text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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MySpace Cuts 300 Jobs Outside U.S.
Website loses two-thirds of foreign workforce
By Ali Jaafar -- Daily Variety
LONDON — MySpace, the social networking site owned by Rupert Murdoch's News Corp., announced Tuesday it was axing 300 jobs outside of the U.S. in an effort to reduce costs.
The cuts amount to two thirds of MySpace's international workforce. The announcement comes a week after the site, which was acquired by Murdoch for $580 million in July 2005, revealed it would be shedding 400 jobs in the U.S., equivalent to 30% of its workforce there.
"As we conducted our review of the company, it was clear that internationally, just as in the U.S., MySpace's staffing had become too big and cumbersome to be sustainable in current market conditions," commented MySpace chief exec Owen Van Natta in a statement.
Click here to read the full story at Variety.com
(Ali Jafaar writes for B&C sister publication Daily Variety)
MySpace Cuts 30% of Staff
Struggling social site issues 420 pink slips
Viacom to Cut 850 Jobs
Restructuring plan to cut 7% of workforce expected to save $200 million to $250 million in 2009
National Public Radio Cuts Jobs, Programs
Public radio operation says it's cutting workforce by 7%
Big Job Cuts Coming to Time Warner Cable
Major restructuring to eliminate 1,250 jobs, save $90 million
Home Shopper QVC To Cut 700 Jobs
Electronic retailer, caught by economic downturn, paring 6% of U.S. workforce
RealNetworks cut 90 jobs
Disney-ABC Cutting Staff
Company is slashing 400 jobs, about 5% of its workforce | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,338 |
Q: Again, the operation of receiving data from a website? I would like to receive data from "http://arsiv.mackolik.com/Program/Program.aspx?st=1" or "http://arsiv.mackolik.com/Program/Program.aspx?st=2". But how can I get the data with "weekac" code. I also want to receive the data with links.
Public Sub Iddaa_Sonuc()
Application.ScreenUpdating = False
Dim i As Long
For i = 3 To 3
Sheets("@").Select
Range("A1").Select
Dim d As WebDriver, clipboard As Object, ele As Object, ws As Worksheet, t As Date, html As HTMLDocument, weeks As Object
Const MAX_WAIT_SEC As Long = 15
Set ws = ThisWorkbook.Worksheets("@")
Set clipboard = GetObject("New:{1C3B4210-F441-11CE-B9EA-00AA006B1A69}")
Set d = New ChromeDriver
Const URL = "http://arsiv.sahadan.com/Iddaa/Program.aspx?st=1"
With d
.Start "Chrome"
.get URL, timeout:=90000
Set weeks = .FindElementsByCss("#weekac option")
.FindElementsByCss("#weekac option")(i).Click
Set html = New HTMLDocument
t = Timer
Do
DoEvents
On Error Resume Next
Set ele = .FindElementByCss("#iddaa-tab-body #resultsList")
On Error GoTo 0
If Timer - t > MAX_WAIT_SEC Then Exit Do
Loop While ele Is Nothing
If Not ele Is Nothing Then
clipboard.SetText ele.Attribute("outerHTML")
clipboard.PutInClipboard
ws.Cells.UnMerge
Application.Wait Now + TimeSerial(0, 0, 1)
ws.Cells(GetLastRow(ws, 1) + 1, 1).PasteSpecial
Application.Wait Now + TimeSerial(0, 0, 3)
End If
Set ele = Nothing
.Quit
End With
Cells.UnMerge
Columns("A:A").Insert
Range("A2").FormulaR1C1 = "=IF(OR(R[-1]C[1]=""Saat"",RC[4]=""Kod""),RC[1],R[-1]C)"
Range("A2").Copy
Range("A2:A" & Range("B1048576").End(xlUp).Row).PasteSpecial Paste:=xlPasteFormulas
Columns("A:A").Copy
Columns("A:A").PasteSpecial Paste:=xlPasteValues
Cells.Replace What:=" ", Replacement:=""
Cells.Replace What:=" ", Replacement:=""
Columns("B:B").Replace What:="PM", Replacement:=" PM"
Columns("B:B").Replace What:="AM", Replacement:=" AM"
Range("D:D,F:F,K:K,O:AB,AF:AZ").Delete
Range("O1").FormulaR1C1 = "=IF(RC[-7]=""v"",""#"",IF(ISNUMBER(RC[-8]),IF(YEAR(RC[-8])=YEAR(TODAY()),CONCATENATE(DAY(RC[-8]),""#"",MONTH(RC[-8])),CONCATENATE(MONTH(RC[-8]),""#"",RIGHT(YEAR(RC[-8]),2))),SUBSTITUTE(RC[-8],""-"",""#"")))"
Range("Q1").FormulaR1C1 = "=IF(RC[-8]=""v"",""#"",IF(ISNUMBER(RC[-9]),IF(YEAR(RC[-9])=YEAR(TODAY()),CONCATENATE(DAY(RC[-9]),""#"",MONTH(RC[-9])),CONCATENATE(MONTH(RC[-9]),""#"",RIGHT(YEAR(RC[-9]),2))),SUBSTITUTE(RC[-9],""-"",""#"")))"
Range("S1").FormulaR1C1 = "=IF(RC[-4]=RC[-3],0,IF(RC[-4]>RC[-3],1,-1))"
Range("T1").FormulaR1C1 = "=IF(RC[-3]=RC[-2],0,IF(RC[-3]>RC[-2],1,-1))"
Range("U1").FormulaR1C1 = "=IF(OR(ISBLANK(RC[-12]),RC[-12]=""-""),0,IF(ISNUMBER(RC[-12]),IF(YEAR(RC[-12])=YEAR(TODAY()),VALUE(CONCATENATE(DAY(RC[-12]),"","",MONTH(RC[-12]))),VALUE(CONCATENATE(MONTH(RC[-12]),"","",RIGHT(YEAR(RC[-12]),2)))),VALUE(SUBSTITUTE(RC[-12],""."","",""))))"
Range("U1").Copy
Range("U1:Z1").PasteSpecial Paste:=xlPasteFormulas
Range("O1:Z1").Copy
Range("O1:O" & Range("A1048576").End(xlUp).Row).PasteSpecial Paste:=xlPasteFormulas
Columns("O:Q").Copy
Columns("O:Q").PasteSpecial Paste:=xlPasteValues
Columns("O:O").TextToColumns Destination:=Range("O1"), DataType:=xlDelimited, TextQualifier:=xlDoubleQuote, ConsecutiveDelimiter:=False, Tab:=True, Semicolon:=False, Comma:=False, Space:=False, Other:=True, OtherChar:="#", FieldInfo:=Array(1, 1), TrailingMinusNumbers:=True
Columns("Q:Q").TextToColumns Destination:=Range("Q1"), DataType:=xlDelimited, TextQualifier:=xlDoubleQuote, ConsecutiveDelimiter:=False, Tab:=True, Semicolon:=False, Comma:=False, Space:=False, Other:=True, OtherChar:="#", FieldInfo:=Array(1, 1), TrailingMinusNumbers:=True
Cells.Copy
Cells.PasteSpecial Paste:=xlPasteValues
Columns("G:N").Delete
ActiveWorkbook.Worksheets("@").Sort.SortFields.Clear
ActiveWorkbook.Worksheets("@").Sort.SortFields.Add Key:=Range("D1:D1048576"), SortOn:=xlSortOnValues, Order:=xlAscending, DataOption:=xlSortNormal
With ActiveWorkbook.Worksheets("@").Sort
.SetRange Range("A1:CC1048576")
.Header = xlGuess
.MatchCase = False
.Orientation = xlTopToBottom
.SortMethod = xlPinYin
.Apply
End With
Range("D" & Columns("D:D").Find(What:="Kod", LookAt:=xlPart).Row & ":D1048576").EntireRow.Delete
Range("A1:R" & Range("A1048576").End(xlUp).Row).Copy
Sheets("Y").Range("A" & Sheets("Y").Range("A1048576").End(xlUp).Row + 1).PasteSpecial Paste:=xlPasteValues
Application.DisplayAlerts = False
Sheets("@").Delete
Sheets.Add.Name = "@"
Sheets("@").Move Before:=Sheets(1)
Application.DisplayAlerts = True
Next
End Sub
It's my code, but it's not working.
When you open "arsiv.mackolik.com/Program/Program.aspx?st=2" in the browser, when you move the mouse over the teams, a link such as "javascript: popBasketTeam (44)" seems to be the id of the 44 teams here. When I have this information, I can get the statistics of the team in the form "arsiv.mackolik.com/Basketball/Team/Default.aspx?id=44" . Make it clear you are after these numbers.
A: I think you can still make your question a lot clearer. From your final point/comments: You can use a css attribute = value selector to gather each of the team links and extract the required number from the href . Below, I gather a nodeList of the required elements and loop that extracting your numbers into an array codes. In the other dimension I place the name associated with that code. You can then loop the first dimension of the array to generate your links by concatenating current array indexed value into URL string.
Integrating with your selenium script:
Dim list As Object, codes()
Set list = d.FindElementsByCss("[href^='javascript:popBasketTeam']")
ReDim codes(1 To list.Count, 1 To 2)
For i = 1 To list.Count
codes(i, 1) = Replace$(Replace$(list(i).Attribute("href"), "javascript:popBasketTeam(", vbNullString), ")", vbNullString)
codes(i, 2) = list(i).Text
Next
Dim newURL As String
'Now loop codes dimension 1 i.e.
For i = LBound(codes, 1) To UBound(codes, 1)
newURL = "http://arsiv.mackolik.com/Basketball/Team/Default.aspx?id=" & codes(i, 1)
' Do something ........
Next
Without selenium:
Option Explicit
Public Sub GetLinks()
Dim sresponse As String, html As HTMLDocument, list As Object, i As Long, codes()
With CreateObject("MSXML2.XMLHTTP")
.Open "GET", "http://arsiv.mackolik.com/Program/Program.aspx?st=2", False
.setRequestHeader "If-Modified-Since", "Sat, 1 Jan 2000 00:00:00 GMT"
.send
sresponse = StrConv(.responseBody, vbUnicode)
End With
Set html = New HTMLDocument
html.body.innerHTML = sresponse
Set list = html.querySelectorAll("[href^='javascript:popBasketTeam']")
ReDim codes(list.Length - 1, 0 To 1)
For i = 0 To list.Length - 1
codes(i, 0) = Replace$(Replace$(list.item(i).href, "javascript:popBasketTeam(", vbNullString), ")", vbNullString)
codes(i, 1) = list.item(i).innerText
Next
Dim newURL As String
'Now loop codes dimension 1 i.e.
For i = LBound(codes, 1) To UBound(codes, 1)
newURL = "http://arsiv.mackolik.com/Basketball/Team/Default.aspx?id=" & codes(i, 1)
' Do something ........
Next
End Sub
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,021 |
$(document).ready(function() {
var timer, delay = 100;
$('#searchBox').bind('keydown blur change', function(e) {
var _this = $(this);
clearTimeout(timer);
timer = setTimeout(queryData, delay );
});
queryData();
});
var queryData = function() {
var searchParam = $("#searchBox").val();
if(searchParam.match(/^\s*$/g)) {
$("#searchResultsTable").children("tbody").html("<tr><td colspan='5'>No Search Results</td></tr>");
}
else {
$.ajax({
type: "POST",
url: "/api/app/search",
data: JSON.stringify({ searchTerm: searchParam }),
contentType: "application/json",
success: function(data, textStatus, jqXHR) {
updateTable(data);
}
});
}
}
var updateTable = function(data) {
if(data.status == "ok") {
$("#searchResultsTable").children("tbody").html(data.apps.map(mapAppToRow).join(""));
console.log("Hit")
}
}
var mapAppToRow = function(app) {
var tdata = "<tr>";
tdata += "<td>" + app.id + "</td>";
tdata += "<td>" + app.appshortkey + "</td>";
tdata += "<td>" + app.name + "</td>";
tdata += "<td>" + app.description + "</td>";
tdata += "<td>" + app.host + "</td>";
tdata += "</tr>";
return tdata;
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 8,223 |
\section{introduction}
The vacuum of electromagnetic field is known \cite{gsa1} to give rise to several
types of very interesting coherence effects. For example it gives rise to
atom-atom correlations and the collective effects. In a single multilevel atom,
it also gives rise to coherences among levels. The coherences are especially
significant if the relevant levels are near by. In Ref. \cite{gsa1} it was
shown that the vacuum of the electromagnetic field can lead to the possibility
of a trapped state in a degenerate V-system. The Ref. \cite{gsa1} also analyzed
the origin of such a trapped state in terms of suitably defined coupled and
uncoupled states. Spontaneous emission produced a coherence between the two excited
states of the V-system.
Further, it was
shown that such a vacuum induced coherence (VIC) can suppress the
steady state resonance fluorescence
\cite{stroud}, and can substantially modify the emission spectrum \cite{agassi}.
Recently, interest in this subject has been revived due to the various
possibilities of manipulating atomic properties using atomic coherence effects
\cite{{plenio},{zhu},{xia},{gsa2},{swain},{imamoglu},{java},{me},{nar},{paspalakis}}.
Using this kind of coherence effect,
Hegerfeldt and Plenio showed that periodic dark states and quantum
beats appear in a near-degenerate V-system \cite{plenio}. The work of Zhu, Scully
and coworkers \cite{zhu}
demonstrate that even
spectral line elimination and spontaneous emission cancellation is possible.
This was observed experimentally by Xia, Ye and Zhu in sodium dimers
\cite{xia}. An appealing physical
picture to explain spontaneous emission cancellation was provided by
Agarwal \cite{gsa2}. The effect of vacuum induced coherence (VIC) on
spontaneous emission has been suggested to achieve gain without
inversion and sub-natural line-widths \cite{swain}. We mention that such a
coherence mechanism is also known to occur in quantum well structures \cite{imamoglu}.
Recent studies have also shown that
vacuum induced coherence effects give rise to phase sensitive
absorption \cite{me} and emission \cite{nar} profiles as well as
to obtain phase control of spontaneous emission in V-systems \cite{paspalakis}.
While much of the work has been in connection with V-systems, other level schemes
like $\Lambda$-systems \cite{{java},{me},{nar}} and $\Xi$-systems \cite{scully}
have also been studied. In case of $\Lambda$-systems,
the coherence is produced in the ground state. We study in this article the origin
of this coherence and the question of a proper probe for such a coherence.
The organization of this paper is as follow. In Sec. II we show the reason
behind the origin of VIC. In Sec. III we derive the spontaneous emission
spectrum in the presence of VIC, and show that the spectrum is independent
of VIC. In Sec. IV we show that absorption of a weak field, as a probe
for VIC, will be uniquely modulated due to VIC. Using second order
perturbation theory we derive the analytical results which well explains
our numerical results. Finally, in Sec. V we present
concluding remarks.
\section{origin of vic in $\Lambda$-systems}
Consider a $\Lambda$-system as shown in Fig. \ref{one}.
Let us take the zero of energy at the state $|\beta\rangle$ and let $\hbar \omega_{ij}$
denote the energy difference between the states $|i\rangle$ and $|j\rangle$.
The Hamiltonian for this system interacting with the vacuum of radiation field is
\begin{equation}
H = H_0 + H_{AV}.
\label{ham1}
\end{equation}
where the unperturbed atom and vacuum field Hamiltonian $H_0$ will be
\begin{equation}
H_0 = \hbar \omega_{1\beta} A_{11} + \hbar\omega_{\alpha\beta}
A_{\alpha\alpha} + \sum_{ks} \hbar\omega_k a^{\dagger}_{ks}a_{ks},
\label{ham2}
\end{equation}
and the interaction Hamiltonian $H_{AV}$ is
\begin{equation}
H_{AV} = - \sum_{ks} \hbar\{(g_{ks} A_{1\alpha} +
f_{ks}A_{1\beta})a_{ks} + {\rm H.c}\}.
\label{ham3}
\end{equation}
Here the operator $A_{ij} = |i\rangle\langle j|$ is the atomic transition operator for
$i \neq j$ and population operator for $i = j$. The field annihilation and creation
operators are $a_{ks}$ and $a^{\dagger}_{ks}$ respectively where the subscript denote the
$k^{th}$ mode of the field with polarization along $\hat{\varepsilon}_{ks}$.
The vacuum
coupling strengths are
$g_{ks} = i(2\pi ck/\hbar L^3)^{1/2} \vec{d}_{1\alpha} \cdot \hat{\varepsilon}_{ks}e^{i\vec{k}\cdot \vec{r}}$ and
$f_{ks} = i(2\pi ck/\hbar L^3)^{1/2} \vec{d}_{1\beta} \cdot \hat{\varepsilon}_{ks}e^{i\vec{k}
\cdot \vec{r}}$, where $\vec{d}_{1 i}$'s ($i = \alpha,\beta$) denote the dipole matrix
elements. We have dropped the anti-resonant terms from (\ref{ham3}).
We work with density matrices and use the standard
master equation technique here. A calculation leads to the following
master equation for the reduced density matrix $\rho$ of the atomic system
\begin{eqnarray}
\dot{\rho} &=& -i [\omega_{1\beta} A_{11} + \omega_{\alpha\beta}
A_{\alpha\alpha}, \rho] -\gamma_{1\alpha}(A_{11}\rho - 2A_{\alpha\alpha} \rho_{11}
+ \rho A_{11}) -\gamma_{1\beta}(A_{11}\rho \nonumber\\
& & - 2A_{\beta\beta}\rho_{11} + \rho A_{11})
+ 2\sqrt{\gamma_{1\alpha} \gamma_{1\beta}} \cos\theta_1
A_{\beta\alpha} \rho_{11} + 2\sqrt{\gamma_{1\alpha}
\gamma_{1\beta}}
\cos\theta_1
A_{\alpha\beta} \rho_{11}.
\label{master}
\end{eqnarray}
Here $2\gamma_{1\alpha}= 4\omega_{1\alpha}^3|d_{1\alpha}|^2/3\hbar c^3$
and $2\gamma_{1\beta} = 4\omega_{1\beta}^3|d_{1\beta}|^2/3\hbar c^3$ denote
the spontaneous emission rates from state $|1\rangle$ to states $|\alpha\rangle$
and $|\beta\rangle$ respectively and
$\theta_1$ is the angle between the two transition dipole moments $\vec{d}_{1i}$ ($i =
\alpha, \beta$).
The last two terms in the above equation are the interference term due to coupling
of the two atomic transition $|1\rangle \rightarrow |\alpha\rangle$,
$|1\rangle \rightarrow |\beta\rangle$ to a common vacuum \cite{harris}
of the electromagnetic
field. The dipole matrix elements should be non-orthogonal for the above
interference to occur. The density matrix elements, $\rho_{ij}$
($\langle i|\rho|j\rangle$), in the Schr\"odinger picture obey equations
\begin{eqnarray}
\dot{\rho}_{11} &=& -2\Gamma_1 \rho_{11},~~~~~~~~~~~~~~~
\dot{\rho}_{1\alpha} = -(\Gamma_1+i\omega_{1\alpha})
\rho_{1\alpha},\nonumber\\
\dot{\rho}_{\alpha\alpha} &=& 2\gamma_{1\alpha}\rho_{11},~~~~~~~~~~~~~~~~~
\dot{\rho}_{1\beta} =
-(\Gamma_1+i\omega_{1\beta})\rho_{1\beta},\nonumber\\
\dot{\rho}_{\beta\beta} &=& 2\gamma_{1\beta} \rho_{11},~~~~~~~~~~~~~~~~~
\dot{\rho}_{\alpha\beta} = -i\omega_{\alpha\beta} \rho_{\alpha\beta} +
2\sqrt{\gamma_{1\alpha}\gamma_{1\beta}} \cos\theta_1 \rho_{11},
\label{elements}
\end{eqnarray}
where $\Gamma_1 = \gamma_{1\alpha} + \gamma_{1\beta}$.
Note that the equation for the ground state coherence $\rho_{\alpha\beta}$ is coupled
to the population of the excited state.
Solving for the coherence $\rho_{\alpha\beta}$ with the initial condition $\rho (0) = A_{11}$ gives,
\begin{equation}
\rho_{\alpha\beta}(t) = \frac{2\sqrt{\gamma_{1\alpha}\gamma_{1\beta}}\cos\theta_1
(e^{-i\omega_{\alpha\beta} t} - e^{-2\Gamma_1 t})}
{(2\Gamma_1 - i\omega_{\alpha\beta})}~.
\label{coh0}
\end{equation}
As the equation reads, this coherence is non-zero as a result of interference term.
Even in the long time limit ($t \gg 1/\Gamma_1$ ) this coherence is finite
and oscillates
with a frequency $\omega_{\alpha\beta}$
\begin{equation}
\rho_{\alpha\beta}(t \rightarrow \infty) = \frac{2\sqrt{\gamma_{1\alpha}
\gamma_{1\beta}}\cos\theta_1e^{-i\omega_{\alpha\beta} t}}
{(2\Gamma_1 - i\omega_{\alpha\beta})}.
\end{equation}
The magnitude of this coherence
is especially significant only if $\omega_{\alpha\beta} \le 2\Gamma_1$ \cite{olga} and if the dipole matrix elements are parallel.
Thus, as mentioned, even vacuum of electromagnetic field can give rise to coherence
in systems with near degenerate levels. We next address the questions: (a) what
leads to the coherence (\ref{coh0}), and (b) how such a coherence can be measured.
In the long time limit the non-zero density matrix in (\ref{elements}) will be
\begin{eqnarray}
\rho_{\alpha\alpha} = \gamma_{1\alpha}/\Gamma_1,~~~ \rho_{\beta\beta} = \gamma_{1\beta}/\Gamma_1,~~~
\rho_{\alpha\beta} = \sqrt{\rho_{\alpha\alpha}\rho_{\beta\beta}}B,~~~
{\rm where} ~~~~~
B = \frac{2\cos\theta_1}{(2-i\omega_{\alpha\beta}/\Gamma_1)}.
\label{elements2}
\end{eqnarray}
The oscillation in $\rho_{\alpha\beta}$
has been removed by writing it in the interaction picture. Thus the density matrix
$\rho$ be reduced to an
effective matrix $\tilde{\rho}$ where
\begin{equation} \tilde{\rho} = \left[
\begin{array}{ccc}
\rho_{\alpha\alpha} & \rho_{\alpha\beta}\\
\rho_{\beta\alpha} & \rho_{\beta\beta}
\end{array}\right] \equiv \left[
\begin{array}{ccc}
\rho_{\alpha\alpha} & \sqrt{\rho_{\alpha\alpha}\rho_{\beta\beta}}B\\
\sqrt{\rho_{\alpha\alpha}\rho_{\beta\beta}}B^* & \rho_{\beta\beta}
\end{array}\right].
\label{matrix}
\end{equation}
A measure of the purity of the state we calculate Tr($\tilde{\rho}^2)$ :
\begin{equation}
{\rm Tr}(\tilde{\rho}^2) = \frac{\gamma_{1\alpha}^2+\gamma_{1\beta}^2}
{\Gamma_1^2} + \frac{8\gamma_{1\alpha}\gamma_{1\beta}\cos^2\theta_1}
{(4\Gamma_1^2 +
\omega^2_{\alpha\beta})}.
\label{trace1}
\end{equation}
For $\omega_{\alpha\beta} = 0$ and $\cos\theta_1 = 1$, $B = 1$, and we get
\begin{equation}
{\rm Tr}(\tilde{\rho}^2) = 1,
\label{trace2}
\end{equation}
which means that the atom would be in a pure state. That is when the VIC
is maximum. Generally, one would
find the system in a mixed state [Tr($\tilde{\rho}^2) < 1$] as $|B| \neq 1$. The
entropy of the final state depends on the parameter $B$.
The case when the atom
is left in a pure state is especially interesting as we can introduce
the coupled ($|c\rangle$) and
uncoupled ($|uc\rangle$) states given by
\begin{equation}
|c\rangle = \frac{|d_{1\alpha}||\alpha\rangle + |d_{1\beta}||\beta\rangle}{|d|},
~~~~~~|uc\rangle = \frac{|d_{1\beta}||\alpha\rangle - |d_{1\alpha}||\beta
\rangle}{|d|},
\end{equation}
where $|d| = \sqrt{|d_{1\alpha}|^2 + |d_{1\beta}|^2}$.
The Hamiltonian (\ref{ham1}) can be written as
\begin{equation}
H = \hbar\omega_{1\beta} |1\rangle\langle 1| +
\sum_{ks}
\omega_k a^{\dagger}_{ks}a_{ks} - \sum_{ks} (g^{\prime}_{ks}|1\rangle\langle c| a_{ks} +
{\rm H.c}),
\label{ham4}
\end{equation}
where $g^{\prime}_{ks} = i(2\pi ck/\hbar L^3)^{1/2}|d|\hat{d}\cdot\hat
{\varepsilon}_{ks} e^{i\vec{k}\cdot\vec{r}}$ is the vacuum coupling between
state $|1\rangle$ and $|c\rangle$ and $\hat{d}$ is the unit vector parallel
to both $\vec{d}_{1\alpha}$ and $\vec{d}_{1\beta}$. Note that state
$|uc\rangle$ is not directly coupled to state $|1\rangle$. Thus $|uc\rangle$
never gets populated if $\rho(0) = A_{11}$. The spontaneous
emission from state $|1\rangle$ occurs to the coherent superposition
state $|c\rangle$ and {\em not just} the individual states $|\alpha\rangle$
and $|\beta\rangle$. Clearly under these conditions the final state will be $|c\rangle$
which agrees with the result (\ref{elements2}) for $\omega_{\alpha\beta} = 0$,
$\theta_1 = 0$.
For $\omega_{\alpha\beta} \neq 0$, the proper basis corresponds to the two
eigenstates $|\psi_{\pm}\rangle$ of (\ref{matrix}) and the steady state will be
an {\em incoherent} mixture of $|\psi_+\rangle$ and $|\psi_-\rangle$.
\section{Emission spectrum}
We now come to the question as to how can one probe the existence of VIC in a
$\Lambda$-system. Thus we naturally think of the spectrum of spontaneous emission.
In a V-system the spontaneous emission is significantly affected
by the presence of VIC
\cite{{gsa1},{stroud},{agassi},{plenio},{zhu},{xia},{gsa2},{swain}}. But for a $\Lambda$-system, as we show, the emission
spectrum is independent of VIC. The emission spectrum corresponds to the normally ordered
two time correlation function of electric field amplitudes \cite{gsa1}. The radiated fields at spacetime points
$\vec{r}_l,t_l$ ($l = 1,2$) will have a correlation given by
\begin{equation}
\langle E^{(-)}(\vec{r}_1,t_1)\cdot E^{(+)}(\vec{r}_2,t_2)\rangle = (r_1r_2)^{-1}
\sum_{i,j = \alpha, \beta} M_{ij}
\langle A_{1i}(t_1) A_{j1}(t_2)\rangle,~~~~~~~~t_1 > t_2,
\label{field1}
\end{equation}
\begin{eqnarray*}
{\rm where}~~~~~~~
M_{ij} = (\frac{\omega_{1i}\omega_{1j}}{c^2})^2
[\hat{r}_1\times(\hat{r}_1\times\vec{d}^*_{1i})]\cdot
[\hat{r}_2\times(\hat{r}_2\times\vec{d}_{1j})],
\end{eqnarray*}
and $r_l$ is much greater than the size of the source.
Using quantum regression theorem and equations (\ref{elements}), it can be shown that
the two time
atomic correlation functions are given by
\begin{equation}
\langle A_{1i}(t_1) A_{i1}(t_2) \rangle = \exp{[(i\omega_{1i} -\Gamma_1)
(t_1 - t_2)]} \exp{(-2\Gamma_1t_2)},~~~~~~t_1 > t_2,
\label{corr1}
\end{equation}
\begin{equation}
{\rm and}~~~~~ \langle A_{1i}(t_1) A_{j1}(t_2) \rangle = 0~~~~~ {\rm for}~~~~~i \neq j.
\end{equation}
Using (\ref{corr1}) in (\ref{field1}) we get the correlation function of the radiated
field
\begin{equation}
\langle E^{(-)}(\vec{r}_1,t_1)\cdot E^{(+)}(\vec{r}_2,t_2)\rangle = (r_1r_2)^{-1}
\sum_{i = \alpha, \beta} M_{ii}
\exp{[(i\omega_{1i} -\Gamma_1)
(t_1 - t_2)]} \exp{(-2\Gamma_1t_2)},~~~~t_1 > t_2.
\end{equation}
This correlation function is
the sum of {\em incoherent} emissions along the two transitions, $|1\rangle \rightarrow
|\alpha\rangle$, $|1\rangle \rightarrow |\beta\rangle$. Thus we conclude that the
spontaneous emission spectrum
in a $\Lambda$-system is {\em not affected} by VIC. Therefore one has to consider other
types of probes to study VIC in such a system.
\section{Modulated absorption as a probe of VIC}
The above result is not surprising because the coherence is created after the
spontaneous emission has occurred. An alternative approach to monitor VIC will be to
study the absorption of a probe field tuned close to some other transition in the
system. In this paper we show that
a unique feature in probe absorption appears
due to the presence of VIC. The model scheme is as
shown in Fig. \ref{two}. Here the spontaneous emission from state $|1\rangle$
creates VIC between the two near-degenerate ground levels $|\alpha\rangle$
and $|\beta\rangle$. We now consider another excited state $|2\rangle$, well separated
from $|1\rangle$. A weak coherent field is tuned between
state $|2\rangle$ and the two ground states
to monitor VIC.
The Hamiltonian in the dipole approximation will be
\begin{equation}
{\cal H} = \hbar\omega_{\alpha\beta} A_{\alpha\alpha} + \hbar\omega_{1\beta} A_{11}
+ \hbar\omega_{2\beta} A_{22} - \{(\vec{d}_{2\beta} A_{2\beta} +
\vec{d}_{2 \alpha} A_{2\alpha})\cdot
\vec{E}_2 e^{-i\omega_2 t} + {\rm H.c}\},
\label{ham}
\end{equation}
where the counter rotating terms in the probe field have been dropped.
The probe field is treated classically here and
has a frequency $\omega_2$
and a complex amplitude $\vec{E}_2$. We use the master equation method to derive
equations for the reduced density matrix of the atomic system. We give the
result of such a calculation,
\begin{mathletters}
\begin{eqnarray}
\dot{\rho}_{11} &=& -2\Gamma_1 \rho_{11},\\
\dot{\rho}_{22} &=& -2\Gamma_2 \rho_{22} + i(G\rho_{\alpha 2}
+ F\rho_{\beta 2})e^{-i\omega_2 t} -i(G^*\rho_{2 \alpha} + F^*\rho_{2 \beta})
e^{i\omega_2 t},
\label{b}\\
\dot{\rho}_{\alpha \alpha} &=& 2\gamma_{1\alpha} \rho_{11} + 2\gamma_{2\alpha} \rho_{22}
-iG e^{-i\omega_2 t}\rho_{\alpha 2} + iG^* e^{i\omega_2 t}\rho_{2 \alpha},
\label{c}\\
\dot{\rho}_{\alpha 2} &=& -(\Gamma_2 - i\omega_{2\alpha})
\rho_{\alpha 2} - iF^*e^{i\omega_2 t}\rho_{\alpha \beta} + iG^*e^{i\omega_2 t}(\rho_{22} -
\rho_{\alpha \alpha}),\label{d}\\
\dot{\rho}_{\beta 2} &=& -(\Gamma_2 - i\omega_{2\beta})
\rho_{\beta 2} - iG^* e^{i\omega_2 t} \rho_{\beta\alpha} + iF^* e^{i\omega_2 t} (2\rho_{22}
+ \rho_{11} + \rho_{\alpha\alpha}- 1),\label{e}\\
\dot{\rho}_{\alpha 1} &=& -(\Gamma_1 - i\omega_{1\alpha})\rho_{\alpha
1}
+ iG^* e^{i\omega_2 t}\rho_{21},\\
\dot{\rho}_{\beta 1} &=& -(\Gamma_1 - i\omega_{1\beta})
\rho_{\beta 1} + iF^* e^{i\omega_2 t}\rho_{21},\\
\dot{\rho}_{21} &=& -(\Gamma_1 + \Gamma_2
-i(\omega_{1\beta}-\omega_{2\beta}))\rho_{21}
+i(G\rho_{\alpha 1} + F\rho_{\beta 1}) e^{-i\omega_2 t},\\
\dot{\rho}_{\alpha \beta} &=& \eta_1 \rho_{11} + \eta_2 \rho_{22} -i\omega_{\alpha\beta}
\rho_{\alpha \beta} - iF e^{-i\omega_2 t} \rho_{\alpha 2} + iG^* e^{i\omega_2 t}\rho_{2 \beta},
\label{density}
\end{eqnarray}
\end{mathletters}
where we have used the trace condition $\sum_i \rho_{ii} = 1$ for (\ref{e}).
Here
\begin{equation}
2\gamma_{2\alpha} = \frac{4\omega_{2\alpha}^3 |d_{1\alpha}|^2}{3\hbar c^3}~~~
{\rm and}~~~
2\gamma_{2\beta} = \frac{4\omega_{2\beta}^3 |d_{1\beta}|^2}{3\hbar c^3}
\end{equation}
define the spontaneous emission rates from $|2\rangle$ to states $|\alpha\rangle$
and $|\beta\rangle$ respectively and we write $\Gamma_2 =
\gamma_{2\alpha} + \gamma_{2\beta}$. The Rabi frequencies
\begin{equation}
2G = 2\vec{E}_2 \cdot \vec{d}_{2 \alpha}/\hbar, ~~~~2F = 2\vec{E}_2 \cdot
\vec{d}_{2 \beta}/\hbar
\end{equation}
are for the probe field acting on transitions $|1\rangle \leftrightarrow |\alpha\rangle$
and $|1\rangle \leftrightarrow |\beta\rangle$ respectively. Further
we can write $G = |G|e^{-i\phi_1}$ and $F = |F|e^{-i\phi_2}$, where the phase
$\phi = \phi_1 - \phi_2$ gives the relative phase
between the complex dipole matrix elements $\vec{d}_{1\alpha}$ and $\vec{d}_{1\beta}$.
The VIC parameters are
\begin{equation}
\eta_1 = 2\sqrt{\gamma_{1\alpha} \gamma_{1\beta}}\cos \theta_1, ~~~~~~
\eta_2 = 2\sqrt{\gamma_{2\alpha} \gamma_{2\beta}}\cos \theta_2.
\end{equation}
We thus include vacuum induced coherence on all possible transitions.
In order to study probe absorption we solve Eqs. (19) perturbatively. We need to know
$\rho_{22}(t)$ to second order in the probe field, assuming that the atom was prepared
in the state $|1\rangle$ at $t = 0$.
Using (\ref{b}) we get
\begin{equation}
\rho_{22}^{(2)} (t) = i\int_0^t d\tau e^{-i\omega_2 \tau} [|G|e^{-i\phi_1}
\rho_{\alpha 2}^{(1)}(\tau) + |F|e^{-i\phi_2} \rho_{\beta 2}^{(1)} (\tau)]
e^{-2\Gamma_2(t-\tau)} + {\rm c.c.}~~.
\label{second1}
\end{equation}
The first order contribution is obtained, for example, by integrating (\ref{d})
\begin{equation}
\rho_{\alpha 2}^{(1)} (t) = -i\int_0^t d\tau e^{i\omega_2 \tau} \{|F|e^{i\phi_2}
\rho_{\alpha \beta}^{(0)} (\tau) - |G|e^{i\phi_1} [\rho_{22}^{(0)}(\tau) -
\rho_{\alpha\alpha}^{(0)}(\tau)]\}
e^{-(\Gamma_2 -i\omega_{2\alpha})(t-\tau)},
\label{first}
\end{equation}
It can be easily show
that $\rho_{22}^{(0)}(t) = 0$ and the other zeroth order terms are known from Sec. I.
The VIC contribution arises from non-zero $\rho_{\alpha\beta}^{(0)}(t)$ in (\ref{first}).
Similarly integrating for $\rho_{\beta 2}^{(1)}(t)$ and combining with Eqs. (\ref{first})
(\ref{second1}), and on simplification we find our key result
\begin{eqnarray}
\rho_{22}^{(2)}(t \gg \Gamma_1^{-1}, \Gamma_2^{-1}) &\equiv& \frac{\eta_1 |F||G| e^{-i(\omega_{\alpha\beta}
t + \phi)}}{(2\Gamma_2-i\omega_{\alpha\beta})(2\Gamma_1-i\omega_{\alpha\beta})(\Gamma_2
-i(\Delta_2+\omega_{\alpha\beta}/2))} \nonumber\\
& &+ \frac{\eta_1 |F||G| e^{i(\omega_{\alpha\beta}
t + \phi)}}{(2\Gamma_2+i\omega_{\alpha\beta})(2\Gamma_1+i\omega_{\alpha\beta})(\Gamma_2
-i(\Delta_2-\omega_{\alpha\beta}/2))} \nonumber\\
& &+ \frac{|G|^2}{4\Gamma_2[\Gamma_2 -
i(\Delta_2 - \omega_{\alpha\beta}/2)]} +
\frac{|F|^2}{4\Gamma_2[\Gamma_2 - i(\Delta_2 + \omega_{\alpha\beta}/2)]} + {\rm c.c}.
\label{second2}
\end{eqnarray}
When $\eta_1 \rightarrow 0$
\begin{equation}
\rho_{22}^{(2)}(t \gg \Gamma_1^{-1}, \Gamma_2^{-1}) \equiv \frac{|G|^2}{4\Gamma_2[\Gamma_2 - i(\Delta_2 -
\omega_{\alpha\beta}/2)]} +
\frac{|F|^2}{4\Gamma_2[\Gamma_2 - i(\Delta_2 + \omega_{\alpha\beta}/2)]} + {\rm c.c},
\label{second3}
\end{equation}
where the last result is the expected result which is the sum of the individual
absorptions corresponding to the transitions $|\alpha\rangle \rightarrow |2\rangle$,
$|\beta\rangle \rightarrow |2\rangle$. The parameter $\Delta_2 = \omega_{1\beta} -
\omega_{\alpha\beta}/2 - \omega_2$ is the probe detuning defined with respect to the
center of level $|\alpha\rangle$ and $|\beta\rangle$. The modulated term in probe
absorption (\ref{second2}) is the
result of VIC. {\em This modulation is the signature of the VIC produced by the
two paths of spontaneous emission} $|1\rangle \rightarrow |\alpha\rangle$,
$|1\rangle \rightarrow |\beta\rangle$. Note the interesting phase dependence
that arises in the probe absorption due to non-zero $\eta_1$. This phase dependence
is another outcome of the presence of VIC in a system. Since the probe is
treated to second
order in its amplitude, the result is independent of the coherence parameter
$\eta_2$ for the transition $|2\rangle\rightarrow |\alpha\rangle$, $|2\rangle
\rightarrow |\beta\rangle$. Needless to say that the Eqs. (19) can
be integrated numerically to obtain the probe absorption for arbitrary times. For
this purpose it is useful to remove the optical frequencies by making
the transformations $\tilde{\rho}_{1i} \equiv \rho_{1i} e^{i\omega_{1i}t}$,
$\tilde{\rho}_{2i} \equiv \rho_{2i} e^{i\omega_{2}t}$ ($i = \alpha, \beta$) and
$\tilde{\rho}_{12} \equiv \rho_{12}e^{i(\omega_2 - \omega_{1\beta})t}$ etc.
We solve these
using fifth-order Runge-Kutta-Verner method with the
initial condition that $\rho_{11}(0) = 1$. We take the probe Rabi frequencies
$F,G$ much smaller than $\Gamma$'s. The numerical results for excited state population
$\rho_{22}(t)$ as a function of time for both the cases when $\eta_1$ is zero and non-zero
are plotted in Fig. 3. Figure 3 shows the significant difference that arises due to
the presence or absence of VIC. The oscillation in the probe absorption is the reflection
of oscillation of in the coherence $\rho_{\alpha\beta}$ (see (\ref{coh0})) and
this confirms the analytical result (\ref{second2}). The numerical result shows
a very slow decay of the envelop of the oscillations. This arises from terms
which are of higher order in probe strength.
Finally we discuss the changes in absorption spectrum that can arise due to VIC.
The modulated component of the population (\ref{second2}) can be written as
\begin{eqnarray}
\rho_{22}^{(2)} &\equiv& \frac{2\eta_1 |F||G|}{D}[\{2\Gamma_1 \Gamma_2^2 + 2\Gamma_1
(\Delta_2^2-
\omega_{\alpha\beta}^2/4) - \Gamma_2 \omega_{\alpha\beta}^2\}\cos(\omega_{\alpha\beta}t
+\phi) \nonumber\\
& &+ \omega_{\alpha\beta}\{2\Gamma_1\Gamma_2 + \Gamma_2^2 + \Delta_2^2-
\omega_{\alpha\beta}^2/4\}\sin(\omega_{\alpha\beta}t + \phi)],
\label{second4}
\end{eqnarray}
where
\begin{eqnarray*}
D = (4\Gamma_1^2+\omega_{\alpha\beta}^2)[\Gamma_2^2+(\Delta_2+\omega_{\alpha\beta}/2)^2]
[\Gamma_2^2+(\Delta_2-\omega_{\alpha\beta}/2)^2].
\end{eqnarray*}
Since it is possible to separate the sine and cosine terms by a phase sensitive
detection we plot these in Fig. 4
as a function of probe detuning. These two components of the absorption spectrum
behave quite differently.
\section{conclusions}
The important criteria for the existence of the vacuum induced coherence between the
close lying levels is the nonorthogonality of the dipole matrix elements. In practice
the nonorthogonality can be achieved by mixing of the energy levels. The mixing
can occur either due to internal fields or due to externally applied fields. For
example in the experiment of Xia {\it et at.} \cite{xia}, spin-orbit interaction
gives rise to mixing. The VIC has also been studied when the level mixing is
produced by using
electromagnetic fields \cite{anil1}, dc fields \cite{berman}, and rf fields
\cite{lenstra}. Special configurations involving cavities can also be utilized to
study VIC \cite{anil2}.
In conclusion we have found that the VIC in a $\Lambda$-system is more difficult
to monitor as it does not show up in the fluorescence spectrum. We have however
demonstrated that the absorption spectrum carries the information on VIC and that the VIC
produces a modulated component in the absorption spectrum.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,321 |
\section{Introduction} \label{sec:intro}
Classical charged particles, confined in a
two-dimensional (2D) layer and interacting via the usual three-dimensional
Coulomb potential, exhibit a crystallization into a Wigner hexagonal structure,
when kinetic energy is small compared to potential energy.\cite{Wi34,GrAd79}
We shall be interested here in bilayer systems, that
describe several properties of real
physical systems in condensed and soft matter, such as
semiconductors,\cite{junctions} quantum dots,\cite{ImMA96}
boron nitride,\cite{boron} laser-cooled trapped ion plasmas,\cite{Mitc98}
dusty plasmas\cite{dusty} and colloids.\cite{colloids}
For a recent review of numerical methods for quasi-2D systems with
long-range interactions, see Ref.\cite{Mazars11}
In addition, the creation of
a bilayer Wigner crystal on two charged plates at some distance is
of primary importance in the study of ``anomalous'' strong-coupling
effects such as like-charge attraction or
overcharging.\cite{LaLP00,Grosberg02,Levin02,Naji05,Samaj11,Samaj12}
In this paper, we study the ground-state properties of a classical
one-component plasma of identical Coulombic particles of the charge $-e$,
evenly distributed between two plates of the same homogeneous
fixed charge density
$\sigma e$ which are at distance $d$.
The total surface density of the particles is $n$, the particle density
in each layer is $n_l=n/2$.
The overall electroneutrality of the system is ensured by the
condition $n_l=\sigma$.
The phase diagram of the system at temperature $T=0$ is determined
by a single dimensionless parameter $\eta=d\sqrt{n/2}=d\sqrt{\sigma}$.
By comparing the static energy of various lattices, five distinct
phases were detected to be stable (providing global minimum of the energy)
in different ranges of $\eta$.\cite{Falk94,EsKa95,Goldoni96,ScSP99,
Weis01,Messina03,Lobaskin07}
In order of increasing $\eta$, these phases are:
a hexagonal lattice (I) for $\eta\in[0,\eta_1^c]$,
a staggered rectangular lattice (II) for $\eta\in[\eta_1^c,\eta_2^c]$,
a staggered square lattice (III) for $\eta\in[\eta_2^c,\eta_3^c]$,
a staggered rhombic lattice (IV) for $\eta\in[\eta_3^c,\eta_4^c]$ and
a staggered hexagonal lattice (V) for $\eta\in[\eta_4^c,\infty]$;
although we use an index $c$ in $\eta^c$, the transition point $\eta^c$
from one structure to the other is not necessarily a critical point.
The structures are pictured in Fig. \ref{fig:Structures}.
The different symbols correspond to particle positions on the opposite
surfaces.
The primitive translation vectors of the Bravais lattice on one of
the surfaces are denoted by $\bm{a}_1$ and $\bm{a}_2$.
\begin{SCfigure}[1.5][t]
\includegraphics[height=10cm,clip]{struct_bis.eps}
\caption{Ground-state structures I--V of counter-ions on two parallel
equivalently and homogeneously charged plates.
Open and filled symbols correspond to particle positions on
the opposite surfaces.
The primitive translation vectors of the Bravais lattice on one of
the surfaces are denoted by $\bm{a}_1$ and $\bm{a}_2$.
For structures I, II and III, we define the aspect ratio as
$\Delta=|\bm{a}_1|/|\bm{a}_2|$, so that $\Delta=\sqrt{3}$
with structure I, $1 < \Delta < \sqrt{3}$ for structure II
and $\Delta=1$ for structure III. The dashed circle for structure
IV is a guide to the eye, for identifying those points that are
equidistant to the ion in the circle center.
For a more detailed description of the structures, see the text.}
\label{fig:Structures}
\end{SCfigure}
The ground-state structures I, III and V are ``rigid'', i.e. they have
fixed ($\eta$-independent) primary cells within their region
of stability.
The structures II and IV are ``soft'', i.e. the shape of their
primary cells is varying with increasing $\eta$, within their region
of stability.
We now outline the basic characteristics of the structures.\cite{Falk94,EsKa95,Goldoni96,ScSP99,
Weis01,Messina03,Lobaskin07}
\begin{itemize}
\item
{\bf Structures I, II and III:}
Within one single layer, the structure corresponds to a rectangular lattice
with the aspect ratio $\Delta=\vert \bm{a}_2\vert/\vert \bm{a}_1\vert$.
The equivalent structures on the two layers are shifted with respect
to one another by a half period, i.e. by $(\bm{a}_1+\bm{a}_2)/2$.
Structure I with $\Delta=\sqrt{3}$ arises naturally in the simple limit
$\eta\to 0$, where the bilayer structure reduces to a single layer,
which is known to crystallize in a hexagonal (equilateral triangular)
lattice.\cite{Bonsall,Jagla}
An open question is whether phase I (with the fixed aspect ratio
$\Delta=\sqrt{3}$) exists only at $\eta=0$ or is stable also in
a finite interval $[0,\eta_1^c]$ with some $\eta_1^c>0$.
Some numerical calculations indicate very small, but nonzero values of
$\eta_1^c=0.006$ (Ewald technique\cite{Goldoni96}) and
$0.028$ (Monte Carlo simulations\cite{Weis01}). On the other hand,
another study for Yukawa bilayers in the limit of infinite screening
length indicates that $\eta_1^c=0$, so that a buckled phase of type II preempts
structure I when $\eta$ is small but non vanishing.\cite{Messina03}
Structure II continuously interpolates between
the rigid structures I and III.
The value of the aspect ratio $\Delta$ then changes smoothly
from $\sqrt{3}$ at $\eta_1^c$ (phase I) to $1$ at the transition point
$\eta_2^c$ to phase III.
It is not clear whether or not $\eta_1^c$, zero or nonzero, is a standard
transition point between phases I and II.
The transition between phases II and III is continuous (of second order).
\item
The structure IV is characterized by an angle $\theta$ between primitive
cell vectors $\bm{a}_1$ and $\bm{a}_2$ of the same length $a$.
Increasing $\eta$, the angle $\varphi$ changes continuously from
$\pi/2$ at $\eta_3^c$ (continuous transition between phases III and IV)
up to $\eta_4^c$, where it drops to $\pi/3$.
Together with an additional shift between the sub-lattices on the two layers,
this corresponds to a discontinuous (first order) transition to phase V.
\item
The presence of structure V is expected for large enough
$\eta\ge \eta_4^c$:
At large separation between the layers, the two particle sub-lattices
are only weakly coupled and so two staggered hexagonal lattices form
the stable structure.
\end{itemize}
In this paper, we derive new series representations of the energy for
all five bilayer structures.
The derivation is based on a sequence of transformations for lattice
sums of Coulomb potentials plus the neutralizing background, having
their origin in the general theory of Jacobi theta functions.\cite{Abra}
The series has excellent convergence properties, which is
convenient for numerical investigations, but is also
conducive to analytical progress.
It will be used to improve the specification
of transition points between the phases and to solve the aforementioned controversy
concerning the stability of phase I.
In particular, it is shown both analytically and numerically that
phase I is realized only at $\eta=0$, i.e. $\eta_1^c=0$.
The expansions of the structure energies around second-order transition
points can be done analytically which enables us to derive the critical
exponents at phase transition points.
The critical behavior is of the Ginzburg-Landau type,\cite{GL} with the mean-field
critical index $\beta=1/2$ for the growth of the order parameters.
A preliminary account of this work has appeared in Ref.\cite{EPL}
The paper is organized as follows.
Sec. \ref{sec:123} is devoted to a detailed derivation of the
series representation of the energy for structures I-III.
The existence of phase I at $\eta=0$ only is established
analytically and illustrated numerically.
The second-order phase transition between phases II and III is then
described.
The energy of structure IV and the second-order phase transition
between phases III and IV are treated in Sec. \ref{sec:4}.
Sec. \ref{sec:5} deals with structure V and the first-order
phase transition between phases IV and V, while our conclusions are finally
presented in Sec.
\ref{sec:conclusion}.
\section{Phases I-III} \label{sec:123}
Structures I, II and III are treated on equal footings by considering
the general case of structure II, see Fig. \ref{fig:Structures}.
For one single layer, the 2D lattice points are indexed by
$j \bm{a}_1 + k \bm{a}_2$, where $j,k$ are any two integers
(positive, negative or zero) and
\begin{equation}
\bm{a}_1 = a (1,0) , \qquad \bm{a}_2 = a (0,\Delta) \qquad
{\rm with}\ a=\frac{1}{\sqrt{\sigma\Delta}}
\end{equation}
are the primitive translation vectors of the Bravais lattice.
The lattice spacing $a$ is determined by the electroneutrality condition
$n_l=\sigma$ with the one-layer particle density $n_l=1/(\Delta a^2)$.
The aspect ratio $\Delta$ is a continuous parameter in the interval
$[1,\sqrt{3}]$; as was already mentioned, the limiting cases $\sqrt{3}$
and $1$ correspond to the phases I and III, respectively.
\subsection{Energy of phases I-III}
The dielectric constant of the medium is set to unity for simplicity,
and we start by a preliminary remark, valid for all phases.
Our goal is to compute the total electrostatic energy,
including particle-particle, particle-plate,
and plate-plate interactions.
The latter two contributions per unit surface
can be derived straightforwardly, and respectively read $4 \pi \sigma^2 e^2 d$
and $-2 \pi \sigma^2 e^2 d$. The sum of both, $2 \pi \sigma^2 e^2 d$,
thus gives $1/2$ of the particle-plate energy, and this is why in the
subsequent analysis, we shall add to the non trivial particle-particle
energy one half of the particle-plate energy (also referred
to as the particle-background term).
The resulting sum provides the full energy of the system.
The energy per particle $E$ of the bilayer system consists of the
intralayer and interlayer contributions,
\begin{equation}
E = E_{\rm intra} + E_{\rm inter} .
\end{equation}
We first consider the intralayer contribution.
It is well known that lattice sums involving the pair Coulomb interactions
exhibit infinities which are canceled exactly by the neutralizing
background term.\cite{Mazars11,dLPS80}
To maintain mathematical rigor, we first restrict ourselves to
a disk of finite radius $R$ around a reference particle localized
at the origin $(0,0)$.
The interaction energy due to the discrete Wigner crystal is given by
\begin{equation} \label{contr1}
\frac{e^2}{2a} \sum_{j,k\atop (j,k)\ne (0,0)}
\frac{1}{\sqrt{j^2 + k^2 \Delta^2}} , \qquad
j^2 +k^2 \Delta^2 \le \left( \frac{R}{a} \right)^2 .
\end{equation}
Hereinafter, the omission of the lower and upper values for integer
indices $j,k$ automatically means a summation from $-\infty$ to $\infty$.
The interaction of the reference particle with the 2D charge background
in the disk is expressed as
\begin{equation} \label{contr2}
- \frac{\sigma e^2}{2} \int_0^R d^2{\bf r} \frac{1}{\vert{\bf r}\vert}
= - \frac{e^2}{2a\Delta} \int_0^{R/a} dr 2\pi r \frac{1}{r} .
\end{equation}
$E_{\rm intra}$ is the sum of (\ref{contr1}) plus (\ref{contr2}).
We intend to rewrite $E_{\rm intra}$ by using the gamma identity
\begin{equation} \label{gamma}
\frac{1}{\sqrt{z}} = \frac{1}{\sqrt{\pi}} \int_0^{\infty}
\frac{dt}{\sqrt{t}} e^{-zt} , \qquad z>0 ,
\end{equation}
a common procedure in the field.\cite{Mazars11,dLPS80}
Each term $1/(j^2 + k^2 \Delta^2)^{1/2}$ in
(\ref{contr1}) can consequently be written as
\begin{equation}
\frac{1}{\sqrt{j^2 + k^2 \Delta^2}} = \frac{1}{\sqrt{\pi}} \int_0^{\infty}
\frac{dt}{\sqrt{t}} e^{-t j^2} e^{-t\Delta^2 k^2} .
\end{equation}
As concerns the background contribution (\ref{contr2}), the application of
the identity (\ref{gamma}) to the term $1/r = 1/\sqrt{r^2}$ under integration
leads to
\begin{equation}
\int_0^{R/a} dr 2\pi r \frac{1}{r} = \int_0^{R/a} dr 2\pi r
\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{dt}{\sqrt{t}} e^{-t r^2} =
\frac{1}{\sqrt{\pi}} \int_0^{\infty}
\frac{dt}{\sqrt{t}} \frac{\pi}{t}\left[ 1 - e^{-t(R/a)^2} \right] .
\end{equation}
Altogether, we get
\begin{equation}
E_{\rm intra} = \frac{e^2}{2a\sqrt{\pi}} \int_0^{\infty} \frac{dt}{\sqrt{t}}
\Bigg\{ \sum_{j,k} e^{-t j^2} e^{-t\Delta^2 k^2}
- 1 - \frac{\pi}{t\Delta} \left[ 1 - e^{-t(R/a)^2} \right] \Bigg\} ,
\qquad j^2 +k^2 \Delta^2 \le \left( \frac{R}{a} \right)^2 .
\end{equation}
Here, the subtraction of unity is due to the absence of the term
$(j,k)=(0,0)$ in the sum (\ref{contr1}).
Having all contributions under the same integration, we are allowed to take
the limit $R/a\to \infty$, which removes the exponentially small term
$\exp[-t(R/a)^2]$ and the disk constraint for lattice indices.
Using the definition of the Jacobi theta function with zero
argument\cite{Gradshteyn} $\theta_3(q)=\sum_j q^{j^2}$
and making the substitution $t \Delta\to t$, we end up with the result
\begin{equation} \label{theta3}
\frac{E_{\rm intra}}{e^2\sqrt{n}} = \frac{1}{2^{3/2}\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}}
\left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right] .
\end{equation}
We shall repeatedly use the Poisson summation formula
\begin{equation} \label{Poisson}
\sum_{j=-\infty}^{\infty} e^{-(j+\phi)^2 t} = \sqrt{\frac{\pi}{t}}
\sum_{j=-\infty}^{\infty} e^{2\pi ij\phi}e^{-(\pi j)^2/t} .
\end{equation}
The asymptotic behaviors
\begin{equation}
\theta_3(e^{-t}) \mathop{\sim}_{t\to 0} \sqrt{\frac{\pi}{t}} \left( 1 +
2 e^{-\pi^2/t} + \cdots \right) , \qquad
\theta_3(e^{-t}) \mathop{\sim}_{t\to\infty} 1 + 2 e^{-t} + \cdots
\end{equation}
follow immediately.
We see that the background charge contribution $-\pi/t$ correctly cancels
the $t\to 0$ singularity of the product of two $\theta_3$ functions inside
the square bracket in (\ref{theta3}) and the integral converges.
The Wigner lattices on the opposite layers are shifted with respect to
one another by the vector $(\bm{a}_1+\bm{a}_2)/2$, see Fig. \ref{fig:Structures}.
To obtain the interlayer contribution to the energy, we first consider
the disk of radius $R$ around the (perpendicular) image of the reference
particle on the opposite layer.
The interaction energy of the Wigner crystal is given by
\begin{equation}
\frac{e^2}{2a} \sum_{j,k} \frac{1}{\sqrt{\left( j-\frac{1}{2}\right)^2 +
\left( k-\frac{1}{2}\right)^2 \Delta^2 +(d/a)^2}} , \qquad
\left( j-\frac{1}{2}\right)^2 +\left( k-\frac{1}{2}\right)^2 \Delta^2
\le \left( \frac{R}{a} \right)^2.
\end{equation}
The interaction with the background charge is described by
\begin{equation}
- \frac{\sigma e^2}{2} \int_0^R d^2{\bf r} \frac{1}{\vert{\bf r}+{\bf d}\vert}
= - \frac{e^2}{2a\Delta} \int_0^{R/a} dr 2\pi r \frac{1}{\sqrt{r^2+(d/a)^2}} .
\end{equation}
Proceeding as in the previous case and taking into account that
$d/a = \eta \sqrt{\Delta}$, we find
\begin{equation}
\frac{E_{\rm inter}}{e^2\sqrt{n}} = \frac{1}{2^{3/2}\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}} e^{-\eta^2 t}
\left[ \theta_2(e^{-t\Delta}) \theta_2(e^{-t/\Delta}) - \frac{\pi}{t} \right]
\label{eq:14}
\end{equation}
with the Jacobi theta function
$\theta_2(q)=\sum_j q^{\left(j-\frac{1}{2}\right)^2}$.
It follows from Eq. (\ref{Poisson}) that
\begin{equation}
\theta_2(e^{-t}) \mathop{\sim}_{t\to 0} \sqrt{\frac{\pi}{t}}
\left( 1 - 2 e^{-\pi^2/t} + \cdots \right) , \qquad
\theta_2(e^{-t}) \mathop{\sim}_{t\to\infty} 2 \, e^{-t/4} + \cdots ,
\end{equation}
so that the integral in Eq. (\ref{eq:14}) converges, as it should.
The total energy per particle $E$ reads
\begin{eqnarray} \label{E}
\frac{E(\Delta,\eta)}{e^2\sqrt{n}} = \frac{1}{2^{3/2}\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}} \Bigg\{
\left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right]
+ e^{-\eta^2 t} \left[ \theta_2(e^{-t\Delta}) \theta_2(e^{-t/\Delta})
- \frac{\pi}{t} \right] \Bigg\} .
\end{eqnarray}
Note the invariance of $E$ with respect to the transformation
$\Delta\to 1/\Delta$, which is physically clear from the configuration
sketched in Fig \ref{fig:Structures} (label exchange of the two Bravais
vectors $\bm{a}_1$ and $\bm{a}_2$).
From a numerical point of view, there are two ``dangerous'' limits:
$t\to 0$ and $t\to\infty$, that jeopardize accuracy.
To simplify the integral representation of $E$, we split the range of
integration into two parts, $[0,\pi]$ and $[\pi,\infty]$, and transform
the integral over $[\pi,\infty]$ to the one over $[0,\pi]$ by using
the Poisson formula (\ref{Poisson}).
For the term containing the product of two $\theta_3$ functions,
the procedure reads
\begin{eqnarray}
\int_{\pi}^{\infty} \frac{dt}{\sqrt{t}}
\left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right]
& \equiv & \int_{\pi}^{\infty} \frac{dt}{\sqrt{t}}
\left[ \sum_j e^{-t j^2\Delta} \sum_k e^{-t k^2/\Delta} - 1 - \frac{\pi}{t} \right]
\nonumber \\
& = & \int_{\pi}^{\infty} \frac{dt}{\sqrt{t}}
\left[ \frac{\pi}{t} \sum_j e^{-(\pi j)^2/(\Delta t)}
\sum_k e^{-(\pi k)^2\Delta/t} - 1 - \frac{\pi}{t} \right] \nonumber \\
& = & \int_0^{\pi} \frac{\pi dt'}{(t')^{3/2}}
\left[ \frac{t'}{\pi} \sum_j e^{-t' j^2/\Delta}
\sum_k e^{-t' k^2\Delta} - 1 - \frac{t'}{\pi} \right] \nonumber \\
& \equiv & \int_0^{\pi} \frac{dt}{\sqrt{t}}
\left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right] .
\end{eqnarray}
Here, going from the third integral to the fourth one, we applied the
substitution $t'=\pi^2/t$.
Similarly, we have
\begin{eqnarray}
\int_{\pi}^{\infty} \frac{dt}{\sqrt{t}} e^{-\eta^2 t}
\left[ \theta_2(e^{-t\Delta}) \theta_2(e^{-t/\Delta}) - \frac{\pi}{t} \right]
& \equiv & \int_{\pi}^{\infty} \frac{dt}{\sqrt{t}} e^{-\eta^2 t}
\left[ \sum_j e^{-t \left( j-\frac{1}{2}\right)^2\Delta}
\sum_k e^{-t \left( k-\frac{1}{2}\right)^2/\Delta} - \frac{\pi}{t} \right]
\nonumber \\
& = & \int_{\pi}^{\infty} \frac{dt}{\sqrt{t}} e^{-\eta^2 t}
\left[ \frac{\pi}{t} \sum_j (-1)^j e^{-(\pi j)^2/(\Delta t)}
\sum_k (-1)^k e^{-(\pi k)^2\Delta/t} - \frac{\pi}{t} \right] \nonumber \\
& = & \int_0^{\pi} \frac{\pi dt'}{(t')^{3/2}} e^{-(\pi\eta)^2/t'}
\left[ \frac{t'}{\pi} \sum_j (-1)^j e^{-t' j^2/\Delta}
\sum_k (-1)^k e^{-t' k^2\Delta} - \frac{t'}{\pi} \right] \nonumber \\
& \equiv & \int_0^{\pi} \frac{dt}{\sqrt{t}} e^{-(\pi\eta)^2/t}
\left[ \theta_4(e^{-t\Delta}) \theta_4(e^{-t/\Delta}) - 1 \right] ,
\end{eqnarray}
where the Jacobi theta function $\theta_4(q) = \sum_j (-1)^j q^{j^2}$.
The asymptotic behaviors
\begin{equation}
\theta_4(e^{-t}) \mathop{\sim}_{t\to 0}
\sqrt{\frac{\pi}{t}} e^{-\pi^2/(4t)} + \cdots ,
\qquad \theta_4(e^{-t}) \mathop{\sim}_{t\to\infty} 1 - 2 e^{-t} + \cdots
\end{equation}
ensure the convergence of the resulting integral.
To summarize this paragraph, the total energy (\ref{E}) can be rewritten
as an integral over the finite interval $[0,\pi]$ as follows
\begin{eqnarray}
\frac{E(\Delta,\eta)}{e^2\sqrt{n}} & = & \frac{1}{2^{3/2}\sqrt{\pi}}
\int_0^{\pi} \frac{dt}{\sqrt{t}} \Bigg\{
2 \left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right]
\nonumber \\ & &
+ e^{-\eta^2 t} \left[ \theta_2(e^{-t\Delta}) \theta_2(e^{-t/\Delta})
- \frac{\pi}{t} \right] + e^{-(\pi\eta)^2/t}
\left[ \theta_4(e^{-t\Delta}) \theta_4(e^{-t/\Delta}) - 1 \right] \Bigg\} .
\label{EE}
\end{eqnarray}
The exact cancellation of singular terms near $t=0$ in this expression for
$E$ represents a numerical obstacle that should be circumvented.
To accomplish the cancellation analytically, we shall consider the series
representations of Jacobi theta functions and apply to them the Poisson
transformation formula (\ref{Poisson});
after subtracting explicitly the singular term, the result appears as a
series of special functions.
In particular, for the first term in the integral (\ref{EE}) we obtain
\begin{eqnarray}
\int_0^{\pi} \frac{dt}{\sqrt{t}}
\left[ \theta_3(e^{-t\Delta}) \theta_3(e^{-t/\Delta}) - 1 - \frac{\pi}{t} \right]
& = & \int_0^{\pi} \frac{dt}{\sqrt{t}} \left[
\sum_{j,k} e^{-t j^2\Delta} e^{-t k^2/\Delta} - \frac{\pi}{t} \right] - 2\sqrt{\pi}
\nonumber \\ & = & \int_0^{\pi} dt \frac{\pi}{t^{3/2}} \left[
\sum_{j,k} e^{-(\pi j)^2/(\Delta t)} e^{-(\pi k)^2\Delta/t} - 1 \right] - 2\sqrt{\pi} .
\label{rep}
\end{eqnarray}
The subtraction of the singularity is equivalent to the omission of
the term $(j,k)=(0,0)$ from the summation.
Using the substitution $t'=t/\pi^2$ and introducing the function
\begin{equation} \label{z}
z_{\nu}(x,y) = \int_0^{1/\pi} \frac{dt}{t^{\nu}} e^{-x t} e^{-y/t} \qquad
\mbox{for $y>0$,}
\end{equation}
the last expression in Eq. (\ref{rep}) can be written as
\begin{equation}
2 \sum_{j=1}^{\infty} \left[ z_{3/2}(0,j^2/\Delta) + z_{3/2}(0,j^2\Delta) \right]
+ 4 \sum_{j,k=1}^{\infty} z_{3/2}(0,j^2/\Delta+k^2\Delta) - 2 \sqrt{\pi} .
\end{equation}
Finally, performing the above procedure for all terms under integration
in (\ref{EE}), we end up with the series representation
\begin{eqnarray}
\frac{E(\Delta,\eta)}{e^2\sqrt{n}} & = & \frac{1}{2^{3/2}\sqrt{\pi}}
\Bigg\{ 4\sum_{j=1}^{\infty}
\left[ z_{3/2}(0,j^2/\Delta) + z_{3/2}(0,j^2\Delta) \right]
+ 8 \sum_{j,k=1}^{\infty} z_{3/2}(0,j^2/\Delta+k^2\Delta) \nonumber \\ & &
+ 2 \sum_{j=1}^{\infty} (-1)^j
\left[ z_{3/2}((\pi\eta)^2,j^2/\Delta) + z_{3/2}((\pi\eta)^2,j^2\Delta) \right]
+ 4 \sum_{j,k=1}^{\infty} (-1)^j (-1)^k z_{3/2}((\pi\eta)^2,j^2/\Delta+k^2\Delta)
\nonumber \\ & &
+ 4 \sum_{j,k=1}^{\infty} z_{3/2}(0,\eta^2+(j-1/2)^2/\Delta+(k-1/2)^2\Delta)
- 4 \sqrt{\pi} - \pi z_{1/2}(0,\eta^2) \Bigg\} . \label{series1}
\end{eqnarray}
The function $z_{\nu}(x,y)$ with $x=0$ is related to the so-called
Misra function\cite{Misra40,Born40} which was extensively used in
single-layer lattice summations.\cite{Borwein88,Bowick06}
For our bilayer system with positive $\eta$ we need the more general
function (\ref{z}) with $x\ge 0$. The convergence properties
of our series can be anticipated from the asymptotic relation
$z_\nu(0,y) \sim e^{-\pi y} \pi^{\nu-2}/y$
In numerical calculations, for a given $\eta$ we have to find such
$\Delta^*$ which provides the minimum value of the energy (\ref{series1}).
In practice, the series (\ref{series1}) must be cut at some $j,k=M$.
We document excellent convergence properties of the series (\ref{series1})
by considering the single-layer case $(\Delta=\sqrt{3},\eta=0)$ for which
the exact\cite{rque50}
\begin{equation}
E(\sqrt{3},0)/(e^2\sqrt{n}) = - 1.96051578931989165\ldots
\label{eq:E0}
\end{equation}
The cut at $M=1,2,3,4$ reproduces this exact value up to $2,5,10,17$
decimal digits, respectively.
A similar accuracy is reached in all considered cases.
To be extremely accurate, we apply the $M=5$ cut everywhere,
and use the {\it Mathematica} software.
Another advantage of the series representation (\ref{series1}) is the
possibility of an explicit expansion of the function
$E(\Delta^*,\eta)/(e^2\sqrt{n})$ around the controversial point $\eta=0$
and around the critical point $\eta_2^c$.
As will be shown, this requires an analogous Taylor expansion of our
$z$-functions.
\subsection{Going from phase I to phase II}
We know\cite{Bonsall,Jagla} that for the single layer, i.e. $\eta=0$, the structure providing
the minimum of the energy is the hexagonal lattice with $\Delta=\sqrt{3}$.
In what follows, we shall investigate the minimum of the energy (\ref{series1})
in the neighborhood of the point $(\Delta=\sqrt{3},\eta=0)$.
We set $\Delta=\sqrt{3}-\epsilon$ and consider $\epsilon$ to be
infinitesimally small.
To derive a small-$\epsilon$ expansion of the energy (\ref{series1}),
we first perform this task for its series components.
From the integral definition (\ref{z}) it is easy to show that
the $z$-functions under consideration exhibit an analytic (Taylor)
expansion in $\epsilon$ of the form
\begin{equation}
z_{3/2}(0,j^2\Delta) = z_{3/2}(0,j^2\sqrt{3})
+ \epsilon j^2 z_{5/2}(0,j^2\sqrt{3})
+ \frac{1}{2} \epsilon^2 j^4 z_{7/2}(0,j^2\sqrt{3}) + {\cal O}(\epsilon^3) ,
\end{equation}
\begin{equation}
z_{3/2}(0,j^2/\Delta) = z_{3/2}(0,j^2/\sqrt{3})
- \epsilon \frac{j^2}{3} z_{5/2}(0,j^2/\sqrt{3})
+ \epsilon^2 \left[ \frac{j^4}{18} z_{7/2}(0,j^2/\sqrt{3}) -
\frac{j^2}{3^{3/2}} z_{5/2}(0,j^2/\sqrt{3}) \right] + {\cal O}(\epsilon^3) ,
\end{equation}
\begin{eqnarray}
z_{3/2}(0,j^2/\Delta+k^2\Delta) & = & z_{3/2}(0,j^2/\sqrt{3}+k^3\sqrt{3})
+ \epsilon \left( k^2 - \frac{j^2}{3} \right) z_{5/2}(0,j^2/\sqrt{3}+k^2\sqrt{3})
\nonumber \\ & & + \epsilon^2 \left[ \frac{1}{2}
\left( k^2 - \frac{j^2}{3} \right)^2 z_{7/2}(0,j^2/\sqrt{3}+k^2\sqrt{3}) -
\frac{j^2}{3^{3/2}} z_{5/2}(0,j^2/\sqrt{3}+k^2\sqrt{3}) \right]
+ {\cal O}(\epsilon^3) .
\end{eqnarray}
Similar expansions can be derived for $z_{3/2}((\pi\eta)^2,j^2\Delta)$,
$z_{3/2}((\pi\eta)^2,j^2/\Delta)$, etc.
Inserting these expansions into (\ref{series1}), we obtain
\begin{equation} \label{epsilon1}
\frac{E(\sqrt{3}-\epsilon,\eta)}{e^2\sqrt{n}}
= \frac{E(\sqrt{3},\eta)}{e^2\sqrt{n}}
+ f_1(\eta) \epsilon + f_2(\eta) \epsilon^2 + {\cal O}(\epsilon^3) ,
\end{equation}
where the explicit form of the prefactor functions $f_1(\eta)$
and $f_2(\eta)$ is written in the Appendix.
\begin{figure} [tb]
\includegraphics[width=0.43\textwidth]{smalleta3.eps}
\hfil
\includegraphics[width=0.43\textwidth]{smalleta2.eps}
\caption{The difference between the dimensionless energies
$[E(\Delta,\eta)-E(\sqrt{3},\eta)]/(e^2\sqrt{n})$ versus
$\epsilon=\sqrt{3}-\Delta$, calculated numerically by using (\ref{series1})
for two small values of $\eta$: a) $\eta=10^{-3}$ and b) $\eta=10^{-2}$.
For the range of aspect ratios chosen, these curves are indistinguishable
from the analytical prediction $-0.5833 \, \eta^2 \epsilon+ 0.0408\, \epsilon^2$,
stemming from Eqs. (\ref{epsilon1}) and (\ref{eq:29}).
The values of $\epsilon^*$, which provide the energy minimum in the asymptotic
limit $\eta\to 0$ according to (\ref{transitionItoII}), are depicted by
the vertical dashed lines for comparison.}
\label{fig:smalleta}
\end{figure}
Being close to the point $(\Delta=\sqrt{3},\eta=0)$, we are interested in
the small-$\eta$ behavior of the functions
$f_1(\eta)$ and $f_2(\eta)$.
The corresponding Taylor expansions in powers of $\eta^2$ can be performed
explicitly, too.
The explicit form of $E(\sqrt{3},\eta)$ is here immaterial.
We find that $f_1(0)=0$, and,
with a high precision,
\begin{equation}
f_1(\eta) = - 0.5833059875\ldots\eta^2 + {\cal O}(\eta^4) , \qquad
f_2(\eta) = 0.0408440789\ldots + {\cal O}(\eta^2) .
\label{eq:29}
\end{equation}
For a fixed $\eta$, the extremum of the energy (\ref{epsilon1}) appears
at $\epsilon^*=\sqrt{3}-\Delta^*$ given by the stationarity condition
\begin{equation}
\frac{\partial}{\partial\epsilon} \frac{E(\sqrt{3}-\epsilon,\eta)}{e^2\sqrt{n}}
\Bigg\vert_{\epsilon=\epsilon^*} = 0 =
f_1(\eta) + 2 f_2(\eta) \epsilon^* .
\end{equation}
Namely,
\begin{equation} \label{transitionItoII}
\sqrt{3} - \Delta^* \equiv \epsilon^*(\eta) =
- \frac{f_1(\eta)}{2 f_2(\eta)} = 7.14064\ldots \eta^2 + {\cal O}(\eta^4) .
\end{equation}
Since $\partial^2_{\epsilon}E(\sqrt{3}-\epsilon,\eta)\big\vert_{\epsilon=\epsilon^*}
= f_2(\eta)>0$, the extremum is the minimum.
The result (\ref{transitionItoII}) tells us that howsoever small
the dimensionless distance $\eta$ is, the buckled structure II with
$\Delta<\sqrt{3}$ takes place.
In other words, the structure I exists only strictly at $\eta_1^c=0$.
The fact that in the previous works\cite{Goldoni96,Weis01} the structure I
was detected also for very small positive values of $\eta$ is probably related
to extremely small values of the deviation $\epsilon^*\propto \eta^2$
for these $\eta$'s, which are ``invisible'' by standard numerical methods.
Like for instance, the structure I border reported
in\cite{Goldoni96} $\eta_1^c=0.006$ corresponds to
$\epsilon^*=0.00026\ldots$.
In Fig. \ref{fig:smalleta}, we present the plots of the difference between
the dimensionless energies $[E(\Delta,\eta)-E(\sqrt{3},\eta)]/(e^2\sqrt{n})$
versus $\epsilon=\sqrt{3}-\Delta$, calculated numerically by using
(\ref{series1}) for two very small values of $\eta$: a) $\eta=10^{-3}$
and b) $\eta=10^{-2}$ which are well below/above the previous estimate
of the phase I threshold,\cite{Goldoni96} respectively.
Alternatively, using the analytical expressions (\ref{epsilon1}) and (\ref{eq:29})
leads to the very same data.
The nonzero values of $\epsilon^*$, which provide the energy minima in
the asymptotic limit $\eta\to 0$ according to formula (\ref{transitionItoII}),
are depicted by the dashed lines for comparison.
We see that the energy minima fit well with the expected $\epsilon^*$
which is clear evidence for the phase I instability.
Note extremely small values $\propto 10^{-12}-10^{-8}$ of the energy difference, in Fig. \ref{fig:smalleta}, which justifies the derivation of an accurate formula for the
Coulombic energy.
In Fig. \ref{fig:zoom12}, the asymptotic relation (\ref{transitionItoII})
(dashed line) is tested against numerical minimization of
the energy (\ref{series1}) (solid curve) for small and intermediary
values of $\eta$, in the logarithmic scale.
The agreement is very good, not only for small $\eta$, but in the
whole range of stability of phase II (it will be shown in the next
subsection that phase II border is given by $\eta_2^c\simeq 0.26276\ldots$).
\subsection{Transition between phases II and III}
Going from phase I to phase II is not a phase transition in the usual sense.
However, the symmetry of the energy $E$ with respect to the transformation
$\Delta\to 1/\Delta$ has the fixed (self dual) point at $\Delta=1$ which is
the critical point of the phase transition from phase II to III.
Let us parameterize $\Delta$ as follows $\Delta=\exp(\epsilon)$.
The symmetry $\Delta\to 1/\Delta$ is now equivalent to
$\epsilon\to -\epsilon$, i.e. the energy $E$ is an even function
of $\epsilon$.
The expansion of $E$ around the critical point $\Delta=1$ $(\epsilon=0)$
in small deviation $\epsilon$ follows from the representation (\ref{series1}):
\begin{equation} \label{phase23}
\frac{E(e^{\epsilon},\eta)}{e^2\sqrt{n}} = \frac{E(1,\eta)}{e^2\sqrt{n}}
+ g_2(\eta) \epsilon^2
+ g_4(\eta) \epsilon^4 + {\cal O}(\epsilon^6) .
\end{equation}
The explicit form of $g_2(\eta)$ is
written in the Appendix and $g_4(\eta)$ is not presented due to lack of space,
but has been derived.
The energy (\ref{phase23}) has the Ginzburg-Landau form, $\epsilon$
being the order parameter.
In contrast to that mean field theory, the expression for our energy is exact.
\begin{figure} [t]
\begin{minipage}{80mm}
\includegraphics[width=1.00\textwidth]{zoom12.eps}
\caption{Going from phase I to II: The test of the asymptotic relation
(\ref{transitionItoII}) (dashed line) against numerical minimization of
the energy (\ref{series1}) (solid curve) for small and intermediary
values of $\eta$, in the logarithmic scale.}
\label{fig:zoom12}
\end{minipage}
\hfil
\begin{minipage}{80mm}
\includegraphics[width=1.00\textwidth]{zoom23.eps}
\caption{Transition between phases II and III: The test of
the asymptotic relation (\ref{transitionIItoIII}) (dashed line) against
numerical minimization of the energy (\ref{series1}) (solid curve),
in the logarithmic scale.}
\label{fig:zoom23}
\end{minipage}
\end{figure}
The critical point is associated with the vanishing of the prefactor
to $\epsilon^2$,
\begin{equation}
g_2(\eta)\Big\vert_{\eta=\eta_2^c} = 0 , \qquad \eta_2^c = 0.2627602682\ldots .
\end{equation}
The values of $\eta_2^c$ obtained in the previous studies were
$0.262$,\cite{Goldoni96}which is remarkably precise, $0.28$\cite{Weis01} and $0.27$.\cite{Lobaskin07}
The functions $g_2(\eta)$ and $g_4(\eta)$ exhibit the following expansions
around the critical $\eta_2^c$:
\begin{equation}
g_2(\eta) = - 0.4620982808\ldots (\eta_2^c-\eta) + {\cal O}((\eta_2^c-\eta)^2) ,
\qquad g_4(\eta) = 0.1054378203\ldots + {\cal O}(\eta_2^c-\eta) .
\end{equation}
The extremum (minimum) of the energy (\ref{phase23}) appears at
$\epsilon^*\simeq \Delta^* - 1$ given by the condition
\begin{equation} \label{eqn}
\frac{\partial}{\partial\epsilon} \frac{E(e^{\epsilon},\eta)}{e^2\sqrt{n}}
\Bigg\vert_{\epsilon=\epsilon^*} = 0 = 2 g_2(\eta) \epsilon^*
+ 4 g_4(\eta) {\epsilon^*}^3 .
\end{equation}
For $\eta<\eta_2^c$ (the ``ordered'' phase II), we have one trivial
solution $\epsilon^*=0$ which however provides the local maximum
of the energy.
There exist two conjugate nontrivial solutions which yield the
needed energy minimum; considering one of these solutions, we arrive at
\begin{equation} \label{transitionIItoIII}
\Delta^* - 1 \simeq \epsilon^* = \left(
- \frac{g_2(\eta)}{2 g_4(\eta)} \right)^{1/2}
\simeq 1.48031 \sqrt{\eta_2^c-\eta} .
\end{equation}
The critical index $\beta$, describing the growth of the order parameter
from its zero critical value via $\epsilon^*\propto (\eta_2^c-\eta)^{\beta}$,
has the mean field value $1/2$.
In Fig. \ref{fig:zoom23}, in the logarithmic scale, the asymptotic
relation (\ref{transitionIItoIII}) (dashed line) is compared with
the numerical minimization of the energy (\ref{series1}) (solid curve).
For $\eta>\eta_2^c$ (the ``disordered'' phase III), we have the only
solution to (\ref{eqn}) $\epsilon^*=0$ (or equivalently $\Delta^*=1$) ,
i.e. the rigid phase III is stable, up to a transition to phase IV
discussed in the next section.
The plot of the lattice aspect ratio $\Delta^*$ versus $\eta$, obtained by
the numerical minimization of the energy (\ref{series1}) in the whole
stability range of the phase II, is pictured by the solid curve
in Fig. \ref{fig:Deltavseta}.
$\Delta^*$ changes from $\sqrt{3}$ at $\eta=0$ to $1$ at $\eta=\eta_2^c$.
Numerical data of Goldoni and Peeters\cite{Goldoni96} (open circles)
are also presented for comparison.
The asymptotic relations (\ref{transitionItoII}) for $\eta\to 0$ and
(\ref{transitionIItoIII}) for $\eta\to \eta_2^c$ are also provided, for completeness.
\begin{figure} [t]
\begin{center}
\includegraphics[width=0.50\textwidth]{Deltavseta.eps}
\caption{The stability range of phase II: The plot of the lattice aspect
ratio $\Delta^*$ versus $\eta$, obtained by the numerical minimization of
the energy (\ref{series1}), is pictured by the solid curve.
Numerical data of Ref.\cite{Goldoni96} are presented
by open circles.
The asymptotic relations (\ref{transitionItoII}) for $\eta\to 0$ and
(\ref{transitionIItoIII}) for $\eta\to \eta_2^c$ are depicted by dashed curves.}
\label{fig:Deltavseta}
\end{center}
\end{figure}
\section{Phase IV} \label{sec:4}
In each of the two layers of phase IV (see Fig. \ref{fig:Structures}),
the elementary cell is the rhombus with angle $\varphi$ between
the primitive translation vectors
\begin{equation}
\bm{a}_1 = a (1,0) , \qquad \bm{a}_2 = a (\cos\varphi,\sin\varphi) \qquad
{\rm with}\ a=\frac{1}{\sqrt{\sigma\sin\varphi}} .
\end{equation}
The lattice spacing $a$ is determined by the electroneutrality condition
$n_l=\sigma$; there is just one particle per rhombus of the surface
$a^2 \sin\varphi$ and so $n_l=1/(a^2\sin\varphi)$.
The special case of $\varphi=\pi/2$ corresponds to phase III.
\subsection{Energy of phase IV}
As before, the energy per particle $E$ of the bilayer structure consists
of the intralayer and interlayer contributions, $E=E_{\rm intra}+E_{\rm inter}$.
As concerns the intralayer part, the 2D lattice vectors on one layer are
indexed with respect to a reference particle on the same layer by
${\bf r}(j,k) = j \bm{a}_1 + k \bm{a}_2$, where $j, k$ are any two integers
except for $(0,0)$.
The square of the lattice vector can be written as
\begin{equation}
\vert {\bf r}(j,k)\vert^2 = a^2 \left( j^2 + k^2 + 2 j k \cos\varphi \right)
= a^2 \left[ (j+k)^2 \cos^2(\varphi/2) + (j-k)^2 \sin^2(\varphi/2) \right] .
\end{equation}
This formula represents a kind of ``diagonalization'' of
$\vert {\bf r}(j,k)\vert^2$ in indices.
If $j+k$ is an even integer, we introduce new indices $n=(j+k)/2$ and
$m=(j-k)/2$ covering all integers except for $(n,m)\ne (0,0)$.
If $j+k$ is an odd integer, we introduce indices $n=(j+k+1)/2$ and
$n=(j-k+1)/2$ covering all integers.
Thus the interaction energy due to the Wigner crystal can be expressed as
\begin{eqnarray}
\frac{e^2}{2a} \sum_{j,k\atop (j,k)\ne (0,0)}
\frac{1}{\vert {\bf r}(j,k)\vert} & = &
\frac{e^2}{4a} \left[ \sum_{n,m\atop (n,m)\ne (0,0)}
\frac{1}{\sqrt{n^2\cos^2(\varphi/2) + m^2\sin^2(\varphi/2)}} \right.
\nonumber \\ & & \left. + \sum_{n,m}
\frac{1}{\sqrt{(n-1/2)^2\cos^2(\varphi/2) + (m-1/2)^2\sin^2(\varphi/2)}}
\right] .
\end{eqnarray}
Adding to this expression the interaction with the neutralizing background
and using the gamma identity (\ref{gamma}) in close analogy with
the previous section, we obtain
\begin{equation} \label{4intra}
\frac{E_{\rm intra}}{e^2\sqrt{n}} = \frac{1}{4\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}} \left\{
\left[ \theta_3(e^{-t\delta}) \theta_3(e^{-t/\delta}) - 1 - \frac{\pi}{t} \right]
+ \left[ \theta_2(e^{-t\delta}) \theta_2(e^{-t/\delta}) - \frac{\pi}{t} \right]
\right\} ,
\end{equation}
where $\delta = \tan(\varphi/2)$.
The Wigner lattices on the opposite layers are shifted with respect to
one another by the vector $(\bm{a}_1+\bm{a}_2)/2$.
To determine the interlayer contribution to the energy, we first consider
the square of the vector between the reference particle on one layer
and the vertices of the Wigner crystal on the other layer at distance $d$:
\begin{eqnarray}
\vert {\bf r}(j,k)\vert^2 & = & a^2 \left[ (j-1/2)^2 + (k-1/2)^2
+ 2 (j-1/2) (k-1/2) \cos\varphi \right] + d^2 \nonumber \\ & = &
a^2 \left[ (j+k-1)^2 \cos^2(\varphi/2) + (j-k)^2 \sin^2(\varphi/2)
+ (d/a)^2 \right] .
\end{eqnarray}
Thus the interaction energy with the Wigner crystal reads
\begin{eqnarray}
\frac{e^2}{2a} \sum_{j,k} \frac{1}{\vert {\bf r}(j,k)\vert} & = &
\frac{e^2}{4a} \left[ \sum_{n,m}
\frac{1}{\sqrt{(n-1/2)^2\cos^2(\varphi/2) + m^2\sin^2(\varphi/2)+d^2/(2a)^2}}
\right. \nonumber \\ & & \left. + \sum_{n,m}
\frac{1}{\sqrt{n^2\cos^2(\varphi/2) + (m-1/2)^2\sin^2(\varphi/2)+d^2/(2a)^2}}
\right] .
\end{eqnarray}
Adding the background term and using the gamma identity, we find that
\begin{equation} \label{4inter}
\frac{E_{\rm inter}}{e^2\sqrt{n}} = \frac{1}{4\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}} e^{-\eta^2 t/2} \left\{
\left[ \theta_3(e^{-t\delta}) \theta_2(e^{-t/\delta}) - \frac{\pi}{t} \right]
+ \left[ \theta_2(e^{-t\delta}) \theta_3(e^{-t/\delta}) - \frac{\pi}{t} \right]
\right\} .
\end{equation}
The total energy per particle $E$ is the sum of (\ref{4intra})
and (\ref{4inter}).
Note the invariance of $E$ with respect to the transformation
$\delta\to 1/\delta$.
With respect to the definition of $\delta = \tan(\varphi/2)$,
this symmetry is equivalent to the obvious one $\varphi\to \pi-\varphi$.
Following subsequently similar lines as in previous sections,
the integral range $[0,\infty]$ can be reduced to $[0,\pi]$ by using
the Poisson formula (\ref{Poisson}),
\begin{eqnarray}
\frac{E(\delta,\eta)}{e^2\sqrt{n}} & = & \frac{1}{4\sqrt{\pi}}
\int_0^{\pi} \frac{dt}{\sqrt{t}} \Bigg\{
2 \left[ \theta_3(e^{-t\delta}) \theta_3(e^{-t/\delta}) - 1 - \frac{\pi}{t} \right]
+ \left[ \theta_2(e^{-t\delta}) \theta_2(e^{-t/\delta}) - \frac{\pi}{t} \right]
+ \left[ \theta_4(e^{-t\delta}) \theta_4(e^{-t/\delta}) - 1 \right]
\nonumber \\ & & + e^{-\eta^2 t/2}
\left[ \theta_3(e^{-t\delta}) \theta_2(e^{-t/\delta}) - \frac{\pi}{t} \right]
+ e^{-(\pi\eta)^2/(2t)}
\left[ \theta_3(e^{-t\delta}) \theta_4(e^{-t/\delta}) - 1 \right]
\nonumber \\ & & + e^{-\eta^2 t/2}
\left[ \theta_2(e^{-t\delta}) \theta_3(e^{-t/\delta}) - \frac{\pi}{t} \right]
+ e^{-(\pi\eta)^2/(2t)}
\left[ \theta_4(e^{-t\delta}) \theta_3(e^{-t/\delta}) - 1 \right] \Bigg\} .
\end{eqnarray}
Applying again the Poisson transformation formula (\ref{Poisson}) to the
series representations of Jacobi theta functions, the singular $t\to 0$
terms are canceled explicitly and we end up with the representation
of the energy per particle $E$ in terms of $z$-functions defined in (\ref{z}):
\begin{eqnarray} \label{energy4}
\frac{E(\delta,\eta)}{e^2\sqrt{n}} & = & \frac{1}{2\sqrt{\pi}}
\Bigg\{ \sum_{j=1}^{\infty} [2+(-1)^j]
\left[ z_{3/2}(0,j^2/\delta) + z_{3/2}(0,j^2\delta) \right]
+ 2 \sum_{j,k=1}^{\infty} [2+(-1)^j (-1)^k] z_{3/2}(0,j^2/\delta+k^2\delta)
\nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} z_{3/2}(0,(j-1/2)^2/\delta+(k-1/2)^2\delta)
+ \sum_{j=1}^{\infty} [1+(-1)^j] \left[
z_{3/2}((\pi\eta)^2/2,j^2/\delta) + z_{3/2}((\pi\eta)^2/2,j^2\delta) \right]
\nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} [(-1)^j+(-1)^k] z_{3/2}((\pi\eta)^2/2,j^2/\delta+k^2\delta)
\nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} \left[ z_{3/2}(0,\eta^2/2+(j-1/2)^2/\delta+k^2\delta)
+ z_{3/2}(0,\eta^2/2+(j-1/2)^2\delta+k^2/\delta) \right]
\nonumber \\ & &
+ \sum_{j=1}^{\infty} \left[ z_{3/2}(0,\eta^2/2+(j-1/2)^2/\delta)
+ z_{3/2}(0,\eta^2/2+(j-1/2)^2\delta) \right]
- 3 \sqrt{\pi} - \pi z_{1/2}(0,\eta^2/2) \Bigg\} . \label{series2}
\end{eqnarray}
\subsection{Transition between phases III and IV}
The symmetry of the energy $E$ with respect to the transformation
$\delta\to 1/\delta$ has the fixed point at $\delta=1$ which is
the critical point of the phase transition between the phases III and IV.
Parameterizing $\delta$ as $\delta\equiv \tan(\varphi/2) = \exp(-\epsilon)$,
the symmetry takes form $\epsilon\to -\epsilon$ and the energy $E$
is an even function of $\epsilon$.
The expansion of $E$ around the critical point $\delta=1$ (equivalent to
$\theta=\pi/2$ or $\epsilon=0$) in small $\epsilon$ follows from
the representation (\ref{series2}):
\begin{equation} \label{phase34}
\frac{E(e^{-\epsilon},\eta)}{e^2\sqrt{n}} = \frac{E(1,\eta)}{e^2\sqrt{n}}
+ h_2(\eta) \epsilon^2
+ h_4(\eta) \epsilon^4 + {\cal O}(\epsilon^6) .
\end{equation}
The explicit form of $h_2(\eta)$ is
presented in the Appendix; The expression for $h_4(\eta)$ is too lengthy
to be given, but is at our disposal.
The critical point is associated with the vanishing of the prefactor
of $\epsilon^2$,
\begin{equation}
h_2(\eta)\Big\vert_{\eta=\eta_3^c} = 0 , \qquad \eta_3^c = 0.6214809246\ldots .
\end{equation}
The values of $\eta_3^c$ obtained in the previous studies are
$0.622$,\cite{Goldoni96} $0.59$\cite{Weis01} and $0.62$.\cite{Lobaskin07}
The functions $h_2(\eta)$ and $h_4(\eta)$ exhibit the following expansions
around the critical $\eta_3^c$:
\begin{equation}
h_2(\eta) = - 0.2675826391\ldots (\eta-\eta_3^c) + {\cal O}((\eta-\eta_3^c)^2) ,
\qquad h_4(\eta) = 0.0863245072\ldots + {\cal O}(\eta-\eta_3^c) .
\end{equation}
The extremum (minimum) of the energy (\ref{phase34}) appears at
$\epsilon^*\simeq \pi/2 - \varphi^*$ given by the condition
\begin{equation}
\frac{\partial}{\partial\epsilon} \frac{E(e^{-\epsilon},\eta)}{e^2\sqrt{n}}
\Bigg\vert_{\epsilon=\epsilon^*} = 0 = 2 h_2(\eta) \epsilon^*
+ 4 h_4(\eta) {\epsilon^*}^3 .
\end{equation}
For $\eta<\eta_3^c$ (``disordered'' phase III), we have the only
solution $\epsilon^*=0$ (or equivalently $\varphi^*=\pi/2$) which provides
the energy minimum, i.e. the rigid phase III is stable.
For $\eta>\eta_3^c$ (``ordered'' phase IV), the trivial solution
$\epsilon^*=0$ becomes unstable.
The couple of conjugate nontrivial solutions, which provide
the energy minimum, implies
\begin{equation} \label{transitionIIItoIV}
1-\delta^* \simeq \epsilon^* \simeq \frac{\pi}{2} - \varphi^* = \left( -
\frac{h_2(\eta)}{2 h_4(\eta)} \right)^{1/2} \simeq 1.24494 \sqrt{\eta-\eta_3^c} .
\end{equation}
The critical index $\beta$ has again the mean field value $1/2$.
In Fig. \ref{fig:zoom34}, in a log-log scale,
the asymptotic relation (\ref{transitionIIItoIV}) (dashed line) is tested
against numerical minimization of the energy (\ref{series2}) (solid curve).
In the upper inset, we show the dependence of the energy on the logarithm
of the angle parameter $\delta=\tan(\varphi/2)$ for $\eta=0.5$, where
the phase III with $\delta=1$ is stable.
In the lower inset, the analogous plot is presented for $\eta=0.7$,
where the phase IV with $\delta\ne 1$ is stable; phase III
corresponds in fact to a local maximum of the energy.
Note the symmetry of the energy with respect to the transformation
$\delta\to 1/\delta$ or, equivalently, $\ln\delta\to -\ln\delta$.
The plot of the angle parameter $\delta^*=\tan(\varphi^*/2)$ versus $\eta$,
obtained by numerical minimization of the energy (\ref{series1}) in the whole
stability range of the phase IV, is displayed by the solid curve
in Fig. \ref{fig:deltavseta}.
$\delta^*$ changes from $1$ at $\eta=\eta_3^c$ (transition point from phase
III to IV) to $\delta^c=0.69334\ldots$ at $\eta=\eta_4^c$ (transition point
from phase IV to V, see the next section).
Numerical data of Goldoni and Peeters\cite{Goldoni96} (open circles)
are presented for comparison.
The asymptotic relation (\ref{transitionIIItoIV}) for $\eta\to \eta_3^c$
is depicted by the dashed curve.
\begin{figure} [t]
\begin{minipage}{80mm}
\includegraphics[width=1.00\textwidth]{zoom34.eps}
\caption{Transition between phases III and IV: The test of
the asymptotic relation (\ref{transitionIIItoIV}) (dashed line) against
a numerical minimization of the energy (\ref{series2}) (solid curve),
in the logarithmic scale.
The content of upper and lower insets is commented in the text.}
\label{fig:zoom34}
\end{minipage}
\hfil
\begin{minipage}{80mm}
\includegraphics[width=1.00\textwidth]{deltavseta.eps}
\caption{
Stability range of phase IV: The plot of the angle parameter $\delta^*$
versus $\eta$, obtained by the numerical minimization of the energy
(\ref{series2}), is shown by the solid curve.
Numerical data of Ref.\cite{Goldoni96} are presented
by open circles.
The asymptotic relation (\ref{transitionIIItoIV}) for $\eta\to \eta_3^c$
is depicted by the dashed curve.}
\label{fig:deltavseta}
\end{minipage}
\end{figure}
\section{Phase V} \label{sec:5}
In a single layer of the phase V (see Fig. \ref{fig:Structures}),
the elementary cell of the hexagonal lattice is the rhombus with
the angle $\pi/3$ between the primitive translation vectors
\begin{equation}
\bm{a}_1 = a (1,0) , \qquad \bm{a}_2 = \frac{a}{2} (1,\sqrt{3}) \qquad
{\rm with}\ a=\frac{\sqrt{2}}{3^{1/4}} \frac{1}{\sqrt{\sigma}} .
\end{equation}
The lattice spacing $a$ follows from the electroneutrality condition
$n_l=\sigma$; there is just one particle per rhombus of surface
$\sqrt{3} a^2/2$, so that $n_l=2/(\sqrt{3} a^2)$.
Note that the images of vertices on the opposite layer are localized in
the center of triangles and not rhombuses, as was the case of phase IV.
There is no continuous way to pass from phase IV to phase V.
\subsection{Energy of phase V}
To study the intralayer contribution to the energy of the reference particle
at the origin, we first consider the Wigner crystal of lattice vectors
${\bf r}(j,k)=j \bm{a}_1 + k \bm{a}_2$, where integers $(j,k)\ne (0,0)$.
The square of the lattice vector is expressible as
\begin{equation}
\vert {\bf r}(j,k)\vert^2 = a^2 \left( j^2 + k^2 + j k \right)
= \frac{a^2}{4} \left[ 3 (j+k)^2 + (j-k)^2 \right] .
\end{equation}
In analogy with phase IV, we introduce new $n,m$ indices for each of
the cases $j+k$ being an even and odd integer.
The interaction energy due to the Wigner crystal then reads
\begin{equation}
\frac{e^2}{2a} \sum_{j,k\atop (j,k)\ne (0,0)} \frac{1}{\vert {\bf r}(j,k)\vert}
= \frac{e^2}{2a} \left[ \sum_{n,m\atop (n,m)\ne (0,0)} \frac{1}{\sqrt{3 n^2+ m^2}}
+ \sum_{n,m} \frac{1}{\sqrt{3(n-1/2)^2 + (m-1/2)^2}} \right] .
\end{equation}
Adding to this expression the interaction with the neutralizing background
and using the gamma identity, we find
\begin{equation} \label{5intra}
\frac{E_{\rm intra}}{e^2\sqrt{n}} = \frac{1}{4\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}} \left\{
\left[ \theta_3(e^{-t/\sqrt{3}}) \theta_3(e^{-t\sqrt{3}}) - 1 - \frac{\pi}{t} \right]
+ \left[ \theta_2(e^{-t/\sqrt{3}}) \theta_2(e^{-t\sqrt{3}}) - \frac{\pi}{t} \right]
\right\} .
\end{equation}
The hexagonal lattices on the opposite layers are shifted with respect to
one another by the vector $(\bm{a}_1+\bm{a}_2)/3$; note that the factor $1/3$
differs from $1/2$ of the previous phases I-IV.
To determine the interlayer contribution to the energy, we first consider
the square of the vector between the reference particle on one layer
and the vertices of the Wigner crystal on the other layer at distance $d$:
\begin{equation}
\vert {\bf r}(j,k)\vert^2 = a^2
\left[ (j+1/3)^2 + (k+1/3)^2 + (j+1/3) (k+1/3) \right] + d^2 =
\frac{a^2}{4} \left[ 3 (j+k+2/3)^2 + (j-k)^2 \right] + d^2 .
\end{equation}
Going from $(j,k)$ to integers $(n,m)$, the interaction energy with
the Wigner crystal on the opposite side takes the form
\begin{equation}
\frac{e^2}{2a} \sum_{j,k} \frac{1}{\vert {\bf r}(j,k)\vert} =
\frac{e^2}{2a} \left[ \sum_{n,m} \frac{1}{\sqrt{3 (n+1/3)^2 + m^2+(d/a)^2}}
+ \sum_{n,m} \frac{1}{\sqrt{3 (n-1/6)^2 + (m-1/2)^2 +(d/a)^2}} \right] .
\end{equation}
Adding the background term, using the gamma identity and the readily
derivable relations
\begin{equation}
\sum_j e^{-3t(j+1/3)^2} = \frac{1}{2} \left[
\theta_3(e^{-t/3}) - \theta_3(e^{-3t}) \right] , \qquad
\sum_j e^{-3t(j-1/6)^2} = \frac{1}{2} \left[
\theta_2(e^{-t/3}) - \theta_2(e^{-3t}) \right] ,
\end{equation}
we find that
\begin{eqnarray}
\frac{E_{\rm inter}}{e^2\sqrt{n}} & = & \frac{1}{4\sqrt{\pi}}
\int_0^{\infty} \frac{dt}{\sqrt{t}}
\left( -\frac{1}{2} e^{-\eta^2 t/2} + \frac{\sqrt{3}}{2} e^{-3\eta^2 t/2} \right)
\nonumber \\ & & \times \left\{
\left[ \theta_3(e^{-t/\sqrt{3}}) \theta_3(e^{-t\sqrt{3}}) - 1 - \frac{\pi}{t} \right]
+ \left[ \theta_2(e^{-t/\sqrt{3}}) \theta_2(e^{-t\sqrt{3}}) - \frac{\pi}{t} \right]
\right\} . \label{5inter}
\end{eqnarray}
The total energy per particle $E$ is given by the sum of (\ref{5intra})
and (\ref{5inter}).
The Poisson formula (\ref{Poisson}) enables us to reduce the integral
range to $[0,\pi]$,
\begin{eqnarray}
\frac{E(\eta)}{e^2\sqrt{n}} & = & \frac{1}{4\sqrt{\pi}}
\int_0^{\pi} \frac{dt}{\sqrt{t}} \Bigg\{
\left( 1 - \frac{1}{2} e^{-\eta^2 t/2} + \frac{\sqrt{3}}{2} e^{-3\eta^2 t/2} \right)
\left[ \theta_3(e^{-t/\sqrt{3}}) \theta_3(e^{-t\sqrt{3}}) - 1 - \frac{\pi}{t} \right]
\nonumber \\ & & +
\left( 1 - \frac{1}{2} e^{-(\pi\eta)^2/(2t)}
+ \frac{\sqrt{3}}{2} e^{-3(\pi\eta)^2/(2t)} \right)
\left[ \theta_3(e^{-t/\sqrt{3}}) \theta_3(e^{-t\sqrt{3}}) - 1 - \frac{\pi}{t} \right]
\nonumber \\ & & +
\left( 1 - \frac{1}{2} e^{-\eta^2 t/2} + \frac{\sqrt{3}}{2} e^{-3\eta^2 t/2} \right)
\left[ \theta_2(e^{-t/\sqrt{3}}) \theta_2(e^{-t\sqrt{3}}) - \frac{\pi}{t} \right]
\nonumber \\ & & +
\left( 1 - \frac{1}{2} e^{-(\pi\eta)^2/(2t)}
+ \frac{\sqrt{3}}{2} e^{-3(\pi\eta)^2/(2t)} \right)
\left[ \theta_4(e^{-t/\sqrt{3}}) \theta_4(e^{-t\sqrt{3}}) - 1 \right] \Bigg\} .
\label{eq:59}
\end{eqnarray}
In terms of the functions
\begin{eqnarray}
I_2(x,y) & \equiv & \int_0^{\pi} \frac{dt}{\sqrt{t}} e^{-x t/\pi^2} e^{-y \pi^2/t}
\left[ \theta_2(e^{-t/\sqrt{3}}) \theta_2(e^{-t\sqrt{3}}) - \frac{\pi}{t} \right]
\nonumber \\ & = & 2 \sum_{j=1}^{\infty} (-1)^j
\left[ z_{3/2}(x,y+j^2/\sqrt{3}) + z_{3/2}(x,y+j^2\sqrt{3}) \right]
+ 4 \sum_{j,k=1}^{\infty} (-1)^j (-1)^k z_{3/2}(x,y+j^2/\sqrt{3}+k^2\sqrt{3}) ,
\phantom{a}
\end{eqnarray}
\begin{eqnarray}
I_3(x,y) & \equiv & \int_0^{\pi} \frac{dt}{\sqrt{t}} e^{-x t/\pi^2} e^{-y \pi^2/t}
\left[ \theta_3(e^{-t/\sqrt{3}}) \theta_3(e^{-t\sqrt{3}}) -1-\frac{\pi}{t} \right]
\nonumber \\ & = & 2 \sum_{j=1}^{\infty}
\left[ z_{3/2}(x,y+j^2/\sqrt{3}) + z_{3/2}(x,y+j^2\sqrt{3}) \right]
+ 4 \sum_{j,k=1}^{\infty} z_{3/2}(x,y+j^2/\sqrt{3}+k^2\sqrt{3}) - \pi z_{1/2}(x,y) ,
\end{eqnarray}
\begin{eqnarray}
I_4(x,y) & \equiv & \int_0^{\pi} \frac{dt}{\sqrt{t}} e^{-x t/\pi^2} e^{-y \pi^2/t}
\left[ \theta_4(e^{-t/\sqrt{3}}) \theta_4(e^{-t\sqrt{3}}) -1 \right]
\nonumber \\ & = & 4 \sum_{j,k=1}^{\infty}
z_{3/2}(x,y+(j-1/2)^2/\sqrt{3}+(k-1/2)^2\sqrt{3}) - \pi z_{1/2}(x,y) ,
\end{eqnarray}
the energy per particle $E$ is expressible as
\begin{eqnarray} \label{energy5}
\frac{E(\eta)}{e^2\sqrt{n}} & = & \frac{1}{4\sqrt{\pi}} \Bigg\{ \left[
2 I_3(0,0) - \frac{1}{2} I_3((\pi\eta)^2/2,0) - \frac{1}{2} I_3(0,\eta^2/2)
+ \frac{\sqrt{3}}{2} I_3(3(\pi\eta)^2/2,0)
+ \frac{\sqrt{3}}{2} I_3(0,3\eta^2/2) \right] \nonumber \\ & &
+ \left[ I_2(0,0) - \frac{1}{2} I_2((\pi\eta)^2/2,0)
+ \frac{\sqrt{3}}{2} I_2(3(\pi\eta)^2/2,0) \right]
+ \left[ I_4(0,0) - \frac{1}{2} I_4(0,\eta^2/2)
+ \frac{\sqrt{3}}{2} I_4(0,3\eta^2/2) \right] \Bigg\} .
\end{eqnarray}
\subsection{Transition between phases IV and V}
Increasing $\eta$ from $\eta_3^c$, phase IV is stable up to the point
$\eta_4^c$ at which the energy of phase IV (\ref{energy4}), evaluated
at $\delta^*$ which minimizes this energy, equals to the energy of
phase V (\ref{energy5}).
Our result is
\begin{equation}
\eta_4^c = 0.73242\ldots .
\end{equation}
The values of $\eta_4^c$ obtained in the previous studies were relatively dispersed:
$0.732$,\cite{Goldoni96} $0.70$\cite{Weis01} and $0.87$.\cite{Lobaskin07}
The phase transition is of first order since the energies of phases IV
and V have as functions of $\eta$ different slopes which causes
the discontinuity of the first derivative of the energy with respect to
$\eta$ at the transition point $\eta_4^c$.
The angle parameter $\delta$, which minimizes the energy of the phase IV
at the critical point $\eta_4^c$, is found to be $\delta^c=0.69334\ldots$.
Since $\delta = \tan(\varphi/2)$, the corresponding angle
$\varphi^c = 69.4702\ldots^{\circ}$; this angle is very close to the
estimate $\varphi^c = 69.48^{\circ}$ of the work.\cite{Goldoni96}
Going from phase IV to V, the angle skips to $60^{\circ}$ as is indicated
in Fig. \ref{fig:deltavseta}.
\subsection{Discussion}
Two different ``sum-rules'' can be derived, that allow for a critical assessment
of the results obtained. The simplest one relies on the geometrical
proximity between structures I (a single hexagonal crystal) and V
(two hexagonal crystals at half density). For large distances, the two
crystals decouple and we have, making use of straightforward notations,
\begin{equation}
E_{I}(\sqrt{3},\eta=0) = \sqrt{2} \, E_{V}(\eta \to \infty).
\end{equation}
With our series representations (\ref{series1}) for $E_{I}$
and (\ref{energy5}) for $E_{V}$, this identity holds.
Another more subtle constraint follows from a combination of elementary geometric considerations,\cite{Rouzina}
which impose that
\begin{equation}
E_{V}(\eta=0) \,=\, \frac{1+\sqrt{3}}{2\sqrt{2}} \, E_{I}\left(\sqrt{3}, \eta=0\right).
\end{equation}
We have also checked that this identity holds with the expressions provided above
(note though that $\eta=0$ lies outside the stability range of structure V).
Finally, it is interesting to consider both the large small and
large distance behavior
of the energy. For small $\eta$, it can be shown that both
structures I and II share the same energy expansion,
up to order $\eta^3$ included:
\begin{equation}
E_{II}(\Delta^*,\eta) = E_I(\eta) + {\cal O}(\eta^4)
\end{equation}
where $\Delta^*$ is the previously introduced optimal
aspect ratio that minimizes $E_{II}(\Delta^*,\eta)$ for a
given $\eta$. Explicit calculation up to order $\eta^2$
shows that
\begin{equation}
\frac{E_{II}(\Delta^*,\eta)}{e^2 \sqrt{n}} \, = \,
\frac{E_{II}(\sqrt{3},0)}{e^2 \sqrt{n}} \,+\,
\frac{\pi \,\eta}{\sqrt{2}} \,- 2.59372\ldots \, \eta^2 +{\cal O}(\eta^3),
\label{eq:smalletaexp}
\end{equation}
where the precise value of $E_{II}(\sqrt{3},0)=E_I(0)$ has been given in Eq.
(\ref{eq:E0}). We note that the linear term in (\ref{eq:smalletaexp})
generates a contact pressure $-2\pi \sigma^2 e^2$
for $\eta\to 0$. A similar term was reported in,\cite{Goldoni96}
where however the term in $\eta^2$ differs from ours by a large factor
(0.2122 instead of 2.5937)
At large distances, the relevant phase is structure V, from which
the inter-plate pressure follows. The large $\eta$ case
is encoded in the small $t$ limit of Eq. (\ref{eq:59}), or equivalently,
Eq. (\ref{5inter}). A saddle point argument leads to
\begin{equation}
\frac{E_{V}(\eta)}{e^2 \sqrt{n}} \, \sim \, \frac{E_{V}(\infty)}{e^2 \sqrt{n}} \,
- \, \frac{3^{5/4}}{4} \, \exp\left( -\frac{4 \pi}{\sqrt{2}\, 3^{1/4}} \, \eta
\right).
\end{equation}
We recover an expression already obtained in,\cite{Goldoni96}
at variance with other
approaches.\cite{EsKa95} Taking the $\eta$ derivative
and remembering that $n=2\sigma$
yields the inter-plate pressure
\begin{equation}
P \, = \, -2\,\sigma \frac{\partial E_{V}}{\partial d}
\,=\, -2\sigma^{3/2} \frac{\partial E_{V}}{\partial \eta}
\,\sim\, -6 \pi (\sigma e)^2 \,
\exp
\left(-\frac{4 \pi}{\sqrt 2 \, 3^{1/4}} \, \eta\right).
\end{equation}
The $\eta$ dependence is well known, since it can be written $\exp(-G_0 d)$,
with $G_0$ the modulus of the first reciprocal lattice vector.
It should be noted though that the prefactor differs from the often
reported $2\pi (\sigma e)^2$ (see e.g. \cite{LaLP00}), by a
factor 3.
\section{Conclusion} \label{sec:conclusion}
The system of classical charged particles, forming a sequence of
bilayer Wigner structures at zero temperature as the distance between
the plates is increasing, has a rather long history.
We have presented here a new method to calculate the Coulomb ground-state
energy of each Wigner structure.
Based on a series of transformations and using general properties of
the Jacobi theta functions, we expressed the energies in terms of
quickly converging series of the functions $z_{\nu}(x,y)$ defined in (\ref{z}).
The presence of the neutralizing background manifests itself simply as
the subtraction of singularities of the Jacobi theta functions under
an auxiliary integration.
\begin{figure} [htb]
\includegraphics[width=0.52\textwidth]{Summary.eps}
\caption{Summary of phase transition scenario. The rounded off values of the
thresholds are mentioned. Structure I is realized at $\eta=0$ only
(vanishing inter-plate separation).}
\label{fig:summary}
\end{figure}
Numerical evaluation of the series requires modest computer and programming
facilities, and at the same time provides extremely accurate estimates of
the energy. We took advantage of this feature, supplemented by analytical
work, to improve and complete previous studies
in three aspects:
\begin{itemize}
\item
There was a relatively large dispersion in the determination of
the transition points between phases; this concerns especially
the first-order phase transition between phases IV and V.
Our method improves significantly the location of all transition points,
that can be worked out with arbitrary precision.
Figure \ref{fig:summary} gives an overview of the sequence
of phases together with the corresponding thresholds.
\item
We resolved, analytically and numerically, a previous
controversy about the stability phase I, thereby corroborating
the findings of Ref.\cite{Messina03}
We found that this phase is stable only at zero distance between the
plates, $\eta=0=d$.
To confirm numerically this result, we worked with extremely small values
of the energy differences $\propto 10^{-12}-10^{-8}$ for distances
$\eta=10^{-3}$ and $10^{-2}$ (see Fig. \ref{fig:smalleta}), which are
``invisible'' by standard numerical methods.
The agreement between the $\eta\to 0$ asymptotic relation
(\ref{transitionItoII}), calculated analytically by using
the Taylor expansions of the functions $z_{\nu}(x,y)$, and
the numerical minimization of the energy, presented in Fig. \ref{fig:zoom12},
is excellent.
\item
The expansions of the structure energies around second-order transition
points can be done analytically which enables us to specify the critical
phenomena at the phase transition points; see the expansions pertaining
to the transitions from phase II to III
(\ref{transitionIItoIII}) and from phase III to IV (\ref{transitionIIItoIV}).
The agreement between these analytic formulas and numerical minimization
of the ground-state energy is very good; It can be appreciated in Figs. \ref{fig:zoom23} and
\ref{fig:zoom34}. Quite expectedly for a zero temperature situation,
the critical behavior is always of the Ginzburg-Landau type,
with the mean-field critical index $\beta=1/2$ for the growth
of the order parameters in the ``ordered'' phases.
\end{itemize}
It is clear that our method can be directly generalized to other problems
concerning the lattice summations over pair interactions, not only
the Coulomb ones.
The bilayers with repulsive Yukawa interactions, extensively studied in
the past,\cite{Messina03,Mazars11} or with inverse power laws,\cite{Mazars10} deserve our attention.
We additionally emphasize that the ground states under consideration here are such that the ions are distributed
evenly (50\% on each plate): in other words, the ionic surface density of
one Wigner crystal on a given plate is $\sigma$, and coincides with the plate
homogeneous surface
density. When dealing with asymmetric plates, this local neutrality
assumption should be relaxed,\cite{Messina00,Messina09}
still enforcing global neutrality.
This makes the asymmetric problem significantly more complex,
and an interesting perspective
for future work. Finally, consideration of dielectric jumps between the walls
and the interstitial slab is also a relevant venue for forthcoming
investigations.
\begin{acknowledgments}
We would like to thank M. Mazars and C. Texier for useful discussions.
L. \v{S}. is grateful to LPTMS for their kind hospitality.
The support received from the grants VEGA No. 2/0049/2012 and CE-SAS QUTE
is acknowledged.
\end{acknowledgments}
\renewcommand{\theequation}{A\arabic{equation}}
\setcounter{equation}{0}
\section*{APPENDIX}
The prefactor functions $f_1(\eta)$ and $f_2(\eta)$ of
the expansion (\ref{epsilon1}) read
\begin{eqnarray}
f_1(\eta) & = & \frac{1}{2^{3/2}\sqrt{\pi}} \Bigg\{ 4 \sum_{j=1}^{\infty} j^2
\left[ z_{5/2}(0,j^2\sqrt{3}) - \frac{1}{3} z_{5/2}(0,j^2/\sqrt{3}) \right]
+ 8 \sum_{j,k=1}^{\infty} \left( k^2 - \frac{j^3}{3} \right)
z_{5/2}(0,j^2/\sqrt{3}+k^2\sqrt{3}) \nonumber \\ & &
+ 2 \sum_{j=1}^{\infty} (-1)^j j^2 \left[ z_{5/2}((\pi\eta)^2,j^2\sqrt{3})
- \frac{1}{3} z_{5/2}((\pi\eta)^2,j^2/\sqrt{3}) \right] \nonumber \\ & &
+ 4 \sum_{j,k=1}^{\infty} (-1)^j (-1)^k \left( k^2 - \frac{j^2}{3} \right)
z_{5/2}((\pi\eta)^2,j^2/\sqrt{3}+k^2\sqrt{3}) \nonumber \\ & &
+ 4 \sum_{j,k=1}^{\infty} \left[ \left( k-\frac{1}{2} \right)^2 -
\frac{1}{3} \left( j-\frac{1}{2} \right)^2 \right]
z_{5/2}(0,\eta^2+(j-1/2)^2/\sqrt{3}+(k-1/2)^2\sqrt{3}) \Bigg\} ,
\end{eqnarray}
\begin{eqnarray}
f_2(\eta) & = & \frac{1}{2^{3/2}\sqrt{\pi}} \Bigg\{ 4 \sum_{j=1}^{\infty}
\left[ \frac{j^4}{2} z_{7/2}(0,j^2\sqrt{3}) +
\frac{j^4}{18} z_{7/2}(0,j^2/\sqrt{3})
- \frac{j^2}{3^{3/2}} z_{5/2}(0,j^2/\sqrt{3}) \right] \nonumber \\ & &
+ 8 \sum_{j,k=1}^{\infty} \left[ \frac{1}{2} \left( k^2 - \frac{j^3}{3} \right)^2
z_{7/2}(0,j^2/\sqrt{3}+k^2\sqrt{3}) - \frac{j^2}{3^{3/2}}
z_{5/2}(0,j^2/\sqrt{3}+k^2\sqrt{3}) \right] \nonumber \\ & &
+ 2 \sum_{j=1}^{\infty} (-1)^j \left[
\frac{j^4}{2} z_{7/2}((\pi\eta)^2,j^2\sqrt{3})
+ \frac{j^4}{18} z_{7/2}((\pi\eta)^2,j^2/\sqrt{3})
- \frac{j^2}{3^{3/2}} z_{5/2}((\pi\eta)^2,j^2/\sqrt{3}) \right] \nonumber \\ & &
+ 4 \sum_{j,k=1}^{\infty} (-1)^j (-1)^k \left[
\frac{1}{2} \left( k^2 - \frac{j^3}{3} \right)^2
z_{7/2}((\pi\eta)^2,j^2/\sqrt{3}+k^2\sqrt{3})
- \frac{j^2}{3^{3/2}} z_{5/2}((\pi\eta)^2,j^2/\sqrt{3}+k^2\sqrt{3}) \right]
\nonumber \\ & & + 4 \sum_{j,k=1}^{\infty}
\Bigg[ \frac{1}{2} \left[ \left( k-\frac{1}{2} \right)^2 -
\frac{1}{3} \left( j-\frac{1}{2} \right)^2 \right]^2
z_{7/2}(0,\eta^2+(j-1/2)^2/\sqrt{3}+(k-1/2)^2\sqrt{3}) \nonumber \\ & &
- \frac{1}{3^{3/2}} \left( j - \frac{1}{2} \right)^2
z_{5/2}(0,\eta^2+(j-1/2)^2/\sqrt{3}+(k-1/2)^2\sqrt{3}) \Bigg] \Bigg\} .
\end{eqnarray}
The prefactor function $g_2(\eta)$ of the expansion (\ref{phase23}) takes the form
\begin{eqnarray}
g_2(\eta) & = & \frac{1}{\sqrt{2\pi}} \Bigg\{ 2 \sum_{j=1}^{\infty}
\left[ j^4 z_{7/2}(0,j^2) - j^2 z_{5/2}(0,j^2) \right] \nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} \left[ (j^2-k^2)^2 z_{7/2}(0,j^2+k^2)
- (j^2+k^2) z_{5/2}(0,j^2+k^2) \right] \nonumber \\ & &
+ \sum_{j=1}^{\infty} (-1)^j \left[ j^4 z_{7/2}((\pi\eta)^2,j^2)
- j^2 z_{5/2}((\pi\eta)^2,j^2) \right] \nonumber \\ & &
+ \sum_{j,k=1}^{\infty} (-1)^j (-1)^k
\left[ (j^2-k^2)^2 z_{7/2}((\pi\eta)^2,j^2+k^2)
- (j^2+k^2) z_{5/2}((\pi\eta)^2,j^2+k^2) \right] \nonumber \\ & &
+ \sum_{j,k=1}^{\infty} \left( \left[ (j-1/2)^2-(k-1/2)^2 \right]^2
z_{7/2}(0,\eta^2+(j-1/2)^2+(k-1/2)^2) \right. \nonumber \\ & & \left.
- \left[ (j-1/2)^2+(k-1/2)^2 \right]
z_{5/2}(0,\eta^2+(j-1/2)^2+(k-1/2)^2) \right) \Bigg\} .
\end{eqnarray}
Finally, the prefactor function $h_2(\eta)$ of the expansion (\ref{phase34})
can be written
\begin{eqnarray}
h_2(\eta) & = & \frac{1}{2\sqrt{\pi}} \Bigg\{
2 \sum_{j=1}^{\infty} \left[ j^4 z_{7/2}(0,j^2) - j^2 z_{5/2}(0,j^2) \right]
+ \sum_{j=1}^{\infty} (-1)^j \left[ j^4 z_{7/2}(0,j^2) - j^2 z_{5/2}(0,j^2) \right]
\nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} \left[ (j^2-k^2)^2 z_{7/2}(0,j^2+k^2)
- (j^2+k^2) z_{5/2}(0,j^2+k^2) \right] \nonumber \\ & &
+ \sum_{j,k=1}^{\infty} (-1)^j (-1)^k \left[ (j^2-k^2)^2 z_{7/2}(0,j^2+k^2)
- (j^2+k^2) z_{5/2}(0,j^2+k^2) \right] \nonumber \\ & &
+ \sum_{j,k=1}^{\infty} \Big( \left[ (j-1/2)^2-(k-1/2)^2 \right]^2
z_{7/2}(0,(j-1/2)^2+(k-1/2)^2) \nonumber \\ & &
- \left[ (j-1/2)^2+(k-1/2)^2 \right]
z_{5/2}(0,(j-1/2)^2+(k-1/2)^2) \Big) \nonumber \\ & &
+ \sum_{j=1}^{\infty} \left[ j^4 z_{7/2}((\pi\eta)^2/2,j^2)
- j^2 z_{5/2}((\pi\eta)^2/2,j^2) \right]
+ \sum_{j=1}^{\infty} (-1)^j \left[ j^4 z_{7/2}((\pi\eta)^2/2,j^2)
- j^2 z_{5/2}((\pi\eta)^2/2,j^2) \right] \nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} (-1)^j \left[ (j^2-k^2)^2 z_{7/2}((\pi\eta)^2/2,j^2+k^2)
- (j^2+k^2) z_{5/2}((\pi\eta)^2/2,j^2+k^2) \right] \nonumber \\ & &
+ \sum_{j=1}^{\infty} \left[ (j-1/2)^4 z_{7/2}(0,\eta^2/2+(j-1/2)^2)
- (j-1/2)^2 z_{5/2}(0,\eta^2/2+(j-1/2)^2) \right] \nonumber \\ & &
+ 2 \sum_{j,k=1}^{\infty} \Big( \left[ (j-1/2)^2-k^2 \right]^2
z_{7/2}(0,\eta^2/2+(j-1/2)^2+k^2) \nonumber \\ & &
- \left[ (j-1/2)^2 +k^2 \right] z_{5/2}(0,\eta^2/2+(j-1/2)^2+k^2) \Big) \Bigg\} .
\end{eqnarray}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,831 |
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\section{#1} \setcounter{equation}{0}}
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\title{\LARGE Wilson Surface Central Charge from Holographic Entanglement Entropy}
\author[1]{John Estes,}
\author[1]{Darya Krym,}
\author[2]{Andy O'Bannon,}
\author[2]{Brandon Robinson,}
\author[2]{and Ronnie Rodgers}
\affiliation[1]{New York City College of Technology, City University of New York,
300 Jay Street, Brooklyn, NY, 11201, USA}
\affiliation[2]{STAG Research Centre, Physics and Astronomy, University of Southampton, Highfield, Southampton SO17 1BJ, UK}
\emailAdd{jestes@citytech.cuny.edu}
\emailAdd{daryakrym@gmail.com}
\emailAdd{a.obannon@soton.ac.uk}
\emailAdd{B.J.Robinson@soton.ac.uk}
\emailAdd{R.J.Rodgers@soton.ac.uk}
\abstract{We use entanglement entropy to define a central charge associated to a two-dimensional defect or boundary in a conformal field theory (CFT). We present holographic calculations of this central charge for several maximally supersymmetric CFTs dual to eleven-dimensional supergravity in Anti-de Sitter space, namely the M5-brane theory with a Wilson surface defect and three-dimensional CFTs related to the M2-brane theory with a boundary. Our results for the central charge depend on a partition of $N$ M2-branes ending on $M$ M5-branes. For the Wilson surface, the partition specifies a representation of the gauge algebra, and we write our result for the central charge in a compact form in terms of the algebra's Weyl vector and the representation's highest weight vector. We explore how the central charge scales with $N$ and $M$ for some examples of partitions. In general the central charge does not scale as $M^3$ or $N^{3/2}$, the number of degrees of freedom of the M5- or M2-brane theory at large $M$ or $N$, respectively.}
\keywords{AdS/CFT correspondence, Gauge/gravity correspondence}
\begin{document}
\maketitle
\flushbottom
\section{Introduction}
\label{sec:intro}
M-theory is currently the leading candidate for an ultra-violet (UV) complete theory of quantum gravity. M-theory is defined as the (presumably unique) UV completion of 11d supergravity (SUGRA), which is a remarkably simple theory~\cite{Becker:2007zj,West:2012vka}: the only bosonic fields are the metric and a three-form, $C_3$. The stable BPS solitons of 11d SUGRA include M2- and M5-branes charged under $C_3$ electrically and magnetically, respectively. A complete formulation of M-theory will necessarily entail understanding these M-branes and other non-perturbative objects of 11d SUGRA more fully.
In a fashion similar to strings ending on D-branes or D-branes ending on other D-branes~\cite{Strominger:1995ac}, M2-branes can end on M2- or M5-branes, and M5-branes can end on other M5-branes. When the distance between two parallel D-branes shrinks to zero, the strings stretched between them become massless point particles and give rise to non-Abelian gauge multiplets. The low-energy worldvolume theory of multiple coincident D-branes is thus a maximally supersymmetric (SUSY) non-Abelian gauge theory~\cite{Witten:1995im}. However, when parallel M2- or M5-branes become coincident, the M2- or M5-branes stretched between them that become massless are extended objects, not point particles---making the low-energy worldvolume theory more challenging to identify.
For M2-branes the low-energy theory turns out to be a conventional quantum field theory (QFT), and in fact a conformal field theory (CFT): the low-energy theory of $N$ coincident M2-branes at a $\mathbb{C}^4/\mathbb{Z}_k$ singularity is a 3d ${\mathcal N}=6$ SUSY $U(N)_k \times U(N)_{-k}$ Chern-Simons-matter theory, called the Aharony-Bergman-Jafferis-Maldacena (ABJM) theory~\cite{Bagger:2006sk,Gustavsson:2007vu,Bagger:2007jr,Bagger:2007vi,Aharony:2008ug}. When the Chern-Simons level $k =1$ or $2$ the SUSY is enhanced to ${\mathcal N}=8$. When $N \gg k^5$ ABJM is holographically dual to 11d SUGRA on $AdS_4 \times S^7$~\cite{Maldacena:1997re}. The CFT and holographic descriptions have provided a wealth of information about M2-branes. For example, at large $N$ the number of worldvolume degrees of freedom scales as $N^{3/2}$~\cite{Klebanov:1996un,Drukker:2010nc}.
For $M$ coincident M5-branes the low-energy worldvolume theory is less well-understood. The 11d SUGRA soliton solution for M5-branes indicates that the worldvolume theory has 6d ${\mathcal N}=(2,0)$ SUSY and that the worldvolume fields should consist of five scalars, a chiral two-form $A_2$, and their fermionic superpartners filling out a tensor multiplet~\cite{Kaplan:1995cp}. The fields in the tensor multiplet arise as Goldstone modes: the scalars from breaking translations in the five normal directions, $A_2$ from breaking the symmetry of shifting $C_3$ by the exterior derivative of a two-form, and the fermions from breaking half the SUSY. The ${\mathcal N}=(2,0)$ SUSY has $SO(5)_R$ R-symmetry, which acts to rotate the five transverse directions, hence the five scalars transform as a vector of $SO(5)_R$. The existence of the chiral $A_2$ in the tensor multiplet means that the three-form field strength, $F_3= dA_2$, is self-dual on the M5-brane worldvolume, i.e. $F_3 = \star_{6d} F_3$. Furthermore the tensor multiplet fields are believed to be valued in a worldvolume $\mathfrak{u}(M)$ gauge algebra~\cite{Strominger:1995ac}. We will henceforth ignore the overall $\mathfrak{u}(1)\subset \mathfrak{u}(M)$ representing the center-of-mass motion of the coincident M5-branes.
The existence of $A_2$ is also intuitive because an M2-brane ending on an M5-brane produces in the M5-brane worldvolume a 2d 1/2-BPS soliton, i.e. a string, which naturally couples to $A_2$~\cite{Howe:1997ue}. Moreover, $F_3 = \star_{6d} F_3$ implies that this string has equal electric and magnetic charges, and hence is called a ``self-dual string.'' When the distance between two parallel M5-branes goes to zero and hence the M2-branes stretched between them become massless, the result should be a 6d theory of massless self-dual strings~\cite{Strominger:1995ac}.
Since strings are extended objects, such a perspective suggests that the M5-brane theory might be non-local. However, compelling evidence has accumulated that in fact the M5-brane theory is not only local, but is a conventional CFT. In particular, when $M\to\infty$ the 11d SUGRA soliton solution for M5-branes has a near-horizon $AdS_7 \times S^4$ geometry in the decoupling limit, suggesting that 11d SUGRA on $AdS_7 \times S^4$ should be holographically dual to the M5-brane theory~\cite{Maldacena:1997re}. The $AdS_7$ factor indicates that all 6d field theory observables computed holographically are consistent with those of a local CFT. Additionally, non-trivial solutions to 6d conformal bootstrap equations have been found that are consistent with ${\mathcal N}=(2,0)$ SUSY and locality~\cite{Beem:2015aoa}.
From the perspective of conventional, perturbative QFT, the existence of any unitary, interacting QFT in $d>4$ is surprising, since power counting rules out any local Lagrangian. Two additional challenges arise in writing a local Lagrangian for the M5-brane theory. First, imposing self-duality at the level of the Lagrangian in any dimension is difficult. Second, generalizing $\mathfrak{u}(M)$ gauge transformations to a higher-form gauge field, such as the two-form $A_2$, is challenging. Remarkably, for the Abelian case, $M=1$, a classical action for the M5-brane worldvolume fields that has $\mathfrak{u}(1)$ gauge invariance, ${\mathcal N}=(2,0)$ superconformal symmetry, and locality, as well as 6d self-duality of $F_3$ enforced by auxiliary scalar field (acting as a Lagrange multiplier), is known~\cite{Pasti:1997gx,Bandos:1997ui}. However, whether a classical action can be written for $M>1$ remains unknown.
Despite the challenges, the M5-brane theory is extremely important to study, not only because M5-branes are key ingredients of M-theory, but also because the M5-brane worldvolume theory holds a uniquely privileged place among QFTs. In 6d, ${\mathcal N}=(2,0)$ is the maximal amount of SUSY possible, and 6d is the maximal dimension in which superconformal symmetry is possible~\cite{Nahm:1977tg}. A 6d ${\mathcal N}=(2,0)$ SUSY CFT cannot be reached as the infra-red (IR) fixed point of a local renormalization group (RG) flow from a free UV fixed point. The M5-brane theory has no known dimensionless parameters besides $M$ that could be tuned to allow a perturbative expansion. In short, the dimensionality, symmetries, and $M$ determine the M5-brane theory completely. The M5-brane theory is thus an isolated, intrinsically strongly-interacting fixed point. Via compactification the M5-brane theory can describe many lower-dimensional SUSY QFTs and dualities among them, and could potentially be a ``master theory'' containing information about all lower-dimensional QFTs.
In this paper, we will use holography to study the M5-brane theory with a 2d conformal defect as well as 3d CFTs with boundaries (BCFTs) related to the ABJM BCFT, using the following 1/4-BPS intersection of M-branes:
\begin{center}
\begin{tabular}{ |c | c || c | c | c | c | c | c | c | c | c | c | c|}
\hline
& & $x_0$ & $x_1$ & $x_2$ & $x_3$ & $x_4$ & $x_5$ & $x_6$ & $x_7$ & $x_8$ & $x_9$ & $x_{10}$ \\
\hline
$N$ & M2 & X & X & X & & & & & & & & \\
$M$ & M5 & X & X & & X & X & X & X & & & & \\ \hline
$M'$ & M5$'$ & X & X & & & & & & X & X & X & X \\ \hline
\end{tabular}
\end{center}
Schematically, this table represents a stack of $N$ coincident M2-branes, a stack of $M$ coincident M5-branes, and another stack of $M'$ coincident M5-branes that we label M5$'$ to distinguish them from the first stack. Most importantly, the M2-branes end on the M5- and M5$'$-branes at $x_2=0$.
Consider first setting $M' =0$, that is, the intersection of $N$ M2-branes ending on $M$ M5-branes, with no M5$'$-branes. Recall that the endpoint of a semi-infinite string ending on a D-brane gives rise to an infinitely massive (s)quark in the D-brane's worldvolume. Integrating out the heavy (s)quark yields a Wilson line, i.e. the holonomy of the D-brane worldvolume gauge field along the (s)quark worldline~\cite{Maldacena:1998im,Rey:1998ik}. Similarly, the end of a semi-infinite M2-brane ending on an M5-brane produces a self-dual string with infinite tension, which can be integrated out to yield a ``Wilson surface'' operator~\cite{Ganor:1996nf}. In the Abelian case, $M=1$, a precise form is known for the 1/2-BPS Wilson surface operator, $\exp\left( i\int A_2 + \ldots \right)$, with the integral over the surface spanned by the self-dual string and the ellipsis denoting terms required by SUSY, involving the M5-branes' worldvolume scalars, see for example ref.~\cite{Bullimore:2014upa}. In the non-Abelian case, $M>1$, a precise form is unknown but is believed to be schematically $\text{tr}_{\cal{R}} \exp \left( i\int A_2 + \ldots\right)$, with the trace in representation $\cal{R}$ of $\mathfrak{su}(M)$. The representation $\cal{R}$ describes how the $N$ M2-branes are partitioned among the $M$ M5-branes, as we discuss in detail below.
We will consider only flat Wilson surfaces, i.e. Wilson surfaces extended along the $\mathbb{R}^{1,1}$ spanned by $x_0$ and $x_1$. From the M5-brane theory's perspective the Wilson surface is a 1/2-BPS 2d superconformal soliton. The M5-brane theory's bosonic symmetry is $SO(6,2) \times SO(5)_R$, where $SO(6,2)$ is the 6d conformal group. The table above shows that the Wilson surface preserves an $SO(2,2) \times SO(4)_R \times SO(4)_R$ subgroup, where the global 2d conformal group $SO(2,2) \subset SO(6,2)$ leaves invariant the Wilson surface, and the first $SO(4)_R$ rotates $(x_3,x_4,x_5,x_6)$ while the second rotates $(x_7,x_8,x_9,x_{10})$. The R subscripts indicate that these act as R-symmetries on the supercharges preserved by the Wilson surface, forming a 2d ``large'' ${\mathcal N}=(4,4)$ SUSY (``small'' ${\mathcal N}=(4,4)$ has a single $SO(4)_R$).
We will compute a central charge associated with the Wilson surface using holography. The holographic description's geometry is an $AdS_3$ and two $S^3$'s fibered over a Riemann surface~\cite{D'Hoker:2008wc,DHoker:2008rje,Estes:2012vm,Bachas:2013vza} and hence has the expected isometry $SO(2,2)\times SO(4)_R \times SO(4)_R$. The holographic description represents a large $M$ limit with arbitrary $N$. The geometry is asymptotically locally $AdS_7 \times S^4$ for all $N$, and becomes precisely $AdS_7 \times S^4$ when $N=0$.
As is well-known, 2d large ${\mathcal N}=(4,4)$ super-groups actually come in a one-parameter family called $D(2,1;\gamma)\times D(2,1;\gamma)$, where $\gamma$ is the free parameter~\cite{Sevrin:1988ew,Frappat:1996pb}. The most general solutions of 11d SUGRA that have super-isometry $D(2,1;\gamma)\times D(2,1;\gamma)$ and locally asymptote to $AdS_7 \times S^4$ are known~\cite{Bachas:2013vza}. The bosonic subgroup of $D(2,1;\gamma)\times D(2,1;\gamma)$ is $SO(2,2)\times SO(4) \times SO(4)$, and hence these solutions all involve an $AdS_3$ and two $S^3$'s fibered over a Riemann surface~\cite{DHoker:2008wvd}. The solutions describing Wilson surfaces in the M5-brane theory at large $M$ and arbitrary $N$ have $\gamma=-1/2$.
Additionally, the most general solutions of 11d SUGRA with super-isometry $D(2,1;\gamma)\times D(2,1;\gamma)$ are known that locally asymptote to ``half'' of $AdS_4 \times S^7$, in a sense we explain below~\cite{Bachas:2013vza}. These solutions are holographically dual to 3d maximally SUSY BCFTs. The exact BCFTs are not yet known, though some properties are clear from the 11d SUGRA solutions. In particular, in these solutions generically both $M$ and $M'$ are non-zero. The solutions thus describe M2-branes ending on M5- and M5$'$-branes, and hence the BCFTs must be cousins of the maximally SUSY ABJM BCFT.\footnote{Solutions of 11d SUGRA that are candidates for the holographic dual of the maximally SUSY ABJM BCFT appear in ref.~\cite{Bachas:2013vza}, but have potentially dangerous singularities.} Presumably these BCFTs are obtained from the ABJM BCFT by couplings to 2d SUSY multiplets at the boundary, and/or by sources or expectation values of scalar operators away from the boundary, similar to the superconformal interfaces between ABJM theories in refs.~\cite{DHoker:2009lky,Bobev:2013yra}. We will henceforth refer to these theories as ``cousins of the ABJM BCFT.''
For $k=1$ or $2$, ABJM's bosonic symmetry is enhanced to $SO(3,2)\times SO(8)_R$, where $SO(3,2)$ is the 3d conformal group, and in the intersection above the $SO(8)_R$ acts on $(x_3,\ldots,x_{10})$. Maximally superconformal boundary conditions at $x_2=0$ preserve the $SO(2,2) \subset SO(3,2)$ that leaves the boundary invariant, and $SO(4)_R \times SO(4)_R \subset SO(8)_R$ R-symmetry. The maximally SUSY ABJM BCFT's bosonic symmetry is thus $SO(2,2) \times SO(4)_R \times SO(4)_R$, and the super-group of the theory is $D(2,1;\gamma)\times D(2,1;\gamma)$ with $\gamma=1$. The cousins of the maximally SUSY ABJM BCFT that we will study have arbitrary $\gamma<0$, and their holographic duals again involve an $AdS_3$ and two $S^3$'s fibered over a Riemann surface~\cite{Bachas:2013vza}. These solutions correspond to a limit with large $N$ and a large number of M5- and M5$'$-branes. However the values of $M$ and $M'$ are in fact undetermined, intuitively because the M2-branes do not ``know'' how many M5- and M5$'$-branes have zero M2-branes ending on them, and so cannot know the total numbers of M5- and M5$'$-branes. Additionally, sending both $M$ and $M'$ to zero produces a singular solution, presumably because removing a BCFT's boundary is a singular operation.
Using these 11d SUGRA solutions, we will compute a central charge associated with the Wilson surface or 3d BCFT boundary. Crucially, the Wilson surface or 2d boundary is not a 2d CFT, but a 2d defect in, or boundary of, an ambient CFT that has $d>2$, which implies in general that the full Virasoro symmetry is not present. To be more precise, the Wilson surface or boundary breaks the ambient $SO(6,2)$ or $SO(3,2)$ conformal symmetry down to $SO(2,2)$, i.e. the global part of the Virasoro symmetry, in which the usual central charge does not appear. We must therefore define a central charge some other way.
We will use entanglement entropy (EE). Following refs.~\cite{Jensen:2013lxa,Estes:2014hka,Gentle:2015jma}, we will compute holographically the EE of a spherical region centered on the 2d defect or of a semi-circle centered on the 2d boundary. This EE has UV divergences, as expected, so we introduce a UV cutoff and subtract the EE of the ambient CFT. What remains is a term logarithmic in the cutoff, a constant term, and terms that vanish as the cutoff is removed. In analogy with the EE of a single interval in a 2d CFT~\cite{Holzhey:1994we,Calabrese:2004eu}, we identify the coefficient of that logarithmic term as $1/3$ times the central charge. In short, we compute the \textit{change} in the coefficient of the EE's logarithmic term due to the 2d defect or boundary, and use the result to define a central charge. We will denote the resulting Wilson surface or 2d boundary central charge as $b_{6d}$ or $b_{3d}$, respectively. Again in analogy with 2d CFT, we will interpret these as counting massless degrees of freedom supported on the Wilson surface or 2d boundary.\footnote{In a 3d BCFT the boundary central charge defined from EE is proportional to a central charge that appears in the trace anomaly~\cite{Fursaev:2013mxa,Fursaev:2016inw} and obeys a c-theorem, strictly decreasing in a boundary RG flow~\cite{Jensen:2015swa}. Our $b_{3d}$ thus counts massless degrees of freedom at the 2d boundary. However, a similar interpretation of $b_{6d}$ may not always be justified. In particular, examples of defects in higher-d CFTs are known in which central charges defined from EE do \textit{not} decrease along defect RG flows~\cite{Kumar:2016jxy,Kumar:2017vjv,Kobayashi:2018lil}. These examples include certain RG flows on a Wilson surface in a totally symmetric representation ${\cal R}$~\cite{Rodgers:2018mvq}.} Whether and how these central charges defined from EE are related to other potential definitions, for example via the thermodynamic entropy, stress tensor correlators, and so on, we leave as an important open question.
Our main result, for $b_{6d}$, takes a remarkably simple form,
\begin{equation}
\label{eq:b6dcompact}
b_{6d} = \frac{3}{5} \left [ 16 \left(\lambda,\varrho\right)-\left(\lambda,\lambda\right)\right],
\end{equation}
where $\lambda$ is the highest weight vector of the Wilson surface's representation $\cal{R}$, $\varrho$ is the $\mathfrak{su}(M)$ Weyl vector, and $(\cdot\,,\,\cdot)$ is defined with the Killing form on the weight space. The inner product $(\cdot,\,\cdot)$ is invariant under the action of the Weyl group, hence so is $b_{6d}$. In particular, $b_{6d}$ is invariant under complex conjugation of a representation, $\cal{R} \to \overline{\cal{R}}$, which acts as a Weyl reflection. Eq.~\eqref{eq:b6dcompact} is also reminiscent of the results for the M5-brane theory's own central charges, $a$ and $c$, which can similarly be written in terms of purely group theoretic data~\cite{Beem:2014kka,Cordova:2015vwa}. However, eq.~\eqref{eq:b6dcompact} obscures $b_{6d}$'s dependence on $N$ and $M$. In sec.~\ref{sec:cc} we present $b_{6d}$ for some specific $\cal{R}$, such as the rank $N$ totally symmetric and anti-symmetric representations, to explore the dependence on $N$ and $M$. Our results for $b_{3d}$ cannot be written so neatly as eq.~\eqref{eq:b6dcompact}, largely because we do not know the total number of M5- and M5$'$-branes and thus do not know $\mathfrak{su}(M)$ or $\mathfrak{su}(M')$.
However, a key observation about our results for both $b_{6d}$ and $b_{3d}$ is that neither naturally scales as $N^{3/2}$, characteristic of M2-branes at large $N$~\cite{Klebanov:1996un,Drukker:2010nc}, or as $M^3$, characteristic of M5-branes at large $M$ ($a \propto c \propto M^3$ at large $M$)~\cite{Freed:1998tg,Harvey:1998bx,Henningson:1998gx}. In other words, in general for the M-brane intersections we consider, the number of massless degrees of freedom on the Wilson surface or 2d boundary does not scale, in any obvious way, with the total number of degrees of freedom of the M2- or M5-brane theory.
This paper is organized as follows. In sec.~\ref{sec:sugra} we review the 11d SUGRA solutions describing M5-branes with Wilson surfaces or cousins of the ABJM BCFT with $\gamma<0$. In sec.~\ref{sec:hc} we derive an integral for the holographic EE, and then evaluate the integral to extract $b_{6d}$ and $b_{3d}$. In sec.~\ref{sec:cc} we summarize our results, including for specific $\cal{R}$, and compare to existing results. (Readers interested only in our results can skip directly to sec.~\ref{sec:cc}.) We conclude in sec~\ref{sec:discussion} with discussion and suggestions for future research.
The companion paper ref.~\cite{Rodgers:2018mvq} reproduces our results for $b_{6d}$ for the fundamental, totally anti-symmetric, and totally symmetric representations, using probe branes in $AdS_7 \times S^4$, namely a probe M2-brane when $N\ll M$ or M5-brane when $N \ll M^2$. Ref.~\cite{Rodgers:2018mvq} also uses the probe M5-branes to explore RG flows on Wilson surfaces.
\textbf{Note added:} After this paper appeared on the arxiv, ref.~\cite{Jensen:2018rxu} clarified how our central charge $b$ obtained from the EE of a (hemi-)sphere is related to central charges in the defect's contribution to the Weyl anomaly. This clarifies the relationship between unitarity and positivity of $b$ discussed in sec.~\ref{sec:wilson-surface}, among other things.
For a CFT in $\mathbb{R}^{1,d-1}$ with $d\geq 3$ and a 2d conformal defect, the trace of the stress tensor splits into two terms, $\left< T_{~\mu}^{\mu} \right> = \left< T_{~\mu}^{\mu}\right>_{\textrm{bulk}} + \delta^{d-2}(x) \left<T_{~\mu}^{\mu}\right>_{\textrm{defect}}$, where $\left<T_{~\mu}^{\mu}\right>_{\textrm{bulk}}$ is the CFT's trace anomaly, which is $0$ for odd $d$ but may be non-zero for even $d$, $\delta^{d-2}(x)$ is a delta function that localizes to the 2d defect, and~\cite{Graham:1999pm,Henningson:1999xi, Schwimmer:2008yh}
\begin{equation}
\label{eq:defecttrace}
\left<T_{~\mu}^{\mu}\right>_{\textrm{defect}} = - \frac{1}{24\pi} \left(
c \, \hat{R} + d_1 \, \Pi_{ab}^\mu \Pi_\mu^{ab} - d_2 \, \hat{g}^{ac} \hat{g}^{bd} W_{abcd}
\right),
\end{equation}
where $\hat{g}_{ab}$ is the defect's induced metric ($a,b=0,1$), $\hat{R}$ is the Ricci scalar built from $\hat{g}_{ab}$, $\Pi^{\mu}_{ab}$ is the defect's traceless extrinsic curvature, $W_{abcd}$ is the Weyl tensor pulled back to the defect, and the dimensionless numbers $c$, $d_1$, and $d_2$ are the defect central charges. In a 3d BCFT the boundary's contribution to the trace anomaly has the form of eq.~\eqref{eq:defecttrace}, but $W_{abcd}=0$ identically, so $d_2$ does not exist. All of $c$, $d_1$, and $d_2$ can depend on boundary conditions imposed on CFT fields at the defect or boundary and on degrees of freedom supported only at the defect or boundary. Indeed, $c$ obeys a $c$-theorem~\cite{Jensen:2015swa}, and may thus serve as a measure of the number of massless degrees of freedom at the defect or boundary. However, unlike a 2d CFT's central charge, unitarity does not require $c \geq 0$, and in fact even simple, unitary theories can have $c<0$, such as a 3d free, massless scalar BCFT with Dirichlet boundary conditions~\cite{Nozaki:2012qd,Jensen:2015swa, Solodukhin:2015eca}. Whether a lower bound exists on $c$ remains unknown, though for 3d BCFTs $c\geq -\frac{2}{3} \, d_1$ is conjectured~\cite{Herzog:2017kkj}, where unitarity requires $d_1\geq 0$~\cite{Herzog:2017xha,Herzog:2017kkj}. Ref.~\cite{Jensen:2018rxu} showed that if the average null energy condition is valid in a CFT with a 2d conformal defect, then $d_2 \geq 0$ also.
Refs.~\cite{Kobayashi:2018lil,Jensen:2018rxu} showed that
\begin{equation}
\label{eq:ee_trace_anomaly}
b = c - \frac{d-3}{d-1} d_2.
\end{equation}
For $d=3$, as is the case for the cousins of ABJM, this reproduces the result \(b = c\)~\cite{Fursaev:2013mxa,Fursaev:2016inw}. For $d > 3$, eq.~\eqref{eq:ee_trace_anomaly} implies that $b$ may be negative even when $c$ and $d_2$ are both positive. Indeed, for Wilson surfaces in the M5-brane theory, ref.~\cite{Jensen:2018rxu} used eq.~\eqref{eq:ee_trace_anomaly}, our result for $b_{6d}$ in eq.~\eqref{eq:b6dcompact}, and the result of ref.~\cite{Gentle:2015jma} for $d_2$ to calculate
\begin{equation}
\label{eq:cd2values}
c = 24\left(\lambda,\varrho\right) + 3\left(\lambda,\lambda\right), \qquad d_2 = 24\left(\lambda,\varrho\right) + 6\left(\lambda,\lambda\right),
\end{equation}
both of which are positive for any ${\cal R}$. However, the linear combination $b$ in eq.~\eqref{eq:ee_trace_anomaly} can be negative for some ${\cal R}$, as we discuss in sec.~\ref{sec:wilson-surface}. These results show that $b<0$ does not signal violation of unitarity.
\section{Review: the SUGRA Solutions}
\label{sec:sugra}
The solutions of 11d SUGRA in ref.~\cite{Bachas:2013vza} that holographically describe Wilson surfaces or cousins of the ABJM BCFT (and which built upon the solutions in refs.~\cite{DHoker:2008rje,D'Hoker:2008wc,Estes:2012vm}) are 1/2-BPS, meaning they support 16 real supercharges, and have super-isometry $D(2,1;\gamma)\times D(2,1;\gamma)$ with $\gamma\in \mathbb{R}$. The super-group $D(2,1;\gamma)\times D(2,1;\gamma)$ has bosonic subgroup $SO(2,1)\times SO(3) \times SO(3)$, where the super-charges anti-commute into a linear combination of the generators of these three bosonic factors. The parameter $\gamma$ determines the relative weights of the coefficients in that linear combination. The super-group $D(2,1;\gamma)\times D(2,1;\gamma)$ is invariant under the simultaneous operations $\gamma \to 1/\gamma$ and swapping the two $SO(3)$'s. Without loss of generality we can thus restrict to $\gamma \in [-1,1]$, and in fact all the solutions we consider below will have $\gamma \in(-1,0)$. These symmetries of the 11d SUGRA solutions match those of 2d large ${\mathcal N}=(4,4)$ SUSY, which admits exactly the same one-parameter family of Lie super-groups~\cite{Sevrin:1988ew,Frappat:1996pb}.\footnote{Ref.~\cite{Sevrin:1988ew} classifies and constructs the Virasoro extension of the super-group.} We leave the full details of these 11d SUGRA solutions to refs.~\cite{Bachas:2013vza,DHoker:2008rje,D'Hoker:2008wc,Estes:2012vm}, and here review only features that we will need in subsequent sections. In particular, we will consider the solutions of refs.~\cite{DHoker:2008rje,Bachas:2013vza}, using the conventions of ref.~\cite{Bachas:2013vza}.\footnote{To clarify, ref.~\cite{D'Hoker:2008wc} constructed solutions for three special values of $\gamma$, ref.~\cite{Estes:2012vm} constructed the solutions for general $\gamma$, and refs.~\cite{DHoker:2008rje,Bachas:2013vza} identified the specific solutions we consider in this paper.}
The $SO(2,2) \times SO(4) \times SO(4)$ isometry implies that the metric involves $AdS_3$ and two $S^3$'s fibered over a Riemann surface, which we take to be the upper half plane,
\begin{equation}
\label{eq:metric2}
ds^2 = f_1^2 \,ds^2_{\text{AdS}_3} + f_2^2 \,ds^2_{S^3} + f_3^2 \,ds^2_{S^3} + 2 \, \Omega^2 \,|dw|^2,
\end{equation}
where $w$ and $\overline{w}$ are the coordinates of the upper half plane (so $\textrm{Im}(w) \geq 0$). The functions $f_1$, $f_2$, $f_3$, and $\Omega$ depend only on $w$ and $\overline{w}$. We denote by $ds^2_{S^3}$ the metric for a unit-radius round $S^3$ and $ds^2_{AdS_3}$ the metric for a unit-radius $AdS_3$, that is,
\begin{equation}
\label{eq:ads3metric}
ds^2_{AdS_3} = \frac{1}{u^2} \left(du^2 -dt^2 + dx_{\parallel}^2\right),
\end{equation}
where $u \in [0,\infty)$, with $AdS_3$ boundary at $u = 0$, $t\in(-\infty,\infty)$ is the time coordinate, and $x_{\parallel}\in(-\infty,\infty)$ is the spatial coordinate parallel to the Wilson surface or 2d boundary.
The solutions in ref.~\cite{Bachas:2013vza} that we will use are completely determined by a triple of data $(h,G,\gamma)$, where $h$ is a harmonic function over the upper half plane, and $G$ is a complex-valued function that obeys
\begin{equation}
\partial_w G = \frac{1}{2} (G + \overline{G}) \, \partial_w \ln h,
\end{equation}
although $G$ is not completely determined by $h$. If we define the complex-valued functions
\begin{align}
\label{eq:wpm}
&W_+ \equiv |G+i|^2 + \gamma (G \overline{G} - 1)\,,&
&W_- \equiv |G-i|^2 + \gamma^{-1} (G \overline{G} -1 ),&
\end{align}
where $\gamma$ is the constant parametrizing the supergroup, then in these solutions $f_1$, $f_2$, $f_3$, and $\Omega$ are given by
\begin{subequations}
\label{eq:metfuncs}
\begin{equation}
f_1^6 = \frac{h^2 W_+ W_-}{c_1^6 (G \overline{G} - 1)^2}, \qquad f_2^6 = \frac{h^2 (G \overline{G} - 1) W_-}{c_2^3 c_3^3 W_+^2}, \qquad f_3^6 = \frac{h^2 (G \overline{G} - 1) W_+}{c_2^3 c_3^3 W_-^2},
\end{equation}
\begin{equation}
\Omega^6 = \frac{1}{8} \frac{|\partial_w h|^6}{c_2^3 c_3^3 h^4} (G \overline{G} - 1) W_+ W_-.
\end{equation}
\end{subequations}
The parameters $c_1$, $c_2$, and $c_3$ are constants that obey $c_1 + c_2 + c_3=0$, so that only two are independent. In fact, a simultaneous re-scaling of $c_1$, $c_2$, and $c_3$ can be absorbed by re-scaling $h$ without changing the solution, so only a single constant is independent. That single constant must map to $\gamma$: the precise relation is $\gamma = c_2/c_3$. As mentioned above, the solutions that we will consider have $\gamma \in (-1,0)$.
When $\gamma<0$, global regularity of these solutions requires that $h$ and $G$ obey $h>0$ and $|G|<1$ everywhere on the interior of the upper half $w$ plane (all $\textrm{Im}\left(w\right)>0$) and that $h=0$ and $G = \pm i$ on the boundary of the upper half $w$ plane ($\textrm{Im}\left(w\right)=0$). All the solutions that we consider below will obey these conditions.
The four-form $F_4 = dC_3$ of these solutions appears in ref.~\cite{Bachas:2013vza}. We will not present the solution for $F_4$ explicitly, but in sec.~\ref{sec:fluxes} we will discuss the M2- and M5-brane charges determined by the solution for $F_4$.
The invariance of $D(2,1;\gamma)\times D(2,1;\gamma)$ under $\gamma \to 1/\gamma$ and swapping of $SO(3)$ sub-groups appears in these 11d SUGRA solutions as invariance under $\gamma \to 1/\gamma$ and exchange of $W_+ \leftrightarrow W_-$. As clear from eq.~\eqref{eq:metfuncs}, that leaves $f_1$ and $\Omega$ invariant but trades $f_2 \leftrightarrow f_3$, thus effectively interchanging the geometry's two $S^3$ factors.
\subsection{Asymptotically Locally $AdS_7 \times S^4$ Solutions}
\label{subsec:ads7metric}
The 11d SUGRA solutions holographically describing Wilson surfaces in the M5-brane theory at large $M$ are of the form in eq.~\eqref{eq:metric2} with
\begin{equation}
\label{eq:ads7metric}
h = - i \left(w-\overline{w}\right), \qquad G = - i\left( 1 + \sum_{j=1}^{2n+2} (-1)^j \frac{w - \xi_j}{|w - \xi_j|} \right),
\end{equation}
where the integer $n \geq 0$ and the $\xi_j$ are $2n+2$ real-valued constants determining $G$'s branch points on the boundary of the upper half plane. More specifically, the $\xi_j$ are points on the real line $\textrm{Im}(w)=0$ where $G$ changes sign from $\pm i$ to $\mp i$.
In general, the upper half plane is invariant under the $SL(2,\mathbb{R})$ group of transformations, which is three-dimensional. Implicitly we fixed two $SL(2,\mathbb{R})$ transformations with our choice of $h$. The third transformation is translations, which we can use to fix one of the $\xi_j$. The remaining $2n+1$ values of the $\xi_j$ determine the M2- and M5-brane charges of the solution, as we discuss in sec.~\ref{sec:fluxes}. The solutions describing M2-branes ending on M5-branes have $\gamma=-1/2$, whereas $\gamma=-2$ describes M2-branes ending on M5$'$-branes. However the latter map to the former via $\gamma \to 1/\gamma$ and swapping the two $S^3$'s, as described above. In what follows, we will consider arbitrary $\gamma \in (-1,0)$ in many intermediate steps but will always set $\gamma = -1/2$ in our final results.
Another form of the $G$ in eq.~\eqref{eq:ads7metric} that will be useful for describing the geometry's asymptotics comes from using polar coordinates in the upper half plane, $w \equiv r e^{i \theta}$ with $r \in [0,\infty)$ and $\theta \in [0,\pi]$. If we expand $G$ in Legendre polynomials, $P_{k}\left(\cos \theta\right)$,
\begin{equation}
\label{eq:ads7legendre}
G = -i\left( 1+ \sum_{k=1}^\infty m_k \frac{\left( e^{i \theta} P_k\left(\cos\theta\right) - P_{k-1}\left(\cos\theta\right) \right)}{r^k} \right), \qquad m_k \equiv \sum_{j=1}^{2 n+2} (-1)^j \left(\xi_j\right)^k,
\end{equation}
then the $\xi_j$ determine the expansion coefficients $m_k$, where $n$ is finite but $k \in [0,\infty)$.
Though not immediately obvious, the solutions in eq.~\eqref{eq:ads7metric} are asymptotically locally $AdS_7 \times S^4$. Indeed, asymptotically we can put the metric of these solutions in the Fefferman-Graham (FG) form of $AdS_7 \times S^4$,
\begin{equation}
\label{eq:ads7asympmet}
ds^2 = \frac{4 L_{S^4}^2}{z^2} \left( dz^2 -dt^2 + dx_{\parallel}^2 + dr_\perp^2 + \frac{\left(1+\gamma\right)^2}{\gamma^2}r_\perp^2 ds^2_{S^3} \right) + L_{S^4}^2 \left(d\phi^2 + \sin^2\phi\, ds^2_{S^3} \right) + \ldots,
\end{equation}
where $z\in[0,\infty)$ is the FG holographic coordinate, with boundary at $z=0$, $r_{\perp}\in[0,\infty)$ is the distance to the defect, $\phi\in[0,\pi]$ is the zenith angle of the asymptotic $S^4$, and the ellipsis represents terms sub-leading in $1/r$. The change of coordinates to FG form admits the following asymptotic expansion,
\begin{subequations}
\label{eq:ads7asympcoords}
\begin{align}
z =& \frac{u}{\sqrt{r}} \left(\frac{1+\gamma}{\sqrt{-\gamma}}\sqrt{2\,m_1} + {\cal O} \left( \frac{1}{r^2} \right) \right), \\
r_\perp =& \, u \left(1 + {\cal O} \left( \frac{1}{r} \right) \right), \\
\phi =& \, \theta \left(1 + {\cal O} \left( \frac{1}{r} \right) \right),
\end{align}
\end{subequations}
which determines the radius of curvature $L_{S^4}$ of the asymptotic $S^4$,
\begin{equation}
\label{eq:ads7rad}
L_{S^4}^6 = \frac{1}{c_1^6}\frac{(1 + \gamma)^6}{\gamma^2}m_1^2.
\end{equation}
The factor $\left(1+\gamma\right)^2/\gamma^2$ in eq.~\eqref{eq:ads7asympmet} indicates that for generic $\gamma$ the dual M5-brane theory lives on a space with a conical singularity at $r_{\perp}=0$. These 11d SUGRA solutions thus suggest that deformations of the M5-brane theory with $D(2,1;\gamma)\times D(2,1;\gamma)$ super-group are only possible, for generic $\gamma$, on spaces with such a conical singularity. Only for $\gamma=-1/2$ does the singularity disappear such that the theory lives on 6d Minkowski space.
Eqs.~\eqref{eq:ads7asympmet} and~\eqref{eq:ads7asympcoords} show clearly that in these solutions we can approach the asymptotic $AdS_7 \times S^4$ boundary in two ways. First, if we fix $u$ and send $r \to \infty$ then $z \to 0$ with fixed $r_{\perp}=u$, which means that we arrive at the boundary a distance $u$ from the defect. Second, if we fix $r$ and send $u \to 0$, then $z \to 0$ but now $r_{\perp} \to 0$, and so we arrive at the boundary precisely on the defect. Recall from eq.~\eqref{eq:ads3metric} that $u \to 0$ is the boundary of the $AdS_3$ factor.
The $AdS_7 \times S^4$ solution is simply the $\gamma=-1/2$ and $n=0$ case, which has two branch points. Using the $SL(2,\mathbb{R})$ translational symmetry we place these two branch points on the real line $\textrm{Im}(w)=0$ symmetrically, at $\textrm{Re}(w)=\pm \xi$, such that the solution has only one free parameter, $\xi$. In this case, in eq.~\eqref{eq:ads7asympcoords} the higher-order powers of $1/r$ vanish, and hence the ellipsis in eq.~\eqref{eq:ads7asympmet} vanishes. We will see below that $\xi$ determines $m_1 = 2 \xi$ and hence $L_{S^4}$, the single free parameter of the $AdS_7 \times S^4$ solution. The cases with $\gamma=-1/2$ and $n>0$ then describe Wilson surfaces in the M5-brane theory.
\subsection{Asymptotically Locally $AdS_4 \times S^7$ Solutions}
\label{subsec:ads4metric}
The 11d SUGRA solutions holographically describing cousins of the ABJM BCFT with $\gamma<0$ are of the form in eq.~\eqref{eq:metric2} with
\begin{equation}
\label{eq:ads4metric}
h = - i \left(w-\overline{w}\right), \qquad G = - i \sum_{j=1}^{2n+1} (-1)^j \frac{w - \xi_j}{|w - \xi_j|},
\end{equation}
where the integer $n\geq 0$ and the $\xi_j$ are $2n+1$ real-valued constants determining $G$'s branch points on the boundary of the upper half plane, i.e. on the real line $\textrm{Im}(w)=0$. As in sec.~\ref{subsec:ads7metric}, we have implicitly fixed two $SL(2,\mathbb{R})$ transformations of the upper half plane with our choice of $h$, and we can use translation symmetry to fix one of the $\xi_j$. The remaining $2n$ values of the $\xi_j$ then determine the solutions' M2- and M5-brane charges, as we discuss in sec.~\ref{sec:fluxes}. We will consider solutions with $\gamma\in(-1,0)$.
As in sec.~\ref{subsec:ads7metric}, we can obtain another form of the $G$ in eq.~\eqref{eq:ads4metric} that will be useful for describing the asymptotics by introducing $w \equiv r e^{i \theta}$ with $r \in [0,\infty)$ and $\theta \in [0,\pi]$. Expanding $G$ in Legendre polynomials then gives
\begin{equation}
\label{eq:ads4legendre}
G = i \left(e^{i \theta} - \sum_{k=1}^\infty m_k \frac{\left( e^{i \theta} P_k\left(\cos\theta\right) - P_{k-1}\left(\cos \theta\right) \right)}{r^k} \right), \qquad m_k \equiv \sum_{j=1}^{2 n+ 1} (-1)^j \left(\xi_j\right)^k,
\end{equation}
where the $\xi_j$ determine the expansion coefficients $m_k$.\footnote{We use the same symbol $m_k$ for the expansion coefficients in both eq.~\eqref{eq:ads7legendre} and~\eqref{eq:ads4legendre}, though the difference should always be clear from the context.}
The solutions in eq.~\eqref{eq:ads7metric} are in fact asymptotically locally ``half'' of $AdS_4 \times S^7$, in the following sense. Asymptotically we can put the metric of these solutions in the FG form of $AdS_4 \times S^7$,
\begin{equation}
\label{eq:ads4asympmet}
ds^2 = \frac{L_{S^7}^2}{4} \left( \frac{dz^2}{z^2} + \frac{-dt^2 + dx_{\parallel}^2 + dx_\perp^2}{z^2} \right) + L_{S^7}^2 \left(d\phi^2 + \cos^2\phi \, ds^2_{S^3} + \sin^2\phi\, ds^2_{S^3} \right)+\ldots,
\end{equation}
where the ellipsis represents terms sub-leading in $1/r$, and where the asymptotic $S^7$ radius, $L_{S^7}$, is given by
\begin{equation}
\label{eq:ads4rad}
L_{S^7}^6 = \frac{16}{c_1^6}\frac{(1 + \gamma)^6}{(-\gamma)^3} (-m_1^2 - m_2).
\end{equation}
The change of coordinates to FG form has the following asymptotic expansion,
\begin{subequations}
\label{eq:ads4asympcoords}
\begin{align}
z =& \,\frac{u}{r} \left(\frac{1 + \gamma}{\sqrt{\gamma}} \frac{\sqrt{m_1^2+m_2}}{2} + {\cal O} \left( \frac{1}{r^2} \right) \right), \\
x_{\perp} =& \, u \left(1 + {\cal O} \left( \frac{1}{r^2} \right) \right), \\
\phi =& \, \frac{\theta}{2} \left(1 + {\cal O} \left( \frac{1}{r^2} \right) \right),
\end{align}
\end{subequations}
where $\theta \in [0,\pi]$ implies $\phi\in [0,\pi/2]$. Eq.~\eqref{eq:ads4asympcoords} shows that $u \in [0,\infty)$ implies the FG holographic coordinate $z\in[0,\infty)$, with boundary at $z=0$, but crucially $x_{\perp}\in [0,\infty)$. The background geometry for the holographically dual field theory thus has coordinates $t \in (-\infty,\infty)$, $x_{\parallel} \in (-\infty,\infty)$, and $x_{\perp}\in[0,\infty)$, i.e. \textit{half} of 3d Minkowski space with a boundary at $x_{\perp}=0$. In this sense, these solutions are locally asymptotic to ``half'' of $AdS_4 \times S^7$, as advertised.
As a result, no continuous limit exists in which these solutions reduce to the exact $AdS_4 \times S^7$ solution. In fact, the exact $AdS_4 \times S^7$ solution differs from these solutions in several ways. For example, in these solutions the $h$ in eq.~\eqref{eq:ads4metric} has a single pole at $r = \infty$, and $\gamma \in (-1,0)$, whereas in the exact $AdS_4 \times S^7$ solution $h$ has two poles and $\gamma=1$. Since we cannot continuously send $\gamma \to 1$, the BCFTs holographically dual to our solutions therefore appear to be cousins of, but not continuously connected to, the ABJM BCFT. Identifying the dual BCFTs is an important question we leave for future research.
Eqs.~\eqref{eq:ads4asympmet} and~\eqref{eq:ads4asympcoords} also show that, as in sec.~\ref{subsec:ads7metric}, we can approach the asymptotic (half) $AdS_4 \times S^7$ boundary in two ways. First, if we fix $u$ and send $r \to \infty$, then $z \to 0$ with fixed $x_{\perp}=u$, so that we arrive at the $AdS_4$ boundary a distance $u$ from the BCFT's boundary. Second, if we fix $r$ and send $u \to 0$, then $z \to 0$ but now $x_{\perp} \to 0$, and so we arrive at the $AdS_4$ boundary precisely on the BCFT's 2d boundary.
\subsection{Partitions and M-brane Charges}
\label{sec:fluxes}
The M-brane intersection of sec.~\ref{sec:intro} describes $N$ coincident M2-branes ending on $M$ coincident M5-branes and $M'$ coincident M5$'$-branes. Such an intersection is characterized by a partition of $N$ describing which M5- or M5$'$-brane each M2-brane ends on. In this section we will briefly sketch how we extract this partition from the 11d SUGRA solutions reviewed above, leaving the full details to ref.~\cite{Bachas:2013vza}.
The asymptotically locally $AdS_7 \times S^4$ solutions of sec.~\ref{subsec:ads7metric} describe M2-branes ending only on M5-branes, not on M5'-branes, and are fully characterized by a partition of $N$ M2-branes among the $M$ M5-branes. The asymptotically locally $AdS_4 \times S^7$ solutions of sec.~\ref{subsec:ads4metric} describe M2-branes ending on both M5 and M5'-branes, but are still fully characterized by a partition of $N$ M2-branes among $M$ M5-branes. The special features of both solutions will be discussed in more detail below. Our discussion of the partitions will applies to both solutions, with one crucial exception: the partitions of the $AdS_7 \times S^4$ solutions include cases in which some M5-branes have no M2-branes ending on them, whereas the $AdS_4 \times S^7$ solutions do not admit such partitions.
We will denote the partition as $\rho$, which we will order with entries decreasing from left to right. To illustrate which M2-brane ends on which M5-brane, we will temporarily imagine separating the M2-branes in one of the directions $(x_3,x_4,x_5,x_6)$ and separating the M5-branes in the $x_2$ direction, as shown in figs.~\ref{fig:part1} and~\ref{fig:part2}. Once $\rho$ is specified we will re-collapse the M2- and M5-branes back to the original intersection of coincident stacks.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.8\textwidth]{fig1}
\caption{\label{fig:part1} The three possible partitions $\rho$ with $N=3$ M2-branes and $M=4$ M5-branes. The horizontal axis represents the $x_2$ direction while the vertical axis represents one of the directions $(x_3,x_4,x_5,x_6)$. The horizontal black lines represent M2-branes while the vertical blue lines represent M5-branes. The M5-branes are ordered such that the number of M2-branes on each M5-brane decreases from left to right. The three possible partitions are then $\rho=\{3\}$ (left), $\rho =\{2,1\}$ (middle), and $\rho=\{1,1,1\}$ (right).}
\end{center}
\end{figure}
We will use two parametrizations of $\rho$. In the first parametrization we label $\ell_q$ as the number of M2-branes ending on the $q^{\textrm{th}}$ M5-brane and write the partition as $\rho=\{\ell_1,\ell_2,\ldots\}$ with integers $\ell_1 \geq \ell_2 \geq \ell_3 \ldots$. In this parametrization the total number $M$ of M5-branes is simply the upper limit of $1 \leq q \leq M$, while the total number $N$ of M2-branes is
\begin{equation}
\label{eq:nellsum}
N = \sum_{q=1}^{M} \ell_q.
\end{equation}
We allow for some $\ell_q=0$, representing M5-branes with zero M2-branes ending on them. However, when writing $\rho=\{\ell_1,\ell_2,...\}$ we will omit all zero entries. As examples, fig.~\ref{fig:part1} shows all $\rho$ for the case with $N=3$ and $M=4$.
In the second parametrization we write the partition as
\begin{equation}
\label{eq:part}
\rho = \{ \underbrace{N_1, N_1, ... N_1}_{M_1}, \underbrace{N_2, N_2, ... N_2}_{M_2}, ..., \underbrace{N_n, N_n, ... N_n}_{M_n},\underbrace{N_{n+1}, N_{n+1}, ... N_{n+1}}_{M_{n+1}}\},
\end{equation}
representing $M_1$ M5-branes each with $N_1$ M2-branes ending on them, $M_2$ M5-branes each with $N_2$ M2-branes ending on them, and so on with $N_1 > N_2 > N_3 \ldots$. In other words, the $N_a$ are $n$ \textit{distinct non-zero} integers, each with degeneracy $M_a$. The final entries in $\rho$ are $N_{n+1}=0$, with degeneracy $M_{n+1}$, representing the number of M5-branes with no M2-branes ending on them. We thus specify $\rho$ by specifying the set of integers $\{N_a\}$ and the set of their degeneracies $\{M_a\}$. The partition $\rho$ has a total of $2n+1$ parameters, the $n$ distinct integers in $\{N_a\}$ plus the $n+1$ distinct integers in $\{M_a\}$. In this parametrization the total numbers $N$ of M2-branes and $M$ of M5-branes are
\begin{equation}
\label{eq:totnm}
N = \sum_{a=1}^{n+1} M_a N_a, \qquad M = \sum_{a=1}^{n+1} M_a.
\end{equation}
As in the first parametrization, when writing $\rho$ we will omit all zero entries, that is, we omit the entries $N_{n+1}$. As examples, fig.~\ref{fig:part2} shows the sets $\{N_a\}$ and $\{M_a\}$ for $\rho=\{2,1\}$ with $M=4$ (fig.~\ref{fig:part2} left) and $\rho=\{3,3\}$ with $M=3$ (fig.~\ref{fig:part2} right).
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.8\textwidth]{fig2}
\caption{\label{fig:part2} Two partitions $\rho$ parametrized as in eq.~\eqref{eq:part}: $\rho=\{2,1\}$ with $M=4$ (left) and $\rho=\{3,3\}$ with $M=3$ (right). The horizontal axis represents the $x_2$ direction while the vertical axis represents one of the directions $(x_3,x_4,x_5,x_6)$. The horizontal black lines represent M2-branes while the vertical blue lines represent M5-branes. The $N_a$ are the distinct integers in $\rho$, each with degeneracy $M_a$. Put differently, $M_a$ is the number of M5-branes with $N_a$ M2-branes ending on them.}
\end{center}
\end{figure}
The ordered partition $\rho$ determines a Young tableau, with the physical interpretation that each box in the Young tableau represents an M2-brane and each row represents an M5-brane. In the first parametrization, $\rho = \{\ell_1,\ell_2,\ell_3,\ldots\}$ with $\ell_1 \geq \ell_2 \geq \ell_3 \ldots$, each $\ell_q$ gives the number of boxes in the $q^{\textrm{th}}$ row of the Young tableau, as shown in fig.~\ref{fig:young} (a). In the second parametrization, eq.~\eqref{eq:part}, $M_1$ is the number of rows with $N_1$ boxes, $M_2$ is the number of rows with $N_2 < N_1$ boxes, and so on, as shown in fig.~\ref{fig:young} (b).
When the M2- and M5-branes re-collapse back to the original intersection of coincident stacks, the gauge algebra on the M5-branes' worldvolume is $\mathfrak{su}(M)$ and the M2-branes' boundary represents the Wilson surface. The partition $\rho$, or the corresponding Young tableau, then determines the Wilson surface's representation $\cal{R}$ of $\mathfrak{su}(M)$. Complex conjugation of the representation, $\cal{R} \to \overline{\cal{R}}$, acts in the first parametrization as $\ell_q \to M - \ell_{M-q}$ and in the second parametrization as
\begin{equation}
\label{eq:conjpart}
M_a \rightarrow M_{n+2-a}, \qquad N_a \rightarrow N_1 - N_{2+n-a}.
\end{equation}
As mentioned above, the asymptotically locally $AdS_4 \times S^7$ solutions are also fully characterized by a partition $\rho$ of $N$. Crucially, however, the partition $\rho$ will have only non-zero entries, as we explain below. In particular, in the parametrization by $\{N_a\}$ and $\{M_a\}$ we will not have a value for $M_{n+1}$. Eq.~\eqref{eq:totnm} will thus give us $N$ but not $M$. Of course, generically the $M_a$ with $a<n+1$ will be non-zero, so we know $M$ must be non-zero, but we will not be able to fix its value. In the parametrization $\rho=\{\ell_1,\ell_2,\ell_3,\ldots\}$ all the $\ell_q$ will be non-zero, and the total number $N$ will still be a sum of the $\ell_q$ as in eq.~\eqref{eq:nellsum}, though the upper limit of the sum will be $\sum_{a=1}^n M_a<M$. In a similar fashion, the asymptotically locally $AdS_4 \times S^7$ solutions will have $M'\neq 0$ whose value we will not be able to fix. Without $M$ or $M'$ we will not be able to identify gauge algebras $\mathfrak{u}(M)$ or $\mathfrak{u}(M')$. The partition $\rho$ will still specify a Young tableau, as described above, but without an algebra we will not be able to identify a representation ${\cal R}$ with the Young tableau.
\subsubsection{Asymptotically Locally $AdS_7 \times S^4$ Solutions}
\label{sec:ads7part}
In the asymoptotically locally $AdS_7 \times S^4$ solutions reviewed in sec.~\ref{subsec:ads7metric}, no explicit M2- or M5-brane sources appear in the 11d SUGRA equations. However, the solutions have non-zero flux of the 11d SUGRA 4- or 7-form wrapping closed, non-contractible 4- or 7-cycles, respectively. Presumably the M2- and M5-branes have been replaced by these fluxes, or ``dissolved into flux,'' similar to how D3-branes are replaced by five-form flux in the $AdS_5 \times S^5$ solution of type IIB SUGRA (i.e. how open string degrees of freedom are replaced by closed string degrees of freedom). As explained in ref.~\cite{Bachas:2013vza}, from these fluxes we can recover an ordered partition $\rho$ and hence a Wilson surface representation $\cal{R}$, as follows.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.8\textwidth]{young}
\caption{\label{fig:young} Extracting the same Young tableau from our two different parametrizations of an ordered partition $\rho$. (a) The parametrization $\rho = \{\ell_1,\ell_2,\ell_3,\ldots\}$ with $\ell_1 \geq \ell_2 \geq \ell_3 \ldots$, where each $\ell_q$ gives the number of boxes in the $q^{\textrm{th}}$ row of the Young tableau, with $1 \leq q \leq M$. In the figure we assumed all the $\ell_q$ are non-zero. (b) The parametrization of eq.~\eqref{eq:part}, where $M_1$ is the number of rows with $N_1$ boxes, $M_2$ is the number of rows with $N_2 < N_1$ boxes, and so on. In physical terms, in the Young tableau each box represents an M2-brane and each row represents an M5-brane.}
\end{center}
\end{figure}
How do we identify closed, non-contractible 4- and 7-cycles in the geometry eq.~\eqref{eq:metric2} with the $h$ and $G$ of eq.~\eqref{eq:ads7metric}? As an example, fig.~\ref{fig:ads7plane} shows the upper-half complex $w$ plane for $n=1$, that is, with $2n+2=4$ branch points on the $\textrm{Re}\left(w\right)$ axis, $\xi_1$, $\xi_2$, $\xi_3$, $\xi_4$. Fig.~\ref{fig:ads7plane} also shows that geometry's three independent non-contractible 4-cycles, the curves ${\cal C}_1$, ${\cal C}_1'$, and ${\cal C}_2$, and a single, non-independent 4-cycle, ${\cal C}_3$, obeying ${\cal C}_3={\cal C}_1+{\cal C}_2$. However, unlike the other 4-cycles, ${\cal C}_3$ can be continuously deformed to $r \to \infty$.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.6\textwidth]{ads7plane}
\caption{\label{fig:ads7plane} Sketch of the upper-half complex $w$ plane for the asymptotically locally $AdS_7 \times S^4$ solutions with metric in eq.~\eqref{eq:metric2} with the $h$ and $G$ of eq.~\eqref{eq:ads7metric}. We show an example in which $G$ has four branch points on the $\textrm{Re}\left(w\right)$ axis, $\xi_1$, $\xi_2$, $\xi_3$, $\xi_4$. The geometry has two $S^3$'s. On the $\textrm{Re}\left(w\right)$ axis, in each red segment one of these $S^3$'s collapses to a point while the other does not, while in each blue segment the behavior of the two $S^3$'s is switched. The geometry thus has three non-contractible 4-cycles, ${\cal C}_1$, ${\cal C}_1'$, and ${\cal C}_2$. The 4-cycle ${\cal C}_3={\cal C}_1+{\cal C}_2$ can be continuously deformed to $r \to \infty$, and indeed, in that region becomes the $S^4$ of the asymptotically locally $AdS_7 \times S^4$ region. Taking the product of a non-contractible 4-cycle in which one $S^3$ collapses with the $S^3$ that does not collapse defines a non-contractible 7-cycle. Each \textit{pair} of branch points thus defines a non-contractible 4- and 7-cycle.}
\end{center}
\end{figure}
To see why the curves ${\cal C}_1$, ${\cal C}_1'$, and ${\cal C}_2$ determine closed, non-contractible 4-cycles, recall from sec.~\ref{sec:sugra} that the geometry has two $S^3$'s, and that regularity requires $h=0$ and $G=\pm i$ on the $\textrm{Re}\left(w\right)$ axis. At each point on the $\textrm{Re}\left(w\right)$ axis one of the two $S^3$'s collapses to a point, while the other does not. Specifically, from eqs.~\eqref{eq:metric2},~\eqref{eq:wpm}, and~\eqref{eq:metfuncs} we see that $h=0$ and $G=+i$ implies $f_2=0$ while $f_3\neq 0$, and so one $S^3$ collapses to a point while the other does not. Conversely, $h=0$ and $G=-i$ implies $f_2 \neq 0$ and $f_3=0$ interchanging the behavior of the two $S^3$'s. When we cross a branch point $\xi_j$, $G$ flips sign from $G = \pm i$ to $G = \mp i$, which implies that different $S^3$'s collapse to a point on the two sides of $\xi_j$. As a result, any curve in the upper-half complex $w$ plane that begins on the $\textrm{Re}\left(w\right)$ axis, jumps over \textit{two} branch points, and then ends on the $\textrm{Re}\left(w\right)$ axis, when combined with the $S^3$ that collapses at the curve's endpoints, forms a closed, non-contractible 4-cycle. In contrast, a curve jumping over an odd number of branch points does not define a closed 4-cycle. Each closed, non-contractible 4-cycle also defines a closed, non-contractible 7-cycle: given a 4-cycle in which one $S^3$ collapses, simply take the product with the $S^3$ that does not collapse. In short, every \textit{consecutive pair} of branch points defines a closed, non-contractible 4-cycle and 7-cycle.\footnote{A word of caution: in ref.~\cite{Bachas:2013vza} the notation ${\cal C}_a$ refers only to a curve in the upper-half $w$ plane, while in this paper it refers to a 4-cycle, i.e. the curve times the $S^3$ that collapses at the curve's endpoints. Correspondingly, in this paper a 7-cycle will be denoted ${\cal C}_a \times S^3$.}
As mentioned above, in fig.~\ref{fig:ads7plane} the 4-cycle ${\cal{C}}_3$ can be decomposed as ${\cal C}_3={\cal C}_1+{\cal C}_2$, and can be continuously deformed to $r \to \infty$. As mentioned below eq.~\eqref{eq:ads7asympcoords}, fixing $u$ and sending $r \to \infty$ takes us to the asymptotic $AdS_7\times S^4$ boundary a distance $u$ from the defect, and so we identify the 4-cycle at $r \to \infty$ as the $S^4$ of the asymptotic local $AdS_7 \times S^4$.
The total number of non-contractible 4-cycles in these geometries is $2n+1$, i.e. the number of branch points minus one. The number of non-contractible cycles thus matches the number of free parameters in the solution: as mentioned in sec.~\ref{subsec:ads7metric}, we can fix one of the $2n+2$ branch points using translations, leaving $2n+1$ free parameters.
Integrating $F_4$ over a non-contractible 4-cycle yields a charge that we interpret as a number of dissolved M5- or M5$'$-branes. We will not write the explicit form for $F_4$, which appears in ref.~\cite{Bachas:2013vza}, but we will need the explicit forms for these charges in terms of the locations of branch points $\xi_j$. In the upper-half complex $w$ plane and starting from the left-most cycle, let ${\cal C}_a$, with $a = 1,2,\ldots,n+1$, denote the independent non-contractible 4-cycles all involving the same collapsing $S^3$. In the $n=1$ example above these are the 4-cycles ${\cal C}_1$ and ${\cal C}_2$. Anticipating their role in a partition $\rho$, we refer to the integrals of the 4-form flux over ${\cal C}_a$ as $M_a$, i.e.
\begin{equation}
\label{eq:madef}
M_a \equiv \frac{1}{2 (4\pi^2 G_{\textrm{N}})^{1/3}} \int_{{\cal C}_a} F_4,
\end{equation}
with $G_{\textrm{N}}$ the 11d Newton's constant. For $\gamma\in(-1,0)$, $M_a$ turns out to be proportional to (minus) the distance between two consecutive branch points~\cite{Bachas:2013vza}:\footnote{As mentioned below eq.~\eqref{eq:metfuncs}, these solutions are invariant under simultaneous re-scaling of $c_1$ and $h = - i (w - \overline{w})$. Re-scaling $h$ clearly means re-scaling $w$ and hence the branch points $\xi_j$. As a result, the charges $M_a$, $M_b'$, and $N_a$ in eqs.~\eqref{eq:m5flux},~\eqref{eq:m5primeflux}, and~\eqref{eq:m2flux}, respectively, are invariant under the re-scaling.}
\begin{equation}
\label{eq:m5flux}
M_a = -\frac{1}{c_1^3} \frac{\left(1+\gamma\right)^3}{\gamma} \frac{\left(2\pi\right)^{4/3}}{G_{\textrm{N}}^{1/3}} \left(\xi_{2a}-\xi_{2a-1}\right).
\end{equation}
We have chosen an orientation so that for positive $c_1$ and $\gamma\in(-1,0)$, the $M_a$ are positive. In the upper-half complex $w$ plane and starting from the left-most cycle, let ${\cal C}_b'$, with $b=1,2,\ldots,n$, denote the independent non-contractible 4-cycles that all involve the same collapsing $S^3$ orthogonal to the $S^3$ that collapses in the ${\cal C}_a$ cycles. In the $n=1$ example above, this is the 4-cycle ${\cal C}_1'$. Since the $S^3$ that collapses in the ${\cal C}_b'$ is orthogonal to that of the ${\cal C}_a$, we refer to the integral of the 4-form flux over ${\cal C}_b'$ as M5$'$-brane charge, which we denote $M_b'$. We define $M_b'$ with the same conventions as eq.~\eqref{eq:madef}. For $\gamma\in(-1,0)$, $M_b'$ also turns out to be proportional to the distance between two consecutive branch points~\cite{Bachas:2013vza},
\begin{equation}
\label{eq:m5primeflux}
M_b' = \frac{1}{c_1^3} \frac{\left(1+\gamma\right)^3}{\gamma^2} \frac{\left(2\pi\right)^{4/3}}{G_{\textrm{N}}^{1/3}} \left(\xi_{2b+1}-\xi_{2b}\right).
\end{equation}
Again, we have chosen the orientation so that for positive $c_1$ and $\gamma\in(-1,0)$, the $M_b'$ are positive.
Such M5$'$-brane charge presumably arises from brane polarization, a.k.a. the Myers effect~\cite{Myers:1999ps}: the M2-branes source $C_3$ in a background with non-zero $F_4$ from the M5-branes, so that the Wess-Zumino (WZ) term $\propto \int C_3 \wedge F_4 \wedge F_4$ in the 11d SUGRA action produces non-trivial $F_4$ flux orthogonal to that of the M5-branes, i.e. M5$'$-brane flux.\footnote{The presence of M5$'$-branes is also related to the fact that totally anti-symmetric representations can be described by a probe M5$'$-brane in $AdS_7 \times S^4$, wrapping $AdS_3 \in AdS_7$ and $S^3 \in S^7$~\cite{Lunin:2007ab}.}
The 4-cycle at $r \to \infty$ identified as the $S^4$ in the asymptotically locally $AdS_7 \times S^4$ region can be decomposed as $\sum_{a=1}^{n+1}{\cal C}_a$. Correspondingly, the 4-form flux through that cycle determines the total number of M5-branes, $M=\sum_{a=1}^{n+1} M_a$. Using eq.~\eqref{eq:m5flux}, we can relate $M$ to the coefficient $m_1$ defined in the Legendre polynomial expansion of eq.~\eqref{eq:ads7legendre},
\begin{equation}
\label{eq:ads7m1exp}
m_1 \equiv \sum_{a=1}^{n+1} (\xi_{2a} - \xi_{2a-1}) = -c_1^3 \, \frac{\gamma}{\left(1+\gamma\right)^3} \frac{G_{\textrm{N}}^{1/3}}{\left(2\pi\right)^{4/3}} \, M.
\end{equation}
Plugging eq.~\eqref{eq:ads7m1exp} into eq.~\eqref{eq:ads7rad} we thus recover the usual relation between $M$ and the radius $L_{S^4}$ of the asymptotic $S^4$,
\begin{equation}\label{eq:ls4-to-m}
L_{S^4}^3 = - \frac{1}{c_1^3 }\frac{(1 + \gamma)^3}{\gamma} \, m_1 = \frac{G_{\textrm{N}}^{1/3}}{\left(2\pi\right)^{4/3}} \, M.
\end{equation}
In contrast to the 4-form flux, the integrated 7-form flux, which we interpret as the number of dissolved M2-branes, is ambiguous: due to the WZ term $\propto \int C_3 \wedge F_4 \wedge F_4$ in the 11d SUGRA action, when $F_4\neq 0$ we cannot define 7-form flux that is simultaneously local, conserved, quantized, and invariant under large gauge transformations of $C_3$~\cite{Marolf:2000cb}. At best we can define 7-form flux that has three of these properties but not the fourth. Following ref.~\cite{Bachas:2013vza} we will use the 7-form flux that gives the Page charge, which is local, conserved, and quantized, but not gauge invariant. Explicitly, we will integrate $\star_{11d} F_4 + \frac{1}{2} C_3 \wedge F_4$ over a closed, non-contractible 7-cycle, where the $C_3 \wedge F_4$ term comes from the WZ term, and clearly produces a M2-brane charge that is not invariant under large gauge transformations of $C_3$.
With that choice, to extract a fixed value of M2-brane charge we must fix a gauge of $C_3$, which we do as follows. In the solutions of ref.~\cite{Bachas:2013vza}, $C_3$ can be written as a sum of three terms, one with legs along $AdS_3$, and two with legs along the two $S^3$'s, respectively. Let $C_3'$ denote the term with legs along the $S^3$ that collapses in a 4-cycle ${\cal C}_b'$ but not in a 4-cycle ${\cal C}_a$. To gauge-fix, we demand that $C_3'=0$ when the $S^3$ collapses at one endpoint of one of the ${\cal C}_b'$.\footnote{Regularity demands that $C_3$ vanish whenever it wraps a vanishing cycle, so our gauge choice simply enforces regularity. However, due to the presence of M5-branes we can make $C_3$ well defined only on a patch. The entire geometry can be covered with $n+1$ patches corresponding to $n+1$ gauge choices.} Since the geometry has $n$ 4-cycles ${\cal C}_b'$, each with two endpoints, we have $n+1$ choices of such endpoints, namely the left end-point of each 4-cycle plus the right endpoint of the right-most 4-cycle. For any such choice, in the solutions of ref.~\cite{Bachas:2013vza} the M2-brane charges are ordered from smallest to largest as we move from left to right in the upper-half complex $w$ plane, but generically include negative values. Of the $n+1$ gauge choices for $C'_3$, only one allows for the M2-brane charges to be all positive: choosing $C_3'=0$ at the right endpoint of ${\cal C}_n'$ produces all positive M2-brane charges, with the right-most M2-brane charge vanishing. To see why, deform all the ${\cal C}_a$ to lie along the real axis and note that $\star_{11d}F_4$ vanishes along the real axis since it wraps a vanishing 7-cycle. The only contribution then comes from the WZ contribution $C_3 \wedge F_4$. Since we have set $C_3' = 0$ along the cycle ${\cal C}_{n+1}$, the WZ contribution also vanishes along this cycle.
With $C_3$ gauge-fixed, and anticipating their role in a partition $\rho$, we denote the integrals of 7-form flux through ${\cal C}_a$ as $M_a N_a$, which we interpret as the total number of M2-branes ending on the M5-branes associated with ${\cal C}_a$,\footnote{Our $M_a N_a$ in eq.~\eqref{eq:manadef} was defined as $N_a$ in ref.~\cite{Bachas:2013vza}.}
\begin{equation}
\label{eq:manadef}
M_a N_a = \frac{1}{4 (4\pi^2 G_{\textrm{N}})^{2/3}} \int_{{\cal C}_a \times S^3} C'_3 \wedge F_4.
\end{equation}
For the solutions in ref.~\cite{Bachas:2013vza}, and using eq.~\eqref{eq:m5primeflux}, $N_a$ turns out to be the sum of M5$'$-brane fluxes on all 4-cycles from ${\cal C}'_a$ to ${\cal C}'_n$,
\begin{equation}
\label{eq:m2flux}
N_a = \sum_{b=a}^n M_b' = \frac{1}{c_1^3} \frac{\left(1+\gamma\right)^3}{\gamma^2} \frac{\left(2\pi\right)^{4/3}}{G_{\textrm{N}}^{1/3}} \sum_{b=a}^n\left(\xi_{2b+1}-\xi_{2b}\right),
\end{equation}
and where the vanishing of the right-most 7-form flux implies $N_{n+1}=0$.
Validity of the SUGRA approximation requires curvatures to be small compared to the Planck scale. This requires the branch points $(\xi_{2b}, \xi_{2b+1})$ to be far apart for all $b$, and therefore $N_{a} \gg N_{a+1}$ for all $a$, and also $N_n \gg 1$ (recall the $N_a$ are ordered). However, with $AdS_7 \times S^4$ asymptotics, when the $N_a$ are of order unity the system should be well-described by probe branes. Taking the probe limit in our results for $b_{6d}$ indeed reproduces calculations using probe branes, so our results seem to apply over a wider regime than might be expected. Similar statements about the range of validity of our results apply to what follows (i.e. the $N_b'$ in eq.~\eqref{eq:m2pflux}).
The total number of dissolved M2-branes is the sum of 7-form fluxes of all the ${\cal C}_a$, $N = \sum_{a=1}^n M_a N_a$. Eq.~\eqref{eq:m2flux} shows that the 7-form fluxes through the non-contractible 7-cycles is fixed by 4-form fluxes, and hence are not independent parameters. The total number of free parameters thus remains $2n+1$, though we have a choice to package some of them as either the $M_b'$ or the $N_a$. In what follows we will choose the latter.
In short, eqs.~\eqref{eq:m5flux} and~\eqref{eq:m2flux} allow us to extract the $2n+1$ free parameters of a partition $\rho$, namely the sets of $n$ distinct non-zero integers $\{N_a\}$ and the $n+1$ distinct degeneracies $\{M_a\}$, from the $2n+1$ free parameters, i.e. the branch points $\xi_j$, of the asymptotically locally $AdS_7 \times S^4$ solutions. We therefore identify the same partition $\rho$ determining the brane intersection, the 11d SUGRA solution, and the Wilson surface's representation ${\cal R}$.
For a Wilson line in Yang-Mills theory all observables are invariant under the combined operations of complex conjugation of the representation, ${\cal R} \to \overline {\cal R}$, and orientation reversal of the Wilson line. We expect the same to be true for a Wilson surface in the M5-brane theory. Indeed, we can demonstrate that these combined operations can be realized in these 11d SUGRA solutions by the combined operations of a gauge transformation of $C_3$, which produces all negative M2-brane charges, followed by an orientation reversal of the 7-cycles, which flips the M2-brane charges back to being positive.
To implement these combined operations in the geometry, we start with a geometry parametrized by a partition $\rho$ corresponding to a representation ${\cal R}$. We perform the unique gauge transformation of $C_3'$ that makes all the M2-brane charges negative, namely requiring $C_3'=0$ at the left endpoint of ${\cal C}_1$. Under this gauge transformation, the $N_a$ are all shifted as $N_a \rightarrow N_a - N_1$. The shifted $N_a$ are thus all negative, and of course the shifted $N_1$ vanishes. To obtain positive M2-brane charges we reverse the orientation of the 7-cycles, thereby flipping the signs of the shifted $N_a$. We can do so for example by reversing the orientation of the two $S^3$'s and the direction of integration in the upper-half $w$ plane (reversing the arrows in fig.~\ref{fig:ads7plane}). Finally, it is convenient, but not necessary, to make the coordinate transformation $\textrm{Re}\left(w\right) \rightarrow - \textrm{Re}\left(w\right)$. This yields a geometry parametrized by a set of branch points $\xi_j^c \equiv \xi_{2n+3-j}$ in the same gauge we defined above eq.~\eqref{eq:m2flux}. Computing the M5- and M2-brane charges via eqs.~\eqref{eq:m5flux} and~\eqref{eq:m2flux}, respectively, then reveals that the charges have been shifted as $M_a \to M_{n+2 - a}$ and $N_a \to N_1 - N_{2+n-a}$. From eq.~\eqref{eq:conjpart} we recognize this shift as complex conjugation of the representation, ${\cal R} \to \overline{{\cal R}}$.
Crucially, the geometry is invariant under these combined operations. As we review in sec.~\ref{sec:hc}, the CFT's EE is proportional to the area of a minimal surface in the dual geometry~\cite{Ryu:2006bv,Ryu:2006ef,Nishioka:2009un} (eq.~\eqref{eq:rt} below). That surface only ``knows'' about the geometry, and hence is invariant under any operations that leave the geometry invariant. Indeed, our results for EE, and in particular $b_{6d}$, are invariant under ${\cal R} \to \overline{{\cal R}}$, as mentioned below eq.~\eqref{eq:b6dcompact}.
The operations leading to complex conjugation in these 11d SUGRA solutions have a simple interpretation in the corresponding brane intersection, as Hanany-Witten moves~\cite{Hanany:1996ie}. As a simple example, consider a brane intersection of the type in figs.~\ref{fig:part1} and~\ref{fig:part2}, with $N$ M2-branes ending on $N$ distinct M5-branes (out of the $M$ total number of M5-branes), producing a totally anti-symmetric representation of rank $N$. Imagine we move an M5$'$-brane from infinity on the left to a finite value of $x_2$, to the left of the M5-branes. If we send this M5$'$-brane to infinity on the right, then when the M5$'$-brane passes through the stack of M5-branes the $N$ M2-branes will be destroyed and $M-N$ anti-M2-branes will be created between the M5$'$-brane and each M5-brane that did not have an M2-brane ending on it. An orientation reversal $x_2 \to -x_2$ then maps the anti-M2-branes to M2-branes, while the M5- and M5$'$-branes are not mapped to anti-branes. Clearly the total number of M2-branes is ambiguous, as in the 11d SUGRA solutions above. Moreover, such a Hanany-Witten move clearly corresponds to complex conjugation of the representation, which for a totally anti-symmetric representation of rank $N$ indeed acts as $N \to M-N$.
\subsubsection{Asymptotically Locally $AdS_4 \times S^7$ Solutions}
\label{sec:ads4part}
Similar to the asymptotically locally $AdS_7 \times S^4$ solutions, in the asymptotically locally $AdS_4 \times S^7$ solutions reviewed in sec.~\ref{subsec:ads4metric} no explicit M2- or M5-brane sources appear. However the solution does have non-zero 4- and 7-form fluxes supported on non-contractible 4- and 7-cycles, respectively, representing dissolved M2-, M5-, and M5$'$-branes. We can extract an ordered partition $\rho$ from these fluxes in a fashion very similar to the asymptotically locally $AdS_7 \times S^4$ solutions in sec.~\ref{sec:ads7part}.
The procedure to identify closed, non-contractible 4- and 7-cycles in the asymptotically locally $AdS_4 \times S^7$ solutions is nearly identical to the asymptotically locally $AdS_7 \times S^4$ solutions in sec.~\ref{sec:ads7part}, with one crucial difference: where the latter solutions had an even number of branch points, the asymptotically locally $AdS_4 \times S^7$ solutions have an odd number. Explicitly, the $G$ in eq.~\eqref{eq:ads7metric} involves a sum over $2n+2$ of the branch points $\xi_j$, while the $G$ in eq.~\eqref{eq:ads4metric} is a sum over $2n+1$ of the $\xi_j$. As an example, fig.~\ref{fig:ads4plane} shows the upper-half complex $w$ plane for $n=2$, meaning $2n+1=5$ branch points on the $\textrm{Re}\left(w\right)$ axis, $\xi_1,\ldots,\xi_5$. Fig.~\ref{fig:ads4plane} also shows the geometry's four independent closed, non-contractible 4-cycles, namely ${\cal C}_1$, ${\cal C}_1'$, ${\cal C}_2$, and ${\cal C}'_2$. By exactly the same arguments as in sec.~\ref{sec:ads7part}, each consecutive pair of branch points defines a closed, non-contractible 4- and 7-cycle.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.6\textwidth]{ads4plane}
\caption{\label{fig:ads4plane} Sketch of the upper-half complex $w$ plane for the asymptotically locally $AdS_4 \times S^7$ solutions with metric in eq.~\eqref{eq:metric2} with the $h$ and $G$ of eq.~\eqref{eq:ads4metric}. We show an example in which $G$ has five branch points on the $\textrm{Re}\left(w\right)$ axis, $\xi_1,\ldots,\xi_5$. Like the asymptotically locally $AdS_7 \times S^4$ solutions (i.e. the example in fig.~\ref{fig:ads7plane}), each \textit{pair} of branch points defines a non-contractible 4- and 7-cycle. In this example we labeled the four independent non-contractible 4-cycles ${\cal C}_1$, ${\cal C}_1'$, ${\cal C}_2$, and ${\cal C}'_2$. However, unlike the asymptotically locally $AdS_7 \times S^4$ solutions, in these solutions at $r \to \infty$ the geometry is asymptotically locally $AdS_4 \times S^7$, which has the non-contractible 7-cycle $S^7$ and no non-contractible 4-cycle except $AdS_4$.}
\end{center}
\end{figure}
Another key difference with the asymptotically locally $AdS_7 \times S^4$ solutions is that none of the non-contractible 4-cycles can be continuously deformed to $r \to \infty$, as obvious in the example of fig.~\ref{fig:ads4plane}. Indeed, as $r \to \infty$ these solutions are asymptotically locally $AdS_4 \times S^7$, which has no non-contractible 4-cycles (besides $AdS_4$) and one non-contractible 7-cycle, $S^7$. The latter cannot be decomposed into the other 7-cycles, ${\cal C}_a \times S^3$ and ${\cal C}_b'\times S^3$, which have the topology of $S^4 \times S^3$ and hence have non-contractible 4- and 3-cycles, unlike $S^7$.
We define the closed, non-contractible 4-cycles as in sec.~\ref{sec:ads7part} by the collapse of one or the other $S^3$, denoted ${\cal C}_a$ and ${\cal C}_b'$ with $a,b=1,2,\ldots,n$. The total number of such ${\cal C}_a$ and ${\cal C}_b'$ in these geometries is $2n$. The number of non-contractible 4-cycles thus matches the number of free parameters in the solution: as mentioned in sec.~\ref{subsec:ads4metric}, we can fix one of the $2n+1$ branch points using translations, leaving $2n$ free parameters.
Integrating $F_4$ over a non-contractible 4-cycle gives us 4-form charge, which we again interpret as a number of dissolved M5- or M5$'$-branes. As in sec.~\ref{sec:ads7part}, $M_a$ is defined as the integral of the 4-form flux over ${\cal C}_a$ in eqs.~\eqref{eq:madef}, and similarly for $M_{b}'$ and ${\cal C}_b'$. In these solutions the expressions for $M_a$ and $M_b'$ in terms of the $\xi_j$ are in fact identical to those in eqs.~\eqref{eq:m5flux} and~\eqref{eq:m5primeflux}, respectively.
As in sec.~\ref{sec:ads7part}, and again following ref.~\cite{Bachas:2013vza}, we will use the 7-form that gives the Page charge, $\star_{11d} F_4 + \frac{1}{2} C_3 \wedge F_4$. We thus need to fix a gauge for $C_3$. Following ref.~\cite{Bachas:2013vza} we make the unique gauge choice in which the gauge potential $C_3$ vanishes in the asymptotic region and produces regular 7-form flux through the $S^7$ at $r \to \infty$, whose integral represents the total number of dissolved M2-branes, $N$. Specifically, with $C_3'$ defined in sec.~\ref{sec:ads7part}, we demand that $C_3'=0$ at the left end-point of ${\cal C}_1'$, and that the part of $C_3$ with legs along the $S^3$ that collapses in a 4-cycle ${\cal C}_a$ vanishes at the right endpoint of ${\cal C}_n$. With these gauge choices the WZ contribution vanishes and
\begin{equation}
N = \frac{1}{2 (4\pi^2 G_{\textrm{N}})^{1/3}} \int_{{\cal S}^7} \star_{11d} F_4.
\end{equation}
Additionally, with these gauge choices the expression for $N_a$, the 7-form flux through a 7-cycle ${\cal C}_a \times S^3$, is identical to eq.~\eqref{eq:m2flux}. We interpret $N_a$ as the number of M2-branes that end on the M5-brane associated with ${\cal C}_a$. A crucial difference from sec.~\ref{sec:ads7part}, however, is that now all of these 7-form charges are non-zero, and in particular there are only $n$ independent charge $N_a$, as opposed to the $n+1$ independent charges of sec.~\ref{sec:ads7part}. Although the $S^7$ is not a sum of the 7-cycles ${\cal C}_a \times S^3$, the total number $N$ of M2-branes nevertheless turns out to be the sum of 7-form charges, $N = \sum_{a=1}^n M_a N_a$~\cite{Bachas:2013vza}.
Using eqs.~\eqref{eq:m5flux},~\eqref{eq:m5primeflux}, and~\eqref{eq:m2flux} for $M_a$, $M_b'$, and $N_a$, respectively, we can relate $N$ to the radius $L_{S^7}$ of the $r \to \infty$ asymptotic $S^7$ in eq.~\eqref{eq:ads4rad}. In particular, we need $-m_1^2-m_2$, with the coefficients $m_1$ and $m_2$ defined via the Legendre polynomial expansion of $G$ in eq.~\eqref{eq:ads4legendre},
\begin{equation}
\label{eq:ads4m1m2exp}
-m_1^2 - m_2 = 2 \sum_{a=1}^n \sum_{b=a}^n (\xi_{2a} - \xi_{2a-1}) (\xi_{2b+1} - \xi_{2b}) = 2 c_1^6\,\frac{\left(-\gamma\right)^3}{\left(1+\gamma\right)^6} \,\frac{G_{\textrm{N}}^{2/3}}{\left(2\pi\right)^{8/3}}\, \sum_{a=1}^n M_a N_a.
\end{equation}
Plugging eq.~\eqref{eq:ads4m1m2exp} into eq.~\eqref{eq:ads4rad} and using $N = \sum_{a=1}^n M_a N_a$, we recover the usual relation between $N$ and the radius $L_{S^7}$ of the asymptotic $S^7$,
\begin{equation}
\label{eq:s7radius}
L_{S^7}^6 = \frac{16}{c_1^6}\frac{(1 + \gamma)^6}{(-\gamma)^3} (-m_1^2 - m_2)
= 32 \,\frac{G_{\textrm{N}}^{2/3}}{\left(2\pi\right)^{8/3}}\, N.
\end{equation}
In short, eqs.~\eqref{eq:m5flux} and~\eqref{eq:m2flux} allow us to extract the $2n$ free parameters of a partition $\rho$, the sets of $n$ distinct non-zero integers $\{N_a\}$ and $n$ degeneracies $\{M_a\}$, from the $2n$ free parameters of the asymptotically locally $AdS_4 \times S^7$ solutions, the branch points $\xi_j$.
In contrast to sec.~\ref{sec:ads7part}, only $n$ degeneracies $\{M_a\}$ appear because the geometry has only $n$ 4-cycles ${\cal C}_a$. In particular, as mentioned below eq.~\eqref{eq:conjpart}, we cannot determine a value for $M_{n+1}$ from these solutions, and hence cannot determine $M = \sum_{a=1}^{n+1} M_a$. However, $\sum_{a=1}^n M_a < M$ will be non-zero, implying $M \neq 0$. Analogous statements apply for the $M_b'$ and $M'$. In 11d SUGRA terms, these solutions have no asymptotically $AdS_7 \times S^4$ regions that would allow us to fix $M_{n+1}$ or $M'_{n+1}$. As a result, we will not be able to identify $\mathfrak{su}(M)$ or $\mathfrak{su}(M')$, so while we will have a partition $\rho$ and corresponding Young tableau, we will not be able to identify a representation ${\cal R}$.
Also in contrast to sec.~\ref{sec:ads7part}, the M5-branes and M5$'$-branes appear here on equal footing: no 4-cycle in the asymptotic $r \to \infty$ region selects one over the other. As a result, instead of labelling the solutions by the partition $\rho$ defined by the charges $N_a$ and $M_a$ coming from the ${\cal C}_a$ cycles, we could have labeled the solutions by another partition $\hat \rho$ defined by the charges $M'_b$ and $N'_b$ coming from the ${\cal C}'_b$ cycles,
\begin{equation}
\label{eq:hatrhodef}
\hat \rho = \{ \underbrace{N'_n, N'_n, ... N'_n}_{M'_n}, \underbrace{N'_{n-1}, N'_{n-1}, ... N'_{n-1}}_{M'_{n-1}}, ... \,\underbrace{N'_1, N'_1, ... N'_1}_{M'_1} \} .
\end{equation}
The $M'_b$ are defined as in eq.~\eqref{eq:m5primeflux} and the $N'_b$ are defined by an analogue of eq.~\eqref{eq:m2flux},
\begin{equation}
\label{eq:m2pflux}
N'_b = \sum_{a=1}^b M_a = -\frac{1}{c_1^3} \frac{\left(1+\gamma\right)^3}{\gamma} \frac{\left(2\pi\right)^{4/3}}{G_{\textrm{N}}^{1/3}} \sum_{a=1}^b \left(\xi_{2a}-\xi_{2a-1}\right).
\end{equation}
Requiring the branch points to be well separated so that the SUGRA approximation is reliable, we find the condition \(N_{b+1}' \gg N_b'\) for all \(b\).
The $2n$ free parameters of these solutions are fully determined by either $\rho$ or $\hat \rho$, hence the two must be related. To see how, we use the fact that we can fix any $2n$ free parameters we like, and hence can determine $\rho$ using $\{M_a\}$ and $\{M_b'\}$. In that parametrization, the $M_b'$ are the degeneracies of \textit{columns}, as shown on the left in fig.~\ref{fig:young2}. In the transposed partition, $\rho^T$, the $M_b'$ become the degeneracies of rows, as in $\hat{\rho}$, so we immediately identify $\rho^T = \hat{\rho}$.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.85\textwidth]{young2alt}
\caption{\label{fig:young2} The asymptotically $AdS_4 \times S^7$ solutions are fully characterized by a partition $\rho$ that can be specified by the sets of M2- and M5-brane charges $\{N_a\}$ and $\{M_a\}$, respectively, with $a = 1,2,\ldots,n$, as shown on the left in fig.~\ref{fig:young} . However, $\rho$ can also be specified by the sets of M5- and M5$'$-brane charges $\{M_a\}$ and $\{M_b'\}$, with $b=1,2,\ldots,n$, as shown on the left above. In that parametrization, the $M_b'$ are degeneracies of columns. In the transposed partition $\rho^T$, shown on the right above, the $M_b'$ are degeneracies of rows, as they are in $\hat{\rho}$ in eq.~\eqref{eq:hatrhodef}, hence we identify $\hat{\rho}=\rho^T$.}
\end{center}
\end{figure}
Indeed, we can show that the solution determined by $\rho$ and $\gamma$ is identical, up to a choice of orientation, to the solution determined by $\hat{\rho}=\rho^T$ and $\gamma^{-1}$, as follows. As mentioned above, taking $\gamma \to 1/\gamma$ leaves $f_1$ and $\Omega$ invariant but maps $f_2 \leftrightarrow f_3$, effectively interchanging the geometry's two $S^3$'s and hence swapping the M5- and M5$'$-branes. (The transformation of the 4-form fluxes under $\gamma \to \gamma^{-1}$ is consistent with this statement.) Performing two operations, namely $\gamma \to \gamma^{-1}$ followed by trading the two $S^3$'s, and hence the M5- and M5$'$-branes, is thus a symmetry of the solution. However, the orientation of the 7-cycles, and hence the sign of the M2-brane fluxes, is reversed in the process. The sign can be reversed by an overall orientation reversal, as in the $AdS_7 \times S^4$ case. The effect on $\rho$ is to map $\rho \to \rho^T$, which because the M5- and M5$'$ were swapped, we identify as $\hat{\rho}$. In short, the solution determined by $\rho$ and $\gamma$ is identical to the solution determined by $\hat{\rho}=\rho^T$ and $\gamma^{-1}$, up to an overall orientation reversal, as advertised. We will see in section \ref{sec:abjm-bcft-cc} that our result for EE, and in particular for $b_{3d}$, is indeed invariant, up to an overall sign, under the simultaneous operations $\gamma \rightarrow \gamma^{-1}$ and $\rho \rightarrow \rho^T = \hat{\rho}$.
This symmetry should appear in the holographically dual BCFT. Of course, as mentioned in sec.~\ref{sec:intro}, the exact BCFTs dual to these 11d SUGRA solutions remain unknown. However, the symmetry relating solutions determined by $\rho$ or $\hat{\rho}$ suggests that the BCFTs arise from M2-branes ending on both M5- and M5$'$-branes, as in the brane intersection of sec.~\ref{sec:intro}. If we separate the M5-branes from one another (the Coulomb branch), then a partition $\rho$ will determine which M5-brane each M2-brane ends on. Alternatively, if we separate the M5$'$-branes from one another (the Higgs branch), then the partition $\hat \rho = \rho^T$ will determine which M5$'$-brane each M2-brane ends on. Simultaneously separating both M5- and M5$'$-branes (moving onto the Coulomb and Higgs branches simultaneously) should not be possible. The BCFT should have a duality that simultaneously swaps the separated M5-branes with separated M5$'$-branes, sends $\gamma \to \gamma^{-1}$, and sends $\rho \to \rho^T = \hat{\rho}$.
\section{Holographic Entanglement Entropy}
\label{sec:hc}
In this section we calculate $b_{6d}$ and $b_{3d}$ holographically from EE, following Ryu and Takayanagi's (RT's) prescription~\cite{Ryu:2006bv,Ryu:2006ef,Nishioka:2009un}. A very similar calculation for the asymptotically locally $AdS_7 \times S^4$ solutions appears in ref.~\cite{Gentle:2015jma}. In this section we will follow ref.~\cite{Gentle:2015jma} very closely.
To compute EE in a QFT vacuum we fix time $t$, separate space into two regions by an ``entangling surface,'' and trace out states outside the entangling surface, thus obtaining a reduced density matrix for the region inside. The EE is this reduced density matrix's von Neumann entropy. Generically EE diverges due to strong UV correlations near the entangling surface, so to extract physical information we must introduce a UV regulator.
Holographically, in an asymptotically AdS geometry, RT's prescription for the EE for a sub-region of the CFT is
\begin{equation}
\label{eq:rt}
S_{\textrm{EE}} = \frac{A}{4 G_{\textrm{N}}},
\end{equation}
where $A$ is the area of the minimal surface in the holographically dual geometry that approaches the entangling surface at the AdS boundary. Computing $S_{\textrm{EE}}$ is thus a two-step process. First, determine the minimal area surface by writing the area functional and solving the associated Euler-Lagrange equations. Second, plug that solution back into the area functional and integrate to obtain $A$. The UV divergences of $S_{\textrm{EE}}$ appear as divergences in $A$ near the AdS boundary. In AdS spacetime, the standard regulator is thus a cutoff on the FG holographic coordinate: rather than integrating to the AdS boundary $z=0$ we integrate only to $z= \varepsilon>0$.
In $AdS_3$ eq.~\eqref{eq:rt} reproduces known results for 2d CFTs~\cite{Ryu:2006bv,Ryu:2006ef,Nishioka:2009un,Holzhey:1994we,Calabrese:2004eu}. For example, when the entangling surface consists of two points a distance $\ell$ apart, the minimal surface in $AdS_3$ is a semi-circle at fixed $t$ with diameter $\ell$ centered on the $AdS_3$ boundary. In that case eq.~\eqref{eq:rt} gives
\begin{equation}
\label{eq:ee2d}
S_{\textrm{EE}}^{2d} = \frac{c}{3} \ln \left(\frac{\ell}{\varepsilon}\right) + \mathcal{O}\left(\varepsilon^0\right),
\end{equation}
with CFT central charge $c$. Henceforth, we will use a superscript to distinguish $S_{\textrm{EE}}$ in different dimensions, such as the superscript $2d$ on $S_{\textrm{EE}}^{2d}$. Crucially, re-scaling the cutoff $\varepsilon$ changes the $\mathcal{O}\left(\varepsilon^0\right)$ terms, while the coefficient of $\ln \left(\frac{\ell}{\varepsilon}\right)$, namely $c/3$, is cutoff-independent and hence physical. In $AdS_4$ eq.~\eqref{eq:rt} produces the form expected for a 3d CFT~\cite{Ryu:2006bv,Ryu:2006ef,Nishioka:2009un}. For example when the entangling surface is a circle of radius $\ell$, the minimal surface in $AdS_4$ is a hemisphere at fixed $t$ with radius $\ell$ centered on the $AdS_4$ boundary. In that case eq.~\eqref{eq:rt} gives
\begin{equation}
\label{eq:ee3d}
S_{\textrm{EE}}^{3d} = c_1 \, \frac{\ell}{\varepsilon} + c_0 + \mathcal{O}\left(\varepsilon\right),
\end{equation}
where $c_1$ and $c_0$ are constants. Re-scalings of the cutoff change $c_1$ but not $c_0$, so only the latter is physical. Indeed $c_0$ is proportional to minus the logarithm of the Euclidean CFT partition function on $S^3$~\cite{Casini:2011kv}. The $AdS_4 \times S^7$ solution of 11d SUGRA gives $c_0 \propto - N^{3/2}$~\cite{Klebanov:1996un,Drukker:2010nc}. In $AdS_7$ eq.~\eqref{eq:rt} produces the form expected for a 6d CFT~\cite{Ryu:2006bv,Ryu:2006ef,Nishioka:2009un}. For example, when the entangling surface is an $S^4$ of radius $\ell$, the minimal surface in $AdS_7$ is a five-dimensional hemisphere at fixed $t$ with radius $\ell$ centered on the $AdS_7$ boundary. In that case eq.~\eqref{eq:rt} gives
\begin{equation}
\label{eq:ee6d}
S_{\textrm{EE}}^{6d} = c_4 \, \frac{\ell^4}{\varepsilon^4} + c_2 \, \frac{\ell^2}{\varepsilon^2} + c_L \ln \left(\frac{\ell}{\varepsilon}\right) + \mathcal{O}\left(\varepsilon^0\right),
\end{equation}
where $c_4$, $c_2$ and $c_L$ are constants. Only $c_L$ is invariant under re-scalings of the cutoff, hence only $c_L$ is physical. Indeed, $c_L\propto - a$, where $a$ is a central charge of the 6d CFT~\cite{Casini:2011kv}. The $AdS_7 \times S^4$ solution of 11d SUGRA gives $c_L \propto - a \propto -M^3$~~\cite{Freed:1998tg,Harvey:1998bx,Henningson:1998gx}.
Following refs.~\cite{Jensen:2013lxa,Estes:2014hka,Gentle:2015jma}, in our (B)CFTs we choose (hemi-)spherical entangling surfaces centered on the 2d defect or boundary, as follows. For the 11d SUGRA solutions reviewed in sec.~\ref{subsec:ads7metric}, dual to the M5-brane theory with a Wilson surface, our entangling surface will be an $S^4$ of radius $\ell$ centered on the Wilson surface, as shown in fig.~\ref{fig:eesurf} (a). For the 11d SUGRA solutions reviewed in sec.~\ref{subsec:ads4metric}, dual to cousins of the ABJM BCFT, our entangling surface will be a semi-circle centered on the CFT's boundary, as shown in fig.~\ref{fig:eesurf} (b).
\begin{figure}[t!]
\begin{center}
\includegraphics[width=.4\textwidth]{eesurf}
\caption{\label{fig:eesurf} Schematics of our entangling surfaces. (a) For the M5-brane theory with a Wilson surface, the blue square represents $\mathbb{R}^5$ at fixed $t$, the vertical black line represents the Wilson surface, and the black circle represents our entangling surface, an $S^4$ of radius $\ell$ centered on the Wilson surface. (b) In cousins of the ABJM BCFT, the blue rectangle represents half of $\mathbb{R}^2$ at fixed $t$, denoted $\mathbb{R}^2_+$, with the boundary at left, and our entangling surface is the semi-circle of radius $\ell$ centered on the boundary.}
\end{center}
\end{figure}
As mentioned above, the first step in the holographic calculation of $S_{\textrm{EE}}$ is to find the minimal surface in the holographically dual geometry at fixed $t$ that approaches our entangling surface at the asymptotic boundary. Luckily, this first step has already been done for us in refs.~\cite{Jensen:2013lxa,Estes:2014hka}. Actually refs.~\cite{Jensen:2013lxa,Estes:2014hka}'s results are much more general: for \textit{any} holographic dual of a CFT with conformal defect or boundary, and for a (hemi-)spherical entangling surface centered on the defect or boundary, refs.~\cite{Jensen:2013lxa,Estes:2014hka} found the solution for the \textit{global} minimum of the area functional.
We can immediately adapt the solution of refs.~\cite{Jensen:2013lxa,Estes:2014hka} to our case: the minimal surface at fixed $t$ wraps both $S^3$'s and the upper-half complex $w$ plane, and in the $AdS_3$ subspace is given by $x_{\parallel}^2 + u^2 = \ell^2$. A quick check of this solution is that for $r \to \infty$ and fixed $u$, which takes us to the asymptotic boundary at $x_{\perp}=u$, the surface becomes $x_{\parallel}^2 + x_{\perp}^2 = \ell^2$, which is indeed the equation for an $S^4$ in $\mathbb{R}^5$ or semi-circle in half of $\mathbb{R}^2$. For fixed $r$ and $u \to 0$, which takes us to the $AdS_3$ boundary, the surface reduces to $x_{\parallel} = \pm \ell$, representing the endpoints of an interval of length $\ell$ on the 2d defect or boundary.
As mentioned above, the second step in the holographic calculation of $S_{\textrm{EE}}$ is to plug the solution for the minimal surface into the area functional and then integrate to obtain the minimal area and hence $S_{\textrm{EE}}$. In our cases, plugging the solution for the minimal surface into the area functional produces
\begin{equation}
\label{eq:integral}
S_{\textrm{EE}} = \frac{2 \,\textrm{Vol}\left(S^3\right)^2}{4 G_{\textrm{N}}} \int dw \, d\overline{w} \left( \frac{\Omega^2}{f_1^2} \, f_1^3 f_2^3 f_3^3 \right) \int \frac{du\,\ell}{u \sqrt{\ell^2 - u^2}},
\end{equation}
where $\textrm{Vol}\left(S^3\right)=2\pi^2$ is the volume of a unit-radius $S^3$, and the integrals are over the upper-half complex $w$ plane and $u \in [0,\ell]$, resdpectively. The latter integration only covers one branch of the minimal surface $x_{\parallel}=\pm \sqrt{\ell^2 - u^2}$, hence the overall factor of $2$.
If we switch to the FG coordinates of eq.~\eqref{eq:ads7asympmet} or~\eqref{eq:ads4asympmet} and introduce an FG cutoff $z=\varepsilon$, then the $S_{\textrm{EE}}$ in eq.~\eqref{eq:integral} exhibits $\varepsilon\to 0$ divergences, as expected. In particular, $S_{\textrm{EE}}$ for the asymptotically locally $AdS_7 \times S^4$ solutions looks like the EE of the 6d CFT, of the form of $S_{\textrm{EE}}^{6d}$ in eq.~\eqref{eq:ee6d}, plus a contribution from the 2d defect that has the form of $S_{\textrm{EE}}^{2d}$ in eq.~\eqref{eq:ee2d}. This structure is common in holographic calculations of EE for CFTs with defects or boundaries~\cite{Jensen:2013lxa,Estes:2014hka,Gentle:2015jma}. As mentioned in sec.~\ref{sec:intro}, following refs.~\cite{Estes:2014hka,Gentle:2015jma} we will subtract the 6d CFT's contribution (using the same cutoff) and then extract the coefficient of any remaining logarithmic term. In other words, we will extract the \textit{change} in the coefficient of the logarithmic term due to the Wilson surface. In SUGRA terms, we will subtract the area of the minimal surface in $AdS_7 \times S^4$ bounded by a sphere of radius $\ell$ on the boundary of $AdS_7$, described above. Stated precisely, we will compute
\begin{equation}
\label{eq:b6ddef}
b_{6d}= 3 \,\ell \, \frac{d}{d\ell}\left(S_{\textrm{EE}}-S_{\textrm{EE}}^{6d}\right),
\end{equation}
where because $S_{\textrm{EE}}-S_{\textrm{EE}}^{6d}$ is of the form of $S_{\textrm{EE}}^{2d}$ in eq.~\eqref{eq:ee2d}, $\ell \, \frac{d}{d\ell}$ extracts the coefficient of the logarithmic term, and the factor of $3$ simply accounts for the normalization of the central charge in eq.~\eqref{eq:ee2d}.
For the asymptotically locally $AdS_4 \times S^7$ solutions our result for $S_{\textrm{EE}}$ looks like \textit{half} the EE of a 3d CFT, meaning $1/2$ times the $S_{\textrm{EE}}^{3d}$ in eq.~\eqref{eq:ee3d}, plus a contribution from the CFT's 2d boundary of the form of $S_{\textrm{EE}}^{2d}$ in eq.~\eqref{eq:ee2d}. Intuitively, in CFT terms the $1/2$ factor appears because introducing the boundary ``cuts off'' half of the 3d CFT, or in SUGRA terms because these solutions are only asymptotic locally to ``half of'' $AdS_4 \times S^7$, as discussed in sec.~\ref{subsec:ads4metric}. Again following refs.~\cite{Estes:2014hka,Gentle:2015jma}, we will subtract the 3d contribution (using the same cutoff), and then extract the coefficient of the logarithmic term. In other words, we will extract the \textit{change} in the coefficient of the logarithmic term due to the CFT's 2d boundary---which of course comes entirely from the 2d boundary, since $S_{\textrm{EE}}^{3d}$ has no logarithmic term. In SUGRA terms, we will subtract $1/2$ the area of the minimal surface in $AdS_4 \times S^7$ that approaches a circle of radius $\ell$ at the boundary, described above. Stated precisely, we will compute
\begin{equation}
\label{eq:b3ddef}
b_{3d} = 3 \,\ell \, \frac{d}{d\ell}\left(S_{\textrm{EE}}-\frac{1}{2}S_{\textrm{EE}}^{3d}\right).
\end{equation}
The FG cutoff $z = \varepsilon$ preserves the symmetry of a Minkowski slice at fixed $z$, dual to the CFT's Poincar\'e symmetry. However in our CFTs the 2d defect or boundary breaks Poincar\'e symmetry to the subgroup that leaves the defect or boundary invariant. In what follows we will thus not use the FG cutoff $z = \varepsilon$, rather we will use the cutoff prescription of refs.~\cite{Estes:2014hka,Gentle:2015jma} for the $u$ and $w$ coordinates, which preserves the reduced symmetry.
The prescription of refs.~\cite{Estes:2014hka,Gentle:2015jma} actually involves two cutoffs. First is an FG cutoff in $AdS_3$, that is, in eq.~\eqref{eq:integral} we perform the $u$ integration from a cutoff $u=\varepsilon_u>0$ to $u=\ell$,
\begin{equation}
\label{eq:uint}
\int_{\varepsilon_u}^{\ell} du \frac{\ell}{u \sqrt{\ell^2 - u^2}} = \ln\left(\frac{2\ell}{\varepsilon_u}\right)+{\cal{O}}\left(\varepsilon_u^2\right),
\end{equation}
so that the integral for $S_{\textrm{EE}}$ in eq.~\eqref{eq:integral} becomes
\begin{equation}
\label{eq:integral2}
S_{\textrm{EE}} = \frac{2 \, \textrm{Vol}\left(S^3\right)^2}{4 G_{\textrm{N}}} \int dw \, d\overline{w} \left( \frac{\Omega^2}{f_1^2} \, f_1^3 f_2^3 f_3^3 \right) \left[\ln\left(\frac{2\ell}{\varepsilon_u}\right)+{\cal{O}}\left(\varepsilon_u^2\right)\right].
\end{equation}
We define ${\cal{I}}$ as the remaining integral,
\begin{equation}\label{eq:Iastintegral}
{\cal{I}}\equiv \frac{2 \, \textrm{Vol}\left(S^3\right)^2}{4 G_{\textrm{N}}} \int dw \, d\overline{w} \left( \frac{\Omega^2}{f_1^2} \, f_1^3 f_2^3 f_3^3 \right).
\end{equation}
To write ${\cal I}$ explicitly we plug in $\textrm{Vol}\left(S^3\right)=2\pi^2$,
\begin{align}
\label{eq:metproducts}
&\frac{\Omega^2}{f_1^2} = \frac{1}{2}
\frac{c_1^2}{c_2 c_3} \frac{|\partial_w h|^2}{h^2} (G \overline{G} - 1) \,, &
&f_1 f_2 f_3 = \pm \frac{h}{c_1 c_2 c_3}\,,&
\end{align}
and in the second equation of eq.~\eqref{eq:metproducts} we choose the sign to guarantee a positive integrand, given that $|G|<1$ as mentioned below eq.~\eqref{eq:wpm}. Eq.~\eqref{eq:Iastintegral} then becomes
\begin{equation}
\label{}
{\cal I} = \frac{2 (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{1}{c_1 c_2^4 c_3^4}\frac{1}{2}\int dw \, d\overline{w} \, |\partial_w h|^2 \, h \left(1-G \overline{G}\right).
\end{equation}
Using $h = - i \left(w-\overline{w}\right)$ from eqs.~\eqref{eq:ads7metric} and~\eqref{eq:ads4metric}, we have $|\partial_w h|^2 = 1$. Introducing polar coordinates $w \equiv r e^{i \theta}$, so that $dw\,d\overline{w} = 2\,d\theta \,dr \, r$ and $h = -i (w - \overline{w})=2r \sin \theta$, we find
\begin{align}
\label{eq:integral3}
{\cal I} =& \frac{2 (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{2}{c_1 c_2^4 c_3^4} \int_0^{\pi} d \theta \sin\theta
\int_0^{r_c} dr \, r^2 (1 - G \overline{G}),
\end{align}
where we have made the endpoints of integration explicit, including a large-$r$ cutoff, $r_c$. The prescription of refs.~\cite{Estes:2014hka,Gentle:2015jma} is to choose $r_c$ in a way that preserves the subgroup of the Poincar\'e group that leaves the 2d defect or boundary invariant. Crucially, a constant $r_c$ does not preserve those symmetries, rather $r_c$ must be a more complicated function whose form depends on the details of the 11d SUGRA solution. In the next two subsections we will compute $r_c$ and then extract ${\cal I}$ in eq.~\eqref{eq:integral3}.
As mentioned above, in principle we would like to extract $b_{6d}$ or $b_{3d}$ from a term in $S_{\textrm{EE}}$ that is $\propto \ln(\ell/\varepsilon)$, with FG cutoff $\varepsilon$. However, how do we do so using the cutoffs $\varepsilon_u$ and $r_c$? The result for the integral ${\cal I}$ will be a sum of terms, including terms with positive powers of $r_c$, a term independent of $r_c$, and terms with negative powers of $r_c$. In eq.~\eqref{eq:integral2} these all multiply $\ln(\ell/\varepsilon_u)$. The terms with positive powers of $r_c$, which are clearly cutoff-dependent and hence unphysical, will turn out to be identical to those of the undeformed $AdS_7 \times S^4$ or (half of) $AdS_4 \times S^7$ solutions, and so will cancel in the background subtraction $S_{\textrm{EE}} - S_{\textrm{EE}}^{6d}$ or $S_{\textrm{EE}} - \frac{1}{2} S_{\textrm{EE}}^{3d}$. The terms with negative powers of $r_c$ clearly vanish as $r_c \to \infty$ and so can be safely ignored. We will thus be left with the term independent of $r_c$, or more precisely what remains of that term after the background subtraction, which still multiplies $\ln(\ell/\varepsilon_u)$. Applying $3 \ell \frac{d}{d\ell}$, as in eqs.~\eqref{eq:b6ddef} and~\eqref{eq:b3ddef}, then extracts this coefficient of $\ln(\ell/\varepsilon_u)$, which is thus our $b_{6d}$ or $b_{3d}$. In short, we will apply eqs.~\eqref{eq:b6ddef} and~\eqref{eq:b3ddef} as advertised, though the form of divergences will look very different in terms of $\varepsilon_u$ and $r_c$ as compared to the usual FG cutoff $\varepsilon$. For a more detailed comparison of these cutoffs, see ref.~\cite{Gentle:2015jma}.
\subsection{Asymptotically Locally $AdS_7 \times S^4$ Solutions}
For the asymptotically locally $AdS_7 \times S^4$ solutions reviewed in sec.~\ref{subsec:ads7metric} we follow ref.~\cite{Gentle:2015jma} very closely, but with three major differences. First, where ref.~\cite{Gentle:2015jma} set $\gamma= -1/2$ we will leave $\gamma$ arbitrary as long as practicable. Second, where ref.~\cite{Gentle:2015jma} set $m_1=2$, we will leave $m_1$ arbitrary. Third and most importantly, we will translate our result into the data $\{N_a\}$ and $\{M_a\}$ of a partition/Young tableau, as described in sec.~\ref{sec:fluxes}.
As discussed above, we implement the double cutoff of refs.~\cite{Estes:2014hka,Gentle:2015jma}. First, in the asymptotic large-$r$ region we put the metric in a form that makes manifest the symmetries of the 2d defect,
\begin{equation}
\label{eq:ads7metriccutoff}
ds^2 = 4 L_{S^4}^2 \left( \frac{dv^2}{v^2}+ \frac{ds^2_{\text{AdS}_3} + \frac{(1+\gamma)^2}{\gamma^2} ds^2_{S^3}}{v^2} \right) + L_{S^4}^2 \left(d\tilde \theta^2 + \sin^2(\tilde \theta) ds^2_{S^3} \right) + \ldots,
\end{equation}
where asymptotically at large $r$,
\begin{subequations}
\begin{align}
\label{eq:AdS7FGtrans}
v =& \, \frac{1+\gamma}{\sqrt{-\gamma}} \frac{\sqrt{2m_1}}{\sqrt{r}} \left(1 + \frac{(1 + 2 \gamma)(-3 + \cos(2\theta))m_1^2 + 12 \gamma m_2 \cos\theta}{48 \gamma m_1} \frac{1}{r} + {\cal O}\left(\frac{1}{r^2}\right) \right),\\
\tilde \theta =& \, \theta + \frac{(1 + 2\gamma)m_1^2 \cos\theta + 3 \gamma m_2}{6 \gamma m_1} \frac{\sin\theta}{r} + {\cal O}\left(\frac{1}{r^2}\right),
\end{align}
\end{subequations}
where $v \in [0,\infty)$, $\tilde{\theta} \in [0,\pi]$, and the ellipsis represent terms orthogonal to $v$ and sub-leading in $1/r$. In these coordinates, we can approach the asymptotic local $AdS_7 \times S^4$ boundary in two ways. First is to fix $u$ and send $v \to 0$, where the latter is equivalent via eq.~\eqref{eq:AdS7FGtrans} to $r \to \infty$, which takes us to the asymptotically local $AdS_7 \times S^4$ boundary at a point away from the 2d defect. For $\gamma = -1/2$ the dual field theory's metric is that of $AdS_3 \times S^3$, which is conformal to 6d Minkowski space, while for $\gamma\neq -1/2$ a conical singularity appears, as in eq.~\eqref{eq:ads7asympmet}. Second is to fix $v$ and send $u \to 0$, which takes us to the $AdS_3$ boundary, i.e. to a point on the 2d defect.
The prescription of refs.~\cite{Estes:2014hka,Gentle:2015jma} is to introduce an FG cutoff for $v$, that is a cutoff $v = \varepsilon_v>0$, which translates to a cutoff $r_c$ that depends on $\varepsilon_v$ and $\theta$. Explicitly, in eq.~\eqref{eq:AdS7FGtrans} we set $v = \varepsilon_v$ and then invert to find $r_c(\varepsilon_v,\theta)$ in a small $\varepsilon_v$ expansion,
\begin{align}
\label{eq:AdS7cutoff}
&r_c(\varepsilon_v,\theta) = \frac{2 (1+ \gamma)^2 m_1}{-\gamma} \frac{1}{\varepsilon_v^2} + \frac{(1+ 2 \gamma)m_1}{-8 \gamma} + \frac{m_2}{2m_1} \cos\theta - \frac{(1+2 \gamma) m_1}{-24 \gamma} \cos(2\theta) \nonumber\\
& + \left\{ \frac{89 m_1^4+212 \gamma(1 + \gamma)m_1^4 + 729 \gamma^2 m_2^2 - 576 \gamma^2 m_1 m_3}{9216 \gamma(1+\gamma)^2 m_1^3}
+ \frac{(1+2\gamma)m_2}{24 (1 + \gamma^2)m_1}\cos\theta \sin^2\theta \right . \nonumber\\&
+ \left . \frac{3m_1^4 + 8 \gamma m_1^4 + 8 \gamma^2 m_1^4 - 36 \gamma^2 m_2^2 + 80 \gamma^2 m_1 m_3}{768 \gamma(1+\gamma)^2 m_1^3} \cos(2\theta) + \frac{(1+2\gamma)^2 m_1}{3072 \gamma(1+\gamma)^2} \cos(4\theta) \right\} \varepsilon_v^2 \nonumber \\ & + {\cal O}(\varepsilon_v^4),
\end{align}
where we have kept an additional order as compared to eq.~\eqref{eq:AdS7FGtrans}, which will be necessary to extract $b_{6d}$ from $S_{\textrm{EE}}$. When $\gamma = - 1/2$, eq.~\eqref{eq:AdS7cutoff} simplifies considerably,
\begin{align}
\label{eq:rcads7m12}
r_c(\varepsilon_v,\theta) =& \frac{m_1}{\varepsilon_v^2} + \frac{m_2}{2m_1} \cos(2\theta) +\left( \frac{4 m_1 m_3 - m_1^4 - 5 m_2^2}{32 m_1^3} + \frac{20 m_1 m_3 - 9 m_2^2 + m_1^4}{96 m_1^3} \cos(2\theta) \right) \varepsilon_v^2 \nonumber \\ & + {\cal O}(\varepsilon_v^4).
\end{align}
Plugging the expression for $G$ in eq.~\eqref{eq:ads7metric} into eq.~\eqref{eq:integral3} gives
\begin{align}
\label{eq:ads7int}
{\cal I} =& \frac{2 (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{2}{c_1 c_2^4 c_3^4} \sum_{j=1}^{2n+2} (-1)^j \int_0^{\pi} d \theta \sin\theta
\int_0^{r_c(\varepsilon_v,\theta)} dr \, r^2 \frac{2(r \cos(\theta) - \xi_j)}{\sqrt{r^2 + \xi_j^2 - 2 r \xi_j \cos(\theta)}} \nonumber\\
&+ \frac{2 (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{2}{c_1 c_2^4 c_3^4} \sum_{j,k=1}^{2n+2} (-1)^{j+k} \int_0^{\pi} d \theta \sin\theta
\int_0^{r_c(\varepsilon_v,\theta)} dr \, r^2 \frac{r e^{i \theta} - \xi_j}{|r e^{i \theta} - \xi_j|}\frac{r e^{-i \theta} - \xi_k}{|r e^{i \theta} - \xi_k|},
\end{align}
where $r_c(\varepsilon_v,\theta)$ is the cutoff in eq.~\eqref{eq:AdS7cutoff}. The integrals in eq.~\eqref{eq:ads7int} are performed in ref.~\cite{Gentle:2015jma},\footnote{In ref.~\cite{Gentle:2015jma} the first and second lines of eq.~\eqref{eq:ads7int} are denoted $J_1$ and $J_2$, respectively.} and are very similar to those in the asymptotically locally $AdS_4 \times S^7$ case performed in the appendix, so here we only quote the result:
\begin{align}
\label{eq:ads7inttemp}
{\cal I} =& -\frac{8(2 \pi^2)^2 L_{S^4}^9}{4 G_{\textrm{N}} \gamma (1 + \gamma)}
\Bigg(\frac{8 (1+\gamma)^4}{3\gamma^2}\frac{1}{\varepsilon_v^4}+\frac{2(1+\gamma)^2(3+16\gamma)}{15\gamma^2}\frac{1}{\varepsilon_v^2} -\frac{85-8\gamma(52+115\gamma)}{5040\gamma^2}\nonumber\\
&+\frac{m_2^2}{10m_1^4}-\frac{2m_3}{15m_1^3} -\frac{1}{3m_1^3}\sum_{\underset{j<k}{j,k=1}}(-1)^{j+k}|\xi_j-\xi_k|^3\Bigg),
\end{align}
where we dropped terms that vanish as $\varepsilon_v \to 0$. We will continue to do so in what follows. If we set $\gamma=-1/2$ then eq.~\eqref{eq:ads7inttemp} becomes
\begin{align}
\label{eq:ads7int2}
{\cal I} = \frac{2^6\pi^4 L_{S^4}^9}{3 \,G_{\textrm{N}}}
\Bigg(\frac{1}{\varepsilon_v^4}-\frac{1}{\varepsilon_v^2} - \frac{3}{40}
+\frac{3 m_2^2 - 4 m_1 m_3}{20 m_1^4}-\frac{1}{2m_1^3}\sum_{\underset{j<k}{j,k=1}}^{2n+2} (-1)^{j+k}|\xi_j-\xi_k|^3\Bigg).
\end{align}
As expected, eq.~\eqref{eq:ads7int2} contains terms that diverge as $\varepsilon_v \to 0$. To extract $b_{6d}$ using eq.~\ref{eq:b6ddef}, we will need the result for ${\cal I}$ for the exact $AdS_7 \times S^4$ solution, which we denote ${\cal I}_{6d}$. As mentioned at the end of sec.~\ref{subsec:ads7metric}, the $AdS_7 \times S^4$ solution has two branch points at $\textrm{Re}\left(w\right)=\pm\xi$. These can be eliminated by conformally mapping the upper half plane to a semi-infinite strip via $w = \xi \cosh Z$, where $Z=X+iY$ with $X\in [0,\infty]$ and $Y\in[0,\pi]$. In the semi-infinite strip coordinates,
\begin{equation}
h=-i\xi\left(\cosh Z-\cosh\overline{Z}\right), \qquad G=-i \left( 1+2\,\frac{\sinh(\frac{Z-\overline{Z}}{2})}{\sinh \overline{Z}} \right).
\end{equation}
In semi-infinite strip coordinates the metric takes the form
\begin{align}
\label{AdS7metric}
ds^2=4L_{S^4}^2\left( \cosh^2 \frac{X}{2} \, ds^2_{AdS_3}+\sinh^2 \frac{X}{2} \, ds^2_{S^3} + \frac{dX^2}{4} \right) + L_{S^4}^2\, ds^2_{S^4}.
\end{align}
Mapping eq.~\eqref{AdS7metric} to the asymptotic form in eq.~\eqref{eq:ads7metriccutoff} gives at leading order $v=2 e^{-X/2}$. As a result, the cutoff $\varepsilon_v$ maps to a cutoff $X_c = -2\log(\varepsilon_v/2)$. Plugging eq.~\eqref{AdS7metric} into eq.~\eqref{eq:Iastintegral} and performing the integration with the cutoff $X_c$, we find
\begin{align}
\label{eq:ads7I}
&{\cal I}_{6d} =\frac{2^6\pi^4 L_{S^4}^9} {3 \,G_{\textrm{N}}} \left(\frac{1}{\varepsilon_v^4}-\frac{1}{\varepsilon_v^2}+\frac{3}{8}\right).
\end{align}
Comparing eqs.~\eqref{eq:ads7int2} and~\eqref{eq:ads7I}, we see that all divergent terms cancel in ${\cal I} - {\cal I}_{6d}$, as advertised. Extracting $b_{6d}$ via eq.~\eqref{eq:b6ddef} then gives
\begin{align}
\label{eq:b6draw}
b_{6d}=&
\frac{2^6\pi^4 L_{S^4}^9}{G_{\textrm{N}}}
\Bigg\{- \frac{9}{20}
+\frac{3\, m_2^2 - 4 \, m_1 m_3}{20 m_1^4}+\frac{1}{2m_1^3}\sum_{\underset{j<k}{j,k=1}}^{2n+2} (-1)^{j+k}(\xi_j-\xi_k)^3\Bigg\}.
\end{align}
\vspace{-0.25cm}If we use the scaling symmetry to set $m_1=2$, then eq.~\eqref{eq:b6draw} reproduces the result of ref.~\cite{Gentle:2015jma}. However, we can go farther, and write $b_{6d}$ of eq.~\eqref{eq:b6draw} in terms of the partition data $\{N_a\}$ and $\{M_a\}$, as follows. On the right-hand-side of eq.~\eqref{eq:b6draw}, we use eq.~\eqref{eq:ads7m1exp} to replace the factors of $m_1$ in the denominators with factors of $M$. Next we consider the combination $3 m_2^2 - 4 m_1 m_3$ in the second term on the right-hand side of eq.~\eqref{eq:b6draw}. While $m_2$ and $m_3$ are not individually invariant under translations of the $\xi_j$, the combination $3 m_2^2 - 4 m_1 m_3$ is invariant. That is important, since such shifts are equivalent to a coordinate transformation, under which our final expression must be invariant. Using the definition of the $m_k$ in terms of the $\xi_j$ in eq.~\eqref{eq:ads7legendre}, we can write
\begin{align}
&3 m_2^2 - 4 m_1 m_3 = - \left(\sum_{a=1}^{n+1} \xi_{2a} - \xi_{2a-1} \right)^4
\nonumber\\
&-12 \left(\sum_{a=1}^{n+1} \xi_{2a} - \xi_{2a-1} \right) \left( \sum_{a=1}^n \sum_{b=a}^{n+1} (\xi_{2b} - \xi_{2b-1}) (\xi_{2a} - \xi_{2a-1}) \sum_{c=a}^n (\xi_{2c+1} - \xi_{2c}) \right)
\nonumber\\
&+12 \left(\sum_{a=1}^{n+1} \xi_{2a} - \xi_{2a-1} \right) \left( \sum_{a=1}^n \sum_{b=1}^{a} (\xi_{2b} - \xi_{2b-1}) (\xi_{2a} - \xi_{2a-1}) \sum_{c=a}^n (\xi_{2c+1} - \xi_{2c}) \right)
\nonumber\\
&+ 12 \left( \sum_{a=1}^n (\xi_{2a} - \xi_{2a-1}) \sum_{b=a}^n (\xi_{2b+1} - \xi_{2b}) \right)^2
\nonumber\\
&-12 \left(\sum_{a=1}^{n+1} \xi_{2a} - \xi_{2a-1} \right) \sum_{a=1}^n
(\xi_{2a} - \xi_{2a-1}) \left( \sum_{b=a}^n (\xi_{2b+1} - \xi_{2b}) \right)^2,
\end{align}
which can be proven using recursion. Using eqs.~\eqref{eq:m5flux} and~\eqref{eq:m5primeflux} to replace the $\xi_j$ with $M_a$ and $M_b'$, and then using eq.~\eqref{eq:m2flux} to replace the $M_b'$ with the $N_a$, we find
\begin{align}
\label{eq:ads7msconv1}
3 m_2^2 - 4 m_1 m_3 =& \, c_1^{12} \frac{G_{\textrm{N}}^{4/3}}{\left(2\pi\right)^{16/3}} \frac{\gamma^4}{\left(1+\gamma\right)^{12}}\bigg[ - M^4 + 12 \gamma M \left( \sum_{a=1}^n \sum_{b=a}^{n+1} M_b M_a N_a \right)\\&
-12 \gamma M \left( \sum_{a=1}^n \sum_{b=1}^{a} M_b M_a N_a \right)+ 12 \gamma^2 \left( \sum_{a=1}^n M_a N_a \right)^2
-12 \gamma^2 M \sum_{a=1}^n M_a N_a^2 \bigg].\nonumber
\end{align}
All that remains in eq.~\eqref{eq:b6draw} is the sum over $(\xi_j- \xi_k)^3$. We decompose this sum as
\begin{align}
\sum_{\underset{j<k}{j,k=1}}^{2n+2} (-1)^{j+k} (\xi_j - \xi_k)^3 =& \bigg[
\left( \sum_{a=1}^{n+1} (\xi_{2a}-\xi_{2a-1}) \right)^3 \\
&+ 6 \left( \sum_{b=1}^{n} (\xi_{2b}-\xi_{2b-1}) \sum_{a=b}^{n} (\xi_{2a+1} - \xi_{2a}) \sum_{c=b}^{n+1} (\xi_{2c}-\xi_{2c-1}) \right) \nonumber\\
&- 6 \left( \sum_{b=1}^{n} (\xi_{2b}-\xi_{2b-1}) \sum_{a=b}^{n} (\xi_{2a+1} - \xi_{2a}) \sum_{c=1}^b (\xi_{2c}-\xi_{2c-1}) \right)
\bigg],\nonumber
\end{align}
which can be proven using recursion. Again using eqs.~\eqref{eq:m5flux} and~\eqref{eq:m5primeflux} to replace the $\xi_j$ with $M_a$ and $M_b'$, and then using eq.~\eqref{eq:m2flux} to replace the $M_b'$ with the $N_a$, we find
\begin{align}
\label{eq:ads7msconv2}
\sum_{\underset{j<k}{j,k=1}}^{2n+2} (-1)^{j+k} (\xi_j - \xi_k)^3 = -c_1^9 \, \frac{G_{\textrm{N}}}{\left(2\pi\right)^4} \frac{\gamma^3}{\left(1+\gamma\right)^9} &\bigg[ M^3 - 6 \gamma \left( \sum_{a=1}^{n} \sum_{b=a}^{n+1} N_a M_a M_b \right)
\nonumber\\ &+ 6 \gamma \left( \sum_{a=1}^{n} \sum_{b=1}^a N_a M_a M_b \right) \bigg].
\end{align}
Plugging eqs.~\eqref{eq:ads7msconv1} and~\eqref{eq:ads7msconv2} into eq.~\eqref{eq:b6draw}, using eq.~\eqref{eq:totnm}, and setting $N_{n+1} = 0$ and $\gamma=-1/2$ then gives
\begin{equation}
b_{6d} = \frac{3}{5}\left[\frac{N^2}{M} -\sum_{a=1}^{n+1} \left(M_a N_a^2 - 8 \sum_{b=a}^{n+1} M_a N_a M_b + 8 \sum_{b=1}^{a} M_a N_a M_b \right)\right],
\end{equation}
which can be simplified by rearranging the summations to give our main result,
\begin{equation}
\label{eq:b6d-sugra-1}
b_{6d} = \frac{3}{5}\left[8MN + \frac{N^2}{M} +\sum_{a=1}^{n} \left(8 M_a^2 N_a -M_a N_a^2 -16\sum_{b=1}^{a}M_b M_a N_a\right)\right].
\end{equation}
\subsection{Asymptotically Locally $AdS_4 \times S^7$ Solutions}
As in the previous case, in the asymptotic large-$r$ region we put the metric in a form that makes manifest the symmetries of the 2d boundary,
\begin{align}\label{AdS4:altFG}
ds^2 = \frac{L_{S^7}^2}{4} \left( \frac{dv^2}{v^2} + \frac{ds^2_{\text{AdS}_3}}{v^2} \right) + L_{S^7}^2 \left(d\tilde{\theta}^2 + \sin^2\tilde{\theta} \, ds^2_{S^3} + \cos^2\tilde{\theta} \, ds^2_{S^3} \right) + \ldots,
\end{align}
where asymptotically at large $r$
\begin{subequations}
\begin{align}
\label{eq:AdS4FGtrans}
v =& \frac{(1+\gamma) \sqrt{-m_1^2-m_2}}{2 \sqrt{-\gamma}} \frac{1}{r} \left(1 + \frac{-m_1^3+m_3}{3(m_1^2 + m_2)} \cos\theta \, \frac{1}{r} + {\cal O}\left(\frac{1}{r^2}\right) \right)\\
\tilde{\theta} =& \frac{\theta}{2} + \frac{-m_1^3+m_3}{6(m_1^2+m_2)} \sin\theta \, \frac{1}{r} + {\cal O}\left(\frac{1}{r^2}\right),
\end{align}
\end{subequations}
where $v \in [0,\infty)$, $\tilde{\theta} \in [0,\pi/2]$, and the ellipsis represent terms orthogonal to $v$ and sub-leading in $1/r$. In these coordinates, we can approach the asymptotic local (half) $AdS_4 \times S^7$ boundary in two ways. First is to fix $u$ and send $v \to 0$, where the latter is equivalent via eq.~\eqref{eq:AdS4FGtrans} to $r \to \infty$, which takes us to the asymptotically local $AdS_4 \times S^7$ boundary at a point away from the 2d boundary. Second is to fix $v$ and send $u \to 0$, which takes us to the $AdS_3$ boundary, i.e. to a point on the 2d boundary.
We introduce an FG cutoff $\varepsilon_v$, which we plug into eq.~\eqref{eq:AdS4FGtrans} and invert to find
\begin{align}
\label{eq:AdS4cutoff}
r_c(\varepsilon_v,\theta) = \frac{(1+\gamma) \sqrt{-m_1^2-m_2}}{2 \sqrt{-\gamma}} \frac{1}{\varepsilon_v} - \frac{-m_1^3+m_3}{3(m_1^2 + m_2)} \cos\theta + {\cal O}\left(\varepsilon_v\right).
\end{align}
Plugging the expression for $G$ given in eq.~\eqref{eq:ads4metric} into eq.~\eqref{eq:integral3} we obtain
\begin{align}
\label{integralcomplex}
{\cal I} =& \frac{ (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{4}{c_1 c_2^4 c_3^4} \int_0^{\pi} d\theta \sin\theta \int_{0}^{r_c(\varepsilon_v,\theta)} dr \, r^2 \left(1 - \sum_{j,k=1}^{2n+1} (-1)^{j+k} \frac{r e^{i\theta} - \xi_j}{|r e^{i\theta} - \xi_j|} \frac{r e^{-i\theta} - \xi_k}{|r e^{i\theta} - \xi_k|} \right).
\end{align}
These integrals are very similar to those in the asymptotically local $AdS_7 \times S^4$ case. We perform the integrals in the appendix, with the result
\begin{equation}\label{eq:ads4int2}
{\cal I} =\frac{(2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{L_{S^7}^9}{24} \frac{1}{\varepsilon_v} - \frac{(2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{8}{3}\frac{1}{ c_1 c_2^4 c_3^4}\sum_{\underset{j<k}{j,k=1}}^{2n+1} (-1)^{j+k}(\xi_j-\xi_k)^3.
\end{equation}
As expected, eq.~\eqref{eq:ads4int2} diverges as $\varepsilon_v \to 0$. To extract $b_{3d}$ using eq.~\ref{eq:b3ddef}, we will need the result for ${\cal I}$ for the exact $AdS_4 \times S^7$ solution, which we denote ${\cal I}_{3d}$. In $AdS_3$ slicing, the $AdS_4 \times S^7$ metric takes the form
\begin{equation}
\label{AdS4metric}
ds^2=L_{S^7}^2 \left(dx^2 + \cosh^2 (2x) \, ds^2_{AdS_3}+dy^2+\sin^2 y\, ds^2_{S^3} + \cos^2 y\, ds^2_{S^3}\right),
\end{equation}
where $x \in (-\infty,\infty)$ and $y \in [0,\pi/2]$. By matching the large-$|x|$ asymptotics of this $AdS_4 \times S^7$ metric to that of eq.~\eqref{AdS4:altFG}, we find $x \in (-x_c,x_c)$ with cutoff $x_ c = -\frac{1}{2}\ln\varepsilon_v$. Plugging eq.~\eqref{AdS4metric} into eq.~\eqref{eq:Iastintegral} and performing the integration with that cutoff, we find
\begin{align}\label{vacuum}
{\cal I}_{3d} = \frac{2 (2 \pi^2)^2}{4 G_{\textrm{N}}} \frac{L_{S^7}^9}{24} \frac{1}{\varepsilon_v}.
\end{align}
Comparing eqs.~\eqref{eq:ads4int2} and~\eqref{vacuum}, we see that the divergence cancels in ${\cal I} - {\cal I}_{3d}$, as advertised. Extracting $b_{3d}$ via eq.~\eqref{eq:b3ddef} then gives
\begin{align}\label{cads4}
b_{3d} = - \frac{ (2 \pi^2)^2}{G_{\textrm{N}}} \frac{2}{c_1 c_2^4 c_3^4}\sum_{\underset{j<k}{j,k=1}}^{2n+1} (-1)^{j+k}(\xi_j-\xi_k)^3.
\end{align}
To rewrite this expression in terms of the partition data $\{N_a\}$ and $\{M_a\}$, we decompose the sum in eq.~\eqref{cads4} as
\begin{align}
\sum_{\underset{j<k}{j,k=1}}^{2n+1} (-1)^{j+k}(\xi_j-\xi_k)^3 = 3 \bigg[
- 2 \left( \sum_{b=1}^n \sum_{c=1}^b (\xi_{2c}-\xi_{2c-1}) (\xi_{2b}-\xi_{2b-1}) \sum_{a=b}^n (\xi_{2a+1} - \xi_{2a}) \right) \nonumber\\
+ \left( \sum_{b=1}^n (\xi_{2b}-\xi_{2b-1})^2 \sum_{a=b}^n (\xi_{2a+1} - \xi_{2a}) \right)
- \sum_{b=1}^n (\xi_{2b}-\xi_{2b-1}) \left( \sum_{a=b}^n (\xi_{2a+1} - \xi_{2a}) \right)^2
\bigg],
\end{align}
which can be proven using recursion. Using eqs.~\eqref{eq:m5flux} and~\eqref{eq:m5primeflux} to replace the $\xi_j$ with $M_a$ and $M_b'$, and then using eq.~\eqref{eq:m2flux} to replace the $M_b'$ with the $N_a$, we find
\begin{align}
\label{eq:ads4cubicconv}
\sum_{\underset{j<k}{j,k=1}}^{2n+1} (-1)^{j+k}(\xi_j-\xi_k)^3= 3 c_1^9 \frac{G_{\textrm{N}}}{\left(2\pi\right)^4} \frac{\gamma^4}{\left(1+\gamma\right)^9}\bigg[
-& 2 \left( \sum_{a=1}^n \sum_{b=1}^a M_b M_a N_a \right) \\
+& \left( \sum_{a=1}^n M_a^2 N_a \right)
+ \gamma \sum_{a=1}^n M_a N_a^2\nonumber
\bigg].
\end{align}
Plugging eq.~\eqref{eq:ads4cubicconv} into eq.~\eqref{cads4} then gives
\begin{equation}
\label{eq:b3dtemp}
b_{3d} = \frac{3}{2} \frac{c^8_1}{c_2^4c_3^4} \frac{\gamma^4}{\left(1+\gamma\right)^9} \left[ - \sum_{a=1}^n M_a^2 N_a- \gamma \sum_{a=1}^n M_a N_a^2+2 \sum_{a=1}^n \sum_{b=1}^a M_b M_a N_a\right].
\end{equation}
Using $c_1+c_2+c_3=0$ and $c_2 = \gamma c_3$, we find
\begin{equation}
\label{eq:celim}
\frac{c^8_1}{c_2^4c_3^4} \frac{\gamma^4}{\left(1+\gamma\right)^9} = \frac{\left(c_2+c_3\right)^8 }{\gamma^4c_3^8}\frac{\gamma^4}{\left(1+\gamma\right)^9} = \frac{1}{1+\gamma},
\end{equation}
and hence we obtain our main result for $b_{3d}$,
\begin{align}
\label{b3dads4}
b_{3d} = \frac{3}{2}\frac{1}{1+\gamma} \left[-\sum_{a=1}^n M_a^2 N_a - \gamma \sum_{a=1}^n M_a N_a^2 +2\sum_{a=1}^n \sum_{b=1}^a M_b M_a N_a \right].
\end{align}
\section{The Central Charge}
\label{sec:cc}
In this section, we explore our results for $b_{6d}$ and $b_{3d}$ in several ways. In sec.~\ref{sec:wilson-surface} we show that our result for $b_{6d}$ in eq.~\eqref{eq:b6d-sugra-1} can be written in the compact form of eq.~\eqref{eq:b6dcompact}, that is, in terms of $\lambda$, the highest weight vector of the representation ${\cal R}$, and $\varrho$, the Weyl vector of $\mathfrak{su}(M)$. We also determine how $b_{6d}$ scales with $M$ and $N$ for some specific choices of ${\cal R}$. Similarly, in sec.~\ref{sec:abjm-bcft-cc} we determine how $b_{3d}$ scales with $N$ for some specific choices of partition $\rho$. In sec.~\ref{sec:other-comparisons} we briefly survey some previous calculations of self-dual string central charges, and discuss how and why these results differ from ours.
\subsection{Wilson Surface}
\label{sec:wilson-surface}
Our first goal is to show that our result for $b_{6d}$ in eq.~\eqref{eq:b6d-sugra-1},
\begin{equation}
\label{eq:b6dsugra}
b_{6d} = \frac{3}{5}\left[8MN + \frac{N^2}{M} +\sum_{a=1}^{n+1} \left(8M_a^2N_a -M_aN_a^2 -16\sum_{b=1}^{a}M_bM_a N_a\right)\right],
\end{equation}
can we re-written in the form of eq.~\eqref{eq:b6dcompact},
\begin{equation}
\label{eq:b6dcompact2}
b_{6d} = \frac{3}{5}\left[16(\lambda,\,\varrho)-(\lambda,\,\lambda)\right],
\end{equation}
where $\lambda$ is the highest weight vector of the representation $\mathcal{R}$, $\varrho$ is the Weyl vector of $\mathfrak{su}(M)$, and $(\cdot ,\,\cdot)$ is the inner product on the weight space. To show the equivalence between the two expressions for $b_{6d}$ we will actually work backwards: starting from eq.~\eqref{eq:b6dcompact2} we will re-write various sums until we reach eq.~\eqref{eq:b6dsugra}.
We start with the parametrization of the partition $\rho=\{\ell_1,\ell_2,\ldots\}$, with integers $\ell_1 \geq \ell_2 \geq \ell_3 \ldots$, where $\ell_q$ with $1 \leq q \leq M$ is the number of M2-branes ending on the $q^{\textrm{th}}$ M5-brane, as illustrated on the left in fig.~\ref{fig:young}. In this parametrization the inner products in eq.~\eqref{eq:b6dcompact2} are simple to write in terms of the Dynkin indices of the representation $\cal{R}$, $\lambda_q = \ell_q - \ell_{q+1}$,
\begin{equation}
\label{eq:b6dsums1}
(\lambda,\,\varrho) = \frac{1}{2}\sum_{q=1}^{M}(M-q)q\lambda_q,\qquad (\lambda,\,\lambda) = \frac{1}{M}\sum_{q=1}^{M}(M-q)\lambda_q\left[- q\lambda_q+2\sum_{p=1}^{q}p\lambda_p \right].
\end{equation}
As is clear from the definition of $\lambda_q$, non-zero contributions to the sums in eq.~\eqref{eq:b6dsums1} only come from cases where the number of boxes in the Young tableau changes from one row to the next. The non-zero contributions are thus more conveniently described using the parametrization of the partition $\rho$ in eq.~\eqref{eq:part}, in terms of the set of distinct integers $\{N_a\}$ with degeneracies $\{M_a\}$ for $a = 1, 2,\ldots,n+1$ and $N_{n+1}=0$. In this parametrization, the non-zero contributions to the sums in eq.~\eqref{eq:b6dsums1} come from $\lambda_a = N_a - N_{a+1}$, and the row number can be written as $q = \sum_{b=1}^a M_b$. Plugging these expressions into eq.~\eqref{eq:b6dsums1} gives
\begin{subequations}
\label{eq:rep-data}
\begin{align}
(\lambda,\,\varrho) &= \frac{1}{2}\sum_{a=1}^{n+1}\sum_{b=1}^a\left(M - \sum_{c=1}^aM_c\right) M_b \left(N_a - N_{a+1}\right),\\
\begin{split}
(\lambda,\,\lambda) &= \frac{1}{M}\sum_{a=1}^{n+1}\sum_{b=1}^a \left(M - \sum_{c=1}^aM_c\right)\left(N_a- N_{a+1}\right)\left[2\sum_{d=1}^b M_d(N_b-N_{b+1}) -M_b(N_a-N_{a+1})\right]\!\!.
\end{split}
\end{align}
\end{subequations}
Expanding the sums in eq.~\eqref{eq:rep-data} then leads to
\begin{subequations}
\label{eq:rep-data2}
\begin{align}
(\lambda,\,\varrho) &= \frac{1}{2}\sum_{a=1}^{n+1} \left(M M_a N_a + M_a^2 N_a - 2 \sum_{b=1}^aM_b M_a N_a\right),\\
(\lambda,\,\lambda) &= \sum_{a=1}^{n+1} M_a N_a^2 - \frac{1}{M}\left(\sum_{a=1}^{n+1} M_a N_a\right)^2,
\end{align}
\end{subequations}
which we simplify using the fact that the total number of M2-branes is $N= \sum_{a=1}^{n+1} M_a N_a$:
\begin{subequations}
\label{eq:rep-data3}
\begin{align}
(\lambda,\,\varrho) &= \frac{1}{2}MN+\frac{1}{2}\sum_{a=1}^{n+1} M_a^2 N_a - \sum_{a=1}^{n+1} \sum_{b=1}^aM_b M_a N_a,\\
(\lambda,\,\lambda) &= \sum_{a=1}^{n+1} M_a N_a^2 - \frac{N^2}{M}.
\end{align}
\end{subequations}
Plugging eq.~\eqref{eq:rep-data3} into eq.~\eqref{eq:b6dcompact2} we find
\begin{eqnarray}
\label{eq:b6dfinal}
\frac{3}{5}\left[16(\lambda,\,\varrho)-(\lambda,\,\lambda)\right] & = & \frac{3}{5}\left[8 MN+8\sum_{a=1}^{n+1} M_a^2 N_a -16\sum_{a=1}^{n+1} \sum_{b=1}^a M_b M_a N_a - \sum_{a=1}^{n+1} M_a N_a^2+\frac{ N^2}{M}\right]\nonumber \\ & = & \frac{3}{5}\left[8MN + \frac{N^2}{M} +\sum_{a=1}^{n+1} \left(8M_a^2N_a -M_aN_a^2 -16\sum_{b=1}^{a}M_bM_a N_a\right)\right]\nonumber \\ & = & b_{6d},
\end{eqnarray}
as advertised. We can alternatively express $b_{6d}$ in terms of the quadratic Casimir of the representation, $\mathcal{Q}_\lambda \equiv 2(\lambda,\,\varrho)+(\lambda,\,\lambda)$,
\begin{equation}
b_{6d} = \frac{24}{5}\left(\mathcal{Q}_\lambda - \frac{9}{8}(\lambda,\,\lambda)\right).
\end{equation}
The inner product $(\cdot,\,\cdot)$ is invariant under the action of the Weyl group. These compact forms of $b_{6d}$ thus make manifest that $b_{6d}$ is invariant under the action of the Weyl group on $\lambda$ and $\varrho$, including in particular the Weyl reflection affected by the complex conjugation of the representation, ${\cal R} \to \overline{{\cal R}}$. Such invariance is expected, given that the 11d SUGRA solutions are invariant under ${\cal R} \to \overline{{\cal R}}$, as explained at the end of sec.~\ref{sec:ads7part}. Such invariance is also expected in the field theory: ${\cal R} \to \overline{{\cal R}}$ combined with orientation reversal of the Wilson surface must leave all observables invariant. The EE of our spherical region is invariant under the orientation reversal alone, and hence must also be invariant under ${\cal R} \to \overline{{\cal R}}$ alone, not just under the combined operation. As a result, $b_{6d}$ is invariant under ${\cal R} \to \overline{{\cal R}}$.
These compact forms of $b_{6d}$ do not make immediately obvious how $b_{6d}$ scales with the total numbers $M$ of M5-branes or $N$ of M2-branes. To see such scaling explicitly, we consider two relatively simple examples of ${\cal R}$. First, consider a totally symmetric representation of rank $N$, which corresponds to a Young tableau with a single row of $N$ boxes. In terms of the partition data, $M_1=1$, $N_1 = N$, and $N_a=0$ for $2\leq a\leq n+1$, which describes all $N$ M2-branes ending on a single M5-brane. In this case we find\footnote{
\textbf{Note added:} A natural question is whether the value $N = 8 M$ at which $b_{6d}$ in eq.~\eqref{eq:b6dsymm} vanishes has any physical significance. In particular, does $b_{6d}=0$ imply that the symmetric-representation Wilson surface with $N=8M$ is somehow equivalent to a trivial-representation object, like a ``baryon vertex''? We suspect not, because as shown in eq.~\eqref{eq:ee_trace_anomaly}, $b_{6d} = c - \frac{3}{5} d_2$ with $c$ and $d_2$ defined as coefficients in the defect trace anomaly eq.~\eqref{eq:defecttrace}, and taking the values in eq.~\eqref{eq:cd2values}. In particular, both $c$ and $d_2$ are positive for the symmetric representation with $N=8M$, and $b_{6d}$ vanishes because of a cancellation between them, whereas we would expect both of them to vanish independently for a trivial-representation object.}
\begin{eqnarray}
\label{eq:b6dsymm}
b_{6d} & = & \frac{3}{5}\left[8 MN + \frac{N^2}{M} - N^2 - 8 N\right] \qquad \textrm{(symmetric)} \\ & = & \frac{3}{5} \frac{N}{M} \left(8M-N\right)\left(M-1\right),\nonumber
\end{eqnarray}
where the second line helps facilitate probe limits. For example, the limit $N \ll M^2$ gives $b_{6d} \approx \frac{24}{5} N \left(M-N/8\right)$. As shown in ref.~\cite{Rodgers:2018mvq}, this probe limit result agrees with a calculation of EE using probe M5-branes in $AdS_7 \times S^4$ wrapping an $AdS_3 \times S^3$ submanifold inside $AdS_7$ and sitting at a point on the $S^4$. This is true even when $N \sim \mathcal{O}(1)$, in which case we would expect the SUGRA approximation to break down, as discussed below eq.~\eqref{eq:m2flux}.
Our second example is a totally anti-symmetric representation of rank $N$, which corresponds to a Young tableau with a single column of $N$ boxes. In terms of the partition data, $M_1=N$ and $N_1=1$, which describes $N$ M5-branes each having one M2-brane ending on them. In this case we find
\begin{eqnarray}
\label{eq:b6dantisymm}
b_{6d} & = & \frac{3}{5}\left[8 MN + \frac{N^2}{M} - 8 N^2 - N\right] \qquad \textrm{(anti-symmetric)} \\ & = & \frac{3}{5}\frac{N}{M}(M-N)(8M-1), \nonumber
\end{eqnarray}
where again the second line helps facilitate probe limits. For example, the limit $N \ll M^2$ gives $b_{6d} \approx \frac{24}{5} N \left(M-N\right)$. As shown in ref.~\cite{Rodgers:2018mvq}, this limit agrees with a calculation of EE using probe M5-branes in $AdS_7 \times S^4$ wrapping an $AdS_3$ submanifold of $AdS_7$ and $S^3$ submanifold of $S^4$, again for all values of $N$. For a totally anti-symmetric representation ${\cal R}$, complex conjugation ${\cal R} \to \overline{\cal R}$ acts as $N \leftrightarrow M-N$. Our result for $b_{6d}$ in eq.~\eqref{eq:b6dantisymm} is clearly invariant under $N \leftrightarrow M-N$, as expected. Furthermore, a totally anti-symmetric representation of rank $M$, meaning $N=M$, is equivalent to a trivial representation, and indeed plugging $N=M$ into eq.~\eqref{eq:b6dantisymm} gives $b_{6d}=0$.\footnote{
\textbf{Note added:} In terms of the trace anomaly coefficients $c$ and $d_2$ defined in eq.~\eqref{eq:defecttrace}, $b_{6d} = c - \frac{3}{5} d_2$ vanishes for an antisymmetric representation with $N = M$ because $c$ and $d_2$ each individually vanish in that case, as expected for a trivial representation.}
Assuming that $b_{6d}$ counts degrees of freedom localized to the Wilson surface, a na\"ive expectation is that in the large-$M$ and/or $N$ limit $b_{6d}$ should scale with the number of degrees of freedom of the M5- or M2-brane theory, $M^3$ or $N^{3/2}$, respectively~\cite{Berman:2007bv}. We can force our result for $b_{6d}$ to scale as $M^3$ or $N^{3/2}$ by choosing a representation such that $N$ scales as a certain power of $M$. For example, for a totally symmetric representation of rank $N \propto M^{3/2}$, in eq.~\eqref{eq:b6dsymm} at large $M$ the term $-8N^2 \propto -8M^3$ with all other terms sub-leading. However, in general none of our results for $b_{6d}$ naturally scale as $M^3$ or $N^{3/2}$. Instead, in our results for $b_{6d}$ most terms scale linearly or quadratically with $M$ or $N$.
As displayed above, our result for $b_{6d}$ can become negative. For example, for a totally symmetric representation of rank $N>8M$, eq.~\eqref{eq:b6dsymm} clearly shows that $b_{6d}<0$. In a standard 2d CFT, unitarity and normalizability of the ground state require the central charge to be non-negative. However, whether unitarity imposes a lower bound on $b_{6d}$ is currently unknown. In similar cases, such as 3d BCFTs, unitarity allows negative values of the boundary central charge. For example, in the free massless scalar 3d BCFT with a Dirichlet boundary condition---a perfectly unitary theory---the boundary central charge is negative~\cite{Nozaki:2012qd,Jensen:2015swa,Fursaev:2016inw}. More generally, for unitary 3d BCFTs ref.~\cite{Herzog:2017kkj} conjectured a lower bound on the boundary central charge that was negative. The fact that Wilson surfaces in the M5-brane theory and their holographic duals have no other known violations of unitarity leads us to suspect strongly that $b_{6d}<0$ does not necessarily signal unitarity violation.
\subsection{Cousins of the ABJM BCFT}
\label{sec:abjm-bcft-cc}
As discussed in sec.~\ref{sec:ads4part}, in the asymptotically locally $AdS_4 \times S^7$ solutions we cannot identify $M$, so while we can identify a partition $\rho$ and corresponding Young tableau, we cannot identify $\mathfrak{su}(M)$ or a representation ${\cal R}$. We can thus write our result for $b_{3d}$ in eq.~\eqref{b3dads4},
\begin{equation}
\label{eq:b3d-cousins}
b_{3d} =\frac{3}{2}\frac{1}{1+\gamma}\sum_{a=1}^n \left(-M_a^2N_a -\gamma M_a N_a^2 +2\sum_{b=1}^a M_b M_a N_a\right),
\end{equation}
only in terms of $N$, and not $M$. As mentioned below eq.~\eqref{eq:ads4metric}, we consider asymptotically locally $AdS_4 \times S^7$ solutions with $\gamma \in (-1,0)$.
As explained at the end of sec.~\ref{sec:ads4part}, the asymptotically locally $AdS_4 \times S^7$ solutions are invariant, up to an orientation reversal, under the combined operations $\gamma \to \gamma^{-1}$ and $\rho \to \hat{\rho} = \rho^T$. To see how these operations act on $b_{3d}$, we re-write eq.~\eqref{eq:b3d-cousins} in terms of the sets $\{N_b'\}$ and $\{M_b'\}$,
\begin{align}
\label{eq:ads4transposeidentity}
b_{3d} =& \frac{3}{2}\frac{1}{1+\gamma}\bigg[
\sum_{b=1}^n M_{n+1-b}' \left( \sum_{a=b}^n M_{n+1-a} \right)^2
- \gamma \sum_{a=1}^n M_a \left(\sum_{b=a}^n M_b'\right)^2
\bigg]
\nonumber\\
=& \frac{3}{2}\frac{1}{1+\gamma} \bigg[
\sum_{b=1}^n M_{b}' N_b'{}^2 - \gamma \sum_{a=1}^n M_a N_a^2
\bigg],
\end{align}
which can be proven using recursion. Eq.~\eqref{eq:ads4transposeidentity} makes clear that the combined operations of $\gamma \to \gamma^{-1}$ and $\rho \to \hat{\rho}=\rho^T$, which acts as $M_a \to M'_{n-a}$ and $N_a \to N'_{n-a}$, sends $b_{3d} \to - b_{3d}$. Equivalently, in the parametrization $\rho=\{\ell_1,\ell_2,\ldots\}$ and $\hat{\rho} = \rho^T = \{\ell_1',\ell_2',\ldots\}$,
\begin{equation}
b_{3d}= \frac{3}{2}\frac{1}{1+\gamma} \left[ - \gamma \sum_{q=1}^n \ell_q^2 + \sum_{\hat{q}=1}^n \ell_{\hat{q}}'{}^2 \right].
\end{equation}
In this parametrization, $\rho \to \hat{\rho}=\rho^T$ acts as $\ell_q \leftrightarrow \ell_{\hat{q}}'$, which when combined with $\gamma \to \gamma^{-1}$ clearly sends $b_{3d} \to - b_{3d}$.
To see how $b_{3d}$ in eq.~\eqref{eq:b3d-cousins} scales with $N$ we consider two examples of partitions analogous to our two examples for $b_{6d}$ in eqs.~\eqref{eq:b6dsymm} and~\eqref{eq:b6dantisymm}. First, we consider the analogue of the rank $N$ totally symmetric representation, which corresponds to a Young tableau with a single row of $N$ boxes. In terms of partition data, this example has $M_1=1$, $N_1 = N$, and $N_a=0$ for $a\geq 2$, which gives
\begin{equation}
\label{eq:b3d-cousins-symm}
b_{3d} = \frac{3}{2(1+\gamma)}N\left(1-\gamma N\right). \qquad \textrm{(``symmetric'')}
\end{equation}
The large-$N$ limit of eq.~\eqref{eq:b3d-cousins-symm} gives $b_{3d}\approx -\frac{3}{2}\frac{\gamma}{1+\gamma} N^2$. As a second example, we consider the analogue of a rank $N$ the totally anti-symmetric representation, which corresponds to a Young tableau with a single column of $N$ boxes. In terms of partition data, this example has $N_{1}=1$ and $M_1 =N$, which gives
\begin{equation}
\label{eq:b3d-cousins-asymm}
b_{3d} = \frac{3}{2(1+\gamma)}N\left(N-\gamma\right). \qquad \textrm{(``anti-symmetric'')}
\end{equation}
The large-$N$ limit of eq.~\eqref{eq:b3d-cousins-asymm} gives $b_{3d}\approx \frac{3}{2}\frac{1}{1+\gamma}N^2$. Clearly in both cases at large $N$ we find that $b_{3d}$ scales as $N^2$ rather than $N^{3/2}$. Of course, as discussed in sec.~\ref{sec:intro}, in these cases the 3d BCFTs are only cousins of the ABJM BCFT, and in particular have a different supergroup from that of the maximally SUSY ABJM BCFT. We therefore have no reason \textit{a priori} to expect $b_{3d}$ to scale as $N^{3/2}$ at large $N$.
As with $b_{6d}$ in sec.~\ref{sec:wilson-surface}, the $b_{3d}$ in eq.~\eqref{eq:b3d-cousins} can become negative. The examples in eqs.~\eqref{eq:b3d-cousins-symm} and~\eqref{eq:b3d-cousins-asymm} never are, because $\gamma\in(-1,0)$. However, as we saw that $\gamma \rightarrow \gamma^{-1}$ combined with $\rho\rightarrow \hat{\rho}=\rho^T$ sends $b_{3d}\rightarrow -b_{3d}$, so any partition $\rho$ giving $b_{3d}$ has a corresponding $\rho^T$ giving $-b_{3d}$ within the same $D(2,1;\gamma)\times D(2,1;\gamma)$ super-group. For the reasons discussed at the end of sec.~\ref{sec:wilson-surface}, even if some choices of $M_a$ and $N_a$ make eq.~\eqref{eq:b3d-cousins} negative, we strongly suspect that $b_{3d}<0$ does not necessarily signal unitarity violation.
\subsection{Comparisons to Other Calculations}
\label{sec:other-comparisons}
In this sub-section we will review two previous calculations of the self-dual string central charge, one by Berman and Harvey (BH)~\cite{Berman:2004ew} in sec.~\ref{sec:bh} and one by Niarchos and Siampos (NS)~\cite{Niarchos:2012cy} in sec.~\ref{sec:ns}. We will denote these as $b_{\textrm{BH}}$ and $b_{\textrm{NS}}$, respectively. In sec.~\ref{sec:freeabjm} we will compute the central charge in the maximally SUSY ABJM BCFT in the free limit, which we will denote as $b_{\textrm{free}}$.
Put simply, we find that none of the calculations of the central charge of the self-dual string or Wilson surface---$b_{6d}$ and $b_{3d}$, BH, NS, and free ABJM---agree perfectly with any of the others, though some share certain features. Indeed, all of these calculations extract a central charge from different quantities (EE, anomalies, and thermodynamic entropy), and use different limits of $M$ and $N$, and so we have little reason \textit{a priori} to expect agreement. Nevertheless, aside from the $M^3$ and $N^{3/2}$ scalings of the ambient CFTs, these are some of the few readily available results that we have for comparison.
\subsubsection{R-symmetry Anomaly}
\label{sec:bh}
In ref.~\cite{Berman:2004ew} BH considered $N$ M2-branes ending on $M$ M5-branes and performed an anomaly inflow analysis. In particular, BH demanded that the total R-symmetry chiral anomaly on the self-dual string's 2d worldsheet vanishes. Let us briefly review their arguments.
The 2d R-symmetry is $SO(4)_{R_T}\times SO(4)_{R_N}$, where $SO(4)_{R_T}$ acts on the directions tangent to the M5-brane, $(x_3,x_4,x_5,x_6)$, and $SO(4)_{R_N}$ acts on the directions normal to the M5-brane, $(x_7,x_8,x_9,x_{10})$. If we gauge each $SO(4) \simeq SU(2)_+ \times SU(2)_-$, where the $\pm$ subscripts indicate 2d chirality, with corresponding field strengths $F_{\pm}=dA_{\pm}$, then the Euler class of the gauge bundle over the 2d self-dual string worldsheet is $\chi= \frac{1}{4\pi^2}\left({\rm Tr} F_+^2 - {\rm Tr} F_-^2\right)$. Written as an exact 4-form, $\chi= d \Phi$ where the 3-form $\Phi$ has the gauge transformation $\delta \Phi = d \phi$ for some 2-form, $\phi$. The 2d chiral R-symmetry has an anomaly that receives contributions from both 2d and 6d degrees of freedom, the latter via anomaly inflow. An anomaly descent calculation shows that the 2d contribution to the anomaly is $2\pi b_{\textrm{BH}} \int (\phi(A_{N}) - \phi (A_{T}))$, where $A_T$ and $A_N$ are the $SO(4)_{R_T}$ and $SO(4)_{R_N}$ connections, respectively, on the normal bundle to the self-dual string. BH define a central charge as the coefficient $b_{\textrm{BH}}$. They determine $b_{\textrm{BH}}$ by demanding that the total 2d chiral R-symmetry anomaly vanishes, i.e. that the 6d anomaly inflow contribution cancels the 2d contribution.
With a single M5-brane, $M=1$, the 2d R-symmetry chiral anomaly is $\pi N\int (\phi(A_{N}) - \phi (A_{T}))$, and so $b_{\textrm{BH}} = N/2$. The $SO(4)_{R_N}$ anomaly, $\pi N \int \phi(A_{N})$, is cancelled by a local counterterm on the M5-brane's wordvolume, $-\int dA_2 \wedge \phi(A_N)$, where as in sec.~\ref{sec:intro}, $A_2$ is the M5-brane's worldvolume 2-form gauge field. The $SO(4)_{R_T}$ anomaly, $-\pi N \int \phi(A_{T})$, is cancelled by the gauge variation of the self-dual string's minimal coupling to $A_2$.
With multiple M5-branes, $M>1$, BH move onto the Coulomb branch, separating some M5-branes out of the stack of $M$ coincident M5-branes. In that case, the M5-branes' effective action includes the Ganor-Intriligator-Motl term~\cite{Ganor:1998ve,Intriligator:2000eq}, $\alpha \int dA_2 \wedge \phi(A_T)$, where $\alpha$ is known. Starting from a stack of M5-branes with ADE Lie algebra $\mathfrak{g}$ and breaking to a subgroup with ADE Lie algebra $\mathfrak{h}$ times $\mathfrak{u}(1)$, $\alpha = \frac{1}{4}(\textrm{dim}\,\mathfrak{g} - \textrm{dim}\,\mathfrak{h}-1)$. The pullback of the Ganor-Intriligator-Motl term to the 2d self-dual string worldsheet produces a term with an $SO(4)_{R_N}$ anomaly that must cancel the 2d contribution, which allows BH to identify
\begin{equation}
b_{\textrm{BH}} = \frac{1}{2} N \alpha.
\end{equation}
If we separate a single M5-brane from the stack then $\mathfrak{g} =\mathfrak{su}(M)$ and $\mathfrak{h}=\mathfrak{su}(M-1)$ (as mentioned in sec.~\ref{sec:intro}, we ignore the overall center-of-mass $\mathfrak{u}(1)$), so $\alpha = \frac{1}{2} \left(M-1\right)$ and hence
\begin{equation}
\label{eq:bbhone}
b_{\textrm{BH}} = \frac{1}{4} N \left(M-1\right) \qquad \textrm{for} \quad \mathfrak{su}(M) \to \mathfrak{su}(M-1) \times \mathfrak{u}(1).
\end{equation}
If we separate all $M$ M5-branes from each other such that $\mathfrak{g} =\mathfrak{su}(M)$ and $\mathfrak{h}=\mathfrak{u}(1)^{M-1}$, then $\alpha = \frac{1}{4} \left(M^2 - M - 1\right)$ and hence
\begin{equation}
\label{eq:bbhmax}
b_{\textrm{BH}} = \frac{1}{8} N \left(M^2-M-1\right) \qquad \textrm{for} \quad \mathfrak{su}(M) \to \mathfrak{u}(1)^{M-1}.
\end{equation}
If we compare $b_{\textrm{BH}}$ in eqs.~\eqref{eq:bbhone} and~\eqref{eq:bbhmax} to $b_{6d}$ of the totally symmetric or anti-symmetric representations in eq.~\eqref{eq:b6dsymm} or~\eqref{eq:b6dantisymm}, respectively, then the only obvious similarities are terms scaling as $MN$ and $N$, with different numerical coefficients. However, we can identify at least four reasons why $b_{6d}$ and $b_{\textrm{BH}}$ need not agree. First, whether $b_{6d}$ and $b_{\textrm{BH}}$ are the same quantity is unclear. In a 2d CFT the chiral R-symmetry anomaly coefficient is proportional to the central charge $c$~\cite{Benini:2012cz}, and BH assume the same remains true for a 2d defect in a higher-d CFT. However, that has not been demonstrated, and moreover, whether and how either quantity is related to the defect's EE is unclear. Second, we calculated $b_{6d}$ for any representation ${\cal R}$. However, $b_{\textrm{BH}}$ seems to involve no data about a representation. Which representation(s) are appropriate in comparing $b_{6d}$ and $b_{\textrm{BH}}$ (if any) is unclear. Third, our $b_{6d}$ was computed at the conformal point where all M5-branes are coincident, whereas $b_{\textrm{BH}}$ was computed on the Coulomb branch. Fourth, we computed $b_{6d}$ in the SUGRA limit of large $M$, whereas the calculation of $b_{\textrm{BH}}$ is in principle valid for any $M$. At best we might expect agreement between $b_{6d}$ and $b_{\textrm{BH}}$'s large-$M$ limit, and indeed, both our results for $b_{6d}$ in eqs.~\eqref{eq:b6dsymm} and~\eqref{eq:b6dantisymm} and the results for $b_{\textrm{BH}}$ in eqs.~\eqref{eq:bbhone} and~\eqref{eq:bbhmax} scale as $MN$ at large $M$ but with different numerical coefficients.
If we compare $b_{3d}$ of the totally ``symmetric'' or ``anti-symmetric'' partitions in eqs.~\eqref{eq:b3d-cousins-symm} or~\eqref{eq:b3d-cousins-asymm}, then the only obvious similarities are terms scaling as $N$, with different numerical coefficients. However, as before, BH compute a different quantity, which appears to involve no information about a partition. We can also identify two further reasons why $b_{3d}$ and $b_{\textrm{BH}}$ need not agree. First, as discussed in secs.~\ref{sec:intro} and~\ref{sec:ads4part}, $b_{3d}$ arises from a case with both M5- and M5$'$-branes, while $b_{\textrm{BH}}$ arises from a case with no M5$'$-branes. Second, as discussed in sec.~\ref{sec:intro} and~\ref{subsec:ads4metric}, $b_{3d}$ arises in a case with super-group $D(2,1;\gamma)\times D(2,1;\gamma)$ with $\gamma\in(-1,0)$, while BH presumably have $\gamma = 1$.
\subsubsection{Blackfolds}
\label{sec:ns}
In refs.~\cite{Niarchos:2012pn,Niarchos:2012cy} NS found a blackfold description~\cite{Emparan:2009at,Emparan:2011hg,Emparan:2011br} of a fully localized intersection of M2- and M5-branes. Specifically, in ref.~\cite{Niarchos:2012pn} in an effective M5-brane theory they found a 1/4-BPS ``spike'' solution representing $N$ M2-branes ending on the $M$ M5-branes. In ref.~\cite{Niarchos:2012cy} they generalized the spike solution to non-zero temperature $T$, in the limit $N \ll M^2$, and computed the thermal entropy density $S$. In the low-$T$ limit, relative to the system's spatial size, they found
\begin{equation}
\label{eq:nsent}
S = \frac{8}{135}\frac{\Gamma(\frac{1}{3})\,\Gamma(\frac{1}{6})\,\sqrt{\pi}}{C^8}\,\frac{N^2}{M}\, T+ {\cal O}(T^4),
\end{equation}
where the constant $C\approx 1.2$ arises as a matching parameter when ``gluing'' the spike to the M5-branes. NS then compare the term $\propto T$ in eq.~\eqref{eq:nsent} with the Cardy entropy density of a 2d CFT, $\frac{\pi}{3} \, c \, T$~\cite{Cardy:1986ie}, and hence identify a central charge,
\begin{equation}
\label{eq:bns}
b_{\textrm{NS}} = \frac{8}{45}\frac{\Gamma(\frac{1}{3})\,\Gamma(\frac{1}{6})}{C^8\,\sqrt{\pi}}\,\frac{N^2}{M} \approx 0.35\,\frac{N^2}{M}.
\end{equation}
Our result for $b_{6d}$ shares at least one superficial similarity with $b_{\textrm{NS}}$, namely $b_{6d}$ in eq.~\eqref{eq:b6dsugra} includes a term $\frac{3}{5} \frac{N^2}{M} \approx 0.6 \frac{N^2}{M}$, similar to $b_{\textrm{NS}}$. However, whether a limit exists---including in particular NS's limit $N \ll M^2$---in which that term dominates over the others in $b_{6d}$ is unclear. Moreover, we can identify at least three reasons why $b_{6d}$ and $b_{\textrm{NS}}$ need not agree. First, whether $b_{6d}$ and $b_{\textrm{NS}}$ are the same quantity is unclear. Specifically, for a 2d defect there appears to be no universal relation between the defect central charge defined from EE and any Cardy-like contribution to $S$~\cite{Jensen:2018rxu}. Second, whether NS's result for $S$ should be interpreted as a Cardy entropy is highly suspect: NS's result is a \textit{low}-$T$ limit, relative to the system's spatial size, whereas Cardy's result is the \textit{high}-$T$ limit. In a 2d CFT modular invariance relates the high- and low-$T$ limits of $S$, so that $c$ determines both limits. Whether that remain true for a 2d defect is unclear---especially since modular invariance is generically absent for a 2d defect. Lastly, $b_{\textrm{NS}}$ seems to involve no data about a representation, which again leaves open which representation(s) to choose in order to compare $b_{6d}$ and $b_{\textrm{NS}}$, if any.
The $b_{3d}$ of the totally ``symmetric'' or ``anti-symmetric'' partitions in eqs.~\eqref{eq:b3d-cousins-symm} or~\eqref{eq:b3d-cousins-asymm} does not appear to have any similarities with $b_{\textrm{NS}}$. As in the comparison to BH in sec.~\ref{sec:bh}, $b_{3d}$ and $b_{\textrm{BH}}$ need not agree: they are different quantities, $b_{\textrm{NS}}$ involves no information about a partition, NS's solution has no M5$'$-branes, and NS's solution at $T=0$ presumably has $\gamma=1$ rather than $\gamma\in(-1,0)$.
Ref.~\cite{Niarchos:2012cy} pointed out that if we define the combination
\begin{equation}
\Lambda \equiv \frac{M^2}{N},
\end{equation}
then the central charge~\eqref{eq:bns} can be written
\begin{equation}
b_\mathrm{NS} \approx 0.35 \frac{N^{3/2}}{\sqrt{\Lambda}} \approx 0.35 \frac{M^3}{\Lambda^2},
\end{equation}
depending on whether $\Lambda$ is substituted for $M$ or $N$, respectively. Hence, for fixed $\Lambda$, $b_\mathrm{NS}$ scales both as $N^{3/2}$, characteristic of M2-branes, and $M^3$, characteristic of M5-branes. However, the same is not true for our results, in general. For example, making the same substitution in result eq.~\eqref{eq:b6dantisymm} for the central charge of a Wilson surface in an antisymmetric representation, we find (using $M \gg 1$)
\begin{equation}
b_{6d} = \frac{3}{5} \left(\sqrt{\Lambda} N^{3/2} - N^2 \right)
= \frac{3}{5} \left(\frac{M^3}{\Lambda} - \frac{M^4}{\Lambda^2} \right).
\end{equation}
which includes a term scaling as $N^{3/2}$ or $M^{3}$, but also another term scaling with a power of $N$ or $M$ that is not necessarily subleading in the SUGRA approximation. Indeed these terms are required to ensure that $b_{6d} = 0$ for the antisymmetric representation with $N = M$, corresponding to a trivial representation.
\subsubsection{Free Limit of the ABJM BCFT}
\label{sec:freeabjm}
A Wilson surface has two equivalent descriptions. The first is from the M5-brane theory perspective, as a non-local operator in the 6d ${\mathcal N}=(2,0)$ SUSY CFT. The second is from the M2-brane perspective, as the boundary of the ABJM BCFT with maximally SUSY boundary conditions~\cite{Berman:2009xd}, and possibly coupled to 2d SUSY multiplets~\cite{Niarchos:2015lla}. In this section we will compute the boundary central charge for the maximally SUSY ABJM BCFT in the free limit, which we denote $b_{\textrm{free}}$. We will assume no 2d SUSY multiplets are present, that is, we will calculate the contributions to $b_{\textrm{free}}$ only from the fields of ABJM.
As mentioned in sec.~\ref{sec:intro}, the ABJM theory~\cite{Aharony:2008ug} is a 3d ${\mathcal N}=6$ SUSY $U(N)_k \times U(N)_{-k}$ Chern-Simons matter CFT, and is the low-energy description of $N$ M2-branes at a $\mathbb{C}^4/\mathbb{Z}_k$ singularity. When $k=1,2$ the SUSY is enhanced to ${\mathcal N}=8$. The field content is two ${\mathcal N}=2$ vector multiplets, two ${\mathcal N}=2$ adjoint hypermultiplets, and two ${\mathcal N}=2$ bi-fundamental hypermultiplets in complex conjugate representations of the gauge group. The on-shell degrees of freedom include, from the adjoint multiplets, the Chern-Simons gauge fields and their fermionic super-partners, and from the bi-fundemental hypermultiplets, eight real scalar fields describing the positions of the M2-branes in the eight transverse directions, and their fermionic super-partners. The theory's 't Hooft coupling is $N/k$, hence the theory becomes weakly-coupled when $k \gg N$ and free when $k \to \infty$ with $N$ fixed.
The maximally SUSY boundary conditions on the ABJM fields that describe M2-branes ending on M5-branes appear for example in ref.~\cite{Berman:2009xd}, which we now briefly review. Four of the scalars have Dirichlet boundary conditions, representing the fact that the M2-branes cannot move away from the M5-branes in the directions $(x_7,x_8,x_9,x_{10})$. The other four scalars obey a Basu-Harvey-type equation~\cite{Basu:2004ed}, which in the free limit $k \to \infty$ with $N$ fixed reduces to a Neumann boundary condition, representing the fact that the M2-branes can move freely along the M5-branes in the directions $(x_3,x_4,x_5,x_6)$. The bi-fundamental gauge indices on the scalars must also encode the partition $\rho$ describing which M5-brane each M2-brane ends on. The boundary conditions on the Chern-Simons gauge fields and fermionic super-partners follow from SUSY.
We also want the boundary conditions to preserve conformal symmetry. In the free limit, for scalar fields Dirichlet and Neumann boundary conditions each preserve conformal symmetry. For Chern-Simons gauge fields conformal invariance requires a boundary term producing a Wess-Zumino-Witten (WZW) model at the 2d boundary. For Dirac fermions conformal symmetry requires acting with a projector that produces a single chiral fermionic mode at the 2d boundary. The maximally superconformal boundary conditions ensure that the WZW and 2d chiral fermion degrees of freedom together preserve parity. For what follows we will need no further details about the boundary conditions.
For 3d BCFTs the central charge defined from EE appears also as a central charge in the trace anomaly~\cite{Fursaev:2013mxa,Fursaev:2016inw}, in essentially the same way as a 2d CFT. To be precise, in a 3d BCFT the only non-zero contribution to the trace anomaly comes from the 2d boundary~\cite{Schwimmer:2008yh}, and includes a term $\frac{b}{24\pi} \int \hat{R}$ localized at the boundary, where $\hat{R}$ is the Ricci scalar of the boundary's induced metric. The coefficient of that Ricci scalar term obeys a $c$-theorem for boundary RG flows, and hence serves as a measure of the number of degrees of freedom localized at the boundary~\cite{Jensen:2015swa}. In 3d the contribution to the trace anomaly's 2d Ricci scalar term is known for Chern-Simons gauge fields, free Dirac fermions, and free scalars with Dirichlet and Neumann boundary conditions~\cite{Jensen:2015swa,Nozaki:2012qd,Fursaev:2016inw}. We can thus calculate $b$ for the maximally SUSY ABJM BCFT in the free limit simply by summing the known results for the on-shell fields in the SUSY multiplets mentioned above.
Crucially, for a free Dirac fermion in 3d, $b=0$, so in the maximally SUSY ABJM BCFT in the free limit the fermion's contribution to $b_{\textrm{free}}$ is zero. For a free real scalar in 3d the values of $b$ for Dirichlet and Neumann boundary conditions are equal and opposite. In the maximally SUSY ABJM BCFT in the free limit, four scalars have Dirichlet boundary conditions and the other four have Neumann, so the scalars' net contribution to $b_{\textrm{free}}$ is also zero. As a result, $b_{\textrm{free}}$ will be blind to the partition $\rho$ describing which M5-brane each M2-brane ends on. We are left with only the $U(N)_k \times U(N)_{-k}$ Chern-Simons gauge fields, whose contribution to $b_{\textrm{free}}$ is two copies of the WZW central charge with the same $N$ but levels $k$ and $-k$. Starting from a Chern-Simons theory with gauge Lie algebra $\mathfrak{g}$ at level $k$, the central charge of the corresponding WZW model is
\begin{equation}
c_{\textrm{WZW}} = \frac{k \, \text{dim} \,\mathfrak{g}}{k +g^{\vee}},
\end{equation}
where $g^{\vee}$ is $\mathfrak{g}$'s dual Coxeter number. For $\mathfrak{g}=\mathfrak{u}(N)$, we have $\text{dim }\mathfrak{g}=N^2$ and $g^{\vee}=N$. The maximally SUSY ABJM BCFT in the free limit therefore has a boundary central charge
\begin{equation}
\label{eq:bfree}
b_{\textrm{free}} = \lim_{k \to \infty} \left[\frac{k \, N^2}{k +N} + \frac{(-k) \, N^2}{(-k) +N}\right] = \lim_{k \to \infty}\frac{2\,k^2 \,N^2}{k^2-N^2} = 2 N^2,
\end{equation}
which indeed contains no information about a partition.
The only similarity between $b_{6d}$ and $b_{\textrm{free}}$ is that $b_{6d}$ contains terms scaling as $N^2$ at large $N$, though generically with different coefficients, as obvious in the examples of eqs.~\eqref{eq:b6dsymm} and~\eqref{eq:b6dantisymm}. In some cases the $N^2$ term in $b_{6d}$ dominates. For example, a totally symmetric representation in the large-$M$ and $N \gg M$ limits has $b_{6d} \approx - \frac{3}{5} N^2$, which however clearly disagrees with $b_{\textrm{free}}$, even in sign.
Superficially, $b_{3d}$ is more similar to $b_{\textrm{free}}$ than $b_{6d}$ is. Indeed, the examples in eqs.~\eqref{eq:b3d-cousins-symm} and~\eqref{eq:b3d-cousins-asymm} clearly scale as $N^2$ at large $N$. In fact, the coefficient even agrees in one case, namely $b_{3d}$ for the ``anti-symmetric'' partition in eq.~\eqref{eq:b3d-cousins-asymm}, with $\gamma = -1/4$.
However, we can identify at least two reasons why $b_{6d}$ or $b_{3d}$ and $b_{\textrm{free}}$ need not agree. First, similar to the previous comparisons, for $b_{3d}$ the 3d BCFT has a different super-group from the maximally SUSY ABJM BCFT. Second, we calculated $b_{6d}$ and $b_{3d}$ via holography, and thus in the limit of large 't Hooft coupling $\lambda = N/k\gg 1$ and hence $N\gg 1$, whereas obviously we calculated $b_{\textrm{free}}$ in the free limit. As a result, any agreement between $b_{6d}$ or $b_{3d}$ and $b_{\textrm{free}}$ is likely just a coincidence, and should probably be treated with skepticism.
\section{Discussion and Outlook}
\label{sec:discussion}
We used the 11d SUGRA solutions of refs.~\cite{Bachas:2013vza,DHoker:2008rje,D'Hoker:2008wc,Estes:2012vm} to compute holographically the EE in two cases. First was a spherical region centered on a Wilson surface in the M5-brane theory, describing $N$ M2-branes ending on $M$ M5-branes at large $M$. Second was the EE of a semi-circular region centered on the 2d boundary in cousins of the maximally SUSY ABJM BCFT, also at large $N$. The Wilson surface or 2d boundary is characterized by a partition $\rho$ of $N$, which for the Wilson surface determines a representation ${\cal R}$ of the M5-branes' worldvolume $\mathfrak{su}(M)$ gauge algebra. From our result for EE we extracted a central charge, $b_{6d}$ or $b_{3d}$, as a contribution from the Wilson surface or 2d boundary to the coefficient of the term logarithmic in the cutoff. Presumably $b_{6d}$ or $b_{3d}$ provides one measure of the number of massless degrees of freedom on the Wilson surface or 2d boundary, respectively.
Our main result for $b_{6d}$ in eq.~\eqref{eq:b6dcompact} is written compactly in terms of ${\cal R}$'s highest weight vector, $\lambda$, and the $\mathfrak{su}(M)$ Weyl vector, $\varrho$, and is manifestly invariant under the action of the Weyl group, including complex conjugation, ${\cal R} \to \overline{\cal R}$. We found that neither $b_{6d}$ nor $b_{3d}$ naturally scales with the number of degrees of freedom of the M5- or M2-brane theories at large $M$ or $N$, namely $M^3$ or $N^{3/2}$, respectively. Instead, for several examples of $\rho$ we found that $b_{6d}$ and $b_{3d}$ typically scale as $M^2$ or $N^2$. Our results also do not meaningfully agree with previous calculations of self-dual string central charges by Berman and Harvey~\cite{Berman:2004ew} and Niarchos and Siampos~\cite{Niarchos:2012cy} (nor do the results in those two references agree with each other), and indeed we provided a long list of reasons why. For example, a chiral anomaly was calculated in ref.~\cite{Berman:2004ew} while ref.~\cite{Niarchos:2012cy} computed thermodynamic entropy. The relation of either of these quantities to EE for a 2d defect or boundary is unclear. Moreover, neither of the previous putative central charges contain any information about the partition of $N$, and so it is unclear which $\rho$ to chose for our comparisons.
Indeed, we know of several more promising possible comparisons, which could provide more details about the origin and implications of our results. Examples include holographic EE in the asymptotically locally $AdS_4 \times S^7$ solutions dual to M2-branes ending on M5-branes~\cite{Bachas:2013vza} (which have potentially dangerous singularities), elliptic genera of M2-branes suspended between M5-branes~\cite{Haghighat:2013gba}, or the superconformal index of the M5-brane theory compactified on $S^1 \times S^5$---where upon dimensional reduction along the $S^1$ the Wilson surface reduces to a Wilson line of the effective 5d maximally SUSY Yang-Mills (SYM) theory on $S^5$~\cite{Bullimore:2014upa}. In some sense all of these should be counting the same massless degrees of freedom that contribute to our $b_{6d}$, though perhaps in different limits.
Our result for $b_{6d}$ in eq.~\eqref{eq:b6dcompact} also suggests a tantalizing potential connection to Toda CFTs. In particular, an $A_{M-1}$ Toda CFT has primaries labeled by representations ${\cal R}$ which have scaling dimensions $\frac{1}{2} \left [ 2Q \left(\lambda,\varrho\right)-\left(\lambda,\lambda\right)\right],$ with background charge $Q$. Our $b_{6d}$ is functionally similar to that with $Q=8$, but with an overall $3/5$ factor instead of $1/2$.
A natural question is thus whether we can reproduce our result for $b_{6d}$ via the AGT correspondence~\cite{Alday:2009aq}. Imagine compactifying M5-branes on a Riemann surface times a squashed $S^4$. The low-energy effective theory on the squashed $S^4$ will be an ${\mathcal N}=2$ SYM theory whose field content depends on the Riemann surface's genus and punctures. The AGT correspondence is the statement that the 4d SYM theory's partition function on the squashed $S^4$, which can be computed via SUSY localization~\cite{Pestun:2007rz,Hama:2012bg}, is equivalent to a certain correlator in a Toda CFT on the Riemann surface, with background charge $Q$ determined by the $S^4$'s squashing parameters. A 2d defect in the M5-brane theory that descends to a 2d defect in the 4d SYM theory appears in the Toda theory as a degenerate operator~\cite{Alday:2009fs,Kozcaz:2010af}. A key question is thus whether and how the dimension of that degenerate operator determines our $b_{6d}$. In other words, can we reproduce our result for $b_{6d}$, in whole or in part, from a calculation either in the Toda CFT with degenerate operator or in the 4d SYM theory with 2d defect~\cite{Bullimore:2014nla,Gomis:2014eya,Gomis:2016ljm}? In particular, does the dimension of the degenerate operator in the Toda CFT determine the EE of the 2d defect in the 4d ${\mathcal N}=2$ SYM? Or is the similarity we found merely a coincidence?
We are currently investigating all of the above possible comparisons. However, more generally we hope our results may help shed light on the mysterious degrees of freedom of self-dual strings, the 6d ${\mathcal N}=(2,0)$ CFT, and most of all, M-theory.
\acknowledgments
We would like to thank Costas Bachas, David Berman, Matthew Buican, Nadav Drukker, Christopher Herzog, Matti J\"arvinen, Kristan Jensen, S.~Prem Kumar, Bruno Le Floch, Vasilis Niarchos, and Kostas Skenderis for useful discussions and correspondence. D.~K. acknowledges support of PSC-CUNY Research Award. A.~O'B. is a Royal Society University Research Fellow. B.~R. and R.~R. acknowledge support from STFC through Consolidated Grant ST/L000296/1. A.~O'B., B.~R., and R.~R. thank the Institut Henri Poincar\'e for hospitality while this work was in progress.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 563 |
Read Next: Jennifer Lopez-Owen Wilson Film 'Marry Me' Moves Back to Universal From STX
August 31, 2018 1:07AM PT
Film Review: 'Reprisal'
A mediocore action-thriller that, not so long ago, would have gone straight to video.
By Joe Leydon
Joe Leydon
Film Critic @joeleydon FOLLOW
Film Review: 'The Other Side of Heaven II: Fire of Faith'
Film Review: 'The Last Whistle'
Film Review: 'The Outsider'
CREDIT: Brian Douglas
Frank Grillo, Bruce Willis, Olivia Culpo, Johnathon Schaech, Natali Yura, Uncle Murda, Natlia Sophie Butler, Tyler Olson, Wass Stevens, Colin Egglesfield.
Rated R 1 hour 29 minutes
In a previous iteration of the movie business, a derivative B movie such as "Reprisal" would have provided gainful employment for actors such as Eric Roberts, Ice-T, and Gary Busey — who, truth to tell, likely would have delivered more than a fair share of down-and-dirty cheap thrills — and premiered on the New Releases shelf at Blockbuster Video stores everywhere. That was then, this is now: Fodder of this kind can land at least a smattering of theatrical engagements on the same day it's made available on demand. And yet, it probably, and deservedly, will have an even shorter shelf life than direct-to-video product of yesteryear, despite the marquee value of Bruce Willis and Frank Grillo in lead roles.
Indeed, with all due respect to Willis, whose admirable professionalism permits him to bring a second dimension to the flat role he's given here, the few curiosity-seekers who bother to track down this trifle in years to come almost certainly will do so because of Grillo's finely calibrated portrayal of a Cincinnati bank manager driven to extremes after a criminal mastermind (Johnathon Schaech) kills a guard during a robbery at his place of employment. Grillo — whose star currently is ascendant after, among other achievements, his larger-than-life villainy in the China-produced smash hit "Wolf Warrior II" and his pedal-to-the-metal anti-heroism in the Netflix neo-noir "Wheelman" — is doubtless bound for bigger and better things. "Reprisal" may someday rank as a footnote in his career, but, really, that's the only reason it will be remembered at all.
"Reprisal" begins with a robbery at the bank managed by Jacob (Grillo), whose devotion to his wife (Olivia Culpo) and young diabetic daughter (Natalie Sophie Butler) is emphasized with all the subtlety of someone dropping a bowling ball out of a fourth-floor window by his insistence on preparing a week's worth of lunch for the pair. After the traumatic holdup, Jacob seeks advice on the subject of revenge from his next-door neighbor, James (Willis), who just happens to be a retired cop — just like Jacob's late father.
One thing leads to another, much more speedily than seems probable or possible, and pretty soon Jacob and James are able to connect dots that somehow remain elusive for the FBI and other law enforcement agencies. After they figure out where the criminal mastermind will pull his next heist, they attempt to intervene. But of course — well, let's just say the movie gives Jacob a wife and daughter for a reason, and his adversary is hardly a novice amateur.
To give fair credit to director Brian A. Miller, two extended shootout sequences — one involving an armored-car robbery, the other depicting Jacob's desperate efforts to save his wife and child — are sufficiently exciting to keep any home viewer distracted from ringing phones or barking dogs. And Willis sporadically suggests (especially when his character recalls trying to offer comfort to the child of a drowning victim) that he still has the chops to do a lot more than has been asked of him lately.
"Reprisal" is not a very good movie, but it leaves you with tantalizing hints that some people involved with it are capable of doing something much better.
Film Review: 'Reprisal'
Reviewed online, Houston, Aug. 30, 2018. MPAA Rating: R. Running time: 89 MIN.
Production: A Lionsgate Premiere release and presentation, in association with Grindstone Entertainment, of an Emmett Furla, Oasis Films production, in association with Checkluck Films, Kind Hearts Entertainment. Producers: Randall Emmett, George Furla. Executive Producers: Henry Winterstern, Arianne Fraser, Delphine Perrier, Ted Fox, Vance Owen, Barry Brooker, Stan Wertlieb.
Crew: Director: Brian A. Miller. Screenplay: Bryce Hammons. Camera (color): Peter A. Holland. Editor: Ryan Dufrene. Music: Giona Ostinelli, Sonya Belousova.
With: Frank Grillo, Bruce Willis, Olivia Culpo, Johnathon Schaech, Natali Yura, Uncle Murda, Natlia Sophie Butler, Tyler Olson, Wass Stevens, Colin Egglesfield. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,172 |
{"url":"https:\/\/byjus.com\/maths\/trigonometric-addition-formulas\/","text":"The trigonometric addition formulas can be applied to simplify a complicated expression or find an exact value when you are with only some trigonometric values. For instance, if you want the Sine of 15 degrees, you can use a subtraction formula to calculate sin(15) as sin(45-30).\n\nFORMULAS\n\n\u2022 sin(a+b) = sin(a)cos(b) + cos(a)sin(b)\n\u2022 sin(a-b)= sin(a)cos(b)-cos(a)sin(b)\n\u2022 cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\n\u2022 cos(a-b)=cos(a)cos(b)+sin(a)sin(b)\n\u2022 $\\tan (a+b) = \\frac{\\tan a + \\tan b}{1 \u2013 \\tan a \\tan b}$\n\u2022 $\\tan (a+b) = \\frac{\\tan a \u2013 \\tan b}{1 + \\tan a \\tan b}$\n\nLet\u2019s solve some examples:\n\nExample 1\n\nVerify the identity\n\n$\\cos (a+b). \\cos (a-b) = \\cos^{2}a-\\sin^{2}b$\n\nSolution\n\ncos(a+b) cos(a-b) = (cos(a)cos(b)-sin(a)sin(b)) (cos(a)cos(b) + sin(a)sin(b))\n\nThis gives,\n\n$\\cos(a+b) \\cos(a-b)= \\cos^{2}a \\cos^{2}b \u2013 \\sin^{2}a \\sin^{2}b$ \u2026\u2026\u2026\u2026(i)\n\nBut,\n\n$\\sin^{2}a \\sin^{2}b = (1-\\cos^{2}a).(1- \\cos^{2}b) = 1-\\cos^{2}a-\\cos^{2}b +\\cos^{2}a\\cos^{2}b$ \u2026\u2026\u2026(ii)\n\nEquating the value of (ii) in (i), we have\n\n$\\cos(a+b) \\cos(a-b)= \\cos^{2}a \\cos^{2}b \u2013 1 + \\cos^{2}a + \\cos^{2}b \u2013 \\cos^{2}a\\cos^{2}b$\n\ncos2(a)cos2(b) \u2013 (1 \u2013 cos2(a) \u2013 cos2(b) + cos2(a) cos2(b)) = cos2(a) + cos2(b) \u2013 1\n\n$\\Rightarrow \\cos(a+b) \\cos(a-b)= \\cos^{2}a + \\cos^{2}b \u2013 1$\n\nWe know, $\\cos^{2}b = 1 \u2013 \\sin^{2}b$\n\nTherefore, we get\n\ncos(a + b) cos(a-b) = cos2(a) \u2013 sin2(b)\n\nExample 2\n\nCalculate the exact value of $\\cos (165^{\\circ})$.\n\nSolution\n\nBy applying the addition formulas for the cosine function, we get\n\ncos(165) = cos(120+45) = cos(120)cos(45) \u2013 sin(120)sin(45)\n\nSince we know that,\n\ncos(120) = -\u00bd\n\nsin(120) = 3\/2\n\ncos(45) = 2\/2\n\nsin(45) = \u221a2\/2\n\nTherefore, we get\n\ncos(165) = $-\\frac{1}{2}.2\\sqrt{2} \u2013 \\frac{\\sqrt{3}}{2}. \\frac{\\sqrt{2}}{2}$\n\n$= -\\frac{\\sqrt{2}+\\sqrt{6}}{4}$<\n\nTo solve more problems and examples on Trigonometry, Trigonometric Addition and subtraction, visit byju\u2019s.com.\n\n#### Practise This Question\n\nWhich of the following cannot be a part of a set of rational numbers?","date":"2019-06-26 06:45:36","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8972529768943787, \"perplexity\": 4238.873868704182}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-26\/segments\/1560628000175.78\/warc\/CC-MAIN-20190626053719-20190626075719-00345.warc.gz\"}"} | null | null |
\section{Introduction}
In this work, we analyse the behaviour of the Laplacian-Heaviside (LapH)
\cite{HadronSpectrum:2009krc} smearing procedure in the presence of open boundary conditions. By
evaluating the pseudoscalar meson two--point functions we determine the portion of
the lattice unaffected by the boundary and the optimal LapH setup.
Our aim is the study of semileptonic decays in this framework, a largely
unexplored setup \cite{Boyle:2019cdl}. The evaluation of three--point
functions with local current insertions gives access to various quantities
relevant to flavour physics. The combination of theory predictions of
semileptonic form factors with experimental data gives access to components
of the Cabibbo-Kobayashi-Maskawa (CKM) matrix \cite{Cabibbo:1963yz, Kobayashi:1973fv}, where an
increase in precision could lead to hints of new physics. Also, the present
tension in lepton flavour universality \cite{London:2021lfn} makes the study of
semileptonic processes particularly interesting.
One source of uncertainty which limits the attainable precision of current
theory predictions arises from the trade-off between the contamination due to
excited states in the two-- and three--point functions and the increased
statistical uncertainty when their separation increases. Short separations
between source and sink make the evaluation of plateau in the relevant
quantities a delicate task and typically require sophisticated fitting
procedures. The LapH smearing procedure ameliorates the effects of excited
states on the relevant correlation functions.
Ensembles with open boundary conditions pave the path towards finer lattice
spacings, which are particularly relevant for heavy--light flavour
phenomenology. However, this requires to control any effects that might arise
from the open boundaries, which we aim to explore in this work.
\section{Computation of meson two--point functions}
Our computations are based on the recently developed LapH \cite{HadronSpectrum:2009krc} smearing
procedure implemented on each time--slice by the smearing matrix
\begin{equation}
\mathcal{S}(\bar{x},\bar{y}) = \Theta \left( \sigma_s^2 + \Delta \right) \simeq V_s V_s^\dagger,
\end{equation}
where $V_s$ is the matrix containing columns of the eigenvectors associated with
the $N_{ev}$ lowest-lying eigenvalues of the 3D Laplacian $\Delta$. The
smeared quark fields thus take the form
\begin{equation}
\tilde{\chi} = \chi\mathcal{S} = \bar{\psi}\gamma_4 \mathcal{S}, \qquad \tilde{\psi} = \mathcal{S} \psi.
\end{equation}
Then, by introducing some $Z(N)$ noise vectors $\rho$ in the LapH eigenspace and
by defining the related dilution projector $\mathcal{P}$ in the time, spin and
$ev$ space \cite{Morningstar:2011ka} one can identify two new objects: quark sinks
$\varphi = \mathcal{S}\Omega^{-1}V_s\mathcal{P}\rho$ and quark sources
$\varrho=V_s\mathcal{P}\rho$, where only the former requires a Dirac matrix
inversion to be computed. With these building blocks, the meson correlation
functions take the form
\begin{align}
C(t - t_0) &= \langle
\tilde{\chi}_{a}(t) \Gamma^A \tilde{\psi}_{a}(t) \
\tilde{\chi}_{b}(t_0) \Gamma^B \tilde{\psi}_{b}(t_0)
\rangle
= \langle \Gamma^B \ Q_{ba}(t_0, t) \ \Gamma^A \ Q_{ab}(t,t_0) \rangle\\
& = \langle \Gamma^B \varphi_{b}(t_0)\varrho_{a}(t)^* \Gamma^A \varphi_{a}(t)\varrho_{b}(t_0)^* \rangle
= \langle \Gamma^B \bar{\varrho}_{b}(t_0)\bar{\varphi}_{a}(t)^* \Gamma^A \varphi_{a}(t)\varrho_{b}(t_0)^* \rangle\\
& = \langle \mathcal{M}^{\Gamma^A}(\bar{\varphi}, \varphi, t) \ \mathcal{M}^{\Gamma^B}(\bar{\varrho}, \varrho, t_0)^* \rangle.
\end{align}
Where $\mathcal{M}$ are the (LapH subspace-sized) diluted meson fields. In this
work we are interested only in evaluating pseudoscalar mesons, therefore we will
study only the $A_{1u}^{(+)}$ channel of the octahedral symmetry
group. Moreover, we are going to employ a trivial $Z(1)$ noise vector, and no
dilution scheme, therefore restricting ourselves to \textit{exact distillation}.
Due to the open boundary conditions, and up to leading order in chiral
perturbation theory, we expect the two--point functions to fall off as
\cite{Luscher:2012av}
\begin{equation}\label{eq:sinh}
C(t) \propto \sinh (m (\tilde{T} - t)),
\end{equation}
where $\tilde{T}$ is a free parameter. In particular, it defines a "virtual"
lattice border inside the lattice itself and can be easily obtained via a fit.
\section{Results}
Measurements are performed on configurations generated by the CLS initiative
\cite{Bruno:2014jqa} with $N_f=2+1$ non-perturbatively improved Wilson fermions with open
boundary conditions, details of which are given in
Table~\ref{tab:ensembleparameters}.
\begin{table}
\begin{tabular}{ccccccccc}
$\mathrm{id}$ & $a[\mathrm{fm}]$ & $N_s$ & $N_t$ & $\kappa_l$ & $\kappa_s$ & $\kappa_c$ \cite{Gerardin:2019rua} & $m_\pi [\mathrm{MeV}]$ & $m_K [\mathrm{MeV}]$\\ \hline
$B105$ & $0.086$ & $32$ & $64$ & $0.136970$ & $0.13634079$ & $0.123244(19)$ & $280$ & $480$
\end{tabular}
\caption{Relevant parameters for the lattice ensemble studied in this work.}
\label{tab:ensembleparameters}
\end{table}
Gauge averages are extracted from a subsample of $100$ configurations and error
estimates are obtained from the bootstrap procedure. On each configuration, we
generate the LapH subspace with $96$ eigenvectors from which we compute the
pseudoscalar meson fields at three source positions $t_0 = 9, 15, 31$.
Since we are using exact distillation, the meson correlation functions with
fewer numbers of eigenvectors can be obtained without any additional
computations by considering the $4 N_{ev} \times 4 N_{ev}$ square sub-matrix
for each of the meson field matrices.
\subsection{Boundary effect on the LapH eigenspace}
As a first step to estimate the effect of the open boundary conditions on the
LapH subspace we consider the gauge average the LapH eigenvalues as a function
of the lattice time coordinate. Periodic boundary conditions guarantee
translational invariance of the LapH subspace, while with open boundary
conditions this might not be the case. As can be seen from the left--hand panel
of figure~\ref{fig:ev}, we observe that the eigenvalues are time--independent
provided they are sufficiently far from the boundary. For the ensemble under
consideration this amounts to approximately 12 time--slices ($\sim
1\,\mathrm{fm}$). Taking the bulk value to be the average of the central 24
time--slices, the right--hand panel of the same figure shows the relative
deviation from this for each eigenvalue. We find that the deviation from the
bulk value is to a good approximation exponential.
\begin{figure}[t]
\centering
\makebox[\textwidth][c]{\includegraphics[width=1.2\textwidth]{img/ev.pdf}}
\caption{\textbf{Left}: eigenvalues of the Laplacian as a function of the
lattice time coordinate. \textbf{Right}: deviation from the bulk value for
each eigenvalue.}
\label{fig:ev}
\end{figure}
\subsection{Considerations on the realness of the meson correlation function}
Another important aspect of the computation of the two--point functions with
open boundary conditions concerns the effects that boundary vacuum states have
on the correlation function. Following~\cite{Guagnelli:1999zf} where the
analogous discussion for the Schr\"odinger functional case is presented, we
write a two--point function with open boundary conditions in the quantum
mechanical representation as
\begin{equation}
C_2(t_1,t_2)= \frac{1}{Z} \langle i_0| P(t_1) P^\dagger(t_2) |i_0\rangle \;, \quad t_1>t_2\;,
\end{equation}
where $| i_0 \rangle$ is the state at the boundaries, with vacuum quantum
numbers. In general
\begin{equation}
| i_0 \rangle = w_0 |0\rangle + w_1|0'\rangle + \cdots
\end{equation}
where $|0\rangle$ is the lowest eigenstate of the Hamiltonian $H$ and
$|0'\rangle$ is the first excited vacuum state. Note that in this formulation
the overlap factors $w_i$ are in general complex and the partition function
reads $\langle i_0 | e^{-TH} | i_0 \rangle$, where $T$ is the temporal extent of
the lattice. Now, up to a constant, the two--point function in the Heisenberg
picture is
\begin{equation}
C_2(t_1,t_2) \propto \langle i_0 |e^{-(T-t_1)H} P e^{-(t_1-t_2)H}P^\dagger e^{-t_2H} |i_0 \rangle\;.
\end{equation}
We consider the different contributions starting with the vacuum-vacuum
contribution to $C_2(t_1,t_2)$
\begin{equation}
w_0^* w_0 \langle 0| P e^{-(t_1-t_2)H}P^\dagger|0\rangle \;,
\end{equation}
which, under the assumptions above, is real. In order to simplify the discussion
we assume that $P$ excites just one state out of $|0\rangle$ and a different one
out of $|0'\rangle$. The two ground state vacuum to first excited state vacuum contributions read
\begin{equation}
w_0 w_1^* \langle 0'|e^{-(T-t_1)E_{0'}} P e^{-(t_1-t_2)H}P^\dagger|0\rangle \;, \textrm{ and} \quad
w_0^*w_1 \langle 0| P e^{-(t_1-t_2)H}P^\dagger e^{-E_{0'}t_2}|0'\rangle \;.
\end{equation}
Even within the strong assumptions above their sum is not real.
However, as long as only pure QCD is taken into account, a single meson state
can be represented equivalently with the two charge conjugated flavour
combinations, and the resulting correlation functions are complex conjugates
of each other. If the boundary contaminations are small, the real part of the
correlation functions still correctly represents the meson state.
\subsection{Boundary effects on the meson correlation functions}
Now we consider the results for the two--point functions computed with the
machinery described above. In the following we restrict ourselves to the
evaluation of the correlation functions for $\pi$, $K$ and $D_s$ pseudoscalar
mesons, and the corresponding effective masses computed as
$m(t + \flatfrac{1}{2}) = \log(\flatfrac{C(t)}{C(t+1)})$.
To ensure that we have full control over all contributions to the correlation
function, we want to understand and disentangle the effects arising from the
LapH parameters and those coming from the boundaries.
In figure~\ref{fig:corr_1} we plot the two--point functions as a function of the
distance from the nearest boundary. The boundary effects are clearly visible at
the right hand edges of the various panels. The fact that the different branches
all coincide close to the boundary shows that these effects are independent of
the source position. They correspond to a tower of excitations arising from the
boundary. In the cases where sufficiently many time slices are available and at
this level of statistical uncertainty, we observe the onset of a plateau at
$\sim 2.0, 1.8, 1.0$ fm from the boundary for $\pi, K$ and $D_s$, respectively.
\begin{figure}
\centering
\includegraphics[width=0.32\textwidth]{img/pi_bshift.pdf}
\includegraphics[width=0.32\textwidth]{img/K_bshift.pdf}
\includegraphics[width=0.32\textwidth]{img/Ds_bshift.pdf}
\caption{Comparison of meson correlation functions for the pion (left), kaon
(middle) and the $D_s$ (right) computed at source times $t_0 = 9, 15, 31$
and with $N_{ev} = 96$. The top panels show the correlation functions, the
bottom panels the effective mass.}
\label{fig:corr_1}
\end{figure}
In figure~\ref{fig:corr_2} we restrict ourselves to a single source position
$t_0=15$ and investigate how quickly the effective mass approaches the plateau
as a function of the number of eigenvectors used in the construction of the LapH
subspace. As expected, a smaller number of lowest-lying eigenvectors results in
a broader smearing and less contamination by excited states near the source.
We observe a moderate increase in the statistical uncertainty as the number of
eigenvectors is reduced. In our setup we estimate the optimal number of
eigenvalues to be around $24$.
\begin{figure}
\centering
\includegraphics[width=0.32\textwidth]{img/pi_ev.pdf}
\includegraphics[width=0.32\textwidth]{img/K_ev.pdf}
\includegraphics[width=0.32\textwidth]{img/Ds_ev.pdf}
\caption{Comparison of the meson effective masses (top) and the relative
uncertainties (bottom) for varying number of eigenvectors $N_{ev} = 6, 24,
96$. The correlation functions are computed for one fixed source at $t_0 =
15$ and we restrict ourselves to the forward branch.}
\label{fig:corr_2}
\end{figure}
Lastly, in figure~\ref{fig:corr_3} we show the forward--time correlation
functions for $N_{ev} = 24$ and compare the short time--separation behaviour for
three source positions. The asymptotic value to which the effective mass
approaches is compatible with the literature~\cite{Bruno:2014jqa} (for the case of $\pi$
and $K$) or in good agreement with the physical value (for the $D_s$). However,
for the case of the pion and kaon at small time separations from the source, we
observe that the $t_0=9$ data set deviates from the other two source
positions. Crucially, this contamination does not arise from the farther
boundary, but rather from the one close to the source. We conclude that this
amounts to a mixing between the excited states generated at the source and those
generated at the boundary.
\begin{figure}
\centering
\includegraphics[width=0.32\textwidth]{img/pi.pdf}
\includegraphics[width=0.32\textwidth]{img/K.pdf}
\includegraphics[width=0.32\textwidth]{img/Ds.pdf}
\caption{Comparison of the forward branch of the meson correlation functions
(top) and effective masses (bottom) computed at different source times
$t_0 = 9, 15, 31$ and with $N_{ev} = 24$.}
\label{fig:corr_3}
\end{figure}
However, as shown in figure~\ref{fig:corr_1}, the boundary effects are a global
characteristic of the lattice to which we have access by considering the large
time separation behaviour of the meson correlation functions. In particular, the
correlation functions are well described by eq.~\ref{eq:sinh} and the parameters
$m$ and $\tilde{T}$ can be determined from a fit. Subsequently, we can define a
continuous estimate of the effective mass of the meson state as
\begin{equation}
m_{\sinh}(t) \equiv \dv{t} \log \left( \sinh (m (\tilde{T} - t)) \right) = m \ \coth(m(\tilde{T} - t)).
\end{equation}
We define the correction term $m_\mathrm{corr}$ by subtracting the fitted meson
mass $m$ from the quantity $m_\mathrm{sinh}(t)$, i.e.
\begin{equation}
m_{\textrm{corr}}(t) = m_{\sinh}(t) - m.
\end{equation}
This quantifies the amount of contamination stemming from the presence of the
open boundary conditions. Since we are interested into the corrections coming
from both open boundaries we correct our numerical estimate for the effective
mass ($m_{\log}(t)$) with $m_{\textrm{corr}}$ and its time reverse
\begin{equation}
m_{\textrm{eff}}(t) = m_{\log}(t) - m_{\textrm{corr}}(t) - m_{\textrm{corr}}(T - t).
\end{equation}
As shown in figure~\ref{fig:meff_corr}, this procedure is able to substantially
reduce the boundary effects on the effective masses. For each meson, we
extracted the two parameters $\tilde{T}$ and $m$ by simultaneously fitting the
two longer branches of the meson correlation functions generated from the
sources at $t_0 = 9$ and $15$. In all cases we obtain values of the reduced
$\chi^2$ lower that unity. While some deviations are still visible close to
$\tilde{T}$, the corrected effective masses show clear plateaux for a wide range
of time separations. This fitting procedure can be easily adapted to the
evaluation of the ratios of two-- and three--point correlation functions
required for the study of semileptonic decays.
\begin{figure}
\centering
\includegraphics[width=0.32\textwidth]{img/pi_meff_corr.pdf}
\includegraphics[width=0.32\textwidth]{img/K_meff_corr.pdf}
\includegraphics[width=0.32\textwidth]{img/Ds_meff_corr.pdf}
\caption{Comparison of the corrected effective masses (top) for $\pi$, $K$
and $D_s$ respectively, and the corresponding cumulative correction
$m_{\textrm{corr}}(t) + m_{\textrm{corr}}(T - t)$ (bottom).}
\label{fig:meff_corr}
\end{figure}
\section{Conclusions and future developments}
We have performed an extensive analysis of the effects of open boundary
conditions on the two--point correlation function computed with the LapH
formalism. On the ensemble we analysed and for our given statistical
uncertainties, we found that the region affected by the boundary extends into
the lattice for $1$ fm (12 time--slices), leaving us with more than half of the
lattice where the time translational invariance is restored. Our two--point
functions analysis lead to the identification of an optimal setup fixing the
source position at $t_0\sim15$ and number of eigenvectors to $N_{ev}\sim24$.
We further devised a fitting procedure that allows for the estimate of the
boundary contribution to the effective masses and a possible correction scheme
that has proven to greatly reduce the boundary effects on the meson effective
masses and that is largely independent of the source/sink positions.
Moreover, we have started generating data for three--point functions with local
current insertions. Using to the distillation workflow this will allow for the
evaluation of the processes needed to study CKM matrix elements and semileptonic
form factors with minimal computational overhead.
\section{Acknowledgments}
The work of J.T.T. has received funding from the European Union's Horizon 2020
research and innovation programme under the Marie Sk{\l}odowska-Curie grant
agreement No 894103. The DFF Research project 1. Grant n. 8021-00122B fully
supports the work of O.F. and partially supports M.D.M and
J.T.T. work. Computations were performed on UCloud interactive HPC system and
ABACUS2.0 supercomputer, which are managed by the eScience Center at the
University of Southern Denmark.
\bibliographystyle{JHEP}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,100 |
\section{CONCLUSION}
\label{sec:conclusion}
In this paper we present a theoretical formulation, and a corresponding numerical algorithm that can find Pontryagin-optimal inputs for general dynamical systems by using a direct method.
The numerical implementation is based on a relaxed-control system and PWM reconstruction.
Our novel approach produces significant improvements both in the quality of the resulting minimizers, and the flexibility of the numerical implementation.
\section{SIMULATION RESULTS}
\label{sec:examples}
\subsection{Constrained LQR}
\label{subsec:example_1}
The first simulation is an input-constrained LQR problem.
The system has 6~states, denoted $x$, and 2~inputs, denoted $u$.
The initial condition is $x(0) = 0$, and the vector field is:
\begin{equation}
\smat{\ddot{x}_1\\ \ddot{x}_2\\ \ddot{x}_3} =
\smat{-\sfrac{d}{m} & 0 & -\gamma\\ 0 & -\sfrac{d}{m} & 0\\ 0 & 0 & -\sfrac{m\, g\, l}{J}}\,
\smat{\dot{x}_1\\ \dot{x}_2\\ x_3} +
\smat{\sfrac{1}{m} & 0\\ 0 & \sfrac{1}{m}\\ \sfrac{r}{J} & 0}\,
\smat{u_1\\ u_2},
\end{equation}
with objective function:
\begin{equation}
\int_0^T \norm{\smats{x_1(t) + 0.3\\ x_2(t) + 0.5\\ x_3(t)}}_2^2 + \eta\, \norm{u(t)}_2^2\, \diff{t}.
\end{equation}
The parameters are $T=2$, $J = 0.0475$, $m = 1.5$, $r = 0.25$, $g = 9.8$, $\gamma = 0.51$, $d = 0.2$, $l = 0.05$, and $\eta = 0.05$.
We use 81~points to sample the vector field, evenly distributed in the control space $U = [-1,1]^2$.
\begin{figure*}
\centering
\subfloat[%
Optimal trajectory.
Final state shown as a triangle, $x(T) = (-0.29, -0.56, -0.06)$.%
]{%
\label{fig:exp1_x}%
\includegraphics[width=.31\linewidth,trim=20 0 0 10,clip]{images/x_lqr.pdf}%
}%
\hfill%
\subfloat[%
PWM projection of the optimal inputs.
Both inputs are identically zero for each $t \geq 0.5$.
]{%
\label{fig:exp1_u}%
\includegraphics[width=.31\linewidth,trim=0 0 0 0,clip]{images/u_lqr.pdf}%
}%
\hfill%
\subfloat[%
Optimality function value per iteration.
]{%
\label{fig:exp1_theta}%
\includegraphics[width=.31\linewidth,trim=0 0 0 0,clip]{images/theta_lqr.pdf}%
}%
\caption{%
Results of the constrained LQR simulation in Sec.~\ref{subsec:example_1}.
}
\label{fig:exp1}
\end{figure*}
Fig.~\ref{fig:exp1_x} shows the trajectory of the first three states of the system.
Fig.~\ref{fig:exp1_u} shows the corresponding controls after wavelet and PWM reconstruction.
Note how the input signal effort is very low, only applying an input for the first $0.1\unit{s}$ approx.
Fig.~\ref{fig:exp1_theta} shows the absolute value of the optimality function $\theta_{N,h}$ at each iteration.
After 62~iterations the optimality function reaches the threshold $\epsilon_{\text{tol}} = 10^{-5}$, validating the convergence of our algorithm.
\subsection{Quadrotor helicopter}
\label{subsec:example_2}
We also consider an application to control a 3-D quadrotor helicopter using a nonlinear model described in~\cite{pounds2006modelling}.
This system has 12~states, denoted $x$, and 4~inputs, denoted $u$.
The inputs are constrained to the set $U = [0, 2]^4$.
The initial condition is $x(0) = 0$, and the vector field is defined as follows:
\begin{equation}
\begin{aligned}
\ddot{x}_1 &= - \sfrac{b}{m}\, \dot{x}_1 + \sfrac{K}{m}\, \sin(x_5)\, \smats{1 & 1 & 1 & 1}\, u,\\
\ddot{x}_2 &= - \sfrac{b}{m}\, \dot{x}_2 + \sfrac{K}{m}\, \sin(x_4)\, \cos(x_5)\, \smats{1 & 1 & 1 & 1}\, u,\\
\ddot{x}_3 &= - \sfrac{b}{m}\, \dot{x}_3 + \sfrac{K}{m}\, \cos(x_4)\, \cos(x_5)\, \smats{1 & 1 & 1 & 1}\, u,\\
\ddot{x}_4 &= - \dot{x}_5\, \dot{x}_6 + \sfrac{K\, L}{I_x}\, \smats{0 & 1 & 0 & -1}\, u,\\
\ddot{x}_5 &= \dot{x}_5\, \dot{x}_6 + \sfrac{K\, L}{I_y}\, \smats{-1 & 0 & 1 & 0}\, u,\\
\ddot{x}_6 &= \sfrac{K}{I_x + I_y}\, \smats{1 & -1 & 1 & -1}\, u,
\end{aligned}
\end{equation}
with objective function:
\begin{equation}
\int_0^T \norm{\smats{x_1(t) - c_1\\ x_2(t) - c_2\\ x_3(t) - c_3}}_2^2 + \sum_{k=4}^6 \sin^2\pb{x_k(t)} + \eta\, \norm{u(t)}\, \diff{t}.
\end{equation}
The states $x_{1, 2, 3}$ represent the $x$, $y$, and $z$ position coordinates, respectively.
The states $x_{4, 5, 6}$ represent the Euler angles of the body.
The parameters are $m = 1.3 \unit{kg}$, $I_{x, y}=0.0605\unit{kg\cdot m^2}$, $g = 9.8\unit{m/s^2}$, $b=0.1$, $K = 1.0$, and $\eta = 0.05$.
We use 625~sampling points for the vector field, and the convergence threshold is $\epsilon_{\text{tol}} = 10^{-4}$.
\begin{figure}[tp]
\centering
\subfloat[%
$p_f = (-1.12, -1.12, -0.98)$.%
]{%
\label{fig:exp2_1}%
\includegraphics[width=.48\linewidth,trim=15 0 0 10,clip]{images/x_hel3d4.pdf}%
}%
\hfill%
\subfloat[%
$p_f = (1.23, 1.23, 1.18)$.%
]{%
\label{fig:exp2_2}%
\includegraphics[width=.48\linewidth,trim=15 0 0 10,clip]{images/x_hel3d5.pdf}%
}%
\caption{%
Results of the quadrotor helicopter simulation in Sec.~\ref{subsec:example_2}.
Final position, $p_f = \sp{x_1(T), x_2(T), x_3(T)}$, is shown as a triangle, and vertical orientation of the helicopter is shown as red arrows.
}
\label{fig:exp2}
\end{figure}
\begin{figure}[tp]
\centering
\includegraphics[width=.67\linewidth,trim=0 0 0 0,clip]{images/theta_hel3d.pdf}%
\caption{%
Optimality function value per iteration for the simulations in Fig.~\ref{fig:exp2_1} (blue solid line) and Fig.~\ref{fig:exp2_2} (orange dashed line).
}
\label{fig:exp2_3}%
\end{figure}
Fig.~\ref{fig:exp2} shows the results of the optimal trajectories for two scenarios: when $c = (-1.2, -1, -1)$ (Fig.~\ref{fig:exp2_1}), and when $c = (1.2, 1, 1)$ (Fig.~\ref{fig:exp2_2}).
Fig.~\ref{fig:exp2_3} shows the absolute value of the optimality function $\theta_{N,h}$ at each iteration in those two simulations.
In both cases we reach the convergence threshold $\epsilon_{\text{tol}}$ in less than 40~iterations.
These results show that our algorithm generates trajectories that move the helicopter to the desired positions while at the same time resulting in minor body-axis displacements.
\section{SYNTHESIS OF RELAXED OPTIMAL INPUTS}
\label{sec:implementation}
Now that we have established a theoretical foundation for the equivalence between first-order derivative and Minimum Principle minimizers for relaxed optimal control problems, we focus our attention on the development of numerical algorithms to synthesize approximated optimal control inputs.
The iteration algorithm in this section is based on solving a sequence of convex optimization problems, even when the dynamical system is nonlinear.
As we show in Fig.~\ref{alg:infinite}, it is possible to formulate an iterative gradient descent method using relaxed inputs that converges to Pontryagin minimizers.
This conceptual algorithm requires solving the optimization problem $\theta_h$, which generates a variation of a stochastic process to locally reduce the cost function of the problem in eq.~\eqref{eq:original2}.
The following propositions enable us to find efficient numerical implementations of our algorithm in Fig.~\ref{alg:infinite}.
\begin{prop}
\label{prop:theta_convex}
$\theta_h$ is a convex optimization problem.
\end{prop}
\begin{prop}
\label{prop:cohull}
For each $x \in \R^n$:
\begin{multline}
\set{\int_{\mathrlap{\R^m}}\, f(x,u)\, \diff{\mu}(u) \mid \mu \in {\cal M}(\R^m),\ \supp(\mu) \subset U}\\
= \cohull\set{f(x,u) \mid u \in U},
\end{multline}
where $\cohull(S)$ is the convex hull of $S \subset \R^n$.
\end{prop}
Using Prop.~\ref{prop:cohull} and eq.~\eqref{eq:ham_optimal_mu} we have almost all the necessary results to develop an implementable version of the algorithm in Fig.~\ref{alg:infinite}.
We now focus our attention on the two remaining problems: how to approximate the convex hull in Prop.~\ref{prop:cohull}, and how to synthesize control inputs given a stochastic process $\mu_t$.
\subsection{Convex Hull Approximation}
\label{subsec:sampling}
\begin{figure}[tp]
\centering
\scalebox{0.8}{\input{images/vector_field2.pdftex_t}}%
\caption{%
Vector field set at $\hat{x}$ (red), and its sampled convex hull (blue), for $f(x,u) = x + \p{u^2 + 1, u}^\tp$, $U = \sp{-1,1}$.
}
\label{fig:convex_hull}
\end{figure}
An effective method to find an inner approximation of a convex set is using the convex hull of samples obtained from the set, as shown in Fig.~\ref{fig:convex_hull}.
In this paper we use a simple Monte Carlo sampling method.
Note that other sampling methods, such as Importance Sampling or MCMC~\cite{andrieu2003introduction}, can also be used and might result in faster computations for large values of the input dimension $m$.
Given an $\mu$ and its trajectory $x^{(\mu)}$, we uniformly obtain $N$ samples from the input set $\set{u_i}_{i=1}^N \subset U$ to create a set of vector field evaluations $\set{f\pb{t, x^{(\mu)}(t), u_i}}_{i=1}^N$.
We build an approximated stochastic process $\mu_N$ from these samples using a set of weight functions $\set{w_i(t)}_{i=1}^N$ such that $\sum_{i=1}^N w_i(t) = 1$ and $w_i(t) \geq 0$ for each $i \in \set{1, \dotsc, N}$.
\begin{prop}
\label{prop:sampling_converge}
Let $\mu \in {\cal M}_p(\R^m)$ be a stochastic process with $\supp(\mu) \subset U$, and let $\set{u_i}_{i=1}^N$ be a set of samples uniformly drawn from $U$.
Then there exists $\set{w_{N,i}(t)}_{i=1}^N$ such that the empirical stochastic process $\mu_{N,t} = \sum_{i=1}^N w_{N,i}(t)\, \one_{u_i} \in {\cal M}_p(\R^m)$ satisfies:
\begin{equation}
\sum_{i = 1}^N w_{N,i}(t)\, f\pb{t, x^{(\mu_N)}(t), u_i}
\to
\int_{\mathrlap{\R^m}}\, f\pb{t, x^{(\mu)}(t), u}\, \diff{\mu_t}(u),
\end{equation}
as $N \to \infty$ for a.e.\ $t \in [0,T]$, where $\one_u$ is the Dirac delta function at $u \in U$.
\end{prop}
Consider the following discretized version of the optimality function $\theta_h$ in eq.~\eqref{eq:ham_optimal_mu}:
\begin{equation}
\label{eq:ham_optimal_muN}
\begin{aligned}
\theta_{N,h}\pb{&x^{(\mu_N)}, \mu_N}\!\\
&= \min_{\delta\mu_N\colon\! [0,T] \to {\cal M}(\R^m)}
\int_0^{\mathrlap{T}} H\pb{t, x^{(\mu_N)}(t), \delta\mu_{N,t}, p(t)} \diff{t},\\
&\text{subject to:}\quad
\supp(\delta \mu_{N,t}) \subset \supp(\mu_{N, t}),\\
&\phantom{\text{subject to:}\quad}
\delta\mu_{N,t} + \mu_{N,t} \geq 0,\\
&\phantom{\text{subject to:}\quad}
\int_{\mathrlap{\R^m}}\ \diff{\delta\mu_{N,t}}(u) = 0,\ \text{for a.e.}\ t \in [0,T],
\end{aligned}
\end{equation}
We can now establish the epi-convergence of $\theta_{N,h}$, which in turn allows us to produce consistent approximations of the optimal solution of the relaxed problem in eq.~\eqref{eq:original2}, as defined by Polak in~\cite[Sec.~3.3]{polak2012optimization}.
\begin{thm}
$\theta_{N,h}$ epi-converges to $\theta_h$ as $N \to \infty$.
\end{thm}
The theorem above follows by Prop.~\ref{prop:sampling_converge}, Thm.~\ref{thm:approx}, and~\cite[Thm.~3.3.2]{polak2012optimization}.
\subsection{Input Signal Synthesis}
\label{subsec:synthesis}
Let $\mu_N^* \in {\cal M}_p(\R^m)$ be an empirical stochastic process built from a fixed set of samples $\set{u_i}_{i=1}^N \subset U$, as explained in Sec.~\ref{subsec:sampling}.
We synthesize a deterministic input signal following the procedure detailed in~\cite{vasudevan2013a,vasudevan2013b}.
Intuitively, given a sampling period $\Delta > 0$, the synthesis procedure generates an input $u^* \in L^2([0,T],\R^m)$ such that, for each integer $k < \frac{T}{\Delta}$:
\begin{equation}
\int_{\mathrlap{k\Delta}}^{\mathrlap{(k+1)\Delta}} f\pb{t, x^{(u^*)}(t), u^*(t)}\, \diff{t} \approx
\int_{\mathrlap{\R^m}} f\pb{t, x^{(\mu_N^*)}(t), u}\, \diff{\mu_{N,k\Delta}^*}(u).
\end{equation}
Denoting $\mu_{N,t}^* = \sum_{i=1}^N w_i^*(t)\, \delta_{u_i}$, the procedure aims to find a new set $\set{\hat{w}_i}_{i=1}^N$ whose values are strictly binary, i.e., $\hat{w}_i(t) \in \set{0,1}$ for each $t$.
We achieve this objective by first applying a Haar wavelet filter to each $w_i$, thus reducing them to piecewise constant functions.
Then, a pulse-width modulation (PWM) transform is applied to each filtered $w_i$, resulting in binary pulses whose widths are proportional to the amplitude of each function at the samples induced by~$\Delta$.
The procedure is explained in detail in~\cite[Sec.~4.4]{vasudevan2013a}.
Moreover, as shown in~\cite[Thm.~5.10]{vasudevan2013a}, $x^{(u^*)}$ converges to $x^{(\mu_N^*)}$ with rate~$\Delta^{1/2}$.
\begin{figure}[tp]
\begin{algorithmic}[1]
\Require $\set{u_i}_{i=1}^N \subset U$, $\set{w_i}_{i=1}^N \subset L^2([0,T],[0,1])$ with $\sum_{i=1}^N w_i(t) = 1$, $\epsilon_{\text{tol}} > 0$, $\alpha, \beta \in (0, 1)$.
\State Let $\mu_{N,t} = \sum_{i=1}^N w_i(t)\, \one_{u_i}$.
\Loop
\State Compute $x^{(\mu_N)}$.
\State Compute $\theta_{N,h}\pb{x^{(\mu_N)}, \mu_N}$ and $\delta \mu_N$ as in~\eqref{eq:ham_optimal_muN}.
\If{$\theta_{N,h}\pb{x^{(\mu_N)}, \mu_N} > -\epsilon_{\text{tol}}$}
\State Go to line~\ref{alg:numerical_break}.
\EndIf
\State Compute $\lambda^*$ as in~\eqref{eq:arimijo_mu}.
\State $\mu_N \gets \mu_N + \lambda^*\, \delta \mu_N$.
\EndLoop
\State \label{alg:numerical_break}%
Construct $\hat{w}_i \in L^2\pb{[0,T],\set{0,1}}$ equal to the PWM transform of a filtered $w_i$ using Haar wavelets, as explained in~\cite[Sec.~4.4]{vasudevan2013a}.
\State \Return $u = \sum_{i=1}^N \hat{w}_i\, \one_{u_i}$.
\end{algorithmic}
\caption{Numerical algorithm to problem~\eqref{eq:original}.}
\label{alg:numerical}
\end{figure}
Fig.~\ref{alg:numerical} summarizes our implementable algorithm.
Note that, in practice, we add an $\ell_1$ regularization term to the objective function of $\theta_{N,h}$, which results in sparse updates to $\mu_{N,t}$ on each iteration.
The sparse updates significantly improve the computation speed of the algorithm.
Moreover, since the $\ell_1$ regularization is strictly convex, it is not hard to show that this modification also results in an optimality function for the problem in eq.~\eqref{eq:original2}.
\section{INTRODUCTION}
\label{sec:introduction}
Optimal control is a theoretical and practical framework that has been widely used to analyze the behavior of controlled dynamical systems~\cite{zhou1996robust, bertsekas1995dynamic}, and to synthesize actuation actions for dynamical systems in the face of safety, robustness, or uncertainty considerations~\cite{bryson1975applied, bellman2015applied}.
While some optimal control problems can be solved using purely analytical or algebraic tools~\cite{boyd1991linear}, modern computers and dynamical systems embedded in changing environments have led to a surge in numerical methods for optimal control~\cite{rao2009survey}.
Numerical methods for optimal control are typically divided into indirect methods, where the optimal solution is found as the solution of a set of equations typically derived from necessary optimality conditions, and direct methods, where the solution is found by iteratively minimizing the cost function at hand.
As shown by Polak~\cite[Sec.~4.2.6]{polak2012optimization} and Schwartz~\cite{schwartz1996theory}, direct numerical methods based on explicit time discretization converge to solutions satisfying derivative necessary optimality condition, which are strictly weaker than Minimum Principle condition.
Although some direct methods result in points satisfying the Minimum Principle, their implementation is usually impractical~\cite{gonzalez2010numerical}.
Some indirect methods also converge to inputs satisfying the Minimum Principle, usually relying on multiple-point boundary-value problems~\cite{betts2010practical}, which only converge for suitably chosen initial guesses~\cite{betts2003initialization}.
This paper presents a theoretical and numerical framework, which results in an algorithm that uses the direct method to converge to Pontryagin Minimum Principle inputs.
Our result is founded on the theory of relaxed control~\cite{warga2014optimal, warga1962relaxed}, which we use to derive a convergent sampling-based numerical method.
Once a relaxed control has been numerically computed, we use a projection operation originally devised for switched dynamical systems~\cite{vasudevan2013a, vasudevan2013b} to synthesize arbitrarily accurate approximations of the trajectories generated by the relaxed inputs.
Our result bridges a significant gap between the formulation of conceptual algorithms and implementable numerical algorithms which converge to Minimum Principle through direct method.
Moreover, our method achieves its goals in a numerically efficient and scalable way.
Our paper is organized as follows.
Sec.~\ref{sec:problem} introduces our notation and the optimal control problem we aim to solve.
A conceptual algorithm is presented in Sec.~\ref{sec:method}, and Sec.~\ref{sec:implementation} describes an implementable sampling-based numerical algorithm.
Finally, simulation results are shown in Sec.~\ref{sec:examples}.
Due to space limitations we omit all the formal proofs in this paper.
\section{OPTIMALITY CONDITIONS FOR OPTIMAL CONTROL}
\label{sec:method}
As shown by the Minimum Principle~\cite{pontryagin1987mathematical}, it is possible to find a necessary condition for optimal points that cannot be formulated using directional derivatives, in contrast of variational-based necessary conditions for optimal points that can be viewed as extensions of finite-dimensional first-order (or KKT) conditions~\cite{nocedal2006}.
Furthermore, necessary conditions for the problem in eq.~\eqref{eq:original} based on directional derivatives are strictly weaker than those based on the Minimum Principle.
Indeed, consider the following optimal control problem:
\begin{equation}
\label{eq:example}
\min\set{x(1) \mid
x(0) = 0,\
\dot{x}(t) = \sfrac{1}{2}\, \abs{u(t)} - \cos u(t)}.
\end{equation}
Note that the costate associated to $x$ is $p(t) = 1$ for each $t$, hence the Hamiltonian of this system is identical to its vector field, as defined in eqs.~\eqref{eq:hamiltonian0} and~\eqref{eq:adjoint0} respectively.
In this example, the Minimum Principle results in a single minimizer $u_g(t) = 0$, while first-order methods will converge to other local minimizers of the form $u_l(t) \in \set{2\, n\, \pi - \sfrac{\pi}{6}, \sfrac{\pi}{6} - 2\, n\, \pi \mid n = 1,2,\dotsc}$ depending on the initialization of the optimization algorithm.
In this section we present the theoretical foundation for our numerical method, including a conceptual infinite-dimensional optimization algorithm.
\subsection{Optimality Functions}
\label{sec:optfunc}
\begin{definition}[Sec.~1.2 in \cite{polak2012optimization}]
Consider an optimization problem defined on ${\cal X}$.
Then $\theta\colon {\cal X} \to (-\infty,0]$ is an optimality function iff $\theta(\hat{x}) = 0$ for each minimizer $\hat{x} \in {\cal X}$.
\end{definition}
Optimality functions are useful in practice since $\theta(x) < 0$ implies $x$ is \emph{not} a minimizer.
Hence, they can be used as numerical tests to check whether a minimizer has been reached.
Given an input $u_0$ and its trajectory $x^{(u_0)}$, consider:
\begin{equation}
\begin{aligned}
\theta_{o, l}\pb{x^{(u_0)}, u_0}
= &\min_{\substack{%
\delta x \in L^2([0,T],\R^n)\\
\delta u \in L^2([0,T],\R^m)
}}\,
\spderiv{\Psi}{x}\pb{x^{(u_0)}(T)}^\tp\, \delta x(T) \\
\text{subject to:} \quad
&\delta x (0) = 0, \\
&\delta \dot{x}(t) = \spderiv{f}{x}\pb{t, x^{(u_0)}(t), u_0(t)}\, \delta x(t)\\
&\phantom{\delta \dot{x}(t) =} + \spderiv{f}{u}\pb{t, x^{(u_0)}(t), u_0(t)}\, \delta u(t), \\
&u_0(t) + \delta u(t) \in U,\ \text{for a.e.}\ t \in [0,T],
\end{aligned}
\label{eq:thetaod}
\end{equation}
and:
\begin{equation}
\label{eq:thetaop}
\begin{aligned}
\theta_{o, h}\pb{x^{(u_0)}, u_0}
= &\hspace{-6pt}\min_{u \in L^2([0,T],\R^m)}
\int_{0}^{\mathrlap{T}}\!\! H_0\pb{t, x^{(u_0)}(t), u(t), p_0(t)}\\
&\hspace{3em} - H_0\pb{t, x^{(u_0)}(t), u_0(t), p_0(t)}\, \diff{t},\\
\text{subject to:} \quad
& u(t) \in U,\ \text{for a.\,e.}\ t \in [0,T],
\end{aligned}
\end{equation}
where $H_0$ is the Hamiltonian of problem~\eqref{eq:original} at time $t$:
\begin{equation}
\label{eq:hamiltonian0}
H_0\pb{t, x^{(u)}(t), u(t), p_0(t)}
= p_0(t)^\tp\, f\pb{t, x^{(u)}(t), u(t)},
\end{equation}
and $p_0(t)$ is the costate of problem~\eqref{eq:original}, defined by:
\begin{equation}
\label{eq:adjoint0}
\begin{aligned}
p_0(T) &= \spderiv{\Psi}{x}\pb{x^{\p{u_0}}(T)},\\
\dot{p}_0(t) &= - \spderiv{f}{x} \pb{t, x^{\p{u_0}} (t), u_0 (t)}^\tp\, p_0(t).
\end{aligned}
\end{equation}
We omit the proofs of the following propositions, but they follow closely the arguments in Thms.~5.6.8 and~5.6.9 in~\cite{polak2012optimization}, and Prop.~4.5 in~\cite{Polak1984}.
\begin{prop}
\label{prop:theta_original}
The functions $\theta_{o,l}$ and $\theta_{o,h}$, defined in eqs.~\eqref{eq:thetaod} and~\eqref{eq:thetaop}, are optimality functions of the problem~\eqref{eq:original}.
\end{prop}
\begin{prop}
\label{prop:thetao_equiv}
Let $u_0$ be an input and $x^{(u_0)}$ its trajectory.
If $\theta_{o, h}\pb{x^{(u_0)}, u_0}$ in eq.~\eqref{eq:thetaop} equals zero, then $\theta_{o, l} \p{x^{\p{u_0}}, u_{0}}$ in eq.~\eqref{eq:thetaod} also equals zero.
\end{prop}
Note that the opposite statement to Prop.~\ref{prop:thetao_equiv} is not true in general~\cite[Sec.~4.2.6]{polak2012optimization}.
This is a significant practical problem, since $\theta_{o,l}$, which captures first-order minimizers, is a convex problem, while $\theta_{o,h}$, which captures the stronger Pontryagin minimizers, is in general a non-convex problem.
Thus, direct numerical methods to compute the optimal control of the original formulation in~\eqref{eq:original} can only provably capture Pontryagin minimizers when the non-convex problem in eq.~\eqref{eq:thetaop} can be simplified or solved analytically.
Now, let us define an optimality function for the relaxed problem in equation~\eqref{eq:original2}.
Given a stochastic process $\mu_{0}$ and its corresponding trajectory $x^{\p{\mu_{0}}}$, the Hamiltonian of problem~\eqref{eq:original2} at time $t$ is:
\begin{equation}
\label{eq:hamiltonian}
H\pb{t, x^{(\mu)}(t), \mu_t, p(t)}
= p(t)^\tp \int_{\mathrlap{\R^m}}\, f\pb{t, x^{(\mu)}(t), u}\, \diff{\mu_t}(u),
\end{equation}
where $p(t)$ is the costate defined by:
\begin{equation}
\label{eq:adjoint}
\begin{aligned}
p(T) &= \spderiv{\Psi}{x}\pb{x^{(\mu)}}(T),\\
\dot{p}(t) &= - \int_{\R^m}\! \spderiv{f}{x}\pb{t, x^{(\mu)}(t), u}^\tp\, \diff{\mu_t}(u)\, p(t).
\end{aligned}
\end{equation}
Given a stochastic process $\mu_0$ and its trajectory $x^{(\mu_0)}$, consider:
\begin{equation}
\label{eq:lin_optimal_mu}
\begin{aligned}
\theta_l\pb{x^{(\mu_0)},\mu_0} =
&\min_{\substack{%
\delta x \in L^2([0,T],\R^n)\\
\delta \mu\colon [0,T] \to {\cal M}(\R^m)
}}\,
\spderiv{\Psi}{x}\pb{x^{(\mu_0)}(T)}^\tp\, \delta x(T)
\\
\text{subject to:}\quad
&\delta\dot{x}(t) =
\int_{\mathrlap{\R^m}}\ \, \spderiv{f}{x}\pb{t, x^{(\mu_0)}(t), u}\, \diff{\mu_{0, t}}(u)\, \delta x(t)\\
&\phantom{\dot{\delta x} \p{t} =}\
+ \int_{\mathrlap{\R^m}}\, f\pb{t, x^{\p{\mu_0}} \p{t}, u}\, \diff{\delta \mu_t}(u), \\
&\delta x \p{0} = 0, \\
&\supp(\delta \mu_{t} + \mu_{0, t}) \subset U,\\
&\delta\mu_t + \mu_{0,t} \geq 0,\\
&\int_{\mathrlap{\R^m}}\ \diff{\delta\mu_t}(u) = 0\ \text{for a.\,e.}\ t \in [0,T],
\end{aligned}
\end{equation}
and:
\begin{equation}
\label{eq:ham_optimal_mu}
\begin{aligned}
\theta_{h}\pb{x^{\p{\mu_0}}, \mu_{0}}\!
=&\hspace{-5pt}
\min_{\delta \mu\colon\! [0,T] \to {\cal M}(\R^m)}\!
\int_0^{\mathrlap{T}} \!\!H\pb{t, x^{(\mu_0)}(t), \delta\mu_t, p(t)} \diff{t},\\
\text{subject to:} \quad
& \supp(\delta \mu_{t} + \mu_{0, t}) \subset U,\\
& \delta\mu_t + \mu_{0,t} \geq 0,\\
& \int_{\mathrlap{\R^m}}\ \diff{\delta\mu_t}(u) = 0,\ \text{for a.e.}\ t \in [0,T],
\end{aligned}
\end{equation}
where $\delta \mu_t$ is a signed measure in $\R^m$.
\begin{prop}
\label{prop:optimality_hamiltonian}
The functions $\theta_l$ and $\theta_h$, defined in eqs.~\eqref{eq:lin_optimal_mu} and~\eqref{eq:ham_optimal_mu}, are optimality functions of the problem~\eqref{eq:original2}.
\end{prop}
Similar to the optimality functions for problem~\eqref{eq:original}, $\theta_l$ in eq.~\eqref{eq:lin_optimal_mu} captures first-order minimizers, while $\theta_h$ in eq.~\eqref{eq:ham_optimal_mu} captures Pontryagin minimizers.
In the next subsection we argue that $\theta_h$ is in fact equivalent to $\theta_l$, thus either can be used in practical and implementable numerical algorithms.
\subsection{Gradient Descent Method for Relaxed Problems}
\label{sec:graddesc}
Now we can show the connection between the directional and Pontryagin optimality functions.
\begin{thm}
\label{thm:eqv_optimal}
The optimality functions $\theta_l$ in eq.~\eqref{eq:lin_optimal_mu} and $\theta_h$ in eq.~\eqref{eq:ham_optimal_mu} are equivalent.
That is, given a pair $\pb{x^{(\mu)},\mu}$, both optimality functions produce the same value and minimizers.
\end{thm}
We omit a detailed proof.
Nonetheless, the argument follows using the costate in eq.~\eqref{eq:adjoint} to derive the Fr\'echet derivative of $\Psi$ as in~\cite[Thm.~5.6.9]{polak2012optimization}, and then rewrite $\theta_l$ using the costate.
A significant feature of optimality functions based on first-order derivatives is that their minimizing arguments are also descent directions for the objective function.
The following proposition shows that this property is preserved by $\theta_l$.
\begin{prop}
Let $\mu_0$ be a stochastic process and $x^{(\mu_0)}$ its corresponding trajectory.
Suppose that $\pb{\delta x^*, \delta \mu_0^*}$ is the minimizing argument of $\theta_l$ in~\eqref{eq:lin_optimal_mu}.
Then there exists a step size $\lambda \in (0, 1)$ such that:
\begin{equation}
\Psi\pb{x^{(\mu_0 + \lambda\, \delta \mu_0^*)}(T)} \leq \Psi\pb{x^{(\mu_0)}(T)}.
\end{equation}
\end{prop}
In practice, given $\alpha, \beta \in \p{0, 1}$, the step size $\lambda$ can be obtained using the following variation of the Armijo algorithm~\cite{Armijo1966minimization}:
\begin{multline}
\label{eq:arimijo_mu}
\lambda^* = \min\setb{\beta^k \mid
\Psi\pb{x^{(\mu_{0} + \beta^k\, \delta \mu_0^*)}(T)} - \Psi\pb{x^{(\mu_0)}(T)}\\
\leq \alpha\, \beta^k\, \theta_h\pb{x^{(\mu_0)}, \mu_0},\ k \in \N},
\end{multline}
\begin{figure}[tp]
\begin{algorithmic}[1]
\Require $\mu\colon [0,T] \to {\cal M}_p(\R^m)$, $\alpha, \beta \in (0, 1)$.
\Loop
\State Compute $x^{(\mu)}$.
\State Compute $\theta_h\pb{x^{(\mu)}, \mu}$ and $\delta \mu$ as in~\eqref{eq:ham_optimal_mu}.
\If{$\theta_h\pb{x^{(\mu)}, \mu} = 0$}
\State \Return $\mu$.
\EndIf
\State Compute $\lambda^*$ as in~\eqref{eq:arimijo_mu}.
\State $\mu \gets \mu + \lambda^*\, \delta \mu$.
\EndLoop
\end{algorithmic}
\caption{Conceptual optimization algorithm to solve problem~\eqref{eq:original2}.}
\label{alg:infinite}
\end{figure}
Fig.~\ref{alg:infinite} shows a conceptual algorithm to solve the relaxed optimal control problem in~\eqref{eq:original2}.
The theorem below shows its convergence, whose proof follows thanks to Thm.~\ref{thm:eqv_optimal} and the argument in the proof of~\cite[Thm.~5.12]{vasudevan2013a}.
\begin{thm}
\label{thm:convegence}
Let $\set{\mu_i}_{i \in \N}$ be a sequence of stochastic processes generated by the algorithm in Fig.~\ref{alg:infinite}, and let $\setn{x^{(\mu_i)}}_{i \in \N}$ be its corresponding sequence of trajectories.
Then $\lim_{i \to \infty} \theta_h\pb{x^{(\mu_i)}, \mu_i} = 0$.
\end{thm}
\section{PROBLEM DESCRIPTION}
\label{sec:problem}
We begin by introducing notation and preliminary results necessary to formulate our theoretical and numerical results in Sec.~\ref{sec:prelim}.
Then, in Sec.~\ref{sec:conceptual}, we formulate the conceptual optimal control problem we will address throughout this paper.
\subsection{Preliminaries}
\label{sec:prelim}
Given $n \in \N$ and $p \geq 1$, we denote the standard finite-dimensional $p$-norm by $\norm{\cdot}_p$, and the induced matrix $p$-norm by $\norm{\cdot}_{i,p}$.
We denote by ${\cal M}(\R^n)$ the set of Radon measures defined over the Borel sets of $\R^n$.
Given $\mu \in {\cal M}(\R^m)$ for $m \in \N$, we say that a function $f\colon \R^m \to \R^n$ is $L^2_\mu$-integrable, denoted $f \in L^2_\mu\p{\R^m, \R^n}$, if there exists $p \geq 1$ such that $\norm{f}_{\mu} = \p{\int_{\R^m} \norm{f(x)}_p^2\, \diff{\mu}(x)}^{\!1/2}$ is finite.
We abuse notation and denote by $L^2\p{\R^m,\R^n}$ the space of Lebesgue square-integrable functions.
Furthermore, we say that $\mu \in {\cal M}(\R^m)$ is a probability measure if $\mu(\R^m) = 1$.
We denote the set of all probability measures by ${\cal M}_p(\R^m)$.
A stochastic process is a function $\mu\colon [0,T] \to {\cal M}_p(\R^m)$, and throughout the paper we simply write $\mu_t$ instead of $\mu(t)$.
Let $\mu_1,\mu_2 \in {\cal M}(\R^m)$ be two Radon measures.
Then the difference $\nu = \mu_1 - \mu_2$ is a signed measure, and we define $L^2_{\nu}(\R^m, \R^n) = L^2_{\mu_1}\p{\R^m,\R^n} \cap L^2_{\mu_2}\p{\R^m,\R^n}$.
Given $f \in L^2_{\nu}\p{\R^m,\R^n}$, its integral with respect to $\nu$ is defined by
$\int_{\R^m} f(x)\, \diff{\nu}(x) = \int_{\R^m} f(x)\, \diff{\mu_1}(x) - \int_{\R^m} f(x)\, \diff{\mu_2}(x)$.
\subsection{Conceptual Optimal Control Problem}
\label{sec:conceptual}
We are interested in solving an input-constrained optimal control problem, formulated as follows:
\begin{equation}
\label{eq:original}
\begin{aligned}
\min_{\substack{%
x \in L^2([0,T],\R^n)\\
u \in L^2([0,T],\R^m)
}}\,
& \Psi\pb{x(T)},\\
\text{subject to:}\quad
& x(0) = \xi,\\
& \dot{x}(t) = f\pb{t, x(t), u(t)},\\
& u(t) \in U,\ \text{for a.\,e.}\ t \in [0,T],
\end{aligned}
\end{equation}
where $U$ is a connected and compact subset of $\R^m$, $\xi \in \R^n$, and the functions $f$ and $\Psi$ are well-defined, each with an appropriate domain and range.
We will say that $f$ is the vector field and $\Psi$ the final cost of the problem in eq.~\eqref{eq:original}.
Note that the optimal control problem in eq.~\eqref{eq:original} is quite general, since other standard formulations, such as those including running cost functions or minimum-time cost functions, can be easily converted to it (see~\cite[Sec.~4.1.2]{polak2012optimization}).
First we give the following assumption to guarantee the uniqueness of the trajectories, as well as the convergence of our numerical method, in this paper.
\begin{assumption}
\label{assump:lipschitz}
The functions $f$ and $\Psi$ are Lipschitz continuously differentiable.
Thus, there exists $L > 0$ such that, for each $t_1, t_2 \in [0,T]$, $x_1, x_2 \in \R^n$, and $u_1, u_2 \in U$:
\begin{align}
&\abs{\Psi(x_1) - \Psi(x_2)} \leq L\, \norm{x_1 - x_2}_2,\\
&\normb{\spderiv{\Psi}{x}(x_1) - \spderiv{\Psi}{x}(x_2)}_2 \leq L\, \norm{x_1 - x_2}_2,\\
&\norm{f(t_1, x_1, u_1) - f(t_2, x_2, u_2)}_2 \notag\\
&\hspace{.5in} \leq L\, \pb{\abs{t_1 - t_2} + \norm{x_1 - x_2}_2 + \norm{u_1 - u_2}_2},\\
&\normb{\spderiv{f}{x}(t_1, x_1, u_1) - \spderiv{f}{x}(t_2, x_2, u_2)}_{i,2} \notag\\
&\hspace{.5in} \leq L\, \pb{\abs{t_1 - t_2} + \norm{x_1 - x_2}_2 + \norm{u_1 - u_2}_2},\\
&\normb{\spderiv{f}{u}(t_1, x_1, u_1) - \spderiv{f}{u}(t_2, x_2, u_2)}_{i,2} \notag\\
&\hspace{.5in} \leq L\, \pb{\abs{t_1 - t_2} + \norm{x_1 - x_2}_2 + \norm{u_1 - u_2}_2}.
\end{align}
\end{assumption}
Now, we relax the problem in eq.~\eqref{eq:original} using the concept of relaxed inputs.
Consider the following relaxed optimal control problem:
\begin{equation}
\label{eq:original2}
\begin{aligned}
\min_{\substack{%
x \in L^2([0,T],\R^n)\\
\mu\colon [0,T] \to {\cal M}_p(\R^m)
}}\,
& \Psi\pb{x(T)},\\
\text{subject to:}\quad
& x(0) = \xi,\\
& \dot{x}(t) = \int_{\mathrlap{\R^m}}\ f\pb{t, x(t), u}\, \diff{\mu_t}(u),\\
& \supp(\mu_t) \subset U,\ \text{for a.\,e.}\ t \in [0,T],
\end{aligned}
\end{equation}
where $\supp(\mu_t)$ is the support of $\mu_t$, i.e., the smallest set $S$ such that $\mu_t(S) = 1$.
In other words, instead of optimizing over the space of $L^2$ functions, we optimize over the space of stochastic processes defined on $U$.
As shown in~\cite[Theorem II.6.5]{warga2014optimal}, if the vector field $f(t, x, u)$ satisfies Asm.~\ref{assump:lipschitz}, the relaxed problem in eq.~\eqref{eq:original2} always has a
solution.
Given a fixed initial condition $x(0) = \xi$, we denote the unique trajectory resulting from the stochastic process $\mu$ by $x^{(\mu)}$.
For simplicity, we also denote the unique trajectory resulting from an input $u$ by $x^{(u)}$.
Note that the problem in eq.~\eqref{eq:original} is a particular case of the problem in eq.~\eqref{eq:original2}.
Indeed, given an arbitrary input $\hat{u} \in L^2([0,T],\R^m)$, the stochastic process defined by $\mu_t(S) = 1$ whenever $\hat{u}(t) \in S$, and $\mu_t(S) = 0$ otherwise, produces exactly the same trajectory
as $\hat{u}$.
This implies that the feasible set of the relaxed problem is strictly larger than that of the original problem.
However, both original and relaxed problems result in the same optimal values.
Their equivalence follows since every point in the feasible set of the relaxed problem can be arbitrarily approximated using points in the feasible set of the original problem.
The following theorem is an extension of the Chattering Lemma~\cite[Thm.~4.1]{berkovitz2013optimal}.
\begin{thm}
\label{thm:approx}
Let $f$ be a vector field satisfying assumption~\ref{assump:lipschitz}, and let $\mu\colon [0,T] \to {\cal M}_p(\R^m)$ be a stochastic process.
Then, for each $\epsilon > 0$ there exists a control signal $\tilde{u}(t)$ such that for each $t \in [0,T]$,
$\normb{x^{(\mu)}(t) - x^{(\tilde{u})}(t)}_2 < \epsilon$.
\end{thm}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,707 |
layout: post
date: 2019-08-07 09:26:01
title: "Running CMSSW with Docker"
categories: [howto]
tags: [CMS, CMSSW, docker, linux]
---
Trying to install CMSSW in an OS that is not officially supported (e.g. Mac OS, Arch Linux, etc) can be very difficult. It's possible to run a virtual machine with [Scientific Linux](http://linux.web.cern.ch/linux/scientific.shtml) or [CentOS](http://linux.web.cern.ch/linux/centos.shtml) as the guest OS instead. But there is also a more lightweight solution, which is to use a [Docker](http://cms-sw.github.io/docker.html) container.
The official Docker images [provided by 'cmssw'](https://hub.docker.com/r/cmssw/cmssw) are huge (5-10 GB). Apparently there is a different Docker image that you can use, which is [provided by 'hepsw'](https://hub.docker.com/r/hepsw/cvmfs-cms), and is smaller (~200 MB). You can find the Dockerfiles used to build the images at <https://github.com/cms-sw/cms-docker> (for 'cmssw') and at <https://github.com/hepsw/docks> (for 'hepsw').
The following instructions should provide a functional CMSSW environment using the [hepsw/cvmfs-cms](https://github.com/hepsw/docks/tree/master/cvmfs-cms) image (tested on Linux Mint 18 as the host OS). Note that the image is based on slc-6.
``` shell
# Install Docker from the repository
sudo apt install docker.io
# Load the Docker image
# '--privileged' option is required because of FUSE
sudo systemctl start docker
sudo docker run -h dev --privileged -i -t hepsw/cvmfs-cms
```
Once the image is loaded, you should be in the interactive session in the guest OS. To setup CMSSW:
``` shell
yum install -y glibc glibc-headers glibc-devel.x86_64
export SCRAM_ARCH=slc6_amd64_gcc700
scramv1 project CMSSW CMSSW_10_4_0
cd CMSSW_10_4_0/src
eval `scramv1 runtime -sh`
scram b
```
To disable the annoying `fastestmirror` plugin for `yum`, open the file `/etc/yum/pluginconf.d/fastestmirror.conf`, and change `enabled=1` to `enabled=0`.
| {
"redpajama_set_name": "RedPajamaGithub"
} | 101 |
\section{Range-Aided SLAM Formulation}
In this section we model the RA-SLAM problem and derive its
corresponding MAP estimation problem.
\subsection{Inference Over a Graph}
We formulate RA-SLAM as inference over a directed graph, with nodes as variables
(i.e. robot poses and landmark positions) and the edges representing
measurements. Without loss of generality, we assume directionality in the edges.
The edge-set of all measurements of relative rigid transformations is denoted
$\TransformEdges$, where $\dedge \in \TransformEdges$ represents a measured
rigid transformation from node $i$ to node $j$. Similarly, the edge-set of all
range measurements is denoted $\RangeEdges$ and each $\dedge \in \RangeEdges$
represents a measured distance from node $i$ to node $j$.
\subsection{Measurement Noise Models}
\label{sec:noise-models}
For relative rigid transformation measurements we follow the formulation of
\cite{rosen19ijrr} in the noise model we assume over our
measurements. We denote the true translation and orientation of node $i$ as
$\ttran_i \in \dvec$ and $\trot_i \in \SOd$, with the true relative rotation from node $i$ to node
$j$ as $\trot_{ij} \triangleq \trot_{i}^\top \trot_{j}$. Similarly, we denote
the true relative translations as $\ttran_{ij} \triangleq \trot_{i}^\top (
\ttran_{j} - \ttran_{i} )$. We indicate noisy measurements with a tilde (e.g.,
$\nrot$):
\begin{equation}
\begin{aligned}
\ntran_{i j} & =\ttran_{i j}+\ptran_{i j}, & \ptran_{i j} & \sim \mathcal{N}\left(0, \tau_{i j}^{-1} I_{d}\right), \\
\nrot_{i j} & =\trot_{i j} \prot_{i j}, & \prot_{i j} & \sim \langevin \left(I_{d}, \kappa_{i j}\right).
\end{aligned}
\end{equation}
We assume the following generative Gaussian model for range measurements:
\begin{equation}
\label{eq:range-measurement-model}
\begin{aligned}
\ndist_{i j} & = \lVert \ttran_i - \ttran_j \rVert_2
+\pdist_{i j}, & \pdist_{i j} & \sim \mathcal{N}\left(0, \sigma_{i j}^2 \right). \\
\end{aligned}
\end{equation}
\subsection{MAP Estimation for RA-SLAM}
We write the MAP problem corresponding to the sensor models of
\cref{sec:noise-models}. The MAP problem is as follows, where $\tran_i \in
\dvec$ and $\rot_i \in \SOd$ denote the variables corresponding to the estimated
translation and rotation of node $i$. Bolded symbols indicate groupings (e.g.,
$\textbf{\tran} \triangleq \{\tran_i~\forall~i=1,\ldots,n \}$):
\begin{equation}
\label{eq:general-MAP}
\begin{array}{r}
\max\limits_{ \bf\tran , \bf\rot } ~ \prob ( \bf \ntran , \bf \nrot, \bf \ndist \mid \bf \tran , \bf \rot).
\end{array}
\end{equation}
This general MAP problem of \cref{eq:general-MAP} is equivalently solved by
minimizing its negative log-likelihood \cite{dellaert2017factor}. Given the
models of \cref{sec:noise-models}, the MAP estimate of the RA-SLAM
problem can be expressed as the following NLS problem:
\begin{equation}
\label{eq:ra-slam-nls}
\begin{aligned}
\min\limits_{\substack{ \tran_i \in \dvec \\ \rot_i \in \SOd}} ~~
& \sum\limits_{\dedge \in \TransformEdges} \kappa_{ij} \lVert \rot_j - \rot_i \nrot_{ij} \rVert_F^2 \\
+ & \sum\limits_{\dedge \in \TransformEdges} \tau_{ij} \left \lVert \tran_j - \tran_i - \rot_i \ntran_{ij} \right \rVert_2^2 \\
+ & \sum\limits_{\dedge \in \RangeEdges} \frac{1}{\sigma^2_{ij}} ( \lVert \tran_i - \tran_j \rVert_2 - \ndist_{ij})^2 ~ .
\end{aligned}
\end{equation}
where $\| \cdot \|_F$ denotes the matrix
Frobenius norm:
In \cref{eq:ra-slam-nls} the first two summands correspond to relative
transformation measurements whereas the third summand corresponds to
range measurements. Derivation of the cost terms in the first two summands can
be found in \cite{rosen19ijrr}. The cost terms in the third summand follow
immediately from the negative log-likelihood of the model in
\cref{eq:range-measurement-model}.
Notice that there are two distinct sources of non-convexity in \cref{eq:ra-slam-nls}. The rotation variables are members of the special orthogonal group,
i.e., $\rot_i \in \SOd$, which is a non-convex set. Additionally, the cost terms
due to range measurements are non-convex due to the $ \lVert \tran_i - \tran_j
\rVert_2 $ component.
\section{Conclusions}
We presented SCORE, a novel SOCP relaxation of the RA-SLAM problem. Through
simulated and real-world experiments we demonstrated SCORE outperforms
state-of-the-art initialization techniques and can obtain high-quality
initializations for RA-SLAM problems. The initializations provided by SCORE
resulted in up to two orders of magnitude reduction in average pose error when
compared to existing odometry-based initializations. Notably, we showed that
SCORE can determine good initializations for the challenging multi-robot RA-SLAM
problem when using only odometry and inter-robot range measurements. Along the
way, we derived a QCQP formulation of the RA-SLAM problem, which is of
independent scientific interest and connects RA-SLAM to a broader body of work.
Finally, we discussed empirically-derived hypotheses as to the conditions under
which SCORE can be expected to return a good initialization and drew connections
to existing theoretical analysis in SNL.
\section{Related Work}
The current state-of-the-art formulation of RA-SLAM is through nonlinear
least-squares (NLS) optimization based upon MAP inference
\cite{dellaert2017factor}. This work specifically focuses on
\textit{initialization} for this MAP formulation. Though we focus on MAP
estimation, there are many important previous formulations of RA-SLAM using:
extended Kalman filters \cite{Newman03icra, menegatti09icra,
Djugash09iros,djugash09springer}, particle filters \cite{gonzalez09ras,
blanco08icra}, mixture models \cite{blanco08iros}, and NLS optimization
\cite{boroson20iros, funabiki21ral, herranz14icra}.
We first survey previous initialization strategies for RA-SLAM problems and then
expand the scope to cover initialization strategies for general pose-graph SLAM
problems. As our initialization approach is a convex relaxation of the RA-SLAM
problem, we discuss related convex relaxations in robotic state-estimation.
Finally, as the key challenge that differentiates RA-SLAM from pose-graph SLAM
is the nature of range measurements, we discuss the closely related field of
sensor network localization (SNL) and convex relaxations developed specifically
for SNL problems.
\subsection{Initialization in RA-SLAM}
Notable work in the single-robot RA-SLAM case used spectral graph clustering to
initially estimate beacon locations \cite{Olson06joe}. Similarly,
\cite{boots13icml} used spectral decomposition to jointly estimate robot poses
and beacon locations. However, these approaches \cite{Olson06joe,boots13icml} do
not account for sensor noise models and do not readily extend to multi-robot
scenarios. Other initializations for multi-robot RA-SLAM used coordinated
movement patterns to find the (single) relative transform between robots
\cite{guo2017ijmav,li20arxiv}.
Finally, a series of works developed increasingly sophisticated methods to
compute relative transforms between two robots using range and odometry
measurements
\cite{zhou08tro,trawny07iros,trawny10rss,trawny10tro,jiang20itaes,li20iros,Li2022ral}.
While these approaches represent great progress, they are only capable of
solving for a single inter-robot relative transform and rely on (noisy)
odometric pose composition to initialize all other poses. In fact,
\cite{jiang20itaes,li20iros,Li2022ral} utilize convex relaxation to solve the
relative transformation problem.
\subsection{Initialization and Convex Relaxation in State-Estimation}
In pose-graph SLAM many existing initializations solve approximations of the MAP
problem. \cite{carlone15icra} leveraged the relationships between rotation and
translation measurements in SLAM and explored the use of several rotation
averaging algorithms
\cite{martinec07cvpr,govindu01cvpr,fredriksson13lnsc,tron14tac} to obtain
initial estimates. The same relationships were used to approximate the SLAM
problem as two sequential linear estimation problems \cite{carlone14ijrr}.
However, these approaches do not consider range measurements and thus do not
extend to RA-SLAM problems.
With exception to \cite{tron14tac}, these initialization procedures represent
convex relaxations of the pose-graph SLAM problem. Subsequent works in
pose-graph SLAM developed convex relaxations which obtained exact
solutions to the non-convex MAP problem \cite{rosen19ijrr,
carlone15iros,briales17ral,tron15rssworkshop, fan20tro
}. Other works developed exact convex relaxations for
the problems of extrinsic calibration \cite{giamou19ral}, two-view relative pose
estimation \cite{briales18cvpr,garcia-salguero21ivc}, and spline-based
trajectory estimation from range measurements with known beacon locations
\cite{pacholska20ral}. With exception to \cite{pacholska20ral}, these convex
relaxations also do not extend to range measurements. However,
\cite{pacholska20ral} required known beacon locations, allowed only for range
measurements, and only considered single-robot localization. SCORE allows for
multi-robot RA-SLAM, and combines measurements of rigid transformations and
ranges without necessitating any information \textit{a priori}.
A number of convex relaxations exist for sensor network localization (SNL). As
SNL centers around point-to-point distance measurements, these works are closely
related to SCORE. Previous works developed and analyzed semidefinite
program (SDP) \cite{biswas06tase,
so07mathematicalprogramming, shamsi13dgo} and SOCP \cite{tseng07siam,naddafzadeh-shirazi14twc,doherty01infocom,
srirangarajan08twc}
relaxations of the SNL problem. Whereas these relaxations only
consider range measurements and can only estimate the Euclidean position of
points, SCORE allows for rigid transformation measurements and estimation of
poses. Additionally, as SCORE is a SOCP, it is substantially more scalable than
these SDP relaxations.
\subsection{Novelty of Our Approach}
SCORE is similar to \cite{martinec07cvpr} in its relaxation of rotations and to
\cite{naddafzadeh-shirazi14twc} in its relaxation of range measurements.
However, it differs from these works in that it jointly considers both relative
pose and range measurements. Notably, SCORE generalizes the
chordal-initialization of \cite{martinec07cvpr} and the SOCP relaxation of
\cite{naddafzadeh-shirazi14twc}.
Additionally, there is much commonality to
\cite{jiang20itaes,li20iros,Li2022ral} in that the convex relaxation of SCORE
considers both range and relative transformation measurements. Unlike these
works, SCORE generalizes to arbitrary dimensions (e.g. 2- or 3-D) and
multi-robot cases. Furthermore, SCORE differs in that the convex relaxation is
derived from the full MAP problem, and thus utilizes all measurements and
jointly solves for an initialization of all RA-SLAM variables.
\section{Notation and Mathematical Preliminaries}
\textbf{Symbols:} This work generalizes across dimensionalities (e.g., 2-D or
3-D). As such the mathematical presentation will assume to be working in
$d$-dimensional space. We denote the $d$-dimensional identity matrix as $I_d$
and the special orthogonal group as $\SOd$. We refer to the set of real values
as $\R$ and the set of nonnegative reals as $\RnonNeg$. Furthermore, we indicate vectors and
scalars with lowercase symbols (e.g., $\tran$) and matrices with uppercase
symbols (e.g., $\rot$). Noisy measurements are indicated with a tilde (e.g.,
$\nrot$). The true (latent) value of a quantity will be underlined (e.g.,
$\trot$)
\textbf{Mathematical Operators:} We denote the matrix trace as $\tr (\cdot)$. A
diagonal matrix with diagonal elements $[a_1, \ldots, a_n]$ is indicated as
$\diag ([a_1, \ldots, a_n])$. The vector 2-norm is indicated as $\|
\cdot \|_2$ and the Frobenius norm is indicated as $\| \cdot \|_F$. We
note the following relationship between the Frobenius norm and trace operators:
$\| A \|_F^2 = \tr(A A^\top) = \tr(A^\top A)$.
\textbf{Probability Distributions:} We indicate a Gaussian probability
distribution over Euclidean space, with mean $\mu \in \dvec$ and variance
$\Sigma \in \dmat$ as $\gaussian (\mu, \Sigma)$. Similarly, we denote the
isotropic Langevin probability distribution\IsotropicLangevinFootnote over
matrix elements of the special orthogonal group, $\SOd$, with mode $\rot_\mu \in
\dmat$ and concentration parameter $\kappa \in \R$ as $\langevin (\rot_\mu,
\kappa)$ \cite{chiuso08infosys}.
This is the distribution whose probability density function is:
\begin{equation}
\label{eq:langevin-distribution}
\begin{array}{ll}
\prot \sim \text{Langevin}(\rot_\mu, \kappa) \\ \prob( \prot )
= \frac{1}{c(\kappa)} \exp\{\kappa \tr (\rot_\mu^\top
\prot)\}
\end{array}
\end{equation}
with normalizing constant $c(\kappa)$.
\section{SCORE: Second Order Conic Relaxation}
\label{sec:ra-slam-score}
In this section we reformulate the MAP problem of \cref{eq:ra-slam-nls} as a
QCQP. We then show how to construct SCORE by relaxing this QCQP to a SOCP.
\subsection{RA-SLAM as a QCQP}
\label{sec:ra-slam-qcqp}
We perform two changes to convert the non-convex NLS problem of
\cref{eq:ra-slam-nls} problem to the QCQP of \cref{eq:ra-slam-qcqp}. First, we
rewrite the $\SOd$ constraint as the orthonormality constraint and remove the
associated determinant constraint. Secondly, we introduce the auxiliary
variables, $\dist_{ij} \in \RnonNeg$ to rewrite the range-measurement cost
terms as quadratics. In this formulation the cost terms are quadratic and
convex with respect to the problem variables and all constraints are quadratic
equality constraints. Thus, \cref{eq:ra-slam-qcqp} is a QCQP.
Previous work \cite{carlone15iros,briales17ral,tron15rssworkshop} demonstrated
that $\det(\rot_i) = 1$ can be written as a set of quadratic equalities and that
the contribution of this determinant constraint is negligible, justifying its
removal. With exception to the determinant constraint, this QCQP is an exact
reformulation of \cref{eq:ra-slam-nls}:
\RaSlamQcqp
\subsection{Relaxing Equation (\ref{eq:ra-slam-qcqp}) to a SOCP}
\label{sec:ra-slam-socp}
We will now introduce the primary contribution of our paper, a \ScoreFull~(SCORE). As
previously mentioned, SCORE naturally arises as a relaxation of the QCQP in
\cref{eq:ra-slam-qcqp}. We emphasize that \cref{eq:ra-slam-qcqp} is not solved
in our approach, as it is not computationally practical to do so.
\cref{eq:ra-slam-qcqp} is important as an intermediate representation to derive
SCORE.
We relax the orthonormality constraints by removing them entirely, eliminating
that specific source of non-convexity. To eliminate the non-convexity from the
quadratic equality constraint on our auxiliary variables ($\dist_{ij}$) we first
recall that $\dist_{ij} \geq 0$, so we can drop the squares on each side of the
inequality without modifying the feasible set. From here, we can relax the
equality constraint to the convex inequality constraint $ \dist_{ij} \geq \lVert
\tran_i - \tran_j \rVert_2$. Importantly, this specific constraint relating the
auxiliary variables and position variables is a (convex) second-order cone
constraint \cite{boyd04book,alizadeh03mathematicalprogramming}. This relaxed
problem takes the following form, where $\CostFunction$ is the same cost
function as in \cref{eq:ra-slam-qcqp}.
\begin{equation}
\label{eq:ra-slam-socp}
\begin{aligned}
\min\limits_{\bf \tran, \bf \rot, \bf \dist} ~~ & \CostFunction \\
\st & \dist_{ij} \geq \lVert \tran_i - \tran_j \rVert_2,~\forall \dedge \in \RangeEdges \\
& \rot_0 = I_d
\end{aligned}
\end{equation}
With this previously mentioned second-order cone constraint on $\dist_{ij}$, it follows that
\cref{eq:ra-slam-socp} is a SOCP, i.e., it is the minimization of quadratic cost
functions over the intersection of an affine subspace with second-order
cones. This is a known form of SOCPs \cite{alizadeh03mathematicalprogramming},
and thus existing solvers (e.g., \cite{gurobi}) can efficiently solve
\cref{eq:ra-slam-socp} to global optimality. Similarly to \cite{martinec07cvpr},
the first rotation ($\rot_0$) is constrained to prevent a trivial solution of
all zeros.
\section{Introduction}
Range-aided simultaneous localization and mapping (RA-SLAM) is a key robotic
task, with applications in planetary \cite{boroson20iros}, subterranean
\cite{funabiki21ral}, and sub-sea \cite{Newman03icra,Bahr06iser,Bahr12iros}
environments. RA-SLAM combines range measurements (e.g., distances between
acoustic beacons) with measurements of relative rigid transformations (e.g.,
odometry) to estimate a set of robot poses and landmark positions.
RA-SLAM differs from pose-graph simultaneous localization and mapping (SLAM) in
that the sensing models of range measurements induce substantial difficulties
for state-estimation. However, ranging is a valuable sensor modality and
further developments in RA-SLAM could substantially advance the state of robot
navigation.
The state-of-the-art formulation of SLAM problems is as \textit{maximum a
posteriori} (MAP) inference, which takes the form of an optimization problem.
However, because robot orientations are a non-convex set, the MAP problem is
non-convex. Additionally, in RA-SLAM, the measurement models of range sensing
introduce further non-convexities. As a result, standard approaches to solving
these MAP problems use local-search techniques and can only guarantee locally
optimal solutions. Thus, good initializations to the MAP problem are key to
obtaining quality RA-SLAM solutions.
\TitleFigure
This work approaches initialization for the non-convex MAP estimation problem
through \textit{convex relaxation} \cite{boyd04book}, the construction of a
convex optimization problem which attempts to approximate the original problem.
Convex problems can be efficiently solved to global optimality, and many convex
relaxations have shown success as initialization strategies in related robotic
problems
\cite{carlone15icra,carlone15iros,giamou19ral,so07mathematicalprogramming}.
This work presents SCORE as a novel methodology for initializing RA-SLAM
problems. SCORE applies to 2D and 3D scenarios with arbitrary numbers of robots
and landmarks. As SCORE is a SOCP, it is easily implemented in existing
convex optimization libraries, and scales gracefully to large problems. We
summarize our contributions as follows:
\begin{itemize}
\item A novel, convex-relaxation approach to initializing RA-SLAM
problems.
\item The first QCQP formulation of RA-SLAM, connecting
RA-SLAM to a broader body of work
\cite{rosen19ijrr,carlone15iros,giamou19ral,so07mathematicalprogramming}.
\item Our implementation of SCORE is open-sourced\footnote{\url{\RepoURL}}.
\end{itemize}
\section{Using SCORE to Initialize Local-Optimization}
\label{sec:score-to-initialize}
Though SCORE can be solved to global optimality, in general the SCORE solution
will differ from that of the original problem due to the relaxed constraints.
This is not a serious limitation, as the intent is that the SCORE solution will
be close to the solution of the original problem. If true, the SCORE solution
can be projected to the feasible set of the original problem, and thereby serve
as a robust and principled initialization procedure.
Projecting the SCORE solution to the original problem's feasible set only
involves the estimated rotations. As SCORE relaxes the $\SOd$ constraints, the
estimated values, $\rot_i$, will generally not be valid rotation matrices. As in
\cite{martinec07cvpr}, we project the approximate rotation estimates to the
nearest rotation matrix via singular value decomposition, as follows:
\begin{align}
& \orot_i = \argmin_{\rot_i}~~\lVert \rot_i - \erot_{i,approx} \rVert_F^2 \label{eq:single-rotation-frob} \\
& U,S,V^\top = \SVD(\erot) \label{eq:SVD} \\
& \orot_i = U ~ \diag([1, 1, \det(U V^\top)]) ~ V^\top \label{eq:SOd-projection}
\end{align}
Projecting all approximate rotations to $\SOd$ produces a valid set of pose and
landmark estimates. We then use this projected solution to initialize the MAP
problem in \cref{eq:ra-slam-nls} and solve the problem with an iterative,
local-search based NLS solver (e.g., \cite{dellaert2012techreport}) to refine
the SCORE initial estimate.
\section{Experiments}
\label{sec:experiments}
In this section we present an experimental evaluation of SCORE's performance on
several RA-SLAM problems, consisting of (1) three real-world trials
involving an autonomous underwater vehicle (AUV) and acoustic beacons, (2)
simulated single-robot experiments with a single static beacon, and (3)
simulated multi-robot scenarios with inter-robot ranging. These experiments were
chosen to
respectively demonstrate (1)
SCORE's utility in real-world settings, (2) the challenge of RA-SLAM in even
seemingly simple cases (single-robot and single-beacon), and (3) SCORE's
performance in the challenging, but useful, scenario in which the relative
transformation between robots must be estimated from odometry and range
measurements.
\subsection{Experimental Evaluation}
\label{sec:experimental-eval}
We compare SCORE to other initialization techniques using the MAP estimate
obtained from the given initialization. Our SCORE implementation follows the
details of \cref{sec:score-to-initialize}, using Drake \cite{drake} and the
Gurobi convex solver \cite{gurobi}.
We emphasize that the initialization strategies compared against represent
either idealistic scenarios (ground truth values) or the existing
state-of-the-art (odometric approaches) for the problems presented in these
experiments. Whereas sophisticated initialization techniques exist for
pose-graph SLAM problems, RA-SLAM lacks such techniques.
Quantitative analysis was in comparison to ground truth values and was based on
absolute pose error (APE) \cite{grupp2017evo}.
\subsection{Real-World AUV Experiments}
The AUV data was collected using four acoustic beacons and a Bluefin Odyssey III
off the coast of Italy. As GPS was not available for most of the collection, we
obtained ground truth AUV positioning via MAP estimation, using GPS for the
initial pose and combining odometry and acoustic ranging with known beacon
locations to localize the remaining trajectory. Ground truth beacon positions
were obtained by surveyed acoustic trilateration prior to the experiment. We
obtained odometry measurements by estimating speed from Doppler velocity log
(DVL) measurements and combining the speed estimate with heading from the INS
magnetometer. We extracted range measurements via two-way time-of-flight ranging
with the acoustic beacons.
\SingleRobotGoatsTrajectoryFigure
\SingleRobotGoatAPEBarPlot
\subsubsection{Single-Robot Odometry Initialization (Odom)}
\label{sec:single-robot-odom-init}
The Odom initialization set the first pose to be the identity matrix and used
composition along the (noisy) odometry chain to obtain the remaining poses. The
static beacon initialization uses a random sample drawn from the environment, as
done in standard techniques \cite{Newman03icra,guo2017ijmav}.
\subsubsection{Evaluation of Single-Robot Experiments}
As seen in \cref{fig:single-robot-trajs}, the SCORE initialized solution closely
matched the ground truth trajectory. The APE results in
\cref{fig:single-trajectory-error} support this argument, showing consistently
lesser error for the SCORE initialized solution in comparison to the Odom
solution.
\subsection{Generation of Simulations}
\label{sec:generation-simulated-experiments}
We simulated RA-SLAM problems in which robots moved along a grid and
measurements used the noise models of \cref{sec:noise-models}. In all simulated
trials the initial position of any robots or beacons were randomly placed within
the simulated environment. At each time step odometry measurements were recorded
for each robot and each robot had a 50\% chance of generating a range
measurement. If generated, the remaining endpoint of the range measurement was
chosen by uniform sampling from the set of all other robots and beacons present.
\SingleRobotTrajectoryFigure
\SingleRobotAPEBoxPlot
To evaluate the effect of different sensor noise levels, we tested low- and
high-noise cases for both the single-robot and multi-robot scenarios. Each set
of noise parameters comes from values found in previous real-world experiments.
All odometry measurements were 1-meter movements, so a standard deviation of
0.01 meters in translation measurements corresponds to 1\% of the distance
traveled.
The low-noise setting uses $\sigma^{2}_{ij} = \lowSigmaSq$, $\tau^{-1}_{ij} =
\lowTauInv$, and $\kappa^{-1}_{ij} = \lowKappaInv$. This corresponds to standard
deviations of $\lowSigma$ meters, $\lowTauInvSqrt$ meters, and 0.002 radians
(0.41 degrees) for distance, translation, and rotation measurements,
respectively. The high-noise setting uses $\sigma^{2}_{ij} = \highSigmaSq$,
$\tau^{-1}_{ij} = \highTauInv$, and $\kappa^{-1}_{ij} = \highKappaInv$.
\subsection{Simulated Single-Robot Experiments}
\label{sec:single-robot-experiments}
In this section we evaluate the single-robot experiments, which considered
single robot and single static ranging beacon. For both the high and low noise
cases we generated 20 trials, with each trial consisting of 400 poses. For each
trial we compare the results of three different initialization strategies:
SCORE, simple odometry composition, and GT-Init (in which the ground-truth is
used as initialization).
\subsubsection{Evaluation of Single-Robot Experiments}
As seen in \cref{fig:single-robot-trajs}, the SCORE initialized solution
appeared to exactly match the GT-Init initialized solution. This is supported by
the quantitative results in \cref{fig:single-trajectory-error}, which suggest
that SCORE and GT-Init initialization effectively resulted in the same APE for
the trials run. In contrast to SCORE and GT-Init, the Odom initialization
obtained notably worse solutions (i.e. substantially higher APE and
qualitatively incorrect). In \cref{fig:single-trajectory-error} it can be seen
that the Odom initialization resulted in an increase in APE of up to two orders
of magnitude.
\subsection{Multi-Robot Experiments}
\label{sec:multi-robot-experiments}
\MultiRobotTrajectoryFourPanelFigure
To test a scenario in which initialization is difficult, we
simulated scenarios with four robots and no ranging beacons. For both the high
and low noise cases we generated 20 trials with 400 poses per robot (1600 poses
per trial). We compare the results of four different initialization strategies:
SCORE, ground truth (GT-Init), odometry with perfect first pose (Odom-P), and
odometry with random first pose (Odom-R). We describe these approaches in detail
in \cref{sec:multi-robot-init}.
\subsubsection{Multi-Robot Initialization}
\label{sec:multi-robot-init}
GT-Init is an idealistic case in which the ground truth state is known \textit{a
priori} and all variables are initialized to their true values. While RA-SLAM is
not needed if the true state is known, GT-Init is meant to serve as an upper
bound for reasonable quality of an initialization technique.
In comparison to the GT-Init initialization, the odometry based initializations
represent more realistic scenarios in which either the initial pose of each
robot is known (Odom-P) or no prior information exists (Odom-R). Odom-P and
Odom-R only differ in choosing the starting pose for each robot. After
determining the starting pose for a given robot both Odom-P and Odom-R compose
odometry measurements to initialize the remaining poses for all robots.
In Odom-P we assume a perfectly known starting pose of each robot and compose
the odometry chain to initialize the other pose variables. Such a situation may
occur in marine robotics, where AUVs begin with GPS before diving underwater.
In contrast to Odom-P, for Odom-R we assume no prior information and choose the
starting pose for each robot randomly. We note that Odom-R is a standard
approach to initialization for multi-robot RA-SLAM \cite{guo2017ijmav,li20arxiv}
and is arguably the fairest comparison for SCORE as both require no prior
information.
\subsubsection{Evaluation of Multi-Robot Experiments}
As seen in the example multi-robot trial displayed in
\cref{fig:multi-robot-trajs}, SCORE provides reliable estimates of the
trajectories of the four robots. In this trial the solution obtained from the
SCORE initialization qualitatively matches the true trajectories. In addition,
the refined SCORE initialization appears to match the trajectories estimated
using the GT-Init initialization and provide a substantially better estimate
than that provided by the Odom-P and Odom-R initializations.
These qualitative observations from \cref{fig:multi-robot-trajs} are supported
by the quantitative analysis of the APE in \cref{fig:multi-trajectory-error}. We
observe that for the high-noise multi-robot experiments the SCORE initialization
strategy resulted in similar APE to the GT-Init initialization and had notably
reduced APE in comparison to Odom-P and Odom-R. For the low-noise multi-robot
experiments SCORE resulted in higher APE than the GT-Init initialization
approach but lesser APE than either of the odometric initialization strategies.
\section{Discussion}
In summary, the results shown in
\cref{sec:experiments} demonstrated
that SCORE works in real-world settings and outperforms state-of-the-art
initialization techniques for a wide range of RA-SLAM problems. We emphasize
the difficulty of the multi-robot problems presented and that SCORE is capable
of providing quality initializations for multi-robot RA-SLAM problems with only
odometry and inter-robot range measurements.
\MultiRobotAPEBoxPlot
Furthermore, in evaluating SCORE's performance on these multi-robot experiments
we emphasize that Odom-R is the only other initialization technique which
assumes the same amount of prior information as SCORE, i.e., no information
beyond the robot's measurements. In comparing SCORE to Odom-R we observe that
the SCORE initialization typically obtains an average APE roughly two orders of
magnitude less than that obtained by Odom-R initialization. In fact, SCORE
appears to obtain more accurate trajectory estimates than those generated by
Odom-P, which assumes knowledge of each robot's first pose. These
observations suggest that SCORE is a more effective initialization strategy than
general odometry-based techniques (the current state-of-the-art initialization
for these problems) even when each robot's first pose is
available.
We note that several instances demonstrated substantial improvement in the
estimate by refining the SCORE initialization. In particular, we observed the
translation estimates were compressed (i.e., the estimated distance traveled was
notably less than the true value of 1 meter). Despite this compression in the
initialized translations, the final MAP estimate appeared high quality. We
hypothesize that good solutions are obtained from these compressed translations
because the SCORE initialization returns good initial estimates of the robot
orientations and beacon locations. This would agree with known results in
pose-graph SLAM \cite{carlone15icra} which, show that the quality of MAP
solutions depends more on good orientation initializations than good
translation initializations.
\subsection{When Does SCORE Return Poor Initializations?}
\label{sec:limitations}
\ScoreDeterminantPlot
Though SCORE generally performed well as an initialization technique, we
observed several multi-robot RA-SLAM instances where SCORE produced poor
initializations. These poor initializations were due to the cost function
tending towards a trivial solution of all zeros (i.e., $\rot_i = 0_{d \times d},
\tran_i = \dist_i = 0_{d}$). While the constraint $\rot_0 = I_{d}$ should
prevent this, in practice certain scenarios demonstrate the
remaining variables rapidly decaying to zero. This phenomenon is unsurprising,
as it is found in similar SLAM relaxations \cite{Rosen15icra,carlone15icra} and
the zeros elements indeed minimizes the problem cost. Importantly, we
note that these failures can be readily detected by observing the determinants
of the estimated rotations, which will quickly decay to less than
$\expnumber{5}{-2}$ (see \cref{fig:rotation-determinants}).
As a result, failures can be identified and alternative initializations used in
this case.
We find that the experiments with failed (poor) initializations resemble the
conditions identified by \cite{tseng07siam} for the SNL SOCP derived. Namely,
the SOCP relaxation returns good solutions when the robot trajectories had
well-constrained poses (via range measurements) within the convex hull of robot
poses connected via odometry to $\rot_0$ (i.e., the trajectory of the robot with
constrained first rotation). Though at present we lack formal analysis,
empirical results suggest the analysis and connections to rigidity theory in
\cite{tseng07siam} are closely linked to SCORE.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,378 |
{"url":"https:\/\/hal.inria.fr\/hal-01215054","text":"HAL will be down for maintenance from Friday, June 10 at 4pm through Monday, June 13 at 9am. More information\n\nA reciprocity approach to computing generating functions for permutations with no pattern matches\n\nAbstract : In this paper, we develop a new method to compute generating functions of the form $NM_\u03c4 (t,x,y) = \\sum\\limits_{n \u22650} {\\frac{t^n} {n!}}\u2211_{\u03c3 \u2208\\mathcal{lNM_{n}(\u03c4 )}} x^{LRMin(\u03c3)} y^{1+des(\u03c3 )}$ where $\u03c4$ is a permutation that starts with $1, \\mathcal{NM_n}(\u03c4 )$ is the set of permutations in the symmetric group $S_n$ with no $\u03c4$ -matches, and for any permutation $\u03c3 \u2208S_n$, $LRMin(\u03c3 )$ is the number of left-to-right minima of $\u03c3$ and $des(\u03c3 )$ is the number of descents of $\u03c3$ . Our method does not compute $NM_\u03c4 (t,x,y)$ directly, but assumes that $NM_\u03c4 (t,x,y) = \\frac{1}{\/ (U_\u03c4 (t,y))^x}$ where $U_\u03c4 (t,y) = \\sum_{n \u22650} U_\u03c4 ,n(y) \\frac{t^n}{ n!}$ so that $U_\u03c4 (t,y) = \\frac{1}{ NM_\u03c4 (t,1,y)}$. We then use the so-called homomorphism method and the combinatorial interpretation of $NM_\u03c4 (t,1,y)$ to develop recursions for the coefficient of $U_\u03c4 (t,y)$.\nKeywords :\nDocument type :\nConference papers\nDomain :\n\nCited literature [12 references]\n\nhttps:\/\/hal.inria.fr\/hal-01215054\nContributor : Coordination Episciences Iam Connect in order to contact the contributor\nSubmitted on : Tuesday, October 13, 2015 - 3:05:46 PM\nLast modification on : Wednesday, August 7, 2019 - 12:18:06 PM\nLong-term archiving on: : Thursday, April 27, 2017 - 12:22:15 AM\n\nFile\n\ndmAO0149.pdf\nPublisher files allowed on an open archive\n\nCitation\n\nMiles Eli Jones, Jeffrey Remmel. A reciprocity approach to computing generating functions for permutations with no pattern matches. 23rd International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2011), 2011, Reykjavik, Iceland. pp.551-562, \u27e810.46298\/dmtcs.2933\u27e9. \u27e8hal-01215054\u27e9\n\nRecord views","date":"2022-05-26 16:35:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5338270664215088, \"perplexity\": 1106.0389418286588}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-21\/segments\/1652662619221.81\/warc\/CC-MAIN-20220526162749-20220526192749-00067.warc.gz\"}"} | null | null |
'great help' Nathan Carter reveals he went to therapist after listening to Bressie's podcast
"That was a big thing, for me to go to a therapist... It was a great help"
Nathan Carter Pic: Fran Veale
Eddie Rowley
December 27 2021 08:30 AM
COUNTRY superstar Nathan Carter revealed how Niall 'Bressie' Breslin personally put him in touch with a therapist as he struggled to cope with his career stalling due to the Covid shutdowns over the last two years.
Carter (31), who first shot to fame with his Wagon Wheel smash hit song, eventually found a lifeline for his anxiety and depression while listening to Bressie discussing mental health issues with Imelda May on his podcast, Where Is My Mind.
"You don't realise how much you miss something until it's taken away," Nathan tells the Sunday World.
"I had often said that I'd love to take three months off, or six months off, but I'll never be saying that again."
The lack of work and uncertainty about the future eventually took its toll on his mental health.
"I was listening to Niall Breslin's podcast and Imelda May was on it, and there were different therapists talking about the issues.
Niall 'Bressie' Breslin
"It sounded so normal, the way they talked about it. So I thought, 'what have I got to lose by trying this?' I contacted Bressie and he gave me the number of the girl that he works with, who is a really qualified therapist in lots of different therapies.
"It was a great help. They don't solve everything, obviously, but it's just a matter of being able to talk, because sometimes you don't want to talk to your family or your closest friends about any of that stuff… you don't want to burden them.
"I would definitely still urge anybody that's feeling down or feeling not in the right place to go and chat to someone.
Nathan Carter dons white suit specially tailored for Joe Dolan during new RTE series
Nathan Carter opens up about turning to therapy to cope with mental health struggles
"I found the lady I went to said hardly anything, and I just literally sat chatting for an hour. You don't realise how much you have to get off your chest and that you need to talk to someone.
"It's having someone to sit there and listen to your problems. It's not about them, it's about you.
Nathan Carter
"That was a big thing, for me to go to a therapist and to tell people that as well because I have been open about it.
"If me speaking out can help anyone else, then that's great. There was a lot of stigma attached to the idea of going to speak to a therapist, particularly for a man because I think we just like to bottle everything up.
"Not that we're all the same, but men in general are not very good at expressing their feelings."
Did he feel the benefits of it? "I did, I could see the bits where I was going wrong and what I should deal with," Nathan says.
"Stress is the thing, and I could see where the stress is coming from. I couldn't pinpoint it before. I didn't know why I was feeling a bit sh*t some days.
"It was the whole thing of not being able to play music.
"Music is a therapy at the end of the day. When that was taken away from me and I couldn't do it, it had a huge impact on me.
"Then the minute we started gigging again my whole ethos just changed, and I was feeling great again. It's crazy.
"I watched I'm A Celebrity and the footballer David Ginola was saying how a lot of footballers deal badly with it after they finish playing in their mid-30s.
"They go from a massive high to stopping overnight when they retire, and that's the way it was for musicians in the lockdowns.
"You miss the buzz you get from seeing people enjoying what you do, whether it's playing in a pub with 50 people dancing and enjoying themselves, or playing in an arena, it doesn't matter, it's the same thing.
"So it's been a tough couple of years for musicians, and everybody else out there, to be honest."
Nathan Carter singing with the Deaf Choir in Croke Park for the Pope
Nathan tells how many musicians and technical crew have either left the business or now need to double job in order to make a living as a result of the devastation of the live entertainment business.
"I was so lucky that out of all of my crew I only lost one guy, but most of them are still double jobbing to make ends meet and it's the same across the industry, which is very sad," he reveals.
"These are seriously talented people who have gone to college and studied for years to do what they're really good at doing, and now they can't actually do that as their profession full-time any more.
"My new year's wish is that that won't be the case and professional people within our industry can go and do what they do best, and do what they have worked so hard at all their lives to create their own careers."
Un-canny
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The jury deliberating the fate of Ghislaine Maxwell at her sex trafficking trial requested a white board and different coloured sticky notes on Monday as it signalled that it had plenty of work to do after the holiday weekend.
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Robbie and Claudine Keane enjoy dinner at lavish 7-star Dubai hotel
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Rosanna Davison a 'proud mama' as twins Hugo and Oscar take first steps
Ciara Kelly reacts to news Leo Varadkar 'wants to end Covid restrictions' on March 31st | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,831 |
\section{General Setup: Expansion in Graviton Exchanges}
\label{general}
AdS/CFT correspondence conjectures that the dynamics of ${\cal N} =4$
$SU(N_c)$ SYM theory in four space-time dimensions is dual to the type
IIB superstring theory on AdS$_5 \times$S$_5$ \cite{Maldacena:1997re}.
In the limit of large number of colors $N_c$ and large 't Hooft
coupling $\lambda = g^2 N_c$ (with $g$ the gauge coupling constant)
such that $N_c \gg \lambda \gg 1$, AdS/CFT correspondence reduced to
the gauge-gravity duality: ${\cal N} =4$ SU($N_c$) SYM theory at $N_c
\gg \lambda \gg 1$ is dual to (weakly coupled) classical supergravity
in AdS$_5$. Hence the gauge theory dynamics at strong coupling, which
includes all-orders quantum effects, is equivalent to the classical
dynamics of supergravity. Instead of summing infinite classes of
Feynman diagrams in the gauge theory or using other non-perturbative
methods, one can simply study classical supergravity in 5 dimensions.
For a review of AdS/CFT correspondence see \cite{Aharony:1999ti}.
Our goal is to describe the isotropization (and thermalization) of the
medium created in heavy ion collisions assuming that the medium is
strongly coupled and using AdS/CFT correspondence to study its
dynamics. We want to construct a metric in AdS$_5$ which is dual to an
ultrarelativistic heavy ion collision as pictured in \fig{spacetime}.
Throughout the discussion we will use Bjorken approximation of the
nuclei having an infinite transverse extent and being homogeneous (on
the average) in the transverse direction, such that nothing in our
problem would depend on the transverse coordinates $x^1$, $x^2$
\cite{Bjorken:1982qr}.
\begin{figure}
\begin{center}
\includegraphics[width=5.cm]{spacetime}
\end{center}
\caption{The space-time picture of the ultrarelativistic heavy ion
collision in the center-of-mass frame. The collision axis is
labeled $x^3$, the time is $x^0$.}
\label{spacetime}
\end{figure}
We start with a metric for a single shock wave moving along a light
cone \cite{Janik:2005zt}:
\begin{equation}\label{1nuc}
ds^2 \, = \, \frac{L^2}{z^2} \, \left\{ -2 \, dx^+ \, dx^- + \frac{2
\, \pi^2}{N_c^2} \, \langle T_{--} (x^-) \rangle \, z^4 \, d
x^{- \, 2} + d x_\perp^2 + d z^2 \right\}.
\end{equation}
Here $x^\pm = \frac{x^0 \pm x^3}{\sqrt{2}}$, $z$ is the coordinate
describing the 5th dimension such that the boundary of the AdS space
is at $z=0$, and $L$ is the radius of $S_5$. According to holographic
renormalization \cite{deHaro:2000xn}, $\langle T_{--} (x^-) \rangle$
is the expectation value of the energy-momentum tensor for a single
ultrarelativistic nucleus moving along the light-cone in the
$x^+$-direction in the gauge theory. We assume that the nucleus is
made out of nucleons consisting of $N_c^2$ ``valence gluons'' each,
such that $\langle T_{--} (x^-) \rangle \propto N_c^2$, and the metric
\peq{1nuc} has no $N_c^2$-suppressed terms in it.
The metric in \eq{1nuc} is an exact solution of Einstein equations in
AdS$_5$:
\begin{align}
\label{ein}
R_{\mu\nu} + \frac{4}{L^2} \, g_{\mu\nu} = 0.
\end{align}
It can also be represented perturbatively as a single graviton
exchange between the source nucleus at the AdS boundary and the
location in the bulk where we measure the metric/graviton field. This
is shown in \fig{1gr}, where the solid line represents the nucleus and
the wavy line is the graviton propagator. Incidentally a single
graviton exchange, while being a first-order perturbation of the empty
AdS space, is also an exact solution of Einstein equations. This means
higher order tree-level graviton diagrams are zero (cf. classical
gluon field of a single nucleus in covariant gauge in the Color Glass
Condensate (CGC) formalism \cite{Kovchegov:1997pc}).
\begin{figure}[h]
\begin{center}
\includegraphics[width=3cm]{1graviton}
\end{center}
\caption{A representation of the metric (\protect\ref{1nuc}) as a graviton
(wavy line) exchange between the nucleus at the boundary of AdS
space (the solid line) and the point in the bulk where the metric
is measured (denoted by a cross). }
\label{1gr}
\end{figure}
Now let us try to find the geometry dual to a collision of two shock
waves with the metrics like that in \eq{1nuc}. Defining $ t_1 (x^-)
\, \equiv \, \frac{2 \, \pi^2}{N_c^2} \, \langle T_{1 \, --} (x^-)
\rangle$ and $t_2 (x^+) \, \equiv \, \frac{2 \, \pi^2}{N_c^2} \,
\langle T_{2 \, ++} (x^+) \rangle$ for the energy-momentum tensors of
the two shock waves in the boundary theory, we write the metric
resulting from such a collision as
\begin{eqnarray}\label{2nuc1}
ds^2 \, = \, \frac{L^2}{z^2} \, \bigg\{ -2 \, dx^+ \, dx^- + d
x_\perp^2 + d z^2 + t_1 (x^-) \, z^4 \, d x^{- \, 2}
+ t_2 (x^+) \, z^4 \, d
x^{+ \, 2} \nonumber \\ + \, \mbox{higher order graviton exchanges}
\bigg\}
\end{eqnarray}
The metric of \eq{2nuc1} is illustrated in \fig{pert}. The first two
terms in \fig{pert} (diagrams A and B) correspond to one-graviton
exchanges which constitute the individual metrics of each of the
nuclei, as shown in \eq{1nuc}. Our goal below is to calculate
higher-order corrections to these terms, which are illustrated by the
diagram C in \fig{pert} and by the ellipsis following it. \fig{pert}
illustrates that construction of dual geometry to a shock wave
collision in AdS$_5$ consists of summing up all tree-level graviton
exchange diagrams, similar diagrammatically to the classical gluon
field formed by heavy ion collisions in CGC
\cite{Kovner:1995ts,Kovchegov:2000hz}. While classical gluon fields
lead to free-streaming final state \cite{Krasnitz:2002mn}, as we will
argue below, their AdS graviton ``dual'' will lead to an ideal
hydrodynamic final state for the gauge theory similar to the one found
in \cite{Janik:2005zt}, though at the same time different from
\cite{Janik:2005zt} due to non-trivial rapidity dependence in the case
at hand.
\begin{figure}
\begin{center}
\includegraphics[width=12cm]{pert}
\end{center}
\caption{Diagrammatic representation of the metric in \protect\eq{2nuc1}.
Wavy lines are graviton propagators between the boundary of the
AdS space and the bulk. Graphs A and B correspond to the metrics
of the first and the second nucleus correspondingly. Diagram C is
an example of the higher order graviton exchange corrections.}
\label{pert}
\end{figure}
\section{Perturbative Solution of Einstein Equations}
\label{gensol}
We begin by parametrizing the metric as
\begin{align}\label{pA_gen}
ds^2 \, = \, \frac{L^2}{z^2} \, \bigg\{ -\left[ 2 + g (x^+, x^-, z)
\right] \, dx^+ \, dx^- + \left[ t_1 (x^-) \, z^4 + f (x^+, x^-, z)
\right] \, d x^{- \, 2} \notag \\ + \left[ t_2 (x^+) \, z^4 +
{\tilde f} (x^+, x^-, z) \right] \, d x^{+ \, 2} + \left[ 1 + h
(x^+, x^-, z) \right] \, d x_\perp^2 + d z^2 \bigg\},
\end{align}
where the functions $f$, $\tilde f$, $g$, and $h$ are zero before the
collision, i.e., for either $x^+ < 0$ and/or $x^- <0$. The exact
Einstein equations \peq{ein} for $f$, $\tilde f$, $g$, and $h$ are
rather complicated and are not going to be presented here. Instead we
are going to solve Einstein equations perturbatively.
To be more specific let us consider in the boundary theory a collision
of two ultrarelativistic nuclei with large light-cone momenta per
nucleon $p_1^+$, $p_2^-$, and atomic numbers $A_1$ and $A_2$. Writing
\begin{align}\label{deltas}
t_1 (x^-) = \mu_1 \, \delta (x^-), \ \ \ t_2 (x^+) = \mu_2 \, \delta
(x^+)
\end{align}
we want to expand Einstein equations in the powers of $\mu_1$ and
$\mu_2$ \cite{Grumiller:2008va,Albacete:2008vs}. The two scales
$\mu_1$ and $\mu_2$ can be expressed terms of physical parameters in
the problem \cite{Albacete:2008vs,Albacete:2008ze}
\begin{align}\label{mus}
\mu_{1} \sim p_{1}^+ \, \Lambda_1^2 \, A_1^{1/3}, \ \ \ \mu_{2} \sim
p_{2}^- \, \Lambda_2^2 \, A_2^{1/3}.
\end{align}
$\Lambda_1$ and $\Lambda_2$ are the typical transverse momentum scales
describing the two nuclei \cite{Albacete:2008vs}, similar to the
saturation scales. Note that $\mu_1$ and $\mu_2$ are independent of
$N_c$. As follows from a simple dimensional analysis, combined with
Lorentz-transformation properties of the relevant quantities, the
expansion parameters in four dimensions would be
\begin{align}
\label{params}
\mu_1 \, ( x^- )^2 \, x^+, \ \ \ \mbox{and} \ \ \ \mu_2 \, ( x^+ )^2
\, x^-,
\end{align}
such that the expansion is valid only at early times, when these
parameters are small.
Linearizing Einstein equations \peq{ein} in $f$, $\tilde f$, $g$, and
$h$ we solve the obtained system of differential equation to obtain
\cite{Albacete:2008vs}
\begin{equation}\label{hsol}
h (x^+, x^-, z) \, = \, h_0 (x^+, x^-) \, z^4 + h_1 (x^+, x^-) \,
z^6
\end{equation}
where $h_0$ and $h_1$ are determined by the causal solutions of the
following equations
\begin{equation}\label{h0eq}
(\partial_+ \, \partial_-)^2 \, h_{0} (x^+, x^-) \, = \, 8 \, t_1
(x^-) \, t_2 (x^+),
\end{equation}
\begin{equation}\label{h1eq2}
\partial_+ \, \partial_- \, h_1 (x^+, x^-) \, = \, \frac{4}{3} \,
t_1 (x^-) \, t_2 (x^+).
\end{equation}
$f$, $\tilde f$, and $g$ are easily expressed in terms of $h (x^+,
x^-, z)$ from \eq{hsol} (see \cite{Albacete:2008vs}).
This lowest-order perturbative solution leads to the energy density of
the produced medium (in the center-of-mass frame)
\cite{Grumiller:2008va,Albacete:2008vs}
\begin{align}
\label{energy}
\epsilon (\tau) \, = \, \frac{N_c^2}{\pi^2} \, \mu_1 \, \mu_2 \,
\tau^2
\end{align}
where $\tau = \sqrt{2 \, x^+ \, x^-}$. The corresponding
center-of-mass energy-momentum tensor is
\begin{equation}\label{emt_ads}
\langle T^{\mu\nu} \rangle \, = \,
\left( \begin{array}{cccc} \epsilon (\tau) & 0 & 0 & 0 \\
0 & 2 \, \epsilon (\tau) & 0 & 0 \\
0 & 0 & 2 \, \epsilon (\tau) & 0 \\
0 & 0 & 0 & - 3 \, \epsilon (\tau) \end{array} \right)
\end{equation}
in terms of the $x^0, x^1, x^2, x^3$ components. One can see that the
longitudinal pressure component in \eq{emt_ads} is large and negative.
Under boosts the $T^{00}$ and $T^{33}$ components of the
energy-momentum tensor mix. This implies that there is a frame in which
the energy density is negative $T'_{00} < 0$. At this point it is not
clear whether this result presents a problem, as there may be nothing
wrong with energy density becoming negative for a short period of time
\cite{Grumiller:2008va}. As we will see below, further evolution of
the energy density with time leads to disappearance of this negative
energy-density problem.
Higher-order perturbative corrections to the energy-momentum tensor
\peq{emt_ads} in powers of $\mu_1$ and $\mu_2$ have been found in
\cite{Grumiller:2008va,Albacete:2008vs} up to the fourth order in
$\mu_i$ (i.e., up to $O(\mu_1^3 \mu_2), O(\mu_1^2 \mu_2^2),O(\mu_1
\mu_2^3)$).
\section{All-Order Resummation in $\mu_2$}
The exact solution of Einstein equations for the collision of two
shock waves involves resummation of both parameters in \eq{params} to
all orders: such calculation appears to be very hard to do. Instead
one can resum all orders of $\mu_2 \, ( x^+ )^2 \, x^-$ while keeping
at the lowest order in $\mu_1 \, ( x^- )^2 \, x^+$. The corresponding
diagram is shown in \fig{pAfig}: in it one resums multiple graviton
exchanges with one shock wave ($t_2$), while exchanging only one
graviton with the other shock wave ($t_1$). By analogy with
perturbative QCD calculations we will refer to these class of diagrams
as to the ``proton-nucleus'' scattering, with the shock wave $t_1$
being the proton and the shock wave $t_2$ the nucleus. The
applicability region of such approximation is defined by
\begin{align}\label{bounds}
\mu_1 \, (x^-)^2 \, x^+ \, \ll \, 1, \ \ \ \mu_2 \, (x^+)^2 \, x^-
\sim 1,
\end{align}
which shows that the resummation is applicable to nucleus-nucleus
collisions, only in the part of the forward light-cone defined by
\eq{bounds}.
\begin{figure}[h]
\begin{center}
\epsfxsize=6cm
\leavevmode
\hbox{\epsffile{pA.eps}}
\end{center}
\caption{A diagram contributing to the metric of an asymmetric
collision of two shock waves.}
\label{pAfig}
\end{figure}
Such resummation was performed in \cite{Albacete:2009ji} using the
eikonal approximation. The result for the expectation value of the
energy-momentum tensor reads \cite{Albacete:2009ji}
\begin{subequations}\label{Tmnd}
\begin{align}
\langle T^{++}\rangle \, & = \, - \frac{N_c^2}{2 \, \pi^2} \,
\frac{4 \, \mu_1 \, \mu_2 \, (x^+)^2 \, \theta (x^+) \, \theta
(x^-)}{\left[ 1 + 8 \, \mu_2 \, (x^+)^2 \, x^- \right]^{3/2}}, \label{T++} \\
\langle T^{--}\rangle \, & = \, \frac{N_c^2}{2 \, \pi^2} \, \theta
(x^+) \, \theta (x^-) \, \frac{\mu_1}{2 \, \mu_2 \, (x^+)^4} \,
\frac{1}{\left[ 1 + 8 \, \mu_2 \, (x^+)^2 \, x^- \right]^{3/2}}
\notag \\ & \times \, \bigg[ 3 - 3 \, \sqrt{1 + 8 \, \mu_2 \,
(x^+)^2 \, x^-} + 4 \, \mu_2 \, (x^+)^2 \, x^- \notag \\ & \times
\, \left( 9 + 16 \, \mu_2 \, (x^+)^2 \, x^- - 6 \, \sqrt{1 + 8 \,
\mu_2 \, (x^+)^2 \, x^-}
\right) \bigg] , \\
\langle T^{+-}\rangle \, & = \, \frac{N_c^2}{2 \, \pi^2} \, \frac{8
\, \mu_1 \, \mu_2 \, x^+ \, x^- \, \theta (x^+) \, \theta
(x^-)}{\left[ 1 + 8 \, \mu_2 \, (x^+)^2 \, x^- \right]^{3/2}}, \\
\langle T^{\, i j}\rangle \, & = \, \delta^{ij} \, \frac{N_c^2}{2
\,\pi^2} \, \frac{8 \, \mu_1 \, \mu_2 \, x^+ \, x^- \, \theta
(x^+) \, \theta (x^-)}{\left[ 1 + 8 \, \mu_2 \, (x^+)^2 \, x^-
\right]^{3/2}}.
\end{align}
\end{subequations}
Provided the complexity of the problem at hand, the resulting formulas
(\ref{Tmnd}) for the energy-momentum tensor are remarkably simple!
Now we can ask a question: what kind of medium is produced in these
strongly coupled proton-nucleus collisions? Is it described by ideal
hydrodynamics, just like Bjorken hydrodynamics was obtained in
\cite{Janik:2005zt}? In our case the produced matter distribution is
obviously rapidity-dependent, so it is slightly more tricky to check
whether Eqs. (\ref{Tmnd}) constitute an ideal hydrodynamics, i.e.,
whether it can be written as
\begin{align}\label{hydro}
T^{\mu\nu} \, = \, (\epsilon + p) \, u^\mu \, u^\nu - p \,
\eta^{\mu\nu}
\end{align}
with the positive energy density $\epsilon$ and pressure $p$.
$\eta^{\mu\nu}$ is the metric of the four-dimensional Minkowski
space-time and $u^\mu$ is the fluid 4-velocity.
For the particular case at hand it is easy to see that the
energy-momentum tensor in \eq{Tmnd} can not be cast in the ideal
hydrodynamics form of (\ref{hydro}). In the case of ideal
hydrodynamics one has
\begin{align}
T^{++} \, = \, (\epsilon + p) \, (u^+)^2 \, > \, 0.
\end{align}
At the same time $\langle T^{++}\rangle$ in \eq{T++} is {\sl negative
definite}. Therefore the ideal hydrodynamic description is not
achieved in the proton-nucleus collisions. We believe this result is
due to limitations of this proton-nucleus approximation. As we will
show below, thermalization (black hole production) is inevitable in
the collision of the two shock waves considered here. Our conclusion
is then that thermalization/isotropization of the medium does not
happen in the space-time region defined by the bounds in \eq{bounds}.
What we found in \eq{Tmnd} is a medium at some intermediate stage, on
its way to thermalization at a later time. It appears that one needs
to solve the full nucleus-nucleus scattering problem to all orders in
both $\mu_1$ and $\mu_2$, to obtain a medium described by ideal
hydrodynamics.
\section{Stopping of Nuclei After Collision}
To better understand dynamics of the shock wave collisions let us
follow one of the shock waves after the interaction. First we
``smear'' the delta-function profile of that shock wave:
\begin{equation}\label{t1s}
t_1 (x^-) \, = \, \frac{\mu_1}{a_1} \, \theta (x^-) \,
\theta (a_1 - x^-).
\end{equation}
Here $a_1 \, \propto \, R_1 \, \frac{\Lambda_1}{p_1^+} \, \propto \,
\frac{A_1^{1/3}}{p_1^+}$, where the nucleus of radius $R_1$ has $A_1$
nucleons in it. The ``$+ +$'' component of the energy-momentum tensor
of a shock wave after the collision at $x^- = a_1 /2$ is
\cite{Albacete:2008vs}
\begin{equation}\label{stop}
\langle T^{+ \, +} (x^+ , x^- = a_1 /2) \rangle \, = \,
\frac{N_c^2}{2 \, \pi^2} \,
\frac{\mu_1}{a_1} \left[ 1 - 2 \, \mu_2 \, x^{+\, 2} \, a_1 \right].
\end{equation}
The first term on the right of \eq{stop} is due to the original shock
wave while the second term describes energy loss due to graviton
emission. \eq{stop} shows that $\langle T^{+ \, +} \rangle$ of a
nucleus becomes {\sl zero} at light-cone times (as $p_1^+ \approx
p_2^-$ in the center-of-mass frame)
\begin{equation}\label{stoptime}
x^+_{stop} \, \sim \, \frac{1}{\sqrt{\mu_2 \, a_1}} \, \sim \,
\frac{1}{\Lambda_2 \, A_1^{1/6} \, A_2^{1/6}}.
\end{equation}
Zero $\langle T^{+ \, +} \rangle$ would mean {\sl stopping} of the
shock wave and the corresponding nucleus. The result can be better
understood by doing all-order resummation of graviton exchanges with
one shock wave performed above for ``proton-nucleus collisions''
\cite{Albacete:2009ji}. The full result for the proton's ``$+ +$''
component of the energy-momentum tensor is
\begin{equation}\label{stop_pA}
\langle T^{++} \rangle \, = \, \frac{N_c^2}{2 \, \pi^2} \,
\frac{\mu_1}{a_1} \, \frac{1}{\sqrt{1 + 8 \, \mu_2 \, (x^+)^2 \,
x^-}}, \ \ \ \mbox{for} \ \ \ 0 < x^- < a_1.
\end{equation}
\eq{stop_pA} is illustrated in \fig{pstop}, in which one can see that
the proton loses all of its light cone momentum over a rather short
time.
\begin{figure}
\begin{center}
\includegraphics[width=7cm]{pstop}
\end{center}
\caption{$T^{++}$ component of the proton's energy-momentum tensor
after the collision as a function of light cone time $x^+$
(arbitrary units).}
\label{pstop}
\end{figure}
We thus conclude that the collision of two nuclei at strong coupling
leads to a necessary stopping of the two nuclei shortly after the
collision. If the nuclei stop completely in the collision, the strong
interactions between them are almost certain to thermalize the system,
probably leading to Landau hydrodynamics \cite{Landau:1953gs}.
(Rapidity-independent Bjorken hydrodynamics \cite{Bjorken:1982qr}
seems to be unlikely after stopping. Even before stopping the
energy-momentum tensor in \eq{Tmnd} is strongly rapidity-dependent.)
\section{Thermalization: Trapped Surface Analysis}
While the exact solution of Einstein equations for the colliding shock
waves remains elusive, one can infer whether a black hole will be
created in the bulk following such collisions by performing a trapped
surface analysis \cite{Penrose,Eardley:2002re}. A trapped surface
analysis for shock waves with sources in the bulk has been carried out
before in \cite{Gubser:2008pc,Lin:2009pn,Gubser:2009sx}. However, with
the resulting trapped surface being centered around the sources, it is
not clear to what extend the trapped surface is the property of these
bulk sources, and what happens to the trapped surface when there are
no bulk sources, as is the case for our shock wave \peq{1nuc}.
Performing a trapped surface analysis for a shock wave without sources
\peq{1nuc} does not allow one to uniquely fix the position of the
trapped surface \cite{Kovchegov:2009du}. We therefore start with a
shock wave having an extended source in the bulk with the only
non-zero component of the bulk energy-momentum tensor being $J_{--} \,
= \, \frac{E}{z_0 \, L} \, \delta (x^-) \, \delta (z - z_0)$. The
corresponding metric is \cite{Lin:2009pn}
\begin{align}\label{nuc1s}
ds^2 \, = \, \frac{L^2}{z^2} \, \left\{ -2 \, dx^+ \, dx^- + \phi
(z) \, \delta (x^-) \, d x^{- \, 2} + d x_\perp^2 + d z^2
\right\}
\end{align}
with
\begin{align}\label{phi_sol}
\phi (z) \, = \, \frac{2 \, \pi^2 \, E \, L^2}{N_c^2} \, \left\{
\begin{array}{c}
\frac{z^4}{z_0^4}, \ z \le z_0 \\~\\
1, \ z > z_0.
\end{array}
\right.
\end{align}
As one can readily see, the metric \peq{nuc1s} reduces to that in
\eq{1nuc} if we send the source to the IR infinity in the bulk by
taking $z_0 \rightarrow \infty$ limit of the metric in Eqs.
(\ref{nuc1s}) and (\ref{phi_sol}) keeping $\mu$ defined by
\begin{align}\label{mu}
\mu \, = \, \frac{2 \, \pi^2}{N_c^2} \, \frac{E \, L^2}{z_0^4}
\end{align}
fixed \cite{Kovchegov:2009du}.
Marginal trapped surface for a collision of two shock waves given by
\eq{nuc1s} was found in \cite{Lin:2009pn}. In the $z_0 \rightarrow
\infty$, $\mu$ fixed limit that trapped surface reduces to
\begin{align}\label{fintr}
x^+ \, = \,0, \ x^- \, = \, - \frac{\mu}{2} \, \left[ z^4 -
\mu^{-4/3} \, 2^{-2/3} \right]
\end{align}
with an analogous expression for the other shock wave obtained by
interchanging $x^+ \leftrightarrow x^-$ in \eq{fintr}. (For simplicity
we work in the center-of-mass frame where $\mu_1 = \mu_2 = \mu$.) The
trapped surface in \eq{fintr} is independent of the shape of the
source being sent to deep IR, as shown in \cite{Kovchegov:2009du}. The
trapped surface from \eq{fintr} is depicted in \fig{surf}.
\begin{figure}[h]
\begin{center}
\epsfxsize=9cm
\leavevmode
\hbox{\epsffile{surf.eps}}
\end{center}
\caption{An illustration of the trapped surface in the collision of
two sourceless shock waves. Vertical axis is the bulk $z$-direction,
the horizontal left-right axis can be thought of either as the
collision axis or as the time direction. The trapped surface is
shaded.}
\label{surf}
\end{figure}
The existence of trapped surface proves that gravitational collapse is
inevitable. That is a black hole will be created in a bulk following a
collision of two sourceless shock waves. In the boundary theory this
means that a thermal medium is created, which is described by ideal
hydrodynamics.
The (lower bound for the) produced entropy per unit transverse area
$A_\perp$ can be found by calculating the area of the trapped surface,
which yields \cite{Kovchegov:2009du}
\begin{align}
\frac{S}{A_\perp} \, = \, \frac{N_c^2}{2 \, \pi^2} \, \left( 2 \,
\mu_1 \, \mu_2 \right)^{1/3}.
\end{align}
Since the trapped surface analysis does not ``know'' anything about
shock wave thickness (e.g. $a_1$), we conclude that the thermalization
time is only a function of $\mu_1$ and $\mu_2$, which gives
\begin{align}
\label{thermtime}
\tau_{th} \, \sim \, \frac{1}{(\mu_1 \, \mu_2)^{1/6}},
\end{align}
in agreement with the thermalization time suggested in
\cite{Grumiller:2008va}. While numerically this thermalization time is
too short to be relevant for RHIC data, it is parametrically shorter
than the stopping time \peq{stoptime}, making our model somewhat more
relevant for description of real-life collisions. As $\mu_1 \, \mu_2
\sim s$ with $s$ the center of mass energy of the collision, we get
\begin{align}
\frac{S}{A_\perp} \, \propto \, s^{1/3}
\end{align}
in agreement with the result obtained in \cite{Gubser:2008pc}. Finally
note that, since, as follows from \eq{mus}, $\mu_1$ and $\mu_2$ are
$N_c$-independent, the produced entropy scales $\propto N_c^2$, in
agreement with $N_c$-counting in a perturbative QCD calculation of
particle production for a collision of two nuclei with $N_c^2$
``valence gluons'' in their nucleons.
\section*{Acknowledgements}
I am grateful to Javier Albacete, Shu Lin, and Anastasios Taliotis for
collaborating with me on various parts of this work
\cite{Albacete:2008vs,Albacete:2009ji,Kovchegov:2009du}.
This research is sponsored in part by the U.S. Department of Energy
under Grant No. {DE-SC0004286}.
\providecommand{\href}[2]{#2}\begingroup\raggedrigh
| {
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{"url":"https:\/\/physics.stackexchange.com\/questions\/533166\/ionizing-radiation-in-thermal-radiation","text":"According to the black-body radiation equation, the spectrum extends to infinitely high frequency (although its intensity gets small quickly towards high frequency).\n\n(1) How do you roughly estimate the ionizing radiation power in common high power thermal sources like a 2KW heater without fan, then make sure it is safe considering it is used for years instead of 0.1 second\/year in the case of DR X-ray scan? (First the portion of power getting radiated instead of convected needs to be estimated, second this seems not to be black-body and will it have a similar radiation curve that extends to infinite frequency?)\n\n(2) In the photoelectric effect, there are things like cut-off frequency; so, why doesn't thermal radiation - which is quantum mechanical in microscopic level - have a cut-off frequency, i.e. it doesn't radiate X-ray and gamma-ray at all?\n\nfor 1) In this link there is the black body formula of Planck's law that gives the\n\nthe power per unit solid angle and per unit of area normal to the propagation\n\nSo for a given temperature of the body one can calculate the power, in your case for high enough frequencies, take a $$\u0394(\u03bd)$$ .\n\nOnce you have the power you will have the energy per second. Divide the power with the average $$h\u03bd$$ of your $$\u0394(\u03bd)$$ and you can estimate how many photons of that approximate frequency radiate per second.\n\nYou will find that one has to wait for a loooo..ng time for an X-ray photon to come our of your radiator. Better worry about the Xrays created by the muons passing your body at the rate of 1 every $$cm^2$$ every minute, much worse when you fly in an airplane.\n\nfor 2) Quantum mechanics gives probabilities for limits. When the probability gets very very small, that is a limit.To get enough energy for an x-ray the material would have to synchronize so that random vibrations would quantum mechanically allow for a quantum level to exist to have enough energy for an x-ray photon to come out.How improbable that is is reflected in the Planck formula.\n\nAlso a limit is given by energy conservation laws , also included in the formula.\n\nThe basic line is that it is very improbable, as the other answers state.\n\n\u2022 Now I'm wondering how much the metal frame of an airplane actually blocks such radiation.... \u2013\u00a0Michael Feb 26 at 16:00\n\u2022 @Michael barely, but if I remember from last time I did the calculation you would have to be at 30,000 feet for about 6 months straight to reach the 20 mSv European annual limit for radiation workers (the American limit is 50 mSv) \u2013\u00a0llama Feb 26 at 16:57\n\u2022 Considering that gold is one of the best X-ray reflectors, and that it only achieves total reflection at less than 3 degree incident angle, you can rest assured that X-rays and above are not the least bit concerned with puny things like atoms. \u2013\u00a0Lawnmower Man Feb 26 at 20:35\n\nPlanck's law says:\n\n$$B_\\nu(\\nu, T) = \\frac{2h\\nu^3}{c^2} \\frac1{\\exp\\left(\\frac{h\\nu}{kT}\\right) - 1}.$$\n\nThis is the power emitted by a black body of temperature $$T$$ at frequency $$\\nu$$. If we consider a black object (thus highest possible purely thermal emission) of temperature $$700\u00b0\\mathrm C$$ (an estimate of temperature of a heater), it'll emit total power of\n\n$$P_\\text{total}(700\u00b0\\mathrm C)=\\pi \\int\\limits_0^\\infty \\mathrm{d}\\nu B_\\nu(\\nu,700\u00b0\\mathrm C)=13.6\\,\\frac{\\mathrm{kW}}{\\mathrm{m}^2}.$$\n\n(This can also be calculated via the Stefan-Boltzmann law.)\n\nIf we now want to find the power of only ionizing radiation, we should instead integrate in the range of $$\\nu\\in[\\nu_0,\\infty)$$, where $$\\nu_0$$ could be taken somewhere in the UV range, e.g. lower border of UVB, i.e. $$950\\,\\mathrm{THz}$$. Then we have:\n\n$$P_\\text{hard}(700\u00b0\\mathrm C)=\\pi \\int\\limits_{950\\,\\mathrm{THz}}^\\infty \\mathrm{d}\\nu B_\\nu(\\nu,700\u00b0\\mathrm C)=3.1\\times 10^{-20}\\,\\frac{\\mathrm{W}}{\\mathrm{m}^2}.$$\n\nSo a surface of $$1\\,\\mathrm{m}^2$$ will emit $$~10^{-20}\\,\\mathrm{W}$$. This is totally negligible, even in the course of a hundred years.\n\n\u2022 Do you have recommendation for some tools to do this calculate? I want to calculate the radiation bellow 700C too. Since the heater burn the skin as Sun, I guess radiation that fall in the range of skin sensing is above 100W. But visible light is not that far away from ionizing radiation, I wonder why the power difference is so much. \u2013\u00a0jw_ Feb 26 at 12:24\n\u2022 @jw_ I did the integration in Wolfram Mathematica, using its builtin NIntegrate function. But if you need a free-software solution, you might want to look at GNU Octave's quadcc function or Scilab's integrate function. \u2013\u00a0Ruslan Feb 26 at 12:47\n\u2022 @undefined well, you can try finding the closed-form indefinite integral (although it's not elementary, see Wolfram|Alpha result), and then use Newton-Leibniz formula on it. You'll need a way to compute polylogarithm though. Or you can make a substitution like $\\nu=\\tan(t)$ to convert the improper integral to a proper one, and then use e.g. trapezoidal quadrature rule to find numerical approximation of the definite integral. \u2013\u00a0Ruslan Feb 26 at 15:30\n\u2022 @jw_ you can also get burns without any ionizing radiation and in the case of your space heater it is not the ionizing radiation you have to worry about. \u2013\u00a0fraxinus Feb 26 at 16:33\n\u2022 Given enough flux, you can incinerate a human with long-wavelength microwaves. This is why you don't want to stand in front of a high-power radar emitter. \u2013\u00a0Lawnmower Man Feb 26 at 20:38\n\n(1) The element in most heaters appears to be a cooler color\/temperature than the sun - and only similarly warm. So, it is safer than the sun.\n\n(2) The photo-electric effect reveals that below a cut-off frequency there is not enough energy to release an electron, so the energy eventually ends up as heat. Above the cut-off an electron may be freed. But all frequencies are involved and can do something.\n\n\u2022 The outer part of the sun is largely in thermal equilibrium and shields you from the inner part of the sun - a classic black body case. And the distance does not matter as long as the warmth you feel is the same - unless you think the atmosphere is shielding you from the sun. \u2013\u00a0Paul Young Feb 26 at 13:22\n\u2022 Good to learn that, and in fact the warmth argument is good, now get it. \u2013\u00a0jw_ Feb 26 at 13:56\n\u2022 Most of the sun is so opaque that it takes a photon thousands of years or more to escape from the core to the surface. Gamma rays from fusion are absorbed by the intermediate layers and re-emitted at lower frequencies, which is why we observe very few of them at Earth. \u2013\u00a0Lawnmower Man Feb 26 at 20:42\n\u2022 \u201cunless you think the atmosphere is shielding you from the sun\u201d This really make the Sun comparison invalid. The atmosphere could shield most high energy radiation. \u2013\u00a0jw_ Feb 27 at 1:34\n\u2022 The atmosphere does shield high-energy radiation, specifically ozone blocks UV (which is right at the edge of ionizing radiation). Makes sense: if the radiation is high enough energy to ionize, why wouldn't it ionize the atoms in the atmosphere? \u2013\u00a0MSalters Feb 27 at 11:09\n\nIf you plug in the values of $$T=1000$$K, $$\\nu=10^{15}$$Hz into the spectral radiance formula of a blackbody (Plank\u2019s law): $$B_{\\nu}(\\nu, T) = \\cfrac{2h \\nu^3}{c^2} \\frac{1}{\\exp\\left(\\frac{h \\nu}{k_BT}\\right) - 1}$$ where $$B_{\\nu}(\\nu, T)$$ is the power delivered per unit area per solid angle at a frequency range centred at $$\\nu$$ by a blackbody at temperature $$T$$ and do the following integral $$P\\left(T,\\nu_0\\right)=\\int_{\\nu_0}^{\\infty} \\cfrac{2h \\nu^3}{c^2} \\frac{d\\nu}{\\exp\\left(\\frac{h \\nu}{k_BT}\\right) - 1}$$ Here since the argument of the exponential is $$\\sim 48$$, we can approximate the integral to\n\n$$P\\left(T,\\nu_0\\right)=\\int_{\\nu_0}^{\\infty} \\cfrac{2h \\nu^3}{c^2}\\exp\\left(-\\frac{h \\nu}{k_BT}\\right)d\\nu$$\n\nThis is a straightforward integral which can be done by parts to get the total ionising radiation power to be $$P\\left(1000\\text K,10^{15}\\text{Hz}\\right)=1.69\\times10^{-16}\\frac{\\text{W}}{\\text{m}^2\\cdot\\text{st}}$$ This as you can see, is minuscule.\n\nIn photoelectric effect, we are knocking off an electron from the bulk using the energy of photons. The excess energy is converted into kinetic energy of the electron. Now consider the reverse. High energy electron is getting captured by the bulk. So now a photon is released depending on how much energy the electron lost. Since the electron before capture can have arbitrary energy, the emitted photon can have arbitrary energy. Thus there is no cutoff for radiation.\n\nThe cutoff comes in the case of photoelectric effect due to the restriction on allowed energies of the bound electron. However there is no such restriction on free electron.\n\n\u2022 (1) some error to fix? Bv(v,t) is per Hz value, it need to be integrated with frequency, this also means \"so all subsequent powers for higher frequencies will be less than the above value\" does not work since the integrage converge but a multiplication will result in infinite. (2) \"Now consider the reverse. High energy electron is getting captured by the bulk\" An electron with high energy, how can it be captured in one step? - the whole reasoning may be right, but this detail is the key to decide whether high energy photon can be emitted right? \u2013\u00a0jw_ Feb 26 at 7:16\n\u2022 Yes! You\u2019re right. Fixed my error. As for the electron capture, it\u2019s a probabilistic process. We\u2019d have to calculate the overlap if the free electron wavefunction with the bound states of the atom. \u2013\u00a0Superfast Jellyfish Feb 26 at 7:28\n\u2022 It seems that you don't want to do the calculation as I did:) , could you make some strict estimation of the upper bound? Why is everybody talking about the Sun, that is not a right comparison! For the second question, I suddenly thing it is just like brake radiation, when electron collide with something and brake or bounce back, during this process photon is emitted - never elastic. \u2013\u00a0jw_ Feb 26 at 7:34\n\u2022 You\u2019re right about the calculation part :P, maybe I\u2019ll try it after my classes. The overlaps only give you the probabilities of capture. But yes, you can have radiation without capture, like brake radiation. That probably is the dominant mechanism for radiation in plasmas (like sun). I compared it with the sun because that\u2019s one constant source of radiation! \u2013\u00a0Superfast Jellyfish Feb 26 at 7:42\n\u2022 Google is not being very helpful, but from en.wikipedia.org\/wiki\/Infrared_heater \"Ceramic elements operate in the temperature of 300 to 700 \u00b0C producing infrared wavelengths in the 2 to 10 \u03bcm range\". I assume the OP is not asking about \"modern\" ceramic heaters, though, and they're more interested in the style of heater shown at the top of the article. IIRC, such heaters radiate at roughly 1000 \u00b0C, and they (obviously) radiate a significant amount of visible light as well as infrared. \u2013\u00a0PM 2Ring Feb 26 at 11:07\n\nHow to roughly estimate the ionizing radiation power in common high power thermal source like 2KW heater without fan then make sure it is safe considering it is used for years instead of 0.1 second\/year in case of DR Xray scan?\n\nA common 2 kW resistance heater does not emit ionizing radiation. It's radiant heat is primarily infra red. The range for infrared radiation is about 430 x 10$$^{12}$$Hz to 300 x 10$$^9$$ Hz. It is considered non-ionizing radiation.\n\nIonizing radiation begins around 3 x 10$$^{15}$$ Hz. X-radiation is in the range of 3 x 10$$^{16}$$ to 2 x 10$$^{19}$$ Hz.\n\nBottom line: Infrared heaters do not pose the health risks associated with ionizing radiation like X-radiation.\n\nHope this helps.\n\n\u2022 It does not, or very little? If doesn not, then what is the theory behind - the radiation curve says it does? If very little, can a roughly reasoning estimation be provided? \u2013\u00a0jw_ Feb 26 at 3:40\n\u2022 @jw_ what \"radiation curve\"? The energy of an infrared photon is insufficient to ionize an atom. \u2013\u00a0Bob D Feb 26 at 3:42\n\u2022 Anything have a radiation curve - the power spectrum of its radition - power vs frequency. \u2013\u00a0jw_ Feb 26 at 3:44\n\u2022 @jw_ Can you provide a reference for the \"radiation curve\" ? The energy of an infrared photon anywhere in the frequency range I cited is insufficient to ionize atoms as are x-rays. \u2013\u00a0Bob D Feb 26 at 3:48\n\u2022 @BobD I believe he is referring to the black body spectra. \u2013\u00a0Cort Ammon Feb 26 at 4:48","date":"2020-04-10 10:29:49","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 30, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7137943506240845, \"perplexity\": 746.5959461602333}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585371893683.94\/warc\/CC-MAIN-20200410075105-20200410105605-00490.warc.gz\"}"} | null | null |
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"Taylor, Imogen (Translator),"
Families -- Germany -- Fiction (1)
Fathers and sons -- Fiction (1)
Germany -- History -- 1933-1945 (1)
Germany -- Politics and government -- 1933-1945 (1)
Germany -- Social conditions -- 1933-1945 (1)
Missing persons -- Fiction (1)
National socialism -- Germany -- Psychological aspects (1)
Suicide -- Germany -- History -- 20th century (1)
World War, 1939-1945 -- Germany -- Psychological aspects (1)
Huber, Florian, 1967 (1)
Kurbjuweit, Dirk, 1962 (1)
Salzmann, Marianna, 1985 (1)
Alternate Languages
Adult New Fiction (1)
» Imogen (Translator) Taylor
Imogen (Translator) Taylor
1) "Promise me you'll shoot yourself": the mass suicide of ordinary Germans in 1945
Huber, Florian
Central - Adult New Nonfiction
940.5316 HUBER
2020. First North American edition. Little, Brown Spark vi, 292 pages : illustrations ; 25 cm English
"By the end of April 1945 in Germany, the Third Reich had fallen and invasion was underway. As the Red Army advanced, horrifying stories spread about the depravity of its soldiers. For many German people, there seemed to be nothing left but disgrace and despair. For tens of thousands of them, the only option was to choose death -- for themselves and for their children. "Promise Me You'll Shoot Yourself" recounts this little-known mass event. Using...
2) Fear: a novel
Kurbjuweit, Dirk
Central - Adult Fiction
F KURBJ
2017. First U.S. edition. Harper, an imprint of HarperCollinsPublishers 257 pages ; 22 cm English
"You'd die for your family. But would you kill for them? I had always believed my father capable of a massacre. Whenever I heard on the news that there had been a killing spree, I would hold my breath, unable to relax until it was clear that it couldn't have been him. Randolph Tiefenthaler insists he had a normal childhood, though he grew up with a father who kept thirty loaded guns in the house. Now he has an attractive, intelligent wife and two...
3) Beside myself
Salzmann, Marianna
Central - Adult New Fiction
F SALZM
[2019] Other Press 317 pages 21 cm. English
"A brilliant literary debut about belonging, family and love, and the enigmatic nature of identity. Beside Myself is the disturbing and exhilarating story of a family across four generations. At its heart is one woman's search for her twin brother. When Anton goes missing and the only clue is a postcard sent from Istanbul, Ali leaves her life in Berlin to find him. Without her twin, the sharer of her memories and the mirror of her own self, Ali is... | {
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\section{Introduction} \label{sec:method}
The Random Coefficients model is often used to model unobserved heterogeneity in a population, an important problem in econometrics and statistics. The model is given by
\begin{equation}\label{eqn:rc_model}
Y_i=\beta_{0i}+\beta_{1i} X_{1i}+\beta_{2i} X_{2i}+\ldots+\beta_{di} X_{di}.
\end{equation}
Where $\mathbf{X}_i = (1, X_{1i}, X_{2i}, \ldots X_{di})^\top$ and ${\boldsymbol{\beta}}_i = (\beta_{0i}, \beta_{1i}, \beta_{2i}, \ldots, \beta_{di})^\top$ are the regressors, and regression coefficients respectively. It is assumed that $\mathbf{X}_i$, $Y_i$, ${\boldsymbol{\beta}}_i$ are i.i.d random variables with ${\boldsymbol{\beta}}_i$ and $\mathbf{X}_i$ being independent, and that $Y_i$ and $\mathbf{X}_i$ are observed while ${\boldsymbol{\beta}}_i$ is unknown.
In this paper we introduce an open source \texttt{Python} module named \pythoninline{PyRMLE} which implemenents regularized maximum likelihood estimation (RMLE) to nonparametrically estimate the density $f_{\boldsymbol{\beta}}$.
Nonparametric estimation of the random coefficients model was first established by \cite{Beran1992} for a single regressor. A kernel method for estimation was developed by \cite{Beran1996}. The optimal rate for design densities with Cauchy-type tails is derived in \cite{hoderlein2010} for a kernel method.
Equality constrained linear regression methods were developed by \cite{fox2011simple, heiss2021nonparametric}. Regularized maximum likelihood methods have been considered for problems other than the random coefficients model specified in \eqref{eqn:rc_model} in \cite{WH:12,HW:13, hohage2016inverse, Dunker2014}. The method implemented by the \pythoninline{PyRMLE} module is the one developed in \cite{dunker2021regularized}.
\texttt{Python} was chosen as the programming language mainly for the extensive scientific and computational libraries at its disposal, namely: \pythoninline{NumPy, SciPy} by \cite{harris2020array,jones2001scipy}. The module takes advantage of the benefits of working with \pythoninline{NumPy} arrays in terms of computational efficiency achieved by doing array-wise computation. Another advantage is that there are no software imposed limits in terms of array size. The maximum array size in \texttt{Python} is solely determined by the amount of RAM available to the user, which allows the user the flexibility to increase the computational complexity of the method to the level that their system allows. The module also uses a trust-region constrained minimization algorithm developed by \cite{byrd1999interior} which is implemented in \pythoninline{SciPy}.
The paper is organized as follows: Section \ref{sec:method} briefly describes the regularized maximum likelihood method developed in \cite{dunker2021regularized}, Section \ref{sec:PythonImplementation} discusses the classes and functions available to the module, and Section \ref{sec:example} discusses examples of the modules usage for general cases.
\section{Regularized Maximum Likelihood} \label{sec:method}
It is assumed that the random coefficients, ${\boldsymbol{\beta}}$, have a Lebesgue density $f_{\boldsymbol{\beta}}$. If the conditional density $f_{Y|\mathbf{X}}$ exists, the two densities are connected by the integral equation
\begin{align*}
f_{Y|\mathbf{X}}(y|\mathbf{X}=\mathbf{x})=\int_{\mathbbm{R}^{d}}\mathbbm{1}\big\{\mathbf{b}^\top \mathbf{x}=y\big\}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b}) = \int_{\mathbf{b}^\top \mathbf{x}=y}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b})
\label{eqn:line_int_f_beta}
\end{align*}
This connection allows us to employ maximum likelihood estimation to nonparametrically identify $f_{\boldsymbol{\beta}}$ as seen in the following expression of the log-likelihood
\begin{equation*}
\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})=\frac{1}{n}\sum_{i=1}^{n}\log\left[\int_{\mathbbm{R}}\mathbbm{1}\big\{{\boldsymbol{\beta}}_i^\top \mathbf{X}_i=Y_i\big\}f_{\boldsymbol{\beta}}(b)d\mu(\mathbf{b})\right].
\label{eqn:avg_lhood}
\end{equation*}
Direct maximization of $\bar\ell(f_{\boldsymbol{\beta}}|Y,\mathbf{X})$ over all densities it not feasible due to ill-posedness and will lead to overfitting. We stabilize the problem by adding a penalty term $\alpha \mathcal{R}(f_{\boldsymbol{\beta}})$ and state the estimator as a minimization problem with negative log-likelihood
\begin{align}\label{eqn:method}
\hat {f_{\boldsymbol{\beta}}}_\alpha = \argmin_{f\ge 0,\, \|f\|_{L^1}=1} -\bar\ell(f|Y,\mathbf{X})+\alpha \mathcal{R}(f).
\end{align}
Here $\alpha \ge 0$ is a regularization parameter that controls a bias variance trade-off. It was poined out in \cite{heiss2021nonparametric} that the constraint $\|f_{\boldsymbol{\beta}}\|_{L^1}=1$ together with a finite difference discretization of $f_{\boldsymbol{\beta}}$ is equivalent to an $\ell^1$ penalty on the discretized values of $f_{\boldsymbol{\beta}}$, which could cause unwanted shrinkage of the estimate. To reduce this LASSO effect, \cite{heiss2021nonparametric} introduced an additional quadratic constraint which turns the method into an elastic net. In \cite{dunker2021regularized} it was stated that the regularization term $\alpha\mathcal{R}(f_{\boldsymbol{\beta}})$ is analogous to this additional constraint but is more flexible as the method is not limited to quadratic $\mathcal{R}$.
The implemented regularization terms $\mathcal{R}$ in this module are: (1) squared $L^2$ norm $\mathcal{R}(f) = \|f\|_{L^2}^2 = \|f\|_{2}^2$, (2) the Sobolev Norm for $H^{1}$ $ \mathcal{R}(f) = \|f_\beta\|_{2}^{2}+\|f^{\prime}_\beta\|_{2}^{2}$, and (3) entropy $\mathcal{R}(f) = \int{f(\mathbf{b})\ln{f(\mathbf{b})}d\mathbf{b}}$.
In addition to the regularization functional, a regularization parameter $\alpha$ also needs to be chosen. In this module we implement two methods of estimating $\alpha$: Lepskii's Balancing principle, and K-fold cross validation.
\section{Python Implementation} \label{sec:PythonImplementation}
The \pythoninline{PyRMLE} module's implementation of regularized maximum likelihood is limited to applications with up to two regressors for the random coefficients model with intercept, and up to three regressors for a model without intercept.
There are two main functions used to implement regularized maximum likelihood estimation using \pythoninline{PyRMLE}, namely: \pythoninline{transmatrix()} and \pythoninline{rmle()}. There are other sub-functions necessary to the implementation, these will be discussed under the appropriate subsections when relevant.
\subsection{The \texttt{transmatrix()} Function}\label{sec:trans_matrix}
The purpose of the function \texttt{transmatrix()} is to construct the discrete version of the linear operator given by
\begin{equation}
T f_\beta = \int_{\mathbf{b}^\top \mathbf{x}=y}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b})
\label{eq:inv_prob}.
\end{equation}
The linear operator above describes the integral of $f_{\boldsymbol{\beta}}$ over the hyperplanes parametrized by the sample points $\mathbf{X}, \text{ and }Y$. The function makes use of a finite-volume method as a discrete analog in evaluating the integral. The function is used as follows
\begin{lstlisting}
trans_matrix = transmatrix(sample,grid)
\end{lstlisting}
The argument \texttt{sample} corresponds to the sample observations. The sample data should be in the following format: \\
\begin{center}
$\begin{bmatrix}
X_{0,1} & X_{1,1} & \hdots & Y_1 \\
X_{0,2} & X_{1,2} & \hdots & Y_2 \\
\vdots & \vdots & \ddots & \vdots \\
X_{0,n} & X_{1,n} & \hdots & Y_n \\
\end{bmatrix}$
\end{center}
In the case of a random coefficients model with intercept the first column would simply be $\mathbf{X}_0 = (1,1,\hdots,1)^{T}$.
The \texttt{grid} argument is a class object generated by the \texttt{grid\_set()} function. It has the following attributes and methods: \{ \texttt{scale}, \texttt{shifts}, \texttt{interval}, \texttt{dim}, \texttt{step}, \texttt{start}, \texttt{end}, \texttt{ks()}, \texttt{numgridpoints()}\}. The \texttt{grid\_set()} function is used as follows:
\begin{lstlisting}
grid_beta = grid_set(num_grid_points=20,dim = 2)
\end{lstlisting}
The base grid that is generated by the \texttt{grid\_set()} function is a symmetric grid that spans $[-5,5]$ in each axis. The user inputs the number of grid points by passing an integer value to the function as \pythoninline{num_grid_points}. This specifies the step size of the grid as $\frac{10}{k}$ where $k$ is the number of grid points along each axis. Additionally, the user can change the range over which each axis is defined by supplying new axes ranges through the arguments: {\pythoninline{B0_range, B1_range, B2_range}} which are passed as lists or arrays that contain the end points of the new range (e.g. \pythoninline{B0_range = [0,10]} ). This is especially useful if the user expects a random coefficient to be significantly larger or smaller than the other random coefficients.
\begin{python}
grid_beta_shifted = grid_set(num_grid_points = 20, \
dim = 2, B0_range = [0,10])
print(grid_beta_shifted.shifts)
[-5,0,0]
\end{python}
The output of the \texttt{transmatrix()} function is the \texttt{`tmatrix'} class object that has the following attributes and methods: \\
\noindent \texttt{Tmat}: returns a $n \times m $ \pythoninline{NumPy}-array that is produced by the function \texttt{transmatrix\_2d()} or \texttt{transmatrix\_3d()}. \\
\noindent \texttt{grid}: returns the \pythoninline{class grid_obj}. \\
\noindent \texttt{scaled\_sample}: returns the scaled and shifted sample.\\
\noindent \texttt{sample}: returns the original sample. \\
\noindent \texttt{n()}: returns the number of sample observations. \\
\noindent \text{m()}: returns the number of grid points $\hat{f}_{\boldsymbol{\beta}}$ is to be estimated over.
\subsubsection{\texttt{transmatrix\_2d()}}
The 2-dimensional implementation of this method works for the random coefficients model with a single regressor and random intercept
\begin{equation}
y_i={\beta_0}_i+{\beta_1}_{i}{x_1}_i,
\label{eqn:rc_model_2d}
\end{equation}
and the model with two regressors and no intercept.
\begin{equation}
y_i={\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i+\epsilon_i
\label{eqn:rc_model_2d_no_intercept}
\end{equation}
The function first initializes an $n \times m$-dimensional array of zeros, where $n$ is the sample size, and $m$ is the number of grid points $f_{\boldsymbol{\beta}}$ is to be estimated over. In the 1-dimensional case, the hyperplanes which $f_{\boldsymbol{\beta}}$ is integrated over reduce to lines which simplifies the problem. A finite-volume type method of estimation is employed to approximate the integral expression as seen in \ref{eq:inv_prob}. This method is akin to the algebraic reconstruction methods used in Computed Tomography. Specifically, it is reminiscent to the discrete Radon Transform methodology laid out by \cite{beylkin1987discrete}.
To implement this finite-volume method, the lines parametrized by the sample points given by equations \eqref{eqn:rc_model_2d} or \eqref{eqn:rc_model_2d_no_intercept} are made to intersect with the grid. The length of each line segment that intersects with the grid is then calculated and stored in an object. The intersection points of these lines with the grid are retrieved and subsequently mapped to their respective array indices. These indices are used to map the length of the line segments to the appropriate entry in the initialized array forming the discretized linear operator $T$. The algorithm is outlined as follows:\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{sample, grid} \\
\Initialization{Initialize \textbf{NumPy} Array of Zeros}\\
\For{s in sample}{
1. get intersection points\;
2. get line segment lengths\;
3. map intersection points to their array indices\;
\For{i in indices}{\
map line segment lengths to initialized array using index i
}
}
\caption{\texttt{transmatrix\_2d()}}
\end{algorithm}
\medskip
The result of this function is a large, sparse array, $\mathbf{T}$ where each row corresponds to a collection of all the lengths of the line segments intersecting the grid for a line parametrized by a sample point, i.e. each $l_{ij}$ corresponds to the length of the line segment intersecting the grid at section $i,j$. The resulting array is sparse because $l_{ij}$ is equal to zero unless a line passes through a section of the grid. The algorithm's implementation is illustrated in figure \ref{fig:1d_Algo}.
The resulting array is then used to evaluate the log-likelihood functional to be optimized
\begin{equation}
\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})= \frac{1}{n}\sum_{i=1}^{n}\log\textbf{T} f_{\boldsymbol{\beta}}^{*},
\label{eqn:discrete_loglikelihood}
\end{equation}
where $f_{\boldsymbol{\beta}}^{*} = (f_{{\boldsymbol{\beta}}_{1}}, f_{{\boldsymbol{\beta}}_{2}}, \hdots, f_{{\boldsymbol{\beta}}_{m}})^T$ is a $m \times 1$-array or vector that serves as the discrete approximation of $f_{\boldsymbol{\beta}}$.
\begin{figure}[!htb]
\begin{tikzpicture}[scale=0.8,every node/.style={minimum size=0.5cm},on grid]
\begin{scope}[
yshift=-50,every node/.append style={
yslant=0.5,xslant=-1.3},yslant=0.5,xslant=-1.3
]
\fill[white,fill opacity=0.9] (-3,-3) rectangle (3,3);
\draw[step=0.5cm, thin, gray] (-3,-3) grid (3,3);
\draw[black,very thick] (-3,-3) rectangle (3,3)
\draw[red,ultra thick,solid] (-1.5,-3.5) -- (2.25,3.5);
\pgfkeys{/pgf/number format/.cd, fixed, zerofill, precision =1}
\coordinate (s1) at (0.9107,1);
\node at (s1) [fill=black,circle,scale=0.1] {$a$};
\coordinate (s2) at (0.64285714285,0.5);
\node at (s2) [fill=black,circle,scale=0.1] {$b$};
\coordinate (s4) at (2.75,2.75);
\node at (s4) [fill=black,circle,scale=0.15] {$c$};
\end{scope},
\coordinate (p1) at (-2.75,1);
\node at (p1) [fill=black,circle,scale=0]{};
\coordinate (p2) at (-2.75,1.5);
\node at (p2) [fill=black,circle,scale=0]{};
\draw[-latex,thick](-2,1)node[left,scale=1]{$P_1$}
to[out=0,in=90] (s1);
\draw[-latex,thick](-2,1.5)node[left,scale=1]{$P_2$}
to[out=0,in=90] (s2);
\draw[-latex,thick](-6,1.5)node[left,scale=1]{}
to[out=0,in=180] (p1);
\draw[-latex,thick](-6,1.5)node[left,scale=1]{$l_{i,j}=\|P_1 P_2\|$}
to[out=0,in=180] (p2);
\coordinate (s3) at (-2.75,1);
\node at (s3) [fill=black,circle,scale=0]{};
\node {} {node (T) at (-4.5,-8) {%
$\begin{aligned}
\mathbf{T}=
\end{aligned}$}};
\node {} {node (line) at (5,-3.75) {%
$\begin{aligned}
\beta_0 = y_i - \beta_1 x_{1_i}
\end{aligned}$}};
\matrix (m1) at (0,-8) [matrix of math nodes,left delimiter=(,right delimiter=)](A) {
l_{1,1} & l_{1,2} & \dots & l_{1,j} & \dots & l_{1,m}\\
l_{2,1} & l_{2,2} & \dots & l_{2,j} & \dots & l_{2,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
l_{i,1} & l_{i,2} & \dots & l_{i,j} & \dots & 0\\
\vdots & \vdots & & \vdots & & \vdots\\
l_{n,1} & l_{n,2} & \dots & l_{n,j} & \dots & l_{n,m}\\
};
\coordinate (lij) at (0.25,-8);
\node at (lij) [fill=black,circle,scale=0]{};
\draw[-latex,dashed](-8.8,1.1)node[left,scale=1]{}
to[out=270,in=90] (lij);
\coordinate (p3) at (2.75,2.75);
\node at (p3) [fill=black,circle,scale=0]{};
\draw[-latex,thick](5,2.75)node[right,scale=1]{$l_{i,m} = 0$}
to[out=200,in=60] (s4);
\coordinate (lim) at (3,-8.4);
\node at (lim) [fill=black,circle,scale=0]{};
\draw[-latex,dashed](7,2.8)node[left,scale=1]{}
to[out=0,in=0] (lim);
\end{tikzpicture}
\caption{2-D Transformation Matrix Algorithm} \label{fig:1d_Algo}
\end{figure}
\subsubsection{\texttt{transmatrix\_3d()}}
The implementation of the function for the 3-dimensional case works for problems with two regressors and a random intercept,
\begin{equation}
y_i={\beta_0}_i+{\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i,
\label{eqn:rc_model_3d}
\end{equation}
and the model with three regressors and no intercept.
\begin{equation}
y_i={\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i+{\beta_3}_{i}{x_3}_i + \epsilon_i
\label{eqn:rc_model_3d_no_intercept}
\end{equation}
Note: for the three-dimensional implementation of the algorithm numerical instabilities occur when the underlying joint density, $f_{\boldsymbol{\beta}}$, has a single mode located at the center of the grid. This problem can occur in two ways: (1) it can occur organically if the mode of the density is located close to or at $(0,0,0)$ with the grid axes ranges at their default values [-5,5], or (2) artificially if the user provides grid ranges such that the mode of $f_{\boldsymbol{\beta}}$ is located at its center.
This problem is overcome by simply imposing a shift onto the resulting density by applying a linear transformation to the sample data, as below:
\begin{equation}
Y_i = (\beta_{0_i} + c) + \beta_{1_i} X_1 + \beta_{2_i} X_2.
\label{eqn:sample_shift}
\end{equation}
This can be achieved in two ways: (1) the more computationally efficient method is to simply supply a grid range for $\beta_0$ that offsets the mode of $\hat{f_{\boldsymbol{\beta}}}$ from the center, or (2) apply the shifting algorithm described in \eqref{eqn:sample_shift} which allows the user to supply any grid range but with an additional computational cost.
Similar to the 2-dimensional implementation, a finite volume type approach is used to create the discrete version of the linear operator $T$. In this higher-dimensional case the hyper-planes parametrized by the sample observations are now 2-dimensional planes, and the grid that $f_{\boldsymbol{\beta}}$ is estimated over is comprised of three axes and is therefore in the shape of a 3-dimensional cube. In this case, the intersection of the plane with the grid are characterized by a collection of points that define a polygon whose areas are used as the 2-dimensional analog of the line segments in the lower dimensional implementation of this finite volume estimation process. The algorithm is outlined below. Figure \ref{fig:2d_Algo} also illustrates the algorithm for a single cuboidal grid section.\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{sample, grid} \\
\Initialization{Initialize \textbf{NumPy} Array of Zeros}\\
\For{s in sample}{
get intersection points\;
\For{p in intersection points}{\
map intersection points to array indices\;
\For{i in indices}{\
assign intersection point to grid location (a point can
belong to more than one grid box)\;}}
\For{m in assigned points}{\
sort points to form polygon and generate the area
}
\For{i in indices}{\
map each polygon's area to the initialized array}
}
\caption{\texttt{transmatrix\_3d()}}
\end{algorithm}
\begin{figure}[!htb]
\begin{tikzpicture}[3d view={120}{15},line join=round, thick,
declare function={a=4;b=2;}]
\draw[dashed] (a,0,-a) -- (0,0,-a)-- (0,a,-a);
\draw[dashed] (0,0,0) -- (0,0,-a);
\draw (a,0,0) -- (a,0,-a) -- (a,a,-a) -- (a,a,-b);
\draw (a,a-b,0) -- (a,0,0) -- (0,0,0) -- (0,a,0) -- (a-b,a,0);
\draw (0,a,0) -- (0,a,-a) -- (a,a,-a);
\draw (a-b,a,0) -- (a,a,0) -- (a,a-b,0);
\draw (a,a,0) -- (a,a,-b);
\fill[cyan, opacity=0.8] (4.5,0.5,1.5) -- (0.5,3.5,1.5) -- (2.75,6.25,-4) -- (6.25,2.75,-4) -- cycle;
\draw[pattern=north east lines, preaction={fill=white, opacity=0.8}] (a,a-0.5*a,-a) -- (a-0.5*a,a,-a) -- (0,a,-a+0.5*a) -- (0,a-0.5*a,0) -- (a,a-0.7*a,0) -- cycle;
\draw (a,0,0) -- (a,0,-a) -- (a,a,-a) -- (a,a,-b);
\draw (a,a-b,0) -- (a,0,0) -- (0,0,0) -- (0,a,0) -- (a-b,a,0);
\draw (0,a,0) -- (0,a,-a) -- (a,a,-a);
\draw (a-b,a,0) -- (a,a,0) -- (a,a-b,0);
\draw (a,a,0) -- (a,a,-b);
\coordinate (p1) at (a,a-0.5*a,-a);
\node at (p1) [fill=red,circle,scale=0.15] {$p1$};
\coordinate (p2) at (a-0.5*a,a,-a);
\node at (p2) [fill=red,circle,scale=0.15] {$p2$};
\coordinate (p3) at (0,a,-a+0.5*a);
\node at (p3) [fill=red,circle,scale=0.15] {$p3$};
\coordinate (p4) at (0,a-0.5*a,0);
\node at (p4) [fill=red,circle,scale=0.15] {$p3$};
\coordinate (p5) at (a,a-0.7*a,0);
\node at (p5) [fill=red,circle,scale=0.15] {$p3$};
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p1);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p2);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p3);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p4);
\draw[-latex,dashed](-8.8,3)node[right,scale=0.8]{$\{p_3,p_2,p_5,p_1,p_4\}$}
to[out=230,in=90] (p5);
\draw[->] (-9,4.5,0) -- (-9,4.5,-0.5);
\coordinate (sort1) at (-9,4.5, -0.9);
\node[draw, scale=0.8] at (sort1) {sort points};
\draw[->] (-9,4.5,-1.25) -- (-9,4.5,-1.75);
\node {} {node[scale=0.8] (T) at (-9,4.5,-2) {%
$\begin{aligned}
\{p_1,p_2,p_3,p_4,p_5\}
\end{aligned}$}};
\draw[->] (-9,4.5,-2.25) -- (-9,4.5,-2.75);
\node {} {node[scale=0.8] (T) at (-9,4.5,-3) {%
$\begin{aligned}
a_{i,j} = Area(\{p_1,p_2,p_3,p_4,p_5\})
\end{aligned}$}};
\matrix (m1) at (-12,4.5,-8) [matrix of math nodes,left delimiter=(,right delimiter=)](A) {
a_{1,1} & a_{1,2} & \dots & a_{1,j} & \dots & a_{1,m}\\
a_{2,1} & a_{2,2} & \dots & a_{2,j} & \dots & a_{2,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
a_{i,1} & a_{i,2} & \dots & a_{i,j} & \dots & a_{i,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
a_{n,1} & a_{n,2} & \dots & a_{n,j} & \dots & a_{n,m}\\
};
\node {} {node (T) at (-12,0,-8.75) {%
$\begin{aligned}
\mathbf{T}=
\end{aligned}$}};
\coordinate (aij) at (-12,4.75,-8);
\draw[-latex,dashed](-6.5,3.5,-2.75)node[right,scale=1]{}
to[out=270,in=90] (aij);
\end{tikzpicture}
\caption{3-D Transformation Matrix Algorithm} \label{fig:2d_Algo}
\end{figure}
The resulting array is in the same form as the one obtained using the 2-dimensional implementation \texttt{transmatrix\_2d()}, and can be used in the following mannner to evaluate the log-likelihood functional in \eqref{eqn:discrete_loglikelihood}
\begin{equation*}
\textbf{T}f_{\boldsymbol{\beta}}^{*} =
\begin{bmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & \hdots & a_{1,m} \\
a_{2,1} & a_{2,2} & a_{2,3} &\hdots & a_{2,m} \\
a_{3,1} & a_{3,2} & a_{3,3} &\hdots & l_{3,m} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n,1} & a_{n,2} &a_{n,3} &\hdots & a_{n,m} \\
\end{bmatrix}
\begin{bmatrix}
{f_\beta}_1 \\
{f_\beta}_2 \\
{f_\beta}_3 \\
\vdots \\
{f_\beta}_m
\end{bmatrix},
\end{equation*}
where in this case each $a_{ij}$ corresponds to the area of the polygonal intersection of the plane with the discrete estimation grid. The resulting product of this matrix and vector is an $n \times 1$ array. Applying an element-wise log-transformation, then taking the sum of this $n \times 1$ array results in the expression in \eqref{eqn:discrete_loglikelihood}.
It is important to note the computational performance of this algorithm. Matrix multiplication parallelizes the computation and eliminates the need to evaluate the log-likelihood functional sequentially for each sample point. However, the process of creating the transformation matrix $\mathbf{T}$ scales linearly with the sample size $n$ and exponentially with the dimensionality of the problem. The function was written in a way that maximizes efficiency as much as possible by making use of Python's parallel computation through array-wise computation when possible.
Solving for the transformation matrix, $\mathbf{T}$ is only one of two steps in the process of estimating the density $f_{\boldsymbol{\beta}}$. The second step is the process of optimizing the regularized log-likelihood functional, the run-time of which scales multiplicatively with respect to the sample size, $n$, and the number of discretezation points, $m$. Therefore, choosing an appropriate sample size and level of discretezation is an important consideration as both primarily determine the total cost of running the algorithm.
\subsection{The \texttt{rmle()} Function}\label{sec:rmle_func}
The \texttt{rmle()} function returns a class named \texttt{`RMLEResult'}. This class stores the solution to the optimization problem, $f_{\boldsymbol{\beta}}^{*}$, and metadata about the solution and the process of optimization. An instance of \texttt{`RMLEResult'} has the following accessible attributes and methods:\\
\noindent \texttt{f}: returns a $m \times 1$ array containing all the estimated function values. It is necessary to reshape the solution before visual representation.\\
\noindent \texttt{f\_shaped}: returns the reshaped array of $f$.\\
\noindent \texttt{dim}: returns an integer which represents how many dimensions $f_{\boldsymbol{\beta}}$ is estimated over.\\
\noindent \texttt{maxval()}: returns a list containing the maximum value of $f_{\boldsymbol{\beta}}^{*}$ and its location. \\
\noindent \texttt{mode()}: returns a list containing possible modes of the density and their locations. \\
\noindent \texttt{ev()}: returns atuple containing the expected value of each $\beta_i$. \\
\noindent \texttt{alpha}: returns a floating-point that specifies the regularization parameter, $\alpha$, used for estimation.\\
\noindent \texttt{alpmth}: returns a string that specifies the method by which the regularization parameter, $\alpha$ was chosen. It can take on the following values: \{`Lepskii', `CV', `User'\}.\\
\noindent \texttt{T}: returns a class object that is created using the \texttt{transmatrix()} function. It contains attributes, methods and subclasses that contain information about the transformation matrix $\mathbf{T}$ and meta-data about it. \\
\noindent \texttt{Tmat}: returns an attribute of the sublcass \texttt{T} but also accessible from the \texttt{RMLEResult} class. It returns the transformation matrix $\mathbf{T}$. \\
\noindent \texttt{grid}: returns a class object that is created using the \texttt{grid\_set()} function. It has attributes and methods that contain information about the grid $\hat{f}_{\boldsymbol{\beta}}$ is estimated over. \\
\noindent \texttt{details}: returns a dictionary containing metadata about the minimization process. \\
The function \texttt{rmle()} serves as a wrapper function for \textbf{SciPy}'s \texttt{minimize} function with \texttt{`trust-const'} set as the minimization algorithm. The choice of this algorithm is crucial as it specializes in large-scale constrained minimization problems, for further details we refer to \cite{byrd1999interior}. The ability to handle large-scale problems is important because depending on the size of the sample, and level of discretezation, the functional evaluations as well as the gradient evaluations could become exceedingly expensive, as it scales with both in a multiplicative manner ($n \times m$). The option to set constraints was also an important consideration. As in equation \eqref{eqn:method} there are two important constraints in estimating $f_{\boldsymbol{\beta}}$, namely: $f\ge 0,\, \text{and } \|f\|_{L^1}=1$. These two constraints ensure the resulting solution satisfies the definition of a density.
Table 1 contains all the arguments for the function \texttt{rmle()} followed by a short description of what they pertain to. More important arguments will be discussed in following subsections. \\
\begin{table}[!h]
\caption{\texttt{rmle()} arguments}
\centering
\begin{tabular}{c ll}
\hline\hline \\ [-1.5ex]
\textbf{Argument} & \textbf{Description} \\\hline
\\ [-1.5ex]
\multicolumn{1}{p{3cm}}{\raggedright functional} & \multicolumn{1}{p{11cm}}{\raggedright Negative likelihood functional with corresponding regularization term} \\
\multicolumn{1}{p{3cm}}{\raggedright alpha} & \multicolumn{1}{p{11cm}}{\raggedright constant $ \geq 0$ that serves as the regularization parameter, or a string matching: `cv' or `lepskii'.}\\
\multicolumn{1}{p{3cm}}{\raggedright tmat} & \multicolumn{1}{p{11cm}}{\raggedright Class object returned by \texttt{transmatrix()} which contains information about the transformation matrix $\mathbf{T}$ and the grid it is estimated over.} \\
\multicolumn{1}{p{3cm}}{\raggedright k} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument: integer which specifies how many folds for modified k-fold cross-validation. Default value is $k = 10$ }\\
\multicolumn{1}{p{3cm}}{\raggedright initial\_guess} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument from the \textbf{SciPy Optimize} minimize function. Used to supply an initial value for the minimization algorithm to start. Default value is set to a \textbf{NumPy} array of values close to zero.} \\
\multicolumn{1}{p{3cm}}{\raggedright hessian\_method} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument from \textbf{SciPy Optimize} minimize function. Default value is set to `2-point'.} \\
\multicolumn{1}{p{3cm}}{\raggedright constraints} & \multicolumn{1}{p{11cm}}{\raggedright Refers to the linear constraints imposed onto the problem. It is set as an optional argument, the default value is set to $\|f\|_{L^1}=1$} \\
\multicolumn{1}{p{3cm}}{\raggedright tolerance} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument for the tolerance criteria for the optimization algorithm's termination. Default value is set to \texttt{1e-6}.} \\
\multicolumn{1}{p{3cm}}{\raggedright max\_iter} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument for the maximum number of iterations. Default value is set to 100.} \\
\multicolumn{1}{p{3cm}}{\raggedright bounds} & \multicolumn{1}{p{11cm}}{\raggedright Refers to the bound constraints of the optimization problem. The default is expressed as $f\ge 0$ } \\
\hline
\end{tabular}
\label{tab:sobolev_args}
\end{table}
\subsubsection{Functionals and Regularization Terms}
Recall the average log-likelihoood functional to be minimized as in equation \eqref{eqn:method}. As specified in section \ref{sec:method} the regularization terms implemented in this module are: the Sobolev norm for $H^1$, the squared $L^2$ norm, and Entropy. The module also includes an option for a functional that has no regularization term if the user wishes to produce a reconstruction of $f_{\boldsymbol{\beta}}$ without any form of regularization. This option will often lead to overfitting, and produce a highly unstable solution.
\subsubsection{Sobolev Norm for $H^1$}
The functional incorporating the Sobolev norm for $H^1$ has the following form,
\begin{align} \label{eqn:h1_penalty}
-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})+\alpha (\|f\|_{2}^{2}+\|f^{\prime}\|_{2}^{2})
\end{align}
where $\|\cdot\|_{2}^{2}$ indicates the squared $L^2$ norm. It is important to note that the choice of the regularization term would typically depend on knowledge about the true solution, as the choice of the regularization term imposes certain assumptions on $f_{\boldsymbol{\beta}}$. In the case of the $H^1$ penalty term, the solution has a square-integrable first derivative in addition to the assumptions of non-negativity and integrability imposed by the constraints of the minimization problem.
The function \pythoninline{sobolev()}, or \pythoninline{sobolev_3d} for the 3-dimensional application returns the value of the discrete implementation of the functional in \eqref{eqn:h1_penalty}. Table 2 provides details on the arguments required for this function. The \textbf{SciPy Optimize} minimize function does not accept array arguments that are greater than one dimension. A necessary step is to unravel the transformation matrix, $\mathbf{T}$, into a one-dimensional array which is passed to the function as \texttt{tm\_long} and simply reshaped into the proper array dimensions. The term $\mathbf{T}f$ is calculated using \textbf{NumPy's} matrix multiplication and sum functions which are much faster alternatives than their non-\textbf{Numpy} counterparts. The regularization term is calculated in \eqref{eqn:h1_penalty}, with the function \texttt{norm\_fprime()} computes for $\|f^{\prime}\|_{2}^{2}$ where $f^{\prime}$ is treated as a total derivative.
\begin{table}[!h]
\caption{\texttt{sobolev(), sobolev\_3d()} arguments}
\centering
\begin{tabular}{c rrrr}
\hline\hline \\ [-1.5ex]
Argument & Notation & Description \\ [0.5ex]
\hline
f & $f_{\boldsymbol{\beta}}^{*}$ & current value of the solution $f_\beta^{*}$ & \\
a & $\alpha$ & constant that serves as the regularization parameter & \\
tm\_long & $\mathbf{T}$ & unraveled form of the transformation matrix, $\mathbf{T}$ &\\
n & n & the sample size & \\
s & $\Delta b$ & the step size of the grid & \\
\hline
\end{tabular}
\label{tab:sobolev_args}
\end{table}
The underlying minimization function is able to approximate the Jacobian of the functional with an additional computational cost. For computational efficiency, we supply an explicit form for the Jacobian of the functional: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+2\alpha (f-f^{\prime\prime})$
The form of this Jacobian implies an additional smoothness assumption on the solution, as it requires $f_\beta$ to be twice differentiable.
\subsubsection{Squared $L^2$ Norm}
The form of the functional in \eqref{eqn:method} that incorporates the squared $L^2$ norm as the regularization term is:
\begin{align*}
-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X}) f+\alpha\|f\|_{2}^{2}
\end{align*}
The arguments for this function are indentical to those listed in Table \ref{tab:sobolev_args}, likewise is true for the Jacobian associated with this regularization term. The Jacobian has the following form: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+2\alpha f$
Choosing this regularization functional imposes less smoothness assumptions on the solution, as the only additional assumption in place is square-integrability. This leads to a typically less smooth reconstruction as compared to using the Sobolev norm for $H^1$ as the regularization functional. The functions in python are coded similarly as with the $H^1$ regularization functional.
\subsubsection{Entropy}
The form of the functional in \eqref{eqn:method} that incorporates the entropy of the function has the following form: $-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})+\alpha \int f \log(f)db$
This functional has the least amount of assumptions on the solution, $f_{\boldsymbol{\beta}}$. It only requires finite entropy which is a weak assumption in addition to the non-negativity and $L^1$ constraints of the minimization problem.
The Jacobian of the entropy functional also does not impose any additional assumptions on the solution, and has the following form: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+\alpha (\log f+1)$
\subsubsection{Parameter Selection}
Recall the minimization problem as in \eqref{eqn:method} where a constant $\alpha \geq 0$ controls the size of the effect of the regularization term. The user can provide the $\alpha$ value directly if the user has a general idea of the level of smoothing necessary for the solution. If the user has no best-guess for the value of the regularization parameter alpha, the module has two options to automatically select the parameter $\alpha$, namely: Lepskii's balancing principle, and k-fold cross-validation. The Lepskii method typically yields a less accurate result relative to k-fold cross-validation; however, its advantage lies in significantly less computational cost.
\subsubsection{Lepskii's Balancing Principle}
Lepskii's principle is an adaptive form of estimation which is popular for inverse problems, e.g. \cite{Tsybakov:00}, \cite{BauHoh:05}, \cite{mathe_2006}, \cite{hohage2016inverse}, \cite{Werner:18}. It is significantly computationally less expensive than other parameter selection methods.
The method works as follows: we compute $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{1}},\ldots,\hat{f_{\boldsymbol{\beta}}}_{\alpha_{m}}$ for $\alpha_1 = c_{L_{n}} \frac{\ln(n)}{\sqrt{n}}$ and $\alpha_{i+1} = r\alpha_i$ with some constants $c_{L_{n}} >0,r>1$. We then select $\alpha_{j}$ as the optimal parameter choice where:
\[
j_{bal}:=\max\{j\leq m, \|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\| \leq 8 r^\frac{1-i}{2}, \text{ for all } i < j\}.
\]
The algorithm is implemented in python as follows: \\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{$\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$} \\
\Initialization{Generate the transformation matrix $\mathbf{T}$}\\
\For{$\alpha_i$ in $\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$}{
Compute for $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{i}}$ using $\mathbf{T}$\;
}
\For{j in $\{1,2,\hdots,m\}$}{\
Set $i = 0$ \;
\While{i $<$ j}{\
Check if $\|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\|\leq 8 r^\frac{1-i}{2}$ \;
\If{$\|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\|> 8 r^\frac{1-i}{2}$}{\
$j_{bal} = j$ \;
\textbf{break}
}
$i+=1$
}
}
\caption{Lepskii's Balancing Principle}
\end{algorithm}
\medskip
The bulk of the computational cost of the Lepskii algorithm implementation can be broken down into two components: the fixed cost of generating $\mathbf{T}$, and the variable cost of computing $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{1}},\ldots,\hat{f_{\boldsymbol{\beta}}}_{\alpha_{m}}$ as the number of $\alpha$ values to be used depends on $c_{L_{n}} >0 \text{, }r>1$, and also the sample size of the data. This implementation of Lepskii algorithm's scales linearly in terms of runtime with the number of $\alpha$ values being tested, $m$.
\subsubsection{K-Fold Cross Validation}
Cross-validation is another popular parameter choice rule. Here we present a modified implementation of k-fold cross-validation with a cost-function that is applicable to our problem. The modification we apply is an algorithm to lessen the computation time by reducing the number of $\alpha$ values it needs to iterate through.
The loss function we considered was,
\begin{align*}
J_{\alpha_{i}} = -\sum_{j=1}^{k} \log \mathbf{T}_{j}\hat{f}_{{\boldsymbol{\beta}}_{-j}}
\end{align*}
Where $\mathbf{T}_{j}$ is the transformation matrix generated from a subsample of the observations, which can be interpreted as the $j$-th fold that is left out in the current iteration, and $\hat{f}_{{\boldsymbol{\beta}}_{-j}}$ is the estimate for $f_{\boldsymbol{\beta}}$ using $\mathbf{T}_{-j}$. The loss function can be interpreted as the negative of the likelihood that the $j$-th fold of the sample used to generate $\mathbf{T}_{-j}$ was drawn from the distribution, as the subsample fold used to produce $\mathbf{T}_{j}$. We aim to choose the $\alpha$ that minimizes this loss function.
The search method for the optimal $\alpha$ value reduces the number of $\alpha$ values tested. The algorithm involves separating the range of $\alpha$ values into two sections $\{\alpha_{1},\ldots,\alpha_j\}$ and $\{\alpha_{j+1},\ldots, \alpha_{m}\}$. Two alpha values $\alpha_a$, and $\alpha_b$ are randomly selected from the respective sections and are used to compute for the corresponding loss function values, $J_{\alpha_{a}}$ and $J_{\alpha_{b}}$. The section from which the $\alpha$ value that produces the smaller $J_{\alpha}$ was drawn from is kept, while the other is discarded. This is repeated until there is a sufficiently small range of $\alpha$ values. Once this range of $\alpha$ values is obtained, the loss function is evaluated over all the remaining $\alpha$ values and the optimal $\alpha$ is chosen as the one which minimizes $J_{\alpha}$. The complete algorithm is implemented as follows:\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{$\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$} \\
\Initialization{Generate the transformation matrix $\mathbf{T}$ and apply a random shuffle, set $\boldsymbol{\alpha}$ = $\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$}\\
\While{\texttt{len}($\boldsymbol{\alpha}$) $>$ 3}{
1. Set $\boldsymbol{\alpha_a}$ = $\{\alpha_1,\alpha_2,\hdots,\alpha_j\}$ \;
2. Set $\boldsymbol{\alpha_b}$ = $\{\alpha_{j+1},\alpha_{j+2},\hdots,\alpha_m\}$ \;
3. Randomly select $\alpha_a$ and $\alpha_b$ from $\boldsymbol{\alpha_a}$ and $\boldsymbol{\alpha_b}$ respectively. \;
4. Evaluate $J_{\alpha_{a}}$ and $J_{\alpha_{b}}$ \;
\If{$J_{\alpha_{a}}$ $<$ $J_{\alpha_{b}}$}{
Set $\boldsymbol{\alpha}$ = $\boldsymbol{\alpha_a}$
}
\Else{
Set $\boldsymbol{\alpha}$ = $\boldsymbol{\alpha_b}$
}
}
\For{$\alpha_i$ in $\boldsymbol{\alpha}$}{\
Compute for $J_{\alpha_{i}}$ \;
}
Choose $\alpha_{cv} = \argmin{J_{\alpha_{i}}}$
\caption{Modified K-fold Cross Validation}
\end{algorithm}
\medskip
The runtime of the unmodified version of k-fold cross-validation scales linearly with the product $k \times m$ where $k$ is the number of folds and $m$ is the number of $\alpha$ values being tested. Applying the modified version reduces the number of $\alpha$ values being tested, $m$, by some logarithmic factor, which is a significant reduction in computational cost which makes cross-validation more computationally feasible.
\section{Examples} \label{sec:example}
The following examples will be demonstrated in this section: the random coefficients model with a single regressor and random intercept for the two-dimensional case, and the two regressors and random intercept for the three-dimensional case. This section will also show how to plot the estimated density using the built in plotting function \pythoninline{plot\_rmle()} which makes use of functions from the \pythoninline{matplotlib} library. It will also show how to plot the density without the use of the built-in function in case the user wishes to explore different plotting options.
\subsection{Example 1: 2-D Case Single Regressor with Random Intercept}
The first example is the case described by \eqref{eqn:rc_model_2d}. The example is demonstrated with simulated data using the function \pythoninline{sim\_sample()}. This function simulates the regressor $X_1 \sim \pazocal{U}(-2,2)$ and the random coefficients $\beta_0$, $\beta_1$ from a bimodal multivariate normal mixture as follows,
\begin{align*}
0.5\mathcal{N}([-0.5,-0.5], 0.01\mathbbm{I}_2) + 0.5\mathcal{N}([0.5,0.5], 0.01\mathbbm{I}_2).
\end{align*}
The general flow of the process of using the module can be broken down in five steps:
\begin{enumerate}[topsep=0pt,itemsep=-1ex,partopsep=1ex,parsep=1ex]
\item Import the necessary modules and functions.
\item Establish the dataset to be used (either real or simulated data).
\item Specify the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is to be estimated over.
\item Generate the transformation matrix $\mathbf{T}$.
\item Run the \pythoninline{rmle()} function.
\end{enumerate}
\begin{python}
from pyrmle import *
sample = sim_sample(n = 10000,dim = 2)
\end{python}
\medskip
The program begins by importing the necessary functions from \pythoninline{pyrmle.py} which contains the high-level functions that the user interacts with. The module \pythoninline{pyrmle_funcs} contains the underlying functions necessary for functions in the main module \pythoninline{pyrmle} to run. The next step is to define the sample to be used in creating the transformation matrix $\mathbf{T}.$ The sample has the same form as described in subsection \ref{sec:trans_matrix}, where the sample has the form $[\mathbf{X}_0, \mathbf{X}_1, \mathbf{Y}]$. In Python it takes the shape of $10000 \times 3$ \pythoninline{NumPy} array as seen below.
\medskip
\begin{python}
[[ 1. , -0.76765455, 2.10802347],
[ 1. , 1.51774991, -0.11337413],
...,
[ 1. , -1.80486996, -3.08120151]]
\end{python}
\medskip
The next step is to generate the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is estimated over. This is done using the \pythoninline{grid\_set()} function, as discussed briefly in \ref{sec:trans_matrix}. In this example we set the ranges of $\beta_0$ and $\beta_1$ to $[-10,10]$ which defines a two-dimensional grid spanning that range in each axis. This function creates an instance of the \pythoninline{class grid_obj} which has attributes and methods enumerated and described in subsection \ref{sec:trans_matrix}. When the \pythoninline{grid} class instance has been created, the user can proceed to generate an instance of the \pythoninline{class tmatrix} using the \pythoninline{transmatrix()} function.
\medskip
\begin{python}
grid_beta = grid_set(num_grid_points = 40, dim = 2)
print(grid_beta.numgridpoints())
1600
T = transmatrix(sample, grid_beta)
\end{python}
\medskip
After the instance of the transformation matrix is produced, the user can then run the \pythoninline{rmle()} function. As stated in subsection \ref{sec:rmle_func}, the \pythoninline{rmle()} function has three essential arguments: \{\texttt{`functional',`alpha',`tmat'}\}.
\medskip
\begin{python}
result = rmle(sobolev,0.25,T)
print(result.ev())
[-0.002865326246726808, -0.010416375635973668]
print(result.mode()[:2])
[[0.39681799850755783, [-0.625, -0.375]],
[0.38831830923870914, [0.625, 0.375]]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_40_grid_base_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 40 grid points}\
\label{fig:cont_40_base}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_40_grid_base_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 40 grid points}
\label{fig:surface_40_base}
\end{minipage}
\end{figure}
As stated in subsection \ref{sec:rmle_func}, the \pythoninline{rmle()} function generates an instance of \pythoninline{class RMLEResult}. This class has a number of useful and informative attributes and methods that describe the estimated density. The \pythoninline{ev()} method returns the expected value of each ${\boldsymbol{\beta}}_j$, while the \pythoninline{mode()} method returns possible maxima of the estimated density. The \pythoninline{mode()} method relies on a naive search method for maxima with no underlying statistical tests.
Figure 3 (reference) shows the contour plot produced by the \pythoninline{plot_rmle()} function. It is clear that there is a large portion of the grid that is essentially unused, and the user could benefit from a reduction in the grid-size in terms of computational costs. This can be done by reducing the number of grid points and shrinking the range of each axis. In terms of tuning the size of the grid, we suggest that the user tries a relatively large range to begin with to ensure that the grid contains the support of $\hat{f_{\boldsymbol{\beta}}}$, and then consider smaller grid ranges. Having a grid range that is too small has a negative effect on the estimate as the optimization algorithm enforces the constraint that $\|\hat{f_{\boldsymbol{\beta}}}\|_{L^1}=1$.
\medskip
\begin{python}
grid_beta_alt = grid_set(20,2,B0_range=[-1.5,1.5],\
B1_range=[-1.5,1.5])
print(grid_obj_alt.numgridpoints)
400
T2 = transmatrix(sample, grid_beta_alt)
result2 = rmle(sobolev,0.15,T2)
print(result2.ev())
[-0.004977073000670898, -0.003964663139211258]
print(result2.mode()[:2])
[[0.3643228046077392, [0.5249999999999, 0.5249999999999]],
[0.3580092260598125, [-0.5249999999999, -0.5249999999999]]]
plot_rmle(result2)
plot_rmle(result2,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_smaller_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on a smaller grid}\
\label{fig:20_smaller_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_smaller_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a smaller grid}
\label{fig:20_smaller_surface}
\end{minipage}
\end{figure}
The reduction in the number of grid points resulted in a significant reduction in the run time of the algorithm while achieving a better estimate for $\hat{f_{\boldsymbol{\beta}}}$. With the reduction in the computational cost of the algorithm makes it more favorable to run an automatic parameter choice method.
\begin{python}
result_cv = rmle(sobolev,'cv',T2)
print(result_cv.alpha)
0.07065193045869372
print(result_cv.ev())
[-0.004977073000670898, -0.003964663139211258]
print(result_cv.mode()[:2])
[[0.3643228046077392, [0.5249999999999, 0.5249999999999]],
[0.3580092260598125, [-0.5249999999999, -0.5249999999999]]]
plot_rmle(result_cv)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_cv_grid_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on a smaller grid using cross-validation}\
\label{fig:20_cv_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_cv_grid_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a smaller grid using cross-validation}
\label{fig:20_cv_surface}
\end{minipage}
\end{figure}
The general workflow that we suggest when tuning the parameters to be used in estimation is as follows:
\begin{enumerate}[topsep=0pt,itemsep=-1ex,partopsep=1ex,parsep=1ex]
\item Establish a relatively large grid range for estimation and generate the transformation matrix $\mathbf{T}$. This should be treated as an exploratory step in terms of analyzing the data.
\item Set $\alpha$ equal to the step size of the grid.
\item Run the \pythoninline{rmle()} function.
\item Plot $\hat{f_{\boldsymbol{\beta}}}$ using the \pythoninline{plot_rmle()} function and determine the necessary grid range.
\item Limit the grid range as well as the grid points to reduce computation costs and genrate the new matrix $\mathbf{T}^*$.
\item Run the \pythoninline{rmle()} function with $\mathbf{T}^*$, and optionally employ one of the two automatic parameter selection methods: \{\pythoninline{'cv','lepskii'}\}.
\end{enumerate}
The following example will demonstrate usage of the \pythoninline{grid_set()} function in terms of supplying a different range $\beta_{1}$. The simulated data in this case will have modes for $\beta_1$ that are significantly larger than that of $\beta_0$ and are not encapsulated by the default range $[-5,5]$. The betas are sampled from the following distribution: $0.5\mathcal{N}([-1.5,6], \mathbbm{I}_2) + 0.5\mathcal{N}([1.5,9], \mathbbm{I}_2).$
\begin{python}
cov = [[[1, 0], [0, 1]],[[1, 0], [0, 1]]]
mu = [[-1.5,6],[1.5,9]]
sample = sim_sample(10000,2,beta_mu = mu,beta_cov = cov))
grid_beta_shifted = grid_set(num_grid_points = 20, \
dim = 2, B1_range=[2,13])
T_shifted = transmatrix(sample, grid_beta_shifted)
result_shifted = rmle(sobolev_norm_penal,0.5,T_shifted)
print(result_shifted.ev())
[0.03520704073478552, 7.524001743037029]
print(result.mode()[0:2])
[[0.07133616078580148, [1.25, 8.875]],
[0.0652140364538059, [-1.75, 5.574999999999999]]]]
plot_rmle(result_shifted)
plot_rmle(result_shifted,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_shifted_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on shifted grid}\
\label{fig:20_shifted_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_shifted_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a shifted grid}
\label{fig:20_shifted_surface}
\end{minipage}
\end{figure}
\subsubsection{Example 2: 3-D Case Two Regressors with Random Intercept}
This second example is the case described by \eqref{eqn:rc_model_3d}. The example is demonstrated likewise with simulated data using the same function \pythoninline{sim_sample()}. The regressors are simulated as follows, $X_1, X_2$ are i.i.d $\pazocal{U}(-2,2)$, and the random coefficients $\beta_0, \beta_1, \text{ and } \beta_2$ are simulated from $\mathcal{N}([2,2,2], 0.01\mathbbm{I}_3)$
\medskip
\begin{python}
from pyrmle import *
sample = sim_sample(n = 10000,dim = 3)
\end{python}
\medskip
As in the previous example, the program begins by importing the necessary modules and functions. The \pythoninline{sim_sample()} function generates sample observations based on the aforementioned distributions. This results in a $10,000 \times 4$ \pythoninline{NumPy} array.
\medskip
\begin{python}
grid_beta = grid_set(10,3,B0_range=[-2,4],\
B1_range=[-2,4],B2_range=[-2,4])
print(grid_beta.numgridpoints())
1000
T = transmatrix(sample,grid_beta)
\end{python}
\medskip
The next step is to establish the number of grid points, and to generate the transformation using the simulated sample observations. In this case, we first consider ten grid points in each axis amounting to a total of 1000 grid points. If the user wishes to estimate $\hat{f_{\boldsymbol{\beta}}}$ over a finer grid it would be more efficient to first determine the smallest possible range of each axis that would fit almost all of the probability mass of $\hat{f_{\boldsymbol{\beta}}}$.
\medskip
\begin{python}
result = rmle(sobolev_norm_penal2d,0.6,T)
print(result.ev())
[2.458436489282234, 2.25500373629305, 1.9058740990043983]
print(result.mode())
[0.08926614291105403, [1.90000000, 1.90000000, 1.90000000]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_10_grid_cont}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_10_grid_surface}
\end{figure}
In the three dimensional application we use the regularization functional \pythoninline{sobolev_3d} and supply it as an argument to the \pythoninline{rmle()} function along with the transformation matrix and $\alpha$. The results show the effect of the level of discretezation on the estimate. It is possible to achieve more accurate estimates with more grid points in conjunction with a narrower grid range, but with a significantly higher computational cost.
\medskip
\begin{python}
grid_beta_alt = grid_set(20,3,B0_range=[0,3],\
B1_range=[0,3],B2_range=[0,3])
print(grid_beta.numgridpoints())
8000
T2 = transmatrix(sample,grid_beta_alt)
result2 = rmle(sobolev_norm_penal2d,0.3,T2)
print(result2.ev())
[2.102855111361121, 2.077365765266784, 1.9697452677152947]
print(result2.mode()[0])
[[0.2008050879224736, [2.025, 2.025, 2.025]]
plot_rmle(result2)
plot_rmle(result2,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_cont}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_surface}
\end{figure}
In this example, the number of grid points $\hat{f_{\boldsymbol{\beta}}}$ is estimated over is set to 20 on each axis which amounts to a total of 8,000 grid points. The additional level of discretezation due to the increased number of grid points and the smaller grid range provided resulted in an estimate $\hat{f_{\boldsymbol{\beta}}}$.
The next example will demonstrate the case when the underlying joint density being reconstructed has a single mode at the center of the grid. Both methods of dealing with this issue described in section \ref{sec:trans_matrix} will be illustrated.
\medskip
\begin{python}
mu = [[0,0,0],[0,0,0],[0,0,0]]
sample = sim_sample(n = 5000,dim = 3, \
beta_mu = mu)
grid_beta = grid_set(20,3,B0_range=[-1,2],\
B1_range=[-1.5,1.5],B2_range=[-1.5,1.5])
T = transmatrix(sample,grid_beta)
result = rmle(sobolev_3d,0.15,T)
print(result.ev())
[-0.19254733276928582, 0.005554898823827626, -0.009788891289943112]
print(result.mode())
.14336637945812933, [-0.024999999999999967, 0.075, -0.075]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_zero_grid_contour}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_zero_grid_surface}
\end{figure}
The following example demonstrates how to apply the shifting algorithm briefly mentioned in section \ref{sec:trans_matrix}. The algorithm is implemented in python by adding $c \sim \pazocal{U}(a,b)$ to the intercept, where $a$ and $b$ are determined by the size of the grid. This applies a transformation that can be back-transformed after estimation by adjusting the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is estimated over. This is repeated across ten different shifted samples, then a simple k-means clustering algorithm is applied using \pythoninline{sklearn.cluster.KMeans()} on the $L2$ penalties of \{$\hat{f_{\boldsymbol{\beta}}}_1, \hdots, \hat{f_{\boldsymbol{\beta}}}_{10}$\}. The reconstruction, $\hat{f_{\boldsymbol{\beta}}}_j$, that is closest to the centroid of the largest cluster is then output as the solution.
\begin{python}
grid_beta = grid_set(20,3,B0_range=[-1.5,1.5],\
B1_range=[-1.5,1.5],B2_range=[-1.5,1.5])
T = transmatrix(sample,grid_beta)
result_shift = rmle(sobolev_3d,0.15,T,shift=True)
print(result_shift.ev())
[-0.12974928815245057, -0.004493337754614205, -0.0052980597609372385]
print(result_shift.mode())
[[0.14624599807641833, [0.009524681155561987, -0.075, -0.075]]
plot_rmle(result)
plot_rmle(result_shift,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_contour_shift}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_surface_shift}
\end{figure}
The final example demonstrates how to use the module on a real dataset, and shows the user how to load the data into Python from a comma-separated value (CSV) format and how to pre-process it, if necessary, into a usable form. The data used here are from the British Family Expenditure Surey that was used in \cite{Dunker2019,dunker2021regularized}.
The model is identified as follows:
\begin{align*}
BS_i = \beta_{0,i} + \beta_{1,i}\ln(TotalExpenditure_i) + \beta_{2,i}\ln(FoodPrices_i) + \epsilon_i
\end{align*}
In the case of this dataset the regressors need to undergo a linear transformation before it generating the transformation matrix to be used in the algorithm. This is to ensure that the grid is sufficiently covered by the hyperplanes generated by the sample observations. For more details (refer to the section of our paper discussing this issue).
The OLS estimate for the random coefficients suggests $f_{\boldsymbol{\beta}}$ is centralized at (0.262755,0.0048,-0.00069) for a subsample size of $n = 5000$, which requires the application of one of the two methods described above to circumvent the problem that arises from a mode close to $(0,0,0)$.
\medskip
\begin{python}
data = np.genfromtxt('filename.csv', delimiter=',')
data = data[1:-1,1:4]
data = data[np.random.randint(1,len(data),5000),:]
data = data[~np.isnan(data).any(axis=1)]
ones = np.repeat(1,len(data))
real_data_sample = np.c_[ones, \
25*data[:,2]-0.3,data[:,1]-5,data[:,0]]
\end{python}
\medskip
The code block above loads the data from the a csv file, selects a random subsample of 5,000 observations, and applies the linear transformations on the data necessary.
\medskip
\begin{python}
grid_beta = grid_set(20,3,B0_range=[-0.75,1.25],\
B1_range=[-1,1],B2_range=[-1,1])
T = transmatrix(real_data_sample,grid_beta)
result = rmle(sobolev_rmle_2d,0.25,T,shift=True)
print(result.ev())
[0.2139255519131858, -0.0022826617746834394, -0.10875379495199464]
print(result.mode()[0:1])
[[0.19533254824055002, [0.3, -0.05, -0.15000000000000002]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_contour}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_surface}
\end{figure}
Included in the module is an option to fit a spline on the estimate $\hat{f_\beta}$. This would be the most computationally feasible option if the user needs a finer grid. The \pythoninline{spline_fit()} function takes two essential arguments. The first positional argument is the \pythoninline{class RMLEResult} object, and second is the number of grid points on each axis.
\begin{python}
spline = spline_fit(result,num_grid_points = 400)
plot_rmle(spline)
plot_rmle(spline,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f12.png}
\caption{Contour plots of the spline interpolated joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_contour_spline}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f12.png}
\caption{Surface plots of the spline interpolated joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_surface_spline}
\end{figure}
\bibliographystyle{apalike}
\section{Introduction} \label{sec:method}
The Random Coefficients model is often used to model unobserved heterogeneity in a population, an important problem in econometrics and statistics. The model is given by
\begin{equation}\label{eqn:rc_model}
Y_i=\beta_{0i}+\beta_{1i} X_{1i}+\beta_{2i} X_{2i}+\ldots+\beta_{di} X_{di}.
\end{equation}
Where $\mathbf{X}_i = (1, X_{1i}, X_{2i}, \ldots X_{di})^\top$ and ${\boldsymbol{\beta}}_i = (\beta_{0i}, \beta_{1i}, \beta_{2i}, \ldots, \beta_{di})^\top$ are the regressors, and regression coefficients respectively. It is assumed that $\mathbf{X}_i$, $Y_i$, ${\boldsymbol{\beta}}_i$ are i.i.d random variables with ${\boldsymbol{\beta}}_i$ and $\mathbf{X}_i$ being independent, and that $Y_i$ and $\mathbf{X}_i$ are observed while ${\boldsymbol{\beta}}_i$ is unknown.
In this paper we introduce an open source \texttt{Python} module named \pythoninline{PyRMLE} which implemenents regularized maximum likelihood estimation (RMLE) to nonparametrically estimate the density $f_{\boldsymbol{\beta}}$.
Nonparametric estimation of the random coefficients model was first established by \cite{Beran1992} for a single regressor. A kernel method for estimation was developed by \cite{Beran1996}. The optimal rate for design densities with Cauchy-type tails is derived in \cite{hoderlein2010} for a kernel method.
Equality constrained linear regression methods were developed by \cite{fox2011simple, heiss2021nonparametric}. Regularized maximum likelihood methods have been considered for problems other than the random coefficients model specified in \eqref{eqn:rc_model} in \cite{WH:12,HW:13, hohage2016inverse, Dunker2014}. The method implemented by the \pythoninline{PyRMLE} module is the one developed in \cite{dunker2021regularized}.
\texttt{Python} was chosen as the programming language mainly for the extensive scientific and computational libraries at its disposal, namely: \pythoninline{NumPy, SciPy} by \cite{harris2020array,jones2001scipy}. The module takes advantage of the benefits of working with \pythoninline{NumPy} arrays in terms of computational efficiency achieved by doing array-wise computation. Another advantage is that there are no software imposed limits in terms of array size. The maximum array size in \texttt{Python} is solely determined by the amount of RAM available to the user, which allows the user the flexibility to increase the computational complexity of the method to the level that their system allows. The module also uses a trust-region constrained minimization algorithm developed by \cite{byrd1999interior} which is implemented in \pythoninline{SciPy}.
The paper is organized as follows: Section \ref{sec:method} briefly describes the regularized maximum likelihood method developed in \cite{dunker2021regularized}, Section \ref{sec:PythonImplementation} discusses the classes and functions available to the module, and Section \ref{sec:example} discusses examples of the modules usage for general cases.
\section{Regularized Maximum Likelihood} \label{sec:method}
It is assumed that the random coefficients, ${\boldsymbol{\beta}}$, have a Lebesgue density $f_{\boldsymbol{\beta}}$. If the conditional density $f_{Y|\mathbf{X}}$ exists, the two densities are connected by the integral equation
\begin{align*}
f_{Y|\mathbf{X}}(y|\mathbf{X}=\mathbf{x})=\int_{\mathbbm{R}^{d}}\mathbbm{1}\big\{\mathbf{b}^\top \mathbf{x}=y\big\}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b}) = \int_{\mathbf{b}^\top \mathbf{x}=y}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b})
\label{eqn:line_int_f_beta}
\end{align*}
This connection allows us to employ maximum likelihood estimation to nonparametrically identify $f_{\boldsymbol{\beta}}$ as seen in the following expression of the log-likelihood
\begin{equation*}
\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})=\frac{1}{n}\sum_{i=1}^{n}\log\left[\int_{\mathbbm{R}}\mathbbm{1}\big\{{\boldsymbol{\beta}}_i^\top \mathbf{X}_i=Y_i\big\}f_{\boldsymbol{\beta}}(b)d\mu(\mathbf{b})\right].
\label{eqn:avg_lhood}
\end{equation*}
Direct maximization of $\bar\ell(f_{\boldsymbol{\beta}}|Y,\mathbf{X})$ over all densities it not feasible due to ill-posedness and will lead to overfitting. We stabilize the problem by adding a penalty term $\alpha \mathcal{R}(f_{\boldsymbol{\beta}})$ and state the estimator as a minimization problem with negative log-likelihood
\begin{align}\label{eqn:method}
\hat {f_{\boldsymbol{\beta}}}_\alpha = \argmin_{f\ge 0,\, \|f\|_{L^1}=1} -\bar\ell(f|Y,\mathbf{X})+\alpha \mathcal{R}(f).
\end{align}
Here $\alpha \ge 0$ is a regularization parameter that controls a bias variance trade-off. It was poined out in \cite{heiss2021nonparametric} that the constraint $\|f_{\boldsymbol{\beta}}\|_{L^1}=1$ together with a finite difference discretization of $f_{\boldsymbol{\beta}}$ is equivalent to an $\ell^1$ penalty on the discretized values of $f_{\boldsymbol{\beta}}$, which could cause unwanted shrinkage of the estimate. To reduce this LASSO effect, \cite{heiss2021nonparametric} introduced an additional quadratic constraint which turns the method into an elastic net. In \cite{dunker2021regularized} it was stated that the regularization term $\alpha\mathcal{R}(f_{\boldsymbol{\beta}})$ is analogous to this additional constraint but is more flexible as the method is not limited to quadratic $\mathcal{R}$.
The implemented regularization terms $\mathcal{R}$ in this module are: (1) squared $L^2$ norm $\mathcal{R}(f) = \|f\|_{L^2}^2 = \|f\|_{2}^2$, (2) the Sobolev Norm for $H^{1}$ $ \mathcal{R}(f) = \|f_\beta\|_{2}^{2}+\|f^{\prime}_\beta\|_{2}^{2}$, and (3) entropy $\mathcal{R}(f) = \int{f(\mathbf{b})\ln{f(\mathbf{b})}d\mathbf{b}}$.
In addition to the regularization functional, a regularization parameter $\alpha$ also needs to be chosen. In this module we implement two methods of estimating $\alpha$: Lepskii's Balancing principle, and K-fold cross validation.
\section{Python Implementation} \label{sec:PythonImplementation}
The \pythoninline{PyRMLE} module's implementation of regularized maximum likelihood is limited to applications with up to two regressors for the random coefficients model with intercept, and up to three regressors for a model without intercept.
There are two main functions used to implement regularized maximum likelihood estimation using \pythoninline{PyRMLE}, namely: \pythoninline{transmatrix()} and \pythoninline{rmle()}. There are other sub-functions necessary to the implementation, these will be discussed under the appropriate subsections when relevant.
\subsection{The \texttt{transmatrix()} Function}\label{sec:trans_matrix}
The purpose of the function \texttt{transmatrix()} is to construct the discrete version of the linear operator given by
\begin{equation}
T f_\beta = \int_{\mathbf{b}^\top \mathbf{x}=y}f_{\boldsymbol{\beta}}(\mathbf{b})d\mu_d(\mathbf{b})
\label{eq:inv_prob}.
\end{equation}
The linear operator above describes the integral of $f_{\boldsymbol{\beta}}$ over the hyperplanes parametrized by the sample points $\mathbf{X}, \text{ and }Y$. The function makes use of a finite-volume method as a discrete analog in evaluating the integral. The function is used as follows
\begin{lstlisting}
trans_matrix = transmatrix(sample,grid)
\end{lstlisting}
The argument \texttt{sample} corresponds to the sample observations. The sample data should be in the following format: \\
\begin{center}
$\begin{bmatrix}
X_{0,1} & X_{1,1} & \hdots & Y_1 \\
X_{0,2} & X_{1,2} & \hdots & Y_2 \\
\vdots & \vdots & \ddots & \vdots \\
X_{0,n} & X_{1,n} & \hdots & Y_n \\
\end{bmatrix}$
\end{center}
In the case of a random coefficients model with intercept the first column would simply be $\mathbf{X}_0 = (1,1,\hdots,1)^{T}$.
The \texttt{grid} argument is a class object generated by the \texttt{grid\_set()} function. It has the following attributes and methods: \{ \texttt{scale}, \texttt{shifts}, \texttt{interval}, \texttt{dim}, \texttt{step}, \texttt{start}, \texttt{end}, \texttt{ks()}, \texttt{numgridpoints()}\}. The \texttt{grid\_set()} function is used as follows:
\begin{lstlisting}
grid_beta = grid_set(num_grid_points=20,dim = 2)
\end{lstlisting}
The base grid that is generated by the \texttt{grid\_set()} function is a symmetric grid that spans $[-5,5]$ in each axis. The user inputs the number of grid points by passing an integer value to the function as \pythoninline{num_grid_points}. This specifies the step size of the grid as $\frac{10}{k}$ where $k$ is the number of grid points along each axis. Additionally, the user can change the range over which each axis is defined by supplying new axes ranges through the arguments: {\pythoninline{B0_range, B1_range, B2_range}} which are passed as lists or arrays that contain the end points of the new range (e.g. \pythoninline{B0_range = [0,10]} ). This is especially useful if the user expects a random coefficient to be significantly larger or smaller than the other random coefficients.
\begin{python}
grid_beta_shifted = grid_set(num_grid_points = 20, \
dim = 2, B0_range = [0,10])
print(grid_beta_shifted.shifts)
[-5,0,0]
\end{python}
The output of the \texttt{transmatrix()} function is the \texttt{`tmatrix'} class object that has the following attributes and methods: \\
\noindent \texttt{Tmat}: returns a $n \times m $ \pythoninline{NumPy}-array that is produced by the function \texttt{transmatrix\_2d()} or \texttt{transmatrix\_3d()}. \\
\noindent \texttt{grid}: returns the \pythoninline{class grid_obj}. \\
\noindent \texttt{scaled\_sample}: returns the scaled and shifted sample.\\
\noindent \texttt{sample}: returns the original sample. \\
\noindent \texttt{n()}: returns the number of sample observations. \\
\noindent \text{m()}: returns the number of grid points $\hat{f}_{\boldsymbol{\beta}}$ is to be estimated over.
\subsubsection{\texttt{transmatrix\_2d()}}
The 2-dimensional implementation of this method works for the random coefficients model with a single regressor and random intercept
\begin{equation}
y_i={\beta_0}_i+{\beta_1}_{i}{x_1}_i,
\label{eqn:rc_model_2d}
\end{equation}
and the model with two regressors and no intercept.
\begin{equation}
y_i={\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i+\epsilon_i
\label{eqn:rc_model_2d_no_intercept}
\end{equation}
The function first initializes an $n \times m$-dimensional array of zeros, where $n$ is the sample size, and $m$ is the number of grid points $f_{\boldsymbol{\beta}}$ is to be estimated over. In the 1-dimensional case, the hyperplanes which $f_{\boldsymbol{\beta}}$ is integrated over reduce to lines which simplifies the problem. A finite-volume type method of estimation is employed to approximate the integral expression as seen in \ref{eq:inv_prob}. This method is akin to the algebraic reconstruction methods used in Computed Tomography. Specifically, it is reminiscent to the discrete Radon Transform methodology laid out by \cite{beylkin1987discrete}.
To implement this finite-volume method, the lines parametrized by the sample points given by equations \eqref{eqn:rc_model_2d} or \eqref{eqn:rc_model_2d_no_intercept} are made to intersect with the grid. The length of each line segment that intersects with the grid is then calculated and stored in an object. The intersection points of these lines with the grid are retrieved and subsequently mapped to their respective array indices. These indices are used to map the length of the line segments to the appropriate entry in the initialized array forming the discretized linear operator $T$. The algorithm is outlined as follows:\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{sample, grid} \\
\Initialization{Initialize \textbf{NumPy} Array of Zeros}\\
\For{s in sample}{
1. get intersection points\;
2. get line segment lengths\;
3. map intersection points to their array indices\;
\For{i in indices}{\
map line segment lengths to initialized array using index i
}
}
\caption{\texttt{transmatrix\_2d()}}
\end{algorithm}
\medskip
The result of this function is a large, sparse array, $\mathbf{T}$ where each row corresponds to a collection of all the lengths of the line segments intersecting the grid for a line parametrized by a sample point, i.e. each $l_{ij}$ corresponds to the length of the line segment intersecting the grid at section $i,j$. The resulting array is sparse because $l_{ij}$ is equal to zero unless a line passes through a section of the grid. The algorithm's implementation is illustrated in figure \ref{fig:1d_Algo}.
The resulting array is then used to evaluate the log-likelihood functional to be optimized
\begin{equation}
\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})= \frac{1}{n}\sum_{i=1}^{n}\log\textbf{T} f_{\boldsymbol{\beta}}^{*},
\label{eqn:discrete_loglikelihood}
\end{equation}
where $f_{\boldsymbol{\beta}}^{*} = (f_{{\boldsymbol{\beta}}_{1}}, f_{{\boldsymbol{\beta}}_{2}}, \hdots, f_{{\boldsymbol{\beta}}_{m}})^T$ is a $m \times 1$-array or vector that serves as the discrete approximation of $f_{\boldsymbol{\beta}}$.
\begin{figure}[!htb]
\begin{tikzpicture}[scale=0.8,every node/.style={minimum size=0.5cm},on grid]
\begin{scope}[
yshift=-50,every node/.append style={
yslant=0.5,xslant=-1.3},yslant=0.5,xslant=-1.3
]
\fill[white,fill opacity=0.9] (-3,-3) rectangle (3,3);
\draw[step=0.5cm, thin, gray] (-3,-3) grid (3,3);
\draw[black,very thick] (-3,-3) rectangle (3,3)
\draw[red,ultra thick,solid] (-1.5,-3.5) -- (2.25,3.5);
\pgfkeys{/pgf/number format/.cd, fixed, zerofill, precision =1}
\coordinate (s1) at (0.9107,1);
\node at (s1) [fill=black,circle,scale=0.1] {$a$};
\coordinate (s2) at (0.64285714285,0.5);
\node at (s2) [fill=black,circle,scale=0.1] {$b$};
\coordinate (s4) at (2.75,2.75);
\node at (s4) [fill=black,circle,scale=0.15] {$c$};
\end{scope},
\coordinate (p1) at (-2.75,1);
\node at (p1) [fill=black,circle,scale=0]{};
\coordinate (p2) at (-2.75,1.5);
\node at (p2) [fill=black,circle,scale=0]{};
\draw[-latex,thick](-2,1)node[left,scale=1]{$P_1$}
to[out=0,in=90] (s1);
\draw[-latex,thick](-2,1.5)node[left,scale=1]{$P_2$}
to[out=0,in=90] (s2);
\draw[-latex,thick](-6,1.5)node[left,scale=1]{}
to[out=0,in=180] (p1);
\draw[-latex,thick](-6,1.5)node[left,scale=1]{$l_{i,j}=\|P_1 P_2\|$}
to[out=0,in=180] (p2);
\coordinate (s3) at (-2.75,1);
\node at (s3) [fill=black,circle,scale=0]{};
\node {} {node (T) at (-4.5,-8) {%
$\begin{aligned}
\mathbf{T}=
\end{aligned}$}};
\node {} {node (line) at (5,-3.75) {%
$\begin{aligned}
\beta_0 = y_i - \beta_1 x_{1_i}
\end{aligned}$}};
\matrix (m1) at (0,-8) [matrix of math nodes,left delimiter=(,right delimiter=)](A) {
l_{1,1} & l_{1,2} & \dots & l_{1,j} & \dots & l_{1,m}\\
l_{2,1} & l_{2,2} & \dots & l_{2,j} & \dots & l_{2,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
l_{i,1} & l_{i,2} & \dots & l_{i,j} & \dots & 0\\
\vdots & \vdots & & \vdots & & \vdots\\
l_{n,1} & l_{n,2} & \dots & l_{n,j} & \dots & l_{n,m}\\
};
\coordinate (lij) at (0.25,-8);
\node at (lij) [fill=black,circle,scale=0]{};
\draw[-latex,dashed](-8.8,1.1)node[left,scale=1]{}
to[out=270,in=90] (lij);
\coordinate (p3) at (2.75,2.75);
\node at (p3) [fill=black,circle,scale=0]{};
\draw[-latex,thick](5,2.75)node[right,scale=1]{$l_{i,m} = 0$}
to[out=200,in=60] (s4);
\coordinate (lim) at (3,-8.4);
\node at (lim) [fill=black,circle,scale=0]{};
\draw[-latex,dashed](7,2.8)node[left,scale=1]{}
to[out=0,in=0] (lim);
\end{tikzpicture}
\caption{2-D Transformation Matrix Algorithm} \label{fig:1d_Algo}
\end{figure}
\subsubsection{\texttt{transmatrix\_3d()}}
The implementation of the function for the 3-dimensional case works for problems with two regressors and a random intercept,
\begin{equation}
y_i={\beta_0}_i+{\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i,
\label{eqn:rc_model_3d}
\end{equation}
and the model with three regressors and no intercept.
\begin{equation}
y_i={\beta_1}_{i}{x_1}_i+{\beta_2}_{i}{x_2}_i+{\beta_3}_{i}{x_3}_i + \epsilon_i
\label{eqn:rc_model_3d_no_intercept}
\end{equation}
Note: for the three-dimensional implementation of the algorithm numerical instabilities occur when the underlying joint density, $f_{\boldsymbol{\beta}}$, has a single mode located at the center of the grid. This problem can occur in two ways: (1) it can occur organically if the mode of the density is located close to or at $(0,0,0)$ with the grid axes ranges at their default values [-5,5], or (2) artificially if the user provides grid ranges such that the mode of $f_{\boldsymbol{\beta}}$ is located at its center.
This problem is overcome by simply imposing a shift onto the resulting density by applying a linear transformation to the sample data, as below:
\begin{equation}
Y_i = (\beta_{0_i} + c) + \beta_{1_i} X_1 + \beta_{2_i} X_2.
\label{eqn:sample_shift}
\end{equation}
This can be achieved in two ways: (1) the more computationally efficient method is to simply supply a grid range for $\beta_0$ that offsets the mode of $\hat{f_{\boldsymbol{\beta}}}$ from the center, or (2) apply the shifting algorithm described in \eqref{eqn:sample_shift} which allows the user to supply any grid range but with an additional computational cost.
Similar to the 2-dimensional implementation, a finite volume type approach is used to create the discrete version of the linear operator $T$. In this higher-dimensional case the hyper-planes parametrized by the sample observations are now 2-dimensional planes, and the grid that $f_{\boldsymbol{\beta}}$ is estimated over is comprised of three axes and is therefore in the shape of a 3-dimensional cube. In this case, the intersection of the plane with the grid are characterized by a collection of points that define a polygon whose areas are used as the 2-dimensional analog of the line segments in the lower dimensional implementation of this finite volume estimation process. The algorithm is outlined below. Figure \ref{fig:2d_Algo} also illustrates the algorithm for a single cuboidal grid section.\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{sample, grid} \\
\Initialization{Initialize \textbf{NumPy} Array of Zeros}\\
\For{s in sample}{
get intersection points\;
\For{p in intersection points}{\
map intersection points to array indices\;
\For{i in indices}{\
assign intersection point to grid location (a point can
belong to more than one grid box)\;}}
\For{m in assigned points}{\
sort points to form polygon and generate the area
}
\For{i in indices}{\
map each polygon's area to the initialized array}
}
\caption{\texttt{transmatrix\_3d()}}
\end{algorithm}
\begin{figure}[!htb]
\begin{tikzpicture}[3d view={120}{15},line join=round, thick,
declare function={a=4;b=2;}]
\draw[dashed] (a,0,-a) -- (0,0,-a)-- (0,a,-a);
\draw[dashed] (0,0,0) -- (0,0,-a);
\draw (a,0,0) -- (a,0,-a) -- (a,a,-a) -- (a,a,-b);
\draw (a,a-b,0) -- (a,0,0) -- (0,0,0) -- (0,a,0) -- (a-b,a,0);
\draw (0,a,0) -- (0,a,-a) -- (a,a,-a);
\draw (a-b,a,0) -- (a,a,0) -- (a,a-b,0);
\draw (a,a,0) -- (a,a,-b);
\fill[cyan, opacity=0.8] (4.5,0.5,1.5) -- (0.5,3.5,1.5) -- (2.75,6.25,-4) -- (6.25,2.75,-4) -- cycle;
\draw[pattern=north east lines, preaction={fill=white, opacity=0.8}] (a,a-0.5*a,-a) -- (a-0.5*a,a,-a) -- (0,a,-a+0.5*a) -- (0,a-0.5*a,0) -- (a,a-0.7*a,0) -- cycle;
\draw (a,0,0) -- (a,0,-a) -- (a,a,-a) -- (a,a,-b);
\draw (a,a-b,0) -- (a,0,0) -- (0,0,0) -- (0,a,0) -- (a-b,a,0);
\draw (0,a,0) -- (0,a,-a) -- (a,a,-a);
\draw (a-b,a,0) -- (a,a,0) -- (a,a-b,0);
\draw (a,a,0) -- (a,a,-b);
\coordinate (p1) at (a,a-0.5*a,-a);
\node at (p1) [fill=red,circle,scale=0.15] {$p1$};
\coordinate (p2) at (a-0.5*a,a,-a);
\node at (p2) [fill=red,circle,scale=0.15] {$p2$};
\coordinate (p3) at (0,a,-a+0.5*a);
\node at (p3) [fill=red,circle,scale=0.15] {$p3$};
\coordinate (p4) at (0,a-0.5*a,0);
\node at (p4) [fill=red,circle,scale=0.15] {$p3$};
\coordinate (p5) at (a,a-0.7*a,0);
\node at (p5) [fill=red,circle,scale=0.15] {$p3$};
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p1);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p2);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p3);
\draw[-latex,dashed](-8.8,3)node[right,scale=1]{}
to[out=230,in=90] (p4);
\draw[-latex,dashed](-8.8,3)node[right,scale=0.8]{$\{p_3,p_2,p_5,p_1,p_4\}$}
to[out=230,in=90] (p5);
\draw[->] (-9,4.5,0) -- (-9,4.5,-0.5);
\coordinate (sort1) at (-9,4.5, -0.9);
\node[draw, scale=0.8] at (sort1) {sort points};
\draw[->] (-9,4.5,-1.25) -- (-9,4.5,-1.75);
\node {} {node[scale=0.8] (T) at (-9,4.5,-2) {%
$\begin{aligned}
\{p_1,p_2,p_3,p_4,p_5\}
\end{aligned}$}};
\draw[->] (-9,4.5,-2.25) -- (-9,4.5,-2.75);
\node {} {node[scale=0.8] (T) at (-9,4.5,-3) {%
$\begin{aligned}
a_{i,j} = Area(\{p_1,p_2,p_3,p_4,p_5\})
\end{aligned}$}};
\matrix (m1) at (-12,4.5,-8) [matrix of math nodes,left delimiter=(,right delimiter=)](A) {
a_{1,1} & a_{1,2} & \dots & a_{1,j} & \dots & a_{1,m}\\
a_{2,1} & a_{2,2} & \dots & a_{2,j} & \dots & a_{2,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
a_{i,1} & a_{i,2} & \dots & a_{i,j} & \dots & a_{i,m}\\
\vdots & \vdots & & \vdots & & \vdots\\
a_{n,1} & a_{n,2} & \dots & a_{n,j} & \dots & a_{n,m}\\
};
\node {} {node (T) at (-12,0,-8.75) {%
$\begin{aligned}
\mathbf{T}=
\end{aligned}$}};
\coordinate (aij) at (-12,4.75,-8);
\draw[-latex,dashed](-6.5,3.5,-2.75)node[right,scale=1]{}
to[out=270,in=90] (aij);
\end{tikzpicture}
\caption{3-D Transformation Matrix Algorithm} \label{fig:2d_Algo}
\end{figure}
The resulting array is in the same form as the one obtained using the 2-dimensional implementation \texttt{transmatrix\_2d()}, and can be used in the following mannner to evaluate the log-likelihood functional in \eqref{eqn:discrete_loglikelihood}
\begin{equation*}
\textbf{T}f_{\boldsymbol{\beta}}^{*} =
\begin{bmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & \hdots & a_{1,m} \\
a_{2,1} & a_{2,2} & a_{2,3} &\hdots & a_{2,m} \\
a_{3,1} & a_{3,2} & a_{3,3} &\hdots & l_{3,m} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n,1} & a_{n,2} &a_{n,3} &\hdots & a_{n,m} \\
\end{bmatrix}
\begin{bmatrix}
{f_\beta}_1 \\
{f_\beta}_2 \\
{f_\beta}_3 \\
\vdots \\
{f_\beta}_m
\end{bmatrix},
\end{equation*}
where in this case each $a_{ij}$ corresponds to the area of the polygonal intersection of the plane with the discrete estimation grid. The resulting product of this matrix and vector is an $n \times 1$ array. Applying an element-wise log-transformation, then taking the sum of this $n \times 1$ array results in the expression in \eqref{eqn:discrete_loglikelihood}.
It is important to note the computational performance of this algorithm. Matrix multiplication parallelizes the computation and eliminates the need to evaluate the log-likelihood functional sequentially for each sample point. However, the process of creating the transformation matrix $\mathbf{T}$ scales linearly with the sample size $n$ and exponentially with the dimensionality of the problem. The function was written in a way that maximizes efficiency as much as possible by making use of Python's parallel computation through array-wise computation when possible.
Solving for the transformation matrix, $\mathbf{T}$ is only one of two steps in the process of estimating the density $f_{\boldsymbol{\beta}}$. The second step is the process of optimizing the regularized log-likelihood functional, the run-time of which scales multiplicatively with respect to the sample size, $n$, and the number of discretezation points, $m$. Therefore, choosing an appropriate sample size and level of discretezation is an important consideration as both primarily determine the total cost of running the algorithm.
\subsection{The \texttt{rmle()} Function}\label{sec:rmle_func}
The \texttt{rmle()} function returns a class named \texttt{`RMLEResult'}. This class stores the solution to the optimization problem, $f_{\boldsymbol{\beta}}^{*}$, and metadata about the solution and the process of optimization. An instance of \texttt{`RMLEResult'} has the following accessible attributes and methods:\\
\noindent \texttt{f}: returns a $m \times 1$ array containing all the estimated function values. It is necessary to reshape the solution before visual representation.\\
\noindent \texttt{f\_shaped}: returns the reshaped array of $f$.\\
\noindent \texttt{dim}: returns an integer which represents how many dimensions $f_{\boldsymbol{\beta}}$ is estimated over.\\
\noindent \texttt{maxval()}: returns a list containing the maximum value of $f_{\boldsymbol{\beta}}^{*}$ and its location. \\
\noindent \texttt{mode()}: returns a list containing possible modes of the density and their locations. \\
\noindent \texttt{ev()}: returns atuple containing the expected value of each $\beta_i$. \\
\noindent \texttt{alpha}: returns a floating-point that specifies the regularization parameter, $\alpha$, used for estimation.\\
\noindent \texttt{alpmth}: returns a string that specifies the method by which the regularization parameter, $\alpha$ was chosen. It can take on the following values: \{`Lepskii', `CV', `User'\}.\\
\noindent \texttt{T}: returns a class object that is created using the \texttt{transmatrix()} function. It contains attributes, methods and subclasses that contain information about the transformation matrix $\mathbf{T}$ and meta-data about it. \\
\noindent \texttt{Tmat}: returns an attribute of the sublcass \texttt{T} but also accessible from the \texttt{RMLEResult} class. It returns the transformation matrix $\mathbf{T}$. \\
\noindent \texttt{grid}: returns a class object that is created using the \texttt{grid\_set()} function. It has attributes and methods that contain information about the grid $\hat{f}_{\boldsymbol{\beta}}$ is estimated over. \\
\noindent \texttt{details}: returns a dictionary containing metadata about the minimization process. \\
The function \texttt{rmle()} serves as a wrapper function for \textbf{SciPy}'s \texttt{minimize} function with \texttt{`trust-const'} set as the minimization algorithm. The choice of this algorithm is crucial as it specializes in large-scale constrained minimization problems, for further details we refer to \cite{byrd1999interior}. The ability to handle large-scale problems is important because depending on the size of the sample, and level of discretezation, the functional evaluations as well as the gradient evaluations could become exceedingly expensive, as it scales with both in a multiplicative manner ($n \times m$). The option to set constraints was also an important consideration. As in equation \eqref{eqn:method} there are two important constraints in estimating $f_{\boldsymbol{\beta}}$, namely: $f\ge 0,\, \text{and } \|f\|_{L^1}=1$. These two constraints ensure the resulting solution satisfies the definition of a density.
Table 1 contains all the arguments for the function \texttt{rmle()} followed by a short description of what they pertain to. More important arguments will be discussed in following subsections. \\
\begin{table}[!h]
\caption{\texttt{rmle()} arguments}
\centering
\begin{tabular}{c ll}
\hline\hline \\ [-1.5ex]
\textbf{Argument} & \textbf{Description} \\\hline
\\ [-1.5ex]
\multicolumn{1}{p{3cm}}{\raggedright functional} & \multicolumn{1}{p{11cm}}{\raggedright Negative likelihood functional with corresponding regularization term} \\
\multicolumn{1}{p{3cm}}{\raggedright alpha} & \multicolumn{1}{p{11cm}}{\raggedright constant $ \geq 0$ that serves as the regularization parameter, or a string matching: `cv' or `lepskii'.}\\
\multicolumn{1}{p{3cm}}{\raggedright tmat} & \multicolumn{1}{p{11cm}}{\raggedright Class object returned by \texttt{transmatrix()} which contains information about the transformation matrix $\mathbf{T}$ and the grid it is estimated over.} \\
\multicolumn{1}{p{3cm}}{\raggedright k} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument: integer which specifies how many folds for modified k-fold cross-validation. Default value is $k = 10$ }\\
\multicolumn{1}{p{3cm}}{\raggedright initial\_guess} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument from the \textbf{SciPy Optimize} minimize function. Used to supply an initial value for the minimization algorithm to start. Default value is set to a \textbf{NumPy} array of values close to zero.} \\
\multicolumn{1}{p{3cm}}{\raggedright hessian\_method} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument from \textbf{SciPy Optimize} minimize function. Default value is set to `2-point'.} \\
\multicolumn{1}{p{3cm}}{\raggedright constraints} & \multicolumn{1}{p{11cm}}{\raggedright Refers to the linear constraints imposed onto the problem. It is set as an optional argument, the default value is set to $\|f\|_{L^1}=1$} \\
\multicolumn{1}{p{3cm}}{\raggedright tolerance} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument for the tolerance criteria for the optimization algorithm's termination. Default value is set to \texttt{1e-6}.} \\
\multicolumn{1}{p{3cm}}{\raggedright max\_iter} & \multicolumn{1}{p{11cm}}{\raggedright Optional argument for the maximum number of iterations. Default value is set to 100.} \\
\multicolumn{1}{p{3cm}}{\raggedright bounds} & \multicolumn{1}{p{11cm}}{\raggedright Refers to the bound constraints of the optimization problem. The default is expressed as $f\ge 0$ } \\
\hline
\end{tabular}
\label{tab:sobolev_args}
\end{table}
\subsubsection{Functionals and Regularization Terms}
Recall the average log-likelihoood functional to be minimized as in equation \eqref{eqn:method}. As specified in section \ref{sec:method} the regularization terms implemented in this module are: the Sobolev norm for $H^1$, the squared $L^2$ norm, and Entropy. The module also includes an option for a functional that has no regularization term if the user wishes to produce a reconstruction of $f_{\boldsymbol{\beta}}$ without any form of regularization. This option will often lead to overfitting, and produce a highly unstable solution.
\subsubsection{Sobolev Norm for $H^1$}
The functional incorporating the Sobolev norm for $H^1$ has the following form,
\begin{align} \label{eqn:h1_penalty}
-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})+\alpha (\|f\|_{2}^{2}+\|f^{\prime}\|_{2}^{2})
\end{align}
where $\|\cdot\|_{2}^{2}$ indicates the squared $L^2$ norm. It is important to note that the choice of the regularization term would typically depend on knowledge about the true solution, as the choice of the regularization term imposes certain assumptions on $f_{\boldsymbol{\beta}}$. In the case of the $H^1$ penalty term, the solution has a square-integrable first derivative in addition to the assumptions of non-negativity and integrability imposed by the constraints of the minimization problem.
The function \pythoninline{sobolev()}, or \pythoninline{sobolev_3d} for the 3-dimensional application returns the value of the discrete implementation of the functional in \eqref{eqn:h1_penalty}. Table 2 provides details on the arguments required for this function. The \textbf{SciPy Optimize} minimize function does not accept array arguments that are greater than one dimension. A necessary step is to unravel the transformation matrix, $\mathbf{T}$, into a one-dimensional array which is passed to the function as \texttt{tm\_long} and simply reshaped into the proper array dimensions. The term $\mathbf{T}f$ is calculated using \textbf{NumPy's} matrix multiplication and sum functions which are much faster alternatives than their non-\textbf{Numpy} counterparts. The regularization term is calculated in \eqref{eqn:h1_penalty}, with the function \texttt{norm\_fprime()} computes for $\|f^{\prime}\|_{2}^{2}$ where $f^{\prime}$ is treated as a total derivative.
\begin{table}[!h]
\caption{\texttt{sobolev(), sobolev\_3d()} arguments}
\centering
\begin{tabular}{c rrrr}
\hline\hline \\ [-1.5ex]
Argument & Notation & Description \\ [0.5ex]
\hline
f & $f_{\boldsymbol{\beta}}^{*}$ & current value of the solution $f_\beta^{*}$ & \\
a & $\alpha$ & constant that serves as the regularization parameter & \\
tm\_long & $\mathbf{T}$ & unraveled form of the transformation matrix, $\mathbf{T}$ &\\
n & n & the sample size & \\
s & $\Delta b$ & the step size of the grid & \\
\hline
\end{tabular}
\label{tab:sobolev_args}
\end{table}
The underlying minimization function is able to approximate the Jacobian of the functional with an additional computational cost. For computational efficiency, we supply an explicit form for the Jacobian of the functional: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+2\alpha (f-f^{\prime\prime})$
The form of this Jacobian implies an additional smoothness assumption on the solution, as it requires $f_\beta$ to be twice differentiable.
\subsubsection{Squared $L^2$ Norm}
The form of the functional in \eqref{eqn:method} that incorporates the squared $L^2$ norm as the regularization term is:
\begin{align*}
-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X}) f+\alpha\|f\|_{2}^{2}
\end{align*}
The arguments for this function are indentical to those listed in Table \ref{tab:sobolev_args}, likewise is true for the Jacobian associated with this regularization term. The Jacobian has the following form: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+2\alpha f$
Choosing this regularization functional imposes less smoothness assumptions on the solution, as the only additional assumption in place is square-integrability. This leads to a typically less smooth reconstruction as compared to using the Sobolev norm for $H^1$ as the regularization functional. The functions in python are coded similarly as with the $H^1$ regularization functional.
\subsubsection{Entropy}
The form of the functional in \eqref{eqn:method} that incorporates the entropy of the function has the following form: $-\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})+\alpha \int f \log(f)db$
This functional has the least amount of assumptions on the solution, $f_{\boldsymbol{\beta}}$. It only requires finite entropy which is a weak assumption in addition to the non-negativity and $L^1$ constraints of the minimization problem.
The Jacobian of the entropy functional also does not impose any additional assumptions on the solution, and has the following form: $-\frac{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})^{\prime}}{\bar{\ell}(f_{\boldsymbol{\beta}}|Y,\mathbf{X})}+\alpha (\log f+1)$
\subsubsection{Parameter Selection}
Recall the minimization problem as in \eqref{eqn:method} where a constant $\alpha \geq 0$ controls the size of the effect of the regularization term. The user can provide the $\alpha$ value directly if the user has a general idea of the level of smoothing necessary for the solution. If the user has no best-guess for the value of the regularization parameter alpha, the module has two options to automatically select the parameter $\alpha$, namely: Lepskii's balancing principle, and k-fold cross-validation. The Lepskii method typically yields a less accurate result relative to k-fold cross-validation; however, its advantage lies in significantly less computational cost.
\subsubsection{Lepskii's Balancing Principle}
Lepskii's principle is an adaptive form of estimation which is popular for inverse problems, e.g. \cite{Tsybakov:00}, \cite{BauHoh:05}, \cite{mathe_2006}, \cite{hohage2016inverse}, \cite{Werner:18}. It is significantly computationally less expensive than other parameter selection methods.
The method works as follows: we compute $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{1}},\ldots,\hat{f_{\boldsymbol{\beta}}}_{\alpha_{m}}$ for $\alpha_1 = c_{L_{n}} \frac{\ln(n)}{\sqrt{n}}$ and $\alpha_{i+1} = r\alpha_i$ with some constants $c_{L_{n}} >0,r>1$. We then select $\alpha_{j}$ as the optimal parameter choice where:
\[
j_{bal}:=\max\{j\leq m, \|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\| \leq 8 r^\frac{1-i}{2}, \text{ for all } i < j\}.
\]
The algorithm is implemented in python as follows: \\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{$\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$} \\
\Initialization{Generate the transformation matrix $\mathbf{T}$}\\
\For{$\alpha_i$ in $\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$}{
Compute for $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{i}}$ using $\mathbf{T}$\;
}
\For{j in $\{1,2,\hdots,m\}$}{\
Set $i = 0$ \;
\While{i $<$ j}{\
Check if $\|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\|\leq 8 r^\frac{1-i}{2}$ \;
\If{$\|\hat{f_{\boldsymbol{\beta}}}_{\alpha_i}-\hat{f_{\boldsymbol{\beta}}}_{\alpha_j}\|> 8 r^\frac{1-i}{2}$}{\
$j_{bal} = j$ \;
\textbf{break}
}
$i+=1$
}
}
\caption{Lepskii's Balancing Principle}
\end{algorithm}
\medskip
The bulk of the computational cost of the Lepskii algorithm implementation can be broken down into two components: the fixed cost of generating $\mathbf{T}$, and the variable cost of computing $\hat{f_{\boldsymbol{\beta}}}_{\alpha_{1}},\ldots,\hat{f_{\boldsymbol{\beta}}}_{\alpha_{m}}$ as the number of $\alpha$ values to be used depends on $c_{L_{n}} >0 \text{, }r>1$, and also the sample size of the data. This implementation of Lepskii algorithm's scales linearly in terms of runtime with the number of $\alpha$ values being tested, $m$.
\subsubsection{K-Fold Cross Validation}
Cross-validation is another popular parameter choice rule. Here we present a modified implementation of k-fold cross-validation with a cost-function that is applicable to our problem. The modification we apply is an algorithm to lessen the computation time by reducing the number of $\alpha$ values it needs to iterate through.
The loss function we considered was,
\begin{align*}
J_{\alpha_{i}} = -\sum_{j=1}^{k} \log \mathbf{T}_{j}\hat{f}_{{\boldsymbol{\beta}}_{-j}}
\end{align*}
Where $\mathbf{T}_{j}$ is the transformation matrix generated from a subsample of the observations, which can be interpreted as the $j$-th fold that is left out in the current iteration, and $\hat{f}_{{\boldsymbol{\beta}}_{-j}}$ is the estimate for $f_{\boldsymbol{\beta}}$ using $\mathbf{T}_{-j}$. The loss function can be interpreted as the negative of the likelihood that the $j$-th fold of the sample used to generate $\mathbf{T}_{-j}$ was drawn from the distribution, as the subsample fold used to produce $\mathbf{T}_{j}$. We aim to choose the $\alpha$ that minimizes this loss function.
The search method for the optimal $\alpha$ value reduces the number of $\alpha$ values tested. The algorithm involves separating the range of $\alpha$ values into two sections $\{\alpha_{1},\ldots,\alpha_j\}$ and $\{\alpha_{j+1},\ldots, \alpha_{m}\}$. Two alpha values $\alpha_a$, and $\alpha_b$ are randomly selected from the respective sections and are used to compute for the corresponding loss function values, $J_{\alpha_{a}}$ and $J_{\alpha_{b}}$. The section from which the $\alpha$ value that produces the smaller $J_{\alpha}$ was drawn from is kept, while the other is discarded. This is repeated until there is a sufficiently small range of $\alpha$ values. Once this range of $\alpha$ values is obtained, the loss function is evaluated over all the remaining $\alpha$ values and the optimal $\alpha$ is chosen as the one which minimizes $J_{\alpha}$. The complete algorithm is implemented as follows:\\
\begin{algorithm}[H]
\SetAlgoLined
\SetKw{Input}{Input:}
\SetKw{Initialization}{Initialization}
\Input{$\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$} \\
\Initialization{Generate the transformation matrix $\mathbf{T}$ and apply a random shuffle, set $\boldsymbol{\alpha}$ = $\{\alpha_1,\alpha_2,\hdots,\alpha_m\}$}\\
\While{\texttt{len}($\boldsymbol{\alpha}$) $>$ 3}{
1. Set $\boldsymbol{\alpha_a}$ = $\{\alpha_1,\alpha_2,\hdots,\alpha_j\}$ \;
2. Set $\boldsymbol{\alpha_b}$ = $\{\alpha_{j+1},\alpha_{j+2},\hdots,\alpha_m\}$ \;
3. Randomly select $\alpha_a$ and $\alpha_b$ from $\boldsymbol{\alpha_a}$ and $\boldsymbol{\alpha_b}$ respectively. \;
4. Evaluate $J_{\alpha_{a}}$ and $J_{\alpha_{b}}$ \;
\If{$J_{\alpha_{a}}$ $<$ $J_{\alpha_{b}}$}{
Set $\boldsymbol{\alpha}$ = $\boldsymbol{\alpha_a}$
}
\Else{
Set $\boldsymbol{\alpha}$ = $\boldsymbol{\alpha_b}$
}
}
\For{$\alpha_i$ in $\boldsymbol{\alpha}$}{\
Compute for $J_{\alpha_{i}}$ \;
}
Choose $\alpha_{cv} = \argmin{J_{\alpha_{i}}}$
\caption{Modified K-fold Cross Validation}
\end{algorithm}
\medskip
The runtime of the unmodified version of k-fold cross-validation scales linearly with the product $k \times m$ where $k$ is the number of folds and $m$ is the number of $\alpha$ values being tested. Applying the modified version reduces the number of $\alpha$ values being tested, $m$, by some logarithmic factor, which is a significant reduction in computational cost which makes cross-validation more computationally feasible.
\section{Examples} \label{sec:example}
The following examples will be demonstrated in this section: the random coefficients model with a single regressor and random intercept for the two-dimensional case, and the two regressors and random intercept for the three-dimensional case. This section will also show how to plot the estimated density using the built in plotting function \pythoninline{plot\_rmle()} which makes use of functions from the \pythoninline{matplotlib} library. It will also show how to plot the density without the use of the built-in function in case the user wishes to explore different plotting options.
\subsection{Example 1: 2-D Case Single Regressor with Random Intercept}
The first example is the case described by \eqref{eqn:rc_model_2d}. The example is demonstrated with simulated data using the function \pythoninline{sim\_sample()}. This function simulates the regressor $X_1 \sim \pazocal{U}(-2,2)$ and the random coefficients $\beta_0$, $\beta_1$ from a bimodal multivariate normal mixture as follows,
\begin{align*}
0.5\mathcal{N}([-0.5,-0.5], 0.01\mathbbm{I}_2) + 0.5\mathcal{N}([0.5,0.5], 0.01\mathbbm{I}_2).
\end{align*}
The general flow of the process of using the module can be broken down in five steps:
\begin{enumerate}[topsep=0pt,itemsep=-1ex,partopsep=1ex,parsep=1ex]
\item Import the necessary modules and functions.
\item Establish the dataset to be used (either real or simulated data).
\item Specify the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is to be estimated over.
\item Generate the transformation matrix $\mathbf{T}$.
\item Run the \pythoninline{rmle()} function.
\end{enumerate}
\begin{python}
from pyrmle import *
sample = sim_sample(n = 10000,dim = 2)
\end{python}
\medskip
The program begins by importing the necessary functions from \pythoninline{pyrmle.py} which contains the high-level functions that the user interacts with. The module \pythoninline{pyrmle_funcs} contains the underlying functions necessary for functions in the main module \pythoninline{pyrmle} to run. The next step is to define the sample to be used in creating the transformation matrix $\mathbf{T}.$ The sample has the same form as described in subsection \ref{sec:trans_matrix}, where the sample has the form $[\mathbf{X}_0, \mathbf{X}_1, \mathbf{Y}]$. In Python it takes the shape of $10000 \times 3$ \pythoninline{NumPy} array as seen below.
\medskip
\begin{python}
[[ 1. , -0.76765455, 2.10802347],
[ 1. , 1.51774991, -0.11337413],
...,
[ 1. , -1.80486996, -3.08120151]]
\end{python}
\medskip
The next step is to generate the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is estimated over. This is done using the \pythoninline{grid\_set()} function, as discussed briefly in \ref{sec:trans_matrix}. In this example we set the ranges of $\beta_0$ and $\beta_1$ to $[-10,10]$ which defines a two-dimensional grid spanning that range in each axis. This function creates an instance of the \pythoninline{class grid_obj} which has attributes and methods enumerated and described in subsection \ref{sec:trans_matrix}. When the \pythoninline{grid} class instance has been created, the user can proceed to generate an instance of the \pythoninline{class tmatrix} using the \pythoninline{transmatrix()} function.
\medskip
\begin{python}
grid_beta = grid_set(num_grid_points = 40, dim = 2)
print(grid_beta.numgridpoints())
1600
T = transmatrix(sample, grid_beta)
\end{python}
\medskip
After the instance of the transformation matrix is produced, the user can then run the \pythoninline{rmle()} function. As stated in subsection \ref{sec:rmle_func}, the \pythoninline{rmle()} function has three essential arguments: \{\texttt{`functional',`alpha',`tmat'}\}.
\medskip
\begin{python}
result = rmle(sobolev,0.25,T)
print(result.ev())
[-0.002865326246726808, -0.010416375635973668]
print(result.mode()[:2])
[[0.39681799850755783, [-0.625, -0.375]],
[0.38831830923870914, [0.625, 0.375]]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_40_grid_base_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 40 grid points}\
\label{fig:cont_40_base}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_40_grid_base_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 40 grid points}
\label{fig:surface_40_base}
\end{minipage}
\end{figure}
As stated in subsection \ref{sec:rmle_func}, the \pythoninline{rmle()} function generates an instance of \pythoninline{class RMLEResult}. This class has a number of useful and informative attributes and methods that describe the estimated density. The \pythoninline{ev()} method returns the expected value of each ${\boldsymbol{\beta}}_j$, while the \pythoninline{mode()} method returns possible maxima of the estimated density. The \pythoninline{mode()} method relies on a naive search method for maxima with no underlying statistical tests.
Figure 3 (reference) shows the contour plot produced by the \pythoninline{plot_rmle()} function. It is clear that there is a large portion of the grid that is essentially unused, and the user could benefit from a reduction in the grid-size in terms of computational costs. This can be done by reducing the number of grid points and shrinking the range of each axis. In terms of tuning the size of the grid, we suggest that the user tries a relatively large range to begin with to ensure that the grid contains the support of $\hat{f_{\boldsymbol{\beta}}}$, and then consider smaller grid ranges. Having a grid range that is too small has a negative effect on the estimate as the optimization algorithm enforces the constraint that $\|\hat{f_{\boldsymbol{\beta}}}\|_{L^1}=1$.
\medskip
\begin{python}
grid_beta_alt = grid_set(20,2,B0_range=[-1.5,1.5],\
B1_range=[-1.5,1.5])
print(grid_obj_alt.numgridpoints)
400
T2 = transmatrix(sample, grid_beta_alt)
result2 = rmle(sobolev,0.15,T2)
print(result2.ev())
[-0.004977073000670898, -0.003964663139211258]
print(result2.mode()[:2])
[[0.3643228046077392, [0.5249999999999, 0.5249999999999]],
[0.3580092260598125, [-0.5249999999999, -0.5249999999999]]]
plot_rmle(result2)
plot_rmle(result2,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_smaller_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on a smaller grid}\
\label{fig:20_smaller_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_smaller_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a smaller grid}
\label{fig:20_smaller_surface}
\end{minipage}
\end{figure}
The reduction in the number of grid points resulted in a significant reduction in the run time of the algorithm while achieving a better estimate for $\hat{f_{\boldsymbol{\beta}}}$. With the reduction in the computational cost of the algorithm makes it more favorable to run an automatic parameter choice method.
\begin{python}
result_cv = rmle(sobolev,'cv',T2)
print(result_cv.alpha)
0.07065193045869372
print(result_cv.ev())
[-0.004977073000670898, -0.003964663139211258]
print(result_cv.mode()[:2])
[[0.3643228046077392, [0.5249999999999, 0.5249999999999]],
[0.3580092260598125, [-0.5249999999999, -0.5249999999999]]]
plot_rmle(result_cv)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_cv_grid_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on a smaller grid using cross-validation}\
\label{fig:20_cv_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_cv_grid_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a smaller grid using cross-validation}
\label{fig:20_cv_surface}
\end{minipage}
\end{figure}
The general workflow that we suggest when tuning the parameters to be used in estimation is as follows:
\begin{enumerate}[topsep=0pt,itemsep=-1ex,partopsep=1ex,parsep=1ex]
\item Establish a relatively large grid range for estimation and generate the transformation matrix $\mathbf{T}$. This should be treated as an exploratory step in terms of analyzing the data.
\item Set $\alpha$ equal to the step size of the grid.
\item Run the \pythoninline{rmle()} function.
\item Plot $\hat{f_{\boldsymbol{\beta}}}$ using the \pythoninline{plot_rmle()} function and determine the necessary grid range.
\item Limit the grid range as well as the grid points to reduce computation costs and genrate the new matrix $\mathbf{T}^*$.
\item Run the \pythoninline{rmle()} function with $\mathbf{T}^*$, and optionally employ one of the two automatic parameter selection methods: \{\pythoninline{'cv','lepskii'}\}.
\end{enumerate}
The following example will demonstrate usage of the \pythoninline{grid_set()} function in terms of supplying a different range $\beta_{1}$. The simulated data in this case will have modes for $\beta_1$ that are significantly larger than that of $\beta_0$ and are not encapsulated by the default range $[-5,5]$. The betas are sampled from the following distribution: $0.5\mathcal{N}([-1.5,6], \mathbbm{I}_2) + 0.5\mathcal{N}([1.5,9], \mathbbm{I}_2).$
\begin{python}
cov = [[[1, 0], [0, 1]],[[1, 0], [0, 1]]]
mu = [[-1.5,6],[1.5,9]]
sample = sim_sample(10000,2,beta_mu = mu,beta_cov = cov))
grid_beta_shifted = grid_set(num_grid_points = 20, \
dim = 2, B1_range=[2,13])
T_shifted = transmatrix(sample, grid_beta_shifted)
result_shifted = rmle(sobolev_norm_penal,0.5,T_shifted)
print(result_shifted.ev())
[0.03520704073478552, 7.524001743037029]
print(result.mode()[0:2])
[[0.07133616078580148, [1.25, 8.875]],
[0.0652140364538059, [-1.75, 5.574999999999999]]]]
plot_rmle(result_shifted)
plot_rmle(result_shifted,plt_type='surface')
\end{python}
\begin{figure}[h]
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_shifted_contour.png}
\caption{$\hat{f_\beta}$ contour plot with 20 grid points on shifted grid}\
\label{fig:20_shifted_cont}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1.1\textwidth]{2d_example_20_grid_shifted_surface.png}
\caption{$\hat{f_\beta}$ surface plot with 20 grid points on a shifted grid}
\label{fig:20_shifted_surface}
\end{minipage}
\end{figure}
\subsubsection{Example 2: 3-D Case Two Regressors with Random Intercept}
This second example is the case described by \eqref{eqn:rc_model_3d}. The example is demonstrated likewise with simulated data using the same function \pythoninline{sim_sample()}. The regressors are simulated as follows, $X_1, X_2$ are i.i.d $\pazocal{U}(-2,2)$, and the random coefficients $\beta_0, \beta_1, \text{ and } \beta_2$ are simulated from $\mathcal{N}([2,2,2], 0.01\mathbbm{I}_3)$
\medskip
\begin{python}
from pyrmle import *
sample = sim_sample(n = 10000,dim = 3)
\end{python}
\medskip
As in the previous example, the program begins by importing the necessary modules and functions. The \pythoninline{sim_sample()} function generates sample observations based on the aforementioned distributions. This results in a $10,000 \times 4$ \pythoninline{NumPy} array.
\medskip
\begin{python}
grid_beta = grid_set(10,3,B0_range=[-2,4],\
B1_range=[-2,4],B2_range=[-2,4])
print(grid_beta.numgridpoints())
1000
T = transmatrix(sample,grid_beta)
\end{python}
\medskip
The next step is to establish the number of grid points, and to generate the transformation using the simulated sample observations. In this case, we first consider ten grid points in each axis amounting to a total of 1000 grid points. If the user wishes to estimate $\hat{f_{\boldsymbol{\beta}}}$ over a finer grid it would be more efficient to first determine the smallest possible range of each axis that would fit almost all of the probability mass of $\hat{f_{\boldsymbol{\beta}}}$.
\medskip
\begin{python}
result = rmle(sobolev_norm_penal2d,0.6,T)
print(result.ev())
[2.458436489282234, 2.25500373629305, 1.9058740990043983]
print(result.mode())
[0.08926614291105403, [1.90000000, 1.90000000, 1.90000000]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_10_grid_cont}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_10_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_10_grid_surface}
\end{figure}
In the three dimensional application we use the regularization functional \pythoninline{sobolev_3d} and supply it as an argument to the \pythoninline{rmle()} function along with the transformation matrix and $\alpha$. The results show the effect of the level of discretezation on the estimate. It is possible to achieve more accurate estimates with more grid points in conjunction with a narrower grid range, but with a significantly higher computational cost.
\medskip
\begin{python}
grid_beta_alt = grid_set(20,3,B0_range=[0,3],\
B1_range=[0,3],B2_range=[0,3])
print(grid_beta.numgridpoints())
8000
T2 = transmatrix(sample,grid_beta_alt)
result2 = rmle(sobolev_norm_penal2d,0.3,T2)
print(result2.ev())
[2.102855111361121, 2.077365765266784, 1.9697452677152947]
print(result2.mode()[0])
[[0.2008050879224736, [2.025, 2.025, 2.025]]
plot_rmle(result2)
plot_rmle(result2,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_cont}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_surface}
\end{figure}
In this example, the number of grid points $\hat{f_{\boldsymbol{\beta}}}$ is estimated over is set to 20 on each axis which amounts to a total of 8,000 grid points. The additional level of discretezation due to the increased number of grid points and the smaller grid range provided resulted in an estimate $\hat{f_{\boldsymbol{\beta}}}$.
The next example will demonstrate the case when the underlying joint density being reconstructed has a single mode at the center of the grid. Both methods of dealing with this issue described in section \ref{sec:trans_matrix} will be illustrated.
\medskip
\begin{python}
mu = [[0,0,0],[0,0,0],[0,0,0]]
sample = sim_sample(n = 5000,dim = 3, \
beta_mu = mu)
grid_beta = grid_set(20,3,B0_range=[-1,2],\
B1_range=[-1.5,1.5],B2_range=[-1.5,1.5])
T = transmatrix(sample,grid_beta)
result = rmle(sobolev_3d,0.15,T)
print(result.ev())
[-0.19254733276928582, 0.005554898823827626, -0.009788891289943112]
print(result.mode())
.14336637945812933, [-0.024999999999999967, 0.075, -0.075]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_zero_grid_contour}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_zero_grid_surface}
\end{figure}
The following example demonstrates how to apply the shifting algorithm briefly mentioned in section \ref{sec:trans_matrix}. The algorithm is implemented in python by adding $c \sim \pazocal{U}(a,b)$ to the intercept, where $a$ and $b$ are determined by the size of the grid. This applies a transformation that can be back-transformed after estimation by adjusting the grid over which $\hat{f_{\boldsymbol{\beta}}}$ is estimated over. This is repeated across ten different shifted samples, then a simple k-means clustering algorithm is applied using \pythoninline{sklearn.cluster.KMeans()} on the $L2$ penalties of \{$\hat{f_{\boldsymbol{\beta}}}_1, \hdots, \hat{f_{\boldsymbol{\beta}}}_{10}$\}. The reconstruction, $\hat{f_{\boldsymbol{\beta}}}_j$, that is closest to the centroid of the largest cluster is then output as the solution.
\begin{python}
grid_beta = grid_set(20,3,B0_range=[-1.5,1.5],\
B1_range=[-1.5,1.5],B2_range=[-1.5,1.5])
T = transmatrix(sample,grid_beta)
result_shift = rmle(sobolev_3d,0.15,T,shift=True)
print(result_shift.ev())
[-0.12974928815245057, -0.004493337754614205, -0.0052980597609372385]
print(result_shift.mode())
[[0.14624599807641833, [0.009524681155561987, -0.075, -0.075]]
plot_rmle(result)
plot_rmle(result_shift,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_contour_shift_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_contour_shift}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_20_zero_grid_surface_shift_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_grid_surface_shift}
\end{figure}
The final example demonstrates how to use the module on a real dataset, and shows the user how to load the data into Python from a comma-separated value (CSV) format and how to pre-process it, if necessary, into a usable form. The data used here are from the British Family Expenditure Surey that was used in \cite{Dunker2019,dunker2021regularized}.
The model is identified as follows:
\begin{align*}
BS_i = \beta_{0,i} + \beta_{1,i}\ln(TotalExpenditure_i) + \beta_{2,i}\ln(FoodPrices_i) + \epsilon_i
\end{align*}
In the case of this dataset the regressors need to undergo a linear transformation before it generating the transformation matrix to be used in the algorithm. This is to ensure that the grid is sufficiently covered by the hyperplanes generated by the sample observations. For more details (refer to the section of our paper discussing this issue).
The OLS estimate for the random coefficients suggests $f_{\boldsymbol{\beta}}$ is centralized at (0.262755,0.0048,-0.00069) for a subsample size of $n = 5000$, which requires the application of one of the two methods described above to circumvent the problem that arises from a mode close to $(0,0,0)$.
\medskip
\begin{python}
data = np.genfromtxt('filename.csv', delimiter=',')
data = data[1:-1,1:4]
data = data[np.random.randint(1,len(data),5000),:]
data = data[~np.isnan(data).any(axis=1)]
ones = np.repeat(1,len(data))
real_data_sample = np.c_[ones, \
25*data[:,2]-0.3,data[:,1]-5,data[:,0]]
\end{python}
\medskip
The code block above loads the data from the a csv file, selects a random subsample of 5,000 observations, and applies the linear transformations on the data necessary.
\medskip
\begin{python}
grid_beta = grid_set(20,3,B0_range=[-0.75,1.25],\
B1_range=[-1,1],B2_range=[-1,1])
T = transmatrix(real_data_sample,grid_beta)
result = rmle(sobolev_rmle_2d,0.25,T,shift=True)
print(result.ev())
[0.2139255519131858, -0.0022826617746834394, -0.10875379495199464]
print(result.mode()[0:1])
[[0.19533254824055002, [0.3, -0.05, -0.15000000000000002]]
plot_rmle(result)
plot_rmle(result,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_contour_f12.png}
\caption{Contour plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_contour}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{3d_example_real_data_20_grid_surface_f12.png}
\caption{Surface plots of joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_surface}
\end{figure}
Included in the module is an option to fit a spline on the estimate $\hat{f_\beta}$. This would be the most computationally feasible option if the user needs a finer grid. The \pythoninline{spline_fit()} function takes two essential arguments. The first positional argument is the \pythoninline{class RMLEResult} object, and second is the number of grid points on each axis.
\begin{python}
spline = spline_fit(result,num_grid_points = 400)
plot_rmle(spline)
plot_rmle(spline,plt_type='surface')
\end{python}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f01.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f02.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_contour_f12.png}
\caption{Contour plots of the spline interpolated joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_contour_spline}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f01.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f02.png}\hfill
\includegraphics[width=.3\textwidth]{real_data_spline_surface_f12.png}
\caption{Surface plots of the spline interpolated joint bivariate marginal distributions of $\hat{f_{\boldsymbol{\beta}}}$: $\hat{f}_{\beta_0,\beta_1}$ (left), $\hat{f}_{\beta_0,\beta_2}$ (middle), and $\hat{f}_{\beta_0,\beta_1}$ (right).}
\label{fig:3d_20_real_grid_surface_spline}
\end{figure}
\bibliographystyle{apalike}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,493 |
Q: Yajra DataTables : Unable to load records in tables I am using Yajra data tables package for the loading of records. But I am getting records on full-page not on a particular table. I am not understanding where is the error.
view file
<table id="clientsTable" class="table table-bordered table-striped dataTable">
<thead>
<tr>
<th>#</th>
<th>First name</th>
<!--<th>Last name</th>-->
<th>Mobile number</th>
<th>Email</th>
<th>Actions</th>
</tr>
</thead>
</table>
script file
$(document).ready(function($) {
$('#menu-clients').addClass('active');
$('#clientsTable').DataTable({
processing: true,
serverSide: true,
"bDestroy": true,
"bAutoWidth": false,
ajax:{
url : '/clients',
method: 'get',
},
columns:[
{ data: 'DT_RowIndex'},
{ data: 'first_name', name: 'first_name'},
{ data: 'mobile_no', name: 'mobile_no'},
{ data: 'email', name: 'email'},
{ data: 'actions', name: 'actions'},
]
});
});
route file
Route::resource('/clients', 'ClientsController');
controller
public function index()
{
$data = DB::table('clients')->orderBy('id', 'desc')->get();
return Datatables::of($data)
->addColumn('actions','buttons.clients')
->rawColumns(['actions'])
->addIndexColumn()
->make(true);
}
response in network tab
{draw: 0, recordsTotal: 166, recordsFiltered: 166, data: [,…], input: []}
draw: 0
recordsTotal: 166
recordsFiltered: 166
data: [,…]
[0 … 99]
[100 … 165]
input: []
A: You need to separate the method for your view and dataTable data.
In your index method call the view of the table.
public function index()
{
return view('yourTableBlade');
}
Also remove the get() method yajraDataTable does the job for you.
public function tableData()
{
$data = DB::table('clients')->orderBy('id', 'desc');
return Datatables::of($data)
->addColumn('actions','buttons.clients')
->rawColumns(['actions'])
->addIndexColumn()
->make(true);
}
Create another route for your datatable.
Route::get('/tableData', 'ClientsController@tableData');
A:
You have to use the Select query Eg:
$data = DB::table('clients')->select('id','first_name','mobile_no','email')->orderBy('id', 'desc')->get();
return Datatables::of($data)
->addColumn('actions','buttons.clients')
->rawColumns(['actions'])
->addIndexColumn()
->make(true);
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 586 |
The units in which metal type is sold, containing relative proportions of letters appropriate for a given language.
References
Typography | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,330 |
{"url":"http:\/\/absoluteozone.com\/half-life-of-ozone.html","text":"ABSOLUTE OZONE\u00ae\n\nOzone Library\n\n# Half Life of Ozone\n\n## Changes of ozone with time\n\n1. General equation\n\nI believe that the equation describing the time evolution of the ozone concentration has the following form:\n\ndn = \u2212kn2 + q (1) dt\n\nwhere, t is the time, k is the reaction rate constant which is independent of concentration, but, probably, depends on the temperature; n is the concentration (number of the molecules per unit volume):\n\nn= M NA (2) \u03bcV\n\n\u03bc is the molar weight of ozone, M is the mass of ozone inside the volume, V , in a given moment of time, NA is the Avogadro number\n\nIn Eq.(1), q is the number of molecules delivered per unit time into the volume and calculate per unit volume::\n\nq= m NA (3) \u03bcV\n\nwhere m is the mass of ozone delivered into volume, V , per unit time.\n\n2. Closed volume\n\nConsider the case of q = 0 . For such a case, integration of Eq. (1) yields n= n0 (4)\n\n1+n0kt where n0 is the initial concentration. Using (2):\n\nn0 =M0 NA (5) \u03bcV\n\nwhere M0 is the initial mass of ozone. Eq.(4) can be rewritten as\n\nn= n0 (6) 1+ t\n\n\u03c4 where we introduced the time scale parameter \u03c4 : when t = \u03c4 the concentration of molecules (and the mass of ozone, for a given volume) decreases by factor of two, see Eq.(6).\n\nComparing Eqs.(4) and (6) one obtains:\n\n\u03c4= 1 (7) n0k\n\nSince k is independent of the concentration, we arrive at an important conclusion: the parameter \u03c4 depends on the initial concentration and, for a given volume and pressure, on the initial mass,\n\nM0 . \u03c4= \u03bc =k1 (8)\n\nNAM0k M0\n\n3. Stationary regime\n\nInspecting Eq.(1) one can conclude that the concentration does not change when q = kn2 (9)\n\nWhen Eq.(9) is satisfied, the rate of concentration changes, dn \/ dt , becomes zero, see Eq.(1).\n\nRealizing that the concentration is maintained to be constant in time, i.e.; n = n0 Eqs.(7) we obtain from (9) that\n\nq = n0 (10) \u03c4\n\nUsing Eqs.(3) and (5) one can rewrite (10) in terms of mass m = M0 (11)\n\nand using\n\n\u03c4\n\n4. Conclusion\n\nThus, to maintain a given mass of ozone, M0 , within a fixed volume at a given temperature, one\n\nshould deliver ozone with the rate m (mass per unit time) given by Eq.(11). The time scale parameter \u03c4 on the right hand side of (11) is determined from an independent experiment. This experiment is to observe a decrease in the ozone mass with time in the same closed volume. The parameter \u03c4 coincides with the time corresponding to the ozone mass M = M 0 \/ 2 . It should be\n\nstressed that, for given volume and temperature, the measured value of \u03c4 is strictly linked to the initial mass of ozone M0 which should be common for both the experiments.\n\nSee in Pdf. Format\n\nRequest Information\n\nShould you have any questions or concerns, please, do not hesitate to contact us either by filling out this form or calling us directly 1-877-486-3761\n\nSubmitting Form...\n\nThe server encountered an error.\n\nCLICK TO VIEW OZONE GENERATORS\n\nOZONE LIBRARY\n\nOZONE APPLICATIONS\n\n10712 181 Street\n\nT5S 1K8\n\nYou need support? Visit our Ozone User support forum on LinkedIn and open tickets for you questions.","date":"2017-08-20 00:05:53","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.814834475517273, \"perplexity\": 1836.4694176569565}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-34\/segments\/1502886105955.66\/warc\/CC-MAIN-20170819235943-20170820015943-00452.warc.gz\"}"} | null | null |
Beate Wischer (* 1969) ist eine deutsche Erziehungswissenschaftlerin.
Leben
Von 1989 bis 1995 studierte sie für das Lehramt der Sekundarstufe I/II in den Fächern Biologie und Pädagogik an der Universität Bielefeld, Abschluss mit dem 1. Staatsexamen. Nach der Promotion 2002 an der Fak. für Pädagogik in Bielefeld (Gutachter: Klaus-Jürgen Tillmann/Wilhelm Heitmeyer) war sie von 2003 bis 2005 Referendarin im Studienseminar für das Lehramt der Sek. I/II in Bielefeld; Ausbildungsschule: Martin-Niemöller-Gesamtschule, Abschluss mit dem 2. Staatsexamen. Von 2008 bis 2016 war sie Professorin für Schulpädagogik mit dem Schwerpunkt Schulforschung/Schultheorie an der Universität Osnabrück. Seit 2016 ist sie Professorin für Erziehungswissenschaft mit dem Schwerpunkt Profession und Organisation im Kontext von Inklusion an der Universität Bielefeld.
Weblinks
ekvv.uni-bielefeld.de
Pädagoge (21. Jahrhundert)
Hochschullehrer (Universität Osnabrück)
Hochschullehrer (Universität Bielefeld)
Deutscher
Geboren 1969
Frau | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,963 |
\section{Introduction} \label{s1}
Symmetries play an important role in quantum field theory.
(For general surveys, one may consult, e.g., Refs.
\cite{Coleman:1985,Itzykson:1980rh,Kaku:1993ym,Peskin:1982mu,
Gasser:1982ap,Manohar:1996cq,Scherer:2002tk,Gasser:2003cg,
Ecker:2006xd,Bernard:2006gx,Kubis:2007iy,Brambilla:2014jmp}.)
They introduce limitations in the choice of possible interactions
for a given physical problem or phenomenon and often they
completely fix the structure of the Lagrangian of the theory.
One distinguishes two types of symmetry, local ones, where the
parameters of the transformations are spacetime dependent, and
global ones, where the latter are spacetime independent.
Local symmetries lead in general to the introduction of gauge
theories, while global symmetries classify particles according to
quantum numbers or predict the existence of massless particles.
\par
QCD, the theory of strong interaction, is a gauge theory with the
local symmetry group $SU(N_c)$, acting in the internal space of
color degrees of freedom. In the real world, $N_c=3$, but leaving
$N_c$ as a free parameter allows one to study the properties of
the theory in more generality. The fundamental fields are the
quarks (matter fields) and the gluons (gauge fields).
\par
Quark fields, with a common mass parameter $m$, belong to the
defining fundamental representation of the color group, which is
$N_c$-dimensional, antiquark fields to the complex conjugate
representation of the latter ($N_c$-dimensional), while the gluon
fields, which are massless, belong to the adjoint representation
($(N_c^2-1)$-dimensional).
\par
\section{Flavor symmetry} \label{s2}
QCD is a confining theory, in the sense that quarks and gluons
are not observed in nature as individual particles. It is admitted
that only gauge invariant objects should be observable. Hadrons
are gauge invariant (color singlet) bound states of quarks and gluons
and are observed in nature as free particles.
\par
From the spectroscopy of hadrons one deduces that there are several
types of quark, currently six, with different masses, having the
same properties with respect to the gluon fields. One distinguishes
them with a new quantum number, called flavor. One thus has the
quarks $u,\ d,\ s$,\ $c,\ b,\ t$.
\par
In the present study, we are mainly interested in the flavor sector
of quarks. To simplify notation, we omit, in general, the color indices
from the quark and gluon fields. The gluons, being flavor singlets, do
not have flavor indices. In gauge invariant quantities, an implicit
summation on color indices is understood. Like the number of quark
colors, $N_c$, the number of quark flavors, designated by $N_f$, will
be considered, for the moment, as a free parameter.
\par
Let us consider the idealized situation where all $N_f$ quarks have
the same mass $m$. We assign the quark and antiquark fields to the
$N_f$-dimensional representation and to its complex conjugate one,
respectively, with respect to the special unitary group $SU(N_f)$
acting in the internal space of flavors. The quark part of the
Lagrangian density becomes
\begin{equation} \label{e2}
\mathcal{L}_q=\overline\psi_a\ \Big(i\gamma^{\mu}
(\partial_{\mu}+igA_{\mu})-m\Big)\ \psi^a.
\end{equation}
(Summation of repeated indices is henceforth assumed: $a$ runs from
1 to $N_f$.)
It is invariant under the continuous global transformations of
the group, with spacetime independent parameters. Designating
the latter by $\alpha^A$ ($A=1,\ldots,N_f^2-1$), one has in
infinitesimal form
\begin{equation} \label{e3}
\delta\psi^a=-i\delta\alpha^A(T^A)_{\ b}^a\psi^b,\ \ \ \ \
\delta\overline\psi_a=i\delta\alpha^A\overline\psi_b(T^A)_{\ a}^{b},
\ \ \ \ \delta A_{\mu}=0,
\end{equation}
where $T^A$ are $N_f\times N_f$ hermitian traceless matrices;
they are the representatives of the generators
of the group and hence satisfy the $SU(N_f)$ algebra
\begin{equation} \label{e4}
[T^A,T^B]=if_{ABC}T^C,\ \ \ \ \ A,B,C=1,\ldots,N_f^2-1.
\end{equation}
The $f$s are called the {structure constants} of the algebra; they
are real and completely antisymmetric in their indices.
\par
Using Noether's theorem, one finds $(N_f^2-1)$ {conserved currents}
\begin{equation} \label{e5}
j_{\mu}^A(x)=-i\frac{\partial \mathcal{L}_q}
{\partial(\partial^{\mu}\psi^a)}(T^A)_{\ b}^a\psi^b
=\overline\psi_a\gamma_{\mu}(T^A)_{\ b}^a\psi^b,\ \ \ \ \ \ \
\partial^{\mu}j_{\mu}^A=0.
\end{equation}
\par
The generators of the group transformations (also called charges) are
obtained from $j_0^A$ by space integration:
\begin{equation} \label{e7}
Q^A=\int d^3x j_0^A(x).
\end{equation}
\par
Because of current conservation, the generators are independent of
time (the fields are assumed to vanish at infinity) and therefore
they commute with the Hamiltonian of the system:
\begin{equation} \label{e8}
[H,Q^A]=0.
\end{equation}
They satisfy, as operators, the $SU(N_f)$ algebra
\begin{equation} \label{e9}
[Q^A,Q^B]=if_{ABC}Q^C,\ \ \ \ \ A,B,C=1,\ldots,N_f^2-1.
\end{equation}
\par
The system under consideration is thus characterized by the existence
of $N_f$ quark fields with {equal} free masses $m$. Since the
interaction term itself is invariant under the group transformations,
the latter property (equality of masses) is preserved after
renormalization.
\par
One might also transcribe the transformation properties of the
fields into similar properties of states.
Ignoring for the moment the confinement problem and introducing
$N_f$ one-particle quark states $|\mathbf{p}, a>$ (spin labels
omitted), created from the vacuum state by the fields $\psi^+_a$,
and assuming the vacuum state is invariant under the transformations
of the group, one obtains
\begin{equation} \label{e10}
Q^A|\mathbf{p}, a>=(T^A)_{\ a}^b|\mathbf{p}, b>.
\end{equation}
Equation (\ref{e8}) implies that the various one-particle states of
the fundamental representation multiplet have equal masses $m$.
\par
This mode of realization of the symmetry is called the
\textit{Wigner-Weyl mode}.
\par
Other types of relationship can be found between observables
involving the multiplet particles, such as form factors or
scattering amplitudes.
\par
\section{Approximate flavor symmetry with hadrons} \label{s3}
Consider now the situation where the quarks have different masses,
which ultimately corresponds to the real world.
The Lagrangian density (\ref{e2}) is replaced by
\begin{equation} \label{e12}
\mathcal{L}_q=\sum_{a=1}^{N_f}\overline\psi_a\ \Big(i\gamma^{\mu}
(\partial_{\mu}+igA_{\mu})-m_a^{}\Big)\ \psi^a.
\end{equation}
The mass term of the Lagrangian density is no longer invariant
under the transformations of the group $SU(N_f)$ and
therefore the currents obtained previously from Noether's theorem
are not conserved. One finds
\begin{equation} \label{e13}
\partial^{\mu}j_{\mu}^A=-i\sum_{a,b=1}^{N_f^{}}(m_a^{}-m_b^{})
\overline\psi_a(T^A)_{\ b}^a\psi^b\neq 0.
\end{equation}
The nonconservation of the currents is thus due to the mass differences
within the representation multiplet.
\par
If, however, the mass differences are much smaller than the
interaction mass scale, which for QCD is of the order of
$\Lambda_{QCD}\sim 300$ MeV, or of a multiple of it,
one is entitled to treat the effects of the mass
differences as perturbations with respect to the symmetric limit
where all the masses are equal. One might write the Lagrangian density
in the form
\begin{equation} \label{e14}
\mathcal{L}=\mathcal{L}_0^{}+\Delta\mathcal{L},
\end{equation}
where $\mathcal{L}_0^{}$ is the Lagrangian density with equal masses
$m$ and $\Delta\mathcal{L}$ corresponds to mass difference terms.
\par
The above procedure does not destabilize the results of the symmetric
theory after renormalization. The reason is that mass operators are
{soft} operators, because their quantized dimension is two for
scalar fields and three for fermion fields, smaller than the dimension
four of the Lagrangian density and in particular of its interaction part.
This has the consequence that mass terms introduce mild effects through
renormalization and at the end the latter remain perturbative.
In particular, they do not affect the currents with anomalous dimensions
\cite{Preparata:1969hg}.
\par
The situation would be different had we introduced the symmetry
breaking through the coupling constants of the interaction terms,
by assigning a different coupling constant to each flavor type quark.
Those have dimension four and their effect on renormalization is
{hard}. At the end, one generally does not find any trace of
approximate symmetry, even if at the beginning the coupling constants
had been only slightly modified.
\par
Since hadrons are bound states of quarks and gluons,
$SU(N_f)$ flavor symmetry, and more generally its
approximate realization, should be reflected in their properties.
In the exact symmetry case, hadrons should be classified in
$SU(N_f)$ multiplets with degenerate masses.
\par
Considering the real world with six flavor quarks, we observe
that they are divided in two groups: the \textit{light quarks}
$u,\ d,\ s$ and the \textit{heavy quarks} $c,\ b,\ t$. The mass
differences between the two categories being large ($>\ 1$ GeV),
an approximate symmetry can be expected only within the space of
the three light quarks. Therefore, flavor symmetry would be concerned
either with $SU(2)$ (isospin symmetry), involving the
quarks $u$ and $d$, or with $SU(3)$, involving the quarks
$u,\ d,\ s$.
\par
Concerning isospin symmetry, one notes the approximate equalities of
the masses of the proton and the neutron, of the charged and neutral
pions, of the kaons and of many other groups of particles.
The nucleons and the kaons could be placed in doublet (fundamental)
representations of $SU(2)$ (isospin 1/2), the pions in
the triplet (adjoint) representation (isospin 1), the
$\Delta$s in the quadruplet representation (isospin 3/2), etc.
\par
Since the mass differences within each multiplet are very small,
of the order of a few MeV, one deduces that the difference
between the masses of the quarks $u$ and $d$ is also of the same
order:
\begin{equation} \label{e15}
|m_d-m_u|\sim \mathrm{a\ few\ MeV}.
\end{equation}
(Precise calculations should also include the contributions of the
electromagnetic interaction, which also are of the same order.)
\par
For the realization of $SU(3)$ symmetry, the quarks $u,\ d,\ s$ are
assigned to the defining representation $3$, while their
antiparticles to the representation $\overline 3$.
\par
Since mesons are made from one quark and one antiquark fields, they
would be classified through the product representation
\begin{equation} \label{e16}
3\otimes\overline{3}\ =\ 8\ \oplus\ 1,
\end{equation}
where $8$ is the octet representation and $1$ the singlet one. On
phenomenological grounds, one notices that for example the vector
mesons $\rho$, $K^*$, $\omega$, $\phi$ can be grouped in an octet
plus a singlet and similarly for the pseudoscalar mesons $\pi$,
$K$, $\eta$, $\eta'$, although for them the mass differences are
much larger than for the vector mesons. (The understanding of
the latter phenomenon comes with chiral symmetry).
\par
Baryons, being made from three quark fields, would be classified
through the product representation
\begin{equation} \label{e17}
3\otimes 3\otimes 3\ =\ 10\ \oplus\ 8\ \oplus 8\ \oplus 1.
\end{equation}
This is the case for example of the baryons $N$, $\Sigma$, $\Xi$ and
$\Lambda$, which may be classified in an octet, of the baryons
$\Delta$, $\Sigma'$, $\Xi'$ and $\Omega$, which may be classified in
a decuplet, etc.
\par
The approximate $SU(3)$ symmetry leads to many relations for the mass
differences of a given multiplet ({Gell-Mann--Okubo formulas}), for the
coupling constants of particles belonging to multiplets, for form
factors, etc.
\par
The mass differences within an $SU(3)$ multiplet being of the order
of 100 MeV, one deduces that the mass difference between the quark
$s$ and the quarks $u$ and $d$ is of the same order of magnitude and
much greater than the mass difference between $u$ and $d$:
\begin{equation} \label{e18}
\big(m_s-(m_u+m_d)/2\big)\sim 100\ \mathrm{MeV}\ \gg |(m_d-m_u)|
\sim \mathrm{a\ few\ MeV}.
\end{equation}
\par
The approximate $SU(3)$ symmetry can also be used for
hadrons containing heavy quarks; these, however, should stand as
backgrounds for the group analysis.
For example one could apply the analysis to mesons made of
$\overline q_hq_{\ell}$, or to baryons made of $q_hq_{\ell}q_{\ell}$,
where $q_h=c,\ b,\ t$ and $q_{\ell}=u,\ d,\ s$.
\par
\section{Chiral symmetry} \label{s4}
Fermion fields may undergo other types of unitary transformation than
those met with the flavor symmetry. They are generated with the inclusion
of the matrix $\gamma_5^{}$ in the former transformations and are called
\textit{axial flavor} transformations, since they change the parity
properties of fields. In infinitesimal form they are of the type
\begin{equation} \label{e19}
\delta\psi^a=-i\delta\alpha^A(T^A)_{\ b}^a\gamma_5^{}\psi^b,
\ \ \ \ \
\delta\overline\psi_a=-i\delta\alpha^A\overline\psi_b\gamma_5^{}
(T^A)_{\ a}^b,\ \ \ \ \delta A_{\mu}=0,
\end{equation}
where the indices $a,\ b,\ A$ refer to the flavor group $SU(N_f)$
representations met before and the $T$s are hermitian traceless
matrices representing the corresponding generators in the fundamental
representation.
\par
We consider the quark part of the QCD Lagrangian density with
equal mass quarks [Eq. (\ref{e2})].
Under the above transformations, this Lagrangian density is not
invariant, because of the presence of the mass terms:
\begin{equation} \label{e20}
\delta\mathcal{L}_q=2im\delta\alpha^A\overline\psi_a
(T^A)_{\ b}^a\gamma_5^{}\psi^b.
\end{equation}
{Thus, invariance under axial flavor transformations requires
vanishing of the quark mass terms.} Contrary to the ordinary flavor
symmetry transformations, equality of masses is no longer sufficient
for ensuring invariance.
\par
In the more general case of unequal masses, the mass terms can be
represented in the form of a {diagonal matrix} $\mathcal{M}$, such
that $\mathcal{M}=\mathrm{diag}(m_1^{},m_2^{},\ldots,m_{N_f}^{})$.
In that case the variation of the Lagrangian density is
\begin{equation} \label{e21}
\delta\mathcal{L}_q=i\delta\alpha^A\overline\psi_a
\{\mathcal{M},T^A\}_{\ b}^a\gamma_5\psi^b,
\end{equation}
where $\{,\}$ is the anticommutator.
\par
We now consider the case of massless quarks. The Lagrangian density
is invariant under both the flavor and axial flavor transformations.
The conserved currents are
\begin{eqnarray}
\label{e22}
& &j_{\mu}^A(x)=\overline\psi_a\gamma_{\mu}(T^A)_{\ b}^a\psi^b,
\ \ \ \ \ \ \ \ \partial^{\mu}j_{\mu}^A=0, \\
\label{e23}
& &j_{5\mu}^A(x)=\overline\psi_a\gamma_{\mu}\gamma_5^{}(T^A)_{\ b}^a
\psi^b,\ \ \ \ \ \ \ \ \partial^{\mu}j_{5\mu}^A=0.
\end{eqnarray}
\par
The corresponding charges are defined from space integration on the
current densities:
\begin{equation} \label{e24}
Q^A=\int d^3x j_0^A(x),\ \ \ \ \ \ Q_5^A=\int d^3x j_{50}^A(x).
\end{equation}
\par
The flavor and axial flavor transformations form the set of
\textit{chiral} transformations. The corresponding charges satisfy
the following algebra:
\begin{eqnarray} \label{e25}
& &[Q^A,Q^B]=if_{ABC}Q^C,\ \ \ \ \ [Q^A,Q_5^B]=if_{ABC}Q_5^C,
\nonumber \\
& &[Q_5^A,Q_5^B]=if_{ABC}Q^C,\ \ \ \ \ A,B,C=1,\ldots,N_f^2-1.
\end{eqnarray}
Note that the axial charges do not form alone an algebra. The previous
algebra can, however, be simplified and become more transparent.
Define
\begin{equation} \label{e26}
Q_L^A=\frac{1}{2}(Q^A-Q_5^A),\ \ \ \ \ \ \ Q_R^A=\frac{1}{2}(Q^A+Q_5^A).
\end{equation}
($L$ for left-handed and $R$ for right-handed.) One obtains
\begin{equation} \label{e27}
[Q_L^A,Q_L^B]=if_{ABC}Q_L^C,\ \ \ \ \ \ \ [Q_R^A,Q_R^B]=if_{ABC}Q_R^C,
\ \ \ \ \ \ \ \ [Q_L^A,Q_R^B]=0.
\end{equation}
Therefore, the left-handed and right-handed charges are decoupled
and operate separately. Each of them generate an $SU(N_f)$
group of transformations. The whole chiral group is then decomposed
into the direct product of two $SU(N_f)$ groups, which will
be labeled with the subscripts $L$ and $R$, respectively:
\begin{equation} \label{e28}
\mathrm{Chiral\ group}\ =\ SU(N_f)_L^{}\ \otimes\ SU(N_f)_R^{}.
\end{equation}
\par
The ordinary flavor transformations form a subgroup of these,
denoted $SU(N_f)_V^{}$:
\begin{equation} \label{e29}
\mathrm{Flavor\ group}\ =\ SU(N_f)_V^{}.
\end{equation}
(The subscript $V$ refers to the vector nature of the corresponding
currents.)
\par
\section{Explicit chiral symmetry breaking} \label{s5}
In nature, quarks have masses. Therefore, chiral symmetry cannot
be an exact symmetry of the QCD Lagrangian. The quark mass terms
introduce an {explicit} chiral symmetry breaking.
\par
The symmetry breaking could be treated as a perturbation only if
the quark masses are much smaller than the QCD mass scale. This
eliminates the heavy quarks $c,\ b,\ t$ from the domain of
investigations. We are left with the sector of light quarks
$u,\ d,\ s$ and the {approximate chiral symmetry}
$SU(3)_L\otimes SU(3)_R$.
\par
What would be the signature of this approximate symmetry in
nature?
The axial charges $Q_5^A$ are pseudoscalar objects. When acting,
in the chiral symmetry limit (massless quarks) and in the Wigner-Weyl
mode, on a massive hadronic state belonging to a multiplet of the
flavor group, they would produce new states with the same mass and spin,
but with opposite parity:
\begin{equation} \label{e30}
Q_5^A\ |M,s,\mathbf{p},+,a>=(T^{\prime A})_{\ a}^b|M,s,\mathbf{p},-,b>.
\end{equation}
The $+$ and $-$ labels refer to the intrinsic parity of the states;
for definiteness, we have assumed above that the initial state has
positive parity. (For chiral singlet representations the matrix
$T'$ is null.)
\par
Thus, if we adopt the Wigner-Weyl mode of realization of chiral
symmetry, we should find \textit{parity doublets} for most of
massive states.
\par
When the light quarks obtain masses, the degeneracy of masses within
the parity doublets would be removed, but the masses would still
remain close to each other.
However, no parity doublets of massive particles with approximately
equal masses are found, in general, in the hadronic world.
This observation forces us to abandon the Wigner-Weyl mode of
realization of chiral symmetry.
\par
The other alternative that remains is the \textit{phenomenon of
spontaneous symmetry breaking}, also called the
\textit{Nambu-Goldstone mode} of realization of chiral symmetry
\cite{Goldstone:1961eq,Goldstone:1962es,Nambu:1960xd,Nambu:1961tp}.
\par
\section{Spontaneous chiral symmetry breaking} \label{s6}
An inherent assumption within the Wigner-Weyl mode is that the
ground state of the theory (the {vacuum state} in QFT) is
{invariant} under the symmetry group of transformations.
This means that the generators of the transformations
annihilate the vacuum state:
\begin{equation} \label{e31}
Q^A\ |0>=0.
\end{equation}
(Here, the specific case of vector flavor charges has been considered.)
\par
Since one-particle states are constructed from the action of the
fields on the vacuum state, the above property guarantees that
one-particle states do transform as elements of irreducible
representations of the symmetry group.
\par
However, the vacuum state (the ground state) may not be invariant
under the symmetry group of transformations, even if the Lagrangian
is. In that case, it is not a symmetric state and the generators of
the symmetry group do not annihilate it:
\begin{equation} \label{e32}
Q_5^A\ |0>\neq 0.
\end{equation}
(Here, the specific case of axial-vector flavor charges has been
considered.)
It is said that the symmetry is \textit{spontaneously broken}.
\par
A similar situation is found in the well-known {sigma model}
\cite{GellMann:1960np} or in the $O(N)$ model of spin-0 fields
\cite{Manohar:1996cq}.
Here, one studies the properties of the potential energy at the classical
level. According to the values of the parameters of theory, the potential
energy may have a non-symmetric ground state.
\par
The absence of parity doublets for massive hadronic states forces us
to explore the possibility of spontaneous breaking of chiral symmetry.
The axial charges, when applied on the vacuum state, would now produce
new states:
\begin{equation} \label{e33}
Q_5^A\ |0>=|\chi^A, ->,\ \ \ \ \ \ A=1,...,8.
\end{equation}
\par
These states have the same quantum properties as their generating
axial charges, the vacuum state being assumed to have a positive
parity and no quantum numbers. In particular, they are pseudoscalar
states. Since in the symmetric limit the charges commute with the
Hamiltonian, the energy of the states $|\chi^A,->$ is null. This is
possible only if there exist \textit{massless pseudoscalar particles},
which might also generate by superposition other many-particle
zero-energy states. This is the content of the \textit{Goldstone theorem}
\cite{Goldstone:1961eq,Goldstone:1962es}.
We note, however, that the states $|\chi^A,->$ are not one-particle
states, but rather a superposition of many-massless-particle states;
their norm is infinite. The fact that these states do not form
new degenerate vacua, but rather represent zero-energy limits of
many-massless-particle states, hinges on the assumption that the
hadronic vacuum state $|0>$ is unique. This is corroborated by the
fact that when the quarks obtain small masses, explicitly breaking
chiral symmetry, these particles become massive, although with small
masses, while the state $|0>$ evolves towards the true vacuum state
of the theory.
\par
Spontaneous chiral symmetry breaking is thus manifested by the
existence of eight pseudoscalar massless particles (mesons), called
\textit{Nambu-Goldstone bosons}.
\par
Spontaneous breaking concerns, however, only the axial sector of the
charges. The ordinary flavor symmetry is still realized with the
Wigner-Weyl mode. Therefore the symmetry group
$SU(3)_L^{}\otimes SU(3)_R^{}$ is spontaneously broken
down to the flavor group $SU(3)_V^{}$:
\begin{equation} \label{e34}
SU(3)_L^{}\otimes SU(3)_R^{}\ \longrightarrow\ SU(3)_V^{}.
\end{equation}
\par
In the real world, where quarks have masses, chiral symmetry will
undergo an additional explicit symmetry breaking. Under this effect,
the eight Nambu-Goldstone bosons will acquire small masses, as compared
to the masses of massive hadrons.
\par
Considering the spectroscopy of mesons, one notices the existence
of eight light pseudoscalar mesons, $\pi,\ K,\ \eta$. They can be
identified, in the chiral limit, with the eight Nambu-Goldstone
bosons.
\par
One would expect that the masses-squared of these particles are
proportional to the masses of the quarks that make their content,
which would explain in turn the mass differences and hierarchies
between them. Thus, one should have (cf. Sect. \ref{ss86})
\begin{equation} \label{e35}
M_P^2=O(\mathcal{M}),
\end{equation}
while for the other hadronic massive states, one should have
the expansion
\begin{equation} \label{e36}
M_h^2\ =\ M_{h0}^2\ +\ O(\mathcal{M}),
\end{equation}
$M_{h0}^2$ being the same for all members of a flavor
multiplet and different from zero in the chiral limit. ($M_{h0}^2$
differs from one flavor multiplet to another.)
This explains why the lightest pseudoscalar meson masses are more
sensitive to the quark masses than those of the massive hadrons.
\par
\section{Properties of Goldstone bosons} \label{s7}
Consider, in the chiral limit, the coupling of the Goldstone
bosons to the axial-vector currents:
\begin{equation} \label{e37}
<0|\ j_{5\mu}^A\ |P^B,p>\ =\ i\delta_{AB}p_{\mu}^{}F_{P0}.
\end{equation}
The coupling $F_{P0}$ is different from zero, since otherwise the axial
charges, which are constructed from $j_{50}^A$, could not create from
the vacuum state zero-energy states.
Using conservation of the axial vector currents and the operator
equation $\partial^{\mu}j_{5\mu}^A=i[P_{\mu},j_{5\mu}^A]$, one
obtains
\begin{equation} \label{e38}
<0|\ \partial^{\mu}j_{5\mu}^A\ |P^B,p>\ =\ 0\ =\
\delta_{AB}p^2F_{P0}\ =\ \delta_{AB}M_{P0}^2F_{P0},
\end{equation}
which implies that $M_{P0}=0$. This is another verification of the
Goldstone theorem.
\par
Consider now the coupling of a {massive} pseudoscalar state
(for example a radial excitation of the Goldstone boson)
to the axial-vector current:
\begin{equation} \label{e39}
<0|\ j_{5\mu}^A\ |P^{\prime B},p>\ =\ i\delta_{AB}p_{\mu}^{}
F_{P'0}.
\end{equation}
Using conservation of the axial-vector currents one again finds
\begin{equation} \label{e40}
<0|\ \partial^{\mu}j_{5\mu}^A\ |P^{\prime B},p>\ =\ 0\ =\
\delta_{AB}p^2F_{P'0}\ =\ \delta_{AB}M_{P'0}^2F_{P'0}.
\end{equation}
However, since $M_{P'0}\neq 0$, one deduces that $F_{P'0}=0$.
This means that the massive pseudoscalar states {decouple} from
the axial-vector currents.
\par
The above results can be summarized as follows:
\begin{eqnarray}
\label{e41}
& &\mathrm{Goldstone\ bosons}:\ \ \ M_{P0}^2=0,\ \ \ F_{P0}\neq 0.\\
\label{e42}
& &\mathrm{Massive\ pseudoscalar\ mesons}:\ \ \ M_{P'0}^2\neq 0,\ \ \
F_{P'0}=0.
\end{eqnarray}
\par
When quarks obtain masses, the above properties are modified
by terms proportional to the quark masses.
\begin{eqnarray}
\label{e43}
& &\mathrm{Goldstone\ bosons}:\ \ \ M_P^2\ =\ O(\mathcal{M}),
\ \ \ \ \ \ F_P^{}\ =\ F_{P0}^{}\ +\ O(\mathcal{M}),\\
\label{e44}
& &\mathrm{Massive\ pseudoscalar\ mesons}:
M_{P'}^2\ =\ M_{P'0}^2\ +\ O(\mathcal{M}),\ \ \ \ \ \
F_{P'}^{}\ =\ O(\mathcal{M}),\\
\label{e45}
& &M_P^2\ll M_{P'}^2,\ \ \ \ \ \ \ \ |F_{P'}|\ll |F_{P}|.
\end{eqnarray}
\par
On experimental grounds, from the leptonic decays of $\pi$ and
$K$ mesons, one has \cite{Agashe:2014kda}
\begin{equation} \label{e46}
F_{\pi}\simeq 92\ \mathrm{MeV},\ \ \ \ \ \ F_K\simeq 110\ \mathrm{MeV}.
\end{equation}
The quantity $(F_K/F_{\pi}-1)\simeq 0.2$ measures the
order of magnitude of flavor $SU(3)$ breaking.
\par
\section{Low-energy theorems} \label{s8}
The decoupling of the massive pseudoscalar mesons from the axial-vector
currents in the chiral limit (massless quarks) allows one to derive
low-energy theorems concerning processes where enters at least
one Goldstone boson. Most of these relations are obtained with the
aid of the Ward-Takahashi identities.
\par
Contrary to to the ordinary flavor symmetry, chiral symmetry does not
yield linear relations between matrix elements of multiplets, but
rather leads to relations between processes involving absorption and/or
emission of Goldstone bosons at low momenta, for example between
the process $\alpha\ \rightarrow\ \beta$ and the process
$\alpha\ +\ n_1P\ \rightarrow\ \beta\ +\ n_2P$, where $n_1$ Goldstone
bosons are absorbed and $n_2$ Goldstone bosons are emitted
($n_1\ge 0$, $n_2\ge 0$).
\par
\textit{Remark.} In QCD, the quark masses appear as free parameters.
Therefore, one expects that all hadronic physical quantities -- masses,
decay couplings, coupling constants, form factors, scattering
amplitudes -- possess analyticity properties in them, up to the
existence of cuts or branching points.
These objects appear in general as residues of Green's functions
at physical particle poles. Therefore, they define
\textit{on-mass shell} quantities. They should not be considered
as functions of the mass-shell variables $p^2,\ p^{\prime 2}$,
etc., but only of the quark mass parameters and of the momentum
transfers or of the Mandelstam variables $s,\ t,\ u$, etc., which,
eventually may take unphysical values by analytic continuation.
Green's functions, on the other hand, may be functions of the
mass-shell variables $p^2,\ p^{\prime 2}$, etc.
\par
\subsection{Goldberger-Treiman relation} \label{ss81}
Consider, in the isospin limit, the matrix element of the axial-vector
current between proton and neutron states \cite{Goldberger:1958vp}:
\begin{equation} \label{e47}
<p(p')|\ j_{5\mu}^{1+i2}\ |n(p)>=\overline u_p(p')\ \Big[\
\gamma_{\mu}\gamma_5^{}g_A^{}(q^2)+q_{\mu}\gamma_5^{}h_A^{}(q^2)\
\Big]\ u_n(p),
\end{equation}
where $q=(p-p')$; $g_A^{}$ and $h_A^{}$ are the axial-vector form
factors of the nucleons.
\par
Taking the divergence of the current, one obtains
\begin{equation} \label{e48}
<p(p')|\ \partial^{\mu} j_{5\mu}^{1+i2}\ |n(p)>=
-i\Big(2M_N^{}g_A^{}(q^2)+q^2h_A^{}(q^2)\Big)\
\overline u_p(p')\gamma_5^{}u_n(p).
\end{equation}
The left-hand side has singularities through the contribution of
pseudoscalar intermediate states.
For simplicity and illustrative purposes, let us assume that the
latter can be saturated by a series of narrow-width particles,
composed of the pion ($\pi$) and of its radial excitations
($\pi^n$, $n=1,2,\ldots$). It can be shown
that multipion states, because of phase space, do not contribute
to the final result. One obtains
\begin{equation} \label{e49}
<p(p')|\ \partial^{\mu} j_{5\mu}^{1+i2}\ |n(p)>=-2i\Big\{
\frac{M_{\pi}^2F_{\pi}}{M_{\pi}^2-q^2}g_{\pi NN}
+\sum_{n=1}^{\infty}\frac{M_{\pi^n}^2F_{\pi^n}}
{M_{\pi^n}^2-q^2}g_{\pi^n NN}\Big\}\ \overline u_p(p')\gamma_5u_n(p),
\end{equation}
where $g_{\pi NN}^{}$ and $g_{\pi^n NN}^{}$ are the coupling constants
of the pseudoscalar mesons with the nucleons. Equation (\ref{e48})
becomes
\begin{equation} \label{e50}
2\frac{M_{\pi}^2F_{\pi}}{M_{\pi}^2-q^2}g_{\pi NN}
+2\sum_{n=1}^{\infty}\frac{M_{\pi^n}^2F_{\pi^n}}
{M_{\pi^n}^2-q^2}g_{\pi^n NN}\ =\ 2M_N^{}g_A^{}(q^2)+q^2h_A^{}(q^2).
\end{equation}
We take the limit $q^2=0$. In the right-hand side,
$h_A(q^2)$ does not have a pole at this value (no massless pseudoscalars
in the real world). This gives
\begin{equation} \label{e51}
F_{\pi}g_{\pi NN}+\sum_{n=1}^{\infty}F_{\pi^n}g_{\pi^n NN}\
=\ M_N^{}g_A^{}(0).
\end{equation}
Consider now the $SU(2)_L^{}\otimes SU(2)_R^{}$ chiral limit
(massless $u$ and $d$ quarks). According to Eq. (\ref{e42}),
all massive pseudoscalar mesons decouple from the axial-vector current.
Equation (\ref{e51}) reduces to the relation
\begin{equation} \label{e52}
g_A^{}(0)\ =\ \frac{F_{\pi}g_{\pi NN}}{M_N^{}}.
\end{equation}
This is an exact result of QCD in the $SU(2)_L^{}\otimes SU(2)_R^{}$
chiral limit (a low-energy theorem). Note that it is independent
of the order of the two limits that were taken. Had we first
considered the chiral limit, then the left-hand side of Eq.
(\ref{e48}) would vanish, but at the same time in the right-hand
side the form factor $h_A^{}(q^2)$ would display a pion pole at
the position $q^2=0$ and would contribute to the equation.
\par
The experimental values are: $g_A^{}\simeq 1.27$ \cite{Agashe:2014kda},
$g_{\pi NN}^{}\simeq 13.15$ \cite{Baru:2010xn}, $F_{\pi}^{}\simeq 92.2$
MeV \cite{Agashe:2014kda}, $M_N^{}=(M_p+M_n)/2=938.92$ MeV. The
right-hand side is then $\simeq 1.29$, to be compared with the left-hand
side, $1.27$. The discrepancy is about $2\%$, which is typical of
the corrections coming from explicit breaking
of $SU(2)_L^{}\otimes SU(2)_R^{}$ symmetry
($M_{\pi}^2/M_N^2\simeq 0.02)$.
\par
\subsection{Ward-Takahashi identities} \label{ss82}
Ward-Takahashi identities (WTI) \cite{Ward:1950xp,Takahashi:1957xn} are
obtained by considering correlation functions of axial-vector currents
with local operators $O(x)$:
\begin{equation} \label{e53}
\int dx e^{iq.x} <\beta\ |Tj_{5\mu}^A(x)O(0)|\ \alpha>,
\end{equation}
where $|\ \alpha>$ and $|\ \beta>$ are hadronic states.
\par
Taking the divergence of the axial-vector current gives
\begin{eqnarray} \label{e54}
& &-iq^{\mu}\int dx e^{iq.x} <\beta\ |Tj_{5\mu}^A(x)O(0)|\ \alpha>\
=\ \int dx e^{iq.x}
<\beta\ |T\partial^{\mu}j_{5\mu}^A(x)O(0)|\ \alpha>\nonumber \\
& &\ \ \ \ \ \ \ \ \ \ \ +\
\int dx e^{iq.x} \delta(x^0) <\beta\ |[j_{50}^A(x),O(0)]|\ \alpha>.
\end{eqnarray}
One notices the presence of the equal-time commutator, which should
be evaluated in the theory. Because of causality, it should
involve a linear combination of $\delta^3(\mathbf{x})$ and a
finite number of its derivatives.
\par
One then takes the limit of low or zero values of $q$
and proceeds with similar methods as in the Goldberger-Treiman case.
\par
The method can also be generalized by considering correlation functions
of the axial-vector currents with a multiple number of local operators.
\par
\subsection{Callan-Treiman relation} \label{ss83}
Choose $|\ \alpha>=|\ K^+>$, $|\ \beta>=|0>$,
$O=j_{\nu}^{4-i5}$ and $j_{5\mu}^A=j_{5\mu}^{3}$ \cite{Callan:1966hu}.
The equal-time commutator of the WTI yields the axial-vector current
$j_{5\nu}^{4-i5}$ plus Schwinger terms that do not contribute.
The matrix element involving the divergence of the
axial-vector current is again saturated by the pion and its radial
excitations. The corresponding residues are proportional to the
$K_{\ell 3}$ form factors with respect to the pion and
to its radial excitations.
\par
One has the definition
\begin{equation} \label{e55}
<\pi^0(p')\ |\ j_{\nu}^{4-i5}\ |\ K^+(p)>=\frac{1}{\sqrt{2}}
\Big[\ (p+p')_{\nu}^{}f_+(t)+(p-p')_{\nu}^{}f_-(t)\ \Big],
\ \ \ \ \ t=(p-p')^2,
\end{equation}
with similar definitions for the radial excitations of the pion.
Take the limit $q\rightarrow 0$ in the WTI [Eq. (\ref{e54})], in which
case the left-hand side vanishes (no poles at $q=0$).
Then the $SU(2)_L\otimes SU(2)_R$ chiral limit is taken
(massless $u$ and $d$ quarks). The massive pseudoscalar states
decouple and one ends up with the relation
\begin{equation} \label{e56}
f_+(M_K^2)+f_-(M_K^2)=\frac{F_K^{}}{F_{\pi}^{}}.
\end{equation}
\par
The physical domain of the $K_{\ell 3}$ form factors
(corresponding to the decay $K\rightarrow \pi\ell\nu$)
being limited by the inequalities
$m_{\ell}^2\leq t\leq (M_K^{}-M_{\pi}^{})^2$, the form
factors appearing in the above relation are evaluated at the
unphysical point $t=M_K^2$. Extrapolations are used from the physical
domain to reach that point. The relation is well satisfied on
experimental grounds with a few percent of discrepancy.
\par
The Callan-Treiman formula establishes a relation between the form
factors of the process $K\rightarrow \pi\ell\nu$ and
the decay coupling of the process $K\rightarrow \ell\nu$.
\par
\subsection{Pion scattering lengths (Weinberg)} \label{ss84}
Choose for $|\ \alpha>$ and $|\ \beta>$ in the WTI [Eq. (\ref{e54})]
target particles like $N$, $K$ or $\pi$, and for $O$ and $j_{5\mu}^A$
axial vector currents with the pion quantum numbers
\cite{Weinberg:1966kf}.
Then calculate the divergences of the two currents. The WTI involves
now two momenta $q$ and $k$ and two equal time commutators. The
momenta squared $q^2$ and $k^2$ are taken to zero, but $q$ and $k$
are maintained nonzero, of the order of $M_{\pi}$.
One ends up with formulas for the S-wave scattering lengths of the
processes $\pi+\alpha\rightarrow \pi+\alpha$, where
$\alpha$ is the target particle, much heavier than the
pion:
\begin{equation} \label{e57}
a_0^I=-\frac{M_{\pi}}{8\pi F_{\pi}^2}(1+\frac{M_{\pi}}{M_{\alpha}})^{-1}
\ [I(I+1)-I_{\alpha}(I_{\alpha}+1)-2],
\end{equation}
where $I$ is the total isospin of the state $|\ \pi\ \alpha>$ and
$I_{\alpha}$ the isospin of the target particle.
This formula is applied to the scattering processes
$\pi+N\rightarrow \pi+N$ and $\pi+K\rightarrow \pi+K$
\cite{Weinberg:1966kf,Tomozawa:1966jm}.
\par
When the target particle is the pion itself, the analysis should be
completed by retaining higher-order terms in the kinematic variables.
Crossing symmetry is also used. One then obtains the $\pi-\pi$
scattering amplitude at low energies:
\begin{equation} \label{e58}
\mathcal{M}_{ac,bd}=\frac{1}{F_{\pi}^2}\{\ \delta_{ac}\delta_{bd}
(s-M_{\pi}^2)+\delta_{ab}\delta_{cd}(t-M_{\pi}^2)
+\delta_{ad}\delta_{bc}(u-M_{\pi}^2)\ \},
\end{equation}
where $a,b,c,d$ are the pion isospin indices and
$s,t,u$ the Mandelstam variables.
\par
The S-wave scattering lengths are
\begin{equation} \label{e59}
a_0^0=\frac{7M_{\pi}}{32\pi F_{\pi}^2}\simeq 0.16\ M_{\pi}^{-1},
\ \ \ \ \ \
a_0^2=-\frac{2M_{\pi}}{32\pi F_{\pi}^2}\simeq -0.046\ M_{\pi}^{-1}.
\end{equation}
The above predictions are well satisfied experimentally within
$10-25\%$ of discrepancy. Direct measurements of the scattering
lengths are, however, not possible because of the instability of
the pion under weak or electromagnetic interactions.
Elaborate extrapolation procedures are used for the extraction of the
scattering lengths from high-energy data.
\par
\subsection{Adler-Weisberger relation} \label{ss85}
The starting point is the same as for the calculation of the
scattering lengths [Sect. \ref{ss84}]. One chooses for
$|\ \alpha>$ and $|\ \beta>$ in the WTI nucleon states and for $O$
and $j_{5\mu}^A$ axial-vector currents with the pion quantum numbers
\cite{Weisberger:1965hp,Adler:1965ka}. The corresponding two momenta
$q$ and $k$ are taken to zero. In this limit, there is in addition
a nucleon pole that contributes, yielding as a residue the
axial-vector form factor of the nucleon at zero momentum transfer.
At the end of the operations, one obtains the isospin antisymmetric
part of the pion-nucleon scattering amplitude at an unphysical point.
The latter is reexpressed by means of a dispersion relation in terms
of an integral over physical pion-nucleon cross sections. The final
formula is
\begin{equation} \label{e60}
g_A^2=1-\frac{2F_{\pi}^2}{\pi}\int_{\nu_0}^{\infty}\ \frac{d\nu}{\nu}\
\Big[\sigma^{\pi^-p}(\nu)-\sigma^{\pi^+p}(\nu)\Big],\ \ \ \ \ \
\nu=q.p\ .
\end{equation}
The right-hand side yields for $g_A$ the value 1.24, to be
compared with its experimental value 1.27.
\par
\subsection{Gell-Mann--Oakes--Renner formulas} \label{ss86}
Choose in the WTI [Eq. (\ref{e54})] $|\ \alpha>=|\ \beta>=|0>$
and for $O$ the divergence of an axial-vector current
\cite{GellMann:1968rz}. One has
\begin{equation} \label{61}
\partial^{\nu}j_{5\nu}^B=i\overline\psi_a
\{\mathcal{M},T^B\}_{\ b}^a\gamma_5\psi^b.
\end{equation}
The operators $v^B=-i\overline\psi_a(T^B)_{\ b}^a
\gamma_5\psi^b$ define the \textit{pseudoscalar densities}.
They transform, in the chiral limit, under the action of the
axial charges as
\begin{equation} \label{e62}
[Q_5^A,v^B]=id_{ABC}^{}u^C+i\frac{1}{3}\delta_{AB}u^0,
\end{equation}
where the $u$s are the \textit{scalar densities}
\begin{equation} \label{e63}
u^C=\overline\psi_a(T^C)_{\ b}^a\psi^b,\ \ \ \ \
C=1,\ldots,8,\ \ \ \ \ u^0=\overline\psi_a \psi^a,
\end{equation}
and the coefficients $d$ are fully symmetric in their
indices; they result from the anticommutators of the matrices
$T$: $\{T^A,T^B\}=d_{ABC}T^C+\frac{1}{3}\delta_{AB}{\mathbf{1}}$.
\par
The WTI takes the form
\begin{eqnarray} \label{e64}
& &-iq^{\mu}\int dx e^{iq.x} <0|T\ j_{5\mu}^A(x)\
\partial^{\nu}j_{5\nu}^B\ |0>\ =\
\int dx e^{iq.x} <0|T\ \partial^{\mu}j_{5\mu}^A(x)\
\partial^{\nu}j_{5\nu}^B\ |0>\nonumber \\
& &\ \ \ \ \ \ \ \ \ \ +\ \int dx e^{iq.x} \delta(x^0)
<0|\ [j_{50}^A(x),\partial^{\nu}j_{5\nu}^B]\ |0>.
\end{eqnarray}
\par
Intermediate states are only pseudoscalar mesons. In the limit
$q=0$ there are no poles to contribute in the left-hand side and
the latter vanishes, yielding
\begin{equation} \label{e65}
\delta_{AB}\Big\{\ M_{P^A}^2F_{P^A}^2+\sum_{n=1}^{\infty}
M_{P^{nA}}^2F_{P^{nA}}^2\ \Big\}\ =\ -\mathrm{tr}
\{T^A,\{\mathcal{M},T^B\}\}\ <0|\frac{1}{3}u^0|0>.
\end{equation}
From Eqs. (\ref{e43}) and (\ref{e44}) we deduce that
$F_{P^{An}}^2\ =\ O(\mathcal{M}^2)$, which is much smaller
than $O(\mathcal{M})$ terms. Keeping only $O(\mathcal{M})$
quantities, one obtains:
\begin{equation} \label{e66}
\delta_{AB}M_{P^A}^2F_{P0}^2=-\frac{}{}\mathrm{tr}
\{T^A,\{\mathcal{M},T^B\}\}<0|\frac{1}{3}u^0|0>.
\end{equation}
\par
The matrix $\mathcal{M}$ is decomposed in the following way along
the matrices $T$:
$\mathcal{M}=(m_u-m_d)T^3-\frac{1}{\sqrt{3}}(2m_s-m_u-m_d)T^8
+\frac{1}{3}(m_u+m_d+m_s)\mathbf{1}$; the relation
tr$(T^AT^B)=\frac{1}{2}\delta_{AB}$ fixes the normalization of the
matrices $T$. Defining
\begin{equation} \label{e67}
B=-\frac{1}{3F_{P0}^2}<0|u^0|0>,\ \ \ \ \ \ \
\hat m=\frac{1}{2}(m_u+m_d),
\end{equation}
and neglecting electromagnetism and $\pi^0-\eta-\eta'$ mixings,
Eq. (\ref{e66}) decomposes into the following equations:
\begin{equation} \label{e68}
M_{\pi^{+}}^2=M_{\pi^{0}}^2=2\hat mB,\ \
M_{K^{+}}^2=(m_s+m_u)B,\ \ M_{K^{0}}^2=(m_s+m_d)B,\ \
M_{\eta}^2=\frac{2}{3}(2m_s+\hat m)B.
\end{equation}
One also verifies here the $SU(3)_V$ Gell-Mann--Okubo formula
(in the isospin limit):
\begin{equation} \label{e69}
4M_{K}^2-3M_{\eta}^2-M_{\pi}^2=0.
\end{equation}
\par
The masses $M_P^{}$ and decay couplings $F_P^{}$ are physical
quantities; threfore, they do not depend on renormalization mass
scales. However, the quark masses and the scalar densities are
renormalized under interaction and depend on the renormalization
mass scale, although their product does not. One must specify,
when providing values for the quark masses, at which scale
they have been evaluated. (Usually, they are chosen at a mass scale
$\mu=2$ GeV.) Also the ratios of quark masses are renormalization
group invariant.
\par
Numerically, one finds from the above formulas
\begin{equation} \label{e70}
\frac{m_s}{\hat m}=26.0,\ \ \ \ \frac{m_u}{m_d}=0.65,\ \ \ \ \
\frac{m_s}{m_d}=21.5.
\end{equation}
One deduces that $m_d>m_u$ and $m_s\gg \hat m$ (cf. also Eq.
(\ref{e18})). More precise results are obtained by incorporating
electromagnetic effects \cite{Gasser:1982ap,Leutwyler:1996qg}.
\par
We consider now the vacuum expectation value of the scalar density
$u^0$: $<0|u^0|0>$. With respect to flavor $SU(3)$,
$u^0$ is a singlet operator. However, with respect to
the chiral group $SU(3)_L^{}\otimes SU(3)_R^{}$, it has a more
complicated structure. To display it, we introduce left-handed and
right-handed quark and antiquark fields:
\begin{equation} \label{e71}
\psi_L^a=\frac{1}{2}(1-\gamma_5)\psi^a,\ \
\psi_R^a=\frac{1}{2}(1+\gamma_5)\psi^a,\ \
\overline \psi_{La}=\frac{1}{2}\overline \psi_a(1+\gamma_5),\ \
\overline \psi_{Ra}=\frac{1}{2}\overline \psi_a(1-\gamma_5).
\end{equation}
In terms of them, $u^0$ is expressed as
\begin{equation} \label{e72}
u^0=\overline \psi_a\psi^a=\overline \psi_{Ra}\psi_L^a
+\overline \psi_{La}\psi_R^a.
\end{equation}
The left-handed and right-handed fields belong to representations of
different groups, $SU(3)_L^{}$ and $SU(3)_R^{}$, respectively.
One finds that $u^0$ belongs to the
$(\overline 3_L,3_R)+(3_L,\overline 3_R)$ representation
of the chiral group. This is not the singlet representation.
If the vacuum were invariant under chiral transformations, then
$<0|u^0|0>$ would be zero, according to the Wigner-Eckart
theorem. Its nonvanishing is a sign that the vacuum is not invariant
under chiral transformations and therefore chiral symmetry is
spontaneously broken. In this case, $<0|u^0|0>$ represents an
\textit{order parameter} of spontaneous chiral symmetry breaking.
One has here an analogy with \textit{ferromagnetism}, where
the magnetization of the atoms has a nonzero value, resulting
from the alignment of their spins along a particular direction,
thus breaking the symmetry of the ground state, which corresponds
to the disordered situation, where the spins are aligned randomly.
\par
\section{Chiral perturbation theory} \label{s9}
After the successes of the predictions of low energy theorems, obtained
in the chiral limit or in leading order of explicit chiral symmetry
breaking, one naturally is interested by the calculation of corrective
terms to the leading-order quantities.
Essentially, there are two types of correction that arise:
1) quark mass term contributions; 2) many-Goldstone boson state
contributions. The latter do not contribute at leading order because
of damping factors coming from phase space coefficients. At nonleading
orders, they are no longer negligible. They produce \textit{unitarity
cuts} with corresponding logarithmic functions of the momenta and the
masses. Early calculations have been done by Li and Pagels
\cite{Li:1971vr,Pagels:1974se}.
\par
Calculation of the nonleading chiral symmetry corrections through a
systematic perturbative approach has been proposed by
Glashow and Weinberg \cite{Glashow:1967rx} and by Dashen
\cite{Dashen:1969eg}.
Initial calculations have been done with the use of the Ward-Takahashi
identities \cite{Dashen:1969bh}. But at higher-orders, this method
becomes rapidly very complicated.
\par
In this context, Weinberg has made several observations
\cite{Weinberg:1968de,Weinberg:1969hw}.
1) In spite of the fact that we are in the domain of strong
interactions, the couplings of the Goldstone bosons with other
particles and with themselves have turned out to be relatively
\textit{weak} in the final results of low-energy theorems.
2) These couplings are reminiscent of \textit{derivative coupling
types}.
3) All results of low-energy theorems could have been obtained (often
more easily) by using phenomenological chiral invariant Lagrangians
involving only the hadrons entering into the concerned processes; it
would be sufficient to do the calculations at the tree level.
Therefore, nonleading contributions to the low-energy theorems could
be evaluated by calculating loop diagrams with the same Lagrangians.
\par
How to construct chiral invariant hadronic Lagrangians?
For this, one should know the chiral transformation properties of
the hadronic fields and in particular of the Goldstone boson fields.
\par
The problem of the realization of chiral symmetry through hadronic
Lagrangians has been systematically studied first by Weinberg
\cite{Weinberg:1968de,Weinberg:1969hw}.
Considering the case of $SU(2)_L^{}\otimes SU(2)_R^{}$ symmetry and
denoting $\pi^A$ ($A=1,2,3$) the pion fields, which belong to the
triplet representation of the flavor group (the isospin), one has
\begin{equation} \label{e73}
[Q^A,\pi^B]=i\epsilon_{ABC}^{}\pi^C,
\end{equation}
while the transformation rule under the axial charges is left in
a general form:
\begin{equation} \label{e74}
[Q_5^A,\pi^B]=-i\Big(\delta_{AB}^{}f(\mbox{\boldmath$\pi$}^2)
+\pi^A\pi^Bg(\mbox{\boldmath$\pi$}^2)\Big).
\end{equation}
($\epsilon_{ABC}^{}=f_{ABC}^{}$ for $A,B,C=1,2,3$.)
Conditions are obtained on the functions $f$ and $g$ from the
action of the chiral algebra [Eqs. (\ref{e25})] on the above
equations and the use of the Jacobi identities. Equations are
obtained for $f$ and $g$ that can be solved. The solution is not
unique, due to the fact that any pseudoscalar field, having the
same quantum numbers as the pion, can be considered as
representing the pion field. These various definitions are related
to each other by field transformations, which should not have any
incidence on physical quantities. The general result is that the
functions $f$ and/or $g$ are nonlinear functions of the pion fields.
The chiral transformation rules of the pion fields are therefore
\textit{nonlinear}.
\par
Similarly, one obtains the chiral transformation properties of
massive hadronic states. The latter transform \textit{linearly},
but the transformation coefficients are \textit{nonlinear} functions
of the pion fields.
\par
To construct chiral invariant Lagrangians, one needs the introduction
of \textit{chiral covariant derivatives}, in which the role of
connections is played by pion field dependent terms proportional to
the derivatives of the pion fields.
One thus finds deductively the property that the couplings of the
Goldstone boson fields are of the derivative type.
\par
Another approach to the construction of chiral invariant Lagrangians
has been developed by Callan, Coleman, Wess and Zumino
\cite{Coleman:1969sm,Callan:1969sn}, which also is
more easily applicable to the $SU(3)_L^{}\otimes SU(3)_R^{}$ symmetry
case. Instead of considering the detailed form of the transformation
of the Goldstone fields with respect to the action of the generators
of the group, one here directly considers the transformation rules
at the group element level. For this, one defines the composite
field
\begin{equation} \label{e75}
U(x)=e^{{\displaystyle 2i\phi^A(x)T^A/F_{P0}}},
\end{equation}
where $\phi^A$ represent the Goldstone boson fields ($A=1,\ldots,8$),
$T^A$ the matrix representatives of the flavor group generators in the
`fundamental representation and $F_{P0}$ the Goldstone boson decay
constants in the chiral limit.
This object is recognized as an element of the
$SU(3)_L^{}\otimes SU(3)_R^{}$ group, obtained from the application of
the group operators on the identity element with parameters identified,
up to multiplicative factors, with the Goldstone boson fields. The
ordinary elements of the group are also of similar type, with parameters
being now c-numbers. One has the transformation law
\begin{equation} \label{e76}
U(x)\ \longrightarrow\ \Omega_R^{}(x)\ U(x)\ \Omega_L^+(x),
\end{equation}
where $\Omega_R$ and $\Omega_L$ are elements of the groups
$SU(3)_R^{}$ and $SU(3)_L^{}$, respectively.
(A detailed derivation of Eq. (\ref{e76}) can be found in Refs.
\cite{Manohar:1996cq,Kubis:2007iy}.) They are considered
as $x$-dependent to allow for the introduction of covariant derivatives.
The latter are defined from Eq. (\ref{e76}) with external currents or
sources playing the role of connections. The chiral invariant
Lagrangian is constructed with the aid of the covariant derivatives,
also including couplings to external scalar and pseudoscalar densities.
This method of approach leads naturally, by means of the external
sources, to the definitions of Green's functions of currents and
densities. It has been adopted by Gasser and Leutwyler in their
formulation of the chiral effective field theory of Goldstone bosons
in the framework of QCD \cite{Gasser:1983yg,Gasser:1984gg}.
\par
Explicit chiral symmetry breaking is further introduced through mass
terms of Goldstone bosons \cite{Weinberg:1968de,Weinberg:1969hw} or
through external sources representing quark masses
\cite{Gasser:1983yg,Gasser:1984gg}.
\par
Next, one calculates loop corrections. Since Goldstone boson couplings
are of the derivative type, each loop introduces, through its vertices,
additional powers of the Goldstone boson momenta. At low energies, the
latter are small, of the same order of magnitude as the Goldstone
boson masses. Increasing the number of loops increases in turn the
powers of the momenta and hence decreases the magnitude of the terms
in comparison with the leading ones \cite{Weinberg:1978kz}.
\par
Therefore, one is led to a systematic power counting rule. The
Goldstone boson masses squared $M_{P}^2$ (or equivalently the quark
masses $\mathcal{M}$) and momenta squared, $q^2,\ k^2$, etc., are
considered as perturbation theory parameters. Their higher powers will
correspond to higher-order terms. Thus, the first-order corrections
will involve at most one-loop diagrams. Chiral perturbation theory
involves therefore expansions in quark masses and Goldstone boson
momenta, rather than in coupling constants.
\par
A crucial point that remains to be dealt with is {renormalization}.
In order for the phenomenological hadronic Lagrangian to describe
correctly the underlying theory (QCD), it is necessary to consider
\textit{the most general} chiral invariant Lagrangian together with
all possible mass terms . Therefore, the latter will involve an
infinite series of terms with an increasing number of derivatives and
powers of the masses. It will involve, from the start
an infinite number of unknown constants. However, the Lagrangian will be
ordered according to the number of derivatives or powers of masses it
involves.
For instance, for the purely Goldstone boson Lagrangian one has the
expansion \cite{Gasser:1983yg,Gasser:1984gg}
\begin{equation} \label{77}
\mathcal{L}=\mathcal{L}_2^{}+\mathcal{L}_4^{}+\cdots
+\mathcal{L}_{2n}^{}+\cdots \ ,
\end{equation}
where the generic subscript $2n$ designates the total number of the
derivatives and of the power of the Goldstone boson masses contained
in the related term. At each finite order of the expansion, there
corresponds a finite number of independent terms. The
leading-order contributions come from $\mathcal{L}_2^{}$, those of
the next-to-leading order from $\mathcal{L}_2^{}+\mathcal{L}_4^{}$, etc.
\par
Calculation of loops introduces divergences. These are then absorbed
by the renormalization of the coupling constants of the higher-order
parts of the Lagrangian. These coupling constants are called
\textit{low energy constants (LEC)}. They also contain information about
the contributions of the high-mass particles which do not appear explicitly
in the Lagrangian.
Thus, in the Lagrangian describing $\pi-\pi$ interaction,
the $\rho$-meson is absent, but since the Lagrangian
is describing only low-energy regions, the $\rho$-meson
field is in some sense integrated out from the total Lagrangian of
the theory and its effects are enclosed in the resulting LECs and
the accompanying expressions.
\par
At each finite order of the perturbation theory, there is only a
finite number of LECs, which should be determined from
experiment. Then the theory becomes predictive at that order.
\par
The phenomenological chiral hadronic Lagrangians that are constructed
from the previous principles define \textit{the chiral effective field
theories} designed to describe definite sectors of hadrons.
Applications have concerned processes involving $\pi,\ K\ ,\eta$
mesons \cite{Gasser:1984ux,Gasser:1984pr,Bernard:1990kw,Colangelo:2001df,
Bijnens:2006zp}, leading to improved precision with respect to the
low-energy theorems. The method has been generalized to include
baryon-Goldstone boson interactions \cite{Gasser:1987rb,Jenkins:1990jv,
Burdman:1992gh,Wise:1992hn,Yan:1992gz,Bernard:1992qa,Cho:1992cf,
Meissner:1997ws,Becher:1999he}.
\par
In conclusion, the chiral effective field theory approach enables one
to convert, at low energies and in the hadronic world, the highly
complicated nonperturbative structure of QCD into a more transparent
and familiar framework, with the sole aid of the symmetry properties
of the theory.
\par
\section{Anomalies} \label{s10}
Invariance properties of Lagrangians established at the classical
level may sometimes not survive quantization. In QFT, the
renormalization process deals with divergent quantities, and some
conventional operations, like interchange of limits or translation of
integration variables, may not be legitimate.
\par
That question has been acutely raised with the study of the decay process
of the neutral pion into two photons ($\pi^0\rightarrow\gamma\gamma$).
Using the usual Ward-Takahashi identities (WTI) of chiral symmetry, one
derives a low-energy theorem stating that the corresponding decay width
should vanish in the chiral limit, while, on experimental grounds, it
is far from vanishing or being small. This paradox has been
analyzed by Adler and Bell and Jackiw
\cite{Adler:1969gk,Bell:1969ts}, who have shown that in QED with
massless fermions, coupled to external axial-vector currents and
pseudoscalar densities, triangle diagrams, corresponding to fermion
loops with two vector vertices and one axial-vector vertex, violate,
through the WTI, the conservation of the axial-vector current.
This result is unchanged, and more precisely unrenormalized, by
higher-order radiative corrections \cite{Adler:1969er}.
Adler has proposed to include the above anomalous contribution in the
divergence of the axial-vector current, which now reads, for massive
fermions, \cite{Adler:1969gk}
\begin{equation} \label{78}
\partial^{\mu}j_{5\mu}=2im\overline\psi\gamma_5\psi
+\frac{e^2}{16\pi^2}\epsilon^{\mu\nu\lambda\sigma}F_{\mu\nu}^{}
F_{\lambda\sigma}^{},
\end{equation}
where $F$ is the electromagnetic field strength and $\epsilon$ the
totally antisymmetric tensor ($\epsilon^{0123}=-\epsilon_{0123}^{}=-1$).
Thus, in the presence of electromagnetism, the axial symmetry is
explicitly broken, irrespective of the fermion masses, a result not
evident at the classical level.
\par
From a more general standpoint, the appearance of anomalies can
be understood as coming from the necessity of renormalizing the
fermionic determinant in the path integral formalism. There are
no regularization schemes that preserve chiral invariance
\cite{Fujikawa:1979ay}.
\par
The general structure of anomalies in the case of non-Abelian currents
has been studied by Bardeen and Wess and Zumino
\cite{Bardeen:1969md,Wess:1971yu}. The anomalous
contributions appear through the presence of an odd number of
axial-vector currents at the vertices of fermion one-loop diagrams,
the number of vertices varying from three to five. Considering the
case of one axial-vector current in the presence of other vector
currents, the expression of its divergence becomes
\begin{eqnarray} \label{79}
\partial^{\mu}j_{5\mu}^A&=&i\overline\psi\{\mathcal{M},T^A\}
\gamma_5\psi
+\frac{N_c}{16\pi^2}\epsilon^{\mu\nu\lambda\sigma}\
\mathrm{tr_f}(T^AT^BT^C)\ F_{\mu\nu}^{B}F_{\lambda\sigma}^{C}
\nonumber\\
& &\ \ \ \ \ +\frac{g^2N_f}{32\pi^2}\epsilon^{\mu\nu\lambda\sigma}
\delta_{A0}\sum_{a=1}^{N_c^2-1}G_{\mu\nu}^{a}G_{\lambda\sigma}^{a},
\ \ \ \ \ A=0,1,\ldots,N_f^2-1,
\end{eqnarray}
where the $F$s are non-Abelian color singlet field strengths
of external vector currents, playing the role of sources, the $G$s
the gluon field strengths, $N_f$ and $N_c$ the numbers of quark
flavors and colors, respectively, $\mathrm{tr_f}$ the trace in flavor
space, and we have also included in the formula the contribtion of
flavor singlet currents, with the convention $T^0=\mathbf{1}$.
\par
We note that in the flavor singlet case, the anomaly receives also a
contribution from the gluon fields. The latter is a dimension-four
operator and, contrary to the usual mass terms, its effect is hard
and could not be treated in general as a perturbation (cf. the comments
after Eq. (\ref{e14})). The presence of the anomaly in the divergence
of the singlet axial-vector current explains the large mass value of
the $\eta'$ meson, which cannot be considered, in the chiral limit,
as a Goldstone boson \cite{'tHooft:1986nc}.
\par
The presence of anomalies destroys the renormalizability and unitarity
properties of field theories. This is not the case of QCD itself, since
it does not contain axial-vector couplings; the anomalies may appear here
only when external currents are considered. However, the situation
is different with the Standard Model, which contains axial-vector
couplings through its weak interaction part. A general theorem states
that the anomalies will intrinsically disappear if the sum of the
charges of all fermions of the theory is equal to zero. This is
indeed the case of the Standard Model, where the sum of the charges
of the quarks and of the leptons cancel to zero, thus confirming the
consistency of the theory
\cite{Itzykson:1980rh,Kaku:1993ym,Bouchiat:1972iq}.
\par
Apart from their implication in filling a hole in the chiral
Ward-Takahashi identities, the anomalies play also an important
role in the analysis of the phases of a theory. Such an attempt has
been undertaken by 't Hooft with the concept of 'naturalness'
\cite{'tHooft:1979bh}.
If physics is described by different effective theories at different
scales, the anomaly structure should be preserved from one scale to
the other when the transition concerns only one part of the particles.
(This is a consequence of the zero-anomaly condition for the global
theory.) This requirement can be applied in particular to QCD. The
latter is described at high energies by elementary quark and gluon
fields, while at low energies it is described by hadrons, leptons
remaining the same. As a consequence of the global zero-anomaly
condition, anomalies resulting from both descriptions should match
each other. This is not a trivial requirement, since quarks belong
to the triplet representation of the flavor group (in the case of
$SU(3)_V^{}$), while hadrons belong to higher representations,
like octets, decuplets, etc. The net result of the investigation shows
that when the number of quark flavors is greater than two, only the phase
of spontaneous chiral symmetry breakdown can fulfill the anomaly
matching condition \cite{'tHooft:1979bh,Frishman:1980dq,Coleman:1982yg,
Coleman:1980mx}. The above analysis brings therefore an
indirect theoretical proof of spontaneous chiral symmetry breaking in
QCD with more than two quark flavors. This result has been completed
with the proof of Vafa and Witten that in vector-like gauge theories,
such as QCD, the vector flavor group and parity cannot be spontaneously
broken \cite{Vafa:1983tf,Vafa:1984xg}.
\par
\vspace{0.5 cm}
\noindent
\textbf{Acknowledgments}. I thank the Organizing Committee of the
Conference Quark Confinement and the Hadron Spectrum XII for
inviting me to give this lecture. I also thank V\'eronique Bernard
and Bachir Moussallam for discussions and useful information.
\par
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,616 |
Q: Pinescript backtest on multiple symbols in a single script? I've a working strategy that I would like to test on different forex pairs but switching between all of this by hand is quite time consuming (also between timeframes).
So I'm trying to look for a way to do it, all in my script.
I've read that I can use something like this : AUDCAD_DAILY = security("OANDA:AUDCAD", "D", close) to get a specific timeframe and a specific symbol.
But now here's come the question : Is it possible to create 28 different lines like this and backtest all of them at the same time (in 1 script) ? If yes, how ?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,520 |
est un maxi-single de Yuki Uchida, portant la mention supported by Chuyan écrite sous son nom. Il est écrit et produit par Hiromasa Ijichi, et sort au nouveau format maxi-CD single de 12 cm le au Japon sur le label King Records, deux mois après une précédente version du single au format mini-CD de 8 cm habituel à l'époque : Rakuen. La version maxi-single contient des titres instrumentaux supplémentaires ; elle atteint la du classement Oricon, et reste classé pendant une semaine.
La chanson-titre a été utilisée comme thème musical pour la séquence régulière Swan no Tabi in The World de l'émission télévisée Susume! Denba Shounen's. Uchida se consacrant désormais à sa carrière d'actrice, elle ne sortira plus d'autre disque ; cette chanson figurera finalement sur sa compilation Perfect Best qui sortira onze ans plus tard en 2010.
Liste des titres
Hard Rock Theme Tune
Liens externes
Fiche du maxi-single sur le site de l'oricon
Chanson interprétée par Yuki Uchida
Single musical sorti en 1999 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,944 |
Q: Django DoesNotExist at /admin/login/ while working with Django forms-builder I was using my http://localhost:8000/admin perfectly before but then I installed django-forms-builder package and created a form. Since then I was getting this error while trying to access /admin.
DoesNotExist at /admin/login/
Site matching query does not exist.
Then I opened setting.py and set SITE_ID = 1. This fixed the /admin problem but now when I go to http://localhost:8000/forms/test it says
Now when I comment the SITE_ID=1 in settings.py this forms works normally but then I can't access /admin. When I use the /admin to access the form using the following way it redirects me to http://example.com/forms/test
Please guide me where I am making trouble?
A: This answer is a continuation of the conversation in the comments of the question.
The issue is connected with Django sites framework, not with django-forms-builder. Just replace default value "example.com" with your domain. For the development you need to use your develoment URL.
You experienced an issue because your newly installed app probably use redirects app or django.contrib.contenttypes.views.shortcut, and they uses django.contrib.sites to works as expected.
Also on the admin framework, the "view on site" link uses the current Site to work out the domain for the site that it will redirect to.
To become more familiar with sites framework visit How Django uses the sites framework.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,011 |
Catocala hariti is a moth in the family Erebidae. It is found in Uzbekistan and Tajikistan.
References
hariti
Moths described in 2002
Moths of Asia | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 316 |
\section{Notation and preliminaries}
All filters will be taken on $\omega$ and be free. For an
infinite set $X$, we let $[X]^{< \omega}=\{A\subseteq X:|A|< \omega\}$ and $[X]^{\omega}=\{A\subseteq X:|A|=\omega\}$. For $A, B\in [\omega]^{\omega}$,
$A\subseteq^*B$ means that $A \setminus B$ is finite. If
$S \in [\omega]^\omega$, we say that $S \to \mathcal{F}$ if $S \subseteq^* F$ for every $F \in \mathcal{F}$.
If $\mathcal{F}$ is a filter, then $C(\mathcal{F}) = \{ S \in [\omega]^\omega : S \to \mathcal{F} \}$ is the set of all sequences converging to $\mathcal{F}$.
For a filter $\mathcal{F}$, we let $\mathcal{I}_\mathcal{F} = \{\omega
\setminus F:F\in \mathcal{F}\}$ ({\it the dual ideal}) and for an ideal $\mathcal{I}$,
we let $\mathcal{F}_\mathcal{I} = \{\omega \setminus I : I\in \mathcal{I}\}$ ({\it the dual
filter}). If $\mathcal{F}$ is a filter and
$f:\omega \to \omega$ is a function, then we define the filter $f[\mathcal{F}]=\{ F :f^{-1}(F)\in \mathcal{F}\}$.
For $A \in [\omega]^\omega$, we define $\mathcal{F}_r(A) = \{ B \subseteq A : |A
\setminus B| < \omega\}$. In particular,
$\mathcal{F}_r(\omega) := \mathcal{F}_r$ is the {\it Fr\'echet filter}.
We say that an infinite family $\mathcal{A} \subseteq [\omega]^{\omega}$ is {\it almost disjoint} ($AD$-family) if $A\cap B$ is finite for distinct $A, B \in \mathcal{A}$.
The ideal generated by an $AD$-family $\mathcal{A}$ is
$\mathcal{I}(\mathcal{A})=\{ X \subseteq \omega: \exists \mathcal{A}'\in[\mathcal{A}]^{< \omega}(X \subseteq^*\bigcup \mathcal{A}')\}$.
An $AD$-family $\mathcal{A}$ is called
{\it maximal almost disjoint} ($MAD$-{\it family}) if it is not contained
properly in another $AD$-family. More general, if $\mathcal{B}
\subseteq [\omega]^\omega$, then we say that a family $\mathcal{A}$ is {\it
maximal} in $\mathcal{B}$ if $\mathcal{A} \subseteq \mathcal{B}$ and
for every $B \in \mathcal{B}$ there is $A \in \mathcal{A}$ such that
$|A \cap B| = \omega$. For a nonempty $\mathcal{A}\subseteq [\omega]^{\omega}$, we define
$\mathcal{A}^{\bot}=\{B \in [\omega]^{\omega} : \forall A\in \mathcal{A}(|A\cap B|<\omega)\}$,
$\mathcal{A}^+ = \{B \in[\omega]^{\omega}:\forall A\in \mathcal{A}(|A\cap B| \neq
\emptyset)\}$ and $\mathcal{A}^* = \{B \in[\omega]^{\omega}: |\{ A \in \mathcal{A} : |A \cap
B| = \omega \}| \geq \omega \}$. Notice that, for a filter $\mathcal{F}$, we have that $\mathcal{F}^+ = \mathcal{P}(\omega) \setminus \mathcal{I}_\mathcal{F}$ and,
for an arbitrary ideal $\mathcal{I}$, $\mathcal{I}^\perp$ is always an ideal and $\mathcal{I}
\subseteq \mathcal{I}^{\perp \perp}$.
For each $X \in[\omega]^{\omega}$ and $\mathcal{A} \subseteq [\omega]^{\omega}$ we let
$\mathcal{A}|_{X}=\{ A \cap X : A \in \mathcal{A} \ \text{and} \ |A \cap X|=\omega\}$.
Observe that if $\mathcal{A}$ is an $AD$-family and
$B \in \mathcal{A}^*$, then $\mathcal{A}|_B = \{ A \cap B : A \in \mathcal{A} \ \text{and} \ |A
\cap B| = \omega \}$ is an $AD$-family on $B$.
\medskip
Now, let $\mathcal{F}$ be a filter on $\omega$ and consider the space $\xi(\mathcal{F}) = \omega\cup \{F\}$ whose topology is defined as follows:
All elements of $\omega$ are isolated and the neighborhoods of $\mathcal{F}$ are of the form $\{\mathcal{F}\} \cup F$ where $F \in \mathcal{F}$.
One class of spaces which has been extensively studied in Topology is the following:
\smallskip
A space $X$ is called a {\it Fr\'echet-Urysohn} space (for short $FU$-space) if for each $x\in X$ such that $x\in cl_X A$, there is a sequence in $A$ converging to $x$.
\smallskip
\begin{definition} A filter $\mathcal{F}$ is called a $FU$-{\it filter} if the space $\xi(\mathcal{F})$ is Fr\'echet-Urysohn.
\end{definition}
The ``smallest'' $FU$-filter is the Fr\'echet filter $\mathcal{F}_r$ and the countable $FAN$-filter is also an example of a $FU$-filter which does not have a countable base. By using $AD$-families, in the paper \cite{gu09}, the authors pointed out the existence of $2^\frak{c}$ pairwise non-equivalent $FU$-filters. In other terms, we have that $\mathcal{F}$ is a $FU$-filter iff for every $A \in \mathcal{F}^+$ there is $S \in [A]^\omega$ such that $S \subseteq^* F$ for all $F \in \mathcal{F}$.
\medskip
Two notions that will help us to distinguish $FU$-filters are the following.
\begin{definition}\label{tu} Let $\mathcal{F}$ and $\mathcal{G}$ be two filters on $\omega$.
\begin{enumerate}
\item $\mathcal{F}\leq_{RK}\mathcal{G}$ if there is a function $f: \omega \to \omega$ such that $f[\mathcal{G}] = \mathcal{F}$ (i. e.,
$F\in \mathcal{F}$ iff $f^{-1}(F)\in \mathcal{G}$).
\item ({\bf \cite{tu05} }) $\mathcal{F}\leq_{TU}\mathcal{G}$ if there are $A\in \mathcal{G}^+$, $B\in \mathcal{F}$ and
a bijection $f:A\to B$ such that $f[\mathcal{G}|_{B}]=\mathcal{F}|_A$.
\end{enumerate}
\end{definition}
We assert that these two relations $\leq_{RK}$ and $\leq_{TU}$ are reflexive and transitive but they are
not anti\-sy\-mmetric.
\begin{definition}
Let $\mathcal{F}$ and $\mathcal{G}$ be filters on $\omega$.
\begin{enumerate}
\item $\mathcal{F}\approx\mathcal{G}$ if there is a bijection function $f: \omega \to \omega$ such that
$f[\mathcal{F}]= \mathcal{G}$.
\item $\mathcal{F}\approx_{RK}\mathcal{G}$ if $\mathcal{F}\leq_{RK}\mathcal{G}$ and $\mathcal{G}\leq_{RK}\mathcal{F}$.
\end{enumerate}
\end{definition}
The definition introduce in $(1)$ can be generalized as:
If $A,B\in[\omega]^{\omega}$ and $\mathcal{F}, \mathcal{G}$ are filters on $A$ and $B$,
respectively, then $\mathcal{F}$ and $\mathcal{G}$ are called {\it equivalent}
if there is a bijection $f: A \to B$ such that $f[\mathcal{F}] =
\mathcal{G}$. It is evident that $\mathcal{F} \approx\mathcal{G}$ implies that $\mathcal{F}\approx_{RK}\mathcal{G}$ for every pair of filters $\mathcal{F}$ and $\mathcal{G}$.
However, we do not know if the inverse implication holds for the class of $FU$-filters:
\begin{question}
Are there two $FU$-filters $\mathcal{F}$ and $\mathcal{G}$ such that $\mathcal{F}\approx_{RK}\mathcal{G}$ and $\mathcal{F} \not\approx\mathcal{G}$ ?
\end{question}
By the symbol $\mathcal{F} <_{RK}\mathcal{G}$ we shall understand that $\mathcal{F}\leq_{RK}\mathcal{G}$ and $\mathcal{F} \not\approx\mathcal{G}$.
\medskip
This paper is a continuation of the work done in the article \cite{gr01}.
The second section is devoted to recall some basic properties of the $FU$-filters. We give combinatorial properties which are equivalent to the
$RK$-order and the $TU$-order. These equivalences will allow to construct $FU$-filters with some interesting properties. In the third section,
we show that if $\mathcal{F}\leq \mathcal{F}_{\mathcal{P}}$, then either $\mathcal{F}$ is relatively equivalent to the Fr\'echet
filter or equivalent to $\mathcal{F}_{\mathcal{P}}$. The pre-orders $\leq_{RK}$ and $\leq_{TU}$ are compared in the forth section. We show that $\leq_{TU} \subseteq \leq_{RK}$ in the category of $FU$-filters.
We also prove that
if $\mathcal{A}$ is a $NMAD$-family of size $\mathfrak{c}$ which is completely separable, then
$S_{\mathcal{P}}\nleq_{TU}S_{\mathcal{A}}$ and $S_{\mathcal{P}} \leq_{RK} S_{\mathcal{A}}$. In the fifth section, we construct an $RK$-chain of size $\mathfrak{c}^+$ which is $RK$-above of each $FU$-filter.
The sixth section is devoted to study the $RK$-incomparability of $FU$-filters. We use the $\alpha_i$ properties to show the $RK$- incomparability of certain $FU$-filters. Besides, we construct an infinite $RK$-antichain of $FU$-filters.
\section{Fr\'echet-Urysohn Filters}
In order to study the $FU$-filters and their relationships we list some useful facts.
\medskip
Let $\mathcal{F}$ be a filter on $\omega$.
\begin{enumerate}
\item If $A\in[\omega]^{\omega}$, then $\mathcal{F}\in cl_{\xi(\mathcal{F})}A$ iff $A\in \mathcal{F}^+$.
\item $S \to \mathcal{F}$ iff $S \in \mathcal{I}_\mathcal{F}^{\perp}$.
\item $\mathcal{F}$ is a $FU$-filter iff $\mathcal{I}_\mathcal{F}^{\perp \perp} = \mathcal{I}_\mathcal{F}$.
\end{enumerate}
It is not hard to prove that if $\emptyset \neq \mathcal{D}\subseteq [\omega]^{\omega}$, then
$$
\mathcal{F}_{\mathcal{D}}=\{F\subseteq \omega:\forall D\in \mathcal{D}(D\subseteq^*F)\}
$$
is a $FU$-filter. By using this kind of filters and $AD$-families it is possible to characterize the $FU$-filters as follows:
\begin{lemma}\label{simon}{\bf \cite{si98}} A filter $\mathcal{F}$ is a $FU$-filter iff there is an $AD$-family $\mathcal{A}$ maximal in $I^\perp_{\mathcal{F}}$ such that $\mathcal{F}=\mathcal{F}_{\mathcal{A}}$.
\end{lemma}
We can see directly from this characterization that $\mathcal{A}$ is a $MAD$-family iff $\mathcal{F}_{\mathcal{A}} = \mathcal{F}_r$. More general,
if $\mathcal{F}_{\mathcal{A}}$ is the filter generated by an $AD$-family $\mathcal{A}$ and $B\to \mathcal{F}_{\mathcal{A}}$, then there is $A\in \mathcal{A}$ such that $|A\cap B|=\omega$.
Observe from Lemma \ref{simon} that for every infinite $\mathcal{D}\subseteq [\omega]^{\omega}$, we can find an $AD$-family $\mathcal{A}$ such that
$\mathcal{F}_{\mathcal{D}} = \mathcal{F}_{\mathcal{A}}$. In what follows, when we write $\mathcal{F}_{\mathcal{A}}$ we shall always assume that $\mathcal{A}$ is an $AD$-family.
\medskip
Given a $FU$-filter $\mathcal{G}$, we say that $\mathcal{G}$ is {\it relatively equivalent} to the Fr\'echet
filter iff $A \to \mathcal{G}$ and $A \in \mathcal{G}$. In particular, we have that
$\mathcal{G} = \{ G \cup E : G \in \mathcal{F}_r(A) \ \text{and} \ E \subseteq \omega \setminus A \}$. If
we do not require that the function $f$ to be onto in the definition of the $RK$-order, then we would have that
$\mathcal{F}_r(A) \leq_{RK} \mathcal{F}_r$ for every $A \in [\omega]^{\omega}$. For our convenience, in the definition of $RK$-order, we shall always require
that the function involved be onto. This convenience is based on the next theorem which was proved in \cite{gr01}.
\begin{theorem}\label{new-RK} Let $\mathcal{F}$ and $\mathcal{G}$ filters such that
$\mathcal{G} \neq \mathcal{F}_r$ and $\mathcal{F}$ is not relatively equivalent to the Fr\'echet
filter. If $\mathcal{F} \leq_{RK} \mathcal{G}$, then there is a surjective function $g: \omega \to \omega$ such that $g[\mathcal{G}] = \mathcal{F}$.
\end{theorem}
In virtue of the previous theorem, we remark that if $\mathcal{F}\leq_{RK}\mathcal{F}_r$, then $\mathcal{F}=\mathcal{F}_r$.
It was pointed out in \cite{gr01} that every filter $\mathcal{G}$
which has a nontrivial convergent sequence satisfies that $\mathcal{F}_r\leq_{RK} \mathcal{G}$.
\medskip
Next, let us describe another useful construction of $FU$-filters which has been very important in the construction of special $FU$-filters (see, for instance, the article \cite{si}):
\medskip
For every $AD$-family $\mathcal{A}$, we define $S_{\mathcal{A}}=\mathcal{F}_{\mathcal{I}(\mathcal{A})}$. In general, $S_{\mathcal{A}}$ is not a $FU$-filter, for instance if $\mathcal{A}$
is a $MAD$-family, then $S_{\mathcal{A}}$ does not have any nontrivial convergent sequence. In order that $S_{\mathcal{A}}$ be a $FU$-filter we need some special $AD$-families:
\medskip
An $AD$-family $\mathcal{A}$ is said to be {\it nowhere $MAD$-family} ($NMAD$-family) if
for every $X \in \mathcal{I}(\mathcal{A})^+$ there is $A \in \mathcal{I}(\mathcal{A})^{\bot}\cap [X]^{\omega}$.
We remark that if $\mathcal{A}$ is a $NMAD$-family, then every infinite
subfamily $\mathcal{B}$ of $\mathcal{A}$ is also a $NMAD$-family.
\begin{theorem}\label{sfu} ({\bf \cite{si}}) Given an $AD$-family $\mathcal{A}$, we have that the filter $S_{\mathcal{A}}$ is a $FU$-filter iff $\mathcal{A}$ is a $NMAD$-family.
\end{theorem}
\medskip
If $\mathcal{P}=\{P_n:n<\omega\}\subseteq [\omega]^{\omega}$ is an infinite partition of $\omega$ in infinite subsets, by Theorem \ref{sfu}, then $\mathcal{F}_{\mathcal{P}}$ is an $FU$-filter which is known as the $FAN$-filter. We also know that $S_{\mathcal{P}}$ and $\mathcal{F}_r$ are the only $FU$-filters with countable base. In what follows, when we write $S_{\mathcal{P}}$ we shall understand that $\mathcal{P}$ is an infinite partition of $\omega$ in infinite subsets. For the filters of the form $S_{\mathcal{A}}$ is very easy to know their characters as we shall see next.
\begin{lemma}\label{char-nmad} If $\mathcal{A}$ is an $AD$-family on $\omega$, then $\chi(S_{\mathcal{A}})=|\mathcal{A}|$.
\end{lemma}
The following combinatorial statements are equivalent to the $RK$-order and were proved in \cite[Th. 3.4]{gr01}.
\begin{theorem}\label{theoeq} Let $\mathcal{A}$ and $\mathcal{B}$ be $AD$-families on $\omega$.
The following statements are equivalent.
\begin{enumerate}
\item $\mathcal{F}_{\mathcal{A}}\leq_{RK}\mathcal{F}_{\mathcal{B}}$ via the function $f: \omega \to \omega$.
\item
\begin{enumerate}
\item $\forall F\in \mathcal{F}_{\mathcal{A}}(f^{-1}(F)\in \mathcal{F}_{\mathcal{B}})$, and
\item $\forall G\in \mathcal{F}_{\mathcal{B}}(f[G]\in \mathcal{F}_{\mathcal{A}})$.
\end{enumerate}
\item
\begin{enumerate}
\item $\forall n<\omega \forall B\in \mathcal{B}(|f^{-1}(n)\cap B|<\omega)$,
\item $\forall B\in \mathcal{B} \forall C\in \mathcal{A}^{\bot}(|f[B]\cap C|<\omega)$, and
\item $\forall S \in C(\mathcal{A}) \exists B\in \mathcal{B}(|f^{-1}(S)\cap B|=\omega)$.
\end{enumerate}
\end{enumerate}
\end{theorem}
The next result can be obtained by a slight modification of the proof of the previous theorem.
\begin{theorem}\label{theoeq2} Let $\mathcal{A}$ and $\mathcal{B}$ be $NMAD$-families on $\omega$.
The following statements are equivalent.
\begin{enumerate}
\item $S_{\mathcal{A}}\leq_{RK}S_{\mathcal{B}}$ via the function $f: \omega \to \omega$.
\item
\begin{enumerate}
\item $\forall F\in S_{\mathcal{A}}(f^{-1}(F)\in S_{\mathcal{B}})$, and
\item $\forall G\in S_{\mathcal{B}}(f[G]\in S_{\mathcal{A}})$.
\end{enumerate}
\item
\begin{enumerate}
\item $\forall I\in \mathcal{I}(\mathcal{A}) (f^{-1}(I)\in \mathcal{I}(\mathcal{B}))$, and
\item $\forall M \in \mathcal{I}(\mathcal{A})^{\bot}\cap[\omega]^{\omega} \exists R\in \mathcal{I}(\mathcal{B})^{\bot}\cap[\omega]^{\omega}(|f^{-1}(M)\cap R|=\omega)$.
\end{enumerate}
\end{enumerate}
\end{theorem}
the last two theorems will be very useful in the construction of spacial filters.
\section{$FAN$-filter}
In what follows, $\mathcal{P}$ will stand for an infinite partition of $\omega$ in infinite subsets. In the paper \cite{gr01}, we proved that the $FAN$-filter $\mathcal{F}_\mathcal{P}$ and the $S_\mathcal{P}$ filter are not $RK$-comparable. We know, by lemma \ref{char-nmad}, that the only filters which are $RK$-predecessors of
$S_{\mathcal{P}}$ are either itself or a filter that is relatively equivalent to the Fr\'echet filter. All these remarks
lead us to ask in \cite[Q. 5.12]{gr01} what are the $RK$-predecessors of the $FAN$-filter ?:
\medskip
{\bf Question.} Is there an $AD$-family $\mathcal{A}$ such that $\mathcal{F}_r <_{RK} \mathcal{F}_\mathcal{A} <_{RK} \mathcal{F}_\mathcal{P}$ ?
\medskip
We shall respond this question in the next theorem.
\begin{lemma}
Let $\mathcal{A}$ be an $AD$-family on $\omega$ and $f: \omega \to \omega$ be a surjective function such that
the restriction $f|_{\mathcal{A}}$ is finite-to-one for each $A\in \mathcal{A}$.
If $\mathcal{D}_f :=\{f[A]:A\in\mathcal{A}\}$, then $\mathcal{F}_{\mathcal{D}_f}\leq_{RK}\mathcal{F}_A$ via $f$.
\end{lemma}
\begin{proof}
We have to prove that $\mathcal{F}_{\mathcal{D}_f}=f[\mathcal{F}_{\mathcal{A}}]$. Let $F\in\mathcal{F}_{\mathcal{D}_f}$ and assume
that $F\notin f[\mathcal{F}_{\mathcal{A}}]$. Then there is $A\in \mathcal{A}$ such that
$A\setminus f^{-1}(F)$ is infinite. Then, we have that $f[A\setminus f^{-1}(F)]$ is infinite
and $f[A\setminus f^{-1}(F)]\to \mathcal{F}_{D_f}$, but $f[A\setminus f^{-1}(F)]\cap F=\emptyset$ which is
impossible. Thus $F\in f[\mathcal{F}_{\mathcal{A}}]$. So, we obtain that $\mathcal{F}_{\mathcal{D}_f} \subseteq f[\mathcal{F}_{\mathcal{A}}]$.
Now fix $H\in f[\mathcal{F}_{\mathcal{A}}]$ and suppose that $H\notin \mathcal{F}_{\mathcal{D}_f}$. Then there is $A\in \mathcal{A}$ such that
$f[A]\setminus H$ is infinite. We know that $A\subseteq^*f^{-1}(H)$ which implies that $f[A]\subseteq^* H$, but this is a contradiction.
Thus $H\in \mathcal{F}_{\mathcal{D}_f}$. This proves that $f[\mathcal{F}_{\mathcal{A}}] \subseteq \mathcal{F}_{\mathcal{D}_f}$. Therefore,
$\mathcal{F}_{\mathcal{D}_f}=f[\mathcal{F}_{\mathcal{A}}]$ and hence $\mathcal{F}_{\mathcal{D}_f}\leq_{RK}\mathcal{F}_A$ via the function $f$.
\end{proof}
\begin{lemma}
If $\mathcal{F}_{\mathcal{A}}\leq_{RK}\mathcal{F}_{\mathcal{B}}$ via the function $f$, then $\mathcal{F}_{\mathcal{D}_f} = \mathcal{F}_{\mathcal{A}}$.
\end{lemma}
\begin{proof}
Observe that for each $B\in \mathcal{B}$, $f[B]\subseteq^* F$ for all $F\in \mathcal{F}_{\mathcal{A}}$. Hence, we have that $\mathcal{F}_{\mathcal{A}}\subseteq \mathcal{F}_{\mathcal{D}_f}$.
Now let $G\in \mathcal{F}_{\mathcal{D}_f}$ and suppose that $G \notin \mathcal{F}_{\mathcal{A}}$. Then we can find $A\in \mathcal{A}$ such that
$|A\setminus G|=\omega$. Since $A\setminus G\to \mathcal{F}_{\mathcal{A}}$, by Theorem \ref{theoeq}, there is $B\in \mathcal{B}$ such that $B\cap (f^{-1}(A)\setminus f^{-1}(G))$ is infinite,
but this contradicts the fact $B\subseteq^* f^{-1}(G)$. Thus, we must have that $G\in \mathcal{F}_A$. Therefore, $\mathcal{F}_{\mathcal{D}_f} = \mathcal{F}_{\mathcal{A}}$.
\end{proof}
\begin{theorem}\label{theofan}
If $\mathcal{F}\leq_{RK} \mathcal{F}_{\mathcal{P}}$, then either $\mathcal{F}$ is relatively equivalent to the Fr\'echet
filter or equivalent to $\mathcal{F}_{\mathcal{P}}$.
\end{theorem}
\begin{proof} It is known that $\mathcal{F}$ is an $FU$-filter (for a proof see \cite{gr01}).
Let $f:\omega \to \omega$ be a function such that $f[\mathcal{F}_{\mathcal{P}}]=\mathcal{F}$ and $\mathcal{P}=\{P_n:n<\omega\}$.
Notice that $\mathcal{D}_f=\{f[P_n]:n<\omega\}$ is a cover of $\omega$ and $\mathcal{F}_{\mathcal{D}_f}=\mathcal{F}$ by the previous lemma.
First assume that there is $n < \omega$ such that $|f[P_{m}] \setminus \big(\bigcup_{i<n} f[P_i] \big)| < \omega$ for all $n < m < \omega$.
It is clearly that $\bigcup_{i<n_k} f[P_i]=A\in \mathcal{F}$ and since $A\to \mathcal{F}$, we also have that
$\mathcal{F}$ is relatively equivalent to the Fr\'echet filter. Now,
set $Q_0=f[P_0]$ and inductively define
a pairwise disjoint family $\mathcal{Q}=\{ Q_k : k < \omega\}$ of infinite subsets of $\omega$, and a strictly increasing sequence $(n_k)_{k < \omega}$ in $\omega$ so that $n_0 = 0$,
$Q_{k}=f[P_{n_{k}}]\setminus \big(\bigcup_{i<n_{k-1}} Q_i\big)$ is infinite and $n_{k}$ is the smallest with this property.
Since each element of $\mathcal{Q}$ is contained in an element of $\mathcal{D}_f$, we have that $\mathcal{F}_{\mathcal{D}_f}\subseteq \mathcal{F}_{\mathcal{Q}}$.
Let $F\in \mathcal{F}_{\mathcal{Q}}$ and fix $P_n\in \mathcal{P}$. Let
$k=min\{ i < \omega : n \leq n_i <\omega\}$. Then, by construction, we obtain that
$f[P_n]\subseteq^*\bigcup_{i\leq n_k} Q_i\subseteq^*F$. Hence, $F \in \mathcal{F}$. Therefore, $\mathcal{F}=\mathcal{F}_{\mathcal{Q}}$.
\end{proof}
To finish this section we pose the following question.
\begin{question} Let $\mathcal{F}$ be a $FU$-filter non-equivalent to the Fr\'echet
filter. If $\mathcal{G} \leq_{RK} \mathcal{F}$ implies that either $\mathcal{G}$ is relatively equivalent to the Fr\'echet
filter or equivalent to $\mathcal{G}$, must $\mathcal{F}$ be equivalent to the $FAN$-filter ?
\end{question}
\section{Todor\v{c}evi\'c-Uzc\'ategui pre-order}
In this section, we shall compare the $FU$-filters by using the pre-order which has been introduced in Definition \ref{tu}. The first goal is to
compare this pre-order with the $RK$-order. To do that we
reformulate the definition of the $TU$-order to make it a little bite easier to handle. Before this reformulation we need to prove a well-known property of certain filters.
\begin{lemma}\label{lemref}
Let $\mathcal{F}$ be a filter non-relatively equivalent to the Fr\'echet filter.
Then for every $A\in \mathcal{F}\setminus \mathcal{F}_r$ there is a bijection $f: A \to \omega$ such that $f[\mathcal{F}|_{A}]=\mathcal{F}$. In particular,
$\mathcal{F} \approx \mathcal{F}|_A$ for all $A \in \mathcal{F} \setminus \mathcal{F}_r$.
\end{lemma}
\begin{proof} Let $A\in\mathcal{F}\setminus \mathcal{F}_r$.
Since $\mathcal{F}$ is not relatively equivalent to the Fr\'echet filter there is
$B\in \mathcal{F}\cap [A]^{\omega}$ such that $A\setminus B\in [A]^{\omega}$. Define
$f:A\to \omega$ so that $f|_B$ is the identity on $B$ and
$f[A\setminus B]=\omega\setminus B$ as a bijection. Let $F\in \mathcal{F}$. Then we have that
$$f^{-1}(F)=f^{-1}(F\cap B)\cup f^{-1}(F\setminus B)=(F\cap B)\cup f^{-1}(F\setminus B).$$
Since $F\cap B\in \mathcal{F}|_A$, we must have that $f^{-1}(F)\in \mathcal{F}|_A$. This shows that $\mathcal{F}\subseteq \mathcal{F}|_A$.
Now, fix $G\in \mathcal{F}|_A$. Notice that
$$f[G]=f[G\cap B]\cup f[G\setminus B]=(G\cap B) \cup f[G\setminus B].$$
As $G\in \mathcal{F}|_{A}$, there is $H\in \mathcal{F}$ such that $G=H\cap A$ and since
that $G\cap B= H\cap A \cap B\in \mathcal{F}$, then $f[G]\in \mathcal{F}$. Thus, $\mathcal{F}|_A \subseteq \mathcal{F}$. Therefore, $f[\mathcal{F}|_{A}]=\mathcal{F}$.
\end{proof}
The next corollary conveniently reformulates the definition of the $TU$-order.
\begin{corollary} Let $\mathcal{F}$ and $\mathcal{G}$ two filters. Then,
$\mathcal{F}\leq_{TU} \mathcal{G}$ iff either
\begin{enumerate}
\item $\mathcal{F}$ is a relatively equivalent to the Fr\'echet filter, or
\item there are $A\in \mathcal{G}^+$ and a bijection $f:A\to \omega$ such that
$f[\mathcal{G}|_A]=\mathcal{F}$.
\end{enumerate}
\end{corollary}
In virtue of the previous corollary, we shall always assume that the element of the filter witnessing being a $TU$-predecessor is $\omega$.
However, the positive element that witnesses being a $TU$-successor cannot be replace by an element of the $FU$-filter as
in the $TU$-order: that is, if we use only members of the filters in the $TU$-order we obtain the $RK$-order as it is shown in the next corollary.
\begin{corollary}
Let $\mathcal{F}$ and $\mathcal{G}$ be two filters non-relatively equivalent to the Fr\'echet filter. Then $\mathcal{F}\leq_{RK}\mathcal{G}$ iff
for each $A\in \mathcal{G}\setminus \mathcal{F}_r$ and $B\in \mathcal{F}\setminus \mathcal{F}_r$ there is a surjection $f:A\to B$ such that
$f[\mathcal{G}|_{A}]=\mathcal{F}|_B$.
\end{corollary}
We will see in the next theorem that the $TU$-order implies the $RK$-order whenever the $TU$-predecessor lies in the category
of the $FU$-filters.
\medskip
Let us remark that if $A\in \mathcal{G}^+\setminus \mathcal{G}$, then $\omega\setminus A\in \mathcal{G}^+\setminus \mathcal{G}$ and
$$
\mathcal{G}=\mathcal{G}|_{A}\oplus\mathcal{G}|_{\omega\setminus A}:=\{F\cup E: F\in \mathcal{G}|_{A} \ {\text and} \ E\in \mathcal{G}|_{\omega\setminus A} \}.
$$
\begin{theorem}
Let $\mathcal{F}$ and $\mathcal{G}$ two filters such that $C(\mathcal{F})\neq \emptyset$. If
$\mathcal{F}\leq_{TU} \mathcal{G}$, then $\mathcal{F}\leq_{RK} \mathcal{G}$.
\end{theorem}
\begin{proof}
Suppose that $A\in\mathcal{G}^+$ and $f:A\to \omega$ witnesses that $f[\mathcal{G}|_A]=\mathcal{F}$. Fix $M\in C(\mathcal{F})$.
Define $g:\omega\to\omega$ so that $g|_A=f$ and $g[\omega\setminus A]=M$
as a bijection. Let $F\in \mathcal{F}$. Since every element of $M$ has
exactly two pre-images, we have that
$g^{-1}(F)=f^{-1}(F)\bigcup g^{-1}(F\cap M)$.
Clearly $f^{-1}(F)\in \mathcal{G}|_A$ and since $M\subseteq^{*} F$, then
$g^{-1}(F\cap M)\cap (\omega\setminus A) \in \mathcal{F}_r(\omega\setminus A)$. By the above remark,
we obtain that $g^{-1}(F)\in \mathcal{G}$. So $\mathcal{F}\subseteq g[\mathcal{G}]$.
Now fix $G\in \mathcal{G}$. Then we have that
$$g[G]=g[G\cap A]\cup g[G\setminus A]=
f[G\cap A]\cup g[G\setminus A].
$$
Since $f[G\cap A]\in \mathcal{F}$, $g[G]\in \mathcal{F}$. Thus, $g[\mathcal{G}]\subseteq \mathcal{F}$. This proves that $g[\mathcal{G}]=\mathcal{F}$.
Therefore, $\mathcal{F}\leq_{RK} \mathcal{G}$.
\end{proof}
\begin{corollary}\label{coroturk}
Let $\mathcal{F}$ and $\mathcal{G}$ two $FU$-filters. If $\mathcal{F} \leq_{TU}\mathcal{G}$, then
$\mathcal{F}\leq_{RK}\mathcal{G}$.
\end{corollary}
In a general context, one may ask what about the implications $\leq_{RK} \Rightarrow \leq_{TU}$ and
$\leq_{TU} \Rightarrow \leq_{RK}$ without any restrictions on the filters ?
We just have seen that the $TU$-order implies the $RK$-order on the class
of $FU$-filters, which is also true for every pair of ultrafilters
$\mathcal{U}, \mathcal{V}$ (i. e., $\mathcal{U}\leq_{TU} \mathcal{V} \Rightarrow \mathcal{U}\leq_{RK} \mathcal{V}$). Let us see, in the next examples,
that both implications could fail in general:
\smallskip
The {\it Arens filter}
$\mathcal{F}_a$ is defined by an infinite partition $\mathcal{P}=\{P_n:n<\omega\}$ of $\omega$
and the Fr\'echet filter on each $P_n$ as follows:
$$\mathcal{F}_a:=\{F\subseteq \omega:\{n<\omega:P_n\cap P\in \mathcal{F}_r(P_n)\}\in \mathcal{F}_r\}.$$
\noindent The filter $\mathcal{F}_a$ is sequential but it is not an $FU$-filter.
Besides, we know that the filter $\mathcal{F}_a|_A$ is a copy of $\mathcal{F}_a$ for every $A\in \mathcal{F}_a^+$.
Thus if $\mathcal{F}\leq_{TU} \mathcal{F}_a$, then $\mathcal{F}\approx \mathcal{F}_a$.
However, the Fr\'echet filter is a $RK$-predecessor of the Arens filter
via the function $f:\omega\to\omega$, defined by $f[P_n]=n$ for each $n<\omega$.
This example shows that the implication $\mathcal{F}\leq_{RK}\mathcal{G}\Rightarrow \mathcal{F}\leq_{TU}\mathcal{G}$ does not hold in general.
Now, we describe an example to show that the implication $\mathcal{F}\leq_{TU}\mathcal{G}\Rightarrow \mathcal{F}\leq_{RK} \mathcal{G}$ could be false.
Choose $M, N\in [\omega]^{\omega}$ so that $M\cap N=\emptyset$ and $M\cup N=\omega$.
We know that $\mathcal{F}_a \leq_{TU} \mathcal{F}_{a}(M)\oplus \mathcal{G} =\{F\cup G: F\in \mathcal{F}_a(M) \ \text{and} \ G\in \mathcal{G}\}$, where $\mathcal{F}_{a}(M)$ is a copy of the Arens filter on $M$ and
$\mathcal{G}$ is a filter on $N$. Let $\mathcal{G}$ be an arbitrary $FU$-filter and suppose that $\mathcal{F}_a\leq_{RK}\mathcal{F}_{a}\oplus \mathcal{G}$ via the function $f$.
Since $\mathcal{G}$ is an $FU$-filter, there is $R\in[N]^{\omega}$ such that $R\to \mathcal{F}_{a}(M)\oplus \mathcal{G}$ and then $f[R]\to \mathcal{F}_a$,
but $\mathcal{F}_a$ does not have any nontrivial convergent sequence. This shows that the implication
$\mathcal{F}\leq_{TU}\mathcal{G}\Rightarrow \mathcal{F}\leq_{RK}\mathcal{G}$ could fail in general.
\medskip
Next, we will use $FU$-filters of the form $S_{\mathcal{A}}$ to show that $\leq_{RK}\nsubseteq \leq_{TU}$.
Before that we need to prove a theorem.
\begin{theorem}\label{theomini}
Let $\mathcal{A}$ and $\mathcal{B}$ be $NMAD$-families such that $S_{\mathcal{A}}\leq_{RK}S_{\mathcal{B}}$ via a function $f:\omega\to \omega$ which satisfies that
$|f^{-1}(n)|=\omega$ for all $n<\omega$. If $\mathcal{C}$ is a $NMAD$ family such that $\mathcal{B}\subseteq \mathcal{C}$,
then $S_{\mathcal{A}}\leq_{RK}S_{\mathcal{C}}$ via the function $f$.
\end{theorem}
\begin{proof} In order to show that $f[S_{\mathcal{C}}] = S_{\mathcal{A}}$, we use clause $(3)$ of Theorem $\ref{theoeq2}$. By using the statement $(3)(a)$, we know that $f^{-1}(n)\in \mathcal{I}(\mathcal{B})$ for all $n<\omega$. Hence, for each $n < \omega$, we have that $|f^{-1}(n)\cap C|<\omega$ for every $C\in \mathcal{C}\setminus \mathcal{B}$. Let $M\in \mathcal{I}(\mathcal{A})^{\bot}$. Notice that the containment $f[S_{\mathcal{B}}]\subseteq S_\mathcal{A}$ implies that there is $N\in \mathcal{I}(\mathcal{B})^{\bot}$ such that $|f^{-1}(M)\cap N|=\omega$.Thus, we obtain that $f^{-1}(M)\in S_\mathcal{C}^{+}=\mathcal{P}(\omega)\setminus \mathcal{I}(\mathcal{C})$. There are two cases to be consider. The first one is when $f^{-1}(M)=M_0\cup M_1$ where $M_0\in \mathcal{I}(\mathcal{C})$ and $M_1\in \mathcal{I}(\mathcal{C})^{\bot}\cap [\omega]^{\omega}$. This case is clearly done since $M_1$ does the job. For the second one assume that $f^{-1}(M)\in \mathcal{C}^{*}$. Since $\mathcal{C}$ is a $NMAD$-family, then there is $N\in \mathcal{I}(\mathcal{C})^{\bot}\cap [\omega]^{\omega}$ such that $|f^{-1}(M)\cap N|=\omega$. Thus, $f[S_{\mathcal{C}}] = S_{\mathcal{A}}$.
Therefore, we conclude that $S_{\mathcal{A}}\leq_{RK}S_{\mathcal{C}}$ via the function $f$.
\end{proof}
Let $\mathcal{A}$ be an $AD$ family on $\omega$.
For each $A\in\mathcal{A}$ choose $E_A\in [\omega]^{< \omega}$ and consider one of the sets either $A'=A\cup E_A$ or $A'=A\setminus E_A$.
It is not difficult to show that $S_{\mathcal{A}}=S_{\mathcal{A}'}$ where $\mathcal{A}'=\{A':A\in \mathcal{A}\}$.
For our convenience, without lose of generality, we shall assume that each $AD$-family $\mathcal{A}$ always contains
a partition $\{A_{n}:n<\omega\}\subseteq \mathcal{A}$ of $\omega$ in infinite subsets. We show next that $S_{\mathcal{P}}$ is an $RK$-minimal filter in the realm of the filters
of the form $S_{\mathcal{A}}$ where $\mathcal{A}$ is a $NMAD$-family.
\begin{corollary}\label{com}
$S_{\mathcal{P}}\leq_{RK}S_{\mathcal{A}}$ for every
$NMAD$-family $\mathcal{A}$.
\end{corollary}
\begin{proof} Let $\mathcal{A}$ be an $AD$-family on $\omega$ such that $\mathcal{A}'=\{A_n:n<\omega\}\subseteq \mathcal{A}$ is
a partition of $\omega$. Enumerate $\mathcal{P}$ as $\{P_n:n<\omega\}$ and
we define $f:\omega\to \omega$ so that
$f[A_n]=P_n$ and $|f^{-1}(n)|$ is infinite for each $n<\omega$. It is straightforward
to prove that $S_{\mathcal{P}}\leq_{RK}S_{\mathcal{A}'}$ via $f$. Therefore, from Theorem \ref{theomini} we deduce that
$S_{\mathcal{P}}\leq_{RK}S_{\mathcal{A}}$ via the function $f$.
\end{proof}
The answer to the following question will be very useful to understand the filter $S_{\mathcal{A}}$.
\begin{question} Is true that
$S_{\mathcal{P}}\leq_{RK}S_{\mathcal{A}}$ for every
$AD$-family $\mathcal{A}$ ?
\end{question}
We recall that an $AD$-family is said to be {\it completely separable} if
for every $M\in \mathcal{B}^*$, there is $B\in \mathcal{B}$ such that $B\subseteq M$.
In the paper \cite{si}, P. Simon showed, in $ZFC$, the existence of a completely separable $NMAD$-family of size $\mathfrak{c}$.
\begin{corollary}
If $\mathcal{A}$ is a $NMAD$-family of size $\mathfrak{c}$ which is completely separable, then
$S_{\mathcal{P}}\nleq_{TU}S_{\mathcal{A}}$ and $S_{\mathcal{P}} \leq_{RK} S_{\mathcal{A}}$.
\end{corollary}
\begin{proof}
Let $\mathcal{A}$ be a completely separable $NMAD$-family of size $\mathfrak{c}$. In the article \cite{si}, the author showed that this family satisfies that
$|\{A\in \mathcal{A}:|M\cap A|=\omega\}|=\mathfrak{c}$ for all $M\in \mathcal{A}^*$.
If $M\in S_{\mathcal{A}}^+$ and $|\{A\in \mathcal{A}:|M\cap A|=\omega\}|< \omega$, then $S_{\mathcal{A}}|_M$ is
a relatively equivalent to the Fr\'echet filter. If $M\in \mathcal{A}^*$, then
$S_{\mathcal{A}}|_M$ has character equal to $\mathfrak{c}$.
Thus, $S_{\mathcal{A}}$ has not a copy of $S_{\mathcal{P}}$.
Therefore, $S_{\mathcal{P}}\nleq_{TU}S_{\mathcal{A}}.$
On the other hand, by Corollary \ref{com}, we have that $S_{\mathcal{P}} \leq_{RK} S_{\mathcal{A}}$.
\end{proof}
The behavior of the filters of the form $S_{\mathcal{A}}$, where $\mathcal{A}$ is completely separable, under the $RK$-order is not well-know yet. For instance, we
do not the answer to the following question.
\begin{question} Are there two completely separable $NMAD$-families $\mathcal{A}$ and $\mathcal{B}$ such that their filters $S_{\mathcal{A}}$ and $S_{\mathcal{B}}$
are $RK$-incomparable ?
\end{question}
In the last section, we will construct two $NMAD$-families of size $\mathfrak{c}$ whose respective filters are $RK$-incomparable.
\section{Chains of $FU$-filters in the $RK$-order and $TU$-order.}
First, we shall describe an
operation of filters that preserves the $FU$-property and produces
$RK$-successors:
\smallskip
Let $I$ be a set, $\mathcal{F}$ a (not necessarily free) filter on $I$ and
$\mathcal{A} = \{ A_i : i \in I \}$ an $AD$-family. For
each $i \in I$, choose a free filter $\mathcal{F}_i$ on the set $A_i$.
Then we define
$$
\sum_{\mathcal{F}}\mathcal{F}_i := \{ F \subseteq \omega :\{i\in I: F\cap A_i \in
\mathcal{F}_i\}\in \mathcal{F}\}
$$
and
$$
\prod_{i \in I}\mathcal{F}_i := \{ F \subseteq \omega : \forall i \in I(F\cap A_i \in
\mathcal{F}_i)\}.
$$
Notice that if the filter on $I$ is the trivial filter $\{I\}$, then
$$
\prod_{i \in I}\mathcal{F}_i := \sum_{\{I\}}\mathcal{F}_i.
$$
The filter $\prod_{i \in I}\mathcal{F}_i$ is referred as the product of the filters $\{ F_i : i \in I\}$. Several interesting properties of this operation of filters are contained in \cite{gr01}.
It is evident that $\sum_{\mathcal{F}}\mathcal{F}_i$ is always a free filter on $\omega$ and
that $F \in (\prod_{i \in I}\mathcal{F}_i)^+$ iff there is $i \in I$ such
that $F \in \mathcal{F}_i^+$. Hence, we deduce that $\prod_{i \in I}\mathcal{F}_i$ is
a $FU$-filter iff $\mathcal{F}_i$ is a $FU$-filter for all $i \in I$. We remark that $\sum_{\mathcal{F}}\mathcal{F}_i $ is not, in general,
an $FU$-filter: for instance the Arens filter $\mathcal{F}_a$. In this context,
the $FAN$-filter is the filter $\prod_{n < \omega}\mathcal{F}_r(P_n)$
where $\{ P_n :n < \omega\}$ is a partition of $\omega$ in infinite subsets.
The product of finitely many filters $\mathcal{F}_0,....., \mathcal{F}_n$ will be denote by $\mathcal{F}_0
\oplus \mathcal{F}_1 \oplus ..... \oplus \mathcal{F}_n$. We point out that if $\mathcal{A} = \{ A_i : i \in I \}$ is an $AD$-family and $\mathcal{A}_i$ is an $AD$-family on $A_i$, for
each $i \in I$, then
$$
\prod_{i \in I}\mathcal{F}_{\mathcal{A}_i} = \mathcal{F}_{\bigcup_{i \in I}\mathcal{A}_i}.
$$
\medskip
To construct $RK$-up-directed chains we need the following lemma from \cite[4.2]{gr01}.
\begin{lemma}\label{many-su} Let $A\in[\omega]^{\omega}$. Suppose that $\mathcal{A} = \{ A_i : i \in I \}\cup \{A\}$ is an $AD$-family
and $\mathcal{A}_i$ is an $AD$-family on $\omega$, for each $i \in I$. If
$f_i: \omega \to A_i$ is a bijection, for every $i \in I$ and, $\mathcal{B}$
is an $AD$-family on $A$, then
$\mathcal{F}_{\mathcal{A}_j} \leq_{RK} \mathcal{F}_{\bigcup_{i \in I}f_i[\mathcal{A}_i]\cup \mathcal{B}}$ for all $j \in
I$.
\end{lemma}
We remark that $\leq_{RK}$ can be replaced by $\leq_{TU}$ in the previous Lemma.
\begin{theorem}
If $\{\mathcal{A}_{\xi}:\xi<\mathfrak{c}\}$ is a collection of $AD$-families, then there is
an $AD$-family $\mathcal{C}$ such that $\mathcal{F}_{\mathcal{A}_{\xi}}<_{RK}\mathcal{F}_{\mathcal{C}}$ for all $\xi<\mathfrak{c}$.
\end{theorem}
\begin{proof} Fix $A\in[\omega]^{\omega}$ so that $\omega\setminus A$ is infinite and let
$\{A_{\xi}:\xi<\mathfrak{c}\}$ be an $AD$-family on $\omega\setminus A$. For each $\xi<\mathfrak{c}$ choose
a bijection $f_{\xi}:\omega\to A_{\xi}$. By the previous lemma we obtain that
$\mathcal{F}_{A_{\xi}}\leq_{RK}\mathcal{F}_{\cup_{\xi<\mathfrak{c}}f[A_{\xi}]\cup \mathcal{B}}$ for all $\xi<\mathfrak{c}$ and for every $AD$-family $\mathcal{B}$ on $A$.
We know that there are $2^{\mathfrak{c}}$
pairwise distinct $AD$-families on $A$, and since every filter $\mathcal{F}_{\mathcal{A}_{\xi}}$ has at most
$\mathfrak{c}$-many $RK$-predecessors, we can find an $AD$-family $\mathcal{B}$ such that $\mathcal{F}_{\mathcal{B}} \nleq_{RK} \mathcal{F}_{\mathcal{A}_{\xi}} $ for all $\xi<\mathfrak{c}$.
Therefore, $\mathcal{F}_{\mathcal{A}_{\xi}}<_{RK}\mathcal{F}_{\mathcal{C}}$ for all $\xi<\mathfrak{c}$, where $\mathcal{C}=\bigcup_{\xi<\mathfrak{c}}f[A_{\xi}]\cup \mathcal{B}$.
\end{proof}
\begin{corollary}
There is a strictly increasing $RK$-chain of $FU$-filters of size $\mathfrak{c}^+$ $RK$-above every $FU$-filter.
\end{corollary}
\section{$RK$-Incomparability of $FU$-filters}
In this section, we construct an $RK$-antichain consisting of $FU$-filters. The authors of \cite{tu05} have proved the existence of a
$TU$-antichain of size $\mathfrak{c}^+$ consisting of $FU$-filters.
\medskip
The next notions introduced by A. V. Arhangel'skii in \cite{ar} will help us to distinguish several $FU$-filters.
\begin{definition}
Let $X$ be an space and $x\in X$. A {\it sheaf} of $x$ is a family of
sequences $\{C_n:n<\omega\}$ in $X$ converging to $x$.
We say that $x$ is an $\alpha_i$-{\it point} (for each $i=1,2,3,4$) if for every sheaf $\{C_n:n<\omega\}$ of $x$
there is a sequence $B$ converging to $x$ such that:
\begin{enumerate}
\item[] ($\alpha_1$) $C_n\subseteq^*B$, for all $n<\omega$.
\item[] ($\alpha_2$) $C_n\subseteq^*B$, for all $n<\omega$.
\item[] ($\alpha_3$) $|C_n\cap B|=\omega$, for infinitely many $n<\omega$.
\item[] ($\alpha_4$) $C_n\cap B\neq\emptyset$, for infinitely many $n<\omega$.
\end{enumerate}
The space $X$ is called $\alpha_i$-{\it space} if every point in $X$ is an $\alpha_i$-point. In particular,
a filter $\mathcal{F}$ is an $\alpha_i$-{\it filter} if its nonisolated point is an
$\alpha_i$ point in the space $\xi(\mathcal{F})$, for every $i=1,2,3,4$.
\end{definition}
It is straightforward to prove the following implications:
$$
first \ countability \Rightarrow \alpha_1 \Rightarrow \alpha_2
\Rightarrow \alpha_3\Rightarrow \alpha_4.
$$
The $FAN$-filter is a canonical example of a $FU$-filter which
is not an $\alpha_4$-filter; indeed, it is well-know that a space is not an
$\alpha_4$-space iff the space contains a copy of $FAN$-space (for a prove see \cite{sw}).
In the article \cite{si}, P. Simon constructed a completely separable $NMAD$-family $\mathcal{A}$ of size $\mathfrak{c}$
such that $S_{\mathcal{A}}$ is an $\alpha_4$-filter which is not an
$\alpha_3$-filter. For this $AD$-family $\mathcal{A}$, it is easy to show that $\mathcal{F}_{\mathcal{A}}$ is also an $\alpha_4$-filter that is not
an $\alpha_3$-filter.
\medskip
In the following, we shall use a standard well-known technic to construct $FU$-filters by using the Cantor tree $2^{<\omega} = \bigcup_{n < \omega}2^n$:
\medskip
For each $x\in 2^{\omega}$ we define $A_x=\{x|n:n<\omega\}\subseteq 2^{<\omega}$. For every
infinite $X\subseteq 2^{\omega}$ we have that $\mathcal{A}_{X}=\{A_x:x\in X\}$ is an $NMAD$-family
on $2^{<\omega}$. By identifying $2^{<\omega}$ with $\omega$, the family $\mathcal{A}_{X}$ can be considered as a family of subsets of $\omega$.
P. Nyikos (\cite{ny1}) proved that $S_{\mathcal{A}_X}$ is an $\alpha_3$-filter
for all infinite $X\subseteq 2^{\omega}$. He also showed that there is $Z\subseteq 2^{\omega}$
for which $S_{\mathcal{A}_Z}$ is an $\alpha_2$-filter, but in general
this assertion could fail; for instance, $S_{\mathcal{A}_{2^{\omega}}}$ is not an $\alpha_2$-filter.
All examples of $FU$-filters given above lie in $ZFC$. Nyikos have proved in \cite{ny2},
under the assumption $\omega_1=\mathfrak{b}$, that there is an $FU$-space that is
$\alpha_2$-space but it fails to be $\alpha_1$-space. In the same paper, it was proved that
if $\omega_1<\mathfrak{b}$, then there is an $FU$-space which is $\alpha_1$ but it is not
a first countable space. Years later, A. Dow (\cite{dow}) proved that the implication ``$\alpha_2\Rightarrow \alpha_1$'' holds
inside of the Lavers Model and together with J. Stepr\={a}ns \cite{dowst} constructed
a model of $ZFC$ in which every $\alpha_1$-space is a first countable space.
The existence of an $\alpha_2$-space
which is not an $\alpha_1$-space, and the existence of an $\alpha_1$-space
which is not an first countable space are still open problems in $ZFC$.
\medskip
Now let us prove that the properties $\alpha_2$, $\alpha_3$ and $\alpha_4$ are preserved
by the $RK$-order down-directed.
\begin{theorem}\label{theoalphas}
Let $\mathcal{F}_{\mathcal{A}}$ and $\mathcal{F}_{\mathcal{B}}$ be two $FU$-filters. If $\mathcal{F}_{\mathcal{B}}$ is an $\alpha_i$-filter
and $\mathcal{F}_{\mathcal{A}}\leq_{RK}\mathcal{F}_{\mathcal{B}}$, then $\mathcal{F}_{\mathcal{A}}$ is also an $\alpha_i$-filter, for each $i=2,3,4$.
\end{theorem}
\begin{proof} We only give a proof for the $\alpha_2$-property since the procedure for $\alpha_3$ and $\alpha_4$ is exactly the same.
Let $\{C_n:n<\omega\}\subset C(\mathcal{F}_{\mathcal{A}})$ be a sheaf of $\mathcal{F}_{\mathcal{A}}$ and $f:\omega\to\omega$
such that $f[\mathcal{F}_{\mathcal{B}}]=\mathcal{F}_{\mathcal{A}}$. By Theorem \ref{theoeq}, we can find
$B_n\in\mathcal{B}$ such that $|f^{-1}(C_n)\cap B_n|=\omega$ for every $n<\omega$.
Notice that $\{f^{-1}(C_n)\cap B_n:n<\omega\}$ is a sheaf of $\mathcal{F}_{\mathcal{B}}$.
Since $\mathcal{F}_{\mathcal{B}}$ is an $\alpha_2$-filter, then there is a sequence $B$ converging
to $\mathcal{F}_{\mathcal{B}}$ such that $|B\cap (f^{-1}(C_n)\cap B_n)|=\omega$ for all
$n<\omega$. We remark that $f[B]\to \mathcal{F}_{\mathcal{A}}$. Fix $n<\omega$. Let us prove that $|f[B]\cap C_n|=\omega$.
Indeed, we have that
$$
f[B\cap (f^{-1}(C_n)\cap B_n)]\subseteq f[B]\cap (C_n\cap f[B_n])
\subseteq f[B]\cap C_n.
$$
Since $B\cap (f^{-1}(C_n)\cap B_n) \to \mathcal{F}_{\mathcal{B}}$, then $|f[B\cap (f^{-1}(C_n)\cap B_n)]|=\omega$
and so $|f[B]\cap C_n|=\omega$. Therefore, $\mathcal{F}_{\mathcal{A}}$ is an
$\alpha_2$-filter.
\end{proof}
For an arbitrary $NMAD$-family $\mathcal{A}$, we know that the filter $\mathcal{F}_{\mathcal{A}}$ cannot be an $\alpha_3$-filter. Thus
we obtain the following corollary.
\begin{corollary}\label{a}
$S_\mathcal{P}$ is not an $RK$-successor of $\mathcal{F}_{\mathcal{A}}$ for any $NMAD$-family $\mathcal{A}$.
\end{corollary}
The next corollary is consequence a from Corollary \ref{a} and Corollary 5.6 from
\cite{gr01}.
\begin{corollary}
$S_\mathcal{P}$ is $RK$-incomparable with every filter $\mathcal{F}_{\mathcal{A}}$ such that $|\mathcal{A}|<\mathfrak{b}$.
\end{corollary}
By using the Corollary \ref{coroturk}, Theorem \ref{theoalphas} and some facts quoted above we obtain the next result.
\begin{corollary}
$\mathcal{F}_{\mathcal{P}}\leq_{RK} \mathcal{F}$ iff $\mathcal{F}_{\mathcal{P}}\leq_{TU} \mathcal{F}$.
\end{corollary}
Now we show that some of the $FU$-filters already described above are $RK$-incomparable.
\begin{theorem}
Let $\mathcal{A}$ be a $NMAD$-family completely separable of size
$\mathfrak{c}$. Then there is a set $X\subseteq 2^{\omega}$ such that $\mathcal{F}_{\mathcal{P}}$, $S_{\mathcal{A}}$ and $S_{\mathcal{A}_X}$ form an
$RK$-antichain.
\end{theorem}
\begin{proof}
Notice that there are $2^{\mathfrak{c}}$ pairwise non-homeomorphic filters of the form $S_{\mathcal{A}_X}$ whit $|X|=\mathfrak{c}$.
We can choose one of them satisfying $S_{\mathcal{A}_X}\nleq_{RK} S_{\mathcal{A}}$.
We know that $S_{\mathcal{A}_X}$ is an $\alpha_3$-filter, $S_{\mathcal{A}}$ is an $\alpha_4$-filter
which is not an $\alpha_3$-filter and the $FAN$-filter $\mathcal{F}_{\mathcal{P}}$ is not an $\alpha_4$-filter. Hence,
by Theorem \ref{theoalphas}, we obtain that
$$
\mathcal{F}_{\mathcal{P}}\nleq_{RK} S_{\mathcal{B}}\nleq_{RK} S_{\mathcal{A}_X} \ \text{and} \ \mathcal{F}_{\mathcal{P}}\nleq_{RK} S_{\mathcal{A}_{X}}.
$$
According to Theorem \ref{theofan}, we have that $S_{\mathcal{A}_X}\nleq_{RK}\mathcal{F}_{\mathcal{P}}$ and
$S_{\mathcal{A}}\nleq_{RK}\mathcal{F}_{\mathcal{P}}$. Therefore, $\mathcal{F}_{\mathcal{P}}$, $S_{\mathcal{A}}$ and $S_{\mathcal{A}_{X}}$ are pairwise $RK$-incomparable.
\end{proof}
Our next task is the construction of an infinite $RK$-antichain consisting of $FU$-filters.
Such filters will be the form $S_{\mathcal{A}_X}$ for suitable sets $X\subseteq 2^{\omega}$. For our purposes
it is important to remark the next characterization of the convergent sequences in $S_{\mathcal{A}_X}$:
\medskip
{\bf Remark.} For $X\in 2^{\omega}$ and $N\in[2^{<\omega}]^{\omega}$, the following statements are equivalents:
\begin{enumerate}
\item $N\to S_{\mathcal{A}_X}$.
\item $N\in \mathcal{I}(\mathcal{A}_X)^{\bot}$.
\item For all $K\in[N]^{\omega}$ there is either:
\begin{enumerate}
\item $x\in 2^{\omega}\setminus X$ such that $|A_x \cap K| = \omega$ or
\item an infinite antichain $M$ such that $|M \cap K| = \omega$.
\end{enumerate}
\end{enumerate}
Thus, we may consider only branches and antichains of $2^{<\omega}$.
The following equivalence is a consequence of the Theorem \ref{theoeq2} and our last remark.
\begin{lemma}\label{lemmi}
Let $X_0,X_1\subseteq 2^{\omega}$ and $f:2^{\omega}\to 2^{\omega}$ a surjection. Then, $f[S_{\mathcal{A}_{X_1}}]\neq S_{\mathcal{A}_{X_0}}$ iff
one of the following conditions is satisfied:
\begin{enumerate}
\item There is $x\in 2^{\omega}\setminus X_1$ such that $f[A_x] \nrightarrow S_{\mathcal{A}_{X_0}}$.
\item There is an infinite antichain $M$ such that $f[M] \nrightarrow S_{\mathcal{A}_{X_0}}$.
\item There is $y\in 2^{\omega}\setminus X_0$ such that $f^{-1}(A_y) \in \mathcal{I}(\mathcal{A}_{X_1})$.
\item There is an infinite antichain $M$ such that $f^{-1}(M) \in \mathcal{I}(\mathcal{A}_{X_1})$.
\end{enumerate}
\end{lemma}
We would like to point out that clauses (1) and (2) imply $S_{\mathcal{A}_{X_0}}\nsubseteq f[S_{\mathcal{A}_{X_1}}]$, and conditions (3) and (4) imply
$f[S_{\mathcal{A}_{X_1}}]\nsubseteq S_{\mathcal{A}_{X_0}}$. If there is an infinite antichain $M$ such that $|f[M]|<\omega$, then we may avoid this kind of
functions, since $f$ cannot be a witness of the $RK$-comparability for any pair of $FU$-filters.
Thus, in what follows, we shall always assume that $f|_M$ is finite-to-one at every antichain $M$.
\medskip
Let us show in the next lemma that we can always extend the sets $X_0$ and $X_1$ in order to
have witnesses for the $RK$-incomparability of their respective $FU$-filters of the extensions.
\begin{lemma}\label{lemant}
Let $X_0$ and $X_1$ be nonempty subsets of $2^{\omega}$ such that $|2^{\omega} \setminus (X_0 \cup X_1)| \geq \omega$ and $f:2^{<\omega}\to 2^{<\omega}$ a surjection such that $f|_M$ is finite-to-one for every infinite antichain $M$. Then, there are $X'_0,X_1', Y_0, Y_1 \subseteq 2^{\omega}$ such that $X_0\subsetneq X_0'$, $0 < |X_0'\setminus X_0|<\omega$,
$X_1\subsetneq X_1'$, $0 < |X_1'\setminus X_1|<\omega$, $0 < |Y_0|, |Y_1|< \omega$, $X_0'\cap Y_0=\emptyset = X_1'\cap Y_1$ and at least one of the following conditions holds:
\begin{enumerate}
\item[(a)] There is $y\in Y_1$ such that $f[A_y] \nrightarrow S_{\mathcal{A}_{X_0'}}$.
\item[(b)] There is an infinite antichain $M$ such that $f[M] \nrightarrow S_{\mathcal{A}_{X_0'}}$.
\item[(c)] There is $x\in Y_0$ such that $f^{-1}(A_x) \in \mathcal{I}(\mathcal{A}_{X_1'})$.
\item[(d)] There is an infinite antichain $M$ such that $f^{-1}(M) \in \mathcal{I}(\mathcal{A}_{X_1'})$.
\end{enumerate}
Thus, by Lemma \ref{lemmi}, we have that $f[S_{\mathcal{A}_{X_1'}}]\neq S_{\mathcal{A}_{X_0'}}$.
\end{lemma}
\begin{proof} We need to consider two cases:
Case I. Suppose that $f[S_{\mathcal{A}_{X_1}}]\neq S_{\mathcal{A}_{X_0}}$. Notice that if the witnesses of the $RK$-incomparability is an antichain satisfying
either (2) or (4) of Lemma \ref{lemmi}, then we can extend arbitrarily $X_0\subseteq X_0'$, $X_1\subseteq X_1'$ and find $Y_0$, $Y_1$ such that $X_0'\cap Y_0=\emptyset$ and $X_1'\cap Y_1=\emptyset$ easily. Hence, either (b) or (d) holds.
Now suppose that the witness is a branch that satisfies (1). There is $y\in 2^{\omega}\setminus X_1$ and $x\in X_0$ such that $|f[A_y]\cap A_x|=\omega$. Define $Y_1=\{y\}$, $X_1'=X_1\cup W$ where $W\subseteq 2^{\omega}\setminus Y_1$, and
$X_0'$, $Y_0$ arbitrarily such that $X_0'\cap Y_0=\emptyset$. Thus we have (a). Assume now that the witness is a branch satisfying (3). There is $v\in 2^{\omega}\setminus X_0$ such that $f^{-1}(A_v)\in \mathcal{I}(\mathcal{A}_{X_1})$. Define $Y_0=\{v\}$, $X_0'=X_0\cup W$ where $W\subseteq 2^{\omega}\setminus Y_0$, and
$X_1'$, $Y_1$ arbitrarily such that $X_1'\cap Y_1=\emptyset$. In this case (c) is satisfied.
In each case we have one of the conditions.
\medskip
Case II. Suppose that $f[S_{\mathcal{A}_{X_1}}]=S_{\mathcal{A}_{X_0}}$. We shall prove that $X_0$ and $X_1$ can be extend and find $Y_0$ and $Y_1$ so that
their extensions will satisfy either (b) or (c).
\begin{enumerate}
\item[(i)] Assume that there are $y\in 2^{\omega}\setminus X_0$ and a nonempty finite set $\mathcal{B}\subseteq \{A_v:v\in 2^{\omega}\}$ such that $f^{-1}(A_z)\subseteq^*\bigcup \mathcal{B}$.
Notice that $X_1\setminus \{v\in 2^{\omega}: A_v\in \mathcal{B}\}\neq \emptyset$. In this case, we set $Y_0=\{y\}$, $X_0'=X_0 \cup W$ where $W\subseteq 2^{\omega}\setminus Y_0$, $X_1'=X_1\cup \{v\in 2^{\omega}: A_v\in \mathcal{B}\}$ and $Y_1$ a finite
nonempty set such that $X_1'\cap Y_1=\emptyset$. Thus, we have (c).
\item[(ii)] Now suppose that there is an antichain $M$ and $x\in 2^{\omega}\setminus X_0$ such that $|f[M]\cap A_x|=\omega$.
In this case, we put $X_0'=X_0\cup \{x\}$, $Y_0$ a finite nonempty set such that $X_0'\cap Y_0=\emptyset$. We can extend $X_1$ arbitrarily and find $Y_1$ a finite
nonempty set such that $X_1'\cap Y_1=\emptyset$. Hence, we have (b0).
\end{enumerate}
If (i) and (ii) fail, then $f^{-1}(A_z)\in \mathcal{I}(\mathcal{A}_{2^{\omega}})^+$, for each $z\in 2^{\omega}\setminus X_0$, and for every infinite antichain $M$ we have that $f[M]\in \mathcal{I}(\mathcal{A}_{2^\omega})^{\bot}$.
Hence $f[M]$ cannot meet any branch $A_y$ in an infinite set, for all $y\in 2^{\omega}$. Fix $z\in 2^{\omega}\setminus X_0$. As
$f^{-1}(A_z)\in \mathcal{I}(\mathcal{A}_{2^{\omega}})^+$, then $f^{-1}(A_z)$ contains an infinite antichain $K$.
Since $f[K]$ is infinite and $f[K]\subseteq f[f^{-1}(A_z)]=A_z$,
we get a contradiction to the negation of (ii). Thus, either (i) or (ii) is satisfied.
\end{proof}
We are ready to construct an infinite $RK$-antichain with $FU$-filters of character equal to $\mathfrak{c}$.
\begin{theorem}
For every infinite cardinal $\kappa<\mathfrak{c}$, there is a family $\{X_{\alpha}:\alpha<\kappa \}\subseteq[2^{\omega}]^{\mathfrak{c}}$
such that $\{S_{X_{\alpha}}:\alpha < \kappa\}$ is an $RK$-antichain.
\end{theorem}
\begin{proof}
Let $\{f_{\beta}:\beta<\mathfrak{c}\}$ be an enumeration of all surjections $f_{\beta}:2^{<\omega}\to 2^{<\omega}$ for which $f_{\beta}|_M$ is finite-to-one for every infinite antichain $M$.
By an inductive procedure, for every $\beta < \mathfrak{c}$ we shall construct, for every $\alpha<\kappa$, sets $X_{\beta}^{\alpha}\subseteq 2^{\omega}$ and $Y_{\beta}^{\alpha}\subseteq 2^{\omega}$ so that:
\begin{enumerate}
\item $X^{\alpha}_{\beta}\cap Y^{\alpha}_{\beta}=\emptyset$ for every $\alpha<\kappa$.
\item $X^{\alpha}_{\mu}\subseteq X^{\alpha}_{\nu}$ and $Y^{\alpha}_{\mu}\subseteq Y^{\alpha}_{\nu}$ if $\mu < \nu < \mathfrak{c}$.
\item For distinct
$\gamma, \delta <\kappa$ one of the following conditions holds:
\begin{enumerate}
\item There is $x\in Y^{\gamma}_{\beta+1}$ such that $f_\beta[A_x] \nrightarrow S_{\mathcal{A}_{X^{\delta}_{\beta+1}}}$.
\item There is an infinite antichain $M$ such that $f_\beta[M] \nrightarrow S_{\mathcal{A}_{X^{\delta}_{\beta+1}}}$.
\item There is $y\in Y^{\delta}_{\beta+1}$ such that $f_\beta^{-1}(A_y) \in \mathcal{I}(\mathcal{A}_{X^{\gamma}_{\beta+1}})$.
\item There is an infinite antichain $M$ such that $f_\beta^{-1}(M) \in \mathcal{I}(\mathcal{A}_{X^{\gamma}_{\beta+1}})$.
\end{enumerate}
\item $|X^{\alpha}_{\beta}|, |Y^{\alpha}_{\beta}|\leq \kappa \cdot |\beta|$ for every $\alpha<\kappa$.
\end{enumerate}
Choose arbitrary distinct elements $x_0, y_0\in 2^{\omega}$ and define $X_{0}^{\alpha}=\{x_0\}$ and $Y_{0}^{\alpha}=\{y_0\}$, for every $\alpha<\kappa$.
Assume that for $\beta<\mathfrak{c}$ the sets
$X_{\theta}^{\alpha}$ and $Y_{\theta}^{\alpha}$ have been defined for all $\theta < \beta$ and $\alpha<\kappa$ so that all of them satisfy the conditions (1), (2), (3) and (4). If $\beta<\mathfrak{c}$ is a limit ordinal, then we define $X_{\beta}^{\alpha}=\bigcup_{\theta<\beta} X_{\theta}^{\alpha}$ and
$Y_{\beta}^{\alpha}=\bigcup_{\theta<\beta} Y_{\theta}^{\alpha}$ for each $\alpha < \kappa$.
Now, suppose that $\beta = \theta +1$. We shall define $X_{\beta+1}^{\alpha}$ and $Y_{\beta+1}^{\alpha}$. Notice from (4) that
$$
|2^{\omega}\setminus \big[\big(\bigcup_{\alpha<\kappa} X_{\theta}^{\alpha}\big) \cup \big(\bigcup_{\alpha<\kappa}
Y_{\theta}^{\alpha}\big) \big]|= \mathfrak{c}.
$$
Fix $\alpha < \kappa$. According to Lemma \ref{lemant}, for every $\gamma < \kappa$ we can find finite nonempty sets $B_{\gamma}^{\alpha}$, $C_{\gamma}^{\alpha}$, $D_{\gamma}^{\alpha}$ and $E_{\gamma}^{\alpha}$
so that the following conditions holds:
\begin{enumerate}
\item[(i)] $B_{\gamma}^{\alpha} \cup C_{\gamma}^{\alpha} \cup D_{\gamma}^{\alpha} \cup E_{\gamma}^{\alpha}\subseteq 2^{\omega}\setminus \big[\big(\bigcup_{\alpha<\kappa} X_{\theta}^{\alpha}\big) \bigcup \big(\bigcup_{\alpha<\kappa}
Y_{\theta}^{\alpha}\big) \big].$
\item[(ii)] One of the following conditions hold
\begin{enumerate}
\item There is $x\in Y^{\alpha}_{\theta}\cup D^{\alpha}_{\gamma}$ such that $f_\beta[A_x] \nrightarrow S_{\mathcal{A}_{X^{\gamma}_{\theta}\cup C^{\alpha}_{\gamma}}}$.
\item There is an infinite antichain $M$ such that $f_\beta[M] \nrightarrow S_{\mathcal{A}_{X^{\gamma}_{\theta}\cup C^{\alpha}_{\gamma}}}$.
\item There is $y\in Y^{\gamma}_{\theta}\cup E^{\alpha}_{\gamma}$ such that $f_\beta^{-1}(A_y) \in \mathcal{I}(\mathcal{A}_{X^{\alpha}_{\theta}\cup B^{\alpha}_{\gamma}})$.
\item There is an infinite antichain $M$ such that $f_\beta^{-1}(M) \in \mathcal{I}(\mathcal{A}_{X^{\alpha}_{\theta}\cup B^{\alpha}_{\gamma}})$.
\end{enumerate}
\item[(iii)] $\big[(\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}B_{\gamma}^{\alpha})\cap
\big[\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}D_{\gamma}^{\alpha}\big]=\emptyset$
\item[(iv)] $\big[(\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}C_{\gamma}^{\alpha})\cap
\big[\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}E_{\gamma}^{\alpha}\big]=\emptyset$
\end{enumerate}
Define
$$
X_{\beta}^{\alpha}=X_{\theta}^{\alpha}\cup (\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}B_{\gamma}^{\alpha})
\cup (\bigcup_{\gamma\in\kappa\setminus \{\alpha\}} C_{\alpha}^{\gamma})
$$ and
$$
Y_{\beta}^{\alpha}=Y_{\theta}^{\alpha}\cup (\bigcup_{\gamma\in \kappa\setminus \{\alpha\}}D_{\gamma}^{\alpha})
\cup (\bigcup_{\gamma\in\kappa\setminus \{\alpha\}} E_{\alpha}^{\gamma}).
$$
Conditions (1), (2), (3) and (4) are clearly satisfied. For every $\alpha<\kappa$ define $X_{\alpha}=\bigcup_{\beta<\mathfrak{c}}X_{\beta}^{\alpha}$. Thus, by the construction and Lemma
\ref{lemant} we have that, for every $\alpha, \gamma <\kappa$,
$$f_{\beta}[S_{\mathcal{A}_{X_{\alpha}}}]\neq S_{\mathcal{A}_{X_{\gamma}}}.$$
Therefore $\{S_{X_{\alpha}}:\alpha<\kappa\}$ is an infinite $RK$-antichain of $FU$-filters which
have character equal to $\mathfrak{c}$.
\end{proof}
We end the paper with the following question that the authors could not solve it.
\begin{question}
Is there an $RK$-antichain of $FU$-filters of size $\mathfrak{c}$ ?
\end{question}
\bibliographystyle{amsplain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,902 |
require 'responders'
class JobsController < ApplicationController
responders :rabl
respond_to :json
def show
respond_with job
end
protected
def job
@job ||= Job.find(params[:id])
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,168 |
{"url":"http:\/\/math.stackexchange.com\/questions\/23045\/non-equality-of-distinct-polynomials-with-the-same-transcendental-argument","text":"# \u201cNon-equality\u201d of distinct polynomials with the same transcendental argument\n\nHow can I show for some transcendental number $\\alpha \\in \\mathbb{C}$, that two distinct polynomials of the form $b_{n}\\alpha^n + \\cdot \\cdot \\cdot + b_{2}\\alpha^2 + b_{1}\\alpha$ and $a_{m}\\alpha^m + \\cdot \\cdot \\cdot + a_{2}\\alpha^2 + a_{1}\\alpha$ can never equal each other?\n\n-\nYou should probably make explicit the assumption that the coefficients $a_i,b_i$ are algebraic numbers, since otherwise the assertion is not true. \u2013\u00a0 Pete L. Clark Feb 21 '11 at 12:43\n@Peter: of course, you're right. But I assume the OP meant transcendental over $\\mathbb Q$, which is what unadorned \"transcendental\" usually means. \u2013\u00a0 lhf Feb 22 '11 at 23:56\nThat follows directly from the definition of transcendental number: if $p(\\alpha)=q(\\alpha)$ then $p(\\alpha)-q(\\alpha)=0$.\nHINT $\\ \\$ By definition, $\\rm\\ \\alpha\\:$ is transcendental over $\\rm\\:F\\ \\iff\\ F[x] \\cong F[\\alpha]\\$ via evaluation $\\rm\\ x\\to \\alpha\\:.\\$ For example, $\\rm\\ \\mathbb Q[\\pi] \\cong \\mathbb Q[x]\\:.\\:$ Therefore, in particular, transcendental evaluation is always injective. Thus any element transcendental over a field $\\rm\\:F\\:$ serves equally well as an \"indeterminate\" over $\\rm\\:F\\:$.","date":"2014-03-12 16:07:20","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9679697155952454, \"perplexity\": 428.08097941513796}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2014-10\/segments\/1394021919409\/warc\/CC-MAIN-20140305121839-00023-ip-10-183-142-35.ec2.internal.warc.gz\"}"} | null | null |
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When you;re thinking that this sounds like something you;ve heard before, you;re almost right. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,892 |
Q: unable to configure ssl for rabbitMQ with web_stomp I have followed the steps mentioned by this link https://www.rabbitmq.com/ssl.html , since i am using web_stomp i have modified the URL of webstomp to wss://my_url:my_web_stomp_port/ws and i have my rabbitmq.config as
[
{rabbit, [
{ loopback_users, [ ] },
{ tcp_listeners, [ 5672 ] },
{ hipe_compile, false },
{ssl_listeners, [5671]},
{ssl_options, [{cacertfile,"/path/to/crt/cacert.pem"},
{certfile,"/path/to/crt/cert.pem"},
{keyfile,"/path/to/crt/key.pem"},
{verify,verify_peer},
{fail_if_no_peer_cert,false}]}
]},
{rabbitmq_web_stomp,
[{ssl_config, [{port, 15671},
{backlog, 1024},
{certfile, "/path/to/crt/cert.pem"},
{keyfile, "/path/to/crt/key.pem"},
{cacertfile, "/path/to/crt/cacert.pem"},
].
to verify my ssl connection is used
openssl s_client -connect localhost:15671 -debug -msg
but the output is just
CONNECTED(00000003)
is waiting for a handshake probably for a very long time
i also get
failed: WebSocket opening handshake timed out in the console
my logs show
=INFO REPORT==== date:time ===
rabbit_web_stomp: listening for HTTPS connections on 0.0.0.0:15671
My rabbitMQ version is 3.6.2 and Erlang version is 19.1
i suspected there is a problem with the TLS version or something else
can anyone give me a hand to solve this problem ?
A: The RabbitMQ team monitors this mailing list and only sometimes answers questions on StackOverflow.
When I have run into the same "waiting for handshake" symptom as you describe it has been because one of the certificate paths is invalid (certfile, keyfile or cacertfile). Double-check that those paths are correct and that the rabbitmq user can read them.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,336 |
Music for Wind Ensemble
C. Alan Publications
Nightmusik.com
"After silence, that which comes nearest to expressing the inexpressible is music."
-Aldous Huxley
Welcome to the website of composer,
Leslie Gilreath
Since I was 7 years old, playing piano in my home church in Anderson, South Carolina, I have been in love with music and music making.
Having a father playing guitar and a mother playing the church organ and piano meant that my childhood was full of music and the love of creating music.
And as is true for so many young people, after finding a home in my school band program, my life was forever positively and wonderfully changed by the music and the experiences therein.
I owe this love of music to my parents, Donnie and Janice Gilreath and to my band directors, Lane Moore and Jay Bocook, whose passion became the fuel for my own lifelong love of creating music. It is to them and to my composition teachers, Michael Hennagin (d. 1993) and Carolyn Bremer (d. 2018), who offered me such wonderful extensions of their gifts, that this site is dedicated.
Thank you for taking an interest in my music! I hope you enjoy it as much as I have enjoyed writing it. And, to those of you who work daily to change the lives of young people through music, BRAVO to you!
Want to get in touch with Leslie Gilreath? Don't hesitate to reach out!
Follow on Social Media or contact via email.
lesliegilreath@me.com
©2018 by lesliegilreath.com. Proudly created with Wix.com | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,165 |
The Temple Owls football statistical leaders are individual statistical leaders of the Temple Owls football program in various categories, including passing, rushing, receiving, total offense, defensive stats, and kicking. Within those areas, the lists identify single-game, single-season, and career leaders. The Owls represent Temple University in the NCAA's American Athletic Conference.
Although Temple began competing in intercollegiate football in 1894, the school's official record book only includes records from after Temple became a Division I-A (now FBS) program in 1971. Even so, these lists tend to be dominated by more recent players for several reasons:
Since the 1970s, seasons have increased from 10 games to 11 and then 12 games in length.
The NCAA didn't allow freshmen to play varsity football until 1972 (with the exception of the World War II years), allowing players to have four-year careers.
Bowl games only began counting toward single-season and career statistics in 2002. The Owls have played in five bowl games since this decision, giving many recent players an extra game to accumulate statistics.
These lists are updated through the end of the 2017 season.
Passing
Passing yards
Passing touchdowns
Rushing
Rushing yards
Rushing touchdowns
Receiving
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Receiving touchdowns
Total offense
Total offense is the sum of passing and rushing statistics. It does not include receiving or returns.
Total offense yards
Total touchdowns
Defense
Interceptions
Tackles
Sacks
Kicking
Field goals
Field goal percentage
References
Temple | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,647 |
Q: How would I parse my own XML..? I made this XML file that I need to try and generate a GUI from. No I am not going to jump over to WPF if you were wondering :-)
Here is the XML that I made:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<gui groupboxlabel="Barnets Stamdata" type="CHILD">
<textbox label="CPR" />
<textbox label="Navn" />
<textbox label="Efternavn" />
<textbox label="Addresse" />
<textbox label="Hus nr." />
<textbox label="Opgang" />
<textbox label="Post Nr." />
<textbox label="By" />
<textbox label="Email" />
<textbox label="Telefon nr." />
<textbox label="Sagsbehandler" />
<textbox label="Konsulent" />
<textbox label="Aflastning" />
<!-- <combobox label="Foranstaltning" /> -->
<!-- <date label="Anbring" /> -->
<!-- <date label="Udskriv" /> -->
</gui>
</root>
I need to find the gui tag first, so I can extract the 2 pieces of information there. Then I have to make a custom textbox control with a certain label name for every textbox child there.
I tried to do something like this at firs to try and print out what it looked like, but the code doesn't work because the child nodes that I find are null:
public void CreateNewLayout(Form parent, String path, String token)
{
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load(path);
XmlNodeList gui = xmlDoc.GetElementsByTagName("gui");
if (gui.Count == 0)
{
MessageBox.Show("XML fil har ingen elementer", "Fejl");
return;
}
while (gui.GetEnumerator().MoveNext())
{
gui.GetEnumerator().Current.ToString();
}
}
Problem is, my XML is pretty rusty...any help?
A: You can use Linq-Xml
var document = XDocument.Parse(inputXmlString);
document
.Root
.Element("gui")
.Elements()
.Select(element =>
new
{
Type = element.Name,
Label = element.Attribute("label").Value,
})
.Dump();
This is a quick example in Linqpad to show how to turn your XML into an anonymous type. It doesn't have to be an anonymous type it can be able type you want...
Also, if you want to filter elements, pass the element name into .Elements(string).
A: I would suggest giving different values for your textbox nodes such as textbox1, textbox2... If not something like the following should work:
System.Xml.XmlDocument doc = new System.Xml.XmlDocument();
List<string> labels = new List<string>();
doc.LoadXml([This would be text from file]);
string groupboxlabel = doc.SelectSingleNode("root/gui").Attributes["groupboxlabel"].Value;
string type = doc.SelectSingleNode("root/gui/textbox").Attributes["type"].Value;
System.XmlNodeList nodeList = doc.SelectNodes("root/gui");
foreach (XmlNode node in nodeList)
{
labels.Add(node.Attributes["label"].Value;);//Now you will have a list of labels
}
Hope this helps
A: Use something like this:
XDocument document = XDocument.Load(@"C:\DOTNET\PRACTICE\XmlTest\XmlTest\XMLFile1.xml");
XElement guiNode = document.Root.Element("gui");
List<XAttribute> attributes = new List<XAttribute>();
foreach(var attribute in guiNode.Attributes())
{
attributes.Add(attribute);
}
This uses the XDocument API instead of the older XMLDocument API. You can add null checks wherever you need them.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,860 |
87-year old Bennett only likes "nice theatres"
Updated / Wednesday, 28 May 2014 14:00
Tony Bennett and daughter Antonia
At 87 years of age, Tony Bennett has understandably cut back on his touring and likes to perform in "nice theaters" only nowadays.
"I always tried to be myself," he recently told New Mexico's Salina Journal, reviewing a career in music that has exceeded 60 years.
"If you try to imitate another singer, you'll just be a member of the chorus. To learn the phrasing of a song, imitate how a musician, an instrumentalist, would do it. Then be yourself, and you'll sound different from anyone. Sinatra, Nat King Cole -- they had different styles, but they were always themselves."
Tony Bennett's daughter Antonia, a singer and pianist, has been opening for the 87-year old crooner for the past few years, most recently in Santa Fe on May 24.
His repertoire, he said, had generally been been based on high quality, intelligent songs. "I never tried to get a hit record, but I've had quite a lot of hit records," he declared.
Bennett first topped the US charts in 1951 with Because of You, but 1962 was the year of his biggest hit around the world, I Left My Heart in San Francisco.
He boasts a healthy lifestyle, regular exercise and daily vocal warm-ups. "People don't think of age when I go out there," he said. "Audiences see me and say, 'Look what this guy is doing at 87.' I'm in top shape. I exercise three times a week and still do a lot of walking. I sleep well, wake up well and I do two things I love every day -- sing and paint."
Bennett paints under his birth name, Anthony Benedetto, and an exhibition of his paintings of legendary jazz musicians will be mounted later this year in Greenwich Village.
"I want to knock the audience out every time I go out there," he said. "I'll never give audiences anything cheap. And when audiences walk out feeling happy, it makes me feel so happy too.
"I just love to make people feel good. That's what I was put on Earth for."
The singer also told the Salina Journal that he still doesn't like rock 'n roll. In March in an interview with BBC Radio 4's Today programme the singer roundly criticised the music industry, saying songs do not have a "lasting quality." He lambasted record companies for their lack respect for the public.
Bennett told Today that there is now such a focus on the young audience that families cannot enjoy music together. "The songs that are written today, most of them are terrible, " he said. "It's a very bad period, musically, throughout the world for popular music."
The crooner argued that the most important thing for the current music industry is money, rather than producing good material. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,049 |
En illusion är en falsk uppfattning av verkligheten, ofta orsakad av flertydiga sinnesintryck. En illusionist är en person som i underhållningssyfte utför trolleri men som medger att dessa är just illusioner, till skillnad från häxor, siare med flera som påstår sig utöva magi med övernaturliga effekter.
Illusioner kommer från något i den verkliga omgivningen medan hallucinationer skapas helt och hållet av hjärnan. Illusion är ett relativt vanligt symtom vid konfusion medan hallucinos är ett centralt symtom vid psykos.
Illusion kan också handla om uppfattning om hur vi kan veta att någon existerar eller att ens världen existerar. Detta är vanligt inom filosofi, existentialismen.
Sinnesintryck som kan ge upphov till illusioner
Hägring
Optisk illusion
Skenperspektiv
Synvilla
Trompe l'œil
Trolleri
Se även
Allusion
Självbedrägeri
Externa länkar
Optical Illusions
Illusion Video.
Termer inom metafysik
Kognitiva symtom
Trolleri
Verklighet | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 381 |
Erricka Bridgeford is an African American activist from Baltimore. She worked to get Maryland's death penalty law repealed in 2013, and founded and co-organizes quarterly 72-hour "Ceasefire" weekends. Baltimore Ceasefire weekends began in 2017, in the hope of reducing violence. Erricka was named 2017's Marylander of the Year by The Baltimore Sun.
Early life
Bridgeford was born on October 9, 1972, the oldest of four children. She grew up in Normount Court, a housing project in West Baltimore. Her family was close-knit, and her parents, she said in an interview, were like "the Huxtables of the ghetto," serving as surrogate parents to other young people in the neighborhood. The area was poor but not plagued with violence until the 1980s and 90s. Since then she has lost a brother, a stepson, two cousins (who were brothers), and several friends to gun violence.
As a result of Amniotic Band Syndrome, Bridgeford was born without a right hand, and with four fingers on her left hand.
Career
Bridgeford began her career working for an organization devoted to infant wellness. In 2001 she trained as a volunteer at the Baltimore Community Mediation Center; two months later she was hired, and in 2002 she became their director of training. In 2005, Erricka became Director of Training at Community Mediation Maryland where she provided training to the 18 community mediation centers in Maryland, as well as to state agencies and organizations. In 2020, she was named Executive Director at Baltimore Community Mediation Center.
On February 3, 2018, she made a guest appearance on the online children's web series, Danny Joe's Tree House.
Activism
Bridgeford has been involved in community activism since the late 1990s. The death of her brother David in 2007 inspired her to focus on ending violence.
Death penalty repeal
Bridgeford began advocating for repeal of the death penalty in Maryland in 2009. Along with several other survivors of homicide victims, she testified before the Maryland legislature, spoke repeatedly to the press, and shared her personal story at rallies and other events. She argued that justice could not be gained through revenge, and that the death penalty was a form of revenge. "It's not justice to me to have another dead body in place of my brother's dead body," she said in an interview. Bridgeford believed the death penalty only brought more pain and failed to honor the victims; she suggested to a CBS reporter that the money Maryland saved by repealing the death penalty could be used to provide services to the victims' families. Senator Bobby Zirkin later cited the testimony of homicide survivors such as Bridgeford as a major factor in his decision to support the repeal. The death penalty was repealed in Maryland in 2013.
Support for survivors of homicide victims
Bridgeford worked with a coalition of murder victims' family members and organizations to support Maryland House Bill 0355 (Programs for Survivors of Homicide Victims). The bill was designed to identify survivors of homicide victims as a group of victims, and to ensure yearly funding. The bill passed and was approved by the governor on April 14, 2015. Maryland became the second state in the United States to provide specific funds and resources to murder victims' family members.
The same year, Senator Lisa Gladden introduced Senate Bill 0512, which called for the Criminal Injuries Compensation Board to include a survivor of homicide. At the hearing, Gladden said that she was inspired by Bridgeford's advocacy for homicide survivors, and believed that someone who had lost a loved one to homicide should be included on the Board that makes decisions about providing resources to victims' families. The bill was signed into law on April 14, 2014.
Baltimore Ceasefire 365
The idea for a ceasefire weekend was first suggested by hip-hop artist Ogun, who was working with Bridgeford on a rally for the anti-violence group 300 Men March. In 2017, with Baltimore's murder rate climbing, Bridgeford revisited the idea with Ogun and several other activists and began planning. The idea was to see if the community could go 72 hours without a shooting, stabbing, or any other kind of violence; their motto was "Nobody kill anybody." In addition, the effort rallied citizens to create and attend life-affirming events for 72 hours, while accepting the Baltimore Peace Challenge.
The first Ceasefire took place the first weekend in August, and drew thousands of residents to special community events and activities. The second Ceasefire weekend, held in November, was even larger. Events included basketball tournaments, a yoga class, art exhibits, a sip-and-paint, a peace walk, and a campout. During the November 2017 ceasefire weekend, Tony Mason was killed, and then, the weekend continued with 48 hours of zero shootings. During the third ceasefire weekend in February 2018, there were zero homicides. This was the first weekend in 2018 in which no Baltimore residents were killed. The February 2018 ceasefire also began a stretch of 11.5 days without murder in Baltimore. This was the first time the city had experienced this since March 2014. Bridgeford gave credit to everyone in the city who had been tirelessly doing good work for years.
When murders occurred during the first two Baltimore Ceasefire weekends, residents were called to show up at the murder locations to pour light and love into the space and into the traumatized community. This began a new campaign, "Don't Be Numb", which encourages Baltimoreans to notice when people are killed, and to remember that each life lost is a loss for all of Baltimore. This also gave birth to Sacred Space Rituals, in which residents go to murder locations to make them sacred ground.
Bridgeford's efforts started a grassroots movement and "awakened a sense that the cycle of killing can be broken and that the power to do it lies in our own hands." She was named Marylander of the Year "for bringing hope to Baltimore in some of its bleakest hours." In 2020, research published by the American Journal for Public Health showed that there was an estimated 52% reduction in gun violence during ceasefire days and no evidence of a postponement effect on either the next 3 days or the next 3-day weekend following each ceasefire weekend.
Awards and distinctions
2017: Marylander of the Year, The Baltimore Sun
2017: Gloria Hertzfelt Unsung Award, Baltimore State's Attorney's Office
2017: Invited to give a TED Talk
2017: Peacemaker of the Year, Baltimore Community Mediation Center
2017: Best Baltimorean, City Paper
2015: Outstanding Volunteer Contribution to Victim's Services, Maryland Governor's Office of Crime Control and Prevention
References
External links
Ted Talk: "How Baltimore called a ceasefire"
Video: Erricka Bridgeford featured on WBAL-TV
Video: Erricka Bridgeford Reflects on Baltimore Ceasefire Weekend. Baltimore Magazine. November 8, 2017.
Video: Documentary featuring Erricka Bridgeford. Maryland Citizens Against State Execution (CASE). March 27, 2013.
Living people
People from Baltimore
American social activists
American community activists
Nonviolence advocates
Activists from Maryland
African-American activists
American anti–death penalty activists
1972 births
21st-century African-American people
21st-century African-American women
20th-century African-American people
20th-century African-American women | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,745 |
Carriera
Giocatore
Cresciuto calcisticamente nel , con cui in cinque stagioni ottenne una promozione in Serie C1, si trasferì nell'estate del 1985 al , in Serie B, dove rimase due per stagioni.
Nel 1987 passò all', società con la quale nella stagione 1987-1988 ottenne la promozione in Serie A e raggiunse la semifinale di Coppa delle Coppe.
Venne quindi ceduto nel 1989 alla per 6 miliardi di lire. A Torino partecipò nella stagione 1989-1990 alla conquista del double continentale composto da Coppa Italia e Coppa UEFA, tuttavia non riuscì a ritagliarsi spazi importanti anche a causa di due brutti infortuni, venendo così dirottato nel 1991 al che lo acquistò per 4,35 miliardi di lire.
Giocò poco anche in Puglia e tornò quindi in Piemonte, stavolta sponda , vincendo con la squadra allenata da Emiliano Mondonico anche la Coppa Italia 1992-1993.
Dalla stagione 1994-1995 tornò a giocare per l'Atalanta, dove ottenne un'altra promozione dai cadetti alla massima serie.
Allenatore
Al termine della carriera ha intrapreso quella da allenatore, inizialmente come vice proprio di Mondonico al e al , poi assumendo incarichi in società di categorie minori, tra cui il e l': a Cuneo è rimasto per tre stagioni (dal 2004 al 2007) riuscendo a centrare una promozione in Serie C2, una semifinale play-off per salire in C1 e una finale di Coppa Italia di Serie C; a Ivrea invece è riuscito a portare la squadra ad un passo dai play-off. Il 9 giugno 2008 è stato ingaggiato come allenatore del , neopromosso in Prima Divisione, per la stagione 2008-2009, venendo però esonerato dall'incarico il 14 ottobre dello stesso anno.
Il 9 luglio 2010 viene ingaggiato dall', ancora come allenatore in seconda di Mondonico; tra il 29 gennaio e il 14 febbraio 2011 gli viene inoltre affidata la guida della prima squadra a causa della temporanea rinuncia di Mondonico per motivi di salute. Il 17 giugno seguente, dopo che Mondonico (a stagione conclusa con la salvezza ai play-out contro il ) lascia definitivamente l'incarico per curarsi, Fortunato viene nominato nuovo allenatore dei lombardi; il 28 gennaio 2012 viene esonerato al termine della sconfitta casalinga contro il Bari (0-2) e sostituito da Sandro Salvioni.
Il 30 gennaio 2014 diventa allenatore del , nella seconda serie portoghese, riuscendo a portare la formazione lusitana alla salvezza con varie giornate d'anticipo. Il 30 giugno 2015 diventa allenatore della formazione Primavera del Vicenza, concludendo la stagione con un 13º posto. Per la stagione 2019-2020 viene chiamato alla guida della formazione Berretti del .
Palmarès
Giocatore
Competizioni nazionali
Legnano: 1982-1983 (girone B)
Juventus: 1989-1990
Torino: 1992-1993
Competizioni internazionali
Juventus: 1989-1990
Allenatore
Competizioni nazionali
Cuneo: 2004-2005 (girone A)
Note
Bibliografia
Collegamenti esterni | {
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} | 6,181 |
\section{Introduction}
The study of universal Banach spaces was initiated by Banach and Mazur \cite{Banach}, who showed that every separable Banach space embeds isometrically into $C([0,1])$ (see also \cite[Corollary 12.14]{Carothers} or \cite[Theorem 8.7.2]{Semadeni}). Motivated by the Rotation Problem of Mazur, Gurarij \cite{Gurarij} constructed a separable Banach space $\mathbb{G}$, now known as the \emph{Gurarij space}, which is universal for separable Banach spaces and which satisfies the following additional property: For every $\varepsilon > 0$, every pair $X \subseteq Y$ of finite-dimensional Banach spaces, and every isometric embedding $f : X \rightarrow \mathbb{G}$, there is an $\varepsilon$-isometric embedding $\tilde{f} : Y \rightarrow \mathbb{G}$ which extends $f$. In other words, $\mathbb{G}$ is \emph{of almost universal disposition} for the class of all finite-dimensional Banach spaces. Lusky \cite{Lusky} later showed that the Gurarij space is unique up to isometry, in the sense that any other space which is of almost universal disposition for all finite-dimensional Banach spaces must be isometrically isomorphic to $\mathbb{G}$. (A simpler proof of the uniqueness and universality of the $\mathbb{G}$ can be found in \cite{KS}.) As a consequence of the fact that $\mathbb{G}$ is of almost universal disposition, one can show that $\mathbb{G}$ is \emph{approximately ultrahomogeneous}: For every finite-dimensional subspace $X \subseteq \mathbb{G}$, every $\varepsilon > 0$ and every isometric embedding $f : X \rightarrow \mathbb{G}$, there is a surjective linear isometry $g : \mathbb{G} \rightarrow \mathbb{G}$ such that $\|g(x) - f(x)\| \leq \varepsilon$ for each $x \in X$. For more information on the Mazur Rotation Problem and related concepts in the context of Banach spaces, we refer the reader to the recent survey \cite{CFR}.
Similar developments have taken place in the setting of Fr\'echet spaces: Mazur and Orlicz (see, e.g., \cite[p. 101]{Rolewicz}) showed that $C(\mathbb{R})$ is homeomorphically universal for the class of all separable Fr\'echet spaces, in the sense that every Fr\'echet space admits a linear homeomorphism into $C(\mathbb{R})$. Moving toward the isometric theory of Fr\'echet spaces, Bargetz, K\k{a}kol and Kubi\'s \cite{BKK} recently proved the existence of an analogue of the Gurarij space in the setting of Fr\'echet spaces equipped with a fundamental system of seminorms; they construct a separable Fr\'echet space $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ with a suitable sequence of seminorms which is approximately ultrahomogeneous (with respect to isometric embeddings which preserve each given seminorm simultaneously) for the class of all finite-dimensional Fr\'echet spaces equipped with an infinite sequence of seminorms. A similar space is also constructed in \cite{BKK} for the class of all finite-dimensional \emph{graded} Fr\'echet spaces, i.e. Fr\'echet spaces equipped with an increasing sequence of seminorms.
Interest in universal spaces such as the Gurarij space is partially motivated by the study of their groups of linear isometries. In particular, the isometry group $\Iso(\mathbb{G})$ of the Gurarij space has been an object of intense investigation in recent years (see, for instance, \cite{BLLM, BY1}). One way to develop a better understanding of Polish groups such as $\Iso(\mathbb{G})$ is via their topological dynamics; a relevant property here is that of \emph{extreme amenability}. Recall that a topological group $G$ is extremely amenable if every continuous action of $G$ on a compact Hausdorff space $X$ has a common fixed point, i.e. a point $x \in X$ such that $g \cdot x = x$ for all $g \in G$. Many examples of such groups have been obtained using a link -- known as the \emph{Kechris-Pestov-Todor\v{c}evi\'c (KPT) correspondence} -- between the Ramsey property and the extreme amenability of automorphism groups of ultrahomogeneous first-order structures. Examples of such structures come from \emph{Fra\"iss\'e theory}, which provides a correspondence between countable ultrahomogeneous structures and certain classes of finitely-generated first-order structures, known as \emph{Fra\"iss\'e classes}. Since Fra\"iss\'e's original paper \cite{Fraisse} (see also \cite{Hodges}), many other versions of the Fra\"iss\'e correspondence have been developed; one of the more prominent instances of this is the development the Fra\"iss\'e theory of \emph{metric structures} (i.e. metric spaces with additional compatible structure) due to Ben Yaacov \cite{BY}. Other presentations of Fra\"iss\'e theory in the context of structures from functional analysis, including ``non-commutative'' structures, have been developed recently in \cite{FLMT, Lupini2018}. Naturally, these events led to the establishment of the KPT correspondence in more general settings, including Banach spaces \cite{BLLM, FLMT}. In fact, the first successful application of the KPT correspondence for metric structures developed in \cite{MT} was recently achieved in \cite{BLLM}, where the authors show that $\Iso(\mathbb{G})$ is extremely amenable by proving an \emph{approximate Ramsey property} for the class of all finite-dimensional Banach spaces. Interestingly, the latter result makes crucial use of the Dual Ramsey Theorem of Graham and Rothschild \cite{GR}. In this paper we obtain analogous results in setting of Fr\'echet spaces and more generally for \emph{multi-seminormed} spaces, i.e. vector spaces equipped with a finite or infinite sequence of seminorms. In particular, we develop Fra\"iss\'e theory for multi-seminormed spaces together with the relevant versions of the KPT correspondence and the approximate Ramsey property. A similar approach was used in \cite{BLLM_2} to study different classes of exact operator spaces, where a sequence of norms is also an essential part of the structure. In the present paper we provide a Fra\"iss\'e-theoretic proof of the existence of the spaces constructed in \cite{BKK}, which were originally defined using properties of a universal (Fra\"iss\'e) operator on $\mathbb{G}$ originally considered in \cite{GK}. Such an operator can be seen as a ``Gurarij'' version of the Rota universal operator on the separable infinite-dimensional Hilbert space (for which we refer the reader to, e.g., \cite[Section 4.1]{Lupini2018}). In addition to the known examples discussed above, we also obtain many new examples of Fr\'echet spaces with strong forms of approximate ultrahomogeneity, including examples which naturally arise from various combinations of Hilbert spaces and $L_p$ spaces. As a result, we obtain many new examples of extremely amenable groups; in particular, we show that the group $\Iso(\mathbb{G}^\omega, (\|\cdot\|_n)_{n<\omega})$ of all surjective linear seminorm-preserving isometries of $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n<\omega})$ is extremely amenable.
The rest of the paper is organized as follows. In Section 2 we consider amalgamation and Fra\"iss\'e properties of finite-dimensional multi-seminormed spaces. In particular, we show that the class $\mathcal M_{<\omega}$ of all finite-dimensional vector spaces equipped with a finite sequence of seminorms is Fra\"iss\'e. Similarly, the classes consisting of all finite-dimensional vector spaces with a sequence of seminorms of length at most $n$ (for various $n\geq 1$) are shown to be Fra\"iss\'e as long as one restricts to \emph{separated} multi-seminormed spaces, where a space is separated if the natural topology induced by the associated sequence of seminorms is a Hausdorff topology. More generally, we show that Fra\"iss\'e classes of finite-dimensional normed spaces naturally give rise to Fra\"iss\'e classes of finite-dimensional multi-seminormed spaces, and we use this to obtain many examples of such classes. In Section 3 we develop a notion of a \emph{Fra\"iss\'e Fr\'echet space} and we obtain examples of such Fr\'echet spaces by taking \emph{Fra\"iss\'e limits} of classes of certain finite-dimensional multi-seminormed spaces. We also show that the separable Fr\'echet spaces of almost universal disposition constructed in \cite{BKK} can be realized as Fra\"iss\'e limits. Section 4 contains a proof of the analogue of the KPT correspondence for multi-seminormed spaces, which involves isolating a useful version of the approximate Ramsey property for such spaces. To conclude, we show how to transfer the approximate Ramsey property of classes of finite-dimensional normed spaces to the corresponding approximate Ramsey property for certain classes of multi-seminormed spaces, thus obtaining new examples of extremely amenable groups by way of the KPT correspondence.
\subsubsection*{Acknowledgments}
We are grateful to Félix Cabello Sánchez for several helpful conversations and remarks. We also thank the referee for many useful comments and corrections, as well as for suggesting Problem \ref{prob3} below.
\section{Fra\"iss\'e classes of finite-dimensional Fr\'echet spaces}
We use standard set-theoretic notation; in particular, $\omega = \{0, 1, \dots\}$ denotes the first countably infinite ordinal and $\mathbb{N}$ will denote $\omega \setminus \{0\}$. When convenient, an integer $n$ will be identified with the set $\{0,\dots, n-1\}$ of its predecessors.
Let $X$ be a Fr\'echet space, i.e. a locally convex Hausdorff topological vector space which is completely metrizable via a translation-invariant metric. A \emph{fundamental system of seminorms} on $X$ is a sequence of seminorms $(\|\cdot\|_{X,n})_{n < \alpha}$ on $X$, where $\alpha \leq \omega$, such that the sets $$B_{n,\varepsilon}(x) = \{y \in X : \max_{m < n} \|x-y\|_{X,m} < \varepsilon\}, \, \, x \in X, n < \alpha, \varepsilon > 0$$ form a basis for the topology of $X$. It is a standard fact that every Fr\'echet space admits a (non-unique) fundamental system of seminorms and so we can view a Fr\'echet space as a \emph{multi-seminormed space} of the above form, i.e. a (finite or infinite) tuple consisting of a topological vector space equipped with a sequence of seminorms. By definition, a fundamental system of seminorms is always \emph{separating}: For every non-zero $x \in X$ there is a seminorm $\|\cdot\|_{X,n}, n < \alpha$ such that $\|x\|_{X,n} \neq 0$; in this case we also say that $X$ is \emph{separated}, which happens precisely when the sequence of seminorms induces a Hausdorff topology on $X$. In this paper we will work more generally with multi-seminormed spaces of the form $\textbf{X} = (X, (\|\cdot\|_{X,n})_{n<\lambda_{\textbf{X}}})$, where $X$ is a vector space, $1 \leq \lambda_{\textbf{X}} \leq \omega$ is the \emph{length} of $\textbf{X}$, which is defined as the length of the associated sequence of seminorms (which we always assume is non-empty), and where $(\|\cdot\|_{X,n})_{n < \lambda_X}$ is a (finite or infinite) sequence of seminorms on $X$. We will always conflate the above tuple $\textbf{X}$ with its underlying space $X$, and so we will always write $\lambda_X$ for the length when the sequence of seminorms is understood. We point out that in general we do not require a multi-seminormed space to be separated. With this in mind, we reserve the term ``Fr\'echet space'' for a separated multi-seminormed space $(X, (\|\cdot\|_{X,n})_{n < \lambda_X})$ such that the topology generated by the seminorms $\|\cdot\|_{X,n}$ is complete. It is clear that in this case the sequence of seminorms $(\|\cdot\|_{X,n})$ forms a fundamental system of seminorms on $X$. Finally, following the terminology in \cite{BKK, vogt1987}, we say that a multi-seminormed space $(X, (\|\cdot\|_{X,n})_{n < \lambda_X})$ is \emph{graded} if, for all $x \in X$, $\|x\|_{X,n} \leq \|x\|_{X,m}$ whenever $n \leq m < \lambda_X$. For more information about (graded) Fr\'echet spaces, we refer the reader to \cite{Hamilton, MV, vogt1987}. The reader is also referred to \cite{Kutateladze, Simon} for more information about the general theory of multi-seminormed spaces (which are also sometimes called seminormed spaces or multinormed spaces in the literature).
Throughout this paper we will work with various subclasses of the class $\mc M$ of finite-dimensional multi-seminormed spaces equipped with a fundamental system of seminorms. Of particular interest will be the subclass $\mathcal G$ consisting of all graded $X \in \mc M$. Given any subclass $\mathcal{K} \subseteq \mc M$, and $\alpha \leq \omega$, we let $\mathcal{K}_{< \alpha}, \mathcal{K}_{\leq \alpha}$ and $\mathcal{K}_{=\alpha}$ denote the collections of all $X \in \mathcal{K}$ such that $\lambda_X < \alpha, \lambda_X \leq \alpha$ and $\lambda_X = \alpha$, respectively. For any such $\mathcal{K}$ we also let $\mathcal{K}^{\mathrm{sep}}$ denote the subclass of $\mathcal{K}$ consisting of all separated members of $\mathcal{K}$.
Given two multi-seminormed spaces $X$ and $Y$ and a linear mapping $f: X\to Y$, we say that a linear mapping $f:X\to Y$ is {\em multi-bounded} when $\lambda_X\le \lambda_Y$ and when for every $m<\lambda_X$ one has $$\nrm{f}_\mr{mb}:= \sup_{m<\lambda_X} \nrm{f}_{(X,\nrm{\cdot}_{X,m}), (Y,\nrm{\cdot}_{Y,m})}<\infty,$$ and where as usual $ \nrm{f}_{(X,\nrm{\cdot}_{X,m}), (Y,\nrm{\cdot}_{Y,m})}$ is the (semi)norm of $f$ defined as $\sup_{\nrm{x}_{X,m}\le 1}\nrm{f(x)}_{Y,m}$. When working with the operator norm, we will usually suppress the notation and omit the reference to the underlying spaces when this is clear from context. Observe that since we are dealing with seminorms, this quantity might be infinite, even when $X$ is finite-dimensional. A {\em multi-isomorphism} $f:X\to Y$ is linear bijection such that both $f$ and $f^{-1}$ are multi-bounded. Observe also that an isomorphism is a Lipschitz mapping, when considering the metrics associated to the fixed seminorms on $X$ and $Y$.
Much like in the class of normed spaces and others, we define a Banach-Mazur-like pseudometric, quantifying distances between isometric types. (The relevant notion of an isometry in this setting will be defined below.) We define
$$d_\mr{BM}(X,Y):=\log\inf_{f}\nrm{f}_\mr{mb} \nrm{f^{-1}}_\mr{mb},$$
where $f$ runs over all multi-isomorphisms (if any) between $X$ and $Y$.
\begin{prop}
${\mc M}_{<\omega}$ and ${\mc G}_{<\omega}$ are $\sigma$-compact, and consequently separable.
\end{prop}
\begin{proof}
${\mc G}_{<\omega}$ is closed in ${\mc M}_{<\omega}$, so it suffices to prove that ${\mc M}_{<\omega}$ is $\sigma$-compact. As in the case of the Banach-Mazur pseudometric on the class of finite-dimensional normed spaces, it is not difficult to see that the closed $d_\mr{BM}$-balls on ${\mc M}_{<\omega}$ are compact. But unlike for the normed space case, where for each dimension $k$ the diameter is finite (hence the Banach-Mazur compactum is compact), this is not the case for multi-seminormed spaces. Given $(X,(\nrm{\cdot}_{X,k})_{k<\lambda_X})\in {\mc M}_{<\omega}$, let $\overline{\alpha}_X=(\alpha^X_s)_{s\subseteq \lambda_X}$ be defined for $s\subseteq \lambda_X$ by $\alpha_s^X:= \dim \bigcap_{k\in s} \ker \nrm{\cdot}_{X,k}$ and $\alpha_\varnothing^X := \dim X$.
\begin{claim}
The classes of multi-seminormed spaces with finite distances (i.e. isomorphic classes) are exactly, given a sequence $\bar\alpha=(\alpha_s)_{s\subseteq n}$ of positive integers, the sets $\mc A_{\bar{\alpha}}:=\{X\in {\mc M}_{<\omega} \,:\,\bar\alpha_X=\bar \alpha\}$.
\end{claim}
This fact easily implies that ${\mc M}_{<\omega}$ is $\sigma$-compact. We prove now the claim. The sets $\mc A_{\bar{\alpha}}$ are closed under isomorphic images because if $f:X\to Y$ is a multi-isomorphism, then $\overline{\alpha}_X= \overline{\alpha}_Y$, because $$f( \bigcap_{k\in s} \ker \nrm{\cdot}_{X,k})= \bigcap_{k\in s} \ker \nrm{\cdot}_{Y,k} \text{ for every $s\subseteq \lambda_X=\lambda_Y$.}$$ This follows from the fact that $f$ is multi-bounded. The converse is also true: for suppose that $\overline{\alpha}_X=\overline{\alpha}_Y$ and let $\lambda = \lambda_X = \lambda_Y$. First, observe that $\alpha_\varnothing^X=\dim X$, so we have that $\dim X=\dim Y$. The argument is by induction on the number $N=N_X=N_Y$ of non-empty $s\subseteq \lambda$ such that $\alpha_s^X>0$. If $N=0$, this means that all seminorms are in fact norms, so the desired result follows from the well-known fact that the norms on finite-dimensional spaces of the same dimension are all equivalent. Suppose now that $N>0$. Let us choose $k_0<\lambda$ such that $\alpha_{\{k_0\}}>0$. For suppose that $X,Y\in \mc A_{\overline{\alpha}}$. Let $X_0:=\ker \nrm{\cdot}_{X,k}$, $Y_0:=\ker \nrm{\cdot}_{Y,k}$, and let $X_1\subseteq X$ and $Y_1\subseteq Y$ be complementary subspaces of $X_0$ and $Y_0$ respectively. Then $N_{X_1}=N_{Y_1}<N$, so there is some multi-isomorphism $f: X_1\to Y_1$, and similarly $X_0':=(X_0, (\nrm{\cdot}_{X, k})_{k<\lambda_X, k\neq k_0})$ and $Y_0':=(Y_0, (\nrm{\cdot}_{Y, k})_{k<\lambda_X, k\neq k_0})$ satisfy that $N_{X_0'}=N_{Y_0'}<N$, so there is a multi-isomorphism $g: X_0'\to Y_0'$. Then $h: X\to Y$ defined by $h(x_0+x_1):= f(x_0)+g(x_1)$ for $x_0\in X_0$ and $x_1\in X_1$ is a multi-isomorphism.
\end{proof}
Given $\varepsilon \geq 0$, let $\Emb_\varepsilon(X,Y)$ denote the set of all injective $\emph{multi-$\varepsilon$-isometric embeddings}$ from $X$ into $Y$, i.e. linear mappings $f: X \rightarrow Y$ such that $$\frac{1}{1+\varepsilon} \|x\|_{X,m} \leq \|f(x)\|_{Y,m} \leq (1+\varepsilon) \|x\|_{X,m}$$ for each $x \in X$ and $m < \lambda_X$. When $\varepsilon = 0$, we simply refer to such a mapping as a \emph{multi-isometric embedding}; the collection of all such mappings $f : X \rightarrow Y$ will be denoted $\Emb(X,Y)$. Note that, unlike in the setting of normed spaces, a seminorm-preserving mapping need not be injective. Also, a seminorm-preserving mapping $f : X \rightarrow Y$ need not be continuous when $\lambda_X < \lambda_Y$. A \emph{multi-isometry} (resp. multi-$\varepsilon$-isometry) is a surjective multi-isometric embedding (resp. multi-$\varepsilon$-isometric embedding).
Below we will make use of the notion of a \emph{modulus} relative to a set $S$, i.e. a function $\varpi: S \times [0,\infty[\to [0,\infty[$ such that, for every $s \in S$, the function $\varpi(s, \cdot)$ is increasing and continuous at $0$ with value $0$. In our case, $S$ will be the product $\mathbb{N} \times (\mathbb{N} \cup \{\omega\})$.
\begin{defi}\label{Fraisse}
Let $\mathcal{K}$ be a class of finite-dimensional multi-seminormed spaces.
\begin{enumerate}[(1)]
\item $\mathcal{K}$ has the \emph{hereditary property} (HP) if $X \in \mathcal{K}$ whenever $Y \in \mathcal{K}$ and $\Emb(X,Y) \neq \varnothing$.
\item $\mathcal{K}$ has the \emph{joint embedding property} (JEP) if for every $X, Y \in \mathcal{K}$ there is $Z \in \mathcal{K}$ such that $\Emb(X,Z)$ and $\Emb(Y,Z)$ are non-empty.
\item $\mathcal{K}$ has the \emph{near amalgamation property} (NAP) if for every $\varepsilon > 0$, every $X,Y,Z \in \mathcal{K}$ and every pair of multi-isometric embeddings $f_0 : X \rightarrow Y, f_1 : X \rightarrow Z$, there is $W \in \mathcal{K}$ together with multi-isometric embeddings $g_0 : Y \rightarrow W, g_1 : Z \rightarrow W$ such that $\|g_0 \circ f_0 - g_1 \circ f_1\|_n \leq \varepsilon$ for each $n < \lambda_X$.
\item $\mathcal{K}$ is an \emph{amalgamation class with modulus of stability $\varpi$} if $(\{0\}, \|\cdot\|) \in \mathcal{K}$ and for every $d \in \mathbb{N}$, $l \in \mathbb{N} \cup \{\omega\}$, $\varepsilon > 0$ and $\delta\ge 0$, every $X,Y,Z \in \mathcal{K}$ such that $\dim X = d$ and $\lambda_X = l$, and every pair of multi-$\delta$-isometric embeddings $f_0 : X \rightarrow Y, f_1 : X \rightarrow Z$, there is $W \in \mathcal{K}$ together with multi-isometric embeddings $g_0 : Y \rightarrow W, g_1 : Z \rightarrow W$ such that:
\begin{enumerate}[(i)]
\item $\|g_0 \circ f_0 - g_1 \circ f_1\|_n \leq \varpi(d,l,\delta)+ \varepsilon$ for each $n < \lambda_X$.
\item $W$ is separated.
\end{enumerate}
\item $\mathcal{K}$ is \emph{Fra\"iss\'e} if it is a hereditary $d_{\mathrm{BM}}$-closed amalgamation class such that $\mathcal{K} \subseteq \mathcal M_{<\omega}$.
\end{enumerate}
\end{defi}
Observe that every amalgamation class $\mathcal{K}$ automatically has the JEP since one can simply amalgamate over the trivial normed space. We remark that condition (4)(ii) does not appear in the usual presentation of metric Fra\"iss\'e theory (and in particular the Fra\"iss\'e theory of Banach spaces developed in \cite{FLMT}). However, it appears to be necessary when working with arbitrary multi-normed spaces in order to guarantee injectivity of certain seminormed-preserving mappings encountered when constructing a Fra\"iss\'e limit. This condition will be trivially satisfied by all collections $\mathcal{K}$ of interest. Conditions (4)(i) and (4)(ii) together imply that for any $X$ belonging to an amalgamation class $\mathcal{K}$, there is some separated $Y \in \mathcal{K}$ such that $\Emb(X,Y) \neq \varnothing$. We will make use of this fact without reference.
We now proceed to show that the classes considered above are amalgamation classes. The following is a particular case of a more general result explained in Proposition \ref{amalgamation}, where it is shown that classes of multi-seminormed spaces defined from amalgamation classes of normed spaces are also amalgamation classes. We present the proof here since a version of the relevant construction (which is essentially a pushout in the category of multi-seminormed spaces) will be used in the sequel.
\begin{prop}\label{NAP}
For each $\mathcal{K} \in \{\mc M,\mathcal G\}$ and each $\alpha \leq \omega$, the classes $\mathcal{K}_{=\alpha}^{\mathrm{sep}}, \mathcal{K}_{\leq \alpha}^{\mathrm{sep}}$ and $\mathcal{K}_{<\alpha}^{\mathrm{sep}}$ are amalgamation classes with modulus of stability $\varpi(d,l,\delta) = 2\delta$. Furthermore, $\mathcal{K}_{<\omega}$ is an amalgamation class with the same modulus of stability.
\end{prop}
\begin{proof}
First we will show that $\mathcal{K}_{\leq \alpha}^{\mathrm{sep}}$ is an amalgamation class with modulus $\varpi(d,l,\delta) = \delta$ with respect to \emph{expansive} multi-$\delta$-isometric embeddings, i.e. multi-$\delta$-isometric embeddings $f: X \rightarrow Y$ which additionally satisfy $\|f(x)\|_{Y,n} \geq \|x\|_{X,n}$ for each $x \in X$ and $n < \lambda_X$. To this end, fix $\varepsilon > 0, \delta \geq 0, d \in \mathbb{N}$ and $l \in \mathbb{N} \cup \{\omega\}$. Let $X, Y, Z \in \langle \bar{\mathcal{K}}\rangle$ where $\dim X = d$ and $\lambda_X = l$, and fix expansive multi-$\delta$-isometric embeddings $f : X \rightarrow Y$ and $g : X \rightarrow Z$. We can suppose, without loss of generality, $\lambda_Y \leq \lambda_Z$. Consider the sum $Y \oplus Z$ together with the canonical inclusion mappings $i : Y \rightarrow Y \oplus Z$ and $j : Z \rightarrow Y \oplus Z$. Equip $Y \oplus Z$ with a sequence $(\|\cdot\|_n)_{n < \lambda_Z}$ of seminorms in the following way:
\begin{enumerate}[(i)]
\item For each $n \in [0, \lambda_X)$, let $$\|(y,z)\|_n := \inf\{\|u\|_{Y,n} + \|v\|_{Z,n} + (\delta+\varepsilon) \|x\|_{X,n} : \, x \in X, u \in Y, v \in Z, y = u + f(x), z = v - g(x)\}.$$
\item Suppose $n \in [\lambda_X, \lambda_Y)$. If $\mathcal{K} \neq \mathcal G$, let $\|(y,z)\|_n := \max\{\|y\|_{Y,n}, \|z\|_{Z,n}\}$. Otherwise, assuming inductively that $\|(y,z)\|_{n-1}$ has been defined, let $\|(y,z)\|_n := \max\{\|(y,z)\|_{n-1}, \|y\|_{Y,n}, \|z\|_{Z,n}\}$.
\item Suppose $n \in [\lambda_Y, \lambda_Z)$. If $\mathcal{K} \neq \mathcal G$, let $\|(y,z)\|_n := \|z\|_{Z,n}$. Otherwise, assuming inductively that $\|(y,z)\|_{n-1}$ has been defined, let $\|(y,z)\|_n := \max\{\|(y,z)\|_{n - 1}, \|z\|_{Z,n}\}$.
\end{enumerate}
Note that the alternative in the last two parts of the above construction is needed to guarantee that $(\|\cdot\|_n)_{n < \lambda_Z}$ is a graded sequence of seminorms whenever $X,Y$ and $Z$ are graded. Furthermore, $Y \oplus Z$ is separated whenever $Y$ and $Z$ are both separated. In the case where we are working in $\mathcal{K}_{<\omega}$, the above space can be turned into a separated space by extending the above sequence of seminorms by a norm.
First we check that $i$ is a multi-isometric embedding. Suppose first that $n < \lambda_X$. Then for each $y \in Y$, $\|i(y)\|_n = \|(y,0)\|_n \leq \|y\|_{Y,n}$ by definition of $\|\cdot\|_n$. On the other hand, given any $x \in X, u \in Y$ and $v \in Z$ such that $y = u+ f(x)$ and $0 = v - g(x)$, we have
\begin{equation*}
\begin{split}
\|u\|_{Y,n} + \|v\|_{Z,n} + (\delta+\varepsilon) \|x\|_{X,n} &= \|u\|_{Y,n} + \|g(x)\|_{Z,n} + (\delta+\varepsilon) \|x\|_{X,n} \\
& \geq \|u\|_{Y,n} + \|x\|_{X,n} + (\delta+\varepsilon) \|x\|_{X,n}\\
& = \|u\|_{Y,n} + (1+\delta+\varepsilon)\|x\|_{X,n}\\
& \geq \|u\|_{Y,n} + \|f(x)\|_{Y,n}\\
& \geq \|u + f(x)\|_{Y,n} = \|y\|_{Y,n}
\end{split}
\end{equation*}
and so $\|i(y)\|_n \geq \|y\|_{Y,n}$. Thus $i$ preserves the first $\lambda_X$ seminorms. If $\lambda_X \leq n < \lambda_Y$, then $\|i(y)\|_n = \|y\|_{Y,n}$ by definition and so $i$ is a multi-isometric embedding.
Next we check that $j$ is a multi-isometric embedding. The case where $n < \lambda_X$ is exactly as before. In the case where $\lambda_X \leq n < \lambda_Y$, we have $\|j(z)\|_n = \|z\|_{Z,n}$ by definition. Otherwise, $n \geq \lambda_Y$. In the graded case we have $$\|j(z)\|_n = \max\{\|(0,z)\|_{\lambda_Y-1}, \|z\|_{Z,n}\} = \max\{\|z\|_{Z,\lambda_Y - 1}, \|z\|_{Z,n}\} = \|z\|_{Z,n}$$ since $Z$ is graded. In the non-graded case we simply have $\|j(z)\|_n = \|z\|_{Z,n}$ by definition, and so $j$ is a multi-isometric embedding.
It remains to check $\|i \circ f - j \circ g\| \leq \delta+\varepsilon$. To see this, note that for each $x \in X$ and $n < \lambda_X$ we have $$\|i(f(x)) - j(g(x))\|_n = \|(f(x), -g(x))\|_n \leq (\delta+\varepsilon) \|x\|_{X,n}$$ by taking $u = 0$ and $v = 0$ in the definition of $\|\cdot\|_n$. Thus $i : Y \rightarrow Y \oplus Z$ and $j : Z \rightarrow Y \oplus Z$ are the desired multi-isometric embeddings.
Now suppose that $f$ and $g$ are merely multi-$\delta$-isometric embeddings. Define a new sequence of seminorms $(\|\cdot\|'_{X,n})_{n<\lambda_X}$ on $X$ by setting $\|x\|'_{X,n} := \frac{1}{1+\delta} \|x\|_{X,n}$. Then $f$ and $g$ become expansive multi-$\delta'$-isometric embeddings from $X$ (equipped with the new sequence of seminorms) into $Y$ and $Z$, respectively, where $\delta' = 2\delta + \delta^2$. Thus we can apply the above argument to find $W \in \mathcal{K}_{\leq \alpha}^{\mathrm{sep}}$ and multi-isometric embeddings $i : Y \rightarrow W, j : Z \rightarrow W$ such that $$\|i(f(x)) - j(g(x))\|_{W,n} \leq (\delta' + \varepsilon) \|x\|'_{X,n} \text{ for all $n < \lambda_X$ and $x \in X$ such that $\|x\|'_{X,n} \leq 1$}.$$ Now fix $n < \lambda_X$ and $x \in X$ such that $\|x\|_{X,n} \leq 1$. Then $\|x\|'_{X,n} \leq \frac{1}{1+\delta} \leq 1$ and so $$\|i(f(x)) - j(g(x))\|_{W,n} \leq (\delta' + \varepsilon) \|x\|'_{X,n} \leq \frac{\delta'}{1+\delta} + \varepsilon = \frac{\delta + \delta(1+\delta)}{1+\delta} + \varepsilon \leq 2\delta + \varepsilon.$$ This shows that $\mathcal{K}_{\leq \alpha}^{\mathrm{sep}}$ is an amalgamation class with modulus $\varpi(d,l,\delta) = 2\delta$. In the case where $\mathcal{K} = \mathcal G$, note that the space $Y \oplus Z$ constructed above is graded whenever $X, Y$ and $Z$ are graded. This proofs for $\mathcal{K}_{<\alpha}^\mathrm{sep}, \mathcal{K}_{=\alpha}^\mathrm{sep}$ and $\mathcal{K}_{<\omega}$ are similar.
\end{proof}
\begin{rem}
The above proof can easily be adapted to prove amalgamation properties for classes of other ``norm-like'' structures from functional analysis. For instance, one can similarly show that the class of all finite-dimensional F-(semi)-normed spaces (as in \cite{K77}) is an amalgamation class with respect to $\delta$-isometric embeddings.
\end{rem}
Another source of examples of amalgamation classes of multi-seminormed spaces will come from the following proposition, which will make use of known amalgamation properties of classes of normed spaces. (The relevant definitions are analogous to our case; the reader is referred to \cite{FLMT} for more details, examples and discussion.) The following is standard notation: Given a seminormed space $X=(X,\nrm{\cdot})$, we denote by $ X_{\nrm{\cdot}}$ the normed space $(X/\ker\nrm{\cdot}, \nrm{\cdot})$ where $\|[x]\| := \|x\|$. Note that this norm is well-defined.
\begin{defi}\label{classes}
Let $\bar{\mathcal{K}}:=\{\mathcal{K}_n\}_{n< \alpha}$, $\alpha\le \omega$, be a collection of classes $\mathcal{K}_n$ of normed spaces. We define $\langle \bar{\mathcal{K}}\rangle_{\le \alpha}$, $\langle \bar{\mathcal{K}}\rangle_{< \alpha}$, $\langle \bar{\mathcal{K}}\rangle_{= \alpha}$ as the classes of separated multi-seminormed spaces $(X,(\nrm{\cdot}_n)_{n<\lambda_X}) \in \mc M_{\le \alpha}$, $\mc M_{< \alpha}$ and in $\mc M_{= \alpha}$, respectively, such that $X_{\|\cdot\|_n} \in \mathcal{K}_n$ for all $n < \lambda_X$.
\end{defi}
\begin{prop}\label{amalgamation}
Let $\{\mathcal{K}_n\}_{n < \omega}$ be a sequence of families of finite-dimensional normed spaces.
\begin{enumerate}[(a)]
\item If each $\mathcal{K}_n$ is hereditary, then so are $\langle \bar{\mathcal{K}}\rangle_{< \alpha}$ and $\langle \bar{\mathcal{K}}\rangle_{\leq \alpha}$ for each $\alpha \leq \omega$.
\item Suppose each $\mathcal{K}_n$ is an amalgamation class with modulus of stability $\varpi^{\mathcal{K}_n} : \mathbb{N} \rightarrow [0,\infty[$. Then for each $\alpha \leq \omega$ the class $\langle \bar{\mathcal{K}}\rangle_{< \alpha}$ is also an amalgamation class with modulus of stability $$\varpi(d,l,\delta) := \max_{n<l} \varpi^{\mathcal{K}_n}(d,\delta).$$ Furthermore, for each $n < \omega$ the classes $\langle \bar{\mathcal{K}}\rangle_{\leq n}$ and $\langle \bar{\mathcal{K}}\rangle_{= n}$ are amalgamation classes with the same modulus of stability.
\end{enumerate}
\end{prop}
\begin{proof}
The proof of (a) is relatively straightforward, so we leave the details to the reader. We only check (b) for $\langle \bar{\mathcal{K}}\rangle_{< \alpha}$ since the proofs for the other classes are similar. Fix $\varepsilon > 0, \delta \geq 0$ together with integers $d, l$. Let $X, Y, Z \in \langle \bar{\mathcal{K}}\rangle_{< \alpha}$ where $\dim X = d$ and $\lambda_X = l$, and fix multi-$\delta$-isometric embeddings $f_0 : X \rightarrow Y$ and $g_0 : X \rightarrow Z$. Suppose, without loss of generality, $\lambda_Y \leq \lambda_Z$. To construct the required element of $\langle \bar{\mathcal{K}}\rangle_{< \alpha}$, we will first define a sequence $(W_i)_{i < \lambda_Z}$ of finite-dimensional normed spaces. For each integer $i < l$, define mappings $f_0^i : X_{\nrm{\cdot}_i} \rightarrow Y_{\nrm{\cdot}_i}$ and $g_0^i : X_{\nrm{\cdot}_i} \rightarrow Z_{\nrm{\cdot}_i}$ by setting $$f_0^i([x]_i) = [f_0(x)]_i \text{ and } g_0^i([x]_i) = [g_0(x)]_i.$$ Then each $f_0^i$ and $g_0^i$ is a $\delta$-isometric embedding between elements of $\mathcal{K}_i$ and so, since $\mathcal{K}_i$ is an amalgamation class for each $i < n$, there are $W_i \in \mathcal{K}_i$ and isometric embeddings $f_1^i : Y_{\nrm{\cdot}_i} \rightarrow W_i, g_1^i : Z_{\nrm{\cdot}_i} \rightarrow W_i$ such that $$\|f_1^i \circ f_0^i - g_1^i \circ g_0^i \|_{X_{\nrm{\cdot}_i}, W_i} \leq \varpi^{\mathcal{K}_i}(d,\delta) + \varepsilon.$$ For $i \in [l, \lambda_Y)$, let $W_i$ be obtained from an application of the JEP to the pair $Y_{\nrm{\cdot}_i}, Z_{\nrm{\cdot}_i}$; let $f_1^i$ and $g_1^i$, respectively, denote the corresponding multi-isometric embeddings. Finally, for $i \in [\lambda_Y, \lambda_Z)$, simply let $W_i = Z_{\nrm{\cdot}_i}$.
Let $W = \prod_{i< \lambda_Z} W_i$ and equip $W$ with a family of seminorms defined by setting $$\|(w_0, \dots, w_{\lambda_Z - 1})\|_i = \|w_i\|_{W_i} \text{ for $i < \lambda_Z$}.$$ Observe that $W \in \langle \bar{\mathcal{K}}\rangle_{< \alpha}$ since $W$ is separated and $W_{\|\cdot\|_i} \cong W_i$ for each $i < \lambda_Z$. Define $f_1 : Y \rightarrow W$ and $g_1 : Z \rightarrow W$ by $$f_1(y) = \left(f_1^0([y]_0), \dots, f_1^{\lambda_Y - 1}([y]_{\lambda_Y - 1}), 0, \dots, 0\right),$$ $$g_1(z) = \left(g_1^0([z]_0), \dots, g_1^{\lambda_Y-1}([z]_{\lambda_Y-1}), [z]_{\lambda_Y}, \dots, [z]_{\lambda_Z-1} \right).$$ Then it is straightforward to check that $f_1$ and $g_1$ are multi-isometric embeddings which witness the amalgamation property for $\mathcal{K}$ for the above parameters. Injectivity follows from the fact that $Y$ and $Z$ are separated.
\end{proof}
Combining our two sources of examples of amalgamation classes yields the following list:
\begin{ex}\label{examples}
\begin{enumerate}[(1)]
\item The class $\mathcal M_{<\omega}$ of all finite-dimensional multi-seminormed spaces $X$ such that $\lambda_X < \omega$ is a Fra\"iss\'e class. In this case, observe that we do not need to restrict to separated multi-seminormed spaces since an arbitrary finite sequence of seminorms can always be extended by a norm. The same is true of the subclass $\mathcal G_{<\omega}$ of all graded $X \in \mathcal M_{<\omega}$.
\item For each $\alpha \leq \omega$, the classes $\mathcal M_{<\alpha}^{\mathrm{sep}}, \mathcal M_{\leq \alpha}^{\mathrm{sep}}$ and $\mathcal M_{=\alpha}^{\mathrm{sep}}$ of all finite-dimensional \emph{separated} multi-seminormed spaces such that $\lambda_X < \alpha, \lambda_X \leq \alpha$ and $\lambda_X = \alpha$, respectively, are amalgamation classes. Note, however, that these are not Fra\"iss\'e classes since it is possible for a non-separated multi-seminormed space to embed into a separated space according to our definition of a multi-isometric embedding. The same is true for the corresponding graded versions $\mathcal G_{<\alpha}^{\mathrm{sep}}, \mathcal G_{\leq \alpha}^{\mathrm{sep}}$ and $\mathcal G_{=\alpha}^{\mathrm{sep}}$.
\item For each infinite $I \subseteq \omega$, let $\mathcal M_{<\omega}^I$ denote the collection of all $(X, (\|\cdot\|_{X,n})_{n<\lambda_X}) \in \mathcal M_{<\omega}$ such that $\|\cdot\|_{X,n}$ is a norm for each $n \in I \cap [0, \lambda_X[$. Then $\mathcal M_{<\omega}^I$ is a Fra\"iss\'e class. (This follows from the proof of Proposition \ref{NAP}.) The amalgamation classes $\mathcal M_{< n}^I$ and $\mathcal M_{=n}^I$ can be defined similarly.
\item Let $\mathcal M_{<\omega}^\mathcal{H}$ denote the class of all finite-dimensional Fr\'echet-Hilbert spaces with a finite sequence of seminorms, where $(X,(\|\cdot\|)_{n<\lambda_X})$ is a \emph{Fr\'echet-Hilbert space} if $\|\cdot\|_n$ is a Hilbertian seminorm, i.e. a seminorm induced by a semi-inner product on $X$. Equivalently, $X$ is Fr\'echet-Hilbert if each quotient space $X_{\|\cdot\|_n}$ is a Hilbert space. Then $\mathcal M_{<\omega}^\mathcal{H}$ is a Fra\"iss\'e class. (We refer the reader to \cite{MV} for more on Fr\'echet-Hilbert spaces and related concepts.)
\item Given a sequence $(p_n)_{n<\omega} \subseteq [1,\infty[$ with $p_n\notin \{4,6,8,\dots\}$, consider the class $\mathcal M_{<\omega}^{(p_n)}$ of all multi-seminormed spaces $(X,(\nrm{\cdot}_n)_{n<\lambda_X})$ such that for each $n<\lambda_X$ the normed space $X_{\nrm{\cdot}_n}$ can be isometrically embedded into $L_{p_n}[0,1]$. Then this is a Fra\"iss\'e class for any such sequence $(p_n)$. This follows from the fact that $\Age(L_p[0,1])$ is a Fra\"iss\'e class of finite-dimensional normed spaces for each $p \in [1,\infty[$ such that $p \not \in \{4,6,8,\dots\}$ (see \cite{FLMT}). In this case, the corresponding modulus of stability $\varpi$ associated to $\Age(L_p[0,1])$ depends on the dimension for each $p$, and hence so does the corresponding modulus of stability associated to $\mathcal M_{<\omega}^{(p_n)}$.
\item In general, one can combine the various classes of normed spaces considered above to form new amalgamation classes of multi-seminormed spaces. For instance, given any sequence $(E_n)_{n<\alpha}$ of Fra\"iss\'e Banach spaces, one can set $\mathcal{K}_n = \Age(E_n)$ for each $n$ (noting that these are all Fra\"iss\'e classes of normed spaces) and form the associated class $\langle \bar{\mathcal{K}}\rangle_{\le \alpha}$ of multi-seminormed spaces, where $\bar{\mathcal{K}} = (\mathcal{K}_n)$.
\end{enumerate}
\end{ex}
In all but the last two examples, the corresponding modulus of stability is always independent of both the ``dimension'' and the ``length'' parameters in the sense that modulus only depends on the third variable. In the setting of normed spaces, such classes have been studied in \cite{Lupini2018} and are called \emph{stable} Fra\"iss\'e classes. Thus, stable Fra\"iss\'e classes of finite-dimensional normed spaces give rise to stable Fra\"iss\'e classes of finite-dimensional multi-seminormed spaces. In the last two families of examples, however, it is unknown if the modulus of stability depends on the dimension. This observation inspires the following:
\begin{que}
Are there any examples of amalgamation classes of finite-dimensional seminormed spaces such that the corresponding modulus of stability depends on all three parameters?
\end{que}
\section{Fra\"iss\'e Fr\'echet spaces}
For a multi-seminormed space $X$, let $\Iso(X)$ be the group of multi-isometries $f: X \rightarrow X$ equipped with the topology generated by basic open sets of the form $$\{g \in \Iso(X) : \max_{m < n} \|(f - g)\restriction Y\|_m < \varepsilon \}$$ where $f \in \Iso(X)$, $Y$ is a finite-dimensional linear subspace of $X$, $\varepsilon > 0$ and $n < \lambda_X$. This is the analogue of the strong operator topology in the setting of multi-seminormed spaces; in particular, a sequence $(f_k)$ in $\Iso(X)$ converges to $f$ if and only if $\|f_k(x) - f(x)\|_n$ converges to $0$ as $k \rightarrow \infty$ for each $x \in X$ and each $n \in \mathbb{N}$. For the sake of brevity, we will simply refer to this topology as the topology of pointwise convergence on $\Iso(X)$. We also equip $\Iso(X)$ with its \emph{left uniform structure}, i.e. the uniformity generated by entourages of the diagonal of the form $$\{(f,g) \in \Iso(X)^2 : f^{-1}\circ g \in U\}$$ where $U$ is a neighbourhood of the identity.
For a multi-seminormed space $(E, (\|\cdot\|_{E,n})_{n < \lambda_E})$, let $\Age(E)$ denote the \emph{age of $E$}, which is defined as the class of all finite-dimensional multi-seminormed spaces of the form $$(X, (\|\cdot\|_{E,n})_{n < \lambda_X})$$ where $X$ is a linear subspace of $E$ and $\lambda_X \leq \lambda_E$. (Warning: This is not the standard definition of the age, which is usually defined as the collection of all finitely-generated substructures of a given structure.) We also let $\Age_\alpha(E) := (\Age(E))_{=\alpha} = \Age(E) \cap \mathcal M_{=\alpha}$ for each $\alpha \leq \omega$. The collections $\Age_{\leq \alpha}(E)$ and $\Age_{\leq \alpha}(E)$ are defined similarly.
\begin{defi}
Let $(E, (\|\cdot\|_n)_{n < \lambda_E})$ be a multi-seminormed space and let $\mathcal{K}$ be a class of finite-dimensional multi-seminormed spaces.
\begin{enumerate}[(a)]
\item $E$ is \emph{$\mathcal{K}$-universal} if every $X \in \mathcal{K}$ embeds into $E$ via a multi-isometric embedding.
\item $E$ is \emph{approximately $\mathcal{K}$-ultrahomogeneous} if for every $\varepsilon > 0, X \in \mathcal{K}$ and $\gamma, \eta \in \Emb(X,E)$, there is $g \in \Iso(E)$ such that $\| g \circ \gamma - \eta \|_n \leq \varepsilon$ for each $n < \lambda_X$.
\item $E$ is \emph{$\mathcal{K}$-Fra{\"i}ss\'e with modulus of stability $\varpi$} if for every $d \in \mathbb{N}, l \in \mathbb{N} \cup \{\omega\}, \varepsilon > 0$ and $\delta \geq 0$, every $X \in \mathcal{K}$ and every $\gamma, \eta \in \Emb_\delta(X,E)$, there is $g \in \Iso(E)$ such that $\| g \circ \gamma - \eta \|_n \leq \varpi(d, l, \delta) + \varepsilon$ for each $n < l$.
\end{enumerate}
When $\mathcal{K} = \Age(E)$, we will occasionally omit the reference to $\mathcal{K}$ in the above definitions. We also omit the reference to the modulus of stability when working with a general Fra\"iss\'e Fr\'echet space.
\end{defi}
In order to transfer properties of an amalgamation class to a relevant Fra\"iss\'e Fr\'echet space, we will need a slight modification of the given modulus. To this end, for a modulus $$\varpi : \mathbb{N} \times (\mathbb{N} \cup \{\omega\}) \times [0,\infty[ \rightarrow [0,\infty[$$ we define a new modulus $$\varpi^*(d,l,\delta) := \inf_{\delta' > \delta} \varpi(d,l,\delta').$$ Note that $\varpi^*$ is indeed a modulus which furthermore satisfies $\varpi(d,l,\delta) \leq \varpi^*(d,l,\delta)$ for any $d \in \mathbb{N}$ and $l \in \mathbb{N} \cup \{\omega\}$.
Our main goal in this section is to show that when $\mathcal{K}$ is a Fra\"iss\'e class with modulus $\varpi$, then there is a unique (up to a multi-isometry) separable, $\mathcal{K}$-universal and $\mathcal{K}$-Fra\"iss\'e Fr\'echet space with modulus $\varpi^*$ whose finite-dimensional subspaces are precisely those from $\mathcal{K}$. Note that any $\mathcal{K}$-Fra\"iss\'e space is automatically approximately $\mathcal{K}$-ultrahomogeneous. Furthermore, if $E$ is Fra\"iss\'e, then $\Age_{<\omega}(E)$ is a Fra\"iss\'e class; the amalgamation property follows from the Fra\"iss\'e property together with the fact that any finite-dimensional subspace $X \in \Age_{<\omega}(E)$ is eventually separated by the sequence of seminorms from $E$.
Below we will need to make use of an inductive limit construction for spaces with a finite sequence of seminorms. Suppose $(X_n, I_n)_{n < \omega}$ is a sequence such that:
\begin{enumerate}[(1)]
\item $X_n \in \mathcal M_{<\omega}$ for each $n$.
\item $(\lambda_{X_n})_{n < \omega}$ is non-decreasing and converges to a fixed $\lambda \leq \omega$.
\item $I_n \in \Emb(X_n, X_{n+1})$ for each $n$.
\end{enumerate}
We define the \emph{inductive limit} $\lim_n (X_n, I_n)$ as follows: First, for each $m \leq n$, define $I_{m,n} \in \Emb(X_m, X_n)$ by setting $I_{m,m} = \id_{X_m}$ and $I_{m,n+1} = I_n \circ I_{m,n}$. Then let $V$ be the linear subspace of the product space $\prod_n X_n$ defined by declaring $(x_n)_{n < \omega} \in V$ if and only if there is some $m$ such that $x_n = I_{m,n}(x_m)$ for all $n \geq m$. For each $k < \lambda$, let $N_k$ be the linear subspace of $V$ consisting of all $(x_n)_n$ such that there is $m$ such that $k < \lambda_{X_m}$ and $\|x_n\|_k = 0$ for all $n \geq m$. Let $N = \bigcap_{k < \omega} N_k$ and let $V_0 = V / N$.
Define a sequence of seminorms $(\|\cdot\|_k)_{k < \lambda}$ on $V_0$ by $$\|(x_n)_n + N\|_k = \|x_m\|_{X_m, k}$$ where $m$ is the least integer such that $k < \lambda_{X_m}$ and $x_n = I_{m,n}(x_m)$ for all $n \geq m$. Note that these seminorms are well-defined and they form an increasing sequence whenever each $X_n$ is graded since each $I_n$ is a multi-isometric embedding.
To take a completion, we proceed as in the case of normed spaces. To this end, define an equivalence relation $\sim$ on the space of all Cauchy sequences in $V_0$ by declaring $$(\sigma_n) \sim (\tau_n) \iff (\forall k < \lambda) \, \lim_n \|\sigma_n - \tau_n\|_k = 0,$$ where a sequence $(\sigma_n) \subseteq V_0$ is Cauchy if it is Cauchy with respect to each seminorm $\|\cdot\|_k$ on $V_0$. Let $\lim_n (X_n, I_n)$ denote the resulting quotient space equipped with the sequence of seminorms of length $\lambda$ given by $$\vertiii{[(\sigma_n)]_\sim}_k = \lim_n \|\sigma_n\|_k \text{ for each $k < \lambda$.}$$ Observe that $(\lim_n (X_n, I_n), (\vertiii{\cdot}_k)_{k<\lambda})$ is a complete, separated multi-seminormed space, and so it is a Fr\'echet space. Given $x \in V_0$, let $\mathcal{C}(x)$ denote the equivalence class (in $\lim_n (X_n, I_n)$) of the Cauchy sequence with constant value $x$. Then for each $m$ the canonical mapping $$I_m^{(\infty)} : X_m \rightarrow \lim_n (X_n, I_n) : x \mapsto \mathcal{C}\big((\overset{(m)}{\overbrace{0,\dots, 0}}, x, I_{m,m+1}(x), I_{m,m+2}(x), \dots)\big)$$ is a seminorm-preserving linear mapping. When $X_m$ is separated, it is routine to check that $I_m^{(\infty)}$ is injective and hence a multi-isometric embedding. Note that union $\bigcup_n I_n^{(\infty)}(X_n)$ is dense in $\lim_n (X_n, I_n)$, in the sense that for every $[\sigma]_\sim \in \lim_n (X_n,I_n)$, there is a sequence $(\sigma_m)$ belonging to the union which converges to $[\sigma]_\sim$ with respect to each $\vertiii{\cdot}_k$. In particular, such a sequence converges to $[\sigma]_\sim$ with respect to the pseudometric induced by $\max_{l < k} \vertiii{\cdot}_l$ for any given $k < \lambda$. From now on, whenever we are working with an inductive limit of separated spaces $X_n$, we will identify $X_n$ with its image $I_n^{(\infty)}(X_n)$ in $\lim_n(X_n, I_n)$. In this way, each $I_n : X_n \rightarrow X_{n+1}$ becomes the corresponding inclusion mapping, so that $(X_n)$ is an increasing sequence of finite-dimensional subspaces of $\lim_n(X_n,I_n)$ and $X_n$ is equipped with the first $\lambda_{X_n}$ seminorms induced by the inductive limit. Furthermore, $\bigcup_n X_n$ is dense in the inductive limit.
We will need the following consequence of the amalgamation property, the proof of which can be found in \cite[Lemma 2.32]{FLMT}.
\begin{lem}\label{AP}
Suppose $\mathcal{K}$ is an amalgamation class with modulus $\varpi$. Fix $\varepsilon > 0$, $\Delta \subseteq \mathbb{R}^+$ finite and $\mathcal A \cup \{Y\} \subseteq \mathcal{K}$ finite. Then there is some $Z \in \mathcal{K}$ and $I \in \Emb(Y,Z)$ such that for every $X \in \mathcal A$, every $\delta \in \Delta$ and every $\gamma, \eta \in \Emb_\delta (X,Y)$, there is $J \in \Emb(Y,Z)$ such that $$\max_{l < \lambda_X} \| I \circ \gamma - J \circ \eta \|_l \leq \varpi(\dim X, \lambda_X, \delta) + \varepsilon.$$
\end{lem}
Before proceeding, we require one more piece of notation. Given two classes $\mathcal{K}$ and $\mathcal{K}'$ of finite-dimensional multi-seminormed spaces, write $\mathcal{K} \preceq \mathcal{K}'$ when for every $X \in \mathcal{K}$ there are $Y \in \mathcal{K}'$ and a multi-isometry of $X$ onto $Y$, and write $\mathcal{K} \equiv \mathcal{K}'$ when $\mathcal{K} \preceq \mathcal{K}' \preceq \mathcal{K}$.
\begin{thm}\label{fraisse}
If $\mathcal{K}$ is an amalgamation class with modulus $\varpi$, then there is a separable $\mathcal{K}$-Fra\"iss\'e Fr\'echet space $E$ with modulus $\varpi^*$ such that $\mathcal{K} \preceq \Age_{<\omega}(E)$. Furthermore, if $\mathcal{K}$ is a Fra\"iss\'e class then $\Age_{<\omega}(E) \equiv \mathcal{K}$.
\end{thm}
\begin{proof}
Let $\{Z_n\}_{n<\omega}$ be a countable $d_{\mathrm{BM}}$-dense subset of $\mathcal{K}$. Fix an enumeration $(\delta_n)$ of $\mathbb{Q} \cap [0,1]$ such that $\delta_0 = 0$. Using Lemma \ref{AP}, we find a sequence $(X_n, I_n)_{n<\omega}$ of separated spaces $X_n \in \mathcal{K}$ and multi-isometric embeddings $I_n \in \Emb(X_n, X_{n+1})$ with the following properties:
\begin{enumerate}[(a)]
\item Let $\lambda_\mathcal{K} := \sup_{X \in \mathcal{K}} \lambda_X$.
\begin{enumerate}
\item[(i)] If $\lambda_\mathcal{K} = \omega$, then $\lambda_{X_n} \geq n$ for all $n$.
\item[(ii)] If $\lambda_\mathcal{K} < \omega$, then $\lambda_{X_n} = \lambda_\mathcal{K}$ for all $n$.
\end{enumerate}
\item For every $k \leq n$, every $X \in \{Z_j\}_{j\leq n} \cup \{X_j\}_{j\leq n}$ and every $\gamma, \eta \in \Emb_{\delta_k}(X, X_n)$, there is $J \in \Emb(X_n, X_{n+1})$ such that $$\max_{l < \lambda_X} \| I_n \circ \gamma - J \circ \eta \|_l \leq \varpi(\dim X, \lambda_X, 2^{-n}) + 2^{-n}.$$
\item $\Emb(Z_m, X_n) \neq \varnothing$ for each pair $m<n$.
\end{enumerate}
Note that (c) can be arranged by applying the JEP of $\mathcal{K}$. Let $E = \lim_n (X_n, I_n)$. For simplicity, we assume from now on that $(X_n)$ is an increasing sequence of subspaces of $E$, $I_n$ is the inclusion mapping $X_n \subseteq X_{n+1}$, and $I_n^{(\infty)}$ is the inclusion mapping $X_n \subseteq E$. In particular, we work exclusively with the sequence of seminorms $(\|\cdot\|_k)_{k<\lambda_\mathcal{K}}$ associated to $E$.
The proof of the theorem will be complete once we prove the following two claims.
\begin{claim}
$E$ is $\mathcal{K}$-Fra\"iss\'e.
\end{claim}
\begin{proof}[Proof of Claim]
Fix $X \in \mathcal{K}$ together with $\varepsilon > 0, \delta' > \delta \geq 0$ and $\gamma, \eta \in \Emb_\delta(X,E)$. Choose a large enough $n$ such that $2^{-(n-1)} < \varepsilon$ and such that there are $j, k \leq n$ and a sufficiently small $\delta'' \geq 0$ with the following properties:
\begin{enumerate}[(i)]
\item $\delta < \delta_j < \delta'$.
\item $(\varpi(\dim X, \lambda_X, \delta_j) + \varepsilon)\delta'' < \varepsilon/3$.
\item There are $\theta \in \Emb_{\delta''}(X,Z_k)$ and $\widetilde{\gamma}, \widetilde{\eta} \in \Emb_{\delta_j}(Z_k, X_n)$ such that $$\max_{l<\lambda_X} \|\widetilde{\gamma} \circ \theta - \gamma\|_l \leq \varepsilon/3 \, \text{ and } \, \max_{l<\lambda_X} \|\widetilde{\eta} \circ \theta - \eta\|_l \leq \varepsilon/3.$$
\end{enumerate}
Using the definition of the sequence $(X_n, I_n)$, recursively construct sequences of embeddings $$J_s \in \Emb(X_{n+2s}, X_{n+2s+1}) \text{ and } L_s \in \Emb(X_{n+2s+1}, X_{n+2s+2})$$ such that:
\begin{enumerate}
\item[(iv)] $\|J_0 \circ \widetilde{\eta} - \widetilde{\gamma} \|_l \leq \varpi(\dim X, \lambda_X,\delta_j) + 2^{-n}$ for each $l < \lambda_X$.
\item[(v)] $\|J_{s+1} \circ L_s - \id_{X_{n+2s+1}}\|_l \leq 2^{-(n+2s)}$ for each $s\geq 0$ and each $l < \lambda_{X_{n+2s+1}}$.
\item[(vi)] $\|L_s \circ J_s - \id_{X_{n+2s}}\|_l \leq 2^{-(n+2s+1)}$ for each $s\geq 0$ and each $l < \lambda_{X_{n+2s}}$.
\end{enumerate}
Letting $\varepsilon_0 = \varpi(\dim X, \lambda_X,\delta_j) + 2^{-n}$ and $\varepsilon_m = 2^{-(n+m)}$ for $m\geq 1$, the above situation can be summarized by the following approximately commutative diagram:
\begin{center}
\begin{tikzcd}[column sep=large]
& & X_n \arrow[r, "I_n", hook] \arrow[dd, phantom, "\scalebox{0.8}{\cca{\varepsilon_0}}" description] & X_{n+1} \arrow[r, "I_{n+1}", hook] \arrow[rdd, "L_0"'] \arrow[dd, phantom,"\scalebox{0.8}{\cca{\varepsilon_{1}}}" description] & X_{n+2} \arrow[r, "I_{n+2}", hook] \arrow[dd, phantom,"\scalebox{0.8}{\cca{\varepsilon_{2}}}" description] & X_{n+3} \arrow[r, "I_{n+3}", hook] \arrow[rdd, "L_1"'] \arrow[dd, phantom,"\scalebox{0.8}{\cca{\varepsilon_{3}}}" description] & X_{n+4} \\
X \arrow[r, "\theta"] & Z_k \arrow[ru, "\widetilde{\gamma}"] \arrow[rd, "\widetilde{\eta}"'] & & & & & \dots \\
& & X_n \arrow[ruu, "J_0"'] \arrow[r, "I_n"', hook] & X_{n+1} \arrow[r, "I_{n+1}"', hook] & X_{n+2} \arrow[r, "I_{n+2}"', hook] \arrow[ruu, "J_1"'] & X_{n+3} \arrow[r, "I_{n+3}"', hook] & X_{n+4}
\end{tikzcd}
\end{center}
Now, according to (v) and (vi), for each fixed $x \in X_{n+2s}$ and $l < \lambda_{X_{n+2s}}$ we have $$\|J_{s+1} L_{s} J_s(x) - J_s(x)\|_l \leq 2^{-(n+2s)}\|J_s(x)\|_l = 2^{-(n+2s)}\|x\|_l.$$ On the other hand, we also have $$\|J_{s+1} L_{s} J_s(x) - J_{s+1}(x)\|_l = \|J_{s+1}(L_{s}J_s(x) - x)\|_l = \|L_{s}J_s(x) - x\|_l \leq 2^{-(n+2s+1)}\|x\|_l$$ and so an application of the triangle inequality then yields $$\|J_{s+1}(x) - J_s(x)\|_l \leq (2^{-(n+2s)} + 2^{-(n+2s+1)})\|x\|_l = 3\cdot 2^{-(n+2s+1)}\|x\|_l.$$ Then for every $s, t < \omega$ and every we have
\begin{equation}
\begin{split}
\|J_{s+t}(x) - J_s(x)\|_l &\leq \sum_{0 \leq i \leq t-1} \|J_{s+i+1}(x) - J_{s+i}(x)\|_l \leq \sum_{0 \leq i \leq t-1} 3\cdot 2^{-(n+2(s+i)+1)} \|x\|_l \\
& = \frac{3}{2^{n+2s+1}} \sum_{0 \leq i \leq t-1} 2^{-(2i)} \|x\|_l \leq \frac{3}{2^{n+2s+1}} \cdot 2 \|x\|_l \leq \frac{3}{2^{n+2s}}\|x\|_l.
\end{split}
\end{equation}
In particular, this implies that for each $x \in \bigcup_{n < \omega} X_n$ the sequence $(J_s(x))_s$ is Cauchy with respect to each seminorm $\|\cdot\|_{E,l}$ for $l < \lambda_E$; indeed, given $l < \lambda_E$, simply choose a large enough $N$ such that $l < \lambda_{X_n}$ and $x \in X_n$ for all $n > N$, noting that inequality (1) holds for all such $n$. Thus $(J_s)_{s<\omega}$ is pointwise Cauchy in $E$, so by completeness we can define a linear mapping $J : \bigcup_{n < \omega} X_n \rightarrow E$ by setting $J(x) = \lim_{s \geq k} J_s(x)$ where $k$ is least such that $x \in X_k$; we then extend $J$ to a mapping $J : E \rightarrow E$. Note that $J$ is a seminorm-preserving linear mapping. To see that $J$ is a bijection, we define as before a seminorm-preserving linear mapping $L : \bigcup_{n < \omega} X_n \rightarrow E$ by $L(y) = \lim_{s \geq k} L_s(y)$ where $k$ is least such that $y \in X_k$, and we extend it to a mapping $L : E \rightarrow E$. Then, since $E$ is separated, (v) and (vi) imply $L \circ J = J \circ L = \id_E$. Thus $J$ and $L$ are both multi-isometries. Finally, note that for each $l < \lambda_X$ we have $$\| J_s \circ \widetilde{\eta} - \widetilde{\gamma}\|_l \leq \varpi(\dim X, \lambda_X, \delta_j) + \sum_{0 \leq i \leq 2s} 2^{-(n+i)} \leq \varpi(\dim X, \lambda_X, \delta_j) + 2^{-(n-1)}.$$ Taking the limit as $s \rightarrow \infty$ we see that $\|J \circ \widetilde{\eta} - \widetilde{\gamma}\|_l \leq \varpi(\dim X, \lambda_X, \delta_j) + 2^{-(n-1)}$ and so
\begin{equation*}
\begin{split}
\|J \circ \eta - \gamma \|_l &\leq \|J \circ \eta - J \circ \widetilde{\eta} \circ \theta\|_l + \|J \circ \widetilde{\eta} \circ \theta - \widetilde{\gamma} \circ \theta\|_l + \|\widetilde{\gamma} \circ \theta - \gamma\|_l \\
& \leq \frac{\varepsilon}{3} + (\varpi(\dim X, \lambda_X, \delta_j) + 2^{-(n-1)})\|\theta\|_l + \frac{\varepsilon}{3} \\
& \leq \varpi(\dim X, \lambda_X, \delta_j) + \varepsilon \leq \varpi(\dim X, \lambda_X, \delta') + \varepsilon.
\end{split}
\end{equation*}
Thus $E$ is $\mathcal{K}$-Fra\"iss\'e with modulus $\varpi^*$.
\end{proof}
\begin{claim}
$\mathcal{K} \preceq \Age_{<\omega}(E)$. Furthermore, $\Age_{<\omega}(E) \preceq \mathcal{K}$ whenever $\mathcal{K}$ is Fra\"isse.
\end{claim}
\begin{proof}[Proof of Claim]
Fix $X \in \mathcal{K}$. Since $\mathcal{K}$ is an amalgamation class, $X$ isometrically embeds into a separated space $X'$. Thus we can assume without loss of generality that $X$ itself is separated. Now, find a decreasing positive sequence $(\delta_n)_n$ such that $\varpi(\dim X,\lambda_X,\delta_n) \leq 2^{-n}$ for each $n$. Using the definition of the sequence $\{Z_n\}$ together with property (c), for each $n$ we find some $\gamma_n \in \Emb_{\delta_n}(X,E)$. Since $E$ is $\mathcal{K}$-Fra\"iss\'e, for each $n$ we can choose $g_n \in \Iso(E)$ such that $\|g_n \circ \gamma_{n+1} - \gamma_n\|_l \leq 2^{-(n-1)}$ (where we take $\varepsilon = 2^{-n}$) for each $l < \lambda_X$. Define a sequence $(\eta_n)_n$ of multi-$\delta_n$-isometric embeddings of $X$ into $E$ by setting $\eta_0 = \gamma_0$ and $$\eta_{n+1} = g_0 \circ \dots \circ g_n \circ \gamma_{n+1} \text{ for each $n > 0$}.$$ Then $(\eta_n)_n$ is pointwise Cauchy, since $$\|\eta_{n+k} - \eta_n\|_l \leq \sum_{j=n}^{n+k-1} 2^{-(j-1)} \leq 2^{-(n-2)}$$ and so the limit $\eta : X \rightarrow E$ is a multi-isometric embedding; injectivity follows from the assumption that $X$ is separated.
Finally, note that the construction of $E$ implies $\Age_{<\omega}(E) \preceq \overline{\mathcal{K}}^{\mathrm{BM}}$, where the latter collection is the $d_{\mathrm{BM}}$-closure of $\mathcal{K}$ in $\mathcal M_{<\omega}$. Thus, if $\mathcal{K}$ is a Fra\"iss\'e class, i.e. a $d_{\mathrm{BM}}$-closed amalgamation class with the hereditary property, then the latter collection is precisely $\mathcal{K}$. Thus $\Age_{<\omega}(E) \preceq \mathcal{K}$ and so $\Age_{<\omega}(E) \equiv \mathcal{K}$ in this case.
\end{proof}
This completes the proof of the two claims and hence of the existence of a separable, $\mathcal{K}$-universal, $\mathcal{K}$-Fra\"iss\'e Fr\'echet space.
\end{proof}
We will henceforth refer to the space $E$ constructed in Theorem \ref{fraisse} as the \emph{Fra\"iss\'e limit} of the class $\mathcal{K}$ and we will denote it by $\Flim(\mathcal{K})$. Next we show that such a space is unique whenever $\mathcal{K}$ is a Fra\"iss\'e class. More generally, we have:
\begin{prop}\label{unique}
Suppose $E$ and $F$ are separable approximately ultrahomogeneous Fr\'echet spaces such that $\lambda_E = \lambda_F$ and $\Age_{<\omega}(E) \equiv \Age_{<\omega}(F)$. Then $E$ and $F$ are multi-isometric.
\end{prop}
\begin{proof}
Recursively define increasing sequences $(X_n)$ and $(Y_n)$ of elements of $\Age(E)$ and $\Age(F)$, respectively, sequences $(k_n)$ and $(l_n)$ of integers, and sequences of multi-isometric embeddings $(\gamma_n : X_n \rightarrow Y_n)$ and $(\eta_n : Y_n \rightarrow X_{n+1})$ such that the following conditions hold:
\begin{enumerate}[(i)]
\item $(k_n)$ and $(l_n)$ are non-decreasing and converge to $\lambda_E = \lambda_F$.
\item $X_n \in \Age_{k_n}(E)$ and $Y_n \in \Age_{l_n}(F)$.
\item $\|\eta_n \circ \gamma_n - \id_{X_n} \|_{E,m} \leq 2^{-n}$ for all $m < k_n$.
\item $\|\gamma_{n+1} \circ \eta_n - \id_{Y_n}\|_{F,m} \leq 2^{-n}$ for all $m < l_n$.
\item $\bigcup_{n < \omega} X_n$ and $\bigcup_{n < \omega} Y_n$ are dense in $E$ and $F$, respectively.
\end{enumerate}
To start, let $X_0 = Y_0 = \{0\}, \gamma_0 = 0$ and $k_0 = 1$. Now assume we have defined $X_n, Y_n, \gamma_n, \eta_{n-1}, k_n$ and $l_n$ for $n \geq 0$. Using the fact that $\Age_{<\omega}(E) \equiv \Age_{<\omega}(F)$, fix $\theta \in \Emb(Y_n, E)$. Since $E$ is approximately ultrahomogeneous, we can find $g \in \Iso(E)$ such that $$\|g \circ \theta \circ \gamma_n - \id_{X_n}\|_{E,m} \leq 2^{-n} \text{ for every $m < k_n$}.$$ Let $\eta_n := g \circ \theta \in \Emb(Y_n, E)$, let $k_{n+1} = \min\{l_n +1, \lambda_E\}$ and let $X_{n+1}$ be a finite-dimensional subspace of $E$ containing $X_n \cup \eta_n(Y_n)$, appropriately enlarged so that condition (v) will eventually hold, and equipped with the first $k_{n+1}$ seminorms from $E$. The construction of $Y_{n+1}, \gamma_{n+1} : Y_{n+1} \rightarrow X_{n+1}$ and $l_{n+1}$ is similar. This completes the inductive construction.
Now, note that for each $x \in \bigcup_{n < \omega} X_n$ the sequence $(\gamma_n(x))_n$ is Cauchy with respect to each seminorm $\|\cdot\|_{F,m}$: Given $m < \omega$, we can choose a large enough $N$ such that $m < \lambda_{X_n}$ and $x \in X_n$ for all $n > N$. Then $\|\gamma_{n+1}(x) - \gamma_n(x)\|_{F,m} \leq 2^{-(n-1)}$, which implies that the sequence $(\gamma_n)_{n<\omega}$ is pointwise Cauchy with respect to $\|\cdot\|_{F,m}$. Thus $(\gamma_n)_{n<\omega}$ is pointwise Cauchy in $F$, so by completeness we can define a linear mapping $\gamma : \bigcup_{n < \omega} X_n \rightarrow F$ by setting $\gamma(x) = \lim_{n \geq k} \gamma_n(x)$ where $k$ is least such that $x \in X_k$. Similarly, we can define a multi-isometric embedding $\eta : \bigcup_{n < \omega} Y_n \rightarrow E$ by $\eta(y) = \lim_{n \geq k} \eta_n(y)$ where $k$ is least such that $y \in Y_k$, and we extend it to a mapping $\eta : F \rightarrow E$. Then, since $E$ and $F$ are both separated, (iii) and (iv) imply $\gamma \circ \eta = \id_F$ and $\eta \circ \gamma = \id_E$, and so $\gamma$ is a multi-isometry.
\end{proof}
\begin{cor}[Fra\"iss\'e correspondence]
Let $\mathcal{K}$ be a class of finite-dimensional multi-seminormed spaces such that $\mathcal{K} \subseteq \mathcal M_{<\omega}$. The following are equivalent:
\begin{enumerate}[(1)]
\item $\mathcal{K}$ is a Fra\"iss\'e class.
\item $\mathcal{K} \equiv \Age_{<\omega}(E)$ for a unique separable Fra\"iss\'e Fr\'echet space $E = \Flim(\mathcal{K})$.
\end{enumerate}
\end{cor}
\begin{proof}
Suppose $\mathcal{K}$ is a Fra\"iss\'e class. By the previous two results, $\Flim(\mathcal{K})$ exists and is unique. By construction, we have $\mathcal{K} \equiv \Age_{<\omega}(\Flim(\mathcal{K}))$. To prove the other direction of the corollary, assume $E$ is s separable Fra\"iss\'e Fr\'echet space such that $\mathcal{K} \equiv \Age_{<\omega}(E)$. Then $\mathcal{K}$ is a hereditary, $d_{\mathrm{BM}}$-closed class. Furthermore, the fact that $E$ is separated implies that the class of separated elements of $\Age_{<\omega}(E)$ is cofinal in $\Age_{<\omega}$. Indeed, given $(X, (\|\cdot\|_n)_{n<m}) \in \Age_{<\omega}(E)$, we can use the fact that $X$ is finite-dimensional to find a sufficiently large $N$ such that $(X, (\|\cdot\|_n)_{n<N})$ becomes a separated subspace of $E$. From this observation, the amalgamation property of $\mathcal{K}$ follows from the Fra\"iss\'e property of $E$. Thus $\mathcal{K}$ is a Fra\"iss\'e class.
\end{proof}
\begin{ex}\label{ex}
\begin{enumerate}[(1)]
\item Let $\mathbb{G}^\omega$ be the product of countably many copies of the Gurarij space $\mathbb{G}$. In \cite{BKK} it is shown that there is a sequence $(\|\cdot\|_n)_{n < \omega}$ of seminorms on $\mathbb{G}^\omega$ such that $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ is a separable graded Fr\'echet space which is Fra\"iss\'e and universal for the class of all finite-dimensional graded multi-seminormed spaces with an infinite sequence of seminorms. It is also shown that there is a sequence $(\|\cdot\|'_n)_{n < \omega}$ of seminorms such that $(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n < \omega})$ is a separable Fr\'echet space which is Fra\"iss\'e and universal for the class of all finite-dimensional multi-seminormed spaces with an infinite sequence of seminorms. (The authors of \cite{BKK} do not use Fra\"iss\'e-theoretic terminology; however, this follows from \cite[Proposition 4.1]{BKK} and \cite[Proposition 5.5]{BKK}, respectively.) Below we will show that these two spaces can be obtained as Fra\"iss\'e limits of the classes $\mathcal G_{<\omega}$ and $\mathcal M_{<\omega}$, respectively.
\item For each $n \geq 1$, the spaces $\Flim(\mathcal M_{\leq n}^{\mathrm{sep}})$ and $\Flim(\mathcal G_{\leq n}^{\mathrm{sep}})$ can be seen as separated $n$-seminormed versions of the spaces considered in the previous example. (See Example \ref{examples} for the relevant notation.) Note that $\Age_{<\omega}(\Flim(\mathcal M_{\leq n}^{\mathrm{sep}}))$ is strictly larger than $\mathcal M_{\leq n}^{\mathrm{sep}}$, since the former collection contains non-separated multi-seminormed spaces. An analogous fact holds for $\Flim(\mathcal G_{\leq n}^{\mathrm{sep}})$.
\item Let $E$ be the space $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ considered in Example \ref{ex}(1). For each $k \in \mathbb{N}$, let $E_k$ be the multi-seminormed space $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < k})$ obtained by truncating the associated sequence of seminorms. Then using the properties of $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ it follows that $E_k$ is universal and Fra\"iss\'e for $\mathcal M_{<k}$. An interesting special case occurs when $n=1$, since $E_1$ can be seen as a seminormed version of the Gurarij space. In fact, if we let $\|\cdot\|$ be the seminorm on $E_1$, then the quotient $E_1 / \ker \|\cdot\|$ equipped with the corresponding quotient norm is separable, universal and approximately ultrahomogeneous for the class of all finite-dimensional Banach spaces, and so it is isometric to the Gurarij space. Note that the spaces $E_k$ are not separated, and so it is unclear if they are unique up to isometry.
\item $E = \Flim(\mathcal M_{<\omega}^{\mathcal{H}})$ is the unique separable Fr\'echet-Hilbert space which is $\mathcal M_{<\omega}^{\mathcal{H}}$-Fra\"iss\'e. To see that $E$ is indeed Fr\'echet-Hilbert, observe that each quotient space $E_{\|\cdot\|}$ (where $\|\cdot\|$ is a seminorm belonging to the sequence of seminorms associated to $E$) is approximately ultrahomogeneous for the class of all finite-dimensional Hilbert spaces, and hence is isometric to a Hilbert space.
\item For each sequence $(p_n) \subseteq [1,\infty[$ with $p_n\notin \{4,6,8,\dots\}$, $\Flim(\mathcal M_{<\omega}^{(p_n)})$ is a Fra\"iss\'e Fr\'echet space. In particular, if $p_n = p$ for each $n$, then $\Flim(\mathcal M_{<\omega}^p)$ can be seen as a \emph{Fr\'echet-$L_p$-space}, which is the $L_p$ analogue of the space considered in the previous example.
\end{enumerate}
\end{ex}
To conclude this section, we will use Proposition \ref{unique} to show that the spaces considered in Example \ref{ex}(1) can be obtained as Fra\"iss\'e limits; we will make use of the fact that $E$ is universal and approximately ultrahomogeneous for the class of all finite-dimensional (graded) multi-seminormed spaces with an infinite sequence of seminorms. From now until the end of the section, $E$ will denote one of these two spaces. Before proceeding, we need some new terminology: Given $n < \omega$ and two multi-seminormed spaces $X$ and $Y$ such that $\lambda_X = \lambda_Y = \omega$, an \emph{multi-isometric $n$-embedding} from $X$ to $Y$ is an injective linear mapping $f : X \rightarrow Y$ such that $\|f(x)\|_{Y,m} = \|x\|_{X,m}$ for each $x \in X$ and each $m < n$.
The following is a version of the near amalgamation property in the context of multi-isometric $n$-embeddings for a fixed $n$.
\begin{lem}
Suppose $X, Y$ and $Z$ are finite-dimensional separated multi-seminormed spaces with $\lambda_X = \lambda_Y = \lambda_Z = \omega$ and $f : X \rightarrow Y, g : X \rightarrow Z$ are multi-isometric $n$-embeddings for a fixed $n < \omega$. Then for every $\varepsilon > 0$ there is a finite-dimensional multi-seminormed space $W$ with $\lambda_W = \omega$ and multi-isometric embeddings $i_Y : Y \rightarrow W$ and $i_Z : Z \rightarrow W$ such that $\| i_Y \circ f - i_Z \circ g \|_m \leq \varepsilon$ for all $m < n$. Furthermore, $W \in \mathcal G$ whenever $X,Y,Z \in \mathcal G$.
\end{lem}
\begin{proof}
Fix all parameters and consider the sum $Y \oplus Z$ together with the canonical inclusion mappings $i_Y : Y \rightarrow Y \oplus Z$ and $i_Z : Z \rightarrow Y \oplus Z$. Equip $Y \oplus Z$ with a sequence $(\|\cdot\|_m)_{m < \omega}$ of seminorms defined by declaring $$\|(y,z)\|_m := \inf\{\|u\|_{Y,m} + \|v\|_{Z,m} + \varepsilon \|x\|_{X,m} : \, x \in X, u \in Y, v \in Z, y = u + f(x), z = v - g(x)\}$$ for each $m < n$, and $\|(y,z)\|_m := \|y\|_{Y,m} + \|z\|_{Z,m}$ for each $m \geq n$. Let $W$ be the sum $Y \oplus Z$ equipped with this sequence of seminorms. Note that $W$ is a graded multi-seminormed space whenever $X,Y$ and $Z$ are graded. As in the proof of Lemma \ref{NAP}, it is straightforward to check that the inclusion mappings $i_Y : Y \rightarrow Y \oplus Z$ and $i_Z : Z \rightarrow Y \oplus Z$ are multi-isometric embeddings which satisfy $\|i_Y \circ f - i_Z \circ g\|_m \leq \varepsilon$ for all $m<n$.
\end{proof}
\begin{lem}
Let $X$ be a finite-dimensional subspace of $E$ such that $\lambda_X = \omega$. For every $\varepsilon > 0, n \in \mathbb{N}$, finite-dimensional multi-seminormed space $Y$ and multi-isometric $n$-embeddings $\eta : X \rightarrow E$ and $\gamma : X \rightarrow Y$, there is a multi-isometric embedding $f : Y \rightarrow E$ such that $\| f \circ \gamma - \eta \|_m \leq \varepsilon$ for each $m < n$.
\end{lem}
\begin{proof}
Apply the previous lemma to $\varepsilon/2$, $\gamma : X \rightarrow Y$ and $\eta : X \rightarrow Z := \eta(X)$ to find the corresponding $W, i_Y$ and $i_Z$. Since $E$ is approximately ultrahomogeneous, there is $g \in \Iso(E)$ such that $\| g(i_Z(z)) - z \|_m \leq \frac{\varepsilon}{2} \|z\|_m$ for all $z \in Z$ and $m \in \mathbb{N}$. Let $f = g\restriction_W \circ \, i_Y$. Then for each $x \in X$ and $m < n$ we have
\begin{equation*}
\begin{split}
\|f(\gamma(x)) - \eta(x)\|_m & \leq \|g(i_Y(\gamma(x))) - g(i_Z(\eta(x)))\|_m + \|g(i_Z(\eta(x))) - \eta(x)\|_m \\
& \leq \|g(i_Y(\gamma(x))-i_Z(\eta(x)))\|_m + \frac{\varepsilon}{2}\|\eta(x)\|_m\\
& \leq \frac{\varepsilon}{2}\|x\|_m + \frac{\varepsilon}{2} \|x\|_m \leq \varepsilon\|x\|_m.
\end{split}
\end{equation*}
Thus $f$ is the desired multi-isometric embedding.
\end{proof}
\begin{cor}
$(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ and $(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n < \omega})$ are approximately ultrahomogeneous for $\mathcal G_{<\omega}$ and $\mathcal M_{<\omega}$, respectively. In particular, $(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega}) = \Flim(\mathcal G_{<\omega})$ and $(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n < \omega}) = \Flim(\mathcal M_{<\omega})$.
\end{cor}
\begin{proof}
We only prove the result for $\mathcal G_{<\omega}$; the proof for $\mathcal M_{<\omega}$ is identical. Fix $X \in \mathcal G_{<\omega}$ together with multi-isometric embeddings $\gamma, \eta : X \rightarrow E$. Extend the sequence of seminorms $(\|\cdot\|_{X, n})_{n < \lambda_X}$ to an infinite sequence in the natural way by declaring $\|\cdot\|_{X,m} = \|\cdot\|_{X,\lambda_X - 1}$ for all $m \geq \lambda_X$. Then $\gamma$ and $\eta$ become multi-isometric $\lambda_X$-embeddings. Let $Y = \gamma(X) \subseteq E$ be equipped with the sequence of seminorms from $E$ and apply the previous lemma to find a multi-isometric embedding $f : Y \rightarrow E$ such that $\|f(\gamma(x)) - \eta(x) \|_m \leq \frac{\varepsilon}{2} \|x\|_m$ for each $x \in X$ and $m < \lambda_X$. By the approximate ultrahomogeneity of $E$, there is $g \in \Iso(E)$ such that $\|g(y) - f(y)\|_m \leq \frac{\varepsilon}{2}\|y\|_m$ for each $y \in Y$ and $m \in \mathbb{N}$. Then $$\|g(\gamma(x)) - \eta(x)\|_m \leq \|g(\gamma(x)) - f(\gamma(x))\|_m + \|f(\gamma(x)) - \eta(x)\|_m \leq \varepsilon \|x\|_m$$ for each $x \in X$ and $m < \lambda_X$, as required.
\end{proof}
\section{The approximate Ramsey property}
As in \cite{BLLM} or \cite{FLMT}, we can characterize the extreme amenability of the group of multi-isometries of certain Fr\'echet spaces in terms of an approximate Ramsey property. Before stating the relevant Ramsey properties, we need some terminology. Given $r \in \mathbb{N}$, an \emph{$r$-colouring} of a set $X$ is simply a mapping $X \rightarrow r$. If $X$ is equipped with a finite sequence of seminorms $(\|\cdot\|_m)_{m < \lambda_X}$ and $n \leq \lambda_X$, an \emph{$n$-continuous colouring} of $X$ is a mapping $c : X \rightarrow [0,1]$ such that $$|c(x) - c(y)| \leq \max_{m < n} \|x-y\|_m \text{ for all $x, y \in X$}.$$ Thus, an $n$-continuous colouring is simply a mapping of $X$ into $[0,1]$ which is 1-Lipschitz with respect to the pseudometric induced by $\max_{m < n} \|\cdot\|_m$.
\begin{defi}
Let $\mathcal{K}$ be a collection of finite-dimensional multi-seminormed spaces.
\begin{enumerate}[(a)]
\item $\mathcal{K}$ has the \emph{discrete approximate Ramsey property} (discrete ARP) if for every $X, Y \in \mathcal{K}, r \in \mathbb{N}$ and $\varepsilon > 0$ there is $Z \in \mathcal{K}$ such that every $r$-colouring of $\Emb(X,Z)$ \emph{$\varepsilon$-stabilizes} on a set of the form $\gamma \circ \Emb(X,Y)$ for some $\gamma \in \Emb(Y,Z)$, i.e. there is $i < r$ such that $\gamma \circ \Emb(X,Y)$ is contained in the set $$(c^{-1}\{i\})_\varepsilon := \{\xi \in \Emb(X,Z) : \exists \eta \, \text{ $c(\eta) = i$ and $\|\xi-\eta\|_m \leq \varepsilon$ for all $m < \lambda_X$}\},$$ where the $m^{\mathrm{th}}$ seminorm $\|\cdot\|_m$ on the space of embeddings $\gamma : X \rightarrow Z$ is defined by setting $\|\gamma\|_m := \sup \{ \|\gamma(x)\|_m : x \in X, \|x\|_m =1 \}.$ In this case we will also say that $\gamma \circ \Emb(X,Y)$ is \emph{$\varepsilon$-monochromatic}.
\item $\mathcal{K}$ has the \emph{continuous approximate Ramsey property} (continuous ARP) if for every $X \in \mathcal{K}_{<\omega}, Y \in \mathcal{K}$ and $\varepsilon > 0$ there is $Z \in \mathcal{K}$ such that for every $\lambda_X$-continuous colouring $c$ of $\Emb(X,Z)$ there is $\gamma \in \Emb(Y,Z)$ such that the \emph{oscillation} of $c$ on $\gamma \circ \Emb(X,Y)$, defined as $$\osc(c \restriction \gamma \circ \Emb(X,Y)) := \sup\{|c(\gamma \circ \eta) - c(\gamma \circ \eta')| : \eta, \eta' \in \Emb(X,Y)\},$$ is at most $\varepsilon$. In this case we will say that $c$ \emph{$\varepsilon$-stabilizes} on $\gamma \circ \Emb(X,Y)$.
\end{enumerate}
\end{defi}
We will abbreviate the conclusions of the discrete and continuous ARP by writing $Z \rightarrow_\varepsilon (Y)^X_r$ and $Z \rightarrow_\varepsilon (Y)^X$, respectively. Exactly as in \cite{BLLM}, it turns out that these two notions are equivalent.
\begin{lem}
Let $\mathcal{K} \subseteq \mathcal M_{<\omega}$. Then $\mathcal{K}$ satisfies the discrete ARP if and only if it satisfies the continuous ARP.
\end{lem}
\begin{proof}
Suppose first that $\mathcal{K}$ has the discrete ARP. Fix $X,Y \in \mathcal{K}$ and $\varepsilon > 0$. Let $D \subseteq [0,1]$ be a finite $\varepsilon$-dense set. Apply the discrete ARP with $|D|$-many colours to find $Z \in \mathcal{K}$ such that $Z \rightarrow_\varepsilon (Y)^X_{|D|}$. We claim that $Z$ witnesses the continuous ARP for the above parameters. Indeed, given a $\lambda_X$-continuous colouring $c : \Emb(X,Z) \rightarrow [0,1]$, define a $|D|$-colouring $\widetilde{c} : \Emb(X,Z) \rightarrow D$ by the condition $|c(\varphi) - \widetilde{c}(\varphi)| \leq \varepsilon$ for every $\varphi \in \Emb(X,Z)$. Then there is $\gamma \in \Emb(Y,Z)$ such that $\widetilde{c}$ $\varepsilon$-stabilizes on $\gamma \circ \Emb(X,Y)$. It follows that $c$ $4\varepsilon$-stabilizes on $\gamma \circ \Emb(X,Y)$.
Now suppose $\mathcal{K}$ has the continuous ARP. We prove that $\mathcal{K}$ has the discrete ARP by induction on $r$, the number of colours. When $r=1$ this is trivial, so suppose inductively that $\mathcal{K}$ satisfies the discrete ARP for $r$-colourings. Fix $X,Y \in \mathcal{K}$ and $\varepsilon > 0$. By the inductive hypothesis, there is $Z_0 \in \mathcal{K}$ such that $Z_0 \rightarrow_\varepsilon (Y)^X_r$. Since $\mathcal{K}$ has the continuous ARP, there is $Z \in \mathcal{K}$ such that $Z \rightarrow_\varepsilon (Z_0)^X$. We claim that $Z$ witnesses the discrete ARP for the parameters $X,Y, \varepsilon$ and $r+1$. Indeed, fix a colouring $c: \Emb(X,Z) \rightarrow r+1$ and define $\widetilde{c} : \Emb(X,Z) \rightarrow [0,1]$ by setting $$\widetilde{c}(\varphi) = \min\left\{1, \inf_{\psi \in c^{-1}\{r\}} \max_{m < \lambda_X} \| \varphi - \psi \|_m \right\}.$$ It is routine to check that $\widetilde{c}$ is an $\lambda_X$-continuous colouring, so there is $\gamma \in \Emb(Z_0, Z)$ such that $\widetilde{c}$ $\varepsilon$-stabilizes on $\gamma \circ \Emb(X,Z_0)$. If there is some $\varphi \in \Emb(X,Z_0)$ such that $c(\gamma \circ \varphi) = r$, then $\gamma \circ \Emb(X, Z_0) \subseteq (c^{-1}\{r\})_\varepsilon$ and so we are done since then $c$ $\varepsilon$-stabilizes on $\gamma \circ \gamma_0 \circ \Emb(X,Y)$ for any choice of $\gamma_0 \in \Emb(Y,Z_0)$. If no such $\varphi$ exists, we can define an $r$-colouring $d$ of $\Emb(X,Z_0)$ by setting $d(\varphi) = c(\gamma \circ \varphi)$. By definition of $Z_0$, there is $\gamma_0 \in \Emb(Y,Z_0)$ such that $d$ $\varepsilon$-stabilizes on $\gamma_0 \circ \Emb(X,Y)$. It then follows that $c$ $\varepsilon$-stabilizes on $\gamma \circ \gamma_0 \circ \Emb(X,Y)$.
\end{proof}
The next lemma is a particular instance of a more general phenomenon which rephrases the Ramsey property of an age in terms of its limit. The proof is similar to that of \cite[Proposition 3.4]{MT}. Recall that $\Age_n(E)$ denotes the collection of all finite-dimensional linear subspaces $X \subseteq E$ equipped with the first $n$ seminorms from $E$.
\begin{lem}
Suppose $E$ is an approximately ultrahomogeneous multi-seminormed space. Then the collection $\Age_n(E)$ has the continuous ARP if and only if for every $X,Y \in \Age_n(E), \varepsilon > 0$ and $\lambda_X$-continuous colouring $c$ of $\Emb(X,E)$, there is $\gamma \in \Emb(Y,E)$ such that $\osc(c\restriction \gamma \circ \Emb(X,Y)) \leq \varepsilon$.
\end{lem}
\begin{proof}
The left-to-right direction is straightforward and will not be used in what follows, so we will only show the right-to-left direction. We will prove the contrapositive. First, let $H = \{\eta_1, \dots, \eta_k\}$ be a finite $\varepsilon/3$-dense subset of $\Emb(X,Y)$, where the latter set is equipped with the pseudometric induced by $\max_{m < \lambda_X} \|\cdot\|_m$. We will need the following claim, the proof of which is routine.
\begin{claim}
Suppose there is $Z \in \Age_n(E)$ such that for every $\lambda_X$-continuous colouring of $\Emb(X,Z)$, there is $\gamma \in \Emb(Y,Z)$ such that $\osc(c \restriction \gamma \circ H) \leq \varepsilon/3$. Then $Z \rightarrow_\varepsilon (Y)^X$.
\end{claim}
Now, if $\Age_n(E)$ does not have the continuous ARP, then there are $X,Y \in \Age_n(E)$ and $\varepsilon > 0$ such that no $Z \in \Age_n(E)$ witnesses $Z \rightarrow_\varepsilon (Y)^X$. Thus, by the Claim, for each such $Z$ we can fix a bad $\lambda_X$-continuous colouring $c_Z$ such that $\osc(c_Z \restriction \gamma \circ H) \geq \varepsilon$ for any choice of $\gamma \in \Emb(Y,Z)$. Fix an ultrafilter $\mathcal{U}$ on $\Age_n(E)$ such that $$\{W \in \Age_n(E) : V \subseteq W\} \in \mathcal{U} \text{ for each $V \in \Age_n(E)$}.$$ Define a mapping $c : \Emb(X, E) \rightarrow [0,1]$ by setting $c(\gamma) = \lim_\mathcal{U} c_Z(\gamma)$. Note that the ultralimit exists (by boundedness) and is well-defined since $\{W \in \Age_n(E) : \gamma(X) \subseteq W\} \in \mathcal{U}$. Furthermore, $c$ is an $\lambda_X$-continuous colouring. We claim that $c$ is a bad colouring of $\Emb(X,E)$. To this end, take any $\rho \in \Emb(Y,E)$ and note that $\{W \in \Age_n(E) : \rho(Y) \subseteq W\} \in \mathcal{U}$. Furthermore, for any such $W$ we have $\rho \in \Emb(Y,W)$ and so by our initial hypothesis we know $$| c_W(\rho \circ \eta_i) - c_W(\rho \circ \eta_j)| > \varepsilon \text{ for some $i, j \in \{1, \dots, k\}$.}$$ Since $\mathcal{U}$ is an ultrafilter, it follows that there are $i, j \in \{1,\dots,k\}$ such that $$\{W \in \Age_n(E) : | c_W(\rho \circ \eta_i) - c_W(\rho \circ \eta_j)| > \varepsilon\} \in \mathcal{U}.$$ It then follows that $|c(\rho \circ \eta_i) - c(\rho \circ \eta_j)| > \varepsilon$. Since $\rho$ was arbitrary, we see that $c$ is a bad colouring of $\Emb(X,E)$.
\end{proof}
The following is the KPT correspondence for multi-seminormed spaces (cf. \cite{KPT, MT}).
\begin{thm}[Kechris-Pestov-Todor\v{c}evi\'c correspondence]\label{KPT}
Suppose $E$ is an infinite-dimensional multi-seminormed space which is approximately ultrahomogeneous. The following are equivalent:
\begin{enumerate}[(i)]
\item $\Age_n(E)$ has the ARP for each $n \in \mathbb{N}$ such that $1\leq n \leq \lambda_E$.
\item $\Iso(E)$ is extremely amenable when endowed with the topology of pointwise convergence.
\end{enumerate}
\end{thm}
\begin{proof}
$(i) \rightarrow (ii)$: Fix an $\Iso(E)$-flow $\Iso(E) \curvearrowright K$ for $K$ compact, $\varepsilon > 0, p \in K$, an entourage $U$ and a finite $F \subseteq \Iso(E)$. By one of the well-known characterizations of extreme amenability (see \cite{Pestov} or \cite[Claim 5.11.2]{FLMT}) it is enough to find $g \in \Iso(E)$ such that $F \cdot (g \cdot p)$ is $U$-small, i.e. such that $$(f_0 \cdot (g \cdot p), f_1 \cdot (g \cdot p)) \in U \text{ for each $f_0, f_1 \in F$}.$$ Before proceeding, we will define a directed family of pseudometrics which generate the left uniformity of $\Iso(E)$: For each $n \in \mathbb{N}$ such that $n \leq \lambda_E$ and each finite-dimensional $X \subseteq E$, define a pseudometric $d_X^n$ on $\Iso(E)$ by $$d_X^n(g,h) = \max_{m< n} \|g\restriction X -h\restriction X\|_m.$$ We can assume without loss of generality that all entourages are symmetric. Fix an entourage $V$ such that $V \circ V \circ V \circ V \subseteq U$. Since the mapping $\Iso(E) \rightarrow K : g \rightarrow g^{-1} \cdot p$ is left uniformly continuous (see, e.g., \cite[Lemma 2.1.5]{Pestov}), there are $n$, $X \subseteq E$ and $\delta > 0$ such that $$d_X^n(g,h) \leq \delta \text { implies } (g^{-1} \cdot p, h^{-1} \cdot p) \in V.$$
Equip $X$ with the first $n$ seminorms induced from $E$, so that $X \in \Age_n(E)$. Let $Y$ be any member of $\Age_n(E)$ containing $\bigcup \{g(X) : g^{-1} \in F \cup \{\id\} \}$. Fix a finite set $\{x_i\}_{i<r} \subseteq K$ such that $K \subseteq \bigcup_{i<r} V[x_i]$, where $V[x] = \{y \in K : (x,y) \in V\}$. Apply the approximate Ramsey property of $\Age(E)$ to the parameters $n, X,Y, \delta/3$ and $r$ to obtain $Z \in \Age_n(E)$ such that $$Z \rightarrow_{\delta/3, n} (Y)^X_r.$$ Define a colouring $c : \Emb(X,Z) \rightarrow r$ by first choosing, for each $\gamma \in \Emb(X,Z)$, some $g_\gamma \in \Iso(E)$ such that $\max_{m < n} \|g_\gamma \restriction X - \gamma\|_m \leq \delta/3$; such a choice is possible by the approximate ultrahomogeneity of $E$. Then define $c(\gamma) = i$ if $i < r$ is the least index such that $g_\gamma^{-1} \cdot p \in V[x_i]$. By definition of $Z$ there are $\rho \in \Emb(Y,Z)$ and $i < r$ such that $$\rho \circ \Emb(X,Y) \subseteq (c^{-1}\{i\})_{\delta/3, n}.$$ In particular, this implies that for each $\eta \in \Emb(X,Y)$ there is $h_\eta \in \Iso(E)$ such that $\max_{m < n} \|\rho \circ \eta - h_\eta\restriction X\|_m \leq 2\delta/3$ and $h_\eta^{-1} \cdot p \in V[x_i]$. Choose $g \in \Iso(E)$ such that $\max_{m < n} \|g \restriction Y - \rho\|_m \leq \delta/3$. Now, given $f_0, f_1 \in F$, let $\eta_j := f_j^{-1} \restriction X$ for $j = 0, 1$ and note $\eta_j \in \Emb(X,Y)$. Then for each $j = 0, 1$,
\begin{equation*}
\begin{split}
d_X^n(g\circ f_j^{-1}, h_{\eta_j}) & = \max_{m < n} \|g \circ \eta_j - h_{\eta_j}\restriction X\|_m \\
& \leq \max_{m < n} \|g \circ \eta_j - \rho \circ \eta_j\|_m + \max_{m < n} \|\rho \circ \eta_j - h_{\eta_j} \restriction X\|_m \\
& \leq \delta/3 + 2\delta/3 = \delta.
\end{split}
\end{equation*}
By choice of $\delta$, this implies $(f_j \circ g^{-1} \cdot p, h_{\eta_j}^{-1} \cdot p) \in V$ for each $j$. Since $(x_i, h_{\eta_j}^{-1} \cdot p) \in V$ for each $j$, our choice of $V$ implies $(f_0 \circ g^{-1} \cdot p, f_1 \circ g^{-1} \cdot p) \in U$. Thus $F \cdot (g^{-1} \cdot p)$ is $U$-small.
\medskip \noindent $(ii) \rightarrow (i)$: To prove the ARP, we use the previous two lemmata together with the following characterization of extreme amenability in terms of the family of pseudometrics defined above. (See \cite{Pestov} or \cite[Proposition 3.9]{MT}.)
\begin{enumerate}
\item[$(\ast)$] $\Iso(E)$ is extremely amenable if, and only if, for every finite $F \subseteq \Iso(E), \varepsilon > 0, X \subseteq E, n \in [1,\lambda_E] \cap \mathbb{N}$ and $1$-Lipschitz map $f : (\Iso(E), d_X^n) \rightarrow [0,1]$, there is $g \in \Iso(E)$ such that $\osc(f \restriction g \circ F) \leq \varepsilon$.
\end{enumerate}
Fix $X, Y \in \Age_n(E)$ together with $\varepsilon > 0$ and an $n$-continuous colouring $c$ of $\Emb(X,E)$. Let $H = \{\eta_1, \dots, \eta_k\}$ be a finite $\varepsilon$-dense subset of $\Emb(X,Y)$ where the latter set is equipped with the pseudometric induced by $\max_{m < n} \|\cdot\|_m$. Apply the approximate ultrahomogeneity of $E$ to find $F = \{g_1, \dots, g_k\} \subseteq \Iso(E)$ such that $\|g_i \restriction X - \eta_i\|_m \leq \varepsilon$ for all $m < n$ and all $i \leq k$. Let $\widetilde{c} : (\Iso(E), d_n^X) \rightarrow [0,1]$ be the 1-Lipschitz mapping defined by $\widetilde{c}(g) = c(g\restriction X)$ and use $(\ast)$ to find $g \in \Iso(E)$ such that $\osc(\widetilde{c} \restriction g \cdot F) \leq \varepsilon$. Then for any $i, j \leq k$, by the triangle inequality the term $|c(g \circ \eta_i) - c(g \circ \eta_j)|$ is bounded above by $$|c(g \circ \eta_i) - c(g \circ g_i\restriction X)| + |c(g \circ g_i \restriction X) - c(g \circ g_j \restriction X)| + |c(g \circ g_j \restriction X) - c(g \circ \eta_j)|.$$ Since $c$ is an $n$-continuous colouring, the first term is bounded above by $$\max_{m < n} \|g \circ \eta_i - g \circ g_i\restriction X\|_m = \max_{m < n} \|\eta_i - g_i\restriction X\|_m \leq \varepsilon$$ by our choice of $g_i$. Similarly, the third term is bounded above by $\varepsilon$. To bound the second term, note that $$|c(g \circ g_i \restriction X) - c(g \circ g_j \restriction X)| = |\widetilde{c}(g \circ g_i) - \widetilde{c}(g \circ g_j)| \leq \varepsilon$$ by our choice of $g$. Thus, if we let $\gamma = g \restriction Y$, we see that the oscillation of $c$ on $\gamma \circ H$ is bounded by $3\varepsilon$. Then, using the definition of the $\eta_i$, it follows that $\osc(\gamma \circ \Emb(X,Y)) \leq 5\varepsilon$.
\end{proof}
Our next goal is to show that various classes of finite-dimensional multi-seminormed spaces have the ARP, which will allow us to apply the KPT correspondence in certain instances to obtain examples of extremely amenable multi-isometry groups. As in the case of the amalgamation property, our main source of examples of such classes will come from classes of finite-dimensional normed spaces which are known to have the ARP in the context of normed spaces. The key lemma is the following:
\begin{lem}\label{ARP1}
Suppose $\mathcal{K}_1, \dots, \mathcal{K}_n$ are classes of finite-dimensional normed spaces with the ARP. Let $X_1,\dots, X_n, Y_1,\dots, Y_n$ be finite-dimensional seminormed spaces such that $(X_i)_{\|\cdot\|}$ and $(Y_i)_{\|\cdot\|}$ belong to $\mathcal{K}_i$ for each $i \leq n$. Let $\varepsilon>0$ and $r < \omega$. Then there are finite-dimensional seminormed spaces $Z_1,\dots, Z_n$ such $(Z_i)_{\|\cdot\|} \in \mathcal{K}_i$ for each $i$ and, for every colouring $c: \prod_{j=1}^n \Emb(X_j, Z_j)\to r$, there are $\rho_j\in \Emb(Y_j,Z_j)$, $j=1,\dots,n$, such that $$\prod_{j=1}^n \rho_j \circ \Emb(X_j,Y_j) \text{ is $\varepsilon$-monochromatic}.$$
\end{lem}
The proof will involve a standard strategy for obtaining product Ramsey properties. First we will need:
\begin{lem}\label{basecase}
Suppose $\mathcal{K}$ is a class of finite-dimensional normed spaces with the ARP. For any finite-dimensional seminormed spaces $X$ and $Y$ such that $X_{\|\cdot\|}$ and $Y_{\|\cdot\|}$ belong to $\mathcal{K}$, any $\varepsilon>0$ and $r\in \mathbb{N}$, there is a finite-dimensional seminormed space $Z$ such that $Z_{\|\cdot\|} \in \mathcal{K}$ and every colouring $c: \Emb(X, Z) \rightarrow r$ $\varepsilon$-stabilizes on a set of the form $\rho \circ \Emb(X,Y)$ for $\rho \in \Emb(Y,Z)$.
\end{lem}
\begin{proof}
Let $X, Y$ be finite-dimensional seminormed spaces, $\varepsilon > 0$ and $r \in \mathbb{N}$. Let $\widetilde{X} = X / \ker \| \cdot \|_X$ and $\widetilde{Y} = Y / \ker \|\cdot \|_Y$ be equipped with the norms $\|[x]\| = \|x\|_X$ and $\|[y]\| = \|y\|_Y$ respectively. By the ARP of $\mathcal{K}$, there is a finite-dimensional normed space $Z \in \mathcal{K}$ such that $$Z \rightarrow_\varepsilon (\widetilde{Y})^{\widetilde{X}}_r.$$ Consider the product $Z \times Z$ equipped with the seminorm $\|(z_1,z_2)\| := \|z_1\|_Z$. We claim that this space witnesses the ARP for the parameters $X, Y, \varepsilon, r$. Note that $Z \times Z \in \mathcal{K}_{\|\cdot\|}$ since the associated quotient is isomorphic to $Z$. Now, fix a colouring $c : \Emb(X,Z) \rightarrow r$ and define $\widetilde{c} : \Emb(\widetilde{X},Z) \rightarrow r$ by $\widetilde{c}(\gamma) = c(\gamma \circ \pi_X)$ where $\pi_X : X \rightarrow \widetilde{X}$ is the canonical surjection. By definition of $Z$, there is $\rho \in \Emb(\widetilde{Y}, Z)$ such that $$\rho \circ \Emb(\widetilde{X}, \widetilde{Y}) \subseteq (\widetilde{c}^{-1}\{i\})_\varepsilon \, \text{ for some $i < r$}.$$ Let $\bar{\rho} : \widetilde{Y} \rightarrow Z \times Z$ be defined by setting $\bar{\rho}(y) = (\rho(y), 0)$. We will show $$(\bar{\rho} \circ \pi_Y) \circ \Emb(X,Y) \subseteq (c^{-1}\{i\})_\varepsilon.$$ To this end, fix $\eta \in \Emb(X,Y)$ and define a mapping $\varphi : \widetilde{X} \rightarrow \widetilde{Y}$ by $\varphi([x]) = \pi_Y(\eta(x))$; it is easy to check that $\varphi$ is a well-defined multi-isometric embedding which satisfies $\varphi \circ \pi_X = \pi_Y \circ \eta$. Then by definition of $\rho$ there is $\theta \in \widetilde{c}^{-1}\{i\}$ such that $\| \rho \circ \varphi - \theta \| \leq \varepsilon.$ Let $\bar{\theta} : \widetilde{X} \rightarrow Z \times Z$ be given by $\bar{\theta}(x) = (\theta(x), 0)$. The situation is summarized by the following diagram:
\begin{center}
\begin{tikzcd}[row sep=large, column sep=large]
X \arrow[r, "\pi_X"] \arrow[d, "\eta"'] & \widetilde{X} \arrow[d, "\varphi"] \arrow[rd, "\bar{\theta}"{name=M}] & \\
Y \arrow[ur, phantom, "\scalebox{1.4}{$\circlearrowleft$}" description]\arrow[r, "\pi_Y"'] & \widetilde{Y} \arrow[r, "\bar{\rho}"'] \arrow[ur, phantom, "\scalebox{0.8}{$\cca{\varepsilon}$}" description, to=M] & Z \times Z
\end{tikzcd}
\end{center}
Then $c(\theta \circ \pi_X) = i$ and, since $$(\rho \circ \pi_Y) \circ \eta = \rho \circ (\pi_Y \circ \eta) = \rho \circ (\varphi \circ \pi_X),$$ we have $$\|(\rho \circ \pi_Y) \circ \eta - \theta \circ \pi_X \| = \|(\rho \circ \varphi) \circ \pi_X - \theta \circ \pi_X\| \leq \|\rho \circ \varphi - \theta \| \leq \varepsilon.$$ Thus $\rho \circ \pi_Y$ is the desired embedding.
\end{proof}
\begin{proof}[Proof of Lemma \ref{ARP1}]
The proof is by induction on $n$. The case $n=1$ follows from the previous lemma, so fix all parameters and suppose that we can find seminormed spaces $Z_1,\dots,Z_n$ such that every $r$-colouring of $\prod_{j=1}^n \Emb(X_j, Z_j)$ has a $\varepsilon/2$-monochromatic set of the form $\prod_{j=1}^n \rho_j\circ \Emb(X_j, Y_j)$. Let $D$ be a finite $\varepsilon/2$-dense subset of $\prod_{j=1}^n \Emb(X_j,Z_j)$, where the seminorm on the product is the maximum of the given seminorms. Apply Lemma \ref{basecase} to $X_{n+1}$ and $Y_{n+1}$ to find a seminormed space $Z$ that works for the error $\varepsilon/2$, and the number of colours being the cardinality of $^{D}{r}$. We claim that $Z_1,\dots,Z_n,Z$ works. For suppose that $c:\prod_{j=1}^{n+1} \Emb(X_j,Z_j)\to r$. We have the induced colouring $\widehat c: \Emb(X_{n+1},Z)\to \, ^{D}r$, $$\widehat c(\xi)(\eta_1,\dots,\eta_n):=c(\eta_1,\dots,\eta_n, \xi)\in r \text{ for every $(\eta_1,\dots,\eta_n)\in D$}.$$ By the choice of $Z$ there is some $\rho\in \Emb(Y_{n+1},Z)$ such that $\rho \circ \Emb(X_{n+1},Y_{n+1})$ is $\varepsilon/2$-monochromatic for $\widehat c$ with colour $\theta\in {}^{D}r$. The mapping $\theta: D\to r$ defines an $r$-colouring $\widehat{\theta}:\prod_{j=1}^n \Emb(X_j, Z_j)\to r$ by $\varepsilon/2$-proximity. For each $j=1,\dots, n$, let $\rho_j\in \Emb(Y_j, Z_j)$ be such that $\prod_{j=1}^n \rho_j \circ \Emb(X_j, Y_j)$ is $\varepsilon/2$-monochromatic for $\widehat\theta$ with colour $s\in r$. Then the set $$\left(\prod_{j=1}^n \rho_j \circ \Emb(X_j, Y_j) \right)\times (\rho \circ \Emb(X_{n+1},Y_{n+1})) \subseteq \prod_{j=1}^{n+1} \Emb(X_j,Z_j)$$ is $\varepsilon$-monochromatic for $c$ with colour $s$: Let $(\gamma_j)_{j=1}^{n+1}\in \prod_{j=1}^{n+1}\Emb(X_j, Y_j)$. There is some $(\mu_{j})_{j=1}^n \in \prod_{j=1}^n \Emb(X_j, Z_j)$ such that $\widehat \theta((\mu_j)_{j=1}^n)=s$ and $(\mu_j)_{j=1}^n$ is $\varepsilon/2$-close to $(\rho_j \circ \gamma_j)_{j=1}^n$. Choose $(\pi_j)_{j=1}^n\in D$ that is $\varepsilon/2$-close to $(\mu_j)_{j=1}^n$ such that $\theta((\pi_j)_{j=1}^n)=s$. Let $\mu\in \Emb(X,Z)$ be such that $\mu$ is $\varepsilon/2$-close to $\rho \circ \gamma$ and $\widehat c(\mu)=\theta$. This last equality means by definition that $$s=\theta((\mu_j)_{j=1}^n)= \widehat c(\mu)((\mu_j)_{j=1}^n)=c(\mu_1,\dots,\mu_n,\mu).$$ Finally, observe that $\mu$ is $\varepsilon/2$-close to $\rho$, and each $\mu_j$ is $\varepsilon$-close to $\rho_j \circ \gamma_j$.
\end{proof}
For the next proposition, recall the definition of the classes $\langle \bar{\mathcal{K}}\rangle_{=n}$ from Definition \ref{classes}.
\begin{prop}\label{ARP}
Let $\bar{\mathcal{K}} = \{\mathcal{K}_n\}_{n<\omega}$ be a collection of finite-dimensional normed spaces such that each $\mathcal{K}_n$ has the ARP. For every $n \in \mathbb{N}, X,Y\in \langle \bar{\mathcal{K}}\rangle_{=n}$, $r\in \mathbb{N}$, and every $\varepsilon>0$ there is $Z\in \langle \bar{\mathcal{K}}\rangle$ such that every $r$-colouring of $\Emb(X,Z)$ has an $\varepsilon$-monochromatic set of the form $\rho \circ \Emb(X,Y)$ for some $\rho\in \Emb(Y,Z)$. Furthermore, if $\mathcal{K}_n = \mathcal{K}$ for each $n$ and $\mathcal{K}$ is closed under $\ell_\infty$-sums, then $Z$ can be chosen to be graded when $X$ and $Y$ are graded.
\end{prop}
\begin{proof}
Fix all parameters and apply the previous lemma to $(X,\nrm{\cdot}_j)_{j=1}^n$, $(Y,\nrm{\cdot}_j)_{j=1}^n$, $\varepsilon$ and $r$ to find the corresponding $Z_1,\dots, Z_n$. Let $Z:= \prod_{j=1}^n Z_j$, and for each $j=1,\dots,n$, let $$\nrm{(z_1,\dots,z_n)}_j:= \nrm{z_j}_{Z_j}.$$ Note that $Z \in \langle \bar{\mathcal{K}}\rangle_{=n}$ by construction. We claim that $Z$ witnesses the ARP for the given parameters. For suppose that $c: \Emb(X,Z)\to r$ is a colouring. Given a sequence $\vec{\gamma} = (\gamma_j)_j \in \prod_{j=1}^n \Emb((X,\nrm{\cdot}_j), Z_j)$, define a mapping $F(\vec{\gamma}) : X \rightarrow Z$ by $$F(\vec{\gamma})(x) = (\gamma_1(x), \dots, \gamma_n(x)).$$ Observe that $F(\vec{\gamma}) \in \Emb(X,Z)$ since each $\gamma_j$ is a multi-isometric embedding. Using this mapping, define an induced colouring $\widehat c: \prod_{j=1}^n \Emb((X,\nrm{\cdot}_j), Z_j)\to r$, by $\widehat{c}(\vec{\gamma}) = c(F(\vec{\gamma}))$ where $\vec{\gamma} = (\gamma_j)_j$. By definition of each $Z_j$, there are $\rho_j \in \Emb((Y,\|\cdot\|_j), Z_j)$ such that $$\prod_{j=1}^n \rho_j \circ \Emb((X,\|\cdot\|_j),(Y_j,\|\cdot\|_j)) \text{ is $\varepsilon$-monochromatic}.$$ Define $\rho \in \Emb(Y,Z)$ by $\rho(y) = (\rho_1(y), \dots, \rho_n(y))$. Note that $\rho \circ \eta = F((\rho_j \circ \eta)_j)$ and so $c(\rho \circ \eta) = \widehat{c}((\rho_j \circ \eta)_j)$. In particular, it follows that $\rho \circ \Emb(X,Y)$ is $\varepsilon$-monochromatic.
In the case where $\mathcal{K}_n = \mathcal{K}$ for all $n$ and $\mathcal{K}$ is closed under $\ell_\infty$-sums, we work with the same underlying space $Z$ but we instead equip it with the sequence of seminorms given by $$\nrm{(z_1,\dots,z_n)}_j:= \max_{i \leq j} \nrm{z_i}_{Z_i}.$$ Then $Z_{\|\cdot\|_j}$ is multi-isometric to the $\ell_\infty$-sum of $Z_i, i \leq j$, and so $Z \in \langle \bar{\mathcal{K}}\rangle_{=n}$ by our additional assumption on $\mathcal{K}$. The rest of the proof is identical to that of the general case.
\end{proof}
By appealing to the various known approximate Ramsey properties of classes of finite-dimensional normed spaces as considered in \cite{BLLM, FLMT}, we can apply Theorem \ref{KPT} and Proposition \ref{ARP} together to obtain the following result. The notation used below corresponds to that of Example \ref{examples}.
\begin{thm}
The following groups are extremely amenable when equipped with the topology of pointwise convergence:
\begin{enumerate}[(1)]
\item $\Iso(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \alpha})$ and $\Iso(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n <\alpha})$ for each $\alpha \leq \omega$. In particular, the multi-isometry group of the separable (graded) Fr\'echet space of almost universal disposition for the class $\mathcal M_\omega$ (resp. $\mathcal G_\omega$) is extremely amenable.
\item $\Iso(\Flim(\mathcal M_{<\omega}^{\mathcal H}))$.
\item $\Iso(\Flim(\mathcal M_{<\omega}^{(p_n)}))$ for any sequence $(p_n) \subseteq [1,\infty[$ with $p_n \not \in \{4,6,8,\dots\}$.
\end{enumerate}
\end{thm}
The extreme amenability of the groups $\Iso(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ and $\Iso(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n <\omega})$ should naturally be compared to the extreme amenability of $\Iso(\mathbb{G})$. The following important questions remain open:
\begin{que}
Are $\Iso(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ and $\Iso(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n <\omega})$ topologically distinguishable from $\Iso(\mathbb{G})$? Are $\Iso(\mathbb{G}^\omega, (\|\cdot\|_n)_{n < \omega})$ and $\Iso(\mathbb{G}^\omega, (\|\cdot\|'_n)_{n <\omega})$ universal Polish groups?
\end{que}
\section{Concluding remarks}
Throughout this paper we have adopted a very particular viewpoint in order to develop Fra\"iss\'e theory for Fr\'echet spaces. Specifically, we have been working with the category of multi-seminormed spaces with morphisms given by seminorm-preserving linear mappings. A natural question is whether or not a similar theory can be developed for more relaxed categories which capture the notion of a Fr\'echet space. For instance, the following general problem remains open:
\begin{prob}
Develop a theory of \emph{metric} Fr\'echet spaces, i.e. pairs of the form $(X,d)$ where $X$ is a vector space and $d$ is a translation-invariant metric inducing a Fr\'echet topology on $X$.
\end{prob}
In the above setting, one may be able to use more general machinery in order to develop a Fra\"iss\'e theory for such spaces, e.g. as in \cite{BY,Kubis}. It is unclear if the use of such machinery is possible in the setting of multi-seminormed spaces, since in general one needs to work with an arbitrarily large (finite) number of seminorms.
One specific corollary of our construction is that the spaces $\Flim(\mathcal M_{<\omega})$ and $\Flim(\mathcal G_{<\omega})$ are -- in addition to their defining properties -- Fra\"iss\'e for the classes $\mathcal M_\omega$ and $\mathcal G_\omega$, respectively, by virtue of being multi-isometric to the spaces constructed in \cite{BKK}. Since the construction of the Fra\"iss\'e limit makes use of the fact that the elements of a given Fra\"iss\'e class are finitely-seminormed, it is not clear that the aforementioned property must be true for general Fra\"iss\'e Fr\'echet spaces. More precisely, given a class $\mathcal{K} \subseteq \mathcal M_{<\omega}$, one can define a class $\mathcal{K}_\omega \subseteq \mathcal M_\omega$ by declaring that $(X,(\|\cdot\|_n)_{n<\omega}) \in \mathcal{K}_\omega$ exactly when
$$(X,(\|\cdot\|_n)_{n<m}) \in \mathcal{K} \text{ for all $m<\omega$}.$$ This motivates the following:
\begin{prob}\label{prob2}
If $E$ is a $\mathcal{K}$-Fra\"iss\'e Fr\'echet space, is $E$ necessarily $\mathcal{K}_\omega$-Fra\"iss\'e?
\end{prob}
The proof that $\mathbb{G}^\omega$ is $\mathcal G_\omega$-Fra\"iss\'e (with an appropriate sequence of seminorms) from \cite{BKK} makes use of the existence of a universal operator $\pi : \mathbb{G} \rightarrow \mathbb{G}$ constructed in \cite{GK}. This operator can essentially be seen as a \emph{Fra\"iss\'e} operator in a precise sense (see, e.g., \cite{Lupini2018}). The operator constructed in \cite{GK} is the Gurarij analogue of Rota's universal operator on $\ell_2$, as in \cite{Caradus, Rota}. Thus, in order to shed light on a possible affirmative answer to Problem \ref{prob2}, one may need to construct similar operators for an arbitrary Fra\"iss\'e Banach space in place of $\mathbb{G}$.
Finally, we mention that throughout the paper we have not assumed any condition on the continuity of the relevant mappings between multi-seminormed spaces. Since seminorm-preserving mappings (as defined above) need not be continuous, the following general problem presents itself:
\begin{prob}\label{prob3}
For various classes $\mathcal{K}$ of finite-dimensional multi-seminormed spaces, study the notions of $\mathcal{K}$-universality and $\mathcal{K}$-Fra\"iss\'e where the embeddings are in addition assumed to be continuous. In particular, is there a version of the Fra\"iss\'e theory for multi-seminormed spaces developed above in which the associated embeddings are also required to be continuous?
\end{prob}
\nocite{BLLM_2}
\bibliographystyle{abbrv}
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"****Space in all the right places***** This property is an end unit and features an enormous great-room with tall 10 ft ceilings and TONS OF NATURAL LIGHT/WINDOWS, plus french door onto a private patio that is perfect for entertaining or just relaxing in your outdoor living space. The entire main level has decorative Tile and hardwood floors for minimal maintenance. The spacious kitchen has stainless appliances and granite counter tops with alder cabinetry, also a large island with a sink. The master bedroom is spacious with an over sized garden tub and separate shower, elaborate granite and a double sink vanity. A true walk in closet with built in shelving/organizers. Come see this property today, it has a total of 5 bedrooms, 4 bathrooms, 2 family-rooms and a 2 car garage!!!!" | {
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extern int weak_symbol __attribute__ ((weak));
int weak_symbol = 1;
// Define a strong symbol with a weak alias.
int strong_aliased = 2;
extern int weak_aliased __attribute__ ((weak, alias ("strong_aliased")));
// And another one.
int strong_aliased_2 = 5;
extern int weak_aliased_2 __attribute__ ((weak, alias ("strong_aliased_2")));
// And a third.
int strong_aliased_3 = 7;
extern int weak_aliased_3 __attribute__ ((weak, alias ("strong_aliased_3")));
// And a fourth.
int strong_aliased_4 = 8;
extern int weak_aliased_4 __attribute__ ((weak, alias ("strong_aliased_4")));
// We want a symbol whose name is the same length as "strong_symbol",
// so that weak_symbol here lines up with weak_symbol in
// weak_alias_test_2.so.
int Strong_Symbol = 101;
| {
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{"url":"https:\/\/math.stackexchange.com\/questions\/2553741\/finding-a-straight-line-equation","text":"# Finding a straight line equation\n\nI am given a point $A =(-1, 0, 1)$. A straight line is defined as the following system: $\\{ y + 3x = 6$ and $z-2 = 0\\}$ . They are one system. This defines a line in 3D space but I don't understand it visually. Let this line be $T$.\n\nLet line $S$ be a line drawn from point $A$ to that line $T$ defined by the given system before so that the angle between them is 90 degrees. Line $S \\perp T$. Find the length of $S$ and straight line equation of $T$.\n\nThe length will probably be $|\\vec AT|$ (line $S$ basicly) calculated by $\\sqrt {x^2 + y ^2 + z ^2}$ or atleast so I think. Maybe this clears up a bit.\n\nHelp appreciated. Thanks.\n\n\u2022 Nothing is clear in your question :-) You don't define whatis point P. As to why there is two lines, you misinterpret the fact that $z=2$ is the equation of a plan, not a line. The intersection of two plans is a line, so the set of 2 equations is defining a line in a 3D space. The rest of your question, to be honest, is not comprehensible to me. Could state very precisely what is the problem at hand, and what are your issues with it, and also what you have tried and where your failed? \u2013\u00a0Martigan Dec 6 '17 at 12:16\n\u2022 $z=2$ is not a straight line, it is a plane. \u2013\u00a05xum Dec 6 '17 at 12:19\n\u2022 @Martigan I cleared up the question a bit, maybe you could take a look now. Thanks. \u2013\u00a0student126 Dec 6 '17 at 13:33\n\nHere's an outline. $T$ is the intersection of two planes. A vector $v$ parallel to $T$ lies in both planes. So $v$ is perpendicular to the normal vectors of both planes. Let $n$ and $m$ be the normal vectors to each plane. Then you can take $v$ to be their cross product. Then find any point on the line. $P=(0,6,2)$ will work. Parametric equations of the line are given by $P+vt.$\n\nThe length of $S$ (a segment, really) is the distance from $A$ to $T$. Pick any point on the line. The same $P$ will do. Let $w$ be the vector from $P$ to $A$. This forms the hypotenuse of a right triangle. If you project $w$ onto $v$, the length of this projection is one leg of the right triangle. The distance you want is the other leg.\n\nFirst you want to understand what the line is (visually). It's nothing but the line $y+3x=6$ in $2d$ in the plane $z=2$. In standard $2d$ coordinate system we assume $z=0$. In this question we simply shifted the line $2$ units in the direction of $z$.\n\nTo find the distance $AT$, first we write the equation of the line $T$:\n\n$$\\frac{x-0}{1}=\\frac{(y-6)}{-3}=\\frac{(z-2)}{0}=r$$\n\nHere $r$ is an arbitrary constant.\n\nSo, a general point on this line $T$ is $(r, 6-3r, 2)$.\n\nLet's call this point $P$.\n\nNow the line joining this point and $A$ i.e., line $AP$ (same as what you call line $S$ in the question) should be perpendicular to line $T$.\n\nDirection ratios of $AP$ are $(r+1,6-3r,1)$\n\nDirection ratios of line $T$ are $(1, -3, 0)$\n\nCondition of perpendicularity:\n\n$(r+1)1 +(6-3r)(-3) + 1(0) =0$\n\nThus, $r = 1.7$\n\nTherefore, point $P$ is $(1.7, 0.9, 2)$\n\nNow just apply distance formula between $A$ and $P$\n\nTherefore required distance $AP = sqrt(9.1)$","date":"2019-07-24 06:41:56","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8656793832778931, \"perplexity\": 179.0063422716168}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-30\/segments\/1563195531106.93\/warc\/CC-MAIN-20190724061728-20190724083728-00183.warc.gz\"}"} | null | null |
Home TV Shows Warrior Season 3: Release Date, Cast, Plot And All Latest News!!
Warrior Season 3: Release Date, Cast, Plot And All Latest News!!
Modified date: Thursday, 21 January 2021, 11:43 EST
Many times, due to a lack of potential storylines and some unavoidable issues, additional series sets are naturally skipped, but there are also some cases in television history where the series got an extension even after being previously canceled. . The Warrior series completed its second chapter recently and the audience is discussing the possibilities of Warrior Season 3.
Although the creators don't have to give any confirmation signals for the show, Warrior Season 3 is one of the most sought after shows of the coming months. Depending on the interesting track of martial arts, including karate and other forms, Warrior's both episodes garnered much love from viewers and also garnered nominations on award shows such as Critic's Choice and Emmy Awards. So all the audience's attention is now on Warrior Season 3.
Release Date: Warrior Season 3
The third set of Warrior may or may not be on the screens, since even after 4 months of the completion of its second installment, the producers of Warrior do not respond about the next chapter. Some say that Warrior Season 3 will not be released to screens due to the current strange situation of the deadly disease of the virus. But if all things went well in the future, then we can definitely all watch Warrior season 3. Still, it could take a general one or a maximum of 2 years to get on television.
Cast: Warrior Season 3
As Warrior Season 3 isn't so sure now, its casting details aren't that clear either. There is a finger crossing situation among fans as they literally want to see the next season of the Warrior series. If Warrior Season 3 hits the screens, then it will have the same stars as its predecessor chapters. Also, if the creators make some changes, then we might witness some new faces being added to the Warrior Season 3 cast. Possible names are Andrew Koji, Jason Tobin, Olivia Cheng, Kieran Bew, Hoon Lee, Dianne Doan, and Joe Taslim.
Plot: Warrior Season 3
In a confusing wind of Warrior season 3's existence, people are getting a lot out about its plot. But we all have to contain our excitement and curiosity as Warrior Season 3 has yet to be confirmed by the producers. The audience is usually very excited when it comes to the news of their favorite shows and the same is true with the Warriors.
So for some relaxing news for fans, we have some unclear details of its plot. The plot may show Ah Sahm's new rage with Jun. To date, the audience has many questions about the Warrior Season 3 plot that has yet to be questioned. But they have no choice but to be patient until the news is out.
Warrior Season 3
Warrior Season 3 Cast
Warrior Season 3 Plot
Warrior Season 3 Release Date | {
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Javan Vidal
By BigAl,
January 16, 2009 in Aberdeen Football Club
http://www.afc.premiumtv.co.uk/page/News/c...1522336,00.html
Welcome on board the good ship AFC loon
Goldie03 7 posts
manc_don 8 posts
TENEMENTFUNSTER 21 posts
Kowalski 7 posts
Wonder if Sone has had some influence in getting the lad to come north ?
Sounds shite.
Boooooooooooooo!
He's a cut above I heard, JC combed through loads of players to find him;
Hopefully he gets a start (sa)soon.
Yep didn't take long...wonder if he'll be hair today, gone tomorrow
Any vids of this guys highlights so far?
Hopefully he manages to Gore some of McGeady's limbs on Sunday.
I had heard a snippet about this but wasn't sure if it was a Gore.
Vidal became the second consecutive Manchester City youngster to sign on loan for Grimsby Town in as many seasons, following the signing of Shaleum Logan at the same time in the previous season.
He signed on an initial one month loan at Grimsby on the 1st September 2008, he was meant to sign 3 days previously but returned to Manchester following a Friday training session in Grimsby.Vidal made his professional debut in the clubs 2-2 draw with Chesterfield in the Football League Trophy 1st Round at Saltergate. He was used as a second half substitute, and the club went on to win 4-1 in the penalty shoot out. He made his league debut on the 6th September against Gillingham as a second half sub for Jamie Clarke. Vidal returned to the Eastlands after playing a total of five games for Grimsby in all competitions.
"Jav reminds me of Earl Barrett in many ways - he's a rubber man who bounces all over the place with great athleticism.. He's played for England under 19s this year and, like the rest of the lads, he will benefit immensely from playing alongside senior players" said Jim Cassell in the Official Manchester City annual 2008
It's fantastic that we are adding to rather than trimming the squad.
glasgowdon
I take it Hodgkiss is away for certain then.
Welcome to AFC!
Heleft a couple of days ago.
Welcome to Aberdeen, Javan.
I hope he's worth it.
Attacking, athletic full back. Sounds decent.
Reckon we're getting him from City at a good time, more than likely to be surplus to requirements even if he is any good.
Hopefully he's not a lightweight and doesn't get brushed aside easily.
It was obvious you were going to join this shower of puns.
Toupee!
minijc
Hope he doesn't take long to gel with the rest of the team.
His speciality is scissor kicks and curlers.
We're also rumoured to be looking at Gianluigi Buffon and Alan Combe.
Biggalloot
Shame about Hodgekiss cause he did look semi decent (in comparison to Foster) but hopefully this new boy will be good. Fingers x'd he makes as big an impact as 'Shaunney' Aluko but doubt that cause he seems a bit thick -
"Scottish football is massive, to be fair, a very good standard.
"Celtic and Rangers: you've just got to look at their European success to know how big it is."
Hmm 1 cup each?
Seems as though he was dying to sign for us.
ntbear
Who we selling in his place then.....
Was only ever on the fringes of the Man City team.
Not a lot on the net but i see he is on facebook
http://en-gb.facebook.com/people/Javan-Noel-Vidal/681735388
If he's any use hopefully we can get him on a permanent contract
If this guy plays a good attacking style of full-back football then we could be looking shit hot on the flanks with him and Mulgrew pinging balls in from deep. Backed up with Sone and Jamie Smith we would be unstoppable.
Is that better ST ? | {
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Q: Does this ROC curve make sense? This code returns and plots the true positive rate, false positive rate, true positive count, false positive count based on predicted and true values :
def get_all_stats(y_true , y_pred) :
def perf_measure(y_true, y_pred):
TP = 0
FP = 0
TN = 0
FN = 0
for i in range(len(y_true)):
if y_true[i] == 1 and y_pred[i] == 1:
TP += 1
if y_pred[i]==1 and y_true[i]!=y_pred[i]:
FP += 1
if y_true[i]== 0 and y_pred[i]==0:
TN += 1
if y_pred[i]==0 and y_true[i] != y_pred[i]:
FN += 1
if(FP == 0) :
FPR = 0;
else :
FPR = FP / (FP + TN)
if(TP == 0) :
TPR = 0
else :
TPR = TP / (TP + FN)
return(TN , FPR, FN , TPR , TP , FP)
tn, fpr, fn, tpr, tp , fp = perf_measure(y_true, y_pred)
return tpr , fpr , tp , fp
tpr1 , fpr1 , tp1 , fp1 = get_all_stats(y_true=[1,1,1] , y_pred=[1,0,0])
tpr2 , fpr2 , tp2 , fp2 = get_all_stats(y_true=[1,0,1] , y_pred=[0,1,0])
tpr3 , fpr3 , tp3 , fp3 = get_all_stats(y_true=[0,0,0] , y_pred=[1,0,0])
plt.figure(figsize=(12,6))
plt.tick_params(labelsize=12)
print(tpr1 , fpr1 , tp1 , fp1)
print(tpr2 , fpr2 , tp2 , fp2)
print(tpr3 , fpr3 , tp3 , fp3)
plt.plot([fpr1,fpr2,fpr3], [tpr1 , tpr2, tpr3], color='blue', label='')
plt.ylabel("TPR",fontsize=16)
plt.xlabel("FPR",fontsize=16)
plt.legend()
The resultant ROC plot produced is :
In order to mimic three different false positive and true positive rates and different thresholds calculate these values by implementing the function get_all_stats three times with different
tpr1 , fpr1 , tp1 , fp1 = get_all_stats(y_true=[1,1,1] , y_pred=[1,0,0])
tpr2 , fpr2 , tp2 , fp2 = get_all_stats(y_true=[1,0,1] , y_pred=[0,1,0])
tpr3 , fpr3 , tp3 , fp3 = get_all_stats(y_true=[0,0,0] , y_pred=[1,0,0])
There are 9 instances to be classified to either 1 or 0 where the truth values are : [1,1,1,1,0,1,0,0,0]
At threshold1 the predicted values are [1,0,0] where the truth values at this threshold are [1,1,1].
At threshold2 the predicted values are [0,1,0] where the truth values at this threshold are [1,0,1].
At threshold3 the predicted values are [1,0,0] where the truth values at this threshold are [0,0,0].
As can see the generated plot of the classifier produced is different from 'typical' ROC curves :
As it first descends and then the false positive and true positive rates decrease causing the line to 'move back'. Have I implemented the ROC curve correctly ? Can the AUC be calculated for this curve ?
A: Okay, motivated to help because you have a lot of rep -> have helped a lot of others. Here we go.
This ROC curve does not make sense. The issue is that you are calculating the FPR/TPR on only subsets of your data at different thresholds. At each threshold, you should use all of the data to calculate FPR and TPR. Thus you seem to have 3 points in your plot, but you should only have had one point with the FPR/TPR for y_true = [1,1,1,1,0,1,0,0,0] and y_pred = [1,0,0,0,1,0,1,0,0]. To ensure that you have an actual ROC curve, though, you also can't just make up the y_pred values at the different thresholds - these need to come from actual predicted probabilities that are then thresholded appropriately. I modified your code a bit because I like to use numpy; here's how you could calculate an ROC curve.
# start with the true labels, as you did
y_true = np.array([1, 1, 1, 1, 0, 1, 0, 0, 0])
# and a predicted probability of each being a "1"
# I just used random numbers for these, but you would get them
# from your classifier
predictions = np.array([
0.07485627, 0.72546085, 0.60287482,
0.90537829, 0.75789236, 0.01852192,
0.85425979, 0.36881312, 0.63893516
])
# now define a set of thresholds (the more thresholds, the better
# the curve will look). There's a smarter way to do this in practice
# (you can sort the predicted probabilities and just have one threshold
# between each), but this is just to help with understanding
thresholds = np.linspace(0, 1, 11) # 0.1, 0.2, ..., 1.0
fprs = []
tprs = []
# we can precompute which inputs are actually 1s/0s and how many of each
true_1_idx = np.where(y_true == 1)[0]
true_0_idx = np.where(y_true == 0)[0]
n_true_1 = len(true_1_idx)
n_true_0 = len(true_0_idx)
for threshold in thresholds:
# now, for each threshold, we use that on the underlying probabilities
# to get the actual predicted classes
pred_classes = predictions >= threshold
# and compute FPR/TPR from those
tprs.append((pred_classes[true_1_idx] == 1).sum() / n_true_1)
fprs.append((pred_classes[true_0_idx] == 1).sum() / n_true_0)
plt.figure(figsize=(12,6))
plt.tick_params(labelsize=12)
plt.plot(fprs, tprs, color='blue')
plt.ylabel("TPR",fontsize=16)
plt.xlabel("FPR",fontsize=16)
Note that an ROC curve is always non-decreasing in TPR (y-axis) as you increase in FPR (x-axis); that is, it goes up as you move to the right. This is clear from the how the thresholding works. At a threshold of 0, all predictions are "1", so we have FPR = TPR = 1. Increasing the threshold gives fewer predictions of "1", so FPR and TPR can only stay the same or decrease.
Note that, even if we used the optimal thresholds, there are still jumps in the curve because we have a finite amount of data, so a finite number of different TPR/FPR pairs we can get with any threshold. If you have enough data, though, then this begins to look smooth. Here I've replaced a few lines in the above code to get a smoother plot:
n_points = 1000
y_true = np.random.randint(0, 2, size=n_points)
predictions = np.random.random(n_points)
thresholds = np.linspace(0, 1, 1000)
If it's not clear, an AUC of 0.5 is the worst possible, and you can see that's what we get with random "predictions". If your AUC is worse than 0.5, you can flip every prediction to be better than 0.5 (and something is probably wrong with your model/training).
If you actually want to plot an ROC curve in practice, not just write it yourself to learn a little more, use sklearn's roc_curve. They also have roc_auc_score to get the AUC for you.
| {
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TAD, 35 YEARS GROWING UP NEXT TO YOU
07 MAY TAD, 35 YEARS GROWING UP NEXT TO YOU
TAD group began its trajectory in May 1986 as a young, diverse, and dynamic company consisted of two partners who also came from the metal sector named Mr. Valentí Sererols and Mr. Juan Huertos, with the business opportunity focused on the Spanish vibratory market.
The company, initially located in a Barcelona's neighbourhood called Poblenou, it is specialised in the design and manufacturing of equipment and own technology systems that feed through vibration to position parts for all kind of industries. Its market at that time was mainly local and national.
Since the partners started with a low capital cost, the beginning of the company was not easy, but with lots of effort, engagement, and commitment, they managed to make a name by themselves in the metal sector.
Little by little TAD group grew, as did the productive needs of their clients, and they decided to expand their workforce and facilities and moved the company's current headquarter to L'Hospitalet de Llobregat in 1996.
It was in the 90s when the company decided to dive into the European exports, taking advantage of the relocation of some national clients.
Currently, approximately 50% of the production is destined for the national market and the rest for global export.
It should be noted that, since its foundation, TAD has managed to adapt to all socioeconomic changes of the moment such as the devaluation of the Spanish currency "la peseta" between the years 1992 and 1995, the globalization which began in 2000, the introduction of the Euro as a single currency in Europe in 2002, the financial crisis of 2008 and the global coronavirus pandemic in 2020.
With the implementation of its own philosophy, a very solid line of principles with a strong productive capacity, a non-stop professional trajectory, a continued improvement of the quality of products and an excellent after-sales service, have been key points of the success of the company.
TAD distinguishes itself in this sector by customizing its product and adapting it to the specific needs of each client. TAD's clients come from many and varied sectors as pharmaceuticals, food, electrical, metallurgical, automotive and any company that needs to automate production processes.
In these past few years, TAD has significant invested on its registered trademark TAD, S.L., advertising, marketing, and social networks, and continues to sell the same product, but always updated and according to the needs of the current market.
Today the TAD group has an operating area of 3.000 m², a team of 60 highly qualified and human professionals specialized in its product and, with permanent commercial delegations in France, Poland, and England for more than ten years.
Being also a forward-thinking company, TAD is currently working on projects that combine the classic product with the intelligent product. In the same way, TAD group explores emerging industrial sectors such as Latin America and North Africa. | {
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Minasolve is a French cosmetic ingredients supplier, with a special focus on multifunctional, nature-derived and sustainable products. Today the personal care industry is striving to find natural ingredients that function as well as their synthetic counterparts. Minasolve, has made great strides in this area by introducing Pentiol Green+, a corncob, sugar cane bagasse based version of Pentylene Glycol. The ingredient has been granted EcoCert and COSMOS certification in the EU and has received the Green Ingredient Silver Award at In-Cosmetics in 2014. This innovation allows customers to use a green variant of a well-known ingredient at competitive price, without any formulation challenges. | {
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Polar Bear Expedition in Churchill, Canada
Collected by Ashley Castle Pittman , AFAR Contributor
I recently went on a week-long polar bear expedition to the polar bear capital of the world, Churchill, Canada. In conjunction with Tauck and the United States Tour Operator Association (USTOA), it was truly a trip of a lifetime.
Tauck Polar Bear Expedition
313 Kelsey Blvd, Churchill, MB R0B 0E0, Canada
Polar bears are some of the most magical creatures in the world, only accessible in remote places of the planet in their natural habitats. On board a Tauck Polar Bear Expedition, we boarded a custom-built crawler that led us through the arctic...
Lazy Bear Lodge
Churchill, Canada, the polar bear capital of the world, is located on the edge of the Hudson Bay and is where the bears come to wait for the ice to form in order to begin their long winter ahead hunting for ring seals. When we arrived to...
Eskimo Museum
242 La Vérendrye Ave, Churchill, MB R0B 0E0, Canada
Step into the stories of the First Nation people of Churchill, Manitoba-- stories that are told through hand-crafted artifacts of the Inuit people, who have made their lives in these lands for several thousand years. Established in Churchill over...
Lazy Bear Lodge & Cafe
313 Kelsey Blvd, PO Box 880, Churchill, MB R0B 0E0, Canada
An extension of the Lazy Bear Lodge, the Lazy Bear Cafe boasts exquisite food and Churchill's only coffee bar. With a vaulted log beam ceiling, glowing candlelight, handmade log chairs and tables, and a crackling fire in a stone fireplace that was...
Inn at The Forks
75 Forks Market Rd, Winnipeg, MB R3C 0A2, Canada
In the heart of Winnipeg's downtown, located across from the Forks Market and just steps from the newly-opened Museum of Human Rights, the Inn at the Forks is modern and chic, but also incorporates materials such as stones, wooden beams and antler...
Smith Restaurant
The first thing that caught my attention in one of Manitoba's newest restaurants and hot spots, Smith, located on the ground floor of the Inn at the Forks hotel, was the impeccably chic decor. A mix of the rustic Canadian landscape with modern and...
321 Kelsey Blvd, Churchill, MB, Canada
Owned and operated by the "top dog," Dave Daley, Wapusk Adventures is a complete immersion experience into the world of Canadian dog sledding. Dave is of the native Maite people in Manitoba, and his passion for dog sledding comes out of his zeal...
The Forks Market
Forks Market Rd, Winnipeg, MB R3C, Canada
The Forks area of Winnipeg gets its name from the union of two rivers, the Red River and the Assiniboine, and for 6,000 years has been a meeting place among people- first, among early Aboriginal peoples, followed by European fur traders. So it's...
Gypsy's Bakery & Restaurant
A popular spot among "Churchillians" and tourists alike, Gypsy's Bakery and Restaurant has an incredible relaxed, diner-kind-of-feel. When you step in, you get the feeling that the same people have been coming here and enjoying their fresh arctic... | {
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press release (1243)
client story (30)
(-) biography (71)
Larry Fickes (3)
Joseph Xiques (2)
Showing 71 results for " grace schools v. burwell"
Emilie Kao
Emilie Kao serves as senior counsel and vice president of advocacy strategy for Alliance Defending Freedom, where she is a member of the U.S. Legal Team. In this role, she supports ADF's legal and legislative objectives through culture shaping initiatives. Before joining ADF, Kao served as director of the DeVos Center for Religion and Civil Society at the Heritage Foundation. She led a team of experts who provided strategy, research, and policy recommendations on life, marriage, and religious freedom consistent with Constitutional principles. She convened coalitions of strategic partners to ...
Type: Biography
Mathew Hoffmann
Mathew Hoffmann serves as legal counsel for Alliance Defending Freedom, where he is a key member of the Center for Academic Freedom.
Hal Frampton
Hal Frampton serves as senior counsel in the Center for Conscience Initiatives at Alliance Defending Freedom.
Frank Chang
Frank Chang serves as legal counsel at Alliance Defending Freedom and plays a key role on the Center for Christian Ministries team.
Cody Barnett
Cody Barnett serves as legal counsel on Alliance Defending Freedom's Appellate Advocacy Team, where he represents various ADF clients before appellate courts across the country.
Andrew D. Graham
Andrew D. Graham serves as senior counsel, academic & professional affairs at Alliance Defending Freedom. In his role, he assists with developing the faculty and curricula for ADF's training and alliance-building programs and works to create professional opportunities for ADF's Blackstone Fellows.
Christy Hirsch
Christy Hirsch serves as legal counsel on the ADF Church and Ministry Alliance Team.
Erin Morrow Hawley
Erin Morrow Hawley serves as senior counsel to the appellate team at Alliance Defending Freedom.
Ali Kilmartin
… mandate which ultimately resulted in the win Zubik v Burwell Kilmartin earned her JD summa cum laude and was first … mandate, which ultimately resulted in the win Zubik v. Burwell . Kilmartin earned her J.D. summa cum laude and …
Tom Minnery
Tom Minnery is chairman emeritus of the governing board of Alliance Defending Freedom. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 252 |
Q: Saving JSON response globally accessible I have integrated one web-service. Parsed the data using GSON and initialized model/pojo class.
Now, I have requirement to access that Model/Pojo class globally all over the application.
I want that class to check various things in separate activities and fragments.
How can I make that model/pojo class accessible all over the app?
Thanks.
A: You should make the DataSource class singleton and use it in your app:
public class JsonDataSource {
private static JsonDataSource instance = new JsonDataSource();
// your json data model
private JsonData data;
// private constructor to prevent creating new instances
private JsonDataSource() {
}
public static JsonDataSource getInstance() {
return instance;
}
public JsonData getData() {
return data;
}
public void setData(JsonData data) {
this.data = data;
}
}
then you can use like this
// set data
JsonDataSource.getInstance().setData(jsonData);
// get data
JsonData data = JsonDataSource.getInstance().getData();
A: Simple and easy way Just create getter and setter in your root Application Class
and access it globally within the application using getter method
private String jsonData;
public String getmJsonData()
{
return jsonData;
}
public void setjsonData(String jsonData) {
this.jsonData= jsonData;
}
//Set Data
((Your_application_class)getApplicationContext()).setjsonData(data);
//get Data
((Your_application_class)getApplicationContext()).getjsondata());
A: save your json to SharedPreferences.
try this:
public class MySessionManager {
public Context context;
private final String PREF_SD_INFO = "Share_Preference";
private final String KEY_DEF = "KEY_PRE";
public MySessionManager(Context context) {
this.context = context;
}
//save your json data
public void setSongs(SongsList songs) {
if (context != null) {
SharedPreferences LoginPref = context.getSharedPreferences(PREF_SD_INFO,
Context.MODE_PRIVATE);
SharedPreferences.Editor editor = LoginPref.edit();
Gson gson = new Gson();
songsLists.add(songs);
String json = gson.toJson(songsLists);
editor.putString(KEY_PRE, json);
editor.apply();
}
//get your json data
public ArrayList<SongsList> getSongs() {
if (context != null) {
SharedPreferences LoginPref = context.getSharedPreferences(PREF_SD_INFO,
Context.MODE_PRIVATE);
Gson gson = new Gson();
String json = LoginPref.getString(KEY_PRE, null);
Type type = new TypeToken<ArrayList<SongsList>>() {
}.getType();
return gson.fromJson(json, type);
} else {
return null;
}
}
}
}
| {
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} | 2,614 |
, més conegut com a Skrillex, és un productor de música electrònica estatunidenc, del gènere electro house, dubstep i glitchstep, a més a més d'haver estat vocalista de la banda de post-hardcore From First to Last, entre els anys 2004 i 2007. Més tard Sonny va haver d'estudiar a casa perquè a l'escola li feien bullying i la seva mare va contractar un professor particular.
Com a productor, destaca la implementació d'Ableton Live. Al novembre del 2011, Skrillex fou nominat a cinc categories dels Premis Grammy, dels quals en va guanyar tres, Millor remix per «Cinema» per a Benny Benassi, Millor gravació de dance per «Scary Monsters and Nice Sprites» i Millor àlbum.
Biografia
Moore, d'ascendència cubana, va néixer a Highland Park, Los Angeles, però després va ser adoptat per un agent d'assegurances i la seva dona, mestressa de casa. Més tard es mudaren a San Francisco, i després la família va tornar a Los Angeles quan ell tenia 12 anys. En aquell moment, es va fer skate-punk, escoltava The Dickies i Subhumans
Moore fou contractat als 16 anys per ser vocalista a la banda "From First to Last". Després de deixar la banda a principis del 2007, amb dos àlbums sota el braç, Sonny es dedicà a la seva carrera solista. Més tard, i com a productor de dubstep, adoptà el nom artístic de "Skrillex".
Carrera musical
2004—2006: From First to Last
A principis del 2004, gairebé amb 16 anys, Sonny fou contactat per Matt Good per a ser guitarrista rítmic a la banda de post-hardcore/emo "From First to Last", Sonny viatjà a Geòrgia i els membres quedaren conformes amb les seves qualitats. Quan, accidentalment, el sentiren cantar Featuring Some of Your Favorite Words quedaren sorpresos per les seves qualitats vocals. El 29 de juny del 2004, la banda llençà el seu primer àlbum, Dear Diary, My Teen Angst Has a Body Count per "Epitaph Records".
Després de diverses gires, la banda començà a produir el seu següent àlbum titulat "Heroine", el qual es va llençar el 21 de març de 2006. Mesos després, durant les gires, Sonny començà a patir problemes amb les cordes vocals, i la banda va haver d'abstenir-se de participar en determinats tours. Després d'un procediment quirúrgic, Moore va informar la banda que els deixaria per a treballar en la seva carrera com a solista.
En la seva estada es generaren controvèrsies, ja que inicialment les lletres de l'àlbum Dear Diary, My Teen Angst Has a Body Count recorrien a la depressió, per la tendència denominada emo. També en varen generar els membre de la banda, en especial Sonny, del qual fou qüestionada la seva sexualitat, ja que en unes fotografies apareix besant a Ben Jorgensen, vocalista de la banda "Armor for Sleep".
2007—2009: Sonny & The Blood Monkeys
Sonny va crear un nou perfil de MySpace presentant tres maquetes: «Signal», «Equinox» i «Glow worm» i el seu primer concert va ser el 7 d'abril amb Carol Robbins. Després de llençar més maquetes, Sonny sortí de gira al "Team Sleep Tour".
El 7 d'abril del 2009, es llençà digitalment Gypsyhook EP, que conté 3 cançons i 4 remescles. També està inclosa la versió japonesa de «Mora» («海水»).
2010—present: Treball solista, èxit i reconeixement als Premis Grammy
El 2010, Moore començà la seva carrera com a productor de música electrònica, sota l'àlias de Skrillex en clubs de Los Angeles. El 7 de juny del 2010, Sonny llençà l'EP My Name is Skrillex] com a descàrrega gratuïta.
A mitjans del 2010, Moore participà com a corista i usant programació per a l'àlbum There Is a Hell, Believe Me I've Seen It. There Is a Heaven, Let's Keep It a Secret de la banda de metalcore Bring Me The Horizon. A finals d'any, Sonny començà una gira nacional amb Deadmau5 després d'haver signat per "mau5trap" llençant son segon EP, Scary Monsters and Nice Sprites, que es posicionà al número dos al Top Heatseekers.
A principis del 2011, Moore començà la gira Project Blue Book Tour amb Porter Robinson i Tommy Lee & DJ Aero. Skrillex donà a conèixer diverses cançons noves en aquesta gira com ara «First Of The Year» (coneguda des del 2007 com a «Equinox»), «Reptile» i «Cinema» (remix d'una cançó de Benny Benassi). «Reptile's Theme» aparegué a l'anunci televisiu de Mortal Kombat 9, i «First Of The Year» fou llençada a la següent edició de Scary Monsters and Nice Sprites, titulada More Monsters and Sprites, EP publicat el 7 de juny, assolint el número tres del Top Heatseekers.
El 28 d'abril de 2011, Skrillex publicà al seu Facebook, el comunicat que en març els seus ordinadors portàtils havien sigut robats a l'hotel on s'allotjava a Milà, Itàlia i que hauria de tornar a gravar el seu àlbum.
Skrillex va estrenar el seu últim àlbum Recess el març de 2014.
Vida personal
No s'ha casat, ni té fills. Gairebé no es coneix la seva vida personal
Discografia
Álbums d'estudi:
Com a solista
2012: Voltage
28 de febrer de 2012 Make It Bun Dem
EP
2010: My Name is Skrillex
2010: Scary Monsters and Nice Sprites
2011: More Monsters and Sprites
2011: Bangarang
2014: Recess
Amb "From First to Last"
2004: Dear Diary, My Teen Angst Has a Body Count
2006: Heroine
Com Sonny Moore
2009: Gypsyhook EP
Senzills
«WEEKENDS!!!»
«Kill EVERYBODY»
«Scary Monsters and Nice Sprites»
«Reptile's Theme»
«Rock n' Roll (Will Take You to the Mountain)»
«First of the Year (Equinox)»
«Ruffneck (FULL Flex)»
«Bangarang»The Devil's Den-(Ewktwk)
Debut de l'Anne-(Ewktwk)
Altres senzills
«Breakn' a Sweat»
«Kyoto»
Col·laboracions
«Get Up!»
«Narcissistic Cannibal»
«Chaos Lives in Everything»
«Still Gettin' It»
«Zoology»
«Lick It»
«Bring Out the Devil»
«Make It Bun Dem»
Remescles importants
«Bad Romance»
«Alejandro»
«Just the Way You Are»
«Rock That Body»
«In for the Kill»
«Cinema»
«Born This Way (Died This Way)»
«Promises»
«Dancing on My Own»
«Levels»
«Welcome to Jamrock»
«Crush on you» (de Nero)
«E.T.»
«Holding on»
Guardons
Premis
2012: Grammy al millor àlbum de música dance/electrònica
2013: Grammy al millor àlbum de música dance/electrònica
Nominacions
2012: Grammy al millor nou artista
Referències
Persones de Los Angeles
Productors musicals estatunidencs
Empresaris californians | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,055 |
Das Ladinium (im deutschen Sprachgebrauch meist verkürzt zu Ladin) ist in der Erdgeschichte die obere chronostratigraphische Stufe der Mittleren Trias, die geochronologisch dem Zeitraum vor ungefähr bis etwa Millionen Jahre entspricht und damit ca. 7 Millionen Jahre dauerte. Die vorhergehende Stufe ist das Anisium, die folgende Stufe das Karnium, das bereits zur Oberen Trias gehört.
Namensgebung und Geschichte
Die Stufe wurde von 1892 Alexander Bittner vorgeschlagen und nach der Volksgruppe der Ladiner benannt.
Definition und GSSP
Der Beginn der Stufe wird durch das Erstauftreten der Ammoniten-Art Eoprotrachyceras curionii definiert. Die Grenze zum Karn ist noch nicht abschließend festgelegt worden. Wahrscheinlich wird die Grenze mit dem Erstauftreten der Ammoniten-Gattungen Daxatina oder Trachyceras, und dem Erstauftreten der Conodonten-Art Metapolygnathus polygnathiformis definiert. Der GSSP (globales Typprofil) des Ladiniums liegt im Tal des Caffaro bei Bagolino (Provinz Brescia, Italien).
Untergliederung
Das Ladinium wird in vier Ammoniten-Zonen unterteilt:
Frechites regoledanus-Zone
Protrachyceras archelaus-Zone
Protrachyceras gredleri-Zone
Eoprotrachyceras curionii-Zone
Regional wird das Ladin in die Unterstufen Fassan (Fassanium = untere zwei Ammonitenzonen) und Longobard (Longobardium = obere zwei Ammonitenzonen) unterteilt.
Einzelnachweise
Literatur
Alexander Bittner: Was ist norisch? Jahrbuch der kaiserlich-königlichen geologischen Reichsanstalt, 42(3): 387–396, Wien 1892 .
Peter Brack, Hans Rieber, Alda Nicora und Roland Mundil: The Global boundary Stratotype Section and Point (GSSP) of the Ladinian Stage (Middle Triassic) at Bagolino (Southern Alps, Northern Italy) and its implications for the Triassic time scale. Episodes, 28(4): 233–244, Beijing 2005 PDF.
Felix M. Gradstein, James G. Ogg, Mark D. Schmitz & Gabi M. Ogg: Geologic Time Scale 2020, Vol. 2. Elsevier 2020 ISBN 978-0-12-824363-3
Hans Hagdorn & Adolf Seilacher (Hrsg.): Muschelkalk : Internationale Muschelkalk-Tagung Schöntal 1991 Sonderbände der Gesellschaft für Naturkunde in Württemberg 2. Stuttgart, Korb : Goldschneck-Verl. Weidert, 1993 ISBN 3-926129-11-5
Hans Murawski & Wilhelm Meyer: Geologisches Wörterbuch. 10., neu bearb. u. erw. Aufl., 278, Enke Verlag, Stuttgart 1998, ISBN 3-432-84100-0.
Weblinks
Kommission für die paläontologische und stratigraphische Erforschung Österreichs der Österreichischen Akademie der Wissenschaften (Hrsg.): Die Stratigraphische Tabelle von Österreich (sedimentäre Schichtfolgen). Wien 2004 (PDF; 1,8 MB)
International Chronostratigraphic Chart 2012 (PDF; 346 kB)
Zeitalter der Trias | {
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All photographs of my work are copyrighted material and may not be reproduced or used in any way without my permission. You may post links to on social media websites, but please be sure to include credit for my work. Thank you for understanding.
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Prices are not accurate, please contact for current price list. Thank you! | {
"redpajama_set_name": "RedPajamaC4"
} | 2,910 |
Question: We live in the flow of time. There are people who waste a lot of energy learning how to plan their time. Does this depend on how much a person has to do?
Answer: It depends on a person's desire. Essentially, the internal self-organization of the person depends on his psychological setup, and everyone has their own.
There are people who are very organized internally: They consider each step from the start, they are compelled to plan their day, sometimes even long before they materialize their plan, and this is how they operate all the time. This is "in their blood." But if there is a desire, it's possible to learn this.
There has to be a clear task set before an individual, the obligation to carry it out, the desire to fulfill it, and the influence of the environment on him or her. All of it has to be organized in a way that it will work within a person.
On the other hand, there is a worldwide trend of evolutionary development, which changes our desires and values. And therefore, even if we obligate a person to work according to the previous values for attaining goals that are no longer important to him today, it will be very difficult for us to get a good work yield out of him.
This also depends on time. A very serious change in man and society is taking place today. Every people, every civilization (today there are six or seven civilizations in the world) has its own sense of time, and therefore we need to take this into account.
For example, the inhabitants of Africa, China, and South America comprehend time completely differently than the inhabitants of Northern Europe and North America. They have an absolutely different approach, internal sensation of time, and they ascribe different value to time. For some time is money, like an immense potential they were given and now must realize, while for others time flows calmly and they flow together with it.
Question: If the representatives of various civilizations feel time differently, does it mean that the amount of activity done by them will also differ?
Answer: Certainly! If I employ a worker from India, with his mentality and with his relationship to time, or I take a person with a Nordic character, they will have completely different requirements for themselves, for those around them, and for everything.
We need to understand that the sensation of time in various modern civilizations is different. In accord with this we have to plan our interrelationships, attitude to general connections, associations, and so on. | {
"redpajama_set_name": "RedPajamaC4"
} | 624 |
{"url":"http:\/\/mothur.org\/wiki\/trim.flows\/","text":"# trim.flows\n\nThe trim.flows command is analogous to the trim.seqs command, except that it uses the flowgram data that comes bundled in the sff file that is generated by 454 sequencing. It\u2019s primary usage is as a preliminary step to running shhh.seqs. Chris Quince has a series of perl scripts that fulfill a similar role 1. This command will allow you to partition your flowgram data by sample based on the barcode, trim the flows to a specified length range, and cull sequences that are too short or have too many mismatches to barcodes and primers. For this page we use a hypothetical sff file - GQY1XT001.sff.\n\n## Preliminaries and defaults\n\nTo get going with trim.flows you first need to have flowgram data. You can generate this in mothur using mothur\u2019s version of sffinfo with the flows option.\n\nmothur\u00a0>\u00a0sffinfo(sff=GQY1XT001.sff,\u00a0flow=T)\n\n\nOpening up GQY1XT001.flow in your favorite text editor will reveal a file that looks like this.\n\n800\nGQY1XT001CQL4K\u00a085\u00a01.04\u00a00.00\u00a01.00\u00a00.02\u00a00.03\u00a01.02\u00a00.05\u00a0...\nGQY1XT001CQIRF\u00a084\u00a01.02\u00a00.06\u00a00.98\u00a00.06\u00a00.09\u00a01.05\u00a00.07\u00a0...\nGQY1XT001CF5YW\u00a088\u00a01.02\u00a00.02\u00a01.01\u00a00.04\u00a00.06\u00a01.02\u00a00.03\u00a0...\nGQY1XT001DB0ZG\u00a085\u00a01.04\u00a00.00\u00a01.01\u00a00.01\u00a00.02\u00a01.01\u00a00.00\u00a0...\nGQY1XT001AKONJ\u00a089\u00a01.03\u00a00.00\u00a01.02\u00a00.01\u00a00.02\u00a01.01\u00a00.00\u00a0...\nGQY1XT001BJEZH\u00a0292\u00a01.03\u00a00.01\u00a01.00\u00a00.02\u00a00.03\u00a01.02\u00a00.00\u00a0...\nGQY1XT001C1YUM\u00a0304\u00a01.04\u00a00.02\u00a01.04\u00a00.02\u00a00.05\u00a00.97\u00a00.02\u00a0...\nGQY1XT001BBRX0\u00a0140\u00a01.02\u00a00.03\u00a00.99\u00a00.03\u00a00.04\u00a01.05\u00a00.04\u00a0...\n\n\nWhat you have on the first row is the total number of flow values - 800 for Titanium data. For GS FLX it would be 400. The second and following lines contain the sequence name, the number of useable flows as defined by 454\u2019s software, and the flow intensity for each base going in the order of TACG. So for example, the first sequence looks like.\u2026\n\nGQY1XT001CQL4K\u00a085\u00a01.04\u00a00.00\u00a01.00\u00a00.02\u00a00.03\u00a01.02\u00a00.05\u00a0...\n\n\nBy default, trim.flows will require each sequence to contain at least 450 flows and will trim each flowgram to 450 flows. We have found that this does the best job of reducing the sequencing noise. However, this assumes that your amplicon is shorter than 450 flows (i.e. Bacterial primers 27F\/519R which amplify V1-3 only requires 350-400 flows to sequence depending on the exact sequence). Chris Quince suggests a minimum number of flows of 360 and a maximum of 720. We use Quince\u2019s settings for denoting the minimum signal (0.50) and noise (0.70). To run trim.flows with the default settings you will do the following:\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow)\n\n\nThis will create a file called GQY1XT001.trim.flow and GQY1XT001.scrap.flow, that contain the trimmed files that passed the criteria and the sequences that didn\u2019t, respectively. In the GQY1XT001.scrap.flow file, you can see appended to the accession numbers the criteria(on) on which each read was culled (l for length, b for barcode, etc). Also created is a file called GQY1XT001.flow.files which lists all of the flow files that were created.\n\nRarely will you really just want the default settings. Below are some options that you are likely to want to use:\n\n## Options\n\n### oligos\n\nThe oligos option takes a file that can contain the sequences of the forward and reverse primers and barcodes and their sample identifier. Each line of the oligos file can start with the key words \u201cforward\u201d, \u201creverse\u201d, and \u201cbarcode\u201d or it can start with a \u201c#\u201d to tell mothur to ignore that line of the oligos file. This same file can be used for running the trim.seqs command. For example, consider a trimmed version of sahl09.oligos:\n\nforward\u00a0\u00a0\u00a0\u00a0CATGCTGCCTCCCGTAGGAGT\u00a0vWhateverF\n#reverse\u00a0\u00a0\u00a0TCAGAGTTTGATCCTGGCTCAG\u00a0vWateverR\nbarcode\u00a0\u00a0\u00a0\u00a0AACCAACC\u00a0\u00a0\u00a0\u00a0ALP50M\nbarcode\u00a0\u00a0\u00a0\u00a0AACCAAGG\u00a0\u00a0\u00a0\u00a0AZAC1\nbarcode\u00a0\u00a0\u00a0\u00a0AACCATCG\u00a0\u00a0\u00a0\u00a0ALP2B\nbarcode\u00a0\u00a0\u00a0\u00a0AACCATGC\u00a0\u00a0\u00a0\u00a0ALP1B\nbarcode\u00a0\u00a0\u00a0\u00a0AACCGCAT\u00a0\u00a0\u00a0\u00a0ALP80M\nbarcode\u00a0\u00a0\u00a0\u00a0AACCGCTA\u00a0\u00a0\u00a0\u00a0ALPG2\nbarcode\u00a0\u00a0\u00a0\u00a0AACCGGAA\u00a0\u00a0\u00a0\u00a0AZ273\n...\n\n\nThe forward primer is best thought of as the forward sequencing primer. So if you are using the 16S rRNA primers 27f and 338r to generate sequencing substrate, but you are sequencing off of the 338r end of the fragment, you would list 338r as the forward primer and 27f as the reverse. Here we are using a \u201c#\u201d for the reverse primer to indicate that we don\u2019t want mothur to screen for the reverse primer. If the \u201c#\u201d were removed, all of the sequences would wind up in the scrap file because the reads are typically not long enough to get to the distal primer. The lines starting with barcode follow the format of \u201cbarcode\u201d - tab - barcode sequence - tab - sample identifier - line break. There is no limit to the number of primers or barcodes that mothur can handle. The forward and reverse primers can also be degenerate using standard IUPAC nomenclature. You can enter your oligos as upper or lowercase letters.\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0oligos=GQY1XT001.oligos)\n\n\nRunning trim.flows with the oligos file will generate a number of new flow files. For example:\n\nGQY1XT001.A01.v35.flow\nGQY1XT001.A02.v35.flow\nGQY1XT001.A03.v35.flow\nGQY1XT001.A04.v35.flow\nGQY1XT001.A05.v35.flow\nGQY1XT001.A06.v35.flow\nGQY1XT001.A07.v35.flow\n...\n\n\nThe GQY1XT001.flow.files file will contain a list of all of these files and will come in handy for the shhh.seqs command.\n\n### bdiffs & pdiffs & ldiffs & sdiffs & tdiffs\n\nThese parameters are used to allow differences in the barcode, primers, linkers and spacers. It has been shown that sequencing errors in the PCR primer region of a sequence correlate highly with poor sequence quality. Therefore, the default is to require an exact match to the primer or barcode sequences that you provide. pdiffs is maximum number of differences to the primer sequence, default=0. bdiffs is maximum number of differences to the barcode sequence, default=0. ldiffs is maximum number of differences to the linker sequence, default=0. sdiffs is maximum number of differences to the spacer sequence, default=0. tdiffs is maximum total number of differences to the barcode, primer, linker and spacer (default to pdiffs + bdiffs + ldiffs + sdiffs).\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0oligos=GQY1XT001.oligos,\u00a0bdiffs=1,\u00a0pdiffs=2)\n\n\n### minflows & maxflows\n\nThe minflows parameter will set the minimum number of flows that each sequence must contain to make it in to a \u201ctrim\u201d file. By default this is set to 450; Chris Quince has preferred 360 in his documentation for processing GSFLX and Titanium data.\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0minflows=450)\n\n\nThe maxflows parameter will set the number of flows after which all other flows should be ignored. For instance, with Titanium data there are 800 flows. Setting maxflows to 450 (the default) will result in trimming the flowgram data to 450 flows. By default this is set to 450; Chris Quince has preferred 360 in his documentation for processing GSFLX data and 720 for Titanium data.\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0maxflows=450)\n\n\nCombining minflows and maxflows allows you to set the range of flows you want. By default you will get flowgrams that are all 450 flows long. The following will give you Quince\u2019s recommended set up for Titanium data:\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0minflows=360,\u00a0maxflows=720)\n\n\n### fasta\n\nThe fasta option allows you to tell trim.flows that you want it to translate the flowgram data to fasta sequence format:\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0fasta=T)\n\n\n### signal & noise\n\nBy default, trim.flows will treat any intensity signal greater than 0.50 as a real signal and any intensity less than 0.70 as noise. If an intensity falls between 0.50 and 0.70, it is treated as ambiguous and set as a trim point. The settings of 0.50 and 0.70 are suggested by Quince and we really see no need to change it, but in case people want to play with the values here is how you\u2019d do it:\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0signal=0.60,\u00a0noise=0.65)\n\n\n### maxhomop\n\nLooking at the summary.seqs output for your dataset you may notice that the longest homopolymer in the dataset is 31 bases long. This is highly suspect as it is well-established that 454 technology struggles with homopolymers. To cap the homopolymer length you use the maxhomop option:\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0maxhomop=9)\n\n\nIt is likely that sequences with longer homopolymers will get culled for other reasons.\n\n### order\n\nThe order parameter is used to select the flow order. Options are A, B and I. Default=A, meaning flow order of TACG.\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0order=A)\n\n\n### processors\n\nThe processors parameter allows you to run the command with multiple processors. Default processors=Autodetect number of available processors and use all available.\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0oligos=GQY1XT001.oligos,\u00a0processors=2)\n\n\n## Putting it together\n\nHere\u2019s how one might mix and match the settings to process their Titanium data...\n\nmothur\u00a0>\u00a0trim.flows(flow=GQY1XT001.flow,\u00a0oligos=GQY1XT001.oligos,\u00a0pdiffs=2,\u00a0bdiffs=1,\u00a0processors=8)\n\n\nThis will generate individual flow files for each barcode\/primer combination, GQY1XT001.trim.flow and GQY1XT001.scrap.flow files and a GQY1XT001.flow.files file.\n\n## Revisions\n\n\u2022 First Introduced - version 1.22.0.\n\u2022 1.23.0 - fixed bug that occurred if you had multiple primers with blank names. - https:\/\/forum.mothur.org\/viewtopic.php?f=4&t=1358\n\u2022 1.24.0 - added linker and spacer option to the oligos file, as well as ldiffs and sdiffs parameters.\n\u2022 1.25.0 - allow for characters other than ATGC in reverse primers.\n\u2022 1.30.0 - added flow orders A, B and I.\n\u2022 1.33.0 - Bug Fix: printing trimmed number of flows to scrap file instead of original number of flows. Caused error if you wanted to read scrapped flow file.\n\u2022 1.40.0 - Rewrite of threaded code. Default processors=Autodetect number of available processors and use all available.\n\u2022 1.45.0 Fixes trim.flows empty *flow.files issue.\n\u2022 1.48.0 Changes default output to count file instead of name file.","date":"2022-07-07 05:29:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.42196765542030334, \"perplexity\": 1418.8735742038139}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656104683683.99\/warc\/CC-MAIN-20220707033101-20220707063101-00044.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.ncatlab.org\/nlab\/show\/free+simplicial+abelian+group","text":"Contents\n\n# Contents\n\n## Definition\n\nThe free simplicial abelian group functor\n\n$\\mathbf{Z}[-]\\colon sSet \\to sAb$\n\nfrom SimplicialSets to SimplicialAbelianGroups is given by the functor\n\n$sSet = Fun(\\Delta^{op}, Set) \\to Fun(\\Delta^{op}, Ab) = sAb,$\n\nwhere the middle functor applies the free abelian group functor\n\n$\\mathbf{Z}[-]\\colon Set \\to Ab.$\n\n## Properties\n\n###### Proposition\n\n(free simplicial abelian group-adjunction)\nThere is a pair of adjoint functors (a free$\\dashv$forgetful-adjunction)\n\n$sAb \\underoverset {\\underset{ frgt }{\\longrightarrow}} {\\overset{\\mathbb{Z}(-)}{\\longleftarrow}} {\\;\\;\\;\\;\\;\\bot\\;\\;\\;\\;\\;} sSet$\n\nbetween SimplicialAbelianGroups and SimplicialSets, where\n\nThis is a Quillen adjunction with respect to the classical model structure on simplicial sets and the projective model structure on simplicial abelian groups.\n\n## Applications\n\nFree simplicial abelian groups are the crucial ingredient of simplicial chains and simplicial cochains, and such also simplicial homology and simplicial cohomology?, in particular, singular homology and singular cohomology. See these articles for more information.\n\nLast revised on July 12, 2021 at 13:35:57. See the history of this page for a list of all contributions to it.","date":"2021-10-27 23:16:20","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 5, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8523460626602173, \"perplexity\": 1051.940848802311}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-43\/segments\/1634323588244.55\/warc\/CC-MAIN-20211027212831-20211028002831-00496.warc.gz\"}"} | null | null |
{"url":"https:\/\/docs.wiris.com\/mathtype\/en\/mathtype-office-tools\/support\/mathtype-support-notices\/support-notices\/mathtype-requires-a-newer-version-of-mt-extra-warning-message.html","text":"# MathType requires a newer version of MT Extra warning message\n\n## Applicability\n\nMathType 6 and later\n\nWindows and Mac\n\n## Issue\n\nWhen launching MathType 7 either by itself or within another application, users sometimes receive the following error message\u2026\n\n\u2026despite having successfully installed MathType 7 and even used it previously without receiving an error. Pressing the OK button will allow MathType 7 to run, but empty boxes will replace several essential characters in equations.\n\n## Reason\n\nThe original version of MT Extra, which is still distributed with many word processing, publishing, and test generation programs, contains only 43 characters:\n\nThe enhanced version of the MT Extra font included with MathType contains 145 characters:\n\nWhen MathType 7 is launched, it checks the version of the installed copy of MT Extra extra. If an older version is installed, the expected characters will be replaced by empty boxes.\n\nIt is essential to understand how Windows and macOS handle fonts in their Fonts folders to understand this problem. In Windows, if multiple files exist for the same font, only one will be displayed in the Fonts Control Panel. Windows may be using an older version of MT Extra, even though the newer version has been properly installed. On the Mac, if Word happens to install all of its fonts into the per-user fonts folder for some reason (maybe an Office reinstallation, etc.), MathType 7 may not find the newer\/correct version of MT Extra, and the above warning will appear. MathType 7 for Mac lists the path to the older MT Extra file, which is helpful as it could be in several places.\n\nIf you reinstall or upgrade your word processor, manually install the equation editor that comes with your word processor, install a new program that contains an equation editor, apply a service release or patch or update to your word processor, or use the \"Detect and Repair\" feature that comes with Microsoft Office, very often these installers will overwrite the version of MT Extra that comes with MathType 7 resulting in an error message when launching MathType 7 because the newer, enhanced version of MT Extra has been replaced with an older version.WS\n\n## Solution\n\nSuppose you are familiar with the Fonts Control Panel and Windows Explorer or the various font locations in macOS. In that case, you could manually install the MathType 7 version of MT Extra after removing extra copies of MT Extra from your system per the instructions below.\n\n### Manually installing the MathType version of MT Extra\n\nYou can manually install MT Extra, but you should first remove all the copies already installed on your computer.\n\nWindows 8 and later\n\nIt's best to have only one copy of a font on your system. In the latest versions of Windows, deleting a font isn't impossible, but doing so is more trouble than necessary. Simply open a File Explorer window, navigate to C:\\Program\u00a0Files\\MathType\\Fonts\\TrueType [for 64-bit Windows it will be C:\\Program\u00a0Files (x86)\\MathType\\Fonts\\TrueType], and double-click mtextra.ttf. In the font preview window, click and answer Yes if it asks you whether you want to replace it.\n\nWindows versions up to Windows 7, 32-bit\n\n### Note\n\nBecause only one copy of MT Extra will be shown in the Fonts Control Panel at a time, you must refresh your fonts list after deleting each copy of it so that if another copy of MT Extra is in your fonts folder, it will appear. It may be necessary to delete and refresh your list several times to delete all copies of MT Extra. If another copy of MT Extra appears in your fonts list, the actual file name for the font will be different.\n\n1. Choose Control Panel from the Start menu.\n\n2. Open Fonts.\n\n3. Look for the MT Extra font and delete it.\n\n4. Close the window and repeat the first three steps. (It's not sufficient to press F5 to refresh your list of fonts. Doing so won't reveal any additional copies of MT Extra that are present.)\n\n5. Look for the MT Extra font again. If it appears, delete it and close the window again.\n\n6. Continue steps 1-5 until MT Extra no longer appears in your list of fonts.\n\n7. From the File menu of the Fonts Control Panel, choose \"Install new font\". There is a copy of the enhanced version of MT Extra in \\MathType\\Fonts\\TrueType.\n\n### Note\n\nThe file you want is mtextra.ttf (the \"ttf\" may not be showing on your computer, but the \"mtextra\" will be). Do not be misled by the similarly-named font \"Mt\u00a0Extra\u00a0Tiger.ttf\". Reinstalling that one won't help.\n\n8. Click MT Extra, and click OK.\n\nMT Extra should be installed into your Fonts folder, and you should no longer receive an error message when launching MathType 7\n\nWindows 7 64-bit\n\n1. Navigate to C:\\Program Files (x86)\\MathType\\Fonts\\TrueType\n\n2. Locate the font MT Extra.\n\n### Note\n\nThe file you want is mtextra.ttf (the \"ttf\" may not be showing on your computer, but the \"mtextra\" will be). Do not be misled by the similarly-named font \"Mt\u00a0Extra\u00a0Tiger.ttf\". Reinstalling that one won't help.\n\n3. Right-click it and from the contextual menu, choose Install New Font.\n\nMT Extra should be installed into your Fonts folder, and you should no longer receive an error message when launching MathType 7\n\nMac\n\n1. From the Applications folder, open Font Book.\n\n2. Look through the list of fonts for MT Extra. If it exists, remove or disable it.","date":"2023-02-05 23:36:17","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3473396897315979, \"perplexity\": 3877.3186500652164}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764500294.64\/warc\/CC-MAIN-20230205224620-20230206014620-00684.warc.gz\"}"} | null | null |
Die Egyptian Football Association () ist der nationale Fußballverband in Ägypten. Er wurde 1921 gegründet und ist seit 1923 Mitglied der FIFA.
Der Verband mit Sitz in Kairo organisiert das nationale Ligasystem, in dem die Egyptian Premier League die höchste Liga des Landes ist. Die zweite Liga ist die Egyptian Second Division. Des Weiteren organisiert der Verband den Ägyptischen Fußballpokal, den nationalen Pokal. Auch die Ägyptische Fußballnationalmannschaft wird von der EFA gestellt sowie alle Jugendnationalmannschaften, die Frauennationalmannschaft und das Futsalnationalteam.
Siehe auch
Ägyptische Premier League
Weblinks
www.efa.com.eg
Fußballverband
Fußball (Ägypten)
Organisation (Kairo)
Gegründet 1921
Sport (Kairo)
Sportverband (Ägypten)
Nationaler Sportverband
Futsal (Ägypten) | {
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2,3-Dibromphenol ist eine chemische Verbindung, die sowohl zu den Phenolen als auch zu den Halogenaromaten zählt.
Darstellung
Die Synthese von 2,3-Dibromphenol geht vom 2,3-Dibromnitrobenzol aus, das mit Eisen in Ethanol zum 2,3-Dibromanilin (Schmelzpunkt: 44–45 °C) reduziert wird. Dieses wird dann durch Verkochen der mit Natriumnitrit entstehenden Diazoniumverbindung zur gewünschten Verbindung umgesetzt.
Einzelnachweise
Bromphenol | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,816 |
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type: 'POST',
data: _data,
cache: false,
success: function(data) {
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_resultField.text('Thanks for your message!');
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_resultField.text(err);
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\section{Introduction}
Let $G$ be a finite group
and $\crp{K}$ a field of characteristic zero.
By Maschke's theorem and Wedderburn-Artin theory,
the group algebra $\crp{K}G$ of $G$ over $\crp{K}$
is a direct product of matrix rings
over division algebras:
\[ \crp{K}G \iso
\mat_{d_1}(D_1)
\times \dotsm \times
\mat_{d_r}(D_r).
\]
A natural question to ask is when each factor in
this decomposition is actually a division ring
(equivalently, the group algebra $\crp{K}G$ contains
no nilpotent elements).
In the classical case where $\crp{K}$ is algebraically closed,
it is well known that
$\crp{K} G$ is a direct product of division rings
if and only if $G$ is abelian.
For $\crp{K}=\mathbb{Q}$, the question was solved
by S.~K.~Seghal~\cite[Theorem~3.5]{Sehgal75}
(see Theorem~\ref{t:qg_divringprod} below).
In this paper, we consider a slightly more general question:
Let $1\neq N \nteq G$ be a normal subgroup.
Then
\[ \crp{K}G \iso \crp{K}[G/N] \times I,
\]
where the (twosided) ideal $I$ is the kernel of the
canonical homomorphism $\crp{K}G \to \crp{K}[G/N]$.
Now we ask:
for which finite groups is there an $N\neq 1$
such that
the ideal $I$ above is a direct product of division rings?
If there is such an $N$, then any nilpotent element
of $\crp{K}G$ has constant coefficients on cosets of $N$.
Also, only twosided ideals of $\crp{K}G$ can distinguish the elements
of $N$.
The following is just a basic observation,
which allows us to state our results more conveniently.
\begin{mainlem}\label{main:introlemma}
For each field $\crp{K}$ (of characteristic zero)
and each finite group $G$, there is a unique
maximal normal subgroup
$ N$, denoted by $\NKer_{\crp{K}}(G)$,
such that the
kernel of the map $\crp{K}G \to \crp{K}[G/N]$
is a direct product of division rings.
\end{mainlem}
We will give a more direct definition of
$\NKer_{\crp{K}}(G)$ in Section~\ref{sec:defnker}
below, before we prove Lemma~\ref{main:introlemma}.
We call $\NKer_{\crp{K}}(G)$ the
\defemph{nonideal kernel} of $G$ (over $\crp{K}$).
We view the zero ideal as an empty
product of division rings, so possibly
$\NKer_{\crp{K}}(G)=1$.
Indeed, this is the case for ``most'' groups,
and we want to classify the groups $G$ for which
$\NKer_{\crp{K}}(G)\neq 1$.
Our first result concerns the field $\mathbb{R}$ of real numbers.
We need to recall a definition:
A nonabelian group $G$
is called \defemph{generalized dicyclic},
if it has an abelian subgroup $A$ of index $2$ and an element
$g \in G \setminus A$
such that $g^2 \neq 1$ and $ a^g = a^{-1} $
for all $a\in A$.
If $A$ is cyclic, then $G$ is called \defemph{dicyclic}
(or generalized quaternion).
Furthermore, $Q_8$ denotes the quaternion group
of order $8$ and $C_n$ a cyclic group of order $n$.
\begin{mainthm}\label{main:reals}
Let $G$ be a finite group.
Then $ \NKer_{\mathbb{R}}(G) > 1 $ if and only if
one of the following holds:
\begin{enumthm}
\item $G $ is abelian and $G\neq \{1\}$.
\item $G$ is generalized dicyclic.
\item $G \iso C_4 \times Q_8 \times (C_2)^r$, $r\in \mathbb{N} $.
\item $G \iso Q_8 \times Q_8 \times (C_2)^r$, $r\in \mathbb{N} $.
\end{enumthm}
\end{mainthm}
The motivation for this work is a question of Babai~\cite{Babai77}.
Babai asked which finite groups
are isomorphic to the affine symmetry group of
an orbit polytope.
(An orbit polytope is a polytope such that its
(affine) symmetry groups acts transitively on the vertices
of the polytope.)
In joint work with Erik Friese~\cite{FrieseLadisch16b}
(continuing our earlier paper~\cite{FrieseLadisch16a}),
we develop a general theory, which shows, among other things,
that $G$ is isomorphic to the affine symmetry group of an orbit
polytope when
$\NKer_{\mathbb{R}}(G) = 1$.
When $\NKer_{\mathbb{R}}(G) > 1$, this may or may not be the case.
Theorem~\ref{main:reals} above is an essential ingredient in our answer
to Babai's question.
Similarly, when $\NKer_{\mathbb{Q}}(G)=1$,
then $G$ can be realized as the affine symmetry group of an
orbit polytope with vertices having rational coordinates.
The classification of groups
with $\NKer_{\mathbb{Q}}(G) > 1$ is more complicated.
To state it, we first describe a special type of such
groups.
\begin{mainlem}\label{main:pq}
Let $p$ and $q$ be primes, let
$P = \erz{g} \times P_0$ be an abelian $p$-group
and $Q$ an abelian $q$-group.
Suppose $P$ acts on $Q$ such that
$x^g = x^k$ for all $x\in Q$
and some integer $k$ independent of $x\in Q$,
and such that $\C_P(Q) = \erz{ g^{p^c} } \times P_0$.
Suppose that $p^d = \ord( g^{p^c} )$ is the exponent
of $\C_P(Q)$,
and that $(q-1)_p$, the $p$-part of $q-1$,
divides $p^d$.
Then for the semidirect product $G=PQ$,
we have $1\neq \NKer_{\mathbb{Q}}(G) \cap \erz{g}$.
\end{mainlem}
Notice that the assumption on the action of $g$ on $Q$
and $\card{P/\C_P(Q)} = p^c$ imply that
$p^c $ divides $q-1$, and that the multiplicative order
of $k$ modulo the exponent of $Q$ is just $p^c$.
One can show that
$\NKer_{\mathbb{Q}}(G) = \erz{g^{p^s}}$,
where $p^s = p^{c-1} (q-1)_p $.
Whenever we mention ``groups as in
Lemma~\ref{main:pq}'', we also use the notation
established in the statement of Lemma~\ref{main:pq}.
\begin{mainthm}\label{main:qclass}
Let $G$ be a finite group.
Then $\NKer_{\mathbb{Q}}(G)\neq 1$
if and only if at least one of the following holds:
\begin{enumthm}
\item $G$ is abelian.
\item \label{it:m_nilp}
$G = S \times A$,
where $S$ is a $2$-group of exponent $4$
which appears on
the list from Theorem~\ref{main:reals},
the group $A$ is abelian of odd order,
and the multiplicative order of\/
$2$ modulo $\card{A}$ is odd.
\item \label{it:m_gendic}
$G$ is generalized dicyclic.
\item \label{it:m_allab}
$G = (PQ) \times B$,
where the subgroups
$P\in \Syl_p(G)$, $Q\in \Syl_q(G)$ and
$B$ are abelian,
$PQ$ is as in Lemma~\ref{main:pq},
and the $p$-part of the multiplicative order
of $q$ modulo $\card{B}$ divides the multiplicative
order of $q$ modulo $p^d$.
\item \label{it:m_dirprod_h}
$G = Q_8 \times (C_2)^r \times H$,
where $H$ is as in \ref{it:m_allab} and has odd order,
and the multiplicative order of\/
$2$ modulo $\card{H}$ is odd.
\end{enumthm}
\end{mainthm}
Case~\ref{it:m_nilp} contains the groups
$G = Q_8 \times( C_2)^r \times A$,
for which $\mathbb{Q} G$ is a direct product of division rings,
as classified by Sehgal~\cite{Sehgal75}.
An important tool in the proofs
of Theorems~\ref{main:reals} and~\ref{main:qclass}
is Blackburn's classification of finite groups
in which all nonnormal subgroups have a nontrivial
intersection~\cite{Blackburn66}.
As we will see below, $\NKer_{\crp{K}}(G)$
is always contained in the intersection
of all nonnormal subgroups of $G$.
While the proof of Theorem~\ref{main:reals} is relatively elementary,
the proof of Theorem~\ref{main:qclass} also depends on some
deep facts about division algebras and Schur indices.
The paper is organized as follows:
In Section~\ref{sec:skewlin}, we review some basic facts
about representations and characters over fields not necessarily
algebraically closed, and in particular Schur indices.
We also introduce the auxiliary concept of
\emph{skew-linear characters}.
In Section~\ref{sec:defnker}, we define $\NKer_{\crp{K}}(G)$
and prove some elementary properties.
In Section~\ref{sec:dedekind}, we consider Dedekind groups
(groups such that all subgroups are normal).
In such groups, we have either
$\NKer_{\crp{K}}(G) = 1$, or $\NKer_{\crp{K}}(G)=G$,
where the latter are exactly the groups such that
$\crp{K}G$ is a direct product of division rings.
Finally, Section~\ref{sec:classr}
contains the proof of Theorem~\ref{main:reals},
and Section~\ref{sec:classq} the (long) proof of
Theorem~\ref{main:qclass}.
\section{Skew-linear characters}
\label{sec:skewlin}
Let $G$ be a finite group.
For simplicity, assume that $\crp{K}\subseteq \mathbb{C}$
and write $\Irr G$ for the set of irreducible complex
characters of $G$.
We begin by reviewing the relation between the representation
theory of $G$ over $\crp{K}$ and over
$\mathbb{C}$~\cite[\S~38]{huppCT}\cite[Chapter~10]{isaCTdov}.
By Maschke's theorem and general Wedderburn-Artin theory,
the group algebra $\crp{K}G$ is the direct product
of simple rings:
\[ \crp{K}G = A_1 \times \dotsm \times A_r.
\]
Each $A_i$ is a simple ideal,
and the set of the $A_i$'s is uniquely determined as the set
of simple ideals of $\crp{K}G$.
The $A_i$'s are called the
\defemph{block ideals of $\crp{K}G$}.
Each $A_i$ is generated by a central primitive idempotent
$e\in \Z(\crp{K}G)$.
By Wedderburn-Artin theory,
each $A_i$ is isomorphic to a matrix ring
over a division ring.
We now relate the above decomposition
to the complex irreducible characters of $G$.
Recall that the \defemph{Schur index}
of $\chi\in \Irr G$ over $\crp{K}$
is the smallest positive integer $m = m_{\crp{K}}(\chi)$
such that
$m\chi $ is afforded by a representation with entries in
$\crp{K}(\chi)$, the field generated by $\crp{K}$
and the values of $\chi$.
\begin{lemma}\label{l:reps_basics}
Let $\chi\in \Irr G$.
\begin{enumthm}
\item \label{it:uniblock}
There is a unique block ideal $A$ of\/ $\crp{K}G$
such that $\chi(A)\neq 0$.
\item \label{it:gal_conj}
Let $\psi\in \Irr G$.
Then
$\psi(A)\neq 0$ if and only if
$\psi$ and $\chi$ are Galois conjugate over $\crp{K}$,
that is, $\psi=\chi^{\alpha}$ for some
$\alpha\in \Gal( \crp{K}(\chi)/\crp{K} )$.
\item \label{it:cent_bl}
Write $A\iso \mat_n(D)$ for some division ring~$D$.
Then $\Z(A)\iso \Z(D)\iso \crp{K}(\chi)$.
\item \label{it:si_deg}
$\fdeg{D:\Z(D)} = m_{\crp{K}}(\chi)^2$
and $\chi(1) = nm_{\crp{K}}(\chi)$.
\end{enumthm}
\end{lemma}
\begin{proof}
This is standard~\cite[Theorems~38.1 and~38.15]{huppCT}.
\end{proof}
It follows from Lemma~\ref{l:reps_basics}
that $A$ is itself a division ring if and only if
$\chi(1)=m_{\crp{K}}(\chi)$.
In this case, the projection
$\crp{K}G\to A$ defines
a homomorphism $\phi$
from $G$ into the multiplicative group of $D$.
Notice also that
$\Ker(\phi)=\Ker(\chi)$.
For this reason,
we call a character $\chi$
\defemph{skew-linear} (over $\crp{K}$),
if $\chi(1) = m_{\crp{K}}(\chi)$.
Thus skew-linear characters generalize linear characters.
Since $m_{\mathbb{C}}(\chi)=1$ for all $\chi$,
skew-linear over $\mathbb{C}$ is the same as linear.
If $\chi\in \Irr(G)$ is linear, then (trivially)
the reduction to any subgroup
is irreducible and linear.
This fact generalizes to skew-linear characters as follows:
\begin{lemma}\label{l:res_skewlin}
Let $\chi\in \Irr(G)$ be skew-linear over
the field\/~$\crp{K}$, and $H\leq G$.
Then the irreducible constituents of $\chi_H$
are skew-linear over $\crp{K}$,
and are Galois conjugate over the field
$\crp{K}(\chi)$.
\end{lemma}
\begin{proof}
Let $\theta\in \Irr(H)$ be a constituent of $\chi_H$.
Then~\cite[Lemma~10.4]{isaCTdov}
\[ m_{\crp{K}}(\chi)
\quad \text{divides} \quad
\ipcf{\chi_H}{\theta}
\card{\crp{K}(\chi, \theta):\crp{K}(\chi)}
m_{\crp{K}}(\theta).
\]
Let $\sigma \in \Gal( \crp{K}(\chi,\theta)/\crp{K}(\chi))$.
Then $\ipcf{\chi_H}{\theta^{\sigma}} = \ipcf{\chi_H}{\theta}$.
Thus each of the
$\card{\crp{K}(\chi,\theta):\crp{K}(\chi)}$
characters $\theta^{\sigma}$ occurs in $\chi_H$ with multiplicity
$\ipcf{\chi_H}{\theta}$.
It follows that
\begin{align*}
\ipcf{\chi_H}{\theta}
\card{\crp{K}(\chi, \theta):\crp{K}(\chi)}
\theta(1)
&\leq \chi(1)
= m_{\crp{K}}(\chi) \\
&\leq \ipcf{\chi_H}{\theta}
\card{\crp{K}(\chi, \theta):\crp{K}(\chi)}
m_{\crp{K}}(\theta).
\end{align*}
This implies that equality holds throughout,
in particular, $\theta(1) = m_{\crp{K}}(\theta)$
and $\chi_H = \ipcf{\chi_H}{\theta} \sum \theta^{\sigma}$,
the sum running over
$\sigma \in \Gal( \crp{K}(\chi,\theta)/\crp{K}(\chi))$.
\end{proof}
In the rest of this section, we record some
(mostly well known) facts about Schur indices
and blocks of group algebras for later reference.
Recall that
\[ e_{\chi}
= \frac{ \chi(1) }{ \card{G} }
\sum_{g\in G} \chi( g^{-1} ) g
\]
is the central primitive idempotent
in $\mathbb{C} G$ corresponding to $\chi\in \Irr G$.
The following simple observation will sometimes be useful.
Notice that it provides an alternative proof of
$\Z(A)\iso \crp{K}(\chi)$.
\begin{lemma}\label{l:charfield_bl}
Let $\chi\in \Irr G$
and let $A$ be the block ideal of\/ $\crp{K}G$
such that $\chi(A)\neq 0$.
Then
\[ A \iso \crp{K}(\chi)Ge_{\chi}
\quad\text{by}\quad
A \ni a \mapsto ae_{\chi}.
\]
\end{lemma}
\begin{proof}
Set
\[ e:=\sum_{\alpha \in \Gal(\crp{K}(\chi)/\crp{K})}
e_{\chi^{\alpha}}.
\]
We claim that $A = \crp{K}Ge$.
We can decompose $1$ into a sum of primitive idempotents
in $\Z(\crp{K}G) $,
and then decompose further in $\Z(\mathbb{C} G)$.
Thus there is a unique primitive idempotent $f$
in $\Z(\crp{K}G)$
such that $fe_{\chi} = e_{\chi}$.
But then also
$fe_{\chi^{\alpha}} = e_{\chi^{\alpha}}$
for all $\alpha \in \Gal(\crp{K}(\chi)/\crp{K})$
and thus $fe=e$.
On the other hand, $e_{\chi}\in \crp{K}(\chi)$ and
$e\in \crp{K}G$, and thus $f=e$.
This shows that $A=\crp{K}Ge$ as claimed.
For $\alpha\in \Gal(\crp{K}(\chi)/\crp{K})$,
\[ b = \sum_g b_g g
\in
\crp{K}(\chi)Ge_{\chi}
\quad \text{implies}\quad
b^{\alpha}:= \sum_{g} b_g^{\alpha} g
\in \crp{K}(\chi)G e_{\chi^{\alpha}}.
\]
Using this, it is straightforward to check that
\[ \crp{K}(\chi) G e_{\chi}
\ni b
\mapsto
\sum_{\alpha\in \Gal( \crp{K}(\chi)/\crp{K} ) }
b^{\alpha} \in \crp{K} G e
\]
yields the inverse of the map
$a\mapsto ae_{\chi}$.
\end{proof}
Since we will often have to consider characters
of direct products of groups,
and the corresponding blocks of the group
algebra, we record the following for later reference.
\begin{lemma}\label{l:blocksdirprod}
Let $G = U \times V$ be a direct product of groups,
$\sigma\in \Irr U$ and $\tau\in \Irr V$.
Then
$\chi = \sigma \times \tau\in \Irr G$
and
$\crp{K}(\chi) = \crp{K}(\sigma,\tau)$.
Let
$ A_{ \crp{K} } ( \chi ) $
be the block ideal of\/ $\crp{K}G$ corresponding to $\chi$,
and
$ A_{ \crp{K} } ( \sigma ) $ and
$ A_{ \crp{K} } ( \tau ) $
the block ideals of\/ $\crp{K}U$ and\/ $\crp{K}V$
corresponding to $\sigma$ and $\tau$.
Then
\[ A_{\crp{K}}(\chi)
\iso \big( A_{\crp{K}}(\sigma)
\otimes_{ \crp{K}(\sigma) }
\crp{K}(\chi)
\big)
\otimes_{ \crp{K}(\chi) }
\big( A_{\crp{K}}(\tau)
\otimes_{ \crp{K}(\tau) }
\crp{K}(\chi)
\big).
\]
\end{lemma}
\begin{proof}
The irreducible characters of $U\times V$
are exactly the characters of the form
$\chi= \sigma \times \tau$,
with $\sigma\in \Irr U$ and
$\tau \in \Irr V$ \cite[Theorem~4.21]{isaCTdov}.
Since $\chi((u,1))= \sigma(u)\tau(1)$ for $u\in U$
and similarly $\chi((1,v))=\sigma(1)\tau(v)$
for $v\in V$,
we see that $\crp{K}(\chi) = \crp{K}(\sigma,\tau)$.
Set $\crp{L}=\crp{K}(\chi)$.
The natural isomorphism
\[ \crp{L}U \otimes_{\crp{L}} \crp{L}V
\to \crp{L}G,
\quad
\sum_u a_u u \otimes \sum_v b_u v
\mapsto \sum_{u,v} a_ub_v(u,v)
\]
sends $e_{\sigma}\otimes e_{\tau}$ to
$e_{\chi}$
and thus induces an isomorphism
\[ \crp{L}Ue_{\sigma}
\otimes_{\crp{L}}
\crp{L}Ve_{\tau}
\iso \crp{L}G e_{\chi}
\]
(by comparing dimensions).
By Lemma~\ref{l:charfield_bl},
the right hand side is isomorphic to
$A_{\crp{K}}(\chi)$,
and on the left hand side
we have
\[ \crp{L}U e_{\sigma}
\iso
\crp{K}(\sigma)U e_{\sigma}
\otimes_{ \crp{K}(\sigma) }
\crp{L}
\iso
A_{\crp{K}}(\sigma)
\otimes_{ \crp{K}(\sigma) }
\crp{L},
\]
and similarly for the other factor.
The result follows.
\end{proof}
In Section~\ref{sec:classq},
we need several deep facts about Schur indices,
which we collect now.
For a prime $q$, we write
$m_q(\chi) := m_{\mathbb{Q}_q}(\chi)$,
where $\mathbb{Q}_q$ denotes the field of $q$-adic numbers.
Sometimes, it will be convenient to use this notation
also for the ``infinite prime'', that is,
$m_{\infty}(\chi):= m_{\mathbb{R}}(\chi)$.
\begin{lemma}\label{l:sifacts}
Let $\chi\in \Irr(G)$.
\begin{enumthm}
\item \label{it:globloc}
$m_{\mathbb{Q}}(\chi)$ is the least common multiple
of the local indices
$m_q(\chi)$, where $q$ runs through all
primes, including the infinite one.
\cite[(32.19)]{reinerMO}
\item \label{it:localsi_div}
$m_{\mathbb{R}}(\chi)$ and $m_2(\chi)$ divide $2$,
and $m_q(\chi)$ divides $q-1$ for odd $q$.
\cite[Theorem~4.3, Corollary~5.5]{Yamada74}
\item \label{it:localsi_brauer}
Let $\phi$ be an irreducible Brauer character
for the prime $q$, and $d_{\chi \phi}$
the decomposition number.
Then
$m_q(\chi)$ divides
$d_{\chi \phi} \card{\mathbb{Q}_q(\chi,\phi):\mathbb{Q}_q(\chi)}$.
\cite[Theorem~IV.9.3]{Feit82}
\item \label{it:localsitriv}
If the finite prime $q$ does not divide $\card{G}$,
then $m_q(\chi) = 1$.
\cite[Corollary~IV.9.5]{Feit82}
\end{enumthm}
\end{lemma}
\begin{cor}\label{c:localindequ}
Let $\chi\in \Irr(G)$ with
$\chi(1) = m_q(\chi)$,
where $q$ is a prime number.
If $H\leq G$ is not divisible by $q$,
then any constituent of $\chi_H$ is linear.
\end{cor}
\begin{proof}
This is immediate from Lemma~\ref{l:res_skewlin}
and Lemma~\ref{l:sifacts}~\ref{it:localsitriv}.
\end{proof}
\section{The nonideal kernel}
\label{sec:defnker}
For every field $\crp{K}$
and any finite group $G$, we define
\[ \NKer_{\crp{K}}(G)
:= \bigcap \{\Ker(\chi) \mid \chi(1) > m_{\crp{K}}(\chi)
\}.
\]
If $m_{\crp{K}}(\chi) = \chi(1)$ for every $\chi\in \Irr(G)$,
we set $\NKer_{\crp{K}}(G):=G$.
We call $\NKer_{\crp{K}}(G)$ the
\defemph{nonideal kernel} of $G$ over $\crp{K}$.
Notice that $\NKer_{\crp{K}}(G)$, for any field $\crp{K}$,
is characteristic in $G$.
\begin{lemma}\label{l:subfields}
Let $\crp{K} \subseteq \crp{L}$ be fields.
Then $\NKer_{\crp{L}}(G) \subseteq \NKer_{\crp{K}}(G)$.
\end{lemma}
\begin{proof}
Since $m_{\crp{L}}(\chi)$ divides
$m_{\crp{K}}(\chi)$ for any $\chi\in \Irr G$,
any character which is skew-linear over $\crp{L}$,
is also skew-linear over $\crp{K}$.
The result follows.
\end{proof}
\begin{lemma}\label{l:kernelnonlin}
Let $G$ be a nonabelian group.
Then
\[ \bigcap_{\substack{\chi\in \Irr G \\ \chi(1)>1}}
\Ker \chi = \{1\}.
\]
\end{lemma}
\begin{proof}
Suppose that $g\neq 1$ is
contained in the kernel of all nonlinear characters.
Then, by the second orthogonality relation~\cite[(2.18)]{isaCTdov},
\begin{align*}
0 &= \sum_{\chi\in \Irr G} \chi(1)\chi(g)
= \sum_{\substack{\chi \in \Irr G \\ \chi(1)>1}}
\chi(1)^2
+ \sum_{\chi\in \Lin G} \chi(1)\chi(g).
\end{align*}
The second sum runs over the irreducible characters of $G/G'$
and has value $\card{G:G'}$ or $0$,
according to whether $g\in G'$ or not.
It follows that
the first sum must be empty.
Thus $G$ has no nonlinear characters,
which means that $G$ is abelian, as claimed.
\end{proof}
\begin{cor}
Let $G$ be a nonabelian group.
Then $\NKer_{\mathbb{C}}(G) = 1$.
\end{cor}
Let us say that a character $\alpha$
(not necessarily irreducible)
is \defemph{strictly nonideal},
if $\ipcf{\alpha}{\chi} < \chi(1)$ for all
$\chi\in \Irr G$.
(Such a character is afforded by a left ideal of the group algebra,
which does not contain any nonzero two-sided ideal.)
If at the same time, $\alpha$ is the character of a
representation with entries in $\crp{K}$, then
$m_{\crp{K}}(\chi)$ divides $\ipcf{\alpha}{\chi}$
for all $\chi\in \Irr G$~\cite[Corollary~10.2(c)]{isaCTdov}.
Thus no constituent
of $\alpha$ can be skew-linear over $\crp{K}$.
Conversely, if $S$ is a set of non-skew-linear characters over $\crp{K}$,
then we may add the characters of the corresponding
irreducible representations
over $\crp{K}$
and get a strictly nonideal character $\alpha$
which is afforded by a $\crp{K}$\nbd representation.
Since $\Ker \alpha = \bigcap \Ker\chi$, where $\chi$ runs through
the constituents of $\alpha$,
it follows that every group $G$ has a strictly nonideal character
$\alpha$ with $\Ker \alpha = \NKer_{\crp{K}}(G)$,
and such that $\alpha$ is afforded by a representation
over $\crp{K}$.
(In the case where $G=\NKer_{\crp{K}}(G)$,
the only such character is $\alpha=0$, however.)
\begin{lemma}\label{l:subgroupskernel}
Let $H\leq G$ with
$N:=\NKer_{\crp{K}}( H ) < H$.
Then
\[ \NKer_{\crp{K}}(G) \leq \bigcap_{g\in G} N^g
\leq \NKer_{\crp{K}}(H).
\]
\end{lemma}
\begin{proof}
Let $\alpha$ be a strictly nonideal character of $H$
with $N = \Ker \alpha$ and which is afforded by a representation
over $\crp{K}$.
Then $0\neq \alpha^G$ is afforded by a representation over $\crp{K}$
and has kernel
$\bigcap_{g\in G} N^g$~\cite[Lemma~5.11]{isaCTdov}.
Let $\rho_G$ be the regular character of $G$.
Notice that a character $\beta$ is strictly nonideal if and only if
$\rho_G-\beta$ is a character and
$\ipcf{\rho_G-\beta}{\chi}> 0$ for \emph{all}
$\chi\in \Irr G$.
Since $\rho_G = (\rho_H)^G$, we have
that $\rho_G -\alpha^G = (\rho_H-\alpha)^G$ is a character, and
\[ \ipcf{\rho_G-\alpha^G}{\chi}
= \ipcf{(\rho_H-\alpha)^G}{\chi}
= \ipcf{\rho_H-\alpha}{\chi_H}_H
> 0
\]
for all $\chi\in \Irr G$.
Thus $\alpha^G$ is strictly nonideal.
\end{proof}
\begin{lemma}\label{l:nt_id}
Let $N$ be a normal subgroup of $G$, and set
\[ e_N = \frac{1}{\card{N}} \sum_{n\in N} n.
\]
Then $\crp{K}G e_N \iso \crp{K}[G/N]$.
If $\chi\in \Irr G$, then
$\chi(e_N)\neq 0 $ if and only if
$N\leq \Ker(\chi)$.
\end{lemma}
\begin{proof}
This is well known:
The canonical epimorphism
$\crp{K}G \to \crp{K}[G/N]$ is split
by the map sending a coset $Ng$ to
$(1/\card{N})\sum Ng = e_N g$.
This proves the first statement.
If $N\leq \Ker(\chi)$, then any representation
affording $\chi$ sends
$e_N$ to the identity map.
If $N\not\leq \Ker(\chi)$,
then any representation affording $\chi$
must send $e_N$ to $0$.
\end{proof}
\begin{lemma}\label{l:idemp}
Let $N:= \NKer_{\crp{K}}(G)$, and
$e_N$ as in Lemma~\ref{l:nt_id}.
Then $\crp{K}G(1-e_N)$ is a direct product of division rings.
In particular, every idempotent $f\in \crp{K}G$ with
$fe_N = 0$ is central.
\end{lemma}
\begin{proof}
By Lemma~\ref{l:nt_id}, it follows that
$ \crp{K} G (1-e_N) $ is the direct product of the block ideals
which correspond to $\chi\in \Irr G$
with $N\not\leq \Ker(\chi)$.
By definition of $N$,
any such $\chi$ is skew-linear over $\crp{K}$,
and thus the corresponding block ideal
is a division ring.
In a direct product of division rings, every idempotent
is central.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{main:introlemma}]
The first part of Lemma~\ref{main:introlemma}
is contained in Lemma~\ref{l:idemp}.
Conversely, if $\crp{K}G(1-e_N)$ is a direct product
of division rings,
then the above considerations yield
that when $m_{\crp{K}}(\chi) < \chi(1)$, we must have
$N \subseteq \Ker(\chi)$, and thus
$N \leq \NKer_{\crp{K}}(G)$.
\end{proof}
Following Blackburn~\cite{Blackburn66},
for any group $G$, we set
\[ \R(G)
:= \bigcap \{U\leq G \mid
U \text{ not normal in } G
\}.
\]
If every subgroup of $G$ is normal, then we set $\R(G) = G$.
Blackburn~\cite{Blackburn66} classified finite groups in which
$\R(G) \neq 1$.
Therefore, a group $G$ with $\R(G) \neq 1$ is called a
\defemph{Blackburn group}.
The following result shows why this is relevant for us:
\begin{lemma}\label{l:many_normals}
For any finite group $G$ and field $\crp{K}$
of characteristic zero,
we have $\NKer_{\crp{K}}(G) \leq \R(G)$.
\end{lemma}
\begin{proof}
Suppose $U\leq G$ is such that
$N:= \NKer_{\crp{K}}(G)\not\subseteq U \leq G$.
We need to show that $U \nteq G$.
Set $f= (1-e_N)e_U$, with $e_N$ as before,
and analogously $e_U:= (1/\card{U})\sum_{u\in U} u$.
Then $f^2 = f \in \crp{K}G(1-e_N)$,
since $e_N$ is central in $\crp{K} G$.
Thus $f$
is central in $\crp{K} G$ by Lemma~\ref{l:idemp}.
We compute
\[ f = \frac{1}{\card{U}} \sum_{u\in U} u
- \frac{1}{\card{NU}} \sum_{x\in NU} x.
\]
As $N$ is not contained in $U$, we have
$U < NU$.
As $g^{-1}fg = f$ for all $g\in G$,
it follows that $U\nteq G$.
\end{proof}
\section{Dedekind groups}
\label{sec:dedekind}
In this section, we compute
$\NKer_{\crp{K}}(G)$ for Dedekind groups,
and determine when $\crp{K}G$ is a direct product
of division rings.
These results are mostly known.
Recall that a \defemph{Dedekind group}
is a finite groups in which all subgroups are
normal.
First, we recall Dedekind's classification of these
groups~\cite[Satz~III.7.12 on p.~308]{huppEG1}.
\begin{thm}[Dedekind 1897]\label{t:dedekind}
Let $G$ be a finite group, such that every
subgroup of $G$ is normal.
Then either $G$ is abelian, or
\[ G \iso Q_8 \times (C_2)^r \times A \quad(r\geq 0),
\]
where $A$ is abelian of odd order.
\end{thm}
Let $\tau\in \Irr(Q_8)$ be the irreducible, faithful
character of degree $2$.
Then $\mathbb{H} := \mathbb{Q} Q_8 e_{\tau}$ is a division ring,
the rational quaternions.
$\mathbb{H}$ can also be described as the
$\mathbb{Q}$-vector space with basis
$\{1, i, j, k\}$ and multiplication
defined by $i^2 = j^2 =-1$, $k= ij = -ji$.
\begin{thm}\label{t:kg_divringprod}
Let $\crp{K}$ be a field and $G$ be a group.
Then $\crp{K}G$ is a direct product of division rings
if and only if either
$G$ is abelian, or
$G\iso Q_8 \times (C_2)^r \times A$,
where $A$ is abelian of odd order,
and\/ $\mathbb{H} \otimes_{\mathbb{Q}}\crp{K}(\lambda)$
is a division ring for all $\lambda\in \Lin(A)$.
\end{thm}
\begin{proof}
Suppose $\crp{K} G$ is a direct product of division rings.
Then all subgroups of $G$ are normal in $G$
by Lemma~\ref{l:many_normals}
(as $\NKer_{\crp{K}}(G) = G$, or directly from the argument in
the proof of Lemma~\ref{l:many_normals}).
It follows that either $G$ is abelian, or
$G \iso Q_8 \times (C_2)^r \times A$
with $A$ abelian of odd order.
In the second case,
let $\tau\in \Irr(Q_8)$ be the irreducible, faithful
character of degree $2$.
Then
\[ \crp{K} Q_8 e_{\tau}
\iso \mathbb{H} \otimes_{\mathbb{Q}} \crp{K},
\]
the quaternions over $\crp{K}$.
Any nonlinear, irreducible character of
$G = Q_8 \times (C_2)^r \times A$
has the form $\chi = \tau \times \sigma \times \lambda$,
where $\sigma \in \Lin(C_2)^r$ and $\lambda\in \Lin A$.
The corresponding block ideal of the rational group algebra
is, by Lemma~\ref{l:blocksdirprod}, isomorphic to
\[
\mathbb{H} \otimes_{\mathbb{Q}} \crp{K}(\lambda).
\]
The result follows.
\end{proof}
\begin{remark}
If $G \iso Q_8 \times (C_2)^r \times A$ with
$A$ abelian of odd order, then either
$\crp{K}G$ is a direct product of division rings,
or $\NKer_{\crp{K}}(G) = 1$.
\end{remark}
\begin{proof}
Suppose that $\crp{K}G$ is not
a direct product of division rings.
Then there is some $\lambda \in \Lin(A)$
such that $\crp{K}(\lambda)$ is a splitting field
for $\mathbb{H}$.
As before, let $\tau\in \Irr(Q_8)$ be the faithful irreducible
character of $Q_8$.
Then
$\Ker(\tau\times 1 \times \lambda)
= 1 \times (C_2)^r \times \Ker(\lambda)$.
It follows that
$\NKer_{\crp{K}}(G) \subseteq 1 \times \Ker(\mu)$ for every
$\mu\in \Lin( (C_2)^r \times A)$ such that
$\ord(\lambda)$ divides the order of $\mu$.
Since $A$ contains elements of order $\ord(\lambda)$, we
see that $\NKer_{\crp{K}}(G) = 1$.
\end{proof}
Notice that for a linear character
$\lambda$, we have
$\crp{K}(\lambda) = \crp{K}(\varepsilon_n)$,
where $\varepsilon_n$ is a primitive $n$-th root of unity
and $n= \ord(\lambda)$.
The following lemma collects some results.
These will be needed also in the proof of
Theorem~\ref{main:qclass}.
\begin{lemma}
\label{l:split_quat}
\hfill
\begin{enumthm}
\item \label{it:squaresum}
$\mathbb{H} \otimes_{\mathbb{Q}} \crp{K}$
is a division ring if and only if
$-1$ is not a sum of two squares in $\crp{K}$.
\item \label{it:qu_local}
$\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{Q}_2$
and\/ $\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{R}$
are division rings,
and\/
$\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{Q}_p$ for $p$ odd
is not a division ring.
(Here $\mathbb{Q}_p$ is the field of $p$\nbd adic numbers.)
\end{enumthm}
Let $\varepsilon_n$ be a primitive $n$\nbd th root of unity,
where $n$ is odd.
Then
\begin{enumthm}[resume]
\item \label{it:qu_cyc}
$\mathbb{H} \otimes_{\mathbb{Q}}{\mathbb{Q}(\varepsilon_n)} $
is a division ring if and only if
the multiplicative order of $2$ in $(\mathbb{Z}/n)^*$
is odd, if and only if
$\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{Q}_2(\varepsilon_n)$
is a division ring.
\item \label{it:qu_qu}
$\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{Q}( \sqrt{2}, \varepsilon_n)$
is a division ring only when $n=1$.
\end{enumthm}
\end{lemma}
\begin{proof}
\ref{it:squaresum} and~\ref{it:qu_local}
are well
known \cite[Example~38.13(a)]{huppCT}
\cite[Ch.~III, Théorème~1]{SerreCA}.
Assertion~\ref{it:qu_cyc} is a result of
Moser~\cite{Moser73}.
(This can be shown without using the
Hasse-Minkowski principle:
If the residue class of $2$ in $(\mathbb{Z}/p)^*$ has even multiplicative
order $2r$,
then $2^r \equiv -1 \mod p$, and thus
$p$ divides $2^r+1$.
Then an elementary argument shows that
$-1$ is a sum of two squares in
$\mathbb{Q}(\varepsilon_p)$~\cite[Example~38.13(d)]{huppCT}.
If $2$ has odd order in $(\mathbb{Z}/n\mathbb{Z})$,
then $\mathbb{H} \otimes_{\mathbb{Q}}\mathbb{Q}_2(\varepsilon_n)$
is also a division ring.)
To see~\ref{it:qu_qu}, assume that $n>1$.
We have to show that
$-1$ is a sum of two squares in
$\crp{K}:=\mathbb{Q}(\sqrt{2}, \varepsilon_n)$.
By the Hasse-Minkowski principle,
it suffices to show that $-1$ is a square
in each possible completion of $\crp{K}$.
Since $n>1$, $\crp{K}$ can not be embedded into $\mathbb{R}$.
If $p$ is odd, then $-1$ is a sum of two squares
in $\mathbb{Q}_p$ already.
Finally, $\mathbb{Q}_2(\sqrt{2})$ is a quadratic extension
of $\mathbb{Q}_2$
and thus a splitting field
of $\mathbb{H}$~\cite[Lemma~VI.2.14]{Lam05_IQFF}.
(We notice that in~\ref{it:qu_qu},
we can replace $\mathbb{Q}(\sqrt{2})$ by any field
such that the completions at all prime ideals over $2$
yield extensions of \emph{even} degree over $\mathbb{Q}_2$.)
\end{proof}
As a consequence, we get the following results.
\begin{thm}[Sehgal 1975~\cite{Sehgal75}]
\label{t:qg_divringprod}
The group algebra $\mathbb{Q} G$ is a direct product of
division rings if and only if one of the following holds:
\begin{enumthm}
\item $G$ is abelian.
\item \label{it:m_q8ab}
$G \iso Q_8 \times (C_2)^r \times A$,
where $r\geq 0$,
and $A$ is abelian of odd order, and
the multiplicative order of $2$ in $(\mathbb{Z}/\card{A})^*$
is odd.
\end{enumthm}
\end{thm}
\begin{thm}\label{t:local_dpdivrings}\hfill
\begin{enumthm}
\item $\mathbb{Q}_2 G$ is a direct product of division rings
if and only if\/ $\mathbb{Q} G$ is a direct product of
division rings.
\item Let $p$ be an odd prime.
Then $\mathbb{Q}_p G$ is a direct product of division rings
if and only if $G$ is abelian.
\item $\mathbb{R} G$ is a direct product of division rings
if and only if either $G$ is abelian,
or $G \iso Q_8 \times (C_2)^r$ for some
$r\geq 0$.
\end{enumthm}
\end{thm}
\section{Classification over the reals}
\label{sec:classr}
In this section, we prove Theorem~\ref{main:reals}.
We begin with the (maybe more interesting) ``only if'' part.
\begin{lemma}\label{l:nn_ord4}
Suppose that $\NKer_{\mathbb{R}}(G) \neq 1$,
and $\erz{g}\not\nteq G$.
Then $g$ has order $4$,
and $\NKer_{\mathbb{R}}(G) = \erz{g^2}$
has order $2$.
\end{lemma}
\begin{proof}
By Lemma~\ref{l:many_normals} and the definition of $\R(G)$,
we have
\[ 1 \neq N:= \NKer_{\mathbb{R}}(G) \leq \R(G) < \erz{g}.
\]
The last inequality is strict since
$\R(G)$ is normal in $G$, but $\erz{g}$ is not.
In particular, the first claim of the lemma implies the second one.
Let $\lambda\in \Lin\erz{g}$ be faithful.
By Lemma~\ref{l:nt_id} applied to
$N \nteq \erz{g}$ and since $N\neq 1$, it follows
$\lambda(e_N)=0$.
Thus
\[ f = e_{\lambda} + e_{\overline{\lambda}}
= \frac{1}{ \card{\erz{g}} }
\sum_{h\in \erz{g}}
( \overline{\lambda}(h) + \lambda(h) ) h
\in \mathbb{R} \erz{g}
\]
is an idempotent with $f e_N = 0$.
It follows from Lemma~\ref{l:idemp}
that $f$ is a central idempotent in
$\mathbb{R} G$, and so
$f^x = f$ for all $x\in G$.
But by assumption, there is some $x\in G$ such that
$g^x \notin \erz{g}$.
It follows that $\overline{\lambda(g)} + \lambda(g) = 0$.
As $\lambda(g)$ is an $n$-th root of unity,
where $n=\ord(g)$, this is only possible when $\ord(g) = 4$.
\end{proof}
\begin{lemma}\label{l:r_not2}
Suppose that $1 < \NKer_{\mathbb{R}}(G) < G$ and that
$G$ is not a $2$-group.
Then $G$ is generalized dicyclic.
\end{lemma}
\begin{proof}
By Lemma~\ref{l:nn_ord4},
we have that $\NKer_{\mathbb{R}}(G) = \R(G) = \erz{z}$,
where $z$ has order~$2$.
Every odd-order subgroup of $G$ is normal in $G$,
and in particular the Sylow $p$-subgroups,
for $p$ odd, generate a normal $2$-complement, $U$,
of $G$.
As $U$ is Dedekind, it follows that $U$ is abelian.
Now set $A= \C_G(U) $, which contains $U$.
There is $g\in G$ such that $\erz{g}\not\nteq G$.
By Lemma~\ref{l:nn_ord4}, we have $g^4 = 1$.
If $gu=ug$ for some $u\in U$,
then $\erz{g}$ is characteristic in
$\erz{gu} = \erz{g} \times \erz{u}$,
and thus $\erz{gu}\not\nteq G$.
Again by Lemma~\ref{l:nn_ord4},
it follows that $(gu)^4 = 1$ and thus $u=1$.
Thus $\C_U(g)=1$ and $g\notin A$.
In particular, $A < G$.
Conversely, let $g\notin A$,
and let $s=g_2$ be the $2$-part of~$g$.
Then $gA = sA$ and thus $s\notin A$.
Thus $u^s \neq u$ for some $u\in U$,
and thus $s^u = s[s,u] \notin \erz{s}$.
It follows that $\erz{s}$ is not normal in $G$,
and thus $\erz{g}$ is not normal in $G$.
By Lemma~\ref{l:nn_ord4}, it follows that
$g^2 = z$ (and $s=g$).
In particular, for $g\in G\setminus A$ and $a\in A$, we have
$g^2 = z = (ga)^2 = g^2 a^g a$,
and thus $a^g = a^{-1}$.
For $u\in U$ and $g$, $h\in G\setminus A$
we have $u^g = u^{-1} = u^h$ and thus
$gh^{-1}\in \C_G(U) = A$, so
$\card{G:A} = 2$.
Thus $G$ is generalized dicyclic.
\end{proof}
To finish the proof of the ``only if'' part
of Theorem~\ref{main:reals}, we use a part of
Blackburn's classification~\cite[Theorem~1]{Blackburn66}:
\begin{thm}[Blackburn 1966]\label{t:blackburn1}
Let $G$ be a $p$-group with $\R(G)\neq 1$.
Then one of the following holds:
\begin{enumthm}
\item $G$ is abelian.
\item $p=2$ and $G$ is generalized dicyclic.
\item $p=2$ and $G \iso C_4 \times Q_8 \times (C_2)^r$, $r\in \mathbb{N} $.
\item $p=2$ and $G \iso Q_8 \times Q_8 \times (C_2)^r$, $r\in \mathbb{N} $.
\end{enumthm}
\end{thm}
Using Theorem~\ref{t:blackburn1},
it is rather straightforward to determine all finite groups
$G$ with $\R(G)\neq 1$, but a rather long list
emerges~\cite[Theorem~2]{Blackburn66}.
However, due to Lemma~\ref{l:r_not2},
we do not need to go through the longer list of
finite groups with $\R(G)\neq 1$.
\begin{proof}[Proof of Theorem~\ref{main:reals}, ``only if'']
Suppose that $\NKer_{\mathbb{R}}(G)\neq 1$.
If $\NKer_{\mathbb{R}}(G) = G$, then
$G$ is abelian or $G\iso Q_8 \times (C_2)^r$,
by Theorem~\ref{t:local_dpdivrings}.
If $1 < \NKer_{\mathbb{R}}(G) <G$ and $G$ is not a $2$-group,
then $G$ is generalized dicyclic, by Lemma~\ref{l:r_not2}.
If $G$ is a $2$-group, then it follows from
Blackburn's classification of $2$-groups
with $\R(G)\neq 1$ (Theorem~\ref{t:blackburn1})
that $G$ appears on the list in Theorem~\ref{main:reals}.
\end{proof}
We now show that conversely, the groups appearing in
Theorem~\ref{main:reals} all have $\NKer_{\mathbb{R}}(G)\neq 1$.
To show that certain characters are skew-linear,
we use the Frobenius-Schur indicator.
Recall that for $\chi\in \Irr G$,
its \defemph{Frobenius-Schur indicator} is defined by
\[ \FS(\chi) := \frac{1}{\card{G}}
\sum_{g\in G} \chi(g^2).
\]
When
$\FS(\chi) = 1$, then
$\chi = \overline{\chi}$ and $\chi$ is afforded by a
representation with entries in $\mathbb{R}$,
so $m_{\mathbb{R}}(\chi) = 1$.
When $\FS(\chi) = 0$, then
$\chi\neq \overline{\chi}$, and again $m_{\mathbb{R}}(\chi)=1$.
Finally, when $\FS(\chi) = -1$, then
$\chi = \overline{\chi}$,
but $m_{\mathbb{R}}(\chi) = 2$.
In the last case,
there is a simple $\mathbb{R} G$-module affording
$2\chi$, and
$\enmo_{\mathbb{R} G}(S) \iso \mathbb{H}$,
the division ring of Hamilton's quaternions
\cite[Theorem~13.12]{huppCT}.
In particular,
$\chi\in \Irr G$ is skew-linear over $\mathbb{R}$,
if and only if either
$\chi(1)=1$ ($\chi$ is linear), or
$\chi(1) = 2$ and $\FS(\chi) = -1$.
We begin by considering generalized dicyclic groups.
\begin{lemma}\label{l:gendickern}
Let $G$ be generalized dicyclic, and
let $g\in G$ and $A\nteq G$ be as in the definition.
Then $\R(G)= \NKer_{\mathbb{R}}(G) = G$ if\/
$G/ \erz{g^2}$ is abelian,
and\/ $\R(G) = \NKer_{\mathbb{R}}(G) = \erz{g^2}$ else.
\end{lemma}
\begin{proof}
First, observe that
$g^2 = (g^2)^g = g^{-2}$ and thus
$g^4 =1$.
Moreover, for any $a\in A$, we have
$(ga)^2 = g^2 a^ga = g^2$.
By assumption, $g^2\neq 1$.
In view of Lemma~\ref{l:nn_ord4}, it suffices to show
that $\erz{g^2} \subseteq \NKer_{\mathbb{R}}(G)$, that is,
all characters $\chi\in \Irr G$ with $g^2\notin \Ker \chi$
are skew-linear.
(In the case when $G/\erz{g^2}$ is abelian,
all characters of $G/\erz{g^2}$
are linear and thus it will follow that all characters of $G$
are skew-linear and $G$ is Dedekind.
Conversely, if $\NKer_{\mathbb{R}}(G)> \erz{g^2}$,
then $\NKer_{\mathbb{R}}(G)=G$ by Lemma~\ref{l:nn_ord4},
and then
$G \iso Q_8 \times (C_2)^r$ by Theorem~\ref{t:local_dpdivrings}
and $G/\erz{g^2}$ is abelian.)
So suppose that $\chi \in \Irr G$ is not linear,
and $g^2 \notin \Ker \chi$.
Let $\lambda\in \Lin A$ be a constituent
of the restriction $\chi_A$.
Then $\chi = \lambda^G$
by Clifford theory~\cite[Corollary~6.19]{isaCTdov}.
As $a^g = a^{-1}$ for all $a\in A$, we have
$\lambda^g = \overline{\lambda}$.
Also, $\lambda(g^2) \neq 1$,
and thus $\lambda(g^2) = -1$ and $\chi(g^2) = -2$.
It follows that
\begin{align*}
\FS(\chi)
= \frac{1}{\card{G}}
\sum_{x\in G} \chi(x^2)
&= \frac{1}{\card{G}}
\sum_{a\in A} (\chi((ga)^2) + \chi(a^2))
\\
&= \frac{1}{\card{G}}
\left( -2\card{A} +
\sum_{a\in A} ( \lambda(a^2) + \overline{\lambda(a^2)} )
\right)
\\
&= \frac{-2\card{A}}{\card{G}}
= -1.
\end{align*}
Here we have used that $(ga)^2 = g^2$ for all $a\in A$,
and that $\sum_{a\in A} \lambda(a^2) = \sum_{a\in A} \lambda^2(a) =0$
since
$\overline{\lambda} \neq \lambda$ and thus $\lambda^2\neq 1$.
Since $\FS(\chi)=-1$ and $\chi(1)=2$,
it follows that $\chi$ is indeed skew-linear, as claimed.
\end{proof}
\begin{lemma}\label{l:type3kernel}
When $G = \erz{u} \times \erz{x,y} \times E$
with $\erz{u} \iso C_4$, $\erz{x,y}\iso Q_8$
and $E\iso (C_2)^r$,
then $\NKer_{\mathbb{R}}(G) = \R(G) = \erz{u^2x^2} \neq 1$.
\end{lemma}
\begin{proof}
As $\erz{ux} \not\nteq G$, we have $\R(G) \leq \erz{u^2x^2}$.
Let $\tau$ be the nonlinear irreducible character of $\erz{x,y}$ and
$\lambda$ a character of $\erz{u}$ with
$\lambda \neq \overline{\lambda}$.
If $\chi$ is a character with
$\chi(u^2x^2) \neq \chi(1)$, then either
$\chi$ is linear, or
$\chi = \lambda^2 \times \tau \times \sigma$, $\sigma\in \Lin E$.
The latter characters all have $\FS(\chi)=-1$.
Thus $ \erz{u^2 x^2} \subseteq \NKer_{\mathbb{R}}(G) $.
\end{proof}
\begin{lemma}
When $G= \erz{u,v} \times \erz{x,y} \times E$
with $\erz{u,v}\iso \erz{x,y} \iso Q_8$ and $E\iso(C_2)^r$,
then $\NKer_{\mathbb{R}}(G) = \R(G)= \erz{u^2x^2} \neq 1$.
\end{lemma}
\begin{proof}
As $\erz{ux} \not\nteq G$, we have $\R(G) \leq \erz{u^2x^2}$.
Let $\tau_1$ and $\tau_2$ be the nonlinear characters of
$\erz{u,v}$ and $\erz{x,y}$, respectively.
If $\chi(u^2x^2) \neq \chi(1)$, then either
$\chi = \tau_1 \times \lambda \times \sigma$ with
$\lambda \in \Lin\erz{x,y}$ and $\sigma\in \Lin(E)$,
or $\chi = \lambda \times \tau_2 \times \sigma$
with $\lambda\in \Lin\erz{u,v}$ and $\sigma\in \Lin(E)$.
In both cases, $\FS(\chi)=-1$
and thus $\erz{u^2x^2} \leq \NKer_{\mathbb{R}}(G)$.
\end{proof}
This lemma finishes the proof of the ``if'' part of
Theorem~\ref{main:reals}.
\begin{comment}
\begin{lemma}\label{l:rg_center}
Suppose $\R(G) < G$.
Then $\R(G) $ is cyclic of prime power order,
and $\R(G) \leq \Z(G)$.
\end{lemma}
\begin{proof}
If $H \leq G$ is not normal in $G$, then
$H$ contains a cyclic subgroup $C$ which is not normal
in $G$,
and $C$ must contain a Sylow $p$-subgroup for some prime $p$
which is not normal in $G$.
Thus $\R(G)$ is cyclic of prime power order.
Now assume that
$\R(G) = \erz{r}$ and let $g\in G$.
We have to show that $[g,r]=1$.
For this, we may assume that $\ord(g)$ is a power
of some prime.
If $\erz{g} \not \nteq G$, then
$\R(G) \leq \erz{g}$ and thus
$rg=gr$.
Otherwise, $\erz{g} \nteq G$,
and of course $\R(G) = \erz{r} \nteq G$.
It follows that
$[g,r] \in \erz{g} \cap \erz{r}$.
Consider $H = \erz{g,r}$.
Then
$H' = \erz{[g,r]} \leq \erz{r} = \R(G) \leq \R(H)$.
Thus every subgroup of $H$ is normal in $H$, and thus
$H$ is a Dedekind group generated by two elements
of prime power order.
If $H$ is nonabelian, then $H\iso Q_8$.
Moreover, $\erz{r}$ and $\erz{g}\leq H$ are normal in $G$,
as is $\erz{rg}$.
It follows that every $x\in G$ induces an inner
automorphism of $H$, and thus
$G = H \C_G(H)$.
Since $R(G)= \erz{r} \not\subseteq \C_G(H)$,
it follows that $\C_G(H)$ is a Dedekind group.
But then either $G$ is itself a Dedekind group,
or $\C_G(H)$ contains elements of order $4$,
and $\R(G) \not\subseteq H$, contradiction.
Thus $H$ must be abelian, as was to be shown.
\end{proof}
\begin{lemma}\label{l:rg_order2}
If $1 < \NKer_{\mathbb{R}}(G) < G$,
then $\card{\NKer_{\mathbb{R}}(G)} = \card{\R(G)}=2$.
\end{lemma}
\begin{proof}
We can assume that $\R(G) = \erz{r} < G$, and thus
$\R(G)$ has prime power order.
Let $z\in \NKer_{\mathbb{R}}(G)$ have prime order $p$.
By Lemma~\ref{l:kernelnonlin}, there is some
$\chi\in \Irr(G)$ with $z \notin \Ker(\chi)$ and
$\chi(1)>1$.
By definition of
$\NKer_{\mathbb{R}}(G)$, we must have
$m_{\mathbb{R}}(\chi) = 2 = \chi(1)$.
In particular, $\chi$ is real-valued.
By Lemma~\ref{l:rg_center}, we have
$\chi_{\R(G)} = \chi(1) \lambda$ for some
$\lambda \in \Lin(\R(G))$.
Since $z \notin \Ker(\chi)$,
the linear character $\lambda$ is faithful.
But since $\lambda$ must have values in $\mathbb{R}$,
we see that $\ord(\lambda) = \card{\R(G)}=2$.
\end{proof}
\end{comment}
\section{Classification over the rational numbers}
\label{sec:classq}
In this section ,we prove Theorem~\ref{main:qclass}.
Throughout, we write
$\NKer(G):=\NKer_{\mathbb{Q}}(G) \neq 1$.
Recall that a \emph{Blackburn group}
is a finite group $G$ such that $\R(G)$, the intersection of
all nonnormal subgroups of $G$, is nontrivial.
For later reference, we record the following observation
(which is part of the argument used by Blackburn to classify
these groups):
\begin{lemma}\label{l:bb_pnilp}
Let $G$ be a Blackburn group and
$p$ a prime dividing $\card{\R(G)}$.
Then $G$ has a normal $p$-complement $A$
such that every subgroup of $A$ is normal in $G$.
If $A$ is nonabelian, then
$G = Q_8 \times (C_2)^r \times H$,
where $H$ is a Blackburn group of odd order.
\end{lemma}
\begin{proof}
By definition of $\R(G)$, all the Sylow $q$-subgroups for
$q\neq p$ are normal in $G$,
and thus generate a normal $p$-complement, $A$.
By definition of $\R(G)$, it follows also that every
subgroup of $A$ is normal in $G$.
In particular, $A$ is a Dedekind group.
If $A$ is nonabelian, then
$S\in \Syl_2(G)$ is isomorphic to
$Q_8\times (C_2)^r$, by Theorem~\ref{t:dedekind}.
As $S\nteq G$, there is a $2$-complement $H$.
Since every subgroup of $S$ is normal in $G$,
it is easy to see that $H$ centralizes $S$
and thus $G = S \times H$
(this is also shown
in~\cite[Proof of Theorem~2(e)]{Blackburn66}).
Any nonnormal subgroup of $H$ is nonnormal in $G$
and thus $\R(G)\leq \R(H)$.
\end{proof}
Thus $A$ is a Dedekind group
and $G = PA$ for any $P\in \Syl_p(G)$.
The classification of Blackburn groups can now be obtained
by considering the different possibilities for $P$ and $A$
(using the fact that $P$ is also a Blackburn group and
Theorem~\ref{t:blackburn1} for $P$,
and Theorem~\ref{t:dedekind} for $A$).
However, in our proof of Theorem~\ref{main:qclass},
we do not have to consider all the cases of Blackburn's
classification separately.
First, we reduce to the case that
$A$ is abelian.
\begin{thm}\label{t:qclass_q8c2h}
Let $G = Q_8 \times (C_2)^r \times H$ with
$H$ of odd order.
Then\/ $\NKer(G)\neq 1$ if and only if\/
$\NKer(H)\neq 1$ and
the multiplicative order of\/ $2$ in $(\mathbb{Z}/\card{H})^*$
is odd.
\end{thm}
\begin{proof}
In view of Theorem~\ref{t:qg_divringprod},
we may assume that $H$ is nonabelian.
Thus $\NKer(G) \leq \NKer(H) \leq \R(H) < H$
by Lemma~\ref{l:subgroupskernel}
and Lemma~\ref{l:many_normals}.
Assume $\NKer(G)\neq 1$ and let $z\in \NKer(G)$
have prime order $p$.
Let $A$ be the abelian $p$-complement of $H$ and
suppose $\lambda\in \Lin(\erz{z,A})$ has maximal possible order.
(This implies $\lambda(z)\neq 1$, in particular.)
Then any $\chi\in \Irr(H\mid \lambda)$ is skew-linear.
For
$\tau\in \Irr(Q_8)$ with $\tau(1)=2$,
we must have that $\tau \times \chi$ is also skew-linear.
Lemma~\ref{l:blocksdirprod} yields in particular,
that $\mathbb{Q}(\chi)$ must not be a splitting field for $\mathbb{H}$
(the quaternions over $\mathbb{Q}$).
On the other hand, we have $\mathbb{Q}(\chi)\subseteq \mathbb{Q}(\varepsilon)$,
where $\varepsilon$ is a primitive $\card{H}$-th root of unity.
Since every prime dividing $\card{H}$ also divides
$\ord(\lambda)$,
we see that
$\card{\mathbb{Q}(\lambda):\mathbb{Q}(\chi_A)}$ is odd.
Therefore, $\card{\mathbb{Q}(\varepsilon):\mathbb{Q}(\chi)}$ is odd as well.
Thus $\mathbb{Q}(\chi)$ is a splitting field for the quaternions,
if and only if $\mathbb{Q}(\varepsilon)$ is a splitting field
for the quaternions.
Now Lemma~\ref{l:split_quat}\ref{it:qu_cyc} yields that the condition
on the order of $2 \mod \card{H}$ holds.
Conversely, assume that this condition holds,
and let $\chi\in \Irr(H)$ be skew-linear over $\mathbb{Q}$,
and $\sigma\in \Irr(S)$, where $S= Q_8 \times (C_2)^r$.
Let $D$ be the block ideal of $\mathbb{Q} H$ corresponding to $\chi$.
This is a division ring with center isomorphic to $\mathbb{Q}(\chi)$.
If $\sigma$ is linear,
then the block ideal corresponding to $\sigma \times \chi$
is again isomorphic to $D$.
If $\sigma$ is nonlinear,
then the block ideal corresponding to $\sigma\times \chi$
is isomorphic to
\[ (\mathbb{H} \otimes_{\mathbb{Q}} \mathbb{Q}(\chi))
\otimes_{\mathbb{Q}(\chi)}
D,
\]
by Lemma~\ref{l:blocksdirprod}.
This is a division ring since both factors are division rings
and one has dimension $4$ over its center,
and the other has odd dimension.
Thus $\sigma \times \chi$ is skew-linear.
This shows that
$\NKer(G) = \NKer(H)$.
The theorem follows.
\end{proof}
Next, we consider nilpotent groups.
\begin{thm}\label{t:qclass_nilp}
Let $G$ be nilpotent.
Then $\NKer(G)\neq 1$ if and only if one of
the following holds.
\begin{enumthm}
\item $G$ is abelian.
\item $G = S \times A$,
where $S\in \Syl_2(G)$ is a nonabelian group
from the list in Theorem~\ref{t:blackburn1}
and has exponent $4$,
and $A$ is abelian of odd order,
and the multiplicative order of $2$ in
$(\mathbb{Z}/\card{A})^*$ is odd.
\item $G$ is a generalized dicyclic $2$-group.
\end{enumthm}
\end{thm}
\begin{proof}
If $G$ is nonabelian, then
the Sylow $2$-subgroup $S$ is nonabelian,
and all other Sylow subgroups are abelian,
by Theorem~\ref{t:blackburn1}.
Thus $G= S\times A$ with $A$ abelian.
If $S$ has exponent $4$,
then the nonlinear, but skew-linear characters
of $S$ yield the quaternions over $\mathbb{Q}$ as block ideal
of the rational group algebra $\mathbb{Q} S$,
and the result follows from
Lemma~\ref{l:split_quat}\ref{it:qu_cyc},
and Lemma~\ref{l:blocksdirprod}.
If $S$ contains elements of order $8$ or greater,
then $S$ is generalized dicyclic,
and there is a skew-linear $\sigma\in \Irr S$ such that
$S/\Ker(\sigma) $ is a dicyclic (=generalized quaternion)
group of order at least $16$.
Then $\mathbb{Q}(\sigma)$ contains $\sqrt{2}$.
As $S/\Ker(\sigma)$ has a subgroup of order~$8$
isomorphic to the quaternion group,
the block ideal of $\mathbb{Q} S$ corresponding to $\sigma$
is isomorphic to the quaternions over
a field containing $\sqrt{2}$.
But since $\mathbb{H} \otimes \mathbb{Q}_2(\sqrt{2})$ splits
(Lemma~\ref{l:split_quat}\ref{it:qu_qu}),
the Schur index of such a character at the
prime $2$ is trivial.
If $G = S\times A$, then any character
$\sigma \times \lambda$ with $1_A \neq \lambda \in \Lin A$
has trivial Schur index over the reals,
and over all other primes anyway.
Thus we can have $\NKer(G)\neq 1$ only if
$G=S$ in this case.
\end{proof}
To prove Theorem~\ref{main:qclass}, we can now
assume that the $p$-complement $A$
in Lemma~\ref{l:bb_pnilp} is abelian,
and that $G=PA$ is not nilpotent.
In other words, $\C_P(A) < P$,
where $P\in \Syl_p(G)$.
It is not difficult to see
that $P$ is then either abelian or generalized dicyclic:
Namely, $\R(P)\neq 1$ and so $P$ occurs on the list
from Theorem~\ref{t:blackburn1}.
If $P\iso C_4 \times Q_8 \times (C_2)^r$
or $P\iso Q_8 \times Q_8 \times (C_2)^r$, however,
then $P$ is generated by elements $u$ such that
$\erz{u}\cap \R(P) =1$ and thus $P$ would centralize
$A$, so this is impossible.
(Alternatively, look at Blackburn's
list~\cite[Theorem~2]{Blackburn66}.)
It remains to show that in this situation,
\ref{it:m_allab} in Theorem~\ref{main:qclass} holds,
or $p=2$ and $G$ is generalized dicyclic.
We begin with some elementary observations,
which were also used in Blackburn's classification.
\begin{lemma}\label{l:specialauts}
Let $Q$ be a finite abelian $q$-group
and suppose that $P$ acts on $Q$ by automorphisms
such that
every subgroup of $Q$ is $P$-invariant,
and $(\card{P}, \card{Q} ) = 1$.
Then $P/\C_P(Q)$ is cyclic of order dividing $q-1$,
and $\C_P(x) = \C_P(Q)= P_{\lambda}$ for every $1\neq x\in Q$
and $1_Q\neq \lambda \in \Lin Q$.
\end{lemma}
\begin{proof}
Take $x\in Q$ of maximal order and $u\in P$.
Since $x^u \in \erz{x}$ by assumption,
we have $x^u = x^k$ for some $k \in \mathbb{N}$.
If $y\in Q$ with $\erz{x}\cap \erz{y} = 1$,
then $y^u = y^k$, since $u$ maps
$\erz{y}$ and $\erz{xy}$ to itself.
It follows that $y^u = y^k $ for all $y\in Q$.
Therefore, $P/\C_P(Q)$ is isomorphic to a $q'$-subgroup
of $\Aut(\erz{x})$, and thus is cyclic of order dividing
$q-1$.
Finally, suppose $1 \neq x\in Q$ and
$x^u = x$ for some $u\in P$.
As we have just seen,
there is $k\in \mathbb{N}$ such that
$y^u = y^k$ for all $y\in Q$.
It follows that $k\equiv 1 \mod q$
(as $q \mid \ord(x)$).
Since $\card{P/\C_P(Q)}$ divides $q-1$,
it follows that
$k^{q-1} \equiv 1 \mod q^n$,
where $q^n$ is the exponent of $Q$.
But this yields that $k\equiv 1 \mod q^n$ and thus
$u\in \C_P(Q)$ as claimed.
The proof for $\lambda\in \Lin Q$ is similar,
using that there is $\ell$ such that
$\mu^u = \mu^{\ell}$ for all $\mu \in \Lin Q$.
\end{proof}
\begin{lemma}\label{l:abbglocal}
Let $G =PA$ be a Blackburn group with
normal abelian $p$-complement $A$ and
$\R(G) \leq P\in \Syl_p(G)$.
Suppose that
$\chi\in \Irr(G)$ is skew-linear over $\mathbb{Q}_q$,
where $q$ is a prime dividing $\card{A}$.
Then $P$ centralizes every
Sylow $r$-subgroup $R$ of $A$
such that $r\neq q$ and $R\not\subseteq \Ker(\chi)$.
\end{lemma}
\begin{proof}
Let $r\neq q$ and $R\in \Syl_r(A)$,
and assume that $R\not\subseteq \Ker(\chi)$.
Let $\lambda\in \Lin(R)$ be a linear constituent
of $\chi_R$, so that $\lambda\neq 1$.
By Lemma~\ref{l:specialauts},
$\C_P(R) = \C_P(x)$ for any $1\neq x\in R$, and thus also
\[\C_P(R) = P_{\lambda} := \{u\in P \mid \lambda^u = \lambda\}.
\]
Consider the subgroup $H=PR$, and choose
a constituent $\theta\in \Irr(H)$ of $\chi_H$
that lies over $\lambda$.
Then $\theta = \psi^H$ for some
$\psi\in \Irr(H_{\lambda})$,
where $H_{\lambda} = \C_P(R) R$.
Thus $\theta(1) \geq \card{P:\C_P(R)}$.
On the other hand, by Corollary~\ref{c:localindequ} we have
that $\theta(1)=1$, and thus
$P=\C_P(R)$ as claimed.
\end{proof}
\begin{lemma}\label{l:abab_1}
Let $G = PA$ be a group
with a normal abelian $p$\nbd complement $A$
and $1 \neq \NKer(G)\leq P\in \Syl_p(G)$.
Suppose that
$\card{P:\C_P(A)} > 2$.
Then $P$ is abelian,
and there is exactly one Sylow subgroup
of $A$ which is not centralized by $P$.
\end{lemma}
\begin{proof}
Let $z\in \NKer(G) \subseteq P$ be an element of order $p$.
Choose $\tau\in \Irr(P)$ with
$z\notin \Ker(\tau)$.
For $\lambda\in \Lin(A)$ arbitrary, we have
\[ \ipcf{ (\tau^G)_A }{ \lambda }_A
= \ipcf{ (\tau_{P\cap A})^A }{ \lambda }_A
= \ipcf{ \tau_{P\cap A} }{ \lambda_{P\cap A} }_{P\cap A}
= \tau(1) > 0,
\]
as $P\cap A = 1$.
Thus for any $\lambda \in \Lin(A)$, there is
$\chi\in \Irr(G)$
such that $\ipcf{\chi}{\tau^G}> 0$ and
$\ipcf{\chi_A}{\lambda}> 0$.
We apply this to a $\lambda$ such that
$\lambda_R \neq 1_R$ for each Sylow subgroup, $R$, of $A$.
Thus there is a $\chi\in \Irr(G)$ lying over $\tau$
and such that $\Ker(\chi)$
contains no Sylow subgroup of $A$.
Notice that $\C_P(A) = P_{\lambda}$ for such a $\lambda$,
by Lemma~\ref{l:specialauts}.
As $\chi$ is induced from a character of $G_{\lambda}$,
it follows that
$\chi(1) \geq \card{G:G_{\lambda}} = \card{P:\C_P(A)} > 2$.
Because $z\not \in \Ker(\chi)$ and $z\in \NKer(G)$,
it follows that $\chi$ is skew-linear over $\mathbb{Q}$
and thus $m_{\mathbb{Q}}(\chi) = \chi(1)$.
By Ito's theorem~\cite[Theorem~6.15]{isaCTdov},
$\chi(1)$ divides $\card{G:A}=\card{P}$
and thus is a power of $p$.
It follows from Lemma~\ref{l:sifacts}\ref{it:globloc}
that there is a prime $q$ (possibly infinite)
such that $m_q(\chi)= m_{\mathbb{Q}}(\chi) = \chi(1)$.
Since $\chi(1)\geq \card{P:\C_P(A)} > 2$,
it follows from Lemma~\ref{l:sifacts}\ref{it:localsi_div}
that the prime $q$, such that
$m_q(\chi)=\chi(1)$, must be a finite, odd prime
and $q\neq p$.
It follows that $q$ divides $\card{A}$.
Now Corollary~\ref{c:localindequ} yields that
$\tau$ is linear.
Since the only assumption on $\tau\in \Irr P$ was
that $z\not\in \Ker(\tau)$,
Lemma~\ref{l:kernelnonlin} yields that $P$ is abelian.
Lemma~\ref{l:abbglocal} yields that
$P$ centralizes all Sylow subgroups of $A$
except the Sylow $q$-subgroup.
\end{proof}
\begin{lemma}\label{l:c2ab}
Let $G = SA$ be a group,
where $A$ is a normal (abelian) $2$-complement
and $1 \neq \NKer(G) \leq S\in \Syl_2(G)$, and suppose
$\card{S:\C_S(A)} = 2$.
Then either $G$ is as in Lemma~\ref{l:abab_1}
(with $p=2$),
or $G$ is generalized dicyclic.
\end{lemma}
(Notice that $A$ is automatically abelian here since
$A$ is a Dedekind group of odd order.)
\begin{proof}[Proof of Lemma~\ref{l:c2ab}]
First we show that $C:= \C_S(A)$ is abelian.
Let $z\in \NKer(G)$ have order $2$.
Then $z\in \Z(G)$ and thus $z\in C$.
If $C$ is not abelian, there is $\tau\in \Irr(C)$
with $\tau(z) \neq \tau(1) > 1$ (Lemma~\ref{l:kernelnonlin}).
Let $t\in S\setminus C$, and
let $\lambda\in \Lin(A)$ be such that
$\lambda^t \neq \lambda$.
Then $\chi:= (\tau \times \lambda)^G \in \Irr(G)$,
and $\chi(1) \geq 2\tau(1) > 2$.
By Lemma~\ref{l:sifacts}~\ref{it:localsi_div},
$\chi$ can not be skew-linear over $\mathbb{R}$
or $\mathbb{Q}_2$.
Since $\chi_C$ has a non-linear constituent,
$\chi$ can not be skew-linear over $\mathbb{Q}_q$
for odd primes $q$, by Corollary~\ref{c:localindequ}.
As $\chi(1) = 2^r$, it follows from
Lemma~\ref{l:sifacts}~\ref{it:globloc} that
$m_{\mathbb{Q}}(\chi) < \chi(1)$,
and thus
$z\notin \NKer(G)$, contradiction.
Thus $C$ is abelian as claimed,
and $G=SA$ has the abelian subgroup $CA$ of index $2$.
Fix $t\in S\setminus C$.
Notice that
$A = [A,t] \times C_A(t)$.
Since every subgroup of $A$ is normal in $G$,
the factors of this decomposition have coprime orders.
Also, we have $[A,t] \neq 1$ by assumption,
and $t$ inverts the elements in $[A,t]$.
Consider first the case $\C_A(t)\neq 1$.
Pick some $\lambda \in \Lin(A)$ such that
$\Ker(\lambda)$ contains no Sylow subgroup of $A$.
Then $\lambda^t \notin \{\lambda, \overline{\lambda}\}$.
Consider extensions $\mu$ to $CA$ with $\mu(z) =-1$,
where $z\in \NKer(G)$ has order $2$ as before.
As $\mu^t \neq \mu$,
we have $\chi = \mu^G \in \Irr(G)$.
Then $\chi$ remains irreducible modulo~$2$, and thus
$m_2(\chi)=1$, by Lemma~\ref{l:sifacts}~\ref{it:localsi_brauer}.
As $\mu^t \neq \overline{\mu}$, we have also
$m_{\mathbb{R}}(\chi)=1$.
But as $z\notin \NKer(G)$,
it follows that $m_q(\chi)=2$ for some odd prime
$q$ dividing $\card{A}$.
Then Lemma~\ref{l:abbglocal} yields
that $S$ centralizes every Sylow subgroup of
$A$ except one.
Also Corollary~\ref{c:localindequ}
yields that $\chi_S$ is a sum of linear characters.
As $\mu$ was an arbitrary extension of $\lambda$
to $CA = C\times A$ with $\mu(z)=-1$,
this means that $\nu^t = \nu$ for
all $\nu\in \Lin(C)$ with $\nu(z)=-1$.
Thus $S$ is abelian and $G$ is as in Lemma~\ref{l:abab_1}
with $p=2$ in this case.
Now assume that $\C_A(t) = 1$.
If $S$ is abelian and $C$ is not just
an elementary abelian $2$-group,
then again we find $\mu$ and $\chi$ as above,
with $m_q(\chi) = 2$ for some odd prime $q$,
and the result follows again.
If $S$ is abelian and $C$ is elementary $2$-abelian,
then $G = S [A,t]$ is generalized dicyclic.
Finally, assume that $\C_A(t) =1$
and that $S$ is nonabelian.
Then $S$ is generalized dicyclic,
and $S$ has an abelian subgroup $D$ of index $2$,
such that $d^s=d^{-1}$ for all $d\in D$ and $s\in S\setminus D$.
If $D=C$, then $G$ is generalized dicyclic.
Thus we may assume that $D\neq C$.
If $S$ is Dedekind, then $S\iso Q_8 \times (C_2)^r$,
and we could choose $D=C$.
So we can assume that $S$ is not Dedekind,
and thus $\R(S) = \erz{z}$ by Lemma~\ref{l:gendickern}.
We may choose $t\in D\setminus C$ and
$s\in C\setminus D$.
Since both $C$ and $D$ are abelian,
it follows that
$s$ centralizes $C\cap D$,
and at the same time inverts the elements in $C\cap D$.
Thus $C\cap D$ has exponent~$2$.
Since $\card{S:C}=\card{S:D}=2$ and
$z \in \erz{s}\cap\erz{t}$,
it follows $s^2 = t^2 = z$.
Since $st\notin D$, we also have $(st)^2 = z$.
It follows that $\erz{s,t}\iso Q_8$ and
$S = \erz{s,t} \times (C\cap D)
\iso Q_8 \times (C_2)^r$.
But then $S$ is Dedekind and $G$ generalized dicyclic,
contradiction.
\end{proof}
The following is part of~\cite[Theorem~2(a)]{Blackburn66}.
\begin{lemma}\label{l:blackburn2a}
Let $G =PA$ be a Blackburn group such that
$\R(G) \leq P\in \Syl_p(G)$ and such that $P$
and the normal $p$-complement $A$ are abelian.
Then we can write
$ P = \erz{g} \times P_0$, such that
$\C_P(A) = \erz{g^{p^c}} \times P_0$ ($c\geq 1$),
and $p^d := \ord( g^{p^c} )$ is the exponent
of $\C_P(A)$.
There is a $k \in \mathbb{N}$ such that
$a^g = a^k$ for all $a\in A$.
\end{lemma}
\begin{lemma}\label{l:localsi_abbg}
Let $G=PA$ be a Blackburn group,
with a normal abelian $p$-complement $A$ and
$P\in \Syl_p(G)$, where $p$ divides $\R(G)$.
Let $q$ be a prime divisor of $\card{A}$,
and $H$ a $q$-complement in $G$.
If $\chi\in \Irr(G)$,
then
$m_q(\chi) = \card{\mathbb{Q}_q(\chi,\theta) : \mathbb{Q}_q(\chi)}$
for any irreducible constituent $\theta\in \Irr(H)$
of $\chi_H$.
\end{lemma}
\begin{proof}
Let $Q\in \Syl_q(G)$.
Notice that $Q\nteq G$ and thus $Q$ has a complement
$H$ in $G$.
Let $\lambda\in \Lin(Q)$ be a constituent of $\chi_Q$.
Then $K=\Ker(\lambda)$ is normal in $G$
(by definition of $\R(G)$) and thus
$K\subseteq \Ker(\chi)$.
We may factor out $K$ and assume without loss of generality
that $K=1$.
This means that $Q$ is cyclic and thus $\chi$ is in a
$q$-block with cyclic defect group.
Thus we can apply Benard's theorem~\cite{Benard76}
to $\chi$
and conclude that
$m_q(\chi) = \card{\mathbb{Q}_q(\chi,\phi):\mathbb{Q}_q(\chi)}$
for any irreducible Brauer constituent $\phi$ of $\chi$.
But an irreducible Brauer character of $G$ contains
the normal $q$-subgroup
$Q$ in its kernel, and thus can be identified
with an ordinary character of the $q'$-group
$H \iso G/Q$.
Thus if $\phi$ is an irreducible Brauer constituent of $\chi$,
then $\phi_H=\theta\in \Irr(H)$ is an
irreducible constituent of $\chi$,
and the result follows from Benard's theorem.
\end{proof}
\begin{lemma}\label{l:qsi_allab}
Let $G =PA$ be a Blackburn group,
with a normal abelian $p$-complement $A$ and
$P\in \Syl_p(G)$, where $p$ divides $\R(G)$.
Assume that $A = Q\times B$,
where $B= \C_A(P)$ and $Q\in \Syl_q(G)$,
and set $C=\C_P(Q)$.
Any nonlinear $\chi\in \Irr(G)$ has the form
$\chi = (\mu \times \lambda)^G$ for some
$\mu\in \Lin(CB)$ and $\lambda\in \Lin(Q)$.
Let $\theta\in \Lin( PB \mid \mu)$.
Then $m_q(\chi) = \ell/k$,
where $\ell$ is the smallest positive integer
such that $\ord(\theta)$ divides $q^{\ell}-1$,
and $k$ is the smallest
positive integer such that $\ord(\mu)$ divides $q^k-1$.
\end{lemma}
(In other words, $\ell$ and $k$ are the multiplicative orders
of $q$ modulo $\ord(\theta)$ and modulo $\ord(\mu)$,
respectively.)
\begin{proof}[Proof of Lemma~\ref{l:qsi_allab}]
Notice that $CB = \Z(G)$, and that
$H = PB$ is an abelian $q$-complement.
Let $\chi\in \Irr(G)$.
If $Q\leq \Ker(\chi)$, then $\chi$ is linear,
since $G/Q = P\times B$ is abelian.
(In fact, $Q=G'$.)
Otherwise, let $\lambda \neq 1$ be a linear constituent of $\chi_Q$.
Then $G_{\lambda} = CBQ =CA$,
and $\chi$ is induced from some linear character of
the abelian group $CBQ$,
say $\chi = (\mu \times \lambda)^G$ with
$\mu \in \Lin(CB)$.
It follows that $\mathbb{Q}_q(\chi) = \crp{K}(\mu)$, where
$\crp{K}\subseteq \mathbb{Q}_q(\lambda)$ is totally ramified over
$\mathbb{Q}_q$,
and the extension $\crp{K}(\mu)/\crp{K}$ is unramified.
By the general form of unramified extensions,
the residue field of $\mathbb{Q}_q(\chi)=\crp{K}(\mu)$ has order $q^k$,
where $k$ is the smallest
positive integer such that $\ord(\mu)$ divides $q^k-1$.
The restriction $\chi_H$ to the $q$-complement $H=PB$
is the sum of all linear characters
$\theta \in \Lin(H)$ lying over $\mu$.
Thus $\mathbb{Q}_q(\chi,\theta)= \crp{K}(\theta)$
is generated by $\mathbb{Q}_q(\chi)$ and
a root of unity of order $\ord(\theta)$.
Since $\ord(\theta)$ is not divisible by $q$,
the extensions
$\mathbb{Q}_q(\chi,\theta)/\mathbb{Q}_q(\chi)$ and
$\crp{K}(\theta)/\crp{K}$ are unramified.
We can thus compute
$\card{\mathbb{Q}_q(\chi,\theta) : \mathbb{Q}_q(\chi)}$
by computing the degrees of the residue fields.
As above, the residue field of
$\mathbb{Q}_q(\chi,\theta) = \crp{K}(\theta)$ has order
$q^{\ell}$, where $\ell$ is the smallest positive integer
such that $\ord{\theta}$ divides $q^{\ell}-1$.
Now the result follows from Lemma~\ref{l:localsi_abbg}.
\end{proof}
\begin{lemma}\label{l:allab_suff}
Let $G =(PQ)\times B$ be as in
Theorem~\ref{main:qclass}~\ref{it:m_allab}.
Then $\NKer_{\mathbb{Q}_q}(G)\neq 1$.
(More precisely, $\NKer_{\mathbb{Q}_q}(G)\cap \erz{g}\neq 1$,
with $g\in P$ as in Lemma~\ref{main:pq}.)
\end{lemma}
Notice that this contains Lemma~\ref{main:pq}
from the introduction.
\begin{proof}
Recall that $P= \erz{g}\times P_0$,
where $g$ has order $p^{c+d}$
and $C:=\C_P(Q)= \erz{g^{p^c}}\times P_0$.
Moreover, we assume that $(q-1)_p$ divides $p^d$.
Let $z\in \erz{g^{p^c}}$ be an element of order $p$.
We claim that $z\in \NKer_{\mathbb{Q}_q}(G)$.
Suppose that $\chi(z)\neq\chi(1)$
for $\chi\in \Irr G$.
If $\chi(1)> 1$, then
$\chi = (\mu \times \lambda)^G$ as in
Lemma~\ref{l:qsi_allab}, with
$\mu\in \Lin(CB)$ and $1\neq \lambda\in \Lin Q$.
To compute the Schur index of such a $\chi$,
we apply Lemma~\ref{l:qsi_allab}.
Since $\mu(z)\neq 1$, it follows that
$\ord(\mu) = p^d n$, where $n$ divides the exponent
of $B$.
For $\theta \in \Lin(PB\mid \mu)$, we have
$\ord(\theta) = p^c \ord(\mu) = p^{c+d}n$.
Let $f$ be the order of $q$ modulo $p^d$,
so that $p^d$ divides $q^f -1$.
As $(q-1)_p$ divides $p^d$,
it follows that
$(q^f -1)_p = p^d$.
Thus the order of $q$ modulo $p^{c+d}$
is $p^c f$.
Let $k$ and $l$ be the multiplicative order of
$q$ modulo $p^d n$ and $p^{c+d}n$, respectively.
The assumption in \ref{it:m_allab} in Theorem~\ref{main:qclass}
yields that $k/f$ is not divisible by $p$.
Thus we must have $l/k \geq p^c$
and thus $m_q(\chi) = p^c = \chi(1)$.
This was to be shown.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{main:qclass}]
By Theorem~\ref{t:qclass_q8c2h},
Theorem~\ref{t:qclass_nilp} and Lemma~\ref{l:allab_suff},
each of the conditions in Theorem~\ref{main:qclass}
ensures that $\NKer(G)\neq 1$.
Conversely, assume that $\NKer(G)\neq 1$.
By Lemma~\ref{l:bb_pnilp},
we have that $G=PA$, where $P\in \Syl_p(G)$
and $A$ is a normal $p$-complement and a Dedekind group.
By Theorem~\ref{t:qclass_q8c2h}, we may assume that
$A$ is abelian.
By Theorem~\ref{t:qclass_nilp}, we can
assume that $G$ is not nilpotent, and thus
$\C_P(A) < P$.
By Lemmas~\ref{l:abab_1} and~\ref{l:c2ab},
we can assume that
$G = (PQ)\times B$,
and that $P$ is also abelian.
Thus Lemma~\ref{l:blackburn2a} applies.
Write $P= \erz{g} \times P_0$ and
$\C_P(A) = \erz{g^{p^c}} \times P_0$,
as in that lemma.
Let $z\in \R(G) \subseteq \erz{g^{p^c}}$ have order $p$
and notice that we must have $z\in \NKer(G)$,
since $1\neq \NKer(G)\leq \R(G)$.
Thus we must have
$m_{\mathbb{Q}}(\chi) = \chi(1)$
for any $\chi\in \Irr(G)$ with $\chi(z)\neq \chi(1)$.
Suppose $\chi = (\mu\times \lambda)^G$ as in
Lemma~\ref{l:qsi_allab}.
By the local-global principle (Lemma~\ref{l:sifacts}~\ref{it:globloc}),
the Schur index of $\chi$ over some local field
must equal $\chi(1)= \card{P:C} = p^c$.
We claim that this local field must be $\mathbb{Q}_q$ at least for
some nonlinear $\chi$.
If $p^c > 2$, then we must have
$m_q(\chi)= \chi(1)$,
by Lemma~\ref{l:abbglocal}.
In the case where $p^c = 2$,
choose $\mu\in \Lin(CB)$ of order greater than $2$.
This is possible since
when the exponent of $C \times B = \Z(G)$ divides $2$,
then $G$ is generalized dicyclic,
and the proof is finished.
We have then $m_{\mathbb{R}}(\chi) = 1$.
Also, $\chi$ is induced from a subgroup of index $2$
(which is not a $2$-group),
and thus $\chi$ remains irreducible after reducing mod~$2$.
Thus $m_2(\chi) =1$ by Lemma~\ref{l:sifacts}~\ref{it:localsi_brauer}.
Thus in every case,
we must have
$m_q(\chi) = \chi(1)$.
We can now apply Lemma~\ref{l:qsi_allab}.
Notice that since $\mu(z)\neq 1$ by assumption,
we have $\ord(\theta) = p^c \ord(\mu)$,
as can be seen from the structure of $P$
(Lemma~\ref{l:blackburn2a}),
and $\ord(\mu) = p^d \cdot n$,
where $n$ divides the exponent of $B$,
and thus is not divisible by $p$.
We may choose $\mu$ such that $n$ equals the exponent of $B$.
Let $f$ be the order of $q \mod p^d$
and let $k'$ be the order of $q^f \mod n$.
Then the order of $q\mod p^d n$ is
$k = fk'$.
If $p^d < (q-1)_p$, or if $p$ divides $k'$,
then certainly $p^d < (q^k-1)_p$.
But this yields that
$q^{kp^{c-1}} \equiv 1 \mod p^{c+d}n$,
and thus
Lemma~\ref{l:qsi_allab}
yields that $m_q(\chi) < p^c$, contradiction.
This means that \ref{it:m_allab} in Theorem~\ref{main:qclass}
holds.
\end{proof}
\section*{Acknowledgment}
I wish to thank Erik Friese for a thorough reading of this paper
and many useful remarks.
\printbibliography
\end{document} | {
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} | 2,523 |
Slopestyle mężczyzn na Zimowej Uniwersjadzie 2023 odbył się w dniach 17–18 stycznia w Gore Mountain.
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Q: Find area of surface enclosed with given curve Find the area of surface with given equation $$\big(x^2 + y^2\big)^2 = a(x^3+y^3)$$
I tried to use polar coordinates
$x=r\cos\alpha$ and $y=r\sin\alpha$
and so $$r = a(\cos\alpha + \sin\alpha)\bigg(1-{\sin2\alpha\over{2}}\bigg)$$
Maybe anyone can suggest better way?
A: Okay i solved, with $\alpha \in [-\pi/4, 3\pi/4]$
and $${a^2\over2}\int_{-\pi\over4}^{3\pi\over4}(\sin\alpha + \cos\alpha)^2 (1-{\sin2\alpha \over 2})^2 d\alpha = \bigg(\dfrac{\cos\left(6x\right)+9\sin\left(4x\right)-9\cos\left(2x\right)-48\sin^4\left(x\right)+48\cos^4\left(x\right)-96\cos^2\left(x\right)+60x}{96}\bigg)_{-\pi\over4}^{3\pi\over4} = {5\pi a^2\over 16}$$
A: Why not directly the double integral (in polar coordinates)?
$$\int_0^{2\pi}\int_0^{a(\cos t + \sin t)\bigg(1-{\sin2t\over{2}}\bigg)}r\,dr\,dt=\frac{a^2}2\int_0^{2\pi}(\cos t+\sin t)^2\left(1-\frac{\sin2t}2\right)^2dt=$$
$$=\frac{a^2}2\int_0^{2\pi}\left(1+\sin2t\right)\left(1-\sin2t+\frac{\sin^22t}4\right)dt=\frac{a^2}2\int_0^{2\pi}\left[1-\frac34\sin^22t+\frac{\sin^32t}4\right]dt=$$
$$=\frac{a^2}4\int_0^{4\pi}\left[1-\frac34\sin^2u+\frac{\sin^3u}4\right]du=$$
$$=\frac{a^2}4\left(4\pi-\frac384\pi+\frac14\overbrace{\int_0^{4\pi}\left(\sin u-\sin u\cos^2u\right)du}^{=0}\right)=\frac{5\pi a^2}8$$
Seeing your solution, at least one of us two is wrong. I wouldn't be surprised at all if it is me...yet in your expression you didn't divide by two $\;\sin2\alpha\;$ ... Check this.
Added. Since it must be $\;t\in\left[-\frac\pi4,\,\frac{3\pi}4\right]\;$ for $\;r\;$ to be positive, the above integral must be changed accordingly to these limits. The outcome indeed is $\;\cfrac{5\pi a^2}{16}\;$
| {
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} | 6,489 |
\section{Introduction}
Low Energy observables sensitive to CP Violation in $b \to s$ transitions constitute excellent probes of possible new CP violating phases in extensions of the Standard Model (SM). Indeed, as CP Violation in such processes is predicted to be tiny in the SM, evidence for sizable CP violation in $b \to s$ transitions would be a clear hint for the presence of New Physics (NP). Examples of such low energy probes are observables that are sensitive to the $B_s$ mixing phase, as the semi-leptonic asymmetry $a_{\rm sl}^s$ in decays of $B_s$ mesons to ``wrong sign leptons'' or the time dependent CP asymmetries in the $B_s \to \psi\phi$ and $B_s \to \psi f_0$ decays.
In the context of generic two Higgs doublet models with Minimal Flavor Violation (MFV), where the CKM matrix is the only source of flavor violation, a large $B_s$ mixing phase can be realized if additional CP violating phases are allowed~\cite{Buras:2010mh,Buras:2010zm,Trott:2010iz,Pich:2009sp}. Simultaneously, these models can also address tensions in fits of the CKM matrix that seem to indicate a sizable NP contribution to the $B_d$ mixing phase~\cite{Lunghi:2008aa,Lenz:2010gu}.
Possible relations between a non-standard $B_s$ mixing phase and the baryon asymmetry of the universe in these models have been studied in~\cite{Cline:2011mm,Liu:2011jh}.
In the context of the Minimal Supersymmetric Standard Model (MSSM) on the other hand, a MFV soft sector is not sufficient to generate sizable NP phases in meson mixing~\cite{Altmannshofer:2008hc,Altmannshofer:2009ne} due to the strong experimental bound on the $B_s \to \mu^+ \mu^-$ branching ratio.
Non SM-like $B_s$ and $B_d$ mixing phases in the MSSM require new sources of flavor violation in addition to the CKM matrix (see e.g.~\cite{Altmannshofer:2009ne,Ko:2008xb} for studies of such frameworks).
Supersymmetric models with MFV {\it can} generate large $B$ mixing phases {\it if} they allow for a strongly reduced muon Yukawa coupling such that the BR$(B_s \to \mu^+ \mu^-)$ constraint can be avoided. Such a situation can be realized for example in the so-called uplifted SUSY Higgs region~\cite{Dobrescu:2010mk}, even though this framework is strongly constrained by other $B$ physics observables and $(g-2)_\mu$~\cite{Altmannshofer:2010zt}.
As studied in detail in~\cite{Altmannshofer:2011rm}, non-negligible corrections to CP violating observables in meson mixing can also be generated if the MSSM with MFV is extended by the two leading higher dimensional operators in the Higgs sector with complex coefficients~\cite{Dine:2007xi}. Still, the stringent bounds on BR$(B_s \to \mu^+ \mu^-)$ only allow for a $B_s$ mixing phase of $S_{\psi\phi} \lesssim 0.15$ in specific regions of parameter space of this model.
In this work we analyze an extension of the MSSM, introducing higher dimensional operators not exclusively in the Higgs sector, but also considering dimension 5 operators that induce non-holomorphic Higgs-fermion couplings already at the tree level~\cite{Antoniadis:2008es}. Such operators are a possible source of flavor and CP violation.
Assuming that these operators obey the Minimal Flavor Violation ansatz~\cite{Chivukula:1987py,Buras:2000dm,D'Ambrosio:2002ex}, we explore to which extent the considered framework allows for non-standard CP violation in $B$ mixing without being in conflict with the bounds on the BR$(B_s \to \mu^+\mu^-)$.
For a study of a similar framework, that however does not consider CP violation in $B$ mixing, see~\cite{Bae:2010ai}.
\section{\boldmath CP Violation in \texorpdfstring{$B$}{B} Meson Mixing}
The $B_q$ mixing amplitude
\begin{equation}
\mathcal{M}^q = M^q_{12} - \frac{i}{2}\Gamma^q_{12}~,
\end{equation}
consists on an dispersive part, $M_{12}^q$, and an absorptive part, $\Gamma_{12}^q$.
The absorptive part is dominated by tree level SM contributions and therefore hardly affected in many NP models. Throughout this work we will assume that $\Gamma_{12}^q$ has no significant NP contributions. The dispersive part $M_{12}^q$ on the other hand is highly sensitive to new heavy degrees of freedom. The effects of NP in $M_{12}^q$ can be parametrized by
\begin{equation}
M_{12}^q = C_q e^{i \phi_q^{\rm NP}} (M_{12}^q)_{\rm SM} ~.
\end{equation}
The main impact of the parameters $C_d$ and $C_s$ is on the mass differences $\Delta M_d$ and $\Delta M_s$, respectively.
The NP phases $\phi_d^{\rm NP}$ and $\phi_s^{\rm NP}$ affect observables that are sensitive to CP violation in B meson mixing, like the semileptonic asymmetries $a_{\rm SL}^d$ and $a_{\rm SL}^s$ as well as the time-dependent CP asymmetries in $B_d \to \psi K_S$, $B_s \to \psi\phi$ and $B_s \to \psi f_0$.
In order to constrain the NP parameters through measurements of these observables, knowledge of the respective SM contributions is required. As several of the observables, in particular $\Delta M_d$ and $S_{\psi K_S}$ -- the time dependent CP asymmetry in $B_d \to \psi K_S$ -- are a key ingredient in the determination of the Unitarity Triangle (UT), a simultaneous fit of the CKM parameters and the NP parameters as performed in~\cite{Lenz:2010gu} is the most consistent approach. Here we focus mainly on the impact of the recent improvements on the determination of the $B_s$ mixing phase at CDF and D0~\cite{Aaltonen:2007he,CDFD0update} and in particular at LHCb~\cite{LHCb}.
As these measurements have no significant effect on the determination of the CKM parameters and the NP parameters other than $\phi_s^{\rm NP}$, we consider a simplified approach and take the CKM parameters as well as $C_d$ and $C_s$ from the generic NP fit in~\cite{Lenz:2010gu}, and fit only the NP parameters $\phi_s^{\rm NP}$ and $\phi_d^{\rm NP}$. We expect this approach to give a good estimate of the allowed region of parameter space compatible with the present experimental data on B meson mixing. In particular we will use
\begin{eqnarray} \label{eq:beta}
\beta &=& \textnormal{Arg}\left[ (V_{tb} V_{td}^*) / (V_{cb} V_{cd}^*)\right] = (27.2^{+1.1}_{-3.1})^\circ \\
\beta_s &=& \textnormal{Arg}\left[ (V_{tb} V_{ts}^*)/(V_{cb} V_{cs}^*)\right] = (-1.3\pm 0.1)^\circ \nonumber
\end{eqnarray}
and the following $2\sigma$ bounds
\begin{equation}
0.62 < C_d < 1.15 ~~,~~~ 0.79 < C_s < 1.23~.
\end{equation}
We now give expressions for the observables that are sensitive to CP violation in B meson mixing.
For the semileptonic asymmetries we obtain~\cite{Lenz:2006hd,Lenz:2010gu}
\begin{eqnarray} \label{eq:asld}
10^4\,a_{\rm SL}^d &\simeq& (55.2 \sin\phi_d^{\rm NP} - 4.80 \cos\phi_d^{\rm NP})/C_d ~~ \\ \label{eq:asls}
10^4\,a_{\rm SL}^s &\simeq& (49.7 \sin\phi_s^{\rm NP} + 0.19 \cos\phi_s^{\rm NP})/C_s ~~
\end{eqnarray}
where the uncertainties on the numerical coefficients are at the level of $\sim 15\%$.
For the time dependent CP asymmetries one has
\begin{eqnarray} \label{eq:SpsiKs}
S_{\psi K_S} &=& \sin(2\beta + \phi_d^{\rm NP}) ~, \\
S_{\psi \phi} = S_{\psi f_0} &=& \sin(2|\beta_s| - \phi_s^{\rm NP}) ~.
\end{eqnarray}
These expressions hold under the usual assumption that the $B_d \to \psi K_S$, $B_s \to \psi\phi$ and $B_s \to \psi f_0$ decays are dominated by the SM tree level amplitudes.
Using~(\ref{eq:beta}) the corresponding SM predictions are
\begin{eqnarray}
S_{\psi K_S}^{\rm SM} = 0.82^{+0.02}_{-0.07} ~,~ S_{\psi \phi}^{\rm SM} = 0.046^{+0.002}_{-0.003} ~.
\end{eqnarray}
The D0 collaboration measured the like-sign dimuon charge asymmetry that is predicted to be composed out of the semi-leptonic asymmetries in the $B_d$ and $B_s$ decays~\cite{Abazov:2011yk}
\begin{eqnarray}
A_{\rm SL}^b &=& 0.59 a_{\rm SL}^d + 0.41 a_{\rm SL}^s \nonumber \\ \label{eq:aslb_D0}
&=& (-78.7 \pm 19.6)10^{-4} ~.
\end{eqnarray}
The corresponding SM prediction $A_{\rm SL}^b(\rm SM) = (-2.8^{+0.5}_{-0.6})10^{-4}$~\cite{Lenz:2010gu} is roughly a factor 25 below the central value in~(\ref{eq:aslb_D0}) and differs from it by $3.9\sigma$. The value~(\ref{eq:aslb_D0}) updates an earlier D0 study~\cite{Abazov:2010hv} that found a $3.2\sigma$ evidence for an anomalous like-sign dimuon charge asymmetry.
A separate extraction of the semileptonic asymmetries results in~\cite{Abazov:2011yk}
\begin{eqnarray} \label{eq:asld_D0}
a_{\rm SL}^d(D0) &=& (-12 \pm 52)10^{-4} ~,\\ \label{eq:asls_D0}
a_{\rm SL}^s(D0) &=& (-181 \pm 106)10^{-4} ~.
\end{eqnarray}
The results in Eqs.~(\ref{eq:aslb_D0}),~(\ref{eq:asld_D0}) and~(\ref{eq:asls_D0}) hint towards large negative values for the NP phases $\phi_d^{\rm NP}$ and, in particular, $\phi_s^{\rm NP}$.
Interestingly enough, there is a (2-3)$\sigma$ tension between the SM prediction of $S_{\psi K_S}$~(\ref{eq:SpsiKs}) and its experimental value~\cite{Asner:2010qj}
\begin{equation}
S_{\psi K_S}^{\rm exp} = 0.67 \pm 0.02 ~,
\end{equation}
that is largely driven by the $B \to \tau \nu$ measurements that prefer a large value of $|V_{ub}|$.
This tension points in the same direction for the NP phase $\phi_d^{\rm NP}$ as the data on the like sign dimuon charge asymmetry.
A small preference for a negative NP phase in $B_s$ mixing was also observed in CDF and D0 data on the time dependent CP asymmetry in $B_s \to \psi\phi$ that give @ 95\%~C.L. ~\cite{Aaltonen:2007he,CDFD0update}
\begin{eqnarray}
-1.36 < &\phi_s^{\rm NP} - 2|\beta_s|& < 0.26 ~~{\rm (CDF)}~, \\
-1.65 < &\phi_s^{\rm NP} - 2|\beta_s|& < 0.24 ~~{\rm (D0)}~.
\end{eqnarray}
Combining the results from~\cite{Aaltonen:2007he} with~\cite{Abazov:2010hv}, global fits to the data found a $B_s$ mixing phase $\phi_s^{\rm NP} = O(-1)$~\cite{Ligeti:2010ia,Lenz:2010gu}.
\begin{figure}[t] \centering
\includegraphics[width=0.47\textwidth]{phid_phis.png}
\caption{
Allowed ranges for the NP phases $\phi_s^{\rm NP}$ and $\phi_d^{\rm NP}$ at the 1 and 2 $\sigma$ level, taking into account the measurements of the time-dependent CP asymmetries in $B_s \to \psi\phi$ at CDF, D0 and LHCb~\cite{Aaltonen:2007he,LHCb}, in $B_s \to \psi f_0$ at LHCb~\cite{LHCb} and in $B_d \to \psi K_S$ at the B factories~\cite{Asner:2010qj}. The measurement of the like-sign dimuon charge asymmetry at D0~\cite{Abazov:2011yk} is included in the black dotted contours but {\it not} in the red solid contours.
}
\label{fig:phi_fit}
\end{figure}
Recently however, LHCb presented results on the time dependent CP asymmetries in $B_s \to \psi\phi$ and $B_s \to \psi f_0$ that are consistent with the tiny SM prediction and that strongly restrict the possible values for a NP phase in $B_s$ mixing~\cite{LHCb}
\begin{equation}
\phi_s^{\rm NP} - 2|\beta_s| = 0.03 \pm 0.16 \pm 0.07 ~.
\end{equation}
In Fig.~\ref{fig:phi_fit} we show the result of a simple fit of the NP phases $\phi_d^{\rm NP}$ and $\phi_s^{\rm NP}$ to the combined LHCb result on the time-dependent CP asymmetries in $B_s \to \psi\phi$ and $B_s \to \psi f_0$~\cite{LHCb}, the results on the time-dependent CP asymmetry in $B_s \to \psi\phi$ from CDF and D0~\cite{Aaltonen:2007he,CDFD0update} as well as the measurement of $S_{\psi K_S}$ at the B factories~\cite{Asner:2010qj}, using the values for $\beta$ and $\beta_s$, $C_d$ and $C_s$ from above.
The allowed region is mainly determined by the measurements of $S_{\psi\phi}$, $S_{\psi f_0}$ at LHCb and $S_{\psi K_S}$ at the B factories, while the measurements of $S_{\psi\phi}$ at CDF and D0 lead to a small shift of the central value of $\phi_s^{\rm NP}$ towards a small negative value. We stress that the very large value of the like-sign dimuon charge asymmetry observed by D0 cannot be explained given the current data on the time dependent CP asymmetries in $B_s \to \psi\phi$ and $B_s \to \psi f_0$. Using our fit results we find a central value of $A_{\rm SL}^b = -11 \cdot 10^{-4}$ and a $2\sigma$ range of $-18 < 10^4\, A_{\rm SL}^b < -2$. This differs from the measured value~(\ref{eq:aslb_D0}) by $\simeq 3 \sigma$.
Including the like-sign dimuon charge asymmetry directly into the fit leads only to a small shift towards slightly larger negative $\phi_s^{\rm NP}$ values. In the following we focus on the fit that does not include the $A_{\rm sl}^b$ measurement.
Due to the small discrepancy between the experimental determination of $S_{\psi K_S}$ and its SM prediction coming from the UT fits~\cite{Lunghi:2008aa,Lenz:2010gu}, the NP phase in $B_d$ mixing shows preference towards a negative value
\begin{equation} \label{eq:phi_d}
\phi_d^{\rm NP} = -0.20^{+0.10}_{-0.04}~,
\end{equation}
that is roughly 2$\sigma$ below 0.
While also for the NP phase in $B_s$ mixing we find a slight preference for a small negative value, $\phi_s^{\rm NP}$ is perfectly consistent with zero
\begin{equation} \label{eq:phi_s}
\phi_s^{\rm NP} = -0.10 \pm 0.15 ~.
\end{equation}
Presently, this still leaves room for NP contributions, but the bound~(\ref{eq:phi_s}) will improve significantly in the near future with more data from LHCb.
\section{\boldmath The \texorpdfstring{$B_q \to \mu^+\mu^-$}{Bq --> mu+mu-} Decays}
The CDF collaboration reported a small excess in $B_s \to \mu^+\mu^-$ candidates~\cite{Aaltonen:2011fi}, leading to
\begin{equation} \label{eq:Bsmm_CDF}
{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm CDF} = (1.8^{+1.1}_{-0.9}) \times 10^{-8}~,
\end{equation}
No excess has been observed by LHCb and CMS that report the following bounds~\cite{Chatrchyan:2011kr}
\begin{eqnarray} \label{eq:Bsmm_LHCb}
{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm LHCb} & < & 1.5 \cdot 10^{-8} ~, \\ \label{eq:Bsmm_CMS}
{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm CMS~} & < & 1.9 \cdot 10^{-8} ~.
\end{eqnarray}
Combining the bounds from LHCb and CMS one finds~\cite{LHCbCMS}
\begin{equation} \label{eq:Bsmm_LHC}
{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm LHC} < 1.1 \times 10^{-8}
\end{equation}
that is only a factor of 3.5 above the SM prediction~\cite{Buras:2003td}
\begin{equation}
{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm SM} = (3.2 \pm 0.2) \times 10^{-9}~.
\end{equation}
The current bounds on the $B_d \to \mu^+ \mu^-$ branching ratio~\cite{Chatrchyan:2011kr} are still a factor 40-50 above the SM expectation and therefore $B_d \to \mu^+ \mu^-$ is much less constraining than $B_s \to \mu^+ \mu^-$ in models with MFV.
Given the strong bound on BR$(B_s \to \mu^+ \mu^-)$, possible neutral Higgs contributions to B mixing in the MSSM with MFV are strongly constrained~\cite{Buras:2002vd,Carena:2006ai}. Also in the BMSSM model considered in~\cite{Altmannshofer:2011rm} non-standard B mixing phases are rather restricted ($S_{\psi\phi} \lesssim 0.15$) and can only be generated in particular corners of parameter space.
In the following we present an extension of the MSSM that respects the MFV principle but allows nonetheless for sizable NP phases in B mixing without being in conflict with the BR$(B_s \to \mu^+ \mu^-)$ constraint.
\section{Effective Higgs - Fermion Couplings Beyond the MSSM}
At the tree level, the MSSM is a 2 Higgs doublet model of type II and the couplings of the neutral Higgs bosons to fermions are flavor conserving. At the loop level on the other hand, non-holomorphic Higgs couplings are generated and have important consequences. Loop induced couplings of down-type quarks and charged leptons to the up-type Higgs can lead to large threshold corrections to the corresponding masses~\cite{Hempfling:1993kv} and modify significantly CKM matrix elements~\cite{Blazek:1995nv} as well as charged Higgs couplings to quarks~\cite{Carena:1999py}. Finally, they also generate flavor changing neutral Higgs couplings that can have a profound impact on flavor phenomenology~\cite{Hamzaoui:1998nu,Buras:2002vd,Carena:2006ai}.
All these effects become relevant for large values of $\tan\beta$ that can compensate for the 1-loop suppression.
We now consider possible extensions of the MSSM with new degrees of freedom at a scale $M \simeq $~several~TeV. As long as the SUSY breaking scale $m_S$ of the new degrees of freedom is small compared to M and as long as M is sufficiently larger than the scale of the MSSM degrees of freedom, one can describe the effects of the Beyond the MSSM (BMSSM) physics by higher dimensional operators suppressed by $1/M$~\cite{Brignole:2003cm,Dine:2007xi}.
An analysis up to order $1/M$ captures the physics of several MSSM UV extensions while the effective description of others need to include $1/M^2$ effects~\cite{Carena:2009gx,Antoniadis:2009rn}. In this work we restrict ourselves to the $1/M$ level.
We consider both the leading higher dimensional superpotential operators that involve only Higgs fields~\cite{Dine:2007xi}
\begin{equation} \label{eq:super}
\mathcal{L} \supset \frac{\omega}{2 M} \int d^2\theta (1 + \alpha Z) (H_u H_d)^2
\end{equation}
and in particular also $1/M$ suppressed K\"ahler potential operators that induce non-holomorphic Higgs-fermion couplings already at the tree level~\cite{Dine:2007xi,Antoniadis:2008es}
\footnote{Possible UV completions that lead to the operators in~(\ref{eq:super}) and~(\ref{eq:kahler}) are discussed e.g. in~\cite{Dine:2007xi,Antoniadis:2008es,Carena:2009gx,Bae:2010ai} and briefly presented in the Appendix.}
\begin{eqnarray} \label{eq:kahler}
\mathcal{L} && \supset \frac{1}{M} \int d^4\theta \left( 1 + Z + Z^\dagger + Z Z^\dagger \right) \\
\times && \left( \lambda_u H_d^\dagger Q U + \lambda_d H_u^\dagger Q D + \lambda_\ell H_u^\dagger L E\right) + {\rm h.c.} ~. \nonumber
\end{eqnarray}
In the above expressions, $Z$ is an auxiliary dimensionless spurion that develops a SUSY breaking F-term
\begin{equation}
Z \to m_S \theta^2~.
\end{equation}
The phenomenological consequences of the operators in~(\ref{eq:super}) have been thoroughly studied in the literature~\cite{Dine:2007xi,Carena:2009gx,Antoniadis:2007xc,Blum:2010by,Altmannshofer:2011rm}. In particular, they can significantly enhance the tree level mass of the lightest Higgs boson of the MSSM and lead to sizable mass splittings between the two heavy neutral Higgs bosons and also the charged Higgs boson. Possible phases of the coefficients $\alpha$ and $\omega$ lead also to scalar-pseudoscalar mixing. Their impact in the context of Higgs and flavor phenomenology has been analyzed in~\cite{Altmannshofer:2011rm}.
The K\"ahler potential operators~(\ref{eq:kahler}) modify the interactions of Higgs bosons, (s)quarks and (s)leptons of the MSSM at the $1/M$ level~\cite{Dine:2007xi,Antoniadis:2008es,Bae:2010ai}.
The supersymmetric part of~(\ref{eq:kahler}) for example leads to corrections of the holomorphic MSSM Yukawa couplings
\begin{equation}
Y_f \to Y_f^\prime = Y_f + \frac{\mu}{M} \lambda_f ~.
\end{equation}
After SUSY breaking also non-holomorphic Higgs-quark couplings are generated
\begin{eqnarray}
\mathcal{L} &\supset& \frac{m_S}{M} (\lambda_u)_{ij} ~H_d^\dagger \bar Q_i U_j + \frac{m_S}{M} (\lambda_d)_{ij} ~H_u^\dagger \bar Q_i D_j \nonumber \\
&& + \frac{m_S}{M} (\lambda_\ell)_{ij} ~H_u^\dagger \bar L_i E_j ~+ {\rm h.c.}~,
\end{eqnarray}
where we now made flavor indices explicit. These terms can lead to flavor changing neutral Higgs vertices and correspondingly to tree level contributions to FCNC processes like $B$ mixing and $B_s \to \mu^+\mu^-$. In the following we will focus on them.
The MFV hypothesis as formulated in~\cite{D'Ambrosio:2002ex} amounts to the assumption that the $SU(3)^3$ quark flavor symmetry of the gauge sector is broken by only two spurions $\mathcal{Y}_u$ and $\mathcal{Y}_d$ that transform as $(3,\bar 3,1)$ and $(3,1,\bar 3)$ respectively. Correspondingly, the couplings $Y_q$ and $\lambda_q$ can be expanded in powers of these spurions. To keep notation simple and concise, we conveniently choose $\mathcal{Y}_q = Y_q^\prime = Y_q + \frac{\mu}{M} \lambda_q$ to be these spurions. Any other linear combination of $Y_q$ and $\lambda_q$ leads to equivalent results. For the non-holomorphic Higgs couplings $\lambda_q$ to the down quarks one then has
\begin{eqnarray} \label{eq:non_holomorphic}
\lambda_d &=& \varepsilon_0 Y_d + \varepsilon_1 Y_d Y_d^\dagger Y_d + \varepsilon_2 Y_u Y_u^\dagger Y_d + \\
&+& \varepsilon_3 Y_u Y_u^\dagger Y_d Y_d^\dagger Y_d + \varepsilon_4 Y_d Y_d^\dagger Y_u Y_u^\dagger Y_d + \dots ~, \nonumber
\end{eqnarray}
where for simplicity we dropped the prime on the corrected Yukawa couplings.
An analogous expression holds for the up quark coupling which is however not relevant for the following discussion.
The coefficients $\varepsilon_i$ are generically of O(1) and complex.
For later convenience we define
\begin{eqnarray}
\bar\varepsilon_0 &=& \frac{m_S}{M} \varepsilon_0 ~~, \nonumber \\
\bar\varepsilon_1 &=& \frac{m_S}{M} y_b^2 \varepsilon_1 ~~,~~~ \bar\varepsilon_3 = \frac{m_S}{M} y_t^2 y_b^2 \varepsilon_3 ~,\nonumber \\
\bar\varepsilon_2 &=& \frac{m_S}{M} y_t^2 \varepsilon_2 ~~,~~~ \bar\varepsilon_4 = \frac{m_S}{M} y_t^2 y_b^2 \varepsilon_4 ~,\nonumber
\end{eqnarray}
\begin{eqnarray}
\bar\varepsilon_5 &=& \bar\varepsilon_0 + \bar\varepsilon_1 + \bar\varepsilon_2 + \bar\varepsilon_3 + \bar\varepsilon_4~, \nonumber \\
\bar\varepsilon_6 &=& \bar\varepsilon_0 + \bar\varepsilon_1 + \bar\varepsilon_4 ~.
\end{eqnarray}
The couplings in~(\ref{eq:non_holomorphic}) modify the relation between the down quark masses $m_q$ and the corresponding Yukawa couplings $y_q$
\begin{eqnarray} \label{eq:masses}
\frac{y_d v}{m_d} = \frac{y_s v}{m_s} &=& \frac{\tan\beta}{1+\bar\varepsilon_0 \tan\beta} \nonumber \\
\frac{y_b v}{m_b} &=& \frac{\tan\beta}{1+ \bar\varepsilon_5 \tan\beta}~,
\end{eqnarray}
with the Higgs vev $v = 174$~GeV and we only show the leading term in a $\tan\beta$ expansion.
Similar to the quark masses, also the CKM matrix receives $\tan\beta$ enhanced corrections. The relations between the affected elements of the bare CKM matrix $V^0$ and the physical CKM matrix $V$ read ($i = 1,2$)
\begin{equation} \label{eq:CKM}
\frac{V_{ti}^0}{V_{ti}} = \left(\frac{V_{ib}^0}{V_{ib}}\right)^* = \frac{1 + \bar\varepsilon_5^* \tan\beta}{1 + \bar\varepsilon_6^* \tan\beta} ~.
\end{equation}
Finally we also give explicit expressions for the corrected flavor changing couplings of right handed down quarks with the Higgs bosons. The leading $\tan\beta$ enhanced terms read ($i \neq j$)
\begin{widetext}
\begin{equation} \label{eq:FCNC_Lagrangian}
\mathcal{L} \supset \bar d_L^i \frac{m_{d_j}}{v} V_{ti}^*V_{tj} ~X_{ij}~ d_R^j \left( c_\alpha H - s_\alpha h + i A\right) ~+~ \bar u_L^i \frac{m_{d_j}}{v} V_{ij} ~Z_{ij}~ d_R^j H^+ ~+ \textnormal{h.c.}
\end{equation}
\begin{eqnarray} \label{eq:Higgs_couplings}
Z_{ib} &=& \frac{t_\beta}{1+\bar\varepsilon_6 t_\beta} ~,~~~~~~~~ X_{ib} = - \frac{(\bar\varepsilon_2 + \bar\varepsilon_3)~t^2_\beta}{(1 + \bar\varepsilon_5 t_\beta)(1 + \bar\varepsilon_6 t_\beta)}~, \\
X_{bi} &=& - \frac{(\bar\varepsilon_2 + \bar\varepsilon_4)~t^2_\beta}{(1 + \bar\varepsilon_5 t_\beta)(1 + \bar\varepsilon_6 t_\beta)}~ \left[ \frac{1+\bar\varepsilon_6 t_\beta}{1+\bar\varepsilon_0 t_\beta} - \frac{1+\bar\varepsilon_6 t_\beta}{1+\bar\varepsilon_6^* t_\beta}~ \frac{\bar\varepsilon_2^* + \bar\varepsilon_3^*}{\bar\varepsilon_2 + \bar\varepsilon_4}~ \frac{(\bar\varepsilon_1 + \bar\varepsilon_3)t_\beta}{1+\bar\varepsilon_0 t_\beta} \right]~. \label{eq:Xbi}
\end{eqnarray}
\end{widetext}
The flavor changing $b_R \to d_L^i$ couplings $X_{ib}$ are generated by the $\varepsilon_2$ and $\varepsilon_3$ terms and they are proportional to $m_b$. The flavor changing $b_L \to d_R^i$ couplings $X_{bi}$ on the other hand are generated by the $\varepsilon_2$ and $\varepsilon_4$ terms and suppressed by light quark masses.
The expression~(\ref{eq:Xbi}) generalizes the results given in~\cite{D'Ambrosio:2002ex,Carena:2006ai,Hofer:2009xb} and, to the best of our knowledge, has not been presented in the literature.
In the MSSM, the $\bar\epsilon$ factors can only be loop induced. Gluino-down squark loops generate for example $\bar\epsilon_0$, while $\bar\epsilon_2$ is generated by chargino-stop loops. Due to the loop suppression, the $\bar\epsilon$ factors are generically of O(0.01) in the MSSM. Correspondingly, the corrections in~(\ref{eq:masses}),~(\ref{eq:CKM}),~(\ref{eq:Higgs_couplings}) and~(\ref{eq:Xbi}) become important only for large values of the ratio of the two Higgs vevs $v_u/v_d = \tan\beta = t_\beta$. On the other hand, in generic 2 Higgs doublet models with MFV as analyzed in~\cite{Buras:2010mh,Buras:2010zm}, where the $\bar\epsilon$ are free parameters, moderate values of $\tan\beta \simeq 5 - 10$ are sufficient to generate O(1) effects. The same is true in the supersymmetric framework considered here as long as the BMSSM scale that controls the size of the $\bar\epsilon$ factors is not too high, i.e. $m_S / M \sim 0.1$.
\section{Higgs Effects in Flavor Physics} \label{sec:flavor}
Observables that are highly sensitive to a non-standard Higgs sector are observables in meson mixing as well as the branching ratios of the rare decays $B_{s,d} \to \mu^+\mu^-$ that receive tree level contributions from flavor changing neutral Higgs exchange.
Charged Higgs effects are relevant in the $B \to X_s \gamma$ decay as well as in the $B \to \tau \nu$, $B \to D \tau \nu$ and $K \to \mu \nu$ decays. The first one is modified only at the loop level while the others receive contributions from tree level charged Higgs exchange. However these tree level decays turn out to give only mild constraints in regions of parameter space with non-standard B meson mixing phases and we do not discuss them in detail here although they are included in our numerical analysis. For a recent study of $B \to \tau \nu$ in the context of multi Higgs doublet models with MFV see~\cite{Blankenburg:2011ca}.
\subsection{\boldmath \texorpdfstring{$B_s$ - $\bar B_s$}{Bs - Bsbar} Mixing}
The flavor changing neutral Higgs couplings in~(\ref{eq:FCNC_Lagrangian}) give rise to tree level contributions to $B_s$ mixing mediated by neutral Higgs exchange.
These contributions can be described by the following effective Hamiltonian
\begin{equation}
\mathcal{H}_\textnormal{eff} = \tilde C_2 (\bar b_R s_L)^2 + C_4 (\bar b_R s_L)(\bar b_L s_R) ~,
\end{equation}
with the Wilson coefficients
\begin{eqnarray}\label{eq:Wilson2}
\tilde C_2 &\simeq& -\frac{1}{2} ~X_{sb}^2 ~(V_{tb}V_{ts}^*)^2~ \frac{m_b^2}{M_A^2} ~ \frac{\alpha \omega m_S}{M} \frac{1}{M_A^2} ~, \\ \label{eq:Wilson4}
C_4 &\simeq& X_{sb} X_{bs}^* ~(V_{tb}V_{ts}^*)^2~ \frac{m_b m_s}{v^2} ~\frac{2}{M_A^2} ~.
\end{eqnarray}
Analogous contributions to $B_d$ mixing can be obtained through the replacements $s \to d$.\footnote{Neutral Higgs contributions to Kaon mixing are proportional to $m_s^2$ or $m_s m_d$ in the considered framework and therefore negligibly small.}
In writing~(\ref{eq:Wilson2}) and~(\ref{eq:Wilson4}), we assume the decoupling limit $M_A \gg M_Z$ and treat the effect of the operators in~(\ref{eq:super}) in a mass insertion approximation. In our numerical analysis instead we work with Higgs mass eigenstates that we derive from the full Higgs potential including MSSM 2-loop corrections~\cite{Carena:1995bx}.
The Wilson coefficient $C_4$ is proportional to $m_b m_q$. Consequently it can only lead to sizable effects in $B_s$ mixing while its impact on $B_d$ mixing is rather restricted. On the other hand, $\tilde C_2$ is proportional to $m_b^2$ and therefore leads to NP contributions of the same size and phase both in $B_s$ and $B_d$ mixing.
We stress that $C_4$ is complex only if higher orders of the bottom Yukawa are considered in the expansion~(\ref{eq:non_holomorphic}) (see also~\cite{Kagan:2009bn}). Indeed, one easily checks that switching off $\bar\varepsilon_1$, $\bar\varepsilon_3$ and $\bar\varepsilon_4$ leads to a real $C_4$.
The Wilson coefficient $\tilde C_2$ however is also highly sensitive to the phases of $\bar\varepsilon_0$, $\bar\varepsilon_2$, $\omega$ and $\alpha$.
From~(\ref{eq:Wilson2}) it is clear that $\tilde C_2$ is only relevant in presence of the higher dimensional operators~(\ref{eq:super}) in the superpotential and for small Higgs masses not far above the electroweak scale.
\subsection{\boldmath The \texorpdfstring{$B_s \to \mu^+\mu^-$}{Bs --> mu+mu-} Decay}
The $B_s \to \mu^+ \mu^-$ decay receives tree level contributions from flavor changing neutral Higgs exchange.
One finds
\begin{eqnarray}
R_{B_s\mu\mu} &=& \frac{{\rm BR}(B_s \to \mu^+ \mu^-)_{\phantom{\rm SM}}}{{\rm BR}(B_s \to \mu^+ \mu^-)_{\rm SM}} \nonumber \\
&\simeq& \left|A\right|^2 + \left|1- A\right|^2 ~,
\end{eqnarray}
\begin{equation} \label{eq:A}
A \simeq - \left(\frac{4\pi}{\alpha_2}\right) \frac{X_{sb}}{4 Y_0(x_t)} \frac{t_\beta}{(1+\bar\epsilon_\ell^* t_\beta)} \frac{m_{B_s}^2}{M_A^2} ~,
\end{equation}
with the SM loop function given by $Y_0(x_t) \simeq 0.96$.
The above expression assumes again the decoupling limit $M_A \gg M_Z$. For small Higgs masses, corrections at the $1/M$ level become important and are included in our numerical analysis.
In writing~(\ref{eq:A}) we also assume that the non-holomorphic lepton-Higgs coupling $\lambda_\ell$ is proportional to the lepton Yukawa
\begin{equation}
\frac{m_S}{M} \lambda_\ell = \frac{m_S}{M} \varepsilon_\ell Y_\ell = \bar\varepsilon_\ell Y_\ell ~.
\end{equation}
As the NP contribution~(\ref{eq:A}) to the $B_s \to \mu^+ \mu^-$ amplitude grows with $\tan^3\beta$, the large $\tan\beta$ regime of the MSSM is strongly constrained by the experimental bound~(\ref{eq:Bsmm_LHC}).
However, as already stressed above, the non-holomorphic tree level Higgs-fermion couplings allow to generate NP contributions to B mixing already for moderate values of $\tan\beta$, where the bound from BR$(B_s \to \mu^+\mu^-)$ is considerably relaxed as long as the muon Yukawa coupling is not largely enhanced by the $\tan\beta$ resummation factors.
\section{Numerical Analysis}
To obtain the $B_q$ mixing amplitudes, we use 2-loop renormalization group running for the Wilson coefficients~\cite{Ciuchini:1997bw} and the hadronic matrix elements from~\cite{Becirevic:2001xt}.
We check compatibility of the model with various constraints.
(i) Vacuum stability: The higher dimensional operators in the superpotential can lead to a second minimum in the Higgs potential. Requiring that the electroweak minimum is stable, gives a lower bound on the charged Higgs mass for given values of the $\mu$ term and the SUSY breaking scale $m_S$~\cite{Blum:2009na,Altmannshofer:2011rm}.
(ii) Electroweak precision observables can constrain regions of parameter space where the dimension 5 superpotential operators lead to a very heavy SM like Higgs boson or a large splitting between the heavy Higgs bosons. We implement the S and T parameter following~\cite{Altmannshofer:2011rm}. Also the $Zbb$ coupling can be modified significantly by Higgs loops~\cite{Cline:2011mm,Degrassi:2010ne}. However the $Zbb$ constraint can be avoided if the Higgs-top coupling is suppressed by non-holomorphic corrections and therefore we do not include it in the numerical analysis.
(iii) Electric Dipole Moments (EDMs) can be induced both by phases of the higher dimensional operators in the K\"ahler potential and the superpotential.
Experimentally accessible EDMs, like the EDMs of Thallium, Mercury or the neutron can be generated by the electron and quark EDMs and chromo EDMs (CEDMs), by CP violating 4 fermion operators~\cite{Barr:1991yx} as well as by the Weinberg 3 gluon operator~\cite{Weinberg:1989dx}.
The fermion (C)EDMs are generated at the 1-loop level by sparticle loops that are sensitive to the phase of the Higgs vev~\cite{Blum:2010by,Altmannshofer:2011rm} as well as to possible complex $1/M$ corrections to the sfermion mass matrices that are induced by modified Higgs-sfermion couplings after electroweak symmetry breaking. The latter corrections can however always be avoided if the parameters $\varepsilon_0$ and $\varepsilon_\ell$ are real.
At the 2-loop level, Barr-Zee diagrams contribute to the (C)EDMs~\cite{Barr:1990vd,Pilaftsis:2002fe,Buras:2010zm}. They are directly sensitive to both the phases in the non-holomorphic Higgs couplings and the scalar-pseudoscalar mixing in the Higgs sector.
The CP violating 4 fermion operators are induced by neutral Higgs exchange at tree level~\cite{Demir:2003js,Buras:2010zm} and, analogously to the 2-loop Barr-Zee contributions, they are sensitive to both the phases in the non-holomorphic Higgs couplings and the scalar-pseudoscalar mixing in the Higgs sector.
Finally, contributions to the Weinberg 3 gluon operator can be induced by 2-loop diagrams that are sensitive to the scalar-pseudoscalar mixing in the Higgs sector and in particular also to the phases in the non-holomorphic Higgs couplings~\cite{Trott:2010iz,Boyd:1990bx}. These contributions can be sizable, but they can be avoided to a large extent if the Higgs-top couplings are suppressed by non-holomorphic corrections. Keeping also in mind the large uncertainties in estimating the contribution of the 3 gluon operator to the neutron EDM, we do not include it in our numerical analysis.
Following this approach, we find that the most important contributions are typically 2-loop Barr-Zee contributions to the mercury EDM and 1-loop Higgsino-Wino-slepton contributions to the Thallium EDM in the regions of parameter space that we consider below.
(iv) Constraints from direct Higgs searches at LEP, Tevatron and LHC are implemented using \verb|HiggsBounds|~\cite{Bechtle:2008jh} as well as the latest updates from Atlas and CMS~\cite{LHChiggs}. Generically direct SM Higgs searches do not lead to strong constraints, as the lightest Higgs boson is usually in the range $120 - 140$~GeV in the regions of parameter space considered below. The two heavier Higgs bosons can be much heavier and, due to the moderate values of $\tan\beta$, SUSY Higgs searches are also not constraining possible large effects in B mixing.
(v) Flavor observables: The main constraints come from $\Delta M_d$ and $\Delta M_s$ as well as the branching ratios of the decays $B \to X_s \gamma$ and $B_s \to \mu^+ \mu^-$. We also implement the constraints from $B \to \tau\nu$, $B \to D\tau\nu$ and $K \to \mu\nu$.
\begin{figure*}[t] \centering
\includegraphics[width=0.48\textwidth]{MA_tanb_1.png} ~~~~~
\includegraphics[width=0.47\textwidth]{MA_tanb_2.png}
\caption{
Possible values for the NP phase $\phi_s^{\rm NP}$ in the $M_{H^\pm}$ - $\tan\beta$ plane in the two example scenarios described in the text. The gray regions are excluded by $\Delta M_s$ (solid line), $\Delta M_d$ (long dashed line), BR$(B \to X_s \gamma)$ (dotted line), BR$(B_s \to \mu^+ \mu^-)$ (short dashed line) and the mercury EDM (dash-dotted line). In the red hatched region of the right plot, the electroweak vacuum is not absolutely stable.
}
\label{fig:MA_tanb}
\end{figure*}
In Fig.~\ref{fig:MA_tanb} we show in two representative scenarios the possible values of the NP phase in $B_s$ mixing in the $M_{H^+}$ - $\tan\beta$ plane together with the above mentioned constraints.
The plot on the left of Fig.~\ref{fig:MA_tanb} shows scenario I, where we chose a common sfermion mass of $\tilde m = 1$~TeV and trilinear couplings $A_t = \tilde m$, $A_b = A_\tau = 0$.\footnote{We remark that in a MFV framework, both the squark soft masses and the trilinear couplings can in general be expanded in powers of the flavor violating spurions $\mathcal{Y}_q = Y_q^\prime$. Higher order terms in the expansion can in principle induce additional flavor changing effects at the loop level. In the case of neutral meson mixing and $B_s \to \mu^+\mu^-$ however, the loop level effects induced by the higher order tems will be subdominant compared to the dominant tree level contributions from the modified Higgs sector. Among the considered flavor observables, only the loop induced $B \to X_s \gamma$ decay can be noticeably affected by possible higher order terms in the trilinear parameters. Considering such higher order terms would therefore add more flexibility in controlling the $B \to X_s \gamma$ constraint. As this constraint turns out to have only a small impact on our analysis, the higher order terms would not change any of our conclusions.} We assume the absence of the higher dimensional operators~(\ref{eq:super}) in the superpotential and set
\begin{eqnarray}
\epsilon_0 &=& -0.8 ~,~ \epsilon_1 = 0 ~,~ \epsilon_2 = 0.3 ~, \nonumber \\
\epsilon_3 &=& 0~,~ \epsilon_4 = -0.5 -0.8i ~,~ \epsilon_\ell = 1 ~,\nonumber \\
\alpha &=& \omega = 0~~, \nonumber \\
\mu &=& 200\,\textnormal{GeV} ~,~ m_S = 1\,\textnormal{TeV} ~, \nonumber \\
M &=& 10\,\textnormal{TeV} ~.
\end{eqnarray}
Correspondingly, NP contribution to the $B$ mixing amplitudes are generated through the Wilson coefficient $C_4$ and therefore effects are much larger in $B_s$ than in $B_d$ mixing. We observe that the $B_s$ mixing phase can easily reach the $2\sigma$ bound given in~(\ref{eq:phi_s}), $\phi_s^{\rm NP} \gtrsim -0.4$, even for moderate values of $\tan\beta \simeq 10$ and very large Higgs masses of $M_{H^\pm} \gtrsim 1$~TeV.
In this region of parameter space, Higgs contributions to the $B_s \to \mu^+\mu^-$ decay and to $B \to X_s \gamma$ are well under control. The strongest EDM constraints in this scenario come from the mercury EDM but due to the large Higgs masses they turn out to be easily fulfilled. As the higher dimensional operators in the superpotential are not present, vacuum stability bounds as well as electroweak precision constraints are always fulfilled.
The plot on the right of Fig.~\ref{fig:MA_tanb} shows scenario II, where we allow both for higher dimensional operators in the superpotential and K\"ahler potential, but consider new sources of CP violation only in the superpotential. We chose third generation squark soft masses of $m_{\tilde q_L} = m_{\tilde t_R} = m_{\tilde b_R} = 700$~GeV, all remaining sfermion masses $\tilde m = 2 m_{\tilde q_L}$ and trilinear couplings $A_t = 2$~TeV, $A_b = A_\tau = 0$. In addition we set
\begin{eqnarray}
\epsilon_0 &=& 0 ~,~ \epsilon_1 = 0 ~,~ \epsilon_2 = -1.8 ~, \nonumber \\
\epsilon_3 &=& -1.6 ~,~ \epsilon_4 = 1.6 ~,~ \epsilon_\ell = 1~, \nonumber \\
\alpha &=& 1.5 - i ~,~ \omega = 1.5 ~, \nonumber \\
\mu &=& 150\,\textnormal{GeV} ~,~ m_S = 500\,\textnormal{GeV} ~, \nonumber \\
M &=& 5\,\textnormal{TeV} ~.
\end{eqnarray}
In this setup the NP phases in $B_s$ and $B_d$ mixing are induced by the Wilson coefficient $\tilde C_2$. Therefore they are of comparable size in both cases and typically well within the $2\sigma$ ranges of~(\ref{eq:phi_d}) and~(\ref{eq:phi_s}). We observe that sizable values for the $B_s$ mixing phase are possible even for small $\tan\beta \simeq 4 $ but require a rather light Higgs spectrum, which agrees with our expectation that $\tilde C_2$ can be important only for small Higgs masses.
Because of the small values of $\tan\beta$, both constraints from $B_s \to \mu^+ \mu^-$ and from the EDMs (that mainly come from the Thallium EDM) are well under control in the considered scenario. Concerning the bound from BR$(B \to X_s \gamma)$ we remark that for the light Higgs masses, there are sizable charged Higgs loop contributions to the $b \to s \gamma$ amplitude that are further enhanced by higher order $\tan\beta$ resummation factors. These contributions can be partly canceled either by $t_\beta m_S/M$ corrections to the couplings of the charged Higgs to the right-handed top quark, or by chargino-stop loops as long as stops are rather light, below 1~TeV. In contrast to scenario I, a very important constraint is now coming from $\Delta M_d$. Also bounds from vacuum stability start to be important. Contrary to the framework discussed in~\cite{Altmannshofer:2011rm} however, vacuum stability bounds can be compatible with a large $B_s$ mixing phase without the need of additional physics that stabilizes the electroweak vacuum.\footnote{We note that in the framework considered in~\cite{Altmannshofer:2011rm} a large negative $\mu$ term is required to generate sizable flavor changing neutral Higgs couplings at the loop level. In the scenario considered here however, small values for $\mu$ are possible that considerably soften the vacuum stability bounds.}
\begin{figure*}[t] \centering
\includegraphics[width=0.47\textwidth]{aSLd_aSLs.png} ~~~~~
\includegraphics[width=0.47\textwidth]{SpsiKs_Spsiphi.png}
\caption{
Correlation between the semi-leptonic asymmetries $a_{\rm SL}^d$ and $a_{\rm SL}^s$ (left) and the time-dependent CP asymmetries $S_{\psi K_S}$ and $S_{\psi\phi}$ (right). Blue (dark gray) points correspond to a scenario with new sources of CP violation in the Yukawa couplings, while green (light gray) points correspond to new CP phases in the Higgs sector. The red thin contours show the allowed regions using the bounds on $\phi_s^{\rm NP}$ and $\phi_d^{\rm NP}$ from Fig.~\ref{fig:phi_fit}. In the left plot, the magenta band (orange ellipse) shows the measurement of $A_{\rm SL}^b$ ($a_{\rm SL}^d$ and $a_{\rm SL}^s$) by D0.
}
\label{fig:aSL}
\end{figure*}
In Fig.~\ref{fig:aSL} we present the results of a parameter scan of the model. The left plot shows the correlation between the semi-leptonic asymmetries $a_{\rm SL}^d$ and $a_{\rm SL}^d$, while the right plot shows the correlation between the time dependent CP asymmetries $S_{\psi K_S}$ and $S_{\psi\phi}$. Blue (dark gray) points correspond to a scenario where new sources of CP violation are arising entirely from the modified Yukawa couplings and the higher dimensional operators in the superpotential~(\ref{eq:super}) are switched off completely. We consider mass scales as in scenario I above and allow $\tan\beta < 15$ as well as epsilon parameters $|\varepsilon_i| < 1$ with $\varepsilon_0$ and $\varepsilon_\ell$ real and arbitrary phases for the remaining $\varepsilon_i$.
As expected, in such a setup the NP effects in $B_s$ mixing are much larger than in $B_d$ mixing and the LHCb bounds on $\phi_s$ exclude a sizable NP phase in $B_d$ mixing.
Green (light gray) points correspond to a scenario where also the higher dimensional operators in the superpotential~(\ref{eq:super}) are considered. We fix mass scales as in scenario II above and allow the parameters $|\alpha|, |\omega| < 2$ with arbitrary phases and real epsilon parameters $|\varepsilon_i| < 2$ as well as $\tan\beta < 15$. In this setup, the NP contributions to CP violation in $B_d$ and $B_s$ mixing are comparable. Consequently, a sizable NP phase in $B_d$ mixing can be compatible with the LHCb constraints on the NP phase in $B_s$ mixing.
Even though new CP phases come entirely from the superpotential operators, we stress that the presence of the K\"ahler potential operators is crucial. They allow for low $\tan\beta$ values and therefore the constraint from $B_s \to \mu^+\mu^-$ can be avoided.
We conclude that in order to generate sizable corrections to CP violation in $B_d$ mixing that are in agreement with the LHCb data on CP violation in $B_s$ mixing, both the higher dimensional operators in the superpotential (that modify the Higgs spectrum) as well as the higher dimensional operators in the K\"ahler potential (that lead to non-holomorphic Higgs-fermion couplings at tree level) are required.
\section{Conclusions}
Recent results from LHCb on the time-dependent CP asymmetries in $B_s \to \psi\phi$ and $B_s \to \psi f_0$ significantly restrict the allowed values for the $B_s$ mixing phase. Combining these results with measurements of CDF and D0 of the time-dependent CP asymmetry in $B_s \to \psi\phi$ as well as the measurements of the time-dependent CP asymmetry in $B_d \to \psi K_S$ at the B factories, we find the following 2$\sigma$ ranges for possible NP phases in $B_s$ and $B_d$ mixing: $-0.28 < \phi_d^{\rm NP} < 0$ and $-0.40 < \phi_s^{\rm NP} < 0.20$.
The preference for a negative NP phase in $B_d$ mixing is driven by tensions in fits of the Unitarity Triangle, while the NP phase in $B_s$ mixing is perfectly consistent with 0. Under the assumptions that neither the absorptive part of the $B_s$ mixing amplitude nor the SM tree-level $B_s \to \psi\phi$ decay amplitude are significantly affected by NP, the anomalous like sign dimuon charge asymmetry observed by D0 cannot be explained given the above bounds.\footnote{The possibility to reconcile the LHCb data on the time dependent CP asymmetries in $B_s \to \psi \phi$ and $B_s \to \psi f_0$ and the D0 measurement of the like sign dimuon charge asymmetry with NP effects in $\Gamma_{12}$ has been analyzed very recently in \cite{Bobeth:2011st}.}
In view of these results we studied the possible impact of higher dimensional operators in the MSSM on B physics.
We considered dimension 5 operators both in the superpotential and in the K\"ahler potential assuming that they are generated at a scale of $M \simeq 5-10$~TeV. The 1/M suppressed operators in the superpotential have important impact on the Higgs spectrum. They can significantly enhance the tree level mass of the lightest Higgs boson and lead to a mass splitting between the two heavy neutral Higgs bosons. With complex coefficients they also lead to scalar-pseudoscalar mixing. The 1/M operators in the K\"ahler potential can induce non-holomorphic Higgs couplings and consequently flavor changing neutral Higgs couplings at the tree level. Assuming that the K\"ahler potential operators follow the minimal flavor violation ansatz, we find a flavor phenomenology that resembles to a large extent the 2 Higgs doublet model with MFV discussed in~\cite{Buras:2010mh,Buras:2010zm}.
In particular large values for $\tan\beta$ are not required to generate sizable corrections to B meson mixing from neutral Higgs exchange and therefore the strong constraint from BR$(B_s \to \mu^+ \mu^-)$ can be significantly relaxed.
We find that both superpotential and K\"ahler potential operators are required to generate non-standard effects in the $B_d$ mixing phase that are in agreement with the current bounds on the $B_s$ mixing phase from LHCb. The corresponding region of parameter space is characterized by rather small values of $\tan\beta \simeq 5$ as well as low masses for the heavy Higgs bosons $M_{H_2}, M_{H_3} \simeq 300 - 400$~GeV.
We stress that in the class of models discussed in this work, a non-standard $B_d$ mixing phase does imply also non-standard effects in the $B_s$ mixing phase at a level that can be tested in the near future by LHCb.
\bigskip
\paragraph{Acknowledgments:}
We thank Michael Trott for useful comments.
Fermilab is operated by Fermi Research Alliance, LLC under Contract No. De-AC02-07CH11359 with the United States Department of Energy.
\section{Appendix}
Here we briefly present UV completions that lead to the higher dimensional operators in~(\ref{eq:super}) and (\ref{eq:kahler}).
The dimension 5 superpotential operator in~(\ref{eq:super}) can be generated by integrating out a heavy singlet S with the following superpotential interactions~\cite{Dine:2007xi}
\begin{equation}
\mathcal{L} \supset \int d^2 \theta \left( \frac{1}{2} M S^2 + \sqrt{\omega} S H_u H_d \right) ~.
\end{equation}
Additional gauge interactions that are broken at a high scale $M$ can be effectively described by dimension 6 operators~\cite{Dine:2007xi,Carena:2009gx}.
One way to generate the K\"ahler potential operators in~(\ref{eq:kahler}) is to introduce two heavy $SU(2)$ doublets $\tilde H_u$, $\tilde H_d$ with hypercharge $+1$ and $-1$ that couple to fermions and mix with the MSSM Higgs doublets $H_u$, $H_d$~\cite{Antoniadis:2008es}.\footnote{To preserve gauge coupling unification in such a case, additional heavy matter fields should be present to complete SU(5) multiplets that include the heavy Higgs doublets.} Neglecting for simplicity gauge interactions, their K\"ahler potential and superpotential read
\begin{eqnarray}
\mathcal{L}&\supset& \int d^4 \theta \left( \tilde H_d^\dagger \tilde H_d + \tilde H_u^\dagger \tilde H_u \right) \nonumber \\
&+& \int d^4 \theta \left( \tilde H_d^\dagger H_d + \tilde H_u^\dagger H_u + {\rm h.c.} \right) \nonumber \\
&+& \int d^2 \theta \Big( M \tilde H_u \tilde H_d + \lambda_\ell \tilde H_d L E \nonumber\\ \label{eq:UV}
& & ~ + \lambda_u \tilde H_u Q U + \lambda_d \tilde H_d Q D + {\rm h.c.} \Big) ~.
\end{eqnarray}
Integrating out the heavy Higgs doublets generates the supersymmetric term in~(\ref{eq:kahler}) and SUSY breaking can be incorporated with the auxiliary spurion $Z$.
In addition to~(\ref{eq:kahler}), integrating out the heavy Higgses from~(\ref{eq:UV}) also generates $1/M$ suppressed terms in the superpotential
\begin{eqnarray}
\mathcal{L} &\supset& \frac{1}{M} \int d^2 \theta \Big( \lambda_u \lambda_d (QU)(QD) \nonumber \\
&& ~~~~~~ + \lambda_u \lambda_\ell (QU)(LE) + {\rm h.c.} \Big)~,
\end{eqnarray}
that in turn generate dimension 5 fermion -- sfermion interactions. Such interactions are however not relevant for the topics discussed in the present work.
| {
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package terrains;
public enum Types {
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Former first lady Michelle Obama has revealed the cover of her new memoir, Becoming, which will be released in the fall.
Mrs. Obama shared the cover photo Thursday on her social media accounts. The image shows her smiling broadly for the camera, and was taken earlier this year in Washington, D.C. by portrait photographer Miller Mobley, the book's publisher said in a press release.
The book will detail the experiences that shaped Obama's life, from her childhood on the South Side of Chicago, to motherhood, her career as a business executive and her time in the White House.
It will be released on Nov. 13.
In the days leading up to the book cover reveal, Obama shared four family photos on her Instagram page that will be included in the memoir.
Among the images are a photo from childhood with her parents, a snapshot from her college years at Princeton, a touching moment from her wedding to Barack Obama in 1992, and a family photo showing the couple with daughters Sasha and Malia.
Earlier this week, President Obama and the former first lady made headlines after signing a multi-year agreement with Netflix to produce films and series for the streaming giant.
The couple will focus on projects that promote "empathy and understanding between peoples" through their newly-formed company, Higher Ground Productions. | {
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A: There may be something wrong with your font cache, located at dompdf/lib/fonts/dompdf_font_family_cache.php (though the exact file name depends on your release and whether you have loaded fonts). This file tells dompdf what fonts are available for use in the PDF. If this file references a font that isn't actually available you can run into major issues on viewing. You may need to re-load your font files. Take a look at the Unicode how-to for an overview of using embedded fonts. This document hasn't been updated to reflect changes implemented in dompdf 0.6.0 beta 3, but the information is still pertinent.
| {
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| {
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} | 8,640 |
# THE
WIKKELING
Steven Arntson
Illustrated by
**Daniela Jaglenka Terrazzini**
For my parents,
**HELEN & JERRY**
© 2011 by Steven Arntson
All rights reserved under the Pan-American and
International Copyright Conventions
Printed in China
_This book may not be reproduced in whole or in part, in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system now known or hereafter invented, without written permission from the publisher._
Books published by Running Press are available at special discounts for bulk purchases in the United States by corporations, institutions, and other organizations. For more information, please contact the Special Markets Department at the Perseus Books Group, 2300 Chestnut Street, Suite 200, Philadelphia, PA 19103, or call (800) 810-4145, ext. 5000, or e-mail special.markets@perseusbooks.com.
ISBN 978-0-7624-3903-4
Library of Congress Control Number: 2010935091
E-book ISBN 978-0-7624-4249-2
9 8 7 6 5 4 3 2 1
Digit on right indicates the numbers of this printing
Cover and interior design by Frances J. Soo Ping Chow
Edited by T. L. Bonaddio
Typography: Blackadder, Blavicke Capitals, Praeton, Requiem, Riva
Published by Running Press Kids an imprint of Running Press Book Publishers
A Member of the Perseus Books Group
2300 Chestnut Street
Philadelphia, PA 19103-4371
Visit us on the web!
www.runningpress.com
## Contents
Acknowledgements
Prologue
Part 1
Efficient Education
Gary
Headaches
The Red Drip
Sunset
The Bestiary
Intentional Detention
Scaredy Gary
The Wikkeling
Through the Windows
Part 2
Spike-Tailed Fish and Flesh-Eating Worms
The Competency Exam
The Department of Insta-Structure
Smashed Sidewalks
The New Route
Trapped!
A Death in the Family
The Escape Plan
The Cat Hall
Frightening Friends
FindEm™
Attacked
Distress Call
The Stink
Two Wild Housecats
Explanations
The Memorial
The Attic Books
Back to the Attic
## Acknowledgements
When a book first sprouts, it feels quite independent, but it is quickly humbled by its needs and indebted for its very existence to those who helped it. Firstly, a thank-you to the members of my writing group for looking at my earliest efforts. My agent, Jenni Ferrari-Adler, went far above and beyond the call of duty in helping me revise and in taking up the cause of the book. My editors Teresa Bonaddio, Marlo Scrimizzi, and Kelli Chipponeri were of tremendous aid in believing in the project and bringing the characters to life through numerous patient rounds of revision. I'd also like to thank Daniela for the beautiful illustrations and Frances Soo Ping Chow for the wonderful cover and book design—they've placed me in the enviable position of having my name attached to something much more beautiful than I'd ever imagined. More than anyone, however, I'm indebted to my wife Anne Mathews, who has, during the course of our marriage and friendship, been the greatest influence on how I think and what I think about. Without our countless conversations about life and the living of it, and without her example, this book would not have been conceived. She also gave the manuscript its first real copyedit, caused the sentences to become readable, and pointed out several directions in which I had forgotten to look (as all who know me know, and she best of all, my eyes are very poor). Thank you, for reading.
Poor kitty,
poor kitty!
The Wikkeling chased you
From city to country
And back again, too.
It won't rest. It won't weary.
It will kill you, poor kitty,
And then all those like you,
And all those you knew.
Jump up to my attic
Poor kitty, and pause–
Rest here. Recover,
And sharpen your claws.
I'll give you refuge
For I understand
What it is to be hunted,
Unwelcome, unwanted,
Pursued and tormented
And fainting from fear
Every night,
Every night,
Every night of the year.
–Anonymous, from Aristotle Alcott's
Riddles and Rhymes of Olden Times
## Prologue
The Old City lies on a long, low hill. It is dangerous and dilapidated. The buildings are crumbling, moss grows in the streets, and garbage festers in the gutters. There are rumors that people live here secretly, breaking into abandoned apartments and living wretched, illegal lives.
Adjoining the Old City is the Addition, which lies on a vast, level plain. The Addition sparkles into the haze, its streets as straight as grocery store aisles, its buildings as shiny as pop cans. The Addition contains countless homes, businesses, schools, and hospitals. Skyscrapers rise in lanky rectangles. Sprawling suburban chessboards meet the broad blocks of industry. The Addition is so large that airports operate to fly people from one part to another.
Between the Old City and the Addition runs a seam, where the decrepit hill meets the youthful plain. Few people know it, but this seam is the kind of place where unexpected things happen. Invisible doors and windows open. Unknown creatures appear. Even now, something strange is afoot.
It's past midnight, but this street, called Boardwalk, is crammed with bumper to bumper traffic. Lilac-scented exhaust fills the air and curls under streetlights and headlights as employees commute to night-shift jobs or return from day-shift jobs, parents take sick children to hospitals, and everyone weaves from lane to lane hoping to dodge the next snarl. Delivery vans deliver new shoes, cell phones, and ready-to-eat dinners to one side of the street while garbage trucks collect old shoes, broken cell phones, and leftovers from the other side. Eighteen wheelers stacked with shipping containers, petroleum tanks, prefab houses, cars, and even new eighteen wheelers pass through on their long-distance routes. In a city as large as this one, there is no time of day or night when such things aren't happening.
Although the Addition seems alive with activity at first glance, it is strangely motionless. Work to home and home to work. Old shoes to new shoes. Delivery and pickup. Repetition becomes stillness, lulling everyone, and this is why no one notices when, in the middle of the street, a shadow the size of a small animal darts from beneath a garbage truck and under a car in the next lane. It dashes across another lane, and then another, camouflaged by a black belch from a tailpipe.
This stretch of Boardwalk is lined with identical, brand-new multi-level houses, constructed of vinyl and glue, sitting behind green plastic lawns. The houses have flat roofs and few windows. They are airtight and soundproof. Someone looking to sneak in would have a tough time. The shadow darts past, slowing at each home and then moving on.
But there is one exception on this block. There's one home made of wood and nails instead of plastic and glue. It's only a single level, and its steeply angled roof rises to a shingled peak, an indication that it was built long ago. Now it's the only one of its kind left here.
The shadow creeps around back, out of the direct glare of the streetlights to a small strip of artificial turf that separates one piece of property from another. It limps, as if injured, but manages a swift, terrific leap to the roof of the old house. It disappears inside through a hole under the eaves.
This old house has a family sleeping within, lying under warm covers, and they are not awakened, soothed as they are by the comforting grind of the endless traffic jam, as monotonous as sheep gliding at intervals over an easy fence. They have no idea that they have a houseguest.
## PART 1
### Efficient Education
"SENSIBLE STUDENTS SUCCEED SPLENDIDLY!" said Ms. Span, a primly dressed teacher sitting behind a computer at the front of the class, her thick, black eyebrows arching over the top of her reading glasses.
"Yes, Ms. Span!" said the students. They sat in neat rows that filled the room, faces lit yellow from the light of their own computers.
On the wall behind Ms. Span, a large projection displayed the sentence she had just recited. "Let's begin today's focus on the letter S," she said, her voice just loud enough to be heard above the whirring of the fans in all of the computers. The students began to type. Next to each child's screen, a plastic cradle held a cell phone hooked to the school's network. The children's practice sentences were instantly graded and transmitted to their parents' phones, ensuring that each parent knew, at each moment, how their child was scoring.
Additionally, all sentences were tabulated, in terms of accuracy and speed, into a data pool describing the class, the school, the district, and the system as a whole. At this moment, every child everywhere was typing "SENSIBLE STUDENTS SUCCEED SPLENDIDLY," allowing every school to be instantly ranked in comparison to every other school. On Competency Exam days these rankings were used to determine whether or not the school was functioning properly, or whether it should be shut down. Today, fortunately, was not the Competency Exam. It was a Practice Test.
Ms. Span flipped through the students' responses on her screen, checking them. She was on edge, even though it was just practice. The thing about Practice Tests was that they led inevitably to the Competency Exam, and if things went poorly then, Ms. Span could be classified a Bad Teacher and lose her job.
"Very good, Andreas," she muttered. "Very good, Sasha."
Ms. Span projected the next sentence.
"SENSIBLE STUDENTS ARE SAFE, SECURE, AND SUPERVISED!" recited the students, some beginning to type as they spoke. Because the exercises were timed, there was little opportunity to fix mistakes. Ms. Span reviewed the responses again. Her computer screen reflected on her glasses, rankings appearing there in columns.
She glanced at the bottom of the stats and winced at the names there, especially at the one at the very bottom. She frowned. She didn't want to stop the lesson, but the whole class was getting dragged down.
"Miss _Gad-Fly_ ," she called out frostily. All the children turned toward the rear of the room to look at the object of their teacher's attention. Some of them snickered. Miss Gad-Fly sat in the back row near the door. She seemed lost in thought, and didn't notice the attentions of the room until Ms. Span said, "Henrietta!"
Being singled out scornfully in class wasn't unusual for Henrietta. Wherever she was and whatever she was doing on any given day, she found herself in a similarly unenviable position. If a poll were to be taken this afternoon by her school, asking all the students in all the grades who was least popular, Henrietta Gad-Fly would win. And that would be the only thing she won all year.
Henrietta looked a little like a brick—her face and body were squat, thick rectangles. Her ruddy skin was prone to pimples, and flushed red when she was embarrassed, like now. Her small, black, beady eyes were set closely together, which lent her a confused and peevish expression that often caused people to explain things to her twice, and then scold her. Her thin eyebrows made her look a little surprised, which didn't help matters.
And yet Henrietta was not a stupid, confused, petulant block. Or at least, she didn't feel like one. Inside, she was just herself—a person to whom she'd scarcely yet been introduced.
Henrietta is the main character of this story. This whole book will be about her—and it's worth mentioning at the outset a few things that aren't going to happen to her.
She will not become beautiful when someone gives her a new hairstyle.
She will not find a miracle cure for her pimples when an angel sees she's a good girl inside.
She will not find out that she's actually a princess, and she won't become happy forever when a prince marries her.
Those books are out there, and your school librarian can help you find one. This isn't it.
"Henrietta," said Ms. Span, "you are _off task_."
With the click of a button, Ms. Span displayed the contents of Henrietta's monitor on the wall screen. Each day contained a moment such as this, in which Henrietta was exposed as an example of the kind of person one should have the common sense not to be. Henrietta, everyone knew, ranked at the bottom of the class in most subjects: writing, reading, math, and even physical safety. Her scores were so low that she was nearly (but not quite) At Risk.
Henrietta's sentence read: SE
What kind of child could so grievously fail in her attempt to type "SENSIBLE STUDENTS ARE SAFE, SECURE, AND SUPERVISED?" The class had a good laugh, but they didn't laugh as hard as they might have. As with any joke, Henrietta's incompetence had grown less funny over time.
"Henrietta, you will stay after class today and retype your practices," said Ms. Span.
"Yes, Ms. Span," said Henrietta, her voice inaudible over the tinny hum of the fans in the computers.
"What?" said Ms. Span.
"YES, MS. SPAN!" said Henrietta. She pressed her lips together in a way that made her look angry, but she was actually feeling humiliated.
"Now, then, everyone," said Ms. Span, "we've fallen behind because of Henrietta. Today is just Practice, but next month is the Competency Exam. If Henrietta delays us then, we'll have to work even faster, with even more accuracy. Remember, we're graded as individuals _and_ as a class. We compete against each other to help each other. Right now, we're in Good Standing. None of you are At Risk, and none are Finished."
A chill ran down the spine of every student when Ms. Span said "Finished." No one wanted that. If the school declared you Finished, that was it. You were kicked out. Your parents would be really, really mad, and you'd become a garbage collector for the rest of your life.
Typing Practice gave way to Composition, and then Math for the remainder of the morning until it was time for History and Nutrition. Ms. Span removed her reading glasses and bade the class stand, and they lined up and followed her into the hallway, quickly completing the short walk past the mural that read: SENSIBLE, EFFICIENT, EDUCATION (S.E.E!).
The History and Nutrition Center was a large room filled with four rows of ten divided study carrels. Each seat in each row was cordoned with yellowish brown walls rising from the desktop. Through the rows ran a conveyor belt used to deliver the day's Approved Nutrition while the students watched various historical videos on the carrel dividers.
Henrietta's class entered just as the kindergartners, whose nutrition came a little earlier, were packing up to return to their room. Henrietta waited at the nearest carrel as its occupant prepared to leave: a little girl, who seemed small even for a kindergartner. Henrietta noticed that she was wearing odd clothes. Instead of a polyester outfit with a yellow safety stripe down the back, she wore a brown shirt made of . . . wool? Henrietta had seen wool in a few old TV shows, where people rode on horses or lived on icebergs. It was said to cause rashes.
The girl's dark-skinned arms looked painfully thin as she stuffed some papers into her backpack, zipped it up, and pulled it with some effort over her shoulders. She ran to join her class, her curly black hair bouncing as she exited.
Henrietta settled into the carrel, noting that the one immediately to her left was occupied by the chubby, bespectacled, Clarence Frederick, and the one to her right by the chubby, bespectacled, Clarice Sodje. Both had bullied Henrietta in the past, and she wasn't thrilled to be between them now.
History and Nutrition period was the day's only non-graded learning experience. History, because it wasn't related to anything, wasn't tested during the Competency Exam. Ms. Span said history had a "Noninstrumental Positive Impact," which meant it was good for you but didn't really matter.
It was also good to watch movies while eating, because watching discouraged talking, and talking with one's mouth full was both dangerous and impolite. The History and Nutrition Center was generally very quiet but for the tinny sounds of the movies playing over the speakers in the carrel dividers.
Henrietta had landed in the HENRIFT ANDI carrel. She'd been in this one a few times before, and had seen the movie already, which was called _Founder, Humanitarian, Forward Thinker_. As she sat, the movie began. Henrift Andi, a tall, clean-shaven man with a stovepipe hat, was giving a speech to a crowd of fascinated onlookers.
"We must be courageous enough to look forward without fear, and sensible enough to fear looking backward!" he said, and the crowd cheered through the little speakers. The title came on:
**H ENRIFT ANDI:
FOUNDER, HUMANITARIAN, FORWARD THINKER**
The end of the title was obscured by a little piece of paper. It was a yellow sticky note attached to the screen. Henrietta leaned forward. The square contained a short message scrawled in a beginner's handwriting.
**henRift and andi**
Henrietta pulled it off of the screen as the movie segued to show Henrift Andi as a little boy, still wearing a stovepipe hat, planting an apple orchard. Henrietta crumpled the note and dropped it in the trash next to the carrel. She thought the kindergartener who had just departed was most certainly the culprit, and she didn't want the tiny girl to get in trouble for vandalism.
Nutrition arrived on the conveyor belt: a cube of corn bread smothered in starchy gravy, some small yellowish carrots with margarine dip, a pile of corn chips, and a glass of apple soda.
"Soon after," said the narrator's calm but engaging voice, "Henrift Andi developed a new breed of apple, which he called Scrumptious!"
Henrietta dipped a corn chip in the margarine. As she brought it to her mouth, a little paper airplane flew over the carrel wall to land squarely atop her corn bread and gravy. On one wing was written:
**DUMBIETTA**
It had been thrown by Clarence or Clarice. Such airplanes were a common part of Henrietta's History and Nutrition periods. Since most of the kids had watched all of the history movies already, and because they weren't getting graded, they made mischief of one kind or another. Henrietta plucked the craft from her lunch and was about to toss it into the garbage when she stopped suddenly. She felt a strange sensation, like someone was standing next to her. It was a creepy feeling, and it made her heart skip a beat: it was the feeling that preceded one of her headaches.
A moment later, the headache began. It wasn't too bad yet. It was still small. But it might get worse. Henrietta understood, though her parents had never said it aloud, that a headache could get bad enough to kill her. She was one of those kids who might not make it. She might not grow up.
Henrietta fished her pills from her pocket and ate three of them. She waited in perfect stillness, one hand still holding the forgotten airplane. After an unknown amount of time, she noticed that the credits for FOUNDER, HUMANITARIAN, FORWARD THINKER were rolling past on the screen. She turned and saw her classmates lining up behind Ms. Span. Carefully, she stood and joined them.
Henrietta had suffered headaches for a few years, since she and her parents had moved into the old house they lived in now—in fact, her mother thought it was the house's fault, and the doctor they'd seen had agreed, saying kids raised in old homes sometimes became House Sick. The cause was unknown, but something about those old places was not good for children. Maybe some kind of toxin. They were told they should move, but unfortunately they didn't have enough money to live anywhere else at present.
Henrietta's headaches, though supposedly caused by her house, seemed to strike her everywhere but at home, and they always followed the same progression. First, it seemed for an instant like someone was standing next to her. Then the headache began behind her eyes, and either dissipated or grew. Henrietta either felt better or went to the hospital.
Today, she felt better. Though she was hungry because she'd eaten nothing during History and Nutrition, relief swept through her when she returned to the classroom. Such moments of recovery constituted the greatest joys of her life. She was so happy that when Ms. Span resumed the typing practices, she produced exactly what she was supposed to, and didn't become distracted.
At the end of the school day, Ms. Span displayed the class's statistics: They were in the top forty percent for the district, which meant they were in Good Standing. If they could keep that up, they'd do fine on the Competency Exam, so Ms. Span ended in a good mood.
Henrietta still had to stay after, though. As the other students left to wait for the buses, Henrietta began to retype her practice sentences.
At the front, Ms. Span previewed the following day's System Approved Lessons (SALs) and leafed through Student Statistical Profiles (SSPs) to see if there was any way to leverage a Competitive Advantage Boost (CAB). She was just starting at the school this year, so none of the students knew her very well. She was generally strict, but Henrietta knew another side of her, which sometimes emerged when the two of them were alone during detention after class.
"Henrietta, your work is looking excellent," Ms. Span said, removing her reading glasses and squinting out across the rows of empty desks. "If only you could perform this well _during class_ , you'd be one of the top students."
"Thank you, Ms. Span," said Henrietta. She did have a knack for schoolwork, and the extra practice she endured in detention had further developed the aptitude.
Ms. Span projected Henrietta's detention and classtime statistics next to one another on the screen. "Why do you have such trouble, Henrietta?" she asked. "It isn't sensible."
"I don't know," said Henrietta. Her detention statistics showed her to be extremely fast and accurate, while her class scores were terrible. Even she was a little confused about it. "I just feel . . . nervous, sometimes."
"Because you get made fun of?" said Ms. Span.
"Yes," said Henrietta. But there was more to it than that. After class, when she and Ms. Span were alone, the work seemed more important, and she felt like she and Ms. Span were a team. This never occurred during class, when everyone competed on a curve. To do better during class meant someone else did worse, and to be singled out meant either you were failing or you were causing others to fail. The best success was to remain unnoticed, right in the middle, and that was pretty boring. Henrietta didn't say any of this to Ms. Span, though. Neither of them could change the way the school worked, and earning detention every day, for Henrietta, was a kind of clever solution to the problem. Not that she felt clever. She felt stupid.
When she completed her extra work, she glanced at her cell phone to check the time. She hadn't missed her bus yet.
"Thank you, Ms. Span," she said, standing.
"Thank you for your good effort, Henrietta. Let's try to do better tomorrow, all right?"
"I'll try, Ms. Span," said Henrietta, and she pulled her backpack onto her shoulders and departed for the bus. Ms. Span clicked a button, and her computer screen read:
STUDENT 3421836 LOG OUT CLASSROOM 7434 16.46.345 [UTC]
Henrietta arrived at the parking lot turnaround just as her bus was preparing to leave. One thing she liked about detention was that she didn't have to stand around with the other kids and wait, friendless. She boarded, found a seat, and began to buckle herself in with a lap belt, two shoulder belts, and a head belt, all designed to assure her survival in the worst imaginable crash.
Her cell phone buzzed in her pocket—her mother calling to see if she needed a ride. Fortunately, as soon as she fastened her last buckle, an automated message appeared on her parents' phones:
STUDENT 3421836 SECURED BUS#1056 ETA 16.46.345 [UTC] STOP 342
The blue warning light over her seat extinguished, indicating to the bus driver that she was fully secure. Around her, other kids talked excitedly and amused themselves, making a racket despite their restraints. The master safety light at the front came on, glowing yellow as the front doors closed and the engine rumbled to life. The bus rolled into the traffic jam.
Gridlock wound unceasingly through nearly every part of the Addition. The cars were packed in the streets, coughing out the city's familiar charred, floral scent of exhaust. Its citizenry succeeded splendidly, labored diligently, recreated briefly, saved sensibly, and spent frugally their short lives in the dense grid of interlocked developments.
The bus crept over perfect asphalt surrounded by a fleet of garbage trucks emitting hazy, yellow-brown streams of lilac-scented particulates. Henrietta looked across the aisle to the seat opposite and saw a boy there she didn't recognize. He was her age, with straight black hair and thick, black eyebrows, locked tightly to his seat by the same network of straps that contained Henrietta and everyone else.
As soon as Henrietta's eyes landed on him, he turned to her, as if on cue. "I hate this!" he exclaimed, and without a pause, he unbuckled all of his safety straps. "Yeah!" he shouted, stretching his arms and sticking his legs out into the aisle.
"Put those back on!" said Henrietta.
There were immediate consequences for him. The yellow safety light went out at the front of the bus, and the blue warning light glared to life above his seat. The bus's engine stopped and then the bus stopped. A thousand messages erupted from outside:
"YOUR PARENTS CAN AFFORD A BIGGER, BETTER HOUSE
FOR YOU AT NEWVIEW ESTATES!"
"YOU'RE NEVER TOO YOUNG FOR YOUR FIRST CAR—NOW AT LURMY'S!"
"IS YOUR CELL PHONE A TINCAN? IT BETTER BE!"
These were Honk Ads, activated by all of the drivers whose vehicles were now stuck behind the bus. Each car horn blared a different advertisement, many of them responding to the presence of the school bus by advertising to the children inside.
"EDIBLE CLEANTASTE CORN SOAP—IT'S
THE CANDY OF SOAP!"
Henrietta heard squealing tires as commuters tried to merge into other lanes, and the horns overlapped into nonsense.
The boy froze. He hadn't realized that this would happen, apparently.
"Don't worry," Henrietta whispered to him through the din. "You'll just get a warning. He doesn't want to be off schedule."
The boy shot Henrietta a worried glance as the driver, a large man wearing a yellow jumpsuit, walked back along the aisle, glaring at every pair of children along the way. When he reached the new boy he frowned down at him and said, "Name."
"I forget!" said the boy, grinning ridiculously.
The driver produced a scanner from his belt, pointed it at the boy, and looked at the screen. "Gary," he said.
"Scary!" quipped a child further back on the bus.
"Scary Gary!" said someone else.
" _Scaredy_ Gary!" said a third student.
Hilarity ensued. Henrietta laughed too, even though she felt bad for the boy. She knew what it was like to be made fun of, but it also felt kind of good to make fun of someone else for a change.
"Okay, Mister," said the driver. "I'm issuing you a warning for releasing your safety belt, and a detention for insubordination." As he said this, he scribbled with a plastic stylus on the small screen of the scanner. Then he reached out, roughly rebuckled Gary's straps, and lumbered back to the front, restarting the bus and entering traffic again. "IT'S TIME FOR A LURMY'S EGG SANDWICH!" one last Honk Ad announced.
"You shouldn't have aggravated him," said Henrietta.
"Thanks for telling me _now_ ," Gary replied.
"Well, I didn't know you were _stupid_ ," Henrietta shot back. That was the end of their conversation.
Henrietta soon recognized the familiar streets of her neighborhood. Because her school was nestled into the Addition's perfect grid of streets, the landmarks Henrietta watched for were older buildings, which signaled the closeness of the Old City. Instead of walls of shiny mirrored windows, the older buildings were dull concrete, stained with traffic exhaust. Henrietta's parents often mentioned how ugly such buildings were, but Henrietta liked them because they meant she was almost home.
Usually she was the only one to disembark at her stop, but today as the bus rolled up, she heard another set of straps release as she released hers. When the door opened and she stepped into the aisle, Gary stood as well.
Henrietta tried not to look at him. She exited the bus and walked to the intersection as the driver stopped Gary and lectured him about the importance of safety. She crossed when the signal turned. Then she heard Gary's voice rise over the din behind.
"I'm sorry!" he yelled. Was he talking to the bus driver or her? She didn't wait to find out.
Henrietta's house stuck out like a sore thumb on her block. It was the only old house, the only house without a front yard, and the only house with a steep, shingled roof. It had originally belonged to Henrietta's grandmother, who had given it to Henrietta's parents two years ago when she'd married her longtime friend Al and moved with him to Sunset Estates retirement community, far into the Addition. The old house was rundown, leaky, cramped, and full of the strong smell of lilac air freshener, which Henrietta's mother used to cover the dusty, mildewy smells of age.
As Henrietta reached the front door, she saw a moving van parked across the street. Several people were carrying boxes from it into a new house that had been erected the previous week. Henrietta saw Gary turn into the drive and enter the house through the garage. He was her across-the-street neighbor. This meant she might see him the following morning at the bus stop, a prospect she didn't choose to dwell on at the moment.
She crossed the narrow petunia border between the sidewalk and her front door, which opened as she approached to reveal her mother, a woman of medium height with Henrietta's closely spaced eyes, ruddy skin, and blockish body. She wore a pair of tan, fitted pants and a loose, white blouse, which made her somewhat resemble a vanilla ice cream cone. She ushered Henrietta inside and closed the door, shutting out the traffic noise.
"How was school?" she asked, giving Henrietta a brief hug.
"It was okay," said Henrietta.
"No makeup work?"
"I did it fast," said Henrietta as they entered the living room.
"That's good," said her mother. "That's an improvement."
"Hello, Henrietta," said her father from the living room couch, where he watched the large, flat-screen television that hung in a picture frame on the wall. He was a bland, unobtrusive man dressed in jeans and a gray sweater.
The sound was muted on the TV, and the screen showed an advertisement for bathtub cleaner. Henrietta and her mother sat on the couch with her father, Henrietta between the two of them.
An animated soap bubble gleefully ate the scum ring from around the inside of a bathtub, and then the news resumed. Her father reactivated the sound to listen to the lead story, in which a family who lived in an old house was crushed when it fell in on them. The next story featured a boy who was scratched across the eyes by a cat. The scratch got infected, and the boy had to have his eyeballs amputated.
The ads returned, and Henrietta's dad muted them.
"Henrietta," said her mother, "stay away from cats."
"I will," said Henrietta. The story had scared her.
"Henrietta," said her father, a note of annoyance in his voice, "why can't you complete your work at school?"
"I'm sorry," said Henrietta. Her parents had asked her this question before, and her inability to answer it, or change her behavior, was a source of constant friction. She restlessly pushed her hands between the pillows of the couch.
"Sorry won't help you if you get Finished," said her father. He pointed at her with the television remote.
"Oh, Tom," said her mother, grabbing the remote from him. "Don't be so hard."
"Well, she needs to think about it," said her father. "I don't want her driving a garbage truck the rest of her life." He looked down at Henrietta. "Do _you_ want to be a garbage truck driver, Henrietta? Is that what you want?"
"No," said Henrietta, staring at the television screen where a magnetic kitchen cabinet door helped a woman lose weight.
"Good," said her father, folding his arms decisively. He turned to the television also, as an animated toothbrush began to dance in a mouth full of happy teeth.
Later, after eating dinner and watching more TV, Henrietta retired to her bedroom, where she typed her homework on her computer and sent it through the school's automated grading program. Henrietta's homework was generally done quickly and with plenty of mistakes because she didn't proofread.
As she worked, a thought occurred to her. The boy from the bus—what was his name? Gary. With a few keystrokes, she opened the school network. Every student at school had a public page that summarized their performance, which was intended to foster healthy competition.
When Henrietta reached the network's front page, she saw two RedAlerts at the top, and stopped to read them:
REDALERT ONE
AN ISSUE OF DISOBEDIENCE ON A SCHOOL BUS HAS NECESSITATED A CHANGE OF SAFETY HARNESS PROTOCOL. BEGINNING TOMORROW, AUGUST 26, STUDENTS WILL NO LONGER HAVE AUTHORIZATION TO UNBUCKLE THEIR SAFETY HARNESSES UNLESS THE VEHICLE IS IN A FULLY STOPPED POSITION.
Henrietta knew this change had come because of Gary unbuckling his straps. He'd revealed a flaw in the system—that was interesting. She went on to read the next alert:
REDALERT TWO
BEGINNING NEXT WEEK, AUGUST 28, TEXTBOOKS WILL NO LONGER BE USED IN CLASSES. ALL CLASSROOM MATERIALS WILL BE ACCESSED THROUGH THE SCHOOL NETWORK. THIS CHANGE IS FACILITATED THROUGH A PUBLIC-PRIVATE PARTNERSHIP WITH TINCAN TELECOMM: **HELPING SCHOOLS HELP CHILDREN HELP THEMSELVES AND US ™**. TEXTBOOKS HAVE MANY DRAWBACKS. THEY CANNOT BE EASILY UPDATED, THEY ARE HEAVY, AND THEY COLLECT MOLD. AS OF THE IMPLEMENTATION DATE, DISPOSE OF ALL TEXTBOOKS IN A SECURE WASTE CONTAINER.
Henrietta looked at her textbook. She wouldn't miss it. It really was pretty heavy.
She clicked around until she found Gary's network page, with his picture at the top. When she saw his performance ratings, her eyes widened. He was number one in the whole class for both reading and math. His behavior on the bus hadn't been suggestive of great intelligence, but there was no arguing with the statistics. Gary was as high up as she was far down. They were opposites.
At the end of the evening, her mother arrived to tuck her in. She supervised as Henrietta donned her bedclothes, brushed her teeth, and got into bed. She pulled the warm blankets up around Henrietta's ears.
"Did you see the alerts on the school page?" her mother asked.
"Yes," said Henrietta. "Actually, I saw why, with the seat belts."
"What do you mean?"
"A boy unbuckled his, and the bus stopped right in traffic!"
Her mom winced. "Why would he do that? Senseless." She shook her head.
"I don't know," said Henrietta. But maybe she did know. Maybe Gary had just been fed up.
"Stay away from that boy. Do you know his name?" said her mother.
"No," said Henrietta. She didn't often lie to her mother, but this seemed like an unusual case. She was curious about Gary, and didn't want to be forbidden from finding out more.
"I'll just check the BedCam quickly," said her mother, stepping over to the wall, where a small camera was mounted, aimed at Henrietta's bed. The BedCam relayed an image to her parents' room so they could keep an eye on Henrietta during the night. Her mother checked the operations light on the underside of the small unit, and looked at the tiny screen on the back. She frowned.
"What is it, Mom?"
"Henrietta, could you wave at the camera?"
Henrietta waved. Her mother knitted her eyebrows. "I'm going to check this. Keep waving." Her mother stepped from the room. She heard her call her father.
"Just now?" said her father.
Henrietta wondered what was going on as she waved at the blank eye of the lens. Soon her parents entered her bedroom again and her father inspected the camera, pressing some of the buttons on its back in various combinations.
"What's happening?" said Henrietta, finally putting her hand down.
Her mother tried to find words for a moment, and then said, "Come look."
She led Henrietta into the master bedroom, which was considerably larger than Henrietta's and featured a wraparound countertop on which sat two computers, two televisions, and countless cell phones plugged into chargers. (Henrietta's father worked for the communications company TinCan TeleComm, so he always had the latest models.)
Also on the countertop was a video screen plugged directly into the output of the live feed from the BedCam. Henrietta looked at the monitor. There on the screen she saw . . . _herself_ , in bed, sleeping. But she wasn't in bed. She was standing right here.
"Is it a recording from last night?" she said.
"It isn't built to record," said Henrietta's father's voice through the wall from her room. "It's a glitch." Henrietta looked at the image of herself on the screen. It was still, like a photo. She was lying on her side, facing the camera, her eyes closed in sleep.
Henrietta and her mother returned to Henrietta's room, and Henrietta looked at the small screen on the camera itself, which showed the same image. After more fruitless button prodding and empty theorizing, Henrietta's parents gave up.
"We all need to get some rest," said her mother. "I'll call the company tomorrow. Would you like to sleep with us tonight, Henrietta?"
"I think I'll just stay in my room," said Henrietta. "I'll be okay."
"Are you sure?" said her mother.
"I'm sure. I can do it."
"If you need anything," said her mother, "even if you're just scared, knock on our door, or call us." She gestured to Henrietta's cell phone, which was charging on her bedside table.
Henrietta climbed back into bed and pulled the covers up, and her mother turned out the light and closed the door. The room glowed yellow from Henrietta's nightlight, a plastic canary with a large round body. Henrietta watched the dark square of the malfunctioning BedCam and eventually fell asleep.
### Gary
She awoke just before her alarm went off, thinking she'd heard a thumping sound somewhere in the house. She listened, but it didn't repeat. She got out of bed and changed drowsily into her school clothes, blue pants and a red shirt with a yellow stripe down the back, designed for good visibility. Her room was lit dimly by her nightlight and her computer screen saver. The screen saver was a counting program. At the moment, it was displaying the number 36,548. When it reached 50,000 (in about a month), the computer would shut down and her parents would replace it.
From the other side of the wall, she heard her parents getting up. They were talking, and although the noise was muffled by the wall, Henrietta understood some of the conversation.
". . . afford to stay if the other houses keep getting bigger," said her mother.
"We've been over that," said her father.
"Maybe it's for the best. Get out of this place. Henrietta's House Sickness—"
"It's pointless. We're stuck. And we don't know it's the house's fault, anyway. It could all be for nothing."
Their voices faded as they left the bedroom and moved down the hall into the kitchen.
After eating breakfast and saying good-bye to her mother, Henrietta walked to the bus stop. The swirls of scented car exhaust dragged at her tired feet. When she reached the crosswalk, she pushed the button and waited for the picture of the dead pedestrian to turn into the picture of the scared pedestrian.
She arrived to find Gary, dressed in matching tan vinyl pants and coat, already there. He was hunched over, kneeling in the artificial turf as she approached, and his short black hair was combed back, plastered to his head and shining. He appeared to be disentangling a piece of trash from where it had become ensnared in the plastic grass blades, muttering to himself under his breath.
When he saw Henrietta approaching, he stood up quickly, looking a little embarrassed. His big eyes were bright black under his thick, black eyebrows, which were the kind that met in the middle. Henrietta didn't speak to him. She felt a little shy and stood a few feet away, watching for the school bus along the line of cars that faded into the hazy distance.
"I'm sorry I was a jerk yesterday!" said Gary loudly. He said it with such drama; it was obvious he'd rehearsed it.
"You were a jerk," said Henrietta. "But I'm sorry you got in trouble."
"Me too." Gary knitted his connected eyebrows for a moment, which made them resemble an inchworm.
"Were your parents mad?" said Henrietta.
"My mom was furious."
"Well, detention is no big deal. I get it almost every day. Oh, did I see you go in the new house yesterday? It's right across the street from me."
"We just moved in," said Gary. "Our house from before got demolished, and we had to move. Which house are you in? Are you in the old one?"
"Yes," she said.
"That's what our house was like. My mom hated it, and so did I. I love our new place. It has heated floors! Our old house was always scary, and it made me sick. At night it creaked and kept me awake."
Henrietta was about to tell him about the mysterious _thump_ that had awakened her that morning, but Gary interrupted her with another question. "Do you have many friends here?" he asked. "Are there nice kids here?"
"I don't know," said Henrietta. "I'm . . . kind of unlikable, I think."
"Me _too_!" said Gary. He clenched his hands excitedly.
"Well, maybe we'll get along," said Henrietta.
The bus rolled up just then and opened its door. Henrietta and Gary boarded as Honk Ads blared around them. "BE FAST AND ACCURATE WITH TINCAN'S NEW SKIPPING-STONE PHONE!"
Henrietta felt an odd sensation. It reminded her of when one of her headaches went away. It was happiness.
### Headaches
That morning the seating arrangement for the remainder of the year was projected on the screen at the front of class. It had been created by integrating high scoring students with low scoring ones in the hope that low scorers would be positively influenced by their proximity to high scorers and improve the class average. Certain studies suggested that this could happen.
As a result of this reshuffling, Henrietta and Gary found themselves seated next to one another—the lowest scoring student and the highest. As everyone found their seats Ms. Span initiated the math session for the day, and a series of problems appeared at the front.
"Be fast and accurate," she said.
The first problem was:
10 + 4 =
Henrietta typed "14," and the next problem appeared:
25 + 13 =
Next to her, Gary grunted, as if physically trying to push through a pile of dirt to get to his answer. It seemed to cost him physically, which was odd, given his excellent grades.
"Good, Hiroki," said Ms. Span, as she watched everyone's progress on her screen.
Henrietta was just about to type "38," when suddenly, she stopped. Her heart skipped a beat as she felt the strange sensation of someone standing next to her. Then the headache appeared.
She instantly forgot what was happening in the classroom as she reached into her pocket for her pill, popped it into her mouth, and swallowed. She clutched the edge of her desk with both hands as the headache grew.
"Are you okay?" said Gary, leaning toward her.
"I'm getting a headache," Henrietta replied. The headache wobbled one way and then another, and then it tipped and fell behind her left eye, which temporarily went blind. The headache was now medium-sized.
"What's going on back there?" said Ms. Span, removing her reading glasses to peer at the back of the room. Henrietta hadn't noticed, but everyone was turned toward her.
"Henrietta has a headache," said Gary.
"She has permission to see the nurse," said Ms. Span. "The rest of you keep working. Don't let her distract you."
The headache was a yellow, pulsing ball. Each pulse got bigger, and she could see its brightness with her otherwise blind left eye. She stood and stumbled down the aisle of desks. As she exited the classroom, Ms. Span logged her out of the test by entering ILLNESS next to her student number. Gary appeared distracted as well, and Ms. Span decided that perhaps it would benefit Henrietta to have a little help. She scrolled to Gary's name and logged him out, too.
"Gary," she said, "please accompany Henrietta to the infirmary."
Gary stood and followed Henrietta out. She seemed entirely oblivious to his presence as she stumbled through the empty hallway past other classrooms to the nurse's office. She tilted her head forward to keep the headache from rolling around in her skull.
The school nurse, Ms. Morse, looked up from her computer as Henrietta and Gary entered. Ms. Morse was a kind, older woman who always seemed sincerely concerned about Henrietta and all the other children who came to her for help. As Henrietta entered, Ms. Morse asked, "How can I help you two?"
"Henrietta has a headache," said Gary.
"How bad is it, Henrietta?" said Ms. Morse.
"Medium," said Henrietta. Ms. Morse led her to a back room that contained several rows of closely spaced cots. Sometimes when Henrietta arrived other students were being treated, usually for turned ankles or bruised elbows, but today there was no one. Henrietta lay face down on the nearest cot, her forehead pressing into the cream-colored plastic pillow.
"Which eye?" said Ms. Morse.
"Left," said Henrietta.
"Should I call your mom?"
"No." Henrietta wanted the questioning to stop. It was distracting, and she needed to concentrate. She tilted her head further forward on the pillow, to try to pin the headache against her blacked-out eye.
"I'll check back," said Ms. Morse, and she left. There was a clatter of keys as she logged Henrietta and Gary into the infirmary.
Henrietta focused on the headache. Of all the abilities she'd acquired in life (walking, speaking, reading, writing) this was her most advanced. Over the past couple years, she'd become aware of every move of her headaches. She studied them with the intensity that a deer studies a mountain lion.
She kept the pulsing ball contained, and eventually it began to subside, shedding its layers. Finally, it melted to nothing.
Her vision returned, although the outlines of everything seemed shaky, and that was how Gary appeared. His edges vibrated.
"Hi," he said. He was sitting opposite her on one of the other cots.
"How did you get in?"
"I came with you," he said. "Ms. Morse said I could stay if I was quiet. School's over now, anyway."
"I didn't even notice you," said Henrietta. "Thanks for coming."
"If we'd been reversed, I wouldn't have noticed you, either. I used to get headaches, too. Like yours. As soon as yours started, I recognized it."
"You don't get them anymore?" said Henrietta.
"Not since my mom and I moved out of our old house. My headaches stopped right away. The doctor says it was House Sickness. I bet that's what you've got, too, because of your old house."
"Did they start behind your forehead?" Henrietta was reluctant to believe Gary had the same problem as she.
"You know what?" Gary leaned toward her conspiratorially.
"What?"
"I knew you were getting it before _you_ did. I was doing the math problems, and then I thought I noticed someone standing next to you. I turned to look, but no one was there. Then you reached for your pills."
Henrietta shook her head. "That's impossible."
"But I _knew_ it," said Gary. "I saw it before you even did anything."
Henrietta shook her head again, but not too hard—she was still a little dizzy. "Who was there?" she said.
"I couldn't tell," said Gary.
"What's really weird," said Henrietta, "is I always feel like someone's there for a second." Henrietta still felt shaky, and she leaned on Gary as they made their way into the hall.
"We have to go back for detention," she said.
"Ms. Span said we could do it tomorrow, since you were sick and I was . . . well, helping out, I guess," said Gary.
As they stepped through the front doors, they immediately saw that something unusual was happening by the turnaround, where a circle of students had formed. At its center, a kindergartner lay on the ground, curled up. Her face, nearly obscured by curly black hair, was screwed into a grimace. Henrietta recognized her as the girl she'd seen in History and Nutrition, who had left the note on the screen.
Henrietta and Gary pushed forward into the ring of students as one of the bus duty supervisors arrived, a large woman named Mason whom all the children were a little afraid of. The bystanders parted at her approach, and she wordlessly plucked the girl from the pavement. "Nothing to see here. Back in your lines," she said, and strode off toward the infirmary.
Gary whispered, "Let's follow." They walked a small distance behind until Mason disappeared inside Ms. Morse's office with the little girl, and then reappeared alone heading back toward the parking lot turnaround.
"Let's ask Ms. Morse," said Henrietta.
Inside, Ms. Morse sat at her computer, typing. "What can I do for you two?" she said.
"We were wondering about that girl," said Henrietta.
"Rose will be all right." Ms. Morse looked quizzically at the two of them. "You all certainly do stick together," she said.
"We all?" said Gary.
"You kids with headaches," said Ms. Morse.
"Rose has a headache?" said Henrietta.
"Can we go sit with her?" asked Gary, glancing at his phone to see how much time they had before the buses came. It would still be a little while.
"I suppose," said Ms. Morse, "but be quiet. She's resting. Her mother is on the way." She led them back to the cots, where Rose lay curled like a baby bird, her eyes squeezed shut in pain. Ms. Morse indicated that Gary and Henrietta should sit on the cot opposite Rose's.
"Is that what I looked like?" whispered Henrietta.
Gary nodded. "Yeah," he said. "I guess it's probably what I looked like, too, when I got them."
Henrietta felt an ache in her heart. At least when the headache was her own, it occupied her to control it. Watching someone else suffer was in some ways worse. She felt Gary's hand on hers, and she firmly grasped his fingers and tried to think positive thoughts.
Eventually, Rose stirred. She opened her eyes.
"Hello," said Henrietta softly.
Rose didn't reply. She appeared disoriented.
"I'm Henrietta." Henrietta pointed to herself.
"I'm Gary," said Gary a little too loudly. "We go to school here."
"We get headaches, too," said Henrietta. "Like yours."
"I'm Rose." The little girl sat up, holding a hand to her head. "I'm all right. It wasn't bad." She looked back and forth between Henrietta and Gary. "You get them too?"
"Gary used to, and I still do," said Henrietta. "It's House Sickness."
"House Sick?" said Rose.
"Do you live in an old house?" Henrietta asked, but Rose didn't answer. She just sat and waited for Henrietta to continue. "Well, um, if you do, sometimes there are poisons that make you sick."
"We should start a _club_ ," said Gary. "We're like superheroes who can get incredible headaches on command!"
"That's a terrible idea," said Henrietta, laughing.
Rose smiled. "I'll be treasurer," she said.
From outside, the three heard Ms. Morse's voice: "Hello, Mrs. Soldottir. She has some friends with her."
Ms. Morse entered the room with Rose's mother, a willowy woman with straight blond hair and a long, pretty face, whose skin was considerably lighter in shade than her daughter's. She wore old-fashioned gray pants and a white button-up dress shirt. She was a little out of breath, as if she'd just been running.
"Hi, Mom," said Rose.
"How are you, Rosie?" said her mother, bending and kissing Rose on the forehead.
"I'm okay."
The two of them hugged.
"Were you two helping her?" her mother asked Gary and Henrietta. Henrietta nodded. "Thank you," she said, and her voice contained none of the suspicion or curtness that Henrietta would have expected from either of her own parents, had they been addressing a stranger. "The buses are here. You two should hurry."
### The Red Drip
Henrietta sat before her computer that evening, but couldn't concentrate on her homework. There was a lot going on that seemed more important than math. She thought about Gary and Rose, and the headaches they all shared. Until today, she'd felt alone. "House Sickness," she mumbled to herself, her hands hanging motionless above her keyboard. She didn't feel satisfied with that explanation, and her doubts surprised her—normally, she accepted what she was told, but this just didn't add up. She looked around her room. If it was the house's fault, why did she never get sick here?
She recalled Gary's claim that he'd seen someone standing next to her at the onset of her headache that day. She, too, had sensed someone. Rather than "House Sickness," it felt to Henrietta like "Outside Sickness," as if something was waiting for them out there.
She surveyed her small room, its bland white walls, bed, and desk. She always complained about the place, but in fact, she felt safe here. She returned to her homework for a few moments, typing out "I will tread water until help arrives," and "It is never too early to buy life insurance."
On the other side of the wall, her parents had begun arguing.
The voices stopped eventually, and her mother entered, looking careworn. "Don't forget we're going to your grandmother's tomorrow," she said. "Set your alarm."
"I will."
"And wear your dress shoes."
"I will." Her uncomfortable black plastic dress shoes were already set out by the bedside table.
"Put your pajamas on, brush your teeth, and go to bed," said her mother. Henrietta wouldn't get tucked in tonight.
"Is the BedCam fixed?" said Henrietta, motioning toward the wall where the black BedCam she used to have had been replaced by a new gray one. Her mother frowned.
"They tried three different models, and all of them had the same problem. We'll get it straightened out. Now, pack up your homework." Her mother disappeared into the hallway, and through the wall Henrietta heard her enter the bedroom again.
Henrietta was named after her grandmother, who was nicknamed Henrie. Henrietta didn't see Grandmother Henrie often, and when she did, the visit was generally short and awkward. Henrietta had always wondered how she ended up with her grandmother's name, because her mother and her grandmother didn't get along very well. If you didn't like your mother, would you name your daughter after her? It made Henrietta wonder if things had been different before she was born.
After thinking on these matters for a few moments, Henrietta's attention drifted back to her textbook. She gazed down at the addition problem she'd been preparing to work out.
It looked . . . strange.
One of the numbers had become a small, perfectly round, red dot.
Henrietta reached out and touched it. It was wet. A little of it came off on her finger. She sniffed it, and the rich smell reminded her of when she skinned her knee once.
It was blood.
She wiped her finger on her pants and checked to see if she was getting a bloody nose, which she wasn't. She stared at the drop. It must have come from somewhere.
She looked up. Above her, emblazoned brightly upon the dingy white ceiling, was a shiny red spot.
Slowly and carefully, Henrietta climbed onto her desk. The drip was seeping through the ceiling along a thin, almost invisible seam. Past the leak, Henrietta followed the seam and saw that it formed a three foot by three foot square. It was a trapdoor.
She visualized her house. She'd never thought about it before, but the pitched roof meant that there was a space, an attic, right over her room. Her first thought was to go to her parents and tell them, but she didn't move. Her mind was racing, and the conclusion to which it sped was that she would not go to them. She stepped down from her desk.
She would look into this herself, right now. From her nightstand, she plucked the flashlight she was supposed to use if there was ever a power outage. Then she climbed back onto the desk.
She pressed gently on the door, and the seam became a dark crack as it tilted upward. She tried to shine her flashlight into the space, but the angle was wrong and her head wasn't high enough to see. She let the door back down, climbed off the desk again, and sat on the edge of her bed. Her heart raced.
It might seem a bit odd that Henrietta, who had been raised to pay such careful attention to safety and the making of sensible decisions, would do something as decidedly unsafe and incautious as investigate something like this alone. It seemed odd to Henrietta, too, and she tried for a moment to convince herself to tell her parents, but she continued to sit, unmoving. She was discovering something about herself that she'd not known before. She was discovering that she was an intensely curious person.
After a few moments she stood, picked up her chair, and placed it quietly on top of her desk. Then she climbed on, balancing precariously, and crouched against the ceiling. She put one hand against the door, knitted her eyebrows, counted to three . . . and stood up.
The attic was larger than she'd imagined. A bank of tall windows to the left admitted the pale light of a greenish full moon, lighting what seemed to be a little living room immediately before her containing a low coffee table with a glass top, a sofa upholstered with a faded flower print, a small end table with a hardcover dictionary placed atop, and two wicker chairs. Behind this stood many tall, full bookcases, which largely blocked the view further into the attic, but Henrietta could see the space stretching back behind them into the shadows, moonbeams glancing upon many obscure, still shapes. Overhead, the peaked rafters diminished into the darkness.
Henrietta was so overwhelmed by it all that for a moment she scarcely noticed what lay immediately before her. When her gaze finally dropped down, she gasped and lunged back against the edge of the door, wobbling uncertainly on the chair atop her desk.
An enormous gray cat lay there, in a puddle of blood.
Ms. Span's voice popped into Henrietta's head instantly, warning her about the risk of bodily fluids, infection, tetanus, dust mites, rabid animals, and falling from heights. But Henrietta didn't budge.
The cat seemed unconscious. Additionally, it wasn't really a cat. It was twice as large as an ordinary housecat. One of its hind legs was splayed out behind its body, as long and thin as a stilt, like nothing Henrietta had ever seen.
She hoisted herself the rest of the way into the attic, keeping a safe distance. The cat's tall ears twitched, and it opened its eyes, which shone in the moonlight an incandescent emerald green with wide, black pupils.
"Hello," Henrietta whispered breathlessly.
The cat didn't make any threatening moves. Henrietta turned on her flashlight. Her eyes were drawn to a patch of wet, matted fur on the cat's hindquarter, where blood pulsed out, thick and slow. She thought back to Physical Safety period at school. This was an arterial wound.
"APPLY DIRECT PRESSURE TO ARTERIAL WOUNDS FOR FIVE MINUTES," she whispered to herself. "Please don't scratch me, kitty." She reached into her pocket and pulled out her cell phone to mark the five-minute interval, but the screen was blank. She shook it, but nothing changed. She removed her sweater, wrapped it around her hand and forearm for protection, and pressed on the gash. The cat's breathing quickened, but to Henrietta's relief it didn't lash out.
She counted out the full five minutes, then slowly removed her hand. Under the matted gray hair, the bleeding seemed to have slowed.
"I'm going to get some things to help you," she said. "Don't go anywhere."
She descended carefully back into her room and tiptoed out into the hallway. As she passed the master bedroom, her mother's voice rang out. "Henrietta? Is that you?"
"Um . . . going to the bathroom."
"Then straight to bed."
"Yes, Mom." When she reached the bathroom, she debated what to take from the medicine cabinet. Veterinary science wasn't emphasized at school beyond "INJURED ANIMALS ARE DANGEROUS." But Henrietta did recall an old television show she'd watched once, in which a boy helped an injured horse, shaving the hair from the wound and bandaging it. Henrietta selected one of her father's safety razors, a bottle of hydrogen peroxide, a package of sterile gauze pads, a tube of antibacterial ointment, and a roll of medical tape.
She headed back along the hallway, but as she passed her parents' bedroom, her mother's voice stopped her again.
" _Henrietta_ ," it said curtly.
"Yes?" said Henrietta, pausing in the hall, trying not to drop any of her supplies.
"You forgot to flush."
"Oh, right!" She quickly returned to the bathroom, flushed the toilet, and headed toward her room again.
"Straight to bed now," said her mother's voice. "And don't forget to set your alarm."
"I won't forget!" said Henrietta, hurrying past.
It was strange to enter her bedroom and see the dark square hole in her ceiling. Two minutes away from it had nearly convinced her it wasn't real. Once again, she entered the greenish glow of the moonlit attic.
The cat remained as she had left it, watching her with its enormous green eyes. Once Henrietta was inside this time, she shut the attic door behind her.
She laid out her supplies in front of the cat. "I'm going to try to help you with these," she said. Hesitantly, she began. She put her hands gently on the cat, and examined the wound. It was tough to see through the fur, and she began to shave carefully. The cat shivered as the blade touched its skin, but Henrietta successfully cleared the wound. It was small, but there was no telling how deep. It looked like a stab. Thick, dark blood pulsed out slowly.
Henrietta applied hydrogen peroxide and antibacterial ointment, and covered the wound with a piece of gauze, which she affixed with medical tape. The cat's tense breathing slowed as she finished.
"I hope it's okay," she whispered. "I have to go. I'll come back tomorrow if I can."
She took one last glance around the attic. The angled walls of exposed rafters glowed in the moonlight, and deep in the space, beyond the bookshelves, stretched a fascinating maze of dim sights: stacked furniture, sealed trunks, towers of boxes.
Reluctantly, she lowered herself into her room and pulled the trapdoor closed. It sealed up as perfectly as ever, leaving only the dot of blood at the edge, which Henrietta wiped away with the arm of her sweater.
In bed, she stared at the ceiling and imagined the animal that was right above her, sitting in the moonlit shadows of the most mysterious place she'd ever seen. And suddenly something occurred to her: when she was in the attic, no one had known she was there. She had been alone.
### Sunset
C _hirp._ The sound came from her cell phone. She'd overslept! When she answered the call, her mother's voice sounded simultaneously over the speaker and through the house from the kitchen. "Henrietta!"
"Sorry!" said Henrietta.
"Hurry up!"
"I'm hurrying." Henrietta rushed into the hallway to find her mother already there. She gasped when she saw her daughter.
"Sweetie!" She reached out and took Henrietta's hand, which was covered in rust-red dried blood. She'd forgotten to wash it off.
She jerked back. "I . . . cut myself," she said. "Picking glass out of my shoe."
"We have to clean it," said her mother, "and bandage it. Let me—"
"I'll do it after I shower. We're late." Henrietta slipped past her mother and closed the door on her. She stepped into the shower, trying to gather her thoughts. Trapdoor. Cat. She looked at the red-brown flakes on her hand. She had really done all of that. "Make it quick!" said her mother from outside—she was already in the bad mood that always possessed her when they visited Henrietta's grandmother and Al. Henrietta's mother had never liked Al for some reason that Henrietta couldn't fathom. To Henrietta, he just seemed like a nice old man, but her mother had been enraged when he and Grandmother Henrie had married two years ago, and since then Henrietta barely ever saw them.
After her shower, Henrietta found her clothes laid out: a long-sleeved shirt printed with yellow leaves, dark green pants, and her dreaded black plastic formal shoes. She threw everything on, and only remembered at the last moment the cut she'd lied about. She went to the medicine cabinet and wrapped a bandage around her thumb.
Because visits to her grandmother's often involved her sitting in a corner somewhere with nothing to do, she grabbed her textbook from her desk as she walked out, hoping she might study a little. She glanced regretfully up at the trapdoor as she left, wishing in vain for a moment to rush up there and check on the cat.
When Henrie and Al married two years ago, they'd moved together to Sunset Estates, a retirement community far out in the Addition, which was an hour's drive out across the endless traffic jam. Along the way they passed Henrietta's school. Henrietta looked at the long, low buildings and wished she was there, which was something she'd never wished before. But she had two friends now, and an amazing story to tell them when she saw them next.
"TURN LEFT AT THE INTERSECTION," suggested the friendly voice of the car's computer. Henrietta's father entered the turn lane.
"What are you smiling about back there?" he said, unexpectedly. Henrietta's eyes snapped into focus. He was watching her in the rearview mirror.
"Nothing," she said. She was a bit shy of her father, partly because she didn't see much of him. He often worked late during the week, and sometimes even on the weekend.
"Are you dreaming about a boy?" he asked impishly, raising one eyebrow to show he wasn't completely serious. But he wasn't completely unserious either.
"No," said Henrietta. She wrinkled her nose as if to say, _yuck_.
"Leave her alone, Tom," said her mother.
"We're just joking," said her father.
"It's embarrassing her."
"No it isn't." Her father's voice lost its levity. Just then, his cell phone rang. Henrietta's father's cell phone was a small, flat oval resembling a polished stone. No one else had anything like it, but her father's employer, TinCan TeleComm, always gave him the latest models early. TinCan was a new company that had formed recently when two other companies merged. Her father's job, as far as Henrietta understood it, was to help the new company communicate with itself. He'd tried once or twice to explain further, with limited success. At the moment, he listened to his polished egg, and then spouted off a string of information Henrietta could scarcely interpret.
"If I.T. can't keep Skipping-Stone's PS for UPC, Marketing just has more time, so it doesn't matter. Right."
"TURN LEFT AT THE NEXT INTERSECTION," said the car's computer. "WHILE YOU DRIVE, WOULD YOU LIKE TO HEAR SOME ADVERTISEMENTS FOR PRODUCTS THAT MIGHT INTEREST YOU?"
"Not right now," said Henrietta's mother.
"I.T. thinks that's important, but it isn't," said Henrietta's father. "Tell them it doesn't matter." This was often what her father said on business calls. Telling people what didn't matter, Henrietta thought, seemed to matter.
The phone conversation continued, as did the directions from the car's computer, until the sign for Sunset Estates appeared:
_**Sunset Estates**_
DINNER 5P BINGO 7P
The car turned into the mazelike complex, composed of single-level row houses with tan vinyl exteriors. Each house was identical to the next, resulting in a pattern as they drove: Garage, porch, front door. Garage, porch, front door. Traffic was considerably less here—it was one of the only places in the Addition where cars were sparse, because it was a dead end.
Henrietta's father concluded his conversation, returned his phone to his pocket, and looked at Henrietta again in the rearview mirror. "Now, about that boyfriend," he said, smiling.
"His name's Gary," said Henrietta, suddenly curious to see what such an admission would bring about.
" _Who?_ " said her mother, turning around and gripping the headrest of her seat with one hand. She had painted her fingernails for the birthday party, and they shone bright pink.
"I knew it!" said her father, triumphantly. He banged one hand on the steering wheel, as if to affix his astuteness there for display.
"He's my friend, not my _boy_ friend," said Henrietta. "He's the best student in our class, actually."
"The highest rank?" said her mother. "That boy Gary?"
"Yes."
"That's wonderful, Henrietta! Maybe he could help you on your homework. It's good to make good friends."
"And you never know, love _could_ blossom," her father jibed.
"Stop it," said her mother, still not amused.
"ARRIVING AT ZERO FIVE, ZERO SEVEN, SIX THREE TWO," said the computer.
"Zero five, zero seven, six three two," said Henrietta to herself, for no particular reason. _If you repeat it a few times, you'll remember it_ , she reflected. Just like the composition sentences at school that stuck in her head, or certain Honk Ads she heard over and over.
Henrietta's father parked, and they exited the car. Henrietta tucked her textbook under one arm and they approached the front door of a home that looked like all the rest, except for its unique address: 0507-632. Henrietta's dress shoes pinched, and she walked strangely, trying to find a way to proceed that didn't hurt too much.
Henrietta's father rang the bell, commencing a computer-generated rendition of "Jingle Bells," and the door opened. There stood Al, a smile lighting his old face. Al was a stooped gentleman, skinny as a stick, wearing black slacks and a green cardigan. Behind him, Henrietta could see and hear the party—guests talking and laughing, holding drinks, sitting or standing. All of them were old.
"Hello, kids!" said Al boisterously, his old voice crackling. "I saw you through the window. It's been too long!" He briskly shook Henrietta's father's hand, saying, "Young Tom!" He hugged Henrietta's mother. "Good to see you, Aline." Then he looked down at Henrietta.
"Henrietta! What occupies your thoughts these days?"
"Oh," said Henrietta, thrown off guard by a real question. Al seemed to sense her discomfort.
"Do you accept hugs or handshakes, young lady?" he asked.
"Handshakes," said Henrietta.
"She accepts _hugs_ , of course," said Henrietta's mother, a note of annoyance in her voice. She prodded Henrietta slightly from behind. It was odd—Henrietta knew she didn't like Al, but for some reason still felt it necessary for Henrietta to hug him.
Al, however, quickly held out his hand, and Henrietta shook it. "I watched you on the porch," he said genially, "and I got the impression your shoes were a little uncomfortable. Am I right?"
"They're super uncomfortable," said Henrietta.
"You know how I could tell? Because you walked like _this_!" said Al, and he limped comically ahead of the three of them, waving for them to accompany him through the crowd. Henrietta's father laughed, a laugh Henrietta recognized as fake, and the three of them followed Al inside.
The house was packed with all kinds of people united by the bonds of their common oldness. In her daily life, Henrietta saw very few old people, and it was daunting to encounter so many at once. Of course, the reason she didn't see many elderly people was because they all lived in Sunset Estates and other similar communities.
Through a sliding glass door on the opposite side of the living room drifted sizzling sounds and the rich, charred smells of a barbecue. Henrietta's mother would never allow an open flame near their house, and barbecue skewers could cause terrible accidents. But there it was, a metal clamshell on the rear patio, tongues of orange flame cooking steaks, hot dogs, and kebabs in sizzling rows. It was so mesmerizing that Henrietta didn't notice she was standing right next to her grandmother until she looked up to see Al, Henrie, and both of her parents watching her.
"Say _hello_ , Henrietta," said her father, a little sharply.
"Would you like a handshake or a hug, Henrietta?" said her grandmother, a short, elderly woman with a pronounced stoop. Henrietta looked up at her and suddenly noted the family resemblance between her grandmother, her mother, and herself, all three of them women with ruddy skin, blockish features, and sturdy frames. This was not, at the moment, a pleasant realization—Henrietta felt strangely trapped by it.
"A handshake," said Henrietta determinedly, though she could feel her mother's withering gaze. She held out her hand, and briefly grasped her grandmother's cool, papery fingers.
They sat with Henrie on a soft, cream-colored couch in the living room. Henrietta thought her grandmother looked tired. In fact, there was something of a forced gaiety to everyone at the party.
The morning melted into a series of loosely connected vignettes of dodging around long legs and overhearing snippets of conversations until the barbecue was ready, and everyone ate. Henrietta was starved because she'd missed breakfast, and she gulped her food ravenously. It tasted excellent, especially the charred bits.
"Don't eat the charred bits," said her mother.
As they all finished their meal, Al turned to Henrietta. "So, Henrietta," he said, "I've heard you're interested in reading."
"Yes, I like reading," she said, though she wondered where Al would have heard about it.
"I've got a few old books in the basement you might enjoy," he said. "You do know what a book is, don't you?" He smiled.
Henrietta held up her textbook silently as an example, a little defensive over being poked fun at.
"Be careful on the stairs," said her father.
"If the books are moldy, hold your breath," said her mother.
"Walk _this_ way!" said Al. As he limped forward, again mocking Henrietta's gait in her uncomfortable shoes, he shot a significant glance at Henrie that terminated in an inscrutable wink.
This time Henrietta imitated his walk, and doing so helped her avoid the pinchiest parts of her shoes. The two of them clowned through a crowded hallway to a narrow white vinyl door where several people lingered in conversation.
"Excuse us," said Al, "we're walking _this way_." He limped through the crowd, opened the door, and headed down a dark, carpeted staircase. Henrietta followed.
A few steps in, Al flipped a light switch, illuminating a series of overhead fluorescent panels. Henrietta had imagined the basement would be like the attic in her house—old, shadowy, and full of cobwebs. But this house was built recently, and its basement was a regular room, full of the stinging scent of new carpet.
Henrietta closed the door behind, but didn't let the latch click shut, remembering Ms. Span's lectures about getting trapped in basements, car trunks, and refrigerators, all of which would result first in suffocation, followed quickly by dehydration, hypothermia, and finally starvation (in that order).
Al descended the stairs slowly, leaning a little on the tan handrail. At the bottom, several plastic folding chairs surrounded a small, vinyl-topped card table. The basement was about the size of the living room upstairs, but instead of being full of people, it was full of books, stacked on shelves and in cases.
"Everything down here is plastic, except the books," said Al as he and Henrietta sat at the table. "These shelves are actually part of the house. It's one molded piece." He gestured to where the tan shelves merged seamlessly with the wall.
On the table before them lay two books, which Al had obviously placed with the intent of showing to Henrietta. "All of the books down here are Henrie's and my collection, which we combined when we moved in together. These two are interesting—one was mine, and one was Henrie's. Can you tell which is older?" he asked.
Both books looked old, and the writing on their covers was a highly ornamented cursive that was difficult to read.
"They have the same title," Henrietta observed.
"They're different editions of the same thing. Look at this one from the side. See how thick it is? Compare." One was considerably thicker than the other. "What do you think?"
"One has more pages."
"If you wrote a book about something, and then wrote it again a few years later, would the new one be shorter?"
"Longer," said Henrietta. "I'd know more."
"Just so," Al said. He opened each book to a random page. "Now look inside, and tell me the difference."
"One is typed, and one is . . . handwriting?" said Henrietta.
"The older one is handwriting. It's so old, they hadn't invented typing yet. The thicker book, the newer one, is typed." Al closed both books, and Henrietta studied the title on their covers, trying to untangle the cursive. "It's fascinating, isn't it, to see how people figure out things? How they learn, and fill more books."
"So, is mine the best of all?" said Henrietta, holding up her textbook.
"What do _you_ think?" said Al. Henrietta looked at the shiny, plastic cover. "I like yours," she said.
"Why?"
"They're just . . . interesting. The handwritten one most of all."
"That's nice to hear," said Al. He smiled. "It's good for an old person like me to know that young people like old things. Look at the title a little more. Have you ever seen cursive?"
"A little," said Henrietta. "B-e-s . . . the word is . . . Bestiary?" She pronounced it BEST-ee-airy.
"That's right," said Al. "But it's pronounced BEAST-ee-airy. Do you know it?"
"No," said Henrietta.
"Say 'not yet,' when someone asks you that," said Al. He winked.
"You winked at Grandma Henrie when we left the living room."
"I'm a winker," said Al.
Henrietta tried to wink in response, but she blinked instead. "Were you sharing a secret?"
"You're an observant girl," said Al. "Henrie and I have a few secrets, some that we've kept a long time. Most haven't been worth it. Do you think that's true?"
"I don't know," said Henrietta. "Some secrets might be important." She thought about the secret hiding in her attic back home.
"Perhaps," said Al. He paused. "Your grandmother asked me to take you down here so she could talk to your parents. She's going to tell them something they won't want you to hear." Henrietta didn't respond. She could tell Al was going to say more. "Henrietta, your grandmother has cancer. She's going to die."
Henrietta suddenly felt very uncomfortable, and she found her eyes drawn to the two old books, covers facing up, resting before Al on the table. The ornate calligraphy was metallic gold.
"This party isn't just a birthday party," Al continued. "It's a farewell party."
"I thought it seemed sad," said Henrietta.
"Your grandmother is eighty today," said Al. "You might not be able to imagine how old that is, but I'll tell you—it's the blink of an eye. That's as old as anyone ever gets."
"I don't know what to say," said Henrietta.
"Say anything or nothing," said Al. "The important thing is that you know. You don't need to be protected from it." A drop of water appeared suddenly right in the middle of the older book's cover. It was a perfect circle, and it came from nowhere. Henrietta thought of the drop of blood she'd seen the night before, and she looked up at the ceiling. She'd once watched a TV news story about a family who drowned when their basement flooded. She looked at Al to see if he'd noticed it.
He was crying. He wiped his eyes with one old hand.
"I don't know grandma very well," said Henrietta. This was difficult for her to admit for some reason, and it felt bad to say.
"Your parents don't come over much," said Al. "It isn't your fault, Henrietta. In fact, if it's anyone's, it's mine."
"Yours?" Henrietta looked curiously into Al's tear-reddened eyes.
He smiled. "It's a strange world," he said.
"I know." Henrietta nodded. In fact, she felt she'd only learned of the world's strangeness recently.
"I assume you're aware that your mother was against Henrie's and my marriage."
Henrietta had never been expressly told, but she'd certainly heard her parents talking about it plenty of times. "Yes," she said.
"Do you know much about your grandfather?" Al asked.
"Mom said he got sick from his work, and died."
"His name was Roy," said Al. "He and I were friends. In fact, I was his best man when he married your grandmother. Did you know that?"
Henrietta shook her head. With regard to this subject, she knew pretty much nothing, but she had long been curious.
"Roy was an agrichemical scientist, and he did grow ill from it. He was bedridden for the better part of two years before he died, and I tried to help him and your grandmother. I was at their house almost every day, tending to your grandfather's needs and helping keep the house in order. Your mother was about your age, then. And, we didn't plan it, Henrietta, but . . . well, your grandmother and I fell in love during that time." Al paused, and picked up the older _Bestiary_ and turned it restlessly in his hands.
"Your mother sensed what was happening, I believe. When Roy died, I think she blamed your grandmother and me. And maybe she was right. Henrie and I tried to end our relationship. We stopped seeing each other. We hoped we'd fall out of love. But we didn't—we couldn't. Even as the years passed. So, once Aline was out on her own and had started her own family, we faced our feelings. I asked Henrie to marry me. You know the rest—we moved out here together, and you and your parents moved into her old house."
"We didn't go to the wedding," said Henrietta.
Al shook his head, studying the book in his hands. "I wish things had been otherwise," he said. Then he suddenly looked up and met Henrietta's gaze and held it. "You know what?" he said. "Maybe they still can be otherwise, Henrietta. After all, we aren't dead yet!" He put the book decisively down on the table with a thump. "Henrietta, by gosh, let me ask you something. I'd like to be your grandfather. What do you think? Do you want a grandfather? Some silly old man?"
Henrietta hesitated. Al was, after all, more or less a stranger to her. She knew her mother disliked him, which made her feel uncomfortable. But at the same time, she saw him clearly—sitting there, his hands clasped nervously on the tabletop, worried she'd reject him. When you yourself are rejected almost every day, it becomes easy to spot in the faces of other people. And perhaps that's one good thing about rejection—it allows you to help others, if you choose to.
"You're not a silly old man," Henrietta said. She looked him up and down, joking a little as if she were studying a product for sale in a store. "I think you'll make a good grandfather," she said.
Al smiled, and released the breath he'd been holding. "Hey," he said, "I haven't given you the tour down here yet." He stood and gestured to Henrietta to follow him, and they walked back among the plastic bookcases. Al pointed out a few volumes as they passed. "There's an old journal—probably my oldest book. And that one's called _How To_ —it has all kinds of instructions in it, like how to build a bird house."
"I saw a bird once," said Henrietta. As she looked at the hundreds, maybe thousands, of titles she thought about how computers had made books obsolete. Even her own textbook, which was practically brand new, was outdated.
On the far wall of the room, past the last set of tall shelves, stood a narrow sliding door. Al pulled it open. "This is the only other room down here. I keep my tools in it. It's pretty jammed full."
The little room was lit with fluorescent lights just like the main room, and the walls were lined with more plastic shelves, which were crowded with old tools instead of old books. Henrietta recognized some of them: a skill saw, a few hammers with different heads, drills, wrenches.
"Have you ever seen tools?" Al asked.
"In safety videos," said Henrietta. "They're dangerous. We don't have any."
"Everything just snaps together nowadays, but there was a time when people had to build from scratch. I keep these back here thinking they'll eventually come in handy."
"You're nostalgic," said Henrietta. She'd seen a movie in class where a father kept some tools out of nostalgia, but then his son died when a hammer fell on him.
"Guilty as charged!" said Al, laughing. He closed up the tool room, and they returned to the card table and plastic chairs by the staircase. Henrietta's mind was carefully circling everything she and Al had talked about. Her eyes played over the covers of the two bestiaries on the table: one ancient and slim, the other old and thick. If people were books, she thought, she'd added a few chapters to herself in the last twelve hours. Finally, she spoke. "I have a secret to tell you, Al," she said.
Al looked at her seriously, and took a seat at the table. "I'm honored to hear it, Henrietta," he said.
Henrietta reached out to the older of the two books, and opened the cover a crack. "I found an attic above my room," she said. "It's full of stuff. And . . . a cat. It was bleeding, and I tried to help it."
Al leaned toward her as Henrietta lowered the cover of the book.
"What did it look like?" he asked.
"Actually, it's not really like a cat," said Henrietta.
"Did it have long legs?" said Al. He held out his hands to indicate roughly how long he meant. "Big green eyes?"
Henrietta was so surprised that it took her a moment to stammer, "Yes!"
"Henrietta," Al said, leaning forward even more, "you've made an _amazing_ discovery. The cat in your attic is a wild housecat!"
Al's excitement was catching. Henrietta felt her heart beat faster. "I don't know what that is," she said, leaning forward a little bit herself. She and Al looked like a pair of conspirators.
"You've heard of domestic housecats," said Al. "They're just like domestic horses, or domestic dogs—they've lived with people so long, they've become used to us. But all of those domestic animals have wild ancestors—wild horses and wild dogs. The same is true of housecats. There are domestic housecats and wild housecats. Now, wild animals are very strange creatures. They don't have anything to do with people. Have you ever seen a wild horse, or a wild dog? Maybe on TV?"
"On history shows," said Henrietta.
"They're all extinct now," said Al. "People made so many buildings and roads that those animals had nowhere left to live. They died out."
"But where do wild housecats live?" said Henrietta.
"In houses," said Al, "that's why they're called housecats. Normally they lived in basements or attics, but the old homes they needed have all been torn down now, except a few. The new plastic houses, like this one, are no good—the cats can't get into them. Also, people misunderstood them, and thought they were dangerous."
"I've heard cats are dangerous."
"Everything is dangerous," said Al, "but not everything is _particularly_ dangerous." He paused. "Henrietta, I'm proud of you. What you did was very brave. And you were smart to keep it a secret. If your parents knew, they'd probably have the cat exterminated."
"Some secrets _are_ worth keeping," said Henrietta.
Al smiled. "Now, as for the rest of what's in that attic—"
Before he could finish, the basement door opened, and Henrietta's mother stuck her head in at the top of the stairwell. "Henrietta," she said, her scowling face hovering over the white ruffles of her blouse. "We're leaving. Now. Your father is in the car."
"All right," said Henrietta. Her mother removed herself and shut the door. Henrietta turned to Al. "The attic," she said.
"Those things belonged to Henrie," said Al, standing. "I wonder if Henrie even remembers it's all up there—inventory from her old store." As Al fell silent, he picked up the older of the two bestiaries from the tabletop. "Henrietta, let me quickly show you this a little more." He opened it. "See how the edges of the pages are dusted in gold? And how the paper is sewn into the binding, and the hand-colored inside cover? Now, if your parents ask, you can tell them what we talked about." He winked at her. "It's a funny coincidence that I pulled these down, actually. When you learn the word _bestiary_ , you'll understand why I decided to give you this." He held out the older of the two editions. "I hope it provides you with good information."
Henrietta didn't know what to say. She carefully took the book in her hands. It was heavy and smelled of old paper, leather, and Al's cologne. The binding was rough and dry.
She looked at her own textbook, which sat on the table between the two of them. "Would you like to trade?" she asked.
"I can't take your schoolbook, Henrietta," said Al.
"It isn't my schoolbook anymore. Starting tomorrow we aren't going to use it. Everything will be on computer."
Al took the plastic book in his hands and looked it over. "All right," he said. "I'll consider it a donation. Thank you, Henrietta."
They ascended the carpeted stairs together. When they reached the top, Henrietta pushed on the door, but it didn't give.
"My mom locked us in!" she said, surprised.
Al laughed, and opened his phone. "I guess we should call for help. Unfortunately, cell phones don't work very well down here. I've been meaning to talk to your father about it." He dialed and handed his phone to Henrietta, who put it to her ear. The line was full of static, but soon her grandmother answered: "Al, are you stuck down there again?"
"Hi, Grandma," said Henrietta. "My mom trapped us by accident."
Henrie laughed. "I'll rescue you," she said.
Henrietta and Al listened through the door to the muted din of the party, Henrietta holding her new book in her arms, until the latch clicked and the door opened to reveal Henrie's amused face. She gazed down at Henrietta searchingly, and again Henrietta noticed the similarities between her and her mother.
"She knows," said Al.
"What does she know?" said Henrie.
"I love you, Grandma," said Henrietta. She held out her arms and hugged her grandmother, and Al joined them.
### The Bestiary
Henrietta had always disliked her house, with its stubby single level and embarrassing peaked roof. The newer houses that surrounded it, two or three levels with nice flat roofs, seemed obviously superior. But now the peaked roof of Henrietta's house was something special—a habitat for a wild housecat, and she was excited to get back to it.
The ride home was quiet. Henrietta's parents both seemed angry, which struck her as odd. Why would finding out that her grandmother had cancer make them mad? They fumed in silence and said nothing to Henrietta by way of explanation.
When the ride finally ended, Henrietta could scarcely contain her desire to get back into the attic. She rushed to the front door of the house and waited impatiently as her father unlocked it. Once inside, she headed immediately in the direction of her room, but her mother stopped her.
"What do you have there?" she asked, pointing at the _Bestiary_ that Henrietta held in one hand.
"Al gave it to me," said Henrietta, hoping her mother wouldn't be too interested.
"It's . . . old?" said her mother, holding out one hand. Henrietta reluctantly gave her the book. Her father looked revolted. Her parents both peered at it, trying to untangle the mess of flourishes that composed the title. "What sort of book is it?"
"I don't know yet," said Henrietta.
"Is it age-appropriate?"
"School-district-approved content?"
"Did you just _wink_ at me, young lady?" said her mother.
"No!" said Henrietta. _Had_ she winked? If so, she was glad she'd finally figured it out.
"I'm worried it will give you a headache," said her father.
"I hadn't thought about that," said her mother, turning the book suspiciously over in her hands.
"It won't," said Henrietta. "I didn't get one when Al showed it to me, and we looked at it for a long time."
Henrietta's mother frowned, but finally returned the book to her. "Put it away if you start to feel ill."
Henrietta moved again toward her room, but again her mother stopped her.
"Before you start your homework, we need to talk about something," she said awkwardly. She began to pick at the pink nail polish on one of her fingers. "Something that happened at the party today." She paused lengthily, formulating a few appropriate euphemisms. Henrietta could see that it would take nearly forever for her to broach the topic. And she just couldn't stand to wait anymore.
"I already know grandma's dying," she said. "Al told me."
Her mother and father both gasped.
"He _what_?" said her mother.
"He said _what_?" said her father.
"He said I didn't need to be protected from it," said Henrietta.
Her father pulled out his cell phone. "I'm calling that man right now," he said.
"You are _grounded_ , young lady," said her mother.
"But what did I—" said Henrietta.
"Just go to your room and think about it!" said her mother, too flustered to substantiate her anger. She quickly peeled all of the nail polish off of one finger and started in on the next.
This was exactly what Henrietta had been hoping to hear, and she left immediately as her father began sputtering a voice message to Al.
Henrietta stopped by the bathroom to grab some sterile gauze before hurrying into her bedroom and closing the door behind. She balanced her chair atop her desk, climbed to the trapdoor, and once again entered the attic, her new book in tow.
The light was much better during the day. The little living room with the couch, coffee table, and wicker chairs looked extremely inviting, and the small table with the dictionary on it reminded Henrietta that she had a word to look up.
Beyond the coffee table and seating, the bookcases towered, covered in an even film of dust. There were several sets, one obscuring the next, largely blocking her view of the rest of the interior, though she could glimpse bits of things back in the shadows: a desk, a sewing table, boxes, a crate, a chest, a dresser, an umbrella stand. The light coming through the windows illuminated more than the moon had the previous night.
She stopped short.
_Windows_? she thought. How odd . . . she'd never noticed them from outside the house before. She shook her head. A mystery to solve later.
The wild house cat had moved away from the trapdoor, and reclined now on the couch. It looked better. It held up its head easily. Its gray-furred ears twitched this way and that, homing in on the small sounds of the attic, and its enormous green eyes with their wide black pupils watched her.
Henrietta placed the _Bestiary_ on the coffee table, and approached the cat. It stood warily on its long, thin legs. This was the first time she'd seen it at its full height, and it was considerably taller than an ordinary housecat.
"I want to change your bandages," said Henrietta, "to prevent infection."
The cat sat, and Henrietta gently removed the tape and gauze she'd applied the night before. The wound looked bad, but better. It was scabbing and didn't appear swollen or infected. She replaced the bloodied gauze with a fresh square.
"I think it's okay," she said. She reached out to pet the cat, hoping to comfort it, but the moment she moved her hand toward it, it retreated to the far end of the couch. "I wish you'd let me pet you," she said. She thought she would never get enough of looking at it. It was the strangest, most wonderful creature she'd ever seen. "I wonder if you're hungry." Her eyes wandered over to the coffee table, and the _Bestiary_. She studied the baroque lettering on the cover, and then took out her phone to look up the word. But the phone was frozen again, just like the other night. She tapped the screen, and then returned it to her pocket as she remembered the dictionary on the table between the wicker chairs. She brought it to the couch and flipped through the _B_ s.
> Bestiary (n) _bes'che-er'e_ A compendium of animals, commonly including those fictitious and those extinct.
She looked up _compendium_ ("a concise collection of detailed information"), and then opened the _Bestiary_. The pages were thick, rough, and discolored into a variety of yellows, unlike the smooth plastic pages she was familiar with.
The book's text was written in a loopy, long cursive. Henrietta marveled at how much time must have gone into the making of it. Early in the year, Ms. Span had shown the class the cursive alphabet, though they hadn't ever practiced it. This book was written more beautifully than the precise, typed examples their class had seen. It flowed like a river. Henrietta touched it with her fingers and followed the lines of the word _Bestiary_ on the title page. Below was another line, which she had to look at for some time before she could unravel its meaning. "Researched and Written by Aristotle Alcott, Henrift, and Many Friends."
She wondered if this Henrift might be Henrift Andi, Humanitarian and Forward Thinker. The movie at school never mentioned him being an author.
Henrietta turned a few more of the brittle pages. Some of the paper crumbled under her fingers. She reached the table of contents and scanned it until she found a section labeled "House Animals," and the subsection "Housecats—Wild."
It seemed unbelievable that this book should have such an entry. She'd never read or heard about wild housecats anywhere before today. Why didn't her teachers ever mention them? She wondered, for the first time, who decided what would be taught at school. Henrietta noted the page number of the chapter on House Animals, and flipped to it.
_Endemic to Attics and Root Cellars. Because of its habitual reclusiveness and a lack of Research (due partly to difficulty of retaining Specimens and partly to poor persistence in captivity), few facts about the animal are known with Certainty._
_The Wild Housecat's diet remains unobserved; despite its probable unreliability, it seems appropriate to report the opinion of Tradition, as a popular Children's Rhyme suggests a subsistence on "Cobwebs and Rat Tails, Dust and Rust."_
_This Animal is considered beneficial to Humankind, as it is held not only to control Rat populations, but also to keep houses free of Spider Webs and Insects. For this reason, many Homes contain so-called "Cat Halls," thought to encourage Ingress and Egress._
_Wild Housecats are thought to possess considerable intelligence, and Tradition holds that, in some respects, they may be the equal of Humans. Such Holdings, also, have sadly not been subjected to verification through the Scientific Method._
_—A.A._
After reading the entry, Henrietta went back over it with the dictionary, looking up the unfamiliar words. She turned to the cat.
"They don't know what you eat," she said. "Maybe cobwebs."
She went to a bookcase, plucked a web from the corner of a shelf, and smeared it onto the sofa cushion near the cat. The cat flicked out one paw and patted the web. Then it yawned widely, and Henrietta saw its long, white teeth. It curled up on the couch, and evinced no further interest in its proposed dinner.
"I'd better go back before my parents miss me," said Henrietta. "I hope you keep getting better."
### Intentional Detention
Henrietta always looked forward to Saturdays, her only day off from school, but every time one arrived, it soon became yesterday. Now, early on Sunday morning, idly watching plumes of exhaust erupt from tailpipes onto the blacktop, the next weekend seemed impossibly distant. Today was a little different from an average Sunday, though. For once something had _really happened_ the day before, and furthermore she had friends to tell about it.
Gary approached from up the block. He was a large boy, and a clumsy walker whose feet frequently tangled with one another or their surroundings. Today, he arrived just as the bus opened its door. Cars honked, annoyed at the holdup even though it happened at precisely the same time every day.
"GET IN SHAPE WITH
LURMY'S NEGATIVE-CALORIE ENCHILADA!"
"TINCAN TELECOMM'S SKIPPING-STONE PHONE IS
PERFECT FOR ATHLETES!"
The children boarded and buckled in. As the bus rolled forward and the sounding horns diminished, Henrietta twisted toward Gary in her network of straps. "I have something to tell you."
"What?" said Gary. He looked interested, but his thick eyebrows loomed tiredly. It seemed to require great effort from him to prop them up in the mornings.
"Um . . . well, you should _see_ it."
Gary grinned sleepily. "You have to tell me you have to show me something?"
"It's at my house. But I can't show you till Wednesday." Henrietta sighed. "I got grounded for finding out something."
"What was it?" said Gary.
"My grandma's dying," said Henrietta. "But my parents wanted it a secret."
"I'm sorry." Gary looked thoughtful. "When my dad died, my mom didn't want me to know either. She told me he was on a cruise."
"That's terrible," said Henrietta. She could scarcely imagine such a brazen lie.
"Yeah," said Gary. "They were divorced, so I didn't see him much anyway." He paused. It seemed obvious from his expression that it had been a terrible secret to discover. "Are you sad about your grandma?"
"Kind of," she said. "I want to be. But I don't see her very much."
"My dad used to send me a card every year for my birthday," said Gary. "That's how I knew he wasn't really on a cruise—because no card came."
Henrietta's grandmother also sent Henrietta a card each year on her birthday, and this struck her suddenly as very sad—not that the cards would stop coming, but that they'd been her grandmother's most consistent presence.
Gary brightened a little bit. "Hey, you know what? I bet I _could_ come over today if you want."
"How?" said Henrietta.
"I just thought of a plan. First, we'll both have to get detention."
"Why?"
"You'll have to see," he said, smiling. His eyes looked more awake as he contemplated his mischief.
This particular Sunday was the first of the month, which meant Physical Safety Period, led by a chubby, pale, balding man named Mr. Safety, who taught his one subject throughout the district, moving from school to school. He never remembered the students' names, and Ms. Span was always present to make sure everyone behaved. It was in Mr. Safety's class that Henrietta had first seen _Watch Out for Pirates_ , one of the most exciting movies ever.
Ms. Span's class walked together in a line down the hall to the Physical Safety room, a small gymnasium with a cushioned floor. As they entered, Mr. Safety blew a whistle that hung from a lanyard around his neck.
"LINES OF FIVE!" he barked. The class formed lines of five students each.
"WE'RE GOING TO DO A JUMPING JACK!" said Mr. Safety. "FEET TOGETHER, ARMS LOOSE. WHEN I WHISTLE, JUMP AND BRING YOUR FEET SHOULDER-WIDTH APART, ARMS OVERHEAD, EXCEPT FOR . . ." Here, he consulted a list on his phone of the names of students whose parents didn't want them to do a jumping jack. "CLARENCE, HIROKI, JOSÉ, AND GARY!"
He whistled. The remainder of the students jumped, put their feet shoulder-width apart, and held their arms overhead.
"COOL DOWN!" said Mr. Safety. "Walk in place." He paced along the lines. "Sit!" he barked. The class sat. "WE'RE GOING TO DO A SIT-UP. LIE ON YOUR BACKS, EXCEPT. . . AMBER, GABRIELLE, AND GARY. EVERYONE ELSE, WHEN I SAY SO, SIT UP. BUT NOT TOO FAST OR YOU'LL GET HERNIAS. YOU DO NOT WANT HERNIAS."
Henrietta sensed Gary in the row next to her, gesturing covertly. "What?" she whispered.
"Look!" Gary pointed to the end of the last row of students. There were five kids there, with Clarence Frederick at the very back. But there was someone else—behind Clarence. Someone who was clearly not a kid.
"Who is it?" Gary said.
Henrietta stared. She blinked and stared again. What she saw there was not a person, though it kind of looked like one. It was the size of an adult, but its face was not a normal adult face. Its skin was pale yellow and even, like pudding smoothed over a tiny nose and an even tinier chin—its small mouth dangled precariously just above. It was dressed in yellow pants and a yellow button-up shirt, and it lay on its back just like the students, its hands by its sides. Its fingers were bizarre, long translucent tapers, like candles. For a moment, it flickered out like a switched-off fluorescent light. When it reappeared, it had changed position: its pale yellow eyes were staring right at Henrietta and Gary.
"SIT UP!" said Mr. Safety. The class did its careful sit-up, and the creature participated obediently.
"STAND!" said Mr. Safety.
The class, and the creature, stood.
"ALL RIGHT EVERYONE. THIS TIME, _TEN_ JUMPING JACKS, EXCEPT . . . GARY, JOSÉ, AND BELINDA. WHEN I WHISTLE, YOU COUNT. READY?"
"Ready, Mr. Safety!" said the class.
_Weeeh!_ went Mr. Safety's whistle, and everyone jumped, including the creature.
"ONE!" said the class. "TWO!"
The creature stopped at Two. It stepped away from its line, and strode purposefully toward Henrietta and Gary. Along the way, as it passed other students, it reached out with its long, waxy index finger and tapped them on their foreheads. No one seemed to notice, or see the creature at all.
"Wh-what is it?" Henrietta stuttered.
"It's coming over here!" said Gary, backing out of his line.
"THREE! FOUR!" said the class.
Then another voice sounded, above everything. It was Ms. Span, and her tone was sharp. "Gary and Henrietta!" she said. "Quit clowning!"
"FIVE! SIX!" said the class.
"Run!" said Henrietta as the creature closed on them.
"I'm . . . not supposed to run," said Gary, his voice shaking.
" _Go!_ " said Henrietta, and she shoved Gary in front of her. They fled.
"Stop!" Ms. Span yelled as they departed their line and zipped through the one adjacent, interrupting the jumping jacks of several students.
"SEVEN!" said the class.
"Oh my!" said Mr. Safety as Henrietta and Gary skidded from line to line, causing considerable confusion.
"EIGHT!" yelled some students. "SIX!" yelled others. The lines began to break up as Henrietta and Gary desperately pushed through, weaving toward the front of the room, toward Mr. Safety.
"Take care now!" said Mr. Safety nervously, holding up one hand as if to ward off their approach. He placed his whistle in his mouth, preparing to blow.
"It's gaining!" yelled Gary, glancing back. The creature wasn't running, exactly—it flickered after them, disappearing and reappearing in a series of approaching snapshots.
Gary's backward glance was poorly timed. He did not see Clarice Sodje looming up before him, right in the middle of her eighth, or maybe ninth, jumping jack. Gary and Clarice were two of the larger kids in the class, and their impact was considerable. Clarice was mid-jump when they collided, and she and Gary crashed to the padded floor in a grand sprawl. Henrietta was right behind, and there was no time for her to stop. She tripped over them both and went flying right into the soft, unprepared stomach of Mr. Safety.
_Weeeeeeeeeeeh!_ shrilled the whistle as she and Mr. Safety tumbled backwards. Mister Safety landed squarely on his back and expelled the remainder of his breath in a rush that popped the whistle from his mouth. It shot to the limit of its lanyard and snapped back, smacking him in the eye. "Weeeeeeeeeeeh!" went the sound again—but it wasn't the whistle this time. It was Mr. Safety himself, and this final falsetto cry was followed by a series of agonized gasps as he struggled for the breath that had been knocked out of him.
Henrietta landed to the side and scrambled forward, looking back to check on Gary, who had now untangled himself from Clarice and was gaining his feet.
"TEN!" said the few members of class who'd managed to keep count during the fracas.
"Gary! Wait!" said Henrietta. Her glance back had revealed that the creature was no longer pursuing them. It was gone—vanished.
Gary turned around, searching. "It must be somewhere," he said, eyes darting this way and that. "Oh, wow, it was . . . that was . . . _the thing_! The one I saw when you got your headache!"
There was no opportunity for further discussion. Ms. Span, who had just called the nurse's office to order some medical assistance for Mr. Safety (who lay splayed and gasping on the ground, both hands covering his injured eye), returned her phone to her pocket and grabbed both Henrietta and Gary by the ears.
"We are returning to our classroom, _now_!" she barked at the class. "Line up and follow!" She dragged Henrietta and Gary with her into the hallway, the class obediently following, tittering animatedly at how exciting it had all been, and what ridiculous weirdos Henrietta and Gary were.
Gary's plan was now, for better or worse, in motion: they'd earned the detention they'd wanted.
At the close of the school day, as the rest of the students filed out to the buses after completing their final typing practice on the subject of autumn ("I WILL NOT JUMP IN A PILE OF LEAVES"), Henrietta and Gary stayed in their seats. Gary winked conspiratorially at Henrietta, who tried to wink back and surprised herself by succeeding.
"Henrietta and Gary, you will spend your detention today sitting quietly and regretting your behavior," said Ms. Span from the front. "Did you know that our class ranked in the thirtieth systemwide percentile today in no small part because of your disobedience?"
"We're sorry," they said in unison.
"Say it to yourselves, not to me," said Ms. Span.
Time crawled by. Ms. Span worked at her computer up front, putting together her materials for the following day. Once the buses had left, Gary gave Henrietta a significant look. Then he turned to the front of the room and said, loudly, "Mom?"
Henrietta's eyes widened. Ms. Span looked up, but didn't remove her glasses—obviously surprised. She squinted at Gary, and then at Henrietta, and then at Gary again.
"Yes, Gary?" she said evenly.
"Can we give Henrietta a ride home? Because she missed her bus, and, you know, we live right across the street from her."
Ms. Span removed her glasses. "Absolutely not," she said. "I don't know her parents at all. I don't want there to be any misunderstandings."
Henrietta was dumbstruck, observing for the first time that Gary and Ms. Span had the exact same thick, black eyebrows that met in the middle.
"We could call her parents and ask," said Gary.
This, then, was it: Gary's plan. Henrietta saw that for it to succeed, she'd have to help.
"My mom would appreciate it," she said. "She's probably working on dinner right now. I can give you my number." Henrietta reached into her pocket for her phone.
"It's already in the school network," said Ms. Span. Without another word, she dialed Henrietta's house. "Hello. Is this Mrs. Gad-Fly?" she said. "Yes, that's right. Henrietta's teacher. My son and your daughter are both here for detention today due to some regretful misbehavior," here she paused to glare out across the room at the two of them, "and they've missed their bus. Would it be convenient for you if I gave Henrietta a ride home? Gary has just pointed out that we live across the street from you." There was a long pause. "Twenty minutes. It will be nice to meet you, too." Ms. Span pocketed the phone and stood from her terminal. "We'd best get going."
"Great!" said Gary. He looked at Henrietta and waggled his eyebrows. The two of them stepped into the hall while Ms. Span gathered her things.
"She's your _mom_?" Henrietta whispered.
"Yeah," said Gary. "No one knows, so don't tell." Their whispered voices echoed in the empty hallway, bouncing off the giant mural image of happy children below the words "SENSIBLE, EFFICIENT, EDUCATION (S.E.E!)"
"I won't," said Henrietta. She shook her head. "It's so weird."
"To you, maybe," said Gary.
Ms. Span emerged from the room and the three of them proceeded down the hall, following a yellow line on the floor through a set of doors and down a flight of stairs to the parking garage. "Henrietta, it will be interesting to meet your parents," said Ms. Span. "I didn't know Gary had told you I was his mother. You two must be good friends."
"We are," said Henrietta.
"Because we sit together," said Gary. "I've been helping Henrietta bring up her scores. We're best friends now."
As they approached Ms. Span's car, a blue station wagon, a recorded voice issued from a speaker on the cement floor of the garage.
"THE SCHOOL DISTRICT IS NOT RESPONSIBLE FOR ANY INJURY RECEIVED WHILE ENTERING, LEAVING, OR TRAVELING THROUGH THE PREMISES."
"Can Henrietta and I sit in the back?" said Gary, as Ms. Span unlocked the doors.
"Buckle all of your seat belts," said Ms. Span. They fastened themselves into the rear seats. The engine started and the car's computer came online.
"WHERE ARE YOU GOING?" it asked.
"Home for Henrietta Gad-Fly," said Ms. Span.
"FOLLOW THE SIGNS TO THE LOT EXIT AND TURN LEFT." Ms. Span navigated accordingly. "WHILE YOU DRIVE, WOULD YOU LIKE TO HEAR SOME ADVERTISEMENTS FOR PRODUCTS THAT MAY INTEREST YOU?"
"No," said Ms. Span.
"THANK YOU," said the car. "THIS THANK-YOU WAS BROUGHT TO YOU BY MIRACLE MEDICAL'S EARHELPER. EARHELPER IS AN AUDIO REFINEMENT DEVICE THAT REDUCES TRAFFIC NOISE WHILE INCREASING VOICE CLARITY. EVER HAVE TROUBLE HEARING WHAT YOUR CAR IS SAYING WHILE YOU DRIVE? WITH EARHELPER, YOU'LL NEVER SAY 'WHAT?' AGAIN."
Ms. Span reached the parking lot exit and passed through the pay station.
"SEVENTY DOLLARS AND SIXTY-SEVEN CENTS HAS BEEN AUTOMATICALLY WITHDRAWN FROM YOUR ACCOUNT," said the car. "THANK YOU FOR USING AUTODEDUCT. YOUR CONVENIENCE IS OUR REWARD. TURN LEFT."
"So, Henrietta," said Ms. Span. "Which house do you live in?"
"The one with the peaked roof."
" _Oh_ ," said Ms. Span. She was silent for a few moments. "Is that . . . very old?"
"It's safe," said Henrietta.
"But, your headaches," said Mrs. Span. "You're getting House Sick, aren't you?"
"Nobody really knows for sure," said Henrietta.
"Gary," said Ms. Span, "I think you should wait in the car when we arrive. I'm worried your House Sickness might come back if you go in."
"Oh, Mom," said Gary, "I haven't had a headache in a long time. Can't I come in just for a minute? _Ple-ease_?" His _please_ utilized a special tone that children have, which can crack almost any parental decree, and Ms. Span reluctantly assented.
They drove in silence for a while, Gary occasionally waggling his eyebrows at Henrietta to call attention to the perfect success of his plan.
When the car reached their block, Ms. Span turned left into Henrietta's driveway, and the engine stopped automatically. Henrietta and Gary extricated themselves from their seat belts and stepped onto the driveway with Ms. Span.
"I can't wait to show—" Henrietta began, but she stopped halfway through the sentence.
"What?" said Gary.
Henrietta pointed toward her roof. There were two dim squares near the top of the house where the siding was a slightly darker color.
"What are you looking at, Henrietta?" said Ms. Span.
Henrietta retracted her hand. "It's a different color," she said. But what she was thinking was, _Where are the attic windows_?
They reached the front door, and Henrietta's mother opened it, wearing a bright yellow blouse and brown polyester slacks.
"Hi, Mom," said Henrietta.
"Hi, honey," she said, looking not at Henrietta, but at Ms. Span. "Hello, Ms. Span," she said, holding out her hand. "I'm Aline, Henrietta's mother."
"It's nice to meet you, Aline. Call me Margaret."
"Please, come in," said Henrietta's mother. Henrietta and Gary entered, and Ms. Span cautiously followed. They stepped into the sitting room, a space Henrietta and her parents didn't use much, though it contained some of their nicest furniture: a shiny vinyl couch, two chairs with faux-leather backs, and a long plastic table with legs shaped like columns of cell phones, which Henrietta's mother had bought for her father when he got his job at TinCan TeleComm. The table was empty but for a single glass vase that contained a fabric rose tastefully adorned with plastic dewdrops.
Ms. Span studied the room, especially the ceiling. "I don't think I've ever been in a house this old," she said.
"Sorry," said Henrietta's mother. "We're moving out soon."
"Not until the end of the school year, though, right?" said Henrietta. Her mother had said this to visitors before, so Henrietta wasn't too alarmed by it, but she still felt the need to speak up.
Her mother smiled a little awkwardly. "No, of course not," she said. In fact, it was unlikely that they would move at all unless they could find someone to buy the place for more than it was worth.
"Gary and I used to live in a place almost this old," said Ms. Span. "The city finally bought it from us and demolished it."
"That sounds wonderful," said Henrietta's mother.
"Mom, can I show Gary my room?" said Henrietta.
"Of course, but Gary should ask his mother."
"Can I, Mom?" said Gary.
"I worry you'll get House Sick," said Ms. Span.
"Does Gary get House Sick?" said Henrietta's mother.
"He used to," said Ms. Span, "until we moved."
" _Ple-ease?_ " said both children at once.
### Scaredy Gary
"What was that thing?" said Gary, as soon as Henrietta shut the door. She shivered, recalling the ghastly, flicking creature.
"I don't know," she said.
"And why _us_?" said Gary.
Henrietta shook her head. She took out her cell phone, thinking to look it up. "Search: Ugly yellow creature," she said into it.
"THANK YOU FOR SEARCHING WITH TINCAN TELECOMM," said a friendly, computerized woman's voice. "YOUR SEARCH FOR UGLY YELLOW CREATURE MAKES ME WONDER IF YOU WOULD LIKE TO BUY SOME YELLOW RAIN BOOTS. WOULD YOU LIKE TO BUY SOME YELLOW RAIN BOOTS?"
Henrietta hung up. She looked at Gary. "I don't know how much time we have before you have to go," she said, "and there's something I have to show you, no matter what." She went to her desk and pulled out her plastic chair. Her computer's counting program was running: 36,548. Henrietta looked twice at it. That was odd—that's where it was the last time she'd looked.
"It was so strange-looking," Gary mused, still thinking about the creature. "Not like a real person at all. And its fingers, tapping everyone!" Gary held his hands out, letting his fingers droop forward limply, and then wiggled his pointer fingers. "Hey, what are you doing?" he said.
Henrietta had placed the chair on top of her desk. "You have to promise not to tell anyone about this," she said.
"I promise," said Gary.
Without further pause, Henrietta climbed onto her desk, and then onto the chair. She put her hand against the ceiling and pushed.
The trapdoor opened.
Gary gasped.
"Shh!" said Henrietta, glaring down at him. "Follow me, but be quiet."
"Follow you?" he said. "But I _can't_." The pitch of his voice rose as he spoke, like air squeaking from a balloon.
Henrietta was a little surprised. On the bus the other day, Gary had seemed fearless. But she'd also noticed he was the only student to be released from every exercise during Physical Safety. "Why not?" she said.
"I . . . could fall."
"We'll be careful," said Henrietta. "This kind of thing is the reason people invented carefulness in the first place." Without allowing him another moment to protest, she grabbed the edge of the trapdoor and pulled herself into the attic.
Gary followed, reluctantly. He wrestled his squat body onto the chair, and then dragged himself up with great effort. As he emerged into the shadowy space, Henrietta whispered, "Welcome to the attic!"
Gary sat up and looked around in amazement, his eyes lighting on the bookcases, the coffee table, the windows, and finally on the wild housecat. He emitted a mousey squeak when he saw it, and held up his hands. The cat stood from its seat on the couch.
"This is a wild housecat," said Henrietta. "It isn't particularly dangerous." She paused. "Gary. Open your eyes." Gary opened them a tiny bit. To the housecat, Henrietta said, "This is my friend Gary. He's not particularly dangerous, either." The cat watched Gary with its huge green eyes.
Gary stepped forward from the trapdoor. " _Wild_ housecat?" he said.
"I found it up here," said Henrietta. "It's hurt, but it's getting better."
"What happened to it?"
"It got stabbed," said Henrietta, "but I don't know how."
Gary continued looking around, his eyes probing back toward the obscure, shadowy interior behind the bookcases. "This place is huge," he said.
"And you know what's really weird?" said Henrietta. "No one knows we're up here."
Gary froze. His black eyebrows bunched together. He opened his mouth and stuck his tongue out, and squinted, and balled his hands up into fists. "Auuughoo . . ." he moaned. The next thing Henrietta knew, he had launched himself to the trapdoor and scrambled down to the chair. In his rush, however, he misstepped, lunged sideways, and fell onto the desk and then to the floor with a powerful thud.
Henrietta hurried after to find him sitting next to her bed, staring intently into her BedCam. Henrietta listened carefully to see if her mother or Ms. Span had heard the commotion. Miraculously, they hadn't.
"I got scared," said Gary, his eyes glued to the BedCam.
"You sure did," said Henrietta. "That's broken by the way." She pointed to the BedCam. "It can't see you."
Gary turned to her. "Really, _really_ scary up there."
"Scaredy Gary," said Henrietta.
Gary frowned. "What?"
"It's what the kids were saying on the bus."
"Well, it isn't true," Gary snapped.
"Then let's go back up," said Henrietta.
Gary set his features determinedly. "Right. Yes. Okay. No problem."
Again, he followed Henrietta through the inky opening, being careful not to stumble off the chair. Henrietta closed the door after them and watched Gary as he turned around in the space, taking everything in.
"Nobody knows," he said to himself. He clenched his fists. Henrietta thought he was going to panic a second time, but he seemed to fight it off. He relaxed. "I'm all right," he said. He strode purposefully over to the windows and looked out through them.
"Remember when we were outside, when I pointed?" said Henrietta. "You can't see the windows outside—they're covered."
"Wow!" said Gary, pointing through them.
"Shh!" Henrietta was still worried the adults would overhear.
" _Look!_ " Gary whispered intently. Henrietta joined him and gazed out, down at the neighborhood outside.
The neighborhood she saw wasn't hers.
### The Wikkeling
Henrietta and Gary's new friend Rose was very smart, but most people—even her parents, who loved her—thought she was a bit dull. She'd often stare vacantly out into space, and she seemed unresponsive at times or said things that didn't make sense. But there was a reason for this. Rose could see something that other people couldn't see—a creature called the Wikkeling.
She had first encountered it when she was very young, before she'd started school. She'd been out for a walk with her father along an old alleyway behind a hospital, to her father's favorite dumpster. The dumpster was made of clean, white plastic, and it was full of unused hospital supplies such as medical tape, unopened boxes of bandages, and even surgical tools—items her father used for repairing books, which was his hobby.
He had opened the top of the dumpster and disappeared inside while Rose kept a lookout (because people don't generally like it when you steal their trash). Once she was alone, however, she felt a strange sensation, like someone was standing next to her. She turned.
Its fine, thin hair was yellow, and its face smooth and young, with a strange small nose and chin. It wore yellow pants and a yellow button-up shirt, and its body flickered like a fluorescent light.
"Who are you?" Rose said, startled.
"I'M WIKKELING," it said. Its lips didn't form the words when it spoke—it opened its round mouth and the sound spilled out, scratchy like an old recording.
It raised one hand revealing five extremely long, waxy fingers, and before Rose could react, it tapped her on the forehead with one of them, and flicked out of sight, disappearing in an instant. Rose immediately felt nauseous. Her head throbbed. She began to cry, and her father heard her and climbed immediately out of the dumpster. As he reached her, she lost consciousness.
She awakened later in her bedroom, her parents kneeling next to her, concern clouding their features. She tried to explain what had happened, but it didn't make much sense, even to her.
After that, Rose saw the Wikkeling often. If she was in the park, it stood near the swing set. If she was in class, it walked past the coatrack. It spent most of its time tapping people on the forehead, just like it had tapped her. But other people didn't seem to mind. They didn't get headaches.
Rose was extremely frightened of the Wikkeling, and quickly learned that for some reason it rarely went into her house, so she spent as much time as she could inside, deep in the interior of the old place, which was filled with old books.
Rose lived far out on the sparkling plain of the Addition, where the homes were constructed entirely of plastic and featured amenities such as disposable bathtubs, antibiotic tabletops, digital windows, and underground parking. But the house Rose lived in with her parents was quite unlike these. Rose's house was very, very old. It was unique in the Addition—an ancient island in a sea of newness.
Rose's parents referred to their house as the Library. The walls all through the place were lined with shelves packed tight with books of all sizes and topics. The house was enormous. The main level and all of the three upper floors were full of reading material. Some of the rooms had hand-painted signs over the doorways describing different genres, such as COOKERY or COMPUTERS. The house had been this way since Rose could remember, and she and her parents were rarely alone there.
Visitors often happened by to borrow or return books, and Rose knew them all by name. Her parents called them the Subscribers. They didn't look like the other people in this story, with new clothes, tidy haircuts, and the latest personal accoutrements. The Subscribers showed up in thick layers of old clothes, often spotted with grease or fringed with dirt. Their shoes looked as though they'd been repaired many times. They carried backpacks, into which they stowed the books they borrowed, and from which they produced the books they returned.
The Subscribers never entered the house through the front door. They arrived from the back, via a narrow alleyway directly abutting the plastic, windowless wall of a skyscraper behind. They were admitted only after supplying a complex secret knock that involved both knocks and scratches, and after Rose or her parents looked through the peephole at them. Once they were recognized and admitted into the kitchen, they'd be offered a cookie.
"Don't let anyone in if you don't know them, or if they don't know the knock," Rose's mother told her once. The Subscribers had all signed a contract after being interviewed by Rose's parents. The Library Use Agreement ran as follows:
_I promise, as a Subscriber to the Library, to:_
_Return what I borrow_
_Donate new materials when possible_
_Conserve the Collection_
_Be a good friend to Rose_
_Conserve the Collection_ meant _Repair the Books_. Once per week the Subscribers gathered to do book maintenance. Rose's father, who had been trained once upon a time as a doctor, was the lead conservator, and as such many of the repaired books somewhat resembled repaired people. Rose's father would often do his work as if in an operating theater, with Rose assisting.
"Thread," he would say, and Rose would hand him thread.
"Glue," he would say, and Rose would hand him the glue.
Because of all of this, Rose knew how to fix books before she knew how to read them. She'd reattached bindings, installed new endpapers, and even affixed new covers to books that arrived stripped, sometimes making her own artwork after her parents told her what the book was about.
Rose's life in the Library was also a surprisingly athletic one. The Library was so large that living in it required a healthy amount of walking, especially going up and down stairs, because the four-level place had neither escalators nor elevators. Two levels, the main level and the second level, had thirty-foot high ceilings with bookcases stretching from top to bottom. These walls had wheeled ladders built into them, which could be rolled from place to place.
Rose's parents encouraged her to use the ladders, and she quickly displayed a natural talent for climbing that was a bit mystifying given how slight she appeared. She could climb before she could walk, and became consequently good at falling without getting hurt.
"She'll swing from her ponytail when she grows up," her mother once predicted as she watched Rose leap from ladder to ladder.
"We'll see whose hair she has when she grows up," said her father, laughing. While her mother's hair was long, blonde, and straight, her father's hair was black and curly. Rose's, so far, was in between.
That was the life of the house's back door: the life of the Library. The house also had a front door, which was seldom used. Visitors at the front door were nicely dressed, and they came from cars they parked in the slim driveway. They were treated very cautiously by Rose's parents, and were never invited in. This was one of the first categories Rose ever understood. _There are two kinds of people: those who come through the kitchen with the secret knock, and those who come to the front._
A few weeks before Rose had started kindergarten, she'd watched her mother answer the front door and talk with a woman whose hair stood in perfect, rigid blond ringlets. The woman wanted to discuss the parent-teacher association at Rose's school. When the woman left, Rose asked, "Do the front door people ever meet the kitchen door people?"
Her mother put her hands on Rose's shoulders. "Rosie," she said, "they don't. And it's very important that they never do. When you start school in the fall, all the kids you'll meet will be front door people. I need you to not ever tell anyone about the Subscribers—not friends, or teachers, or anyone. Do you understand?"
"It's not alright for us to have all of these books?" said Rose.
"The books are fine," said her mother. "The problem is us. You, your father, and I live here secretly. No one, except for the Subscribers, knows we're here. This house is supposed to be empty. If you tell anyone that we're living here, we'll have to leave forever."
"But people come and talk to you at the front door all the time," said Rose.
"We're tricking them for now," said her mother. "We said we own the place. Hopefully, they won't look into it any further."
This issue proved so important that her parents came to her bedside that night and repeated the whole conversation. Then, at breakfast the next morning, they talked it over a third time.
At the moment, Rose sat at the small kitchen table on the ground floor of the Library, eating a bowl of corn cereal. The kitchen door opened behind her, and a waft of lilac scented air entered from the alleyway as her mother stepped into the house and closed the door behind. She took a chair opposite Rose, and set her bags on the floor: a cloth sack filled with groceries and a backpack constructed of stitched-together inner tubes from old bicycle tires.
"Hello, Rosie," she said. She kissed Rose on the forehead. "Is your father home?"
"He's in Political Science," said Rose around a mouthful of cereal. Political Science was a small section in the northwest corner of the fourth floor, a cozy wood-lined room with a stained-glass window depicting a boy and girl walking along a dirt road, carrying schoolbooks in their hands.
Her mother's cell phone rang, a chirping sound like a cricket. She dug it from her pocket and saw GAD-FLY on the screen. "Hello?" she said. "Oh, yes—of course I remember you." She paused. "No, Rose doesn't have her own phone. But she's here, if you'd like to talk to her." She held out the phone to Rose. "It's for you."
### Through the Windows
Henrietta and Gary gazed down in wonder at a street they'd never seen before—a broad, one-lane brick boulevard planted on both sides with enormous, leafy maples whose branches stretched out to touch Henrietta's house.
People strolled up and down paths on either side—a couple arm in arm, a man walking a dog, a group of grandparents with grandchildren. Everyone's clothes were of a strange style that Henrietta thought she'd seen once or twice in old pictures. Men and women alike were dressed in wool coats, buttoned across the front over wide lapels, and sported hats of a variety of styles, some with brightly colored feathers protruding jauntily from hat bands. Everyone, from little boys to old women, wore brown or black dress shoes.
Gary was the first to master his surprise sufficiently to say something. "The houses look like yours." Across the street they could see a row of homes with pitched, shingled roofs. The general style was the same for all, but each house was different from its neighbors. It was as if ten people had each been asked to draw a circle three inches across, and the ten circles that resulted were similar, but also different depending on who made each. A few of the houses were one story, like Henrietta's, but some were two. Several had open front porches, one with a porch swing, and some had no porch at all. From the chimney of one house billowed dark smoke, illuminated occasionally by tiny embers. Inside the front room of another, Henrietta saw several candles burning.
"Our house is old," said Henrietta. "It was my grandmother's. All of _those_ houses . . ." she trailed off.
"They became like mine," said Gary.
They were both thinking the same thought, but neither spoke it right away because it sounded ridiculous.
"This is the _past_ ," Henrietta whispered finally. "It sort of makes sense, I mean, why the windows are blocked off when we're outside, but not here. If we're looking into the past—the windows weren't blocked yet."
Golden-green light filtered through the tops of the trees. Below, the brick street turned the sunlight that reached it into cinnamon. As the two continued to stare, their eyes fell upon an object in the middle of the boulevard in front of Henrietta's house. It resembled an enormous, irregularly shaped table.
"That's the biggest picnic bench I've ever seen," said Gary.
" _Is_ it a picnic bench?" said Henrietta. A couple of passersby stopped and sat on its edge, conversing. "I think it's . . . a stump!"
"You mean, from a tree?" said Gary. "Like in the history unit at school?" Henrietta knew what Gary was referring to—a movie called _They Built with Trees_ , about how people used to make things out of wood. "It can't be a stump," Gary mused. "Stumps aren't that big. You could park four cars on that!"
"But it is," said Henrietta.
"Wow," said Gary, and he looked up, obviously trying to imagine the enormity of the tree that once grew there. "Hey," he said. "What do you suppose will happen if we go outside now? Will it be now, or then?"
They opened the trapdoor. Henrietta looked at the housecat, who silently watched from the couch. Before it on the coffee table sat the open _Bestiary_.
"It looks like the cat's been reading," Henrietta joked.
Gary laughed. "Studying for the Competency Exam!"
They dropped into Henrietta's room and returned to the living room, where their mothers were still talking, sitting on the couch.
"Ready to go already?" said Ms. Span as Gary entered.
"Um . . . actually, if it's okay, Henrietta and I are going to work on homework together. I thought I'd help her with math."
"That would be wonderful, Gary," said Henrietta's mother. "I'm sure Henrietta would appreciate it."
"I will," said Henrietta. "And we were just going to also . . . "
"Go outside for a second!" said Gary.
The two children, nodding in unison, rushed to the front door and exited onto the sidewalk.
Before them stretched the same scene they saw every day: a four-lane asphalt road crammed with cars. Traffic lights winked. Enormous plastic houses squatted behind green squares of fake turf.
"I wonder when they widened the street," said Gary.
"I guess things had to get bigger," said Henrietta. "Did the clothes those people wore remind you of anyone?"
"Through the windows?" said Gary. He shook his head.
"Rose—with the headache. She wears a wool shirt sometimes."
"You're right," said Gary.
"We should call her," said Henrietta, stepping back inside. "We could invite her over."
They entered the living room to find Gary's mother preparing to leave.
"Gary, I'm glad you're going to help Henrietta. But be on time for dinner, and be _careful_ when you cross the street."
"I will, Mom," said Gary.
Ms. Span turned to Henrietta's mother. "It was a pleasure to meet you, Aline."
"And you, Margaret."
Ms. Span departed.
"Mom, we were wondering," said Henrietta, "if we could invite our other friend to study with us. She's in kindergarten. She was sick yesterday at school, and Gary and I helped her."
"Sick?" said her mother. "Is she contagious?"
"Sick from headaches, like me. Can we invite her?"
"Your father will be home soon, dinner is on the way, and I'm still finishing up some work . . ." Her mother trailed off when she realized how nice it was that Henrietta was gaining some friends. "Oh, all right," she said.
Henrietta and Gary returned to Henrietta's room, where they looked up Rose's number on the school network. Rose's mother answered and agreed that Rose's father could bring their daughter over and that, of course, he would enjoy seeing Henrietta's house and meeting her mom.
They waited out on the sidewalk. Traffic crawled through the lilac haze, and they thought about the street in the past. Henrietta wondered if the red bricks had been buried under the asphalt. Gary wondered about the trees, and the giant stump—were the roots still growing underground, even now? The thought made him feel claustrophobic.
"I wonder what kind of car Rose's dad drives," said Henrietta. "Hey, did you see any cars when we looked out the attic window?"
"I didn't," said Gary. Just then, a small puff of exhaust blew a candy-wrapper past their feet. Gary bent somewhat reflexively to grab it, and when he straightened he saw two figures approaching along the sidewalk.
He squinted at them. "There they are," he said. "They're . . . _walking_?"
Rose and her father were indeed walking along the sidewalk, holding hands. Henrietta and Gary had scarcely ever seen adults walking anywhere except to and from their cars.
"On foot they're going faster than the traffic," Henrietta observed.
Rose's father was tall, slender, and darker-skinned than Rose, with curly black hair. He wore khaki cotton pants, which Henrietta could identify easily because they weren't shiny like most pants. Rose was wearing the same wool shirt Henrietta had seen her in the other day.
"You _walked_ ," Gary said as they arrived.
"It's good to meet you both," said Rose's father. His voice struck Henrietta as quite friendly. "Rose's mother told me you looked after Rose during her headache. That was very kind."
"I'm Henrietta, and this is Gary," said Henrietta. "Come in and meet my mom." She opened the front door and ushered everyone in.
Henrietta's mother was waiting in the sitting room, and she stood as the guests entered.
"It's nice to meet you," said Rose's father. "Call me Sid."
"Aline," said Henrietta's mother. "Won't you have a seat? Sorry about this old house. We're planning to move soon."
"Oh, I like old houses," said Rose's father. "Some things about them can be quite nice."
"You'll have to tell me what those things are," said Henrietta's mother with a laugh.
"I don't think you've met my daughter, Rose," said Sid.
"It's nice to meet you, Rose," said Henrietta's mother. "Are you in kindergarten? You seem a little young, I must say."
"I'm old," said Rose.
"Can I make everyone some instant lavender tea?" She gestured toward the kitchen.
"Thank you, that would be nice," said Sid.
"Um, can we go study now?" Henrietta blurted.
As soon as they entered Henrietta's room and closed the door, they saw that Henrietta and Gary had accidentally left the chair sitting atop her desk. Henrietta winced at the oversight. They would need to be more careful.
"Why is that up there?" said Rose.
"Rose," said Henrietta, "we're going to show you a secret. But you have to promise not to tell."
Rose, used to keeping secrets, nodded.
Henrietta gave Gary an expectant look, and he clambered onto the chair, opened the trapdoor, and pulled himself up easily, as if he'd been doing it for months.
"An attic," Rose observed, intrigued but not apparently amazed.
"When we're up there," said Henrietta, "you have to be quiet so our parents don't hear."
Henrietta was going to help Rose, because she seemed too small to climb up by herself, but to her surprise Rose zipped onto the desk and the chair, jumped to catch the frame of the attic door, and pulled herself inside, all in about half the time it normally took Henrietta, who followed as quickly as she could, feeling oafish in comparison.
The wild housecat stood and stretched on the couch as they entered.
"Rose," said Henrietta, "this is a wild housecat. I found it up here." Then to the cat she said, "This is my other friend, Rose."
The cat curled its tail over its feet and yawned.
"What's its name?" said Rose.
"I . . . don't know!" said Henrietta, a little taken aback.
"Yeah, we _should_ call it something other than just 'the cat,'" said Gary.
"What's a good name for a wild housecat?" said Henrietta.
"Is it a boy or a girl?" said Rose.
"I don't know that, either," said Henrietta. "My grandpa said it's supposed to be extinct, but I don't know much else."
They all looked at the cat. It did not announce its gender.
"We should pick something that would work either way," said Henrietta.
"How about Mister Lady?" said Rose.
Gary and Henrietta grinned. "Perfect!" said Gary.
"Do you mind if we call you Mister Lady?" Henrietta asked.
The cat narrowed its eyes a little and cocked its head. It didn't seem thrilled. "It doesn't _not_ like it," said Gary. He turned away from the couch and walked over toward the windows.
"If you ever want us to call you something different," said Henrietta, "just . . . let us know." That cat perhaps nodded a little bit, though Henrietta couldn't be sure. It lay down and curled up on the couch, closing its eyes.
"I've only been up here a few times," said Henrietta to Rose. "I don't know what any of the stuff back there is." She gestured past the bookcases at the many shadowy objects behind in the deep recesses. "It used to belong to my grandmother."
"Rose, come look out the windows," said Gary. Rose and Henrietta joined him.
On the street below, a woman with a cart full of fruit gestured toward her apples, several adults on the sidewalk carried on an animated discussion, and a group of children skipped past, joking with one another and playing tag. One of them jumped onto the giant stump and ran across it.
"I don't understand," said Rose.
"This is the past," said Gary. "That's what the street outside used to look like here, a long time ago."
"We don't know why we can see it, though," said Henrietta. She and Gary were waiting for the moment when Rose understood, and became as thrilled as they had been. Instead of seeming surprised, though, Rose said, "Their school gets out a little after ours." Gary and Henrietta saw that the children walking past below were all carrying books and book bags.
Henrietta glanced over at the couch then to see Mister Lady hop down from it and limp over with some difficulty to stand next to the three of them, looking down on the old town with a gaze that seemed somehow sad. After a moment, the cat put a paw gently up against the glass.
"I think Mister Lady might be from there," Rose speculated. "Maybe she's wishing she could go back."
Henrietta glanced at the couch again, and noticed something. The cobweb she'd earlier placed there was gone. "Hey!" she said to Mister Lady. "Did you eat that?"
Mister Lady limped back over to the couch and laboriously climbed onto it, her injured leg dangling awkwardly. She looked at the three of them, and her expression, if Henrietta interpreted it correctly, said _It's about time you noticed._
Henrietta said, "I've been trying to figure out what it eats! I was reading this—" she pointed to the _Bestiary_ , open on the coffee table. She flipped to the page on wild housecats as Gary and Rose looked on.
"That's a drawing of one!" said Gary, pointing at the illustration in amazement. "Right there, like in a textbook, like it's real!"
"It is real," said Henrietta. "And read the entry." She gestured to the cursive.
"I can't read cursive," said Rose.
Henrietta looked to Gary.
"Uh . . ." he said. His eyes darted around suddenly, evasively, and he backed away from the book. He turned partway toward the trapdoor, but stopped. A grim determination settled on his features, and he faced his friends. "I have to tell you both something," he said. "My secret." He looked up at the rafters and then down at his shoes. "It's . . . well . . ." He hesitated, and seemed to brace himself. "I can't read," he said finally. He spat the words.
In the midst of the strangeness of the attic, this really was a surprise. Henrietta had half expected Gary to admit that he could make things float with his mind, or become invisible. It took her a moment to absorb his admission.
"But you're the teacher's kid," she said. Gary's face turned scarlet, and he glanced at Rose. Henrietta clapped one hand over her mouth—she'd been so shocked, she'd forgotten that this was also a secret.
"It's okay," Gary sighed. "Rose, my mom is our teacher." Rose nodded, still somehow immune to surprise.
"But you pass the practice tests!" Henrietta protested. "And the Competency Exams! You're the best in class."
"I know how to _type_ ," said Gary. "I just can't read what I type."
"What about when we write compositions?"
"I ask my mom what it's going to be, and then I copy stuff the night before and memorize the letters."
"That seems harder than reading," said Henrietta.
"Well, it isn't," said Gary. "And don't tell my mom. If she knew, she'd ground me forever."
"You'd be Finished," said Henrietta.
"Yeah," said Gary. "But the thing is, too—" Now that he'd started talking, he wasn't about to stop. From the quickness with which the words spilled forth, it was obvious he'd wanted to confess this for a long time. "—I think I'd _like_ being a garbage collector. I wouldn't have to fake anything anymore. And, and, actually . . ." Gary put his hand in his pocket and pulled it out to reveal a small, crumpled up piece of paper. He sat at the table and began to unfold it. "I kind of like garbage. I have a collection of it."
Henrietta and Rose stared at him as he continued to attentively smooth the small sheet.
"Trash?" said Henrietta.
"Interesting trash," said Gary. "Like this," he said. He finished smoothing out the little piece of paper. It was a sticky note which had the words **henRift and andi** scrawled upon it.
"Where did you get that?" Henrietta and Rose said simultaneously. For the first time, Rose looked completely surprised.
"History Nutrition room trash can," said Gary. "I saw Henrietta throw it out, and I wondered what it was. I've looked at it for awhile. I think someone made a mistake on it. I think it says 'And and I.' I'm not sure," he admitted, still obviously embarrassed by his poor reading skills.
"Rose wrote it," said Henrietta.
"That movie was wrong," said Rose. "There's no one named Henrift Andi. It was two people. Henrift and Andi."
"How do you know?"
"A book my dad read me," said Rose. "It said there were two people. They were scientists. A man and a woman."
"That's weird," said Henrietta. "Why would the movie be wrong?"
"I wish I could read," said Gary, folding the sticky note and returning it to his pocket.
"Well, we did tell my mom we'd study," said Henrietta.
Gary looked doubtful. "I'm pretty stupid," he said. "Maybe we could try later. Weren't you about to tell us something?"
Henrietta let the issue drop. She pointed at the _Bestiary_. "This says wild housecats eat cobwebs, and so I put a cobweb on the couch, and . . ."
"And Mister Lady ate it?" said Gary. "Let's get more!" There was a relieved note in his voice.
For the next several minutes, they forgot themselves in a flurry of gathering. Henrietta and Gary stayed close to the main area, both still a little nervous about being on their own. The dark lanes of old stuff behind the bookshelves, though full of magnificent cobwebs, seemed a bit scary. But this didn't deter Rose. The attic reminded her of home. The moment she'd arrived and taken a breath she smelled the wonderful rich pages of all the old books.
She went straight behind the bookcases, going far back into the dusty depths, where there were old dressers, tables, locked chests, stacked boxes, a record player, and many other fascinating antiques.
Encouraged by Rose's confidence, Henrietta and Gary ventured a little further as well, and soon the three of them had amassed a considerable ball of cobwebs, which they deposited next to Mister Lady on the couch. As the children watched, the cat batted the ball lightly with one velvet paw, sending it rolling a few inches across the cushion, and then looked up at them. Henrietta thought its expression might have said, _This could have been a little bigger, but thanks._
"We should probably go back down," said Henrietta, "in case my mom comes to check."
She opened the trapdoor, and the three returned to her room. They heard Rose's father's voice from elsewhere in the house, and proceeded to the sitting room, where Henrietta's mother and he were conversing.
"We have instant lavender or instant peppermint," Henrietta's mother said. Henrietta turned to Gary and Rose, and motioned for them to step back into the hall.
"They're _just_ sitting down," she whispered. "That's what they were saying when we left!"
"I don't get it," said Gary.
"I don't think any time has passed," Henrietta said.
## PART 2
### Spike-Tailed Fish and Flesh-Eating Worms
Time passed—or didn't—happily for a few weeks after these many discoveries. Henrietta visited the attic twice each day: afternoons after school with Gary and Rose, and at night before bed, alone, when she changed Mister Lady's bandages. The wound was almost entirely healed, although for some reason it seemed reluctant to close completely.
Henrietta liked being in the attic at night. It reminded her of the evening she'd first discovered it, and she'd begun to enjoy being by herself. She found she could think more clearly, and she sometimes sat on the couch in silence with Mister Lady, not doing anything other than rolling the events of the day around in her head.
When the moon shone through the windows it brought a haunting glow to the little living room, the dusty bookshelves, and the deep interior. It was both eerie and beautiful.
Afternoons when school finished, Rose, Gary, and Henrietta went up together. They'd spend hours talking, looking out the windows, reading old books, playing games, and studying. One afternoon Gary brought some of his most prized trash objects, which included a strip of fabric from an old chair that had an image of an acorn stitched on it, and a box from a brand of TV dinner that didn't exist anymore. Henrietta and Rose couldn't see exactly what Gary found so fascinating, but he spoke in low tones, saying, "Now _this_ is of _particular_ interest. . . ." before revealing a tan Styrofoam packing peanut. Henrietta began to wonder if maybe Gary really _should_ let himself get Finished from school. If anyone would make a good garbage collector, it would be him.
Another afternoon, Rose surprised Henrietta and Gary when she arrived in the attic armed with linen thread, glue, a ruler, a razor, and a metal spike with a wooden handle, all of which she laid on the coffee table with great solemnity.
"What is all of this?" said Gary. "It looks dangerous. What's that?" He pointed to the spike.
"An awl," said Rose. "These are for repairing books. Like the _Bestiary_." She opened the front cover and showed Henrietta and Gary that the endpapers were partly unglued, and a tear had started down the hinge of the back cover.
"I didn't know books could be repaired," said Gary. "Isn't it more sensible to just get a new one?"
"A new _Bestiary_?" said Rose.
"Oh, right," said Gary. When it's irreplaceable, it makes sense to take care of it.
There followed a most illuminating conservation lecture in which Rose described the parts of a book, its materials, and common construction methods. Then she effected a simple repair of the _Bestiary_ , much to the amazement of Henrietta and Gary. "My dad is better at it," she said.
"Where did he learn?" said Henrietta.
". . . Nowhere," said Rose. Further questioning produced only silence.
Mister Lady grew increasingly active as days passed, often chasing after motes of dust in the main area. One afternoon Henrietta found a tuft of fur beside the couch, and concluded that the cat was also doing some hunting.
Through the windows, Henrietta, Gary, and Rose continued their fascination with the old world outside. Usually when they arrived they'd see the children coming home from school, but sometimes the two eras came unstuck from one another and they'd arrive to find midnight through the windows, or sunrise.
Whenever the sun was out, so were people, crossing the boulevard on errands, chatting on street-side benches, selling groceries and other goods from carts. One afternoon a whole picnic took place atop the gigantic stump, and thirty people ate and drank there. Occasionally, a car passed, strange and large, and people stopped what they were doing to look. Cars were a curiosity.
At sunset, a man would stroll the boulevard bearing a long metal stick with a flame at the end, lighting the streetlamps. A real flame burned in every one, flickering and casting wavering shadows along the trunks of the maple trees.
All three children spent considerable time reading up in the attic, even Gary. One of the things that had always kept him from reading was that he couldn't see the point. Why read when you could learn from TV or the radio, or ask a computer, or your cell phone? But in the attic, reading was quite helpful.
One afternoon, while exploring behind the bookshelves, Gary happened upon a large glass jar full of thick, glaucous liquid. Floating in it was some kind of preserved creature. He carefully picked up the jar and carried it out past a pile of luggage, an old dresser, a sewing table, and the bookcases to the living room, where he placed it on the coffee table.
Henrietta jumped when she saw it. The thing looked like a gray bird, but it was covered in scales like a fish. It sloshed gently back and forth in the cloudy solution, which Gary's transport had agitated.
If Gary had seen something like this at school he would have taken a picture of it with his phone, and the phone would have told him what he was looking at. Or he could have spoken a few descriptive words, like "Part bird, part fish," and his phone would have sorted some search results based on those keywords.
And, indeed, Gary did produce his phone with the intent of taking a picture . . . but the phone was dead. "It's broken," he said, shaking it a little.
"That happened up here to me, too," said Henrietta. She took out her own phone. The screen was blank.
"Rose, does yours work?" said Henrietta. Rose, over by the windows, was watching a group of children kick an empty can along the brick street.
"I don't have one," she said.
"I forgot," said Henrietta. "Why don't your parents get you one? What if there's an emergency?"
Rose shrugged. She came over to look at Henrietta's phone.
"It's like at my house," said Rose.
"Your house?" said Henrietta.
Rose's mouth snapped shut so fast her teeth clicked.
"What is it, Rose?" said Gary, sensing her hesitation.
"Nothing." Rose sat by the coffee table and looked into the glass jar at the strange preserved creature.
"Does your house have a lot of books in it?" said Henrietta. "Is that why you know how to repair them?"
Rose continued to look silently at the creature. At the end of its tail was a long, threateningly curved hook.
"Let's see if it's in the _Bestiary_." She slid the book toward Gary. "Do you want to try it?"
"You know I can't," he said, scowling.
"I didn't mean it like that," said Henrietta. "I was just thinking—it doesn't matter how long it takes when we're up here. If you want to try."
"I guess that's true," said Gary. He brightened a little. The lack of time pressure made the situation seem a little more encouraging. He opened the Bestiary, and Henrietta and Rose sat on either side of him. Even Mister Lady approached to watch.
"This might be kind of hard," said Henrietta, "since we don't know what it's called."
"But there are pictures," said Gary, flipping through the brittle old pages. "Let's just look for it."
"I didn't think of that," said Henrietta.
"Because you can read," said Gary, smiling. It took a while, but they finally found an illustration that resembled the creature in the jar.
Henrietta read the title silently. She looked at Gary.
"Do you know how to sound things out?" she asked.
"A little," said Gary.
"That's a Q," said Rose, pointing at the first letter.
"Kw . . ." Gary said. "And the next letter is a _U_ , then an _A_ . . . _A_ is like 'apple,' right?"
"It depends," said Henrietta.
"That's what I hate about reading!" said Gary, instantly exasperated. "It's like they make it hard on purpose."
If the alphabet had been invented in the Addition, all _A_ s would probably sound like the _A_ in _apple_. In the Addition, streets ran on a grid, and the houses were identical. The "number one" at one restaurant was the same as the "number one" at every other restaurant, and it would only make sense to have an equally regular alphabet.
But the alphabet was irregular. The rules all had exceptions, and some of the exceptions even had exceptions. The alphabet was like the Old City. The pitched roof of Henrietta's house, for instance, was an exception to the rule of flat roofs. Maybe Henrietta's house should be remodeled to have a flat roof, and maybe the alphabet should be remodeled so all the _A_ s sounded like the _A_ in _apple_.
But if Henrietta's house had a flat roof, this story wouldn't have happened. Mister Lady would not have been able to sneak in. Mysterious jars would not be hidden away among strange artifacts. There would be no windows looking into the past. The alphabet's different _A_ s had caused Gary some problems, but the attic was giving him the time he needed to sort them out.
With occasional guidance from Henrietta, Gary and Rose sounded out the text of the description of the quaverly (a word in which the _A_ sounds like the _A_ in " _Safe_ "). Here's what they read:
**_A beach-dwelling, nocturnal Carnivore, Quaverly live in Schools of up to one hundred thousand individuals. Roosting and sleeping during the Day under driftwood logs, suspended by a chitinous Hook, this placid Creature drops from its roost at night to the Sand, and enters tidal pools and shallows to provender upon Shrimp, Sand Fleas, and other small Fauna._**
_**Though edible, Quaverly is rarely prepared owing to its bitter Taste, want of Meat, and nearly imperturbable Hide. However, its Abundance has made it useful to Humans in times of severe Lack—especially in Wintertime, when its Population swells after the Autumnal mating season.
—Henrift**_
They didn't figure it all out at once. Even Henrietta had to use the dictionary here and there for words like _chitinous_ (hard, like a beetle shell) and _provender_ (to feed). When they finished, Gary slumped. "That was the hardest thing I've ever done."
"Way harder than what we read in school," said Henrietta.
Rose peered at the page. "What's that?" she asked, pointing to the word "Henrift" at the very end.
"That's the name of the person who wrote this entry," said Henrietta. "Look here." She flipped to the front of the book. "See, _Aristotle Alcott_ , that's the _A.A._ at the end of some of the entries. I wonder if this _Henrift_ is the same one from History in school."
"I can't believe they wrote this whole thing out by _hand_ ," said Gary.
"My grandfather said it was made before people typed. He has a newer _Bestiary_ , too. It's old, but not this old. It's typed, and it's longer, because they'd learned more."
"It's weird, as we flipped through," said Gary. "I've never heard of any of the animals."
They looked again, skimming past the illustrations. The diversity of pictured life was fascinating: mesmer vole, airship whale, springer, tail fox, candlefly, statium, pulchritude hound, pif, greater pif, paf. . . .
"I guess they're all extinct," said Henrietta. "That's what my grandfather said about wild housecats, too."
As if on cue, Mister Lady took a brief experimental swipe at the quaverly in the jar, as if to ascertain whether or not it might be chased.
"I think Mister Lady is a girl," Rose said abruptly.
Henrietta and Gary agreed, though they weren't sure why.
At school, the lessons slowly progressed day to day and the Competency Exam grew inexorably nearer. Gary continued to cheat as always, and so maintained his position at the top of the class. Henrietta never cheated, but she found her work improving considerably, even though school seemed more awful and boring than ever. She wasn't sure why she was doing better. Partly, it was that she didn't want to miss her bus after school, because she could go into the attic with her friends. Partly it was because she was feeling happier than she had in a long time.
And partly, it was because her reading skills were improving. Every day, she learned new words, and encountered more difficult sentences, and she waded through them with increasing expertise. In fact, she was becoming quite an excellent reader, and she had an excellent memory for new words, which seemed to stick in her head like flies to flypaper. She was ensnaring herself a superlative vocabulary.
After school on the day before the Competency Exam, while Henrietta ducked around with Rose behind the book cases collecting cobwebs, they ran across a book whose title caught Henrietta's eye: _Early Town_. She slid it from the shelf and returned to the couch, laying her harvest of webs next to Mister Lady. She opened the book and turned to the title page.
**EARLY TOWN:**
**A Book of Records Including Maps and Services**
The next page, printed in black, blue, and red ink, folded out to become twice as wide as the book. The legend at the top read CITY MAP. Henrietta had never seen a map before, except in the movie _Watch Out for Pirates_ , when some buccaneers had used one to find buried treasure.
Gary and Rose looked over Henrietta's shoulder at the folded-out page. Mister Lady approached as well, curious as she always was whenever anyone read something. She often peered over their shoulders while they waded through the old books, and Henrietta had begun to wonder about it—maybe the cat really could read.
"What is it?" Gary asked.
"A map," said Henrietta.
"Like in _Watch Out for Pirates_?"
"But not a treasure map," said Henrietta. "Just a regular map. It shows how the city used to be."
"I don't get it," said Gary.
"Pretend we're floating over the buildings," said Henrietta. She pointed at a straight red line. "This is a road."
"And that's a river," said Rose, pointing to a meandering blue line.
"And the squares are buildings?" said Gary. "We don't really need these anymore I guess, since your car or your phone just tells you."
"This seems kind of better, though," said Henrietta.
"It seems complicated," said Gary.
"But when your car tells you, you don't really know where you're going."
"Yeah, I guess so," said Gary. "If the phone says 'turn left' . . . I can see how a map is better."
"Also, I like how it looks," said Henrietta. "It's like a painting."
"If I found it in the trash, I'd definitely keep it," said Gary.
"It doesn't show what's here now," said Rose.
"It's what used to be, back then, I bet," said Henrietta, gesturing out the attic windows.
It took some puzzling out because the map contained many streets, houses, and streams. Labels for everything were crammed in at different angles, but they eventually found a street with the same name as theirs: _Boardwalk_. Strangely, it was right at the far left edge of the map, like it was the last street, period.
"It looks like we're on the edge of the world," said Henrietta. "I wonder what's beyond it."
"The Addition," said Rose.
"But why isn't it on here?"
"It wasn't built yet."
For a few moments the children contemplated the fact that the city they lived in, every building and every street, hadn't always existed.
"I want to know everything that happened between then and now," said Gary.
"I wonder how long ago it was," said Henrietta.
"If any of those people are still alive," said Rose.
The three of them walked to the windows and looked down at the narrow brick road. At the moment, the boulevard was nearly empty. Two young men sat on the edge of the giant stump, arguing earnestly and passing a steaming thermos back and forth between them.
"They should see their street now," said Gary. "I wonder if they'd like it."
"Maybe some of them _have_ seen it both ways," said Henrietta. "Like my grandparents."
"I want to tell those people not to chop down the trees," said Gary. He looked out at the soaring boughs and wished he could somehow prevent them from disappearing. The leaves were full of the gold and red of autumn, and some had fallen in stiff breezes and littered the ground, skittering here and there along the bricks and across the great stump.
"Hey, look!" said Henrietta, pointing high into the branches of one of the trees. "There!" She kept pointing, following the form as it moved from branch to branch within the deep orange foliage. Because of its orange fur, it was tough to see. "It's . . . I think it's a wild housecat!"
Suddenly, from behind them, Mister Lady leaped down from the couch. Before Henrietta could turn, the cat was beside them, pressing her front paws against the windowsill. Her green eyes were wide.
"I see it!" said Gary.
"Me too," said Rose.
The cat was huge, a tabby even larger than Mister Lady. It left the shadows for a moment to run across a thick branch right by the windows. As it passed, Mister Lady let out a long, plaintive meow. She pawed at the glass, but the tabby neither saw nor heard her. In another instant it reentered the shadows of the heavy foliage, and disappeared as it leaped effortlessly on its long legs from one tree to another and continued down the boulevard.
Mister Lady stared after it, her eyes seeking this way and that for another glimpse. Finally she turned and walked back to the couch. Her limp was nearly gone now, but her gait was slow, dejected.
"I think she really is from out there," said Gary.
"I wonder how much longer she'll stay here," said Henrietta. "She's almost better." This was a tough thought, because none of the three wanted to lose Mister Lady, who was as much a part of their shared friendship as anyone.
They looked down as a few more schoolchildren ran by, chasing after leaves and playing. Seeing them turned Henrietta's thoughts back toward class. "Gary, do you think you'll cheat on the Competency Exam tomorrow?" she asked.
"I'm just not good enough yet to do it on my own," he replied. "What if I got Finished? I'd never see you guys again. Say, do _you_ want to cheat?" he asked. "I could help you."
"Actually, I'm kind of looking forward to it," said Henrietta, "after all the practice we've done up here."
"These books are different from the computer questions," said Gary.
"It's all reading, though," said Henrietta. "People have been doing it since forever."
That night, as Henrietta changed Mister Lady's bandages and sterilized the wound, she made a disturbing discovery. In the near-darkness of the attic, she shone her flashlight on the cat's injured leg and saw several small, white worms crawling.
" _Ugh!_ " She simultaneously recoiled from them and reached out to brush them off. Mister Lady wrenched away and leapt onto a bookshelf several feet above Henrietta's head. It was the first such leap Henrietta had ever seen the cat make. The ease and strength of the jump was astounding, and Henrietta felt a small resurgence of the fear she'd felt when she and Mister Lady first met.
"I'm sorry," said Henrietta. "Please come down."
The sight of the worms squirmed awfully in her brain. How could that have happened? She'd been so careful to keep the wound clean. Her eyes fell on the _Bestiary_ , which lay as always atop the glass coffee table.
Henrietta flipped to the index. Scanning through, she noticed an entry for "Worm—Flesh-Eating."
**_Worm, Flesh-Eating:_**
**_Though disturbing on first encounter, and often erroneously associated with Uncleanliness and Disease, the Flesh-Eating Worm is beneficial to Humans. The presence of Flesh-Eating Worms in a Wound is indicative of the Restoration of an Injury to Health, as the creatures consume only dead and diseased Flesh. Their attentions are a boon to Healing. It is most unfortunate that many souls benighted by Ignorance view these worms incorrectly as the Cause and Continuation of the Necrosis they in fact hasten to eradicate.
—Recorded and observed by E.S._**
It went against good sense, but somehow it also made sense. Henrietta turned to the bookcase, from which Mr. Lady glared down. "All right," she said. "I'll leave them. But it's gross."
### The Competency Exam
After weeks of practice tests, detention, and homework, the morning of the Competency Exam arrived to find Henrietta waiting at the bus stop for Gary, bursting to inform him about flesh-eating worms. Gary was a good person to tell disgusting things to because he could be relied upon to make revolted faces. Hopefully the diversion would take both of their minds off the test for a few minutes.
Unfortunately, Gary was also the kind of person to show up a little late for things, and he trudged up just as the bus arrived. As it slowed for the children, the cars behind it began honking.
"MIRACLE MEDICAL WISHES ALL STUDENTS GOOD LUCK
ON THEIR COMPETENCY EXAMS!"
"GOT AN A ON YOUR EXAM? CELEBRATE WITH A SKIPPING-STONE
PHONE FROM TINCAN TELECOMM!"
Henrietta said hello to Gary, and he said hello back, but neither could hear the other over the racket. They strapped themselves into their seats, the blue warning lights turned off, and the yellow safety light turned on. As the bus picked up speed and the Honk Ads diminished, Henrietta said, "Are you nervous?"
"I know all the questions," said Gary. "I'm ready."
"You really just ask your mom?"
"Yeah."
"Do you feel guilty about it?" Henrietta couldn't imagine taking similar advantage, were she in his position.
"Maybe a little," he said, shrugging as well as he could in his harness. "But I also think the tests are dumb."
"I never thought about it that way," said Henrietta. She always felt bad about her grades, but what if grades were the problem, and not her?
Suddenly, something happened that didn't normally happen.
From the empty seat in front of them, a head rose up and peered back over the headrest. The face had a tiny nose, a weak chin, and thin yellow hair partly covering a tall forehead. It flickered briefly as its pale yellow eyes focused on the two of them.
The bus's security system didn't respond to the infraction: No lights went on or off, and the engine didn't shut down.
Gary gasped. "You!" he said.
Henrietta froze. Her heart gulped.
"Get away!" said Gary. He wrestled against his restraints, but after his encounter with the driver weeks back, the school district had retrofitted the buses with automated seat belts. The children were trapped.
The thing's mouth opened into a round hole with a guttering yellow light glowing up from its pink throat. Its voice leaked out, slow and scratchy, like a recording. Its lips didn't move to form the words. "WHERE DO YOU GO?" it said.
Henrietta wanted to look over at Gary, but her eyes wouldn't respond to her urging. "WHERE DO YOU GO?" the voice came again. Its inflections were exactly the same as the first time. The thing brought an arm up over the back of the seat and one long, waxen index finger reached toward Gary.
"Don't!" Henrietta whispered.
The finger lightly tapped Gary's forehead, and the creature winked out, disappearing like a switched-off light.
"I need my medicine," said Gary, his voice trembling and small. He was pale and sweating. He squeezed his eyes shut.
Henrietta racked her brain, trying to figure out if there was any way to get out of her straps. Finally, she did the only thing she could think of: she screamed.
Once they arrived at school, a crumpled Gary was whisked to Ms. Morse's office by the bus duty supervisor. Henrietta, after a brief trip to the principal's office to receive a Behavioral Citation and a lecture about Being Disruptive on the Bus, had to proceed straight to class to take the Competency Exam.
It was no longer practice. This was what they'd all been preparing for. Statistics from this test would be tabulated for all districts, schools, classes, and students. Schools could lose funding, teachers could lose jobs, and students could be classified as At Risk or even Finished. Henrietta entered the room and sat silently at the back, next to Gary's empty chair.
At the front, Ms. Span stood and silently watched her nervous students. She was dressed in a black sweater and black slacks, scrupulously devoid of any piece of lint. Her black hair was pulled back into a perfect bun, and her normally thick eyebrows were plucked into precise arcs. Despite her severe appearance, Henrietta could tell she was worried. Not only was the exam itself stressful, and her best student sidelined, but that student was her son, and she couldn't go visit him to see how he was doing.
When Ms. Span had everyone's undivided attention, she spoke. "Read every question twice," she said, speaking slowly and articulating carefully. "Be fast and accurate. Never leave a blank. It is vitally important that you do well today, for yourself, and for me, and for the school. Failure will have real consequences, up to and including being Finished." She paused. "Don't be nervous," she added. She smiled nervously. "Don't be anxious."
The class hung on her every word. Finally, she donned her reading glasses and spent a few moments tilting them back and forth on her nose to make sure they were perfectly straight. Meanwhile, in homes and workplaces all across the Addition, parents held their cell phones before them in sweating hands, watching anxiously for news.
Ms. Span sat at her terminal. "Get ready," she said as the clock approached the hour. In seconds, every student in every school in every district everywhere would be presented with their first short response question; the same question for everyone, every child responding in parallel with every other child.
The clock turned. "Begin!" said Ms. Span as everyone's screen went momentarily blank, and then lit up with the prompt.
WHERE DO YOU GO?
"Don't start!" said Ms. Span, her voice shrill. She typed at her terminal. "It must be a glitch. . . ." She frowned. "No, this is it. This is the question! Go!"
"What does it mean?" said one student.
"It doesn't matter," said Ms. Span. "Just answer it!"
"But what do we say?"
"Where you go! Write about where you go!"
The students wrote, though the question was unlike the ones they were accustomed to. Henrietta studied her own screen a little longer than everyone else, half expecting the face of the yellow-haired creature to loom up and look back at her. Her cursor blinked impatiently. Time was clicking by. She slouched a bit in concentration, and began.
I NEVER WEARY OF PERAMBULATING TO LURMY'S, BECAUSE IT ALWAYS HAS HIGH-QUALITY, AFFORDABLE PROVENDER.
As she typed, the word _perambulating_ (which she knew meant to walk about leisurely) showed up underlined in red—a mistake. She had learned the word from the _Bestiary_ , though, and had even looked it up in the attic's dusty dictionary, so she knew she was using it correctly. Another red underline appeared under _provender_ , but she kept writing—there was no time for second thoughts.
Soon, the two minutes were up, and the terminal froze. The data was sent to the district for processing. "Don't worry about how you did," Ms. Span chirped from the front. "It's too late. Math is next!"
Many of the students looked mystified and nervous as the seconds ticked down toward the appearance of the first multiple-choice math question. "Remember, C is most common," said Ms. Span quietly. Henrietta looked up at her, a little surprised. Giving directions after the test began was against the rules. The class's composition grades must not have been good. The first math question appeared.
20 + 5 =
A) 25
B) 25
C) 25
D) 25
"Don't respond!" Ms. Span said. She typed for a few moments and watched her screen. "That's the question!" she called out. "Answer and move on!"
"But which is it?" said one student.
"Just _answer_ ," barked Ms. Span.
Henrietta clicked C), and her computer made a soft _ding_ sound to indicate that she'd answered correctly. The same _ding_ sounded around her as other students submitted their answers, and here and there a _clunk_ rang out, the sound of a garbage can's contents being dumped into a truck. Henrietta saw some students sitting with their eyes squeezed shut. The next problem appeared. This one was fill-in-the-blank.
30 - 10 =
Henrietta typed "20" and submitted it. To her surprise, though, her computer responded with a heart-stopping _clunk_.
Around her, other computers _clunk_ ed. At the front of the room, Ms. Span's neat bun had begun to come loose. A strand of hair hung in front of her face, and her reading glasses were askew. She typed madly at her terminal.
The next question appeared.
20 - 5 =
"Don't respond!" she said, typing furiously. "All right, everyone. _Silence._ Pay attention. Math is different now. Minus means plus now. When you see minus, think _plus_."
"But—" began several students.
" _Minus means plus_!" Ms. Span shouted. Her voice cracked.
Henrietta looked at the problem. All right, then. "20 - 5" should be "20 + 5." Henrietta entered "25." Her computer _ding_ ed—correct. Around her, other computers _ding_ ed, although some _clunk_ ed. Some of the students were evidently having difficulty thinking of "minus" as "plus."
The next problem:
10 + 10 =
"Ms. Span, if minus is plus, is plus minus?" said Bernard Faust, a large boy whose scores were generally near the bottom of the class. He sat next to Clarence Frederick, and both of them were shooting worried glances all around, unsure what to do.
"I don't know," said Ms. Span. Her voice had become very calm. "Answer and move on," she said. She held up her arms, like someone making a plea for help.
Henrietta typed "20" and submitted it. _Ding_.
That was how the remainder of the math portion worked out. Minus had become plus, but plus was still plus. In other words, the test contained no subtraction problems. Once the students got the hang of it, Ms. Span seemed to recover some degree of her composure, straightening her glasses and running a hand over her black hair to smooth it. The exam ended and the bell rang for History and Nutrition.
"Everyone, Mason the bus supervisor is waiting outside to escort you today, while I stay here to collate the results."
The students stood, some a little shakily, others wiping tears from red cheeks, and walked silently out the door. Henrietta was preparing to follow when Ms. Span's voice stopped her in her tracks. "Henrietta." She did not sound pleased.
Henrietta approached the front as the last few students exited.
"You did well on the math today," Ms. Span said, removing her reading glasses.
"Thank you," said Henrietta. She could tell this wasn't going to be good news.
"But your composition. It was . . . inexcusable." Ms. Span turned her computer screen to face Henrietta and displayed Henrietta's essay, with FAIL written across the top. Below, the many words Henrietta had learned in the attic were underlined in red, one after the other.
"What is this supposed to mean?" said Ms. Span, donning her glasses again to peer at the screen and then removing them as if the sight caused her physical pain. "Why are you making up words, Henrietta?"
"I didn't," said Henrietta.
Ms. Span shook her head. "This essay decreased our class's aggregated statistic by two percent."
"I'm sorry, Ms. Span," said Henrietta.
"Henrietta, this essay, plus your Behavioral Citation this morning, has forced me to classify you as At Risk for the remainder of the year. If you perform like this again, you will be Finished."
"But—" said Henrietta.
"If I don't declare you At Risk, the whole class will suffer from having your scores included. Do you want that?"
"No," said Henrietta.
Ms. Span sighed. She massaged her plucked eyebrows with one hand. "Henrietta, I don't want to do it. I like you. Gary likes you. And I want to help you."
"I understand," said Henrietta, numbly. She wasn't really following the conversation anymore. Her mind had stopped at the words At Risk.
"We'll get through it if we commit to working hard. Gary will help you, too, I'm sure."
"Okay," said Henrietta. She looked up at the wall clock above, and observed the seconds clicking past.
"If you'd like to visit him at the nurse's office, Henrietta, I'll release you from History and Nutrition today. Ms. Morse just sent me a message that he's recovering."
"Thank you, Ms. Span."
Henrietta entered the hallway with her stomach clenched in a knot. Detention she could handle, but At Risk was something else entirely. Just a step away from a lifetime in a dingy apartment in the crime-ridden Old City, collecting garbage. A step from never seeing her friends again. She blinked furiously and wiped away shameful tears as she walked to Ms. Morse's office.
Her cell phone rang, and her mother's name appeared on the screen. She knew she should answer, but she didn't. Once it stopped ringing, it rang again—her father. By now they would both have received the news that she'd been reclassified.
When she opened the door to the infirmary, Ms. Morse was behind her desk, and wasn't surprised to see her.
"Rose is already with him," she said, gesturing.
Henrietta entered the recovery room. Gary lay on a cot, curled up with his hands loosely covering his face. He looked small. Rose sat across from him. "He'll be all right," she said as Henrietta sat next to her. They watched Gary's still form for awhile. "Rose, I'm At Risk," said Henrietta. "My parents are going to ground me forever. You probably shouldn't come over today."
"I'm sorry," said Rose. Her small face was full of sympathy, which made Henrietta feel a little better.
Henrietta's Behavioral Citation had also earned her detention, during which she typed out a long list of District-Approved Vocabulary words provided by Ms. Span that included terms such as _bucket_ , _grunt_ , and _rug_. By the end, Henrietta had missed her bus and had to call her mother.
Her parents were both in the car when it arrived, which was almost unheard of. They received her with disappointed faces, told her she was grounded, supplied many unpleasant scenarios of her future life that would occur if she were Finished, and emphasized the importance of using District-Approved Vocabulary, glancing at Ms. Span's report recommendations on their phones as they spoke. Henrietta wished she could shrink into nothing. As her parents lectured, she dropped her chin to her chest.
When they arrived home, dinner was served in near-silence, and her father curtly assured Henrietta that she would get no dessert. Other grim facts were aired:
– The amount of money Henrietta's father lost by coming home early.
– The exact layout of the tiny, rat-infested Old City apartment where Henrietta would spend the rest of her days as a garbage collector, which did not feature a private bathroom.
– The shame that would be heaped forever upon the name of Gad-Fly if
Henrietta were to become Finished.
Henrietta nodded when it seemed appropriate. Everything had piled on top of everything until she felt nothing. Finally, she was sent to her room to think about what she'd done. As soon as she arrived there, she climbed past the unseeing eye of the still-broken BedCam into the attic.
A luxuriant, greenish full moon shone through the large windows, reflecting perfectly in the glass top of the coffee table. The couch, ornamented with brocades of deep shadow, looked like a stone sculpture in the pale light.
On it Mister Lady reclined casually, reading _Early Town_ , turning a page with a single sharp claw.
Henrietta froze when she saw this, and Mister Lady looked up abruptly. It was obvious she hadn't been expecting Henrietta at this moment.
"I'm sorry," said Henrietta. "I didn't mean to interrupt. I just can't do _anything_ right."
The black pupils of the cat's green eyes were so large Henrietta felt swallowed by them. Mister Lady dropped the page, stood, and leaped easily to the top of a nearby bookcase, from which she looked down at Henrietta.
"I'm sorry," said Henrietta again. She felt like she should leave, but couldn't bear to go back down. Everything was too terrible.
When she blinked, she must have missed something. She must have because suddenly Mister Lady was gone. The place she'd just occupied atop the bookcase was empty.
"Hello?" said Henrietta quietly. There was no response. The silence of the attic was oppressive.
After wearying herself with searching, Henrietta descended back to her bedroom. The day could scarcely get worse. She shoved her feet and arms into the legs and sleeves of her polyester pajamas and slipped under her bedcovers. Disheartened, she fell asleep almost immediately, retreating from it all.
It seemed like the end of an awful day, but it wasn't over yet. The world didn't stop turning just because Henrietta had gone to sleep.
At that moment, a massive, yellow truck turned onto the street outside of her house, blocking both northbound lanes of traffic. A yellow car followed it at a walking pace. A worker emerged from the passenger side of the car with a can of spray paint in one hand, and made marks on the road—numbers and symbols of obscure meaning.
Behind the yellow car drove a yellow van with a hole in its roof, through which protruded a tall swiveling platform where a worker stood, holding a large remote control covered with buttons. When the van approached a stoplight, the worker pushed a few buttons on the control, and the light went out.
Behind the yellow van drove another yellow truck, this one dropping sawhorses at intersections. Each sawhorse featured two bright yellow, flashing lights with a sign between that read: ROAD CLOSED. Eventually, this caravan of northbound vehicles passed an identical caravan traveling southbound. After they passed one another, the street was empty.
For the first time in many years, there was silence on Henrietta's block.
### The Department of Insta-Structure
The next morning, after a depressing breakfast of cornslaw and further remonstrances from both of her parents, Henrietta left the house to find Gary waiting outside. "Look!" he said, pointing to the street.
It was empty. There was no layer of smog curling on the ground. No horns blared. No engines rumbled. Henrietta stepped onto the sidewalk, looking up and down the street. Several blocks distant, she saw some cross-traffic out past the ROAD CLOSED signs. She scratched her head and turned to Gary. "How are you feeling?" she asked. She hadn't talked to him since his headache.
Instead of answering, Gary took a few comical steps out into the road and did a little dance. He waved his arms, jiggling them like rubber bands. "I just ran across to your house!" he said.
"Don't stand out there," said Henrietta. "The cars could come back." She looked up the street nervously.
"Oh, right," said Gary.
As they walked to the bus stop, Gary's levity diminished. He grabbed Henrietta's hand when they arrived. "I'm scared," he said. "I hope the bus doesn't come."
"Me, too," said Henrietta.
They hoped in vain. The crowded bus arrived, and they reluctantly boarded. No cars honked in annoyance, and no targeted advertisements were deployed.
"Let's sit somewhere different," said Gary quietly.
They buckled into a pair of seats a few rows further back than usual. Gary closed his eyes.
Not long after Henrietta departed for school, her mother Aline sat at the living room computer and checked her mail.
The first thing she saw was an advertisement for a Halloween trick-or-treating event at a nearby mall. Aline didn't like Halloween. It was dangerous, and she was perturbed that it endured year after year. Henrietta would insist on going out, she knew. Was it the ads? Something they were learning at school? Aline couldn't understand why kids liked Halloween. (She'd forgotten that she liked it herself when she was young.)
The next mail item was from the city. It read:
"DEAR HOMEOWNER,"
YOUR HOUSEHOLD HAS BECOME NONCOMPLIANT WITH THE MINIMUM STREETSIDE OFFSET ALLOWANCE (MSOA) SUBSEQUENT TO AN EMINENT DOMAIN APPROPRIATION BY THE CITY. NONCOMPLIANT STRUCTURES ARE DEMOLISHED IN ACCORDANCE TO THE DEMOLITION AND RESTRUCTURE ACT (DRA). MITIGATION FEES ARE EQUIVALENT MARKET VALUE (EMV) PLUS EXPENSES. THE EMV OF YOUR DRA AWARD IS:
**$1,000,000**
THE DATE OF IMPLEMENTATION OF THE REQUIREMENTS OF THE DRA PURSUANT TO BRINGING THE STRUCTURE(S) ON THIS PROPERTY INTO COMPLIANCE WITH THE MSOA IS:
**OCTOBER 30**
AND THE PROPERTY HAS BEEN SCHEDULED FOR DEMOLITION ON:
**OCTOBER 31**
PLEASE MAKE NECESSARY ARRANGEMENTS, AND RELOCATE OR LIQUIDATE PERSONAL POSSESSIONS IN ADVANCE OF THE FORMER DATE.
SIGNED,
THE DEPARTMENT OF INSTA-STRUCTURE AND HOUSING AFFAIRS (DIHA)
ADDITION DISTRICT 002
The letter didn't make much sense to Aline. Minimum street-side offset? Eminent domain appropriation? Her eyes fell again on the figure in the middle of the screen:
**$1,000,000**
Aline and Tom had tried to sell the place several times. It was the topic of many of their fights. For Aline, living in her rickety childhood home was a chronic aggravation, and she felt sure it was the cause of Henrietta's House Sickness. But it was valued at $900,000, and the cheapest new houses were a million dollars.
This letter changed everything. The city would pay them more to tear down the crumbling place than they could have gotten from a buyer.
Tom entered the living room with his work clothes on, his cereal bowl in one hand, chewing. Aline gestured to the screen, and he peered at it over her shoulder. She watched his eyes.
"Can you believe it?" she said.
Tom took a bite of cereal. "Barely," he said, as he chewed. "They don't give us a ton of time. Is today September 30th?"
"A month," said Aline.
"We'll have to talk more about it," said Tom, "but it seems like good news."
"It seems like great news," said Aline.
They were silent then, and both of them had the same secret thought: they imagined, for a moment, moving out on their own. Not being together anymore. If they split the money, they could each get a nice condominium.
Tom was running a little late, and he walked into the master bedroom and grabbed his coat, noticing as he did that Henrietta's BedCam seemed to be working again. He approached the viewing screen. It showed Henrietta's empty, rumpled bed.
"Finally," he muttered, not giving the matter further thought. He certainly did not speculate about whether the departure of a wild housecat from the attic had anything to do with it. Back in Henrietta's room, the counting program on her computer switched from 36,565 to 36,566.
Outside Tom was surprised to see the empty, silent street. Several blocks had been closed off, and he concluded that some road work would occur later. He entered his car, a compact red two-door, started the engine, and backed onto the vacant asphalt.
"HELLO, AND THANK YOU FOR DRIVING," said the onboard computer.
"Work," said Tom.
"TURN LEFT," said the car. "WOULD YOU LIKE TO HEAR SOME ADVERTISEMENTS FOR PRODUCTS THAT MAY INTEREST YOU WHILE YOU DRIVE?"
"Yes," said Tom, and then, "sixty percent volume, double speed. Dial work, full volume, priority."
"THANK YOU FOR USING THE ADVANCED FEATURES," said the car. A stream of advertisements issued from the car's speakers, sped up and smashed together, at sixty percent of normal volume. It formed a linguistic wallpaper against which Tom's phone dialed his work.
HAS SOMEONE YOU HAVE KNOWN DIED RECENTLY? WITH PERFUME CREMATION, ASHES ARE TRANSFORMED INTO PERFUME—THE SCENT OF LOVE. IF YOU ARE PLAGUED BY RATS TRY RAT-B-GON! RATS EAT IT, DIE, AND BECOME PENCIL ERASERS. WERE YOU RECENTLY FINISHED FROM SCHOOL? WONDERING WHAT'S NEXT? ATTEND THE GARBAGE ELIMINATION INSTITUTE—WHEN SCHOOL ENDS, THIS BEGINS.
"Hello, Tom," said a voice exactly forty percent louder than the advertisements.
"Elton," said Tom. "What's the Intermediary Technology Report?"
"Are you inbound?"
"I'll be there in—" Tom stopped to allow the car's computer to answer for him. The computer monitored his conversations and automatically filled in information that seemed appropriate. "THIRTY-THREE MINUTES."
"The report is rendering in the System Manager now."
"Stats?" said Tom.
"They're—" said Elton, and he paused to allow his own auto-complete program to fill in:
"INTEROPERABILITY 75 PERCENT, SCALABILITY 78 PERCENT, PREDICTED POTENTIAL SATURATION 35 PERCENT, MODULAR COMPONENT CROSS-MARKET INDEX 7.5, EARLY ADOPTION INDEX 5.7, PERCEIVED OBSOLESCENCE VELOCITY 85 PERCENT."
"TURN RIGHT," said the car. Tom turned right.
—CONSIDER BUYING A RING FOR THAT SPECIAL SOMEONE! THE CHERISHMENT RING IS MADE OF SPECIAL PLASTIC—PLASTIC, LIKE LOVE, LASTS FOREVER. SEASIDE HOSPITAL'S TEAM OF SURGICAL PRACTITIONERS CAN HELP YOU LOOK VIRTUALLY YOUNGER! I NEVER THOUGHT I WOULD BE OUT OF MONEY, BUT FOR THE PRICE OF ONE PAYCHECK I GOT AN ADVANCE FROM GAME OVER PAYDAY LOANS. BEEFCRAFT: LOOKS LIKE BEEF, TASTES LIKE—
"Those ratings sound good," said Tom.
"Potential saturation is a little low," said Elton.
"It doesn't matter. When the render reaches upgrade potential, we should see high numbers. Anything else?" said Tom, as if it were Elton who had called him, and not he who had called Elton.
"The DBAs have been troubleshooting inflated valuations in the System Manager. It's not a huge deal, but you may want to interface with the Virtual Operator when you get here."
"Inflated valuations?" said Tom.
"So far it's in accounts payable and interest."
"Oh, _well_ ," said Tom, laughing—if customers were footing the bill for a mistake, that wasn't so bad.
"TURN RIGHT," said the car. Tom turned right, merging with the flow of traffic on a cross street, which angered someone behind him, and they honked.
"BUY THE NEW SKIPPING-STONE PHONE FROM TINCAN TELE -COMM!" said the ad. Tom honked back, and his horn also said, "BUY THE NEW SKIPPING-STONE PHONE FROM TINCAN TELECOMM!"
"See you shortly, Elton," said Tom, and disconnected.
—FOR THE MAN IN CHARGE, PROFORMA PANTS SHOW EVERYONE YOU APPRECIATE THE GOOD THINGS IN LIFE. PARENTS, WHEN THE KIDS ARE GROWN, MOVE TO ADDEDGE AND ENJOY A VIEW OF THE OPEN OCEAN, BREEZES, AND THE LEISURE THAT EVERY PARENT DESERVES—
"Computer," said Tom. "Purchase two pairs ProForma pants, color gray, waist thirty-four, inseam thirty-three. Also, solicit information from AddEdge—buyin price, resalability, location."
AUTODEDUCTION OF FOUR HUNDRED THIRTY-THREE DOLLARS FOR TWO PAIRS OF PROFORMA PANTS. INFORMATION ON ADDEDGE REQUESTED. THANK YOU FOR USING AUTODEDUCT. YOUR CONVENIENCE IS OUR REWARD.
"Computer," he said, "send information about the Garbage Elimination Institute to the following phones: me, Aline Gad-Fly, and Henrietta Gad-Fly."
"DISTRIBUTED. MERGE ONTO THE HIGHWAY." Tom merged.
"BUY THE NEW SKIPPING-STONE PHONE FROM TINCAN TELE -COMM!" honked someone behind him. Tom smiled and slowed down a little, to get them to honk again.
When he reached work, things were not in the same good state they'd been in minutes before. As he exited his car in the underground parking lot, his cell rang, and he saw Elton standing a hundred feet away next to the elevators, a tall, skinny man about Tom's age with short brown hair, wearing blue jeans and a T-shirt with an artificially faded slogan on it. Elton was holding his own phone to his ear. Tom answered. "Elton," he said.
"Tom, we have a situation. I need you to interface with the Virtual Operator ASAP."
"The accounts receivable glitch?" said Tom.
"It's spreading."
Tom approached Elton as they spoke, and they stepped into the elevator together. Though they were right next to each other, they continued to communicate via their phones, because the sound quality was better than face-to-face.
"Floor sixty," said Elton. A subtle lurch followed as the room rocketed skyward.
"What's the DBA report?"
"Not a storage issue—it's in the Intelligence. Anyway, the system is adding only."
" _Adding only?_ " said Tom, speaking into his phone and looking blankly at Elton.
"Right—no subtraction."
The elevator doors opened, and Tom and Elton strode into a high-ceilinged hallway with one full glass wall that looked out over the Addition. From here, above the surrounding buildings, one could see the haze of pollution that covered everything in a yellow-grey cloud, through which poked some of the taller buildings, like this one.
Tom and Elton had no time to enjoy the view. They passed through a pair of automatic doors into a conference room containing a large table surrounded by twenty beige swivel-chairs.
On the wall hung a large screen, and on the screen was a face—the Virtual Operator of the System Manager. Tom had never liked the image, which had been designed before he worked at the company. The face had scarcely any nose or chin, and featured a pale haze of thin hair and light yellow eyes. The face flickered occasionally, a glitch that no one ever seemed to be able to fix. The Virtual Operator was the graphical interface for the System Manager, a program that interacted with the Intelligence Kernal, which directed all General Subsidiary Applications. Tom, though he could recite this, didn't fully understand what it meant, and this was one reason he never adequately explained his job to his daughter: he couldn't explain it to himself.
"GOOD MORNING TOM," said the Operator. When it spoke, its mouth hung open and the words emerged scratchily from the speakers at the side of the screen.
"There's a glitch in the Intelligence," said Tom. "Diagnose and repair."
"DIAGNOSING." The Operator's face flickered out of sight for a moment, and then returned. "THE GLITCH YOU REFER TO IS A GRADUATED ADJUSTMENT ALGORITHM, CURRENTLY ENGAGED TO COUNTER MALICIOUS SYSTEM CONTENT REMOVAL."
"Malicious removal?" said Tom. "What's the nature of the attack?"
The Operator looked at Tom with an expression he'd never seen it make before. It appeared . . . confused. "TOM," it said, "WHERE DO HENRIETTA AND HER FRIENDS GO AFTER SCHOOL?"
### Smashed Sidewalks
While Henrietta, at school, wrote compositions and figured math problems, and while Aline, at home, generated accounting statistics for her clients, and while Tom, at work, tried to repair the glitch in the Intelligence Kernal, developments continued on the empty street outside their house.
More yellow vehicles arrived, in various shapes and sizes. Some looked like crustaceans, others like giraffes. Some carried massive rollers, others massive lungs. They lined up one after another and revved their engines.
The spectacle that followed was as precise as a parade. Each machine moved over the street in succession, performing the task for which it was designed. First, a massive caterpillar with jackhammer legs hammered the street into shards, and then the shards were sucked up by an enormous vacuum cleaner. Underneath the shards, a narrow old road of red bricks was revealed, which shone in the yellow autumn light just as Henrietta, Gary, and Rose had often seen through the attic windows.
The next row of vehicles used gigantic knife attachments to slice away the edge of the present street and widen it by removing the sidewalks, as well as some of the front yard of every house.
What was the purpose of all of this? The city had recently concluded, through an extensive systemwide analysis, that busy roads operate at higher efficiency if sidewalks and lawns (which scarcely anyone ever used anyway) were turned into lanes. There was one problem with this plan, however: Henrietta's house.
Henrietta's house had no front lawn. Aside from two rows of flowers and the sidewalk, it already abutted the street. The machines did what they could, lapping up the sidewalk and the flowers and leaving Henrietta's front door opening right onto what would shortly be a lane of traffic.
Next, massive dump trucks spilled mounds of hot asphalt over the bricks, up and down the block. Then a paver built to the exact width of the new street lowered its enormous roller and squashed the mounds into a steaming black plain, and a sprinkler at the rear sprayed atomized chemicals that instantly set the new surface, releasing a tremendous flowery stench. Lastly, a paint truck rolled through, squirting streams of quick-drying yellow paint—some solid lines, some dotted, some double—that defined the lanes.
Once upon a time this block had been a sleepy brick boulevard bordered on both sides by maple trees. Then it grew to a busy four-lane road with sidewalks and no trees. Now it was a six-lane highway.
Workers ascended the utility poles along the street and added the appropriate traffic signals before reconnecting the electricity and turning everything back on. The crosswalk signs were all removed, as there were no more sidewalks. The entire street had become permanently DON'T WALK.
The construction vehicles retreated, and the street reopened. Traffic flooded into the new lanes, producing instantaneous gridlock.
If it weren't for Henrietta's house, the road could have been wider. If a map were made of the street now, it would show Henrietta's house protruding forward from the even line of identical houses until it nearly stumbled into the street. One might wonder, looking at it from that perspective, just how long it would be before Henrietta's house went the way of the sidewalks—the way of the maple trees—the way of anything that pinched the growth of the road.
### The New Route05
The school day was largely uneventful. Ms. Span was mostly pleased with the results of the Competency Exam, and she indulged a relaxed morning, smiling, nodding and dispensing universal appreciations. Tomorrow, her attentions would anxiously redirect themselves toward the next Exam, but for now she was looking back from the safe side of the hurdle they'd jumped together. Even Henrietta, though reclassified, was ensconced within the glow of the class's success.
After school, while Henrietta and Gary stood in line at the turnaround waiting for the bus, Henrietta told him she was At Risk. Gary immediately suggested that they both get Finished on purpose and become a garbage-collecting team, a proposal that didn't seem too terrible—but the incalculable amount of grief Henrietta would endure from her parents kept her from seriously considering the idea. She had to pull up her grades.
"You could cheat," said Gary.
"I can't," said Henrietta, shaking her head.
"Why not?"
"I don't know. I'm just not that kind of person."
"Well, I am," said Gary. He puffed out his chest a little.
"That's one of the things I like about you, actually," said Henrietta.
Gary punched her arm, and she punched his arm, and then he punched his own arm as Rose approached.
"Henrietta and I are going to become a garbage-collecting team," said Gary.
"Can I be on it?" said Rose.
"Definitely," said Gary.
"I have something else to tell you guys," said Henrietta as the buses slowly approached the school from far up the clogged street. She launched into the story of Mister Lady's disappearance the previous evening, which she'd been wanting to tell them since it happened.
"You saw Mister Lady _reading_?" said Gary, shaking his head in disbelief. "We always joked."
"She turned the page with a claw."
"Do you think she was angry that you caught her?"
"I don't know. I was so upset already that I wasn't paying attention. Then, before I knew it, she was gone."
"She wanted to go home," said Rose. "Back to the old town."
The buses arrived, and the children boarded together and strapped themselves in. The ride was noisy as always, what with other children talking and the constant eruption of ads from outside. They sat silently.
As the bus approached Henrietta's and Gary's stop, Henrietta noticed it wasn't slowing. She craned her neck and looked out the window.
"We're missing our stop," she said. "We're—" She saw the massive new street. There were no sidewalks anymore. If the bus dropped them off, they'd be standing right in traffic.
"The road!" said Henrietta.
"What is it?" said Gary, straining to see.
The bus rolled up to Henrietta's house and opened its door at the entrance to Henrietta's driveway. From behind came the predictable litany:
**"GOT FINISHED?**
ATTEND THE GARBAGE ELIMINATION INSTITUTE.
WHEN SCHOOL ENDS, THIS BEGINS!"
"DINNER COOKIES ARE DINNER AND DESSERT!
**TOTALLY EFFICIENT!"**
"Let's go," Henrietta shouted over the din. They unbuckled themselves and disembarked, and the bus computer logged them out, messaging their whereabouts to their parents' phones. They walked up the driveway to the carport.
"Look at my front door!" said Henrietta, pointing. The door now stood right at the edge of the freshly paved road, with a plank nailed across to keep it from opening.
"How will we get in?" said Gary.
"Over here," said Henrietta, going to the side door near the rear of the house, next to the narrow strip of plastic grass that separated Henrietta's house from the neighbors behind.
Her mother opened the door and looked down at the three of them. Henrietta expected her to be unhappy. After all, their house was now inches from six roaring lanes of traffic, and Henrietta had just brought her friends over even though she was grounded. But, surprisingly, her mother looked cheerful.
"Come in!" she said to them, and they filed into the kitchen to find a paper plate on the table piled high with Dinner Cookies.
"I thought you might need some fuel for studying," said Henrietta's mother.
"Thanks," said Gary. He immediately crammed one cookie in his mouth and another in his pants pocket.
"Mom," said Henrietta, a little uncertain how to proceed in light of her mother's unexpected good mood, "I know I'm grounded, but Gary said he'd help Rose and I study."
"Study hard!" said her mother, smiling.
Henrietta hesitated a little, still perplexed.
"Go ahead, silly!" said her mother, shooing the three of them amiably toward Henrietta's room.
### Trapped!
They climbed directly into the attic and began searching for Mister Lady. Despite the considerable hours they'd spent up there during the past weeks, the network of paths through the antiques behind the bookshelves were still only partly explored, even by the intrepid Rose. Some places they just couldn't get to—spaces that had been completely boxed off by walls of old furniture.
After hunting for awhile to no avail, they regrouped glumly at the couch.
"I guess she decided she'd gotten better," said Henrietta.
"That's good, right?" said Gary.
"It's just that I liked her. And I thought she liked us."
"She did like us," said Rose. "But she had things to do."
"If I was lost and injured, I'd want to get back home," said Gary.
"I wonder if she has a family," Henrietta mused. "Do you suppose that other cat we saw is her mate?"
They went over to the windows, and looked down on the old town. The street was crowded, as it usually was at this time, with warmly dressed children returning home from school, laughing and talking, some of them even walking down the middle of the empty brick street. Mister Lady was nowhere to be seen, but Henrietta saw something else, quite unexpectedly, that caught her attention.
"Look," she said. "By the stump." The stump was currently crawling with children playing a game of tag, scurrying around the edges and across the top in pursuit and flight.
"What?" said Gary.
"That kid, there." Henrietta pointed. There was one child standing at the base of the stump who didn't look quite right. Dressed in a yellow button-up shirt and yellow pants, its face was not a person's face.
"It's . . ." said Gary. He didn't need to finish the sentence. Clearly, none of the other children could see it. As they ran past, it occasionally reached out with one long finger and tapped their foreheads.
"It's not grown up yet," Henrietta murmured.
"It isn't flickering," said Rose. She was right—it looked strangely solid compared to what Gary and Henrietta had seen the previous day.
"Have you seen it before?" said Gary, turning to Rose.
"My whole life," said Rose. "It tries to tap me, just like out there. But those kids don't mind."
"Nobody minds but us," said Henrietta, "because we're House Sick."
"What's House Sick?" said Rose.
"Nobody knows," said Henrietta. As she spoke, though, several ideas began to come together in her mind. "Wait . . ." she said, holding up one hand. She knitted her eyebrows. "We _do_ know. It's that thing. I think . . . I think it doesn't like our old houses, because it can't get in!"
"I've seen it in my house," said Rose, "but you're right. It doesn't like it. It never stays."
"Maybe it's making us sick so we'll move," said Henrietta.
"But I did move," said Gary. "My mom and I left our old house, so why would it attack me on the bus?"
"Because you've been up here in the attic," said Henrietta. She clenched her hands, feeling like she'd finally grabbed something that had been hovering just beyond her reach. "On the bus it asked us where we went. And then during the Competency Exam, that was the composition question."
"That was the _Exam question_?" said Gary, aghast. He still needed to take the makeup test. "My mom said it was 'Why is it dangerous to swim?'"
Henrietta smiled despite herself—it would have been funny, in an awful way, to see Gary's careful cheating method go awry. "It can go almost anywhere—even into computers," she said. "But not into old houses. Not up here."
"But _why_?" said Gary, still not satisfied.
Their conversation ended abruptly, however, when something surprising transpired down below. The strange child stopped tapping kids on the forehead. It stepped away from the edge of the giant stump, and looked up—right at Henrietta's house. Right at the three of them looking down.
They stumbled back from the window. "It saw us!" said Gary.
Cautiously, they crawled forward and peeked over the sill. It was still looking, its yellow eyes locked on them. They fled back to the couch.
"But nobody out there _ever_ sees us!" said Gary.
"If that version of it knows..." said Henrietta, reluctant to finish the sentence. Filled with trepidation, they approached the trapdoor, and Henrietta cautiously opened it.
There was the chair sitting atop the desk, and the desk on the tan carpet, as always. There was the bedside table, and Henrietta's canary night-light, and the windows with the shades pulled. There was the bed.
And standing next to the bed, feet together, arms at its side, was the creature. It looked frozen, like a photo, and its empty yellow eyes stared up at the three of them. Henrietta dropped the door.
"Oh, _no_ ," said Gary.
"Maybe it'll leave," said Henrietta. "Let's just wait."
They returned to the couch and sat together.
"Do you think it's gone yet?" said Gary immediately. He reflexively pulled his phone from his pocket and looked at it. "Hey," he said. "It's . . . well, it's not really working, but it's kind of working." He showed them the screen, which was lit, and the digital numbers that indicated the time were there, but crawling at a fraction of their normal pace. "Uh oh," he said, and they all simultaneously reached the same conclusion: When you're up in this attic, where time is frozen, it doesn't matter how long you wait. The thing down below will be there, from your perspective, forever.
"But it arrived after we climbed up," said Henrietta. "It came later. So time must be passing a little, or just passing for it, or . . ."
"Let's keep waiting," said Rose.
They sat for another few moments in silence. "We should try to distract ourselves," said Henrietta. Gary picked up an old deck of cards he'd found some weeks ago, and started laying them out with Rose for Solitaire, a game they'd learned from one of the dusty attic books.
Henrietta picked up _Early Town_ from the coffee table. It was still open to the page Mister Lady had been reading, which was an odd-looking, close-up map that showed Boardwalk, the street right outside Henrietta's house. It appeared to be two maps, one laid on top of the other. The bottom map was the one Henrietta had seen before, of the single-lane brick boulevard. The other map, printed on transparent paper and laid atop the first, showed the road Henrietta had grown up with—four lanes with sidewalks.
Henrietta tried to puzzle the pieces together. Was this a map of the future—or what _was_ the future, once upon a time?
"Hey," she said to Gary and Rose, but the moment she spoke, a loud _bang!_ sounded from outside. The children started and looked at the windows.
Their view of the enormous trees was obscured by a young man standing outside atop a ladder. He wore a red checked shirt and a blue cotton cap. In one hand he held a hammer, and in the other a two-by-four plank. He looked in but didn't appear to see the children.
"Who is that?" said Gary.
"I don't know . . ." said Henrietta. Then she did know. "It's Al! My grandfather!" But it wasn't the Al she knew. It was Al as he had once been, decades before Henrietta was born—a young man. He lifted the plank and began nailing it to the outside of the house, across the windows.
"What's he doing?" said Gary.
"He's making the house like it is now," said Henrietta.
"We've got to stop him!" said Gary.
"We can't," said Henrietta. "He has to do it. The city's about to knock down the trees and turn the road into more traffic lanes. Look." She held up _Early Town_ so Gary and Rose could see the overlaid maps.
"But that just happened!" said Gary.
"It just happened _again_ , you mean. It happened the first time then, and again now," said Henrietta.
"The same thing all over," said Gary.
"Do you think Mister Lady left the book open there on purpose?" said Rose. Outside, Al nailed another board on the window, blocking out more light.
The attic began to darken.
"Why's he doing that?" said Gary.
"So the trees don't shatter the windows when they fall," said Henrietta.
Al disappeared down the ladder for a moment and then returned with more boards. The light diminished further, the shadows deepened, and soon Al's face was no longer visible. The last board was in place, leaving the attic in pitch darkness. They listened to the final strokes of the hammer. Then, silence.
"You guys," said Gary's voice, strained and a little high-pitched. Henrietta could imagine his eyebrows bunching together.
"What is it?" said Henrietta.
"I'm afraid of the dark. Could we open the trapdoor again, just for a second?" There was a cold tremor in his voice, and Henrietta recognized that he was starting to panic. She recalled his disastrous descent the first time she'd shown him the place.
"Let's all go together," said Henrietta. She took Gary's and Rose's hands, and they shuffled through the darkness to the trapdoor.
Henrietta pulled it open a crack. The light from the bedroom flooded in, and Gary began a relieved sigh that caught almost immediately in his throat as he looked down. The thing was still waiting below, fixing them with its steady stare. Henrietta slowly closed the door again.
"You know," Henrietta said, "I remember seeing some candles on one of the bookshelves, once. Did you guys see those?"
"Yeah," said Gary. "And a box next to them—matches!"
"Remember when we watched a movie about them in class?" said Henrietta. " _Don't Strike Those Matches_. A kid burns his house down."
"That movie was really good," breathed Gary.
"You two stay here," said Henrietta. "I'm going to find them." She released her friends' hands and felt her way across the floor. She maneuvered behind the coffee table to the base of the nearest bookshelf and slid her hands up the spines of the old books until she reached the top shelf, letting her fingers skim the outlines of the items there. A bronze baby shoe. A glass ashtray. A bookend. And . . . a candelabra, loaded with seven long tapers. She curled her fingers around the metal base. With her other hand, she continued searching until she found the small cardboard box.
She crawled back to the coffee table, set the candelabra on it, and opened the box. Inside it felt like a bunch of sticks. "Do you remember how that kid lit them in the movie?" she said.
"There should be some red stuff on one end," said Gary, "and you swipe it against the side."
"I can't see anything," said Henrietta.
"The bigger end," said Rose, who dealt with matches frequently in the Library.
She felt inside the box. Each stick had a little bulb on one end. She swiped one against the side of the box. "It didn't work."
"Feel around on the box," said Gary. "One side should have a rough strip. I have a couple in my trash collection, and I think they're all the same."
He was right. Henrietta tried again, and— _Fiss!_ The match ignited. She held it to the wick of one of the long candles, which caught easily and glowed with a steady, swelling light. She lit two more candles, and then dropped the match on the floor and stomped it out thoroughly.
"You did it!" said Gary. It was comforting to have a little light, even if their situation was the same as before.
"What now?" said Gary.
"Keep waiting, I guess," said Henrietta.
"We should sleep soon," said Rose
Gary rubbed his eyes. "Yeah," he said. "I'm almost as tired as I am scared."
"It does seem pretty late," said Henrietta.
"We could camp up here," said Gary, and for the first time in a while there was a hopeful note in his voice. He'd always wanted to camp.
"There are probably blankets somewhere," said Henrietta.
"In the dresser," said Rose. Henrietta handed her the candelabra, and Rose led the way back behind the bookcases to an old wooden dresser. Henrietta pulled open the top drawer to reveal a pile of neatly folded blankets. She and Gary scooped up several of them, and they all returned to the main room, where they made up three little beds on the floor next to the couch.
"I don't think I've ever gone to sleep without brushing my teeth," said Henrietta. "I hope they don't rot."
"They won't," said Gary. "I've done it."
"Doesn't your mom make you?"
"Sometimes I fake it." He opened his mouth wide, showing two rows of reasonably clean-looking teeth in the candlelight.
"When I first found Mister Lady up here, I told my mom I was using the bathroom, but I was really getting some bandages," said Henrietta. The two of them grinned at one another, pleased about their shared history of mischievousness.
"I've lied, too," said Rose. "To everyone. About everything." And then, finally, she told the story she had been forbidden to tell.
She told it because she trusted her two friends and because they had trusted her, been kind to her, and helped her. She began with her home, the Library, with its thousands of old books stretching to high ceilings. She told them about the Subscribers who arrived in the dead of night; about elaborate secret knocks; and about thieves who read historical romances, dumpster diving scientists obsessed with narrative poetry, and former telecommunications employees who loved cookbooks. Once she got started, she kept talking until it was all out, and when she finished, Henrietta and Gary were dumbstruck.
Henrietta was first to form a question: "You have friends from the _Old City_?" she said. "What are they like? Are they scary?"
"The Subscribers are nice," said Rose. "They come over and fix books with my dad."
"How do they live?" said Gary. "Do they kidnap people?"
"They find things," said Rose. "Like in trash cans."
"They _steal trash_?" said Gary, his eyes widening.
"Sometimes it isn't really trash," said Rose. "All of my clothes I'm wearing. My shoes."
Henrietta and Gary looked at Rose's shoes. In the low light of the candles, they looked like regular shoes, white plastic with Velcro straps.
"Those were in a dumpster?" said Gary. He reached out and touched them gingerly.
"Someone outgrew them," said Rose.
"This is amazing," said Gary. "Amazing." Everything he'd ever thought about the world was turned on its head in an instant. Trash wasn't trash. Criminals weren't criminals. He frowned, and then smiled, and then bit his lip.
"Why didn't you tell us this before?" said Henrietta.
"My parents said if I ever told we'd have to leave the Library, and I'd have to quit school."
" _I_ won't tell," said Gary. "Not a soul. Not ever. Unless it's by accident," he corrected. "Sometimes I say things by accident."
"I'll remind you not to," said Henrietta. She turned back to Rose. "But how do your parents keep the whole thing secret? And why is an old library sitting out in the middle of the Addition?"
"I don't know," said Rose to both questions.
"I can hardly believe it," said Henrietta. She stared at Rose as if looking at a complete stranger. This little kid was a never ending font of surprises, and Henrietta felt proud to have her as a friend.
Gary said, "My only secret is that I can't read."
"But you can read now," said Rose.
"That's true." Gary grinned. "I guess I don't have any secrets."
"You cheat on tests and collect garbage," said Henrietta, and she leaned over and blew out the candles. In the darkness, they crawled under the covers they'd laid out, and rested their heads on the couch cushions.
After talking with his friends, Gary felt a little less scared to be in the dark, and he drifted off quickly. Henrietta and Rose followed soon after, both exhausted and scared, but, for the moment, safe.
### A Death in the Family
After the children departed to Henrietta's room to study, Aline returned to her work, but even as she prepared a lengthy accounting report, part of her mind continued to ponder the mail from the city. That money could make her life into something she'd begun to think it would never be: happy. Perhaps she could escape the feeling, one she'd had for so long, of being trapped. She wondered how Henrietta would react to the news, and realized that she had no idea at all.
Aline didn't feel very close to her daughter. As Henrietta had grown into childhood and revealed her personality, the two of them had begun to clash. Henrietta couldn't seem to do anything the normal way, and not because she lacked the ability. Rather, she seemed to intentionally avoid fitting in and doing what people required of her. Aline had wondered, on occasion, if Henrietta wasn't her real daughter. Maybe her real daughter had been switched with someone else's at the hospital, and some set of sloppy, willful, uncharming parents were wondering how they'd managed to produce a beautiful, obedient, tidy little girl.
Aline's phone rang. She looked at the screen and frowned. It was Al. For a moment, she became angry. Why would he be calling her? She'd told him in no uncertain terms when he'd married her mother that she wanted nothing to do with him. It had been the last straw, and she was still, even now, furious that her mother had—
Her heart sank as she realized why Al was calling. There was only one possible reason. She turned away from her computer and answered the call. "Mother," she whispered into the phone.
"Aline, I'm sorry," said Al's scratchy old voice. "She passed away peacefully in her sleep. This morning."
And the worst of all: Aline hadn't seen her mother since the birthday party, over a month ago. She'd been so angry, about . . . something.
"I'd like to talk to you a bit about memorial arrangements," Al said stiffly.
"I need to go," said Aline. "I'll . . ." she didn't even finish the sentence, just disconnected the call. She turned to the windows where the slow mass of traffic passed within inches of her house. The cars seemed miles away.
### The Escape Plan
Henrietta's eyes opened in the pitch darkness. It took her a moment to remember where she was—the attic, with the windows blocked off, and Gary and Rose beside her.
"What was that?" said Gary's voice.
Something had awakened them both.
"I don't know," said Henrietta.
Then came a whooshing sound, like a broom scraping across a floor. It came from the blocked windows.
"It's the trees," said Henrietta. "They're getting chopped down!" The sound continued, and Henrietta imagined the grand limbs toppling, their red and gold leaves brushing the house as they fell.
"I wonder how much time has passed," said Henrietta. She sat up, felt for the box of matches and the candelabra on the table, and lit the candles.
Gary and Rose sat up in their piles of blankets. Rose's black hair was frizzed out from tossing and turning, and Gary's was a spiky thicket. They certainly looked as though they'd been asleep for a while.
They went to the trapdoor and cracked it open, although they all had a feeling about what they'd find. They were right. The creature hadn't moved. Its eyes were still fixed on the entrance.
"I'm kind of hungry," said Gary once Henrietta replaced the trapdoor.
"Me, too," said Henrietta. Her stomach growled.
"I wish we could eat cobwebs, like Mister Lady," said Gary. He plucked one from an empty bookshelf and eyed it critically.
Rose tapped the _Bestiary_ 's cover, on the table. "Have you ever looked up the Wikkeling in here?" she asked. Gary and Henrietta looked at her blankly.
"The what?" said Henrietta.
"The Wikkeling. In your room," said Rose, gesturing toward the trapdoor.
"How do you—" said Gary, and then stopped himself. "Never mind," he said. "You just know everything. I accept that."
They sat at the coffee table, and Gary and Rose looked on as Henrietta flipped to the table of contents and skimmed it fruitlessly. She went to the index. Nothing. Page by page, they examined all the pictures.
"I guess they didn't know about it," she said as she turned past "Alphabeetle," "False Apple," and "Tree Goat." Then a thought occurred. "You know what, though. My grandfather's version might be better. It's a later edition."
"I wish our phones worked," said Gary.
Henrietta continued flipping through the ancient book a page at a time, until she landed on the very first thing she'd ever looked up: _Housecats—Wild_. She remembered sitting with Mister Lady, studying the strange words, combing the mildewy attic dictionary for the meanings of terms like _ingress_ and _egress_. She contemplated those terms now, still locked in their same sentence.
**_. . . many homes contain so-called "Cat Halls," thought to encourage Ingress and Egress._**
"Hey," she said, putting her finger on the sentence.
"What is it?" said Gary.
"Cat Halls," said Henrietta. "Ingress and egress!"
"You're right!" said Rose.
"What? What?" said Gary, looking from one of them to the other. "What are they?"
Henrietta grabbed the candelabra.
"Coming and going," said Rose.
"People used to put doors in their attics for wild housecats," said Henrietta. "That's how Mister Lady got in here, and it's how we're going to get out."
### The Cat Hall
They split up, each taking a candle and moving behind the bookshelves into the various deep corners of the attic. Now that they had an idea to work toward, the attic seemed a little less scary. As she headed down a dark aisle bordered by stacks of old paintings of the ocean, Henrietta heard Gary and Rose rooting around elsewhere, moving boxes, squeezing between stacked chairs. Suddenly, she found herself smiling. They were going to figure this out!
She balanced her candle on a tabletop and started moving some of the paintings to see if the hall might be hidden there when a shout stopped her, echoing through the attic.
" _I found it!_ "
Henrietta and Gary followed Rose's triumphant exclamation along a row of ceramic plant containers sitting on narrow plant stands and wooden magazine racks until they reached a large stack of luggage.
Rose's voice seemed to emerge from under the pile. Henrietta held out her candle and saw a narrow, dark space framed by an alligator-skin purse, a wooden hatbox, and a denim-sided briefcase.
"Are you _under_?" said Henrietta.
"It goes through," said Rose's muted voice.
Henrietta and Gary knelt and squirmed one after the other into the luggage tunnel. Henrietta held her candle carefully, remembering the part in _Don't Strike Those Matches_ when a flame ignited a boy's hair. The tunnel ran for about fifteen feet, and they emerged in a little room formed of the backs of several bookshelves and one wall of the attic. Rose jumped up and down in triumph when they arrived, pointing toward her find.
The cat hall was right there, at the intersection between wall and floor, a miniature doorway just large enough for a wild housecat to squeeze through—or a kid. Its frame was carved with intricate designs, and a message had been engraved over the lintel. Gary leaned down to read it.
**EAT OUR COBWEBS, CHASE OUR RATS.
IN THIS HOUSE WE LOVE OUR CATS.**
Henrietta peered through the cat hall. "It's dark outside. Time must be passing out there—it looks like the middle of the night."
"I don't get it," said Gary.
"It's unpredictable," said Rose.
Henrietta knelt and stuck her head through the cat hall. Below lay the narrow lane of plastic turf that ran between the back of her house and the back of the neighbors'. Neither house had any windows on this side—it was an unused area. Henrietta felt dizzy as she contemplated the drop. She pulled her head back in. "It's really far," she said.
"It's not too bad," said Rose. "We just need a landing pad."
Before Henrietta could formulate a follow-up question, Rose took her candle and crawled back out through the luggage, into the attic proper. Henrietta and Gary looked at one another, and Gary shrugged. Rose soon returned, dragging the blankets and pillows they'd slept on.
"Landing pad," she said again, and pushed the blankets through the cat hall, letting them fall to the ground outside. She stuck her head through and looked down. "We'll need a lot more."
They all went into the attic and located as many soft objects as they could find—bedding and ornamental pillows, towels, a few old teddy bears, two bath mats, seven throw rugs, pants, shirts, a fluffy down parka. . . . Henrietta quickly lost track of it all as they dropped one item after another onto the growing pad below.
"Good enough," Rose declared finally, peering down through the cat hall at the prodigious pile. She pulled her head back in and stood. "Gary, you first."
"M-me?" said Gary.
"You're tough, but sometimes you don't realize it," said Rose. Gary opened his mouth to make a retort, but closed it again. He blinked several times, realizing that Rose was correct. In fact, her assessment was one of the truest things anyone had ever said about him. With a thoughtful, determined frown, he blew out his candle and set it aside. He lay on his belly and entered the cat hall feet first. He began wiggling through. "It's chilly outside!" he said as he progressively disappeared. "Okay, this is as far as I can go, so . . . I'm going to drop. Wish me luck!"
"Good lu—" said Henrietta, and Gary was gone. Henrietta and Rose scrambled forward and looked out after him. He was lying on the landing pad. For a moment, he was still, and then he rolled onto the ground, got up, and showily dusted himself off. "I'm okay!" he called up. "It was . . . fun! You next, Henrietta!"
Henrietta blew out her candle. "Rose, do you think I can make it?"
"Definitely," said Rose.
Henrietta put her feet out, and squirmed through the hall, feeling the cool night air on her ankles. Soon, she'd gone as far as she could. She'd have to drop.
"Wish me luck," she said.
"Good luck," said Rose, and she wiggled her fingers to indicate the luck flying out.
With a sick feeling in her stomach, Henrietta let go, and she saw, in a blur, the attic disappear, the door shrink away, and the siding of the house flying past along with cold air and dizziness. . . .
_Foomp!_
She hit the pile, sinking deep down and then rolling off.
Gary hovered over her.
"Are you okay?" he said.
Henrietta stood and took a couple of shaky steps. She looked up at the house, and saw, far above, the cat hall, and Rose's face looking down through it. She smiled. She waved up to Rose. "It _was_ kind of fun!" she said.
The yellow glow of Rose's candle went out. Then her feet appeared through the door. Soon, she was as far as she could go, and she dropped. The speed of the plummet was a little stomach-churning, but Rose sank into the soft pile and emerged quickly, bouncing right onto her feet.
All three of them laughed from nervousness and relief as they looked up at the cat hall far above.
"What now?" said Rose.
"Call my grandfather," said Henrietta, taking out her phone. She looked at the screen, about to search for Al's listing, but she froze.
"What is it?" said Gary.
Wordlessly, Henrietta turned the phone around and showed it.
WHERE DO YOU GO?
She placed it on the ground and stepped on it firmly, crushing it. Gary perused his own phone for a moment, frowned, put it next to Henrietta's, and smashed it, too.
"We've got to get out of here," said Henrietta. "We should try to get to my grandfather's."
"How?" said Gary.
"I don't know. It's called Sunset Estates. I know the address."
"Maybe we can use my phone," said Rose.
"Your phone?" said Henrietta. "I thought—"
"I lied," said Rose. She reached into her pocket and pulled out her phone. Gary raised his eyebrows when he saw it. It was last year's model and looked a bit ridiculous. Instead of the curvy oval shape of this year's phones it was rectangular, with sharp corners.
Rose opened it and held it out so Henrietta and Gary could see the screen. It was blank except for a blinking green cursor.
"It's broken," said Gary.
"It isn't," said Rose. "It's cracked."
"Cracked?" said Gary.
"My mom cracked it. It's off the system."
"I don't understand," said Henrietta.
"You said the Wikkeling was in the school computer," said Rose. "Now it's on your phones. But mine is manual." She started typing numbers onto her screen.
"What are you doing?" said Henrietta.
"Calling some friends."
"You . . . you just _know_ the number?" said Henrietta.
"I memorized it," said Rose. She finished typing. She held the phone to her ear and stepped away as she began a whispered conversation.
A moment later, she pocketed the phone and turned to Henrietta and Gary. "They're coming," she said.
"Who?" said Gary.
"Friends from the Old City," said Rose.
A chill ran through Henrietta's spine as she heard the words. She wondered if Rose's "friends" would make things better or worse.
All three children jumped nearly out of their skins as a voice sounded from behind the landing pad they'd assembled.
"WHERE DO YOU GO?" it said.
### Frightening Friends
The Wikkeling had appeared from around the front corner of the house. "WHERE DO YOU GO?"it asked again, repeating the words exactly, the scratchy echo resounding down its pink throat as its mouth hung slack. Its body flickered out of sight for a moment, and when it returned it was a step closer.
"You want to know where we were?" said Henrietta, hoping to perhaps reason with it. "We were in the attic. Up there." She pointed toward the cat hall. The Wikkeling looked up.
"Why does it matter?" said Henrietta. "Why don't you leave us alone?"
The Wikkeling flicked off and on again, returning to sight with one hand raised, its long, waxen index finger pointed at them.
Then another voice sounded, from behind the children at the opposite corner of the house. "Rose!" it lisped. Gary clutched his heart reflexively, wondering if he could possibly be any more nervous that he already was.
The children turned to see two strangers, and to Henrietta and Gary they looked more terrifying even than the Wikkeling. The one who'd spoken was wiry and bald with a pale face and a streaked gray beard. He wore a black long-sleeved shirt, grimy gray pants, and a pair of dark green knee-high rubber boots. His pants were held up with suspenders that appeared to be laminated strips cut from plastic grocery bags.
The second man was equally dirty, and a mountain—at least six and a half feet tall. His dark red hair was wrapped into a loose bun, and his face was scruffy with stubble. A streak of grease was smudged across his forehead, as if he'd scratched an itch with a dirty hand.
Rose leaped forward and threw herself into the arms of the bald man. "That was fast!" she said. "We need help. We're going to Sunset Estates. We're in danger." She pointed at the Wikkeling, which had stopped its advance for a moment, perhaps to assess the developing situation. It was immediately apparent that the two visitors couldn't see it, however, and Rose continued, "We need to go _now_." Her voice filled with a tremendous authority that made her seem much older.
The bald man shrugged. "Jump on, then," he said, kneeling. Rose scrambled onto his back, putting her arms around his neck, and he stood.
The enormous man knelt also, and indicated that Henrietta and Gary should follow Rose's example. "Nice to meet you," he said in a scratchy voice. "My name is Oak."
Henrietta hesitated.
"Hurry," said Rose from her perch atop the bald man. Henrietta and Gary both pulled themselves onto Oak's impossibly broad back. Henrietta grabbed his left shoulder, and Gary his right.
The two strangers began to run, and Oak's gait almost shook Henrietta from her purchase. She grabbed tighter, curling her hands into his thick clothing as she looked ahead, wondering where they could possibly go. The only option she knew of was the sidewalk—and that was gone now. But if you're the kind of person who can't be seen on sidewalks without getting arrested, you learn other ways. Henrietta had never noticed, for instance, that the rear alley behind her house also ran behind the house next door, and the one after that, continuing all the way up the block if you knew how to follow it: a little to the left, between two fences, and behind a narrow shed.
Henrietta and Gary held on for dear life against the jolts of Oak's long strides. She glanced back briefly—the Wikkeling was already out of sight behind them. She breathed a sigh of relief. Next to her, Gary said, "G-g-g—d-d-d," which meant, "Good riddance!" His words were pounded down to two consonants by Oak's gait.
They made another turn down a narrow, pitch-dark alley between two tall buildings. Behind them, the sounds of traffic disappeared, replaced with sounds of splashing—they were running through water. Oak's stride slowed a little, and the jouncing settled enough for Henrietta to speak.
"Where are we?" she said, still holding on tightly.
"You won't understand the route," said Oak."We're . . . behind the scenes, you might say." Ahead of them, the bald man turned suddenly right, disappearing. Oak followed, and they emerged into a wider alley with a few streetlamps overhead. Henrietta saw that they were running over bricks.
"Bricks," she said to Gary. "Just like through the windows."
"This is incredible!" said Gary. "I can't believe I never knew about this."
"We're in the Old City now," said Oak. "School buses never go back here—there's no children." They took another turn into an open alley square at the intersection of several buildings. The bald man had stopped ahead, and Rose was climbing off. Oak knelt and let Henrietta and Gary dismount. They both looked around in wonderment at the old buildings rising into the darkness around them. "Wow, wow," Gary muttered under his breath as he took it all in.
"Rosie, you better explain yourself," the bald man lisped, a little out of breath. "I'm of a mind to call your parents. The whole gosh-dang Old City is looking for you right now. And _these two_ ," he gestured toward Henrietta and Gary. "The whole gosh-dang Addition is looking for them. And blaming us, I might add." He snorted. Then he nodded at Henrietta and Gary. "Nice to meet you both, by the way," he said, loping forward and extending his hand. "The name's OK."
"OK?" said Henrietta.
"All right," said OK. He smiled, and Henrietta saw he was missing his front teeth, which explained his lisp. "Now go ahead, Rose, with that explanation."
"We can see an invisible creature," said Rose. "It's called the Wikkeling. It's after us."
"You're _sure_ this ain't a game," said OK sternly.
"I'm sure," said Rose. "We think Henrietta's grandfather can help us. He's got a _Bestiary_."
"An Alcott _Bestiary_?" said OK. Rose looked at Henrietta.
"That's right," said Henrietta. "Do you know it?"
"Saw one once," said OK. "A beautiful old book. You think your Wick-ling will be in there, do you?"
Just then a low rumbling sound entered the alleyway. "Here comes yer ride," said OK. Then he pointed to the greasy dumpster. "Jump in here, and the truck'll scoop you up with the trash." He picked up Rose, carried her to the lip, and dropped her inside. "C'mon," he said to Henrietta and Gary.
"You want us to get in _there_?" said Henrietta, spreading her fingers in revulsion.
"Wow!" said Gary. He immediately clambered over the top and flipped inside, disappearing. "Wow!" he said again, his voice echoing within.
When Henrietta didn't move, OK gestured to Oak, who scooped her up and heaved her gracelessly through the opening. " _Augh_ ," said Henrietta as the terrible thick smell entered her nose.
OK's face appeared above them. "Pay attention, and make sure you get off at the right stop." He grinned at Henrietta. "Pretend you're on the school bus." He disappeared, and the rumbling grew quickly until Henrietta could feel it vibrating in the trash bags all around her. The dumpster began to tilt.
"Look out!" yelled Gary as the trash shifted, tumbling over itself. Henrietta tried to keep her balance, but before she knew it she was midair.
She landed on a soft pile, and a black trash bag flopped down on her head.
Gary rushed to his feet, his eyes wide. "We're riding in a garbage truck!" he said, obviously thrilled.
"This is gross," said Henrietta, pinching her nose with one smelly hand.
"Are you kidding?" said Gary. He looked out from the open rear of the truck as it pulled from the alleyway and into traffic, the roar of the engine mixing with the sounds of other cars and trucks all around. "I've wanted to do this for my _entire life_!" He held his hands up in the air, striking a triumphant pose. " _GARBAGE TRUCK_!" he yelled.
Surrounding the truck were lanes of traffic, and the children peeked out at the cars.
"Stay down," said Rose. Henrietta tackled Gary, and they both fell into the trash. Gary was laughing hysterically.
"I can't believe you wanted to do this," said Henrietta.
"I thought _everyone_ wanted to do this," said Gary.
The truck took a slow turn onto a larger road, and the children peeked out. They were in one of the middle lanes of a ten-lane thoroughfare, high enough to look down through the windshields and see the blank faces of all the people driving.
"Watch for Sunset Estates," said Rose. "If we miss it, we'll get compacted."
"Compacted?" said Henrietta.
"Squashed," said Rose, "when the truck reaches the dump."
They began looking in earnest.
### FindEm™
The news of her mother's death shook Aline, wiping her daughter and her studying friends entirely from her mind. She started to get dinner ready in a daze, threw out the breakfast dishes, and worked halfheartedly on a project for one of her clients.
Tom returned late from work. He entered the kitchen holding his cell phone in front of his face, reading something on the small screen. He pocketed the phone and gave Aline an exhausted look. "How was your day?" he asked.
"My mother passed," said Aline.
Tom stopped short, caught off guard. Then he put his arms around Aline, who held him tightly.
"I knew she didn't have long, but . . . I never called. She raised me here, in this house." Aline pulled away from Tom and looked disconsolately around.
"Did you tell Henrietta?" said Tom.
"Not yet. She's studying with her friends." At that moment, Aline realized that Henrietta's room had been uncommonly quiet for a very long time. Without another word, she walked to Henrietta's closed door and entered.
Henrietta, Gary, and Rose were not there. Aline yelled, and Tom raced in from the living room. He was already on his cell, and Aline activated hers also. She accessed FindEm™, a tracking program parents used to locate kids.
HELLO! LET US HELP YOU FINDEM™!
Aline scrolled down the Find menu, which contained herself, her mother, Tom, and Henrietta. She selected Henrietta and clicked FINDEM™.
FINDING . . .
Tom was similarly occupied. Aline sat on the edge of Henrietta's bed, holding the phone tightly in both hands. After a few moments, the word "Finding" was replaced on both their phones with
SORRY. CONTACT THE POLICE? Y/N
Aline selected _γ_ and held the phone to her ear.
"Seaside Precinct. Missing persons," said a disinterested woman's voice on Aline's phone.
"Henrietta Gad-Fly," said Aline.
"Scanning for the phone," said the officer. "I don't see it active."
"What does that mean?" said Aline. "Is the battery down?"
"The phone was destroyed. When did you last see your daughter?"
"A couple of hours ago."
"Where?"
"In her bedroom, with her friends," said Aline.
"Have you searched the house to ensure she and her friends aren't hiding?"
"Well, no, but if her phone is destroyed—" She looked around as she said it. Tom knelt and peered under Henrietta's bed.
"Sometimes children destroy phones and hide because they're angry at their parents," said the officer. "Maybe she's angry at you."
"I . . . don't think so," said Aline.
"I suggest you search the house. All I can do here is scan her cell records to see if a malfunction message was logged. What are her friends' names?"
"Gary Span, and Rose Soldottir," said Aline.
"We'll check their phones as well."
"Thank you," said Aline. She hung up.
"Why is Henrietta's chair sitting on her desk?" said Tom. There was no immediate answer to this question, and the two of them split up and searched the house. Aline opened the kitchen cupboards and moved couches away from the walls. She felt a terrible, constricting ache in her heart.
Tom went through the master bedroom and the bathroom, and everywhere else he could think of, and eventually collapsed on the living room couch. Aline sat next to him, and her phone rang. The officer informed her that Henrietta had been classified as a Person of Unknown Whereabouts, as were Gary and Rose, whose parents were also being contacted.
Tom and Aline were directed to search the house again. Had they looked in the freezer? The toilet tank? The oven?
Tom, fed up, seized the phone. "She's obviously been kidnapped!" he shouted.
### Attacked
The garbage truck turned off the ten-lane arterial into a residential neighborhood.
Gary stood tall, the wind whipping his hair, a wide smile on his face. Henrietta, nauseated from the stench of the garbage, focused on scanning the buildings around her for anything familiar, and soon spotted the sign for Sunset Estates.
The truck pulled into the Estates, passed a long stretch of identical homes, and paused at a row of dumpsters.
" _Now!_ " said Rose.
They extricated themselves from the trash, jumped from the stinking truck onto the asphalt drive, and beelined for the shadows at the side of the nearest town home, even as one of the garbage collectors hopped from the truck's cab.
"I wish my mom drove a garbage truck instead of a car," Gary whispered, panting a little from the sprint. "So," he said, taking in the line of identical houses. "How do we figure out which is your grandpa's?"
"His address," said Henrietta. "Zero five, zero seven, six three two."
The sidewalks running through Sunset Estates' honeycomb of streets were silent. All of the old people were inside, probably sleeping, and no one but the garbage collector was driving through. The sound of the truck's grumbling engine diminished behind them as they walked.
They figured out the pattern of addresses and headed toward Al's.
"I hope he's home," said Rose.
"I hope he's awake," said Gary.
All of the town homes followed the same architectural pattern: Garage, porch, front door. Garage, porch, front door.
"Just a few more," said Henrietta.
"WHERE DO YOU GO?" said a voice. The children all turned, eyes wide, to find the Wikkeling out in the street, pacing them. Its haunting, flickering face gaped. A sound of static pulsed from its open mouth.
" _Run_ ," said Henrietta.
The Wikkeling blipped along, appearing and disappearing, easily in step with them as they fled. Henrietta glanced at the address they were rushing past: Al's was next.
Garage.
Porch.
Front door. The Wikkeling closed in perilously as Henrietta pressed the doorbell, which began its cheery rendition of "Jingle Bells."
"Grandpa!" Henrietta shouted.
This time, the Wikkeling went for Rose.
Its long finger stretched out and tapped her, but instead of disappearing as usual, the Wikkeling's finger stuck to Rose's head, as if it were glued there.
Rose collapsed as the front door opened to reveal Al, an old book in one hand and a questioning look on his face as he took in the scene.
"Henrietta?" he said. "What—"
"Let us in!" said Henrietta.
"Who—" said Al.
Henrietta grabbed Rose's arm to support her. "Hurry!" she said. Gary grabbed Rose's other arm.
Al stepped aside to make way, and Henrietta and Gary stumbled in with Rose in tow. The Wikkeling, still attached to Rose by its index finger, dragged with Rose across the threshold.
The door swung closed behind them. Rose and the Wikkeling lay in a heap on the floor.
"Stop it!" Henrietta shouted.
"The _Wikkeling_?" said Al, his brow furrowing.
"You can see it?" said Gary.
"Release her!" said Al, whacking at it with the old book he held. Surprisingly, the Wikkeling recoiled as it was struck, and it swatted at the book, knocking it out of Al's hand and ripping a few pages from the binding.
"Books!" said Henrietta, grabbing Al's arm.
"What?" said Al.
"Books! Old books!" She turned to Gary. "Gary—it's not just that the houses are old. Rose's house is a library, and so is the attic! It's the books, too!"
"You're right," said Gary.
Henrietta turned to Al. "We're taking Rose into your basement."
Al asked no further questions, but knelt painfully on his old knees and picked up Rose. The Wikkeling stood, its finger still stuck to Rose's forehead, and walked along beside.
"Don't let it touch you," said Henrietta. She ran ahead and opened the basement door, ushering Gary in first. "Get the light," she said. Gary flipped the switch, and the fluorescent lights flickered on.
Henrietta descended ahead of Al, Rose, and the Wikkeling.
As Al went, the Wikkeling didn't follow—it rooted itself to the spot on the top step and its arm began to stretch, unwinding from its body like a garden hose.
Once Al and Rose reached the bottom, the Wikkeling's finger disengaged with a _pop_ and its arm recoiled up the stairs. It moaned out a sound like twisting metal, and flickered out for a moment. When it reappeared a second later, it was sitting on the top step, holding its hands to its head.
Al laid Rose on the basement couch, and the three gathered around her.
"Is she breathing?" said Henrietta as Gary knelt and listened.
"Yes," he said.
Henrietta glanced up the stairwell at the Wikkeling. "Look at it," she said. It sat there, head in its hands, as if in pain.
"Tell me what's going on," said Al.
"We need to see your _Bestiary_ ," said Henrietta.
Al ducked into one of the rows of bookshelves and brought out the copy he'd shown Henrietta on her previous visit.
"Is there an entry on the Wikkeling?" said Henrietta. "You knew its name when you saw it."
"When I was young," said Al, "I used to see it. Everybody did."
"Just . . . walking around?"
"There was a story that it was supposed to protect people. But it disappeared as time went on. I'd started to think it was a dream."
"We're trapped again," said Gary, looking up the stairs where the Wikkeling was still seated miserably on the top step, between them and the exit.
"It has never protected us from anything," said Henrietta.
Al flipped to the index of the _Bestiary_.
"Here it is," he said. "Under _Nonliving Creatures_."
`Wikkeling, The:`
`It is argued in recent years whether the Wikkeling exists or is simply a story, though this author is compelled to assert his conviction that this unique creature is real, and was once seen by many. As time has gone on, and sightings have diminished, its existence has become increasingly mythologized.`
`The Wikkeling was engineered by two scientists named Henrift and Andi in the early days of civilization to destroy the Draageling (page 345) and to generally harness the power of nature toward human industry. It served faithfully in the outset, but grew maleficent, even killing its creators. The Wikkeling became increasingly reclusive as years wore on, shrinking from sight until finally absenting itself altogether from human contact. Its whereabouts are no longer known. Of those who believe in its existence, some maintain that it moved away, some that it died, and some that it remains invisibly among humans.`
`It is said to be violently repelled by highly time-resistant objects such as old books, old trees, and wild housecats (whose shrinking population is sometimes attributed to the Wikkeling).`
**—**`Observed by H.G-F. Recorded by A.A.`
Henrietta was thinking hard. "Al," she said, "remember the cat I told you about?"
"Yes."
"She got better, and finally left. I wonder—if the Wikkeling doesn't like wild housecats, is there any way. . . ."
"Right," said Al, and he turned and disappeared again into one of the long rows of bookshelves.
"Al," Henrietta called after him, "is time ever weird down here?"
"Time?" said Al, rummaging through the stacks.
"Does it seem not to pass down here?"
"Sometimes," Al's voice came, "I'm down here for what seems like hours with these old books, and I'll get hungry for dinner, and head upstairs, and find that your grandmother—"
His voice cut off suddenly. Henrietta ran back to see what was the matter. She found him stooped over, having just removed a book from a low shelf. He was holding one hand to his forehead, and his eyes were squeezed shut. Henrietta's heart skipped a beat. Al turned and Henrietta saw tears on his cheeks. "I'm sorry, Henrietta. I'm grieving."
"Grieving?" said Henrietta.
Al paused. "Did your parents not tell you?" he asked, his brow furrowing. "Your grandmother—my Henrie—passed away."
"Oh," said Henrietta. She felt a lump in her throat.
"Henrietta . . ." Gary's voice from the main room sounded urgent. "Henrietta!" he shrieked. Then, there was a soft thud.
Henrietta and Al both raced out to see that the Wikkeling had somehow managed to reach the bottom of the staircase. Its arm had snapped out again, and its finger was now locked to Gary's forehead. Gary lay unconscious on the floor, his face pale.
### Distress Call
Henrietta didn't know she was rushing at the Wikkeling until Al restrained her. "Let him go!" she screamed at it.
"Henrietta, stop," said Al.
"It's going to kill Gary," she said.
"Look," said Al. He opened the book he'd just retrieved from the stacks. _How To: A Book of Instructions_.
Al paged through the table of contents ( _How to Fix a Leaky Faucet; How to Carve a Turkey; How to Stop Time_ ) and then flipped forward to the chapter he was looking for: How to Attract a Wild Housecat.
They skimmed the entry.
`Wild housecats are fascinating, intelligent creatures, but tend to be wary of people. The easiest way to attract a wild housecat is to construct a cat hall **—** a wild-housecat-sized doorway built into an exterior wall. Some report advantages from framing the hall, writing a welcome message over the lintel, or otherwise giving the hall an inviting appearance. Once the hall is built, wait six to six hundred weeks, during which time it is likely that a wild housecat will appear.`
"Six to six _hundred_?" Henrietta exclaimed.
"Keep reading," said Al.
`If you are in immediate need, there is an additional method available, although a cat hall must still be installed. Procure a small amount of dried wild housecat blood (obtainable from many Friends of Wild Housecats societies). Dampen the blood with water and apply it to the entrance of the cat hall. When the wind carries the smell of the blood outside, any wild housecat in the vicinity should smell it and come quickly, assuming one of its kind is in danger. Since this method involves attracting these very intelligent and easily perturbed animals through false pretense, you'd best be ready with a good explanation for your actions when the cat arrives.`
"We can make a hall," Al said, "but what about the blood?"
Henrietta thought of the blood on the attic floor, in her house . . . and her eyes lit up. "Al!" she said. "Do you still have the textbook I gave you?"
Al loped back into the shelves and emerged with the book in his hands. Henrietta opened it.
There it was: the drop of blood, dry and brown now, that had fallen onto her math problem the night she'd found Mister Lady.
"Amazing!" said Al. There was no time to enjoy their good fortune. Al hurried off into the rear room he'd shown Henrietta on her first visit, where he kept his old tools.
Henrietta looked over at Gary. His lips were closed tight, his face screwed up in pain. The Wikkeling's finger was still attached to his forehead, its long arm snaking back to its body at the base of the stairs.
"We've got to hurry," said Henrietta as Al returned.
"Won't be attractive, but it'll do," said Al. He handed Henrietta a hammer with a large metal head, then plugged a small power saw into a nearby wall socket. "I'll saw through the wall, and you hit it with the hammer to knock it out."
The saw whined loudly as Al revved it, dust flying off the long-unused blade. He pressed it into the wall, at ground level outside, and an awful-looking plume of black smoke poured from the slice. He steered the saw through a neat square cut, and then backed away. "Whack it!" Henrietta struck the square with the hammer and knocked it out onto Al's front yard. Evening air flooded in.
"We've got to dampen the blood," said Al, "and there's no running water down here!"
Henrietta grabbed the textbook, spat on the page, mixed the spit and blood with her finger, and quickly daubed it around the edges of the cat hall.
She and Al turned around just in time to see the Wikkeling's arm retract into its body with a snap, like a measuring tape. It flickered out for a moment, reappeared lying on the ground, blinked out again, and then returned, standing. The proximity to the books seemed to be taking a toll on it.
Henrietta ran to Gary. He was unconscious, but now that the connection to the Wikkeling was broken, he seemed to relax a little bit.
The Wikkeling visibly gathered its strength. It straightened and solidified. The vague circle of its gaping mouth stretched into an empty smile.
### The Stink
If anyone had happened to go strolling past Al's house that night, they would have observed a small chunk of the house's outer wall falling into the lawn in a cloud of black smoke; then, a child's hand appearing and smearing something around the edges.
That mixture of blood and spit began to stink, though too subtly for a human nose to detect. The stink wafted across the plastic green grass to the street. It ascended above rooftops and gusted along highways.
It drifted past OK and Oak, deep in the Old City. After leaving Rose, Henrietta, and Gary in the dumpster, OK had called Rose's parents and told them everything, including the children's destination. Rose's parents called Henrietta's parents and asked them if they knew anyone at Sunset Estates. They did. They called Al, but Al didn't answer. They called Gary's mother.
In this manner it came to pass that Rose's parents Sid and Sigrid, Henrietta's parents Aline and Tom, and Ms. Span all headed toward Al's town house in Sunset Estates, along with several police officers and a fire truck that had been summoned when some black smoke had drifted through Al's neighbor's window.
The stink traveled onward on the wind, past the place you've reached in this story and into the next, and the one after, until it finally encountered a nose sensitive enough to smell it.
### Two Wild Housecats
The Wikkeling's smile was the most frightening thing Henrietta had ever seen. Its flickering pink mouth and yellow eyes widened as it raised its arm and pointed its long index finger at her. Its arm began to extend.
"Don't!" Al shouted.
Henrietta scrambled behind the couch. She braced herself.
Then a shadow fell across the cat hall.
A familiar gray head appeared, followed by a pair of shoulders and long legs. Mister Lady leapt silently to the carpeted floor. Then another cat entered. A fat, orange-striped tabby, considerably larger than Mister Lady, squeezed its bulk through, looking strangely dignified as it tumbled down with a graceless _thud_.
The directions in _How To_ had said you'd better be ready to explain yourself when the wild housecats show up, but Mister Lady and her ample friend seemed to know exactly why they'd been summoned. Mister Lady crouched and then sprang in a fantastic leap, landing right on the Wikkeling's shoulder. The Wikkeling jerked back, but she clung tenaciously and raised a paw, exposing a set of long, razor-sharp claws. She slashed its face, and the Wikkeling's smile wilted as four pale gashes appeared across its cheek and nose. It winked out of existence for a moment, and Mister Lady fell through to the stairs even as it reappeared a few steps above her. The orange tabby stood above it near the top landing, arching his back and hissing. Mister Lady, from below, yowled low in her throat, exposing her long, white teeth.
Henrietta had always heard that cats were dangerous, but when she befriended Mister Lady she'd somewhat forgotten those warnings. Now she saw that they could be dangerous indeed when the situation called for it.
The Wikkeling cast about, trapped between the cats. Its face was bleeding profusely, a clear liquid like corn syrup. With one long finger, it stabbed out at the cats as they closed in.
"Look at its legs," said Al. At first, Henrietta wasn't sure what he meant, but then she saw that the Wikkeling's legs had stopped flickering. The yellow fabric of its pants was cracking at the folds, like clay.
The Wikkeling began to sweat as it turned upstairs and down, desperately seeking escape, and its face shone in the fluorescent lights. Soon its clothing was wet as well, as if slicked in mud.
"It's getting younger," said Henrietta. Except for its long hands and fingers, which dangled enormously, the Wikkeling was no longer full-grown. Even as she watched, its stature shrank until it appeared as she'd seen it through the attic windows: a monstrous child.
Next to Henrietta, Rose stirred and sat up. Henrietta held her hand, and together they watched the awful reduction of their tormentor. The child Wikkeling's mouth opened and closed, but no sound emerged, just the flickering light in the back of its throat. It looked terrified, desperate.
"What's it saying?" said Henrietta.
"I can't hear," said Al.
Whatever it was, the Wikkeling repeated it over and over. As it continued to shrink, its features dislodged from their positions on its face. Its mouth drifted up onto its cheek, and its teeth dribbled down its neck. The rest of its body lost cohesion and spread out over the stairs, mixing with its clothes into a yellow soup. In mere moments, only a lumpy puddle remained, with a pair of eyes sinking into it. Then, nothing.
The wild housecats stepped toward one another and sat together on the now empty stair. Mister Lady licked the ear of the tabby and straightened his fur fussily with one paw.
Rose went to Gary. His hands, which had been balled in tight fists, loosened. His eyes opened. "What happened?" he gasped, sitting up quickly.
"The Wikkeling evaporated," said Rose.
Henrietta turned toward the cats to thank them, but saw only the flick of a gray tail at the edge of the cat door as Mister Lady disappeared after her friend. It was as if they'd stopped by on their way to another engagement—which may indeed have been the case.
Then a knock sounded from upstairs, and a cheerful rendition of "Jingle Bells" began.
"Someone's here," said Al.
Lots of someones, in fact: Henrietta's parents, Rose's parents, Gary's mother, ten police officers, two police dogs, and five firefighters were crowded on the porch, waiting for some explanations.
### Explanations
The children realized that an honest account of their experiences wouldn't be well-received, so they chimed in on an improvised medley of conflicting fabrications (each confirmed by a bemused but willing Al), and the following story emerged:
Henrietta, Gary, and Rose had wanted to borrow some books from Al. Because Henrietta was grounded, they decided to sneak out her bedroom window. They caught a public bus, but accidentally boarded the wrong one and also accidently dropped their phones while trying to call for help. Humiliated and phoneless, they'd ended up at Sunset Estates just as Al was sawing through his wall to . . . deal with . . . an electrical problem.
Of the crowd assembled at Al's door, the firefighters were the first to leave, since there was no fire. The police officers departed next, since there was no crime.
Rose's parents followed, with Rose in tow. They were upset at her, but they knew there was more to the story and that Rose would tell them the details once they'd returned to the Library.
Ms. Span and Gary made their exit. Gary would be grounded for a week. Unlike Rose, he would not be explaining the real story to his mother. To her, the truth would sound more like a lie than the lies did. There was one thing Gary did want to come clean about, however. As he sat next to his mother while she drove them home, he clenched his hands, collecting his courage. "Mom, I have to tell you something."
"Are you going to explain why you smell like a garbage pile?" his mother said, turning to eye his pockets as she spoke. "You haven't been collecting again, have you? You know I've expressly forbidden it."
"Oh," Gary said, putting his hands around his waist. "I think I sat in something." In fact, his pockets were bulging with some amazing pieces he'd discovered during the garbage truck ride, which he could hardly wait to catalog.
"You'll scrub the seat when you get home, and shower after that," said her mother briskly.
"Mom, what I wanted to say is . . ." Gary cleared his throat. "I had a secret. I've been keeping it a long time, but it isn't a secret anymore. So I want to tell you."
"Oh," said Ms. Span, and her face showed some concern.
"I can't . . . I mean, I _couldn't_ . . . read."
"What?" she said. "Don't joke, Gary. Of course you can read."
"I can now," said Gary, "but not until this year. I used to cheat. But I'm learning—Henrietta and Rose have been helping me."
" _Henrietta_ has been helping _you_?" said Ms. Span, aghast.
Gary insisted on his story, and somehow—through exhaustion, relief, or too much worry in one day—his mother finally believed him. "Beginning tomorrow, you and I are going to have supervised study sessions after dinner," she said crisply.
"Okay," said Gary happily, flicking a bit of eggshell from the leg of his pants and onto the pristine car floor.
After Gary and his mother left, a long, awkward silence ensued between Henrietta, her parents, and Al.
"Mom," said Henrietta. "I'm sorry."
"Sweetie," her mother sighed. She looked worn-out and sad. "That was a terrible, dangerous thing to do. I have no idea what you were thinking. But I'm glad you're all right."
"You're still grounded, though," said her father. Then he almost smiled. He gathered Henrietta in for a hug, but just as quickly released her, waving a hand in front of his face. "You smell awful," he said.
Henrietta's mother sniffed the air. "That's coming from you?"
Henrietta shrugged.
Then Al spoke up, looking uncomfortable. "Aline," he said, "we need to talk."
Henrietta's mother tensed up. She folded her arms across her chest and looked impatient. "I'm not sure what you mean," she said.
Al's face clouded with doubt. "Aline," he said slowly, "you and your parents meant the world to me. They were all I lived for." He paused, took a deep breath and let it out and then continued. "But that world . . . well, it's passing away. I hope it isn't too late. I want things to change. I want . . ." He struggled for words for a moment. "I want to be your father, Aline. For as long as I have left. If you'll have me."
Another long silence fell on the room.
Aline looked at her husband, and then at Henrietta. She closed her eyes, and her face relaxed. Then she stood, crossed the room to Al, and hugged him. As her arms closed around his old shoulders, a grieving sob suddenly escaped her. Al's eyes reddened, and his old, wrinkled face grimaced, which is the only expression you can make when you're feeling too many emotions at once.
### The Memorial
The memorial service for Grandmother Henrie was held at the Sunset Estates community center. Henrietta dressed in her nicest pants and wore her uncomfortable dress shoes.
Al and Henrie's elderly friends were there, and they were all terribly nice, and terribly sad—they were often attending funerals these days. During the service, Al stood next to Henrietta and her parents, and it felt strange and new to all of them to be a family.
After the service, everyone gathered at Al's house. For the first time in Henrietta's memory, she and her parents stayed until the very end. They didn't look for an excuse to leave early, and they didn't stand awkwardly and act like they wanted to be elsewhere. Henrietta's father was unexpectedly charming, and her mother was a whiz at learning people's names.
Near the end of the evening, when the remaining guests were chatting in the living room, Henrietta approached Al in the kitchen.
"Grandpa?" she said. Al turned from the counter and smiled as he saw her.
"It's nice to hear you call me that," he said.
"I was wondering if you'd start a book club with me. We aren't using books at school anymore, but I thought we could read some, with my friends. When we aren't grounded anymore."
"I'd like that very much," said Al.
Henrietta and her parents were the last guests to depart, and they helped Al clean up. As they said their good-byes on the front porch, Aline and Al hugged silently. Henrietta heard a sound coming from up the street and turned just in time to see the garbage truck pull through the parking lot. She smiled, even though the sight also made her a little nauseous.
On the drive home, they reflected on the responsibilities that awaited them the next morning. Henrietta would return to school, and her father and mother both had work to do. Of course, life wouldn't be quite so regular as it once was. "Henrietta," said her mother, "there's something I should have mentioned."
Henrietta was thinking about how wonderful it would be to take off her horrid, too tight shoes when she got home. Her feet throbbed. "What is it?" she asked.
"We found out last week that the city wants us to move. Our house is going to be demolished. It's good news, actually. We can go anywhere. I know you've never liked that old place. It made you sick."
"It didn't make me sick, actually," said Henrietta, but she stopped herself quickly from explaining further—the story would be far too strange for her parents to accept.
Her mother continued: "We'll move someplace new. A fancy house like Ms. Span's."
"I want to stay in our house," said Henrietta, firmly. "Can we tell them we don't want to leave?"
"I'm afraid their minds are made up, Henrietta," said her father.
"Could we find another old house, then?" said Henrietta.
Her mother looked over at her father, and they shared a puzzled glance. Their daughter was quite an inexplicable creature. Just then, Henrietta's father's cell phone rang, and he hooked it to his ear by a little ear clip.
"Tom here," he said. "Yes, I—what? I'm sorry, say that again. Elton, am I hearing you right?" Tom frowned. "Elton, I'm going to have to call you back. I'm spending some time with my family right now." He disconnected and returned the phone to his pocket.
"What is it, Tom?" said Aline.
"Just work," said Tom.
"But what?" said Henrietta.
"Tough to explain," he said. This was what he always said, and Henrietta pushed a little further.
" _Try_ ," she said insistently.
"Well, there's a program called the System Manager. It connects other programs together, you could say. It makes them run efficiently. Anyway, it crashed yesterday, and we've been trying to repair it. Now I just heard that apparently it has deleted itself."
"Deleted?" said Aline.
"That's what I've been told. Needless to say, I'll be going in early tomorrow. But I'm not going to worry about it now."
Henrietta leaned back in her seat and looked out the window at the other lanes of traffic and the buildings that walled the street, shining in the strong glow of the streetlights. She thought about Al's _How To_ book. If she did have to move, she thought, the first thing she'd do in the new house would be to build a beautiful cat hall with a painted lintel that read ALL CATS WELCOME across the top.
### The Attic Books
Gary and Henrietta were grounded for a week. In Henrietta's case, this involved having no friends over, supervised homework sessions after school, and no personal use of her brand new Skipping-Stone Phone.
It wasn't so bad, though. Henrietta's mother was quite supportive of her efforts to improve her schoolwork, and sitting with her after dinner to complete compositions and math problems was kind of fun. Henrietta hadn't known that her mother was a whiz at math, and she even talked about her own accounting work a little bit, using examples from her job to illustrate some of the problems Henrietta had to work.
But Henrietta immensely missed going into the attic with Gary and Rose. Although they spent time together at school, they couldn't really relax there, Henrietta especially, since she was AT RISK.
Henrietta hadn't been back to the attic since the three of them escaped through the cat hall. Now that the BedCam was working again, and her parents were watching her every move, she hadn't had even a second alone—a deplorable situation, since she'd come to depend on the solitude of the attic to give her time to think about things.
When the day finally dawned that signaled the end of Henrietta's punishment, she awakened not to the sound of her alarm, but to the sound of her phone ringing. She reached out groggily and checked the screen, brightening immediately when she saw the name.
"Grandpa!" she said as she answered.
"Hello, Henrietta," said Al. "Remember how you said you wanted to start a book club? How about tonight?"
Henrietta beamed. "I want to! But I have to ask—"
"I asked already," said Al. "In fact, your friend Rose's mother suggested we meet at their house. I've arranged to pick up you and Gary, if you're interested."
"Yes!" Henrietta shouted.
School was a blur of typing practices and advice about Halloween (a topic that would continue to wax for weeks in anticipation of the dreaded holiday, including cautionary movies about Jack-O'-Lantern disasters, poison candy, and dangerous strangers). Henrietta worried over all of the Practice Tests because the next Competency Exam could spell the end of her scholastic career.
Soon enough though, the day ended, and Henrietta, Gary, and Rose headed home on the bus. Henrietta quizzed Rose about what her house was like, only to find her mysterious, evasive, and more than a little amused. When the bus arrived at Rose's stop, Rose smiled as she disembarked, saying, "See you later!"
Henrietta ate a distracted dinner in front of the TV with her parents, fielding a few questions about school and apologizing for a mistype she'd made on a composition when she had written "affect" instead of "effect." Nonetheless, her parents seemed pleased with her progress.
When the evening news began, a knock sounded at the side door. Henrietta jumped up from the couch.
"Have a good time," said her mother.
"I will!" said Henrietta. She exited to the kitchen, opened the door, and hugged her grandfather.
They walked to the car and Henrietta opened the rear passenger door to find Gary already inside, waggling his eyebrows delightedly.
The car crawled through traffic toward Rose's house, Al following the lead of the car's computer. As they drove, they heard an ad about a kind of ice that stayed cold even in ovens. Imagine: a broiled milkshake.
"NOW ARRIVING AT YOUR DESTINATION," said the computer. "TURN LEFT, AND PARK IN THE DRIVEWAY. THESE INSTRUCTIONS BROUGHT TO YOU BY EARHELPERS. WITH EARHELPERS—," Al turned off the engine, and they all stepped onto the driveway, Al holding a few books in one hand that he'd brought for the meeting.
Before them stood an enormous mansion. But its size wasn't the most impressive thing about it. Rather, it was the look of the place. It resembled the old houses Henrietta and Gary had seen through the attic windows, but much, much bigger. The roof wasn't one roof, but many small roofs, sloping down multiple planes and along gables. A turret rose from the middle that appeared constructed from a single old tree trunk.
"It's hu-uge," said Gary, extending the word because "huge" wasn't quite huge enough.
A grand staircase led to a pair of immense double doors, where Al knocked with his free hand. A latch clicked on the other side, and the heavy wood creaked on old hinges. They all took a step back, a little nervous about the imposing place, but their nerves quieted when Rose peeked out, smiling.
"Come in!" she said.
They entered a gigantic sitting room that contained two warmly burning fireplaces, a variety of upholstered chairs scattered in groups, reading desks, and even a study carrel along one wall. At the center of the room stood a long table covered with old maps. Every wall in the place was clothed in bookshelves, which reached to the ceiling far overhead. As the three looked around, awestruck, Rose's parents approached to greet them.
"I'm Sigrid," said Rose's mother. "And I'm Sid," said Rose's father. "It's nice to meet you all officially. Things were a bit rushed the other night."
"It's nice to meet you both," said Al.
"Me, too," said Gary.
"When Rose told us you were starting a book club," said Sigrid, "it seemed sensible to have it here, since we call this house the Library. And as you can see, that's what it is."
"But it didn't used to be," said Al.
"Hm?" said Sid.
"This . . . was . . . ." Al trailed off. He seemed lost in a memory, and he turned in a slow circle, taking everything in. Then he strode to a far corner of the room where two bookcases connected. He knelt and began pulling books from the bottom shelf and peering into the shadows. He laughed. "I can't believe it!" he said.
Everyone joined him, kneeling and looking at the little shadowed spot he indicated. There, clumsily carved into the wood of the shelf, were three words. ALBERT WUS HERE
"Albert?" said Henrietta.
"That's me," said Al. "A long, long time ago. Back when this place wasn't a library. It was a _school_. So long ago I'd nearly forgotten, if you can believe that."
"You are pretty old," said Henrietta.
"But it's strange. The school wasn't here in the city. We took a boat to it."
"When we found this house, it was abandoned," said Sid. "You're the first person I've met who knows anything about it."
"I remember sitting here, and talking with a teacher there. I was scolded once over there."
"We should start the meeting," said Rose. "The other members are waiting."
"Other members?" said Gary.
Instead of answering, Rose turned to a wide staircase that ascended from the first level of the house to the second in a broad rightward sweep.
"Have a good time," said Sigrid, as she and Sid disappeared off in another direction.
"My parents will talk to you about this when you leave," said Rose, "but don't tell anyone you were here. And from now on, you'll use the back door. There's a secret knock."
Al's eyebrows raised, but he said nothing. Lately, anything seemed possible. The world was swiftly becoming quite interesting, he reflected. He wished Henrie was alive to see it.
They climbed another staircase once they reached the second floor, walked along a hallway lined with hardback history books to a third staircase, ascended that, turned left, walked through a reading room where several people read books and drank tea from antique cups, and finally arrived in a small corner that contained a nondescript, narrow door.
"I decided to have the meetings up here," said Rose.
"Is this the part that sticks up, outside?" said Gary.
"The turret," said Rose. "I call it the attic. I thought we could call our book club The Attic Books."
She opened the door, revealing an exceedingly narrow, ascending spiral staircase. It was made all of stone.
"This is weird," said Gary, touching his hand to the stone wall. "It looks like wood, but it's rock." The stone was composed of many distinct minerals pressed together in layers that resembled wood grain, some translucent and others opaque.
"It's petrified," said Rose, mounting the steep case. The spiral wound twice and they emerged in a small, circular room lined with books and lit by candles. The walls were made of petrified wood, and it was apparent that the room was the interior of an ancient tree. They'd walked right up the inside of the trunk.
High up on the walls several small windows let in the early evening light. In the middle of the room, a group of six chairs were arranged in a circle. Two were occupied, and their inhabitants stood as the group entered.
"Oak!" said Gary.
"OK!" said Henrietta. And then, "This is my grandfather, Al. Al, this is Oak and OK. They're friends who helped us, even though they look scary."
"Do I look scary?" said OK from behind his gray-streaked beard, his bald head and grocery bag suspenders glinting in the candlelight. Oak couldn't even stand fully upright in the space, he was so massive. His shoulders bulked against the lowest rafters.
"It's nice to meet you both," said Al. He shifted the books he was carrying from his right hand to his left, so he could shake their hands.
"What's that?" said Gary, pointing up at one of the walls. In the space between the tops of the bookshelves and the bottom of the high windows a small wooden sign was set into the stone, which read,
**IYCHMN EON**
"I don't know," said Rose. "This place is really old."
"Well," said Al, "it's very nice to have everyone together, finally ungrounded! I brought a few books to show you all, if you'd like to start the meeting." Everyone found seats in the circle, and Al put the three books he'd brought on his lap. He held up the first one. "This is Henrietta's textbook from school," he said. He flipped through the plastic pages. "I'd call this a modern book. It's unusual because it has a spot of wild housecat blood between two pages. It was probably only printed a year or so ago, but it already has an interesting history." He handed Henrietta's textbook to Oak. "The second book I brought is very old. It's one of Aristotle Alcott's journals." Al showed the old, softcover book. Its pages were crumbling, the leather spine cracked. OK immediately held out his hands for it, and Al gave it to him.
"This is the same Alcott who made the _Bestiary_?" OK asked, opening the journal carefully and looking into the handwritten interior.
"The same," said Al. "It's the oldest book I own. I would call it ancient."
OK passed the book to Rose, who turned it gently over in her hands.
"This could use conservation," she said. Al gave her a questioning look. "Repair."
"Rose can fix any book that ever was," said Gary. "She's an expert."
"I didn't know books could be fixed," said Al.
"I didn't either," said Henrietta, "but Rose even repaired the _Bestiary_."
"I'll do it this week," said Rose.
"The final book I brought today," said Al, "was one I'd almost completely forgotten about. It's neither modern nor ancient. I'll call it 'old' because I think it's about my age. I'd like to nominate it as our first book club assignment."
Al showed everyone this final, slim volume, which immediately reminded Henrietta of _Early Town_. She read the title in the shifting candlelight:
**THE WIKKELING:**
**Fact or Fairy Tale?**
"Where did you _get_ that?" said Gary, aghast.
"I don't remember," said Al. "I've owned it for many years. In fact, I have two copies, so we can share it."
"Maybe we have one at the Library, too," said Rose. "I'll ask my mom and dad."
"There are three different times," Henrietta mused. "The present, the past, and the ancient past."
"So I'm in the present," Gary said, "and the past is through the windows, and I think about the _ancient_ past when I see the big stump."
Al looked at him curiously. "What are you referring to?"
"Well...." said Henrietta. And, as it seemed as good a time as any, she told the story. Gary and Rose both chimed in too, adding details here and there and occasionally contradicting one another a little bit—but not much. The facts were plain. As the tale unwound, Al, Oak, and OK listened raptly.
"Things are coming together," said Al. "You say that the Wikkeling started making you sick when it couldn't find you. In this book," Al indicated _Wikkeling: Fact or Fairy Tale_ , "it says that the Wikkeling was originally created to destroy a dangerous animal, called the Draageling. To bring the Wikkeling to life, the scientists placed a slip of paper in its mouth with a secret word written on it. Do you know what the word was?" Al paused dramatically, and when no one conjectured he said, "It was _grow_."
"I don't understand," said Gary.
"I wonder if that's what it was repeating as it melted," said Rose.
"I think I see now," said Henrietta. "When it taps people, it's feeding from them somehow, and it's always hungry because it's growing."
"And it couldn't reach us in our old houses, or in the attic," said Gary. "So it got mad?"
"Maybe it didn't even mean to make us sick," said Henrietta. "It was just upset—like how our parents grounded us when they found us at Al's. Even the Competency Exam makes sense! All the subtraction problems turned into addition."
"It would hate subtraction, if it only wants to grow," said Gary.
"And it would love addition," said Henrietta.
Then Rose, who had been silent for some time, spoke up. "This whole city is called the Addition," she said. This statement stopped the conversation in its tracks for a few moments.
"Well, we've certainly got plenty to think about," said Al. "But we also have the tools to do it. Oh, and here's something else." He opened the book and showed everyone the inside of the front cover. At the top of the page was written,
**T HIS BOOK HAS BEEN READ BY:**
The rest of the page was ruled with lines, like notebook paper. The first seven or so had names written on them, some in pen, some in pencil, some cursive, some printed. The remainder were blank. "I have several books like this," said Al. "People used to write their names in when they finished reading, and you could see who'd had it before you. Since I read this one last week, I added my name at the bottom." He handed the book to Henrietta. "You'll write yours next," he said.
### Back to the Attic
After the meeting, Al drove Henrietta home, and she entered the living room to find her mother watching television alone. The late news was on, and Henrietta sat next to her.
In the top story, a man had been eaten by a bear, deep in the Old City. The only thing the bear didn't swallow, the solemn shellac-haired newscaster told the camera, was the man's mouth. "The police report that the mouth was still screaming when they found it," he said. An ad came on after, and Henrietta's mother muted it.
"Henrietta," she said, "if you ever see a bear, call an adult."
"I will," said Henrietta.
"How was your book club?"
"It was fun," said Henrietta. "We decided on a book to read."
"Does it contain District-Approved Vocabulary?" said her mother.
"Yes," said Henrietta.
The news returned for a final story about research being done on a new cell phone that would be implanted directly into the subscriber's brain. "We'll all have instant access to everyone, all the time," the newscaster chirped gleefully, twirling his finger at his own head.
"We just bought that spot today," said Henrietta's father, emerging from the hallway.
"What's the phone going to be called?" said Henrietta's mother.
"We're leaning toward _Perpetuality_."
"That's got a nice ring to it."
"Pun intended?"
Henrietta saw beads of sweat on her father's forehead, as if he'd been working on something.
"What are you doing, dad?" she asked.
"Your BedCam is broken again," said her father. "I swear, I'm looking forward to this old place getting torn down!"
Henrietta's eyes widened. "I have homework to do!" she blurted, and she ran down the hallway, her mother calling after her, "It's almost bedtime!"
Henrietta closed her bedroom door and lifted her chair onto her desk. She climbed up past the broken eye of the BedCam, pressed her hands against the seam, and at long last entered the attic.
It was perfectly quiet and dark now that the windows were sealed off. Henrietta lit the candles in the candelabra and closed the trapdoor. She wondered if the moon was shining right now on that other town, lighting the bricks of its old street, or if those bricks had already been paved over.
Her eyes fell on the brown stain next to the trapdoor. It seemed like forever ago when she'd found this place. It seemed, furthermore, like a story about someone else. And she _was_ a different person then. Everything that had happened since had changed that girl into the person she was now.
A small noise came from behind her, interrupting her thoughts. She turned and searched the deep shadows cast by the candles. Her eyes traveled up the tallest bookcase. A familiar form was illuminated there by the candlelight.
Mister Lady jumped down and alighted easily on the armrest on the far end of the couch. She looked at Henrietta with her enormous eyes—clear green, like new leaves.
Henrietta's heart leapt. She wanted to spring from her seat and throw her arms around the cat. She wanted to rub her ears and kiss her between the eyes, and hold her soft paws in her hands. But instead she sat quietly.
If you or I had seen her, we might have thought Henrietta was afraid.
She wasn't, though. She was simply a considerate person—one who knew that wild animals don't like to be petted, even if you and they are friends.
| {
"redpajama_set_name": "RedPajamaBook"
} | 3,338 |
Google Maps shows a list of the people that contributed with the most customized maps, reviews and edits related to a location. In some cases, you'll find helpful maps that show hidden parks, interesting museums or landmarks.
Google has also released a customized maps editor for Android. "With this application you can create, edit, share, and view personalized maps on your Android powered phone synchronized with the My Maps tab on Google Maps. Create a map on your desktop computer using Google Maps and then take it with you on the go and update it on location."
Does Google Map Editor allow me to give my contributed map data away with my own licences? For instance giving it to OpenStreetmap?
If there was one thing I had to choose that I liked best about google it would be the Google Map Editor.
Birth place of Buddha is in fact, Kapilvastu, Lumbini, NEPAL But the Google map search showed UttarPradesh, INDIA, which is not correct. Though, the search showed the location within the boundary of NEPAL.
Now I cannot see Top Contributors list in google map. Why? | {
"redpajama_set_name": "RedPajamaC4"
} | 2,458 |
TiVo TV Talk
Now Playing - TV Show Talk
Game Show prizes misrepresented
getreal
postcrastinator
NBC promotes their generosity by awarding $1m prize to winners, but the 42-year old winner would be 82 years old before he actually receives the full amount in the year 2047 for his win in 2007. And how much would $25,000/yr. be worth in 2047 dollars? Imagine winning the $10,000 Pyramid in 1973, and they paid it off over 40 years in increments of $250/yr. You'd finally be getting your last check for $250 in six years -- in the year 2013! How much more could you have done with that $250 cash back in '73 compared to today in 2007 or in 2013?
As America's Got Talent 2's winner, Fator ... received a $1,000,000 prize and the show's "Best New Act In America" title. However unfortunately for Fator, unlike the winners of most other reality shows, he won't actually be receiving his full cash prize anytime soon.
Instead, similar to what NBC (which appears to have established a lock on the "America's Cheapest Reality TV Broadcast Network" title) did with For Love Or Money's $1,000,000 prize and Treasure Hunters' $3,000,000 prize, Fator's $1,000,000 prize will be paid over via a lottery prize-like 40-year period. Fator can (and probably will) opt to pass on the $25,000 annual payments and instead receive a lump sum, however should he do so, the present cash value of the prize will be no more than several hundred thousand dollars.
After revealing Fator's victory, Springer also revealed that Fator would also be receiving one more "surprise" prize. "We've got one more surprise for you," Springer told Fator. "We know your dream is to play Las Vegas, (so) we're going to make that dream come true. You're going to be playing the Jubilee Theater at Bally's Resort." Springer didn't reveal whether, based on NBC's "fine print" prize practices, Fator might actually have to pay for the privilege.
http://www.realitytvworld.com/news/terry-fator-crowned-america-got-talent-second-season-winner-5668.php
Stay Tooned!
"Misrepresented"? In what way?
The guy's getting $1M. That's what was promised. Did they say that they'd get a single check for $1M?
Didn't the contestants get to read the rules on payouts before they signed up?
If I lose, I get my Turtle Wax and Rice-a-Roni up front, right?
"I'm Buddy Rich when I fly off the handle"
IndyJones1023
anom said:
You get one grain of rice a day for the next 40 years.
Common sense is a gift, not a given.
Amnesia said:
Well, I just watched the last 10 minutes of the show, but Springer clearly said that "the winner will wake up tomorrow with $1m and the title of America's Got Talent champ." So that implies that the winner would receive $1m that night before he goes to sleep. If it's clarified in the contestant's fine print contract when they sign on to do the show (and it should be), that's okay for the contestants, but it's certainly misrepresented to the viewing audience, IMHO.
doom1701
Time for a new Title
That's how most million dollar contests work. Heck, the lottery has been doing that for ages. I'm sure it's well represented in the paperwork you sign when you want to be on one of those shows.
Fator can (and probably will) opt to pass on the $25,000 annual payments and instead receive a lump sum, however should he do so, the present cash value of the prize will be no more than several hundred thousand dollars.
Man, only several hundred thousand dollars? How will he eat?
TAsunder
Debates Ghee vs Gi
I don't think that's how most of them work. The lottery does, but not game shows.
Well, in Canada, when you win $10m in the lottery, you get a check (cheque) for $10m. Tax free.
Jebberwocky!
getreal said:
And how much would $25,000/yr. be worth in 2047 dollars?
Using 8% discount rate, the value of $25,000 a year for 40 years in 2047 dollars is approximately $7 million
Using 8% discount rate, the value of $25,000 a year for 40 years in 2007 dollars is approximately $325 thousand
XBOX360 Gamertag- jeffebbe (new)
TAsunder said:
There's, what, 3 or 4 "Million Dollar" shows? I'm asking--I don't watch much TV.
But every million dollar giveaway run by something like McDonalds, or Doritos, or any other company that would do a contest like that is done as either a 20 or a 40 year payout (well, maybe there's some 30's as well).
dswallow
Save the Moderatоr
Wow, I never expected that. In this day of production costs for an episode of a TV series being in the $2 million to $4 million range, I would never have expected these prizes to be paid as annuities. That's extraordinarily cheap of the producers.
Doug Swallow
dswallow said:
cheap or really smart?
ID-10-T
Yeah it is very common for almost all sweepstakes to pay out in annuities.
I agree with Dswallow though, the networks are cheap. It is ridiculous that the prize for 13 weeks of Survivor is 1 million dollars or for close to 50 shows for Big Brother is $500,000.
That is why I think anyone who goes on these shows are foolish. They are seriously being taken advantage of...
A nobody on a sitcom who is in every episode is going to make as much as someone who wins a show like Big Brother.
The biggest stars will make more than the winners every episode.
All that being said, I don't know that paying out a million dollars in an annuity is misleading. It is cheap, but not really misleading.
I Support Global Warming!
The difference is that you need talent (or at least a good agent) to get a job on a sitcom. You don't need anything to be on a reality show...
brnscofrnld
where is this place?
I'd take the 25k per year, but then again, I'm only 24.
Great supplement for your current salary no matter what your current income is.
That Don Guy
Now with more GB
Actually, every AGT episode this season has mentioned that the prize was "paid from an annuity over 40 years" in the closing credits, and that the winner has the option of taking the current value of the annuity instead. (At least they mentioned it this year; no one has said if Bianca Ryan's prize from last year was handled the same way or not.)
Nothing new about paying large game show prizes over time; back in 1976, the top prize on $100,000 Name That Tune was $10,000 a year (they mentioned this every time someone had a chance to win it), and a few years later, the prize on $1,000,000 Chance of a Lifetime was also paid out over time (I think it was 25 payments of $40,000, but I'm not sure). (On the other hand, the two grand prize winners on The $128,000 Question in the mid-1970s were paid all at once.)
For the record, the closing credits on Power of 10 say that "some prizes are paid out over time."
(And speaking of Terry Fator's "bonus" prize on AGT, while they said he would be playing the Jubilee Room at Bally's, they didn't say if he would be headlining his own show or just be part of the existing "Jubilee" seven-act cabaret show...)
martinp13
Roller coaster addict :)
Jebberwocky! said:
smartly cheap!
Doesn't bother me much (I stopped watching a few episodes ago. I might watch the finale if I'm bored), but somebody should probably put a spoiler warning in the title or tag the spoilers.
"Of all the dogs in the world, I never would have expected this goofy one here to know the Heimlich"
MikeMar
isn't this the EXACT same as the lottery?
even if you get hit with like 50% tax or whatever, it's better to take the lump sum and invest it then the 40 year thing.
do not see at ALL what the big deal is.
Take the lump sum $1 million, uncle sam hits you for a lot of.
Do you want NBC to pay $1 million PLUS taxes??
LetterBoxd Movie Profile
dagap
So where's the line? It'd be much cheaper to stretch it out 50 years or more. Why not 100 years?
"And here's America's Newest Millionaire!!! ****"
(**** paid in 1,200 monthly installments of $833.33)
That Don Guy Aug 24, 2007 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,601 |
\section{Introduction} \label{sec:intro}
With the appearance of data from a variety of direct, indirect and collider
search strategies, there has been great interest in formalisms which allow one
to, for any particular model, relate the expected observables which can be probed
with each detection strategy. One such formalism is that of simplified models~\cite{SimplifiedModels}, in
which one focusses on a effective Lagrangian involving only the small number
of fields and couplings relevant for dark matter interactions with the Standard Model. In its
most basic form, a simplified model could describe dark matter interactions with
the Standard Model particles through
exchange of a single mediating particle. For such a simplified model, one can describe
the full set of dark matter-Standard Model interactions by specifying only a
few quantities, such as the mass and spin of the dark matter and mediating particle,
the strength of the couplings, and the exchange channel ($s$, $t$ or $u$). A single
simplified model can thus capture the essential physics of dark matter-Standard Model
interactions for a variety of new physics scenarios.
We consider a class of simplified models in which dark matter is a scalar, and
dark matter annihilation to light charged leptons ($XX \rightarrow e^+ e^-, \mu^+ \mu^-$)
proceeds through $t$-channel exchange of a single
mediating particle, $f'$ (see fig.~\ref{fig:FeynmanDiagrams}, left panel).
Examples of models which can realize this scenario include WIMPless dark matter~\cite{WIMPless},
and this scenario can be realized in models of leptophilic dark matter~\cite{LeptophilicDarkMatter}.
This particular scenario is of interest because it allows one to decouple the dark matter annihilation
cross section from bounds arising from direct detection and collider searches.
These latter search strategies place tight constraints on dark matter interactions with
quarks. In some (though by no means all) models, these bounds are satisfied by requiring the
particles mediating interactions with quarks to be very
heavy, suppressing the annihilation cross section to quarks. But if the particles mediating
interactions with leptons are light, then the cross section for dark matter to annihilate
to leptons could still be sizable. In this light, it is important to consider the case
where dark matter is a scalar whose annihilation cross section does not suffer from the
chirality/$p$-wave suppression which typically arises for Majorana fermions.
\begin{figure}[b]
\raisebox{-0.5\height}{ \includegraphics[scale=0.8]{annihilation.pdf} }\
\hspace{0.05 \textwidth}
\raisebox{-0.5\height}{ \includegraphics[scale=0.8]{loop.pdf} }
\label{fig:FeynmanDiagrams}
\caption{Feynman diagrams for the annihilation
process $XX \rightarrow \bar f f$ (left) and for the lepton dipole moment corrections (right).}
\end{figure}
Because the dark matter has vanishing electric charge,
the mediating particle must necessarily be electrically charged. As a result,
the dark matter-lepton interaction vertices can contribute to the electric or
magnetic dipole moments of the electron and muon (see fig.~\ref{fig:FeynmanDiagrams}, right panel).
Both electric and magnetic dipole moments have long been used a probes of new physics (see, for
example,~\cite{NewPhysics,ElectronMagMoment,MuonMagMoment,Roberts:2011fq,ElecMoment}).
In particular, the current measurement of the muon magnetic dipole moment differs from the Standard
Model prediction by $\sim 3.5\sigma$~\cite{MuonMomentAnomaly}, and there have been attempts to
interpret this anomaly in terms of a signal for new physics~\cite{MuonMagMomentNewPhysics}.
We will consider the connection between corrections to the dipole moments and the dark matter
annihilation cross section.
We will consider the case where dark matter is a scalar, with the most general
allowed Yukawa coupling to light charged leptons and an exotic
charged lepton. We will find that the annihilation cross section depends on
two terms, one of which is $CP$-conserving and the other of which is $CP$-violating.
The $CP$-conserving term will be constrained by the new physics contribution to the
magnetic dipole moment of the charged lepton, while the $CP$-violating term will constrained
by the electric dipole moment. This will allow us to constrain the the $XX \rightarrow \bar f f$
annihilation cross-section with experimental measurements of the fermion dipole moments.
We thus focus on annihilation to light charged leptons (electrons or muons), whose dipole moments
are most tightly constrained.
The outline of this paper is as follows. In section II, we describe the simplified
model, and the scaling of $CP$-conserving and $CP$-violating terms. In section III we
compute the $XX \rightarrow \bar f f$ annihilation cross section and the corrections
to the electric and magnetic dipole moments. In section IV, we describe experimental
constraints on the dipole moments, and the implied constraints on the dark matter annihilation
cross section. We conclude with a discussion of our results in section V.
\section{The Simplified Model}
If dark matter is a scalar, which couples to Standard Model leptons and an
exotic fermion, then the most generic renormalizable interaction can be
written in terms of the effective Lagrangian:
\begin{eqnarray}
L_{X} = X \bar f' (\lambda_L P_L + \lambda_R P_R) f + X^* \bar f (\lambda_L^* P_R + \lambda_R^* P_L) f' ,
\end{eqnarray}
where $f$ is the Standard Model lepton, $f'$ is an exotic lepton, and $X$ is a scalar dark matter field, which
in general can be real or complex. We also assume that $X$ and $f'$ are charged under
some unbroken symmetry which stabilizes the dark matter, and under which $f$ is neutral. We thus require
$m_X < m_{f'}$.
If $X$ is an $SU(2)_L \times U(1)_Y$ singlet, then gauge-invariance implies that
$f'$ must be chiral under $SU(2)_L$ and $U(1)_Y$. As a result, $f'$ will act like an exotic lepton and will get mass
through a coupling to the higgs; it cannot be arbitrarily
heavy, though it is certainly allowed by data. Note, however, that if $f'$ is part of a full generation (to
cancel the hypercharge mixed anomaly), then this would imply the existence of a 4th generation quark, which is
more tightly constrained~\cite{4thGenQuarks}. However, the anomaly can be cancelled in other ways (say, the presence of another
mirror lepton). For our purposes, we simply assume that the hypercharge mixed anomaly is cancelled.
Note that it is also possible for $f'$ to be completely vector-like, in which case its mass is unconstrained.
But in this case, gauge-invariance implies that $X$ must be a linear combination of fields with different
charges under $SU(2)_L$ and $U(1)_Y$ (though electrically neutral). The coefficients $\lambda_{L,R}$ then
also include the relevant mixing angles.
We will thus treat the mass of $f'$ as unconstrained (aside from $m_{f'}>m_X$), though for specific models
$m_{f'}$ can be constrained both from above and from below by data.
We can choose to write the propagators for $f$ and $f'$ with real mass terms. With this choice, the only remaining
allowed phase rotation of the fields is an overall rotation of $X$, $f$ and $f'$ (the relative phase between
the left-handed and right-handed fermion components are fixed by the real mass condition). Any
overall phase for the coefficients $\lambda_L$ and $\lambda_R$ can be absorbed by a phase rotation of the fields,
but a relative phase between $\lambda_L$ and $\lambda_R$ cannot be absorbed by any field redefinition. We thus
see that $CP$-violating
terms will be proportional to $Im(\lambda_L \lambda_R^*)$.
\section{Connecting Dark Matter Annihilation to Electric/Magnetic Dipole Moments}
\subsection{Dark Matter Annihilation}
If $X$ is real, then the $XX \rightarrow \bar f f$ annihilation process proceeds
though $t$- and $u-$channel exchange of $f'$. If $X$ is complex, then the $X^* X \rightarrow \bar f f$
annihilation process proceeds through only one of these channels.
For scalar dark matter, the $XX \rightarrow \bar f f$ process can proceed from an $s$-wave initial state,
and with no chirality suppression. As a result, although velocity-dependent terms can be relevant for
the annihilation rate at freeze-out, they will be largely irrelevant in the current epoch. We can thus
limit ourselves to the the non-relativistic limit. We will also focus on the limit $m_X ,m_{f'} \gg
m_f$.\footnote{However, if either $\lambda_L$ or $\lambda_R$ is sufficiently small, then the leading
terms in the cross section will scale as $m_f^2$ or $v^4$~\cite{3body}. Since these terms will be small,
internal bremsstrahlung can also
be important. The annihilation cross section is suppressed in this limit, so we will not
discuss it further. }
We then find
\begin{eqnarray}
\sigma(XX \rightarrow \bar f f) v &\simeq & \left[ Re(\lambda_L \lambda_R^*)^2 + Im(\lambda_L \lambda_R^*)^2 \right]
{ m_{f'}^2\over \pi ( m_{f'}^2 + m_{X}^2)^2 } ,
\nonumber\\
\sigma(X^* X \rightarrow \bar f f) v &\simeq & \left[ Re(\lambda_L \lambda_R^*)^2 + Im(\lambda_L \lambda_R^*)^2 \right]
{ m_{f'}^2\over 4\pi ( m_{f'}^2 + m_{X}^2)^2 } ,
\end{eqnarray}
as $v \rightarrow 0$.
It is useful to consider the limit $m_{f'} \gg m_X \gg m_f$.
In this limit, the annihilation matrix element
is generated by the dimension 5 effective operator:
\begin{eqnarray}
{\cal O}_{eff} &=& {Re (\lambda_L \lambda_R^*) \over m_{f'}} (X^* X )(\bar f f)
- {Im (\lambda_L \lambda_R^*) \over m_{f'}} (X^* X)(\imath \bar f \gamma^5 f) .
\end{eqnarray}
Note that each of the two terms permits annihilation from an $S=0$, $L=0$, $J=0$ $CP$-even initial state. However, the
first term annihilates to a $S=1$, $L=1$ $CP$-even final state, while the second term annihilates to a
$S=0$, $L=0$ $CP$-odd final state. The annihilation matrix elements generated by these two terms thus do not interfere,
and each matrix element is neither chirality nor $p$-wave suppressed~\cite{Kumar:2013iva}.
As expected, the terms in the cross section proportional to $Re (\lambda_L \lambda_R^*)$ arise from
the $CP$-invariant part of the operator, while the terms proportional to $Im (\lambda_L \lambda_R^*)$ arise
from the $CP$-violating part.
\subsection{Corrections to the Electric/Magnetic Dipole Moment}
The most general fermion-fermion-photon coupling consistent with gauge-invariance is
given by
\begin{eqnarray}
\Gamma^\mu &= \gamma^\mu F_1 (q^2) +{\imath \sigma^{\mu \nu} q_\nu \over 2m} F_2 (q^2)
+{\imath \sigma^{\mu \nu} q_\nu \gamma^5 \over 2m} F_3 (q^2)
+(\gamma^\mu q^2 - \Dsl q q^\mu) \gamma^5 F_A (q^2) ,
\end{eqnarray}
where $F_{1,2,3,A}(q^2)$ are form factors which depend on the momentum $q$ of the photon.
Gauge-invariance requires $F_1 (0) =1$. $F_A (0)$ is the anapole moment, which will
not concern us here. But the electric and magnetic dipole moments are determined by
$-\imath F_3 (0)$ and $F_2 (0)$. These moments are denoted by $d$ and $\mu = g (e /2m)S$,
respectively. At the classical level, one finds a vanishing electric dipole moment, $d_{cl}=0$,
and a magnetic dipole moment given by $g_{cl}=2$. The one-loop quantum corrections to these
dipoles are given by the quantities
\begin{eqnarray}
a \equiv {g-2 \over 2} &=& F_2 (0),
\nonumber\\
{d \over |e|} &=& {\imath F_3(0) \over 2m}.
\end{eqnarray}
These corrections can be computed at lowest order from the one-loop correction to the
vertex. Note that, at lowest order, neither correction is affected by the fermion wavefunction
renormalization, so a computation of the vertex correction is sufficient.
In the limit $m_f \ll m_X, m_{f'}$, we then find (see also~\cite{Cheung:2009fc})
\begin{eqnarray}
\Delta a &=& {Re( \lambda_L \lambda_R^* ) \over (4 \pi)^2 }
{m_{f}m_{f'}\left( (m_{f'}^2-m_X^2)(m_{f'}^2-3m_X^2) + 2 m_X^4 \log{m_{f'}^2 \over m_X^2} \right) \over (m_{f'}^2-m_X^2)^3} ,
\nonumber\\
{d \over |e|} &=& {Im(\lambda_L \lambda_R^* ) \over 2(4 \pi)^2 }
{m_{f'}\left( (m_{f'}^2-m_X^2)(m_{f'}^2-3m_X^2) + 2 m_X^4 \log{m_{f'}^2 \over m_X^2} \right) \over (m_{f'}^2-m_X^2)^3} .
\end{eqnarray}
As expected, the correction to the magnetic dipole moment depends on $Re(\lambda_L \lambda_R^*)$, while the
electric dipole moment, which arises from $CP$-violation, depends on $Im(\lambda_L \lambda_R^*)$. Note that the dipole
moment corrections do not depend on whether $X$ is a real or complex scalar.
It is also worth noting that, in general, the dipole moment operators will run under renormalization group (RG) flow
between the energy scale at which one integrates out the $X$ and $f'$ fields ($\sim m_X$) and the energy scale
relevant for the vertex corrections ($\sim m_f$). However,
since the $\lambda_{L,R}$ we consider are typically small and since the only other relevant couplings are electroweak couplings,
the effects of running are not significant~\cite{Jung:2008it} and we will ignore them. If we had considered the coupling of dark matter to
quarks instead, RG-running of the operators would be non-trivial.
\section{Constraints and Prospects}
We focus on dark matter couplings to electrons or muons, whose dipole moments
are most tightly constrained.
We can compare the experimental measurement and theoretical calculation of $a=(g-2)/2$~\cite{muon,mm,electron}
for the electron and muon:
\begin{eqnarray}
a_{e(exp)} = && 1159652180.76 (0.27) \times 10^{-12} ,
\nonumber\\
a_{e(theory)} = && 1159652181.13 (0.11)(0.37)(0.77) \times 10^{-12} ,
\nonumber\\
a_{e (exp)} - a_{e(theory)} = \Delta a_e = && -0.37 (0.82) \times 10^{-12} ;
\\
a_{\mu (exp)} = && 116592089.(54)(33) \times 10^{-11} ,
\nonumber\\
a_{\mu (SM)} = && 116591802(2)(42)(26) \times 10^{-11} ,
\nonumber\\
a_{\mu (exp)} - a_{\mu (SM)} = \Delta a_{\mu} = && 287(63)(49) \times 10^{-11} .
\end{eqnarray}
The Standard Model prediction for the electric dipole moment $d$ of the electron and
muon is negligible.
The experimental measurement~\cite{electron,Moyotl:2011yv} is:
\begin{eqnarray}
\left| {d_{\text{e}} \over e} \right| &<& 10.5 \times 10^{-28} \text{cm} ,
\nonumber\\
2 m_e \left| {d_{\text{e}} \over e} \right| &<& 5.5 \times 10^{-17} ;
\\
\left| {d_\mu \over e} \right| &< & 1.5 \times 10^{-19} \text{cm} ,
\nonumber\\
2 m_\mu \left| {d_\mu \over e} \right| &< & 1.6 \times 10^{-6} .
\end{eqnarray}
We see that constraints on the corrections to $\Delta a$ and $d$ imply constraints
on $Re(\lambda_L \lambda_R^*)$ and $Im(\lambda_L \lambda_R^*)$, which in turn imply a bound on
the annihilation cross section.
But since there can be additional new physics which contribute to both $\Delta a$ and $d$, a contribution
to these moments from dark matter interactions alone which exceeds the difference between theory and the measured
value cannot be strictly excluded.
Instead, we will consider as ``disfavored" a model for which the magnitude of the correction from dark matter interactions
is larger than
the magnitude of the total correction allowed by the data (regardless of sign). Although such models can be allowed
by data if there is additional new physics, those contributions would necessarily have to be fine-tuned against the
dark matter contribution.
Clearly, this characterization is somewhat ambiguous, and there is no clear answer as to how much fine-tuning is
acceptable; this analysis merely provides a benchmark.
Using the analysis above, we plot in fig.~\ref{fig:CrossSecBound} regions of parameter-space in the
$(\langle \sigma_{a}v \rangle, m_{f'} )$ plane which are disfavored by the dipole moment data,
for $m_X = 10, 100~\text{GeV}$ and $m_f \ll m_X$. The annihilation processes considered are $X^* X \rightarrow e^+ e^-$
and $X^* X \rightarrow \mu^+ \mu^-$. In each panel the labeled shaded regions correspond to annihilation cross sections due either to
$CP$-conserving or $CP$-violating interactions, and are constrained by the appropriate
dipole moment. Note that $m_{f'}$ is constrained by direct searches for charged particles at
LEP. The precise exclusion contours depend on the spin of the new particle as well as its decay chain, but LEP searches
roughly exclude new charged particles which are
lighter than $\sim 100~\text{GeV}$~\cite{pdg,Carpenter:2011wb}.
LHC constraints on new charged leptons can be up to a factor of 3 tighter if the charged
lepton couples to $SU(2)_L$, but can be much weaker if they do not~\cite{lhcslepton,lhcsleptonrh}. As these constraints are somewhat
model-dependent, we plot the full range of $m_{f'}$ for completeness.
\begin{figure}[h]
\includegraphics*[width=3in]{e10.pdf}
\includegraphics[width=3in]{e100.pdf}
\includegraphics*[width=3in]{u10.pdf}
\includegraphics[width=3in]{u100.pdf}
\caption{Bounds on the annihilation cross section of complex scalar dark matter to $e^+ e^-$ (top panels)
and $\mu^+ \mu^-$ (bottom panels) from experimental
measurement of the lepton magnetic and electric dipole moments in the $m_f \to 0$
limit for $m_X =10~\text{GeV}$ (left panels) and $m_X =100~\text{GeV}$ (right panels). Each disfavored region is
shaded and demarcated by a line
which indicates if it is disfavored by magnetic dipole moment bounds, electric dipole moment bounds, or by
perturbativity of the coupling constant (perturbativity is not constraining for $X^*X \rightarrow e^+ e^-$).
Note that the annihilation cross section will be enhanced by a factor of 4
if $X$ is a real scalar. }
\label{fig:CrossSecBound}
\end{figure}
It is interesting to note that, for annihilation to the electron channel, the bounds on both $CP$-conserving
and $CP$-violating contributions to the annihilation cross section are tight enough to exclude any models which
could be observed at current or upcoming indirect detection experiments, although bounds on $CP$-violating
contributions are tighter than bounds on the $CP$-conserving contribution. On the other hand, for annihilation
to the muon channel, bounds from $g_\mu-2$ on the $CP$-conserving contribution are much tighter than
electric dipole moment bounds on the $CP$-violating contribution. Indeed, if this simplified model describes
$X^* X \rightarrow \mu^+ \mu^-$ annihilation which can be detected at indirect detection experiments, it must
arise from $CP$-violating contributions.
One can compare the annihilation cross sections and dipole moment corrections more directly in the limit
$m_{f'} \gg m_X \gg m_f$. In this limit, we find
\begin{eqnarray}
\Delta a_f &\simeq & {Re( \lambda_L \lambda_R^* ) \over (4 \pi)^2 } {m_f \over m_{f'}}
\nonumber\\
2m_f (d_f /|e|) &\simeq & {Im( \lambda_L \lambda_R^* ) \over (4 \pi)^2 } {m_f \over m_{f'}}
\nonumber\\
\sigma(X^* X \rightarrow \bar f f ) v &\simeq & \left[ Re(\lambda_L \lambda_R^*)^2 + Im(\lambda_L \lambda_R^*)^2 \right]
{ 1\over 4 \pi m_{f'}^2 } \sim 64\pi^3 [\Delta a_f^2+ (2m_f (d_f /|e|))^2] m_{f}^{-2}
\nonumber\\
&\simeq & (7.7 \times 10^{11}~{\rm pb} )[\Delta a_f^2 + (2m_f (d_f /|e|))^2]\left({m_f \over {\rm GeV}} \right)^{-2}
\end{eqnarray}
In this limit, the annihilation cross section depends on $m_X$ and $m_{f'}$ only through the dipole moment
corrections.
As expected, we see that for $f=e$, the $CP$-violating contribution is more highly
suppressed than the $CP$-conserving contribution. On the other hand, for $f=\mu$,
it is the $CP$-conserving contribution which is more highly suppressed.
With $\Delta a_\mu \sim 3 \times 10^{-9}$ and $d_\mu =0$ ($Im (\lambda_L \lambda_R^*)=0$),
we find $\sigma(X^*X \rightarrow \bar \mu \mu ) v \sim 6\times 10^{-4}~{\rm pb}$. This implies that
a $CP$-conserving interaction producing a $g_\mu-2$ correction large enough to explain the
data would still produce a annihilation cross section of only $\sim 10^{-3}~\text{pb}$. If a larger
cross section is observed in indirect detection experiments, then there must be additional fine-tuning
between corrections to the muon magnetic dipole moment, or there must be $CP$-violating interactions
which also contribute to the muon electric dipole moment.
Note that, if this simplified model describes
the physics underlying the $g_\mu -2$ anomaly, $m_{f'}$ could be as large as $\sim 35~\text{TeV}$ while still
having perturbative couplings. For the case of annihilation to electrons, perturbativity constraints
are not relevant for any of the mass scales considered.
\section{Conclusions}
\label{sec:conclusions}
We have studied dipole moment constraints on simplified models in which scalar dark matter
annihilates to light charged leptons through $t$-channel exchange of a heavy
mediating fermion. In particular, we have found that such simplified models are tightly
constrained by the magnitude of experimentally allowed corrections to electric and magnetic
dipole moments. Interestingly, the tightest constraints on electron interactions arise from
electric dipole moment measurements, while the tightest constraints on muon interactions arise
from magnetic dipole moment interactions. If such a simplified model describes dark matter
interactions in nature, then an observable $XX \rightarrow \mu^+ \mu^-$ annihilation cross-section
must be accompanied by large $CP$-violation. It is also worth noting that, if dark matter is a
thermal relic, then the $XX \rightarrow e^+ e^-$ process can be only a small part of total annihilation
at freeze-out, while $XX \rightarrow \mu^+ \mu^-$ can be the dominant annihilation process at freeze-out.
These bounds can be weakened if there are other new physics effects whose contribution to the
dipole moments cancels that from loop diagrams involving dark matter. However, models with
cross sections far larger than these constraints can only be consistent if there is significant fine-tuning
between dipole moment corrections.
In the limit $m_{f'} \gg m_X \gg m_f$, both $\Delta a$ and
$m_f (d/e)$ scale as $m_f / m_{f'}$.
One could instead consider the case where $X$ is a fermion and the $t$-channel mediator $f'$
is a scalar. This simplified model would again yield corrections to the electric and magnetic dipole
moments, but these moments would instead scale as
$(m_f m_X / m_{f'}^2) \times [(m_f / m_X)\,{\rm or}\,(\sin \alpha)]$, where $\alpha$ is the
scalar mixing angle~\cite{Cheung:2009fc}.
This suppression is due to the fact that dipole moment interactions require a helicity-flip for the
Standard Model lepton, which can be provided by the fermion mass term or by scalar mixing.
However, if $X$ is a fermion, then the $s$-wave annihilation cross section is suppressed
by a factor $m_X^2 / m_{f'}^2$ (if dark matter is a Majorana fermion and there is negligible scalar mixing,
the annihilation cross section is suppressed by an additional factor $m_f^2 / m_X^2$). As a result, the bounds
on dark matter annihilation to light charged leptons for the case of fermionic dark matter with
large scalar mixing may be expected to be roughly the same as those shown here for the scalar dark matter case.
The same is true if dark matter is a Majorana fermion, and there is negligible scalar mixing.
But if dark matter is a
Dirac fermion with negligible scalar mixing, then the constraints on dark matter annihilation from
dipole moments would be much weaker~\cite{Buckley:2013sca}.
We can thus compare the analysis of the simplified model considered here with that of an MSSM neutralino. The correction
to the muon magnetic moment due to loop diagrams has been well studied~\cite{Cheung:2009fc},
and for the reasons stated above,
we expect the bounds on the cross section for $\chi \chi \rightarrow \mu^+ \mu^-$ to be similar to those
found here in the $CP$-conserving case, up to ${\cal O}(1)$ factors. This amounts to the statement that,
even if superparticle corrections are large enough to be responsible for the deviation between the measured
$g_\mu -2$ and the Standard Model prediction, the branching fraction for dark matter annihilation to muons
in the current epoch will still be quite small. This is not surprising, given that the neutralino is a Majorana
fermion. One might have thought that a larger annihilation cross section would be allowed for scalar dark matter,
since the required helicity-flip is provided by the mediating fermion. However, the mediating fermion provides
the same helicity-flip for the magnetic moment diagram, enhancing the correction to $g_\mu-2$. Thus a $CP$-conserving
interaction involving scalar dark matter which provides the needed correction to $g_\mu-2$ would still yield
an annihilation cross section to muons which is very small. But for a fixed coupling, the
mediating particle mass required to match the $g_\mu -2$ anomaly in the scalar dark matter case
will be much larger than in the neutralino dark matter case.
Moreover, large $CP$-violation is possible in the simplified model considered
here, allowing significant annihilation to muons which is unconstrained by $g_\mu-2$ bounds.
For the case of Dirac neutralinos, dark matter can annihilate from a $S=1$, $L=0$, $J=1$ initial state. In this
case, no helicity-flip is needed for the final state. A helicity-flip is still needed for the correction to $g_\mu -2$,
implying that the correction to $g_\mu -2$ can be quite small even if the cross section for annihilation to muons is
large~\cite{Buckley:2013sca}.
As we have seen, for Majorana fermion dark matter, the spin-flip required for an $s$-wave annihilation
cross section is correlated with the spin-flip needed for a contribution to the dipole moments. But it
is worth noting that even dark matter with $p$-wave suppressed annihilation can still have the correct
thermal relic density, because a $p$-wave annihilation cross section is only suppressed by a factor $\sim 10$
at the time of freeze-out.
As a result, Majorana fermion dark matter with largely $p$-wave annihilation could arise as a thermal relic,
yet be unconstrained by dipole moment bounds.
In this work, we have only focussed on annihilation to light charged leptons. Annihilation to
$\bar \tau \tau$ or $\bar q q$ can also be constrained, but experimental constraints on the dipole
moments are much weaker. It would be interesting to revisit constraints on annihilation to those
channels from dipole moment corrections. Finally, it is worth noting that all of these bounds on
the annihilation cross section can be avoided by models with $s$-channel annihilation, where the mediating
particle is neutral. The observation
of dark matter annihilation to $e^+ e^-$ pairs would then provide interesting information regarding
the nature of the mediating particle.
\vskip .2in
\textbf{Acknowledgments}
We are grateful to M.~Buckley, D.~Hooper and X.~Tata for useful discussions.
J.~K.~thanks the Center for Theoretical Underground Physics and
Related Areas (CETUP* 2013) in South Dakota for their support and
hospitality during the completion of this work.
The work of J.~K.~before June 1, 2013 is supported in part by Department of Energy
grant DE-FG02-04ER41291. The work of J.~K.~after June 15, 2013 is supported in
part by NSF CAREER Award PHY-1250573.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,315 |
\section{Introduction}
A \emph{track} layout of a graph consists of a vertex $k$ colouring, and a total order of each vertex colour class, such that between each pair of colour classes no two edges cross.
A \emph{queue} layout of a graph consists of a total order of the vertices, and a partition of the edges into sets (called queues) such that no two edges that are in the same set are nested with respect to the vertex ordering. The minimum number of queues in a queue layout of a graph is its \emph{queue number}.
Track layouts have been extensively studied in \cite{D15, DFJW12, DMW05, DMW13, DPW04, DW04, DW05, FLW02}.
Queue layouts have been introduced by Heath, Leighton, and Rosenberg \cite{HLR92, HR92} and have been extensively studied from \cite{BFP10, DMW05, DMW13, DPW04, DW05, EI71, H04, HLR92, HR92, P92, RM95, SS00, T72, W05, W08}. Both track and queue layouts have applications in parallel process scheduling, fault-tolerant processing, matrix computations, and sorting networks (see \cite{P92} for a survey).
Queue layouts of directed acyclic graphs \cite{BCLR96, HP99, HPT99, P92} and posets \cite{HP97, P92} have also been investigated.
The question in Heath et al. \cite{HLR92, HR92}, whether the queue number of a planar graphs is constant-bound (it also leads to constant-bound track number), remains open. Heath et al. \cite{HLR92, HR92} conjectured that the question has an affirmative answer.
However, Pemmaraju \cite{P92} conjectured that every planar graph has $O(\log n)$ queue number. Also,
he conjectured that this is the correct lower bound.
Up to now, the best known lower bound is still constant-bound.
On the other hand, the well-known upper bound for the queue number of planar graphs had remained stagnant as $O(\sqrt{n})$ roughly two decades. This upper bound utilizes the fact that planar graphs have path width at most $O(\sqrt{n})$.
Recently, the upper bounds of queue and track numbers for planar graphs were reduced to $O(\log^2 n)$, by Di Battista, Frati and Pach \cite{BFP10} and $O(\log n)$, by Vida Djumovic \cite{D15}, respectively.
In this paper, we provide a layout on constant number of tracks for a plane graph.
Our result attempts to break Pemmaraju's conjecture in a positive direction.
The proof that a plane graph has constant-bound track number is simple.
It utilizes a novel graph partition technique.
In particular, our main result states that every $n$-vertex plane graph has such a graph partition and it leads to $O(1)$-track layouts for plane graphs.
One of the most important motivations for studying queue layouts is 3D crossing-free straight-line grid drawing in a small volume.
Particularly, a 3D crossing-free straight-line grid drawing of a graph is a placement of the vertices at distinct points in a 3D grid, and the straight-line representing the edges are pairwise non-crossing.
One of the most important open problems that Felsner et al. \cite{FLW02} present in graph drawing questions is whether planar graphs have 3D crossing-free straight-line grid drawings in a linear volume.
A 3D crossing-free straight-line grid drawing with volume $X \times Y \times Z$ is an $X \times Y \times Z$ drawing that fits in an axis-aligned box with side lengths $X-1$, $Y-1$, and $Z-1$.
The following theorem has been established in \cite{DMW05, DPW04}.
\begin{theorem}
An $n$-vertex graph $G$ has a 3D crossing-free straight grid drawing in an $O(1) \times O(1) \times O(n)$ volume, if and only if $G$ has a constant-bound queue number. (constant-bound track number.)
\end{theorem}
The road map for this paper is as follows:
the first half part from Sections \ref{sec:prelim} to \ref{sec:cons-tracks}
explain the basic framework and ideas for this article.
The second half part explain more details in the first half part of this article.
\section{Preliminaries}\label{sec:prelim}
In this section, Some definitions and important preliminary results are given.
Definitions not mentioned here are standard. A graph $G=(V,E)$ is called
{\em planar} if it can be drawn on the plane with no edge
crossings. Such a drawing is called a {\em plane embedding} of $G$.
A {\em plane graph} is a planar graph with a fixed plane embedding.
A \emph{layerlike} graph $\Pi$ is a graph whose vertices are partitioned and placed on contiguous layers such that
no edge is placed between any two non-contiguous layers and no edges are crossing.
Given a layerlike graph $\Pi$, a \emph{down-pointing} triangle $\triangledown$ is a a cycle $(l, \cdots, r, m)$
that vertices on the cycle $(l, \cdots, r, m)$
are on the two contiguous layers
where the path from $l$ to $r$ are on the upper layer and the vertex $m$ is on the lower layer.
A \emph{bowl} $\heartsuit$ is a cycle $(l, \cdots, r)$
that the cycle are on the same layer
where each vertex of the cycle is on the same layer. In Fig. \ref{fig:perfect-layer-graph},
vertices $(b_8, b_9, b_{10}, b_{11}, b_{12}, b_{13})$ form a bowl in the composite-layerlike graph $\mathcal{G}$.
\begin{definition}
A \emph{composite-layerlike} graph $\mathcal{G}$ can be recursively defined as follows:
$\mathcal{G}$ consists of a layerlike graph $\Pi$ such that
each bowl $\heartsuit$ of $\mathcal{G}$ has a smaller composite-layerlike graph $\mathcal{G}_1$
where $\mathcal{G}_1$'s first layer is the bowl $\heartsuit$,
and each down-pointing triangle $\triangledown$ has a composite-layerlike graph $\mathcal{G}_2$
where the first layer of $\mathcal{G}_2$ is the upper layer of $\triangledown$.
\end{definition}
An edge $e=(u, v)$ is called a \emph{chord} if
both end-vertices $u$ and $v$
are on the same layer in a composite-layerlike graph $\mathcal{G}$.
A \emph{region} $\mathcal{W}$, rooted at a vertex $r$ in a composite-layerlike graph $\mathcal{G}$, consists of a left boundary $\mathcal{B}^L$ and a right boundary $\mathcal{B}^R$ such that $\mathcal{W}$ satisfies (1): $B^L$ and $\mathcal{B}^R$ are two paths walking along contiguous layers from the vertex $r$ to two different vertices on lower layers in $\mathcal{G}$, and
(2) $\mathcal{W}$ is a separator of the composite-layerlike graph $\mathcal{G}$. Also, we denote the left and right boundaries of a region $\mathcal{W}$ as $\mathcal{B}^L(\mathcal{W})$ and $\mathcal{B}^R(\mathcal{W})$, respectively.
Consider a region $\mathcal{W}$ rooted at a vertex $r$ in a composite-layerlike graph $\mathcal{G}$.
A composite-layerlike graph $\mathcal{G}(\mathcal{W})$ is \emph{induced} by $\mathcal{W}$ if
$\mathcal{G}(\mathcal{W})$ is a subgraph of $\mathcal{G}$ inside by the two boundaries $\mathcal{B}^L(\mathcal{W})$ and $\mathcal{B}^R(\mathcal{W})$.
Also,
we denote $\mathcal{W}^{\mathcal{M}}$ as the maximum region bounded by the leftmost and rightmost boundaries of $\mathcal{G}$.
Obviously, a composite-layerlike graph $\mathcal{G}$ is a maximum composite-layerlike graph $\mathcal{G}(\mathcal{W}^{\mathcal{M}})$
induced by the maximum region $\mathcal{W}^{\mathcal{M}}$.
A \emph{ladder} $\mathcal{H}$ is defined to
consist of contiguous tracks.
A \emph{layout} of a composite-layerlike graph $\mathcal{G}$ in a ladder $\mathcal{H}$ is defined to be an arbitrary vertices's partition of $\mathcal{G}$ on tracks of $\mathcal{H}$.
For a layout of a composite-layerlike graph $\mathcal{G}$ in a ladder $\mathcal{H}$,
a set of chords $\{e_1=(u_1, v_1), e_2=(u_2, v_2), \cdots, e_q=(u_q, v_q)\}$ are called \emph{nest} if $\{e_1, e_2, \cdots, e_q\}$ are placed on a track in $\mathcal{H}$ as the order:
$(u_1,$ $u_2,$ $\cdots,$ $u_q,$ $v_q,$ $\cdots,$ $v_2,$ $v_1)$.
A set of edges $\{e_1=(u_1, v_1),$ $e_2=(u_2, v_2),$ $\cdots,$ $e_q=(u_q, v_q)\}$ are called \emph{$X$-cross} if
$(u_1,$ $u_2,$ $\cdots,$ $u_q)$ are orderly placed as $(u_1,$ $u_2,$ $\cdots,$ $u_q)$ on a track and
$(v_1,$ $v_2,$ $\cdots,$ $v_q)$ are reversely placed as
$(v_q,$ $\cdots,$ $v_2,$ $v_1)$ on another track in $\mathcal{H}$.
Given an edge $e=(u, v)$ a layout in $\mathcal{H}$, let $\mathcal{L}_{\mathcal{H}}(u)$ and $\mathcal{L}_{\mathcal{H}}(v)$ be the track numbers where the vertices $u$ and $v$ placed in $\mathcal{H}(\mathcal{G})$, respectively. The \emph{gap} of an edge $e=(u, v)$ is the absolute difference $|\mathcal{L}_{\mathcal{H}}(u)-\mathcal{L}_{\mathcal{H}}(v)|$.
Also, the \emph{queue number} on a track is defined as the maximum size of edges nest on the track, and
the $X$-\emph{crossing number} for any two tracks in $\mathcal{H}$ is defined as the maximum size of edges $X$-cross between the two track.
The \emph{distance-number} of a layout in a ladder $\mathcal{H}$ is defined as the maximum gaps among all edges.
\begin{figure}[t]
\begin{center}
\includegraphics[width=1\textwidth, angle =0]{perfect-layer-graph}
\centering
\caption{(1): $\mathcal{G}$ is a composite-layerlike graph;
(2) is a layerlike graph $\Pi$ of $\mathcal{G}$ with three layers $(\mathcal{L}_1, \mathcal{L}_2, \mathcal{L}_3)$ and $\Pi$ has three down-pointing triangles and one bowl
$\triangledown_1=(a_2, a_3, a_4, a_5, a_6, b_3)$,
$\triangledown_2=(b_1, b_2, c_1)$,
$\triangledown_3=(b_2, b_3, b_4, c_3)$ and
$\heartsuit_1=(b_8, b_9, b_{10}, b_{11}, b_{12}, b_{13})$;
(3) $\triangledown_1$'s inner vertices can be placed on a layerlike graph $\Pi_1$ which has the same first layer $\mathcal{L}_1$ with $\Pi$;
(4) the inner vertices of
$\{\triangledown_2, \triangledown_3\}$ and $\heartsuit_1$ can be placed on a $\Pi_2$ which has the same first layer $\mathcal{L}_2$ with $\Pi$.
}
\label{fig:perfect-layer-graph}
\end{center}
\vspace{-0.2in}
\end{figure}
\begin{definition}\label{def:well-placed}
A layout of a composite-layerlike graph $\mathcal{G}$ in
a ladder $\mathcal{H}$ is called $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-\emph{well-placed} if they can be placed in $\mathcal{H}$ such that
\begin{itemize}
\item each track's queue number is less than $\mathcal{Q}$,
\item the $X$-crossing number between any two tracks is less than $\mathcal{X}$,
\item the distance-number in $\mathcal{H}$ is less than $\mathcal{D}$, and
\item $\mathcal{G}$ can be placed as sequential regions in $\mathcal{H}$;
The sequential regions are denoted as $\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{G})$.
\end{itemize}
\end{definition}
\begin{theorem}\label{thm:wrap}
If a composite-layerlike graph $\mathcal{G}$ is an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}'$, then $\mathcal{G}$ can be placed as an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout on $2\mathcal{D}$ tracks in $\mathcal{H}$.
\end{theorem}
\begin{proof}
Assume that a composite-layerlike graph $\mathcal{G}$ can
be placed in a ladder $\mathcal{H}'$
such that
\begin{enumerate}
\item the queue number of each track in $\mathcal{H}'$ is less than or
equal to $\mathcal{Q}$,
\item the $X$-crossing number of between any two tracks in $\mathcal{H}'$ is less than or equal to $\mathcal{X}$, and
\item the difference $|\mathcal{L}_{\mathcal{H}}(u)-\mathcal{L}_{\mathcal{H}}(v)|$ is less than or equal to $\mathcal{D}$ for any edge $e=(u, v)$ in $\mathcal{H}'$.
\end{enumerate}
Since the total tracks in $\mathcal{H}'$ could grow beyond constant bound,
we need to wrap $\mathcal{H}'$ as follows:
move vertices on $(i\times 2\mathcal{D} + j)$-th track to right of
vertices on $((i-1)\times 2\mathcal{D} + j)$-th track on the wrapped $\mathcal{H}$'s $j$-th track.
Now each track $i$ in the wrapped ladder $\mathcal{H}$,
vertices are from tracks
$(0\times 2\mathcal{D} + j)$, $(1\times 2\mathcal{D} + i)$, $(2\times 2\mathcal{D} + j)$, $\cdots$ in the unwrapped ladder $\mathcal{H}'$.
And, for each edge $e=(u, v)$ in the wrapped ladder $\mathcal{H}$,
the edge $e$ comes from pair of tracks
$(0\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 0\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$(1\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 1\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$(2\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 2\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$\cdots$
in the unwrapped ladder $\mathcal{H}'$.
Because for any edge $e=(u, v)$ in the unwrapped ladder $\mathcal{H}'$, the difference $|\mathcal{L}_{\mathcal{H}}(u)-\mathcal{L}_{\mathcal{H}}(v)|$ is at most $\mathcal{D}$,
only edges on pair tracks
$(0\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 0\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$(1\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 1\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$(2\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), 2\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$,
$\cdots$
in the unwrapped ladder $\mathcal{H}'$
can be placed on the pair tracks $(\mathcal{L}_{\mathcal{H}}(u), \mathcal{L}_{\mathcal{H}}(v))$ in the wrapped ladder $\mathcal{H}$.
Also,
for a track $(\mathcal{L}_{\mathcal{H}}(u))$ ($\mathcal{L}_{\mathcal{H}}(v)$, respectively)
on the wrapped ladder $\mathcal{H}$, we know that vertices on a track
$i\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u)$
($i\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v)$, respectively)
from the unwrapped ladder $\mathcal{H}'$ are placed at left of
vertices on a track
$(i+1) \times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v)$
($(i+1)\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u)$, respectively)
from the unwrapped ladder $\mathcal{H}'$.
Hence
there is no any $X$-crossing edge
between edges from pair tracks
$(i\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), i\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$
and
pair tracks
$((i+1)\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(u), (i+1)\times 2\mathcal{D} + \mathcal{L}_{\mathcal{H}}(v))$.
Finally, We can conclude that a composite-layerlike graph $\mathcal{G}$ can be placed in the wrapped laddder graph $\mathcal{H}$ such that
\begin{enumerate}
\item the queue number of each track in the wrapped ladder $\mathcal{H}$ is less than or
equal to $\mathcal{Q}$,
\item the $X$-crossing number of between any two track in the wrapped ladder $\mathcal{H}$ is less than or equal to $\mathcal{X}$, and
\item the number of tracks in the wrapped ladder $\mathcal{H}$ is $2\mathcal{D}$.
\end{enumerate}
\end{proof}
\section{A Framework to Construct an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-Well-Placed Layout on Constant Number of Tracks for a Composite-Layerlike Graph $\mathcal{G}$}\label{sec:perfect-layer-graph-layout}
\begin{algorithm}[ht]
\caption{A Framework to Place an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-Well-Placed Layout on Constant Number Tracks in a Ladder $\mathcal{H}$ for a Composite-Layerlike Graph $\mathcal{G}(\mathcal{W}^{\mathcal{M}})$}
\label{alg:framework}
\KwIn{A composite-layerlike graph $\mathcal{G}(\mathcal{W}^{\mathcal{M}})$.}
Place the $\mathcal{W}^{\mathcal{M}}$'s root on the first track in $\mathcal{H}$\;
Place the contiguous layers $(2, 3, \cdots)$ of $\mathcal{W}^{\mathcal{M}}$ at right of $\mathcal{W}^{\mathcal{M}}$'s root on the contiguous $(\mathcal{Z}+2, \mathcal{Z}+3, \cdots)$ tracks in $\mathcal{H}$\;
Add the maximum region $\mathcal{W}^{\mathcal{M}}$ into the empty first-in-first-out queue $\tilde{\mathcal{Y}}$\;
\While{$\tilde{\mathcal{Y}}$ is not empty}
{
Let $\mathcal{W}$ rooted at $r$ be the first region in $\tilde{\mathcal{Y}}$\;
Find sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ inside the region $\mathcal{W}$\;
Place the sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ orderly at the rightmost part in $\mathcal{H}$ and
from the $(\mathcal{L}_{\mathcal{H}}(r)+2\mathcal{Z})$-th track in $\mathcal{H}$
where $\mathcal{L}_{\mathcal{H}}(r)$ is the track number of the vertex $r$ in $\mathcal{H}$\;
Add the maximum subsequential regions of $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}))$ into $\tilde{\mathcal{Y}}$
such that each region of the maximum subsequential regions does not root at $r$\;
Remove the region $\mathcal{W}$ from $\tilde{\mathcal{Y}}$\;
}
Wrap $\mathcal{H}$\;
\end{algorithm}
In this section, we provide a framework in Algorithm \ref{alg:framework} to place a composite-layerlike graph $\mathcal{G}$ as $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}$ on constant number of tracks.
Before we describe the framework, we introduce a structure \emph{skeleton}
as follows:
\begin{definition}\label{def:skeleton}
Consider a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$ rooted at a vertex $r$. A subgraph $\Psi$ of a composite-layerlike graph $\mathcal{G}(\mathcal{W})$ is called a \emph{skeleton} of $\mathcal{W}$ if
$\Psi$ consists of
\begin{enumerate}
\item the $\mathcal{W}$'s root $r$,
\item sequential regions $\tilde{\mathcal{W}}(\Psi)$ such that
there are subsequential regions $(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots,$ $\mathcal{W}^M_1,$ $\cdots,$ $\mathcal{W}^M_m,$ $\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots)$ $\subseteq \tilde{\mathcal{W}}(\Psi)$
where
(1) for each vertex $u\in \mathcal{B}^L$, each $u$'s child is inside a region in $(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots)$,
(2) for each vertex $v\in \mathcal{B}^R$, each $v$'s child is inside a region in $(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots)$, and
(3) each $r$'s child is inside a region in $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$.
\end{enumerate}
\end{definition}
\begin{figure}[t]
\includegraphics[width=1\textwidth, angle =0]{skeleton}
\caption{(1) shows an example of a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)=((a_1, b_1, c_1, \cdots), (a_1, b_6, c_{12}, \cdots))$} and the subgraph induced by $\mathcal{W}$;
(2) shows a $\mathcal{W}$'s skeleton that consists of sequential regions
$\mathcal{W}_1=((b_1, c_1, \cdots), (b_1, c_3, \cdots))$,
$\mathcal{W}_2=((b_1, c_3, \cdots), (b_1, c_5, \cdots))$,
$\mathcal{W}_3=((a_1, b_1, c_5, \cdots), (a_1, b_2, c_6, \cdots))$,
$\mathcal{W}_4=((a_1, b_2, c_6), (a_1, b_2, c_6))$,
$\mathcal{W}_5=((a_1, b_4, c_7), (a_1, b_5, c_7))$,
$\mathcal{W}_6=((a_1, b_5, c_7, \cdots), (a_1, b_6, c_9, \cdots))$,
$\mathcal{W}_7=((b_6, c_9, \cdots), (b_6, c_{12}, \cdots))$.
In addition, the region $\mathcal{W}_1$ consists of two subregions
$\mathcal{W}'_1=((b_1, c_1, \cdots), (b_1, c_2, \cdots))$ and
$\mathcal{W}'_2=((b_1, c_2, \cdots), (b_1, c_3, \cdots))$.
The region $\mathcal{W}_2$ consists of two subregions
$\mathcal{W}'_3=((b_1, c_3, \cdots), (b_1, c_4, \cdots))$ and
$\mathcal{W}'_4=((b_1, c_4, \cdots), (b_1, c_5, \cdots))$.
the region $\mathcal{W}_7$ consists of three subregions
$\mathcal{W}'_5=((b_6, c_9, \cdots), (b_6, c_{10}, \cdots))$,
$\mathcal{W}'_6=((b_6, c_{10}, \cdots), (b_6, c_{11}, \cdots))$ and
$\mathcal{W}'_7=((b_6, c_{11}, \cdots), (b_6, c_{12}, \cdots))$;
(3): the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \mathcal{W}_3, \mathcal{W}_4, \mathcal{W}_5, \mathcal{W}_6, \mathcal{W}_7)$ are placed
as:
$(\mathcal{W}_1, \mathcal{W}_2, \mathcal{W}_3, \mathcal{W}_4, \mathcal{W}_7, \mathcal{W}_6, \mathcal{W}_5)$
in a ladder $\mathcal{H}$;
\label{fig:skeleton}
\vspace{-0.2in}
\end{figure}
\begin{figure}[t]
\includegraphics[width=1\textwidth, angle =0]{framework}
\caption{(1), (2) and (3) show an example to place a region $\mathcal{W}$ and $\mathcal{W}$'s skeleton
of Fig. \ref{fig:skeleton} in a ladder $\mathcal{H}$ by Algorithm \ref{alg:framework}.
The $\mathcal{W}$'s skeleton is placed at right of the region $\mathcal{W}$ in $\mathcal{H}$ and
starts to place $\mathcal{W}$'s skeleton from the track three
in $\mathcal{H}$.
Because $\mathcal{W}$'s distance-number is one, place the $\mathcal{W}$'s skeleton from the track three
doesn't make any $X$-crossing edge
between $\mathcal{W}$ and $\mathcal{W}$'s skeleton.
After the $\mathcal{W}$'s skeleton is placed,
the subgraph inside each region of $\mathcal{W}$'s skeleton
is placed from the track five.
Moreover,
the placement cannot
make $X$-crossing edges with the previous one
in a ladder $\mathcal{H}$.
}
\label{fig:framework}
\vspace{-0.2in}
\end{figure}
Firstly, this framework place $\mathcal{W}^M$ on the contiguous tracks from the first track in $\mathcal{H}$.
Next, this framework also consists of a loop and each iteration of the loop in Algorithm \ref{alg:framework}
starts from the first region $\mathcal{W}$ rooted at $r$ of the first-in-first-out queue $\tilde{\mathcal{Y}}$
and executes the following steps:
find sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ such that
each skeleton $\Psi_i, i\geq 1,$ roots at the vertex $r(\mathcal{W})$ and
each $r(\mathcal{W})$'s child inside $\mathcal{W}$ is a vertex in a skeleton of $(\Psi_1, \Psi_2, \cdots)$.
Also, the sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ are placed orderly at the rightmost part in $\mathcal{H}$
and starts from the track $(\mathcal{L}_{\mathcal{H}}(r(\mathcal{W}))+2\mathcal{Z})$ in $\mathcal{H}$.
Before we prove the correctness of Algorithm \ref{alg:framework} in Subsection \ref{subsec:correctness-framework},
we assume the following conjecture in advance.
This conjecture is proved in Section \ref{sec:skeleton}.
\begin{conjecture}\label{conj:skeleton}
Given a region $\mathcal{W}$ rooted at a vertex $r$,
we have a skeleton $\Psi$ that
$\Psi$ can have sequential regions $\tilde{\mathcal{W}}(\Psi)$. And, the skeleton $\Psi$ can be placed in $\mathcal{H}$ as $\Psi_{\mathcal{H}}$ such that
$\Psi_{\mathcal{H}}$ is $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed in $\mathcal{H}$.
\end{conjecture}
\subsection{Sequential Skeletons $(\Psi_1, \Psi_2, \cdots)$ inside a Region $\mathcal{W}$ Rooted at a Vertex $r$}\label{subsec:correctness-framework}
In this subsection, we start to show two lemmas, the first one shows how to find sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ inside a region $\mathcal{W}$ rooted at $r$ such that
each $r$'s child in $\mathcal{W}$ is a vertex of a skeleton of $(\Psi_1, \Psi_2, \cdots)$,
And the second one shows how to place the sequential skeletons
$(\Psi_1, \Psi_2, \cdots)$ in $\mathcal{H}$ and
$(\Psi_1, \Psi_2, \cdots)$ can partition the region $\mathcal{W}$ into sequential regions such that
$(\Psi_1, \Psi_2, \cdots)$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed in $\mathcal{H}$.
In the following lemma, we give a constructive proof to find sequential skeletons $(\Psi_1, \Psi_2, \cdots)$ consisting of all $r$'s children inside a region $\mathcal{W}$.
\begin{lemma}\label{lem:partition}
For a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$ rooted at a vertex $r$, sequential skeletons $\tilde{\Psi}(\mathcal{W})=$
$(\Psi_1, \Psi_2, \cdots)$ can be constructed such that
each $r$'s child is a vertex of some skeleton $\Psi$ of $\tilde{\Psi}(\mathcal{W})$.
\end{lemma}
\begin{proof}
From Theorem \ref{thm:skeleton}, there exists a skeleton $\Psi$ for the region $\mathcal{W}$ and
$\Psi$ can partition the region $\mathcal{W}$ into sequential regions
$\tilde{\mathcal{W}}(\Psi)=(\mathcal{W}_1, \mathcal{W}_2, \cdots)$.
From Definition \ref{def:skeleton}, each $r$'s children not included into $\Psi$ is inside in a region of $\tilde{\mathcal{W}}(\Psi)$.
Let $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$ be the maximal subsequential regions of
$\tilde{\mathcal{W}}(\Psi)$ such that each $r$'s child not included in $\Psi$ is inside
a region of $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$ rooted at the vertex $r$.
Now for each region $\mathcal{W}^M_i\in (\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$,
a skeleton $\Psi_i$ for the region $\mathcal{W}_i$ can be found.
Then by the above discussion,
for each skeleton $\Psi_i, 1\leq i\leq m$, the skeleton $\Psi_i$ can partition the region $\mathcal{W}^M_i$
into sequential regions $\tilde{\mathcal{W}}(\Psi_i)$. And,
each $r$'s children inside $\mathcal{W}^M_i$ not included into $\Psi_i$ is inside
a region of $\tilde{\mathcal{W}}(\Psi_i)$.
the same partition can be repeatedly executed till each $r$'s child is included into a skeleton.
Hence we can conclude that given a region $\mathcal{W}$ rooted at a vertex $r$,
we can have sequential skeletons $\tilde{\Psi}(\mathcal{W})$
such that each $r$'s child is a vertex of some skeleton $\Psi$ in $\tilde{\Psi}(\mathcal{W})$.
\end{proof}
Consider sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ in $\mathcal{H}$.
The sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ and the sequential subgraphs $(\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_1), \tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_2), \cdots)$ are called an \emph{ordered} layout in $\mathcal{H}$ if the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ are at left of the sequential subgraphs $(\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_1), \tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_2), \cdots)$ in $\mathcal{H}$, and for any two regions $\mathcal{W}_i$ and $\mathcal{W}_j$ in $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$, the region $\mathcal{W}_i$ is at left of the region $\mathcal{W}_j$ if and only if the subgraph $\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_i)$ is at left of the subgraph $\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_j)$ in $\mathcal{H}$.
The task of this subsection is to show that if the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ and the sequential subgraphs $(\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_1), \tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_2), \cdots)$ in $\mathcal{H}$ are an ordered layout in $\mathcal{H}$, then the layout is also $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
Now consider sequential edges $(\mathcal{E}_1, \mathcal{E}_2, \cdots)$ where each $\mathcal{E}_i, i\geq 1,$ are edges connected between the region $\mathcal{W}_i$ and the subgraph $\tilde{\Psi}_{\mathcal{H}}(\mathcal{W}_i)$.
Now we prove that the sequential skeletons $\tilde{\Psi}(\mathcal{W})$
can be $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed on contiguous tracks in $\mathcal{H}$.
From Theorem \ref{thm:skeleton}, we can have a skeleton $\Psi$ for the region $\mathcal{W}$
such that
$\Psi$ can be $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed in $\mathcal{H}$ as $\Psi_{\mathcal{H}}$ and
have sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi)$ in $\mathcal{H}$.
Then we can pick the maximum subsequential regions $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$ of $\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi)$
such that each $r$'s child is inside a region of $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$.
And, for the sequential regions $(\mathcal{W}^M_1, \mathcal{W}^M_2, \cdots, \mathcal{W}^M_m)$,
we can have corresponding sequential skeletons $(\Psi_1, \Psi_2, \cdots, \Psi_m)$
such that $(\Psi_1, \Psi_2, \cdots, \Psi_m)$ can be $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed as
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1), \tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_2), \cdots, \tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_m))$ in $\mathcal{H}$.
Now we have placed $(\Psi, \Psi_1, \Psi_2, \cdots, \Psi_m)$ in $\mathcal{H}$ as sequential regions
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_m))$.
Since regions in $(\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_m))$ are mutually disjoint and
$(\mathcal{W}^M_1,$
$\mathcal{W}^M_2,$
$\cdots,$
$\mathcal{W}^M_m,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_1),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_m))$ are an ordered layout,
edges $\mathcal{E}_i$ between $\mathcal{W}^M_i$ and $\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_i)$ and
edges $\mathcal{E}_j$ between $\mathcal{W}^M_j$ and $\tilde{\mathcal{W}}_{\mathcal{H}}(\Psi_j)$ don't nest in $\mathcal{H}$.
Hence the layout in $\mathcal{H}$ is $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed.
We can repeat the above steps until each $r$'s child inside the region $\mathcal{W}$ is a vertex of some skeleton of
the sequential skeletons $\tilde{\Psi}(\mathcal{W})$ and have sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}))$ in $\mathcal{H}$.
From the above discussion, we immediately have the following lemma.
\begin{lemma}\label{lem:skeleton-wrap}
For a region $\mathcal{W}$ rooted at a vertex $r$,
sequential skeletons $\tilde{\Psi}(\mathcal{W})$ can be placed as
the new order $\tilde{\Psi}_{\mathcal{H}}(\mathcal{W})$
on contiguous tracks in $\mathcal{H}$
such that $\tilde{\Psi}_{\mathcal{H}}(\mathcal{W})$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{J})$-well-placed in $\mathcal{H}$ where
the three numbers $\mathcal{Q}$, $\mathcal{X}$ and $\mathcal{J}$ are constant-bound.
\end{lemma}
In the next lemma,
we prove that when a sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}))$ are placed in $\mathcal{H}$ in Algorithm \ref{alg:framework},
all edges connecting between $\mathcal{W}$ and $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}))$ cannot make $X$-crossing with the existing layout in $\mathcal{H}$.
\begin{lemma}\label{lem:noncrossing}
Given sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ in $\mathcal{H}$
where each region $\mathcal{W}_i, i\geq 1,$ roots at a vertex $r_i$.
If each sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_i)), i\geq 1,$ is placed at right of $(\mathcal{W}_1, \mathcal{W}_2, \cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_1)),$ $\cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_{i-1}))$ and
from the track $\mathcal{L}_{\mathcal{H}}(r_i)+2\mathcal{Z}$ in $\mathcal{H}$,
then (1): the edges set
$\mathcal{E}_i$ connecting between $\mathcal{W}_i$ and $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_i))$ don't have $X$-crossing edges with the layout
$(\mathcal{W}_1, \mathcal{W}_2, \cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_1)),$ $\cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_i))$ in $\mathcal{H}$.
\end{lemma}
\begin{proof}
Because the region $\mathcal{W}_i$ is placed at right of the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots, \mathcal{W}_{i-1})$ and $(\mathcal{W}_1, \mathcal{W}_2, \cdots, \mathcal{W}_i)$ are mutually disjoint,
the edges sets $\mathcal{E}_i$ and $\mathcal{E}_j, 1\leq i-1$ don't have any $X$-crossing edge in $\mathcal{H}$
where $\mathcal{E}_j$ are all edges connecting between $\mathcal{W}_j$ and $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_j))$.
From Lemma \ref{lem:skeleton-wrap}, we can assume that each edge's gap number on each region $\mathcal{W}_i$ is less than or equal to $\mathcal{J}$.
Because the gap numbers of edges in each edges set $\mathcal{E}_i$ is greater than or equal to $\mathcal{Z}$ and $\mathcal{Z}$ is greater than $\mathcal{J}$,
each region of $(\mathcal{W}_1, \mathcal{W}_2, \cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_1)),$ $\cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\Psi}(\mathcal{W}_{i-1}))$ and $\mathcal{E}_j$ cannot make $X$-crossing in $\mathcal{H}$.
\end{proof}
\begin{lemma}\label{lem:gap_number}
For each edge $e$ placed in Algorithm \ref{alg:framework},
$e$'s gap number is at most $2\mathcal{Z}$.
\end{lemma}
\begin{proof}
Initially,
Each edge of the maximum region $\mathcal{W}^{\mathcal{M}}$ except the edges connecting to $\mathcal{W}^{\mathcal{M}}$'s root can be placed in $\mathcal{H}$ with gap number = 1.
Each Edge connecting to $\mathcal{W}^{\mathcal{M}}$'s root has gap number = $\mathcal{Z}$.
Each edge $e$ connecting to $\mathcal{W}^{\mathcal{M}}$'s root inside $\mathcal{W}^{\mathcal{M}}$ is an
edge of a skeleton of $\tilde{\Psi}(\mathcal{W}^{\mathcal{M}})$ and $\tilde{\Psi}(\mathcal{W}^{\mathcal{M}})$ is placed in $\mathcal{H}$ from the track $2\mathcal{Z}+1$ in $\mathcal{H}$. It leads to $e$'s gap number = $2\mathcal{Z}$ in $\mathcal{H}$.
For a region $\mathcal{W}$ in $\mathcal{H}$, the region $\mathcal{W}$ is called \emph{processed} if $\mathcal{W}$ has been partitioned by sequential skeletons $\tilde{\Psi}(\mathcal{W})$ and each edge $e$ connecting to a boundary of $\mathcal{W}$,
either $e$ is an edge of a skeleton $\Psi$ for $\mathcal{W}^{\mathcal{M}}$ or $e$ is inside a region in $\tilde{\mathcal{W}}(\Psi)$ in $\mathcal{H}$.
In each iteration of Algorithm \ref{alg:framework}, we pick the leftmost unprocessed region $\mathcal{W}$ in $\mathcal{H}$ and place its sequential skeletons $\tilde{\Psi}(\mathcal{W})$ into $\mathcal{H}$. From Lemma \ref{lem:skeleton-wrap}, for each region $\mathcal{W}' \in \tilde{\Psi}(\mathcal{W})$, each edge of $\mathcal{W}'$ except connecting to $\mathcal{W}'$'s root has gap number = $\mathcal{J}$.
Now for each vertex $v$ on a boundary of the leftmost unprocessed region $\mathcal{W}$, all edges connecting to the vertex $v$ inside the region $\mathcal{W}$ are processed by two consecutive steps:
(1): some edges connecting to the vertex $v$ are edges of a skeleton of $\tilde{\Psi}(\mathcal{W})$ and have gap number $\mathcal{Z}$ in $\mathcal{H}$. And,
(2): each remaining edge connecting to the vertex $v$ is inside a region $\mathcal{W}'$ rooted at the vertex $v$ of $\tilde{\mathcal{W}}(\tilde{\Psi}(\mathcal{W}))$.
Then the region $\mathcal{W}'$ has sequential skeletons $\tilde{\Psi}(\mathcal{W}')$ such that each remaining edge $e'$ is an edge of some skeleton of $\tilde{\Psi}(\mathcal{W}')$ and has gap number $2\mathcal{Z}$.
Hence we can conclude that
each edge placed in Algorithm \ref{alg:framework} has gap number at most $2\mathcal{Z}$.
\end{proof}
From Lemmas \ref{lem:partition}, \ref{lem:skeleton-wrap} and \ref{lem:noncrossing},
we know that each current step in Algorithm \ref{alg:framework}, the placement of sequential skeletons for a region $\mathcal{W}$ in $\mathcal{H}$ cannot make $\mathcal{X}$-crossing with the placement of previous steps in $\mathcal{H}$
because each edge's gap number in each skeleton is at most $\mathcal{J}$.
Also, from Lemma \ref{lem:gap_number},
we know that the distance number of the layout in Algorithm \ref{alg:framework} is at most $2\mathcal{Z}$.
Hence we can immediately prove that the framework in Algorithm \ref{alg:framework} can place a composite-layerlike graph $\mathcal{G}$ as an $(\mathcal{Q}, \mathcal{X}, \mathcal{D}=2\mathcal{Z})$-well-placed layout on $2\mathcal{D}$ tracks of a ladder $\mathcal{H}$
in Theorem \ref{thm:cons-track}.
\begin{theorem}\label{thm:cons-track}
If Conjecture \ref{conj:skeleton} can be proven,
then by Algorithm \ref{alg:framework}, every composite-layerlike graph $\mathcal{G}$ can be an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout on $2\mathcal{D}$ tracks in a ladder $\mathcal{H}$.
\end{theorem}
\section{From a Plane Graph $G$ To a Composite-Layerlike Graph $\mathcal{G}$}\label{sec:transformation}
\begin{figure}[t]
\begin{center}
\includegraphics[width=1\textwidth, angle =0]{reform-1}
\centering
\caption{It shows a procedure to transform a plane graph $G$ of (1) to a composite-layerlike graph $\mathcal{G}$;
(2) The outer boundary of $G$ is orderly placed from the vertices $a_1$ to $a_{10}$.
And, the second layer consists of three
sequential vertices $(b_1, b_2, b_3, b_4)$,
$(b_5, b_6, b_7, b_8, b_9, b_{10})$ and
$(b_{11}, b_{12}, b_{13}, b_{14})$.
The dash edges are wires.
(3) The wires
$\{(a_1, b_3),$
$(a_1, b_4),$
$(a_1, b_5),$
$(a_1, b_8),$
$(a_1, b_9),$
$(a_1, b_{10}),$
$(a_1, b_{11}),$
$(a_1, b_{13}),$
$(a_1, b_{14}))\}$ and the bridges
$\{(b_3, b_5),$
$(b_8, b_{11})\}$ are removed.
The dummy edges
$\{(a_3, b_4),$
$(a_4, b_5),$
$(a_8, b_{10}),$
$(a_9, b_{11}),$
$(a_9, b_{13})\}$ are added;
(4) shows how to reform a inner cycle into a composite-layerlike graph.
The outer boundary of the inner cycle is placed on a layer in $\mathcal{G}$
from the vertex $b_5$ and follows the clockwise order as $(b_5, b_6, b_7, b_8, b_9, b_{10})$.
Inside the inner cycle, we have two smaller inner cycles
$(d_1, d_2, d_3, d_4)$ and $(d_5, d_6, d_7, d_8)$.
The two smaller inner cycles are placed in $\mathcal{H}$ from the vertices $d_1$ and $d_8$ by clockwise order, respectively.
Next, the four wires
$\{(b_5, d_3), (b_5, d_4), (b_5, d_5), (b_5, d_8)\}$ and the bridge $(d_3, d_5)$ are removed,
and the two dummy edges $\{(b_8, d_4), (b_{10}, d_8)\}$ are added into $\mathcal{G}$;
(5) shows how to reform a down-pointing triangle in a composite-layerlike graph.
The six piles
$\{(b_6, c_1), (b_6, c_3), (b_6, c_4), (b_6, c_5), (b_6, c_6), (b_6, c_7)\}$ and a bridge $(b_7, b_9)$ are removed,
and the dummy edges $\{(a_6, c_4), (a_6, c_6), (a_6, c_7)\}$ are added into a composite-layerlike graph $\mathcal{G}$.}
\label{fig:reform}
\end{center}
\vspace{-0.2in}
\end{figure}
In this section,
we show how to reform a plane graph $G$ to a composite-layerlike graph $\mathcal{G}$.
Let $G$ be a plane graph and ${\cal O}(G)$ be its outer boundary.
Each layer in a composite-layerlike graph $\mathcal{G}$ can be recursively defined as follows:
the first layer is ${\cal O}(G)$ and ${\cal O}_{G}$ is placed as clockwise order $(m, u_1, u_2, \cdots)$.
Then $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$ are the sequential maximal inner cycles inside ${\cal O}$ such that
for each maximal inner cycle ${\cal O}_i, 1\leq i\leq p$,
there are some vertices on ${\cal O}_i$ connecting to the vertex $m$.
Now for each maximal inner cycle ${\cal O}_i, 1\leq i\leq p$,
we can walk around the cycle ${\cal O}_i$ by clockwise order
to get two contiguous vertices $(L^U({\cal O}_i)=(v^i_1=u^i_y, v^i_2, \cdots, v^i_x=u^i_1), L^B({\cal O}_i)=(u^i_1=v^i_x, u^i_2, \cdots, u^i_y=v^i_1) )$
where each vertex of $L^U({\cal O}_i)$ don't connect to the vertex $m$ except the first and last vertices $\{v^i_1, v^i_y\}$ of $L^U({\cal O}_i)$ and each vertex of $L^B({\cal O}_i)$ connects to the vertex $m$.
All maximal inner cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$ can be placed on the second layer as the order:
$({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$.
Moreover,
for each maximal inner cycle ${\cal O}_i$,
${\cal O}_i$ is placed on the second layer from the vertex $u^i_1$ by
clockwise order as:
$(v^i_1=u^i_y, v^i_2, \cdots, v^i_x=u^i_y, u^i_{y-1}, \cdots, u^i_2)$.
By the above placement, we can have sequential induced subgraphs
$(G|\triangledown_1, G|\triangledown_2, \cdots, G|\triangledown_q)$
that each induced subgraph $G|\triangledown_i, 1\leq i\leq q$,
is a subgraph induced by a maximal down-pointing triangle $(l_i, \cdots, r_i, m_i)$
where the path from vertices $l_i$ and $r_i$ are on the outer boundary ${\cal O}(G)$ and $m_i$ is a vertex on some inner cycle ${\cal O}_i \in ({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$.
In the followings, we discuss how to place
each induced subgraphs $G|\triangledown_i, 1\leq i\leq q,$
on the subsequent layers in a composite-layerlike graph $\mathcal{G}$.
For a subgraph induced by a maximal down-pointing triangle $\triangledown=(l, \cdots, r, m)$.
Sequential maximal inner cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_q)$ inside $\triangledown=(l, \cdots, r, m)$ can be found
such that each maximal inner cycle ${\cal O}_i, 1\leq i\leq q$ inside $\triangledown$ connects to the vertex $m$ and
each maximal inner cycle ${\cal O}_i$ can be partitioned into two contiguous vertices $(L^U({\cal O}_i), L^B({\cal O}_i))$ by clockwise order
where each vertex of $L^U({\cal O}'_i)$ doesn't connect the vertex $m$ except the first and last vertices of $L^U({\cal O}_i)$ and
each vertex of $L^B({\cal O}'_i)$ connects to the vertex $m$.
Also,
(1) the sequential cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_q)$ is placed orderly,
(2) the contiguous vertices $\mathcal{L}^U({\cal O}_i)$ are placed orderly and
the contiguous vertices $\mathcal{L}^B({\cal O}_i)$ are
placed reversely, and
(3) $\mathcal{L}^U({\cal O}_i)$ is placed at left of $\mathcal{L}^B({\cal O}_i)$
on the subsequent layer of
a new frame $\Pi$ in a composite-layerlike graph $\mathcal{G}$
where the $\Pi$'s first layer is the same as the down-pointing triangle $\triangledown$'s upper layer.
From the above description,
we assume there is no any chord $(u, v)$ on each cycle ${\cal O}$.
Next we start to explain how to eliminate each chord on each cycle.
Now we plan to remove edges from the above construction such that there is no any $X$-crossing edge in a composite-layerlike graph.
For each maximal inner cycle ${\cal O}$ that the cycle ${\cal O}$ is placed as the clockwise order: $(m, u_1, \cdots)$ on a layerlike graph $\Pi$,
and the sequential maximal inner cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$ inside ${\cal O}$ such that
each maximal inner cycle ${\cal O}_i, 1\leq i\leq q,$ connects to the vertex $u_1$.
we remove edges between the vertex $m$ and all vertices on the $(L^B({\cal O}'_1), L^B({\cal O}'_2), \cdots, L^B({\cal O}'_q))$ from the frame $\Pi$.
Moreover,
let $v_i$ be the vertex on the cycle ${\cal O}$ such that
$v_i$ connects to the both cycles ${\cal O}_i$ and ${\cal O}_{i+1}$.
Then, we add each sequential edges between $v_i$ and $L^B({\cal O}_i), 1\leq i\leq q,$ in the frame $\Pi$.
Similarly, for each maximal down-pointing triangle $\triangledown=(l, \cdots, r, m)$,
we execute the above procedure for the vertex $m$ of the down-pointing triangle $\triangledown$.
During the above transformation, how to process that if there exists a chord
on a cycle ${\cal O}$ is neglected to discuss. The reason is explained below.
A $d$-subdivision of a graph $G$ is a graph obtained by
replacing each edge of $G$ with a path having at most $2 + d$ vertices.
For each chord $(u, v)$ on a cycle ${\cal O}$, a 1-subdivision plane graph $G^1$ without any chord on a cycle ${\cal O}$ can be constructed by the following steps:
(1): find another
vertex $w\notin {\cal O}$ such that the triple vertices $(u, v, w)$ form a triangle face in a plane graph $G$,
(2): a vertex $w'$
can be added inside the face $(u, v, w)$,
(3): the chord $(u, v)$ can be replaced with two edges
$\{(u, w'), (w', v)\}$, and
(4): a dummy edge $(w, w')$ can be added to satisfy triangulation property.
Hence how to process a chord on a cycle ${\cal O}$ found
during the above transformation can be neglected by replaced a chord with two edges.
Now we immediately have the following theorem.
\begin{theorem}\label{thm:reform}
For each plane graph $G$, a $1$-subdivision $G^1$ of $G$ can be reformed into a composite-layerlike graph $\mathcal{G}$.
\end{theorem}
\section{A Track Layout for a Plane Graph on Constant Number of Tracks}\label{sec:cons-tracks}
We explain how a plane graph $G$ can be placed as a track layout on constant number of tracks.
\begin{theorem}\label{thm:subdivision}\emph{\cite{DW05}}
Suppose a graph $G$ has a $d$-subdivision $k$-track layout.
If the two numbers $k$ and $d$
are constant-bound,
$G$ also have a track layout on constant number of tracks.
\end{theorem}
\begin{theorem}
Every plane graph $G$ has a track layout on constant number of tracks.
\end{theorem}
\begin{proof}
From Theorem \ref{thm:G1-well-placed}, a $1$-subdivision plane graph $G^1$ can have a track layout on constant number of tracks $\mathcal{H}$.
Because $G^1$'s track number is constant-bound,
by Theorem \ref{thm:subdivision},
$G$ can also have a track layout on constant number of tracks.
\end{proof}
\section{An $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-Well-Placed Layout for a Raising Fan $\tilde{\mathcal{F}}$ in a Ladder $\mathcal{H}$}\label{sec:layout-fan-path}
\begin{figure}[t]
\begin{center}
\includegraphics[width=1\textwidth, angle =0]{fan-path-layout}
\centering
\caption{(1): shows a raising fan $\tilde{\mathcal{F}}$ consists of six fans
$\mathcal{F}_1=(a_3, m_1, a_4, m_2)$,
$\mathcal{F}_2=(a_2, a_5, m_3)$,
$\mathcal{F}_3=(a_1, a_6, m_4)$,
$\mathcal{F}_4=(d_2, d_3, m_5)$,
$\mathcal{F}_5=(d_1, d_4, m_6)$, and
$\mathcal{F}_6=(d_1, d_4, m_7)$.
Also, $\tilde{\mathcal{F}}$'s middle path is $\mathcal{M}=(m_1, m_2, m_3, m_4, m_5, m_6, m_7)$.
For the raising fan $\mathcal{F}$,
$\mathcal{L}_1=\{(a_1, a_2, a_3)\}$,
$\mathcal{R}_1=\{(a_4, a_5, a_6)\}$,
$\mathcal{M}_1=\{m_1\}$,
$\mathcal{L}_2=\{(b_1, b_2, b_3), (c_1, c_2), (d_1, d_2)\}$,
$\mathcal{R}_2=\{(b_4, b_5, b_6), (d_3, d_4)\}$,
$\mathcal{M}_2=(m_2, m_3, m_4)$,
$\mathcal{L}_3=\{(e_1, e_2, e_3)\}$,
$\mathcal{R}_3=\{((e_4, e_5, e_6)\}$,
$\mathcal{M}_3=(m_5, m_6, m_7)$;
(2): shows an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout for $\tilde{\mathcal{F}}$ in $\mathcal{H}$
by orderly placing the left wing of $\tilde{\mathcal{F}}$ and
reversely placing the right wing of $\tilde{\mathcal{F}}$.
}
\label{fig:fan-path-layout}
\end{center}
\vspace{-0.2in}
\end{figure}
In this section, we present an approach to have a specified type of sequential fans $\tilde{\mathcal{F}}$ in a composite-layerlike graph $\mathcal{G}(\mathcal{W}^{\mathcal{M}})$ called a \emph{raising-fan} path and defined later and
show hot to place it as an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in a ladder $\mathcal{H}$.
Also, an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout for a \emph{raising-fan} path $\tilde{\mathcal{F}}$ can partition the region
$\mathcal{W}^{\mathcal{M}}$ into sequential regions in $\mathcal{H}$.
A \emph{fan} $\mathcal{F}$ consists of sequential vertices $(u_1, \cdots, u_a, m)$
such that $(u_1, \cdots, u_a)$ and $w$ are on two contiguous
layers $i$ and $i+1$ of a layerlike graph, and $w$ connects each vertex in $(u_1, \cdots, u_a$.
The sequential vertices $(u_1, \cdots, u_a)$ are called upper vertices of the fan $\mathcal{F}$.
Also, the vertex $m$ are called the lower vertex of the fan $\mathcal{F}$.
A \emph{raising-fan} path $\tilde{\mathcal{F}}$ consists of sequential fans $(\mathcal{F}_1, \mathcal{F}_2, \cdots)$
such that for each fan $\mathcal{F}_i, i\geq 1$,
there is a down-pointing triangle in $\mathcal{F}_{i+1}$ bounds $\mathcal{F}_i$.
Also, the lower vertices of the raising fan $\tilde{\mathcal{F}}$ form
sequential vertices $\mathcal{M}=(m_1, m_2, \cdots)$ and is called a \emph{middle path}
where each vertex $m_i, i\geq 1$, is the lower vertex of the fan $\mathcal{F}_i$.
For a fan $\mathcal{F}\in \tilde{\mathcal{F}}$ with the middle path $\mathcal{M}$,
edges of the fan $\mathcal{F}$ at left and right of the middle path $\mathcal{M}$ are called
a \emph{left arm} and \emph{right arm} of the fan $\mathcal{F}$, respectively
A path $\Re(v)$ is called a \emph{raising-path} from a vertex $v$ in a composite-layerlike graph $\mathcal{G}$
if the path $\Re(v)$ starts from the vertex $v$ along edges from the lower layer to the upper layer
till the first layer in $\mathcal{G}$,.
Any two raising-paths $\Re(v_1)$ and $\Re(v_2)$,
$\Re(v_1)$ and $\Re(v_2)$ are called \emph{upward-merging} if either
the two raising-paths $\Re(v_1)$ and $\Re(v_2)$ are vertex-disjoint or
the intersection of the two raising-paths $\Re(v_1) \cap \Re(v_2)$ is a subpath from some vertex to a vertex on the first layer in $\mathcal{G}$.
A set of raising-paths $\tilde{\Re}=\{\Re(v)|v\in V(\mathcal{G})\}$ in a composite-layerlike graph $\mathcal{G}$ are called \emph{upward-merging} to $\mathcal{G}$
if for any two raising-paths $\Re(v_1)$ and $\Re(v_2)$ in $\tilde{\Re}$ are upward-merging and each vertex in $\mathcal{G}$
is in a raising-path $\Re(v)$ in $\tilde{\Re}$.
From now on, when mention a raising-path $\Re(v)$, it always means that
the path $\Re(v)$ is a raising-path in some specific upward-merging raising-path set $\tilde{\Re}$ for a composite-layerlike graph $\mathcal{G}$.
Assume that $\triangledown$ is a down-pointing triangle inside a fan $\mathcal{F}$ of a raising fan $\tilde{\mathcal{F}}$ with the middle path $\mathcal{M}$.
\begin{itemize}
\item A \emph{left spine} of $\triangledown$ with respect to the $\tilde{\mathcal{F}}$'s middle path $\mathcal{M}$ consists of
sequential maximal cycles $LS(\triangledown)=({\cal O}^L_1, {\cal O}^L_2, \cdots, {\cal O}^L_p)$ at left of the middle path $\mathcal{M}$
such that for each cycle ${\cal O}^L_i, 1\leq i\leq p$, there is an edge connecting between the upper vertices of $\triangledown$ and the boundary of ${\cal O}^L_i$.
\item A \emph{right spine} of $\triangledown$ with respect to the $\tilde{\mathcal{F}}$'s middle path $\mathcal{M}$ consists of
sequential maximal cycles $RS(\triangledown)=({\cal O}^R_1, {\cal O}^R_2, \cdots, {\cal O}^R_q)$ at right of the middle path $\mathcal{M}$
such that for each cycle ${\cal O}^R_i, 1\leq i\leq q$, there is an edge connecting between the upper vertices of $\triangledown$ and the boundary of ${\cal O}^R_i$.
\item the $i$-th \emph{joint} of a spine inside $\mathcal{F}'$ consists of the $i$-th cycle's the leftmost and rightmost vertices in the spine of $\triangledown$.
\end{itemize}
If a down-pointing triangle $\triangledown\in \mathcal{F}$ which is not passed through the middle path $\mathcal{M}$, then the sequential maximal cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_t)$ is called a \emph{spine} in $\triangledown$
because the spine is not partitioned by the middle path $\mathcal{M}$.
Let $\mathcal{F}$ be a fan in a raising fan $\tilde{\mathcal{F}}$ with the middle path $\mathcal{M}$ that
Then the fan $\mathcal{F}$ are partitioned by the middle path $\mathcal{M}$ into the following sequential down-pointing triangles
$(\triangledown^L_1,$
$\triangledown^L_2,$
$\cdots,$
$\triangledown^L_a),$
$\triangledown^M,$
$(\triangledown^R_1,$
$\triangledown^R_2,$
$\cdots,$
$\triangledown^R_b)$
where each down-pointing triangle $\triangledown^L_i, 1\leq i\leq a,$ is at left of the middle path $\mathcal{M}$,
the down-pointing triangle $\triangledown^M$ consists of the middle path $\mathcal{M}$ and
each down-pointing triangle $\triangledown^R_i, 1\leq i\leq b,$ is at right of the middle path $\mathcal{M}$.
\begin{itemize}
\item A \emph{left wing} $\mathcal{L}(\mathcal{F})$ of a fan $\mathcal{F}$ is the union of raising-paths consisting of all joints
in the sequential left down-pointing triangles $\triangledown^L_i, 1\leq i\leq a$ of the fan $\mathcal{F}$,
and the down-pointing triangle $\triangledown^M$'s left joints with respect to the middle path $\mathcal{M}$. And
\item A \emph{right wing} $\mathcal{R}(\mathcal{F})$ of a fan $\mathcal{F}$ is the union of raising-paths consisting of all joints
in the sequential right down-pointing triangles $\triangledown^R_i, 1\leq i\leq b$ of the fan $\mathcal{F}$,
and the down-pointing triangle $\triangledown^M$'s right joints with respect to the middle path $\mathcal{M}$.
\end{itemize}
From the definition of a raising-path, two contiguous raising-paths $(\Re(u), \Re(u')) \in \mathcal{L}$,
$\Re(u)$ and $\Re(u')$ form
a disjoint region $\mathcal{W}^L$, and $\Re(u)$ and $\Re(u')$ are the
left and right boundaries of $\mathcal{W}^L$, respectively.
Similarly,
two raising-paths $(\Re(u), \Re(u')) \in \mathcal{R}$ form
a region $\mathcal{W}^R$, and $\Re(u)$ and $\Re(u')$ are the
left and right boundaries of $\mathcal{W}^R$, respectively.
Hence we have two sequential regions for a fan $\mathcal{F}$: $\tilde{\mathcal{W}}^L(\mathcal{F})$ and $\tilde{\mathcal{W}}^R(\mathcal{F})$ that are at left and right of the middle path $\mathcal{M}$, respectively.
From the definitions of a raising fan $\tilde{\mathcal{F}}$ and a composite-layerlike graph $\mathcal{G}$,
we can give another representation for a raising fan $\tilde{\mathcal{F}}$ as
$(\mathcal{F}_{1, 1}, \mathcal{F}_{1, 2}, \cdots, \mathcal{F}_{1, a_1},$ $\mathcal{F}_{2, 1}, \mathcal{F}_{2, 2}, \cdots, \mathcal{F}_{2, a_2},$ $\cdots)$
where for each subsequential fans $(\mathcal{F}_{i, 1}, \mathcal{F}_{i, 2}, \cdots, \mathcal{F}_{i, a_i}), i\geq 1$,
all upper vertices of each fan $\mathcal{F}_{i, j}, 1\leq j\leq a_i,$ are on the $i$-th layer of the layerlike graph $\Pi$ of $\mathcal{G}$.
Also, subsequential fans
$\tilde{\mathcal{F}}^C=(\mathcal{F}_{1, 1}, \mathcal{F}_{1, 2}, \cdots,$ $\mathcal{F}_{1, a_1}, \mathcal{F}_{2, 1}, \mathcal{F}_{2, 2},$ $\cdots, \mathcal{F}_{2, a_2}, \cdots)$
are called \emph{characteristic-fans} of $\tilde{\mathcal{F}}$
if for each $i\geq 1$, all upper vertices of $(\mathcal{F}_{i, 1}, \mathcal{F}_{i, 2}, \cdots, \mathcal{F}_{i, a_i})$
are on the $i$-th layer of the $\mathcal{G}$'s layerlike graph $\Pi$.
Also, the union of left wings and right wings of all fans in a characteristic-raising fan
on the $i$-th layer are denoted as $\mathcal{L}_i(\tilde{\mathcal{F}}^C)$ and $\mathcal{R}_i(\tilde{\mathcal{F}}^C)$, respectively. And,
the sequential lower vertices $(m_{i, 1}, m_{i, 2}, \cdots, m_{i, a_i})$
of the sequential fans $(\mathcal{F}_{i, 1}, \mathcal{F}_{i, 2}, \cdots, \mathcal{F}_{i, a_i})$
are denoted as $\mathcal{M}_i(\tilde{\mathcal{F}}^C)$.
Let $\tilde{\mathcal{F}}^C$
$=(\mathcal{F}_{1, 1}, \mathcal{F}_{1, 2},$
$\cdots,$
$\mathcal{F}_{1, a_1},$
$\mathcal{F}_{2, 1},$
$\mathcal{F}_{2, 2},$
$\cdots,$
$\mathcal{F}_{2, a_2},$
$\cdots)$ be the characteristic-raising fan of $\tilde{\mathcal{F}}$.
The characteristic-raising fan $\tilde{\mathcal{F}}^C$ can form sequential regions as follows:
initially, let the two term $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)$ be empty sequence.
Then repeatedly find the last fan $\mathcal{F}\in \tilde{\mathcal{F}}^C$
(the last fan $\mathcal{F}\in \tilde{\mathcal{F}}^C$ is inside a region between the rightmost and leftmost boundaries of $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)$, respectively),
remove the fan $\mathcal{F}$ from $\tilde{\mathcal{F}}^C$, and
add two sequential regions $\tilde{\mathcal{W}}^L(\mathcal{F})$ and $\tilde{\mathcal{W}}^R(\mathcal{F})$
into $\tilde{\mathcal{F}}^L(\tilde{\mathcal{F}}^C)$ and $\tilde{\mathcal{F}}^L(\tilde{\mathcal{F}}^C)$, respectively
where the two sequential regions $\tilde{\mathcal{W}}^L(\mathcal{F})$ and $\tilde{\mathcal{W}}^R(\mathcal{F})$ are formed by the fan $\mathcal{F}$
that are at left and at right of the middle path $\mathcal{M}$, respectively.
After all fan are removed from the characteristic-raising fan $\tilde{\mathcal{F}}^C$,
we can have two sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ and
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})$ that are at left and right of the middle path $\mathcal{M}$, respectively.
For each subsequential fans $\tilde{\mathcal{F}}_{i, j}$ from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}, i, j\geq1$,
we know that the unions of left wings $\mathcal{L}_{i, j}$ and right wings $\mathcal{R}_{i, j}$ for all fans $\mathcal{F}$ from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}$
are inside the rightmost region $\mathcal{W}^L_{i, j}$ and leftmost regions $\mathcal{W}^R_{i, j}$ of $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)_{i, j}$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)_{i, j}$, respectively.
Hence we can recursively partition the regions $\mathcal{W}^L_{i, j}$ and $\mathcal{W}^R_{i, j}$ by the two left wings $\mathcal{L}_{i, j}$ and right wings $\mathcal{R}_{i, j}$ and replace the two regions
$\tilde{\mathcal{W}}^L_{i, j}$ and $\tilde{\mathcal{W}}^R_{i, j}$
by the two sequential regions $\tilde{\mathcal{W}}^L_{i, j}$
and $\tilde{\mathcal{W}}^R_{i, j}$, respectively.
Now we can conclude that
the union of the left wings and right wings of the raising fan $\tilde{\mathcal{F}}$
form two sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})=(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$ and
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})=(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$.
We can conclude that a raising fan $\tilde{\mathcal{F}}=(\mathcal{F}_1, \mathcal{F}_2, \cdots)$ can form sequential regions as
$(\cdots,$
$\tilde{\mathcal{W}}^L_2,$ $\tilde{\mathcal{W}}^L_1,$ $\tilde{\mathcal{W}}^R_1,$ $\tilde{\mathcal{W}}^R_2,$
$\cdots)$
where $\tilde{\mathcal{W}}^L_i$
and $\tilde{\mathcal{W}}^R_i$
are sequential regions at left and right of the middle path $\mathcal{M}$ for a fan $\mathcal{F}_i$ of a raising fan $\tilde{\mathcal{F}}$, respectively.
From the above discussion, we immediately have the following lemma.
\begin{lemma}\label{lemma:fan-path-partition}
Given a raising fan $\tilde{\mathcal{F}}$ and its middle path $\mathcal{M}$,
the unions of $\tilde{\mathcal{F}}$'s left wings and right wings can form sequential regions $(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p,$ $\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$
where the two subsequential regions $(\mathcal{W}^L_1, \mathcal{W}^L_2,$ $\cdots,$ $\mathcal{W}^L_p)$ and $(\mathcal{W}^R_1, \mathcal{W}^R_2,$ $\cdots,$ $\mathcal{W}^R_q)$ are
at left and right of the middle path $\mathcal{M}$, respectively.
Moreover, $(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$ and $(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$
are called left and right sequential regions of $\tilde{\mathcal{F}}$, and denoted to $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})$, respectively.
\end{lemma}
Now we can consider how to have an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout for a raising fan $\tilde{\mathcal{F}}=(\mathcal{F}_1, \mathcal{F}_2, \cdots)$ in a ladder $\mathcal{H}$.
When we have a raising fan $\tilde{\mathcal{F}}$, it implies that we have two sequential
regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})$ from Lemma \ref{lemma:fan-path-partition}.
The basic idea to have an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}$ is that
we orderly place the sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})$ and reversely place the sequential regions $\mathcal{W}^R(\tilde{\mathcal{F}})$ in $\mathcal{H}$.
To place the left wing and right wing of a raising fan $\tilde{\mathcal{F}}$,
we can start to place the $\tilde{\mathcal{F}}$'s characteristic-fans
$(\mathcal{F}_{1, 1}, \mathcal{F}_{1, 2}, \cdots, \mathcal{F}_{1, a_1},$ $\cdots,$ $\mathcal{F}_{2, 1}, \mathcal{F}_{2, 2}, \cdots, \mathcal{F}_{2, a_2}, \cdots)$
and their union of left wings $\mathcal{L}_i$, union of right wings $\mathcal{R}_i$ and
their sequential lower vertices $\mathcal{M}_i=(m_{i, 1}, m_{i, 2}, \cdots, m_{i, a_i})$ as the order:
for each $i\geq 1$, we place $\mathcal{L}_i$ orderly,
$\mathcal{M}_i$ reversely and $\mathcal{R}_i$ reversely on the $i$-th layer of $\mathcal{H}$.
Since the sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C_i)$ are placed orderly in $\mathcal{H}$,
we must place $\mathcal{M}_i(\tilde{\mathcal{F}}^C)=(m_{i, 1},$ $m_{i, 2},$ $\cdots,$ $m_{i, a_i})$ reversely in $\mathcal{H}$ as
$(m_{i, a_i},$ $\cdots,$ $m_{i, 2},$ $m_{i, 1})$ to avoid $X$-crossing edges in $\mathcal{H}$.
For each raising fan $\tilde{\mathcal{F}}^C_i=(\mathcal{F}^C_{i, 1}, \mathcal{F}^C_{i, 2}, \cdots, \mathcal{F}^C_{i, a_i})$, the sequential lower vertices $\mathcal{M}_i=(m_{i, 1}, m_{i, 2}, \cdots, m_{i, a_i})$ are reversely placed as
$(m_{i, a_i},$ $\cdots,$ $m_{i, 2},$ $m_{i, 1})$ in $\mathcal{H}$,
the sequential regions $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C_i)$ need to be placed reversely in $\mathcal{H}$ to avoid that
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C_i)$ make $X$-crossing edges in $\mathcal{H}$.
\begin{lemma}\label{lem:fan-path-region-order}
Given the characteristic-raising fan $\tilde{\mathcal{F}}^C$ of a raising fan $\tilde{\mathcal{F}}$,
the unions of left wings and right wings of $\tilde{\mathcal{F}}^C$
form two sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ and
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)$,
the sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ are orderly placed
as $\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}^C)$ and
the sequential regions
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)$ are reversely placed
as $\tilde{\mathcal{W}}^R_{\mathcal{H}}(\tilde{\mathcal{F}}^C)$.
Also,
the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^C)=(\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}^C),
\tilde{\mathcal{W}}^R_{\mathcal{H}}(\tilde{\mathcal{F}}^C))$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{lemma}
In the next theorem, we prove that a raising fan $\tilde{\mathcal{F}}$ can be $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in a ladder $\mathcal{H}$.
\begin{theorem}\label{thm:fan-path-region-order}
Given a raising fan $\tilde{\mathcal{F}}$, the left wings and right wings of $\tilde{\mathcal{F}}$
form two sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})=(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$ and
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})=(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$.
If the sequential regions $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}})=(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$ are sequentially placed in $\mathcal{H}$ as the order:
$\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}})=(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$
and
the sequential regions
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}})=(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$ are reversely placed
as the order in $\mathcal{H}$:
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})=$
$(\mathcal{W}^L_1, \mathcal{W}^L_2,$ $\cdots, \mathcal{W}^R_p,$
$\mathcal{W}^R_q, \cdots,$ $\mathcal{W}^R_2, \mathcal{W}^R_1)$,
then the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})=(\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}), \tilde{\mathcal{W}}^R_{\mathcal{H}}(\tilde{\mathcal{F}}))$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{theorem}
\begin{proof}
From Lemma \ref{lemma:fan-path-partition}, we know that
a raising fan $\tilde{\mathcal{F}}^C$ can form two sequential regions $(\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}),$
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}))$
that are at left and right of the middle path $\mathcal{M}$, respectively.
For a characteristic-raising fan $\tilde{\mathcal{F}}^C$,
we know that
$(\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}^C), \tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}^C))$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in a ladder $\mathcal{H}$ by Lemma \ref{lem:fan-path-region-order}.
Moreover,
the sequential regions
$\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)$ are orderly placed in $\mathcal{H}$ as$\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}^C)$ and
the sequential regions
$\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)$ are reversely placed in $\mathcal{H}$ as $\tilde{\mathcal{W}}^R_{\mathcal{H}}(\tilde{\mathcal{F}}^C)$.
For each subsequential fans $\tilde{\mathcal{F}}_{i, j}$ from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}, i, j\geq1$,
we know that the unions of left wings $\mathcal{L}_{i, j}$ and right wings $\mathcal{R}_{i, j}$ of all fans from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}$
are inside the rightmost region $\mathcal{W}^L_{i, j}$ and leftmost regions $\mathcal{W}^R_{i, j}$ of $\tilde{\mathcal{W}}^L(\tilde{\mathcal{F}}^C)_{i, j}$ and $\tilde{\mathcal{W}}^R(\tilde{\mathcal{F}}^C)_{i, j}$, respectively.
Hence we can recursively partition the regions $\mathcal{W}^L_{i, j}$ and $\mathcal{W}^R_{i, j}$ by the two left wings $\mathcal{L}_{i, j}$ and right wings $\mathcal{R}_{i, j}$ to
have two sequential regions
$\tilde{\mathcal{W}}(\mathcal{L}_{i, j})$ and
$\tilde{\mathcal{W}}(\mathcal{R}_{i, j})$ that can be
$(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed as
$\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{L}_{i, j})$ and
$\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{R}_{i, j})$
in $\mathcal{H}$.
Because the regions $\mathcal{W}^L_{i, j}$ and $\mathcal{W}^R_{i, j}$ are orderly and reversely placed in $\mathcal{H}$, respectively,
we can place the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{L}_{i, j})$ orderly
and the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{R}_{i, j})$ reversely
inside the two regions
$\tilde{\mathcal{W}}^L_{i, j}$ and $\tilde{\mathcal{W}}^R_{i, j}$, respectively
to have an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}$.
Moreover, the middle path from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}$ is placed reversely between the two vertices $m_{(i, j)}$ and
$m_{(i, j+1)-1}$.
Because (1) $\mathcal{L}_{i, j}$ are placed between $\mathcal{F}_{i, j}$ and $\mathcal{F}_{i, j+1}$, and
(2) the middle path from $\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{(i, j+1)-1}$ is placed reversely between the two vertices $m_{(i, j)}$ and
$m_{(i, j+1)-1}$,
each left arm from
$\mathcal{F}_{(i, j)+1}$ to $\mathcal{F}_{i, j+1}-1$ is placed between tracks $i$ and $i+1$ cannot make $X$-crossing with
left arms of the characteristic-raising fan $\tilde{\mathcal{F}}^C$ between tracks $i$ and $i+1$in $\mathcal{H}$.
Similarly, each right arm of $\mathcal{R}_{i, j}$ is placed between
tracks $i$ and $i+1$ cannot
make $X$-crossing with
right arms of the characteristic-raising fan $\tilde{\mathcal{F}}^C$ between tracks $i$
and $i+1$ in $\mathcal{H}$.
Finally, we conclude that
the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})=(\tilde{\mathcal{W}}^L_{\mathcal{H}}(\tilde{\mathcal{F}}), \tilde{\mathcal{W}}^R_{\mathcal{H}}(\tilde{\mathcal{F}}))$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{proof}
\section{How to Find a Skeleton $\Psi$ in a Region $\mathcal{W}$ to Have an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-Well-Placed Layout $\Psi_{\mathcal{H}}$ in a Ladder $\mathcal{H}$?}\label{sec:skeleton}
In this section, we prove that a skeleton $\Psi$ of a regions $\mathcal{W}$ can be obtained from all \emph{rightward-outer} and \emph{leftward-outer} fans in $\mathcal{W}$ and can have an
$(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}$.
\subsection{A Forest-Like Representation $\clubsuit_{\mathcal{B}}$ of Raising Fans from a Boundary $\mathcal{B}$}\label{sec:forest-like}
In this subsection, we show how to build a forest-like representation $\clubsuit_{\mathcal{B}}$
from all \emph{rightward-outer} fans of all vertices on a boundary $\mathcal{B}$.
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$,
let $v$ be a vertex on the boundary $\mathcal{B}^L$ and $(v_1, v_2, \cdots)$ be the sequential children of $v$ inside the region $\mathcal{W}$.
A fan $\mathcal{F}=(u_1, u_2, \cdots, m)$ is called a \emph{rightward} for $v$' children on a left boundary of a region $\mathcal{W}'$
if $\mathcal{F}$'s upper vertices can be
partitioned into two contiguous subsequences that the first contiguous subsequence is
among $v$'s children and
the second contiguous subsequence
is not among $v$'s children.
A fan $\mathcal{F}=(u_1, u_2, \cdots, m)$ is called a \emph{rightward} fan for a vertex $v$ on a left boundary of a region $\mathcal{W}'$
if $\mathcal{F}$'s the first upper vertex $u_1$ is the vertex $v$.
Given a subregion $\mathcal{W}'$ inside a region $\mathcal{W}$ and the vertex $v$ is on the lowest layer of the intersection of the $\mathcal{W}$'s left boundary $\mathcal{B}^L$ and $\mathcal{W}'$'s left boundary,
a \emph{leftmost-raising-fan} path $\tilde{\mathcal{F}}$ inside the subregion $\mathcal{W}'$ of $\mathcal{W}$ is a maximal raising fan consisting of all $v$'s rightward-outer fans.
Intuitively, the layer of any rightward-outer fan not included in $\tilde{\mathcal{F}}$ can not be lower than the lowest layer of a rightward-outer fan included in $\tilde{\mathcal{F}}$.
Similarly,
a \emph{leftward-outer} fan and
a \emph{leftward} fan of a vertex $v$ in a region $\mathcal{W}'$ also can be defined symmetrically to a rightward fan and a rightward-outer fan of a vertex $v\in \mathcal{B}^R$, respectively. And,
a \emph{rightmost-raising-fan} path inside a region $\mathcal{W}'$ also can be defined symmetrically to a leftmost raising fan inside a region $\mathcal{W}'$.
Without loss of generality,
assume that $\mathcal{B}$ is $\mathcal{W}$'s
the left boundary,
we plan to find a set of raising fans from $\mathcal{B}$'s all rightward-outer fans and
represent them as a forest-like representation $\clubsuit_{\mathcal{B}}$
where each vertex of $\clubsuit_{\mathcal{B}}$ represents a raising fan.
Initially $\clubsuit_{\mathcal{B}}$'s root is a raising fan $\tilde{\mathcal{F}}$ consisting the only region $\mathcal{W}$
and add the region $\mathcal{W}$ into $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}})$.
Then $\clubsuit_{\mathcal{B}}$ can be constructed as follows:
repeatedly find a region $\mathcal{W}'$ from the sequential regions having uncolored rightward-outer fans inside the region $\mathcal{W}'$
and $\mathcal{W}'$ is a region of a raising fan $\tilde{\mathcal{F}}$
, to
\begin{enumerate}
\item find a maximal raising fan $\tilde{\mathcal{F}}'$ consisting of all leftmost and uncolored fans inside the region $\mathcal{W}'$,
\item assign the raising fan $\tilde{\mathcal{F}}'$ to a child of $\tilde{\mathcal{F}}$ consisting of the region $\mathcal{W}'$ in $\clubsuit_{\mathcal{B}}$ and color the raising fan $\tilde{\mathcal{F}}'$,
\item partition the region $\mathcal{W}'$ into sequential regions $\tilde{\mathcal{W}}'$ by the raising fan $\tilde{\mathcal{F}}'$ and replace the region $\mathcal{W}'$ by the sequential regions $\tilde{\mathcal{W}}'$ in $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}})$, and
\end{enumerate}
till no any such region can be found. Now we immediately have the following two lemmas:
\begin{lemma}
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$ with the left and right boundaries $\mathcal{B}^L$ and $\mathcal{B}^R$, the collection of all rightward-outer and leftward-outer fans of the boundary $\mathcal{B}^L$ and
$\mathcal{B}^R$ form sequential regions $(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p)$ and $(\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$, respectively.
Also, the two sequential regions partition the region $\mathcal{W}$ into sequential regions
$(\mathcal{W}^L_1, \mathcal{W}^L_2, \cdots, \mathcal{W}^L_p,$ $\mathcal{W}^M,$ $\mathcal{W}^R_1, \mathcal{W}^R_2, \cdots, \mathcal{W}^R_q)$ where $\mathcal{W}^M$
is a region bounded by the right boundary of $\mathcal{W}^L_p$ and the left boundary of $\mathcal{W}^R_1$.
\end{lemma}
\begin{lemma}\label{lem:forest-like}
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$,
we can have two forest-like representations $\clubsuit_{\mathcal{B}^L}$ and $\clubsuit_{\mathcal{B}^R}$
from the collection of all rightward-outer and leftward-outer fans of $\mathcal{B}^L$ and $\mathcal{B}^R$, respectively.
\end{lemma}
A forest-like structure $\clubsuit_{\mathcal{B}}$ is called \emph{left-forest-like} structure, if $\mathcal{B}$ is the left boundary of a region $\mathcal{W}$.
Recall that a boundary $\mathcal{B}$'s forest-like structure $\clubsuit_{\mathcal{B}}$ is the union of all rightward-outer fans of $(u_1, u_2, \cdots, u_l)$
where each $u_i, i\geq 1,$ is a vertex on the boundary $\mathcal{B}$.
Observe that for each vertex $u\in \mathcal{B}$,
every $u$'s child $u'$ is inside a region of $u$'s rightward-outer fan $\mathcal{F}$ and $\mathcal{F}$ is a fan in a raising fan $\tilde{\mathcal{F}}\in (\tilde{\mathcal{F}}_1, \tilde{\mathcal{F}}_2, \cdots)$.
Hence $u'$ is inside a region of $\tilde{\mathcal{W}}(\tilde{\mathcal{F}})$ and $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}})$.
From the above discussion, we have the following lemma:
\begin{lemma}\label{lem:region-partition}
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$ and a boundary $\mathcal{B}\in (\mathcal{B}^L, \mathcal{B}^R)$, for each vertex $u\in \mathcal{B}$, every $u$'s child is inside a region of sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}})$.
\end{lemma}
From the above lemma, we know that the union of the two forest-like structures
$\clubsuit_{\mathcal{B}^L}$ and $\clubsuit_{\mathcal{B}^R}$ form a skeleton of a region $\mathcal{W}$.
$\clubsuit_{\mathcal{B}^L}$ and $\clubsuit_{\mathcal{B}^R}$ are called \emph{left skeleton} $\Psi^L(\mathcal{W})$ and \emph{right skeleton} $\Psi^R(\mathcal{W})$, respectively.
\begin{theorem}\label{thm:skeleton}
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$, the union of the left skeleton and right skeleton $\Psi^L(\mathcal{W})$ and $\Psi^R(\mathcal{W})$ is a skeleton of $\mathcal{W}$.
\end{theorem}
\subsection{An $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-Well-Placed Layout in a Ladder $\mathcal{H}$ for a Forest-Like Structure $\clubsuit_{\mathcal{B}}$}\label{sec:placement-forest-like}
\begin{algorithm}[t]
\caption{A Framework to Place Forest-Like Raising-Fan Paths $\clubsuit_{\mathcal{B}}$ in a Ladder $\mathcal{H}$}
\label{alg:framework-boundary}
\KwIn{A forest-like structure $\clubsuit_{\mathcal{B}}$ consists of the only one root $r$.}
\KwOut{sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(r)$ in $\mathcal{H}$.}
Let $\tilde{\mathcal{F}}$ be the raising fan at the $\clubsuit_{\mathcal{B}}$'s root $r$\;
Orderly place the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})$ in $\mathcal{H}$\;
Orderly add the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})$ into the first-in-first-out queue $\tilde{\mathcal{W}}$\;
\While{$\tilde{\mathcal{W}}$ is not empty}
{
Remove the first region $\mathcal{W}'$ in $\tilde{\mathcal{W}}$\;
Place the subsequential regions
$\tilde{\mathcal{W}}'_{\mathcal{H}}$ of
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}'))$ at rightmost in $\mathcal{H}$
where each region $\mathcal{W}''$ in $\tilde{\mathcal{W}}'_{\mathcal{H}}$ consists of some raising fans in $\clubsuit_{\mathcal{B}}$\;
Orderly add $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}'))$ into the first-in-first-out queue $\tilde{\mathcal{W}}$\;
}
\end{algorithm}
Before describing Algorithm \ref{alg:framework-boundary},
we need to roughly define the term $\clubsuit_{\mathcal{B}}(\mathcal{W}_i)$ as a subset of raising fans in $\clubsuit_{\mathcal{B}}$ which consist of all raising fans inside $\clubsuit_{\mathcal{B}}(\mathcal{W}_i)$ meet at a vertex on the right boundary of $\mathcal{W}_i$.
Moreover, the subset of raising fans $\clubsuit(\mathcal{W}_i)$ partition $\mathcal{W}_i$ into sequential regions and can be a $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout in $\mathcal{H}$. (They are proven in Lemmas \ref{lem:partition-boundary} and \ref{lem:layout-boundary}, respectively.)
Initially, in Algorithm \ref{alg:framework-boundary}, the root $r$'s raising fan $\tilde{\mathcal{F}}$ in a forest-like structure $\clubsuit_{\mathcal{B}}$ form sequential
regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})$ in $\mathcal{H}$ and we put them into the first-in-first-out queue orderly.
Next each iteration $i$ of Algorithm \ref{alg:framework-boundary} picks up
the first region $\mathcal{W}_i$ in the first-in-first-out queue $\tilde{\mathcal{W}}$
(it means that the region $\mathcal{W}_i$ at the leftmost region
such that the region $\mathcal{W}_i$ consists of raising fans in $\clubsuit_{\mathcal{B}}$ in $\mathcal{H}$;
$|\clubsuit_{\mathcal{B}}(\mathcal{W}_i)|>0$)
to place the sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ (it can be proven in Lemma \ref{lem:partition-boundary}) at the rightmost side in $\mathcal{H}$ as $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$. (It can be proven in Lemma \ref{lem:layout-boundary}.)
It means that all regions $\mathcal{W}_j$ at left of $\mathcal{W}_i$ in $\mathcal{H}$ doesn't consist any raising fan in $\clubsuit_{\mathcal{B}}$ ($|\clubsuit_{\mathcal{B}}(\mathcal{W}_j)|=0$) and the chords between $\mathcal{W}_j$ and $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ don't nest with
the chords between $\mathcal{W}_i$ and
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$
because $\mathcal{W}_j$ is at left of $\mathcal{W}_i$ in $\mathcal{H}$ and
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are at left of
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ in $\mathcal{H}$.
It means that
the left end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are placed at left of
the left end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$, and
the right end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are placed at left of
the right end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ in $\mathcal{H}$.
Next, we give a precise definition of $\clubsuit_{\mathcal{B}}(\mathcal{W}_i)$ as follows:
\begin{definition}
Given a region $\mathcal{W}_i=(\mathcal{B}^L, \mathcal{B}^R)$ and
a left forest-like structure $\clubsuit_{\mathcal{B}}$,
$\clubsuit_{\mathcal{B}}(\mathcal{W}_i)$ is defined to consist of two sequential raising fans
$\clubsuit^L=(\tilde{\mathcal{F}}^L_1, \tilde{\mathcal{F}}^L_2, \cdots, \tilde{\mathcal{F}}^L_p)$ and
$\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$
where
\begin{itemize}
\item
$\clubsuit^L=(\tilde{\mathcal{F}}^L_1, \tilde{\mathcal{F}}^L_2, \cdots, \tilde{\mathcal{F}}^L_p)$
are the maximal subsequential raising fans in $\clubsuit_{\mathcal{B}}$
such that for each raising fan $\tilde{\mathcal{F}}^L_i, 1\leq i\leq p$,
(1) $\tilde{\mathcal{F}}^L_i$ does not touch any vertex on the right boundary $\mathcal{B}^R$,
(2) $\tilde{\mathcal{F}}^L_i$ is not a descendant of any raising fan $\tilde{\mathcal{F}}^R_j, 1\leq j\leq q$, and
(3) $\tilde{\mathcal{F}}^L_i$ is a right sibling of $\tilde{\mathcal{F}}^L_{i-1}$ in $\clubsuit_{\mathcal{B}}$. And,
\item
$\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$
are the maximal subsequential raising fans in $\clubsuit_{\mathcal{B}}$
such that for each raising fan $\tilde{\mathcal{F}}^R_i, 1\leq i\leq q$,
$\tilde{\mathcal{F}}^R_i$'s right wings touch the right boundary $\mathcal{B}^R$.
\end{itemize}
\end{definition}
In the followings, we prove three properties: the first one proves that sequential raising fans $\clubsuit^R$ form a contiguous path in $\clubsuit_{\mathcal{B}}$. Properties \ref{prop:left-contiguous} and \ref{prop:right-contiguous} state that for each vertex $v$, all rightward-outer fans for children of a vertex $v\notin \mathcal{B}$ form a contiguous subsequence in $\clubsuit^R$.
Also, all rightward fans for a vertex $v\notin \mathcal{B}$ also form contiguous subsequence in $\clubsuit^R$.
For any region $\mathcal{W}_i\in \tilde{\mathcal{W}}(\tilde{\mathcal{F}^R_i})$,
$\mathcal{W}_i$ has three different types: the type one is the $\mathcal{W}_i$'s root is at a vertex on the left boundary of $\mathcal{W}_{i-1}$. The type two is that $\mathcal{B}^L(\mathcal{W}_i)\cap \mathcal{B}^L(\mathcal{W}_{i-1})$ and $\mathcal{B}^R(\mathcal{W}_i)\cap \mathcal{B}^R(\mathcal{W}_{i-1})$ are sub-paths of $\mathcal{B}^L(\mathcal{W}_{i-1})$ and $\mathcal{B}^R(\mathcal{W}_{i-1})$ from the $\mathcal{W}_{i-1}$'s root, respectively. The type three is that the $\mathcal{W}_i$'s is at a vertex on the right boundary of $\mathcal{W}_{i-1}$.
In the sequential regions $\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_i)$, types one, two and three are orderly appeared from left to right.
Notes that the third type of region can bound contiguous raising fans of $\clubsuit^R$.
Given a region $\mathcal{W}_i\in \tilde{\mathcal{F}}^R_i$, the region $\mathcal{W}_i$ is called a \emph{black-hole} if $\mathcal{W}_i$ is the type two.
Intuitively, a black-hole is a region such that it can bound a contiguous subsequence $(\tilde{\mathcal{F}}^R_{i+1}, \tilde{\mathcal{F}}^R_{i+2}, \cdots, \tilde{\mathcal{F}}^R_q)$ in $\clubsuit^R$.
The following observation states that for each raising fan in $\clubsuit^R$, there is the only one region which can be a black-hole.
Note that for a region $\mathcal{W}_i$ in a raising fan $\tilde{\mathcal{F}}^R_i$.
If the $\mathcal{W}_i$'s root is at a vertex on the left boundary of $\mathcal{W}_{i-1}$ or on the right boundary of $\mathcal{W}_{i-1}$, then $\mathcal{W}_i$ cannot consist of any raising fan in $\clubsuit^R$.
Hence the only region in $\tilde{\mathcal{F}}_i$ which can consist of contiguous raising fans $(\tilde{\mathcal{F}}^R_{i+1}, \tilde{\mathcal{F}}^R_{i+2}, \cdots, \tilde{\mathcal{F}}^R_q)$ in $\clubsuit^R$ is the $\tilde{\mathcal{F}}$'s black-hole.
\begin{observation}\label{obs:black-hole}
If a region $\mathcal{W}_i\in \tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_i)$ is a black-hole, then the region $\mathcal{W}_i$ bounds subsequential raising fans $(\tilde{\mathcal{F}}^R_{i+1}, \tilde{\mathcal{F}}^R_{i+2}, \cdots, \tilde{\mathcal{F}}^R_q)\subseteq \clubsuit^R$
where $(\tilde{\mathcal{F}}^R_{i+1}, \tilde{\mathcal{F}}^R_{i+2}, \cdots, \tilde{\mathcal{F}}^R_q)\subseteq \clubsuit^R$
form a contiguous subpath in $\clubsuit^R$ and $\clubsuit_{\mathcal{B}}$.
And, $\mathcal{W}_i$ is the only one black-hole in the raising fan $\tilde{\mathcal{F}}^R_i$.
\end{observation}
\begin{property}
Sequential raising fans $\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$
are an ancestor-descendant path in $\clubsuit_{\mathcal{B}}$
such that
for each $1\leq i\leq q-1$,
a raising fan $\tilde{\mathcal{F}}^R_i$ is the parent
of $\tilde{\mathcal{F}}^R_{i+1}$ in $\clubsuit_{\mathcal{B}}$.
($\tilde{\mathcal{F}}^R_i$ bounds $\tilde{\mathcal{F}}^R_{i+1}$.)
\end{property}
\begin{proof}
From Observation \ref{obs:black-hole},
we know that for each raising fan $\tilde{\mathcal{F}}^R_i, 1\leq i\leq q$, there is the only one black-hole in the raising fan $\tilde{\mathcal{F}}^R_i$.
Then
for each raising fan $\tilde{\mathcal{F}}^R_i, 1\leq i\leq q-1$,
$\tilde{\mathcal{F}}^R_i$ bounds the raising fan $\tilde{\mathcal{F}}^R_{i+1}$ in $\clubsuit^R$ and $\tilde{\mathcal{F}}^R_i$ is the parent of $\tilde{\mathcal{F}}^R_{i+1}$ in $\clubsuit_{\mathcal{B}}$.
Hence we can prove that $\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)
$ is a path in $\clubsuit_{\mathcal{B}}$.
\end{proof}
\begin{property}\label{prop:left-contiguous}
Sequential raising fans $\clubsuit^R$ have the following property:
for each vertex $v \notin \mathcal{B}$ in the right wing of a raising fan $\tilde{\mathcal{F}} \in \clubsuit^R$,
\begin{enumerate}
\item all rightward fans for the vertex $v$ form at most one raising fan in $\clubsuit^R$ and
\item all rightward fans for $v$'s children form at most one raising fan in $\clubsuit^R$.
\end{enumerate}
\end{property}
\begin{proof}
Let $v$ be a vertex at the right boundary of $\tilde{\mathcal{F}}$'s right wing.
All rightward fans for the vertex $v$ and rightward-outer fans for $v$'s children are consisted in at most one raising fan $\tilde{\mathcal{F}}' \in \clubsuit^R$.
We know that no any raising fan in $\clubsuit^R$ can be inside a region formed by $\tilde{\mathcal{F}}'$'s left wing and it leads that the right boundary of $\tilde{\mathcal{F}}$'s left wing cannot overlap $\tilde{\mathcal{F}}'$'s right wing.
Hence we can conclude that for each vertex $v\notin \mathcal{B}$ that $v$ is a vertex of the right wing of $\tilde{\mathcal{F}} \in \clubsuit^R$,
all rightward fans for the vertex $v$ and all rightward-outer fans for $v$'s children form at most one raising fan $\tilde{\mathcal{F}}'$ in $\clubsuit^R$.
\end{proof}
\begin{property}\label{prop:right-contiguous}
Sequential raising fans $\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$ have the following property:
for each vertex $v\notin \mathcal{B}$ in the right wing of a raising fan $\tilde{\mathcal{F}}^R_i\in \clubsuit^R$,
all $u$'s leftward-outer fans form contiguous raising fans $(\tilde{\mathcal{F}}^R_i, \tilde{\mathcal{F}}^R_{i+1}, \cdots, \tilde{\mathcal{F}}^R_a) \in \clubsuit^R$.
\end{property}
\begin{proof}
Let $\mathcal{W}'$ be a black-hole passes through a vertex $u$ where the vertex $u$ is on the right boundary $\mathcal{B}^R(\mathcal{W}')$ of $\mathcal{W}'$.
Let $\tilde{\mathcal{F}}^R_{a_1}$ be the first raising fan in $\clubsuit^R$ such that the black-hole $\mathcal{W}_{a_1} \in \tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_{a_1})$ passes through the vertex $u$.
Observe that all $u$'s leftward-outer fans are shared by a maximal contiguous black-holes $(\mathcal{W}_{a_1}, \mathcal{W}_{a_1+2}, \cdots, \mathcal{W}_{a_2})$ where
each black-hole $\mathcal{W}_j, a_1\leq j\leq a_2$, passes through the vertex $u$ till the black-hole $\mathcal{W}_{a_2+1}\in \tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_{a_2+1})$ doesn't pass through the vertex $u$.
Now we know that there are sequential raising fans $(\tilde{\mathcal{F}}^R_{a_1}, \tilde{\mathcal{F}}^R_{a_1+1}, \cdots, \tilde{\mathcal{F}}^R_{a_2}) \subseteq \clubsuit^R$ such that
each raising fan $\tilde{\mathcal{F}}^R_i, a_1\leq i\leq a_2,$ consists of a black-hole $\mathcal{W}_i$.
When the black-hole in
$\tilde{\mathcal{F}}^R_{a_2+1}$ doesn't passes through the vertex $u$,
each raising fan $\tilde{\mathcal{F}}^R_j, a_2+1\leq j\leq q$, cannot consists of any $u$'s leftward-outer fan.
Hence we can prove that $u$'s leftward-outer fans are shared by contiguous raising fans $(\tilde{\mathcal{F}}^R_{a_1}, \tilde{\mathcal{F}}^R_{a_1+1}, \cdots, \tilde{\mathcal{F}}^R_{a_2}) \subseteq \clubsuit^R$.
\end{proof}
Now we can construct sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$ from the sequential raising fans $(\clubsuit^L, \clubsuit^R)$ in a region $\mathcal{W}$ as follows:
firstly, because the sequential raising fans $(\tilde{\mathcal{F}}^L_1, \tilde{\mathcal{F}}^L_2, \cdots, \tilde{\mathcal{F}}^L_p, \tilde{\mathcal{F}}^R_1)$ are mutually disjoint (they don't have any ancestor-descendant relation in $\clubsuit_{\mathcal{B}}$), the sequential raising fans $(\tilde{\mathcal{F}}^L_1, \tilde{\mathcal{F}}^L_2, \cdots, \tilde{\mathcal{F}}^L_p, \tilde{\mathcal{F}}^R_1)$ partition $\mathcal{W}$ into sequential disjoint regions $(\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^L_1),$ $ \tilde{\mathcal{W}}(\tilde{\mathcal{F}}^L_2),$ $\cdots,$ $\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^L_p),$ $\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_1))$.
Secondly, for each raising fan $\tilde{\mathcal{F}}^R_i, 2\leq i\leq q$, process the following steps:
let $\mathcal{W}_i$ be a region in $\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_i)$ bounds a raising fan $\tilde{\mathcal{F}}^R_{i+1}$.
Then replace $\mathcal{W}_i$ by $\tilde{\mathcal{W}}(\tilde{\mathcal{F}}^R_{i+1})$.
After the last raising fan $\tilde{\mathcal{F}}^R_q$ is processed,
we get sequential disjoint regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$.
Now we show how to place sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$ in $\mathcal{H}$ as $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$ and
utilize it to place sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}})$ in $\mathcal{H}$ as $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}})$
such that $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}})$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\begin{enumerate}
\item Firstly, we place the sequential regions $(\tilde{\mathcal{W}}(\mathcal{F}^L_1), \tilde{\mathcal{W}}(\mathcal{F}^L_2), \cdots, \tilde{\mathcal{W}}(\mathcal{F}^L_p), \tilde{\mathcal{W}}(\mathcal{F}^R_1))$ as $(\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{F}^L_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{F}^L_2),$ $\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{F}^L_p),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\mathcal{F}^R_1))$ in $\mathcal{H}$.
\item Because each raising fan $\tilde{\mathcal{F}}^R_i, 2\leq i\leq q$, is inside a region $\mathcal{W}_i$ of $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_{i-1})$, we place the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_i)$ at right of the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_{i-1})$ in $\mathcal{H}$.
\end{enumerate}
Now we have new sequential regions in $\mathcal{H}$ as follows: $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))=$
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_p),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$.
And, we immediately have the following lemma.
\begin{lemma}\label{lem:partition-boundary}
Regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$
are sequential in $\mathcal{H}$.
\end{lemma}
The next lemma proves that
the sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))=$
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_p),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$
are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\begin{lemma}\label{lem:layout-boundary}
Sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}))$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{lemma}
\begin{proof}
Recall that $\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$ are sequentially placed in $\mathcal{H}$
as $(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1)$, $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$ $\cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$
such that each raising fan $\tilde{\mathcal{F}}^R_i, 2\leq i\leq q$, is at right of $\tilde{\mathcal{F}}^R_{i-1}$ in $\mathcal{H}$.
In the followings, we utilize
Property \ref{prop:left-contiguous} to prove that
for each vertex $v\notin \mathcal{B}$,
edges between $v$ and $v$'s children have constant
$X$-crossing edges with other edges in $(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1)$, $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$ $\cdots,$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$.
Because all $v$' rightward-outer fans form contiguous raising fans $(\tilde{\mathcal{F}}_i, \tilde{\mathcal{F}}_{i+1})$
with length at most two in $(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$.
The edges between $v$ and $v$'s children make $X$-crossing edges with
the only raising fan $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}_{i+1})$ in $\mathcal{H}$.
Hence the edges between $v$ and $v$'s children have $X$-crossing number at most one.
As we place $\clubsuit^R=(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$ as
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$
in $\mathcal{H}$,
we have sequential edges
$(\tilde{e}_1, \tilde{e}_2, \cdots, \tilde{e}_{q-1})$ that
each edge $e\in \tilde{e}_i, 1\leq i\leq q-1$, connects between two raising fans $\tilde{\mathcal{F}}^R_i$ and
$\tilde{\mathcal{F}}^R_{i+1}$ except for
edges between $v$ and $v$'s children,
Observe that
$(\tilde{e}_1, \tilde{e}_2, \cdots, \tilde{e}_{q-1})$
are orderly placed in $\mathcal{H}$.
So, there is no any $X$-crossing edge among
$(\tilde{e}_1, \tilde{e}_2, \cdots, \tilde{e}_{q-1})$.
Since any two raising fans in $(\tilde{\mathcal{F}}^L_1,$ $\tilde{\mathcal{F}}^L_2,$
$\cdots,$
$\tilde{\mathcal{F}}^L_p)$ are siblings in $\clubsuit_{\mathcal{B}}$ (two raising fans $\tilde{\mathcal{F}}$ and $\tilde{\mathcal{F}}'$ are siblings in $\clubsuit_{\mathcal{B}}$, $\tilde{\mathcal{F}}$ and $\tilde{\mathcal{F}}'$ are not bounded to each other),
the placement: $(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_1),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_2),$ $\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_p))$
are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
Also,
because $(\tilde{\mathcal{F}}^L_1, \tilde{\mathcal{F}}^L_2, \cdots, \tilde{\mathcal{F}}^L_p)$ and $(\tilde{\mathcal{F}}^R_1, \tilde{\mathcal{F}}^R_2, \cdots, \tilde{\mathcal{F}}^R_q)$ are not bounded to each other,
except for the edges between a
vertex on a black-hole,
the placement: $(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_1), \tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_2), \cdots, \tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_p))$
and
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$ $\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$
are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
In the followings, we utilize Property \ref{prop:right-contiguous} to prove that
for each vertex $v\notin \mathcal{B}$,
all $v$'s leftward-outer fans have constant number of
$X$-crossing edges in $\mathcal{H}$.
Let $v$ be a vertex on the right boundary of a black-hole.
By Property \ref{prop:right-contiguous},
there exists contiguous raising fans $(\tilde{\mathcal{F}}^R_i, \tilde{\mathcal{F}}^R_{i+1}, \cdots, \tilde{\mathcal{F}}^R_a) \subseteq \clubsuit^R$ which consists of $v$'s leftward-outer fans.
All edges between $v$ and $v$'s children cross from the sequential raising fans
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_i),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_{i+1}),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_a))$ in $\mathcal{H}$.
Let the sequential vertices $(v_1, v_2, \cdots, v_h)$ be orderly on a track in $\mathcal{H}$
such that
each vertex $v_j, j\geq 1,$ consists of some leftward-outer fans.
Let $(\mathcal{W}_1, \mathcal{W}_2, \cdots, \mathcal{W}_h)$ be sequential black-holes in $\mathcal{H}$ such that
each vertex $v_j, 1\leq j\leq h$, is on the right boundary of $\mathcal{W}_j$.
We know that
for each vertex $v_j, 1\leq j\leq h$, there are contiguous raising fans
$(\tilde{\mathcal{F}}_{a_j},$ $\tilde{\mathcal{F}}_{a_j+1},$
$\cdots$
$\tilde{\mathcal{F}}_{b_j})$
passing through $v_j$.
Also, since the sequential black-holes $(\mathcal{W}_1, \mathcal{W}_2, \cdots, \mathcal{W}_h)$ are disjoint, we have the following ordered relation
$(a_1 < a_2 \cdots < a_h)$.
Now,
the sequential vertices $(v_1, v_2, \cdots, v_h)$ are orderly placed on a track in $\mathcal{H}$ and
the sequential children $(\mathcal{C}_{v_1}, \mathcal{C}_{v_2}, \cdots, \mathcal{C}_{v_h})$ of $(v_1, v_2, \cdots, v_h)$
are also orderly placed on other track in $\mathcal{H}$
because the order relation
$(a_1\leq b_1 < a_2 \leq b_2 \cdots < a_h \leq b_h)$.
Hence for all sequential edges
$(\tilde{e}(v_1),$
$\tilde{e}(v_2),$
$\cdots,$
$\tilde{e}(v_h))$
where each edges $\tilde{e}(v_j), 1\leq j\leq h$, are edges between
the vertex $v_j$ and and $v_j$'s children $\mathcal{C}_{v_j}$,
$(\tilde{e}(v_1),$
$\tilde{e}(v_2),$
$\cdots,$
$\tilde{e}(v_h))$ are not $X$-crossing in $\mathcal{H}$.
Finally, when we add the sequential edges
$(\tilde{e}(v_1), \tilde{e}(v_2), \cdots, \tilde{e}(v_h))$ into $\mathcal{H}$,
the $X$-crossing number in $\mathcal{H}$ increase one in $\mathcal{H}$
because the sequential edges $(\tilde{e}(v_1), \tilde{e}(v_2), \cdots, \tilde{e}(v_h))$
make $X$-crossing edges with the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots, \mathcal{W}_h)$ in $\mathcal{H}$.
and
the layout:
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_2),$
$cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^L_p),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_1),$ $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_2),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}}^R_q))$
are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{proof}
\begin{lemma}\label{lem:full-layout-boundary}
Sequential regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}})$ are $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{lemma}
\begin{proof}
Initially, the root $r$'s raising fan $\tilde{\mathcal{F}}$ in a forest-like structure $\clubsuit_{\mathcal{B}}$ form sequential
regions $\tilde{\mathcal{W}}_{\mathcal{H}}(\tilde{\mathcal{F}})$ in $\mathcal{H}$.
Next each iteration of Algorithm \ref{alg:framework-boundary} picks up
the first region $\mathcal{W}_i$ in the first-in-first-out queue $\tilde{\mathcal{W}}$
to place the sequential regions $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ at the rightmost side in $\mathcal{H}$; The region $\mathcal{W}_i$ at the leftmost region consists of raising fans in $\clubsuit_{\mathcal{B}}$ in $\mathcal{H}$.
($|\clubsuit_{\mathcal{B}}(\mathcal{W}_i)|>0$.)
All regions $\mathcal{W}_j$ at left of $\mathcal{W}_i$ in $\mathcal{H}$ doesn't consist of any raising fan in $\clubsuit_{\mathcal{B}}$ ($|\clubsuit_{\mathcal{B}}(\mathcal{W}_j)|=0$) and the chords between $\mathcal{W}_j$ and $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ don't nest with
the chords between $\mathcal{W}_i$ and
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$
because $\mathcal{W}_j$ is at left of $\mathcal{W}_i$ in $\mathcal{H}$ and
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are at left of
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ in $\mathcal{H}$.
The above description also implies that
the left end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are placed at left of
the left end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$, and
the right end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_j))$ are placed at left of
the right end-vertices of the chords $\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ in $\mathcal{H}$.
Observe that if sequential regions
$(\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_1)),$
$\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_2)),$
$\cdots)$ are orderly placed in
$\mathcal{H}$ as
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_1)),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_2)),$
$\cdots)$ in Algorithm \ref{alg:framework-boundary}, then
the sequential regions $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$ are also orderly
placed in $\mathcal{H}$.
Let $(\mathcal{W}_1, \mathcal{W}_2, \cdots)$
be sequential regions in $\mathcal{H}$
and $(\tilde{e}_1, \tilde{e}_2, \cdots)$ be
sequential chords in $\mathcal{H}$
where $\tilde{e}_i, i\geq 1,$ are edges between $\mathcal{W}_i$ and $\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$.
When we place a new sequential regions
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i))$ at right of
the sequential regions
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_1)),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_2)),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_{i-1})))$,
the sequential chords $(\tilde{e}_1,$
$\tilde{e}_2,$
$\cdots,$
$\tilde{e}_i)$
don't nest to each other in $\mathcal{H}$
because
the order of $(\mathcal{W}_1,$
$\mathcal{W}_2,$
$\cdots,$
$\mathcal{W}_i)$ in $\mathcal{H}$ is the same as the order of
$(\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_1)),$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_2)),$
$\cdots,$
$\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}}(\mathcal{W}_i)))$ in $\mathcal{H}$.
Hence the layout in Algorithm \ref{alg:framework-boundary}
is $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$.
\end{proof}
In Algorithm \ref{alg:framework-boundary}, we place a forest-like raising fans $\clubsuit_{\mathcal{B}}$ of a boundary $\mathcal{B}$ in a ladder $\mathcal{H}$ where the input $\clubsuit_{\mathcal{B}}$ consists of the only one root.
However, $\clubsuit_{\mathcal{B}}$ would be a forest with sequential roots $(r_1, r_2, \cdots, r_s)$,
we can slightly modified Algorithm \ref{alg:framework-boundary} as follows:
if $\clubsuit_{\mathcal{B}}$ is a forest with the sequential roots $(r_1, r_2, \cdots, r_s)$,
we can orderly place the sequential regions
$(\tilde{\mathcal{W}}_{\mathcal{H}}(r_1), \tilde{\mathcal{W}}_{\mathcal{H}}(r_2), \cdots, \tilde{\mathcal{W}}_{\mathcal{H}}(r_s))$ in $\mathcal{H}$.
\begin{theorem}
Given a region $\mathcal{W}=(\mathcal{B}^L, \mathcal{B}^R)$, the collection of all rightward-outer and leftward-outer fans from the boundaries $\mathcal{B}^L$ and $\mathcal{B}^R$
can be $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in $\mathcal{H}$ and
there are sequential regions $(\tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}^L}), \mathcal{W}^M, \tilde{\mathcal{W}}_{\mathcal{H}}(\clubsuit_{\mathcal{B}^R}))$ in $\mathcal{H}$
where $\mathcal{W}^M$ is a region between
the rightmost boundary of $\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}^L})$
and the leftmost boundary of
$\tilde{\mathcal{W}}(\clubsuit_{\mathcal{B}^R})$.
Moreover, the union of the two forest-like structures
$\clubsuit_{\mathcal{B}^L}$ and $\clubsuit_{\mathcal{B}^R}$ is a skeleton $\Psi(\mathcal{W})$ of $\mathcal{W}$.
\end{theorem}
\begin{figure}[t]
\begin{center}
\includegraphics[width=1\textwidth, angle =0]{ugly-edges-layout}
\centering
\caption{(1): $(\triangledown_1, \triangledown_2, \triangledown_3)$ are sequential down-pointing triangles that $\triangledown_1=(a_3, m_0, a_4, m_1)$,
$\triangledown_2=(a_2, a_6 m_2)$ and $\triangledown_3=(a_1, a_7, m_3)$.
The middle path $\mathcal{M}=(m_0, m_1, m_2, m_3)$ partition $(\triangledown_1, \triangledown_2, \triangledown_3)$ into two disjoint parts.
By our algorithm, $(b_1, \cdots, b_4, m_1)$, $(c_1, \cdots, c_4, m_2)$ and $(d_1, \cdots, d_4, m_3)$ are placed on a track as the order: $(d_1, \cdots, d_4, m_3, c_1, \cdots, c_4, m_2, b_1, \cdots, b_4, m_1)$.
The three sequential vertices $(b_5, \cdots, b_7)$, $(c_5, \cdots, c_8)$ and $(d_5, \cdots, d_8)$ are placed on a track as the order: $(d_8, \cdots, d_5, c_8, \cdots, c_5, b_7, \cdots, d_5)$;
(2): $m_3$'s left piles are
$\{(m_3, c_1), (m_3, c_2), (m_3, c_3), (m_3, c_4)\}$ and
$m_2$'s left piles are $\{(m_2, b_1), (m_2, b_2), (m_2, b_3),$ $(m_2, b_4)\}$. Because $(m_3, m_2, m_1)$ are orderly placed in $\mathcal{H}$ and the sequential vertices $(c_1, c_2, c_3, c_3, c_4)$ are placed at left of the sequential vertices $(b_1, b_2, b_3, b_4)$, $m_2$'s left piles cannot make nested with $m_3$'s left piles in $\mathcal{H}$.
Similarly,
$m_3$'s right piles are $\{(m_3, c_5), (m_3, c_6), (m_3, c_7), (m_3, c_8)\}$, and
$m_2$'s right piles are $\{(m_2, b_5), (m_2, b_6), (m_2, b_7)\}$.
Because the sequential vertices $(c_5, c_6, c_7, c_8)$ are placed at left of the sequential vertices $(b_5, b_6, b_7)$ in $\mathcal{H}$,
$m_2$'s right piles cannot make nested with $m_3$'s right piles in $\mathcal{H}$;
(3): The bridges $((d_2, d_4), (c_2, c_4), (b_2, b_4))$ at left of the middle path $\mathcal{M}$ are orderly placed on a track in $\mathcal{H}$. Hence
$((d_2, d_4), (c_2, c_4), (b_2, b_4))$ cannot form nested chords on a track in $\mathcal{H}$.
The bridges $((m_3, d_7), (m_2, c_7), (m_1, b_6))$ that are connecting to middle path $\mathcal{M}$ are also orderly placed, so $((m_3, d_7), (m_2, c_7), (m_1, b_6))$ cannot form nested chords on a track in $\mathcal{H}$;
(4): The chord $((d_1, d_3), (c_1, c_3), (b_1, b_3))$ at left of the middle path $\mathcal{M}$ are orderly placed on a track in $\mathcal{H}$.
The right chord $((d_7, d_8), (c_7, c_8), (b_6, b_7))$ at right of middle path are also orderly placed on a track in $\mathcal{H}$.
The chord $((d_4, d_6), (c_4, c_6), (b_4, b_5))$ across the middle path $\mathcal{M}$ are not nested on a track in $\mathcal{H}$ because both of the sequential vertices $(d_4, c_4, b_4)$ and
$(d_6, c_6, b_5)$ are orderly placed on a track in $\mathcal{H}$.
}
\label{fig:ugly-edges-layout}
\end{center}
\vspace{-0.2in}
\end{figure}
\section{Deleted Edges Increase $X$-Crossing, Queue and Gap Numbers Sightly}
In this section, we explain why our layout is still $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed after deleted edges are re-added into our layout.
Recall that in the Section \ref{sec:transformation}, for a cycle ${\cal O}$,
the cycle ${\cal O}$ is clockwisely placed from the vertex $m$ of ${\cal O}$ in a composite-layerlike graph $\mathcal{G}$.
And,
for a down-pointing triangle $\triangledown$,
the down-pointing triagle $\triangledown$ is clockwisely placed from the lower vertex $m$ of $\triangledown$ where
all vertices except the lower vertex $m$ of $\triangledown$ are placed at a upper layer
and the lower vertex $m$ is placed at a lower layer of a composite-layerlike graph $\mathcal{G}$.
Also, there exist sequential maximal inner cycles $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$ inside ${\cal O}$ or $\triangledown$;
The spines of the cycle ${\cal O}$ or
the down-pointing triangle $\triangledown$.
And, the leftmost and rightmost vertices of the $i$-th cycle ${\cal O}_i \in ({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$ are the $i$-th joint of the spine.
Moreover,
for the $i$-th cycle ${\cal O}_i \in ({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$,
the $i$-th \emph{hoop}
$(u_i, u'_i)$ on the cycle ${\cal O}$ is defined that the vertices $u_i$ and $u'_i$ are the parents of the $i$-th joint of the spine of the cycle ${\cal O}$ or the down-pointing triangle $\triangledown$.
The vertex $m$ is called the \emph{bad} vertex of a cycle ${\cal O}$
or a down-pointing triangle $\triangledown$.
\begin{definition}
For a cycle ${\cal O}$,
a deleted edge on the bad vertex $m$ of the cycle ${\cal O}$ are called a \emph{wire} of $m$ inside ${\cal O}$ denoted as $\curlywedge(m)$. Also, a removed edge $e_i, 1\leq i\leq p-1,$ which connects between two contiguous maximal inner cycles ${\cal O}_i$ and ${\cal O}_{i+1}$ of the spine is called a \emph{bridge}.
\end{definition}
Recall that in Algorithm \ref{alg:framework},
we simultaneously
pick all joints of ${\cal O}_i, 1\leq i\leq p$ and their hoops
$\{(u_1, u'_1), (u_2, u'_2), \cdots, (u_p, u'_p)\}$
on the two contiguous tracks in $\mathcal{H}$.
Also, we order all hoops $\{(u_1, u'_1),$
$(u_2, u'_2),$
$\cdots,$
$(u_p, u'_p)\}$ as ordering the lower boundary $\{L^B({\cal O}_1), L^B({\cal O}_2), \cdots, L^B({\cal O}_p)\}$ of the ${\cal O}$'s spine.
Moreover, for all hoops $((u_1, u'_1),$
$(u_2, u'_2),$
$\cdots,$
$(u_p, u'_p))$ and the lower boundary
$\{L^B({\cal O}_1),$
$L^B({\cal O}_2),$
$\cdots,$
$L^B({\cal O}_p)\}$ of the spine, we place the two sequential vertices contiguously on any track in $\mathcal{H}$.
Hence we have the following observations that state the key reasons why wires cannot make $X$-crossing in our layout.
\begin{observation}
Given a bad vertex $m$ on a cycle ${\cal O}$,
let $\{{\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p\}$ be ${\cal O}$'s spine and $\{(u_1, u'_1),$
$(u_2, u'_2),$
$\cdots,$
$(u_p, u'_p)\}$
be corresponding hoops on ${\cal O}$, the layout in $\mathcal{H}$ has the following properties:
\begin{enumerate}
\item all hoops are placed contiguously on the same track in $\mathcal{H}$,
\item all joints are placed contiguously on a track in $\mathcal{H}$,
\item the order of all joints on a track is the same as the order of all hoops on a track in $\mathcal{H}$. And,
\item all hoops and all joints are placed at two contiguous tracks in $\mathcal{H}$.
\end{enumerate}
\end{observation}
\begin{observation}
Let $m$ and $m'$ be bad vertices on cycles ${\cal O}$
and ${\cal O}'$, respectively.
The bad vertex $m$ is placed at left of the bad vertex $m'$ on a track in $\mathcal{H}$ if and only if
the spine of ${\cal O}$ is placed at
left of the spine of ${\cal O}'$ at any track in $\mathcal{H}$.
\end{observation}
\begin{observation}
For a cycle ${\cal O}$ with the bad vertex $m$,
\begin{enumerate}
\item the gap number between the bad vertex $m$ and any vertex on the lower boundary $\{L^B({\cal O}_1),$
$L^B({\cal O}_2),$
$\cdots,$
$L^B({\cal O}_p)\}$ of the ${\cal O}$'s spine is at most $2\mathcal{J}$, and
\item each lower boundary $L^B({\cal O}_i),$
$1\leq i\leq p$
except its joint
is placed contiguously on a track in $\mathcal{H}$.
\end{enumerate}
\end{observation}
For any two bad vertices $m_1$ and $m_2$ that $m_1$ is at left of $m_2$ on a track in $\mathcal{H}$,
the wires $\curlywedge(m_1)$ are placed at left of the wires $\curlywedge(m_2)$ in $\mathcal{H}$.
So, there is no any $X$-crossing edge between $\curlywedge(m_1)$ and $\curlywedge(m_2)$.
From the above observations,
we have the following lemma:
\begin{lemma}\label{lem:vertical-X-crossing}
Suppose sequential bad vertices $(m_1, m_2, \cdots, m_p)$ are placed from left to right on a track in $\mathcal{H}$,
the sequential wires $(\curlywedge(m_1),$ $\curlywedge(m_2),$ $\cdots,$ $\curlywedge(m_p))$ are not $X$-crossing in $\mathcal{H}$.
\end{lemma}
\begin{definition}
Suppose (1) $\triangledown$ is a down-pointing triangle with the bad vertex $m$ and
(2) the down-pointing triangle $\triangledown$ is in a fan $\mathcal{F}$ of a raising fan $\tilde{\mathcal{F}}$ with the middle path $\mathcal{M}$,
\begin{enumerate}
\item a left (right, respectively) \emph{pile} of the bad vertex $m$
is defined as an edge connecting between the bad vertex $m$ and a vertex in the lower boundary $\{L^B({\cal O}_1), L^B({\cal O}_2), \cdots, L^B({\cal O}_p)\}$ of the spine of $\triangledown$ that
is at left (right, respectively) of the middle path $\mathcal{M}$.
A left (right, respectively) pile with respect to the middle path $\mathcal{M}$ is denoted to $\curlyvee^L(m)$ ($\curlyvee^R(m)$, respectively).
\item
A left (right, respectively) spine of the down-pointing triangle $\triangledown$ with respect to the middle path $\mathcal{M}$ is the subsequential spine of the down-pointing triangle $\triangledown$ at left (right, respectively) of the middle path $\mathcal{M}$.
\item
A left (right, respectively) hoop with respect to the middle path $\mathcal{M}$ is a hoop of the down-pointing triangle $\triangledown$ at left (right, respectively)
of the middle path $\mathcal{M}$.
\end{enumerate}
\end{definition}
Suppose $\triangledown'$ is a down-pointing triangle with the bad vertex $m'$
inside the down-pointing triangle $\triangledown$.
Then
the sequential regions $\tilde{\mathcal{W}}(m)$ consisting of all $m$'s left joints are placed at left of the sequential regions $\tilde{\mathcal{W}}(m')$ consisting of all $m'$'s left joints.
Because we place the bad vertex $m$ at left of the bad vertex $m'$ on a track in $\mathcal{H}$ and
all $m$'s joints at left of $m'$'s hoops on a track in $\mathcal{H}$,
we can have that the left piles $\curlyvee^L(m)$ are not nested with the left piles $\curlyvee^L(m')$ on any track in $\mathcal{H}$.
Similarly,
the sequential regions $\tilde{\mathcal{W}}(m)$ consisting of all $m$'s right joints are at left of the sequential regions $\tilde{\mathcal{W}}(m')$ consisting of all $m'$'s right joints.
Because we place the bad vertex $m$ at left of
bad vertex $m'$ on a track in $\mathcal{H}$ and
place
all $m$'s right joints at left of
all $m'$'s right joints on a track in $\mathcal{H}$,
there is no any nested edge between
all right piles $\curlyvee^R(m)$ of the bad vertex $m$ and all right piles $\curlyvee^R(m')$ of the bad vertex $m'$.
Now we can the following observations:
\begin{observation}
Suppose raising down-pointing triangles $(\triangledown_1, \triangledown_2, \cdots, \triangledown_p)$ and their sequential bad vertices
$\mathcal{M}=(m_1, m_2, \cdots, m_p)$ are placed on the same track in $\mathcal{H}$,
\begin{enumerate}
\item the sequential left joints of the sequential down-pointing triangles $(\triangledown_1, \triangledown_2, \cdots, \triangledown_p)$ are orderly placed at a track in $\mathcal{H}$. And,
\item the sequential right joints of the sequential down-pointing triangles $(\triangledown_1, \triangledown_2, \cdots, \triangledown_p)$ are orderly placed at a track in $\mathcal{H}$.
\end{enumerate}
\end{observation}
\begin{observation}
For a down-pointing triangle $\triangledown$ with the bad vertex $m$,
\begin{enumerate}
\item the gap number between the bad vertex $m$ and any vertex on the lower boundary
$\{L^B({\cal O}_1),$
$L^B({\cal O}_2),$
$\cdots,$
$L^B({\cal O}_p)\}$ of the $\triangledown$'s spine is at most $2\mathcal{J}$, and
\item each lower boundary $L^B({\cal O}_i), 1\leq i\leq p$
except its joint is placed contiguously on a track in $\mathcal{H}$.
\end{enumerate}
\end{observation}
From the above observations, we can have that the right piles $\curlyvee^R(m_i)$ are not nested with the left piles $\curlyvee^R(m_j)$ on any track in $\mathcal{H}$ in the following lemma:
\begin{lemma}\label{lem:pile-nested}
Given sequential bad vertices $(m_1, m_2, \cdots, m_p)$ orderly placed on a track in $\mathcal{H}$,
their sequential left and right piles $(\curlyvee^L(m_1), \curlyvee^L(m_2), \cdots, \curlyvee^L(m_p))$ and $(\curlyvee^R(m_1), \curlyvee^R(m_2), \cdots, \curlyvee^R(m_p))$ are not nested on any track in $\mathcal{H}$.
\end{lemma}
For a cycle ${\cal O}$,
all joints of the spine of the cycle are orderly placed on any track
in $\mathcal{H}$,
all bridges of the spine cannot have nested chords on any track in $\mathcal{H}$.
Similarly,
for a down-pointing triangle $\triangledown$,
all left and right joints of the left and right spines of the cycle are orderly placed on any track in $\mathcal{H}$, respectively,
all bridges of the spine cannot have nested chords on any track in $\mathcal{H}$.
From the above fact, we can have the following lemma:
\begin{lemma}\label{lem:bridge-nested}
Given a cycle ${\cal O}$ or a down-pointing triangle $\triangledown$ with their spine $({\cal O}_1, {\cal O}_2, \cdots, {\cal O}_p)$,
their sequential bridges $(e_1, e_2, \cdots, e_{p-1})$ are not nested on any track in $\mathcal{H}$
where $e_i, 1\leq i\leq p-1,$ is the bridge between
the cycles ${\cal O}_i$ and ${\cal O}_{i+1}$.
\end{lemma}
\begin{theorem}\label{thm:G1-well-placed}
Every $1$-subdivision plane graph $G^1$ can have an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout on constant number of tracks.
\end{theorem}
\begin{proof}
From Theorem \ref{thm:reform}, we know a plane graph $G$ can be reformed into a composite-layerlike graph $\mathcal{G}$.
From Lemmas \ref{lem:vertical-X-crossing}, \ref{lem:pile-nested}
and \ref{lem:bridge-nested}, deleted edges slightly increase $X$-crossing number in any two tracks and queue number in any track in $\mathcal{H}$.
Hence we conclude that
a plane graph $G$ can be $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed in a ladder $\mathcal{H}$.
Also, from Theorem \ref{thm:cons-track},
an $(\mathcal{Q}, \mathcal{X}, \mathcal{D})$-well-placed layout can
be wrapped into a ladder $\mathcal{H}$ on constant number of tracks.
\end{proof}
\bibliographystyle{plain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,195 |
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\section{Introduction}
The Hurewicz map
\begin{eqnarray}
\label{eqn:hurewicz}
\pi_* (\Omega^\infty
\Sigma^\infty X)
\rightarrow
H_* (\Omega^\infty \Sigma^\infty X; \field)
\end{eqnarray}
from the stable homotopy groups of a pointed space $X$ to the mod $2$ homology
of
the infinite loop space $QX := \Omega ^\infty \Sigma^\infty X$ is of
significant
interest; for
example, for $X=S^0$, so that $ \pi_* (Q S^0)$ gives the stable homotopy of the
sphere spectrum,
the spherical class conjecture predicts that the image in
positive degree consists only of the images of elements of Hopf invariant one
and those of Kervaire invariant one (see
\cite{H_spherical}, \cite{Curtis} and \cite{Wellington}, for
example). This conjecture is usually attributed either to Madsen or to Curtis.
The first named author (cf. \cite[Conjecture 1.1]{HT}) proposed the following
generalization
(which is related to a conjecture due to Peter Eccles cited in \cite{Z}).
Write
$Q_0X$ for the basepoint
component of $QX$ and henceforth always take homology (and cohomology) with
$\field$ coefficients.
\begin{conj}[The generalized spherical class conjecture]
\label{Conjecture_SC}
Let $X$ be a pointed CW-complex. Then the Hurewicz homomorphism $H:
\pi_{*}(Q_0
X) \to H_{*}(Q_0 X)$
vanishes on classes of $\pi_* (Q_0 X)$ of
Adams filtration greater than $2$.
\end{conj}
This paper is motivated by an algebraic version of the generalized spherical
class conjecture, stated below as Conjecture \ref{Conjecture_Hung}; the
main result, Theorem \ref{main_intro}, proves a weak algebraic version,
Conjecture \ref{weak_conjecture_intro}. These conjectures are due to the first
named author; to state them, we recall the Singer transfer and the
Lannes-Zarati morphism.
An algebraic model for the $\mathbb{Z}/2$-transfer $\Sigma^\infty \mathbb{R}P^\infty_+
\rightarrow \Sigma^\infty S^0$ (see \cite{MMM}, for example)
is given by the non-trivial extension class $e \in
\mathrm{Ext}^1_{\mathscr{A}} (\Sigma^{-1} \field, P_1)$, where $P_1 = H^* (\mathbb{R}P^\infty) \cong
\field [x]$.
This class is represented by the short exact sequence of $\mathscr{A}$-modules
\[
0 \longrightarrow P_1 \longrightarrow \widehat{P}_1
\longrightarrow\Sigma^{-1}\mathbb{F}_2 \longrightarrow 0,
\]
where $\widehat{P}_1$ denotes the submodule of elements of degree
$\geq -1$ in the algebra $\field [x^{\pm 1}]$ equipped with the structure of
$\mathscr{A}$-algebra extending that on $P_1$.
For a positive integer $s$, the $s$-fold iterated smash product of
the transfer is represented by the class $e^s \in
\mathrm{Ext}^s_{\mathscr{A}} (\Sigma^{-s} \field, P_s)$, given by the tensor product of $s$
copies of $e$, where $P_s \cong H^* (BV_s)
\cong P_1^{\otimes s}$.
Recall that the destabilization functor ${\mathcal{D}}$ from $\mathscr{A}$-modules to
unstable modules gives ${\mathcal{D}} N$, the
largest unstable quotient of the $\mathscr{A}$-module $N$.
There is a natural transformation ${\mathcal{D}} N \to \field \otimes_{\mathscr{A}} N$
for functors from $\mathscr{A}$-modules to $\mathscr{A}$-modules, where $ \field
\otimes_{\mathscr{A}} N$ has
trivial $\mathscr{A}$-module structure. This is obtained
by applying ${\mathcal{D}}$ to the quotient
$N \to \field \otimes_{\mathscr{A}} N$ and then composing with the canonical
inclusion:
\[
{\mathcal{D}} N \to {\mathcal{D}}(\field \otimes_{\mathscr{A}} N) \equiv (\field \otimes_{\mathscr{A}}
N)^{\geq 0}
\stackrel{\subset}{\longrightarrow}
\field \otimes_{\mathscr{A}} N.
\]
This passes to derived functors to give
$$
{\mathcal{D}}_s N \rightarrow \tor_s^{\mathscr{A}} (\field, N).
$$
The cap product with the algebraic transfer class $e^s$ induces a natural
transformation
$\cap e^s : {\mathcal{D}}_s (\Sigma^{-s} N) \rightarrow {\mathcal{D}} (P_s \otimes N)$ for any
$\mathscr{A}$-module $N$ and hence, for $M$ an unstable module,
\[
\alpha^M_s : {\mathcal{D}}_s (\Sigma^{-s} M) \rightarrow P_s \otimes M
.\]
Lannes and Zarati \cite{LZ} used this to relate the functors ${\mathcal{D}}_s$ to the
Singer
functors $R_s$. (As recalled in Section \ref{sect:background}, the Singer
functor $R_s$ is an exact
functor defined on the category of unstable modules and $R_s M$ is canonically
a
submodule of $P_s \otimes M$,
for $M$ an unstable module.) Namely, by \cite[Théorème 2.5]{LZ}, for $M$ an
unstable module, $\alpha^{\Sigma M}_s$ induces an isomorphism
\[
{\mathcal{D}}_s (\Sigma ^{1-s } M)
\stackrel{\cong}{\longrightarrow}
\Sigma R_s M .
\]
Hence, as in \cite{LZ}, there is a
natural homomorphism of $\mathscr{A}$-modules
$\Sigma R_s M \rightarrow \tor_s^\mathscr{A} (\field,
\Sigma^{1-s} M) \cong \Sigma \tor_s^\mathscr{A} (\field,
\Sigma^{-s} M) $ and thus
\[
\field \otimes_\mathscr{A} R_s M
\rightarrow
\tor_s^\mathscr{A} (\field,
\Sigma^{-s} M).
\]
The linear dual of this map:
\begin{eqnarray}
\label{eqn:dualLZ}
\mathrm{Ext}_{\mathscr{A}}^{s}(\Sigma^{-s} M, \mathbb{F}_2) \to (\field \otimes_\mathscr{A} R_s
M)^*
\end{eqnarray}
is called the Lannes-Zarati homomorphism. It corresponds to an associated
graded of the Hurewicz map (\ref{eqn:hurewicz})
when $M$ is the reduced cohomology of a pointed space $X$. (The proof of this
assertion is sketched in \cite{Lannes} and \cite{Goerss}; the Singer functors
arise in identifying $H^* (QX) $ in terms of $H^* (X)$.)
The first named author proposed the following algebraic version of the
generalized spherical class conjecture for $M=\field$ in \cite{H_spherical}
and, for any
unstable module $M$, in his seminar at the Vietnam National University (cf.
\cite[Conjecture 1.2]{HT}).
\begin{conj}
[The generalized algebraic spherical class conjecture]
\label{Conjecture_Hung}
The Lannes-Zarati homomorphism (\ref{eqn:dualLZ}) vanishes in positive stem,
for $s>2$ and any unstable $\mathscr{A}$-module $M$.
\end{conj}
The conjecture was established by the first named author and his co-workers
for
the case $M=\field$ for $s=3,4$, and $5$ respectively in
\cite{H_spherical} and \cite{Hung1999}, \cite{Hung2003}, and \cite{HQT}. That
the Lannes-Zarati morphism for $M=\field$ vanishes for $s>2$ on
the algebra decomposables of $\mathrm{Ext}_{\mathscr{A}}^s(\field,\field)$ was proved in
\cite{HP}.
The definition of Singer's algebraic transfer \cite{S} is similar to that of
the
Lannes-Zarati morphism, replacing the derived functors of
destabilization by $\tor^\mathscr{A}_s$. Namely, for an $\mathscr{A}$-module $N$, the cap
product with $e^s$ induces a natural transformation:
\[
\psi_s ^N : \tor_s^\mathscr{A} (\field, \Sigma^{-s} N)
\rightarrow
\field \otimes_\mathscr{A} (P_s \otimes N) .
\]
The dual of this map is called the Singer transfer.
By naturality of the cap product, this fits into the commutative diagram:
\[
\xymatrix{
{\mathcal{D}}_s (\Sigma^{-s} N)
\ar[r]^{\cap e^s}
\ar[d]
&
{\mathcal{D}}( P_s \otimes N)
\ar[d]
\\
\tor_s^\mathscr{A} (\field, \Sigma^{-s} N)
\ar[r]_{\psi_s^N}
&
\field \otimes_\mathscr{A} (P_s \otimes N)
}
\]
in which the vertical morphisms are the natural transformations.
In particular, for $N= \Sigma M$, where $M$ is an unstable module, the dual of
the composition of the Singer transfer with the Lannes-Zarati
morphism identifies (up to suspension) with the composite:
\[
R_s M \hookrightarrow P_s \otimes M \twoheadrightarrow \field \otimes_\mathscr{A}
(P_s \otimes M)
\]
of the canonical inclusion with the projection to $\mathscr{A}$-indecomposables.
A weak form of Conjecture~\ref{Conjecture_Hung} asserts that the $s$-th
Lannes-Zarati homomorphism vanishes for $s>2$ in any positive stem on the image
of the Singer transfer. This is equivalent to the following (see also
\cite[Conjecture 1.6]{HT}):
\begin{conj}
[The weak generalized algebraic spherical class conjecture]
\label{weak_conjecture_intro}
Let $M$ be an unstable $\mathscr{A}$-module and $s>2$ be an integer. Then every
positive degree element of
the Singer construction $R_sM$ is $\mathscr{A}$-decomposable in
$P_s\otimes M$.
\end{conj}
The conjecture was proved for $M=\widetilde{H}^*(S^0)$ and
$M=\widetilde{H}^*(BV_k)$ (for $k$ a non-negative integer) by
Tr{\`\acircumflex}n N. Nam and the first named author in \cite{HN} and
\cite{HN2} respectively.
Conjecture \ref{weak_conjecture_intro} is proved by the main result of the
paper:
\begin{Theorem}\label{main_intro}
If $s>2$, the morphism
\[
R_s M \rightarrow \field \otimes_{\mathscr{A}} (P_s \otimes M)
\]
is trivial on elements of positive degree, for
any unstable $\mathscr{A}$-module $M$.
\end{Theorem}
The condition $s > 2$ is necessary: in the case $M = \field$, the map is
non-zero for $s\in \{1, 2 \}$ (see \cite{H_spherical}). This phenomenon arises
from the existence of the spherical classes of Hopf invariant one and those of
Kervaire invariant one.
\begin{exam}
\label{exam:known}
\
\begin{enumerate}
\item
Taking $M=\field$ recovers the main result of \cite{HN};
\item
taking $M= P_k$ for $k$ a non-negative integer recovers the
main result of \cite{HN2}.
\end{enumerate}
\end{exam}
The proof of Theorem \ref{main_intro} builds upon the methods of
Tr{\`\acircumflex}n N. Nam and the first named author.
An important new ingredient is the usage of the Milnor operations $Q_0$,
$Q_1$,
which is a key element of the inductive strategy used in Section
\ref{sect:proofs}.
Moreover, the general result proved is stronger than obtained hitherto in the
cases of Example \ref{exam:known}, giving information on the Steenrod
operations which are required to hit elements (see Theorem
\ref{thm:hyper_refined} below).
\bigskip
\noindent
{\bf Organization:} background and references are provided in
Section \ref{sect:background} and
the results are stated in Section \ref{sect:results}, where the standard
reduction
arguments are explained. The first
result, concerning the image of the Steenrod total power, is proved
in Section \ref{sect:image_sts}; the
precise form of the main result is stated and proved in Section
\ref{sect:proofs}.
\section{Background}
\label{sect:background}
Fix $\field$ the prime field of characteristic two and write $\mathscr{A}$ for the
mod $2$ Steenrod algebra.
Let $\mathscr{A}(n)$ denote the subalgebra of $\mathscr{A}$
generated by $\{Sq^1, \ldots , Sq^{2^n} \}$
for $n \in \nat$, where $\nat$ denotes the set of non-negative integers.
Recall that
$\Phi $ denotes the doubling functor on the category $\mathscr{U}$ of unstable
modules (respectively unstable algebras,
$\mathscr{K}$) \cite{schwartz_book}.
Let $P_s=\field[x_1,\dots,x_s]$ denote the polynomial algebra on $s$ generators
of degree $1$, equipped with the usual actions of the
general linear group $GL_s=GL(s,\field)$ and of the Steenrod algebra. The rank
$s$ Dickson algebra,
$D_s$, is the algebra of invariants
$
\field[x_1,\dots,x_s]^{GL_s},
$
which is an unstable $\mathscr{A}$-algebra.
The underlying algebra is polynomial
\[
D_s \cong \field[c_{s,0}, \dots, c_{s,s-1}],
\]
where $c_{s,i}$ denotes the Dickson invariant of degree $2^s-2^i$
(for the action of the Steenrod algebra, see \cite{W}, \cite{H}).
The Dickson algebra $D_s$ occurs as the unstable
algebra $R_s \field$, where $R_s$ is the version of the Singer functor used by
Lannes and Zarati
\cite{LZ}. This is a functor
\[
R_s : \mathscr{U} \rightarrow D_s \mbox{-} \mathscr{U} \stackrel{\mathrm{forget}}{\rightarrow}
\mathscr{U}
\]
from unstable modules to the category $D_s\mbox{-}\mathscr{U}$ of $D_s$-modules in
unstable modules or, by forgetting the $D_s$-action, to unstable modules. For an
unstable module $M$,
there is a canonical inclusion $R_s M \hookrightarrow D_s \otimes M$ in
$D_s\mbox{-}\mathscr{U}$ and the image identifies with the $D_s$-module generated by the
image of
$\mathrm{St}_s : \Phi^s M \dashrightarrow P_s \otimes M$, the total Steenrod power map
\cite{LZ} (the dashed arrow indicates that this is linear but not $\mathscr{A}$-linear
in general and the iterated double $\Phi^s$ ensures that the degree is
preserved). The total Steenrod power is defined recursively using an
isomorphism
$P_s \cong (P_1)^{\otimes s}$ of unstable algebras, starting from
$\mathrm{St}_1 (x) = \sum u^{|x|-i} \otimes Sq^i x \in P_1 \otimes M$ and setting
$\mathrm{St}_s
= \mathrm{St}_{1} \circ \mathrm{St}_{s-1}$, interpreted
in the appropriate manner. The map $\mathrm{St}_s$ takes values in $D_s \otimes M$, in
particular is independent of the isomorphism $P_s \cong (P_1)^{\otimes s}$ used
in its construction.
The Singer functor $R_s$ has many good properties \cite{LZ,Psinger}; in
particular, it
is exact and commutes with tensor products: $R_s (M_1 \otimes M_2) \cong R_s M_1
\otimes_{D_s} R_s M_2$. Moreover,
$R_s$ restricts to a functor $R_s : \mathscr{K} \rightarrow D_s \downarrow
\mathscr{K}$ from unstable algebras to $D_s$-algebras in unstable algebras.
An important example is given by the action of $R_s$ on the polynomial algebra
$P_k$:
\begin{prop}
\cite[Section 5.4.7]{LZ}, \cite[Theorem 1.2]{HN2}.
\label{prop:invt_ring_Singer}
For positive integers $k, s$, the unstable algebra $R_s P_k$ is isomorphic to
the algebra of invariants
\[
R_s P_k \cong P_{k+s}^{\mathrm{GL}_{s,k}}
\]
where $\mathrm{GL}_{s,k}\subset \mathrm{GL}_{k+s}$ denotes the subgroup which induces the
identity on $\field^k$ under the projection $\field^{k+s} \cong\field^s \oplus
\field^k \twoheadrightarrow \field^k$.
Moreover, there is an isomorphism of algebras
\begin{eqnarray}
\label{eqn:RsPk}
R_s P_k \cong \field [c_{s,i} , V_s(j) ~|~ 0\leq i < s, \ 1 \leq j \leq k]
\end{eqnarray}
where $V_s (j):= \mathrm{St}_s x_j$, in particular $|V_s (j)|= 2^s$.
\end{prop}
\begin{rem}
\
\begin{enumerate}
\item
The identification of the underlying algebra of $R_s P_k$ is not explicit
in \cite{LZ}.
\item
The group $\mathrm{GL}_{s,k}$ is denoted $\mathrm{GL}_s \bullet \mathbf{1}_k$ in \cite{HN2}.
\item
As an element of $P_{k+s}$, $V_s (j)$ is the Mui invariant
$
\prod_{y \in \field^s} (x_j + y)
$,
where $y$ ranges over elements of $\field^s \subset \field^{k+s}$.
\end{enumerate}
\end{rem}
For the proof of the main result, it is sufficient to consider the case
$s=3$, hence the following notation is adopted:
\begin{nota}
Write $V(j)$ for $V_3(j) \in R_3 P_k$ for $1 \leq j \leq k$.
\end{nota}
Since $R_3P_k$ is a polynomial algebra on specified generators
(\ref{eqn:RsPk}),
it is equipped
with a length grading:
\begin{nota}
\label{nota:length}
Write $\mathsf{l} (\mathfrak{m})$ for the length of a monomial $\mathfrak{m}$ in $\{ c_{s,i}
, V(j) ~|~ 0\leq i < s, \ 1 \leq j \leq k \}$ (that is the length of $\mathfrak{m}$
as a
word).
\end{nota}
\begin{defn}
\label{def:length}
A non-zero element of $R_3 P_k$ is {\em length homogeneous} if it is the sum
of monomials $\sum_i \mathfrak{m}_i $ of the same length; its length is $\mathsf{l}
(\mathfrak{m}_i)$ for any monomial appearing.
\end{defn}
Recall that the first two Milnor operations are $Q_0 = Sq^1$ and $Q_1= [Sq^2,
Sq^1]$.
\begin{prop}
\label{prop:Sq_invts}
\cite{H}
The $\mathscr{A}$-action on
$R_3 P_k$ is determined by
\[
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline
Sq^0&Sq^1&Sq^2&Sq^3&Sq^4&Sq^5&Sq^6&Sq^7&Sq^8\\
\hline\hline
c_{3,2} & &c_{3,1}& c_{3,0} & c_{3,2}^2 &&&&\\
\hline
c_{3,1} & c_{3,0} & && c_{3,1}c_{3,2} & c_{3,0}c_{3,2} & c_{3,1}^2 &&\\
\hline
c_{3,0}&& &&c_{3,0}c_{3,2} & & c_{3,0}c_{3,1}& c_{3,0}^2 &\\
\hline
V(j) &&&&V(j)c_{3,2} & & V(j)c_{3,1} & V(j) c_{3,0} & V(j)^2\\
\hline
\end{array}
\]
In particular, the Milnor operations $Q_0, Q_1$ act trivially on $V(j)$
and $c_{3,0}$, $Q_0 c_{3,2} = Q_1 c_{3,2}= 0$ whereas
$Q_1 c_{3,2} = Q_0 c_{3,1}= c_{3,0}$.
\end{prop}
Recall that the Milnor operation $Q_i \in \mathscr{A}$ satisfies $Q_i^2=0$ and
the Margolis cohomology
groups of an unstable module $M$ are defined as
$$H^* (M; Q_i)
:=
\ker Q_i / \mathrm{Im} Q_i
$$
with grading inherited from $M$. Moreover, since $Q_i \in \mathscr{A}$ is
primitive
with respect to the Hopf algebra structure, it acts as a derivation on
unstable
algebras.
\begin{nota}
Let $\overline{P}_k$ denotes the augmentation ideal of
$P_k$.
\end{nota}
\begin{lem}
\label{lem:Milnor_observation}
For $i, k \in \nat$,
\begin{enumerate}
\item
\label{item:MO_1}
$\overline{P}_k$ is $Q_0$-acyclic, that is $\kerQ_0 = \mathrm{Im} Q_0$ on
$\overline{P}_k$;
\item
\label{item:MO_2}
$\Phi P_k \subset P_k$ lies in $\ker Q_i$
and induces a
surjection
$$
\Phi P_k\cong
\field [x_j^2 ~|~ 1 \leq j \leq k] \twoheadrightarrow H^* (P_k; Q_i)
\cong
\field [x_j^2 ~|~ 1 \leq j \leq k]/ (x_j^{2^{i+1}});
$$
\item
\label{item:MO_3}
$\ker (Q_i|_{\overline{P}_k})= \Phi \overline{P}_k + \mathrm{Im}
Q_i \subset
\mathrm{Im} Q_0 + \mathrm{Im} Q_i.
$
\end{enumerate}
In particular
$$
\ker (Q_0|_{\overline{P}_k}) + \ker (Q_1|_{\overline{P}_k}) \subset \big( \mathrm{Im} Sq^1 +
\mathrm{Im}
Sq^2 \big) = \overline{\mathscr{A} (1)} \ \overline{P}_k,
$$
where $\overline{\mathscr{A} (1)}$ is the augmentation ideal of $\mathscr{A}(1)$.
\end{lem}
\begin{proof}
The first two points are standard; for instance, to
calculate the Margolis cohomology, by the Künneth theorem one reduces to the
case $k=1$, which is elementary.
For \ref{item:MO_3}, the case of $Q_0$ is \ref{item:MO_1}. For $Q_1$, it follows from \ref{item:MO_2}
that $\ker (Q_1|_{\overline{P}_k})= \Phi \overline{P}_k + \mathrm{Im}
(Q_1)$; now $\Phi \overline{P}_k \subset \ker Q_0 = \mathrm{Im} Q_0$,
establishing this case.
Since $Q_0 = Sq^1$ and $Q_1 = [Sq^2, Sq^1]$, the final statement holds.
\end{proof}
Recall that $F(n)$ denotes the
free unstable module on a generator $\iota_n$ of degree $n$, for $n \in \nat$
(see e.g. \cite{StE}, \cite{schwartz_book}); these
modules
form a set
of projective generators of $\mathscr{U}$. One identifies $F(0)= \field$ and
$F(1) =
\langle x^{2^i} | i = 0, 1, 2, \dots \rangle \subset P_1 \cong \field [x]$.
Moreover, $F(n)$ is isomorphic to the invariants for the action of the symmetric
group by place permutations on $F(1) ^{\otimes n}$:
\[
F(n) \cong \big(F(1) ^{\otimes n} \big) ^{\mathfrak{S}_n}.
\]
Hence $F(n)$ embeds in $P_n$ as the submodule generated by the class $\prod
_{i=1}^n x_i$; in particular $F(n) \subset P_n ^{\mathfrak{S}_n}$, for the
action permuting the generators.
\begin{nota}
For $M$ an unstable module,
write $\mathrm{St}_s M $ for the image of the linear map $\mathrm{St}_s : \Phi^s M
\dashrightarrow D_s
\otimes M$.
\end{nota}
\begin{rem}
As graded vector spaces, there is a natural isomorphism $\mathrm{St}_s M \cong \Phi ^s
M$, hence $\mathrm{St}_s$ can be considered as an exact functor.
\end{rem}
Recall (see \cite{StE}, for example) that the total Steenrod power is
multiplicative: for example, if $K$ is an unstable algebra and $x, y \in K$,
then $\mathrm{St}_s (xy)= \mathrm{St}_s (x) \mathrm{St}_s (y)$, where the product is formed in $R_s K
\subset P_s \otimes K $.
The following result is stated for $s=3$ for notational simplicity, but holds
for any positive integer $s$.
\begin{lem}
\label{lem:length_homog}
For $ k$ a non-negative integer, there is an embedding in $D_3 \mbox{-} \mathscr{U}$:
\[
R_3 F(k)
\hookrightarrow
R_3 P_k.
\]
Moreover, as a graded vector space, $R_3 F(k)$ identifies as the free
$D_3$-submodule of $D_3 \otimes F(k) \hookrightarrow D_3 \otimes P_k$ generated
by $\mathrm{St}_3 F(k)$ and
$\mathrm{St}_3 F(k) \subset R_3 P_k$ has a basis of length homogeneous elements.
\end{lem}
\begin{proof}
The only point which is not immediate from the definitions is that $\mathrm{St}_3 F(k)$
has a basis of length homogeneous elements.
For $k=1$ this is clear: $\mathrm{St}_3 x = V$ (by definition) and the
multiplicativity of $\mathrm{St}_3$ gives $\mathrm{St}_3 (x^{2^i})= V^{2^i}$.
For the general case, multiplicativity of $\mathrm{St}_3$ leads to the identification
of $\mathrm{St}_3 F(k) \subset R_3 P_k$ with the invariants
$\big(\mathrm{St}_3(F(1))^{\otimes k} \big)^{\mathfrak{S}_k}$. It follows that
there is a basis of $\mathrm{St}_3 F(k)$ given by symmetric monomials from
$$\big(\field [V(1), \ldots , V(k)]\big)^{\mathfrak{S}_k} \subset \field [V(1),
\ldots , V(k)];$$
in particular, these are length homogeneous.
\end{proof}
\section{Reduction arguments}
\label{sect:results}
For a graded module $N$, let $N^{>0}$ denote the submodule of elements of
positive degree. The main result of the paper (Theorem~\ref{main_intro} of the
introduction) can be stated as follows:
\begin{thm}
\label{thm:main}
For an unstable module $M$ and an integer $s\geq 3$,
the natural map
\[
(R_s M) ^{>0}
\rightarrow
\field \otimes _\mathscr{A} (P_s \otimes M)
\]
induced by $R_s M \hookrightarrow D_s \otimes M \hookrightarrow P_s \otimes M$
and passage to $\mathscr{A}$-indecomposables is zero.
\end{thm}
The following standard argument reduces to the case $s=3$:
\begin{prop}
\label{prop:reduction_s=3}
Suppose that $
(R_3 M) ^{>0}
\rightarrow
\field \otimes _\mathscr{A} (P_3 \otimes M)
$ is zero for each unstable module $M$, then
\[
(R_s M) ^{>0}
\rightarrow
\field \otimes _\mathscr{A} (P_s \otimes M)
\]
is zero for each unstable module $M$ and all $s \geq 3$.
\end{prop}
\begin{proof}
For $s\geq 3$, by construction (see \cite{LZ}) there is a natural inclusion
$R_s
M \hookrightarrow R_3
R_{s-3} M $ which fits into the commutative diagram
\[
\xymatrix{
R_s M
\ar[r]
\ar@{^(->}[d]
&
\field \otimes _\mathscr{A} (P_s \otimes M)
\\
R_3 R_{s-3} M
\ar[r]
&
\field \otimes _\mathscr{A} (P_3 \otimes R_{s-3}M),
\ar[u]
}
\]
where the upwards arrow is induced by $R_{s-3} M \hookrightarrow
P_{s-3} \otimes M$ and the isomorphism $P_s \cong P_3 \otimes P_{s-3}$.
The lower horizontal arrow is the natural morphism $R_3 N \rightarrow \field
\otimes _\mathscr{A} (P_3 \otimes N)$ with $N= R_{s-3}M$.
By hypothesis, this is zero on positive degree elements, hence so is the upper.
\end{proof}
\begin{nota}
For $s$ a positive integer and $M$ an unstable module, write $\rtilde M$ for
$\overline{D}_s \otimes_{D_s} R_s M$.
\end{nota}
Recall \cite{LZ} that, for $s$ a positive integer,
there is a canonical surjection in
$\mathscr{U}$:
\begin{eqnarray}
\label{eqn:proj_Phi}
R_s M \twoheadrightarrow \Phi^s M \cong \field \otimes_{D_s} R_s M.
\end{eqnarray}
\begin{lem}
\label{lem:rtilde:properties}
For $s$ a positive integer
and $M$ an unstable module,
\begin{enumerate}
\item
$\rtilde M $ is the kernel of the surjection (\ref{eqn:proj_Phi});
\item
there is a natural isomorphism of graded vector spaces $R_s M \cong \rtilde M
\oplus \mathrm{St}_s M$;
\item
the functor $\rtilde : \mathscr{U} \rightarrow \mathscr{U}$ is exact and commutes with
direct sums.
\end{enumerate}
\end{lem}
\begin{proof}
The first statement follows by applying the exact functor $- \otimes _{D_s}
R_s
M$ ($R_s M$ is free as a $D_s$-module)
to the short exact sequence of $D_s$-modules:
\[
0
\rightarrow
\overline{D}_s
\rightarrow
D_s
\rightarrow
\field
\rightarrow
0.
\]
The remaining statements are straightforward.
\end{proof}
The following is an immediate consequence of the fact that $P_s$ is free as a
$D_s$-module:
\begin{lem}
\label{lem:induct_include}
For $s$ a positive integer
and $M$ an unstable module, the canonical inclusion $R_sM
\hookrightarrow D_s \otimes M$ induces an inclusion
\[
P_s \otimes _{D_s} R_s M \hookrightarrow P_s \otimes M.
\]
As a $P_s$-module, the image is free on $\mathrm{St}_s M$.
\end{lem}
The natural map $R_s M \rightarrow \field \otimes _\mathscr{A} (P_s \otimes M) $
factors:
\[
\xymatrix{
R_s M
\ar[r]
\ar[rd]
&
\field \otimes _\mathscr{A} (P_s \otimes_{D_s} R_s M)
\ar[d]
\\
& \field \otimes _\mathscr{A} (P_s \otimes M)
}
\]
and
restricts to a linear map
$
\Phi^s M \cong \mathrm{St}_s M \dashrightarrow \field \otimes _\mathscr{A} (P_s \otimes M).
$
\begin{rem}
Since $(\mathrm{St}_s M)^{>0}=\mathrm{St}_s (M^{>0})$, parentheses can be omitted.
\end{rem}
Theorem \ref{thm:main} follows (using Lemma \ref{lem:rtilde:properties}) from
the refined statement:
\begin{thm}
\label{thm:refined}
For an unstable module $M$ and an integer $s>2$,
\begin{enumerate}
\item
$ \mathrm{St}_s M^{>0} \dashrightarrow \field \otimes _\mathscr{A} (P_s \otimes M)$ is
zero;
\item
the composite morphism $\rtilde[s] M \hookrightarrow R_s M \rightarrow
\field \otimes
_\mathscr{A} (P_s \otimes_{D_s} R_s M) $ is zero.
\end{enumerate}
\end{thm}
\begin{rem}
By Proposition \ref{prop:reduction_s=3}, it suffices to treat the case $s=3$.
For this, a more refined result is stated in Theorem \ref{thm:hyper_refined}.
\end{rem}
The theorem can be proved by considering the projective generators of $\mathscr{U}$:
\begin{lem}
\label{lem:F(n)-reduction}
Theorem \ref{thm:refined} holds for all unstable modules $M$ if and only if it
holds for the free unstable module $F (n)$, for any non-negative integer $n$.
\end{lem}
\begin{proof}
This is a formal consequence of the fact that the set of all $F(n)$, $n$
a non-negative integer, forms a set of projective generators of $\mathscr{U}$ and
that
$\mathrm{St}_3$, $\rtilde[3]$
are exact and commute with direct sums, by Lemma \ref{lem:rtilde:properties}.
\end{proof}
\section{The image of $\mathrm{St}_s$}
\label{sect:image_sts}
Recall that, for $M$ an unstable module, $\mathrm{St}_3$ denotes the linear map
$$\Phi^3
M \dashrightarrow D_3 \otimes M \subset P_3 \otimes M.$$
The aim of this section is to prove Proposition \ref{prop:st3_hit} below,
which
corresponds to the first statement of Theorem \ref{thm:refined}.
\begin{lem}
\label{lem:mil_trivial}
For $x \in M$, where $M$ is an unstable module, $Q_0, Q_1, Sq^2 $ act
trivially on
$\mathrm{St}_3
(x)$.
\end{lem}
\begin{proof}
It is straightforward to reduce to the universal example $x = \iota_n \in
F(n)$, the fundamental class of the free unstable module $F (n)$ (see Section
\ref{sect:background}).
Now $F(n) \hookrightarrow P_n= \field [x_1, \ldots , x_n]$ under
the map induced by $\iota_n \mapsto \prod_{i=1}^n x_i$, so that $\mathrm{St}_3
(\iota_n) \in
\mathrm{St}_3 (F(n)) \subset R_3 P_n$; here one has $\mathrm{St}_3 (\iota_n) = \prod_{j=1}^n
V(j)$. Since $Sq^1 , Sq^2$ (and
hence $Q_1$) act trivially on $V(j)$ by Proposition
\ref{prop:Sq_invts}, the result follows from the Cartan formula.
\end{proof}
Consider the subalgebra $\mathscr{A} (0) \subset \mathscr{A}$ generated by $Q_0$, which
identifies with the exterior algebra $\Lambda (Q_0)$.
\begin{lem}
\label{lem:ker_image}
As an $\mathscr{A} (0)$-module:
$$D_3 \cong (\field [c_{3,2}], Q_0c_{3,2} =0) \otimes
(\field [c_{3,1}, c_{3,0}], Q_0 c_{3,1} = c_{3,0}).
$$
Hence
\begin{enumerate}
\item
$H_* (D_3;Q_0) \cong \field [c_{3,2}, c_{3,1}^2]$;
\item
a monomial in $\{c_{3,0}, c_{3,1}, c_{3,2} \}$
of $D_3$ lies in $\ker Q_0 $ if and only if it has the
form $ c_{3,2}^{i_2} c_{3,1}^{2n} c_{3,0} ^{t}$ for $i_2,n, t\in \nat$; if $t>0$
then it lies in $\mathrm{Im} Q_0$.
\end{enumerate}
\end{lem}
\begin{proof}
Straightforward.
\end{proof}
\begin{nota}
Index the monomial basis of $D_3$ by $I = (i_2,i_1, i_0) \in \nat^{ 3}$ with
\begin{eqnarray}
\label{eqn:monomial_basis}
c^I = c^{i_2,i_1,i_0} := c_{3,2}^{i_2}c_{3,1}^{i_1}c_{3,0}^{i_0}.
\end{eqnarray}
\end{nota}
\begin{lem}
\label{lem:mil0_reduction}
Let $M$ be an $\mathscr{A}(0)$-module and write an element $y\in
\overline{D}_3
\otimes M$ as $y= \sum c^I \otimes y_I $, where $c^I$ runs over the monomial
basis (\ref{eqn:monomial_basis}) of $\overline{D}_3$.
Then $y$ lies in $ \ker Q_0$ if and only if both
\begin{enumerate}
\item
$
Q_0 y_{i_2, 2u, t+1} = y_{i_2, 2u+1, t}
$ for all $(i_2, u, t) \in \nat^{\times 3}$ and
\item
$
Q_0 y_{i_2, 2u,0} =0.
$
\end{enumerate}
If these conditions are satisfied,
\[
y = \sum c^{i_2, 2u,0} \otimes y_{i_2, 2u,0} + \sum Q_0 (c^{i_2 , 2u+1, t}
\otimes y_{i_2, 2u, t+1}).
\]
\end{lem}
\begin{proof}
Clearly $Q_0 y = \sum \{ (Q_0 c^I) \otimes y_I + c^I \otimes Q_0 y_I
\}$. The result follows by identifying coefficients in the expansion in terms
of the monomial basis of $D_3$.
\end{proof}
\begin{prop}
\label{prop:st3_hit}
For $M$ an unstable module, $\mathrm{St}_3 M^{>0} \subset \overline{\mathscr{A}} (P_3
\otimes M)$. More precisely,
$
\mathrm{St}_3 (x) \in \mathrm{Im} Sq^1 + \mathrm{Im} Sq^{4|x|}
$
for $x \in M^{>0}$.
\end{prop}
\begin{proof}
As in Lemma \ref{lem:mil_trivial}, one reduces to the universal example
$\iota_n
\in F(n)$: it suffices
to show that $\mathrm{St}_3 \iota_n \in \overline{\mathscr{A}} (P_3 \otimes F(n))$ for each
$0<n \in \nat$.
Lemma \ref{lem:mil_trivial} implies that $\mathrm{St}_3 \iota_n \in \ker Q_0$;
Lemma \ref{lem:mil0_reduction} shows that it suffices to consider the terms
\[
c_{3,2}^{i_2} c_{3,1} ^{2u} \otimes y_{i_2,2u}
\]
which appear in the expansion of $\mathrm{St}_3 \iota_n$ (for simplicity, writing
$y_{i_2, 2u}$ for $y_{i_2, 2u, 0}$). By Lemma
\ref{lem:mil0_reduction}, both $y_{i_2, 2u}$ and
$c_{3,2}^{i_2} c_{3,1} ^{2u}$
lie in $\ker Q_0$.
Clearly $y_{0,0} = (Sq_0)^3 \iota_n \in F(n)$ (recall that $Sq_0 (x) :=
Sq^{|x|}x$); in the other
cases, $c_{3,2}^{i_2} c_{3,1} ^{2u}$ has positive degree so that there exists
$a_{i_2,2u} \in P_3$ such that $Q_0 a_{i_2,2u} = c_{3,2}^{i_2} c_{3,1}
^{2u}$,
by Lemma \ref{lem:Milnor_observation}. Thus $Q_0 (a_{i_2,2u} \otimes
y_{i_2,2u}) = c_{3,2}^{i_2} c_{3,1} ^{2u} \otimes y_{i_2,2u}$ in $P_3 \otimes
F(n)$.
This establishes that, for $x \in M^{>0}$, $\mathrm{St}_3 (x) \equiv 1
\otimes (Sq_0)^3 (x) = (Sq_0)^3 (1 \otimes x) \mod \mathrm{Im} Q_0$ in $P_3 \otimes
M$.
Since $(Sq_0)^3 (x) = Sq^{4|x|} (Sq_0)^2(x)$, the final statement follows.
\end{proof}
\begin{cor}
\label{cor:sts_hit}
For an integer $s \geq 3$ and an unstable module $M$,
$$
\mathrm{St}_s M^{>0} \subset
\overline{\mathscr{A}} (P_s \otimes
M).
$$
\end{cor}
\begin{proof}
Use a reduction argument similar to the proof of Proposition
\ref{prop:reduction_s=3}.
\end{proof}
\section{Proof of the main theorem}
\label{sect:proofs}
To complete the proof of Theorem \ref{thm:refined}, by combining Lemma
\ref{lem:F(n)-reduction} with Proposition \ref{prop:st3_hit}, it suffices
to show that
\[
\rtilde[3] F(n) \subset \overline{\mathscr{A}} (P_3 \otimes_{D_3} R_3 F(n))
\]
for each $n \in \nat$.
Recall that $P_3 \otimes_{D_3} R_3 M \cong P_3 \otimes
\mathrm{St}_3 M$ as a $P_3$-module (forgetting the $\mathscr{A}$-action). The inclusion
$\overline{P}_3 \hookrightarrow P_3$
of $D_3$-modules induces
\[
\overline{P}_3 \otimes \mathrm{St}_3 M \cong \overline{P}_3 \otimes _{D_3}R_3 M
\hookrightarrow
P_3 \otimes_{D_3} R_3 M \cong P_3 \otimes
\mathrm{St}_3 M.
\]
\begin{lem}
\label{lem:basic_Milnor_reduction}
For $M$ an unstable module, there is a natural isomorphism of
$\mathscr{A}(1)$-modules
\[
P_3 \otimes_{D_3} R_3 M \cong P_3 \otimes
\mathrm{St}_3 M,
\]
where $\mathscr{A}(1)$ acts trivially on $\mathrm{St}_3 M$.
This restricts to an isomorphism of $\mathscr{A}(1)$-modules
$
\overline{P}_3 \otimes_{D_3} R_3 M \cong \overline{P}_3 \otimes
\mathrm{St}_3 M.
$
Hence
\[
\big(\overline{P}_3 \otimes \mathrm{St}_3 M \big) \cap \big(\ker Q_0 + \ker Q_1 \big)
\subset \big( \mathrm{Im} Sq^1 + \mathrm{Im} Sq^2 \big) \subset P_3 \otimes _{D_3} R_3 M.
\]
\end{lem}
\begin{proof}
Lemma \ref{lem:mil_trivial} implies that $\mathrm{St}_3M \subset R_3 M$ is a trivial
$\mathscr{A}(1)$-submodule.
The $P_3$-module structure of $ P_3 \otimes_{D_3} R_3 M$ then leads to the
isomorphism of $P_3$-modules
\[
P_3 \otimes \mathrm{St}_3 M \rightarrow P_3 \otimes_{D_3} R_3 M
\]
which is $\mathscr{A}(1)$-linear. The case of the restriction to $\overline{P}_3$ is
straightforward.
The final statement follows from Lemma \ref{lem:Milnor_observation}.
\end{proof}
Recall that $\rtilde[3] F(n) \subset R_3 F(n) \subset R_3
P_n$ and that $R_3 F(n)$ is the free $D_3$-module on $\mathrm{St}_3 F(n)$.
Moreover, Lemma \ref{lem:length_homog} implies that $\mathrm{St}_3 F(n) \subset R_3 P_n$
has a basis given by length homogeneous polynomials in
$\field [V(1), \ldots , V(n)]$.
\begin{lem}
\label{lem:generators}
For $n$ a non-negative integer,
$ \rtilde[3] F(n) \subset R_3 F(n)$ has a basis of elements
of
the form
\begin{eqnarray}
\label{eqn:standard_basis}
c^I \mathrm{St}_3 y_i
\end{eqnarray}
where $y_i \in F(n)$ ranges over a basis such that $\mathrm{St}_3 y_i$ is length
homogeneous and $c^I\neq 1$ ranges over the monomial basis.
\end{lem}
\begin{proof}
Straightforward.
\end{proof}
\begin{nota}
A {\em term} of an element of $ \rtilde[3] F(n)$ is a basis element (as in
equation (\ref{eqn:standard_basis}) of Lemma
\ref{lem:generators}) which appears with non-zero coefficient.
\end{nota}
To prove Theorem \ref{thm:refined}, it is sufficient to consider
the basis elements (\ref{eqn:standard_basis}) of Lemma \ref{lem:generators}. The
inductive proof
is based upon the submonomial of $c^I$ in the generators $c_{3,2}$ and
$c_{3,1}$; the
behaviour is analysed using Lemmas \ref{lem:full_reduction} and
\ref{lem:B} below.
\begin{lem}
\label{lem:full_reduction}
Let $\mathfrak{m}= c^I \mathrm{St}_3 y \in \rtilde[3] F(n)$ where $c^I \neq 1$, $\mathrm{St}_3 y$ is
length homogeneous
and let $v \in P_3$.
If $i_1 i_2 \equiv 0\mod 2$ then $\mathfrak{m} v^2 \in \big( \mathrm{Im} Sq^1 +
\mathrm{Im} Sq^2\big) \subset P_3 \otimes _{D_3} R_3 M$.
\end{lem}
\begin{proof}
If $i_1 \equiv 0 \mod 2$, then $Q_0 \mathfrak{m}=0$ and if $i_2 \equiv 0 \mod 2$,
then $Q_1 \mathfrak{m}=0$, by Proposition \ref{prop:Sq_invts}. Since $Q_i$ ($i \in
\{0,1 \}$) is a derivation, if $Q_i x=0$, then $Q_i x v^2 =0$; hence,
under the hypothesis,
$\mathfrak{m} v^2 \in \ker Q_0 +\ker Q_1$. The result thus follows from Lemma
\ref{lem:basic_Milnor_reduction}.
\end{proof}
This result is complemented by the following special case of \cite[Lemma
B]{HN}.
\begin{lem}
\label{lem:B}
The element $c_{3,2} c_{3,1} \in D_3$ lies in $\mathrm{Im} Sq^1 +
\mathrm{Im} Sq^2 \subset P_3$.
\end{lem}
Lemmas \ref{lem:full_reduction} and
\ref{lem:B} motivate the following:
\begin{defn}
For $\mathfrak{m} = c^I \mathrm{St}_3 y$ as above:
\begin{enumerate}
\item
the standard form for $\mathfrak{m}$ is
\begin{eqnarray}
\label{eqn:c_monomial}
\mathfrak{m} = (c_{3,2}c_{3,1})^{2^f-1} (c_{3,2}^{j_2} c_{3,1}^{j_1}) ^{2^f}
c_{3,0}^{i_0} \mathrm{St}_3 y,
\end{eqnarray}
where $f\in \nat$ is determined by $j_1 j_2 \equiv 0 \mod 2$;
\item
the fullness of $\mathfrak{m}$, $\mathsf{f}(\mathfrak{m}) \in \nat$, is the integer $f$
appearing in the standard form;
\item
$\mathfrak{m}$ is full if $j_1 =0= j_2$.
\end{enumerate}
\end{defn}
\begin{rem}
The significance of the notion of {\em fullness} is suggested by Lemma
\ref{lem:full_reduction} and is important in the primary
induction of the proof of the main theorem below. A secondary argument uses
the
notion of {\em length} inherited from $R_3
P_n$ (see Definition \ref{def:length} and Lemma \ref{lem:length_homog}).
\end{rem}
\begin{rem}
\label{rem:length_inductive_scheme}
The length argument uses Lemma \ref{lem:exclude} below (which is inspired by
\cite[Lemma 3.4]{HN2}), which allows monomials which satisfy
\[
\mathsf{l} (\mathfrak{m}) \not \equiv 2^{\mathsf{f}(\mathfrak{m})}-1 \mod 2^{\mathsf{f}({\mathfrak{m}})}
\]
to be treated by using an induction upon the fullness.
The remaining case is studied by using Lemma \ref{lem:parity_case} (which is
inspired by \cite[Lemma 3.5]{HN2}); this reduces
to terms of smaller fullness or terms which can be treated by Lemma
\ref{lem:exclude}.
\end{rem}
The following elementary observation is used:
\begin{lem}
\label{lem:2-adic_binomial}
Let $f$ and $\ell$ be non-negative integers. Then $\ell \equiv 2^{f}-1 \mod
2^{f}$ if and only if
\[
\binom {\ell}{2^i} \equiv 1 \mod 2,
\]
for each $0 \leq i < f $.
\end{lem}
\begin{lem}
\label{lem:exclude}
For $\mathfrak{m} = c^I \mathrm{St}_3 y \in \rtilde[3] F(n)$ with $\mathrm{St}_3 y$ length
homogeneous of length $\mathsf{l} (\mathfrak{m})$, and $\mathsf{f}
(\mathfrak{m})= f>0$, suppose that, for some $0 \leq i < f$,
\[
\binom {\mathsf{l} (\mathfrak{m})}{2^i} \equiv 0 \mod 2.
\]
Then, $\mathfrak{m} c_{3,2}^{-2^i} \in \rtilde[3]F(n)$ and, for any element $v \in
P_3$, there is an equality in $P_3
\otimes_{D_3} R_3 F(n)$:
\[
Sq^{4\cdot 2^i} (\mathfrak{m} c_{3,2}^{-2^i} v ^{2^{f}})
=
\mathfrak{m} v ^{2^{f}}
+
\sum_l \mathfrak{s}_l v_l ^{2^{f}}
\]
where $v_l \in P_3$ and $\mathfrak{s}_l \in R_3 F(n)$ is an element of the form
$c^{I_l} \mathrm{St}_3 (y_l)$ such that
\begin{enumerate}
\item
$\mathfrak{s}_l \in \rtilde[3] F(n)$;
\item
$\mathsf{f}(\mathfrak{s}_l)\leq i<f $.
\end{enumerate}
\end{lem}
\begin{proof}
The fact that $\mathfrak{m} c_{3,2}^{-2^i} \in \rtilde[3]F(n)$ is clear.
The remainder of the argument uses the action of the Steenrod
algebra (given in Proposition \ref{prop:Sq_invts}) together with the Cartan
formula. In particular, $Sq^4$ acts on the generators of $R_3 P_n$ as
multiplication by $c_{3,2}$.
A monomial appearing in the expansion of $Sq^{4\cdot 2^i} (\mathfrak{m}
c_{3,2}^{-2^i} v ^{2^{f}}) $ can only have fullness $> i$ by the
application of $Sq^4$ to $2^i$ generators. The hypothesis on the binomial
coefficient ensures that this contributes the term $ \mathfrak{m} v ^{2^{f}}$.
The remaining terms have fullness $\leq i$ and can be written in the required
form $ \mathfrak{s}_l v_l ^{2^{f}}$. The fact that $\mathfrak{s}_l\in
\rtilde[3] F(n)$ is clear, since it is impossible to
destroy a contribution from $\overline{D}_3$.
\end{proof}
\begin{lem}
\label{lem:parity_case}
Let $\mathfrak{m} = c^I \mathrm{St}_3 y \in \rtilde[3] F(n)$ with $\mathrm{St}_3 y$ length
homogeneous and $\mathfrak{m}$ full with $\mathsf{f}(\mathfrak{m}) = f$.
Suppose that $\mathsf{l} (\mathfrak{m}) \equiv 2^{f}-1 \mod 2^{f}$ and consider a Steenrod
operation $\theta \in \mathscr{A}$ such that $|\theta|\leq 2^{f+1}$.
For $\mathfrak{t}$ a term of $\theta \mathfrak{m}$,
\begin{enumerate}
\item
if $\mathsf{l}(\mathfrak{t})=\mathsf{l} (\mathfrak{m})$, then $\mathsf{f}(\mathfrak{t}) < f$;
\item
if $\mathsf{l}(\mathfrak{t})> \mathsf{l} (\mathfrak{m})$, then $\mathsf{l} (\mathfrak{t})\not \equiv 2^f
-1 \mod 2^f$.
\end{enumerate}
In particular, one of the following holds:
\begin{enumerate}
\item
$ \mathsf{f}( \mathfrak{t}) < f$;
\item
$ \mathsf{f}( \mathfrak{t}) \geq f$ and $\mathsf{l} (\mathfrak{t})\not \equiv
2^{\mathsf{f}(\mathfrak{t})} -1 \mod 2^{\mathsf{f}(\mathfrak{t})}$.
\end{enumerate}
\end{lem}
\begin{proof}
The hypothesis that $\mathfrak{m}$ is full of fullness $f$ means that $$\mathfrak{m} =
(c_{3,2}c_{3,1})^{2^{f}-1} c_{3,0}^{i_0} \mathrm{St}_3 y.$$
In the case $\mathsf{l} (\mathfrak{t})= \mathsf{l} (\mathfrak{m})$, $\mathfrak{t}$ arises from the
action of the operation $\theta$ upon the factor $(c_{3,2}c_{3,1})^{2^{f}-1}$
of $\mathfrak{m}$ since
Steenrod operations increase the length when operating non-trivially on the
other
terms (this is where fullness is used). It follows by inspection
that
$\mathsf{f}(\mathfrak{t}) < f$.
In the remaining case, write $\mathsf{l} (\mathfrak{m}) = 2^{f}-1 + k 2^f$, for some $k \in
\nat$. Then, by inspection of the action given by Proposition
\ref{prop:Sq_invts},
\[
\mathsf{l} (\mathfrak{m}) = 2^{f}-1 + k 2^{f} < \mathsf{l} (\mathfrak{t}) \leq 2^{f}-1 + k 2^{f}
+
|\theta|/4 \leq 2^{f}-1 + k 2^{f} + 2^{f-1}.
\]
The conclusion follows by elementary arithmetic.
\end{proof}
Theorem \ref{thm:refined} is implied by the following, more precise,
result. The main case of interest is when $v=1$ in the statement; the
inductive proof requires the stronger statement.
\begin{thm}
\label{thm:hyper_refined}
Let $\mathfrak{m} = c^I \mathrm{St}_3 y \in \rtilde[3] F(n)$ be a basis element with $\mathsf{f}
(\mathfrak{m})= f$ and let $v \in P_3$.
Then
\begin{enumerate}
\item
if $\mathfrak{m}$ is full (hence $f>0$)
\[
\mathfrak{m} v^{2^{f+1}} \in \overline{\mathscr{A} (f)} \big(P_3 \otimes_{D_3} R_3
F(n)\big);
\]
\item
otherwise
\[
\mathfrak{m} v^{2^{f+1}} \in \overline{\mathscr{A} (f+1)} \big(P_3 \otimes_{D_3} R_3
F(n)\big).
\]
\end{enumerate}
\end{thm}
\begin{proof}
The result is proved by increasing induction upon $f$.
The initial case of the induction for $f=0$ (only the non-full case occurs),
is
established by Lemma \ref{lem:full_reduction} since,
by definition of fullness, the hypothesis $i_1 i_2 \equiv 0 \mod 2$ is
satisfied.
The initial case of the induction for the full case is for $f=1$; here one can
write:
\[
\mathfrak{m} v^4 = c_{3,2} c_{3,1} \mathfrak{m}' v^4
\]
and $Sq^1, Sq^2$ act trivially on $\mathfrak{m}' v^4$ (since $\mathfrak{m}'$ contains
neither $c_{3,2}$ nor $c_{3,1}$). The result
therefore follows from Lemma \ref{lem:B} and the Cartan formula.
There are two inductive steps to consider. In order to present a unified proof
of the final step, the
induction first treats the non-full case with $\mathsf{f} (\mathfrak{m})= f$ and then the
full case with $\mathsf{f} (\mathfrak{m})= f+1$.
\begin{enumerate}
\item
In the non-full case with $\mathsf{f} (\mathfrak{m})= f$, one can write
\[
\mathfrak{m} v^{2^{f+1}} = (c_{3,2}^{j_2}c_{3,1}^{j_1} v^2)^{2^{f}}
\mathfrak{m}_{\mathrm{full}}
\]
where $\mathfrak{m}_{\mathrm{full}}$ is full with $\mathsf{f} (\mathfrak{m}_{\mathrm{full}}) =
f$, $j_1 + j_2 >0$ and $j_1j_2 \equiv 0 \mod 2$. Lemma \ref{lem:full_reduction}
implies that $c_{3,2}^{j_2}c_{3,1}^{j_1} v^2 \in \big( \mathrm{Im} Sq^1 + \mathrm{Im}
Sq^2 \big) \subset \overline{P}_3$, say is $Sq^1 u_1 + Sq^2 u_2$, so that
$ (c_{3,2}^{j_2}c_{3,1}^{j_1} v^2)^{2^{f}} = Sq^{2^f} u_1^{2^f} + Sq^{2^{f+1}}
u_2 ^{2^f} $.
Hence it is sufficient to consider elements of the form
\[
(Sq^{2^f} u_1^{2^f} + Sq^{2^{f+1}} u_2 ^{2^f} ) \mathfrak{m}_{\mathrm{full}} ,
\]
where $u_1, u_2 \in \overline{P}_3$. For such elements, see below.
\item
In the full case with $\mathsf{f} (\mathfrak{m})= f+1$, first observe that it suffices to
prove the case for $v=1$, since
if $\mathfrak{m} \in \mathrm{Im} ( Sq^1, \ldots , Sq^{2^{f+1}} )$, the Cartan formula
shows that
$\mathfrak{m} v^{2^{f+2}}$ is also.
Now write
\[
\mathfrak{m} = (c_{3,2}c_{3,1})^{2^{f}} \mathfrak{m}_{\mathrm{full}}
\]
where $\mathfrak{m}_{\mathrm{full}}$ is full with $\mathsf{f} (\mathfrak{m}_{\mathrm{full}}) =
f$. By Lemma \ref{lem:B},
$c_{3,2}c_{3,1}$, considered as an element of $P_3$, lies in $\mathrm{Im} Sq^1 +
\mathrm{Im} Sq^2$. Hence
$(c_{3,2}c_{3,1})^{2^f}$ lies in $\mathrm{Im} Sq^{2^f} + \mathrm{Im} Sq^{2^{f+1}}$, thus
it is sufficient to consider elements of the form:
\[
(Sq^{2^{f}} u_1^{2^{f}} + Sq^{2^{f+1}} u_2 ^{2^{f}} ) \mathfrak{m}_{\mathrm{full}},
\]
where $u_1, u_2 \in \overline{P}_3$. This expression has the same form as in
the non-full case.
\end{enumerate}
The indexing was chosen so that $\mathsf{f} (\mathfrak{m}_{{\mathrm{full}}}) = f$ in both cases.
To establish the inductive steps, it is sufficient to show, for
$u_1, u_2 \in \overline{P}_3$ and $\mathsf{f} (\mathfrak{m}_{\mathrm{full}}) =f$, that
\[
(Sq^{2^{f}} u_1^{2^{f}} + Sq^{2^{f+1}} u_2 ^{2^{f}} ) \mathfrak{m}_{\mathrm{full}}
\in
\mathrm{Im} (Sq^1, \ldots , Sq^{2^{f+1}}) \subset P_3 \otimes_{D_3} R_3 F(n).
\]
(Note that the appropriate subalgebra of $\mathscr{A}$ is $\mathscr{A} (f+1)$ in both
cases,
by the choice of indexing.)
Hence consider elements of the form
\[
(Sq^{\delta} u)^{2^f} \mathfrak{m}_{\mathrm{full}} = Sq^{2^{f -1 +\delta}} (u^{2^f})
\mathfrak{m}_{\mathrm{full}}
\]
for $u \in \overline{P}_3$ and $\delta \in \{1, 2 \}$ and $\mathsf{f} (\mathfrak{m}_{\mathrm{full}}) =f$.
If $\mathsf{l} (\mathfrak{m}_{\mathrm{full}}) \not \equiv 2^f -1 \mod 2^f$, then Lemma
\ref{lem:exclude},
using $Sq^{2^{i+2}}$ for the appropriate $i$, $0 \leq i < f$,
allows reduction to terms earlier in the inductive scheme, since
\begin{enumerate}
\item
$i+2 < f+2$, so that $i+2 \leq f+1$ and hence $Sq^{2^{i+2}} \in \mathscr{A}
(f+1)$;
\item
the terms $\mathfrak{s}_l v_l^{2^f}$ (in the notation of Lemma
\ref{lem:exclude}) which occur have fullness $\mathsf{f}(\mathfrak{s}_l) < f$, hence
are
treated by the inductive hypothesis.
\end{enumerate}
Finally, if $\mathsf{l} (\mathfrak{m}_{\mathrm{full}}) \equiv 2^f -1 \mod 2^f$, consider
\[
Sq^{2^{f -1 +\delta}} (u^{2^f} \mathfrak{m}_{\mathrm{full}} )
\]
using the Cartan formula and Lemma \ref{lem:parity_case} to understand the
terms
which occur from applying a Steenrod square
to $\mathfrak{m}_{\mathrm{full}}$, which are of the form $Sq^j \mathfrak{m}_{\mathrm{full}}$ with $j \leq
2^{f+1}$. (In fact, only $Sq^j \mathfrak{m}_{\mathrm{full}}$ with $j \in \{2^f , 2^{f+1} \}$
have to be considered, but this precision is not required.)
Note that the operation $ Sq^{2^{f -1 +\delta}}$ lies in $\mathscr{A} (f+1)$.
By the final statement of Lemma \ref{lem:parity_case}, the terms which arise
either have
fullness $<f$ or can be treated as above by using Lemma \ref{lem:exclude}.
This completes the proof of the inductive steps.
\end{proof}
\smallskip
\noindent
{\bf Acknowledgement.}
This research was carried out when the first named author was a CNRS invited
researcher at the LAREMA, Angers in the autumn of 2014. The public manuscript
was
prepared in the winter of 2015-16, when the first named
author visited the Vietnam Institute for Advanced Study in Mathematics
(VIASM), Hanoi; he would like to express his warmest thanks to the CNRS and to
the VIASM for hospitality and for the wonderful working conditions.
\smallskip
The authors are grateful to Bill Singer for comments on an earlier version
and to Hadi Zare for his remarks on the relationship between the generalized
spherical class
conjecture and a conjecture due to Peter Eccles.
\bigskip
The first named author was partially funded by the National Foundation for
Science and Technology Development
(NAFOSTED) of Vietnam under grant number 101.04-2014.19.
The second named author was partially supported by the project {\em Nouvelle
Équipe}, convention No. 2013-10203/10204 between the Région des Pays de la
Loire and the Université d'Angers.
\bigskip
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,315 |
Q: Margin doesn't work properly in IE11 I have HTML code:
h1{
margin-bottom: 2px;
margin-top: 3px;
}
<div class="header col-md-7 col-sm-8 col-xs-12">
<h1><span class="title-pink">FIRST TITLE</span></h1>
<h1><span class="title-pink">SECOND TITLE</span></h1>
<h1><span class="title-pink">THIRD TITLE</span></h1>
</div>
On Chrome, this margin works properly, but when I open the website on IE 11, these margins don't appear at all. Why? Can you help me?
A: Your code snippet runs same on IE 11 and chrome, so style was applied correctly.
You used bootstrap, and it will override h1 style, so IE11 won't apply your style. On chrome, bootstrap may not be loaded because of CORS problem.
Overall, adding !important to style attributes will solve problem.
h1{
margin-bottom: 2px !important;
margin-top: 3px !important;
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,372 |
Nathan is a licensed Landscape Architect holding national CLARB certification with registration in Idaho, Montana and Wyoming. He is also certified as a golf irrigation auditor by the Irrigation Association. Nathan's focus is planning and design of projects ranging from intimate memorials and small public spaces to regional parks and multi-sport athletic facilities.
His design of improvements for cemeteries, including full development of new cemeteries and expansions, are regionally recognized. Nathan's background and experience gives him the ability to effectively manage projects as well as lead multi-disciplinary design teams on larger projects for both private and public clients. His design and technical skills allow him to envision the big picture yet focus on the key details that are important to sensitive design execution.
Bachelor of Landscape Architecture – University of Idaho, 1997. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,094 |
Q: Apache access log, strange post requests Getting lot strange requests in my access log:
ip login:"-" - - [24/May/2017:01:26:30 +0700] "POST /3A348409-DD98-D443-96A4-D712F51D8B11/D89B1EDB-4CED-D145-9246-16243451D23D/from HTTP/1.0" 404 1346 Time:"2s" pid:23050 Mem:"2097152
ip login:"-" - - [24/May/2017:00:48:35 +0700] "POST /3A348409-DD98-D443-96A4-D712F51D8B11/E970DBFE-0DB1-A749-9392-CF1704CC81FD/from HTTP/1.0" 404 1348 Time:"0s" pid:22893 Mem:"4194304"
ip login:"-" - - [23/May/2017:00:33:08 +0700] "POST /CE92AFB2-2FDE-8742-B5ED-0629F2B9B622/2D682DC1-D8C5-574F-8A0E-AC62EB96CBD8/from HTTP/1.0" 404 1348 Time:"0s" pid:6695 Mem:"4194304"
...
Also, sometimes (not so frequently), getting another type of logs records containing parts of my HTML pages:
ip login:"-" - - [23/May/2017:14:00:49 +0700] "GET /static/legacy/js/ion%20value=201602>%D4%E5%E2%F0%E0%EB%FC%202016</option><option%20value=201601>%DF%ED%E2%E0%F0%FC%202016</option><option%20value=201512>%C4%E5%EA%E0%E1%F0%FC%202015</option><option%20value=201511>%CD%EE%FF%E1%F0%FC%202015</option><option%20value=201510>%CE%EA%F2%FF%E1%F0%FC%202015</option><option%20value=201509>%D1%E5%ED%F2%FF%E1%F0%FC%202015</option><option%20value=201508>%C0%E2%E3%F3%F1%F2%202015</option><option%20value=201507>%C8%FE%EB%FC%202015</option><option%20value=201506>%C8%FE%ED%FC%202015</option><option%20value=201505>%CC%E0%E9%202015</option><option%20value=201504>%C0%EF%F0%E5%EB%FC%202015</option><option%20value=201503>%CC%E0%F0%F2%202015</option><option%20value=201502>%D4%E5%E2%F0%E0%EB%FC%202015</option><option%20value=201501>%DF%ED%E2%E0%F0%FC%202015</option><option%20value=201412>%C4%E5%EA%E0%E1%F0%FC%202014</option><option%20value=201411>%CD%EE%FF%E1%F0%FC%202014</option><option%20value=201410>%CE%EA%F2%FF%E1%F0%FC%202014</option><option%20value=201409>%D1%E5%ED%F2%FF%E1%F0%FC%202014</option><option%20value=201408>%C0%E2%E3%F3%F1%F2%202014</option><option%20value=201407>%C8%FE%EB%FC%202014</option><option%20value=201406>%C8%FE%ED%FC%202014</option><option%20value=201405>%CC%E0%E9%202014</option><option%20value=201404>%C0%EF%F0%E5%EB%FC%202014</option><option%20value=201403>%CC%E0%F0%F2%202014</option><option%20value=201402>%D4%E5%E2%F0%E0%EB%FC%202014</option><option%20value=201401>%DF%ED%E2%E0%F0%FC%202014</option><option%20value=201312>%C4%E5%EA%E0%E1%F0%FC%202013</option><option%20value=201311>%CD%EE%FF%E1%F0%FC%202013</option></select></td></tr><script%20type= HTTP/1.0" 404 1347 Time:"0s" pid:15377 Mem:"4194304"
Anyone know something about it?
OS: ubuntu 15.10 x64
Apache: v 2.4.24
A: Looks to me like someone found a cross-site scripting (XSS) vulnerability somewhere in your code.
Without seeing the code found in the file found (presumably) at /static/legacy/js/ion, it's almost impossible to offer any advice or answers as to what needs to be done.
Generally speaking though, somewhere along the line there's code that exists which is producing output without first being sanitized. It could be inside that file, or maybe even inside the file that produces the output that writes that line.
Either way, it would probably be best to search for things like $_POST, $_GET, $_REQUEST, etc., that are producing output provided by the user without first being sanitized.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,969 |
De Lacets de Montvernier is een tot 782 meter hoogte stijgende bergroute met haarspeldbochten in de Alpen.
De weg is gelegen in het departement Savoie en verbindt de gemeenten Pontamafrey-Montpascal in het dal met Montvernier boven op de alp. De weg met de haarspeldbochten is de D77B.
Wielrennen
De helling werd in de Ronde van Frankrijk 2015 voor het eerst beklommen. Het circuit wordt gelijkgesteld met de beklimming van een col van tweede categorie. Romain Bardet kwam eerst boven gevolgd door Pierre Rolland en Jakob Fuglsang. De Ronde van Frankrijk 2018 herhaalde de beklimming, toen reed Pierre Rolland alleen naar boven. Als tweede passeerde Julian Alaphilippe, als derde Serge Pauwels. In de Ronde van Frankrijk 2022 was het podium voor (in dalende volgorde) Pierre Latour, Simon Geschke en Warren Barguil.
Bergpas in Savoie
Beklimming in de wielersport in Frankrijk | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 135 |
{% include header.html %}
{% include body.html %}
<div id="content">
{% include banner.html %}
{% include navigation.html %}
<h1>{{ page.title }}</h1>
<h5>{{ page.date | date_to_long_string }}</h5>
{% include blog.html %}
</div>
{% include footer.html %}
</body>
</html>
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,604 |
28" 14" 16" 18" 22"
3 YRS Xuzhou ZhiWalk New Power Technology Co., Ltd.
adult electric bike dal Technicial parameters: Tire size:16inches Power:350W 48V Brushless motor Battery:Lead acid 12Ah Removable Max speed:45km/h Max mileage:45KM Max loading:300KG Brake:drum brake Warranty:2years for motors,15months for battery Packing size:L160*W35*H77CM Other products: Contact person:Thomas Wang Whatsapp:008615162153068 XUZHOU ZHIWALK NEW POWER TECHNOLOGY.,LTD.
Alibaba.com offers 258,087 electric bicycle products. About 10% of these are electric bicycle, 4% are electric scooters, and 4% are other electric bicycle parts.
A wide variety of electric bicycle options are available to you, such as ce, eec, and dot. You can also choose from 48v, > 60v, and 36v. As well as from > 500w, 200 - 250w, and 251 - 350w. And whether electric bicycle is 28", 14", or 16".
There are 258,087 electric bicycle suppliers, mainly located in Asia. The top supplying country is China (Mainland), which supply 100% of electric bicycle respectively.
Electric bicycle products are most popular in North America, Western Europe, and Domestic Market.
You can ensure product safety by selecting from certified suppliers, including 80,286 with ISO9001, 22,819 with Other, and 19,143 with ISO14001 certification. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,888 |
US Men's Clay Court Championships 2007 – tenisowy turniej ATP rangi ATP International Series z cyklu US Men's Clay Court Championships rozgrywany w dniach 9–15 kwietnia 2007 roku w Houston na kortach ziemnych o puli nagród 416 000 dolarów amerykańskich.
Gra pojedyncza
Zawodnicy rozstawieni
Drabinka
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Pula nagród
Gra podwójna
Zawodnicy rozstawieni
Drabinka
Pula nagród
Eliminacje gry pojedynczej
Drabinka eliminacji
Przypisy
Bibliografia
2007 w tenisie ziemnym
2007
2007 w Stanach Zjednoczonych | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,372 |
package sk.stuba.fiit.perconik.ivda.server;
import sk.stuba.fiit.perconik.ivda.activity.dto.BashCommandEventDto;
import sk.stuba.fiit.perconik.ivda.activity.dto.EventDto;
import sk.stuba.fiit.perconik.ivda.activity.dto.ProcessesChangedSinceCheckEventDto;
import sk.stuba.fiit.perconik.ivda.activity.dto.ide.IdeEventDto;
import sk.stuba.fiit.perconik.ivda.activity.dto.web.WebEventDto;
import sk.stuba.fiit.perconik.ivda.server.metrics.Loc;
import sk.stuba.fiit.perconik.ivda.server.metrics.SourceCodeMetric;
import javax.annotation.concurrent.ThreadSafe;
import javax.validation.constraints.NotNull;
/**
* Created by Seky on 16. 8. 2014.
* <p/>
* Pomocna trieda pre pracovanie s EventDto
*/
@ThreadSafe
public final class EventsUtil {
private static final SourceCodeMetric metric = new Loc();
@NotNull
public static String event2name(EventDto event) {
//noinspection IfStatementWithTooManyBranches
if (event instanceof WebEventDto) {
return "Web";
} else if (event instanceof IdeEventDto) {
return "Ide";
} else if (event instanceof BashCommandEventDto) {
return "Bash";
} else if (event instanceof ProcessesChangedSinceCheckEventDto) {
return "Iny proces";
} else {
return "Unknown";
}
}
public static int codeWritten(String txt) {
return metric.eval(txt);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 5,379 |
{"url":"https:\/\/cs.stackexchange.com\/questions\/22939\/if-i-solve-hard-instance-therefore-i-prove-np-p?noredirect=1","text":"# If I solve hard instance, therefore I prove NP=P? [duplicate]\n\nIf someone (off-topic) asks a question (on-topic) like this:\n\nSuppose that he claims that $\\mathcal{P=NP}$. Suppose that someone else (on-topic) gives him an instance of an NP-complete problem that cannot be solved by any computer optimally, i.e., to get the optimal solution, one must run an algorithm for very long time (the age of the universe for example).\n\nIf this someone (off-topic) solves this instance very fast optimally. Because the problem is NP-complete, we know that it can be easily verified.\n\nCan we verify easily that it is the optimal solution or we can only verify that it is just a solution?\n\nIn the other hand, if this someone (off-topic) solves every instance (hard instances) of an NP-hard problem very fast. Can we claim that he proved that $\\mathcal{P=NP}$?\n\nI want an answer, not a vote up\/down or on\/off-topic.\n\n\u2022 There is no such thing as a \"hard instance\". Any single instance of a problem can be solved in linear time: just check for that instance and output the correct answer, which is either \"yes\" or \"no\" and can be pre-computed. Since the answer to any problem in NP is either \"yes\" or \"no\", there is no such thing as an optimal solution. Mar 22 '14 at 19:54\n\u2022 If you give me an instance of sat and always I give you back yes or no correctly. Is this a proof? That's my question. Mar 22 '14 at 20:07\n\u2022 Related question. But suspect you need this one.\n\u2013\u00a0Raphael\nMar 22 '14 at 23:04\n\u2022 No, demonstrating that you can solve individual SAT instances says nothing about P=NP. Mar 26 '14 at 4:34\n\n$$\\mathrm{P}$$ and $$\\mathrm{NP}$$ are both sets of languages. A language $$A \\subseteq \\Sigma^{\\star}$$ is in $$\\mathrm{P}$$ ($$\\mathrm{NP}$$), if and only if there exists a deterministic (nondeterministic) turing machine $$M$$, that can decide in polynomial time for all $$x \\in \\Sigma^\\star$$, whether $$x \\in A$$.\n\nThere is an alternative definition of $$\\mathrm{NP}$$: $$A \\subseteq \\Sigma^\\star$$ is in $$NP$$, if there is an language $$L_0 \\in \\mathrm{P}$$ and an polynom $$p(x)$$, such that $$A = \\left\\{ x \\mid \\exists y \\text{ with } |y| \\leq p(|x|) \\text{ and } x\\#y \\in L_0 \\right\\}$$ That $$y$$ is called proof (or witness) that $$x$$ is in $$A$$. For example, the satisfiability problem is to check whether a given boolean formula is satisfiable or not. $$\\mathrm{SAT}$$ is the language of all satisfiable boolean formula. Given a boolean formula $$\\phi$$ (a instance of the satisfiability problem), we want to check whether $$\\phi$$ is satisfiable (if $$\\phi \\in \\mathrm{SAT}$$). A witness for $$\\phi$$ is an interpretation. We can determine in polynomial time, whether the interpretation of that formula is true or false.\n\nSuppose that someone else (on-topic) gives him an instance of an NP-complete problem that cannot be solved by any computer optimally, i.e., to get the optimal solution, one must run an algorithm for very long time (the age of the universe for example)\n\nFor descition problems, there no such thing as an optimal solution. The answer is yes or no (since the question is whether a string over an alphabet is in the language or not). Also, just because an algorithm runs for a very long time does not imply that the problem is in $$\\mathrm{NP}$$: Consider the worst case time complexity of something like $$n^{100}$$, which is even for small input sizes very high, but in $$\\mathrm{P}$$.\n\nBut lets consider we give him an instance of an $$\\mathrm{NP}$$-complete problem, for example of the satisfiability problem:\n\nIf this someone (off-topic) solves this instance very fast optimally. Because the problem is NP-complete, we know that it can be easily verified.\n\nCan we verify easily that it is the optimal solution or we can only verify that it is just a solution?\n\nAgain, for descition problems there is no such thing as an optimal solution. Therefore, we can verify an solution in polynomial time.\n\nIn the other hand, if this someone (off-topic) solves every instance (hard instances) of an NP-hard problem very fast. Can we claim that he proved that P=NP?\n\nIf he presents us an (correct) deterministic polynomial time algorithm, the answer is yes. Theoretically, he could just got lucky everytime and guess a correct solution, or own a non-deterministic turing machine. Since there is an infinite amount of $$\\mathrm{NP}$$-complete problems and only limited time, there is no way of solving all instances. Also, for only a finite amount of instances, one could use an (huge) lookup table.\n\nFor futher information regarding the $$\\mathrm{P}$$ vs $$\\mathrm{NP}$$ problem, you could look at the official problem description of the Clay Mathematics Institute.\n\n\u2022 Can you give me a good reference in P vs. NP definitions and related topics, ...? Like the definitions you used. Books, Tutorial, ... Mar 22 '14 at 20:09\n\u2022 @user3439590 Most of my knowledge and definitions are from lectures at my university. One source is he book Introduction to Automata Theory, Languages, and Computation by Hopcroft et. al., which is very comprehensive. Mar 22 '14 at 20:41\n\u2022 Your effort is appreciated, but I think your answer (and the question, for that matter) is redundant given the reference question we have. If you disagree, feel free to drop by in Computer Science Chat and discuss the matter!\n\u2013\u00a0Raphael\nMar 22 '14 at 23:05\n\u2022 Ok. It is off-topic and duplicate. I will delete it to not discuss the matter. Mar 22 '14 at 23:22","date":"2021-11-30 10:31:37","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 29, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6611390113830566, \"perplexity\": 306.2163448467013}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964358966.62\/warc\/CC-MAIN-20211130080511-20211130110511-00194.warc.gz\"}"} | null | null |
Q: How to use a value imported from a database Hey Everyone here is my question .
The code below gets data from my database and displays it both in an input field and a button. I want it to be in such a way that if i click the button it should get the value(which is imported from the db).But the problem i am facing is that all the inputs and buttons have the same ids so it only captures the value of the first button(or so i think). How can i make it in such a way that for every button i click it should have its own separate value.
<?php
$dbcon=mysqli_connect("localhost","root","");
mysqli_select_db($dbcon,"codex");
require('config/server.php');
?>
<table class="table table-striped">
<th>ID</th>
<?php
$view_users_query="select * from users";//select query for viewing
users.
$run=mysqli_query($dbcon,$view_users_query);//here run the sql
query.
while($row=mysqli_fetch_array($run))//while look to fetch the result
and store in a array $row.
{
?>
<!--here showing results in the table -->
<form id="loginForm" method="" action="" novalidate>
<tr>
<div class="panel2">
<main class="content">
<td><input name="hanis" id="hanis" type="text" value="<?php echo
$row['email']?>" autofocus /></td>
<td><button type="button" class="btn btn-success btn-block"
name="hanis" id="hanis" onclick="hanisdata()" value="<?php echo
$row['email']?>" ><?php echo $row['email']?></button><</td>
</main></div>
</div>
</div>
</form>
<?php } ?>
</tr>
<script type="text/javascript">
function hanisdata() {
var hanis=$("hanis").val();
alert(hanis);
// AJAX code to send data to php file.
$.ajax({
type: "POST",
url: "hanis.php",
data: {hanis:hanis},
dataType: "JSON",
success: function(data) {
$("#message").html(data);
$("p").addClass("alert alert-success");
},
error: function(err) {
alert(err);
}
});
}
</script>
A: NOTE :- Don't use same id for elements
You can get values by passing this with onclick function like onclick="hanisdata(this)"
Example
<button type="button" class="btn btn-success btn-block"
name="hanis" id="hanis" onclick="hanisdata(this)" value="<?php echo
$row['email']?>" ><?php echo $row['email']?></button>
Then you can get specific element in js and then can find parent and search for input field in that like below example.
JS CODE
<script type="text/javascript">
function hanisdata(el) {
var hanis=$(el).parent().find("input").val();
alert(hanis);
// AJAX code to send data to php file.
$.ajax({
type: "POST",
url: "hanis.php",
data: {hanis:hanis},
dataType: "JSON",
success: function(data) {
$("#message").html(data);
$("p").addClass("alert alert-success");
},
error: function(err) {
alert(err);
}
});
}
</script>
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,860 |
Dr. Buckley is a gynecologist in the Canton area. She is affiliated with Graham Hospital Association and Methodist Medical Center of Illinois.
Dr. Buckley saw pregnancy patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley treated pelvic surgery patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for pelvic surgery in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated vaginal delivery patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for vaginal delivery in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated more genetic testing patients than 85% of similar doctors in the Peoria area for 2014–2018. Learn more.
Dr. Buckley treated glucose challenge test patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for glucose challenge test in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated urine test patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for urine test in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated complete blood count (CBC) patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for complete blood count (CBC) in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated screenings patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for screenings in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated more ultrasound patients than 75% of similar doctors in the Peoria area for 2014–2018. Learn more.
Dr. Buckley saw pain patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley treated preventive care patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for preventive care in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated ovarian surgery patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for ovarian surgery in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated laparoscopic myomectomy patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for laparoscopic myomectomy in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated pain management patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for pain management in the Peoria area during 2014–2018. Learn more.
Dr. Buckley treated transvaginal ultrasound patients in 2014–2018, but we couldn't determine from our database whether she was was a top doctor for transvaginal ultrasound in the Peoria area during 2014–2018. Learn more.
Dr. Buckley saw morning sickness patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw digestive tract conditions patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw urinary incontinence patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw menopause patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw hormonal imbalance patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw ovarian cyst patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw high blood pressure (hypertension) patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw diabetes patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw vaginitis patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley saw fibroids patients in 2014–2018, but we couldn't determine whether she was a top doctor for the condition in the Peoria area. Learn more.
Dr. Buckley's C-section rate is similar to the rate we predicted for her mix of cases and patients.
We first measure the actual C-section rate for the pregnant patients Dr. Buckley treats.
We then analyze the typical practices of all other doctors in the United States who deliver babies, using a data-driven model to predict the C-section rate for patients like Dr. Buckley's.
By comparing these two numbers, you can see whether Dr. Buckley's C-section rate is lower than, similar to, or higher than predicted.
Dr. Buckley sees patients and can be reached at the following locations.
Dr. Buckley is able to admit and treat patients at the following hospitals. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,622 |
Reine Enchanteur ( 1967) was a Thoroughbred racehorse out of Sea Bird who was sold for a world-record $405,000 ($ million inflation adjusted) in 1968.
She earned a total career purse of $9,305. Owner Wendell P. Rosso used her for breeding, including with 1971 Blue Grass Stakes winner Impetuosity.
References
1967 racehorse births
Racehorses bred in Kentucky
Racehorses trained in the United States
Thoroughbred family 4-c | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,990 |
Amazonas Province was one of the provinces of the Empire of Brazil. It was created in 1850 from territory of Grão-Pará Province.
In 1889 it became Amazonas (Brazilian state).
1850 establishments in Brazil
1889 disestablishments in Brazil
Former countries in South America
Provinces of Brazil | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,085 |
ACCEPTED
#### According to
Index Fungorum
#### Published in
Sydowia 8(1-6): 67 (1954)
#### Original name
Gibbera lepontina E. Müll.
### Remarks
null | {
"redpajama_set_name": "RedPajamaGithub"
} | 346 |
{"url":"https:\/\/gmatclub.com\/forum\/if-h-x-2x-3-3-and-h-m-19-what-is-the-value-of-m-271710.html","text":"GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Aug 2018, 06:02\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47967\nIf h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Jul 2018, 01:10\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n93% (00:41) correct 7% (01:39) wrong based on 46 sessions\n\n### HideShow timer Statistics\n\nIf $$h(x) = 2x^3 - 3$$ and $$h(m) = -19$$, what is the value of m?\n\n(A) \u20133\n(B) \u20132\n(C) 2\n(D) 6,856\n(E) 6,862\n\n_________________\nManager\nJoined: 19 Nov 2017\nPosts: 146\nLocation: India\nSchools: ISB\nGMAT 1: 670 Q49 V32\nGPA: 4\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Jul 2018, 01:16\nBunuel wrote:\nIf $$h(x) = 2x^3 - 3$$ and $$h(m) = -19$$, what is the value of m?\n\n(A) \u20133\n(B) \u20132\n(C) 2\n(D) 6,856\n(E) 6,862\n\nIf\n$$h(x) = 2x^3 - 3$$\nthen\n$$h(m) = 2m^3 - 3$$\nand\n$$h(m) = -19$$\n\nSo\n$$2m^3 - 3 = -19$$\n$$m=-2$$ fits the equation\n\n$$2*(-8) - 3 = -19$$\n$$-16-3 = -19$$\n$$-19 = -19$$\n\nThus, B\n_________________\n\nRegards,\n\nVaibhav\n\nSky is the limit. 800 is the limit.\n\n~GMAC\n\nDirector\nJoined: 30 Jan 2015\nPosts: 525\nLocation: India\nConcentration: Operations, Marketing\nGPA: 3.5\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Jul 2018, 02:25\nh(x) = 2*x^3 \u2212 3 and h(m) = \u221219\nm = ?\n\nNow,\n-19 = 2*m^3 \u2212 3\n-16 = 2*m^3\n-8 = m^3\nm = -2\n\nHence, B.\n_________________\n\nThe few, the fearless !\n\nThanks\n\nRC Moderator\nStatus: Perfecting myself for GMAT\nJoined: 22 May 2017\nPosts: 432\nConcentration: Nonprofit\nSchools: Haas '21\nGPA: 4\nWE: Engineering (Computer Software)\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Jul 2018, 03:10\nBunuel wrote:\nIf $$h(x) = 2x^3 - 3$$ and $$h(m) = -19$$, what is the value of m?\n\n(A) \u20133\n(B) \u20132\n(C) 2\n(D) 6,856\n(E) 6,862\n\n$$h(x) = 2x^3 - 3$$ and $$h(m) = -19$$\n\n=> $$2m^3 - 3 = -19$$\n\n=> $$2m^3 = -19 + 3 = -16$$\n\n=> $$m^3 = -8$$\n\n=> m = -2\n\nHence option B\n_________________\n\nIf you like my post press kudos +1\n\nNew - RC Butler - 2 RC's everyday\n\nTag me in RC questions if you need help. Please provide your analysis of the question in the post along with the tag.\n\nIntern\nJoined: 24 Dec 2017\nPosts: 18\nLocation: India\nConcentration: Strategy, Real Estate\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Jul 2018, 04:50\nh(x) = 2x^3-3\nh(m) = -19\n\nh(m) = 2m^3-3\n2m^3-3 = -19\n2m^3 = -19+3\n2m^3 = -16\nm^3 = -16\/2\nm^3 = -8\nm = -2\n\nHence IMO B\nIntern\nJoined: 23 Jun 2018\nPosts: 14\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m?\u00a0 [#permalink]\n\n### Show Tags\n\n28 Jul 2018, 10:29\nBunuel wrote:\nIf $$h(x) = 2x^3 - 3$$ and $$h(m) = -19$$, what is the value of m?\n\n(A) \u20133\n(B) \u20132\n(C) 2\n(D) 6,856\n(E) 6,862\n\n2m^3-3 = -19\n2m^3 = -19+3 = -16\nm^3 = -8\nm=-2\n\nIMO B\nRe: If h(x) = 2x^3 \u2013 3 and h(m) = \u201319, what is the value of m? &nbs [#permalink] 28 Jul 2018, 10:29\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.","date":"2018-08-17 13:02:38","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5139427185058594, \"perplexity\": 11278.746160309143}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-34\/segments\/1534221212323.62\/warc\/CC-MAIN-20180817123838-20180817143838-00537.warc.gz\"}"} | null | null |
\section{Introduction} \label{sec:intro}
Interstellar dust grains location are a potential
diagnostic of the interaction between the solar wind and the interstellar
medium. The relative velocity between the Sun and the surrounding
interstellar cloud is 26.3 \hbox{km s$^{-1}$}, and neutral interstellar
gas and large dust grains flow through the
raindrop-shaped solar wind bubble known as the heliosphere. Interstellar ions and the
small interstellar dust grains that dominate interstellar polarization
are trapped in the
interstellar magnetic field that is deflected around the heliosphere.
The fundamental interstellar properties that dominate the
shape of the heliosphere are the direction and strength of the
magnetic field ${B_\mathrm{IS}}$, the ionization level, and the thermal and ram
pressure of gas \citep[e.g.][]{Holzer:1989}.
The upstream directions of interstellar H$^{\rm o}$\ and He$^{\rm o}$\ inside the
heliosphere differ by 5 \hbox{$^{\rm o}$}, serving as a diagnostic of
the interstellar magnetic field at the Sun
\citep{WellerMeier:1974,Quemeraisetal:2000,Lallementetal:2005}.
A second diagnostic of
${B_\mathrm{IS}}$\ at the heliosphere is the polarization of light from nearby
stars in the galactic center hemisphere, discovered by
\citep[][T82]{Tinbergen:1982} and now interpreted as partly caused by
interstellar dust grains trapped in the outer heliospheath regions
\citep[][Paper I]{Frisch:2005L}. The polarization maximum occurs
for stars near the ecliptic plane and located at ecliptic longitudes of
about $\sim +35^\circ$ from the upstream
directions given by interstellar dust and He$^{\rm o}$.
Three independent sets of data indicate that the heliosphere
shape is not symmetric around the upwind direction of interstellar
material (ISM) flowing into the heliosphere. Voyager 1 crossed the
solar wind termination shock (TS) at 94 AU in December 2004, and
Voyager 2 crossed the TS at $\sim 84$ AU in August 2007, indicating the TS
is closer by $\sim 10$ AU in the southern ecliptic compared to northern
ecliptic regions
\citep{Stone:2007,Stoneetal:2005,Burlagaetal:2005,Deckeretal:2005}.
These asymmetries result from an interstellar magnetic field inclined
to the velocity of inflowing ISM \citep{Pogorelovetal:2004,Opheretal:2007}.
Observations of low frequency radio emissions (1.8--3.6 kHz) by the
plasma wave instruments on Voyager 1 and 2 during 1992-1994 showed
that these events arise $\sim 130$ AU from the Sun at the edge of the
heliosphere, and that the events are consistent with an asymmetric
heliosphere \citep[][Appendix
\ref{app:khz}]{KurthGurnett:2003,GurnettKurthetal:2006}. The third
data set consists of measurements of $\sim 1$ keV energetic neutral atoms
(ENAs) near the heliosphere nose, that have an origin
offset to positive ecliptic longitudes compared to the heliosphere
nose direction of \hbox{$\lambda$}$\sim 10^\circ - 40^\circ$
\citep{Collieretal:2004,Wurzetal:2004}. ENAs are formed beyond the termination shock
by charge exchange between interstellar H$^{\rm o}$\ and the solar wind;
the ENAs show the bulges in the solar wind such as the
heliotail and the neutral current sheet wrapping around the
heliosphere nose \citep{Heerikhuisenetal:2007}.
The asymmetry of the outer heliosphere may vary with time
since the solar magnetic activity cycle causes the
solar and interstellar magnetic field lines to alternate between parallel and
antiparallel modes
\cite[e.g.][]{NerneySuess:1995,Macek:1990,WashimiTanaka:1996,Zurbuchen:2007}.
The Tinbergen polarization data and 3 kHz data were collected during
north-pole-positive polarities, with field lines emerging at the solar
north pole. The Voyager satellites crossing of the termination shock
occurred during solar cycle phases when the south pole had a
positive polarity.
The structure of this paper is as follows.
The polarization data that yield the local magnetic field
direction are discussed in \S \ref{sec:data}, with more
details in Appendix \ref{app:pa}.. The polarization data used in Paper I are
augmented with additional data acquired during the 1970's.
The question as to whether
this weak polarization originates at the Sun or in nearby ISM in the upwind
direction is discussed in \S \ref{sec:origin}.
The magnetic field direction of nearby and distant stars in
the ecliptic nose region for near and far stars are found to be similar
(\S \ref{sec:blocal}), and yield a magnetic field direction at the Sun
that is independent of whether the polarization is formed at the heliosphere or in the nearest upwind
ISM (\S \ref{sec:bfield}). Several characteristics of
the low-$\ell$ components of the cosmic
microwave background that are related to the heliosphere are discussed
in \S \ref{sec:cmb}, including the area vectors that occur
in the direction of the 3 kHz emissions,
the relation of the hot and cold poles of the CMB Doppler dipole
to the heliosphere, and the distribution of nearby ISM.
The influence on the local magnetic field direction of the S1 magnetic bubble
is discussed in \S \ref{sec:loop}.
Conclusions are presented in \S
\ref{sec:conclusions}. Appendix \ref{app:pa} presents additional
details about the optical polarization data, and discusses the
coincidental correlation between the position angles of distant stars
and ecliptic latitude. Appendix \ref{app:khz} briefly discusses primary
versus alternate locations found for the 3 kHz emissions observed by
the Voyager satellites.
The conclusions in the following discussions are sensitive to the physical
properties of the ISM surrounding the Sun. The ISM at the solar location has a temperature of 6300$\pm$340 K,
a heliocentric velocity of --26.3 \hbox{km s$^{-1}$}, and an upstream direction near the ecliptic
plane at \hbox{$\lambda$}=254.7$\pm$0.5\hbox{$^{\rm o}$}, \hbox{$\beta$}=5.1$\pm$0.2\hbox{$^{\rm o}$}\ (corresponding
to $\ell$=3.5\hbox{$^{\rm o}$}, $b$=15.2\hbox{$^{\rm o}$}), based on Ulysses interstellar
He$^{\rm o}$\ data \citep[][W04]{Moebiusetal:2004,Witte:2004}.
The ISM flow through the
heliosphere contains dust grains, which are observed at the same
velocity and flow direction as the gas
\citep[][]{Gruenetal:1993,Baguhletal:1996,Gruenetal:2005,Frischetal:1999}.
The dust-to-gas mass ratio for the ISM flowing into the
heliosphere is $<120$, compared to $>180$ for the LIC (SF07).
Models that evaluate the ionization gradient in the surrounding ISM
yield interstellar H$^{\rm o}$\ densities at the Sun of $n \mathrm{(H^\circ)}$=0.19 cm$^{-3}$, and
electron densities $n \mathrm{(e)}$$=0.07 \pm 0.01$ cm$^{-3}$\ \citep[e.g. Model 26
in][SF07]{SlavinFrisch:2007}. The plasma oscillation frequency of ISM
at the Sun is then $\sim 2.4$ kHz, but will rise to higher values in
the interstellar plasma pile-up in the outer heliosheath. If the
partially ionized gas close to the Sun is in pressure equilibrium with
${B_\mathrm{IS}}$, then $P_{ \rm B} \sim P_{ \rm th} \sim 0.2$ eV and ${B_\mathrm{IS}}$$\sim
2.8$ $\mu $G. If the gas and
interstellar magnetic field are in pressure equilibrium the
heliosphere has a Mach$\sim 1$ bowshock.
\section{Local Magnetic Field Direction from Interstellar Polarization Data}\label{sec:polarization}
The direction of the interstellar magnetic field at the Sun is
measured from starlight polarization.
Weak but systematic polarization of light from stars within 40 pc
in a region in the galactic center hemisphere was discovered by
\citet{Tinbergen:1982}. Tinbergen attributed this weak polarization,
where detections are at the $3 \sigma -5 \sigma$ level,
to a patch of ISM close to the Sun.
Tinbergen's data were acquired during solar minimum conditions,
in the south during 1974 and in the north
during 1973 (J. Tinbergen, private communication).
\footnote{During the mid-1970's, the solar magnetic polarity was north
pole positive (A$>$0, field lines emerging at the north pole).}
Paper I showed that this polarization reaches a maximum
at a location offset by $ \sim +35^\circ$
from the heliosphere nose. Below it is shown that this
maximum also occurs in the ecliptic plane, and that the weak polarizations
extend to negative ecliptic latitudes
in the eastern hemisphere of the heliosphere (\S \ref{sec:data}).
The weak polarization observed by Tinbergen appears to arise
from interstellar dust interacting with the heliosphere (\S \ref{sec:heliosphere}).
Data are sparse and additional observations are needed.
\subsection{Polarization Data for Nearby Stars}\label{sec:data}
The catalog of polarizations of nearby stars used here consists of
data from \citet{Tinbergen:1982} and \citet[][acquired in
1974]{Piirola:1977}. This combined Tinbergen-Piirola catalog (TPC)
has 202 stars, 86\% of which are within 40 pc (for Hipparcos distances).
The two data sets have 45 stars in common; the noisiest data from
the original studies are omitted from this discussion. The errors on
the mean Stokes parameters are discussed by Tinbergen, and are $6 - 17
\times 10^{-5}$; combining different measurements yielded $1
\sigma \sim 6 \times 10^{-5}$ \citep[Table 1 of][]{Tinbergen:1982}.
The unit $P_5$$=10^{-5}$ degree of polarization will be used. The
linear Stokes parameters U and Q give polarization $P$ =
(Q*Q+U*U)$^{1/2}$.
The polarizations of the TPC are plotted in Fig. \ref{fig:aitoff} in
an Aitoff projection. A region within 30\hbox{$^{\rm o}$}\ of the ecliptic meridian
$\lambda \sim 280^\circ$ shows the strongest polarizations down to latitudes of
$\beta \sim - 45^\circ$. The polarization for stars close to
the ecliptic plane reaches a maximum at a
location offset by $\Delta \lambda \sim +35^\circ$ from the heliosphere nose.
The average polarization as a function of $\lambda$ is plotted in Fig. \ref{fig:correl},
with averages smoothed over $\pm
20^\circ$ around the central ecliptic longitude \hbox{$\lambda_\mathrm{0}$}\ because
of the small number of stars. Averages are shown for stars within 20\hbox{$^{\rm o}$},
and within 50\hbox{$^{\rm o}$}, of the ecliptic plane. The polarization peaks at $\lambda \sim 295^\circ$
for stars with $| \beta | < 20^\circ$.
The polarization peak is offset by $\sim +30^\circ$ and $\sim +40^\circ$ from the
upwind directions of interstellar dust and He$^{\rm o}$, respectively.
The uncertainties on the polarization data have been carefully
carefully discussed by Tinbergen (1982) and Piirola (1977).
Paper I used only the Tinbergen (1982) data; the addition here of the Piirola (1977) data
improves the statistics for many regions of the sky, but not in the
heliosphere nose direction.
Additional information on the smoothing process
is given in Appendix \ref{app:pa}.
These polarizations are weak, $\sim 0.02$\%, and have been detected at the $3 \sigma -5
\sigma$ levels.
The polarization maximum regions coincides with
upstream direction of the ENA flux observed from the outer heliosphere
(\S \ref{sec:heliosphere}).
Fig. \ref{fig:aitoff} also shows the upstream direction of
dust grains flowing into the heliosphere
\citep[from Ulysses and Galileo data,][]{Frischetal:1999}
and of interstellar He$^{\rm o}$\ flowing into the heliosphere
\citep{Witte:2004,Moebiusetal:2004}.
The weak polarization becomes unobservable when the angle between the
sightline and the direction of the magnetic field, $\hat{k_\mathrm{B}}$, deviates
significantly from 90$^\circ$. This condition can be estimated
because the observed polarization can not be
identified when $P < 2.5 \sigma$, so the polarization is only detected
in stars within $\sim 40^\circ$ of the perpendicular direction to
${B_\mathrm{IS}}$.
\subsection{Is Polarization Origin at Heliosphere or in Nearby Upwind ISM?} \label{sec:origin}
\subsubsection{Polarization Origin at Heliosphere} \label{sec:heliosphere}
Several characteristics support the origin of this weak polarization
in the outer heliosheath region (defined as the region between the
heliopause and bow shock). The first is that the upwind polarizations
are strongest in the ecliptic plane (\S
\ref{sec:data}). The five stars that dominate the upstream
polarization maximum in the ecliptic plane have the same polarization
position angles (\S
\ref{sec:blocal}, Table \ref{tab:pa}, Appendix \ref{app:pa}).
The polarization maximum region is elongated along the ecliptic plane
towards positive longitudes.
This may indicate either a blunt heliosphere or the sideways
deflection of the grains following ${B_\mathrm{IS}}$\ around the heliosphere.
The second argument supporting an outer heliosheath origin of the
polarization is that the small grains capable of polarizing starlight
are tied by small gyroradii to the interstellar magnetic field
deflected around the heliosphere. The magnetic field upstream of the
heliopause filters out grains with large charge-to-mass ratios, Q/M,
and gyroradii that are smaller than the characteristic lengths between
the heliosphere bow shock and heliopause
\citep[e.g.,][]{Frischetal:1999,CzechowskiMann:2003,LindeGombosi:2000}.
Dust grains in the LIC are silicates (SF07). The radii of the
magnetically excluded grains are $\lesssim 0.1 - 0.25 $
$\mu $G, and are comparable to the sizes of interstellar polarizing grains \citep[e.g.][]{Mathis:1986}.
The coincidence of the polarization maximum location with the ENA flux
from the outer heliosphere \citep{Collieretal:2004,Wurzetal:2004} may
indicate that maximum charge densities in the outer heliosheath occur
where the strongest compression of the interstellar magnetic field
occurs. The photoejection of electrons gives
positively charged dust grains \citep{Weingartner:2004}, and
ENAs form where interstellar neutrals charge exchange with protons, indicating
ENAs and small dust grains in the outer heliosheath may have
similar distributions.
The properties of the outer heliosheath regions are not yet understood,
and IBEX data on ENAs will elucidate the processes of the outer heliosphere
\citep{McComasetal:2005sw11}.
The polarization increases by $\sim 46$\% between the $3
\sigma$ detection towards 36 Oph AB at 6 pc, and HD 161892 at 39 pc
and 5\hbox{$^{\rm o}$}\ away from 36 Oph AB. This suggests the observed
polarization is partly formed beyond the heliosphere in
the cloud, but that the polarization level is boosted to detectable
levels in the ecliptic plane where the interstellar magnetic field
is compressed between the bow shock and heliopause and
grain charging rates differ from those in the ISM.
Interstellar dust grains traversing the outer heliosheath
experience a gradient in magnetic field direction and strength, plasma
density, and grain charging rates, so that the grain gyroradii vary
with position. \citet{CzechowskiMann:2003} have modeled the passage of
interstellar dust grains across the transition region between the ISM and the heliopause
and found that in the presence of a strong shock small grains develop
velocity differences compared to the plasma velocity, and are
deflected to the heliosphere flanks. Such a process may explain
the $\sim 35^\circ$ offset in the polarization maximum from the upstream
direction. For some magnetic field directions and grain masses, grains are reflected back away from the
heliopause and form density waves. The following section shows that
collisional disruption of grain alignment is slow. This indicates that polarization
position angles will not vary if the ${B_\mathrm{IS}}$\ component in the plane of the
sky is relatively constant as the field is initially displaced
by the heliosphere.
The polarizations appear to be strong for the small
($\sim 150$ AU) distance between the heliopause and bow shock.
Comparisons between 36 Oph AB and HD 161892 suggests
that only half of the polarization must originate in the outer heliosheath
to boost the polarization strength to detectable levels (\S \ref{sec:data}).
There are no comparable data sets of high-sensitivity very weak optical
polarizations with which to compare the Tinbergen data.
\subsubsection{Polarization Origin in Upwind ISM}\label{sec:upwind}
The Sun is located in an interstellar cloud that is part of a
decelerating cluster of local interstellar cloudlets (CLIC) flowing away
from the direction of the Scorpius-Centaurus Association and past the
Sun \citep{Frisch:1995,FGW:2002}. Figure \ref{fig:localfluff1} shows
the distribution of neutral ISM within 15 pc for sightlines with
$N$(H$^{\rm o}$)$>1.0 \times 10^{18}$ cm$^{-2}$, based on data in \citet{Woodetal:2005}
and \citet{RLII}, and for some stars using the assumption D$^{\rm o}$/H$^{\rm o}$$=1.5 \times 10^{-5}$.
Interstellar H$^{\rm o}$\ within 15 pc is seen to coincide with the
eastern hemisphere of the heliosphere. The initial interpretation of the
Tinbergen data was that the polarizations were formed in this nearby
patch of ISM in the galactic center hemisphere
\citep[T82;][]{FrischYork:1983,Bruhweiler:1984}. Comparisons between
the polarizations of 36 Oph AB and HD 161892 suggest that the starlight
polarization observed in the ecliptic plane maximum also includes a contribution
from ISM upstream of the heliosphere nose (\S \ref{sec:data}).
This ISM within 15 pc appears to be the closest part of a magnetic shell
from Loop I that has expanded to the solar location (see \S \ref{sec:loop}).
The polarization maximum in the ecliptic plane is seen to occur
in a sightline through the boundary of the ISM within 15 pc.
The possibility that the CLIC is the exclusive origin of the upwind
polarizations concentrated in the ecliptic plane can not be ruled out.
Most of the polarization is formed near the ecliptic meridian $\lambda \sim 280^\circ$,
which cuts through the low column density boundary of the ISM shown in Fig. \ref{fig:localfluff1}.
Ionization levels and magnetic field strengths may be higher
towards the boundary regions than in the cloud interior, yielding stronger
polarizations.
Stars in the CLIC that are located between \hbox{$\lambda$}$\sim 330^\circ$ and \hbox{$\lambda$}$\sim 60^\circ$
are unpolarized.
For stars towards the polarization maximum near the ecliptic plane,
polarization position angles from interstellar dust grains in the outer
heliosheath and from grains in the upstream ISM should be the same,
since polarization position angles do not change significantly
over the nearest $\sim 200$ pc in this direction (\S \ref{sec:bheiles}),
Grain alignment persists over long time scales in low density gas such
as is upwind of the Sun, when compared to the denser clouds. Low
densities significantly reduce the collisional disalignment of the
dust grains when compared to denser clouds. The minimum and maximum
mean densities for the H$^{\rm o}$\ within 15 pc are $<$$n \mathrm{(H^\circ)}$$>$=0.01 cm$^{-3}$ and
0.12 cm$^{-3}$, respectively, and the average density for stars within 15 pc
is 0.05 atoms cm$^{-3}$. The LIC is less dense by a factor of $\sim$65,
warmer by a factor of $\sim$100, with proton densities a factor of
$\sim$10 larger than typical cold ISM. Grain alignment is
disrupted when grains accumulate their own mass in thermal collisions
with the gas, which occurs over time scales of $\tau_\mathrm{mass} \sim
10^6 a \rho_\mathrm{gr} / n
\sqrt{T_\mathrm{gas} }$ Myrs = 0.7--1.2 Myrs in the LIC. In the LIC
the collision rate is down by a factor of $\sim n / \sqrt{T} \sim 600
$ compared to denser clouds, so that even though ${B_\mathrm{IS}}$\ is weaker in
the LIC compared to dense clouds, it is fundamentally easier to
magnetically align ISDGs in the LIC than in denser clouds. Although
precise spatial densities are not known for the upwind ISM, the low
average densities indicate similar arguments will apply.
\section{Magnetic Field Direction at Polarization Maximum in Ecliptic Plane}\label{sec:bfield}
The direction of the local interstellar magnetic field is established
by polarization position angles of starlight. The magnetic field
direction towards the polarization maximum in the ecliptic plane is the same
for local stars (\S \ref{sec:blocal}) and distant stars in the same
region (\S \ref{sec:bheiles}). The polarization position angle in
celestial coordinates of the plane of vibration of the electric vector
is given by $\theta_{\rm C}$=0.5 arctan(U/Q) (also see Appendix
\ref{app:pa}). The interstellar magnetic field is parallel to the
polarization, as is shown by synchrotron emission that is polarized
perpendicular to the field direction \citep[e.g.][]{Heiles:1976araa}.
\subsection{Magnetic Field Direction near the Sun}\label{sec:blocal}
The stars in the polarization maximum near the heliosphere nose have
a mean position angle in ecliptic coordinates of PA$_\mathrm{E}$ $=-25^\circ \pm 4^\circ$ (Table \ref{tab:pa}).
The polarization position angles
do not vary significantly with ecliptic longitude in the
polarization maximum region (see Appendix \ref{app:pa}).
The interstellar magnetic field direction is parallel to the polarization vector
\citep{Spitzer:1978}.
The polarization position angles indicate that the
magnetic field direction close to the Sun
is inclined by $\sim 55^\circ$ with respect to the
galactic plane and $\sim 65^\circ$ with respect to the ecliptic plane,
regardless of whether the polarization originates in the outer
heliosheath or in the very local ISM.
The 5\hbox{$^{\rm o}$}\ angular offset between the interstellar He$^{\rm o}$\ and H$^{\rm o}$\
flowing through the heliosphere, discovered by
\citet{WellerMeier:1974} and now accurately determined using Ulysses
He$^{\rm o}$\ data and SOHO H$^{\rm o}$\ \hbox{Ly$\alpha$}\ florescence data, is also a diagnostic
of the interstellar magnetic field direction at the Sun
\citep{Witte:2004,Quemeraisetal:2000,Lallementetal:2005}. The H$^{\rm o}$\ and
He$^{\rm o}$\ upstream directions (Table \ref{tab:direct}) can be used to
define a position angle for comparison with the polarization data.
This position angle is the H$^{\rm o}$\ upwind direction with respect to a great
circle meridian passing through the He$^{\rm o}$\ upwind direction, and is
PA$_\mathrm{E,HHe} = -33^\circ \pm 14^\circ$ in ecliptic coordinates
(Table \ref{tab:pa}).
The magnetic field directions defined by starlight polarization in the
heliosphere nose, and by the 5\hbox{$^{\rm o}$}\ offset between H$^{\rm o}$\ and
He$^{\rm o}$\ flowing into the heliosphere, therefore give the same result to within the uncertainties.
However a component of the magnetic field that is parallel to the sightline
would not be detected by these methods.
\subsection{Magnetic Field Direction from Distant Stars} \label{sec:bheiles}
The upwind direction of the LIC corresponds to an astronomically
interesting direction, dominated by the North Polar Spur
\citep[][]{Frisch:1981}, which is the radio-intense section of the
Loop I superbubble shell that is traced by both polarized synchrotron
emission and magnetically aligned dust grains
\citep[e.g.][]{Heiles:1976araa}. As a test of the relation between
the nearby and distant magnetic field in the direction of the
heliosphere nose, the polarization position angles for stars in the
TPC are compared with those of distant stars in the Heiles
Polarization Catalog \citep[HPC,][available on Vizier as catalog
II/226]{Heiles:2000pol}. The scaled polarizations for stars 40--100
pc distant and within 20\hbox{$^{\rm o}$}\ of the ecliptic plane are plotted
versus \hbox{$\lambda$}\ in Fig. \ref{fig:correl}, where the polarizations are
smoothed over ecliptic intervals of $\pm 5^\circ$ from the central
longitude. Locations where the ecliptic crosses Loop I are seen as
two polarization peaks at \hbox{$\lambda$}$\sim 264^\circ$ and $\lambda \sim
220^\circ$, and the strongest peak near $\ell$$\sim 4^\circ$ and
$b$$\sim 4^\circ$ is within 10\hbox{$^{\rm o}$}\ of the ecliptic nose direction.
The polarizations of stars in the HPC within 250 pc are displayed in
Fig. \ref{fig:loop}, using only objects with HD numbers and
parallaxes in the Hipparcos catalog \citep{Perrymanetal:1997}. No
screening is made for discrepancies between the celestial and galactic
positional angles listed in the HPC. The familiar pattern of magnetic
filaments traced by polarization vectors is evident.
The comparison between the position angles of near and far stars in
the heliosphere nose region of maximum polarization has two distinct
properties. First, for stars in the heliosphere nose region the
average position angle of the distant stars in the HPC is
PA$_\mathrm{E}$$=-20^\circ \pm 20^\circ$, in agreement with the TPC data (Table \ref{tab:pa},
Fig. \ref{fig:loop}). Slightly different definitions of the
heliosphere nose region are used for the TPC versus the HPC stars
because of the small number of TPC stars (see Table \ref{tab:pa} for
region specifications). These position angles for the nearby TPC
stars agree to within uncertainties with those of the distant HPC
stars in the region of maximum nearby polarization. For example, the
nearby star HD 155885 (6 pc) and distant star HD 157236 (189 pc) in
the same direction towards the heliosphere nose both have the same
polarization position angle (PA$_\mathrm{G}$=39\hbox{$^{\rm o}$}\ in galactic coordinates), although the
polarization of HD 157236 is stronger than for HD
155885. Second, for a more extended spatial interval around
the heliosphere nose the position angles of the distant stars vary
systematically with \emph{ecliptic} latitude, due to the chance
alignment of a Loop I magnetic filament (this is shown in Appendix
\ref{app:pa}).
Close to the ecliptic plane
in the polarization maximum region the distant and nearby stars
show the same mean polarization position angles. The similarity in the polarization position angles of
nearby stars within 40 pc, and distant stars (140--240 pc) that trace
the Loop I magnetic field, indicates that
the interstellar magnetic field is
well-ordered over several hundred parsecs in this direction,
and interstellar densities are low, minimizing
the collisional disruption of the aligned grains.
It is suprising that the polarization data indicate the interstellar
magnetic is well-ordered over a few hundred parsecs, especially since
the distant region includes the Riegel-Crutcher cold cloud where
magnetic filaments are seen
\citep{RiegelCrutcher:1972,McClurGriffiths:2006}. However
the Local Bubble surroundings of the Sun
provide a minimal barrier to the expansion of Loop I \citep{Frisch:1981}.
Superbubbles formed by stellar evolution in
the Sco-Cen Association during the past $\sim 5$ Myrs evidently
dominate morphology of the local interstellar magnetic field.
\section{Evidence for Unrecognized Foreground Contamination of the Cosmic Microwave Background }\label{sec:cmb}
Paper I suggested that the interstellar dust grains trapped in the
outer heliosheath and deflected around the heliosphere may contribute
a weak unrecognized foreground to the cosmic microwave background.
The large scale CMB power spectrum contains poorly understood
anomalies related to ecliptic geometry, indicating that the low-$\ell$
CMB may have unrecognized very local foreground contamination related
to the ecliptic geometry
\citep[e.g.][]{Eriksenetal:2004,SchwarzStarkmanetal:2004}. These
anomalies appear in the Year 1, and remain in the Years 1-3 WMAP data
after instrumental corrections were improved
\citep[][]{HinshawNoltaetal:2007,Copietal:2007}.
The interstellar magnetic field near the Sun (\S \ref{sec:bfield}) and
small interstellar dust grains excluded from the heliosphere (\S
\ref{sec:data}) both provide possible foreground contaminants of the
CMB radiation.
In this section the geometrical properties of the CMB low-$\ell$
moments are compared to the geometrical properties of heliospheric
phenomena. Several coincidences are found. The geometrical
properties of what has been identified as the Doppler contribution to
the CMB dipole moment \citep[e.g.][]{Matheretal:1994} are related to
both the heliosphere geometry and the distribution of ISM within 35
pc. CMB low-$\ell$ multipole moments, including the non-cosmological
Doppler shift to the CMB dipole moment, and the quadrupole and octopole
multipoles, trace important heliospheric quantities.
The interstellar magnetic field at the Sun may influence these apparently independent
phenomena, and this field direction appears to be determined
by the S1 subshell of Loop I (\S \ref{sec:loop}).
This qualitive discussion focuses on the spatial coincidences
between the CMB and heliosphere properties. The source of possible
heliospheric contamination of the CMB microwave radiation is unknown.
The outer heliosphere region is only recently explored for the first
time by Voyager 1 and Voyager 2, which have crossed the solar wind
termination shock at 94 AU and 85 AU in December 2004 and August 2007,
respectively. Since the heliosphere itself varies with the magnetic
solar cycle of the Sun, any possible heliospheric foreground to the
CMB radiation may vary with solar cycle phase.
\subsection{CMB Features Related to Heliosphere and Local ISM }\label{sec:cmbhelio}
Figures \ref{fig:copi} and Fig. \ref{fig:ilc3} summarize the geometrical
coincidences between the CMB low-$\ell$ multipole moments and
heliospheric properties.
Structure in the Cosmic Microwave Background (CMB) signal over large
angular scales was detected in the COBE maps
\citep[][]{Smootetal:1992,KogutSmootetal:1992,Matheretal:1994}, and
persists in the high signal-to-noise Wilkinson Microwave Analyzer
Polarimeter (WMAP) data
\citep[e.g.][]{BennettHilletal:2003,HinshawNoltaetal:2007}. The
anisotropies of the CMB at large angular scales are dominated by
characteristic anisotropies which include a dipole moment of amplitude
$\Delta T/T \sim 10^{-3}$ due to the energy shift induced by the
Doppler motion through the CMB rest frame, a weaker quadrupole moment
with amplitude $\Delta T/T \sim 10^{-5.2}$, and weak higher order
asymmetries
\citep{Smootetal:1977,Smootetal:1992,KogutLineweaveretal:1993}.
The plane that separates the hot and cold hemispheres of the CMB
dipole moment traverses the heliosphere nose $\sim 6^\circ$ from the
upstream direction defined by interstellar He$^{\rm o}$\ flowing into the
heliosphere (Table \ref{tab:direct}). The poles of the CMB Doppler
motion are therefore nearly equidistant from the inflow direction of
ISM into the heliosphere, and the plane separating the hot and cold
hemispheres of the CMB dipole moment separates the eastern and western
parts of the heliosphere (Fig. \ref{fig:copi}).
\footnote{This fact does not appear to have been discussed
previously in the literature.
\citet{Smootetal:1977} report that Peebles referred to the
CMB kinematic Doppler dipole as the ``new aether drift''.} The
position angle defined by this plane, at the point of closest approach
to the heliosphere nose, is within 15\hbox{$^{\rm o}$}\ of the position angles of
the polarizations in the ecliptic plane maximum (Table
\ref{tab:pa}).
\footnote{By coincidence the heliocentric upstream direction of ISM
flowing through the heliosphere, which represents the vector sum of
the solar and LIC motions through the LSR, is within $\sim 5^\circ$ of
the ecliptic plane and $\sim 15^\circ$ of the galactic center.}
\nocite{Copietal:2007}
The plane separating the hot and cold hemispheres of the CMB dipole
moment therefore appears to be within $ \sim 15^\circ$ of the
interstellar magnetic field direction at the Sun found from optical
polarizations. For comparison, the WMAP ILC3 map data
\citep{HinshawNoltaetal:2007} are plotted in Fig. \ref{fig:ilc3} in
ecliptic coordinates, with the same features as shown as in
Fig. \ref{fig:copi}.
Anomalies also appear in the low-$\ell$ quadrupole and octopole
moments. Using ``multipole vectors'' to decompose the Internal Linear
Combination map, Years 1-3 (ILC3) data, Copi et al. (2007) found that
there is an alignment between the quadrupole and octopole planes, and
that the ecliptic traces a zero between the extremes in the quadrupole
and octopole moments. The multipole vectors carry the angular
distribution information of the power for each $\ell$-value. Copi et
al. showed that the $\stackrel{\rightarrow}{w}$ area vectors (normals
to planes defined by multipole pairs) coincide with the ecliptic plane
for southern galactic latitudes.
In Fig. \ref{fig:copi} the positions of the $\ell=$2 and 3 multipole
vectors $\hat{v}$, and $\stackrel{\rightarrow}{w}$ area vectors, are
plotted in ecliptic coordinates \citep[locations from ][Table 1,
ILC123 data]{Copietal:2007}. The $v^{22}$ component is directed
towards the heliosphere nose. The four area vectors $w^{323}$,
$w^{212}$, $w^{312}$, and $w^{331}$ are all directed towards or close
to the strip on the sky containing the 3 kHz emissions (Appendix
\ref{app:khz}) and the hot pole of the dipole. This strip is nearly
perpendicular to the ecliptic plane, in the sidewind direction
corresponding to ecliptic longitude $\lambda \sim 180^\circ$
(Fig. \ref{fig:aitoff}). The 3 kHz emissions are formed where outward
propagating global merged interaction regions interact with the
magnetically-shaped heliopause \citep{MitchellCairnsetal:2004}. The
low-$\ell$ moments moments of the CMB therefore appear to respond to
the ecliptic coordinate system partly because of the shaping of the
heliosphere by the interstellar magnetic field.
Separate analysis of the WMAP data as a function of solar cycle phase
may yield clues about the origin of the new foreground. Copi et
al. (2007) pointed out that there is a discrepancy between the Year 1
and Year 1--3 quadrupole moments. WMAP was launched in June 2001, and
the Year 1 data represent data acquired between Sept. 2001 and
Sept. 2002 during solar maximum conditions, versus between Sept. 2001
and Sept. 2004 for the Year 1--3 period that includes the solar
minimum conditions of 2004. The shift of the $v^{22}$ multipole from
\hbox{$\lambda$}=268.9\hbox{$^{\rm o}$}, \hbox{$\beta$}=15.6\hbox{$^{\rm o}$}\ during Year 1 (17.6\hbox{$^{\rm o}$}\ away from the
upwind direction) to \hbox{$\lambda$}=252.6\hbox{$^{\rm o}$}, \hbox{$\beta$}=8.2\hbox{$^{\rm o}$}\ during Years 1--3
(3.7\hbox{$^{\rm o}$}\ from the upwind direction) indicates that the process
contributing CMB foreground may be solar cycle dependent. An
analogous shift has not been reported for the upstream direction of
inflowing He$^{\rm o}$, which has been measured up to 2003 by Ulysses
\citep{Witte:2004,Moebiusetal:2004}.
\subsection{The CMB Dipole Moment and Local Interstellar Matter}\label{sec:cmbclic}
The cold hemisphere of the CMB dipole moment coincides
with the location of neutral ISM within 15 pc with $N$(H$^{\rm o}$)$>10^{-18}$ cm$^{-2}$\ (Fig. \ref{fig:localfluff1}).
When the highest column densities of interstellar H$^{\rm o}$\ within 35 pc
are plotted, the hot pole of the CMB dipole
moment is seen to coincide with distant parts of the S1 magnetic
shell (Fig. \ref{fig:localfluff2}, \S \ref{sec:loop}).
Together with the polarization data,
this indicates that the S1 magnetic bubble has expanded to the
solar location. CMB foreground templates calculated for synchrotron emissivity
do not include contributions from a local magnetic field
or this magnetic bubble.
\section{S1 Subshell of Loop I Superbubble as Explanation of Local Magnetic Field }\label{sec:loop}
The relation between the ISM at the Sun
and the Loop I superbubble formed by stellar evolution in the Scorpius-Centaurus
Association was discovered from the kinematics and abundances seen
in ISM upwind of the Sun \citep{Frisch:1981}, and has now been investigated
in a number of studies
\citep[e.g.][]{Crutcher:1982,Bochkarev:1984,Crawford:1991,deGeus:1992,Frisch:1995,Breitschwerdtetal:2000,MaizApel:2001}.
Recently the Loop I magnetic superbubble has been modeled as the
superposition of two magnetic bubbles with centers offset from each
other by $\sim 37^\circ$ \citep{Wolleben:2007}. The nearest of these
structures, S1, is centered at $\ell$$=346^\circ \pm 5^\circ$,
$b$$=3^\circ \pm 5^\circ$, distance $68 \pm 10$ pc, with radii for
the inner and outer rims of $72 \pm 10$ pc and $91 \pm 10$ pc,
respectively. The inner and outer tangential rims
of S1 are approximated as circles and plotted in
ecliptic coordinates in Figs. \ref{fig:localfluff1} and \ref{fig:localfluff2}, and in galactic coordinates in Fig. \ref{fig:loop}.
The nearest CLIC gas, $d < 15$ pc,
coincides roughly with the eastern part of the S1 shell (Fig. \ref{fig:localfluff1}).
Interstellar H$^{\rm o}$\ with $N$(H$^{\rm o}$)$> 2 \times 10^{-18}$ traces the tangential regions through
the western part of the shell (Fig. \ref{fig:localfluff2}). The
upwind direction of the CLIC in the local standard of rest
is centered near the center of the S1
shell.
The best fit bulk flow velocity of interstellar cloudlets within 30 pc is
$-17 $ \hbox{km s$^{-1}$}\ from $\ell$,$b$ = 0\hbox{$^{\rm o}$}, --5\hbox{$^{\rm o}$}\ \citep{FGW:2002},
using the Hipparcos solar apex motion \citep{DehnenBinney:1998}.
The CLIC H$^{\rm o}$\ densities decrease towards the shell center
as if the CLIC is part of the expanding shell. The H$^{\rm o}$\ minimum is
centered near $\ell$$\sim 330^\circ$ and $b$$\sim +8^\circ$, where
the radiation field from hot stars from the Upper Centaurus Lupus
subgroup of the Centaurus association appears to ionized upwind CLIC
gas in the solar vicinity. An additional source of
ionizing radiation is the white dwarf star WD1634-573, located
at $\ell$=330\hbox{$^{\rm o}$}, $b$=--7\hbox{$^{\rm o}$}, and where 60\% of the gas is ionized
\citep{Woodetal:2002}.
The locations of the two poles of the CMB dipole moment both fall just outside of the S1
shell, and coincide with the H$^{\rm o}$\ gas surrounding the S1 magnetic shell (Fig. \ref{fig:localfluff2}).
Partially ionized ISM is tied to the interstellar magnetic field so
that gas and dust are trapped in the expanding S1 magnetic shell. If
the LIC gas and magnetic field at the Sun are in pressure equilibrium,
then $|$${B_\mathrm{IS}}$$| \sim 2.8$ $\mu $G\ (Model 26, SF07). For the warm LIC, $T
> 5000$ K, the proton gyroradius is $< 0.4$ pc and the ISM will couple
to the magnetic field in the moving shell unless cloud densities are
$n \mathrm{(H^\circ)}$$> 2$ cm$^{-3}$\ \citep[which is possible but not
likely,][]{Frisch:2003apex}. However there is a well-known
discontinuity in the velocity of the LIC versus the upstream gas
towards the nearest star $\alpha$ Cen
\citep[e.g.][]{LinskyWood:1996,Landsmanetal:1984}, indicating possible
substructure in the S1 shell induced by higher field strengths. The
position angle data indicate that the magnetic field direction does
not vary significantly over several hundred parsecs in the upstream
direction in the region of the ecliptic-plane-polarization-maximum, so there must be empty regions of space with minimal dust
between the near and far fields, such as expected for the interior of
a magnetic bubble.
Rotation measures of pulsars and extragalactic radio sources show that
the uniform component of the interstellar magnetic field in the
interarm region around Sun is directed towards $\ell$$\sim 82^\circ$.
The strength of the large-scale galactic regular magnetic field at the
solar radius is $B_\mathrm{IS} \sim 2.1 \pm 0.3 $ $\mu $G, found from
rotation and dispersion measures towards an assembly of pulsars within
several kiloparsecs of the Sun \citep{Han:2006}. Faraday rotation data
show that the large-scale magnetic field reverses polarity around the
galactic $\ell$=0\hbox{$^{\rm o}$}\ meridian at the galactic center and also
around the galactic plane. In the part of the CLIC within 15 pc,
$b$$<0^\circ$ and \hbox{$\lambda$}$>270^\circ$, the global field polarity is directed
away from the Sun.
The polarization data establish the direction of the magnetic field
at the Sun (Table \ref{tab:pa}). Radio synchrotron polarization data
indicate that the optical polarization vector is parallel to the
magnetic field direction in the Loop I filament, and there is no
reason to assume otherwise for the local ISM, especially given the
coincidence in position angles for near and far polarization.
However, there may be a component of ${B_\mathrm{IS}}$\ that is perpendicular to
the plane of the sky that is not sampled by optical polarization. For
a high-latitude direction through the distant Loop I, the ${B_\mathrm{IS}}$\ component parallel
to the sightline is small. At the location $\ell$$= 34^\circ$,
$b$$=42^\circ$, where a $\sim 70 \pm 30$ pc tangential sightline
through Loop I is sampled, the volume-averaged field strength is ${B_\mathrm{IS}}$$
\sim 4$ $\mu $G\ \citep{Heilesetal:1980}. For this sightline, the
Faraday rotation of extragalactic radio sources indicate the ${B_\mathrm{IS}}$\
component parallel to the sightline is small, with an average value of
$ B_\mathrm{||} = 0.9 \pm 0.3 $ $\mu $G\ \citep{Fricketal:2001}.
\section{Conclusions} \label{sec:conclusions}
\begin{itemize}
\item
The polarizations of nearby stars, $< 45 $ pc, form a maximum
in the ecliptic plane that peaks at an ecliptic longitude shifted by
$\sim 35^\circ$ from the upstream direction of interstellar gas and
dust flowing into the heliosphere. Half of this polarization is
formed within $\sim 6$ pc of the Sun, and the polarization maximum
indicates a significant contribution from
small interstellar dust grains interacting with the outer
heliosheath regions.
\item
The polarization position angles of nearby and distant stars in this
polarization peak agree to within the uncertainties (Table
\ref{tab:pa}). The position angle defined by the offset between the
upstream directions of interstellar He$^{\rm o}$\ and H$^{\rm o}$\ in the heliosphere
also agrees with the position angles found from the polarization-peak
stars, to within the uncertainties.
\item
These polarization data give a direction for the
interstellar magnetic field at the Sun that is inclined by $\sim
65$\hbox{$^{\rm o}$}\ with respect to the ecliptic plane, and $\sim 55$\hbox{$^{\rm o}$}\ with
respect to the galactic plane. An additional component into the plane
of the sky may also be present. If the interstellar magnetic field
direction nearby is similar to the global field in the solar vicinity,
then the field is directed away from the Sun east of the heliosphere
nose in ecliptic coordinates.
\item
The plane that separates the cold and hot hemispheres of the CMB
dipole moment also divides the east and west hemispheres of the
heliosphere, and also encloses most sightlines within 15 pc that have
interstellar $N$(H$^{\rm o}$)$>1.0 \times 10^{18}$ cm$^{-2}$.
\item The CMB quadrupole area vectors of \citet{Copietal:2007} line up
with the positions of the locations of the 3 kHz events detected by
Voyagers 1 and 2. A mix of primary and alternate locations for the 3
kHz emissions are used for this comparison (Appendix \ref{app:khz}).
\item
A plane can be formed from points that are equidistant between the
two poles of the CMB dipole.
At the closest point to the heliosphere nose
direction defined by inflowing interstellar He$^{\rm o}$, the position angle of this plane is within $\sim 15^\circ$
of the position angles of the optical
polarizations that form the ecliptic-plane-polarization-maximum.
\item
\citet{Wolleben:2007} has modeled the Loop I magnetic field in terms
of two bubbles, one of which (S1) has expanded to the solar location.
The distribution of interstellar H$^{\rm o}$\ within 35 pc shows evidence of a
shell morphology coinciding with the rims of S1. The eastern part of
the shell, in ecliptic coordinates, is closest to the Sun. The
upstream direction of the ISM flow past the Sun is towards the center
of the S1 shell. The polarizations seen by Tinbergen are somewhat concentric
with the curvature of the S1 shell for southern ecliptic regions.
\item
The hot and cold CMB Doppler dipole moment poles are both directed
toward the boundaries of the S1 shell.
\item
These results suggest that the magnetic field associated with
the S1 subshell of the Loop I supernova remnant
creates the interstellar magnetic field at the heliosphere nose.
The S1 shell dominates the polarization of stars close to the Sun,
the distribution of ISM within 15 pc, the heliosphere
configuration that yielded the 3 kHz emissions, and the hot
and cold poles of the CMB Doppler dipole moments.
\item
The CMB foreground that mimics the spatial
distribution of the 3 kHz emissions and the east-west heliosphere
symmetry may arise from the heliosphere itself,
or from another property that is also affected by the
interstellar magnetic field in the S1 shell, such as the spectrum of synchrotron emission or dust grains trapped and heated in the shell.
\item
The polarization maximum in the ecliptic plane is dominated by five stars
with the same polarization position angles. The polarization detections
by Tinbergen (1982) for these stars are $3 \sigma - 5 \sigma$, and
additional data on the polarization of nearby stars are desirable.
Tinbergen's data were collected during solar minimum in the mid-1970's,
and there may be a solar cycle dependence to the polarization strength.
\end{itemize}
\acknowledgements This research has been supported by the NASA the
grants NAG5-13107 and NNG05GD36G to the University of Chicago. I
would like to thank Dragan Huterer for important comments on the CMB
multipoles, Carl Heiles for helpful suggestions, and Hiranya Peiris,
Cora Dvorkin, and Bruce Winstein for providing the ILC map data for
Years 1-3 in {\sc ascii} form.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,649 |
class MyFirstCairoPlugin : public IPlug
{
public:
MyFirstCairoPlugin(IPlugInstanceInfo instanceInfo);
~MyFirstCairoPlugin();
void Reset();
void OnParamChange(int paramIdx);
void ProcessDoubleReplacing(double** inputs, double** outputs, int nFrames);
private:
double mGain;
};
#endif
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,518 |
\section{Introduction}
Statistical channel models are needed for system-level network simulations and air interface design. Current radio-systems use the 3GPP and WINNER spatial channel models, valid for 1 - 6 GHz, and for RF signal bandwidths up to 100 MHz~\cite{3GPP:1,WinnerII}. The COST 2100 models utilize circular regions (e.g., visibility regions) whose size vary as a user equipment (UE) physically moves and interacts with more or less scattering objects to model realistic links~\cite{Liu12}. This modeling approach supports temporal and spatial channel correlations over large-scale distances, allowing for smooth channel transitions without discontinuities between closely separated UEs (e.g., spatial consistency)~\cite{Liu12,metis}.
Recently, new channel modeling frameworks have been developed to characterize millimeter-wave (mmWave) bands. The MiWEBA models use a 60 GHz quasi-deterministic channel model, where strong deterministic components are modeled using Friis' free space path loss equation and path-length geometry, while the properties of weaker clusters of multipaths are generated from measurement-based statistical distributions~\cite{miweba}. The METIS models, in contrast, use a combination of map-based and geometry-based stochastic models to generate channel coefficients~\cite{metis}.
In this paper, the original statistics from a measurement-based channel impulse response (CIR) model~\cite{Samimi15_2} are extended to include multiple frequencies and arbitrary antenna beamwidths to recreate directional CIRs. A multi-frequency 3-dimensional (3-D) CIR model is presented here, based on 28 GHz and 73 GHz ultrawideband channel measurements in New York City~\cite{Rap13:2, Rap15_3}. The model supports arbitrary carrier frequency, RF signal bandwidth, and antenna beamwidth, following a 3GPP-like modeling approach. The model uses time clusters and spatial lobes to represent the propagation channel~\cite{Samimi15_2}. Note that the 28 GHz and 73 GHz frequency bands for outdoor communications are attractive, as the Federal Communications Commission (FCC) and other governments are about to issue rulemaking, bringing these bands into service~\cite{Rap15_3,MacCartney15_2}.
The measurements~\cite{Rap15_3} provided over 12,000 measured power delay profiles (PDPs) obtained at unique transmitter (TX) - receiver (RX) pointing angles to construct an accurate statistical spatial channel model (SSCM) at both 28 GHz and 73 GHz separately in non-line of sight (NLOS), and in line of sight (LOS) conditions where the 28 GHz and 73 GHz statistics were combined. The LOS statistics from 28 and 73 GHz were lumped into one data set to extract combined 28 GHz and 73 GHz statistics, motivated by the nearly identical channel statistics for LOS environments at both frequencies. For example, the LOS path loss exponents (PLE) are both nearly identical to free space ($n=2$)~\cite{MacCartney14:2,Rap15_3}, and have virtually the same values of average number of multipath components~\cite{Rap15_3}. Thus, one early finding is that LOS mmWave channels behave very similarly in terms of path loss, temporal, and spatial properties in~\cite{Rap15_2,MacCartney15_2} as long as there is no oxygen absorption. Thus, it makes sense to generate simplified LOS models that pool data at different frequencies when creating CIRs.
\section{Measurement Description}
28 GHz and 73 GHz wideband propagation measurements were performed at 74 and 36 RX locations, and for three and five distinct TX sites, respectively, with TX-RX distances ranging from 31 m to 425 m on the streets of New York City~\cite{Rap13:2,Rap15_3}. A 400 megachips-per-second broadband sliding correlator channel sounder and highly directional horn antennas were used to recover angle of departure (AOD) and angle of arrival (AOA) statistics. The directional, steerable horn antennas were exhaustively rotated in azimuth and elevation in half-power beamwidth (HPBW) step increments, and many thousands of PDPs were collected at distinct azimuth and elevation unique pointing angles, which provided the means to develop the necessary statistical channel models~\cite{Rap15_3}. Note that impact of sidelobes was minor, with -20 dB sidelobe levels, and a cross-polarization discrimination factor of 21 dB and 25.4 dB for the 28 GHz and 73 GHz outdoor measurements~\cite{Rap15_2}, respectively. The measurement system provided excess delays of the multipath arrivals at the many recorded unique TX-RX pointing angles, and complementary 3-D ray-tracing recreated absolute timing of multipath arrivals from TX to RX~\cite{Samimi15_2}. Table I of~\cite{Rap15_3} gives details about the campaigns and equipment used.
\section{3-D Impulse Response Channel Model}
The SSCM presented here uses temporal clusters and spatial lobes to model the mmWave channel, motivated by observations of the collected New York City measurements~\cite{Rap13:2,Rap15_3}. A temporal cluster represents a group of traveling multipath components arriving closely spaced in time from arbitrary angular directions, while a spatial lobe represents a main directional of arrival where many temporal clusters can arrive at different time delays~\cite{Samimi15_2}. In the 3GPP model, a path is defined as a time-delayed multipath copy of the transmitted signal, that is sub-divided into $M = 20$ subpaths, whose delays are identical to the path delay~\cite{3GPP:1,Calcev07:1}, where each multipath component in the channel is a part of the CIR as shown in~(\ref{IREq}). The WINNER model defines a cluster as a propagation path diffused in space, either or both in delay and angle domains, and is sub-divided into 20 rays, where the two strongest clusters are sub-divided into three sub-clusters with delay offsets of 0 ns, 5 ns, and 10 ns~\cite{WinnerII}. In the COST 2100 model, the time delay of a multipath component is the sum of three delays: the base station (BS)-to-scatterer delay, the mobile station (MS)-to-scatterer delay, and the cluster-link delay~\cite{Liu12}. Ultrawideband PDP measurements at mmWave frequencies, obtained with greater temporal (2.5 ns) and narrower spatial ($7^{\circ},10^{\circ}$) resolution over models in~\cite{3GPP:1,WinnerII,Liu12} indicate that temporal clusters are composed of many intra-cluster subpaths with different random delays, as shown in Fig. 13 of~\cite{Rap15_3}. Thus, the approach presented here offers an improvement over existing models developed for lower frequencies and bandwidths.
The double-directional omnidirectional CIR is commonly used to represent the radio propagation channel between a transmitter and receiver, and can be expressed as in~(\ref{IREq})~\cite{Steinbauer01,Samimi15_2},
\begin{equation}\label{IREq}\begin{split}
h_{omni}(t,\overrightarrow{\mathbf{\Theta}},\overrightarrow{\mathbf{\Phi}}) =& \sum_{n=1}^{N} \sum_{m=1}^{M_n} a_{m,n} e^{j\varphi_{m,n}} \cdot \delta(t - \tau_{m,n}) \\
& \cdot \delta(\overrightarrow{\mathbf{\Theta}}-\overrightarrow{\mathbf{\Theta}}_{m,n}) \cdot \delta(\overrightarrow{\mathbf{\Phi}}-\overrightarrow{\mathbf{\Phi}}_{m,n})
\end{split}\end{equation}
\noindent where $t$ denotes absolute propagation time, $\overrightarrow{\mathbf{\Theta}}=(\theta,\phi)_{TX}$ and $\overrightarrow{\mathbf{\Phi}}=(\theta,\phi)_{RX}$ are the vectors of azimuth/elevation AODs and AOAs, respectively; $N$ and $M_n$ denote the number of time clusters (defined in~\cite{Samimi15_2}), and the number of cluster subpaths, respectively; $a_{m,n}$ is the amplitude of the $m$\textsuperscript{th} subpath belonging to the $n$\textsuperscript{th} time cluster; $\varphi_{m,n}$ and $\tau_{m,n}$ are the phases and propagation time delays, respectively; $\overrightarrow{\mathbf{\Theta}}_{m,n}$ and $\overrightarrow{\mathbf{\Phi}}_{m,n}$ are the azimuth/elevation AODs, and azimuth/elevation AOAs, respectively, of each multipath component.
The statistical channel model presented here also produces the joint AOD-AOA power spectra $P(\overrightarrow{\mathbf{\Theta}},\overrightarrow{\mathbf{\Phi}})$ in 3-D obtained by integrating the magnitude squared of~(\ref{IREq}) over the propagation time dimension,
\begin{align}\label{3DEq1}
P(\overrightarrow{\mathbf{\Theta}},\overrightarrow{\mathbf{\Phi}}) &= \int_{0}^{\infty} |h(t,\overrightarrow{\mathbf{\Theta}},\overrightarrow{\mathbf{\Phi}}) |^2 dt\\
\begin{split}P(\overrightarrow{\mathbf{\Theta}},\overrightarrow{\mathbf{\Phi}}) &= \sum_{n=1}^{N} \sum_{m=1}^{M_n} |a_{m,n}|^2 \\ \label{3DEq2}
&\cdot \delta(\overrightarrow{\mathbf{\Theta}}-\overrightarrow{\mathbf{\Theta}}_{m,n}) \cdot \delta(\overrightarrow{\mathbf{\Phi}}-\overrightarrow{\mathbf{\Phi}}_{m,n})
\end{split}
\end{align}
\noindent Current channel models~\cite{3GPP:1,WinnerII,Liu12} use \textit{global} azimuth and elevation spreads to quantify the degree of angular dispersion over the $4\pi$ steradian sphere, using~(\ref{3DEq1}) and the equations in Annex A of~\cite{3GPP:1}. The RMS \textit{lobe} angular spread~\cite{Samimi15_2} is different from the global angular spread, as it only considers the strongest measured lobe directions (as opposed to the entire $4\pi$ steradian power spectrum over space). Typical spatial lobes have absolute and RMS lobe azimuth spreads of $30^{^\circ}$ and $6^{\circ}$~\cite{Samimi15_2}, respectively, thereby quantifying spatial directionality for realistic multi-element antenna simulations, to emulate beamforming of future directional mmWave systems in the strongest angular directions.
The omnidirectional CIR can further be partitioned to yield \textit{directional} PDPs at a desired TX-RX unique antenna pointing angle, and for arbitrary TX and RX antenna patterns,
\begin{equation}\label{DIREq}\begin{split}
h_{dir}&(t,\overrightarrow{\mathbf{\Theta}_d},\overrightarrow{\mathbf{\Phi}_d}) = \sum_{n=1}^{N} \sum_{m=1}^{M_n} a_{m,n} e^{j\varphi_{m,n}} \cdot \delta(t - \tau_{m,n}) \\
& \cdot g_{TX}(\overrightarrow{\mathbf{\Theta}_d}-\overrightarrow{\mathbf{\Theta}}_{m,n}) \cdot g_{RX}(\overrightarrow{\mathbf{\Phi}_d}-\overrightarrow{\mathbf{\Phi}}_{m,n})
\end{split}\end{equation}
\noindent where ($\overrightarrow{\mathbf{\Theta}_d},\overrightarrow{\mathbf{\Phi}_d})$ are the desired TX-RX antenna pointing angles, $g_{TX}(\overrightarrow{\mathbf{\Theta}})$ and $g_{RX}(\overrightarrow{\mathbf{\Phi}})$ are the arbitrary 3-D (azimuth and elevation) TX and RX complex amplitude antenna patterns of multi-element antenna arrays, respectively. In~(\ref{DIREq}), the TX and RX antenna patterns amplify the power levels of all multipath components lying close to the desired pointing direction, while effectively setting the power levels of multipath components lying far away from the desired pointing direction to 0.
\subsection{Step Procedures for Generating Channel Coefficients}
\label{sec:mmWaveProc}
The step procedure for generating temporal and spatial mmWave channel coefficients is outlined below. In the following steps, $DU$ corresponds to the \textit{discrete uniform} distribution, and the notation $[x]$ denotes the closest integer to $x$. Steps 11 and 12 apply to both AOD and AOA spatial lobes.
\noindent \textit{Step 1: Generate the TX-RX separation distance d (in 3-D) ranging from 30 - 60 m in LOS, and 60 - 200 m in NLOS (based on our field measurements, and may be modified)}:
\begin{equation}\label{step1}
d \sim U(d_{min},d_{max})
\end{equation}
\noindent where, \[
\begin{cases}
d_{min}=30 \text{ m}, d_{max} = 60 \text{ m} ,& LOS\\
d_{min}=60 \text{ m}, d_{max} = 200 \text{ m} ,& NLOS\\
\end{cases}\]
\noindent To validate our simulation, we used the distance ranges in Step 1, but for standards work other distances are likely to be valid. Users located near BSs (i.e., small TX-RX separation) will be power-controlled in the near field~\cite{MacCartney15_2,Thomas15}.
\noindent \textit{Step 2: Generate the total received omnidirectional power Pr (dBm) at the RX location according to the environment type:}
\begin{align}
&P_r (d)[dBm] = P_t[dBm] - PL(d)[dB]\\
&PL[dB](d) = PL(d_0) + 10\overline{n}\log_{10}\left( \frac{4\pi d}{\lambda} \right)+\chi_{\sigma}\\
&PL(d_0) = 20 \times \log_{10}\left(\frac{4\pi d_0}{\lambda}\right)
\end{align}
\noindent where $P_t$ is the transmit power in dBm, $d_0 = 1$ m, $\lambda$ is the carrier wavelength, $\overline{n}$ is the path loss exponent (PLE) for omnidirectional TX and RX antennas, given in Table~\ref{tbl:50} for 28 GHz or 73 GHz in both LOS and NLOS environments, and $\chi_{\sigma}$ is the lognormal random variable with 0 dB mean and standard deviation $\sigma$~\cite{Rap15_3}. The $d_0=1$ m close-in (CI) free space reference path loss model is a simple physically-based one-parameter (PLE) model~\cite{Rap15_3,Sun15}, that is more stable across frequencies and environments, than the traditional floating-intercept (FI) least-squares regression equation line~\cite{Rap15_3,Thomas15,MacCartney15_2}. Further, the CI and FI models perform similarly over identical data sets, with differences in standard deviations that are within a fraction of a dB~\cite{Rap15_3,MacCartney15_2,Sun15}. Also, the CI model allows the pooling of LOS power statistics at multiple mmWave frequencies without any change in model coefficients (this is not the case for the FI model). Note that we used $n=2$ to simulate free space propagation in LOS.
\noindent \textit{Step 3: Generate the number of time clusters N and the number of AOD and AOA spatial lobes $(L_{AOD}, L_{AOA})$ at the RX location:}
\begin{align}
&N \sim DU[1,6]\\
& L_{AOD} \sim \min \bigg\{ L_{max},\max \Big\{ 1, \text{Poisson} \big( \mu_{AOD} \big) \Big\} \bigg\}\\
& L_{AOA} \sim \min \bigg\{ L_{max},\max \Big\{ 1, \text{Poisson} \big( \mu_{AOA} \big) \Big\} \bigg\}
\end{align}
\noindent where $L_{max}=5$ is the maximum allowable number of spatial lobes, $\mu_{AOD}$ and $\mu_{AOA}$ are the empirical mean number of AOD and AOA spatial lobes, respectively (see Table~\ref{tbl:softTable}). At 28 GHz in NLOS, the maximum number of time clusters observed was 5, while it was 6 at 73 GHz, using a -10 dB threshold based on work in~\cite{Samimi15_2}. We therefore choose 6 to simplify the model across frequency bands. Note that in~\cite{Samimi15_2}, ($L_{AOD},L_{AOA}$) were conditioned upon $N$, but since subpaths from the same time cluster can arrive and depart from arbitrary directions, the number of spatial lobes is here generalized to be independent of the number of time clusters.
\noindent \textit{Step 4: Generate the number of cluster subpaths (SP) $M_n$ in each time cluster:}
\begin{align}
&M_n \sim DU[1,30]\hspace{.3cm},\hspace{.3cm} n = 1, 2, ... N
\end{align}
\noindent At 28 GHz in NLOS, the maximum and second to maximum number of cluster subpaths were found to be 53 and 30, respectively, over all locations, while at 73 GHz the maximum was 30 in NLOS, therefore 30 is chosen as the upper bound of the uniform distribution for all frequencies. Subpath components were identified using a peak detection algorithm.
\noindent \textit{Step 5: Generate the intra-cluster subpath excess delays $\rho_{m,n}$}:
\begin{align}
&\rho_{m,n}(B_{bb}) =\left\{ \frac{1}{B_{bb}}\times (m-1) \right\}^{1+X}\\
&m = 1, 2, ..., M_n \hspace{.3cm},\hspace{.3cm} n = 1, 2, ..., N
\end{align}
\noindent where $B_{bb}=400$ MHz is the baseband bandwidth of the transmitted PN sequence (but can be modified for different baseband bandwidths less than 400 MHz), and $X$ is uniformly distributed between 0 and $X_{max}$. This step ensures a bandwidth-independent channel model, while reflecting observations that intra-cluster subpath delay intervals tend to increase with delay (through the random variable $X$). The upper bound $X_{max}$ is easily adjustable to field measurements (see Table~\ref{tbl:softTable}).
\noindent \textit{Step 6: Generate the cluster excess delays $\tau_n$:}
\begin{align}\label{mu_tau}
&\tau^{\prime\prime}_n \sim \text{Exp}(\mu_{\tau})\\
&\Delta \tau_n = \text{sort}(\tau^{\prime\prime}_n)-\min(\tau^{\prime\prime}_n)
\end{align}
\vspace{-.6cm}
\begin{align}
&\tau_n =
\begin{cases}
0, &n = 1\\
\tau_{n-1}+\rho_{M_{n-1},n-1}+\Delta \tau_n+ 25, &n = 2, ..., N
\end{cases}
\end{align}
\noindent where $sort()$ orders the delay elements $\tau^{\prime\prime}_n$ from smallest to largest, and where $\mu_{\tau}$ is given in Table~\ref{tbl:softTable}. This step assures no temporal cluster overlap with a 25 ns minimum inter-cluster void interval. The value of 25 ns for minimum inter-cluster void interval was found to match the measured data, and makes sense from a physical standpoint, since multipath components tend to arrive in clusters at different time delays~\cite{Saleh87} over many angular directions, most likely due to the free space air gaps between reflectors (buildings, lampposts, streets, etc). The narrowest streets have a typical spatial width of 8 m (25 ns in propagation delay) in New York City, thus physically describing the regularly observed minimum void interval for arriving energy.
\noindent \textit{Step 7: Generate the time cluster powers $P_n$ (mW):}
\begin{align}\label{eqCluster2}
&P^{\prime}_n = \overline{P}_0 e^{-\frac{\tau_n}{\Gamma}} 10^{\frac{Z_n}{10}}\\ \label{eqCluster}
&P_n = \frac{P^{\prime}_n}{\sum^{k=N}_{k=1} P^{\prime}_k}\times P_r [mW]\\
&Z_n \sim N(0, \sigma_Z )\hspace{.2cm},\hspace{.2cm}n = 1, 2, ... N
\end{align}
\noindent where $\overline{P}_0$ is the average power in the first arriving time cluster, $\Gamma$ is the cluster decay time constant, and $Z_n$ is a lognormal random variable with 0 dB mean and standard deviation $\sigma_Z$ (see Table~\ref{tbl:softTable}). ~(\ref{eqCluster}) ensures that the sum of cluster powers adds up to the total omnidirectional received power $P_r$. Note that $\overline{P}_0$ cancels out in~(\ref{eqCluster}) using~(\ref{eqCluster2}), but can be used as a secondary statistic to validate the channel model~\cite{Samimi15_2}. The 3GPP, WINNER, COST, and METIS models also parameterize an exponential function over delay, as in~(\ref{eqCluster2}), to estimate mean cluster power levels~\cite{3GPP:1,WinnerII,Liu12,metis}.
\noindent \textit{Step 8: Generate the cluster subpath powers $\Pi_{m,n}$ (mW)}:
\begin{align}
&\Pi^{\prime}_{m,n} = \overline{\Pi}_0 e^{-\frac{\rho_{m,n}}{\gamma}} 10^{\frac{U_{m,n}}{10}}\\ \label{eqSP}
&\Pi_{m,n} = \frac{\Pi^{\prime}_{m,n}}{\sum^{k=N}_{k=1} \Pi^{\prime}_{k,n}}\times P_n [mW]\\
& U_{m,n} \sim N(0, \sigma_U)
\end{align}
\noindent where $\overline{\Pi}_0$ is the average power in the first received intra-cluster subpath, $\gamma$ is the subpath decay time constant, and $U_{m,n}$ is a lognormal random variable with 0 dB mean and standard deviation $\sigma_U$ (see Table~\ref{tbl:softTable}); $m=1, 2, ..., M_n$ and $n=1,2,...,N$. ~(\ref{eqSP}) ensures that the sum of subpath powers adds up to the cluster power. For model validation, the subpath path losses were thresholded at \text{180 dB} (maximum measurable path loss~\cite{Rap15_3}). Note: the measurements have much greater temporal and spatial resolution than previous models. Intra-cluster power levels were observed to fall off exponentially over intra-cluster time delay (see Fig. 4 in~\cite{Samimi15_2}).
\noindent \textit{Step 9: Generate the subpath phases $\varphi_{m,n}$ (rad)}:
\begin{align}
&\varphi_{m,n} \sim U(0,2\pi)
\end{align}
\noindent where $m=1, ..., M_n$ and $n=1,2,...,N$. Different from~\cite{Samimi15_2} where phases are estimated from frequency and delays, here the subpath phases are assumed independently and identically distributed (i.i.d), and uniform between 0 and $2\pi$~\cite{Saleh87} since each subpath may experience a different scattering environment.
\noindent \textit{Step 10: Recover absolute time delays $t_{m,n}$ of cluster subpaths using the TX-RX separation distance $d$ (Step 1):}
\begin{equation}\label{step10}
t_{m,n} = t_0 + \tau_n + \rho_{m,n} \hspace{.3cm},\hspace{.3cm}t_0 = \frac{d}{c}
\end{equation}
\noindent where $m = 1,2,...M_n, n = 1,2,...N$, and $c = 3\times 10^8 \text{ } m/s$ is the speed of light in free space.
\noindent \textit{Step 11a: Generate the mean AOA and AOD azimuth angles $\theta_i(^{\circ})$ of the 3-D spatial lobes to avoid overlap of lobe angles:}
\begin{align}
&\theta_{i} \sim U(\theta_{min},\theta_{max})\hspace{.3cm},\hspace{.3cm} i=1, 2, ..., L\\
& \theta_{min} = \frac{360(i-1)}{L}, \theta_{max} = \frac{360i}{L}
\end{align}
\noindent \textit{Step 11b: Generate the mean AOA and AOD elevation angles $\phi_i(^{\circ})$ of the 3-D spatial lobes:}
\begin{align}
&\phi_{i} \sim N(\mu,\sigma), i = 1, 2,..., L.
\end{align}
\noindent Values of $\phi_i$ are defined with respect to horizon, namely, a positive and negative value indicate a direction above and below horizon, respectively. While the 28 GHz measurements used a fixed 10$^{\circ}$ downtilt at the TX, and considered elevation planes of $0^{\circ}$, and $\pm 20^{\circ}$ at the RX, mmWave transceivers will most likely beamform in the strongest directions, as emulated in the 73 GHz measurements~\cite{MacCartney15_2}. Consequently, the provided elevation angle distributions for all frequency scenarios are extracted from the 73 GHz measurements (see Table~\ref{tbl:softTable}).
\noindent \textit{Step 12: Generate the AOD angles $(\theta_{m,n,AOD},\phi_{m,n,AOD})$ and AOA angles $(\theta_{m,n,AOA},\phi_{m,n,AOA})$ of each subpath component using the spatial lobe angles found in Step 11:}
\begin{align}
&\theta_{m,n,AOD} = \theta_i+(\Delta \theta_i)_{m,n,AOD}\\
&\phi_{m,n,AOD} = \phi_i +(\Delta \phi_i)_{m,n,AOD}\\ \label{eq37}
&\theta_{m,n,AOA} = \theta_j+(\Delta \theta_j)_{m,n,AOA}\\ \label{eq38}
&\phi_{m,n,AOA} = \phi_j +(\Delta \phi_j)_{m,n,AOA}
\end{align}
\begin{align} \text{where:}\hspace{.2cm}
&i \sim DU[1,L_{AOD}] \hspace{.2cm}, \hspace{.2cm}j \sim DU[1,L_{AOA}] \\ \label{norm}
& (\Delta \theta_i)_{m,n,AOD} \sim N(0,\sigma_{\theta,AOD})\\
&(\Delta \phi_i)_{m,n,AOD} \sim N(0,\sigma_{\phi,AOD})\\ \label{eq42}
& (\Delta \theta_j)_{m,n,AOA} \sim N(0,\sigma_{\theta,AOA})\\ \label{Laplace}
&(\Delta \phi_j)_{m,n,AOA} \sim \text{Laplace}(\sigma_{\phi,AOA})
\end{align}
\noindent This step assigns to each multipath component a single spatial AOD and AOA lobe in a uniform random fashion, in addition to a random angular offset within the spatial lobe with distributions specified in\text{~(\ref{norm}) - (\ref{Laplace})}. Note that the Laplace distribution in~(\ref{Laplace}) provided a better fit to all data across frequencies and environments than a normal distribution. The 3GPP model uses a uniform distribution from $-40^{\circ}$ to $+40^{\circ}$ to generate path azimuth AODs, and for path azimuth AOAs uses a zero-mean normal distribution whose variance is a function of path powers for the UMi scenario~\cite{3GPP:1}. The WINNER models use a wrapped Gaussian distribution that is a function of path powers and delays to generate path AODs and AOAs~\cite{WinnerII}.
\subsection{Implementing the step procedures}
To facilitate the implementation of this SSCM, Table~\ref{tbl:50} and Table~\ref{tbl:softTable} provide the necessary parameters required in Steps 2, 3, 5, 6, 7, 8, 11b, and 12 as a function of the frequency-scenarios considered in this work.
\begin{table}[b!]
\centering
\caption{Measured path loss exponents and shadow factors~\cite{MacCartney14:2,Rap15_3}, used to generate the omnidirectional received power in Step 2 of Section~\ref{sec:mmWaveProc}. }
\begin{tabular}{|c|c|c|c|}
\hline
\textbf{Step \#} & \textbf{Frequency} & \textbf{Environment} & \textbf{Measured}$\bm{(\overline{n},\sigma)}$ \\ \hline
\multirow{4}{*}{Step 2} & \multirow{2}{*}{28 GHz} & LOS & (2.1, 3.6 dB) \\ \cline{3-4}
& & NLOS & (3.4, 9.7 dB) \\ \cline{2-4}
& \multirow{2}{*}{73 GHz} & LOS & (2.0, 5.2 dB) \\ \cline{3-4}
& & NLOS & (3.3, 7.6 dB) \\ \hline
\end{tabular}
\label{tbl:50}
\end{table}
\begin{table*}
\centering
\captionsetup{width=\textwidth}
\caption{Key frequency-dependent parameters that reproduce the measured statistics for the combined 28 - 73 GHz LOS, 28 GHz NLOS, 73 GHz NLOS, and combined 28 - 73 GHz NLOS frequency scenarios.}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{\textbf{Step \#}} & \multirow{2}{*}{\textbf{Input Parameters}} & \multicolumn{4}{c|}{\textbf{Frequency Scenario}} \tabularnewline \cline{3-6}
& & \textbf{28 - 73 GHz LOS} & \textbf{28 GHz NLOS} & \textbf{73 GHz NLOS} & \textbf{28 - 73 GHz NLOS} \tabularnewline \hline
Step 3 & $\mu_{AOD},\mu_{AOA}$ & 1.9, 1.8 & 1.6, 1.6 & 1.5, 2.5 & 1.5, 2.1 \tabularnewline \hline
Step 5 & $X_{max}$ & 0.2 & 0.5 & 0.5 &0.5 \tabularnewline \hline
Step 6 & $\mu_{\tau} [ns]$ & 123 & 83 &83 & 83 \tabularnewline \hline
\multirow{1}{*}{Step 7} & $\Gamma [ns], \sigma_Z [dB]$ & 25.9, 1 & 49.4, 3 & 56.0, 3 & 51.0, 3 \tabularnewline \hline
Step 8 & $\gamma [ns], \sigma_U [dB]$ & 16.9, 6 & 16.9, 6 & 15.3, 6 & 15.5, 6 \tabularnewline \hline
\multirow{2}{*}{Step 11b} & $\mu_{AOD}[^{\circ}], \sigma_{AOD}[^{\circ}]$ & -12.6, 5.9 & -4.9, 4.5 & -4.9, 4.5 & -4.9, 4.5 \tabularnewline \cline{2-6}
& $\mu_{AOA}[^{\circ}], \sigma_{AOA}[^{\circ}]$ & 10.8, 5.3 & 3.6, 4.8 &3.6, 4.8 & 3.6, 4.8 \tabularnewline \hline
\multirow{2}{*}{Step 12} & $\sigma_{\theta,AOD}[^{\circ}],\sigma_{\phi,AOD}[^{\circ}]$ & 8.5, 2.5 & 9.0, 2.5 & 7.0, 3.5 & 11.0, 3.0 \tabularnewline \cline{2-6}
& $\sigma_{\theta,AOA}[^{\circ}],\sigma_{\phi,AOA}[^{\circ}]$ & 10.5, 11.5 & 10.1, 10.5 & 6.0, 3.5 & 7.5, 6.0 \tabularnewline \hline
\end{tabular}
\label{tbl:softTable}
\end{table*}
\subsection{Sample Output Functions}
Figs.~\ref{fig:SamplePDP_28GHzNLOS} and \ref{fig:SampleAOD_28GHzNLOS} show output functions of a 28 GHz NLOS omnidirectional PDP, and corresponding AOA 3-D power spectrum, obtained from a MATLAB-based statistical simulator that implemented the channel models given \text{in~(\ref{step1}) - (\ref{Laplace})}. The generated PDP in Fig.~\ref{fig:SamplePDP_28GHzNLOS} is composed of four multipath taps, grouped into two time clusters with exponentially decaying amplitudes with cluster decay constant $\Gamma=49.4$ ns and intra-cluster subpath decay constant $\gamma=16.9$ ns (see Section~\ref{sec:mmWaveProc}, Steps 6 and 7). Here, the simulated PDP has a total path loss of 120 dB with TX-RX separation distance of 112 m, and RMS delay spread of 50 ns, with 0 dBi TX and RX antenna gains. The AOA spectrum (Fig.~\ref{fig:SampleAOD_28GHzNLOS}) shows the four multipath grouped into two AOA spatial lobes according to~(\ref{eq37}),~(\ref{eq38}),~(\ref{eq42}), and~(\ref{Laplace}).
\begin{figure}[t]
\begin{center}
\includegraphics[width=3.5in]{IR_Plot_3.eps}
\end{center}
\caption{Example of simulated 28 GHz NLOS PDP, showing four multipaths, obtained from the MATLAB-based statistical simulator, assuming 0 dBi TX and RX antenna gains. Temporal cluster powers and intra-cluster subpath powers decay with increasing delay according to empirical time decay constants, $\Gamma = 49.4$ ns and $\gamma=16.9$ ns, respectively.}
\vspace{-0.4cm}
\label{fig:SamplePDP_28GHzNLOS}
\end{figure}
\begin{figure}[t]
\begin{center}
\includegraphics[width=3.5in]{AOA_Plot_WithArrows_3.eps}
\end{center}
\caption{Example of simulated 28 GHz NLOS 3-D AOA power spectrum (top view of the azimuth plane) of Fig.~\ref{fig:SamplePDP_28GHzNLOS} obtained from the MATLAB-based statistical simulator, showing four multipath components grouped into two AOA spatial lobes.}
\vspace{-0.4cm}
\label{fig:SampleAOD_28GHzNLOS}
\end{figure}
\section{Simulation Results}
The statistical channel models presented in~(\ref{step1}) - (\ref{Laplace}) were implemented in a MATLAB-based statistical simulator to confirm the accuracy of the SSCM against the measured statistics. A large simulation was carried out in which 10,000 omnidirectional PDPs for omnidirectional TX and RX, and 3-D AOD and AOA power spectra were generated according to~(\ref{IREq}) and~(\ref{3DEq1}). Simple number generators were utilized to obtain the number of time clusters, the number of AOD and AOA spatial lobes, cluster and subpath delays, and cluster and subpath powers, as described in Section~\ref{sec:mmWaveProc}. To remain faithful to the measurements, the total dynamic range was set to 180 dB, and the TX and RX 3-dB antenna beamwidths were set to 10$^{\circ}$ and 7$^{\circ}$ (in azimuth and elevation)~\cite{MacCartney15_2}, when performing directional simulations.
\subsection{Simulated RMS Delay Spreads}
Fig.~\ref{fig:JointOmni_DS} compares the omnidirectional simulated RMS delay spreads and empirical values from omnidirectional PDPs~\cite{Samimi15_2}, at both 28 GHz and 73 GHz in LOS and NLOS scenarios. The empirical and simulated medians were 18 ns and 16 ns, respectively, for the combined 28-73 GHz LOS scenario, and 32 ns and 35 ns for the empirical and simulated medians, respectively, for the combined 28-73 GHz NLOS scenario. The few measured data samples considerably skewed the empirical distributions, so the median (instead of the mean) was selected to represent the distribution trend. The empirical and simulated medians in NLOS were 31 ns and 32 ns at 28 GHz (see Fig. 5 in~\cite{Samimi15_2}), respectively, and 47 ns and 39 ns at 73 GHz, respectively, providing good agreement to empirical values
\begin{figure}
\begin{center}
\includegraphics[width=3.5in]{Joint_28_73_Omni_CDF.eps}
\end{center}
\caption{Combined 28 - 73 GHz LOS and NLOS omnidirectional RMS delay spreads synthesized from absolute timing PDPs, superimposed with 10,000 simulated RMS delay spreads from generated omnidirectional PDPs~\cite{Samimi15_2}. }
\vspace{-0.4cm}
\label{fig:JointOmni_DS}
\end{figure}
The model also produces \textit{directional} PDPs at arbitrary TX-RX pointing angle combination, and reconstructs the temporal statistics of arbitrary antenna beamwidths. This is achieved by weighting the multipath component power levels with a user-defined complex amplitude antenna pattern, such that the multipath components closest to a desired direction are amplified, while those farthest away are effectively set to 0, as shown in~(\ref{DIREq}). Using~(\ref{DIREq}), the multipath component power levels were weighted by $|g_{TX}(\theta,\phi)|^2$ and $|g_{RX}(\theta,\phi)|^2$, the TX and RX horn antenna patterns, respectively, commonly parameterized as follows~\cite{Balanis05},
\begin{align}
&|g(\theta,\phi)|^2 = \max \big( G_0 e^{\alpha \theta^2+\beta \phi^2}, \frac{G_0}{100} \big) \\
& \alpha = \frac{4 \ln(2)}{\theta_{3dB}^2}, \beta = \frac{4 \ln(2)}{\phi_{3dB}^2}, G_0 = \frac{41253 \eta }{\theta_{3dB}\phi_{3dB}}
\end{align}
\noindent where $(\theta,\phi)$ are the azimuth and elevation angle offsets from the boresight direction in degrees, $G_0$ is the maximum directive gain (boresight gain) in linear units, $(\theta_{3dB},\phi_{3dB})$ are the azimuth and elevation HPBWs in degrees, $\alpha,\beta$ are parameters that depend on the HPBW values, and $\eta = 0.7$ is a typical average antenna efficiency.
Fig.~\ref{fig:RMSDS_Sim} shows simulated directional RMS delay spreads obtained from the 28 GHz and 73 GHz channel models presented here, obtained from~(\ref{DIREq}), in comparison to reported values in the literature for the 10\%, 50\%, and 90\% CDF points of measured directional RMS delay spreads at 28 GHz, 38 GHz, 60 GHz, and 73 GHz~\cite{MacCartney15_2}, for antenna beamwidths of 7.3$^{\circ}$, 10.9$^{\circ}$, 28.8$^{\circ}$, and 49.4$^{\circ}$. To test~(\ref{DIREq}), we generated sample functions and computed directional RMS delay spreads from directional PDPs for 20 random TX-RX separation distances using antenna HPBW of 10$^{\circ}$, $7^{\circ}$, and $30^{\circ}$ in azimuth/elevation at 28 GHz and 73 GHz to emulate the NYC measurements. Note that the empirical CDFs consist of all available data, while the simulated directional RMS delay spreads were generated from channel models obtained exclusively from up to four strongest AOD and AOA PDP data. Fig.~\ref{fig:RMSDS_Sim} indicates that the simulated and measured omnidirectional and directional RMS delay spread distributions match relatively well across antenna beamwidths and many mmWave bands.
\begin{figure}
\begin{center}
\includegraphics[width=3.5in]{Dir_RMS_DS_Sim_Meas.eps}
\end{center}
\caption{Simulated directional RMS delay spread CDFs (lines) for various frequencies and for antenna beamwidths, obtained from directional PDPs generated using (\ref{DIREq}). Values reported from the literature are shown as points~\cite{Rap13:2,MacCartney14:2}. In the figure legend, 'C' stands for 'Cellular', 'P2P' stands for Peer-to-Peer', and 'Sim' stands for 'Simulated'.}
\vspace{-0.4cm}
\label{fig:RMSDS_Sim}
\end{figure}
\subsection{Simulated RMS Angular Spreads}
The omnidirectional azimuth and elevation spreads describe the degree of angular dispersion at a BS or MS over the entire $4\pi$ steradian sphere~\cite{3GPP:1,WinnerII}, also termed \textit{global} angular spreads in~\cite{Liu12}. The AOD and AOA global angular spreads were computed from all available 28 GHz and 73 GHz NLOS measured data, using the total (integrated over delay) received power at unique azimuth/elevation pointing angles, but not requiring absolute multipath time delays, and the equations in Annex A of~\cite{3GPP:1}. These were compared with the simulated angular spreads using the 3-D SSCM, where the SSCM was developed from the statistics of up to four strong measured angles. The simulated and measured mean global angular spreads match relatively well at 28 GHz and 73 GHz, as can be seen from Fig.~\ref{fig:GlobalAngularSpreads_AOA_28NLOS} and Fig.~\ref{fig:GlobalAngularSpreads_AOA_73NLOS}. The slight differences (slight under- or over-estimation of global azimuth spreads using Step 12 in Section~\ref{sec:mmWaveProc}) may be due to the model focusing only on the multipath components contained in the strongest several spatial lobes measured at every location, which will in actuality be the strongest components in a practical wireless system.
\begin{figure}
\begin{center}
\includegraphics[width=3.5in]{28_AOA_NLOS_Global_Angular_Spreads.eps}
\end{center}
\caption{28 GHz NLOS global azimuth and elevation spreads, obtained from the NYC measurements~\cite{Rap15_3} and the 3-D SSCM presented here.}
\vspace{-0.4cm}
\label{fig:GlobalAngularSpreads_AOA_28NLOS}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=3.5in]{73_AOA_NLOS_Global_Angular_Spreads2.eps}
\end{center}
\caption{73 GHz NLOS global azimuth and elevation spreads, obtained from the NYC measurements~\cite{Rap15_3} and the 3-D SSCM presented here.}
\vspace{-0.4cm}
\label{fig:GlobalAngularSpreads_AOA_73NLOS}
\end{figure}
The directional AOD and AOA RMS \textit{lobe} azimuth and elevation spreads were also computed based on a -10 dB lobe threshold~\cite{Samimi15_2} from the 28 GHz and 73 GHz data, and compared with simulated values using the 3-D SSCM. Fig.~\ref{fig:AngularSpreads_AOA_73NLOS} shows typical measured against simulated AOA RMS lobe angular spreads for the 73 GHz NLOS scenario, showing an excellent match over empirical and simulated means of 4$^{\circ}$ and $2^{\circ}$ in azimuth and elevation, respectively, over all measured spatial lobes. Similar agreement was found for the 28 GHz NLOS and LOS datasets, across AOD and AOA spatial lobes, indicating that the model accurately recreates the empirical spatial statistics.
\begin{figure}
\begin{center}
\includegraphics[width=3.5in]{AngularSpreads_AOA_73NLOS.eps}
\end{center}
\caption{73 GHz NLOS AOA RMS lobe azimuth and elevation spreads, measured and simulated, showing good agreement~\cite{Samimi15_2}.}
\vspace{-0.4cm}
\label{fig:AngularSpreads_AOA_73NLOS}
\end{figure}
\section{Conclusion}
This paper presented a 3-D statistical model of the channel impulse response for mmWave LOS and NLOS mobile access communications, that supports arbitrary carrier frequency, signal bandwidth, and antenna beamwidth, with good agreement between the model and published directional RMS delay spreads. The RMS \textit{lobe} angular spreads~\cite{Samimi15_2} provide a realistic representation of directional angular spreads for future multi-antenna mmWave systems, where the strongest multipath components contained in spatial lobes are most important for characterizing system behavior. A MATLAB-based statistical simulator was implemented to generate a large ensemble of PDPs and 3-D power spectra, showing good agreement to field measurements, allowing for future millimeter-wave system and bit-error rate simulations, as well 5G wireless network capacity analyses.
\bibliographystyle{IEEEtran}
| {
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{"url":"https:\/\/homework.cpm.org\/category\/CCI_CT\/textbook\/int3\/chapter\/7\/lesson\/7.1.4\/problem\/7-48","text":"### Home > INT3 > Chapter 7 > Lesson 7.1.4 > Problem7-48\n\n7-48.\n1. What is the inverse of each of the functions below? Write your answers using function notation. Homework Help \u270e\n\n1. p(x) = 3(x3 + 6)\n\n2. k(x) = 3x3 + 6\n\n3. h(x) =\n\n4. j(x) =\n\nLet p(x) = y, switch the x and the y in the equation and solve for y.\n\n$p^{-1}(x)=\\sqrt[3]{\\frac{x}{3}-6}$\n\nSee part (a).\n\nLet h(x) = y . Switch x and y. Multiply both sides of the equation by (y \u2212 1) to remove the fraction.\n\nDistribute and then rewrite the equation so that all the y-terms are on the left side and everything else is on the right side.\n\nYour equation should look like this: xyy = x + 1. Factor the left side of the equation and divide both sides by (x \u2212 1) to get y by itself.\n\nGraph the equation on a graphing calculator to see why this equation is its own inverse.\n\nSee part (a).","date":"2019-11-13 16:20:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 1, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7366674542427063, \"perplexity\": 618.9518821625398}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-47\/segments\/1573496667262.54\/warc\/CC-MAIN-20191113140725-20191113164725-00486.warc.gz\"}"} | null | null |
Judge Tosses Child Sexual Assault Case, Citing Prosecutor's 'Shocking' Conduct
A California judge cited "shocking" prosecutorial conduct in dismissing a child sexual assault case last week, saying that a deputy district attorney hadn't been truthful after learning there was a videotaped medical examination of the alleged victim that hadn't been turned over. The prosecutor's conduct, the judge said, "strikes at the foundation of our legal system and is so grossly shocking and outrageous that it offends the universal sense of justice to allow prosecution in this matter to proceed."
By Kate Moser | January 11, 2010 at 12:00 AM
A Santa Clara County, Calif., Superior Court judge cited "grossly shocking" prosecutorial conduct in dismissing a child sexual assault case Wednesday. | {
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