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TITLE: Is every set class generic over a given inner model?
QUESTION [4 upvotes]: In a paper by B. Mitchell, I stumbled into the following sentence:
"In the summer of 1986 Woodin discovered the second of the forcing orders
associated with a Woodin cardinal, the extender algebra. This forcing goes back
to the class forcing of Vopěnka [Vopěnka and Hájek, 1972], by which any set is generic, by a class forcing, over any given class model of set theory."
If I interpreted the result correctly, it means that for every inner model $M$ and for every set $x\in V$, there is a class forcing notion $\mathbb{P}$, definable over $M$, s.t. $x$ is $\mathbb{P}$-generic over $M$.
I looked up the reference, which was the book "The theory of semisets", but it was really hard to figure out anything because of the uncommon symbolization.
My questions are the following:
1) Is my formulation of this result correct?
2) Is there another, more approachable, reference where I could find its proof (and maybe more information on class forcing)?
REPLY [3 votes]: What you are referring to it, is called the "extender algebra". It is a Boolean algebra and so can be considered as a forcing notion.
Then the result you have stated might be the following theorem of Woodin:
Theorem (Woodin). Assume $(M;\vec{E})$ is fully iterable and
$\vec{E}$ witnesses a countable ordinal $\delta$ is a Woodin cardinal in $M$. Then for every set of ordinals $x$
there is a (well-founded) iteration $j: M \to M^*$
of length < $card(x)^+$
such that $x$ is $j(W_{\delta,\delta}(\vec{E}))-$generic over $M^*$.
For undefined notions and a proof of this theorem see the following papers:
Farah, "The extender algebra and $\Sigma^2_1 $-absoluteness".
Schindelr-Doebler, "The extender algebra and vagaries of $\Sigma^2_1 $-absoluteness''.
There is also another result of Woodin which is related to your question.
Theorem (Woodin) Work in $ZF$. Suppose the $HOD$ Conjecture is $\Omega-$valid from
$ZFC$ + “There is a supercompact cardinal”,
$\delta_0$ is a supercompact cardinal, and that there is a supercompact cardinal below $\delta_0$.
Then there exists a transitive class $N ⊂ V$ and $X ∈ V_{\delta_0}$ such that the following hold:
(1) $N \models ZFC$.
(2) $N$ is $Σ_2-$definable from $X$.
(3) There exists a partial order $P ∈ N ∩ V_{\delta_0}$ such that for all $A ⊂ Ord, A ∈ N[G]$
for some $N-$generic filter $G ⊂ P.$
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- Jun 13, 2019 · ...
Outlook add people to meeting without sending update
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Outlook add people to meeting without sending update
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Outlook add people to meeting without sending update
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Outlook add people to meeting without sending update
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Click the To-> button to add the selected name to the list. Repeat steps 3 and 4 if you want to select several names. Click OK when you've finished selecting the names. This returns you to the invitation message window. If you want to add some expanatory words to your share invitation, type them in the message box just like in an ordinary email.
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Hide Details of Outlook Appointment and Meetings Sometime we have appointments and events which we want to keep private. Now you can easily hide your personal appointments from regular calendar view through Custom Setting.Oct 30, 2018 ·;
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01.11.2016 · To add people to an existing meeting without sending the invitation to all existing members, we suggest you follow the Based on my test on version 1710 and 1708, we can open up the Meeting from Calendar, and field the new attendee in the To, then click on Send Update, we can...
How to unlock toshiba laptop forgot passwordIt appears that in the latest update of Outlook for Office 365, an important feature was dropped - The ability to change a meeting without sending updates to participants. Yes, I know that attendees must know if the time or place of a meeting change, but they don't need to know if I shorten the name of.. ...
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Supermax sm9200 ca hd manualOutlook 2011 for Mac In Outlook 2011, you must open the Delegates window and add the resource you manage. This will cause the shared calendar for the resource to appear in your Shared Calendars list on the left side of the Outlook 2011 window. To view the room calendar: 1. Click Calendar on the left side of the window. 2. Click Tools > Accounts ... In order to add Zoom to your Outlook email client, you'll need to download the Zoom add-on. This is available for free in Microsoft's Apps store. Just click on this link and then click on the blue ...
Apr 03, 2019 · If you’ve never sent a meeting invitation in Outlook, you might think it’s a complicated process but it’s only slightly different than sending an email. Meeting invitations in Outlook. Open Outlook and click the New Items dropdown on the Home tab. From the menu, select Meeting. A new window will open and it will, for the most part ...
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TITLE: What is the difference between the equations $x^2=4y^2$ and $x^2=2yx$
QUESTION [2 upvotes]: Suppose I am given the equation $x=2y$.
My interpretation of this equation is $A=\{\langle x,y \rangle \in \mathbb R \times \mathbb R |\ x=2y\ \}$
Given this interpretation, I want to know the difference between the equations $x^2=4y^2$ and $x^2=2yx$.
For the equation $x^2=2yx$, we have a single line that completely overlaps with $x=2y$.
For the equation $x^2=4y^2$, we have two intersecting lines, only one of which overlaps with $x=2y$.
In either case, this is satisfying because it explains the logical concepts of:
$$x=2y \implies x^2=4y^2$$ $$\text{and}$$ $$x=2y \implies x^2=2yx$$ which I feel as though I am implicitly using during different algebraic proofs. That is to say, the set of solutions for $x=2y$ also satisfies (intersects with) the equations (sets) $x^2=4y^2$ and $x^2=2yx$.
In line with our earlier set interpretation, letting $x^2=2yx$ represent the set $B=\{\langle x,y \rangle \in \mathbb R \times \mathbb R | x^2=2yx\}$ and letting $x^2=4y^2$ represent the set $C=\{\langle x,y \rangle \in \mathbb R \times \mathbb R | x^2=4y^2\}$, the aforementioned implication suggests $A \subseteq B$ and $A\subseteq C$.
However, given my descriptions of the graphs, it is quite clear that we can be more specific...namely:
$A=B$ and $A \subsetneqq C$
I see that for the equation $x^2=4y^2$, square rooting both sides yields $\pm x=2y$, which explains its graph of the two intersecting lines. But I am wondering if there is a more fundamental reason that explains this observation. In particular, the quality of inverse exponents versus the quality of inverse multiplication.
To simplify $x^2=2yx$, I am effectively multiplying each side by $x^{-1}$. Comparatively, to simplify $x^2=4y^2$ (perhaps better written as $x^2=(2y)^2$), I am taking the inverse operation of squaring. It seems like the inverse operation of squaring yields "more solutions" than the inverse operation of multiplication.
Any added insight would be greatly appreciated!
REPLY [0 votes]: Suppose
$$x = 2y \tag{1}\label{eq1}$$
Then certainly
$$x^2 = 4y^2 \tag{2}\label{eq2}$$
Now, if all we know is $x^2 = 4y^2$, then we would know either $x = 2y$ or $x = -2y$. However, we know more than this. We can eliminate $x = -2y$ because of $\eqref{eq1}$.
$x = 2y$ doesn't stop being true even after squaring both sides. \eqref{eq2} implies two posibilies, one of which we can eliminate because of \eqref{eq1}.
(To be more precise, $x=-2y$ is eliminated for $x, y \neq 0$. The solution $x = y = 0$ is fine, and is already covered by $x = 2y$.)
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Financial News
Here you can find ad-hoc releases according to § 15 of the Securities Trading Act (WpHG) of Logwin AG as well as current financial news of the Logwin Group.))
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DOLCE&GABBANA
DOMENICO DOLCE AND STEFANO GABBANA IN LONDON
Sicilian Inspiration, Spring/Summer 2013 (Photo courtesy of Dolce&Gabbana)
LONDON, UK - London Collections: Men’s calendar during which the Summer 2014 Tailoring Collection for men will be presented.
“We are thrilled to welcome Dolce&Gabbana to London Collections: Men, and we are delighted that they are opening this season with such a high-profile event. London welcomes Domenico and Stefano and hope they decide to have an event here every season.” - commented Dylan Jones, Editor in Chief – British GQ.
Italian and English sartorial traditions meet and blend in a new, one-of-a-kind space, designed and created exclusively for London and located at 53-55 New Bond Street.
Milan Fashion Week SS 2013 (Photo courtesy of MFW)
Milan Fashion Week SS 2013 (Photo courtesy of MFW).
“London has always been one of the cities we love the most. It’s not by chance that we have decided to open this particular store here. British sense of style is part of their DNA, it’s been part of their culture for ages. Just like in Italy. When Dylan invited us to come, we couldn’t say no!” - said Domenico Dolce.
Dolce&Gabbana’s boutiques in London are located at 6 (menswear) and 175 (womenswear) Sloane Street and at 6/8 Old Bond Street (menswear and womenswear).
Source: BFC
Spring/Summer 2013
Photos by Getty Images
Photos by Getty Images
AUTUMN/WINTER 2013-14
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Trial Of Accused $300 Million Ponzi Scheme Masterminds Ends In Acquittal, Mistrial
In a rare loss for the government in a Ponzi scheme trial, the trial of a husband-and-wife duo accused of defrauding thousands of investors out of hundreds of millions of dollars in an alleged timeshare Ponzi scheme ended with a full acquittal for the wife and a mistrial declared for her husband. After a federal jury acquitted Cristal Clark of all fraud charges, U.S. District Judge Jose E. Martinez declared a mistrial after a federal jury deadlocked for the fourth day of negotiations on the fate of Fred "Dave" Clark. The government has indicated that it intends to retry Fred Clark, who faces up to thirty years in federal prison for each bank fraud charge..
In the wake of the charges against the Clarks related to their operation of Cay Clubs, authorities targeted former sales directors Barry Graham and Ricky Lynn Stokes and charged the pair with conspiracy to commit bank fraud. The charges resulted in guilty pleas and identical five-year sentences, and each of the men reportedly testified at the Clarks' trial..
Reader Comments
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\subsection{Persistence Modules as Multigraded Modules}
\label{sec:multigraded}
As the name suggests, persistence modules can be seen as modules over a ring, and sometimes this perspective is very useful.
We explain this just in the case where $P=\Z^n$, though the story is much the same when $P=\R^n$ \cite{lesnick2015theory}.
\begin{definition}
Let $\mathbf e_i$ denote the $i^{\mathrm{th}}$ standard basis vector in $\Z^n$. An \emph{$n$-grading} on a $k[x_1,\dots,x_n]$-module $M$ is a vector space decomposition \[M=\oplus_{z\in \Z^n} M_z\] such that $x_i M_z\subset M_{z+\mathbf e_i}$ for all $z\in \Z^n$ and $i\in \{1,\ldots,n\}$. A $k[x_1,\dots,x_n]$-module $M$ is said to be \emph{$n$-graded} if it comes equipped with an $n$-grading.
\end{definition}
A \emph{morphism} $f:M\to N$ of $n$-graded modules is a module homomorphism (in the usual sense) such that $f(M_z)\subset N_z$ for all $z\in \N^n$. With these morphisms, the $n$-graded modules form a category $\dmod$.
\begin{proposition}[Carlsson, Zomorodian \cite{carlsson2009theory,corbet2018representation}]\label{Prop:ModDiagramEquivalence}
The categories $\Vect^{\Z^n}$ and $\dmod$ are equivalent.
\end{proposition}
$k[x_1,\dots,x_n]$-modules are the basic objects of study in commutative algebra, and thus have a rich and highly developed theory \cite{eisenbud1995commutative}.
Proposition \ref{Prop:ModDiagramEquivalence} allows us to adapt standard language and constructions for $k[x_1,\dots,x_n]$-modules to the study of persistence modules, provided those constructions make sense in the $n$-graded setting. Fortunately, as a rule of thumb, definitions and results about $k[x_1,\dots,x_n]$-modules do tend to carry over to the $n$-graded setting, and often become simpler there.
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.
At Comic-Con 2016, ESET’s Anna Keeve dives deep into the “real” world of Mr. Robot, has a VR adventure, and looks for cybersecurity clues in Elliot’s room.
When it comes to APAC cybersecurity, there is room for improvement. In this feature we take a look at key trends across this region.
Restaurant chain Cicis has released a statement informing customers of a data breach experienced in some of its restaurants across the US.
One of the surprises of last year for everyone working in IT security was, without a doubt, the hit TV show Mr. Robot, explains ESET’s Josep Albors.
Make sure that all your Apple devices are patched before online criminals attempt to take advantage of this flaw.
The Library of Congress in the US has experienced a distributed denial of service (DDoS) attack, it revealed via social media. Also affected was the US Copyright Office.
Beware jackware, malicious software that seeks to take control of a device, the primary purpose of which is not data processing or digital communications.
Malicious scripts are gaining prevalence in Brazil, reports ESET’s Matías Porolli.
A major data breach on the Ubuntu Forums has not compromised the passwords of its affected users. Usernames, emails addresses and IPs have been exposed.
Both the OurMine and PoodleCorp hacking gangs appear to be taking credit for Pokémon GO being offline over the weekend. But might there be a more down-to-earth explanation?
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Welcome back to Bonza’s Daily Puzzle answers and today’s is a great one as it’s to do with food and who can’t deny a cook meal at Mc Donalds? Today’s puzzle is everything to do with cooking (Cook It!). March 2nd 2015 puzzle is an easy one but a great one, Ever heard of that saying ‘The simple the better’ well this is your perfect match.
Now lets just wait till tomorrow and see what Bonza has to offer, But we’re focused on today 2nd March’s puzzle and we’ve got you covered incase your stuck. So tie your shoes, Buckle your belt and lets get started.
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TITLE: Evaluating $\int\frac{e^{\tan^{-1} x}}{(1+x^2)^{\frac32}}dx$
QUESTION [5 upvotes]: I have $$\int\dfrac{e^{\tan^{-1} x}}{(1+x^2)^{\tfrac32}}dx$$
I tried using Integral By Parts:
$$\int \dfrac{1}{\sqrt{1+x^2}}\times\dfrac{e^{\tan^{-1} x}}{(1+x^2)}dx=e^{\tan^{-1}x}\times\dfrac1{\sqrt{x^2+1}}+\int x(1+x^2)^{\tfrac{-3}2}e^{\tan^{-1}x}dx$$
$$=\dfrac{e^{\tan^{-1}x}}{\sqrt{x^2+1}}+\int\dfrac{{xe^{\tan^{-1}x}}}{(1+x^2)^{\frac32}}dx$$
But I don't know how to continue
REPLY [2 votes]: I wanted to figure out how the exponent of $1+x^2$ is resolved to be $-\dfrac32$
$$\dfrac{d \left(\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}\right)}{dx}=\dfrac{e^{\tan^{-1}x}}{(1+x^2)^{n+1}}-2n\cdot\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^{n+1}}$$
Integrate both sides with respect to $x,$
$$\implies\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}+K=I_{n+1}-2nJ_{n+1}\ \ \ \ (1)$$ where $\displaystyle I_m=\int\dfrac{e^{\tan^{-1}x}}{(1+x^2)^m}\ dx\text{ and } J_m=\int\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^m}\ dx$
$$\dfrac{d \left(\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^n}\right)}{dx}=-(2n-1)\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}+\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^{n+1}}+2n\cdot\dfrac{e^{\tan^{-1}x}}{(1+x^2)^{n+1}}$$
Integrate both sides with respect to $x,$
$$\implies\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^n}+C=-(2n-1)I_n+J_{n+1}+2nI_{n+1}\ \ \ \ (2)$$
If $2n-1=0,$ $(1),(2)$ becomes simultaneous equations for $$I_{3/2},J_{3/2}$$
Can you take it from here?
Had $2n-1$ been $\ne0,$ the problem would not have remained so elementary.
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\begin{document}
\maketitle
\vspace{-0.3in}
\vspace*{0.3in}
\noindent
\textbf{Abstract.} Two signed graphs are called switching isomorphic to each other if one is isomorphic to a switching of the other. The wheel $W_n$ is the join of the cycle $C_n$ and a vertex. For $0 \leq p \leq n$, $\psi_{p}(n)$ is defined to be the number of switching non-isomorphic signed $W_n$ with exactly $p$ negative edges on $C_n$. The number of switching non-isomorphic signed $W_n$ is denoted by $\psi(n)$. In this paper, we compute the values of $\psi_{p}(n)$ for $p=0,1,2,3,4,n-4,n-3,n-2,n-1,n$ and of $\psi(n)$ for $n=4,5,...,10$. Our method of obtaining $\psi_{p}(n)$ not only count the switching non-isomorphic signed wheels but also generates them.
\section{Introduction}\label{introduction}
A \textit{k-ary necklace} of length $n$ is an equivalence class of necklaces, under rotation, formed with $n$ beads which have $k$ available colors. It is known~\cite{Weisstein} that the number $N(n,k)$ of non-equivalent $k$-ary necklaces of length $n$ is given by
\begin{equation}\label{equ for no of necklaces}
N(n,k) = \frac{1}{n} \sum_{d~ |~ n}\phi(d) k^{\frac{n}{d}} = \frac{1}{n} \sum_{i=1}^n k^{\gcd(n,i)},
\end{equation}
where $\phi$ is Euler's totient function.
Two necklaces are said to be \textit{isomorphic} if one can be obtained from the other by (cyclic) rotation or reflection. A \textit{k-ary bracelet} of length $n$ is an equivalence class, up to isomorphism, of necklaces of length $n$ with $k$ colors. It is also known~\cite{Weisstein} that the number $N'(n,k)$ of non-equivalent $k$-ary bracelets of length $n$ is given by
\begin{equation}\label{equ for no of braceletes}
N'(n,k) = \frac{N(n,k) + R(n,k)}{2},
\end{equation}
where
\begin{equation}
R(n,k) =
\begin{cases} k^{\frac{n+1}{2}} &\text{if $n$ is odd} \\
(\frac{k+1}{2})k^{\frac{n}{2}} &\text{if $n$ is even}.
\end{cases}
\end{equation}
George P\'{o}lya~\cite{Polya} was the first who discovered a powerful method for enumerating the number of orbits of a group on particular configurations. This method became known as the P\'{o}lya Enumeration Theorem. The numbers $N(n,k)$ and $N'(n,k)$ were determined by finding the number of orbits of the \textit{cyclic group} $Z_n$ and the \textit{dihedral group} $D_{2n}$ on $k$-ary $n$-tuples, respectively.
Fredricksen and Kessler~\cite{Fredri1} and Fredricksen and Maiorana~\cite{Fredri2} firstly developed an algorithm for generating necklaces. An algorithm for generating $k$-ary bracelets was developed by Joe Sawada~\cite{Sawada}. In best of our knowledge, no other method than algorithm is known for generating bracelets.
We will see that counting isomorphism type (non-equivalent) of 2-ary bracelets is equivalent to counting of isomorphism types of signed wheels. Our approach for enumerating isomorphism type signed wheels also generates them and does not depend upon any algorithm.
A \textit{signed graph}, denoted by $\Sigma=(G,\sigma)$, is a graph consisting of an ordinary graph $G$ and a sign function $\sigma : E(G) \rightarrow \{+1,-1\}$ which labels each edge of $G$ as positive or negative. In $\Sigma=(G,\sigma)$, $G$ is called the \textit{underlying graph} of $\Sigma$ and the set $\sigma^{-1}(-1)=\{e\in E(G)~|~\sigma(e)=-1\}$ is called the \textit{signature} of $\Sigma.$ \textit{Switching} $\Sigma$ by a vertex $u$ changes the sign of each edge incident to $u$. Two signed graph $\Sigma_1 =(G,\sigma_1)$ and $\Sigma_2 = (G,\sigma_2)$ are called \textit{switching equivalent} if one can be obtained by a sequence of switchings from the other. If the number of negative edges in a cycle is even then we call the cycle \textit{positive} and \textit{negative}, otherwise.
The following characterization for two signed graphs to be switching equivalent is given by Zaslavsky~\cite{Zaslavsky}.
\begin{lemm}\label{two SGs equivalence}
Two signed graphs $\Sigma_1,\Sigma_2$ are switching equivalent if and only if they have the same set of negative cycles.
\end{lemm}
Given a graph $G$ on $n$ vertices and $m$ edges, there are $2^m$ ways of constructing signed graphs on $G$.
\begin{lemm}\cite{Reza}\label{lemma of switching non-equivalent SGs}
There are $2^{m-n+1}$ switching non-equivalent signed graphs on a connected graph $G$ on $n$ vertices and $m$ edges.
\end{lemm}
We say the signed graphs $\Sigma_1 =(G,\sigma_1)$ and $\Sigma_2 = (H,\sigma_2)$ are \textit{isomorphic} if there exists a graph isomorphism between $G$ and $H$ preserving the edge signs. Two signed graphs are \textit{switching isomorphic} if one is isomorphic to a switching of the other.
Up to switching isomorphism, it is known that there are two signed $K_3$, three signed $K_4$, and seven signed $K_5$. In \cite{Deepak}, the authors classified all sixteen switching non-isomorphic signed $K_6$. Mallows and Sloane~\cite{Mallows} proved that the number of switching non-isomorphic signed complete graphs on $n$ vertices is equal to the number of Euler graphs on $n$ vertices. In~\cite{Zaslavsky}, Zaslavsky proved that there are only six signed Petersen graphs, up to switching isomorphism.
Recently, Y. Bagheri et al.~\cite{Ramezani} proved that the number of mutually switching non-isomorphic signed graphs associated with a given graph $G$ is equal to the number of orbits of the automorphism group of $G$ acting on the set of all possible signed graphs with underlying graph $G$. In this paper, we have used a different technique to determine the number of switching non-isomorphic signed wheels of some particular orders.
A \textit{wheel}, denoted by $W_n$, is the join of the cycle $C_n$ and a vertex. Let $V(W_n)=\{v,v_1,v_2,...,v_n\}$ and $E(W_n)=\{vv_i, v_iv_{i+1}~|~i=1,2,...,n\}$, where the subscripts are read modulo $n$. For $1 \leq i \leq n$, the edges $vv_i$ are said to be the \textit{spokes} of $W_n$, and the cycle induced by all the edges of form $v_iv_{i+1}$ is said to be the \textit{outer cycle}, denoted by $C_n$, of $W_n$. For $n=3$, the graph $W_3$ is the complete graph $K_4$. It is known that the number of switching non-isomorphic signed graphs over $K_4$ is 3. Thus, in the subsequent discussion, we consider the wheels $W_n$ for $n \geq 4$.
If a spoke $vv_{j}$, for some $1\leq j \leq n$, is negative in $(W_n, \sigma)$ then one can make it positive by switching $v_j$. Thus for any $(W_n, \sigma)$ there is an equivalent $(W_n, \sigma_1)$ such that $\sigma_1^{-1}(-1) \subseteq E(C_{n})$. The signed wheels whose signatures are subsets of the edges of the outer cycle $C_n$ will be denoted by $(W_n, \sigma)^o$. Also two signatures of $C_n$, with no switching, are isomorphic if and only if the corresponding signed wheels are isomorphic. Therefore, counting of isomorphism types of signed wheels is equivalent to counting isomorphism types of 2-ary bracelets, say bracelets of beads having colors blue and red.
For a fixed $0 \leq p \leq n$, $\psi_{p}(n)$ denotes the number of switching non-isomorphic signed wheels of the form $(W_n, \sigma)^o$ with exactly $p$ negative edges. In other words, $\psi_{p}(n)$ denotes the number of non-equivalent 2-ary bracelets with exactly $p$ red beads and $n-p$ blue beads. By $\psi(n)$, we denote the number of switching non-isomorphic signed wheels on $n+1$ vertices. Thus, $ \psi(n) = \sum\limits_{p=0}^{n} \psi_{p}(n)$.
The values of $\psi_{p}(n)$ for $p=0,1,2,3,4,n-4,n-3,n-2,n-1,n$ are determined in Section~\ref{main results}. Using these values, the values of $\psi(n)$ for $n \leq 10$ are obtained in Section~\ref{exact values}.
\section{Terminology and Methodology}\label{methodology}
Our approach to enumerate the switching non-isomorphic signed wheels is to put $p$ negative edges on $C_n$ at different distances that generate all mutually switching non-isomorphic signed wheels.
By $G_n$, we denote a $regular~n\text{-}gon$ having vertex set $V(G_n)=\{v_{1},v_{2},v_{3},...,v_{n}\}$ and edge set $E(G_n) = \{v_{i}v_{i+1}~|~i=1,2,...,n\}$, where the subscripts are read modulo $n$.
The \textit{distance} between two vertices $u~\text{and}~v$, denoted $d(u,v)$, in a graph $G$ is defined to be the number of edges in a shortest path between $u~\text{and}~v$. The \textit{distance} between two edges $e_1 = u_1u_2$ and $e_2 = v_1v_2$ in a graph $G$, denoted by $d(e_1, e_2)$, is $\min\{d(u_i, v_j)~ : ~ i \in \{1,2\},~ j \in \{1,2\}\}$. In $G_n$, it is clear that $1 \leq d(v_i,v_j) \leq \lfloor \frac{n}{2} \rfloor$ for all $i \neq j$. Further, if we measure the distance along one particular direction (clockwise or anticlockwise), then we have $1 \leq d(v_i,v_j) \leq n-1$ for all $i \neq j$.
If $n$ is an even number then the vertices $v_{i}$ and $v_{i+ \frac{n}{2}}$ of $G_n$ are called \textit{diagonally opposite vertices} and the edges $v_{i}v_{i+1}$ and $ v_{i+ \frac{n}{2}}v_{i+1+\frac{n}{2}}$ are called the \textit{opposite edges}. On the other hand, if $n$ is an odd number, the edge $v_{i+\lfloor \frac{n}{2} \rfloor}v_{i+(\lfloor \frac{n}{2} \rfloor+1)}$ is called the \textit{opposite edge} of $v_{i}$ for $1 \leq i \leq n$.
Clearly, $G_n$ features $n$ axes of symmetry. A common point at which all these axes meet is called the \textit{center} of $G_n$. Observe that if $n$ is an even number then half of the axes pass through diagonally opposite vertices and the remaining axes pass through the midpoints of opposite edges. On the other hand, if $n$ is an odd number, all the axes pass through a vertex and the midpoint of its opposite edge.
Let Aut($G$) denotes the automorphism group of a graph $G$. It is well known that Aut($W_n$) = Aut($G_n$) = $<\alpha, \beta~|~ \alpha^n = \beta^2 = 1, \alpha\beta = \beta\alpha^{-1}>$, the dihedral group $D_n$.
The following result will be helpful to examine whether two signed wheels are switching equivalent.
\begin{lemm}\label{switching non-equi wheels}
Two signed wheels of the form $(W_n, \sigma)^o$ with different signatures are always switching non-equivalent.
\end{lemm}
\begin{proof}
Let $\Sigma_1=(W_n,\sigma_1)^o$ and $\Sigma_2=(W_n,\sigma_2)^o$ be two signed wheels such that $\sigma_{1}^{-1}(-1) \neq \sigma_{2}^{-1}(-1)$, where $\sigma_{1}^{-1}(-1)$ and $\sigma_{2}^{-1}(-1)$ are subsets of $E(C_n)$. Since each negative edge makes exactly one triangle negative, the result follows from Lemma~\ref{two SGs equivalence}.
\end{proof}
Let $\Sigma=(W_n,\sigma)^o$ be a signed wheel with $p$ negative edges. Corresponding to $\Sigma$, we associate an ordered \textit{distance tuple} $D(\Sigma)=(r_{0},r_{1},r_{2},r_{3},...,r_{\lfloor \frac{n}{2} \rfloor})$, where $r_{l}$ denotes the number of distinct pairs of negative edges which are at distance $l$ and $r_{0}+r_{1}+r_{2}+r_{3}+\cdots +r_{\lfloor \frac{n}{2} \rfloor} = \binom{p}{2}$.
\begin{ex}
\rm{Consider $\Sigma = (W_8,\sigma)^o$, as depicted in Figure~\ref{signed wheel W_8}. Let $e_1= v_1v_2, e_2=v_2v_3, e_3=v_4v_5, e_4=v_7v_8$ so that $\sigma^{-1}(-1)= \{e_1,e_2,e_3,e_4\}$. It is easy to see that $d(e_1,e_2)=0,~d(e_1,e_4)=d(e_2,e_3)=1,~\text{and}~d(e_1,e_3)=d(e_2,e_4)=d(e_3,e_4)=2$. Therefore, $r_{0}=1,~r_{1}=2,~r_{2}=3,~r_{3}=0,~r_{4}=0$. Hence we have $D(\Sigma) = (1,2,3,0,0)$.}
\end{ex}
\begin{figure}[ht]
\begin{center}
\begin{tikzpicture}[scale=0.7]
\node[vertex] (v1) at (10,0) {};
\node[vertex] (v2) at (11.50,0.90) {};
\node[vertex] (v3) at (12.5,2.5) {};
\node[vertex] (v4) at (11.50,4.15) {};
\node[vertex] (v5) at (10,5) {};
\node[vertex] (v6) at (8.25,4.15) {};
\node[vertex] (v7) at (7.5,2.5) {};
\node[vertex] (v8) at (8.50,0.90) {};
\node[vertex] (v9) at (10,2.5) {};
\node [above] at (v5) {$v_{1}$};
\node [xshift=0.25cm, yshift=0.15cm] at (v4) {$v_{2}$};
\node [xshift=0.3cm, yshift=-0.05cm] at (v3) {$v_{3}$};
\node [xshift=0.25cm, yshift=-0.2cm] at (v2) {$v_{4}$};
\node [xshift=0.05cm, yshift=-0.3cm] at (v1) {$v_{5}$};
\node [xshift=-0.25cm, yshift=-0.2cm] at (v8) {$v_{6}$};
\node [left] at (v7) {$v_{7}$};
\node [xshift=-0.25cm, yshift=0.15cm] at (v6) {$v_{8}$};
\node at (10.2,3) {$v$};
\foreach \from/\to in {v2/v3,v5/v6,v7/v8,v8/v1,v9/v1,v9/v2,v9/v3,v9/v4,v9/v5,v9/v6,v9/v7,v9/v8} \draw (\from) -- (\to);
\draw[dotted] (10,0) -- (11.50,0.90);
\draw[dotted] (10,5) -- (11.50,4.15);
\draw[dotted] (12.5,2.5) -- (11.50,4.15);
\draw[dotted] (7.5,2.5) -- (8.25,4.15);
\end{tikzpicture}
\end{center}
\caption{A signed $W_8$, where dark lines denote positive edges and dotted lines denote negative edges.} \label{signed wheel W_8}
\end{figure}
The following lemma will help us in deciding whether two signed wheels of the form $(W_n,\sigma)^o$ with $p$ negative edges are isomorphic to each other.
\begin{lemm}\label{key lemma}
Two signed wheels $\Sigma_1=(W_n,\sigma_1)^o$ and $\Sigma_2= (W_n,\sigma_2)^o$ with $p$ negative edges are isomorphic to each other if and only if $D(\Sigma_1) = D(\Sigma_2)$.
\end{lemm}
\begin{proof}
Let $\Sigma_1$ and $\Sigma_2$ be isomorphic to each other. Since an isomorphism preserve the distance, it follows that $D(\Sigma_1) = D(\Sigma_2)$.
\noindent
Conversely, let $\Sigma_1=(W_n,\sigma_1)^o$ and $\Sigma_2= (W_n,\sigma_2)^o$, with $p$ negative edges, satisfy $D(\Sigma_1) = D(\Sigma_2)$. We need to show that $\Sigma_1$ and $\Sigma_2$ are isomorphic to each other. To establish an isomorphism that maps $\Sigma_1$ onto $\Sigma_2$, we first fix the position of $p$ negative edges of $\Sigma_1$ in clockwise direction, say, at $v_{1_1}v_{1_1 +1}, v_{1_2}v_{1_2 +1},v_{1_3}v_{1_3 +1},...,v_{1_p}v_{1_p +1}$ such that $1 \leq 1_i < 1_j \leq n$ for $1 \leq i < j \leq p$.
Since $D(\Sigma_1) = D(\Sigma_2)$, the positions of $p$ negative edges of $\Sigma_2$ can also be fixed in clockwise direction say, at $v_{2_1}v_{2_1+1}, v_{2_2}v_{2_2+1},v_{2_3}v_{2_3 +1},...,v_{2_p}v_{2_p +1}$, where $1 \leq 2_{i} \leq n$ so that
\begin{equation}\label{equation 1}
d(v_{1_i}v_{1_i +1},v_{1_j}v_{1_j +1})=d(v_{2_i}v_{2_i+1},v_{2_j}v_{2_j +1}),~\text{ for all}~ i,j\in \{1,2,...,p\}.
\end{equation}
Define $\phi : V(W_n) \rightarrow V(W_n)$ by
$$\phi(x) = \begin{cases} v & \text{if $x=v$}, \\
v_{2_1 + t} & \text{if $x=v_{1_1 + t}$ for $t=0,1,2,...,n-1$}.
\end{cases}$$
It is easy to verify that $\phi$ is an isomorphism that maps $\Sigma_1$ onto $\Sigma_2$. Hence if $D(\Sigma_1) = D(\Sigma_2)$ then $\Sigma_1$ and $\Sigma_2$ are isomorphic to each other.
\end{proof}
Lemma~\ref{switching non-equi wheels} and Lemma~\ref{key lemma} together yield the following result.
\begin{lemm}\label{main lemma}
Let $\Sigma_1=(W_n,\sigma_1)^o$ and $\Sigma_2= (W_n,\sigma_2)^o$ be two signed wheels with $p$ negative edges such that $D(\Sigma_1) \neq D(\Sigma_2)$. Then $\Sigma_1$ and $\Sigma_2$ are switching non-isomorphic.
\end{lemm}
\begin{lemm}\label{lemma having bounds on dist when p=4}
Among any four edges $e_1, e_2, e_3~\text{and}~e_4$ of $C_n$, there exist two edges $e_i$ and $e_j$ such that $d(e_i,e_j) \leq \lfloor \frac{n-4}{4} \rfloor$.
\end{lemm}
\begin{proof}
For a fix $n$, let if possible
\begin{equation}
d(e_i,e_j) \geq \Big \lfloor \frac{n-4}{4} \Big \rfloor + 1,~\text{for all}~ i,j \in \{1,2,3,4\},~ i \neq j.
\end{equation}
If the distance between $e_i~\text{and}~e_j$ is $k$ then there are at least $k-1$ vertices between end vertices of $e_i~\text{and}~e_j$. Therefore there are at least $\lfloor \frac{n-4}{4} \rfloor$ vertices between $e_i~\text{and}~e_j~\text{for all}~ i,j \in \{1,2,3,4\}$. This means there are at least $4\lfloor \frac{n-4}{4} \rfloor +8$ vertices in $C_n$, a contradiction. Hence the result follows.
\end{proof}
Let us place the vertices of $W_n$ in such a way that the outer cycle $C_n$ becomes the regular $n$-gon $G_n$. Let $e_1,e_2,e_3~\text{and}~e_4$ be four negative edges of $(W_n,\sigma)^o$. We place these four edges $e_1,e_2,e_3~\text{and}~e_4$ in such a way that if $i < j$ and $e_i=v_{r}v_{r+1}, e_j=v_{l}v_{l+1}$ then $r+1 \leq l$. Further, in light of Lemma~\ref{lemma having bounds on dist when p=4}, we can always assume that $\text{d}(e_1,e_2) \leq \lfloor \frac{n-4}{4} \rfloor$. Without loss of generality, let $e_1=v_1v_2$. To calculate the value of $\psi_{4}(n)$, we will count different signatures of size four by applying the following strategies.
\begin{enumerate}
\item[S1.] Take $d(e_1,e_2)=0$ and count the different possibilities for $e_3~\text{and}~e_4$ up to isomorphism. This is carried out in Lemma~\ref{lemma 1 of S1}, Lemma~\ref{lemma 2 of S1}, Lemma~\ref{lemma 3 of S1} and Lemma~\ref{lemma 4 of S1}.
\item[S2.] Take $d(e_1,e_2)=1$ and count the choices for $e_3~\text{and}~e_4$ under the following conditions:\\
(i) $d(e_2,e_3) \geq 1$;\\
(ii) $d(e_3,e_4) \geq 1$;\\
(iii) $d(e_4,e_1) \geq 1$.\\
Note that if any one of $d(e_2,e_3),d(e_3,e_4),d(e_4,e_1)$ is zero then replacement of those two edges with $e_1~\text{and}~e_2$ will give us a signature which is already encountered in S1.
\item[S3.] For $d(e_1,e_2)= r$, where $1 \leq r < \lfloor \frac{n-4}{4} \rfloor$, count different choices of $e_3~\text{and}~e_4$.
\item[S4.] If $d(e_1,e_2)= r+1$, where $2 \leq r+1 \leq \lfloor \frac{n-4}{4} \rfloor$, count different choices for $e_3~\text{and}~e_4$ under the following conditions:\\
(i) $d(e_2,e_3) \geq r+1$;\\
(ii) $d(e_3,e_4) \geq r+1$;\\
(iii) $d(e_4,e_1) \geq r+1$.\\
Note that if any one of $d(e_2,e_3),d(e_3,e_4),d(e_4,e_1)$ is less than $r+1$ then replacement of those two edges with $e_1~\text{and}~e_2$ will give us a signature which is already encountered in S3.
\end{enumerate}
\section{Computation}\label{main results}
In this section, we compute the value of $\psi_{p}(n)$ for $p=0,1,2,3,4,n-4,n-3,n-2,n-1~\text{and}~n$, where $n \geq 4$. To count the number of switching non-isomorphic signed wheels with $p$ negative edges, it is enough to count the different choices of $p$ edges from $E(C_n)$ up to isomorphism (rotations as well as reflections). Note that the counting of different $p$ edges on $C_n$ is same as the counting of different $n-p$ edges. Thus for any $0 \leq p \leq n$, we have
\begin{equation}\label{main equation}
\psi_{p}(n) = \psi_{n-p}(n).
\end{equation}
The following lemma is trivial.
\begin{lemm}\label{case p=0}
For each $n \geq 4$, $\psi_{0}(n)= \psi_{n}(n) =1$.
\end{lemm}
Any two signed wheels of the form $(W_n, \sigma)^o$ with exactly one negative edge are isomorphic (rotationally equivalent) to each other. Therefore, in the view of Equation~\ref{main equation}, the following lemma is immediate.
\begin{lemm}\label{case p=1}
For each $n \geq 4$, $\psi_{1}(n)=\psi_{n-1}(n)=1$.
\end{lemm}
We now determine the value of $\psi_{n-2}(n)~\text{and}~\psi_{2}(n)$.
\begin{lemm}\label{case p=n-2}
For each $n \geq 4$, $\psi_{2}(n)=\psi_{n-2}(n)=1+\lfloor \frac{n-2}{2} \rfloor$.
\end{lemm}
\begin{proof}
We classify it into two cases.
\begin{enumerate}[wide=0pt]
\item[\textit{Case 1.}] If two edges form a path $P_{3}$ then there is only one possibility up to rotation. One such path is $P_{3}= v_{1}v_{2}v_{3}$.
\item[\textit{Case 2.}] If two edges are disjoint, then the number of choices of two edges among $E(C_n)$ is $\lfloor \frac{n-2}{2} \rfloor$ up to isomorphism.
\end{enumerate}
Each choice of two edges in Case 1 and Case 2 produces a signed wheel $(W_n, \sigma)^o$ with two negative edges. In light of Lemma~\ref{main lemma}, all these signed wheels are mutually switching non-isomorphic. This proves that $\psi_{2}(n)=\psi_{n-2}(n)=1+\lfloor \frac{n-2}{2} \rfloor$.
\end{proof}
A number $n$ is said to have a $k\text{-}partition$ if $n=\lambda_1+\lambda_{2}+...+\lambda_{k}$, where $\lambda_{1} \geq \lambda_{2} \geq \lambda_{3} \geq ...\geq\lambda_{k} \geq 1$. Par($n;k$) denotes the set of all $k$-partitions of $n$ with $p(n;k) = |\text{Par}(n;k)|$. Clearly, the number $p(n;k)$ is zero if $n<k$. The numbers $p(n-3;2)$ and $p(n-3;3)$ are used to compute $\psi_{n-3}(n)$. It is well known that $p(n;3)= \bm{[} \frac{1}{12} n^2 \bm{]}$, where $\bm{[} x \bm{]}$ is the nearest integer function. See \cite{Honsberger} for details.
\begin{lemm}\label{case p=n-3}
For each $n \geq 4$, $\psi_{3}(n)=\psi_{n-3}(n)=1+\lfloor \frac{n-3}{2} \rfloor +\bm{[} \frac{1}{12} (n-3)^2 \bm{]}$.
\end{lemm}
\begin{proof}
Since $n-3$ edges are to be chosen from $C_n$, only following three cases are possible:
\begin{itemize}
\item[(i)] all $n-3$ edges form a path;
\item[(ii)] $n-3$ edges form two different paths;
\item[(iii)] $n-3$ edges form three different paths.
\end{itemize}
Clearly, there is only one possibility, up to rotation, if $n-3$ edges form a path. For case (ii), the number of two different paths comprising $n-3$ edges is same as the number of partitions of $n-3$ with exactly two parts. Therefore, the number of two such different paths is $\lfloor \frac{n-3}{2} \rfloor$.
For case (iii), let three distinct paths formed by $n-3$ edges be $P_{t}, P_{t^\prime},~\text{and}~P_{t^{\prime\prime}}$ such that $t \geq t^{\prime}\geq t^{\prime\prime} \geq 2$. For each $t \geq t^{\prime}\geq t^{\prime\prime} \geq 2$, it is easy to see that there is a unique possibility for three such paths, up to rotation. Thus the number of three such paths is same as the number of partitions of $n-3$ with exactly three parts. Hence there are $p(n-3;3)$ different choices for three such paths.
Each different possibility of $n-3$ edges in cases (i), (ii) and (iii) produces a signed wheel $(W_n, \sigma)^o$ with $n-3$ negative edges and in light of Lemma~\ref{main lemma}, all these signed wheels are mutually switching non-isomorphic. Hence $\psi_{n-3}(n)=\psi_{3}(n)=1+\lfloor \frac{n-3}{2} \rfloor +p(n-3;3)=1+\lfloor \frac{n-3}{2} \rfloor +\bm{[} \frac{1}{12} (n-3)^2 \bm{]}$, as desired.
\end{proof}
Let $\Sigma=(W_n, \sigma)^o$ be a signed wheel with exactly four negative edges, say $e_1, e_2, e_3,~\text{and}~e_4$. By Lemma~\ref{lemma having bounds on dist when p=4}, it is possible to choose two edges $e_i$ and $e_j$ so that $\text{d}(e_i,e_j) \leq \lfloor \frac{n-4}{4} \rfloor$. Again, a rotation permits us to choose these two edges as $e_1$ and $e_2$ so that $\text{d}(e_1,e_2) \leq \lfloor \frac{n-4}{4} \rfloor$. We now proceed to compute $\psi_{4}(n)$, and to do so we will make use of S1, S2, S3 and S4 .
\begin{lemm}\label{lemma 1 of S1} If edges $e_1, e_2, e_3~\text{and}~e_4$ form a path on $C_n$, then there is only one signed wheel up to rotation.
\end{lemm}
\begin{lemm}\label{lemma 2 of S1} If edges $e_1, e_2~\text{and}~e_3$ form a path $P_4$ and the edge $e_4$ is at distance at least one from $P_4$, then the number of non-isomorphic signed wheels is $ \lfloor \frac{n}{2} \rfloor -2$.
\end{lemm}
\begin{proof}
Let $e_1=v_1v_2, e_2=v_2v_3~\text{and}~e_3=v_3v_4$. Due to the reflection passing through the mid point of $e_2$, the edge $e_4$ can be $v_{5}v_{6}, v_{6}v_{7},..., v_{\lfloor \frac{n}{2} \rfloor+2}v_{\lfloor \frac{n}{2} \rfloor +3}$ for a total of $\lfloor \frac{n}{2} \rfloor -2$.
\end{proof}
\begin{lemm}\label{lemma 3 of S1} If the edges $e_1, e_2$ form a path $P_3$ and $e_3,e_4$ form an another path on three vertices disjoint from $P_3$, then the number of non-isomorphic signed wheels is $ \lfloor \frac{n}{2} \rfloor -2$.
\end{lemm}
\begin{proof}
Let $e_1=v_1v_2, e_2=v_2v_3$ and $P_3= v_1v_2v_3$. Let $e_3~\text{and}~e_4$ form an another path $P_3^{\prime}$ disjoint from $P_3$. Due to the reflection passing through $v_2$, the path $P_3^{\prime}$ can be $v_{4}v_{5}v_{6}, v_{5}v_{6}v_{7},...,v_{\lfloor \frac{n}{2} \rfloor +1}v_{\lfloor \frac{n}{2} \rfloor +2}v_{\lfloor \frac{n}{2} \rfloor +3}$ for a total of $\lfloor \frac{n}{2} \rfloor -2$.
\end{proof}
\begin{lemm}\label{lemma 4 of S1} Let the edges $e_1, e_2$ form a path $P_3$ and $e_3,e_4$ be non-adjacent with each other as well as with $P_3$. Then the number of non-isomorphic signed wheels is $(k-2)^2$ and $(k-3)(k-2)$ when $n=2k+1$ and $n=2k$, respectively.
\end{lemm}
\begin{proof}
Let $e_1=v_1v_2, e_2=v_2v_3$ and $P_3= v_1v_2v_3$. We classify $n$ into two cases.
\begin{enumerate}[wide=0pt]
\item[Case 1.] \textit{Let $n=2k+1$.} If $e_3=v_4v_5$ then due to the reflection passing through $v_{2}$, the edge $e_4$ can be $v_{6}v_{7}, v_{7}v_{8},...,v_{2k}v_{2k+1}$ for a total of $2k-5$.
If $e_3=v_{l}v_{l+1}$ for $5 \leq l \leq k+1$, then the edge $e_4$ can be $v_{l+2}v_{l+3},...,v_{2k-l+4}v_{2k-l+5}$ for a total of $2k-2l+3$. Thus the number of different choices of $e_3$ and $e_4$ is
\vspace*{-0.05in}
\begin{align*}
& (2k-5)+ \sum_{l=5}^{k+1}[2k-2l+3] \\
=& (2k-5)+2k(k-3)-2 \Big [\frac{(k+1)(k+2)}{2}- 10 \Big ]+3(k-3)\\
=& (k-2)^2.
\end{align*}
\item[Case 2.] \textit{Let $n=2k$.} If $e_3=v_{l}v_{l+1}$ for $4 \leq l \leq k$, then the edge $e_4$ can be $v_{l+2}v_{l+3},...,v_{2k-l+3}v_{2k-l+4}$ for a total of $2k-2l+2$. Thus the number of different choices of $e_3$ and $e_4$ is
\vspace*{-0.05in}
\begin{align*}
& \sum_{l=4}^{k}[2k-2l+2] \\
=& 2k(k-3)-2 \Big [\frac{k(k+1)}{2}- 6 \Big ]+2(k-3)\\
=& (k-3)(k-2).
\end{align*}
\end{enumerate}
In Case 1 and Case 2, each choice of $e_3$ and $e_4$ along with $P_3$ produces a signed wheel with four negative edges. By Lemma~\ref{key lemma}, all these signed wheels are pairwise non-isomorphic. This completes the proof.
\end{proof}
\begin{lemm}\label{lemma1 p=4}
Let $(W_{2k+1}, \sigma)^o$ be a signed wheel with four negative edges in which $d(e_1,e_2)=r$, where $1 \leq r \leq \lfloor \frac{2k-3}{4} \rfloor$. Then the number of non-isomorphic signed wheels is $[k-(2r+1)]^2$.
\end{lemm}
\begin{proof}
Let $e_1=v_1v_2$ and $e_2=v_{r+2}v_{r+3}$ such that $d(e_1,e_2)=r$, where $1 \leq r \leq \lfloor \frac{2k-3}{4} \rfloor$. We count the choices for $e_3$ and $e_4$ in the following two cases.
\begin{enumerate}
\item[(i)] If $e_3=v_{2r+3}v_{2r+4}$, then due to the reflection passing through the mid-point of $e_2$, the edge $e_4$ can be $v_{3r+4}v_{3r+5},...,v_{k+r+2}v_{k+r+3}$ for a total of $k-(2r+1).$
\item[(ii)] If $e_3=v_{l}v_{l+1}$, then $e_4$ can be $v_{l+1+r}v_{l+1+r+1},...,v_{(2k+1)-(l-r-3)}v_{(2k+1)-(l-r-3)+1}$ for a total of $2k-2l+4$, where $2r+4 \leq l \leq k+1$.
\end{enumerate}
Thus if $e_1=v_1v_2$, $e_2=v_{r+2}v_{r+3}$, then the number of choices for $e_3$ and $e_4$ is the sum of all choices obtained in (i) and (ii). Each such choice produces a signed wheel with four negative edges, and by Lemma~\ref{key lemma}, all these signed wheels are mutually non-isomorphic. Hence the number of non-isomorphic signed wheels is
\begin{align*}
& k-(2r+1)+ \sum_{l=2r+4}^{k+1} (2k-2l+4) \\
=& \{k-(2r+1)\}+ \left[ 2k(k+1-2r-3)-2 \left\{\frac{(k+1)(k+2)}{2}-\frac{(2r+3)(2r+4)}{2} \right\}+4(k+1-2r-3) \right]\\
=&[k-(2r+1)]^2.
\end{align*}
This completes the proof.
\end{proof}
\begin{lemm}\label{lemma2 p=4}
Let $(W_{2k}, \sigma)^o$ be a signed wheel with four negative edges in which $d(e_1,e_2)=r$, where $1 \leq r \leq \lfloor \frac{2k-4}{4} \rfloor$. Then the number of non-isomorphic signed wheels is $[k-(2r+1)]+[k-(2r+2)]^2$.
\end{lemm}
\begin{proof}
Let $e_1=v_1v_2$ and $e_2=v_{r+2}v_{r+3}$ such that $d(e_1,e_2)=r$, where $1 \leq r \leq \lfloor \frac{2k-4}{4} \rfloor$. We count the different choices for $e_3$ and $e_4$ in the following two cases:
\begin{enumerate}
\item[(i)] If $e_3=v_{2r+3}v_{2r+4}$, then due to the reflection passing through the mid-point of $e_2$, the edge $e_4$ can be $v_{3r+4}v_{3r+5},...,v_{k+r+2}v_{k+r+3}$ for a total of $k-(2r+1).$
\item[(ii)] If $e_3=v_{l}v_{l+1}$, then $e_4$ can be $v_{l+1+r}v_{l+1+r+1},...,v_{(2k)-(l-r-3)}v_{(2k)-(l-r-3)+1}$ for a total of $2k-2l+3$, where $2r+4 \leq l \leq k+1$.
\end{enumerate}
Thus the number of non-isomorphic signed wheels is
\begin{align*}
& k-(2r+1)+ \sum_{l=2r+4}^{k+1} (2k-2l+3) \\
=& \{k-(2r+1)\}+ \left[ 2k(k-2r-2)-2 \left\{\frac{(k+1)(k+2)}{2}-\frac{(2r+3)(2r+4)}{2} \right\}+3(k-2r-2) \right]\\
=& [k-(2r+1)]+[k-(2r+2)]^2.
\end{align*}
This proves the lemma.
\end{proof}
Note that, in light of Lemma~\ref{main lemma}, all the signed wheels counted in Lemma~\ref{lemma 1 of S1} to Lemma~\ref{lemma2 p=4} are switching non-isomorphic. We now compute $\psi_{4}(n)$ by classifying $n$ into two cases depending upon whether $n$ is odd or even. In the following two theorems, we put $ \lfloor \frac{n-4}{4} \rfloor=l$.
\begin{theorem}\label{Total arrangements of size 4 when n is even}
Let $n = 2k$ for some $k \geq 2$. Then
\begin{equation}\label{n even and n/4 even}
\psi_{4}(2k) = (l+1)k^{2} - (2l+3)(l+1)k + \frac{4l^3+15l^2+20l+9}{3}.
\end{equation}
\end{theorem}
\begin{proof}
Let $\psi_{i}$ be the number of non-isomorphic signed wheels with four negative edges $e_1,e_2,e_3,e_4$ such that $d(e_1,e_2) = i$, where $0 \leq i \leq l$. We have
\begin{align*}
\psi_{4}(2k) &= \sum_{i=0}^{l} \psi_{i} \\
&= \{1+(k-2)+(k-2)+(k-3)(k-2)\} + \sum_{i=1}^{l} \psi_{i}\\
&= \{k^2-3k+3\} + \sum_{i=1}^{l} [k-(2i+1)]+[k-(2i+2)]^2 \\
&= \{k^2-3k+3\} + \sum_{i=1}^{l} [k^2-3k+6i-4ki+4i^2+3]\\
&= \{k^2-3k+3\} + \Big \{ lk^2-3kl+6\frac{l(l+1)}{2}-4k \frac{l(l+1)}{2}+4 \frac{l(l+1)(2l+1)}{6} +3l \Big \} \\
&= (l+1)k^2-(2l+3)(l+1)k + \frac{4l^3+15l^2+20l+9}{3}.
\end{align*}
This completes the proof.
\end{proof}
Note that the value of $\psi_{0}$ is the sum of all the values obtained in Lemma~\ref{lemma 1 of S1},~\ref{lemma 2 of S1},~\ref{lemma 3 of S1} and Lemma~\ref{lemma 4 of S1}. For each $1 \leq i \leq l$, the value of $\psi_{i}$ is given in Lemma~\ref{lemma2 p=4}.
\begin{theorem}\label{Total arrangements of size 4 when n is odd}
Let $n = 2k+1$ for some $k \geq 2$. Then
\begin{equation}\label{n odd and n/4 even}
\psi_{4}(2k+1) = (l+1)k^{2} - 2(l+1)^2k + \frac{(2l+1)(2l+3)(l+1)}{3}.
\end{equation}
\end{theorem}
\begin{proof}
Let $\psi_{i}$ be the number defined in the proof of Theorem~\ref{Total arrangements of size 4 when n is even}. We have
\begin{align*}
\psi_{4}(2k+1) &= \sum_{i=0}^{l} \psi_{i} \\
&= \{1+(k-2)+(k-2)+(k-2)^2\} + \sum_{i=1}^{l} \psi_{i}\\
&= \{k^2-2k+1\} + \sum_{i=1}^{l} [k-(2i+1)]^2 \\
&= \{k^2-2k+1\} + \sum_{i=1}^{l} [k^2-2k+4i-4ki+4i^2+1]\\
&= \{k^2-2k+1\} + \Big \{ lk^2-2kl+4\frac{l(l+1)}{2}-4k \frac{l(l+1)}{2}+4 \frac{l(l+1)(2l+1)}{6} +l \Big \} \\
&= (l+1)k^{2} - 2(l+1)^2k + \frac{(2l+1)(2l+3)(l+1)}{3}.
\end{align*}
This completes the proof.
\end{proof}
\section{Main Results}\label{exact values}
In this section, we compute the number of switching non-isomorphic signed wheels $W_{n}$, for $4 \leq n \leq 10$.
\begin{lemm}\label{remaining values}
The value of $\psi_{5}(10)$ is 16.
\end{lemm}
\begin{proof}
To count $\psi_{5}(10)$, the different choices for five edges on $C_{10}$ are considered in the following cases.
\begin{enumerate}
\item If the five edges form a path $P_6$, then there is only one choice for such a path, up to rotation.
\item If the set of five edges is a disjoint union of $P_5~\text{and}~P_2$ then we can assume that $P_5=v_1v_2v_3v_4v_5$. Due to the reflection passing through $v_3$ and $v_8$, the possibilities for $P_2$ are $v_6v_7$ and $v_7v_8$. Therefore there are only two such choices.
\item If the set of five edges is a disjoint union of $P_4~\text{and}~P_3$, assume that $P_4=v_1v_2v_3v_4$. Due to the reflection passing through the mid point of $v_2v_3$ and its opposite edge $v_7v_8$, the choices for $P_3$ are $v_5v_6v_7$ or $v_6v_7v_8$. Thus there are only two such choices.
\item If the set of five edges is a disjoint union of $P_4,P_2^{1}~\text{and}~P_2^{2}$, where $P_2^{1}~\text{and}~P_2^{2}$ are paths on two vertices, assume that $P_4=v_1v_2v_3v_4$. Further, if $P_2^{1}=v_5v_6$, then $P_2^{2}$ can be $v_{7}v_{8},v_{8}v_{9},v_{9}v_{10}$. If $P_2^{1}=v_6v_7$, then $P_2^{2}$ must be $v_8v_9$. Hence there are four such choices.
\item If the set of five edges is a disjoint union of $P_3^{1},P_3^{2}~\text{and}~P_2$, where $P_3^{1},P_3^{2}$ are paths on three vertices, assume that $P_3^{1}=v_1v_2v_3$. If $P_3^{2}=v_4v_5v_6$ then due to the reflection passing through the mid point of $v_3v_4$ and its opposite edge $v_8v_9$, $P_2$ can be $v_7v_8~\text{or}~ v_8v_9$. If $P_3^{2}=v_5v_6v_7$ then due to the reflection passing through $v_4$ and its opposite vertex $v_9$, $P_2$ must be $v_8v_9$. Finally, if $P_3^{2}=v_6v_7v_8$ then due to the reflection passing through mid point of $v_4v_5$ and its opposite edge $v_9v_{10}$, $P_2$ must be either $v_4v_5$ or $v_9v_{10}$. Thus there are four choices for this case.
\item If the set of five edges is a disjoint union of $P_3,P_2^{1},P_2^{2}~\text{and}~P_2^{3}$, where $P_2^{1},P_2^{2}~\text{and}~P_2^{3}$ are paths on two vertices, then there are two such choices, up to automorphism.
\item If all five edges are mutually disjoint then there is only one choice, up to rotation.
\end{enumerate}
From all these cases, we find that $\psi_{5}(10)=16$. These 16 signed $W_{10}$ are shown in Figure~\ref{psi_5(10)}.
\end{proof}
\begin{lemm}
For $4 \leq n \leq 10$ and $0 \leq p \leq 10$, the values of $\psi_{p}(n)$ are those listed in Table~\ref{table-2}.
\end{lemm}
\begin{proof}
In Table~\ref{table-2}, entries of row $i$, for $i=2,3,4,~\text{and}~5$, are computed from Lemma~\ref{case p=0},~\ref{case p=1},~\ref{case p=n-2}, and Lemma~\ref{case p=n-3} respectively. The values of $\psi_{r}(s)$ for $r=s$ are computed from Lemma~\ref{case p=0} and of $\psi_{r}(s)$ for $r=s-1$ are computed from Lemma~\ref{case p=1}. The values of $\psi_{r-2}(r)$ and $\psi_{r-3}(r)$ for $r=7,8,9,~\text{and}~10$ are computed from Lemma~\ref{case p=n-2} and Lemma~\ref{case p=n-3}, respectively. The values of $\psi_{4}(8)$ and $\psi_{4}(10)(=\psi_{6}(10))$ are computed from Theorem~\ref{Total arrangements of size 4 when n is even} and of $\psi_{4}(9)(=\psi_{5}(9))$ is computed from Theorem~\ref{Total arrangements of size 4 when n is odd}. The value of $\psi_{5}(10)$ is obtained in Lemma~\ref{remaining values}. This proves the lemma.
\end{proof}
\begin{theorem}\label{exact no of signed wheels}
For $n=4,5,6,7,8,9,10$, the number of switching non-isomorphic signed wheels on $W_n$ are those given in Table~\ref{table-3}.
\end{theorem}
\begin{proof}
The values of Table~\ref{table-3} are obtained by respective columns sums of Table~\ref{table-2}.
\end{proof}
\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
\diaghead(-1,1){aaa}
{$p$}{$n$} & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
\hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
\hline
2 & 2 & 2 & 3 & 3 & 4 & 4 & 5\\
\hline
3 & 1 & 2 & 3 & 4 & 5 & 7 & 8\\
\hline
4 & 1 & 1 & 3 & 4 & 8 & 10 & 16\\
\hline
5 & & 1 & 1 & 3 & 5 & 10 & 16\\
\hline
6 & & & 1 & 1 & 4 & 7 & 16\\
\hline
7 & & & & 1 & 1 & 4 & 8\\
\hline
8 & & & & & 1 & 1 & 5\\
\hline
9 & & & & & & 1 & 1\\
\hline
10 & & & & & & & 1\\
\hline
\end{tabular}
\end{center}
\caption{The number $\psi_{p}(n)$ for $n=4,5,...,10$ and $0 \leq p \leq 10$} \label{table-2}
\end{table}
\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
$n$ & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
$\psi(n)$ & 6 & 8 & 13 & 18 & 30 & 46 & 78\\
\hline
\end{tabular}
\end{center}
\caption{The number of switching non-isomorphic signed $W_n$ for $n=4,5,...,10$} \label{table-3}
\end{table}
\section{Conclusion}
Recall from Lemma~\ref{lemma of switching non-equivalent SGs} that the number of switching non-equivalent signed wheels is $2^{n}$. Another way of getting this number is the following.
It was already noticed that any signed wheel is switching equivalent to a signed wheel whose signature is a subset of $E(C_n)$. Also, by Lemma~\ref{two SGs equivalence}, any two signed wheels whose signatures are different subsets of $E(C_n)$ are switching non-equivalent. As the total number of subsets of $E(C_n)$ are $2^{n}$, there are $2^{n}$ switching non-equivalent signed wheels on $n+1$ vertices. However many of these $2^{n}$ signed wheels are isomorphic to each other. For this purpose, we have determined the value of $\psi_{p}(n)$, for $p=0,1,2,3,4,n-4,n-3,n-2,n-1,n$ and the value of $\psi(n)$, for $n=4,5,6,7,8,9,10$. The values of $\psi_{p}(n)$, for $p=5,6,...,n-5$ are still unknown.
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\begin{tikzpicture}[scale=0.22]
\begin{scope}[rotate=18]
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\end{scope}
\end{tikzpicture}
\end{minipage}
\caption{Switching non-isomorphic signed $W_{10}$ with exactly five negative edges}\label{psi_5(10)}
\end{figure}
| 28,897
|
\begin{document}
\begin{abstract}
Let $X$ be an Abelian surface and $C$ a holomorphic curve in $X$ representing a
primitive homology class. The space of genus $g$ curves in the class of
$C$ is $g$ dimensional. We count the number of such curves that pass
through $g$ generic points and we also count the number of curves in the
fixed linear system $|C|$ passing through $g-2$ generic points. These two
numbers, (defined appropriately) only depend on $n$ and $g$ where
$n=\frac{C\cdot C}{2} +1-g$ and not on the particular $X$ or $C$ ($n$ is
the number of nodes when a curve is nodal and reduced).
G\"ottsche conjectured that certain quasi-modular forms are the generating
functions for the number of curves in a fixed linear system
\cite{Gott-conj}. Our theorem proves his
formulas and shows that (a different) modular form also arises in the problem of
counting curves without fixing a linear system. We use techniques that
were developed in \cite{Br-Le} for similar questions on $K3$ surfaces. The
techniques include Gromov-Witten invariants for families and a degeneration
to an elliptic fibration. One new feature of the Abelian surface case is
the presence of
non-trivial $\pic^{0} (X)$. We show that for any surface $S$ the cycle in the
moduli space of stable maps defined by requiring that the image of the map
lies in a fixed linear system is homologous to the cycle defined by
requiring the image of the map meets $b_{1}$ generic loops in $S$
representing the generators of $H_{1}(S;\znums )/\operatorname{Tor}$.
\end{abstract}
\thanks{The first author is supported by a grant from the Ford Foundation
and the second author is supported by NSF grant DMS-9626689.}
\maketitle
\markboth{Curves on Abelian surfaces}
{Curves on Abelian surfaces}
\renewcommand{\sectionmark}[1]{}
\tableofcontents
\pagebreak
\section{Introduction}\label{sec: intro}
Let $X$ be an Abelian surface and let $C$ be a holomorphic curve in $X$
representing a primitive homology class. For $n$ and $g$ satisfying $C\cdot
C=2g-2+2n$, there is a $g$ dimensional space of curves of genus $g$ in the
class of $C$. To define an enumerative problem, one must impose $g$
constraints on the curves. There are two natural ways to do this. One way
is to count the number of curves passing through $g$ generic points which
we denote $N_{g,n}(X,C)$. The second way is to count the number of curves
in the fixed linear system $|C|$
passing through $g-2$ generic points which we denote $N_{g,n}^{FLS}(X,C)$. We
define (modified) Gromov-Witten invariants that compute the numbers
$N_{g,n}(X,C)$ and $N_{g,n}^{FLS}(X,C)$
and we prove that they do not depend on $X$ or $C$ but are universal
numbers henceforth denoted $N_{g,n}$ and $N^{FLS}_{g,n}$. Our main theorem
computes these numbers as the Fourier coefficients of quasi-modular
forms. Note that $X$ does not contain any genus 0 curves
so implicitly $g>0$ throughout.
\begin{thm}[Main theorem]\label{thm: main thm}
The universal numbers $N^{FLS}_{g,n}$ and $N_{g,n}$ are given by the
following generating functions:
\begin{eqnarray*}
\sum_{n=0}^{\infty }N_{g,n}q^{n+g-1}&=&g(DG_{2})^{g-1},\\
\sum_{n=0}^{\infty }N_{g,n}^{FLS}q^{n+g-1}&=&(DG_{2})^{g-2}D^{2}G_{2}\\
&=&(g-1)^{-1}D((DG_{2})^{g-1})
\end{eqnarray*}
where $D$ is the operator $q\frac{d}{dq}$ and $G_{2}$ is the Eisenstein
series, \ie
$$
G_{2}(q)=-\frac{1}{24}+\sum_{k=1}^{\infty }(\sum_{d|k}d)q^{k}.
$$
\end{thm}
Note that the right hand sides are
quasi-modular forms (quasi-modular forms are an algebra generated by
modular forms, $G_{2}$, and $D$, c.f. \cite{Gott-conj}) and the left hand
sides are reminiscent of theta series since the power of $q$ is
$\frac{C\cdot C}{2}=n+g-1$.
The formula for $N_{g,n }^{FLS}$ in Theorem \ref{thm: main thm} was
first conjectured by Lothar
G\"ottsche \cite{Gott-conj} who proved the genus 2 case (see also
\cite{DeB}). More generally, G\"ottsche gave conjectural generating
functions for
the number of curves with $n$ nodes in any $n$-dimensional sublinear system of
$|C|$ for any sufficiently ample divisor $C$ on any surface $S$. His
formulas involve only
$c_{2}(S)$, $C\cdot C$, $C\cdot K$, $K\cdot K$ and universal functions
($K$ is the canonical class). For surfaces with numerically trivial
canonical class, his formulas reduce to quasi-modular forms.
For concreteness, we expand the equations of the theorem to write:
\begin{eqnarray*}
\sum_{n=0}^{\infty }N_{g,n}q^{n+g-1}&=&g\left(\sum_{k=1}^{\infty
}k(\sum_{d|k}d)q^{k} \right)^{g-1},\\
\sum_{n=0}^{\infty }N_{g,n }^{FLS}q^{n+g-1}&=&\left(\sum_{k=1}^{\infty
}k(\sum_{d|k}d)q^{k}
\right)^{g-2}\left(\sum_{k=1}^{\infty }k^{2}(\sum_{d|k}d)q^{k} \right).\\
\end{eqnarray*}
So for example, for small values of $n $ and $g$, $N_{g,n}^{FLS}$
and $N_{g,n}$ are given in the following tables:
\vskip .25in
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}\hline
$N_{g,n}^{FLS}$& n= 0& n=1& n=2& n=3& n=4& n=5&
n=6& n=7\\ \hline \hline
g=2 & 1& 12& 36& 112& 150& 432& 392& 960\\ \hline
g=3 & 1& 18& 120& 500& 1620& 4116& 9920& 19440\\ \hline
g=4 & 1& 24& 240& 1464& 6594& 23808& 73008& 198480\\ \hline
g=5 & 1& 30& 396& 3220& 18960& 88452& 344960& 1169520 \\ \hline
\end{tabular}
\vskip .25in
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}\hline
$N_{g,n}$& n= 0& n=1& n=2& n=3& n=4& n=5&
n=6& n=7\\ \hline \hline
g=2 & 2& 12& 24& 56& 60& 144& 112&
240\\ \hline
g=3& 3& 36& 180& 600& 1620& 3528& 7440& 12960\\
\hline
g=4& 4& 72& 576& 2928& 11304& 35712& 97344& 238176\\
\hline
g=5& 5& 120& 1320& 9200& 47400& 196560& 689920& 2126400\\
\hline
\end{tabular}
\vskip .25in
Gromov-Witten invariants have been remarkably effective in answering many
questions in enumerative geometry for certain varieties such as $\P ^{n}$;
however, the ordinary Gromov-Witten invariants are not very effective for
counting
curves on most surfaces. One basic reason is that the moduli space of
stable maps often fails to be a good model for a linear system (and the
corresponding Severi varieties) for dimensional
reasons. For an bundle $L$ such that $L-K$ is ample, the dimension of the
Severi variety
$V_{g}(L)$ (the closure of the set of geometric genus $g$ curves
in the complete linear system $|L|$) is
$$
\dim _{\cnums }V_{g}(L)=-K\cdot L +g-1+p_{g}-q
$$
where $K$ is the canonical class, $p_{g}=\dim H^{0}(X,K)$, and $q=\dim
H^{1}(X,\mathcal{O})$. On the other hand, the virtual dimension of the
moduli space $\M _{g,L}(X)$ of stable maps of genus $g$ in the class
dual to $c_{1}(L)$ is
$$
\virdim _{\cnums }\M _{g,L}(X)=-K\cdot L+g-1.
$$
The discrepancy $p_{g}-q$ arises from two sources. Since the image of maps
in $\M _{g,L}(X)$ are divisors not only in $|L|$ but also potentially in
every linear
system in $\pic ^{c_{1}(L)}(X)$, one would expect $\dim \M _{g,L}(X)$ to
exceed $\dim V_{g}(L)$ by $q=\dim \pic ^{c_{1}(L)}(X)$ (we use $\pic
^{c}(X)$ to denote the component of $\pic (X)$ with Chern class
$c$). We show that this
discrepancy can be accounted for within the framework of the usual
Gromov-Witten invariants (see Theorem \ref{thm: pullback of class on pic is
ft*ev*}).
However, even if we consider $\M _{g,L}$ as a model for the parametrized
Severi varieties
$$
V_{g}(c_{1}(L))\equiv \bigcup _{L'\in \pic^{c_{1}(L)}(X)}V_{g}(L'),
$$
there is still a $p_{g}$ dimensional discrepancy (see also \cite{Do}).
The reason is the following. The virtual dimension
of $\M _{g,L}(X) $ is the dimension of the space of curves that persist as
{\em pseudo-holomorphic} curves when we perturb the K\"ahler structure to a
generic almost K\"ahler structure. The difference of $p_{g}$ in the
dimensions of $\M _{g,L}$ and $V_{g}(c_{1}(L))$ means that only a
codimension $p_{g}$ subspace of $V_{g}(c_{1}(L))$ persists as
pseudo-holomorphic curves when we perturb the
K\"ahler structure. One way to rectify this situation is to find a compact
$p_{g}$-dimensional\footnote{By this we mean a real $2p_{g}$
dimensional family. Note that the parameter space for the family need not
have an
almost complex structure.} {\em family} of almost K\"ahler structures that
has the
property that the only almost K\"ahler structure in the family that
supports pseudo-holomorphic curves in the class $c_{1}(L)$ is the original
K\"ahler structure. If $T$ is such a family, then the moduli space $\M
_{g,L}(X,T)$ of stable maps for the family $T$ is a better model for the
space $V_{g}(c_{1}(L))$ in the sense that its dimension is stable under
generic perturbations of the family $T\mapsto T'$.
It is straight-forward to extend the notion of the ordinary Gromov-Witten
invariants to invariants
for families of almost K\"ahler structures. The invariants will only depend
on the deformation class of the underlying family of symplectic structures.
Given the existence of a $p_{g}$-dimensional family as described above,
these invariants can be used to answer enumerative geometry questions for
the corresponding surface and linear system.
In general, it is not clear when such a family will exist; however,
if $X$ has a hyperk\"ahler metric $g$ (\ie $X$ is an Abelian or $K3$ surface),
then there is a natural candidate for $T$, namely the hyperk\"ahler family
of K\"ahler structures. We call this family the {\em twistor family}
associated to the metric $g$ and we denote it $T_{g}$. It is parameterized
by a 2-sphere and so $\dim _{\rnums }T_{g}=2=2p_{g}$ as it should.
Furthermore, the property that all the curves in $\M _{g,L}(X,T_{g})$ are
holomorphic for the original complex structure can be proved with Hodge
theory (of course this need no longer be the case for a perturbation of
$T_{g}$ to a generic family of almost K\"ahler structures).
We will define the numbers $N_{g,n}^{FLS}$ and $N_{g,n}$ as certain
Gromov-Witten invariants
for the twistor family $T_{g}$ associated to a hyperk\"ahler metric on an
Abelian surface $X$ (see our previous paper \cite{Br-Le} for the $K3$
case). We show that the invariants only depend on $g$ and $n$
(not $X$ or $C$) and they count each irreducible geometric genus $g$ curve
with positive integral multiplicity. Furthermore, the multiplicity is 1 if
additionally the curve is nodal. There are additional Gromov-Witten
invariants for the twistor family that we also compute. These invariants
are easier to describe with the notation of the Gromov-Witten invariants so
we will postpone the statement of the result until section \ref{sec:
twistor family} (Defintion \ref{def: definition of Ngn and NngFLS} and
Theorem \ref{thm: restatement of main thm including Nij}). The
enumerative problem these additional invariants correspond to is counting
curves that pass through $g-1$ points and lie in a certain one dimensional
family of linear systems.
Section \ref{sec: GW for families} reviews Gromov-Witten invariants for
families and formulates
our result that equates the subset of $\M _{g,L}(X)$ consisting of maps
whose image have fixed divisor class with a cycle more familiar in ordinary
Gromov-Witten theory (see Theorem \ref{thm: pullback of class on pic is
ft*ev*}). In section \ref{sec: twistor family}
we discuss properties
of the twistor family, define $N_{g,n}^{FLS}$ and $N_{g,n}$, and prove they
have the enumerative
properties discussed above. In section \ref{sec: computing Ng(n)} we
compute the invariants to
complete the proof of our main theorem and its generalization. We conclude
with an appendix
containing the proof of Theorem \ref{thm: pullback of class on pic is
ft*ev*} that was postponed in the main exposition.
The authors would like to thank O. DeBarre, A Givental, L. G\"ottsche, K.
Kedlaya, and C. Taubes for helpful discussions and correspondence.
\section{Gromov-Witten invariants for families}\label{sec: GW for families}
We review Gromov-Witten invariants for families and we refer the reader to
\cite{Br-Le} for details.
Let $X$ be any compact symplectic manifold with an almost K\"ahler
structure.
Recall that an $n$-marked, genus $g$ {\em stable map} of degree $C\in
H_{2}(X,\znums )$ is a (pseudo)-holomorphic map $f:\Sigma \to X$ from an
$n$-marked nodal curve $(\Sigma ,x_{1},\ldots,x_{n})$ of geometric genus
$g$ to $X$ with $f_{*}([\Sigma ])=C$ that has no infinitesimal
automorphisms. Two
stable maps $f:\Sigma \to X$ and $f':\Sigma' \to X $ are equivalent if
there is a biholomorphism $h:\Sigma \to \Sigma '$ such that $f=f'\circ h$.
We write $\M _{g,n,C}(X,\omega )$ for the moduli space of equivalence
classes of genus $g$, $n$-marked, stable maps of degree $C$ to $X$. We will
often drop the $\omega $ or $X$ from the notation if they are understood
and we sometimes will drop the $n$ from the notation when it is $0$. If $B$
is a family of almost K\"ahler structures, we denote parameterized version
of the moduli space:
$$
\M _{g,n,C}(X,B)=\bigcup_{t\in B}\M _{g,n,C}(X,\omega _{t}).
$$
If $B$ is a compact, connected, oriented manifold then $\M _{g,n,C}(X,B)$
has a fiduciary cycle $[\M _{g,n,C}(X,B)]^{vir}$ called the virtual
fundamental cycle (see \cite{Br-Le} and the fundamental papers of Li and
Tian \cite{Li-Tian2}\cite{Li-Tian}\cite{Li-Tian3}). The dimension of the
cycle is
$$
\dim_{\rnums }[\M _{g,n,C}(X,B)]^{vir}=2c_{1}(X)(C)+(6-\dim _{\rnums
}X)(g-1)+ 2n+\dim _{\rnums }B.
$$
The invariants are defined by evaluating cohomology classes of $\M _{g,n,C}$
on the virtual fundamental cycle. The cohomology classes are defined via
incidence relations of the maps with cycles in $X$. The framework is as
follows. There are maps
$$
\begin{CD}
\M _{g,1,C}@>{ev}>>X\\
@VV{ft}V\\
\M _{g,C}
\end{CD}
$$
called the {\em evaluation} and {\em forgetful} maps defined by
$ev(\{f:(\Sigma ,x_{1})\to X \})=f(x_{1})$ and $ft(\{f:(\Sigma ,x_{1})\to X
\})=\{f:\Sigma \to X \}$.\footnote{There is some subtlety to making this
definition rigorous since forgetting the point may make a stable map
unstable, but it can be done.} The diagram should be regarded as the
universal map over $\M _{g,C}$.
Given geometric cycles $\alpha _{1},\ldots,\alpha _{l}$ in $X$ representing
classes $[\alpha _{1}],\ldots,[\alpha _{l}]\in H_{*}(X,\znums )$ with
Poincar\'e duals $[\alpha _{1}]^{\vee},\ldots,[\alpha _{l}]^{\vee}$, we can
define the Gromov-Witten invariant
\begin{equation*}
\Phi _{g,C}^{(X,B)}(\alpha _{1},\ldots,\alpha _{l})=\int_{[\M
_{g,C}(X,B)]^{vir}} ft_{*}ev^{*}([\alpha _{1}]^{\vee })\cup \cdots \cup
ft_{*}ev^{*}([\alpha _{l}]^{\vee }) .
\end{equation*}
$\Phi _{g,C}^{(X,B)}(\alpha _{1},\ldots,\alpha _{l})$ counts the number of
genus $g$, degree $C$ maps which are pseudo-holomorphic with respect to
some almost K\"ahler structure in $B$ and such that the image of the map
intersects each of the cycles $\alpha _{1},\ldots,\alpha
_{l}$.\footnote{The integral is defined to be 0 if the integrand is not a
class of the correct degree.} The Gromov-Witten invariants are multi-linear
in the $\alpha $'s and they are symmetric for $\alpha $'s of even degree
and skew symmetric for $\alpha $'s of odd degree. If $p_{1},\ldots,p_{k }$
are points in a path-connected $X$, we will use the shorthand
$$
\Phi^{(X,B)}_{g,C}(\point ^{k},\alpha _{k+1},\ldots,\alpha _{l}):=\Phi
^{(X,B)}_{g,C}(p_{1},\ldots,p_{k},\alpha _{k+1},\ldots,\alpha_{l} ).
$$
Now suppose that $X$ is a K\"ahler surface.
In order to count curves in a fixed linear system $|L|$ with
$c_{1}(L)=[C]^{\vee }$ one would like to restrict the above integral
to the cycle defined by $\Psi_{\Sigma _{0}} ^{-1}(0)$ where $\Psi_{\Sigma
_{0}} $ is the map
$$
\Psi_{\Sigma _{0}} :\M _{g,C}(X,\omega )\to \pic ^{0}(X)
$$
given by $f\mapsto \mathcal{O}(\im (f)-\Sigma _{0})$ where $\Sigma _{0}\in
|L|$ is a fixed divisor. Dually, one can add the pullback by
$\Psi_{\Sigma _{0}} $ of the volume form on $\pic ^{0}(X)$ to the integrand
defining the invariant:
$$
\int_{[\M _{g,C}(X)]^{vir}}\Psi_{\Sigma _{0}} ^{*}([\point ]^{\vee })\cup
ft_{*}ev^{*}([\alpha
_{1}]^{\vee }) \cup \cdots \cup ft_{*}ev^{*}([\alpha _{l}]^{\vee }).
$$
We show that $\Psi_{\Sigma _{0}} ^{*}([\point ]^{\vee })$ can be expressed
in the usual Gromov-Witten framework.
\begin{thm}\label{thm: pullback of class on pic is ft*ev*}
Let $X$ be a K\"ahler surface and let $[\gamma]\in H_{1}(X,\znums ) $ and
let $\til{\gamma }$ be the
corresponding class in $H^{1}(\pic ^{0}(X),\znums )$ induced by the
identification $\pic ^{0}(X)\cong H^{1}(X,\rnums )/H^{1}(X,\znums )$. Then
$$
\Psi _{\Sigma _{0}}^{*}(\til{\gamma})=ft_{*}ev^{*}([{\gamma}]^{\vee }).
$$
\end{thm}
\begin{cor}
Let $[\gamma _{1}],\ldots,[\gamma _{b_{1}}]$ be an oriented integral basis
for $H_{1}(X;\znums )$. Then
$\Psi_{\Sigma _{0}} ^{*}([\point ]^{\vee })=ft_{*}ev^{*}([\gamma
_{1}]^{\vee })\cup \cdots \cup
ft_{*}ev^{*}([\gamma _{b_{1}}]^{\vee }).$
\end{cor}
\pf We defer the proof of Theorem \ref{thm: pullback of class on pic is
ft*ev*} to the appendix. The corollary follows immediately.
The upshot is that we count curves in a fixed linear system using the usual
constraints from Gromov-Witten theory. Namely, the invariant:
$$
\Phi _{g,C}(\gamma _{1},\ldots,\gamma _{b_{1}},\alpha _{1},\ldots,\alpha _{l})
$$
counts the number of genus $g$ maps whose image lie in a fixed linear
system $|L|$ with $c_{1}(L)=[C]^{\vee }$ and hit the cycles $\alpha
_{1},\ldots,\alpha _{l}$.
\section{The twistor family and the definition of
$N_{g,n}$ and $N^{FLS}_{g,n}$}\label{sec: twistor family}
The discussion in this section is very similar to the corresponding
discussion for $K3$ surfaces given in section 3 of \cite{Br-Le}. We show
that there is a unique family $T$ (up to deformation) corresponding to the
twistor family. We define $N_{g,n}^{FLS}$ and $N_{g,n}$ using a suitable
set of Gromov-Witten invariants for the family $T$. We show that the
invariants solve the enumerative problems we are interested in. We use this
framework to prove that the invariants $N^{FLS}_{g,n}$ and $N_{g,n}$ are
universal numbers independent of the choice of the Abelian surface $X$ and
the linear system $|L|$.
Let $(X,\omega )$ be an Abelian surface and let $g$ be the unique
hyperk\"ahler metric given by Yau's theorem \cite{Yau}. Define $T_{g}$ to
be the family of K\"ahler structures given by the unit sphere in
$\mathcal{H}^{2}_{+,g}$, the space of self-dual, harmonic 2-forms.
\begin{prop}\label{prop: twistor family is unique}
For any two hyperk\"ahler metrics $g$ and $g'$, the corresponding twistor
families $T_{g}$ and $T_{g'}$ are deformation equivalent. There is
therefore a well-defined deformation class which we denote by $T$.
\end{prop}
\pf The moduli space of complex structures on the 4-torus is connected and
the space of hyperk\"ahler structures for a fixed complex torus is
contractible (it is the K\"ahler cone). Therefore the space parametrizing
hyperk\"ahler 4-tori is connected (in fact it is just the space of flat
metrics). We can thus find a path of hyperk\"ahler metrics connecting $g$
to $g'$ and by associating the twistor family to each metric, we obtain a
continuous deformation of $T_{g}$ to $T_{g'}$. \qed
An important observation concerning the twistor family is the following
corollary.
\begin{cor}
Let $f:X\to X$ be an orientation preserving diffeomorphism, then $f^{*}(T)$
is deformation equivalent to $T$.
\end{cor}
\pf Let $T_{g}$ be the twistor family for a hyperk\"ahler metric $g$. Then
$f^{*}(T_{g})=T_{f^{*}(g)}$ where $f^{*}(g)$ is the pullback metric which
is also hyperk\"ahler. \qed
The corollary has the consequence that for any orientation preserving
diffeomorphism $f$ the Gromov-Witten invariants for the twistor family
satisfy
$$
\Phi ^{X,T}_{g,C}(\alpha _{1},\ldots,\alpha _{l})=\Phi
^{X,T}_{g,f_{*}(C)}(f_{*}\alpha _{1},\ldots,f_{*}\alpha _{l}).
$$
There is an orientation diffeomorphism of the 4-torus for every element of
$Sl_{4}(\znums )$ given by the descent of the linear action on the
universal cover $\rnums ^{4}$. It follows from the elementary divisor
theorem that for any
two classes $C $ and $C' $ in $H_{2}(X,\znums )$ with the same divisibility
and square there is a linear diffeomorphism $f$ such that $f_{*}(C)=C'$
(\cite{abelian} pg. 47). This means that there is a lot of symmetry among
the Gromov-Witten invariants for the twistor family so that (for primitive
classes) they can be encompassed by $N_{g,n} $ and $N_{g,n}^{FLS}$ and
the other invariants which we define below.
\begin{definition}\label{def: definition of Ngn and NngFLS}
Let $\gamma _{1},\ldots,\gamma _{4}$ be loops in $X$ representing an
oriented basis for
$H_{1}(X,\znums )$ and let $C\in H_{2}(X,\znums )$ be a primitive class
with $C\cdot C=2g-2+2n$. We define:
\begin{eqnarray*}
N_{g,n}&=&\Phi ^{(X,T)}_{g,C}(\point ^{g})\\
N_{g,n}^{FLS}&=&\Phi ^{(X,T)}_{g,C}(\gamma _{1},\gamma _{2},\gamma
_{3},\gamma _{4},\point ^{g-2}).
\end{eqnarray*}
Now let $C=[\gamma _{1}]\wedge [\gamma _{2}]+(n+g-1)[\gamma _{3}]\wedge
[\gamma _{4}] $ and define\footnote{Here we use the fact that on any torus
there is a
natural identification $$\Lambda ^{2}H_{1}(X;\znums )\cong H_{2}(X,\znums
).$$ }
$$
N^{ij}_{g,n}=\Phi ^{(X,T)}_{g,C}(\gamma _{i},\gamma _{j},\point ^{g-1})
$$
for $i<j$.
\end{definition}
The invariants $N_{g,n}$, $N_{g,n}^{FLS}$, and $N_{g,n}^{ij}$ encompass all
possible Gromov-Witten invariants for the twistor family and primitive
homology classes. Since $X$ has real dimension four, the only non-trivial
constraints come
from the point class and $H_{1}(X;\znums )$. Since the point class is
invariant under orientation preserving diffeomorphisms, $\Phi
^{(X,T)}_{g,C}(\point ^{g})$ only depends on the square (and divisibility)
of $C$. Similarly, since $\Phi $ is skew-symmetric in the classes $\gamma
_{i}$, the only possibility with four
$\gamma $-constraints are the invariants $\Phi _{g,C}^{(X,T)}(\gamma
_{1},\gamma _{2},\gamma _{3},\gamma _{4},\point ^{g-2})$ which also only
depends on the square (and divisibility) of $C$.
\footnote{This invariant only depends on the square and divisibility of $C$
since for any orientation preserving diffeomorphism $f$ we have:
\begin{eqnarray*}
\Phi ^{(X,T)}_{g,C}(\gamma _{1},\ldots,\gamma _{4},\point ^{g-2})&=&
\Phi _{g,f_{*}(C)}^{(X,T)}(f_{*}\gamma _{1},\ldots,f_{*}\gamma _{4},\point
^{g-2})\\
&=&\det (f_{*}:H_{1}\to H_{1})\Phi _{g,f_{*}(C)}^{(X,T)}(\gamma
_{1},\ldots,\gamma _{4},\point ^{g-2})\\
&=&\Phi _{g,f_{*}(C)}^{(X,T)}(\gamma
_{1},\ldots,\gamma _{4},\point ^{g-2}).
\end{eqnarray*}
}
For dimensional reasons, the invariants with one or three $\gamma $'s
are zero. The invariants $N_{g,n}^{ij}$ encompass the remaining invariants
since for any primitive $C$ with $C^{2}=2g-2+2n$ we can first move $C$ to
$[\gamma _{1}]\wedge [\gamma _{2}]+(n+g-1)[\gamma _{3}]\wedge
[\gamma _{4}] $ by an orientation preserving diffeomorphism.
We now wish to show that $N_{g,n}$, $N_{g,n}^{FLS}$, and $N_{g,n}^{ij}$
enumerate holomorphic curves as we want.
\begin{lemma}\label{lem: twistor family has only one member supposrting
curves in C}
Suppose that $X$ is an Abelian surface with a hyperk\"ahler metric $g$ and
suppose that $C\subset X$ is a holomorphic curve. Then the only K\"ahler
structure in $T_{g}$ that has a holomorphic curve in the class $[C]\in
H_{2}(X;\znums )$ is the original K\"ahler structure for which $C$ is holomorphic.
\end{lemma}
\pf A necessary condition for the class $[C]$ to contain holomorphic curves
is that $[C]^{\vee }\in H^{1,1}(X,\rnums )$ and $[C]$ pairs positively with
the K\"ahler form. Since the class orthogonal to $\mathcal{H}^{2}_{+,g}$
are always of type $(1,1)$ we just need to see when the projection of
$[C]^{\vee }$
to $\mathcal{H}^{2}_{+,g}$ is type $(1,1)$.
We may assume $[C]^{2}$ is non-negative since $X$ has no rational curves
and so the projection of $[C]^{\vee }$ onto $\mathcal{H}^{2}_{+,g}$ is
non-zero. By
definition, the complex structure associated to a form $\omega \in
T_{g}\subset \mathcal{H}^{2}_{+,g}$ defines the orthogonal splitting
$\mathcal{H}^{2}_{+,g}\cong \omega \rnums \oplus (H^{2,0}\oplus
H^{0,2})_{\rnums }$. Therefore, the only $\omega \in T_{g}$ for which
$[C]^{\vee }$
is type $(1,1)$ and pairs positively with $\omega $ is when $\omega $ is a
positive multiple of the projection of $[C]^{\vee }$ to
$\mathcal{H}^{2}_{+,g}$.
This is unique and so must be the original K\"ahler structure for which
$C\subset X$ is holomorphic. \qed
In general, Gromov-Witten type invariants may give a different count from
the corresponding enumerative problem because Gromov-Witten invariants
count maps instead of curves and the image of a map may have geometric
genus smaller than its domain. We show that in the case at hand this does
not happen.
\begin{lemma}
Suppose $X$ is generic among those Abelian surfaces admitting a curve in
the class of $C$. Then the invariants $N_{g,n}$, $N_{g,n}^{FLS}$, and
$N_{g,n}^{ij}$ count only maps whose image has geometric genus $g$ and they
are counted with positive integral multiplicity.
\end{lemma}
\pf The assumption guarantees that $C$ generates $\pic (X)/\pic ^{0}(X)$ so
that all the curves in the class $[C]$ are reduced and irreducible. Then a
dimension count shows that maps with contracting components of genus one or
greater are not counted. The details are the same as in the proof of
Theorem 3.5 in \cite{Br-Le}. \qed
We summarize the results of this section in the following theorem.
\begin{theorem}\label{thm: summary of properties of invariants}
Let $C\subset X$ be a holomorphic curve $C$ in an Abelians surface $X$ with
$C^{2}=2g-2+2n$. Then the invariants $N_{g,n}$, $N_{g,n}^{FLS}$, and
$N^{ij}_{g,n}$ defined in Definition \ref{def: definition of Ngn and
NngFLS} count the number of genus $g$ curves in the class $[C]$
that:
\begin{enumerate}
\item pass through $g$ generic points ($N_{g,n}$);
\item pass through $g-2$ generic points and are confined
to the linear system $|C|$ ($N_{g,n}^{FLS}$).
\item pass though $g-1$ generic points and are confined to lie in a certain
1-dimensional family of linear systems determined by $ij\in
\{12,13,14,23,24,34 \}$ ($N_{g,n}^{ij}$).
\end{enumerate}
Furthermore, the invariants do not depend on the particular Abelian surface
$X$ or the curve $C$ and the invariants count all irreducible, reduced
curves with positive multiplicity that is one for nodal curves.
\end{theorem}
\section{Computing the invariants}\label{sec: computing Ng(n)}
In this section, we compute the invariants $N_{g,n}$, $N_{g,n}^{FLS}$ and
$N_{g,n}^{ij}$ by using a particular choice for $X$ and $C$. Namely, we
will choose $X$ to be a product of elliptic curves and $C$ to be a section
together with a multiple of the fiber. The computations
give our main theorem as well as the generalization to the $N_{g,n}^{ij}$
case. For convienence we state the results below:
\begin{theorem}\label{thm: restatement of main thm including Nij}
The invariants $N_{g,n}$, $N_{g,n}^{FLS}$ and
$N_{g,n}^{ij}$ (defined in Definiton \ref{def:
definition of Ngn and NngFLS}) are given by the following generating
functions:
\begin{eqnarray}
\label{eqn: formula for Ngn}
\sum_{n=0}^\infty N_{g,n}q^{n+g-1} &=&g\left( DG_2\right) ^{g-1}\\
\label{eqn: formula for
NFLS}
\sum_{n=0}^{\infty }N_{g,n}^{FLS}q^{n+g-1}
&=&(DG_{2})^{g-2}D^{2}G_{2}\\
&=&(g-1)^{-1}D((DG_{2})^{g-1})\nonumber\\
\label{eqn: formula for N12}
\sum_{n=0}^\infty N_{g,n}^{12}q^{n+g-1}&=&D\left((DG_{2})^{g-1} \right)\\
\label{eqn: formula for N34}
\sum_{n=0}^\infty N_{g,n}^{34}q^{n+g-1}&=&\left( DG_2\right) ^{g-1}\\
\label{eqn: N13,N14,N23 are 0}
\sum_{n=0}^{\infty }N_{g,n}^{13}q^{n+g-1}&=&0\\
\sum_{n=0}^{\infty
}N_{g,n}^{14}q^{n+g-1}&=&0\nonumber\\
\sum_{n=0}^{\infty
}N_{g,n}^{23}q^{n+g-1}&=&0\nonumber\\
\sum_{n=0}^{\infty
}N_{g,n}^{24}q^{n+g-1}&=&0.\nonumber
\end{eqnarray}
where $D=q\frac{d}{dq}$ and $G_{2}=-\frac{1}{24}+\sum_{k=1}^{\infty
}(\sum_{d|k}d)q^{k}$.
\end{theorem}
From Theorem \ref{thm: summary of properties of invariants} we are free to
compute the invariants for any choice of $X$ and $C$. Let $X$ be the
product of two generic elliptic
curves $S\times F$ and let $C$ be the primitive homology class $S+\left(
g+n-1\right) F\in H_2\left( X,
\mathbf{Z}\right)$. We write $X=S^1\times S^1\times S^1\times S^1$ by
choosing
a diffeomorphism between $S^1\times S^1$ with $S$ and also one between $
S^1\times S^1$ with $F$ . Next we choose representatives for the loops and
points on $X.$ We consider following four loops generating $H_1\left( X,
\mathbf{Z}\right) :$
\begin{equation*}
\begin{array}{ll}
\gamma _1:S^1\rightarrow X, & \gamma _1\left( e^{it}\right) =\left(
e^{it},b_1,c_1,d_1\right) , \\
\gamma _2:S^1\rightarrow X, & \gamma _2\left( e^{it}\right) =\left(
a_2,e^{it},c_2,d_2\right) , \\
\gamma _3:S^1\rightarrow X, & \gamma _3\left( e^{it}\right) =\left(
a_3,b_3,e^{it},d_3\right) , \\
\gamma _4:S^1\rightarrow X, & \gamma _4\left( e^{it}\right) =\left(
a_4,b_4,c_4,e^{it}\right) .
\end{array}
\end{equation*}
where the $a_{i}$'s, $b_{i}$'s, $c_{i}$'s, and $d_{i}$'s are distinct
points on $S^1.$ We also choose $g$
generic points $p_1=\left( s_1,f_1\right) ,...,p_g=\left( s_g,f_g\right) $
on $X=S\times F.$ For any $f\in F$ and $s\in S$ we call $S_f=S\times \left\{
f\right\} \subset X$ a section curve and $F_s=\left\{ s\right\} \times F$ a
fiber curve of $X.$
{\sc Proof of Equation \ref{eqn: formula for Ngn}:}
We suppose that $\phi :D\rightarrow X$ is a stable map from a genus $g$
curve $D$ with $g$ marked points to $X$ representing the class $C=S+\left(
g+n-1\right) F$ and sending corresponding marked points to the $p_{i}$'s.
First we observe that the image of $\phi $ consists of one section curve
and some fiber curves. This is because the projection of any irreducible
component of $\im (\phi ) $ to $S$ is of degree zero except for
one which has degree one. On the other hand, the projection of the degree
one component to $F$ must have degree
zero because there is no nontrivial morphism between two generic elliptic
curves $S$ and $F$. Therefore $\im (\phi )$ consists of a single
section curve and a number of fibers.
In order for $\phi $ to pass through the $g$ generic points
$p_{1},\ldots,p_{g}$
on $X,$ we need at least $\left( g-1\right) $ fiber curves in the image of $
\phi .$ On the other hand the geometric genus of $D\ $is $g$ and covering
each section or fiber curve will take up at least one genus of $D$.
Therefore $\im \left( \phi \right) $ consists of exactly $g-1$ fiber curves
(possibly with multiplicity) and one section curve.
In fact, $D$ has to have precisely $g$ irreducible components
$D_1,\ldots,D_g$ and
each component is a genus one curve and contains one marked point. The
restriction of $\phi $ to each $D_i$ is either (i) a covering of some fiber
curve containing one of the $p_{i}$'s or (ii) an isomorphism to a
section curve containing one of the $p_{i}$'s.
We can assume that $\phi $ restricted to $D_g$ is an isomorphism onto one of
the section curves containing some $p_i=\left( s_i,f_i\right) .$ There are
$g$
choices of such $p_{i}$'s. Without loss of generality we assume $\phi
\left( D_g\right) $ contains $p_g$ or equivalently $\phi \left( D_g\right)
=S_{f_g}.$ In this case the marked point on $D_g$ must be the unique point
$\phi ^{-1}\left( p_g\right)$.
For $i<g,$ we label the component of $D$ covering $F_{s_i}$ as $D_i.$ Then
$\phi :D_i\rightarrow F_{s_i}$ is an unbranched cover by the Hurwitz
theorem
because both $D_i$ and $F_{s_i}$ are genus one curves. We denote
$k_i=\operatorname{deg}\left( \phi :D_i\rightarrow F_{s_i}\right) >0$ then
we have
$\sum_{i=1}^{g-1}k_i=g+n-1$ since $\phi $ represents the homology class of
$S+\left( g+n-1\right) F$. The number of elliptic curves that admit a
degree $k$ homomorphism to a fixed
elliptic curve is classically known to be $\sum_{d|k}d$. We
fix the origin of the elliptic curve $D_i$ (and $F_{s_i}$) to be its
intersection point with $D_g$ (and $S_{f_g}$ respectively), so that $\phi
:D_{i}\to F_{s_{i}}$ is a homomorphism and thus the
number of choices of $\left( \phi :D_i\rightarrow F_{s_i}\right) $ is given
by $\sum_{d|k_i}d$. Since the marked point on $D_i$ could be any one of the
$k_i$ points in $\phi ^{-1}\left( p_i\right)$, there are a total of
$k_{i}\sum_{d|k_{i}}d$ choices for each marked curve $D_{i}$.
We denote the $(g-1)$-tuple $k_{1},\ldots,k_{g-1}$ by $\mathbf{k}$ and we
write $|\mathbf{k}|$ for $\sum_{i}k_{i}$.
From the preceding discussion, the number of stable maps $\phi $ of
geometric genus $g$ and $g
$ marked points to $X$ in the class of $S+\left( g+n-1\right) F$ is given by
\begin{equation*}
g\sum_{\mathbf{k}:|\mathbf{k}|=n+g-1} \prod_{i=1}^{g-1}k_i\left(
\sum_{d|k_i}d\right).
\end{equation*}
It is
not difficult to see that each such stable map $\phi $ contributes one to
the family Gromov-Witten invariant (c.f. \cite{Br-Le}).
The formula is then proved by summing over $n$ and rearranging:
\begin{eqnarray*}
\sum_{n=0}^\infty N_{g,n}q^{n+g-1} &=&\sum_{n=0}^\infty \left(
g\sum_{|\mathbf{k}|=n+g-1}\prod_{i=1}^{g-1}k_i\left(
\sum_{d|k_i}d\right) \right) q^{n+g-1} \\
&=&g\sum_{n=0}^\infty \left(
\sum_{|\mathbf{k}|=n+g-1}\prod_{i=1}^{g-1}\left(
k_i\sum_{d|k_i}dq^{k_{i}}\right)\right) \\
&=&g\left( \sum_{k=1}^\infty k\sum_{d|k}dq^k\right) ^{g-1}\\
&=&g(DG_{2})^{g-1}.
\end{eqnarray*}\qed
{\sc Proof of Equations \ref{eqn: formula for N12}, \ref{eqn: formula for
N34}, and \ref{eqn: N13,N14,N23 are 0}:}
For these computations we count the number of stable maps $\phi
:D\rightarrow X$ of genus $g$ with $g+1$ marked points that represent the
homology class $C=S+\left( g+n-1\right) F$. The maps are constrained by
requiring that the first $g-1$ marked points are mapped to the points
$p_{1},\ldots,p_{g-1}$ and the image of the remaining two points must lie
on $\gamma _{i}$ and $\gamma _{j}$ respectively. We argue as before that
$D$ has $g$ irreducible components $D_1,...,D_g$ where
each $D_i$ is a genus one curve and it contains one marked point except $D_g$
which contains two marked points. Moreover, the image of the two marked
points on $D_g$ must lie on the two loops $\gamma _{i}$ and $\gamma _{j}$.
It is easy
to check that when the points $p_{i},a_{i},\ldots,d_{i}$ are in general
positions, then no single section or fiber curve can pass through both one
of the loops $ \gamma_{i}$ and one of the points $p_{i}$. We can also verify
directly the following lemma.
\begin{lemma}
No fiber or section curve can pass through two different
$\gamma_{i}$'s unless they are $\left( \gamma _1,\gamma _2\right) $ or
$\left( \gamma _3,\gamma _4\right)$.
No section curve can pass through both $\gamma _1$ and
$\gamma _2.$ The only fiber curve passing through $\gamma _{1}$ and $\gamma
_{2}$ is $F_{12}=F_{\left( a_2,b_1\right) }$.
Moreover, $F_{12}\cap \gamma _1 =\left(
a_2,b_1,c_1,d_1\right) $ and $F_{12}\cap \gamma _2 =\left(
a_2,b_1,c_2,d_2\right) .$
Similarly, no fiber curve can pass through both $\gamma _3$
and $\gamma _4.$ The
only section curve passing through $\gamma _{3}$ and $\gamma _{4}$ is
$S_{34}=S_{\left( c_4,d_3\right) }$. Moreover, $S_{34}\cap \gamma _3
=\left( a_3,b_3,c_4,d_3\right) $ and $S_{34}\cap \gamma _4 =\left(
a_4,b_4,c_4,d_3\right)$.
\end{lemma}
From the lemma the moduli space of stable maps is empty in all cases except $
\left( \gamma _i,\gamma _j\right) =\left( \gamma _1,\gamma _2\right) $ or $
\left( \gamma _3,\gamma _4\right) $ up to permutations. Therefore the
corresponding family Gromov-Witten invariants vanish (proving Equation
\ref{eqn: N13,N14,N23 are 0})
\begin{equation*}
N_{g,n}^{13}=N_{g,n}^{14}=N_{g,n}^{23}=N_{g,n}^{24}=0.
\end{equation*}
Next we compute $N_{g,n}^{34}.$ Let $\phi $ be any stable map in the
corresponding moduli space, we have $\phi \left( D_g\right) =S_{34}$ by the
above lemma. Moreover the restriction of $\phi $ to $D_g$ is an isomorphism
and $\phi $ must send the two marked points on $D_g$ to $\left(
a_3,b_3,c_4,d_3\right) $ and $\left( a_4,b_4,c_4,d_3\right) .$ It is not
difficult to check that the orientation of the moduli space is compatible
with the one induced from $\left( \gamma _3,\gamma _4\right) .$ The
calculation for the contribution from the other $D_i$'s are identical
with the earlier computation and we obtain
$$
\sum_{n=0}^\infty N_{g,n}^{34}q^{n+g-1}=\left( \sum_{k=1}^\infty
k\sum_{d|k}dq^k\right) ^{g-1}=(DG_{2})^{g-1}
$$
thus proving Equation \ref{eqn: formula for N34}.
Now we compute $N_{g,n}^{12}$. From the lemma again we have $\phi \left(
D_g\right) =F_{12}$ for any stable map $\phi $ in the corresponding moduli
space. Let us denote the degree of $\phi $ restricted to $D_g$ by $k_0.$
Then the number of possible $\left( \phi :D_g\rightarrow F_{12}\right) $ is
given by $\sum_{d|k_0}d$ as before. The image of the two marked points on $
D_g$ are $\left( a_2,b_1,c_1,d_1\right) $ and $\left( a_2,b_1,c_2,d_2\right)
.$ Since $\phi $ is an unbranched covering map, there are $k_0$ choices for
the location of each marked point. There are then a total of
$k_0^2\sum_{d|k_0}d$ different choices associated the component $D_g$.
There are $g-1$ choices for which curve $D_{1},\ldots,D_{g-1}$ is mapped to
a section. After making such a choice we may assume without loss of
generality that $D_{g-1}$ is mapped to the section curve passing through
$p_{g-1}$ (since we can relabel the points). Then since $\phi $ is an
isomorphism on $D_{g-1}$, the location of the
marked point on it is determined. The analysis for the other fibers is
the same as before and the issue of the orientation is the same as above.
In conclusion we have
\begin{equation*}
N_{g,n}^{12}=\left( g-1\right) \sum_{\mathbf{k}:|\mathbf{k}|=n+g-1} \left(
k_0^2\sum_{d|k_0}d\prod_{i=1}^{g-2}k_i\sum_{d|k_i}d\right) ,
\end{equation*}
where the summation is over $(g-1)$-tuples
$\mathbf{k}=\{k_{0},\ldots,k_{g-2} \}$ with $k_i>0$ and
$|\mathbf{k}|=\sum_{i=0}^{g-2}k_i=g+n-1$. Summing over $n$, we prove
Equation \ref{eqn: formula for N12}:
\begin{eqnarray*}
\sum_{n=0}^\infty N_{g,n}^{12}q^{n+g-1} &=&\left( g-1\right) \left(
\sum_{k=1}^\infty k^2\sum_{d|k}dq^k\right) \left( \sum_{k=1}^\infty
k\sum_{d|k}dq^k\right) ^{g-2} \\
&=&\left( g-1\right) \left( D^2G_2\right) \left( DG_2\right) ^{g-2}\\
&=&D\left((DG_{2})^{g-1} \right).
\end{eqnarray*}\qed
{\sc Proof of Equation \ref{eqn: formula for
NFLS}:}
Suppose that $\phi :D\rightarrow X$ is a stable map of genus $g$ with $g+2$
marked points and represents the homology class $C=S+\left( g+n-1\right) F$.
We argue as before that $D$ has $g$ irreducible components $D_1,...,D_g$.
The curves $D_{1},\ldots,D_{g-2}$ each have one marked point mapping to
$p_{1},\ldots,p_{g-2}$ respectively. The curves $D_{g-1}$
and $D_g$ contain the remaining four marked points which must lie on the
four loops $\gamma
_1,...,\gamma _4$. The image of each $D_i$ must either be a section curve or
a fiber curve. By the previous lemma the only fiber curve that intersects
two $\gamma_{i}$'s must be $F_{12}$ and it intersects $\gamma _1$ and
$\gamma _2$
at $\left( a_2,b_1,c_1,d_1\right) $ and $\left( a_2,b_1,c_2,d_2\right) $
respectively.
Similarly, the only section that intersects two $\gamma$'s is $S_{34} $ and
it must intersect $\gamma _3$ and $\gamma _4$ at $\left(
a_3,b_3,c_4,d_3\right) $ and $\left( a_4,b_4,c_4,d_3\right) $
respectively.
Hence the image of $D_{g-1}$ and $D_g$ must be $F_{12}$ and $S_{34}$
respectively. For $i\leq g-2,$ we have $\phi \left( D_i\right) =F_{s_i}.$
Let
the degree of $\phi $ on $D_i$ be $k_i$ where $1\leq i\leq g-1.$ Then $\sum
k_i=g+n-1.$ The number of choices of $\left( \phi :D_i\rightarrow
F_{s_i}\right) $ is again given by $\sum_{d|k_i}d.$ For $i<g-1$ there are $
k_i$ choices of the location of the marked point on $D_i.$ For $i=g-1,$
there are two marked points on $D_{g-1}$ and each has $k_{g-1}$ possible
locations. Again it is not difficult to check that such maps each
contribute $+1$ to the family Gromov-Witten invariants. In conclusion we have
\begin{equation*}
N_{g,n}^{FLS}=\sum_{\mathbf{k}:|\mathbf{k}|=n+g-1} \left(
k_{g-1}^2\sum_{d|k_{g-1}}d\right)
\prod_{i=1}^{g-2}\left( k_i\sum_{d|k_i}d\right)
\end{equation*}
and so summing over $n$ we have
\begin{eqnarray*}
\sum_{n=0}^\infty N_{g,n}^{FLS}q^{n+g-1} &=&\left( \sum_{k=1}^\infty
k\sum_{d|k}dq^k\right) ^{g-2}\left( \sum_{k=1}^\infty
k^2\sum_{d|k}dq^k\right) \\
&=&\left( DG_2\right) ^{g-1}\left( D^2G_2\right)
\end{eqnarray*}
proving Equation \ref{eqn: formula for
NFLS}.\qed
\appendix
\section{Proof of Theorem \ref{thm: pullback of class on pic is
ft*ev*}}\label{appendix}
First we determine a more explicit formulation of the map
$$
\Psi _{\Sigma _{0}}:\M _{g,C}\to \pic ^{0}(X)=H^{1}(X,\rnums
)/H^{1}(X,\znums ).
$$
Throughout this section we will abuse notation and refer to a stable map
$f:\Sigma \to X$ by its image so that when we say $\Sigma \in \M_{g,C}$ we
mean the curve $\Sigma \subset X$ given by the image of the map $f:\Sigma
\to X$. By definition, $\Psi_{\Sigma _{0}} (\Sigma )=\mathcal{O}(\Sigma
-\Sigma _{0})$. We wish to get a 1-form representative for
$\mathcal{O}(\Sigma -\Sigma _{0})$. First, choose a $C^{\infty }$
trivialization of the bundle $\mathcal{O}(\Sigma -\Sigma _{0})$. Let
$s_{0}$ and $s$ be defining sections of $\mathcal{O}(\Sigma _{0})$ and
$\mathcal{O}(\Sigma )$. Then via the trivialization, $h=s_{0}/s$ is a
non-vanishing $C^{\infty }$ function on $X^{o}=X-\{\Sigma _{0}\cup \Sigma
\}$. Define an integrable real 1-form $a_{h}$ by
$$
a_{h}=\frac{1}{2\pi i}\left(\frac{\dbar h}{h}-\frac{\del
\overline{h}}{\overline{h}} \right).
$$
The form $a_{h}$ defines a real 1-current and let $a_{\Sigma _{0}}^{\Sigma
}$ be the 1-form
representing the harmonic projection of $a_{h}$. We wish to define a map to
$H^{1}(X,\rnums )/H^{1}(X,\znums )$ using $a_{\Sigma _{0}}^{\Sigma }$ and
we need the following lemma:
\begin{lemma}
The harmonic 1-form $a_{\Sigma _{0}}^{\Sigma }$ is independent of the
choices of $s_{0}$, $s$, and the trivialization up to translation by an
integral harmonic 1-form.
\end{lemma}
\pf Let $h'=s_{0}'/s'$ be defined using a possibly different
trivialization. Then $h'=gh $ for some $C^{\infty }$ map $g:X\to \cnums
^{*}$ and so
$$
a_{h}-a_{h'} = \frac{1}{2\pi i}\left(\frac{\dbar g}{g}-\frac{\del
\overline{g}}{\overline{g}} \right) = a_{g}
$$
is a smooth 1-form. We need to show that the harmonic projection of $a_{g}$
is integral. Write $g=e^{t}\theta $ where $t:X\to \rnums $ and $\theta
:X\to S^{1}\subset \cnums $ are $C^{\infty }$. A direct computation shows that
$$
a_{g}=d^{c}t+\frac{1}{2\pi i}\theta ^{-1}d\theta =d^{*}(t\omega
)+\frac{1}{2\pi i}\theta ^{-1}d\theta
$$
where the last equality uses the K\"ahler identities and so we see that
the harmonic projection of $a_{g}$ is the integral class $\frac{1}{2\pi
i}\theta ^{-1}d\theta $.\qed
The lemma shows that
$$
\Sigma \mapsto a_{\Sigma _{0}}^{\Sigma }
$$
determines a well-defined map
$$
\M _{g,C}\to H^{1}(X,\rnums )/H^{1}(X,\znums ).
$$
This is the same map as $\Psi _{\Sigma _{0}}$ since
\begin{enumerate}
\item if $\Sigma $ is linearly equivalent to $\Sigma _{0}$ then $a_{\Sigma
_{0}}^{\Sigma }\equiv 0$ and
\item the sum of divisors corresponds to the sum of forms, \ie $a_{2\Sigma
_{0}}^{\Sigma +\Sigma '}=a_{\Sigma _{0}}^{\Sigma
}+a_{\Sigma _{0}}^{\Sigma '}$.
\end{enumerate}
The first property holds since if $\Sigma \sim \Sigma _{0}$ then $h$ can be
chosen to be a meromorphic function and so $a_{h}\equiv 0$. The second
property follows from $a_{h h'}=a_{h}+a_{h'}$.
To prove Theorem \ref{thm: pullback of class on pic is ft*ev*} we need to
show that $\Psi ^{*}_{\Sigma _{0}}(\til{\gamma
} )=ft_{*}ev^{*}([\gamma ]^{\vee })$. Recall that $\gamma $ is a loop in
$X$, $[\gamma ]^{\vee}$ is the Poincare dual of the 1-cycle $[\gamma ]$,
and $\til{\gamma }\in H^{1}(\pic ^{0}(X),\znums )$ is the natural class
associated to $[\gamma ]$ arising from the identification $\pic
^{0}(X)\cong H^{1}(X,\rnums )/H^{1}(X,\znums )$.
We use the general correspondence
between elements of $H^{1}(M,\znums )$ and homotopy classes of maps $M\to
S^{1}$. One can
get a circle valued function on $M$ from a class $\phi \in H^{1}(M;\znums
)$ by choosing a base point $x_{0}\in M$ and defining $f_{\phi }:M\to
\rnums /\znums $ by
$$
f_{\phi }(x)=\int_{\Gamma _{x_{0}}^{x}}\phi \mod \znums
$$
where $\Gamma _{x_{0}}^{x}$ is a path from $x_{0}$ to $x$. Since $\phi $ is
an integral class, $f_{\phi }$ does not depend on the choice of the path
(mod $\znums $).
Using $\Sigma _{0}$ as the base point for $\M _{g,C}$, the class
$ft_{*}ev^{*}([\gamma ]^{\vee })\in H^{1}(\M _{g,C},\znums )$ is given by the
$S^{1}$-valued map
\begin{eqnarray*}
\Sigma &\mapsto &\int_{\Gamma _{\Sigma _{0}}^{\Sigma
}}ft_{*}ev^{*}([\gamma ]^{\vee }) \mod \znums \\
&=&\int_{ev(ft^{-1}(\Gamma _{\Sigma _{0}}^{\Sigma }))}[\gamma ]^{\vee } \mod
\znums \\
&=&\int_{W_{\Sigma _{0}}^{\Sigma }}[\gamma ]^{\vee }\mod \znums
\end{eqnarray*}
where $W_{\Sigma _{0}}^{\Sigma }$ is the 3-chain in $X$ swept out by the
curves in the path $\Gamma _{\Sigma _{0}}^{\Sigma }$. Note that $\del
W_{\Sigma _{0}}^{\Sigma }=\Sigma _{0}-\Sigma $ and that the map
$$
\Sigma \mapsto \int_{W}[\gamma ]^{\vee }\mod \znums
$$
is the same for any 3-chain $W$ such that $\del W=\Sigma _{0}-\Sigma $
(since the difference $W_{\Sigma _{0}}^{\Sigma }-W$ is a 3-cycle and
$[\gamma ]^{\vee }$ is an integral class).
On the other hand, the class $\til{\gamma } \in H^{1}(\pic ^{0},\znums
)$ is by definition given by the $S^{1}$-valued function on $\pic ^{0}$
defined by
$$
[a]\mapsto \int_{X}a\wedge [\gamma ]^{\vee } \mod \znums
$$
where $a\in H^{1}(X,\rnums )$ and $[a]$ is the corresponding equivalence
class in
$$H^{1}(X,\rnums )/H^{1}(X,\znums ).$$
The class $\Phi _{\Sigma
_{0}}^{*}(\til{\gamma } )$ is therefore represented by the
$S^{1}$-valued function
$$
\Sigma \mapsto \int_{X}a_{\Sigma _{0}}^{\Sigma }\wedge [\gamma ]^{\vee } \mod
\znums
$$
and to prove theorem \ref{thm: pullback of class on pic is ft*ev*} then we need
to show that
$$
\int_{X}a_{\Sigma _{0}}^{\Sigma }\wedge [\gamma ]^{\vee }=\int_{W_{\Sigma
_{0}}^{\Sigma }} [\gamma ]^{\vee }\mod \znums .
$$
Recall that $a_{\Sigma _{0}}^{\Sigma }$ is the harmonic projection of
$a_{h}$. Let $\zeta $ be the harmonic representative for $[\gamma
]^{\vee }$. Then
$$
\int_{X}a_{\Sigma _{0}}^{\Sigma }\wedge [\gamma ]^{\vee }=\int_{X}a_{h}\wedge
\zeta
$$
so we need to show that
\begin{equation}\label{eqn: int over Xo is same as in over W mod Z}
\int_{X^{o}}\frac{1}{2\pi i}\left(\frac{\dbar h}{h}-\frac{\del
\overline{h}}{\overline{h}} \right)\wedge \zeta =\int_{W_{\Sigma
_{0}}^{\Sigma }}\zeta \mod \znums .
\end{equation}
Recall that $h$ is a $\cnums ^{*}$-valued function on $X^{o}=X-\{\Sigma
_{0}\cup \Sigma \}$ and so writing $h=e^{t}\theta $ for smooth functions
$t:X^{o}\to \rnums $ and $\theta :X^{o}\to S^{1}$ we can rewrite the left
hand side of Equation \ref{eqn: int over Xo is same as in over W mod Z} as
$$
LHS=\int_{X^{o}}d^{*}(t\omega )\wedge \zeta +\int_{X^{o}}\frac{1}{2\pi
i}\theta ^{-1}d\theta \wedge \zeta .
$$
We can do the first of these integrals by first integrating along the
fibers of $t:X^{o}\to \rnums $:
\begin{eqnarray*}
t_{*}(d^{*}(t\omega )\wedge \zeta )&=&-t_{*}(*d(*(t\omega ))\wedge \zeta
) \\
&=&t_{*}(d(t\omega )\wedge *\zeta )\\
&=&t_{*}(dt\wedge \omega \wedge *\zeta )\\
&=&dt\wedge t_{*}(\omega \wedge *\zeta )
\end{eqnarray*}
but $t_{*}(\omega \wedge *\zeta )=0$ because $\omega \wedge *\zeta
$ is a closed form defined on {\em all} of $X$ and $t^{-1}(\point )$ is
a boundary 3-chain in $X$.
We perform the remaining integral by integrating first along the fibers of
$\theta :X^{o}\to S^{1}$. Since $\frac{1}{2\pi i}\theta ^{-1}d\theta=\theta
^{*}(dvol_{S^{1}})$, we can write the remaining integral as
$$
\int_{X^{o}}\theta ^{*}(dvol_{S^{1}})\wedge \zeta =\int_{S^{1}}\theta
_{*}(\zeta ) dvol_{S^{1}}.
$$
Now $\theta _{*}(\zeta )=\int_{\theta ^{-1}(c)}\zeta $ is
independent of $c$ (mod $\znums $) since $\theta ^{-1}(c)$ is a 3-chain in
$X$ with boundary $\Sigma _{0}-\Sigma $ and so (mod $\znums $) we have
\begin{eqnarray*}
\text{LHS of Eqn \ref{eqn: int over Xo is same as in over W mod Z} }
&=&\left(\int_{W}\zeta \right)\int_{S^{1}}dvol_{S^{1}}\\
&=&\int_{W}\zeta \\
&=&\text{RHS of Eqn \ref{eqn: int over Xo is same as in over W mod Z} }
\end{eqnarray*}
which proves the theorem. \qed
| 45,810
|
Investopedia (11/18, Hawley) explains flexible spending accounts, stating “medical and dental expenses that exceed the benefits provided by medical and dental insurance policies can be paid using funds from the FSA.”
Patient education library includes information on various topics ranging from dental care to oral health problems.
Freiberg Family Dentistry 255 Union Blvd, Suite 495 Lakewood, Colorado, 80228, United States 303-989-1423
| 354,488
|
ELSMERE PTA
Class registration
Class descriptions
Schedule
Elsmere After-school Enrichment Program (ASEP)
2020 Registration
It’s time to register for 2020 After-school Enrichment Program (ASEP) classes! Each class takes place at Elsmere from 3:30-4:30pm. Classes are held once a week for six weeks, with exceptions as noted. You can register your child(ren) by clicking on "CLASS REGISTRATION" on the toolbar above.
Online registration begins on
Wednesday, 1/22/2020
.
The deadline is Thursday, 1/30/2020
. Classes will be filled on a first-come, first-served basis at the end of each day. We are not able to accept late registrations, so please be sure to submit your enrollment forms on time!
Your registration is not complete until we have received THREE ITEMS:
* Online registration form
*
Payment
(you may pay by PayPal or credit card through the registration form online)
*
Parent volunteer commitment date
There are two ways to
guarantee
your child’s first choice class AND to have your child attend the program at
no cost
.
* Serve as a
p
ermanent volunteer
:
You may opt to volunteer for all of the six classes of a session.
* Serve as a
per weekday volunteer
:
This parent will be present for each of the six weeks on a specific day (i.e. every Monday, every Tuesday, etc.). Your role would be to manage the attendance and make sure each child is accounted for after school as a helper to the class parents, be available until each child has left the building at 4:30pm dismissal, and to be a “floater” in case there is backup needed for any of the classes during session time.
There are five spots available for this post
. If you would like to serve this very important function, please email me when registration begins and let me know what day would be best for you.
Important Items:
* Students may register for more than one class.
Please complete a
separate online form for each child
(although you can register for more than one class per student on the form).
* If a class reaches capacity, students will be chosen by lottery at the end of each day for that day’s registrants.
* Please note that if we have a snow day, or if school closes early, the ASEP class will be cancelled. If that happens, we will notify you regarding the date of the make-up class.
*
Students MUST be picked up promptly at 4:30pm
following ASEP classes. We cannot accommodate late pick-ups. If your child attends School’s Out at Elsmere, the SOI staff will pick him/her up at class. If your child has permission to walk home unaccompanied, you may indicate that on the registration form.
* We will
notify you
by email
to confirm your child(ren)’s class registration could require any medication or has an allergy that could require an EpiPen, we request that you plan to be in school during ASEP classes. There will not be an adult in the building who is trained to administer an EpiPen, and there is no access to the nurse’s medication closet during that time. Parents will be asked to send nut-free snacks for children attending ASEP classes.
* Some ASEP classes reach capacity. Please be mindful that if your child signs up for a no-cost class, s/he is taking a spot from another child. If there’s a possibility that your child won’t end up taking the class, please don’t sign up. Also, if your child took a class last year that was full, we hope that s/he will consider other classes this year in order to leave room for students who’ve not yet had a chance to take the more popular options. We will try to give priority to students who have not taken a particular class in the past, but cannot make guarantees.
* The PTA would like to make the program available to every student who wishes to participate. If you need financial assistance to cover the cost of a class, please contact Mrs. Kloss confidentially at
kkloss@bethlehemschools.org
and she will help make arrangements.
Elsmere School PTA ASEP Behavior Expectations
We need your assistance to make this program successful. Please review these behavior expectations with your child(ren) before the program begins and for the duration. Your electronic signature on the online registration form indicates that you have read this material, discussed with your children, and understand that a students failure to adhere to this code of conduct could result in his/her being removed from the program. The PTA is unable to refund fees in the event that a child is removed from a class.
Rules will, in general, follow the rules of Elsmere School. Students should be familiar with these rules and will be reminded of them by our instructors and volunteers. We ask that you remind your students to be on his/her best school behavior during this program. Students are required to be respectful of other students, instructors, and parent volunteers during the program (which includes check-in, snack, instruction, and dismissal). If a student demonstrates behavior that could cause concern for his or her own safety or the safety of another person, the parent will be notified and the student might be asked to leave the class. If a students behavior is disruptive to the class, the instructor or parent volunteer will ask the student to sit out for a period of time. If it happens more than once, the student will be asked to leave the program. If a students behavior could result in his/her being removed from the class, the PTA president or ASEP coordinator will notify the parent.
A designated adult is expected to be at the school promptly at 4:30pm to pick up each child, unless the student has permission to walk home or is released to School’s Out. If the adult is consistently late picking up a student, the PTA reserves the right to require that the student leave the ASEP. We cannot be responsible for children after 4:30pm, and they cannot be left alone in the school building or outside. If a child needs to be removed from the program for any reason, the PTA is unable to issue refunds. These behavior expectations have been created to ensure a safe, productive, and fun learning environment for all students. Please feel free to contact the ASEP coordinator or PTA president with any questions. With your help and with the support of our volunteers, we hope to run a program that is safe, fun and educational for all students.
We’re looking forward to another fabulous Elsmere After-school Enrichment Program in 2020! If you have any questions, please don’t hesitate to email us at
elsmereasep@gmail.com.
Warm regards,
Becki Ida, Jenny Horn, and Suzie Ko, ASEP coordinators
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| 158,459
|
First of all, cement’s plasticity increases when it is prepared using magnetic water.
Plasticity levels depend on the qualities of cement glue and since magnetic treatment influences glue’s qualities, then the level of cement’s plasticity changes, as shown on fig. 1.
Fig. 1. Change in the plasticity level of cement.
When using magnetic water for kneading of cement, its strength increases by 10-20%, cement consumption decreases by 9-12%, and its mobility increases in some cases by more than 1,5 times.
Products made from cement, which is prepared using magnetic water, have higher resistance to outside forces, such as drastic temperature changes, humidity, salty and acidic environments, mechanical overload, etc.; as shown on figures 2, 3, 4, 5, 6.
Fig. 2. Samples of cement after repeated freezing and de-freezing.
Fig. 3. Change in the appearance of cement sea parapet after 5 years since it was installed.
Black Sea, Sochi, Russia.
Fig. 4. Highway precast pavement, 3 years old. Rostov- na-Donu, Russia.
Fig. 5. Cement slabs on the highway, exposure to a wide range of temperature changes
(summer – up to 40 C, winter – up to -40 C). One year after laying, Siberia, Russia.
Fig. 6. Change in the structure of cement pavements, 5 years after it was made, Moscow, Russia.
| 118,381
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\begin{document}
\input epsf
{\abstract In this paper we develop an integer-affine
classification of three-dimen\-sional multistory completely empty
convex marked pyramids. We apply it to obtain the complete lists
of compact two-dimensional faces of multidimensional continued
fractions lying in planes at integer distances to the origin equal
$2$, $3$, $4$, $\ldots$ The faces are considered up to the action
of the group of integer-linear transformations. In conclusion we
formulate some actual unsolved problems associated with the
generalizations for $n$-dimensional faces and more complicated
face configurations.}}
\maketitle \sloppy \normalsize
\tableofcontents
\section*{Introduction and background}
The main purpose of the present paper is to develop an
integer-affine classification of three-dimensional multistory
completely empty convex marked pyramids. We apply it to deduce an
integer-linear classification of compact two-dimensional faces of
multidimensional continued fractions in the sense of Klein lying
in planes at integer distances to the origin greater than unity.
Classification of two-dimensional faces leads to new algorithms
of two-dimensional continued fraction calculations. It is also
the first step in studying the combinatorial structure of
multidimensional continued fractions.
\subsection{General definitions}
Consider a vector space $\r^{n+1}$ for some $n\ge
1$ over $\r$. A point or vector of $\r^{n+1}$ is called {\it
integer} if all its coordinates are integers.
Consider some $k$-dimensional plane of $\r^{n+1}$. The
intersection of a finite number of closed $k$-dimensional
half-planes of the plane is said to be a {\it convex $($solid$)$
$k$-dimensional polyhedron} if it is homeomorphic to
$k$-dimensional closed disk. For $k=2$, $1$, or $0$ we have a {\it
convex polygon}, a {\it segment}, or a {\it point}. Here we
consider polyhedra as convex hulls (with all their inner points).
A convex polyhedron is said to be a {\it convex marked pyramid}
with some marked face and a vertex outside the plane containing
the face if it coincides with the union of all segments joining
the marked vertex with each point of the marked face. The marked
face is called the {\it base} of the marked convex pyramid and
the marked vertex --- the {\it vertex} of the marked convex
pyramid. A polyhedron is called a {\it convex pyramid} if some
structure of convex marked pyramid can be introduced for it.
A convex polyhedron (polygon, segment) is said to be {\it
integer} if all its vertices are all integer points. A convex
(marked) pyramid is said to be {\it integer} if it is an integer
convex polyhedron.
\begin{definition}
An integer convex polyhedron is called {\it empty} if it does not
contain integer points different from the vertices of the
polyhedron. An integer convex marked pyramid is called {\it
completely empty} if it does not contain integer points different
from the marked vertex and from the integer points of the base.
\end{definition}
Two sets in $\r ^{n+1}$ are said to be {\it integer-affine
equivalent} (or have the same {\it integer-affine type}), if
there exists an affine transformation of $\r^{n+1}$ preserving
the set of all integer points, and transforming the first set to
the second. Two sets in $\r ^{n+1}$ are said to be {\it
integer-linear equivalent} (or have the same {\it integer-linear
type}), if there exists a linear transformation of $\r^{n+1}$
preserving the set of all integer points, and transforming the
first set to the second.
\begin{definition}
A plane is called {\it integer} if it is integer-affine equivalent
to some plane passing through the origin and containing a
sublattice of the integer lattice, and the rank of the sublattice
is equivalent to the dimension of the plane.
\end{definition}
Consider some integer $(k-1)$-dimensional plane and an integer
point in the complement to this plane. Let the Euclidean distance
from the given point to the given plane equal $l$. The minimal
value of nonzero Euclidean distances from all integer points of
the ($k$-dimensional) span of the the given plane and the given
point to the plane is denoted by $l_0$. Note that $l_0$ is always
greater than zero and can be obtained for some integer point of
the described span. The ratio $l/l_0$ is said to be the {\it
integer distance} from the given integer point to the given
integer plane.
\begin{definition}
An integer convex marked pyramid is called {\it $l$-story} for
some positive integer $l$ if the integer distance from the vertex
of this pyramid to its base plane is equal to $l$. An integer
convex marked pyramid is called {\it multistory}/{\it
single-story} if the integer distance from the vertex of this
pyramid to its base plane is greater than one/equal to one. (See
example on Figure~\ref{pyramid}.)
\end{definition}
\begin{figure}
$$
\epsfbox{pyramid.1}
$$
\caption{Two images of a completely empty three-story marked
pyramid with vertex $O$ and base $ABC$.}\label{pyramid}
\end{figure}
For any convex polygon there exist a single-story integer
three-dimensional convex marked pyramid with the given polygon as
the base (since any single-story integer convex marked pyramid is
completely empty). Two single-story three-dimensional convex
marked pyramids are integer-affine equivalent iff their bases are
integer-affine equivalent.
It turns out that the case of multistory convex marked pyramids
is essentially different from the single-story case. Only
polygons of a few integer-affine types can be bases of multistory
convex marked completely empty pyramids. For example, the
parallelogram with vertices $(0,0)$, $(0,1)$, $(1,1)$ and $(1,0)$
is not of that type. Besides, there exist integer-affine
nonequivalent multistory convex marked completely empty pyramids
whose bases are integer-affine equivalent.
In Section~1 of the present paper, we give the complete list of
integer-affine types of integer multistory convex marked
completely empty pyramids. To classify the pyramids, we study
arrangements of integer sublattices on the planes parallel to the
bases of the pyramids.
\subsection{Definition of multidimensional continued fra\-c\-ti\-ons in the
sense of Klein}
The problem of generalizing
ordinary continued fractions to the higher-dimensional case was
posed by C.~Hermite~\cite{Herm} in 1839. A large number of
attempts to solve this problem lead to the birth of several
different remarkable theories of multidimensional continued
fractions. In this paper we consider the geometrical
generalization of ordinary continued fractions to the
multidimensional case presented by F.~Klein in~1895 and published
by him in~\cite{Kle1} and~\cite{Kle2}.
Consider a set of $n{+}1$ hyperplanes of $\r^{n+1}$ passing
through the origin in general position. The complement to the
union of these hyperplanes consists of $2^{n+1}$ open orthants.
Let us choose an arbitrary orthant.
\begin{definition}
The boundary of the convex hull of all integer points except the
origin in the closure of the orthant is called the {\it sail}.
The set of all $2^{n+1}$ sails of the space $\r^{n+1}$ is called
the {\it $n$-dimensional continued fraction} associated to the
given $n{+}1$ hyperplanes in general position in
$(n{+}1)$-dimensional space.
\end{definition}
Note that the union of all sails of any continued fraction is
centrally symmetric.
\begin{figure}
$$
\epsfbox{cfrac.1}
$$
\caption{A one-dimensional continued fraction.}\label{cfrac}
\end{figure}
On Figure~\ref{cfrac} we show an example of one-di\-men\-si\-onal
continued fraction. This continued fraction contains four sails
(four broken lines on Picture~\ref{cfrac}). A description of
connections between ordinary continued fractions and geometrical
one-dimensional continued fractions can be found
in~\cite{KarItrig}, \cite{Hir}, and~\cite{Jun}.
Two $n$-dimensional continued fractions are said to be {\it equivalent}
if there exists a linear transformation that preserves the integer lattice
of the $(n+1)$-dimensional space and maps the sails of the first
continued fraction to the sails of the other.
Multidimensional continued fractions in the sense of Klein have
many relations with other branches of mathematics. For example,
J.-O.~Moussafir~\cite{Mou1} and O.~N.~German~\cite{Ger} studied
the connection between the sails of multidimensional continued
fractions and Hilbert bases. In~\cite{Tsu} H.~Tsuchihashi found
the relationship between periodic multidimensional continued
fractions and multidimensional cusp singularities, which
generalizes the relationship between ordinary continued fractions
and two-dimensional cusp singularities. M.~L.~Kontsevich and
Yu.~M.~Suhov discussed the statistical properties of the boundary
of a random multidimensional continued fraction in~\cite{Kon}.
The combinatorial topological generalization of Lagrange theorem
was obtained by E.~I.~Korkina in~\cite{Kor1} and its algebraic
generalization by G.~Lachaud~\cite{Lac}.
Theory of ordinary continued fractions was described, for example,
by A.~Ya.~Hinchin in~\cite{Hin}. V.~I.~Arnold presented a survey
of geometrical problems and theorems associated with
one-dimensional and multidimensional continued fractions in his
article~\cite{ArnPT} and his book~\cite{Arn2}). For the
algorithms of constructing multidimensional continued fractions,
see the papers of R.~Okazaki~\cite{Oka},
J.-O.~Moussafir~\cite{Mou2} and the author~\cite{Kar4}.
E.~Korkina in~\cite{Kor0},~\cite{Kor2},~\cite{Kor3}
and G.~Lachaud in~\cite{Lac},~\cite{Lac2},
A.~D.~Bruno and V.~I.~Parusnikov in~\cite{BP}, \cite{Par1},
\cite{Par1.1}, \cite{Par1.2} and~\cite{Par2}, the author in~\cite{Kar1}
and~\cite{Kar2} produced a large number of fundamental domains
for periodic algebraic two-dimensional continued fractions.
A nice collection of two-dimensional continued fractions
is given in the work~\cite{site} by K.~Briggs.
Besides the multidimensional continued fractions in the sense of Klein,
there exist several different generalizations of continued fractions
to the multidimensional case.
Some other well-known generalizations of continued fractions can
be found in the works of H.~Minkowski~\cite{Min}, G.~F.~Voronoi~\cite{Voro},
A.~K.~Mittal and A.~K.~Gupta~\cite{Mit1}, A.~D.~Bryuno and V.~I.~Parusnikov~\cite{BP2},
V.~Ya.~Skorobogat'ko~\cite{Sko}, and V.~I.~Shmoilov~\cite{Shm}.
\subsection{Two-dimensional faces of multidi\-men\-si\-o\-nal continued fractions}
Many classical papers were dedicated to studying algebraic and algorithmic properties
of multidimensional continued fractions.
The interest to geometrical properties of multidimensional continued fractions
was revived by V.~I.~Arnold's work~\cite{Arn1}
and the subsequent paper of E.~I.~Korkina~\cite{Kor0} on the classification of
$A$-algebras with three generators.
In~1989 and later, V.~I.~Arnold formulated a series of problems and conjectures
associated to the geometrical and topological properties of sails
of multidimensional continued fractions. The majority of these problems
are still open. The geometry of sails has not been sufficiently studied.
In the present work, we make the first steps in the investigation of
geometric properties of sails.
One of the first natural questions here is the following:
{\it what compact faces can sails of multidimensional continued fractions have?
$($these objects are usually studied up to the integer-linear equivalence relation$)$?}
The complete answer to this question was known only for
one-dimensional continued fractions.
{\it For any non-negative integer number $n$ there exists a one-dimensional face
with exactly $n$ integer points inside.
Two compact faces with the same numbers of integer points inside are integer-linear equivalent.}
In the two-dimensional case the original question decomposes into two questions.
{\it What compact faces contained in planes at integer distances
from the origin equal to one can sails of multidimensional
continued fractions have $($up to integer-linear equivalence$)$?}
{\it What compact faces contained in planes at integer distances
from the origin greater than one can sails of multidimensional
continued fractions have $($up to integer-linear equivalence$)$?}
The answer to the first question is quite straightforward. For
any convex polygon $P$ at the unit integer distance from the
origin, there exist an integer positive $k$, a $k$-dimensional
continued fraction, and some face $F$ of a sail of this continued
fraction, such that $F$ is integer-affine equivalent to $P$.
Furthermore, two two-dimensional faces in the planes at the unit
integer distance from the origin are integer-linear equivalent iff
the corresponding polygons are integer-affine equivalent.
Note that up to this moment the following statement on compact
two-dimensional faces (of sails of multidimensional continued
fractions) contained in planes at integer distances from the
origin greater than one was known. {\it Such faces are either
triangles or quadrangles} (see the work~[3] by J.-O.~Moussafir).
In the present work we classify compact two-dimensional faces
contained in planes at integer distances from the origin greater
than one up to integer-linear equivalence. This result was
announced in~\cite{Kar5}. We give the complete lists for continued
fractions of any dimension. This result is based on the
classification of three-dimensional multistory completely empty
convex marked pyramids.
\subsection{Description of the paper}
We start in Section~1 with introducing Theorem~A on
integer-affine classification of three-dimensional multistory
completely empty convex marked pyramids. In this section we also
formulate Theorem~B on integer-linear classification of
two-dimensional faces of the sails at integer distance greater
than one. The integer-affine classification of two-dimensional
faces contained in planes at integer distances from the origin
greater than one (Corollary~C) directly follows from the
integer-linear classification of two-dimensional faces contained
in planes at integer distances from the origin greater than one.
In Sections~2 and~3 we prove Theorem~A and Theorem~B respectively.
And, finally, in Section~4 we give a list of unsolved problems
associated with main theorems of this work.
\section{Formulation of main results}
\subsection{Classification of two-dimensional multistory completely empty pyramids}
By $(a_1, \ldots , a_k)$ in $\r ^n$ for $k<n$ we denote the point
$(a_1, \ldots,a_k, 0, \ldots , 0)$.
Denote the marked pyramid with vertex at the origin and
quadrangular base $(2,-1,0)$, $(2,-a-1,1)$, $(2,-1,2)$, $(2,b-1,1)$,
where $b\ge a \ge 1$, by $M_{a,b}$.
Denote the marked pyramid with vertex at the origin and
triangular base \\
$(\xi,r-1,-r)$, $(a+\xi,r-1,-r)$, $(\xi,r,-r)$,
where $a \ge 1$, $r\ge 1$, by $T_{a,r}^{\xi}$;\\
$(2,1,b-1)$, $(2,2,-1)$, $(2,0,-1)$, where $b\ge 1$, by $U_b$;\\
$(2,-2,1)$, $(2,-1,-1)$, $(2,1,2)$ by $V$;\\
$(3,0,2)$, $(3,1,1)$, $(3,2,3)$ by $W$ (pyramid $W$ is shown on
Figure~\ref{pyramid}).
The integer-affine types of the bases of the described above triangular
and quadrangular pyramids are shown on Figure~1.
\begin{figure}
$$
\epsfbox{mtheorem.1}
$$
\caption{The integer-affine types of the bases of the marked pyramids
of List ``M-W''.}\label{osnovaniya}
\end{figure}
\vspace{1mm} {\bf Theorem A.} {\it Any multistory completely
empty convex three-dimensional marked py\-ra\-mid is
integer-affine equivalent exactly to one of the marked pyra\-mids
from the following list.
{\bf List ``M-W'':}\label{sp_M-W}
\nopagebreak
--- the quadrangular marked pyramids $M_{a,b}$, with integers $b\ge a \ge 1$;
--- the triangular marked pyramids $T_{a,r,}^{\xi}$,
where $a \ge 1$, and $\xi$ and $r$ are relatively prime, and
$r\ge 2$ and $0<\xi\le r/2$;
--- the triangular marked pyramids $U_b$, where $b\ge 1$;
--- the triangular marked pyramid $V$;
--- the triangular marked pyramid $W$.
} \vspace{1mm}
We give the proof of Theorem~A in Section~\ref{pA}.
\subsection{Compact two-dimensional faces at distance greater than one}
Note that up to this moment the following statement on compact
two-dimensional faces contained in planes at the integer distance
from the origin greater than one was known.
\vspace{1mm} {\bf Theorem (J.-O.~Moussafir~\cite{Mou2}.)} {\it Let
$F$ be a two-dimensional compact face of some sail of a
two-dimensional continued fraction. Let $r$ be the integer
distance from the origin to the plane, containing the face.
1. If $r=1$, $F$ may have arbitrary many vertices.
2. If $r=2$, $F$ has at most 4 vertices.
3. If $r\ge 3$, $F$ has three vertices. \qed
}\vspace{1mm}
Here we present a new theorem on integer-linear classification
and its corollary on integer-affine classification of
two-dimensional faces of multidimensional sails (the faces are
again contained in the planes at integer distances greater than
one from the origin). Note that from this theorem and its
corollary it follows that the second item of Moussafir's theorem
can be strengthened:
{\it 2 $'$. If $r=2$, $F$ has three vertices.}
Quadrangular faces for the case of $r=2$ are possible only for
$n$-dimensional continued fractions where $n\ge 3$.
\vspace{1mm} {\it {\bf Theorem~B.}
Any compact two-dimensional face of a sails of a two-dimensional
continued fraction contained in a plane at an integer distance
from the origin greater than one is integer-linear equivalent
exactly to one of the faces of the following list.
{\bf List ``$\alpha_{2}$'':}\label{sp_alpha}
--- triangle with vertices $(\xi,r-1,-r)$, $(a+\xi,r-1,-r)$,
$(\xi,r,-r)$, where $a \ge 1$, integers $\xi$ and $r$ are
relatively prime and satisfy the following inequalities $r\ge 2$
and $0<\xi\le r/2$;
--- triangle with vertices $(2,1,b{-}1)$, $(2,2,-1)$, and $(2,0,-1)$ for $b\ge 1$;
--- triangle with vertices $(2,-2,1)$, $(2,-1,-1)$, and $(2,1,2)$;
--- triangle with vertices $(3,0,2)$, $(3,1,1)$, and $(3,2,3)$.\\
All triangular faces of List ``$\alpha_2$'' are realizable by
sails of dimension two and integer-linear nonequivalent to each
other.
Any compact two-dimensional face of a sails of a $n$-dimensional
$(n\ge 3)$ continued fraction contained in a plane at an integer
distance from the origin greater than one is integer-linear
equivalent exactly to one of the faces of the following list.
{\bf List ``$\alpha_n$'', $n\ge 3$:}
--- quadrangle with vertices $(2,-1,0)$, $(2,-a-1,1)$, $(2,-1,2)$,
$(2,b-1,1)$, where $b\ge a \ge 1$,
--- triangle with vertices $(\xi,r-1,-r)$, $(a+\xi,r-1,-r)$,
$(\xi,r,-r)$, where $a \ge 1$, integers $\xi$ and $r$ are
relatively prime and satisfy the following inequalities $r\ge 2$
and $0<\xi\le r/2$;
--- triangle with vertices $(2,1,b-1)$, $(2,2,-1)$, and $(2,0,-1)$ for $b\ge 1$;
--- triangle with vertices $(2,-2,1)$, $(2,1,2)$, and $(2,-1,-1)$;
--- triangle with vertices $(3,0,2)$, $(3,1,1)$, and $(3,2,3)$.\\
All faces of List ``$\alpha_n$'' are realizable by sails of any
dimension greater than two and integer-linear nonequivalent to
each other. } \vspace{1mm}
\begin{remark}
Note that for any compact face of a sail we can associate an
integer completely empty convex marked pyramid with marked vertex
at the origin and this face as base. Therefore integer-affine
types of such marked pyramids are in one-to-one correspondence
with integer-linear types of faces (see lemma~\ref{l81} below).
\end{remark}
We give the proof of Theorem~B in Section~\ref{pB}.
{\bf Corollary~C.} {\it
Any compact two-dimensional face of a sails of a multidimensional
continued fraction contained in a plane at integer distance from
the origin equals $r$ is integer-affine equivalent exactly to one
of the polygons of the list $\beta_r$ shown below.
\begin{figure}
$$
\begin{array}{c}
\epsfbox{theorem1.11}\\
\epsfbox{theorem1.12}\\
\end{array}
$$
\caption{Integer-affine types of faces of List
``$\beta_2$''.}\label{beta2}
\end{figure}
{\bf List ``$\beta_2$'':}\label{sp_beta}
\nopagebreak
--- quadrangle with vertices $(-1,0)$, $(-a-1,1)$, $(-1,2)$,
$(b-1,1)$, where $b\ge a \ge 1$ $($see the case of $a=2$, $b=3$
on Figure~\ref{beta2}a$)$$)$; quadrangular faces are possible
only for $n$-dimensional continued fractions where $n\ge 3$;
--- single triangle $(-1,0)$, $(0,-2)$,
$(2,1)$ $($see Figure~\ref{beta2}b$)$$)$;
--- triangle $(0,-1)$, $(0,1)$, $(b,0)$, for $b\ge 1$
$($see the case of $b=5$ on Figure~\ref{beta2}c$)$$)$;
--- triangle
$(0,0)$, $(a,0)$, $(0,1)$, for $a\ge 1$ $($see the case of $a=5$
on Figure~\ref{beta2}d$)$$)$.
\begin{figure}[ht]
$$\epsfbox{theorem1.13}$$
\caption{Integer-affine types of faces of List
``$\beta_3$''.}\label{beta3}
\end{figure}
{\bf List ``$\beta_3$'':}
\nopagebreak
--- single triangle $(-1,-1)$, $(1,0)$,
$(0,1)$ $($see Figure~\ref{beta3}a$)$$)$;
--- triangle
$(0,0)$, $(a,0)$, $(0,1)$, for $a\ge 1$ $($see the case of $a=5$
on Figure~\ref{beta3}b$)$$)$.
\begin{figure}
$$\epsfbox{theorem1.3}$$
\caption{Integer-affine types of faces of List ``$\beta_r$'', for
$r \ge 4$.}\label{betar}
\end{figure}
{\bf List ``$\beta_r$'', ($r\ge 3$):}
\nopagebreak
--- triangle with vertices $(0,0)$, $(a,0)$, and $(0,1)$, for some $a\ge 1$
$($see the case of $a=6$ on Figure~\ref{betar}$)$, the
corresponding convex marked pyramid is integer-affine equivalent
to~$T_{a,r}^\xi$, where the integers $\xi$ and $r$ are relatively
prime and satisfy $0<\xi\le r/2$. For different $\xi$ the
corresponding faces are integer-linear nonequivalent but
integer-affine equivalent.
For any integer $r$ the faces of List $\beta_r$ are
integer-affine nonequivalent to each other; List $\beta_r$ is
irredundant. \qed }
\vspace{1mm}
The integer-affine and the integer-linear classifications
coincide, for $r<5$. For $r\ge 5$, the integer-linear
classification contains the integer-affine classification.
For any integers $n\ge 3$ and $r\ge 2$, the integer-linear
classification of compact two-dimensional faces contained in
planes at integer distances from the origin greater than one of
sails of $n$-dimensional continued fractions coincides with the
integer-affine classification of completely empty $r$-story
three-dimensional convex marked pyramids.
\section{Proof of Theorem~A}\label{pA}
\subsection{Preliminary definitions and statements}
Before proving the main theorem, we give several definitions and fix the notation,
and also formulate some general statements
that we will further use in the proof of the main statements.
For an integer polygon in some two-dimensional subspace
the ratio of its Euclidean volume
to the minimal possible Euclidean volume of an integer triangle in the same
two-dimensional subspace is called the{\it integer volume} of this polygon.
\begin{remark}
Our integer volume is a positive integer
(for a parallelogram, the usual volume will be two times less).
The integer volume of a triangle is equal to the index of the lattice
generated by its edges.
\end{remark}
An integer polyhedron (polygon) is called {\it empty},
if it does not contain integer points in its interior, and the set of integer points
of the faces coincides with the set of vertices of the polyhedron (polygon).
Let $ABCD$ be a tetrahedron with an ordered set of vertices $A$, $B$, $C$
and $D$. Denote by $P(ABCD)$ the following parallelepiped:
$$
\{A+ \alpha \bar{AB}+ \beta \bar{AC} +\gamma \bar{AD}|
\makebox[.2cm]{} 0\le\alpha\le1, 0\le\beta\le1, 0\le\gamma\le1\}.
$$
\begin{definition}
Now we specify some useful coordinates in the three-dimensional subspace
containing $P(ABCD)$ of $\r ^n$.
Let $b$, $c$, and $d$ be the distances from $B$, $C$, and $D$
to the two-dimensional planes containing the faces $ACD$, $ABD$, and $ACD$
respectively.
Let us define the coordinates of $A$, $B$, $C$, and $D$
as follows: $(0,0,0)$, $(b,0,0)$, $(0,c,0)$, and $(0,0,d)$ respectively.
The coordinates of all other points in this three-dimensional subspace are
uniquely defined by means of linearity. We call them the {\it integer-distance
coordinates} with respect to $P(ABCD)$.
\end{definition}
\begin{remark}
For any set of vertices $A$, $B$, $C$, and $D$ with the order as in $P(ABCD)$,
the integer-distance coordinates are uniquely defined.
\end{remark}
Using integer-distance coordinates by {\it integer lattice nodes
of $\r ^n$} (or, for short, {\it lattice nodes})
we mean integer points in the original coordinates in $\r ^n$.
\begin{remark}
Note that any lattice node of the three-dimensional space described above
has integer coordinates in the new integer-distance system of coordinates.
The inverse is not true. There exists an integer-distance system of coordinates
and such a point in the corresponding three-dimensional space
with integer coordinates which is not a lattice node.
For lattice nodes, the absolute values of their new coordinates coincide with the
integer distances from these lattice nodes to the planes containing the corresponding faces
of the parallelepiped.
\end{remark}
Let us continue with the following definition.
\begin{definition}
Two points $P$ and $Q$ are said to be
{\it equivalent with respect to some integer parallelogram} $ABCD$,
if there exist such integers $\lambda$ and $\beta$
that $P=Q+\lambda \bar{AB}+\beta \bar{AC}$.
The set of all equivalence classes of the integer lattice
with respect to the integer parallelogram $ABCD$
is called the {\it quotient-lattice} of the space by this integer parallelogram.
\end{definition}
Note that any equivalence class is contained in one of two-dimen\-sio\-nal planes
parallel to the plane of the parallelogram.
\begin{proposition}\label{m_st}
Consider an integer parallelepiped $ABCDA'B'C'D'$ in $\r^3$
and some integer plane $\pi$ parallel to the face $ABCD$.
Let $\pi$ intersect the parallelepiped $($along a parallelogram$)$.
Then the following two statements hold.
First, $\pi$ contains only finitely many equivalence classes of the integer lattice
with respect to the integer parallelogram $ABCD$.
Their number is equivalent to the index of the sublattice generated by the edges
$\bar {AB}$ and $\bar{AC}$ in the integer lattice of the plane containing
$GBCD$.
Secondly, for any yquivalence class of the integer lattice contained in $\pi$
with respect to the integer parallelogram $ABCD$ exactly one of the following conditions holds.
\\
a$)$ only one point of the equivalence class is in the parallelogram,
it is an inner point of the parallelogram;
\\
b$)$ two points of the equivalence class are in the parallelogram,
they are contained in $($open$)$ opposite edges of the parallelogram;
\\
c$)$ four points of the equivalence class are in the
parallelogram, they coincide with vertices of the parallelogram.
\end{proposition}
We skip the proof of Proposition~\ref{m_st}.
It is straightforward and is based on the following easy lemma.
\begin{lemma}\label{l1}
Consider an integer parallelepiped with an empty face.
Let some parallel to this face plane intersect the parallelepiped $($along a parallelogram$)$.
Then exactly one the following statements holds.
\\
a$)$ only one integer point is in the parallelogram,
it is an inner point of the parallelogram;
\\
b$)$ two integer points are in the parallelogram,
they are contained in $($open$)$ opposite edges of the parallelogram;
\\
c$)$ four integer points are in the parallelogram, they coincide
with vertices of the parallelogram. \qed
\end{lemma}
\subsection{First results on empty integer tetrahedra}
In this subsection we present the corollary of White's
theorem~\cite{Whi} (see also~\cite{Ger}). Here without lose of
generality we consider only the three-dimensional space.
\begin{theorem}{\bf (G.~K.~White, 1964~\cite{Whi}.)}
Let $\Delta \subset \r ^3$ be an integer three-dimensional
simplex, let $E_i=\{ \sigma_i, \sigma_i '\}$, $i=1,2,3$ be a set
of points of two opposite edges $ \sigma_i, \sigma_i '$ for
$\Delta$. Then $(\Delta \setminus E_i)\cap \z ^3$ is empty iff
there exist a $j$ and two neighboring planes $\pi_j$, $\pi_j '$
$($by neighbor we mean that there is no integer points
``between'' these planes $\pi_j$ and $\pi_j '$$)$, such that
$\sigma_i \subset \pi_j$ and $\sigma_i ' \subset \pi_j '$. \qed
\end{theorem}
We will use the following corollary on empty integer tetrahedra
for the classification of empty convex multistory tetrahedra and
also further in the proof of Theorem~A.
\begin{corollary}\label{symplex}
Let $ADBA'$ be some empty integer tetrahedron.
Then all integer points of the parallelepiped $P(ADBA')$
are in the plane passing trough
two centrally-symmetric edges of the parallelepiped.
This two edges are not the edges of the tetrahedron $ADBA'$.
\qed
\end{corollary}
\begin{remark}
The number of planes passing through two centrally-symmetric edges of the parallelepiped
equals six, but only three of them do not contain the edges of the tetrahedron.
\end{remark}
For the proofs see~\cite{Whi}.
\subsubsection{Classification of empty triangular marked
pyra\-mids}
Corollary~\ref{symplex} allows to describe all integer-affine
types of empty triangular marked pyramids (i.e. tetrahedra with
one marked vertex each).
Let $r$ be some positive integer, and $\xi$ be nonnegative integer.
Denote by $P_{r}^{\xi}$ the marked pyramid with vertex at $(0,0,0)$
and the triangular base $(0,1,0)$, $(1,0,0)$, $(\xi,r-\xi,r)$.
\begin{corollary}\label{col_pyr}
Any integer empty triangular marked pyramid is inte\-ger-affine equivalent to exactly
one of the pyramids of the
{\bf List ``P'':}
--- $P_1^0$;
--- $P_r^\xi$, where $\xi$ and $r$ are relatively prime, and $r{\ge} 2$, and $0{<}\xi{\le} r/2$.
All triangular marked pyramids of List~``P'' are empty and
integer-affine nonequivalent to each other.
\end{corollary}
\begin{proof}
{\bf 1. Completeness of List ``P''.}
Let us show that an arbitrary empty integer marked pyramid $ADBA'$
(with a vertex $A$) is integer-affine equivalent to one of the marked pyramids of ``P''.
Suppose that, the integer distance from its marked vertex to the plane containing the marked base
equals some positive integer $r$.
If $r=1$, then the vertices of the marked pyramid generate the three-dimensional
integer lattice, and therefore such a marked pyramid is integer-affine equivalent to $P_1^0$
(here $A$ corresponds to the marked vertex of $P_1^0$).
Suppose now that $r>1$. By Corollary~\ref{symplex},
all integer points of the parallelepiped $P(ADBA')$ are contained
exactly in one of the three planes passing through centrally-symmetric
edges of the parallelepiped and not containing the edges of the tetrahedron $ADBA'$.
Denote the vertices of the marked base $DBA'$ by $\bar B$,
$\bar D$, and $\bar A'$ in a such way that all inner integer points of the parallelepiped
$P(A \bar D \bar B \bar A')$ are contained in the plane passing
through $\bar B\bar D$ and the centrally-symmetric edge.
Consider the integer-distance coordinates with respect to the parallelepiped
$P(A \bar D \bar B \bar A')$.
Take the intersection of the parallelepiped with the plane $z=1$ in these coordinates.
There is only one lattice node in the intersection, by Corollary~\ref{symplex}
its coordinates are $(r-\xi,\xi,1)$.
Denote this lattice node by $K$.
If the integers $\xi$ and $r$ have some common integer divisor $c\ge 1$,
then the point with the coordinates $(\frac{r-\xi}{c}r,\frac{\xi}{c}r,c)$
is a lattice node.
Hence the point $(0,0,c)$ is also a lattice node.
And then the marked pyramid $A \bar D \bar B \bar A'$ is not empty.
Thus the integers $\xi$ and $r$ are relatively prime.
Since the integer distance from $K$ to the two-dimensional plane
containing the face $A \bar D \bar B$
equals one,
there exists an integer-affine transformation taking the tetrahedron $A\bar B\bar D K$
to the tetrahedron with vertices $(0,0,0)$,
$(0,1,0)$, $(1,0,0)$, and $(1,1,1)$.
Here the point $\bar A'$ maps to $(\xi,r-\xi,r)$.
Hence the integer-affine type of the marked pyramid $ABDA'$
coincides with the integer-affine type of the marked pyramid $A \bar B \bar D \bar A'$,
and it in turn coincides with the integer-affine type of the marked pyramid
$P_r^\xi$, where $0<\xi< r$, and $\xi$ and $r$ are relatively prime.
It remains to say that the marked pyramids $P_r^\xi$ and $P_r^{r-\xi}$
can be mapped one to another by the symmetry about the plane $x=y$
(which preserves the integer lattice).
Therefore the marked pyramids $P_r^\xi$ and $P_r^{r-\xi}$
are integer-affine equivalent.
\medskip
{\bf 2. Emptiness of the marked pyramids of List ``P''.}
Let us show that all listed marked pyramids $P_\xi^r$ are empty.
The intersection of the plane $z=b$ (for $1\le b \le (r-1)$) and marked pyramid $P_\xi^r$
is the triangle $A_kB_kD_k$ with the following coordinates of the vertices:
$$
\left( \frac{b}{r}\xi,\frac{b}{r}(r{-}\xi),b \right), \quad
\left( \frac{b}{r},\frac{b}{r}(r{-}\xi){+}\frac{r{-}b}{r},b
\right), \quad \left(
\frac{b}{r}\xi{+}\frac{r{-}b}{r},\frac{b}{r}(r{-}a),b \right).
$$
The triangle $A_kB_kD_k$ is contained in the band $b\le x+y \le b+\frac{r-b}{r}$, $z=b$.
This band contains only integer points with coordinates $(t,b-t,b)$ for integer $t$.
Hence it remains to check if $A_k$ is integer.
Since $\xi$ and $r$ are relatively prime and $d<r$, the first coordinate of
$A_k$ is not integer.
Therefore all marked pyramids $P_\xi^r$ of List ``P'' are empty.
\medskip
{\bf 3. Irredundance of List ``P''.}
We will show now that all marked pyramids $P_\xi^r$ of List ``P'' are integer-affine
nonequivalent to each other.
Note that the integer distance from the marked vertex to the plane containing the base
is an integer-affine invariant. Therefore the pyramids with nonequivalent parameter $r$
are integer-affine nonequivalent.
To distinguish the marked pyramids with the same $r$,
we construct the following integer-affine invariant.
Consider an arbitrary empty mar\-ked pyramid $ABDA'$ with a marked vertex $A$
and the corresponding trihedral angle also with vertex $A$ and triangle
$DBA'$ as its base.
By White's theorem, exactly one lattice node of the trihedral angle (we denote this lattice node by $K$)
is contained in the two-dimensional plane parallel to the face $DBA'$
and at the integer distance $r+1$ from~$A$.
By Corollary~\ref{symplex}, the integer distances from $K$
to two-dimensional planes of the angle are equal to $1$, $\xi$, $r-\xi$
(for some integer $\xi$).
The trihedral angle and $K$ are uniquely defined by the marked pyramid
up to the symmetries of the marked pyramid preserving the marked vertex.
The group of such symmetries permutes all integer distances
from $K$ to the planes containing the faces of the angle.
Hence, the unordered system of integers $[1, \xi, r-\xi]$
is an invariant.
This invariant distinguishes all marked pyramids $P_\xi^r$
with the same integer distance $r$.
\end{proof}
\begin{proposition}
Let relatively prime integers $\xi$ and $r$ satisfy the following
inequalities: $r\ge 2$, $0<\xi\le r/2$.
Then the marked pyramid $P_r^{\xi}$ is integer-affine equivalent
to the marked pyramid $T_{1,r}^{\xi}$.
\end{proposition}
\begin{proof}
The marked pyramid $T_{1,r}^{\xi}$ is the image of $P_r^{\xi}$ under
the integer-linear transformation
$$
\left(
\begin{array}{ccc}
\xi +1 & \xi & -\xi \\
r-1 & r-1 & 2-r \\
-r & -r & r-1 \\
\end{array}
\right)
.
$$
\end{proof}
\begin{corollary}\label{col_pyr2}
Any integer empty $r$-story $($$r\ge 2$$)$ triangular marked pyramid
is integer-affine equivalent exactly to one of the marked pyramids
$T_{1,r}^\xi$ for relatively prime integers $\xi$ and $r$ satisfying $0<\xi\le r/2$.
All such pyramids $T_{1,r}^\xi$ are empty
$($and integer-affine nonequivalent if the corresponding parameters $r$ and $\xi$
do not coincide$)$.
\qed
\end{corollary}
\subsubsection{Classification of integer empty tetrahedra}
A certain difference between the integer-affine classification of
integer empty triangular marked pyramids (with marked vertex) and
the integer-affine classification of integer empty tetrahedra
(without mar\-ked vertices) occurs. The first steps in the
integer-affine classifications of integer empty tetrahedra were
made by J.-O.~Moussafir in~\cite{Mou2}.
\begin{theorem}{\bf (J.-O.~Moussafir~\cite{Mou2}.)}
Any integer empty tetrahedron is integer-affine equivalent
to the tetrahedron with vertices $(0,0,0)$,
$(1,0,0)$, $(0,1,0)$, and $(u,v,d)$, for some integers $u$, $v$ and $d$,
where $u$, $v$ and $u+v-1$ are relatively prime with $d$,
and one of the integers $u+v$, $u-1$, $v-1$ is divisible by $d$.
$($These tetrahedra are sometimes called
{\it Hermitian normal forms of the simplexes}.$)$
\end{theorem}
Note that many of such Hermitian normal forms are integer-affine equivalent
to each other.
The following consequence of Corollary~\ref{symplex} improves
Moussafir's theorem.
\begin{corollary}\label{col_symp}
Any integer empty tetrahedron is in\-te\-ger-affine equivalent
exactly to one of the following tetrahedra:
--- $P_1^0$;
--- $P_r^\xi$, where $r \ge 2$, $0<\xi< r$,
and the element $(\xi \mod r)$ of the additive group $\z/m\z$
is also contained in the associated multiplicative group $(\z/m\z)^*$
$($i.e. integers $\xi$ and $r$ are relatively prime$)$.
All listed integer tetrahedra are empty.
Two tetrahedra $P_{r_1}^\xi$ and $P_{r_2}^{\nu}$ are integer-affine
equivalent iff $r_1=r_2$ and $($for $r_1>1$$)$ one of the following
equalities in $(\z/m\z)^*$ holds:
$$
(\xi \mod r_1)=(\pm 1)\cdot (\nu \mod r_1)^{\pm 1}.
$$
\end{corollary}
\begin{proof}
{\bf 1. Completeness of the list.}
By Corollary~\ref{col_pyr}, any empty integer tetrahedron is
integer-affine equivalent to some tetrahedron of the list of
Corollary~\ref{col_symp}.
\medskip
{\bf 2. Emptiness of the tetrahedra of the list.}
By Corollary~\ref{col_pyr}, the tetrahedron $P_\xi^{r}$
is empty for relatively prime integers $r$ and $\xi$ satisfying $r \ge 2$ and $\xi\le r/2$.
Since $P_\xi^{r}$ and $P_\xi^{r-\xi}$ are integer-affine equivalent
and $P_1^0$ is empty, all tetrahedra of the list of Corollary~\ref{col_symp} are empty.
\medskip
{\bf 3. Proof of the last statement of Corollary~\ref{col_symp}.}
The (affine) symmetry group of the right tetrahedron ($S_4$) includes
the (affine) symmetry group of the right tetrahedron with marked vertex ($S_3$).
Now the list of the four trihedral angles associated with all four vertices
of the tetrahedron is uniquely defined.
We chose one lattice node for each of these angles as we did in the proof of the previous corollary.
Direct calculations show that the integer distances from these points to the
four two-dimensional planes containing the faces of the tetrahedron are
$$
(1,1,\xi, r-\xi), \quad (1,1,\xi,r-\xi), \quad (\nu,r-\nu,1,1),
\quad \mbox{and} \quad (\nu,r-\nu,1,1),
$$
where $(\xi \mod r)\cdot (\nu \mod r)=1$ in $(\z/m\z)^*$. The
set of these numbers up to the group $S^4$ of permutations action
(for all points at the same time) is an integer-affine invariant.
Therefore, the tetrahedra $P_{r_1}^\xi$, $P_{r_2}^\nu$,
$P_r^{r-\xi}$, and $P_r^{r-\nu}$ are integer-affine equivalent
and the invariant distinguishes all other tetrahedra.
\end{proof}
\begin{remark} The integer-affine classifications of integer empty
triangular marked pyramids and of integer empty tetrahedra are
coincide only for $r=1,2,3,4,5,6,8,10,12,24$.
\end{remark}
\subsection{Proof of Theorem~A for the case of polygonal marked
pyramids}
In this subsection we study the case of marked pyramids with
polygonal bases (containing more than three angles distinct from
the straight angle). In the next subsection we will study
triangular marked pyramids.
\subsubsection{Integer parallelograms contained in some integer
polyhedron}
First of all we show that the integer convex polygons under
consideration contain an integer parallelogram.
\begin{proposition}\label{st1}
Let some four vertices of a convex polygon be integer points.
Then this polygon contains some integer parallelogram that is
integer-affine equivalent either to the standard unit
parallelogram shown on Figure~\ref{2xPar}a$)$, or to the
parallelogram with vertices $(1,0)$, $(0,1)$, $(-1,0)$, and
$(0,-1)$ shown on Figure~\ref{2xPar}b$)$:
\begin{figure}[h]
$$\epsfbox{mnogoug.1}$$
\caption{Any integer polygon contains an integer parallelogram that is integer-affine
equivalent to one of this two parallelograms.}\label{2xPar}
\end{figure}
\end{proposition}
\begin{proof}
Suppose that an integer polygon contains four integer vertices.
Consider the quadrangle generated by these four vertices and denote it by $KLMN$.
Let us prove that the quadrangle contains some integer parallelogram.
Consider the parallelogram $P(KLN)$ and denote it by $KLM'N$.
The vertex $M$ can be in any of the four octants with respect to the lines
containing $M'N$ and $M'L$.
For any of these four cases, we explicitly
construct an integer parallelogram contained in the quadrangle on Figure~\ref{4-gons}
(we draw the quadrangle $KLMN$ with thick line, the corresponding parallelogram is
shaded).
\begin{figure}
$$\epsfbox{other.2}$$
\caption{The possible positions of the convex quadrangle $KLMN$
with respect to $P(KNL)$ (i.e. the quadrangle
$KLM'N$).}\label{4-gons}
\end{figure}
Further we use the following statement.
Let some point of an integer parallelogram be integer.
Consider the point which is centrally-symmetric about the intersection point
of the diagonals of this parallelogram. This point is also
in the parallelogram and is integer.
Denote the integer parallelogram in the polygon by $ABCD$.
{\bf 1. Integer empty parallelogram.} Suppose $ABCD$ is empty.
Then it generates the whole integer lattice and hence is integer-affine
equivalent to the standard one.
{\bf 2. Integer parallelogram with the only one integer point inside.}
Suppose $ABCD$ contains only one integer point $O$ in its interior.
Then this point coincides with the centrally-symmetric point
about the intersection point of the diagonals of this parallelogram.
And hence it coincides with the intersection point of the diagonals.
Therefore the integer triangle $OAB$ is empty. Hence it is
integer-affine equivalent to the standard unit triangle.
Now the integer-affine type of $ABCD$ is uniquely defined and is just
the integer-affine type of the parallelogram with vertices
$(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
{\bf 3. Remaining cases.} Let the parallelogram $ABCD$ contain
more than one integer point except the vertices.
Then there exists a points among these points such that it is different
from the intersection point of the diagonals of this parallelogram.
We denote it by$~O$.
Denote the centrally-symmetric point about the intersection point
of the diagonals of this parallelogram by~$O'$.
Without loss of generality, we suppose that~$OO'$ is not a subset of $AC$
(otherwise $OO'$ is not a subset of~$BD$).
Therefore $AOCO'$ (or $AO'CO$) is an integer parallelogram contained in $ABCD$.
The number of integer points of $AOCO'$ is smaller than
the number of integer points of $ABCD$ at least by two.
Since the initial parallelogram contains only a finite number of integer points,
we iteratively come to one of the cases of item {\bf 1.} or {\bf 2.}
Therefore any convex polygon with four integer vertices contains a parallelogram
integer-affine equivalent to one of the parallelograms of Proposition~\ref{st1}.
\end{proof}
\subsubsection{The case of an empty marked pyramid with empty
parallelogram as base}
\begin{proposition}\label{lt1}
Let an empty integer parallelogram be a base of some marked pyramid.
If this pyramid is empty, then it is single-story.
\end{proposition}
\begin{proof}
We prove this proposition by contradiction.
Let $A'ABCD$ be an empty marked pyramid with marked vertex $A'$ and an empty
parallelogram $ABCD$ as its base.
Suppose that the integer distance from the point $A'$ to the plane containing
$ABCD$ equals $r>1$.
Consider the parallelepiped $P(AA'BC)$ and the
integer-distance coordinates corresponding to it
(we denote such coordinates in the following way: $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $A'$, $B$, and $C$
equal to $(r,0,0)$, $(0,r,0)$, and $(0,0,r)$ respectively.
Note that coordinates of lattice nodes (of the integer lattice in the old coordinates)
are integers.
Let us find the lattice node of the parallelepiped
at the unit integer distance to the plane containing $ABC$,
i.e. the lattice node with coordinates $(1,y,z)$, where $0\le y \le r$,
$0 \le z \le r$.
On one hand, it does not contain in the marked pyramid $A'ABCD$,
and hence $y+1>r$ or $z+1>r$.
On the other hand, by Corollary~\ref{l1}, the two-dimensional faces of $P(AA'BC)$
do not contain integer points different from vertices since $AA'BC$ is empty.
Therefore $y$ and $z$ are not equal to $r$.
Hence there are no lattice nodes in the plane containing $ABC$.
We come to the contradiction with Lemma~\ref{l1}.
\end{proof}
\subsubsection{The case of a completely empty marked pyramid whose
base is an integer parallelogram containing a unique integer
point in its interior}
\begin{lemma}\label{mn1}
Consider an integer marked pyramid with vertex $O$ and parallelogram
$ABCD$ as base. Let $ABCD$ be integer-affine equivalent to the parallelogram with
vertices $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$ $($see Figure~\ref{mnogoug.3}$)$.
If the marked pyramid $OABCD$ is completely empty and multistory, then
it is two-story. The integer-affine type of such pyramid coincides
with the integer-affine type of the pyramid with vertex $(0,0,0)$
and base $(2,-1,0)$, $(2,-2,1)$, $(2,-1,2)$, $(2,0,1)$.
\begin{figure}[h]
$$\epsfbox{mnogoug.3}$$
\caption{Quadrangle with vertices $(1,0)$, $(0,1)$, $(-1,0)$, and
$(0,-1)$.}\label{mnogoug.3}
\end{figure}
\end{lemma}
\begin{proof}
Let the integer base $ABCD$ of the completely empty $r$-story
integer marked pyramid $OABCD$ ($r\ge 2$) be integer-affine equivalent to the
parallelogram with vertices $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
Consider the parallelepiped $P(AOBC)$ and the
integer-distance coordinates corresponding to it
(we denote these coordinates as $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $O$, $B$, $C$, and $D$
equal $(r,0,0)$, $(0,2r,0)$, $(0,0,2r)$, and $(0,2r,2r)$ respectively.
Let us consider the parallelogram of intersection of $P(AOBC)$
with
the plane $x=1$.
Now we will find all lattice nodes in this parallelogram.
By Proposition~\ref{m_st}, there are exactly two lattice nodes in the parallelogram
of intersection.
Let us describe all possible positions of these nodes in the intersection of
$P(AOBC)$ and the plane $x=1$.
First, there are no lattice nodes in the intersection of
the marked pyramid $AOBCD$ and the plane $x=1$,
i.e. in the closed parallelogram with vertices
$(1,0,0)$, $(1,0,2r-2)$, $(1,2r-2, 2r-2)$, and $(1,2r-2,0)$.
Secondly, there are no lattice nodes in all parallelograms
obtained from the given one by applying translations by the vectors
$\lambda (0,2r,0) + \mu (0,r,r)$, where $\lambda$
and $\mu$ are integers.
In Figure~\ref{lemmas.10}, we show some parallelograms
that do not contain any lattice nodes.
These parallelograms are painted shaded.
So, the lattice nodes of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ can be only in integer points of
open parallelograms obtained from the parallelogram with vertices
$K(1,r-2,2r-2)$, $L(1,r,2r-2)$, $M(1,r,2r)$, and $N(1,r-2,2r)$
by the symmetry with respect to the plane $y=z$ and translations by the vectors
$\lambda (0,2r,0) + \mu (0,r,r)$, where $\lambda$ and $\mu$ are some integers.
The parallelogram $KLMN$ contains exactly one integer point
$(1,r-1,2r-1)$, see Figure~\ref{lemmas.10}.
\begin{figure}[ht]
$$\epsfbox{lemmas.10}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ and the plane $x=1$.}\label{lemmas.10}
\end{figure}
Suppose that this point is a lattice node.
Since the intersection parallelogram contains exactly two lattice nodes,
the point symmetric to the point $(1,r-1,2r-1)$ with respect to the plane $y=z$ is also a lattice node
(there are no other integer points in the intersection parallelogram).
Therefore $(2,2r-2, 4r-2)$ is a lattice node.
Hence $(2,2r-2,2r-2)$ is a lattice node,
and hence $(2,r-2,r-2)$ is also a lattice node.
However, for $r\ge 3$ the point $(2,r-2,r-2)$ is contained in the closed parallelogram
of intersection of $P(AOBC)$ with the plane $x=2$.
The vertices of this parallelogram are the following:
$(2,0,0)$, $(1,0,2r-4)$, $(1,2r-4, 2r-4)$, and $(1,2r-4,0)$.
Thus there are no pyramids satisfying all the conditions of Lemma~\ref{mn1} for $r\ge 3$.
Now consider the case $r=2$.
The integer points $A$, $B$, $C$, and $(1,1,3)$ define the integer lattice in a unique way.
This implies that all marked pyramids satisfying all the conditions of Lemma~\ref{mn1}
are of the same integer-affine type, and it coincides
with the integer-affine type of the marked pyramid with vertex $(0,0,0)$
and base $(2,-1,0)$, $(2,-2,1)$,
$(2,-1,2)$, $(2,0,1)$ (in the old coordinates).
\end{proof}
\subsubsection{General case}
Now we study the general case of integer completely empty marked
pyramids with convex polygonal bases.
\begin{lemma}\label{mn2}
Consider an integer marked pyramid with vertex $O$ and convex polygonal base $M$.
If this marked pyramid is completely empty and multistory, then
it is two-story. The base of the marked pyramid is integer-affine
equivalent to the quadrangle $(b,0)$, $(0,1)$, $(-a,0)$, $(0,-1)$ where $b\ge a \ge 1$
$($see the case $a=2$, $b=3$ on Figure~\ref{mnogoug.4}$)$.
\begin{figure}[ht]
$$\epsfbox{mnogoug.4}$$
\caption{Quadrangle with vertices $(b,0)$, $(0,1)$, $(-a,0)$, and
$(0,-1)$ where $b\ge a \ge 1$.}\label{mnogoug.4}
\end{figure}
The integer-affine type of the pyramid is uniquely determined
by the integers $a$ and $b$ $($for $b \ge a \ge 1$$)$
and coincides with the integer-affine type of the marked pyramid $M_{a,b}$.
Two marked pyramids $M_{a,b}$ and $M_{a',b'}$ $($$b \ge a \ge 1$, $b' \ge a' \ge 1$$)$
are integer-affine equivalent iff $a=a'$ and $b=b'$.
\end{lemma}
\begin{proof}
Under the assumptions of the lemma, the integer distance from the two-dimensional plane
containing the parallelogram $M$ to the vertex $O$ is greater than one.
From Proposition~\ref{st1} it follows that the parallelogram $M$ contains either
some empty parallelogram or a parallelogram with exactly one integer point in its interior
(and different from the vertices).
By Proposition~\ref{lt1}, the case of an empty parallelogram is eliminated.
Consider an empty parallelogram~$P$ or a parallelogram with exactly one integer point inside.
Choose coordinates on the plane containing the base $M$ so that
the vertices of $P$ have the following coordinates:
$(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
Note that all the coordinates of a point of this plane are integers iff
this point is integer (with respect to the old system of coordinates).
Let an integer point with coordinates $(x,y)$ for some $x,y > 0$ be in the base $M$.
Since $M$ is convex, the point $(1,1)$ is also in $M$.
This implies that the empty integer parallelogram with vertices
$(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ is contained in $M$.
Therefore by, Proposition~\ref{lt1}, the distance from the vertex of the pyramid
to the two-dimensional plane containing
the polygon $M$ equals one.
The cases $x>0$, $y<0$; $x<0$, $y>0$; and $x,y<0$ are similar.
Let the integer points with coordinates $(x,0)$ and $(0,y)$, where $|x|>1$ and $|y|>1$,
be in the base $M$.
Then $M$ contains one of the points: $(1,1)$, $(1,-1)$, $(-1,1)$, or $(-1,-1)$.
And for the same reason, the distance from the vertex of the pyramid
to the two-dimensional plane containing $M$ equals one.
Without loss of generality we suppose that $M$ does not contain points with coordinates
$(0,y)$ for $|y|>1$.
Then $M$ is integer-affine equivalent to the quadrangle with vertices
$(b,0)$, $(0,1)$, $(-a,0)$, $(0,-1)$, where $b\ge a \ge 1$.
Since the polygon $M$ contains the parallelogram $P$,
by Lemma~\ref{mn1} the integer distance from the vertex $O$ of the marked pyramid
to the two-dimensional plane containing the base $M$ equals two.
The parallelogram $P$ is uniquely defined by the quadrangle with vertices
$(b,0)$, $(0,1)$, $(-a,0)$, $(0,-1)$, where $b\ge a \ge 1$
(such a quadrangle contains the unique integer parallelogram
with exactly one integer point different to the vertices).
Therefore, by Lemma~\ref{mn1}, the marked pyramid is integer-affine equivalent
to the marked pyramid with vertex $(0,0,0)$ and base
$(2,-1,0)$, $(2,-a-1,1)$, $(2,-1,2)$, $(2,b-1,1)$.
The point of intersections of the diagonals of the base
quadrangle divides the diagonals into four segments with integer
lengths $1$, $1$, $a$ and $b$. Therefore the (unordered) pair of
integers $[a,b]$ is an integer-affine invariant of the marked
pyramids.
\end{proof}
\subsection{Proof of Theorem~A for the case of triangular marked pyramids}
We continue the proof by exhausting some special cases.
Throughout this subsection we denote by $OABC$ a triangular marked
pyramid with vertex $O$ and base $ABC$.
\subsubsection{Case 1: the base contains an integer polygon}
Suppose that the triangle $ABC$ contains such two integer points
$D$ and $E$, that the line $DE$ intersects the edges of the
triangle $ABC$ and does not contain any vertex of the triangle.
Without loss of generality we suppose that the open ray $DE$ with
vertex at $D$ intersects $AB$, and the open ray $ED$ with vertex
at $E$ intersects $BC$. Hence the triangle $ABC$ contains some
integer convex quadrangle $AEDC$. By Proposition~\ref{st1}, the
triangle $ABC$ contains either an integer empty parallelogram or
a parallelogram integer-affine equivalent to the parallelogram
with vertices $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
If the triangle $ABC$ contains an integer empty parallelogram, then by Proposition~\ref{lt1}
the marked pyramid $OABC$ is single-story.
Suppose that the triangle $ABC$ does not contain an integer empty parallelogram
and contains a parallelogram integer-affine equivalent to the parallelogram with vertices
$(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
Consider coordinates on the plane containing the base such that
the vertices of the above-mentioned parallelogram have the following coordinates:
$(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
If the points $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$ are not contained
in $ABC$, then the marked pyramid is no longer triangular.
Therefore any marked pyramid of Case~1 contains some empty parallelogram,
and, by Proposition~\ref{lt1}, is single-story.
\subsubsection{Case 2: the integer points of the base different
from the vertices are not contained in one line}
Now suppose, that there are two integer points $G$ and $H$ such
that the line $GH$ intersects the edges of the triangle $ABC$ and
does not contain any vertex of the triangle. Here we consider the
case of integer points of the base different from the vertices
and not contained in one line. The only possible affine type is
shown on Figure~\ref{theorem1.4}.
\begin{figure}[h]
$$\epsfbox{theorem1.4}$$
\caption{The affine type of triangles of Case~2.}\label{theorem1.4}
\end{figure}
Let us find all possible integer-affine types of such a triangle.
Since the triangle $FED$ (see Fig.~\ref{theorem1.4}) is empty,
it is integer-affine equivalent to the triangle $(1,0)$, $(0,0)$, and $(0,1)$.
The points $A$, $B$, and $C$ correspond to $(-1,0)$, $(2,1)$, and
$(0,-2)$ respectively.
Hence the integer-affine type is determined in a unique way.
\begin{lemma}
Consider an integer multistory marked pyramid with vertex $O$ and triangular base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(-2,1)$, $(-1,-1)$, and $(1,2)$,
shown on Figure~\ref{theorem1.5}.
Then the marked pyramid~$OABC$ is two-story and integer-affine equivalent to
the marked pyramid $V$ of List ``M-W''.
\begin{figure}[ht]
$$\epsfbox{theorem1.5}$$
\caption{The triangles with the following vertices $(-2,1)$,
$(-1,-1)$, and $(1,2)$.}\label{theorem1.5}
\end{figure}
\end{lemma}
\begin{proof}
Let the base of an $r$-story ($r \ge 2$) completely empty marked pyramid
$OABC$ be integer-affine equivalent to the
triangle with vertices $(-2,1)$, $(-1,-1)$, and $(1,2)$.
Consider the parallelepiped $P(AOBC)$ and the
integer-distance coordinates corresponding to it
(we denote these coordinates by: $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of the vertices $O$, $B$, and $C$
are $(r,0,0)$, $(0,7r,0)$, and $(0,0,7r)$ respectively.
Let us consider the intersection parallelogram of $P(AOBC)$ with
the plane $x=1$.
Now we will find all lattice nodes in this parallelogram. By
Proposition~\ref{m_st}, there are exactly seven lattice nodes in
the parallelogram of intersection. Let us describe all possible
positions of these nodes in the intersection of $P(AOBC)$ with
the plane $x=1$. First, there are no lattice nodes in the
intersection of the marked pyramid $AOBC$ with the plane $x=1$,
i.e. in the closed triangle with vertices $(1,0,0)$,
$(1,0,7r{-}7)$, and $(1,7r{-}7,0)$. Secondly, there are no
lattice nodes in all triangles obtained from the given one by
applying translations by vectors $\lambda (0,r,2r) + \mu
(0,4r,r)$ for all integers $\lambda$ and $\mu$. In
Figure~\ref{lemmas.1}, ($r\ge 4$) and Figure~\ref{lemmas.2}
($r=2,3$) we show some triangles that do not contain any lattice
nodes. These triangles are shaded.
So the lattice nodes of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ can be only at integer points of
open triangles obtained from two triangles by translations by the vectors
$\lambda (0,r,2r) + \mu (0,4r,r)$ for all integers $\lambda$ and $\mu$.
The vertices of the first triangle are
$K(1,3r,4r{-}7)$, $L(1,3r,2r)$, and $M(1,5r{-}7,2r)$.
Here the points $(1,0,0)$ and $L$ should be in different
half-planes with respect to the plane $LM$.
This condition is satisfied only if $2r>4r-7$, i.e. $r< 7/2$.
The vertices of the second triangle are
$P(1,4r-7,3r)$, $Q(1,r,3r)$, and $R(1,r,6r-7)$.
And again the points $(1,0,0)$ and $Q$ should be in different
half-planes with respect to the plane $PR$.
This condition is satisfied only if $(4r-7<r)$, i.e. $r< 7/3$.
So for $r>3$ all points of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ are covered, see Figure~\ref{lemmas.1}.
\begin{figure}[ht]
$$\epsfbox{lemmas.1}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=1$ (for $r>3$).}\label{lemmas.1}
\end{figure}
If $r=2$, then the triangle $KLM$ contains only one integer point
with coordinates $(1,5,3)$, see Figure~\ref{lemmas.2}a). If
$r=3$, then the triangle $KLM$ does not contain any integer
point, see Figure~\ref{lemmas.2}b).
\begin{figure}[ht]
$$
\epsfbox{lemmas.2}
$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with
the plane $x=1$: a) $r=2$; b) $r=3$.}\label{lemmas.2}
\end{figure}
Since the intersection parallelogram of the plane $x=1$ with the open parallelepiped
must contain seven lattice nodes, the only possible case is the case $r=2$.
There are exactly seven integer points in the complement of the union of the described triangles
in the parallelogram. Hence all these points are lattice nodes.
Therefore, the marked pyramid~$OABC$ is two-story and integer-affine equivalent to
the marked pyramid with vertex $(0,0,0)$ and base $(2,-2,1)$, $(2,-1,-1)$, $(2,1,2)$
(i.e. to the pyramid $V$ of List ``M-W'').
\end{proof}
It remains to study the cases of triangular pyramids with the following property.
All integer points of the base of such a pyramid different from the vertices of the pyramid are
contained in some straight line passing through one of the vertices of the base triangle.
\subsubsection{Case 3: all integer points of the base different
from vertices are contained in a straight line --- {\bf I}}
Suppose that all integer points of the triangle $ABC$ are
contained in a ray with vertex at $A$. Let the number of points
be equal to $c$ ($c\ge 1$), and also suppose all these points are
inner. Denote the inner points by $D_1, \ldots, D_c$, starting
from the point closest to $A$ and increasing the indexing in the
direction from $A$. It turns out that for any positive integer
$c$ there exist exactly one integer-affine type of such pyramid.
Since the triangle $BD_cC$ is empty there exists an integer-affine transformation
that maps the triangle to any other empty triangle.
Let us transform the triangle $BD_cC$ to the triangle $\tilde B\tilde D_c\tilde C$
with vertices $(0,1)$, $(0,0)$, and $(1,0)$ respectively.
Now we determine the image of $A$.
Since the point $\tilde D_{c}(0,0)$ is an integer point of the triangle,
the point $\tilde A$ is in the third orthant ($x<0$, $y<0$).
Since $(-1,0)$ is not in the triangle,
the point $\tilde A$ is in the half-plane defined by $y<x+1$.
Since $(0,-1)$ is not in the triangle,
the point $\tilde A$ is in the half-plane defined by $y>x-1$.
Since $\tilde A$ is integer, its coordinates are $(-t,-t)$ for some positive integer $t$.
Since there are exactly $c$ inner integer points in the triangle $\tilde B\tilde D_c\tilde C$,
we obtain $t=c$.
Therefore the triangle $\tilde A \tilde B \tilde C$ is integer-affine equivalent to the triangle
with vertices $(1,0)$, $(0,1)$, and $(-c,-c)$
(the case of $c=4$ is shown on Figure~\ref{theorem1.6}).
\begin{figure}[h]
$$\epsfbox{theorem1.6}$$
\caption{The triangle with vertices $(1,0)$, $(0,1)$, and $(-c,-c)$.}\label{theorem1.6}
\end{figure}
First we study the case $c=1$.
\begin{lemma}\label{l3}
Consider an integer multistory marked pyramid with vertex $O$ and triangular base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(-1,-1)$, $(0,1)$, and $(1,0)$
shown on Figure~\ref{theorem1.7}.
Then the marked pyramid~$OABC$ is three-story and integer-affine equivalent to
the marked pyramid $W$ of List ``M-W''.
\begin{figure}[h]
$$\epsfbox{theorem1.7}$$
\caption{The triangle with vertices $(-1,-1)$, $(0,1)$, and $(1,0)$.}\label{theorem1.7}
\end{figure}
\end{lemma}
\begin{proof}
Suppose that the base of $r$-story ($r {\ge} 2$) completely empty
marked pyramid $OABC$ be integer-affine equivalent to the
triangle with the following vertices $(-1,-1)$, $(0,1)$, and
$(1,0)$.
Consider the parallelepiped $P(AOBC)$ and the
integer-distance coordinates corresponding to it
(we denote such coordinates as $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $O$, $B$, and $C$
equal $(r,0,0)$, $(0,3r,0)$, and $(0,0,3r)$ respectively.
Let us consider the parallelogram at intersection of $P(AOBC)$ and
the plane $x=1$.
Now we will find all lattice nodes in this parallelogram.
By Proposition~\ref{m_st}, there are exactly three lattice nodes in the parallelogram
at intersection.
Let us describe all possible positions of these nodes in the intersection
of $P(AOBC)$ with the plane $x=1$.
First there are no lattice nodes in the intersection of
the marked pyramid $AOBC$ with the plane $x=1$,
i.e. in the closed triangle with vertices
$(1,0,0)$, $(1,0,3r{-}3)$, and $(1,3r{-}3,0)$.
Secondly, there are no lattice nodes in all triangles
obtained from the given one by applying translations by vectors
$\lambda (0,3r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$.
In Figure~\ref{lemmas.3},
we show some triangles that do not contain any lattice nodes.
These triangles are shaded.
So the lattice nodes of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ can be only at integer points of
open triangle obtained from the triangle $K(1,3r,r{-}3)$, $L(1,3r,r)$, $M(1,3r{-}3,r)$
by translations by vectors
$\lambda (0,3r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$.
Only one point with integer coefficients $(1,3r{-}1,r{-}1)$ is in the triangle $KLM$,
see Figure~\ref{lemmas.3}.
\begin{figure}[ht]
$$\epsfbox{lemmas.3}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=1$.}\label{lemmas.3}
\end{figure}
Shaded triangles covers almost all integer points of the
intersection parallelogram of $P(AOBC)$ with the plane $x=1$.
Only two three-tuples of integer points are still uncovered:\\
{\bf 1)} $(1,3r{-}1,r{-}1)$, $(1,r{-}1,2r{-}1)$, $(1,2r{-}1,3r{-}1)$;\\
{\bf 2)} $(1,r{-}1,3r{-}1)$, $(1,2r{-}1,r{-}1)$, $(1,3r{-}1,2r{-}1)$.\\
So the lattice nodes are either the points of the first
three-tuples or the points of the second one.
Suppose $(1,3r{-}1,r{-}1)$ is a lattice node. (If no, then the
point $(1,r{-}1,3r{-}1)$ is a lattice node. Since the
transformation that maps $(x,y,z)$ to $(x,z,y)$ is integer-affine
and it preserves the parallelepiped $P(AOBC)$ and the marked
pyramid $OABC$, this case is similar.) Then the point
$(r,(3r{-}1)r, (r{-}1)r)$ is a lattice node. Hence
$(3r-1)r-(r-1)r$ is divisible by three, and hence $2r^2$ is also
divisible by three. Therefore $r$ is divisible by three.
Suppose $r=3$, then the marked pyramid exists
and is integer-affine equivalent to $W$.
Let us study the case of $r=3k$, for $k\ge 2$.
Consider the parallelogram at intersection of $P(AOBC)$ and
the plane $x=3$.
Now we will find all lattice nodes in this parallelogram.
By Proposition~\ref{m_st}, there are exactly three lattice nodes in the parallelogram
of intersection.
Let us describe all possible positions of these nodes in the intersection of
of $P(AOBC)$ with the plane $x=3$.
First, there are no lattice nodes in the intersection of
the marked pyramid $AOBC$ with the plane $x=3$,
i.e. in the closed triangle with vertices
$(3,0,0)$, $(3,3r-9,0)$, and $(3,3r-9,0)$.
Secondly, there are no lattice nodes in all triangles
obtained from the given one by applying translations by vectors
$\lambda (0,3r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$.
This includes the triangle with vertices $P(3,2r,2r)$,
$Q(3,5r-9,2r)$, and $R(3,2r,5r-9)$ shown on Figure~\ref{lemmas.4}
\begin{figure}[ht]
$$\epsfbox{lemmas.4}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=3$.}\label{lemmas.4}
\end{figure}
Since $(1,3r-1,r-1)$ is a lattice node,
the point $(3,9r-3,3r-3)$ is a lattice node.
Thus $(3,3r-3,3r-3)$ is a lattice node.
However, this point is in $KLM$ (for $r>1$) and hence $(1,3r-1,r-1)$ is not a lattice node.
We come to the contradiction, the case of $r=3k$ for $k\ge 2$ is empty.
\end{proof}
\begin{lemma}
Consider an integer multistory marked pyramid with vertex $O$ and triangular base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(-c,-c)$, $(0,-1)$ , and $(-1,0)$, for $c\ge 2$.
Then the marked pyramid~$OABC$ is not completely empty.
\end{lemma}
\begin{proof}
We prove by reductio ad absurdum.
Let the base of $r$-story ($r \ge 2$) completely empty marked pyramid
$OABC$ be integer-affine equivalent to the
triangle with vertices $(-c,-c)$, $(0,-1)$, and $(-1,0)$, for $c\ge 2$.
Since the triangle with vertices
$(-c,-c)$, $(1,0)$, and $(0,1)$ contains the triangle with vertices
$(-1,-1)$, $(1,0)$, and $(0,1)$, the marked pyramid $OABC$ contains
a marked subpyramid integer-affine equivalent to the pyramid
of Lemma~\ref{l3}.
(By {\it marked subpyramid $P$} of some marked pyramid $Q$ we call such convex
pyramid $P$ that the vertices of $P$ and $Q$ coincides and
the base of $Q$ contains the base of $P$.)
Therefore by Lemma~\ref{l3} we have $r=3$.
Let us show that $r\ne 3$.
Suppose $r=3$.
Since $c\ge 2$, the marked pyramid $OABC$ contains some
marked subpyramid $OA'BC$ with base $A'BC$ integer-affine equivalent to
the triangle with vertices $(-2,-2)$, $(1,0)$, and $(0,1)$.
We show now that $OA'BC$ is not completely empty.
Consider the parallelepiped $P(A'OBC)$ and the
integer-distance coordinates corresponding to it
(we denote such coordinates as $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $O$, $B$, and $C$
equal $(3,0,0)$, $(0,15,0)$, and $(0,0,15)$ respectively.
Let us consider the parallelogram at intersection of $P(A'OBC)$
and the plane $x=1$. Now we will find all lattice nodes in this
parallelogram. First, there are no lattice nodes in the
intersection of the marked pyramid $A'OBC$ with the plane $x=1$,
i.e. in the closed triangle with vertices $(1,0,0)$, $(1,0,12)$,
and $(1,12,0)$. Secondly, there are no lattice nodes in all
triangles obtained from the given one by applying translations by
vectors $\lambda (0,15,0) + \mu (0,3,3)$ for all integers
$\lambda$ and $\mu$. This triangles contains all integer points
of the intersection of $P(A'OBC)$ with the plane $x=1$, see
Figure~\ref{lemmas.5}.
\begin{figure}[h]
$$\epsfbox{lemmas.5}$$
\caption{The intersection of the parallelepiped $P(A'OBC)$ with the plane $x=1$.}\label{lemmas.5}
\end{figure}
So, the marked pyramid $OA'BC$ is not completely empty.
Hence the marked pyramid $OABC$ is not completely empty.
Thus $r\ne 3$.
Therefore, for any $r \ge 2$, the base of any $r$-story completely empty
pyramid $OABC$ is not integer-affine equivalent to the
triangle with vertices $(-c,-c)$, $(0,-1)$, and $(-1,0)$, for $c\ge 2$.
We come to the contradiction.
\end{proof}
\subsubsection{Case 4: all integer points of the base different
from vertices are contained in a straight line --- {\bf II}}
Suppose that all integer points of the triangle $ABC$ are contained in
the ray with vertex $A$. Let the number of points be equal to $b$ ($b\ge 1$),
and the last point is in the edge $BC$.
Denote these points by $D_1, \ldots, D_c$, starting
from the point closest to $A$ and increasing the indexing in the direction from $A$.
It turns out that for any $b$ there exist exactly one integer-affine type of such
pyramid.
Since the triangle $D_b D_{b-1} B$ is empty there exists an integer-affine transformation
that maps the triangle to any other empty triangle.
We transform the triangle $D_b D_{b-1} B$ to the triangle with vertices
$(0,0)$, $(1,0)$, and $(0,-1)$ respectively.
Then $C$ maps to $(0,1)$, and $A$ maps to $(b,0)$.
Therefore the triangle $ABC$ is integer-affine equivalent to the triangle
with vertices $(0,-1)$, $(b,0)$, and $(0,1)$.
(the case of $b=5$ is shown on Figure~\ref{theorem1.8}).
\begin{figure}[ht]
$$\epsfbox{theorem1.8}$$
\caption{The triangle with vertices $(0,-1)$, $(b,0)$, and $(0,1)$.}\label{theorem1.8}
\end{figure}
First we study the case $b=2$.
\begin{lemma}\label{l5}
Consider an integer multistory marked pyramid with vertex $O$ and triangular base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(2,0)$, $(0,-1)$, and $(0,1)$
shown on Figure~\ref{theorem1.9}.
Then the marked pyramid~$OABC$ is two-story and integer-affine equivalent to
the marked pyramid $U_2$ of List ``M-W''.
\begin{figure}[h]
$$\epsfbox{theorem1.9}$$
\caption{The triangle with vertices $(2,0)$, $(0,-1)$, and $(0,1)$.}\label{theorem1.9}
\end{figure}
\end{lemma}
\begin{proof}
Suppose that the base of $r$-story ($r \ge 2$) completely empty
marked pyramid $OABC$ be integer-affine equivalent to the
triangle with vertices $(2,0)$, $(0,-1)$, and $(0,1)$.
Consider the parallelepiped $P(AOBC)$ and the
integer-distance coordinates corresponding to it
(we denote such coordinates as $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $O$, $B$, and $C$
equal $(r,0,0)$, $(0,4r,0)$, and $(0,0,4r)$ respectively.
Now consider the parallelogram at intersection of $P(AOBC)$ and
the plane $x=1$.
Now we will find all lattice nodes in this parallelogram.
By Proposition~\ref{m_st}, there are exactly three lattice nodes in the parallelogram
at intersection.
Let us describe all possible positions of these nodes in the intersection
of $P(AOBC)$ with the plane $x=1$.
First, there are no lattice nodes in the intersection of
the marked pyramid $AOBC$ with the plane $x=1$,
i.e. in the closed triangle with vertices
$(1,0,0)$, $(1,0,4r-4)$, and $(1,4r-4,0)$.
Secondly, there are no lattice nodes in all triangles
obtained from the given one by applying translations vectors
$\lambda (0,4r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$.
We show (shaded) triangles that do not contain any lattice nodes on Figure~\ref{lemmas.7}.
So the lattice nodes of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ can be only at integer points of
open triangle obtained from the triangle $K(1,4r,2r-3)$, $L(1,4r,2r)$, $M(1,4r-3,2r)$
by translations by vectors
$\lambda (0,4r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$
and the symmetry about the plane $y=z$.
Only the points with integer coefficients $(1,4r-2,2r-1)$, $(1,4r-1,2r-1)$, and $(1,4r-1,2r-2)$
are in the triangle $KLM$, see Figure~\ref{lemmas.7}.
\begin{figure}[ht]
$$\epsfbox{lemmas.7}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=1$.}\label{lemmas.7}
\end{figure}
We prove that one of these points is a lattice node by reductio
ad absurdum. Suppose that the triangle $KLM$ does not contain a
lattice node. Then there are no lattice nodes in all triangles
obtained from $KLM$ by applying translations by vectors of the
form $\lambda (0,4r,0) + \mu (0,r,r)$ for all integers $\lambda$
and $\mu$. Hence the intersection of the parallelepiped $P(AOBC)$
with the plane $x=1$ does not contain integer nodes. We come to
the contradiction. So one of the points $(1,4r-2,2r-1)$,
$(1,4r-1,2r-1)$, and $(1,4r-1,2r-2)$ is a lattice node.
Suppose that $r \ge 3$ and consider the plane $x=2$.
First, there are no lattice nodes in the intersection of
the marked pyramid $AOBC$ with the plane $x=2$,
i.e. in the closed triangle with vertices
$(1,0,0)$, $(1,0,4r-8)$, and $(1,4r-8,0)$.
Secondly, there are no lattice nodes in all triangles
obtained from the given one by applying translations by vectors
$\lambda (0,4r,0) + \mu (0,r,r)$ for all integers $\lambda$ and $\mu$.
In particular, there are no lattice nodes in the triangle
with vertices $P(2,3r,3r)$,
$Q(2,7r-8,3r)$, and $R(2,3r,7r-8)$.
\begin{figure}[ht]
$$\epsfbox{lemmas.8}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=2$.}\label{lemmas.8}
\end{figure}
Suppose that the point
$(1,4r-2,2r-1)$, $(1,4r-1,2r-1)$, or $(1,4r-1,2r-2)$ is a lattice node,
then
$(2,8r-4,4r-2)$, $(2,8r-2,4r-2)$, or $(2,8r-2,4r-4)$ respectively is also a lattice node.
Hence the point
$(2,4r-4,4r-2)$, $(2,4r-2,4r-2)$, or $(2,4r-2,4r-4)$ respectively is a lattice node.
The last three points are contained in the triangle $PQR$
with vertices
$P(2,3r,3r)$, $Q(2,7r-8,3r)$, and $R(2,3r,7r-8)$, for $r>3$ (see Figure~\ref{lemmas.8}),
and hence these points are not lattice nodes.
For $r=3$, the point $(1,11,5)$ is not a lattice node by the same reason.
The points $(1,10,5)$ and $(1,11,4)$ are not lattice nodes,
since the points $(3,30,15)$ and $(3,33,12)$ are not lattice nodes of the plane $x=3$
(all such node coordinates are $(3,4m,4n)$ for some integers $m$ and $n$).
From the above we conclude that $r\le 2$.
Suppose now that $r{=}2$ and consider the points $(1,6,4)$,
$(1,7,3)$, and $(1,7,4)$. The points $(1,6,4)$ and $(1,7,3)$ are
not lattice nodes since the points $(2,12,6)$ and $(2,14,8)$ are
not lattice nodes of the plane $x{=}2$ (all such nodes coordinates
are $(2,4m,4n)$ for some integers $m$ and $n$). The point
$(1,7,4)$ defines a unique-possible integer-affine type of marked
pyramids with such base --- the integer-affine type of the marked
pyramid $U_2$.
\end{proof}
Now we will study the general case ($b\ge 2$).
\begin{lemma}
Consider an integer multistory marked pyramid with vertex $O$ and triangle base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(b,0)$, $(0,-1)$, and $(0,1)$, for $b\ge 2$.
Then the marked pyramid~$OABC$ is two-story and integer-affine equivalent to
the marked pyramid $U_b$ of List ``M-W''.
\end{lemma}
\begin{proof}
Let the base of $r$-story ($r {\ge} 2$) completely em\-pty marked
pyramid $OABC$ be integer-affine equi\-valent to the triangle with
vertices $(b,0)$, $(0,-1)$, and $(0,1)$.
Since the triangle with vertices $(b,0)$, $(0,-1)$, and $(0,1)$
contains the triangle with vertices $(2,0)$, $(0,-1)$, and $(0,1)$,
the marked pyramid $OABC$ contains a marked subpyramid that is
integer-affine equivalent to the marked pyramid of Lemma~\ref{l5}.
Since the subpyramid is completely empty, by Lemma~\ref{l5}
we have that it is two-story.
Suppose now $r{=}2$. Consider the parallelepiped $P(AOBC)$ and the
integer-distance coordinates corresponding to it (let us denote
such coordinates as $(x,y,z)$). By Proposition~\ref{m_st}, the
coordinates of $O$, $B$, and $C$ equal $(2,0,0)$, $(0,4b,0)$, and
$(0,0,4b)$ respectively.
Consider the parallelogram at the intersection of $P(AOBC)$ and
the plane $x=1$.
Now we will find all lattice nodes in this parallelogram.
By Proposition~\ref{m_st}, there are exactly $2b$ lattice nodes in the parallelogram
at intersection.
Let us describe all possible positions of these nodes in the intersection
of $P(AOBC)$ with the plane $x=1$.
First, there are no lattice nodes in the intersection of
the marked pyramid $AOBC$ with the plane $x=1$,
i.e. in the closed triangle with vertices
$(1,0,0)$, $(1,0,2b)$, and $(1,2b,0)$.
Secondly, there are no lattice nodes in all triangles
obtained from the given one by applying translations by vectors
$\lambda (0,4b,0) + \mu (0,2,2)$ for all integers $\lambda$ and $\mu$.
We show some (shaded) triangles that do not contain any lattice nodes on Figure~\ref{lemmas.9}.
So the lattice nodes of the intersection parallelogram
of $P(AOBC)$ with the plane $x=1$ can be only at integer points of
open triangle obtained from the triangle $K(1,4b,2b-4)$, $L(1,4b,2b)$, $M(1,4b-4,2b)$
by translations by vectors
$\lambda (0,4b,0) + \mu (0,2,2)$ for all integers $\lambda$ and $\mu$
and the symmetry about the plane $y=z$.
Only the points with integer coefficients $(1,4b-2,2b-1)$, $(1,4b-1,2b-1)$, and $(1,4b-1,2b-2)$
are in the triangle $KLM$ (the case $b=3$ is shown on Figure~\ref{lemmas.9}).
\begin{figure}[ht]
$$\epsfbox{lemmas.9}$$
\caption{The intersection of the parallelepiped $P(AOBC)$ with the plane $x=1$.}\label{lemmas.9}
\end{figure}
One of the integer points of this triangle is a lattice node (the
other uncovered parts of the section can be obtained by
translations by vectors $\lambda (0,4b,0) + \mu (0,2,2)$ for all
integers $\lambda$ and $\mu$).
Consider the plane $x=2$.
The point $(2,y,z)$ is a lattice node iff there exist such integers
$m$ and $n$ that $z=2m$, and $y=2m+2bn$.
We show that the point $(1,4b{-}2,2b{-}1)$ is not a lattice node by reductio ad absurdum.
Let this point be a lattice node.
Then the point $(2,8b{-}4,4b{-}2)$ is also a lattice node.
Let us find the such integers $m$ and $n$ that $4b-2=2m$ and $8b-4=2m+2bn$.
Then $m=2b-1$, $n=\frac{2b{-}1}{b}$. For $b \ge 2$, the number $n$ is not integer.
We come to the contradiction. Therefore the point $(1,4b{-}2,2b{-}1)$
is not a lattice node.
By the same reasons the point $(1,4b-1,2b-2)$ is not a lattice node.
The last point of the triangle $(1,4b-1,2b-1)$ determines the pyramid of the
integer-affine type $U_b$.
\end{proof}
\subsubsection{Case 5: integer points of the base different from
vertices are contained in one edge of the base}
It remains to study the case of the last most simple series of triangular
marked pyramids.
Suppose that all integer points of the base $ABC$ different from the vertices
are contained in $AC$, and the integer length of $AC$ is $a-1$, for some $a\ge 2$.
The case of $a=1$ is the case of empty marked pyramid was studied before in
Corollary~\ref{col_pyr2}.
Denote these points by $D_1, \ldots, D_{a-1}$ starting
from the point closest to $A$ and increasing the indexing in the direction to $C$.
(See Figure~\ref{theorem1.10}.)
\begin{figure}[ht]
$$\epsfbox{theorem1.10}$$
\caption{The triangle with vertices $(0,0)$, $(0,1)$, and $(a,0)$.}\label{theorem1.10}
\end{figure}
Consider an integer multistory marked pyramid with vertex $O$ and triangular base $ABC$.
Let the triangle $ABC$ be integer-affine equivalent to the
triangle with vertices $(0,0)$, $(0,1)$, and $(a,0)$, for $a\ge 2$.
Then the marked pyramid~$OABC$ is two-story and integer-affine equivalent to
the marked pyramid $U_b$ of List ``M-W''.
\begin{lemma}\label{zzzzz}
The marked pyramid~$OABC$ is integer-affine equivalent to
the marked pyramid of the following list.
{\bf List ``T'':}
--- $T_{a,1}^0$;
--- $T_{a,r}^\xi$, where
$\xi$ and $r$ are relatively prime and satisfy: $r \ge 2$ and
$0<\xi \le r/2$.
All integer marked pyramids listed in $``T''$ are completely empty and
integer-linear nonequivalent to each other.
\end{lemma}
\begin{proof}
{\bf 1. Preliminary statement.}
Let us show that the marked pyramid $OABC$ is integer-affine equivalent
to the marked pyramid $T_{a,r}^{\xi}$, for some positive integer $\xi \le r/2$.
First of all two single-story marked pyramids with the same $a$
are integer-affine equivalent, since the integer points of the edges of the pyramid
generates all integer lattice.
Let the base of $r$-story ($r \ge 2$) completely empty marked pyramid
$OABC$ be integer-affine equivalent to the
triangle with vertices $(0,0)$, $(0,1)$, and $(a,0)$.
Consider the parallelepiped $P(AOBD_1)$ and the
integer-distance coordinates corresponding to it
(we denote such coordinates as $(x,y,z)$).
By Proposition~\ref{m_st}, the coordinates of $O$, $B$, and $C$
equal $(r,0,0)$, $(0,r,0)$, and $(0,0,r)$ respectively.
By Corollary~\ref{symplex} (since the tetrahedron $AOBD_1$ is empty)
all inner lattice nodes are contained in one of three diagonal planes:
$x+z=r$, $y+z=r$, or $x+y=r$. Examine all the cases.
Let all inner lattice nodes are contained in the plane $x+z=r$.
By Lemma~\ref{l1} there exist exactly one lattice node $K$ contained in the
plane $x=1$.
So, $K$ is in the intersection of these two planes,
and its coordinates are $(1,\xi,r-1)$, where $0<\xi<r$.
Now we come back to the old coordinates associated with the lattice.
Since the integer distance from $K$ to the two-dimensional plane
containing the face $AD_1B$ equals one,
the tetrahedron $AD_1BK$ can be mapped by some integer-affine transformation
to the tetrahedron with vertices $(0,0,0)$,
$(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.
By such transformation the vertex $O$ maps to $(-\xi,1-r,r)$, and $C$
maps to $(a,0,0)$.
Let us translate the obtained pyramid by the integer vector $(\xi , r-1,r)$.
Finally we get the marked pyramid $T_{a,r}^{\xi}$.
Hence the marked pyramid $OACB$ is integer-affine equivalent to the marked pyramid
$T_{a,r}^\xi$, where $0<\xi< r$.
Consider the integer-affine transformation mapping the points
$O$, $A$, $B$, $C$ to the points $O$, $C$, $B$, $A$ respectively,
then the point $K$ maps to the point $(r-\xi,1-r,r)$.
Chose the smallest one of $\xi$ and $r-\xi$. Obviously,
this number is less then $r/2$.
Let all inner lattice nodes be contained in the plane $y+z=r$
in the integer-distance coordinate system.
By Lemma~\ref{l1} there exists exactly one lattice node $K$ contained in the
plane $x=1$.
So, $K$ is in the intersection of these two planes,
and its coordinates are $(1,\xi,r-\xi)$, where $0<\xi<r$.
The intersection of the marked pyramid $OABC$ with the plane $x=1$
is a triangle with vertices $(1,0,0)$, $(1,ar-a,0)$, and $(1,0,r-1)$.
This triangle contains all integer points $(1,t,r-t)$, for $2\le t\le r$. Hence
$\xi=1$.
Therefore the point $K$ is in the plane $x+z=r$,
so, we are in the position of the previous case.
Let all inner lattice nodes be contained in the plane $x+y=r$
in the integer-distance coordinate system.
By Lemma~\ref{l1} there exist exactly one lattice node $K$ contained in the
plane $z=1$.
So, $K$ is in the intersection of these two planes,
and its coordinates are $(\xi,r-\xi,1)$, where $0<\xi<r$.
The intersection of the marked pyramid $OABC$ with the plane $z=1$
is a triangle with vertices $(0,0,1)$, $(r-1,0,1)$, and $(0,ar-a,1)$.
This triangle contains all integer points $(t,r-t,1)$, for $1\le t\le r-1$. Hence
$\xi=r-1$.
Therefore the point $K$ is in the plane $x+z=r$,
so, we are in the position of the previous case.
So, the marked pyramid~$OABC$ is integer-affine equivalent to
a marked pyramid $T_{a,r}^{\xi}$, for some positive integer $\xi \le r/2$.
{\bf 2. Completeness of List ``T'' and completely emptiness
of the marked pyramids of ``T''.}
Let us show that the marked pyramids $T_{a,r}^{\xi}$ of the list ``T''
are completely empty.
Denote the vertices of the marked pyramids by $O$, $A$, $B$, $C$,
and the integer points of $AC$ by $D_i$.
Denote also the point $A$ by $D_0$, and the point $C$ by $D_a$.
Note that the marked pyramid $OD_{i}D_{i+1}B$ is integer-affine equivalent
to the marked pyramid $P_r^{\xi}$, for any positive integer $i\le a$,
since the marked pyramid $OD_{i}D_{i+1}B$ can be obtained from the
pyramid $P_r^{\xi}$ by applying the compositions of
the integer-linear transformation defined by the following matrix
$$
\left(
\begin{array}{ccc}
\xi +i+1& \xi +i& -\xi -i\\
r-1& r-1& 2-r\\
-r&-r&r-1\\
\end{array}
\right)
,
$$
and the translation by the integer vector $(-\xi, 1-r,r)$.
By corollary~\ref{col_pyr}, if $\xi$ and $r$ are relatively prime, then
the marked pyramids $OAD_1B$, $OD_1D_2B$,~$\ldots$, $OD_{a-1}CB$
are empty, and hence their union $OABC$ is completely empty.
By the same reasons the marked pyramids $T_{a,r}^{\xi}$ with
relatively prime $\xi$ and $r$ are completely empty.
Therefore List ``T'' is complete, and all integer pyramids of the list
are completely empty.
{\bf 3. Irredundance of List~``T''.}
Now we prove that all marked pyramids $T^\xi_{a,r}$ of List~``T''
are integer-affine nonequivalent to each other.
Obviously, that the marked pyramids with different $a$ are nonequivalent.
Since the integer distance from the marked vertex to the two-dimensional
plane of the marked base is an integer-affine invariant,
the marked pyramids with different $r$ are nonequivalent.
For the case of pyramids with the same integers $a{>}1$ and $r$,
we construct the follo\-wing integer-linear invariant. Consider an
arbitrary marked pyramid $OABC$, where all its integer points are
contained in the edge $AC$. As it was shown before the empty
marked pyramids $OAD_1B$, $OD_1D_2B$,~$\ldots$, $OD_{a-1}CB$ are
integer-affine equivalent to the mar\-ked pyramid $P_r^\xi$ with
$0\le\xi\le r/2$. Since the col\-lection of this marked pyramids
is defined in a unique way and by Corollary~\ref{col_pyr}, the
type of such $P_r^\xi$ is an invariant. This invariant
distinguishes different marked pyramids of List~``T''.
\end{proof}
So, we have studied all possible cases of integer-affine types of
multistory completely empty convex three-dimensional marked
pyramids. It remains to say a few words about the irredundance of
List ``M-W'' of Theorem~A.
\subsubsection{Irredundance of List ``M-W''}
If two marked pyramids have integer-affine nonequivalent bases, then
these pyramids are also integer-affine nonequivalent.
The integer-affine types of the base distinguish almost all marked
pyramids of List ``M-W''.
This does not work only for pyramids $T_{a,r}^{\xi}$ with the same $a$ and $r$,
and distinct $\xi$ from List~``M-W''.
Such pyramids $T_{a,r}^{\xi}$ are integer-affine nonequivalent by Lemma~\ref{zzzzz}
(see List ``T'').
The proof of the main theorem is completed.
\qed
\section{Proof of Theorem~B}\label{pB}
\subsection{Completeness of Lists ``$\alpha_n$'' for $n \ge 2$ of Theorem~B }
Note that polygonal faces of any sail are faces of the boundary of the convex.
Hence all faces are convex.
Consider some marked pyramid with marked vertex at the origin
and some compact two-dimensional face of a sail as base.
By the definition of multidimensional continued fractions it follows
that such pyramid is completely empty.
\begin{lemma}\label{l81}
Two two-dimensional faces are integer-linear equivalent iff
the corresponding completely empty marked pyramids are integer-affine equivalent.
\end{lemma}
\begin{proof}
If two two-dimensional faces are integer-linear equ\-iva\-lent,
then one of them maps to the other with some integer-line\-ar transformation.
The marked pyramid corresponding to the first face maps
to the marked pyramid corresponding to the second face at that.
Suppose now that the corresponding completely empty marked
pyramids are in\-te\-ger-affine equivalent. Then one of them maps
to the other with some integer-affine transformation. Since the
marked vertices of both pyramids are at the origin, the origin is
a fixed point of the transformation. Hence the transformation is
integer-linear. Since the base of the first pyramid maps to the
base of the second, the first face maps to the second also. Hence
these two-dimensional faces are integer-linear equivalent.
\end{proof}
So, for any $r\ge 2$, the following is true. Any integer-linear
type of compact two-dimensional faces contained in the
two-dimensional planes at integer distances equal $r$ from the
origin is uniquely defined by the corresponding integer-affine
type of $r$-story completely empty convex marked pyramids. Hence
by Theorem~A (see List~``M-W'') Lists~``$\alpha_n$'' of theorem~B
are complete if $n>2$.
Now we study the case of two-dimensional continued fractions. By
Theorem~A the list of all triangular faces in List~``$\alpha_2$''
is complete. It remains to show that there are no faces of sails
integer-linear equivalent to the quadrangle with vertices
$(2,-1,0)$, $(2,-a-1,1)$, $(2,-1,2)$, $(2,b-1,1)$, for some $b
\ge a \ge 1$.
Let us prove the following lemma.
\begin{lemma}\label{zapret}
{\bf (On some restrictions for two-di\-men\-sional continued
fractions.)} Let some compact two-dimensional face $F$ of some
two-dimensional continued fraction contains some integer
parallelog\-ram~$P$ integer-affine equivalent to the
parallelogram the with vertices
$(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
Then the integer distance from the origin to the plane containing the face $F$
equals one.
\end{lemma}
\begin{proof}
Consider such coordinates on the plane containing $F$ that
the vertices of $P$ become $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.
Note that the point in this plane is integer iff
its new coordinates are integers.
Suppose that the point $(1,1)$ is in $F$. Then the empty
parallelogram with vertices $(0,0)$, $(1,0)$, $(1,1)$, and
$(0,1)$ is contained in $F$. Therefore, by Proposition~\ref{lt1},
the distance from the plane containing $F$ equals one.
By the same reasons if $(1,-1)$, or $(-1,1)$, or $(-1,-1)$ is in
$F$, then the distance from the plane containing $F$ equals one.
Now we show that $F$ contains one of the listed four points by reduction ad absurdum.
Suppose the points $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$
are in the complement to $F$.
Three planes of two-dimensional continued fraction intersect with the plane
containing $F$ at three lines. The face $F$ is in the interior the triangle $T$
generated by the intersection lines.
The triangle $T$ contains $F$, and the set $T\setminus F$ does not
contain any integer point.
Notice that the point $(1,0)$ is in $F$, and the points $(1,1)$ and $(1,-1)$
are not in $F$. Note also that the points $(1,0)$, $(1,1)$, and $(1,-1)$
are in one straight line.
Then the open angle with vertex $(0,0)$ and edges passing through
the points $(1,1)$, and $(1,-1)$,
contains some vertex of the triangle $T$, see Figure~\ref{mnogoug.2}.
\begin{figure}[ht]
$$\epsfbox{mnogoug.2}$$
\caption{One of the vertices of $T$ is in shaded
(open) angle.}\label{mnogoug.2}
\end{figure}
The same holds for two adjacent angles and for the opposite angle.
Therefore the triangle $T$ has at least four vertices.
We come to the contradiction.
So, we have studied all the cases. Lemma~\ref{zapret} is proven.
\end{proof}
\begin{corollary}
Any two-dimensional continued fractions does not contain faces
that are integer-linear equivalent to the quadrangle with vertices
$(2,-1,0)$, $(2,-a{-}1,1)$, $(2,-1,2)$, $(2,b{-}1,1)$ for $b \ge a \ge 1$.
\end{corollary}
\begin{proof}
Take the polygon with the following vertices $(2,-1,0)$,
$(2,-a-1,1)$, $(2,-1,2)$, $(2,b-1,1)$. It contains the
parallelogram integer-affine equivalent to the parallelogram with
vertices $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$. Hence by
Lemma~\ref{zapret} the distance between the origin and the plane
containing such face equals one. But the integer distances from
the origin to the faces of the corollary equal two. Therefore
two-dimensional continued fractions do not contain faces of the
corollary.
\end{proof}
Therefore all Lists ``$\alpha_n$'' of Theorem~B (for
$n$-dimensional continued fractions) are completed for all $n\ge
2$.
\subsection{The completion of proof of Theorem~B}
First we show that all triangular faces of List ``$\alpha_n$''
are realizable for $n\ge 2$. We prove more general statement for
the triangles, and then generalize it to the case of polygons.
\subsubsection{All triangular faces are realizable}
Consider some completely empty triangular marked pyramid $OABC$
with marked vertex $O$ and base $ABC$. Let $\Sigma_{ABC}(3)$ be
the configuration space of ordered 3-tuples of points of the
plane containing the triangle $ABC$. The configuration space
$\Sigma_{ABC}(3)$ is homeomorphic to $\r ^6$. Consider the
standard topology of this space. Denote by $U_3 \subset
\Sigma_{ABC}(3)$ the set of such 3-tuples $A'B'C'$, that the open
marked pyramid $OA'B'C'$ contains the marked pyramid $OABC$
(except the point $O$) and the set $OA'B'C'\setminus OABC$ does
not contain integer points.
\begin{lemma}\label{last}
The set $U_3$ is open and nonempty.
Any point of the set $U_3$ defines such two-dimensional continued fraction,
that this fraction contains
the triangle $ABC$ as a two-dimensional face.
\end{lemma}
\begin{proof}
First we prove that $U_3$ is open. Consider such integer planes
parallel to the plane $ABC$ that the origin and the pane $ABC$
are in different half-planes (in the three-dimensional space
spanning the points $O$, $A$, $B$, and $C$). The number of such
planes is finite and equals $r-1$, where $r$ is an integer
distance between $O$ and the plane $ABC$. Denote by $\pi_i$, for
$i\le r$, one of the described planes at the integer distance
from $O$ equal~$i$. The marked pyramid intersects $OABC$ with
$\pi_i$ by the triangle, we denote it by $T_i$. The triangle
$T_i$ does not contain integer points for $i<r$. Consider all
open triangular angles centered at $O$, that intersect with
$\pi_i$ by some triangles that contain closed $T_i$ and do not
contain other integer points (different from the integer points
of $T_r$ for the case of $i=r$). Any such angle defines three
points in the plane containing the triangle $ABC$ (i.e. in
$\pi_r$). These points determine six ordered 3-tuples points of
$U_3$. The set of all such triangular angles determines the
nonempty open subset of $U_3$ (denoted by $U_{3,i}$). As it is
easy to see, the set $U_3$ coincides with the intersection of
the sets $U_{3,i}$ for all positive integers $i\le r$. Therefore
$U_3$ is open.
Secondly we prove that $U_3$ is nonempty.
We denote by $u_0$ the point of $U_3$ corresponding to the ordered 3-tuple
points $A$, $B$, $C$.
On one hand, there exist a neighborhood of $u_0$
containing the points with the following property:
if $A'B'C'$ in the neighborhood, then the set of
integer points of the marked pyramid $OA'B'C'$
is contained in the marked pyramid $OABC$.
On the other hand, any neighborhood of $u_0$ (and also the described one)
contains such point $A''B''C''$ that the closed marked pyramid $OABC$ is contained
in the open triangular angle $OA''B''C''$ except the point $O$.
From these two facts it follows that $U_{3}$ is not empty.
Now consider an arbitrary triangle $A'B'C'$ corresponding to some
point of $U_3$. Chose the planes $OA'B'$, $OA'C'$, and $OB'C'$.
The two-di\-men\-si\-onal continued fraction defined by these
planes contains the triangle $ABC$ as a face.
\end{proof}
\subsubsection{Realizability of polygonal faces}
Let us now generalize Lemma~\ref{last}. Consider some completely
empty convex polygonal marked pyramid $OA_1\ldots A_n$ with
marked vertex $O$ and base $A_1\ldots A_n$. Let $\Sigma_{A_1
\ldots A_n}(n)$ be the configuration space of ordered $n$-tuples
of points of the plane containing the polygon $A_1\ldots A_n$.
The configuration space $\Sigma_{A_1 \ldots A_n}(n)$ is
homeomorphic to $\r ^{2n}$. Consider the standard topology of
this space. Denote by $U_n \subset \Sigma_{A_1 \ldots A_n}(n)$
the set of such $n$-tuples $A'_1\ldots A'_n$, that the open
marked pyramid $OA'_1\ldots A'_n$ contains the marked pyramid
$OA_1\ldots A_n$ (except the point $O$) and the set $OA'_1\ldots
A'_n\setminus OA_1\ldots A_n$ does not contain integer points.
\begin{lemma}\label{last2}
The set $U_n$ is open and nonempty.
Any point of the set $U_n$ defines such two-dimensional continued fraction,
that this fraction contains
the polygon $A_1\ldots A_n$ as a two-dimensional face.
\qed
\end{lemma}
For any convex $k$-gon $P$ in $\r^{n+1}$ for $k\le n+1$
whose two-dimensional plane does not contain the origin,
there exists a $n$-tuple of hyperplanes that divides the two-dimensional
plane containing $P$ onto connected components, such that
one of these components coincides with $P$.
Further proof of Lemma~\ref{last2} repeats the proof of Lemma~\ref{last}.
\subsubsection{Realizability of faces}
\begin{proposition}
For any $n\ge 2$, any two-dimensional face of List~``$\alpha_n$''
is realizable as a face of some $n$-dimensional continued fraction.
\end{proposition}
\begin{proof}
Since all faces of List~``$\alpha_n$'' ($n \ge 2$)
are triangular or quadrangular (and the corresponding marked pyramids with vertices
at the origin and bases in these faces are completely empty),
Lemmas~\ref{last} and~\ref{last2} can be applied.
\end{proof}
\subsubsection{Nonequivalence of faces}
\begin{lemma}\label{xxxx}
For any $n\ge 2$, any two different faces of List~``$\alpha_n$''
are integer-linear nonequivalent to each other.
\end{lemma}
\begin{proof}
Lemma~\ref{xxxx} follows directly from Theorem~A (see
List~``M-W'').
\end{proof}
\subsubsection{On polygonal faces of two-dimensional continued fractions}
\begin{corollary}
Let some compact two-dimensional face of some two-dimensional
continued fraction is an integer polygon containing more that three
angles less than the right angle.
Then the integer distance between the origin and the plane containing this face equals one.
\end{corollary}
\begin{proof}
The corollary follows from Lemma~\ref{zapret}.
\end{proof}
\section{Unsolved questions and problems}
In conclusion of this work we outline some arising natural problems here.
First of all let us make the following remark.
By ``classification problems'' for some subset in some other set
through this section we mean the study of the following questions:\\
{\bf a)} which elements of the set are in the subset;\\
{\bf b)} which elements of the set are not in the subset;\\
{\bf c)} which infinite series of elements of the set
are in the subset, how many such series exist;\\
{\bf d)} which infinite series of elements of the set
are not in the subset, how many such series exist;\\
{\bf e)} describe properties of the elements of the subset;\\
{\bf f)} describe properties of the elements of the complement of the
subset in the set;\\
{\bf g)} is the problem of verification weather the given element of the set
is in the subset or not in it algorithmicaly solvable (find the corresponding algorithms);\\
{\bf h)} give the complete list of elements and series of the subset explicitly.
For instance, in this paper we solve the ``classification
problem'' ${\bf h)}$ for the subset of integer-linear or
integer-affine equivalence classes of compact two-dimensional
sails faces of multidimensional continued fractions at the integer
distances from the origin great than one (in the set of
integer-linear/affine equivalence classes of all polygons). Here
the answer to Question~${\bf h)}$ also implies the answers to
Questions~${\bf a)}$, ${\bf b)}$, ${\bf c)}$. Question~${\bf d)}$
becomes meaningless. Also we partially get answers to
Questions~${\bf e)}$ and~${\bf f)}$. Question~${\bf g)}$ is also
closely related to Question~${\bf h)}$ and also was studied by
the author, but it does not appear in the present paper by volume
reasons. The result of this question can be also applied in
algorithms of constructing two-dimensional continuous fractions.
\begin{problem}
Solve the ``classification problems'' for the subset of
integer-linear or integer-affine equivalence classes of compact
three-dimensional $($multi\-di\-men\-sional$)$ sail faces
contained in three-dimensional $($multidimensional$)$ planes at
the integer distance from the origin greater than one $($in the
set of integer-linear/affine equivalence classes of all
polyhedra$)$.
\end{problem}
In connection with the last problem the following question about
marked (compact by definition) pyramids is natural.
\begin{problem}\label{3Dproblem}
Solve all the ``classification problems'' for the subset of
integer-linear or integer-affine equ\-i\-va\-lence clas\-ses of
four-dimensional $($multi\-di\-men\-si\-onal$)$ multistory
com\-pletely empty convex marked pyramids $($in the set of
integer-linear/affine equ\-iv\-a\-lence classes of all convex
marked pyramids of the same dimension$)$.
\end{problem}
The geometrical contents of the next problem is extremely different
from the above two ones.
\begin{problem}
Solve the ``classification problems'' for the subset of
integer-linear or integer-affine equi\-valence clas\-ses of
two/three-dimensional $($multi\-di\-men\-si\-o\-nal$)$ sail faces
contained in the two/three-dimensional
$($multi\-di\-men\-si\-onal$)$ planes at the unit distance from
the origin.
\end{problem}
As a matter of fact this problem can be reduced to the
``classification problems'' for the integer-affine classes of
convex hulls of all integer points in some polygons
$($polyhedra$)$ with bounded number of faces of maximal
dimension. The following question about the polygons and
polyhedra is in its place here.
\begin{problem}
Solve the ``classification problems'' e$)$, f$)$, and h$)$ for the
subset of in\-te\-ger-af\-fine equ\-i\-va\-len\-ce classes of
integer polygons/polyhedra $($in the set of classes of all
polygons/polyhedra$)$.
\end{problem}
The following result on this subject is known.
Denote by $H(\mu)$ the logarithm of the number of integer-affine
equivalence classes of integer polygons of volume
$\mu /2$ in the plane, for some integer $\mu$.
\begin{theorem}{\bf (V.~Arnold~\cite{Arn5}.)}
For sufficiently large $\mu$ the following holds
$$
c_1 \mu ^ {1/3}\le H(\mu) \le c_2 \mu^{1/3}\ln \mu.
$$
\end{theorem}
Further investigations lead to the problems of classifications
of some face arrangements.
We give the simplest intensional example of such problems.
\begin{problem}{\bf(V.~Arnold.)}
Solve the ``classification problems'' for 1-stars of vertex
$($i.e. the union of a vertex and all adjacent edges to this
vertex$)$ for sails of two$($multi$)$-dimensional continued
fractions up to the integer-linear/affine equivalence.
\end{problem}
Here is another problem of this series.
\begin{problem}
Solve the ``classification problems'' for two-tuples of
two-dimensional adjacent faces for the sails of
two$($multi$)$-dimensional continued fractions up to the
integer-linear/affine equivalence.
\end{problem}
The last two problems can be naturally generalized to the case of
more complicated arrangements of faces.
Now we formulate the following problem on so-called {\it stable
integer-affine types} of polyhedra.
\begin{definition}
The integer-affine type of some polyhedron $($polygon$)$ is called
{\it stable in dimension $k$} if for any positive integer $r$
there exists such $k$-dimensional continued fraction that one of
the sails of this fraction contains the face with the given
integer-affine type in the plane at the integer distance equal $r$
to the origin.
\\
The integer-affine type of some polyhedron $($polygon$)$ is called
{\it almost stable in dimension $k$} if for any positive integer
$N$ there exist such $r>N$ and such $k$-dimensional continued
fraction that one of the sails of this fraction contains the face
with the given integer-affine type in the plane at the integer
distance equal $r$ to the origin.
\end{definition}
\begin{problem}
Which integer-affine types of polyhedra are $($almost$)$ stable in
dimension $3$ $($in dimension $k>3)$?
\end{problem}
We can answer on the similar question for the case of polygons.
\begin{corollary}
For any positive integer $k\ge 2$, and any positive integer $a\ge 1$
the integer-affine type of the triangle $(0,0)$, $(a,0)$, $(0,1)$
is stable in dimension $k$ $($see the case $a=6$ on Figure~\ref{qqq}$)$.
\begin{figure}[h]
$$\epsfbox{theorem1.3}$$
\caption{Stable polygons ($a\ge 1$).}\label{qqq}
\end{figure}
There are no other integer-affine types of integer polygons stable or almost stable
in dimension~$k$.
\qed
\end{corollary}
Even the answer to the following question is unknown to the author.
\begin{problem}
Is it true that the set of all stable in dimension $3$ $($in
dimensiona $k$$)$ integer-affine types of polyhedra coincide with
the set of all almost stable in dimension $3$ $($in dimensiona
$k$$)$ integer-affine types of polyhedra?
\end{problem}
Except the series of problems listed before the problems similar to the
following one are very important and interesting.
\begin{problem}
Do there exist three-dimensional polyhedra that appear
as a faces of sails of $k$-dimensional continued fractions
contained in three-dimensional planes at integer distances
greater than one only for $k>3$, and do not
appear for $k=3$?
In the case of positive answer solve the corresponding ``classification problems''
for them.
\end{problem}
Now we formulate some problems on statistical properties
of sail faces for multidimensional continued fractions.
Denote the set of all integer $(n+1)$-dimensional operators with
real rational distinct eigenvalues by $\Lambda_{n+1}$.
Denote by $\Lambda_{n+1,r}^I$
a subset of $\Lambda_{n+1}$ of operators with the norm not greater than $r$.
Denote by $\Lambda_{n+1,r}^{II}$ a subset of $\Lambda_{n+1}$ of operators with
the norm not greater than $r$ and the square root of the sum of squares of all
characteristic polynomial coefficients not greater than $r$.
(The operator norm here is the square root of the sum of squares of all
its matrix coefficients in some fixed basis.)
Since all eigenvalues of some operator $A$ in the set $\Lambda_{n+1}$
are real and distinct, the number of eigen hyperspaces
for $A$ (in $\r^{n+1}$) equals $n+1$.
The continued fraction is uniquely defined by these hyperspaces.
Since all eigenvalues of $A$ are rational, the sails consist of
finite number of compact faces.
Inversely, if all sails of multidimensional continued fraction consist of
finite number of compact faces, then the continued fraction corresponds to
some operator of the set $\Lambda_{n+1}$.
Let $\Gamma$ be some set of integer-linear types of faces
of $n$-dimensional continued fractions.
By $\sharp_{n+1,r}^I(\Gamma)$ (by $\sharp_{n+1,r}^{II}(\Gamma)$)
we denote the total number of faces with integer-linear type of the set $\Gamma$
for the continued fractions of the set $\Lambda_{n+1,r}^{I}$ ($\Lambda_{n+1,r}^{II}$ respectively).
\begin{problem}{\bf (V.~Arnold.)}
Does there exist a statistics of triangular faces for general
sails of finite multidimensional continued fractions?
Find this statistics in the case of positive answer.
\end{problem}
In other worlds, we have to study the existence of the following limit:
$$
\lim_{r\to \infty}\left(
\frac{\sharp_{n+1,r}^I(\triangle)}{\sharp_{n+1,r}^I(*)}\right)
\quad \left(
\mbox{or } \lim_{r\to \infty}\left( \frac{\sharp_{n+1,r}^{II}(\triangle)}{\sharp_{n+1,r}^{II}(*)}\right)
\right),
$$
if the limit exists, it is extremely important to find the limit (or even its approximation).
By $\triangle$ we denote the set of all integer-linear types of triangles,
the set of all integer-linear types of faces is denoted by $*$.
Is it true that the limits (for $I$ and $II$) equal to each other?
The similar questions are interesting for the cases of polygons
with $n>3$ vertices, and also for single cases of integer-affine types.
Besides that, the similar questions exist and are interesting for three-dimensional
and multidimensional polyhedra.
Note that nonexistence of the statistics for some sets of types
does not imply nonexistence of ``relative'' statistics for these sets.
\begin{problem}{\bf (V.~Arnold.)}
Does there exist a ``relative'' statistics of triangular faces
and quadrangular faces for general sails of finite multidimensional continued fractions?
Find this ``relative'' statistics in the case of positive answer.
\end{problem}
As in the previous case we have to study the existence of the following limit (and find it):
$$
\lim_{r\to \infty}\left(
\frac{\sharp_{n+1,r}^I(\triangle)}{\sharp_{n+1,r}^I(\diamondsuit)}\right)
\quad \left(
\mbox{or } \lim_{r\to \infty}\left( \frac{\sharp_{n+1,r}^{II}(\triangle)}{\sharp_{n+1,r}^{II}(\diamondsuit)}\right)
\right).
$$
Here by $\diamondsuit$ we denote the set of all integer-linear types of quadrangles.
Is it true that the limits (for $I$ and $II$) equal to each other?
\begin{remark}
It is also possible to consider some other exhaustions of $\Lambda_{n+1}$ (except $I$ and $II$)
for calculating the corresponding statistics.
(For more information see, for instance, the work of V.~Arnold~\cite{Arn2}.)
\end{remark}
In the papers~\cite{Arn2},~\cite{Arn4} and the book~\cite{Arn3}
(see problem~1993-11) by V.~Arnold
he discusses notions of statistics for types of sail faces of
multidimensional continued fractions
more detailed and formulates many interesting and actual statistical
problems and conjectures.
For one-dimensional continued fractions some of the conjectures of
V.~Arnold were completely studied by M.~Avdeeva and
V.~Bikovskii~\cite{Avd1} and~\cite{Avd2}. Denote by $\mbox{``}k
\mbox{''}$ a unique integer-linear type of the segment of integer
length $k>0$. In the works~\cite{Avd1} and~\cite{Avd2} for any
$k>0$ M.~Avdeeva and V.~Bikovskii proved the existence and found
the following limits:
$$
\lim_{r\to \infty}\left( \frac{\sharp_{3,r}^I( \mbox{``}k \mbox{''} )}{\sharp_{3,r}^I(*)}\right)
=
\frac{1}{\ln (2)}\ln\left( 1+ \frac{1}{k(k+2)} \right).
$$
and also the authors gave the estimate for the convergence rate of these limits.
It turns out that the limits coincide with the statistics of theorem of Gauss-Kuzmin-L\'evi
(for more information see the works of A.~Wiman~\cite{Wim} and R.~O.~Kuzmin~\cite{Kuz}).
M.~Kontsevich and Yu.~Suhov~\cite{Kon} proved the existence
of an average number of a polyhedron with the prescribed number of integer
points for almost all sails of multidimensional continued fractions
(except some zero Lebesque measure set).
These statistics are not calculated yet, and the methods of their calculation
are not yet developed.
The first statistical data for the periodic two-dimensional continued fractions
is given by the author in the work~\cite{Kar3}.
In conclusion of this section it remains to note that all the problems listed above
can be posed also for the case of sails of periodic algebraic
continued fractions.
We give the following problem as an example.
\begin{problem}{\bf (V.~Arnold.)}
Solve the ``classification problems'' for inte\-ger-affine
equivalence classes of compact two/three$($multi$)$-dimensional
faces of sails of periodic algebraic continued fractions.
\end{problem}
In the last problem it is also useful to study cases of faces contained
in the planes at distances equal/greater than one to the origin.
\begin{remark}
All the statistical questions (similar to the questions
for finite multidimensional continued fractions described above)
can be posed also for the case of periodic multidimensional continued fractions.
\end{remark}
\vspace{10pt} \noindent {\bf Acknowledgements:} \ The author is
grateful to professor V.~I.~Arnold, professor A.~B.~Sossinsky, and
E.~I.~Korkina for constant attention to this work and useful
remarks and discussions, and Universit\'e Paris-Dauphine
--- CEREMADE for the hospitality and excellent working
conditions.
| 13,799
|
Gentle reader, do not be concerned at the silence from this particular corner of paradise. In truth, I've been a bit busy with affairs of the heart, or at least, the fallout when they go sour.
Those of you who know me well, will be aware that I've been in mid-divorce for about three years now. The decree absolute came through two days before the General Election, which would have been immaculate timing on the part of my ex, had it not been for that fact that I didn't find out for six weeks...
And so the financial wrangling began, leading to a hearing just five weeks before the local elections this year (are you detecting a pattern here?). Even then, I thought that things could be wrapped up and I could carry on with my life. Sadly, I was wrong... and the whole charade carried on for another seven months... until this week, which saw the final court hearing. Or not, as the case might be. We'll probably know next week what the judge has decided, I'm told.
So I've been a bit occupied, all in all, with lawyers, estate agents, pension advisors and the entire legion of people who you might want to have a beer with occasionally but not necessarily want to deal with professionally. It could all be rather depressing if you let it.
Curiously, though, I'm not depressed, although I am a little saddened. I am a hopeless romantic, albeit not the most expressive of that ilk, and I had rather hoped to mate for life, optimistic though that might sound.
The curious thing though, is that the chorus to 'Simon Smith and His Amazing Dancing Bear' kept echoing through my head afterwards. "Oh who will think a boy and bear could be well accepted everywhere? It's just amazing how fair people can be..."
1 comment:
Keep your pecker up, Valladares.
| 38,467
|
TITLE: The integer $m$ is odd if and only if there exists $q \in \mathbb{Z}$ such that $m = 2q + 1$
QUESTION [0 upvotes]: The integer $m$ is odd if and only if there exists $q \in \mathbb{Z}$ such that $m = 2q + 1$.
Proof.
We have to prove both ways.
Suppose $m$ is odd, then by definition of odd number, $m = 2q+1$ for some $q \in \mathbb{Z}$.
Suppose $m = 2q + 1$, then $m$ is not divisible by $2$ so it's odd.
This is the way I proved it but not sure if I did it right. Any help please?
REPLY [1 votes]: It seems to me that you are assuming what you want to prove.
Definition. An integer $n$ is said to be even if there exists an integer $k$ such that $n = 2k$. An integer that is not even is said to be odd.
Division Algorithm. Let $n, d \in \mathbb{Z}$, with $d \neq 0$. Then there exist integers $q$ (the quotient) and $r$ (the remainder) such that $n = dq + r$, where $0 \leq r < |d|$.
Assume $m$ is odd. By the Division Algorithm, there exist integers $q$ and $r$, with $0 \leq r < 2$ such that $m = 2q + r$. There are only two non-negative integers less than $2$. They are $0$ and $1$. If $r = 0$, then $m = 2q$, so $m$ is even, contrary to our hypothesis that $m$ is odd. Hence, $r = 1$. Therefore, $m = 2q + 1$.
Assume there exists $q \in \mathbb{Z}$ such that $m = 2q + 1$. Since the integers are closed under multiplication, $2q$ is an integer. Since the integers are closed under addition, $m = 2q + 1$ is an integer. Observe that
$$m = 2\left(q + \frac{1}{2}\right)$$
Since $q$ is an integer, $q + 1/2$ is not an integer. Thus, $m \neq 2k$ for some integer $k$. Hence, $m$ is not even, so it is odd.
| 145,227
|
TITLE: Why A Basic Solution Is A Vertex?
QUESTION [0 upvotes]: A basic solution is a vector with $n$ components ($n=$ the number of variables), in which $m$ of them ($m=$ the number of functional constraints) are positive and the remaining ($n-m$) components are zero.
How does this implies that it must be a vertex point?
REPLY [1 votes]: Assuming the LP is written in standard form (all functional constraints are $\le$, all decision variables are $\ge 0$), then there are a total of $m+n$ constraints ($m$ functional and $n$ non-negativity), and each constraint corresponds to a variable (either a slack variable or a "regular" variable).
A basic solution is the solution you obtain by:
choosing $m$ of the constraints (out of the total of $m+n$ functional and non-negativity constraints)
setting them to equalities
forcing the $n-m$ variables that do not correspond to your $m$ constraints to 0, and
solving the resulting system of equations for the $m$ variables that do correspond to your $m$ constraints.
Since you are solving a system with $m$ equations and $m$ unknowns, there is a single unique solution (assuming there's nothing funny going on like degeneracy), and that is your basic solution. Graphically, the basic solution is a vertex of the feasible region because it is the single point that is on the boundary of all $m$ of those constraints.
| 166,304
|
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| 335,761
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TITLE: Irreducible representations and Hilbert spaces
QUESTION [3 upvotes]: I am reading Howard Georgi's book "Lie Algebras in Particle Physics" where he writes the following (chapter 1.14:eigenstates):
"... if some irreducible representation appears only once
in the Hilbert space, then the states in that representation must be eigenstates of $H$ (and any other invariant operator)."
The irreducible representation here, as far as I can tell, is meant to be part of a representation $D(g)$ on the full Hilbert space and we assume $H$ to commute with $D(g): [H, D(g)] = 0.$
My question is: what is meant by "appearing only once" in the Hilbert space?
Does it mean that, when I write the full representation D(g) as a direct sum of irreps, it appears only once in this direct sum?
To motivate why I think this is the case:
in this work explaining Schur's Lemma it is stated that, if the Hamiltonian commutes with $D(g) = \begin{pmatrix}\pi(g) & 0 \\ 0 & \pi(g) \end{pmatrix}$ where $\pi(g)$ is an irrep, then Schur's lemma does not apply but we can say that $H = \begin{pmatrix}A \mathbb{I} & B \mathbb{I} \\C\mathbb{I} & D\mathbb{I} \end{pmatrix}$.
So my questions are: 1) is my assumption correct? and 2) can you point me to an example for the two different cases (an irrep appearing once and more than once) that may potentially clarify my confusion?
REPLY [3 votes]: If we can decompose
$$
{\mathcal H}=\bigoplus_{{\rm irreps}\, J} {\mathcal H}_J
$$
into $\hat H$-invariant irreps of $G$ then Schur's lemma tells us that in each ${\mathcal H}_J$ the hamiltonian $\hat H$ will act as a multiple of the identity operator. In other words every state in ${\mathcal H}_J$ will be an eigenstate of $\hat H$ with a common energy $E_J$.
If an irrep $J$ occurs only once in the decomposition of ${\mathcal H}$ then it is automatically an $H$ invariant subspace and we can find the eigenstates directly by applying projection to vectors in the total Hilbert space ${\mathcal H}$. If the irrep occurs $n_J$ times in the decomposition, then we can project onto the reducible subspace
$$
\underbrace{{\mathcal H}_J\oplus {\mathcal H}_J\oplus\cdots {\mathcal H}_J}_{n_J\, {\rm copies}}={\mathcal M}\otimes {\mathcal H}_J.
$$
Here ${\mathcal M}$ is an $n_J$ dimensional multiplicity space.
The hamiltonian $\hat H$ will act in ${\mathcal M}$ as an $n_J$-by-$n_J$ matrix. In other words, if the vectors
$$
|n,i\rangle \equiv |n\rangle \otimes |i \rangle \in {\mathcal M}\otimes {\mathcal H}_J
$$
form a basis for ${\mathcal M}\otimes {\mathcal H}_J$, with $n$ labelling which copy of ${\mathcal H}_J$ the vector $|n,i\rangle $ lies in, then the hamiltonian and the symmetry group act as
$$
\hat H |n,i\rangle = |m,i\rangle H^J_{mn},\nonumber\\
D(g)|n,i\rangle = |n,j\rangle D^J_{ji}(g),
$$
where $D^J_{ji}(g)$ is the representation matrix in representation $J$.
Diagonalizing $H^J_{nm}$ provides us with $n_j$ $\hat H$-invariant copies of ${\mathcal H}_J$ and gives us the energy eigenstates.
| 109,136
|
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| 168,394
|
Notice
Disadvantaged Business Enterprise (DBE) Proposed Goal for the three years ending September 30, 2018.
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| 176,131
|
What is Quantserve?
Quantserve.com creates web beacons and cookies operated by audience research and behavioural advertising company Quantcast.
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| 371,118
|
TITLE: Show that the vector space $F((-\infty, 0)\cup (0, \infty))$ can be written as a direct sum in two different ways.
QUESTION [0 upvotes]: Prove that the vector space $F((-\infty, 0)\cup (0, \infty))$ can be written as a direct sum in two different ways using these pairs of subspaces:
(a) even functions and odd functions
(b) functions $f :(-\infty, 0)\cup (0, \infty)$ such that $f(x) = 0$ for all $x < 0$ and
functions g : $g :(-\infty, 0)\cup (0, \infty)$ such that $g(x) = 0$ for all $x > 0$.
For this question, why the direct sum of even and odd functions does not include 0 in vector? And why $f(x) = 0$, $g(x) = 0$ is nessessary in (b)?
The question can be understood in graph, but can someone give me hints about how to write a formal proof? Thx.
REPLY [4 votes]: For (a):
Observe that any function $h \in F(\Bbb{R} \setminus\{0\})=V$ can be written as
$$h(x)=\underbrace{\frac{h(x)+h(-x)}{2}}_{\in E}+\underbrace{\frac{h(x)-h(-x)}{2}}_{\in D},$$
where $E$ is the subspace of even functions and $D$ is the subspace of odd functions. Observe that for a direct sum we want two things: $V=E+D$ and $E \cap D=\{\mathbf{0}\}$ (only the zero function should be in both). By above we have demonstrated the first condition. The second condition is easy to verify as only the zero function is both even and odd.
For (b).
For any function $h \in F(\Bbb{R} \setminus\{0\})=V$, let us define
$$f(x)=\begin{cases}h(x), & \text{ if } x >0\\ 0, & \text{ if } x < 0\end{cases} \qquad
g(x)=\begin{cases}0, & \text{ if } x >0\\ h(x), & \text{ if } x < 0\end{cases}.$$
Now $h$ can be written as
$$h(x)=\underbrace{f(x)}_{\in R}+\underbrace{g(x)}_{\in L},$$
where $R$ is the subspace of all functions which are zero for $x<0$ and $L$ is the subspace of all the functions which are $0$ for $x>0$.
The reason we want to have $L$ and $R$ to have functions which disappear on a certain interval is to make sure when functions are added then they won't interfere with each other (once again this is to ensure that both the properties $V=R+L$ and $R \cap L=\{\mathbf{0}\}$ associated with the direct sum can be achieved).
| 61,962
|
TITLE: Rolling a fair die
QUESTION [1 upvotes]: You and your friend play a game in which you and your friend take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. Play continues until either player wins if, after the player rolls, the number
on the running tally is a multiple of 7. Should you start first or it is better if you let your friend rolls the die first.
REPLY [0 votes]: I thought I'd answer this question as another opportunity to learn some math that's way out of my league in hopes that some of my usual favorites will evaluate it and see if I'm on the right track.
The first thing I did was consider a few trials:
Player 1 has a 0/6 chance of winning
Player 2 has a 1/6 chance of winning
Player 1 has a 1/6 chance of winning
Player 2 has a 1/6 chance of winning
The probability of player 1 winning a 1 round game is 0. The probability of player 1 winning a 2 round game is $\mathbb{P} = \mathbb{P}(\text{Winning Round 1}) + \mathbb{P}(\text{Rounds 1 and 2 Losses}) \cdot \mathbb{P}(\text{Winning Round 3}))$.
I figured that continues infinitely with the outcomes becoming less and less likely as the number of trials approaches infinity.
So, I came up with this:
$$\mathbb{P}(\text{Player 1}) = \dfrac{0}{6} + \dfrac{6}{6} \cdot \dfrac{5}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots $$
$$\mathbb{P}(\text{Player 1}) = 0 + \dfrac{5}{6} \cdot \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots = \dfrac{1}{6} \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $$
So, then I had to figure out how to sum an infinite series: $\left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $
I did some research on how exactly that's done and the best I could come up with is to multiply the entire series by $\left(\dfrac{5}{6}\right)^2$ and subtract that from the original series. So, I think this is valid:
$$\left(\dfrac{5}{6}\right)^2 \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) = \left( \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right)$$
If I subtract the new series from the original, all of the elements but the first should cancel out and I should have this:
$$\mathbb{S} - \left(\dfrac{5}{6}\right)^2 \cdot \mathbb{S} = \dfrac{5}{6}$$
$$\mathbb{S} \cdot \left[1 - \left(\dfrac{5}{6}\right)^2\right] = \dfrac{5}{6}$$
$$\mathbb{S} = \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]}$$
If I plug that into my original formula, I get this:
$$\mathbb{P}(\text{Player 1}) = \dfrac{1}{6} \cdot \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]} = \overline{.45}$$
Because there are only two players this should be true: $\mathbb{P}(\text{Player 2}) = 1 - \mathbb{P}(\text{Player 1}) = \overline{.54} $
Unsure of my original conclusion, I thought I'd calculate the probability of Player 2 winning to verify.
$$\mathbb{P}(\text{Player 2}) = \dfrac{6}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^2 \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^4 \cdot \dfrac{1}{6} + \ldots $$
$$\mathbb{P}(\text{Player 2}) = \dfrac{1}{6} + \dfrac{1}{6} \cdot \left(\left(\dfrac{5}{6}\right)^2 \cdot \left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^6 + \ldots \right) $$
To sum that series I thought: $\mathbb{S} = \dfrac{25}{36} + \left(\dfrac{25}{36}\right)^2 + \left(\dfrac{25}{36}\right)^3 + \ldots $
So, $\mathbb{S} - \dfrac{25}{36}\mathbb{S} = \dfrac{25}{36}$ and $\mathbb{S} = \dfrac{\dfrac{25}{36}}{1 - \dfrac{25}{36}}$
I plug that into the original formula and find that $\mathbb{P}(\text{Player 2}) = \overline{.54}$ which is what I expected from before.
| 85,108
|
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| 371,468
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Description
Condition: Certified Refurbished - Brown/White Box
Warranty Term: 90 Days
Warranty Provider: Distributor
Apple MacBook Pro MGX92LL/A 13.3" 16GB 256GB Intel Core i5-4308U, Silver (Certified Refurbished). 2.8 GHz Dual-Core Intel Core i5 Processor (Turbo Boost up to 3.1 GHz, 3 MB shared L3 cache) 16 GB 1600 MHz DDR3L RAM; 512 GB PCIe-based Flash Storage 13.3-inch IPS Retina Display, 2560-by-1600 resolution Intel Iris Graphics-4308U Processor (Dual-Core, 2.8GHz, 3MB Cache)
- 16GB DDR3 System Memory
- 256GB Solid State Drive (SSD)
- Integrated Intel Iris Graphics 6100
- No Optical Drive
- 2 x USB 3.0, 1 x HDMI, 2 x DisplayPort
- Backlit Standard Keyboard and Touchpad
- 1MP Webcam and Integrated Microphone
- No Ethernet, IEEE 802.11a/b/g/n and Bluetooth 4.0 + HS
- 6-cell Lithium-Polymer Battery with up to 7 hours of use
- Operating System: macOS 10.14
- 0.75" x 12.35" x 8.62" (HxWxD); 3.5lbs
- Energy Star Compliant
Whats in the Box?
- Apple MacBook Pro 13.3" Laptop
- Power Cord
- Charging Cable
Specifications:
General Information
- Color (Exact): Silver
- Model Number: MGX92LL/A
- Energy Star Compliant: YES
- Release Year: 2014
- Product Line (Notebook): MacBook Pro
- Lifestyle: Entertainment
- Screen Refresh Rate (Hz):
- Form Factor: Notebook
- EPEAT Level: Not Applicable
Storage
- Hard Drive Type: Solid State Drive (SSD)
- Hard Drive Speed (RPM): 0
- Hard Drive Interface: SSD
- Hard Drive Size (GB): 256
Inputs/Outputs
- Fingerprint Reader: NO
- Mouse Type: Touchpad
- Total Headphone Outputs: 1
- Total USB 2.0 Ports: 0
- Total DVI Outputs: 0
- Integrated Microphone: YES
- Optical Drive Type: Not Applicable
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TITLE: How to calculate the voltage induced in a toroid by a charged particle moving through the center?
QUESTION [0 upvotes]: I'm trying to derive the expression for the voltage induced in a toroid by a charged particle moving perpendicularly through the opening/center of the toroid? For the purposes of clarification, the center of the toroid is where the magnetic field would be zero and the electric field lines would be parallel and unbending.
I've gotten to the expression that the induced voltage is
$\varepsilon = -N \frac{\Delta \phi_B}{\Delta t}$
$\phi_B = \vec{B} \cdot \vec{A}$
$\vec{B} = \frac{\mu _0 q_e \vec{v}}{4\pi r^2}$
$d\phi = |\vec{B}||\vec{A}|cos(\theta)d\theta$
$\theta = tan^{-1}(\frac{r_{tor}}{z})$
$d\theta = -\frac{r_{tor}}{r_{tor}^2 + z^2} dz$
These expressions are used in the first equation to give
$\varepsilon = -N |\vec{B}||\vec{A}|cos(tan^{-1}(\frac{r_{tor}}{z})) \frac{\Delta z}{\Delta t}$
$\varepsilon = -N |\vec{B}||\vec{A}|cos(tan^{-1}(\frac{r_{tor}}{z})) |\vec{v}|$
This doesn't seem completely correct since the distance between the moving particle and the toroid center, $z$, is changing in time. I imagine it would be more accurate to have an integral of this expression where the total voltage is the integral of the EMF as a function of time, but I'm just not convinced by my math. I also read that here that the charged particle would lose energy as it streams through the toroid. I don't know where to start with this problem. It's been many years since my advanced E&M class in college.
REPLY [0 votes]: Starting from the assumption that the electron beam is actually a line source is the key to solving my problem. The magnetic potential is defined as
$\vec{M} = \frac{\mu_0 \mu_r}{4\pi} \int _{\Omega} \frac{J(\vec{r}',t)}{|\vec{r} - \vec{r}'|} d^3\vec{r}'$
then the magnetic field can be determined using Maxwell's equations
$\vec{B} = \nabla \times \vec{M} $
$\vec{B} = \frac{\mu_0 \mu _r Q_e |\vec{v}_e|}{2\pi\vec{r}}$
Finally for Faraday's Law the induced voltage in the toroid is
$\varepsilon = -N \frac{\Delta \phi_B}{\Delta t}$
$\phi_b = \int_{r_1}^{r_2} \mu_0 \mu_r \vec{B} \cdot d\vec{A}$
$\Delta t = \frac{w}{|\vec{v}_e|}$
where $w$ is the width of the toroid.
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Police will soon be able to tell what a crook ate and drank and what drugs they took before they committed a crime, thanks to pioneering fingerprint technology developed at Sheffield Hallam University.
Researchers have already been able to determine the sex of a criminal with the new crime-fighting weapon.
Tests show it can also detect certain drinks and foods such as coffee and garlic.
The team are now working to see if it is able to test for medical conditions, as well as being able to tell how long the fingerprint has been at the crime scene.
The microscopic technology is called Matrix Assisted Laser Desorption Ionisation Mass Spectrometry Imaging - MALDI-MSI - which traces drugs, hair and cleaning products in fingerprints.
It has been developed by researchers at Hallam’s biomedical research centre led by Dr Simona Francese.
Dr Francese said: “MALDI enables you to detect the chemistry of the finger marks so essentially what chemicals are present.
“In one example we found the presence of cocaine in traces, we also found the presence of cocaine metabolite, that is very good because it actually tells us immediately the person who left the mark has not just touched cocaine but actually ingested it.
“So that changes the forensic scenario very quickly. With food, I’ve tried that on my own fingerprints, I’ve drunk a cup of coffee, then looked at my own finger marks at a certain time, after ten minutes, and you could see a very clear signal for caffeine.”
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TITLE: Find the generator of Markov Process
QUESTION [1 upvotes]: Homework question:
Consider the Markov process $X_t=B_t-t^2+t$ where $B_t$ is the Brownian motion. Find the generator $Q$ of this process.
I am completely confused how to find the generator for this.
Any help please?
REPLY [1 votes]: First of all, we note that
$$X_t := B_t-t^2+t = \int_0^t \, dB_s + \int_0^t (-2s+1) \, ds.$$
Therefore, we find by Itô's formula for any $f \in C^2$
$$\begin{align*} f(X_t)-f(X_{t_0}) &= \int_{t_0}^t f'(X_s) \, dX_s + \frac{1}{2} \int_{t_0}^t f''(X_s) \, d\langle X \rangle_s \\ &= \int_{t_0}^t f(X_s) \, dB_s + \int_{t_0}^t f'(X_s) (-2s+1) \, ds + \frac{1}{2} \int_{t_0}^t f''(X_s) \, ds. \end{align*}$$
As $\int_{t_0}^t f(X_s) \, dB_s$ is a stochastic integral, hence a martingale, we see that
$$\begin{align*} &\frac{1}{t-t_0} \bigg(\mathbb{E}^{t_0,x}f(X_t)-f(x)\bigg) \\ &= \frac{1}{t-t_0} \int_{t_0}^t \mathbb{E}^{t_0,x}(f'(X_s)) (-2s+1) \, ds + \frac{1}{2(t-t_0)} \int_0^t \mathbb{E}^{t_0,x}(f''(X_s)) \, ds. \end{align*}$$
Since both $s \mapsto \mathbb{E}^{t_0,x}(f'(X_s))$ and $s \mapsto \mathbb{E}^{t_0,x}(f''(X_s))$ are continuous functions, the fundamental theorem of calculus yields
$$\begin{align*} \lim_{t \to t_0} \frac{1}{t-t_0} \int_{t_0}^{t} \mathbb{E}^{t_0,x}(f'(X_s)) (-2s+1) \, ds &= \mathbb{E}^{t_0,x}(f'(X_{t_0})) (-2 \cdot t_0+1) \\ &= f'(x)(-2t_0+1) \\ \lim_{t \to t_0} \frac{1}{2t-t_0} \int_{t_0}^t \mathbb{E}^{t_0,x}(f''(X_s))&= \frac{1}{2} \mathbb{E}^{t_0,x}(f''(X_{t_0})) \\ &= \frac{1}{2} f''(x) \end{align*}$$
Hence,
$$Q_{t_0}f(x) = \lim_{t \downarrow t_0} \frac{1}{t-t_0} \bigg(\mathbb{E}^{t_0,x}f(X_t)-f(x)\bigg) \to f'(x)(-2t_0+1)+ \frac{1}{2} f''(x).$$
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Caribbean No Frills Vacations in Barbados
Barbados on a budget
Need a break but don't want to break the bank? Well then a budget vacation to a hotel or resort in the Caribbean is perfect for you. Why not visit the Caribbean on the cheap? The weather and beaches are free!
The Caribbean island of Barbados has a reputation as a playground for the rich and famous but, no frills cheap vacations are also available, so you can enjoy a Caribbean vacation without spending a fortune. And 'no frills' does not mean low quality. It is comfortable hotels and resorts at an affordable price. Several Barbados properties are ideally suited for a cheap Caribbean hotel or resort vacation.
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TITLE: $ L(P)= (1+X^2)P''(X)-2XP'(X)$
QUESTION [3 upvotes]: $E=\mathbb{R}[X]$
$L$ is an endomorphism of $E$ and $L(P)= (1+X^2)P''(X)-2XP'(X)$
What are the possible eigenvalues of $L$ and the dimension of the eigenspace ?
Let's $p(x)=\sum_{k=0}^{n} a_k X^k$ be an eigenvector associated to the eigenvalue $\lambda$.
$a_{n+1} = a_{n+2} = .. =0$
$
\begin{cases}
p(x) &= \sum_{k=0}^{+\infty} a_k x^k \\
p'(x) &= \sum_{k=1}^{+\infty} a_k x^{k-1} k \\
p''(x) &= \sum_{k=2}^{+\infty} a_k x^{k-2} k (k-1) \\
\end{cases}
$
$
\begin{cases}
-2x p'(x) &= \sum_{k=1}^{+\infty} -2 a_k x^k \\
x^2 p''(x) &= \sum_{k=2} a_k k (k-1) x^k \\
p''(x) & = \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) \\
\end{cases}
$
$
\begin{align*}
(1+x^2)P''(x) -2xP'(x) - x P'(x) &= \lambda P(x) \\
\sum_{k=1}^{+\infty} -2 k a_k x^k +\sum_{k=2} a_k k (k-1) x^k + \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) &= \lambda \sum_{k=0}^{+\infty} a_k x^k \\
\end{align*}
$
$
\begin{cases}
-2 k a_k + a_k k (k-1) + a_{k+2} (k+2) (k+1) &= \lambda a_k ~~ \text{for} ~ k \geq 2 \\
a_2 \times 2 &= \lambda a_0 \\
-2 a_1 +a_3 \times 6 &= \lambda a_1 \\
\end{cases}
$
\begin{cases}
(k^2-3k-\lambda)a_k &=-(k+2)(k+1)a_{k+2}\tag{1} \\
a_2 \times 2 &= \lambda a_0 \\
-2 a_1 +a_3 \times 6 &= \lambda a_1 \\
\end{cases}
$(1)$ is available only for $k \geq 2$.
$
\begin{align*}
\psi : \mathbb{N} &\to \mathbb{N}\\
n & \mapsto n^2-3n \\
\end{align*}
$
$\psi(n)-\psi(m) = (n-m)(n+m-3)$
REPLY [2 votes]: Let $\lambda$ be an eigenvalue of $L$ and $p(x)=\sum_{k=0}^n a_kx^k=\sum_{k=0}^\infty a_kx^k$ with $a_n=1$ and $0=a_{n+1}=a_{n+2}=\cdots$ be a corresponding eigenvector. From $(1+x^2)p''(x)-2xp'(x)=\lambda p(x)$, we obtain the recurrence relation
$$
\left[(k+2)(k+1)a_{k+2}+k(k-1)a_k\right]-2ka_k=\lambda a_k
$$
or equivalently
$$
(k^2-3k-\lambda)a_k=-(k+2)(k+1)a_{k+2}\tag{1}
$$
for each $k\ge0$. It follows that $\lambda=n^2-3n$.
Since $(n^2-3n)-(m^2-3m)=(n-m)(n+m-3)$, if $\lambda=n^2-3n$ for some $n>3$, then $n$ is uniquely determined and the LHS of $(1)$ is nonzero for every $0\le k<n$. Thus $a_{n-1},a_{n-2},\ldots,a_0$ can be uniquely and recursively determined from $(1)$ using the boundary conditions that $a_{n+1}=0$ and $a_n=1$. Hence the eigenspace for $\lambda$ exists and is one-dimensional.
When $\lambda=n(n-3)$ for some $0\le n\le3$, complications arise because $\lambda$ has two factorisations $n(n-3)$ and $m(m-3)$, where $m=3-n$.
When $\lambda=0$, the equation $\lambda=n(n-3)=0$ has two nonnegative integer solutions $n=3$ and $n=0$. The null space of $L$ therefore consists of only cubic polynomials and constant polynomials. Suppose $p$ be a monic cubic polynomial in the null space of $L$. The recurrence relation $(1)$ then gives $a_2=0$ and $a_1=3$, but $a_0$ is undetermined because $(1)$ reduces to $0a_0=0$ when $k=0$. Hence the null space of $L$ exists and it is a two-dimensional subspace spanned by $x^3+3x$ and $1$.
When $\lambda=-2$, the equation $\lambda=n(n-3)$ has two nonnegative integer solutions $n=2$ and $n=1$. The corresponding eigenspace thus consists of (apart from the zero polynomial) only polynomials of degrees $1$ and $2$. Let $p$ be a monic quadratic polynomial in the eigenspace. The recurrence relation $(1)$ gives $a_0=1$ but $a_1$ is undetermined because $(1)$ reduces to $0a_1=0$ when $k=1$. Therefore the eigenspace for this eigenvalue exists and it is a two-dimensional subspace spanned by $x^2-1$ and $x$.
In short, the eigenvalues of $L$ are $\lambda=n^2-3n$ for each $n\ge2$. The corresponding eigenspace is two-dimensional when $n=2,3$ and one-dimensional when $ n>3$.
| 145,096
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TITLE: Quadratic subfields of Quartic extensions
QUESTION [1 upvotes]: In this article, on page 18, the author describes quadratic subfields of quartic extensions. However, the proofs are somewhat obscure (at least to me). Let $L/K$ be a quartic extension containing a quadratic extension $M$, $K$ not characteristic $2$. The aim is to prove that $L=K(\theta)$, where $\theta$ has minimum polynomial over $K$ of the form $X^4 + AX^2 + B$.
$M=K(\sqrt{a})$ with $a \in K^{\times}$, obviously $a$ not square in $K$. Also $L=M(\sqrt{\theta})$, with $\theta \in M^{\times}$, obviously $\theta$ not square in $M$. The author puts it immediately like this: $L=M(\sqrt{u + v\sqrt{a}})$, $u,v \in K$, $v \neq 0$ (does something similar in characteristic $2$). The fact that $v \neq 0$ is key, because then $L=M(\sqrt{u + v\sqrt{a}})=K(\sqrt{u + v\sqrt{a}})$. Without justification, to me, it seems that the author uses what he is about to prove as a result. So, why $L=M(\sqrt{u + v\sqrt{a}})$, $u,v \in K$, $v \neq 0$ ? As counterexample, where this is not immediately obvious, consider e.g. $\mathbb{Q}(\sqrt{2})(\sqrt{-1})$.
REPLY [1 votes]: Right, the case with $v=0$ must be settled. I keep all the previous notations. If $v=0$, one has $L=K(\sqrt \theta)=K(\sqrt a, \sqrt u)$. The case $K(\sqrt a)=K(\sqrt u)$, which corresponds to $a$ mod ${K^*}^2 = u$ mod ${K^*}^2 $, leads to a contradiction between the degrees. In the other case, $L/K$ is a biquadratic extension, i.e. a galois extension with group $G\cong C_2\times C_2$, hence contains exactly 3 quadratic subextensions which are $K(\sqrt a),K(\sqrt u), K(\sqrt {ua})$: this is an immediate consequence of Kummer theory, but can also be shown "by hand".
Let us introduce the sum $s=\sqrt a + \sqrt u\in L$. Since $s^2=a+u+2\sqrt a\sqrt u$ does not belong to $K$, the degree of $s$ over $K$ must be 4, and its minimal polynomial has the form $X^4+AX^2+B$. This can be checked by computing $s^4$ and eliminating $\sqrt a\sqrt u$. So the announced property holds even if $v=0$. I guess the author neglected this abelian case because he implicitly considered it as well known in the abelian case.
| 200,865
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Abstract kA/cm2 Nb/AlOx/Nb process. The operation of a fabricated neuron circuit is experimentally demonstrated. Network performance of a neural network using proposed neuron circuits is also estimated by numerical dynamic simulations.
Keywords
- Combinatorial optimization problem
- Neural network
- SQUID
- Superconducting circuit
ASJC Scopus subject areas
- Electronic, Optical and Magnetic Materials
- Electrical and Electronic Engineering
| 353,648
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Sweater: Anthropologie | Jeans: Nordstrom | Booties: Here, Other Leopard Shade | Bag: LV, Similar | Earrings: Beehive, Similar | Lipstick: Candy Yum Yum | Sunglasses: Dupe | Tank: BP Photography by: Angie Garcia Happy Halloween! Besides…
| 103,196
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Pitching class tv writing: the new york film academy’s new york screenwriting school offers classes year round 12-week evening screenwriting workshop. Creative writing schools in new york new york has 25 accredited creative writing schools where creative writing faculty who teach creative writing classes can find. Get information about creative writing courses at new york university in new york, ny including enrollment information and other schools that may offer a creative. Creative writing school offers classes in new york city and online.
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I held off as long as I could. Really I did. But those Memorandum scraps have been shouting Christmas! at me and I was finding them hard to resist.
I decided to make some gift tags with the help of my Slice.
I layered the different tag shapes up to make them feel a bit more special
and even added a bit of Christmas sparkle. Then I hung them all on my new display board and turned it into this:
| 296,952
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TITLE: Groups all of whose extensions are split
QUESTION [9 upvotes]: Is there a sensible characterization of groups $G$ with the following property?
Every extension of groups $1\to G\to H\to K\to 1$ is split.
A complete group $G$ has that property and in fact such a group has a normal complement in every group that contains it as a normal subgroup (moreover, completeness is characterized by this)
For comparison, a group $K$ has the property that every extension $1\to G\to H\to K\to 1$ is split iff it is free (there is such an extension with $H$ free, and the splitting map $K\to H$ is injective, so gives an isomorphism of $K$ with a subgroup of $H$, which is free by the Nielsen–Schreier theorem)
NB: the title does use the old meaning of «extension of a group»...
REPLY [7 votes]: Since my late comment to YCor's answer is easily overlooked, I allow myself to repeat it here: The question was answered in [J. S. Rose, Splitting properties of group extensions, Proc. London Math. Soc. (3) 22 (1971), 1–23] with exactly the same outcome as in YCor's answer.
| 148,258
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Posted by Willy Franzen on February 25, 2011. Jobs updated daily.
When I was a kid, I’d read almost anything. When we had free reading time in elementary school, I’d often grab the dictionary out of my desk and start reading (I always forgot to bring another book to read). Not surprisingly, kids made fun of me for this. I’d also often pick up whatever was lying around my house. That often meant that I’d end up with a copy of The New Yorker in my little hands. The illustrations on the cover would draw me in, and then I’d get frustrated as I rustled through page after page of black text. Occasionally, I’d come across one of the cartoons, read it, not get it, and continue on. But for some reason, I’d keep picking up copies of The New Yorker in hope of finding a cartoon that I could understand (or an Absolut ad, which was the hot thing to collect in middle school). Since The New Yorker doesn’t really say anything about themselves on their About page (besides the fact that they’ve been published since February 21, 19250, I went to Wikipedia which says, “The New Yorker is an American magazine of reportage, commentary, criticism, essays, fiction, satire, cartoons and poetry published by Condé Nast Publications.”
Now, our approach with The New Yorker is going to be a bit different from how we usually look at companies and their entry level jobs. The New Yorker is a well known brand that tons of people want to work for, but they’re super mysterious about what kind of opportunities they offer for recent college graduates. The Careers page that they link to is just CondeNastCareers.com, and that doesn’t actually have any information on positions with The New Yorker. It may have job postings when they’re current, but right now they’ve got nothing. Googling “The New Yorker jobs” or “The New Yorker entry level jobs” doesn’t turn up much at all. In fact, the most interesting thing that I could find was the Wikipedia article on Fact checker, which says, “fact-checking is an entry-level publishing job at major magazines; fact-checker jobs at The New Yorker are considered prestigious and can lead to higher-level positions, usually at other magazines.” So there are in fact entry level opportunities at The New Yorker. So, how do you find them? Go to The New Yorker’s LinkedIn Company Profile, take a look at the types of positions that appear to be entry level and backgrounds of the people who hold them, then network, network, network. It’s low probability, but if you want to work at prestigious name like The New Yorker, it’s probably your only shot.
Links to Help You Begin Your Research
Do you read The New Yorker?
We've identified The New Yorker as having career opportunities in the following categories:
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Game Crusade is recently released WordPress Theme from Shape5 Club for the month of August 2011. Game Crusade WordPress Theme is professionally designed for gaming websites and it is 100% transparent core. This theme comes packed with lots of great features! Choose between a fading or non-fading background.
You can rotate up to six images in the site’s background or completely disable this feature. Choose between a fixed or scrolling background images as well. With an amazing selection of fully collapsible sidebar positions and custom page and columns widths you will never run out of space for your content and widgets!
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- Sliced PSDs included
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- Site Shaper available (WordPress install that includes demo data)
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- Lytebox enabled
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- 5 Menu systems: Choose style: Drop Down, Fading, Scroll Down, Suckerfish, None
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Get instant access to demo or download: Game Crusade WordPress Theme.
| 155,342
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In "Give Me Some Sugar," Emily Hilliard introduces us to some of the South's most talented female pastry chefs. They do right by the classics while developing a new canon of their own. Check back every Monday to meet a reason to save room for dessert..
The work of pastry chef Carla Cabrera-Tomasko is part of this picture of the Global South. Growing up in Ecuador, she started selling cakes at age 16. “My aunt taught me to make cakes and flan,” says Carla. “I always liked being in the kitchen.” As her cake business grew, she wanted to learn more about the culinary arts. Her cousin told her about the Atlanta Art Institute’s Culinary Arts program, and in 2000 she moved to Atlanta to enroll.
Now the pastry chef at Anne Quatrano and Clifford Harrison’s Bacchanalia, Carla’s desserts often feature locally sourced ingredients and typical Southern products, interpreted through her memories of what she grew up eating in Ecuador. Earlier this month, for example, the Bacchanalia dessert menu featured a tres leches cake with fresh Florida strawberries. Last summer she made blueberry hand pies, which simultaneously recalled South American empanadas and traditional Southern fried pies.
“I try not to be too crazy or copy something that’s already been done. I don’t want to recreate, but I want to inspire the memory,” she says. She explains a dessert she recently made for the Southern Foodways Alliance’s New South Family Supper: her take on the Tango, a popular packaged cookie in Ecuador. “It’s two vanilla wafers sandwiched with vanilla cream and covered in chocolate. It’s like a Moon Pie,” Carla says. “When I was little, I would buy them all the time.”
Carla likes finding these intersections between the cuisines of her two homes, and says it’s a way for her to feel connected to the South while expressing her own heritage. “There’s a lot of overlap between Ecuador and the American South,” she says. “For instance, I grew up eating grits, but in a different way. It’s prepared differently, but the grain itself is the same. It’s from this region, but also reminds me of home.”
| 32,420
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- Walk / Run
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- SeaWorld - 5 Minute Drive
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| 5,684
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TITLE: Are localization functors always essentially surjective?
QUESTION [6 upvotes]: Let $\mathcal{C}$ be a category and $\mathcal{W} \subseteq \text{Arr}(\mathcal{C})$ a set (or class) of arrows.
There are (at least) two notions of localization of $\mathcal{C}$ with respect to $\mathcal{W}$, a strict and a weak one.
In the strict one, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is a localization iff $\lambda$ inverts all arrows in $\mathcal{W}$ and if for every functor $G: \mathcal{C} \rightarrow \mathcal{E}$ with the same property there exists a unique functor $G':\mathcal{D} \rightarrow \mathcal{E}$ with
$$ G = G'\circ{}\lambda $$
It is not hard to see, that such a strict localization functor $\lambda$ must actually be surjective on objects.
In the weak version of localization, one merely asks for the existence of a functor $G'$ together with an isomorphism
$$ \eta: G \stackrel{\sim}{\rightarrow} G'\circ{}\lambda$$
such that for every other such pair $(G'',\eta')$ there exists a unique isomorphism $\kappa:G' \stackrel{\sim}{\rightarrow} G''$ with
$$ (\kappa \circ{} \lambda) \eta = \eta' $$
Now, if such a weak localization functor would always be essentially surjective, then I (think I) can prove that the notions of strict and weak localization are essentially equivalent, in the sense that the existence of one implies that of the other.
So, are all weak localization functors essentially surjective?
REPLY [4 votes]: The question has already been answered by Fernando and Mike, but I think it might be useful to expand on the details and provide another (partial) argument as to why localizations are essentially surjective.
As Fernando suggested, a weak localization functor
$$ \lambda: \mathcal{C} \rightarrow \mathcal{D}$$
can be factorized (like any other functor) as the composition $\lambda = \lambda_2 \circ{} \lambda_1$ of an essentially surjective followed by a fully faithful one:
$$ \begin{array}{ccccc} \mathcal{C} & & \xrightarrow{\lambda} & & \mathcal{D} \\
& \searrow && \nearrow \\
&& \mathcal{C}'\end{array} $$
One simply takes $\mathcal{C}'$ to be the category with the same objects as $\mathcal{C}$ and puts $\text{Hom}_{\mathcal{C'}}(X,Y) := \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. The functors $\lambda_1$ and $\lambda_2$ are then defined in the obvious way, with $\lambda_1$ being the identity on objects.
From the construction it's clear that
$$ \lambda \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ is an equivalence} $$
This has nothing to do with $\lambda$ being a localization. However since $\lambda$ is a localization (in the following, "localization" refers to the weak notion), it follows by easy arguments that
$$ \lambda_2 \text{ is an equivalence } \Leftrightarrow \lambda_1 \text{ is a localization } \Leftrightarrow \lambda_1 \text{ is a strong localization}$$
The first equivalence follows by observing that $\lambda_1$ already inverts all arrows in $\mathcal{W}$, which implies that for all categories $\mathcal{E}$ the diagram
$$ \begin{array}{ccccc} [\mathcal{D},\mathcal{E}] & & \xrightarrow{\lambda^\ast} & & \{ F \in [\mathcal{C},\mathcal{E}] : F(\mathcal{W}) \subseteq \text{Isos}(\mathcal{E}) \} \\
& \searrow && \nearrow \\
&& [\mathcal{C}', \mathcal{E}] \end{array} $$
of functor categories is well-defined (and commutative). Because $\lambda$ is a localization, the top arrow is an equivalence. If $\lambda_1$ is a localization, it follows that $\lambda_1^\ast$ and therefore $\lambda_2^\ast$ is an equivalence for all categories $\mathcal{E}$, which implies that $\lambda_2$ itself must be an equivalence. This shows the direction $\Leftarrow$ of the first equivalence, and the direction $\Rightarrow$ is obvious.
For the second equivalence the direction $\Leftarrow$ is obvious, and the direction $\Rightarrow$ follows easily because given any weak factorization
$$\eta:\widehat{F}\circ{}\lambda_1 \stackrel{\simeq}{\rightarrow} F$$
of a functor $F: \mathcal{C} \rightarrow \mathcal{E}$ inverting the arrows in $\mathcal{W}$, we can simply twist $\widehat{F}$ via $\eta$ to obtain a functor $\widehat{F}'$ giving an on-the-nose-factorization $F = \widehat{F}'\circ{}\lambda_1$:
$$ \widehat{F}'(X \stackrel{\phi}{\rightarrow} Y) := \eta(Y) \circ{} \widehat{F}(\phi) \circ{} \eta(X)^{-1} $$
Given these preliminary remarks, let us prove that $\lambda$ is essentially surjective, using
Mike's argument
Fernando's argument
My own (partial) argument
1. Essential surjectivity via left orthogonality to fully faithful functors
According to this nLab entry, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is essentially surjective iff it is left orthogonal to fully faithful functors, meaning that for every fully faithful functor $\mu: \mathcal{A} \rightarrow \mathcal{B}$ the induced diagram
$$ \require{AMScd} \begin{CD} [\mathcal{D},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{D},\mathcal{B}] \\ @V{\lambda^\ast}VV @V{\lambda^\ast}VV \\ [\mathcal{C},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{C}, \mathcal{B}] \end{CD} $$
is a 2-categorical pullback. Now whatever this means, it should certainly imply the following: given functors $\phi \in [\mathcal{C},\mathcal{A}]$, $\psi \in [\mathcal{D},\mathcal{B}]$ and an isomorphism $\eta: \mu\circ{} \phi \stackrel{\sim}{\rightarrow} \psi\circ{}\lambda$, there exists a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ together with isomorphism $\varepsilon_1: \alpha\circ{}\lambda \stackrel{\sim}{\rightarrow} \phi$ and $\varepsilon_2: \mu\circ{}\alpha \stackrel{\sim}{\rightarrow} \psi$.
This last property implies essential surjectivity of $\lambda$ as follows. Factorize $\lambda = \lambda_2 \circ{} \lambda_1$ as above and consider $\mathcal{A} = \mathcal{C}'$, $\mathcal{B} = \mathcal{C}$, $\mu = \lambda_2$ (fully faithful by construction), $\phi = \lambda_1$, and $\psi = \text{id}$. Because $\psi\circ{}\lambda = \lambda = \lambda_2\circ{}\lambda_1 = \mu \circ{} \phi$ on the nose, we get $\alpha \in [\mathcal{D},\mathcal{C}']$ and isomorphisms $\alpha \circ{} \lambda \simeq \lambda_1$ and $\lambda_2\circ{}\alpha \simeq \text{id}$, showing that $\lambda_2$ is essentially surjective and hence an equivalence, implying that $\lambda = \lambda_2\circ{}\lambda_1$ is essentially surjective (because $\lambda_1$ is by construction).
Therefore it only remains to see that a localization functor has the property above. But given $\mu$, $\phi$, $\psi$, and an isomorphism $\eta: \mu\circ \phi \stackrel{\sim}{\rightarrow} \psi\circ \lambda$ as in the statement of the property, because $\lambda$ inverts the arrows in $\mathcal{W}$, so must $\mu \circ{} \phi$; since $\mu$ is fully faitful and therefore detects isomorphisms, $\phi$ itself must already $\mathcal{W}$. The universal property of the localization then gives a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ and an isomorphism $\varepsilon_1: \alpha \circ{} \lambda \stackrel{\sim}{\rightarrow} \phi$ out-of-the-box. Moreover the universal property also states that $\lambda^\ast$ defines a bijection
$$ \text{Nat}(\mu\circ \alpha, \psi) \stackrel{\sim}{\rightarrow} \text{Nat}(\mu\circ \alpha \circ \lambda, \psi \circ \lambda)$$
which we can use to lift the isomorphism given by the composition
$$ \mu\circ\alpha\circ\lambda \stackrel{\mu\varepsilon_1}{\rightarrow} \mu\circ\phi \stackrel{\eta}{\rightarrow} \psi\circ\lambda$$
This lift is again an isomorphism (because of the fully faithfulness of $\lambda^\ast$), giving the desired second isomorphism $\varepsilon_2$.
2. Constructing a quasi-inverse to $\lambda_2$
Because $\lambda_1$ inverts the arrows in $\mathcal{W}$ as we've seen already, we can apply the universal property of $\lambda$ to get a weak factorization of $\lambda_1$:
$$\eta: \mu \circ\lambda \stackrel{\sim}{\rightarrow} \lambda_1$$
From this we deduce an isomorphism
$$\lambda_2\eta: \lambda_2\circ\mu\circ\lambda \stackrel{\sim}{\rightarrow} \lambda_2 \circ \lambda_1 = \lambda$$
Applying the second part of the universal property of $\lambda$, the fully faithfulness of $\lambda^\ast$:
$$\text{Nat}(\lambda_2\circ\mu , \text{id}) \stackrel{\sim}{\rightarrow} \text{Nat}(\lambda_2\circ\mu\circ\lambda , \lambda)$$
we deduce a natural transformation $\kappa: \lambda_2\circ\mu \rightarrow \text{id}$ that must be an isomorphism for the same argument used in the previous section.
3. Using the description of morphisms in localized categories
One can verify directly that $\lambda_1$ is a localization, therefore proving the essential surjectivity of $\lambda_2$, by simply writing down the objects asked for by the universal property. The verification that these objects have the required properties is not obvious, but follows easily if one knows that all morphisms between objects $\lambda(X)$, $\lambda(Y)$ in the localized category are given by images of morphisms in $\mathcal{C}$ and inverses of morphisms in $\mathcal{W}$.
Let $F: \mathcal{C} \rightarrow \mathcal{E}$ be a functor inverting $\mathcal{W}$. By the universal property of $\lambda$, we get a weak factorization
$$ \eta: \widehat{F}\circ\lambda \stackrel{\sim}{\rightarrow} F$$
which we can also read as an equivalence $\widehat{F}'\circ\lambda_1 \simeq F$ if we put $\widehat{F}' := \widehat{F}\circ\lambda_2$, proving one half of the universal property for $\lambda_1$. To show the other half, we must prove that for any two functors $F,G: \mathcal{C}' \rightarrow \mathcal{E}$ we get a bijection
$$\lambda_1^\ast: \text{Nat}(F,G) \stackrel{\sim}{\longrightarrow} \text{Nat}(F\circ\lambda_1 , G\circ\lambda_1 )$$
Since $\lambda_1$ is the identity on objects, this map is obviously injective. To see surjectivity, we would have to know that if for an $\eta:F\circ\lambda_1 \rightarrow G\circ\lambda_1$ the diagrams
$$ \require{AMScd} \begin{CD} F(X) @>\eta(X)>> G(X) \\ @VF(\phi)VV @VG(\phi)VV \\ F(Y) @>\eta(Y)>> G(Y)\end{CD} $$
commuted for all $\phi \in \lambda(\text{Hom}_{\mathcal{C}}(X,Y)) \subseteq \text{Hom}_{\mathcal{C}'}(X,Y) = \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$, then they commuted for all $\phi \in \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. But given a description of the morphisms in $\mathcal{D}$ as compositions of images of morphisms in $\mathcal{C}$ and inverses of images of morphisms in $\mathcal{W}$, this would be immediate.
Such as description would follow for instance from the explicit construction of the localized category via generators and relations (which may not exist due to size issues).
| 110,314
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Flexible heat pipes can be installed in either gravity-aiding or against-gravity orientations. The against-gravity orientation is possible because of Thermacore products."
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Wedding and event design, hire + styling servicing Noosa and Sunshine Coast surrounds
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Chegg does not guarantee CDs, access codes, or lab manuals with this book.
PUBLISHER: Bethany House Publishers
AUTHORS: Tracie Peterson, Judith Miller, Judith McCoy-Miller
A Tapestry of Hope:... Show more»
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Is it, self-sabotage?
Change is scary even when it is planned. The unknown of what lies ahead, gives our imagination an opportunity to show us just how creative you can really be. When I am scared but excited, I tend to imagine a beneficial outcome. But when I am just plain scared, the imagined outcome tends to be not so great. I do think that the more energy I put into imagining a ‘good’ result, the more likely it is that I get one.
Do you know why? I do. It is because my thoughts direct my actions. When I am fearful, I tend to make reserved choices and keep myself small, while at the same time being ready to fight at any given moment. I am on edge with myself and the people I encounter in my life when I am fearful. I have realized this is not the energy I wish to share with the world.
Instead I have chosen to look at life as an adventure in which my attitude, my imagination and my state of mind really do manifest the outcome I desire. When I am scared, yet excited about the direction I am headed, I can decide to make the best of the day and choose to bring good energy to my experiences. I can decide with my everyday choices where I direct my energy and what kind of energy I am emitting.
I can choose to be kind and inquisitive when I encounter people or situations unfamiliar to me. I can choose to believe in the good in the world around me. I can choose to know that the path that lies ahead of me is exactly where I am supposed to be. I can choose to do my best through my everyday actions to make the day enjoyable or not…
So, even though I tend to be somewhat resistant to change, I know that nothing is permanent in our lives. We can get comfortable and stuck in our ruts that do not challenge us, which is a nice rest from constantly striving or doing. Yet, when you hear the call of your heart, those ruts can be hard to get out of. It takes a great deal of effort, persistence and determination to get going. Which is a lot of work; gosh darn it! Why can’t change just be easy?
I believe change can be easy when I accept it and let it be the divine guidance it was meant to be. But, no… that’s just not how I roll somedays. Somedays I get upset about the changes forced upon my daily life that may require more effort and energy then I really want to give at that particular time or day. I can be as stubborn as my beloved bulldog Gus about change and I know it. Especially if I had planned to do something and life has different ideas for me.
Stacy and I had been talking about our common goals, that we are trying to accomplish together, yet we do them in our own ways. Publishing our book series is one of our goals, but we also share our life goals together. When I am thinking of making a change to my diet, or my lifestyle, I call my friend Stacy and we talk about it like sisters do. We process outcomes together, we offer support and personal experiences that we think will help each other.
Recently I had told her of a personal goal that I was not sure I would be able to accomplish but I know in my heart would be for the best. She said “Okay! Let’s do this together so we can support each!” this showing of sisterly support made me feel like “Yes. We can do this!” Because we will have each other to talk to, when the test of life and time come around like it always does.
The very next day after this conversation we saw a statement that stated “Resistance to change, is really self-sabotage”
When I think about it, I think it is true. Can you imagine if we all still acted like we did in high school while well into your forties?! I know a few people like this. People who like how it was, people who have learned all they needed to learn (or knew it all at age 18) so they feel they are just fine, thanks. Some folks just choose not to grow or change! We must learn to accept that, but we do not have to do the same.
Although, in my experience, the people who refuse to grow get very uncomfortable watching you grow. And that’s okay. I have decided I will not dim my light because it is shining in their eyes. I have experienced more of what life has to offer by taking risks and allowing myself to grow and change. I would not change any of the challenges of change that life has asked of me. It has made me into who I am today.
You know what? I really like who I am! I love that I have been changed by adventure, challenges, heartbreak, motherhood, friendship, and even my career path that all got me here today. My journey has softened my edges, given me patience and wisdom to know that I always have something to learn. I have learned that my resistance to change is futile and just makes the journey harder on me then it has to be! I have learned that if I want something new in my life I must do and accomplish new things and apply new effort.
So, I agree, resistance to change can be self-sabotage that keeps me from learning the lessons I need to grow. I don’t have to change all at once, I can usually choose to change in my own way, at my own pace. What makes change easier for me is putting my creative imagination to good use by expecting and acting in ways that will foster a favorable outcome. I can choose to make the most of it or go kicking and screaming the whole way. But, purposely putting out good vibes into the direction I am going, seems to make it a more comfortable journey.
Shine your light and vibe on, my adventurous friends! You CAN do this, and you GOT this! I believe in you.
Adventure awaits! Don’t be scared, be excited!
Wishing you an abundance of joyful blessings,
Emy Minzel ~ Adventure Sister
EmyforHouse.com
@EmyforHouse15A
Follow the Adventure Sisters on Facebook!
Stacy Crep ~ Adventure Sister
Photo credit by International Impact / Google Images
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Slim Fast Reviews – Diet & Shakes Does Slim Fast Work?
Slim-Fast 3-2-1 Plan, Ready To Drink Shake, Cappuccino Delight, 11-Ounce Cans (Pack of 24)
- Safety/build quality: 97/100
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Does Slim Fast Work?
Slim Fast Low Carb is a ready to go chocolate flavor drink that’s a great meal replacement for dieters who are looking to slash their carbohydrate intake and lose weight. This drink is creamy and sweet and tastes much better (in my opinion) than other leading low carb drinks, such as the Atkins protein shake, or EAS diet shakes.
Slim Fast Diet Effectiveness:
Each can comes in an 11 ounce size that is a perfect grab and go meal. Slim fast low carb drink is quite filling and is made of natural and artificial flavors with two grams of net carbohydrates per serving. Each can has 190 calories with 80 calories from fat, and has 20 grams of protein.
Slim-Fast Low-Carb Drink is designed for a low carb diet and is a Kosher food. The Creamy Chocolate Slim Fast comes in 11-Ounce Cans in 4-Count Boxes and can be purchased in a six pack.
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Here are some of the best features of the Slim Fast Review – Chocolate Low Carb Drink:
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Slim Fast Review – Nutrition Facts:
Fat Free Milk, Water, Sugar, Canola Oil, Gum Arabic, Calcium Caseinate, Cellulose Gel, Coffee Powder, Natural and Artificial Flavor, Potassium Phosphate, Soybean Lecithin, Cellulose Gum, High Fructose Corn Syrup, Mono and Diglycerides, Carrageenan, Maltodextrin, Sucralose (a Nonnutritive Sweetener), Dextrose, Citric Acid and Sodium Citrate.) and Cholecalciferol (Vitamin D3).
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Can I really get balanced nutrition from the Slim·Fast 3·2·1 Plan?
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\section{Stream Decoding}
\label{sec:streamdecoding}
We have so far considered the case where frame boundaries are known exactly.
While there are practical cases involving single-frame transmission where
this is true, exact frame boundaries are often unknown.
The MAP decoder can handle such cases by changing the initial conditions
for \eqref{eqn:alpha} and \eqref{eqn:beta} and choosing appropriate state
space limits.
This obviates the need for explicit frame-synchronization markers as used
in conventional communication systems, and can therefore reduce this overhead.
The approach presented here is in principle similar to that used in
\cite{dm01ids} for `sliding window' decoding.
However, there are some critical differences which we explore further in
Section~\ref{sec:stream_comparison}.
\subsection{Choosing End-of-Frame Priors}
\label{sec:eof_priors}
Consider first the common case where a sequence of frames is transmitted in
a stream.
The usual practice in communication systems is for the receiver to decode
one frame at a time, starting the decoding process as soon as all the data
related to the current frame is obtained from the channel.
In this case, the current received frame is considered to be
$\vec{Y}\sw{m_\tau^{-}}{\tau+m_\tau^{+}}$,
which may include some bits from the end of the previous frame and start of
the next frame.
The end-state boundary condition for \eqref{eqn:beta} can be obtained by
convolving the expected end-of-frame drift probability distribution with
the start-state distribution:
\begin{equation}
\beta_N(m) = \sum_{m'} \alpha_0(m') \Phi_\tau(m-m')
\text{.}
\end{equation}
Note that in general this distribution $\beta_N(m)$ has a wider spread than
$\Phi_\tau(m)$.
As discussed in Sections~\ref{sec:drift_limits} and \ref{sec:summation_limits},
the choice of state space limits depends on the expected distribution of drift.
For limits involving the whole frame, the distribution used is $\Phi_\tau(m)$,
which assumes that the initial drift is zero.
The assumption does not hold under stream decoding conditions, where the
initial drift is not known \emph{a priori}, although its distribution can
be estimated.
The uncertainty in locating the start-of-frame position increases the
uncertainty in locating the end-of-frame position, resulting in a wider
prior distribution for the end-state boundary condition $\beta_N(m)$.
Therefore, any limits on state space determined using $\Phi_\tau(m)$ will
be underestimated.
The severity of this error depends on the difference between $\beta_N(m)$
and $\Phi_\tau(m)$, which increases as channel conditions get worse.
For stream decoding, therefore, it is sensible to recompute the state space
limit $M_\tau$ at the onset of decoding a given frame, using $\beta_N(m)$
in lieu of $\Phi_\tau(m)$.
Doing so avoids underestimating the required state space, and implies that
for stream decoding, the state space size will change depending on how
well-determined the frame boundaries are.
After decoding the current frame, we obtain the posterior probability
distribution for the drift at end-of-frame, given by:
\ifdraft
\begin{align}
\prob{ S_{\tau}=m \;\middle\vert\; \vec{Y}\sw{m_\tau^{-}}{\tau+m_\tau^{+}} }
& = \lambda_N(m) / \prob{ \vec{Y}\sw{m_\tau^{-}}{\tau+m_\tau^{+}} }
= \frac{ \lambda_N(m) }{ \sum_{m'} \lambda_N(m') }
\text{.}
\end{align}
\else
\begin{align}
\prob{ S_{\tau}=m \;\middle\vert\; \vec{Y}\sw{m_\tau^{-}}{\tau+m_\tau^{+}} }
& = \lambda_N(m) / \prob{ \vec{Y}\sw{m_\tau^{-}}{\tau+m_\tau^{+}} } \nonumber\\
& = \frac{ \lambda_N(m) }{ \sum_{m'} \lambda_N(m') }
\text{.}
\end{align}
\fi
The most likely drift at end-of-frame can be found by:
\begin{align}
\hat S_{\tau}
& = \argmax_{m} \frac{ \lambda_N(m) }{ \sum_{m'} \lambda_N(m') }
= \argmax_{m} \lambda_N(m)
\text{.}
\end{align}
As in \cite{dm01ids}, we determine the nominal start position of the next
frame by shifting the received stream by $\tau + \hat S_{\tau}$ positions.
The initial condition for the forward metric for the next frame,
$\hat\alpha_0(m)$, is set to:
\begin{align}
\hat\alpha_0(m)
& = \frac{ \lambda_N(m + \hat S_{\tau}) }{ \sum_{m'} \lambda_N(m') }
\text{,}
\end{align}
replacing the initial condition for \eqref{eqn:alpha} reflecting a known
frame boundary.
\subsection{Stream Look-Ahead}
\label{sec:lookahead}
Taking advantage of the different constituent encodings in TVB codes, the
MAP decoder can make use of information from the following frame to improve
the determination of the end-of-frame position.
We augment the current block of $N$ symbols with the first $\nu$ symbols from
the following block (or blocks, when $\nu > N$), for an augmented block size
$N' = N + \nu$.
The MAP decoder is applied to the corresponding augmented frame.
After decoding, only the posteriors for the initial $N$ symbols are kept;
the start of the next frame is determined from the drift posteriors at the end
of the first $N$ symbols, and the process is repeated.
Consider the latency of the MAP decoder to be the time from when the first
bit of a frame enters the channel to when the decoded frame is available.
The cost of look-ahead is an increase in decoding complexity and latency
corresponding to the change in block size from $N$ to $N'$.
The effect on complexity is seen by using terms corresponding to the augmented
block size in the expressions of Table~\ref{tab:complexity}.
The latency is equal to the time it takes to receive the complete frame and
decode it.
Look-ahead increases the time to receive the augmented frame linearly with
$\nu$ and the decoding time according to the increase in complexity.
The required look-ahead $\nu$ depends on the channel conditions and the
code construction.
In general, a larger value is required as the channel error rate increases.
We show how an appropriate value for $\nu$ can be chosen for a given code
under specific channel conditions in Section~\ref{sec:results_stream}.
Typical values for $\nu$ are small ($\nu < 10$) for good to moderate channels
($\Pri,\Prd < 10^{-2}$).
The required look-ahead increases significantly for poor channels:
the example in Section~\ref{sec:results_stream} requires $\nu=1000$ at
$\Pri=\Prd = 2\times10^{-1}$.
\subsection{Comparison with Davey-MacKay Decoder}
\label{sec:stream_comparison}
A key feature of the Davey-MacKay construction is the presence of a known
distributed marker sequence that is independent of the encoded message.
This allows the decoder, in principle, to compute the forward and backward
passes over the complete stream.
However, to reduce decoding delay, the decoder of \cite{dm01ids} performs
frame-by-frame decoding using a `sliding window' mechanism.
The `sliding window' mechanism seems intended to approximate the computation
of the forward and backward passes over all received data at once.
This approach is similar in principle to ours when stream look-ahead is used;
however, there are some critical differences which we discuss below.
In \cite{dm01ids}, the starting index for a given frame is taken to be the
most likely end position of the previous frame, as determined by the Markov
model posteriors.
This is the same as the approach we use in Section~\ref{sec:eof_priors}.
However, in \cite{dm01ids}, the initial conditions of the forward pass are
simply copied from the final values of the forward pass for the previous frame.
This is consistent with the view that the `sliding window' mechanism
approximates the computation over all received data at once, but contrasts
with our method.
In Section~\ref{sec:eof_priors} the initial conditions of the forward pass
are determined from the posterior probabilities of the drift at the end of
the previous frame.
These drift posteriors include information from the look-ahead region and
from the priors at the end of the augmented frame, which were determined
analytically from the channel parameters.
Observe that in the `sliding window' mechanism of \cite{dm01ids}, the backward
pass values cannot be computed exactly as for the complete stream.
Instead, the decoder of \cite{dm01ids} computes the forward pass for some
distance beyond the expected end of frame position, and initializes the
backward pass from that point.
The suggested distance by which to exceed the expected end of frame position
is `several (e.g.\ five) multiples of $x_\textrm{max}$', where $x_\textrm{max}$
is the largest drift considered.
The concept is the same as the stream look-ahead of
Section~\ref{sec:lookahead}.
However, we recommend choosing the look-ahead quantity $\nu$ based on
empirical evidence (c.f.\ Section~\ref{sec:results_stream}).
It is claimed in \cite{dm01ids} that the backward pass is initialized from the
final forward pass values; the reasoning behind this is unclear, and does not
seem to have a theoretical justification.
We initialize the backward pass with the prior probabilities for the drift
at the end of frame, as explained in Section~\ref{sec:eof_priors}.
\subsection{Initial Synchronization}
The only remaining problem is to determine start-of-frame synchronization
at the onset of decoding a stream.
This can be obtained by choosing state space limits $M_\tau$ large enough to
encompass the initial desynchronization and by setting equiprobable initial
conditions:
$\alpha_0(m) = \beta_N(m) = \frac{1}{M_\tau} \quad \forall m$.
Previous experimental results \cite{dm01ids} have assumed a known start for
the first frame, with the decoder responsible for maintaining synchronization
from that point onwards.
We adopt the same strategy in the following.
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TITLE: Identity matrix
QUESTION [0 upvotes]: Edited: Suppose $M$ represents a linear transformation from $\mathbb R^3\to \mathbb R^3$, where $M = \left( \begin{smallmatrix} a&1&0\\ 0&a&1 \\ 0&0&a \end{smallmatrix} \right)$. What are the bases for the domain and range s.t. $M$ is the identity?
(Sorry about the previous edition.)
REPLY [2 votes]: Thanks for improving the formulation of the question. It's still not really clear, since $M$ is a given matrix and it makes no sense to ask when it is the identity. A clearer formulation of what I think you mean would be:
Suppose a linear transformation from $\mathbb R^3$ to $\mathbb R^3$ is represented by the matrix $M = \left( \begin{smallmatrix} a&1&0\\ 0&a&1 \\ 0&0&a \end{smallmatrix} \right)$ if the canonical basis is used for both the domain and the codomain. What are bases of the domain and the codomain such that the matrix representing the transformation with respect to these bases is the identity matrix? (Note the lack of a definite article on "bases", which in your version seems to imply that there is a unique solution, which is not the case.)
The are infinitely many pairs of bases with this property. Two particularly simple ones are obtained if you use the canonical basis for one of the spaces and only look for a new one for the second one. This is particularly straightforward if you only change the basis of the codomain: Since the columns of the matrix are the images of the basis vectors of the domain, represented in the basis of the codomain, the matrix will be the identity matrix if you choose the column vectors of $M$ as the basis vectors of the codomain.
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Higher ed cheerleader
Robert Caret, the new University of Massachusetts president, wants to increase investment in the state's university system--both emotional and monetary--of taxpayers, elected officials and alumni
The University of Massachusetts matters. It matters for a lot of reasons and to a lot of people—parents, students, lawmakers, government officials, and taxpayers with no connections except that their money supports the system and they all want a bang for their buck.
More than 68,000 students, full- and part-time, mostly Massachusetts kids, attend one of UMass’s five campuses, and nearly 14,000 graduated last year. Nearly 300,000 UMass alumni live in Massachusetts. You’d think with those kinds of numbers, fiscal, physical, and civic support for the system would be easy to find. But when you’re sharing the education spotlight with the likes of Harvard, MIT, Tufts, and other world-renowned institutions, you struggle for attention and support and carry something of an inferiority complex.
Robert Caret knows that. Caret, who took over as UMass president from Jack Wilson on July 1, has spent his entire professional life in the public higher education world: as a professor and administrator at Towson State University in Maryland for two decades, followed by nearly a decade as president of San Jose State University in California, and then the last eight years back at Towson as president.
Caret, who earned a PhD in organic chemistry at the University of New Hampshire, has a booming voice that complements his oversized persona and assertive and confident manner. He’s part politician, like past UMass president William Bulger, and part academic, like Wilson, the man he is replacing. He’s been recognized for his success in growing Towson and San Jose State in size, academic scope, and diversity, while at the same time increasing minority graduation and retention rates. At Towson, Caret eliminated the gap between minority and white graduation rates, one of only 11 colleges or universities nationwide that can claim such success. It’s that national stature that made him Gov. Deval Patrick’s choice as Wilson’s replacement. The perception that Caret can also till the rocky terrain of Massachusetts politics like a pol—without having the baggage of actually being one—is a bonus.
Caret, 63, hails from Biddeford, Maine, a hardscrabble industrial town 20 miles south of Portland. He was one of the first members of his family —indeed, he says one of the first in his town—to go to college. His journey has made him a passionate advocate for the role publicly funded colleges and universities play in educating the country’s students. It’s that passion that UMass trustees and Patrick hope can tap the minds, souls, and wallets of potential supporters, while competing against some of the world’s most highly regarded private institutions.
One area Caret sees as essential as a top administrator is ensuring the school blends in with the community where it resides. In San Jose, he developed numerous “town-gown” committees to address the needs of the area’s diverse residents as well as those of a school going through growing pains. His approach upon returning to Towson was similar, trying to become a magnet for the majority-minority population in Baltimore.
Sitting in his temporary office in Post Office Square one morning in early May prior to taking the presidential reins, Caret marvels at his water view in a city he first became familiar with as an undergraduate at Suffolk University. More than four decades later, Caret is coming back to Boston on a mission to energize support for five very different campuses that make up one system that’s never received the kind of recognition and support its backers think it deserves.
Recent years have been particularly brutal, with student fees soaring to offset declining state aid. Tuition has remained virtually flat over the last decade, a little over $1,600 per year. But fees have gone up more than 300 percent in that same period, from $3,069 in 2001 to nearly $9,500 this year. With the fiscal 2012 budget of $429 million the same as last year—but down from nearly $480 million just a few years ago—and with the loss of federal money, the system is facing a $55 million deficit. To cover half that gap, UMass trustees just approved another 7.5 percent hike in fees for the upcoming school year, an average increase of more than $800. Layoffs and other reductions are being eyed to cover the remainder.
Where the state once picked up nearly half the cost of educating UMass students, the burden now falls squarely on the student. That is one major area that Caret emphatically says he will change. He says it in a way that precludes doubt. But Caret is no Pollyanna, and he did not come into this situation blind. He knew the state of the economy nationally and familiarized himself with UMass’s funding struggles before accepting the job. He says he’s had general conversations with Patrick and Education Secretary Paul Reville about “shared responsibility,” the idea that the burden for educating students be spread more equally among many shareholders as a public obligation.
I sat down with Caret to talk about his educational philosophy, football, his knowledge of UMass, and his goals. Over the course of our conversation, a number of subjects were addressed two and three times. I’ve rearranged the order of some of his answers to avoid jumping around too much for the reader. We began by talking about why he’s remained committed to public education all these years. (Suffolk was his only foray at a private school.) This is an edited transcript of our conversation.
—JACK SULLIVAN
commonwealth: When you look at public versus private higher education, outside of the positive affordability factor that most state schools have, is there a broader value to public institutions?
caret: You hit the obvious point—the price differential. The less obvious is the quality of the educational environment, or the climate, the ambience, the environment itself. If you go to a public institution, in general, you will find an institution that is much more diverse because it has many more pathways into it, and it is often taking many more students from many more walks of life. So the peer group is a much broader peer group, much like the peer group you will be dealing with in the real world when you go to work, as opposed to, sometimes, a subset of that peer group that all group up in the same private, preparatory schools and wind up in the same country clubs and the same board rooms. And there are plusses to that, I think, in terms of the society we are trying to create. I don’t say all privates are one way and all publics are the other way. But just in a general sense, you are going to get a much more diverse peer group [in public higher ed institutions] that represents the diversity of ideas, race, ethnicity, and sexual orientation that you find in society. There’s an excitement to that. Our job as a public university is to provide the alternative choice to the private universities. Public universities are not just chosen because of affordability but are chosen because of quality and a different ambience than you would get at a private school. We don’t want to compete with them, but to complement them in terms of civic society.
cw: When you first went to San Jose State as president and then Towson State, did you find that kind of diversity in those schools or was that something that you had to create? You said that at both of those schools one of your goals was to increase the minority graduation rate and retention rate, and at Towson State, one of the ways you did that was to recruit more high-performing, high-achieving minorities.
caret: Well, they are very different schools. San Jose was an extremely diverse environment when I arrived. So you had an extremely complex ethnic, racial, cultural background there, and we had to put in structures to really deal with the complexity of that population. In California, you have this liberal, open society, but the African-American students at San Jose State, because they’re only about 4 to 5 percent [of the student population], really felt, to some extent, disenfranchised by the campus and the community because they just didn’t have enough voice. So what you do in those kinds of situations is make sure these organizations, these student groups, have a voice. And I did that through creating ethnic town-and-gown committees. I had an African-American town-and-gown committee, an Asian-American, and a Hispanic-American committee.
Towson was, when I arrived, probably 90 percent, 95 percent white. We kept opening up the doors more and more and now we’re probably only 80 percent white, but it’s still largely a white campus. When I first got back, I’m asking a thousand questions and one of the questions was, “We just took in 2,200 freshmen, how many came from the Baltimore city public schools?” Out of 2,200, 25 came from the Baltimore public schools.
cw: Not 25 percent?
caret: No, 25 students. We had a lot from Catholic schools and religious schools in the city, private schools, but not the publics. We’re only a quarter mile from the city line and only four miles from downtown [Baltimore], and I said, “That’s crazy. Is anybody doing anything about this?” They said, “Well, we don’t need to. We’ve got 20,000 applicants. We don’t do town halls down there, we don’t do college fairs.” I said, “It’s the nucleus of the state. It’s the political hub of the state. We’ve got to reach out to the city.” So we opened up and we did a top 10 percent program—we guaranteed the top 10 percent of the high school kids in the city and in the county no charge to apply and up to a $4,000 scholarship, if you needed it, to match the financial aid. And guaranteed admission, regardless of your grades —if you were in the top 10 percent, you were in. And we went from 25 to 200 in one year, and that process has continued up until today. If you’re in a region that’s 30 percent, 40 percent African-American, and you’ve only got an 8 percent African-American campus, you’re not doing your job. We need to get ready for the world that’s coming as opposed to the world that’s here.
cw: What was the draw for you with public versus private higher ed?
caret: When I got out of grad school—every grad school in the country has niches that it’s developed—and UNH in chemistry had two large niches: one was people went on and became faculty someplace, and the other one was the pharmaceutical industry. I had taught high school, I liked teaching, so I decided to go the faculty route. Most of the positions I was offered were non-tenure track, so I wound up taking the position at Towson, which was also non-tenure track. But Towson was growing like crazy at the time and I just felt I had a really good shot there as a tenure track. The next year, they had two positions open up and I applied and got one of them, so I just started off there, and 21 years later, I’m still there. So I just wound up being in the public sector. It wasn’t necessarily planned.
cw: So you are not one of those people who thinks that tenure is a bad thing?
caret: I am one of those people who can argue both sides. There’s a part of me that understands, because of all the legal protections we have, that tenure is not as necessary as it was 100 years ago. But on the other side—and I unequivocally can give you examples to prove this—what tenure does is protect universities from outside intrusion. The example I like to use is when [Ronald] Reagan was governor of California, and Clark Kerr, who is considered one of the gurus of higher education in all of history, was in charge of Berkeley and the students were protesting the Vietnam War. Reagan fired Kerr because he didn’t think he was hard enough on the students. Kerr left, but Berkeley never changed because you couldn’t fire the faculty and you’re not going to change a university by just changing the president. I can assure you that if Reagan could have fired the faculty, he would have, and Berkeley would have changed. We need a place in society where government can be criticized without fear of retribution, and universities have often played that role. But I do believe if we continue tenure, which I think is important because of the protection it provides us, we need to be substantively involved to make sure that faculty are living up to what we expect them to do, both pre-tenure and post-tenure. So post-tenure review, on a periodic basis with substantive outcomes, is something that I think is necessary and desirable.
cw: So what would be the value of tenure if you do that?
caret: Well, [dismissal of a tenured faculty member would] require a substantive negative [finding]. It’s not just that you don’t like the person’s personality or you don’t like their political point of view, but that it’s actually destructive to the educational environment, in the classroom, and in our buildings. I think the main reason for tenure, from an institutional perspective, is to protect the institution, but the way you are protecting the institution, what you’re doing there, really, is protecting the academic freedom to speak your mind within the framework of your discipline. And that’s where it’s most important.
cw: Your two prior jobs were at places that were, for the most part, single campuses. How do you apply the singular vision that you’ve had for the universities you previously ran to this much wider, multi-campus audience?
caret: I think each of them will be somewhat unique, so that, together, the system is serving a variety of needs as well. I do think both Lowell and Boston, and Boston in particular, want some housing and want to have some critical mass of students so it’s not just everybody comes [to campus each day] and everybody leaves [at the end of the day].
cw: From the beginning, UMass Boston was opened for the minority student and the low-income urban student who can’t afford room and board, who can’t afford going away to school, who has to work nearby. And also for the single parent who wants to finish their degree, for the older student who wants to return to college. It would seem to me that the building of residential halls is going to take away from that mission.
caret: No, I don’t think so. I think what will happen will be much like what happened at San Jose—it was a bicameral campus. Half the population was traditional age, the other half was average age of 22, 23, largely commuter, largely working, coming to school sometimes full-time, sometimes part-time. I think you’ll see at UMass Boston the same kind of thing. They unequivocally understand their history, if you talk to the faculty and staff there. I don’t think you’ll ever see them give up on their original mission because they all too intimately believe in it, which is the access mission for those broad populations you just described.
cw: You have a history at San Jose State and Towson State, especially at Towson, of being a very big supporter of the athletic department and the sports teams. UMass Amherst has just announced that it is going to be going 1A with its football program, moving up from the championship subdivision to the bowl subdivision. Is it coincidental that this is happening with your arrival?
caret: Yeah, it’s just coincidental, but I am supportive of it. I’m really a social animal—I’m not a jock per se. I go to a lot of athletic events just like I go to lots of other events where I socialize with people. I like athletics, I don’t live and breathe it. I can’t quote you names and numbers. That’s not my thing. I like going to games. Amherst is the flagship campus, and people hate that term, but it is the flagship campus and it is the primary public research university in the state. We need to develop it so that it’s better and better at what it does. And part of that is going to be characteristics of the campus, including what sports it plays and at what level and what peer group is it in. The move to the FBS [Football Bowl Subdivision] level from FCS [Football Championship Subdivision] is a move that I fully understand for that school.
cw: But if they’re going to play their home games, as was announced, at Gillette Stadium in Foxborough, which is 95 miles from Amherst, how does that benefit the campus and how does that make a connection to the campus?
caret: Everybody would prefer to have a 30,000-seat stadium on the campus. If you made that a criterion for doing this, it wouldn’t happen because there’s no way we could afford to do that—because that is done largely with auxiliary money and, hence, largely on the backs of student fees. We’re just not in a situation, budget-wise, to take on that kind of debt. So the reality of the situation is, Gillette becomes a necessary partner in the mix in order to get the kind of stadium we need. Lots of the students at Amherst go home on weekends, so a lot of them are already in the Boston suburbs on weekends. There’s a huge alumni base, 265,000 alumni, many of them in the Boston metropolitan region. So what the school is doing is drawing on the alumni base, drawing on students who are already in the region on weekends, and we’ll use busing as we have in the past to get students from the campus back and forth. If it’s successful, maybe a stadium comes someday in the future. But it can’t be a fixed variable today because it would just end the discussion, so I think it will either be successful or it won’t.
cw: When you spoke with Gov. Patrick about the budget, were you given any commitments?
caret: No, no commitments. I’ve talked to the governor a couple of times, I’ve talked to [the education] secretary, Paul Reville, several times, and there are no commitments. I don’t think you’re going to see, in Massachusetts, any precipitous funding change in where we are. My job is to continue to educate the governor and the Legislature as to how important UMass is. I’ve been using the term, what is our “shared responsibility” to the citizens of Massachusetts and to the Commonwealth itself. By shared responsibility I mean, what do we think is an appropriate mix of expense to the student and to his or her family, to the state, and to the federal government. I don’t think Massachusetts is where it needs to be. I don’t think 20 cents on the dollar, for the UMass system, is an appropriate level of funding from the state. My goal for the last 10 years has been to get the states back to at least 50 cents on the dollar and then we can move from there. But I’m open to that debate. What is the appropriate percentage? The three primary funding sources for a student are the state, the student himself or herself with their tuition, and the federal government behind the scenes providing financial aid to the student that comes to us. Those are the three big pieces of revenue that we use. The feds have been, even though it’s diluted by inflation, pretty constant, the state has decreased and therefore tuition has increased, but that’s pretty much the mix we have today. I have pointed out that I do think we need to look at that shared responsibility piece and see where Massachusetts used to be, where it would like to be, and can we get there. I’m pretty sure, other than because of the budget crisis, Massachusetts is not where it thinks it should be. It’s where it has to be at the moment, and we need to get it back to where it should be.
cw: Do you believe that college campuses, as one way to raise funds, should be in competition with the private sector, for instance, in research and development and patents? UMass Worcester has a nonprofit corporation, Commonwealth Medical, that does audits and analysis of pharmaceutical purchases for Medicaid and Medicare. They also provide prison health care, and they are in competition with private companies that are bidding for it. Is that the purview of a college, to get state funds to go into a state entity?
caret: I’ll answer that two ways. One is, I think absolutely we need to be involved in research and consulting. The state invests $500 to $600 million in us every year on the operating side, not counting on the capital side, and there’s no reason they shouldn’t benefit from the expertise on the campuses. I think having us there as a research arm, as a consulting arm, is an important tool. It gets touchier when you start to compete with private business. I understand if I had a private business and I was competing with a state entity that is being subsidized and so their costs are much lower, that I would have trouble competing. So what I’ve tried to do—and I haven’t had this dialogue in Massachusetts, this is one of those things where people could start yelling at me—but what we’ve always tried to do in the states that I’ve been involved in is that when we get involved with that kind of enterprise, we do it with an educational mix to it, too. As one of my mentors used to say, just because you can make money selling cars on a campus, it doesn’t mean you are going to make money selling cars on a campus. Because our job is not to make money, we need to be doing things on campus, whether they are for profit, or not for profit, for the benefit of the education and the research mix that goes on on campus.
cw: Where does the endowment picture fit into this? UMass has the 142nd largest endowment in the country [$520 million]. For a system this size, it is undersized. What do you do to engage UMass alumni, because they’re really not an engaged group?caret: If you go back and really look 40 years ago, 50 years ago, when people started fundraising [at public colleges and universities] for the first time, states didn’t want us to do it because they felt companies and people are paying their taxes and those taxes are underwriting these campuses at 70 cents to the dollar, and we don’t want you to go hassle them for more money. A lot of the companies in those days, because I was raising money in those days, actually had policies in place where they didn’t give to publics, because they felt that their tax dollars were going to the publics and their endowment dollars or their foundation dollars should go to the privates. What you’ve got to do is, from the day you bring [students] in, develop strong orientations, a strong belief and understanding of the campus that they’re becoming part of, and, for the traditional students at least, tell them about the history of the campus, teach them the fight song. The most important thing is to get them attached to the campus, whether it’s through a student club or a major, or through volunteerism. What we need is to ramp that up by several thousand percent and get more kids engaged so that when they do leave its not just, “Hey, I paid you, you gave me a decent education, I liked a couple of my faculty, I’m off with my life,” but that when it’s homecoming time, or when you get a call to come back to the campus for some kind of an alumni event, they want to be there. And that’s the beginning. The biggest piece is that 4- to 6-year period that they’re with you, to give them the kind of experience that they reflect back and say “That was fun, I’d love to do that again, I’d love to be part of that again.”
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| 274,887
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SVAY RIENG
The Safe House, a women's shelter at an undisclosed address in Svay Rieng, battles rising domestic violence by giving hope to the province's
most at-risk victims of rape and battery
Photo by:
THOMAS GAM NIELSEN
Educational boards and posters are handed out in Svay Rieng to change people's attitude towards domestic violence.
FOR more than a year, a shelter simply called the Safe House has been helping the most vulnerable victims of rape and domestic violence in Svay Rieng province, where, like most areas outside the capital, social services are limited.
"There has been a big difference [since we opened]. Now, we can help these women with full security, and they do not have to feel afraid," said Nget Thy, project manager at the Cambodian Centre for the Protection of Children's Rights, the organisation that runs the Safe House.
The idea for the Safe House, which is located at an undisclosed address, was born after an original open-door shelter proved unable to protect victims from their abusers.
"When we received victims of rape or domestic violence, the perpetrator would sometimes come and complain and want to take the victim from our shelter," Nget Thy said.
"Also, brothel owners would pass by and want to get back their former sex workers," he added.
Since moving to their new address, the organisation has been able to provide security for at-risk victims of domestic violence or rape who have been targeted by perpetrators trying to track them down.
According to Lourdes Autencio, coordinator of counter-trafficking in persons at the Asia Foundation - which is a major donor to the Safe House - the centre is particularly impressive as it evolved in response to local needs rather than because of an explicit donor request.
"It was born out of the need of those high-risk individuals," she said, adding that the Asia Foundation gives technical support to the Safe House.
Growing need
A report released Tuesday by the Women's Affairs Ministry said that a quarter of Cambodian women were the victims of abuse, while young girls have increasingly become targets of sexual assault.
In Svay Rieng, the rights group Licadho has reported 14 cases of rape this year, including those involving seven children, and 21 cases of abuse.
At the moment, five females between ages 12 and 25 years are living at the Safe House, referred to the program by local authorities and other NGOs.
The organisation counsels the women and tries to create an atmosphere that will allow them to take back control of their lives. "These are very sensitive cases, and the main point is to make them feel safe," Nget Thy said.
Clients can live at the Safe House for up to six months, and afterwards are given vocational training or help finding a job.
At the Cambodian Women's Crisis Centre, general director Nop Sarin Sreyroth welcomed Safe House initiative.
"It is good that they have the Safe House in Svay Rieng, because even though they could refer some of victims to us, we cannot take all cases," she told the Post.
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gant)..
Taken from the ACC website.
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TITLE: A property regarding intervals
QUESTION [0 upvotes]: While I was solving a problem on TopCoder I used the following assumption. I have n intervals: $ [a_1,b_1], [a_2,b_2],...,[a_n,b_n]$ and a number $T$ such that:
$$ a_1 + a_2 + ... + a_n \leq T \leq b_1 + b_2 + ... + b_n $$
Then there is a choice of numbers $x_1,x_2,...,x_n$, with $x_i \in [a_i,b_i] $, such that:
$$x_1 + x_2 + ... + x_n = T $$
For example, let's say I have to intervals: [1.6 , 2.3] and [1.7 , 2.7]. If T = 4.5 ,
with 1.6 + 1.7 < 4.5 < 2.3 + 2.7, then I have 4.5 = 2.2 + 2.3 , with 2.2 belonging to the first interval, and 2.3 to the second. The solution is not unique.
Now, I know that this the claim I stated is pretty obvious and intuitive. One can easily find a greedy algorithm that can find a choice of x's. We can start with $x_1 = a_1, x_2 = a_2, ... , x_n = a_n $, and then increment each x as needed to attain the $T$ value.
For example, I begin with T1 = 1.6 + 1.7 = 3.3. I didn't attain the T value! I look at the first x, and I try to increase it. T2 = 2.3 + 1.7 = 4.0. I haven't reached the T value and I can no more increase the first x. So I proceed to the second x. By increasing the second x to 2.2, I now have 2.3 + 2.2 = 4.5.
Intuitively this algorithm works every time. But how can I prove its correctness? Note that I don't have much experience in proving this type of claim. My 'common sense' got me to this statement. It would be great if somebody could show me how to prove it. Thank you!!
REPLY [0 votes]: To prove the correctness you can use an invariant. This is a statement that holds true throughout the execution of the algorithm.
In your case, the invariant expresses that the partial sum $S$ that you form never exceeds $T$. (When you try to increase some $x_i$, either the sum reaches $T$ and you stop, or $x_i$ reaches its upper bound $b_i$ before.)
So the invariant guarantees that either you have reached $T$ (and you are done), or you can still increase the sum (by increasing the next $x_i$).
A second aspect of the proof is that the solution will be found after a finite number of steps: as long as the sum has not reached $T$, you can increase the next $x_i$. Indeed, $S<T$ implies that you can add a finite amount to the sum, and this is always possible unless all $x_i$ are saturated. But if all $x_i$ are saturated, $S=T=$the sum of all $b_i$.
S= sum of Ai; { S <= T }
i= 1;
while i <= n and S < T:
{ S < T }
if S + Bi - Ai < T
then S+= Bi - Ai; i++; { We saturated Xi and still have S < T}
else S= S + T - S; { The last increment was T-S > 0, and now S==T }
i++;
{ When arrived here, we cannot have i>n as this would imply T > sum of Bi }
For completeness, you should also express in the invariant that all $x_j$ for $j<i$ are saturated.
If you really want to gently but rigorously learn how to write proofs of algorithms, this is a must.
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Requirements
A minimum of 36 physics units, 17 mathematics units, 4 chemistry units, and 4 computer science units are required for the physics major.
*Meets a General Studies core (or elective) requirement.
Note: This information is current for the 2012–13 academic year; however, all stated academic information is subject to change. Please refer to the current Academic Catalog for more information. For additional information, please contact the appropriate office.
| 326,829
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TITLE: Relating the signs of $f$, $f'$ and $f''$
QUESTION [4 upvotes]: Let $f:\Bbb R \to \Bbb R$ be a function such that $f$ and $f'$ are differentiable for all $x$. If $f(x) \gt 0$ for all $x\ge 0$ and $f''(x)\lt 0$ for all $x\ge 0$, prove that $f'(x)\ge 0$ for all $x\ge 0$.
REPLY [0 votes]: Let's assume that $f'(a) < 0$ for some $a \geq 0$. Then since $f''(x) < 0$ for all $x \geq 0$, it follows that $f'(x) < f'(a) < 0$ for all $x > a$. It follows that $f(x)$ is strictly decreasing in $[a, \infty)$ and since $f(x) > 0$ for all $x \in [a, \infty)$, it follows that $\lim_{x \to \infty}f(x) = A$ exists and $A \geq 0$.
Next we can see that $f''(x) < 0$ so that $f'(x)$ is strictly decreasing for all $x$ and hence either $\lim_{x \to \infty}f'(x) = B$ or $f'(x) \to -\infty$ as $x \to \infty$. In the first case if $B$ exists then we must have $B \leq f'(a) < 0$ (because $f'(x) < f'(a)$ for all $x \in (a, \infty)$). By Mean Value theorem we have $f(x) - f(x/2) = (x/2)f'(c)$ for some $c$ satisfying $x/2 < c < x$. Taking limits as $x \to \infty$ we get a contradiction as LHS tends to $A - A = 0$ and RHS diverges to $-\infty$ because $f'(c)$ tends to a negative value. Same contradiction holds if $f'(x) \to -\infty$.
It follows that our assumption of existence of $a$ with $f'(a) < 0$ is wrong. Thus $f'(x) \geq 0$ for all $x \geq 0$.
Update: From the above proof it is easy to see that the result in question is true even if we assume $f(x) > K$ for some constant $K$. It is not necessary for $f(x)$ to be positive (like in the original question), it just needs to be bounded below.
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Many of you have probably already read about how a single tweet led to two people getting fired and plenty of backlash across the internet. In a nutshell, the author of the tweet called out two fellow conference-goers for a comment she overheard and felt was sexist. She didn't just share her outrage, though. She shared a photo of the men who made the comment. Ultimately, two people were fired, including the author of the tweet. If you want more background, take a look at this Forbes story on the incident.
This story caught my eye for two reasons.
First, this tweet had significant real-world costs. Two people lost their jobs and a company was targeted by a cyber-attack as a result of the tweet. I hope those involved were able to learn from the experience and move forward with their lives.
Second, it raises a lot of tough questions about how we approach social media. As a company that helps future students (mostly high-school kids) connect in a virtual environment, these are questions that we think about often.
Public vs. Private
It's amazing (and unsettling, at times) to think of the expanded reach every person has access to because of the internet and social media and the increased velocity with which information is spread. And it's all happened so fast that I think everyone is still learning how to navigate these new tools.
As part of this, there's clearly been a blurring of the line between public and private. In this particular case, the author of the tweet overheard a comment made by two people having a "private" conversation. I put "private" in quotes because it was happening at a tech conference where they were surrounded by dozens of people who were likely within earshot. Are we at the point with the spread of smart phones equipped with cameras and easy access to broadcasting tools like Twitter, YouTube, and Facebook where conversations in public spaces should always be considered public?
Look, if you're a public figure, making comments on a radio show or on your Twitter account with 100,000 followers, I think you're fair game. But should you be fair game if you're talking to a friend in a cafe? What about a post in a private Class of 2017 Facebook Group? Anyone in that group can copy-and-paste a post and share it?
I know I approach my posts on my personal Facebook profile with the thought that they might be shared publicly at some point, even though I'm typically only sharing them with my friends. There has been more than one instance where I didn't post something as a result. But am I losing out on sharing memories or having virtual conversations with friends because of this caution? Probably.
Now, imagine you're an 18-year-old senior and everyone is sharing every detail of their life on Twitter, YouTube, and Facebook. What are you likely to be sharing? Are you equipped to appreciate the implications of sharing information and content about yourself? We only need to look at the recent events in Steubenville, Ohio, to realize many are not.
In-Person Conversations vs. Virtual Conversations
While I've had some really interesting and valuable back-and-forth conversations on social media tools like Twitter or Facebook, let's be honest... it's not the same as having an in-person conversation with a friend.
One of the first things to go in a virtual conversation is nuance. Trying to pare an idea down to a 140-character tweet necessitates the removal of words that often soften a stance. I can't use "sometimes" or "in this particular instance" because those are precious characters I'm giving up. In this case, the tweet didn't include the full comment, just references to "big dongles" and jokes about "forking repo's in a sexual way." No real sense for what made this offensive vs. juvenile humor.
On top of that, you often lose the context and history behind a comment. When I'm talking to my friend, we have shared experiences. We know some of each other's background and have an understanding of where the other person is coming from. I also have visual cues to help me glean a better understanding of these comments. If someone retweets something I said in a back-and-forth exchange, there's no context and history there, no visual cues. Was the comment meant sarcastically or in earnest? Was it referencing a previous conversation? Was it some sort of inside joke?
How do we balance our desire to share our ideas and have virtual conversations with the risk that anything we share can be taken out of context, removed from the conversation and evaluated in isolation (think, retweet)? How do we educate high school students who text hundreds of times a day with friends that a comment on YouTube or Facebook may carry additional risks?
Law of Unintended Consequences
One of the saddest parts of this story is that it seemed to snowball into a much bigger deal than I'm guessing anyone involved could ever have imagined. When you release something into social media, it often takes on a life of its own, for better or for worse.
While I believe she shouldn't have shared a photo of the "offenders"—sharing the comment itself probably would have been enough and funneled any resulting outrage toward the comment and not this person whom she did not seem to know—I doubt her goal was to get anyone fired from their jobs. I doubt she ever could have imagined such a strong backlash against herself, let alone the company she works for.
The whole incident almost leaves my head spinning a bit. Should I never share my thoughts and opinions for fear of some unimaginable consequence? Is this just the new reality for participating in the world of social media? Should we all consider ourselves having been Mirandized when it comes to social media "Anything you say, can and will be used against you in the court of social media and in ways you can't even anticipate"?
Again, bringing it back to the typical target audience for most of our partner schools... how do you teach an 18-year-old high school student to appreciate these risks and to answer these difficult questions for themselves?
Labels: Social Media, Teens, Twitter
Note: Only a member of this blog may post a comment.
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\begin{document}
\title{On the Classification of Extremal Doubly Even
Self-Dual Codes with $2$-Transitive Automorphism Groups}
\author{
Naoki Chigira\thanks{
Department of Mathematics,
Kumamoto University,
Kumamoto 860--8555, Japan.
email: chigira@kumamoto-u.ac.jp},
Masaaki Harada\thanks{
Department of Mathematical Sciences,
Yamagata University,
Yamagata 990--8560, Japan.
email: mharada@sci.kj.yamagata-u.ac.jp}
and
Masaaki Kitazume\thanks{
Department of Mathematics and Informatics,
Chiba University,
Chiba 263--8522, Japan.
email: kitazume@math.s.chiba-u.ac.jp}
}
\maketitle
\begin{abstract}
In this note, we complete the classification of extremal
doubly even self-dual codes with $2$-transitive automorphism
groups.
\end{abstract}
\noindent
{\bf Keywords} extremal doubly even self-dual code,
automorphism group, 2-transitive group
\noindent
{\bf Mathematics Subject Classification} 94B05, 20B25
\section{Introduction}
As described in~\cite{RS-Handbook},
self-dual codes are an important class of linear codes for both
theoretical and practical reasons.
It is a fundamental problem to classify self-dual codes
of modest lengths
and determine the largest minimum weight among self-dual codes
of that length (see~\cite{Huffman05, RS-Handbook}).
It was shown in~\cite{MS73} that
the minimum weight $d$ of a doubly even
self-dual code of length $n$ is bounded by
$d \le 4 \lfloor{\frac {n}{24}} \rfloor + 4$.
A doubly even self-dual code meeting the bound is called {\em extremal}.
A common strategy for the problem whether there is an
extremal doubly even self-dual code for a given length
is to classify extremal doubly even
self-dual codes with a given nontrivial automorphism group
(see~\cite{Huffman05, RS-Handbook}).
Recently, Malevich and Willems~\cite{MW} have shown that
if $C$ is an extremal doubly even self-dual code with a $2$-transitive
automorphism group then
$C$ is equivalent to one of the extended quadratic residue codes of
lengths $8,24,32$, $48,80,104$, the second-order Reed--Muller
code of length $32$ or a putative
extremal doubly even self-dual code of length $1024$
invariant under the group $T \rtimes \SL(2,2^5)$, where
$T$ is an elementary abelian group of order $1024$.
The aim of this note is to complete the classification
of extremal doubly even self-dual codes with $2$-transitive
automorphism groups.
This is completed by excluding the open case in the above
characterization~\cite{MW}, using Theorem~A in~\cite{CHK}.
\begin{thm}\label{main}
Let $C$ be an extremal doubly even self-dual code with a $2$-transitive
automorphism group.
Then $C$ is equivalent to one of the the extended quadratic
residue codes of lengths $8,24,32,48,80,104$ or
the second-order Reed--Muller code of length $32$.
\end{thm}
\section{Proof of Theorem~\ref{main}}
For an $n$-element set $\Omega$,
the power set ${\cal P}(\Omega)$
-- the family of all subsets of $\Omega$ --
is regarded as an $n$-dimensional binary vector space
with the inner product $(X, Y) \equiv |X \cap Y| \pmod 2$
for $X,Y \in {\cal P}(\Omega)$.
The {\em weight} of $X$ is defined to be the integer $|X|$.
A subspace $C$ of ${\cal P}(\Omega)$
is called a {\em code} of length $n$.
Note that all codes in this note are binary.
The {\em dual code} $C^\perp$ of $C$ is
the set of all $X \in {\cal P}(\Omega)$
satisfying $(X, Y)=0$ for all $Y\in C$.
A code $C$ is said to be
{\em self-orthogonal} if $C \subset C^\perp$, and
{\em self-dual} if $C = C^\perp$.
A {\em doubly even} code is a code
whose codewords have weight a multiple of $4$.
Let $G$ be a
permutation group on an $n$-element set $\Omega$.
We define the code $C(G,\Omega)$ by
\[
C(G,\Omega)=
\langle \Fix(\sigma) \mid \sigma \in I(G)\rangle^\perp,
\]
where
$I(G)$ denotes the set of involutions of $G$
and $\Fix(\sigma)$ is the set of fixed points of $\sigma$ on $\Omega$.
\begin{thm}[Chigira, Harada and Kitazume~\cite{CHK}]
\label{thm:CHK}
Let $C$ be a binary self-orthogonal code of length $n$
invariant under the group $G$.
Then $C \subset C(G,\Omega)$.
\end{thm}
By using Theorem~\ref{thm:CHK},
some self-dual codes invariant under sporadic almost
simple groups were constructed in~\cite{CHK}.
In this note, we apply Theorem~\ref{thm:CHK} to
a family of 2-transitive groups containing the group
$(2^{10}) \rtimes \SL(2,2^5)$.
Let $r, s$ be positive integers.
We consider the following group $G$
$$ G = T \rtimes H \quad
(T = (2^r)^{2s}, H = \SL(2s, 2^r)), $$
where the group $T$ is regarded as the natural module $GF(2^r)^{2s}$ of $H$.
Here $T$ acts regularly on $T$ itself and $H$ acts on $T$ as
the stabilizer of the unit of $T$, which is regarded as
the zero vector of $GF(2^r)^{2s}$.
Then $G$ naturally acts 2-transitively on $T$.
\begin{lem}
There is no self-dual code of length $2^{2rs}$
invariant under $G = T \rtimes H$.
\end{lem}
\begin{proof}
By the fundamental theory of Jordan canonical forms in
basic linear algebra,
the dimension of the subspace of $GF(2^r)^{2s}$
spanned by the vectors fixed by
an involution in $H = \SL(2s, 2^r)$ is equal to or greater than $s$.
Then it is easily seen that
there exist two involutions $\sigma, \tau$ in $H$
such that each of them fixes some $s$-dimensional subspace
of $GF(2^r)^{2s}$, and
the zero vector is the only vector fixed by both of them
(i.e.\ $T = \Fix(\sigma)\oplus\Fix(\tau)$).
As codewords in $C(G,\Omega)^\perp$, the inner product
$(\Fix(\sigma), \Fix(\tau))$ is equal to $1$,
since $|\Fix(\sigma)\cap \Fix(\tau)|=1$.
This yields that $C(G, T)^\perp$ is not self-orthogonal.
Suppose that
$B$ is a self-dual code invariant under $G$.
By Theorem~\ref{thm:CHK}, $B \subset C(G, T)$.
Since $B^\perp \supset C(G, T)^\perp$ and $B=B^\perp$,
$C(G, T)^\perp$ is self-orthogonal.
This is a contradiction.
\end{proof}
The case $(r,s)=(5,1)$ in the above lemma completes the
proof of Theorem~\ref{main}.
\begin{rem}
In the above proof, the cardinality of the fixed subspace of dimension $s$
is $2^{rs}$, which is smaller than the value
$ 4 \lfloor{\frac {2^{2rs}}{24}} \rfloor + 4$,
except for the cases $(r, s) = (1, 2), (2, 1)$.
This shows immediately that
there is no extremal doubly even self-dual code of length $2^{2rs}$
invariant under the group $G = T \rtimes \SL(2s, 2^r)$
if $rs >2$.
On the other hand,
the smallest cardinality of the fixed subspace of an involution
in $\SL(2s-1, 2^r)$ is $2^{rs}$.
If $s >1$ then this number is smaller than the value
$ 4 \lfloor{\frac {2^{(2s-1)r}}{24}} \rfloor + 4$,
except for the small cases
$(r, s) = (1, 2), (1,3), (2, 2)$.
When $(r, s) = (1, 2)$ or $(1,3)$, the code $C(G, T)$,
for $G = T \rtimes \SL(2s-1, 2^r)$ where $T= (2^r)^{2s-1}$,
is equivalent to
the extended Hamming code of length $8$, or
the second-order Reed--Muller code of length $32$
(see~\cite[Example 2.10]{CHK}), respectively.
For the remaining case $(r, s) = (2, 2)$
(i.e.\ $G = T \rtimes \SL(3, 2^2)$, $T=2^6$),
the smallest cardinality of the fixed subspace of an involution
is $16\ (>12)$,
and so such an argument does not work.
(Indeed the code $C(G, T)^\perp$ is self-orthogonal
with minimum weight $16$.)
\end{rem}
\bigskip
\noindent {\bf Acknowledgment.}
This work is supported by
JSPS KAKENHI Grant Numbers 23340021, 24340002, 24540024.
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Central Okanagan News
Crime by the numbers
Kelowna RCMP have sent out the first crime stats information of 2014.
These numbers encompass the one month period between Dec 16, 2013 – Jan 12, 2014.
From Dec 16 to 22, 2013 Theft of a Motor Vehicle was higher than average in West Kelowna coming in at four vehicles stolen, while theft of a bicycle was higher than average in Kelowna coming in at four bikes stolen.
From Dec 23 to 29, 2013 we saw higher than average crime numbers in several categories. In Kelowna theft from a vehicle was at a three year high with 31 cars broken into.
In Lake Country, theft from a motor vehicle and theft of a motor vehicle were also at a three year high seeing one car stolen and three break-ins.
In West Kelowna, three year highs were also seen in residential break and enters and theft from a motor vehicle. Five homes and 10 cars were broken into.
For the final two week period running from Dec 30, 2013 to Jan 12, 2014 overall numbers remained within yearly averages with the exception of residential break and enters in West Kelowna and Lake Country, which each saw three year highs.
West Kelowna experienced five break-ins and Lake Country saw two.
Check out our interactive map above & below to see whether crime is happening in your neighbourhood, and what kind of crime is most common.<<
| 358,959
|
\begin{document}
\maketitle
\begin{abstract}
Concentration of measure is a principle that informally states that in some spaces any Lipschitz function is essentially constant on a set of almost full measure. From a geometric point of view, it is very important to find some structured subsets on which this phenomenon occurs. In this paper, I generalize a well-known result on the sphere due to Milman to a class of Riemannian manifolds. I prove that any Lipschitz function on a compact, positively curved, homogeneous space is almost constant on a high dimensional submanifold.
\end{abstract}
\section*{Introduction}
The celebrated Lévy's concentration of measure inequality for the sphere, together with the work of V. Milman on the asymptotic behavior of Banach spaces, put forward the concentration of measure phenomenon in high dimensional spaces. Among the several results achieved in this field, let me mention the work on Banach spaces (\cite{Ledoux},\cite{Milman}), infinite-dimensional groups (\cite{Pestov1}), Riemannian manifolds (\cite{Gromov2},\cite{Ledoux}) or even general metric measure spaces (\cite{Burago},\cite{Gromov}).
The fundamental idea underlying these results is that a Lipschitz function tends to asymptotically concentrate near a single value. This type of results is usually stated in the sense of the measure, meaning that the probability of the subset where the function is almost constant tends to $1$ when the dimension of the space approaches infinity. Nevertheless, especially from the geometric point of view, it is important to find more structured subsets on which the function is concentrated. A well-known result in this direction, due to Milman (\cite{Milman1}), roughly states that every Lipschitz function on $\mathbb{S}^n$ is almost constant on a sufficiently high-dimensional sphere $\mathbb{S}^k\subset \mathbb{S}^n$. In this note I generalize this result for a class of positively curved Riemannian manifolds. I want to highlight that recently, Faifman, Klartag, and Milman (\cite{Faifman}) have found out that a similar result also holds on the torus, where the strong concentration property is not available due to the flatness of the space.
I also want to mention two others papers of Milman (\cite{Milman2},\cite{Milman3}), in which he extends the idea of concentration to some homogeneous structure like Stiefel and Grassmann manifolds of an infinite dimensional Hilbert space. Some very clever applications of these results are proved by Gromov and Milman in (\cite{Gromov1}) and (\cite{Milman4}).
I now briefly recall some elementary facts from Riemannian geometry. In the course of this paper I will use the overline notation for a quantity defined on the ambient manifold. The same quantity intrisically defined on a submanifold will not have the bar.
A Riemannian manifold $(\bar{M},\bar{g})$ is a real, smooth manifold $\bar{M}$ with a symmetric, positive definite $(0,2)$ tensor $\bar{g}$. I always assume that $\bar{M}$ is connected. Given a 2-plane $P$ spanned by $(v,w)$ on the tangent space $T_x\bar{M}$, the sectional curvature at the point $x$ is defined by $$\overline{Sec}_x(P)=\frac{\bar{R}_x(v,w,v,w)}{\bar{g}(v,v)\bar{g}(w,w)-(\bar{g}(v,w))^2},$$ where $\bar{R}_x$ is the Riemannian $(0,4)$ curvature tensor.
Suppose now $L$ is a $m$-plane section of $T_x\bar{M}$ and $v$ a unit vector in $L$. Take an orthonormal basis $\{e_1,...,e_m\}$ of $L$ such that $e_1=v$. The $m$-Ricci curvature of the subspace $L\subset T_x\bar{M}$ and the vector $v$ at the point $x$ is given by the formula $$\textit{m-}\overline{Ric}_x^{L}(v)=\sum_{j=2}^m\overline{Sec}_x(P_{1j})$$ where $P_{1j}$ denotes the $2$-plane spanned by $e_1,e_j$. The $n$-Ricci curvature of the vector $v$, $n=\dim(\bar{M})$, does not depend on the subspace and it is simply indicated by $\overline{Ric}_x(v)$. It coincides with the usual notion of Ricci curvature.
By writing $\overline{Sec}(\bar{M})\geq K$ I mean that $\overline{Sec}_x(P)\geq K$ for every $x\in \bar{M}$ and every 2-plane $P$ of $T_x\bar{M}$.
By writing $m$-$\overline{Ric}(\bar{M})\geq K$ I mean that $m$-$\overline{Ric}_x^L(v)\geq K$ for every $x\in \bar{M}$, every $m$-dimensional subspace $L$ of $T_x\bar{M}$ and every unit vector $v\in L$.
In view of the inductive identity $$\sum_{j=2}^{m+1}\overline{Sec}(P_{1j})=\frac{1}{m-1}\sum_{j=2}^{m+1}\sum_{i\neq j}\overline{Sec}(P_{1i}),$$ it follows the implication $$\textit{m-}\overline{Ric}(\bar{M})\geq (m-1)K\Rightarrow (\textit{m}+1)\textit{-}\overline{Ric}(\bar{M})\geq mK.$$
Recall that a median for a random variable $T:(\Omega,P)\rightarrow \mathbb{R}$ is a value $m_T$ such that $P(T\leq m_T)\geq \frac{1}{2}$ and $P(T\geq m_T)\geq \frac{1}{2}$.
To state the main result, let $(\bar{M},\bar{g})$ be a compact, $n$-dimensional homogeneous space with $m$-Ricci curvature $m$-$\overline{Ric}(\bar{M})\geq (m-1)K>0$ for a sufficiently high $m\leq n$. Let $r$ be the maximal dimension of the totally geodesic submanifolds of $\bar{M}$. Assume $\bar{M}$ is equipped with the geodesic distance $d_{\bar{g}}$ and the normalized Riemannian measure $\bar{\mu}$.
\begin{theorem}
\label{teorema}
For every $1$-Lipschitz function $T:\bar{M}\rightarrow \mathbb{R}$ and for every $\epsilon>0$, it exists an $s$-dimensional submanifold $S\subset \bar{M}$ such that $|T(z)-m_f|\leq\epsilon$ for every $z\in S$, $m_T$ the median of $T$, with $s$ of order $$ s\sim r\wedge\frac{\epsilon^2K(n-1)}{8\ln(\frac{12}{\epsilon\sqrt{K}})}$$
\end{theorem}
\begin{remark}
The condition on the curvature of the manifolds is certainly satisfied if $\overline{Sec}(\bar{M})\geq K>0$. Neverthless, I decide to state the theorem with the hypothesis on the $m$-Ricci curvature to stress the fact that we do not need positive curvature in every direction in order to have a high dimensional submanifold with positive intrinsic Ricci curvature. What is really needed is $m$-$\overline{Ric}(\bar{M})>0$ for $m$ at least as high as the dimension of the submanifold where the function is concentrated.
\end{remark}
\begin{remark}
Basically every bound in this paper, included the one in theorem $\ref{Levy-Gromov}$, is not sharp at all. I simply try to provide the correct order of the terms. Specifically, if we assume that the constant $K$ does not depend on the dimension of the manifold, the order of the dimension of the submanifold $S$ is the same achieved by Milman in the case of the sphere (\cite{Milman}, theorem $2.4$).
\end{remark}
This paper is organized as follows: firstly I introduce the notions of Riemannian distance and measure, and I state the standard concentration result on Riemannian manifold. Secondly, in section $2$, I show how to put "the right" measure on the isometry group of a homogeneous space. After that in section 3 I prove theorem $\ref{teorema}$ and I make some remarks about the statement and the proof (following once again $\cite{Milman}$, remarks 2.8 and 2.9). In section 4 I present a class of manifolds with sufficiently high dimension totally geodesic submanifolds: the symmetric spaces. Finally, in the last section, I give some explicit examples of spaces where the result can be applied.
\section{Riemannian manifold as a metric measure space and the standard concentration theorem}
Let $(\bar{M},\bar{g})$ be a Riemannian manifold and let $\gamma:[a,b]\rightarrow \bar{M}$ be a smooth curve in $\bar{M}$. I define the lenght of $\gamma$ as $$L(\gamma)=\int_a^b\sqrt{\bar{g}(\dot{\gamma},\dot{\gamma})}dt.$$
Every Riemannian manifold $(\bar{M},\bar{g})$ is equipped with a Riemannian distance $$d_{\bar{g}}(x_0,x_1):=\inf\{L(\gamma)|\gamma:[a,b]\rightarrow\bar{M}\hspace{1mm}\textrm{smooth}\hspace{1mm}\gamma(a)=x_0,\gamma(b)=x_1\}.$$ It can be seen that the Riemannian distance induces the same topology of the manifold.
Furthermore, every Riemannian manifold has a Riemannian measure $\bar{\nu}$. It is a Borel measure and if $\bar{M}$ is compact, it has finite volume $V=\bar{\nu}(\bar{M})<\infty$ and the probability measure $\bar{\mu}=\frac{\bar{\nu}}{V}$ is called the normalizied Riemannian measure.
We are now ready to state the standard concentration theorem for Riemannian manifold:
\begin{theorem}(Levy-Gromov)
\label{Levy-Gromov}
Let $(\bar{M},\bar{g})$ be a compact, connected Riemannian manifold equipped with the Riemannian distance $d_{\bar{g}}$ and the normalized Riemannian measure $\bar{\mu}$. Suppose that $\overline{Ric}(\bar{M})\geq K>0$. Then, given a 1-Lipschitz function $T$, it exists $m_T$ such that, for every $\epsilon>0$ $$\mu(\{|T-m_T|>\epsilon\})<e^{-\frac{K\epsilon^2}{2}}.$$
\end{theorem}
\section{Isometry group, homogeneous spaces and Haar measure.}
Let $(\bar{M},\bar{g})$ be a Riemannian manifold. I denote with $Iso(\bar{M})$ the isometry group of $\bar{M}$, i.e. the set of the maps $f:\bar{M}\rightarrow \bar{M}$ such that $\bar{g}(X,Y)=\bar{g}(f_{*}X,f_{*}Y)$ for every vector fields $X,Y$, togheter with the operation of composition. Every isometry is a metric isometry in respect to the geodesic distance. In 1938, Myers and Steenrod proved that $Iso(\bar{M})$ is a Lie group (\cite{Myers}).
$Iso(\bar{M})$ acts naturally on $\bar{M}$ via the map $f\cdot x=f(x)$. If this action is transitive, $\bar{M}$ is called an homogeneous space.
In case $\bar{M}$ is compact, $Iso(\bar{M})$ is a compact Lie group and we can equipped it with the unique bi-invariant Haar probability measure $\theta$. We can transport this measure on $\bar{M}$ with a classical construction: let $x$ be a point in $\bar{M}$, I denote with $h^x:Iso(\bar{M})\rightarrow \bar{M}$ the map $h^x(f)=f(x)$. Moreover I denote with $\bar{\mu}^x=h^x_{\sharp}\theta$ the pushforward of $\theta$ by $h^x$, i.e. $\bar{\mu}^x(A)=\theta((h^x)^{-1}(A))$ for any Borel set $A$. Therefore, it seems there are several measures on a compact, homogeneus space $\bar{M}$: $\bar{\mu}^x$ for any $x\in \bar{M}$ and $\bar{\mu}$, the normalized Riemannian measure. Theorem 1.3 in $\cite{Milman}$ shows us they are the same measure: to see this, note that $\bar{\mu}$ and $\bar{\mu}^x$ are probability measure invariant by the action of the isometry group on $\bar{M}$.
To sum up I state the next lemma:
\begin{lemma}
\label{lemma1}
Let $\bar{M}$ be a compact homogeneous space and let $\mu$ be the normalized Riemannian measure. Then, for any $x\in \bar{M}$ and for any Borel set $A$, we have $$\bar{\mu}(A)=\theta(\{f \in Iso(\bar{M}): f(x)\in A\}),$$ where $\theta$ is the unique Haar probability measure on the compact Lie group $Iso(\bar{M})$.
\end{lemma}
\section{Totally geodesic submanifolds}
Let $(\bar{M},\bar{g})$ be a smooth Riemannian manifold. Let $M$ be a Riemannian submanifold of $\bar{M}$, i.e. a submanifold endowed with the first fundamental form $g=i^{*}\bar{g}$. $M$ is called a totally geodesic submanifold if any geodesic on $M$ is also a geodesic on $(\bar{M},\bar{g})$.
I write $\bar{\nabla}$ and $\nabla$ for the Levi-Civita connenction on $\bar{M}$ and $M$, respectively. I recall that the second fundamental form is the symmetric tensor field defined by $$II(X,Y)=\bar{\nabla}_XY-\nabla_XY,$$ where $X,Y$ are vector fields on the submanifold $M$.
$M$ is totally geodesic if and only if the second fundamental form $II$ vanishes, i.e
\begin{equation}
\label{curvatura}
\bar{\nabla}_XY=\nabla_XY.
\end{equation}
From $\eqref{curvatura}$ follows that the Riemannian curvature tensors $\bar{R}$ and $R$ agree on the domain of $R$. In particular, it is easy to note that $\overline{Ric}_x^{T_xM}(v)=Ric_x(v)$ (see $\cite{Abedi}$ for a more general result).
\section{Proof of the theorem $\ref{teorema}$}
Let's first state a lemma on an upper bound on the cardinality of a $\delta$-net in a totally geodesic submanifold.
\begin{lemma}
\label{lemma3}
Let $\bar{M}$ be a compact symmetric space with $$\textit{m-}\overline{Ric}(\bar{M})\geq (m-1)K>0$$ and let $M$ be an $m$-dimensional, totally geodesic submanifold of $\bar{M}$. Then it exists a constant $C$ such that we can find on $M$ a $\delta$-net, in respect to the extrinsic distance, of cardinality less than $(\frac{6}{\delta\sqrt{K}})^m$
\end{lemma}
\begin{proof}
Recall that a subset $\{y_j\}_{j=1}^N$ of points in $M$ is a $\delta$-net for $(M,d_g)$ (where $d_g$ is the Riemannian intrinsic distance on $M$) if, for every $y\in M$, it exists a $j\in \{1,...,N\}$ such that $d_g(y,y_j)<\delta$. Note that a $\delta$-net for the intrinsic distance $d_g$ is also a $\delta$-net for the extrinsic distance $d_{\bar{g}}$, thanks to the obvious inequality $d_{\bar{g}}\leq d_g$.
Let $\{y_j\}_{j=1}^N$ be a maximal set of points in $M$ such that $d_g(y_i,y_j)\geq\delta$ for every $i\neq j$. It is easy to see that $\{y_j\}_{j=1}^N$ is a $\delta$-net.
I denote by $B(x,r)$ the geodesic ball centered in $x$ of radius $r$ and by $\mu$ the normalized Riemannian measure on $M$. The measure of the ball $\mu(B(x,r))$ doesn't depend on $x$ because $M$ is homogeneus. Moreover
$B(y_i,\delta/2)\cap B(y_j,\delta/2)=\emptyset$, which implies that $$N\mu(B(\delta/2))\leq \mu(M).$$
We know that $Ric(M)\geq (m-1)K$ because $M$ is a totally geodesic submanifold of a positively curved manifold, so we can apply the Bishop-Gromov theorem (theorem 4.19 in $\cite{Gallot}$) and we obtain that $$\phi(r)=\frac{\mu(B(r))}{Vol(B_K(r))}$$ is a non-increasing function in $(0,+\infty)$, where $B_K(r)$ is the ball of radius $r$ in the simply connected $m$-dimensional space of costant sectional curvature $K$.
Let $D=diam(M)$ be the diameter of the submanifold, it follows that $$\frac{\mu(B(\delta/2))}{Vol(B_K(\delta/2))}\geq \frac{\mu(M)}{Vol(B_K(D/2))}$$ and
$$\frac{\mu(M)}{Vol(B_K(D/2))}\geq N\frac{\mu(B(\delta/2))}{Vol(B_K(D/2))}.$$
So $$N\leq \frac{Vol(B_K(D/2))}{Vol(B_K(\delta/2))}.$$
From Myers theorem (theorem 3.85 in $\cite{Gallot}$), it follows that $D\leq \frac{\pi}{\sqrt{k}}$, so we can have an upper bound on $N$ that depends only on $K$, $m$ and $\delta$.
In particular, we see that $$N\leq \frac{\int_0^{\frac{\pi}{2\sqrt{K}}}(\sin(\sqrt{K}t))^{m-1}dt}{\int_0^{\delta/2}(\sin(\sqrt{K}t))^{m-1}dt}=$$ $$=\frac{\int_0^{\frac{\pi}{2}}(\sin(t))^{m-1}dt}{\int_0^{\frac{\delta\sqrt{k}}{2}}(\sin(t))^{m-1}dt}\leq \frac{\sqrt{\pi}\Gamma({\frac{m}{2}})m2^{2m-1}}{2\Gamma({\frac{m+1}{2})}\delta^{m}K^{m/2}}<\frac{\pi}{4}m(\frac{4}{\delta\sqrt{K}})^m<(\frac{6}{\delta\sqrt{K}})^m$$ where we use the inequality $\frac{t}{2}<\sin(t)$ to bound the denominator.
\end{proof}
We are now ready to prove the main theorem:
\begin{proof}(Main theorem)
\label{proof}
Let $T:\bar{M}\rightarrow \mathbb{R}$ be a $1$-Lipschitz function. By Levy-Gromov theorem we know that it exists $m_T$ such that, for every $\epsilon>0$, we have $$\bar{\mu}(\{|T-m_T|>\frac{\epsilon}{2}\})<e^{-\frac{\epsilon^2K(n-1)}{8}}.$$
Let $M$ be a totally geodesic submanifold with an extrinsic $\delta$-net $\{y_j\}_{j=1}^N$. Lemma $\ref{lemma1}$ implies that $$\theta(\{f\in H:|T(f(y_j))-m_T|\leq\frac{\epsilon}{2}, j=1,...,N\})>1-N\exp^{-\frac{\epsilon^2K(n-1)}{8}},$$ and, if ($\star$) $N\exp^{-\frac{\epsilon^2K(n-1)}{8}}<1$, it exists at least one $f$ such that the function $T$ is almost constant on $z_j:=f(y_j)$ for every $j$. I highlight the fact that $\{z_j\}_{j=1}^N$ is an extrinsic $\delta$-net on the submanifold $S:=f(M)$, simply because $f$ is an isometry.
From the Lipschitz condition of $T$ it follows that this function is almost constant for every point $z$ in the submanifold $S$. To see this let $z$ be a point on $S$; we know that an index $j$ exists such that $d_{\bar{g}}(z,z_j)<\delta$ and $$|T(z)-m_T|\leq \frac{\epsilon}{2}.$$ Hence
$$T(z)=T(z)-T(z_j)+T(z_j)\leq |T(z)-T(z_j)|+T(z_j)\leq \delta + m_T+\frac{\epsilon}{2}$$
$$T(z)=T(z)-T(z_j)+T(z_j)\geq -|T(z)-T(z_j)|+T(z_j)\geq -\delta + m_T-\frac{\epsilon}{2}$$
and we have the concentration property $$|T(z)-m_T|\leq \delta+\frac{\epsilon}{2}$$ for every $z$ in $S$ as claimed. Put $\delta=\frac{\epsilon}{2}$ to conclude.
The condition ($\star$), thanks to lemma $\ref{lemma3}$, is certainly satisfied if $s$, the dimension of the submanifold, satisfies $$(\frac{6}{\delta\sqrt{K}})^s<e^{\frac{\epsilon^2K(n-1)}{8}}$$ i.e. $$s<\frac{\epsilon^2K(n-1)}{8\ln(\frac{12}{\epsilon\sqrt{K}})}.$$
\end{proof}
\begin{remark}
Note that, taking a slightly smaller $s$, the theorem $\ref{teorema}$ is still true for two different Lipschitz functions on the same submanifold.
\end{remark}
\begin{remark}
I give here another proof to theorem $\ref{teorema}$: let $x_0$ be a point in $\bar{M}$, let $M_0$ be a $s$-dimensional totally geodesic submanifold of $\bar{M}$ cointaining $x_0$ and let $Y$ be the set of submanifolds $M=f(M_0)$, $f\in Iso(\bar{M})$. The set $Y=Iso(\bar{M})/H$ can be endowed with a structure of manifold, where $H$ is the closed Lie subgroup of all $f\in Iso(\bar{M})$ such that $f(M_0)=M_0$. Let $dy$ the $Iso(\bar{M})$-invariant probability measure on $Y$. Then for every continuous function $u:\bar{M}\rightarrow \mathbb{R}$ it follows
\begin{equation}
\label{disintegration}
\int_{\bar{M}}u(x)d\bar{\mu}(x)=\int_Y\left(\int_Mu(x)d\mu_M(x)\right) dy,
\end{equation}
where $\bar{\mu}$ is the normalized Riemannian measure on $\bar{M}$ and $\mu_M$ is the normalized Riemannian measure on $M$ (see $\cite{Rouviere}$, section 2). Let $I_{A^{\epsilon}}$ be the indicator function of the set $A^{\epsilon}:=\{|T(x)-m_T|\leq \frac{\epsilon}{2}\}$. Use now the equation $\eqref{disintegration}$ with $u=I_{A^{\epsilon}}$ (obviously $u$ is not continuous, but it is bounded and measurable, so it is a bounded pointwise limit of continuos functions) and the theorem $\ref{Levy-Gromov}$ to obtain that there exists a submanifold $M$ such that $$\mu_M(M\cap A^{\frac{\epsilon}{2}})\geq 1-e^{-\frac{\epsilon^2K(n-1)}{8}}.$$
Note also that, arguing as in the proof of lemma $\ref{lemma3}$, $$\mu_M(B(\frac{\epsilon}{2}))\geq (\frac{\epsilon\sqrt{K}}{6})^s.$$
Finally impose that $$(\frac{\epsilon\sqrt{K}}{6})^s+1-e^{-\frac{\epsilon^2K(n-1)}{8}}\geq 1$$ in order to get a bound on $s$ such that any ball of radius $\frac{\epsilon}{2}$ in the $s$-dimensional submanifold $M$ intersects $A^{\frac{\epsilon}{2}}$, i.e $$|T(x)-m_T|\leq \epsilon$$ for every $x\in M$.
\end{remark}
\section{Symmetric spaces}
An isometry $f:\bar{M}\rightarrow \bar{M}$ is called involutive if $f\circ f=Id$, the identity isometry. $\bar{M}$ is a symmetric space if, for each point $x\in \bar{M}$, there exists an involutive isometry $f_x$ such that $x$ is an isolated fixed point of $f_x$.
A symmetric space is a particular homogeneous space (\cite{Kobayashi}, pag. 223).
A lot is known about totally geodesic submanifolds in symmetric spaces (see $\cite{ChenSurvey}$ for an excellent exposition). For the purpose of this paper I need two facts that I state in a lemma:
\begin{lemma}
\label{lemma2}
\ \\
\begin{itemize}
\item A complete totally geodesic submanifold of a symmetric space is a symmetric space.
\item Let $\bar{M}$ be a symmetric space, $dim(\bar{M})\geq 2$, then there exists a complete totally geodesic submanifold $M$ whose dimension satisfies: $$\frac{1}{2}dim(\bar{M})\leq dim(M) <dim(\bar{M}).$$
\end{itemize}
\end{lemma}
The first assertion is a standard result, which can be found in $\cite{Kobayashi}$. The second statement was proven for the first time by Chen and Nagano in $\cite{Chen}$.
Now, it is easy to deduce from lemma $\ref{lemma2}$ that the submanifolds $S$, whose existence is stated in theorem $\ref{teorema}$, can be choosen so that $dim(S)\sim \frac{\epsilon^2K(n-1)}{8\ln(\frac{12}{\delta\sqrt{K}})}$.
\section{Examples: the sphere and the projective spaces}
The standard sphere $\mathbb{S}^n$ is a compact $n-$dimensional symmetric space with constant sectional curvature equal to 1. $\mathbb{S}^r\subset \mathbb{S}^n$ is a totally geodesic submanifold for every $1\leq r\leq n$, so theorem $\ref{teorema}$ applies with $s\sim \frac{\epsilon^2(n-1)}{8\ln(\frac{12}{\epsilon})}$.
The real projective space $\mathbb{RP}^n$ is a compact $n-$dimensional manifold. As a quotient of $\mathbb{S}^n$, it is a symmetric space of constant sectional curvature equal to 1. $\mathbb{RP}^r\subset \mathbb{RP}^n$ is a totally geodesic submanifold for every $1\leq r\leq n$, so theorem $\ref{teorema}$ applies with $s\sim \frac{\epsilon^2(n-1)}{8\ln(\frac{12}{\epsilon})}$.
The complex projective space $\mathbb{CP}^n$ is a compact $2n-$dimensional manifold. Equipped with the Fubini–Study metric, it is a symmetric space with sectional curvature $Sec(\mathbb{CP}^n)\geq \frac{1}{4}$. $\mathbb{RP}^n$ and $\mathbb{CP}^{n-1}$ are the maximal totally geodesic submanifolds, so theorem $\ref{teorema}$ applies with $s\sim \frac{\epsilon^2(2n-1)}{32\ln(\frac{24}{\epsilon})}$.
The quaternionic projective space $\mathbb{HP}^n$ is a compact $4n-$ dimensional manifold. Equipped with the Fubini–Study metric, it is a symmetric space with sectional curvature $Sec(\mathbb{HP}^n)\geq \frac{1}{4}$. $\mathbb{CP}^{n}$ and $\mathbb{HP}^{n-1}$ are the maximal totally geodesic submanifolds, so theorem $\ref{teorema}$ applies with $s\sim \frac{\epsilon^2(4n-1)}{32\ln(\frac{24}{\epsilon})}$.
\section{Acknowledgements}
The author would like to thank Vitali Milman for the precious discussion that led to an improvement of the introduction and the bibliography.
The author is also thankful to Emanuele Casini, Fabio Cavalletti, Alessandro Ghigi, Bo'az Klartag, Stefano Pigola and Giuseppe Savaré, for reading an earlier version of this paper and suggesting improvements. This work was partially supported by GNAMPA–INDAM.
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Harrison Ford talks Blade Runner sequel and sets record straight on Star Wars/Indiana Jones classic moments
Story by Jack Foley
HARRISON Ford has been speaking of his excitement at the prospect of a Blade Runner sequel as well as setting the record straight on speculation surrounding two of his most iconic moments in, respectively, Star Wars and Raiders of the Lost Ark.
Answering questions as part of Reddit’s Ask Me Anything series, the iconic star – best known as Han Solo and Indiana Jones – said he was most excited about the opportunity of reuniting with Ridley Scott if the script for the proposed Blade Runner sequel is good enough.
He wrote: .”
And that’s despite the fact that Ford cites Blade Runner as being one of his toughest roles, given the fact that it was “always raining” and he worked 50 nights in a row.
Further gems from Ford on Reddit included revelations concerning iconic moments from both Stars Wars and Raiders of the Lost Ark – and, in particular, scenes that were supposedly ad libbed or spontaneous.
Addressing the Star Wars rumour first, Ford denied that his “I know” line (in response to Princess Leia’s declaration of “I love you”) was not – as widely reported – ad libbed.
“It’s not really an ad lib, it was a suggestion, and movie making’s a real collaborative process at its best,” he wrote. .”
Moving on to Raiders, Ford gave a detailed run-down of the events that led up to his decision to shoot a swordsman instead of indulging in another lengthy fight… most of which were dictated by the fact that the actor was suffering from dysentery at the time.
He wrote: ‘I.”
And on another forthcoming note, Ford explained his reasons for agreeing to take part in Sylvester Stallone’s forthcoming Expendables 3, stating: ?”
Next story: Guardians of the Galaxy will feed into Avengers 3
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In this episode of The Chrome Cast, we talk the biggest news in the Chromebook and Assistant space, including:
- Google At CES
- Play Store Uncertified Devices
- Acer Chromebook 15
- CrossOver on Chrome OS
- Samsung Chromebook Pro With 64GB
The podcast is live on iTunes, Stitcher and Google Play Music, so find us there with a quick search for Chrome Unboxed. You can also, of course, just add our feed straight to your podcast catcher of choice!
Enjoy!
| 232,383
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TITLE: Stability of fixed points in this example from a dynamical systems textbook
QUESTION [1 upvotes]: I hope people will indulge me a bit of a beginner Dynamical Systems question.
The following example was taken from Nonlinear Dynamics and Chaos (Second Edition) by Steven Strogatz (example 2.2.1).
Find all fixed points for $\dot{x}=x^2 - 1$ and classify their stability.
Solution: Here $f(x) = x^2 - 1$. To find the fixed points, we set $f(x*) = 0$ and solve for $x*$. Thus, $x* = \pm1$. To determine stability, we plot $x^2 - 1$ and then sketch the vector field (see figure below). The flow is to the right where $x^2 - 1 > 0$ and to the left where $x^2 - 1 < 0$. Thus, $x* = -1$ is stable, and $x* = 1$ is unstable.
The solid dot in the picture below (at $x = -1$) represents a stable equilibrium, whereas the open dot (at $x = +1$) represents an unstable equilibrium. The arrows represent the direction of "local flow."
So I understand that the fixed points are the roots of $f(x)=x^2 + 1$, which is obviously $\pm1$, and that the fixed points are equilibrium solutions. I understand that the equilibrium at $x = -1$ is stable because the local flow is "towards" it and the fact that the equilibrium at $x = +1$ is unstable because the equilibrium is "away from" it, but I'm confused as to how they came up with the direction of the local flow in the first place. Can someone explain why the local flow is in those directions in this case?
REPLY [1 votes]: The local flow going to the right means that $x(t)$ is increasing, which happens when $\dot x>0$.
The equation says that $$\dot x=x^2-1,$$ so $\dot x > 0$ if and only if $x^2-1 > 0$, which is true in the part of the figure where the curve $y=x^2-1$ lies above the $x$ axis.
| 211,390
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TITLE: Does multivariate polynomial over a finite field always have a solution (in the field)?
QUESTION [0 upvotes]: Let $K = F_{p^e}$ be a finite field. Say I have a single polynomial $f \in K[x_1,\ldots, x_n]$ of degree $d$.
Under what conditions on $n$ and $d$ can I claim that a root to $f$ always exists? In other words, do there exist polynomials over finite fields with an arbitrary number of variables and bounded degree that have no roots in the field?
If a root exists, can I compute it efficiently?
REPLY [2 votes]: You can't do it with $n$ and $d$ alone, unless $d=1$ (which is a hyperplane and so is easy to find all solutions).
For example, we know $\mathbb{F}_{p^{2e}}$ is a degree 2 extension of $\mathbb{F}_{p^e}$, so there is an irreducible quadratic $x^2+ax+b$ in $\mathbb{F}_{p^e}[x]$. Consequently,
$$
(a_1x_1+\dots+a_nx_n)^2+a(a_1x_1+\dots+a_nx_n)+b=0
$$
has no solutions for all $(a_1,\dots,a_n)\in\mathbb{F}_{p^e}^n-\{(0,\dots,0\}$. Similar construction gives no $d>1$ can guarantees the existence of a zero regardless of how large $n$ is.
Note that this doesn't contradict Chavelley-Warning theorem, since the number of zeros, $0$, is divisible by $p$.
| 154,807
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Normally.
The full rumored (and partially guessed) specs are below, but a few things stand out. First is the relatively low maximum ISO: At just 6,400, this is behind modern day specs.
On the other hand, face detection in DSLRs is pretty new (the D4 does it, for example) and weather sealing is certainly a high-end feature. And if this thing comes in at under a grand and offers full-frame video shooting (as it seems that it will) then it could be a complete Canon 5D MkII killer, despite being in a different range altogether.
If this turns out to be real, I see it as the beginning of the end for crop-sensor SLRs. Some will still be around (sports and wildlife shooters like them because they make their telephoto lenses half as long again, for free), but everyone else will enjoy the shallower depth-of-field and greater range of second-hand lenses available to full-frame cameras.
Remember back when DSLRs were all over $1,000 and only bought by pros? Full frame might see the same journey to the mass market as those.
Source: Nikon Rumors
Via: Photography Bay
| 398,885
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TITLE: How fast must the ball be launched to land 450 feet away?
QUESTION [0 upvotes]: A golf ball is hit in a horizontal direction off the top edge of a building that is $100$ feet tall. How fast must the ball be launched to land 450 feet away?
I have some of the equations that I have been working with here:
$$
\begin{align*}
\mathbf{s}(t) &= \langle v_0t \cos(\theta),v_0 t \sin(\theta) - \tfrac{1}{2}g t^2 \rangle \tag{1} \\
\mathbf{v}(t) &= \mathbf{s}^{\prime} (t) = \langle v_0 \cos(\theta),v_0 \sin(\theta) -gt \rangle \tag{2}\\
\mathopen|\mathbf{s}^{\prime}(t)\mathclose| &=\sqrt{v_0^2-gt} \tag{3}
\end{align*}$$
where $(1)$ is the position of the object at time $t$, $(2)$ is the velocity of the object at time $t$, and $(3)$ is the speed of the object at time $t$ and $g$ is the gravitation constant equaling $32$ feet per second squared.
With the given information I'm not sure how to proceed with the problem. Ideally I would want to start use $(3)$, but I don't know the value of $v_0$. I know that I want the object to travel a horizontal distance of $450$ feet, so the horizontal component of $\mathbf{s}$ is equal to $450$, however, again, I confound my problem with unknown values for $v_0$ and $\theta$.
Any hints or guidance for this seemingly trivial problem would be greatly appreciated.
REPLY [0 votes]: Take $g = 32 \,\text{feet/sec}^2$. $S = \frac{1}{2}gt^2$. Thus, it takes time $\sqrt{\frac{2S}{g}} = \sqrt{\frac{200}{g}} = 2.5\,\text{secs}$ for the ball to reach the ground. Now, in 2.5 seconds, the ball must travel 450 feet. Thus, the speed must be $\frac{450}{2.5} = 180 \,\text{feet/s}$.
EDIT: The original poster asked for a visualization of what I did, hence this edit.
Forgive me for the crudely drawn diagram, but this is what I visualized. Since the ball is hit horizontally, the velocity component is completely along the horizontal direction. There is no component along the vertical direction.
Now, the matter of whether to use + or - with the $\frac{1}{2}at^2$ term is just a matter of where you are looking from. If you are standing at the bottom and throwing a ball up in the air, then you have to use the negative sign.
So, given this visualization, what you have to realize is that since the velocity is $u$ feet per second along the horizontal direction, in time $t$ seconds, the ball will travel a total of $ut$ feet. Thus is because I used $S = ut + \frac{1}{2}at^2$. Note that the acceleration along the horizontal direction is zero (the problem could be more complicated if someone says that the ball was hit with some $x$ Newton amount of force. Then you would calculate acceleration appropriately), so $S = ut$. Since you know $S$, and what you want to calculate is $u$, you just need $t$, i.e. the time for which the ball travels horizontally to complete your calculation.
The time for which the ball travels horizontally is obviously determined by how long the ball takes to reach the ground. We know that the ball has 100 feet to travel in the vertical direction and we know that there is no initial velocity along the vertical direction. Thus, the only thing that is making it move vertically is the gravitational acceleration. That is why I used $+\frac{1}{2}gt^2$ to determine how much time it will take to reach the ground.
Hope this helps.
| 148,235
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Downtime triggers millions of dollars well worth of harm to computer system systems all over the world. Power associated concerns as it ends up, generally induces the unusual downtime, time-outs, and various other networking glitches. The source of power must be secure as well as secure. Ensure that the planet connection is properly working and the polarities of live and also neutral are in order. A certified electrical expert could effortlessly check this and also the little financial investment would conserve you from all future significant repair services. If the planet’s line falls short as well as the stray present on the circuit board lack appropriate grounding, then it causes element failure. Later this could become a costly repair service.
Feed power to your computer system though a line conditioner with continuous power supply unit. computer repair This would certainly guarantee that in the event of a blown fuse and variations in supply, you manage to shut down your computer system without damaging the operating system.
Place the Central Processing Device in a well-ventilated area. The CPU creates a great deal of warmth and the cooling fans must dissipate this warmth into the setting. If hot air could not get away and cold air could not get in the system, overheating would bring about element failing. There need to not be excessive moisture around because there are components called “capacitors” that fail if way too much of wetness is prevalent in the air. Dust is an opponent. It settles all over as well as forms a slim layer on the motherboard. The dust layer combined with moisture comes to be a great conductor of electrical power that could damage the elements leading to failing. Periodically impact air via a vacuum cleaner to remove all that dust choosing the motherboard, Switching-Mode Power Supply (SMPS), as well as hard drive published circuit board.
Avoid closing on and off the computer as well frequently. Every single time you put on the computer, it heats up and also when you shut off they cool. This inevitably brings about thermal deterioration of parts because of constant growth and also tightening. If you intend to go with supper while working with your PC let it operate on minimum of resources. This would certainly make it possible for the device to keep its optimal temperature, as opposed to cooling down as well as again reheating resulting in thermal wear and tear.
Precautionary maintenance is the best remedy for the optimum utilization of component life. Modern equipment consists of couple of serviceable components inside the computer system. The rapid development of modern technology obsoletes the serviceable spares if any. It is suggested that the customer take care of the vital elements that could lead to failure of the equipment.
| 175,167
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TITLE: Geometric Intuition for "Right-Veering" Property of $f$ in MCG(S)?
QUESTION [1 upvotes]: let $S$ be a compact surface with non-empty boundary, let $\alpha : [0,1] \rightarrow S$ be a Properly-embedded arc (meaning both endpoints of the arc are in $\partial S$) and let $f$ be an element in $MCG(S)$, the mapping class group of $S$. Let $\alpha(0)=x$. I am trying to understand the geometric motivation behind the property of "right-veeringness" of $f$. We say that $f$ is right-veering if the orientation given at the point $x$ by the pair $(f'(\alpha(0)) ,\alpha'(0))$ agrees with the orientation of the surface $S$ at the point $x$. Any ideas ?
REPLY [1 votes]: For the overall definition: You need to either work in the universal cover and homotop lifts of $\alpha$ and $f(\alpha)$ so they don't intersect before checking the condition or work on the surface and homotop them so they intersect minimally. Then, $f$ is right veering if for all $\alpha$ the condition is true.
The condition: The condition is that the arc to $f(\alpha)$ starts out to the right of $\alpha$ at the boundary (i.e. at $0$). (That is, the initial tangent directions $(f \circ \alpha)'(0)$ and $\alpha'(0)$ form an oriented basis.) But this isn't robust under homotopy (or rather, isotopy of $f$ fixing the boundary), so the correct condition is to first homotop $f(\alpha)$ to intersect $\alpha$ minimally, and then see if it starts out heading rightward. (It would have to come across the arc an extra time if you change which direction it starts, so this is a well-defined concept.)
This first showed up in Honda-Kazez-Matic relating to tight contact structures. The basic idea is that $f$ not right-veering allows you to find an overtwisted disk in the contact manifold associated to the open book with monodromy $f$. Very roughly, this is because the way the contact structure associated to the open book works involves "continuing to twist further" in the right-veering way as you go closer to the binding. So non-right-veering means there'd have to be "extra twisting" and an overtwisted disk, meaning the contact structure is not tight.
| 87,737
|
Out of nowhere, website will not load. I have a Cisco ASA 5505 doing PPPoE authentication for a Qwest Actiontec M1000 modem.
If you open up your web browser it just runs and runs and nothing loads. If you do a nslookup for a domain you get the dns servers. I've tried changing the DNS settings on the ASA to point to Google, etc and I get the same result.
I did a packet trace from an inside IP (Private) to a website IP (global)
nat (inside) 1 0.0.0.0 0.0.0.0 match ip inside any outside any dynamic translation to pool 1 (71.33.87.24 [Interface PAT]) translate_hits = 502, untranslate_hits = 4to a website IP and it says it's dropped at NAT.
Everything worked earlier this morning and all of a sudden it stopped.
3 Replies
Jun 13, 2011 at 7:17 UTC
Need a little more info........
When you say "website" will not load, I am assuming you mean any external website isn't loading?
If that's the case, can you get to a external website by it's IP address?
You said when you do an nslookup, you get the DNS servers. Are you talking about doing an internal DNS lookup or an external, and "what" DNS server does it show when you do an nslookup, internal, external?
-Jay
Jun 13, 2011 at 7:39 UTC
I don't get it. I was tired of doing it remotely so I came down to the site, everything works.
And yes, the external website wouldn't load (google.com, msn.com, etc). I even tried via the IP and it wouldn't work. I tested it out on my home machine and I could get to the websites via the IP (I was doing the troubleshooting on a different network via TeamViewer). I was down at the client's side earlier in the day and noticed it was extremely slow loading a site if it would at all. Found it odd because no one ever complained.
When I did a nslookup I would do an external DNS like google, etc.
Here is the config file for the ASA. Let me know if you see anything that should be changed. I'm using an ACL rule that makes it so you have to use OpenDNS' servers in order to surf the net.
Jun 13, 2011 at 8:13 UTC
Not sure what your speeds are at the client so hard to say what is slow. If you do a tracert from a client machine to an internet address, do the hops show the same basic route you are trying to achieve in that config?
The only thing I see in your config that might make a difference is where you ahve the...
dhcpd address 10.50.1.100-10.50.1.250 inside
dhcpd auto_config outside interface inside
dhcpd enable inside
Just wondering why you didn't add the...
dhcpd dns 205.171.3.65 205.171.2.65
Give that a try and see if it improves.
-Jay
| 100,640
|
TITLE: Relationship between cohomology and higher-homotopy
QUESTION [4 upvotes]: Let $M$ be a connected, compact, and orientable 3-manifold ($H^3(M)\cong\mathbb{Z}$), and let $G$ be a simple Lie group satisfying $\pi_1(G)=\pi_2(G)=0$. Let
$\pi_M(G)$ denote the set of homotopy classes of maps from $M\to G$
$\pi_3(G)$ denote the set of homotopy classes of maps from $\mathbb{S}^3\to G$
Suppose I have a map $f:M\to \mathbb{S}^3$ which induces an isomorphism on $H^3$, and for which $f_*:\pi_3(G)\to \pi_M(G)$ is surjective.
$$f_*[g]:=[g\circ f]$$
Is this enough information to infer that $f_*$ is an isomorphism?
REPLY [1 votes]: If $G$ is a simple Lie group, then $\pi_3(G) = \Bbb Z$ (see here). So the only way this homomorphism could fail to be an isomorphism is if $\pi_M(G)$ were finite. Now $G$ is homotopy equivalent to a space $G'$ with a single 0-cell, a single 3-cell, 4-cells whose boundary maps are the constant maps to the basepoint, and cells of higher dimension; this follows from theorem 4C.1 in Hatcher.
From this, cellular approximation shows that every map $M \to G'$ is homotopic to one lying entirely in the copy of $S^3$; Hopf's theorem says that these are classified by their degree; and therefore even after wedging on the extra copies of $S^4$ they are still not homotopic. (This is because the inclusion $S^3 \hookrightarrow G'$ induces an isomorphism on $H^3$.)
So there are infinitely many homotopy classes of maps $M \to G$ as desired. Your homomorphism is an isomorphism.
| 157,710
|
\begin{document}
\date{}
\maketitle
\begin{center}
{\bf Ualbai Umirbaev}\footnote{Supported by an MES grant 1226/GF3; Eurasian National University,
Astana, Kazakhstan and
Wayne State University,
Detroit, MI 48202, USA,
e-mail: {\em umirbaev@math.wayne.edu}}
\end{center}
\begin{abstract} We prove that the ideal membership problem and the subalgebra membership problem are algorithmically undecidable for differential polynomial algebras with at least two basic derivation operators.
\end{abstract}
\noindent {\bf Mathematics Subject Classification (2010):} Primary 12H05; Secondary 12L05, 13P10, 16E45, 16Z05.
\noindent
{\bf Key words:} Differential polynomial algebras, the ideal membership problem, the subalgebra membership problem, Minsky machines, Gr\"obner bases.
\section{Introduction}
\hspace*{\parindent}
Let $P_n=k[x_1,x_2,\ldots,x_n]$ be the polynomial algebra in the variables $x_1,x_2,\ldots,x_n$ over a constructive field $k$. One of the first applications of Gr\"obner bases (see, for example \cite{CLO}) gives the decidability of the ideal membership problem for $P_n$, i.e., there exists an effective algorithm which for any finite sequence of elements $f,f_1,\ldots,f_m\in P_n$ determines whether $f$ belongs to the ideal $(f_1,\ldots,f_m)$ or not. Another application of Gr\"obner bases \cite{SS} gives the decidability of the subalgebra membership problem for $P_n$, i.e., there exists an effective algorithm which for any finite sequence of elements $f,f_1,\ldots,f_m\in P_n$ determines whether $f$ belongs to the subalgebra $\langle f_1,\ldots,f_m\rangle$ or not. The subalgebra membership problem in characteristic zero was also solved in \cite{Noskov} without Gr\"obner bases.
Traditionally, the ideal membership problem for free algebras is called the word problem for corresponding variety of algebras. The word problem is undecidable for many subvarities of semigroups \cite{Gurevich}, groups, and associative and Lie algebras \cite{Kharlampovich81,SK}. More details on this classical problem can be found in a survey paper \cite{KS}. The decidability of the word problem, in general, is related to the study of Gr\"obner-Shirshov bases \cite{BC}. The word problem is decidable for polynilpotent $\mathfrak{N}_2\mathfrak{A}$-groups \cite{Kharlampovich87} and for polynilpotent $\mathfrak{N}_2\mathfrak{N}_c$-Lie algebras \cite{BU}.
A well known Nielsen-Schreier Theorem states that the subgroups of free groups are free \cite{KMS} and a Shirshov-Witt Theorem states that the subalgebras of free Lie algebras are free \cite{Shir1,Witt}. These results easily imply the decidability of the subalgebra membership problem for free groups and free Lie algebras. The subalgebra membership problem is decidable also for free metabelian groups \cite{Romanovskii} and free metabelian Lie algebras \cite{Zaicev}. It is undecidable for free associative algebras \cite{Um7} and for free solvable Lie algebras \cite{Um8} and for free solvable groups \cite{Um14} of solvability index $\geq 3$. The subalgebra membership problem for free metanilpotent Lie algebras, i.e., $\mathfrak{N}_s\mathfrak{N}_t$-Lie algebras, is decidable \cite{GU1,GU2}.
The basic concepts of differential algebras can be found in \cite{Kolchin,Ritt,PS03}.
Let $\Delta = \{\delta_1,\ldots, \delta_m\}$ be a basic set of derivation operators and
let $\Phi\{x_1,x_2,\ldots,x_n\}$ be the differential polynomial ring in free differential variables $x_1,x_2,\ldots,x_n$ over an arbitrary commutative ring $\Phi$ with unity such that $\delta_i(\Phi)=0$ for all $i$.
There are several approaches to define analogues of the Gr\"obner bases for differential polynomial algebras \cite{CF89,Man91,Ol91} and some recent results can be found in \cite{KLMP}. The differential ideal membership problem is solved positively only in some interesting particular cases (see, for example in \cite{KZ,Z}). At the moment the membership problem for differential ideals generated by a single polynomial is still open. It is negatively solved for recursively generated differential ideals \cite{GMO}. The membership problem for finitely generated differential ideals of differential polynomial algebras was formulated by J.F. Ritt in \cite[p. 177, Question 2]{Ritt}.
In this paper we prove that the membership problem for finitely generated differential ideals is algorithmically undecidable, i.e., the word problem for differential algebras is undecidable. The main instrument of proving this is an interpretation of Minsky machines. The proof uses the fact that every recursive function can be calculated by Minsky machines without cycles \cite{Um7}. Using a method of interpreting the ideal membership problem from \cite{U95A}, we also prove that the membership problem for finitely generated differential subalgebras is undecidable.
Our proofs need at least two derivation operators. Thus, these problems are still open for ordinary differential polynomial algebras.
The rest of the paper is organized as follows. In Section 2 we fix some standard notations and recall some definitions on differential algebras. In Section 3, we give an interpretation of the Minsky machines and prove the undecidability of the ideal membership problem. In Section 4 we give an interpretation of the ideal membership problem and prove the undecidability of the subalgebra membership problem.
\section{Differential polynomial algebras}
\hspace*{\parindent}
All our rings are assumed to be commutative and with unity. Let $R$ be an arbitrary ring. A mapping $d: R\rightarrow R$ is called a {\em derivation} if
\bes
d(s+t)=d(s)+d(t)
\ees
\bes
d(st)=d(s)t+sd(t)
\ees
holds for all $s,t\in R$.
Let $\Delta = \{\delta_1,\ldots, \delta_m\}$ be a basic set of derivation operators.
A ring $R$ is said to be a {\em differential} ring or $\Delta$-ring if all elements of $\Delta$ act on $R$ as a commuting set of derivations, i.e., the derivations $\delta_i: R\rightarrow R$ are defined for all $i$ and $\delta_i\delta_j=\delta_j\delta_i$ for all $i,j$.
Let $\Theta$ be the free commutative monoid on the set $\Delta = \{\delta_1,\ldots, \delta_m\}$ of
derivation operators. The elements
\bes
\theta = \delta_1^{i_1}\ldots \delta_m^{i_m}
\ees
of the monoid $\Theta$ are called {\em derivative}
operators. The {\em order} of $\theta$ is defined as $|\theta|=i_1+\ldots+i_m$.
Let $R$ be a differential ring. Denote by $R^e$ the free left $R$-module with a basis $\Theta$. Every element $u\in R^e$ can be uniquely written in the form
\bes
u=\sum_{\theta\in \Theta} r_{\theta}\theta
\ees
with a finite number of nonzero $r_{\theta}\in R$. We turn $R^e$ to a ring by
\bes
\delta_i r=r \delta_i+ \delta_i(r)
\ees
for all $i$ and $r\in R$. It is well known \cite{KLMP} that these relations uniquely define a structure of a ring on $R^e$ and every left module over $R^e$ is called a {\em differential module} over $R$. In particular, $R$ is a left $R^e$ and every $I\subseteq R$ is a differential ideal of $R$ if and only if $I$ is an $R^e$-submodule of $R$. The ring $R^e$ is called the {\em universal enveloping} ring of $R$.
Let $x^\Theta=\{x^\theta | \theta\in \Theta\}$ be a set of symbols enumerated by the elements of $\Theta$. Consider the polynomial algebra $R[x^\Theta]$ over $R$ generated by the set of (polynomially) independent variables $x^\Theta$. It is easy to check that the derivations $\delta_i$ can be uniquely extended to a derivation of $R[x^\Theta]$ by $\delta_i(x^\theta)=x^{\delta_i\theta}$. Denote this differential ring
by $R\{x\}$; it is called the {\em ring of differential polynomials} in $x$ over $R$.
By adjoining more variables, we can obtain the differential ring $R\{x_1,x_2,\ldots,x_n\}$ of the differential polynomials in $x_1,x_2,\ldots,x_n$ over $R$. Let $M$ be the free commutative monoid generated by all elements $x_i^{\theta}$, where $1\leq i\leq n$ and $\theta\in \Theta$. The elements of $M$ are called {\em monomials} of $R\{x_1,x_2,\ldots,x_n\}$. Every element
$a\in R\{x_1,x_2,\ldots,x_n\}$ can be uniquely written in the form
\bes
a=\sum_{m\in M} r_m m
\ees
with a finite number of nonzero $r_m\in R$.
Every ring can be considered as a differential ring under the trivial action of all derivation operators. If all differential operators act as zeroes on $R$, then $R\{x_1,x_2,\ldots,x_n\}$ becomes an $R$-algebra. In studying of Gr\"obner bases, we usually assume that $R$ is a constructive field $k$ or the ring of integers $\mathbb{Z}$.
\section{The ideal membership problem}
\hspace*{\parindent}
Minsky machines are multi-tape Turing machines \cite{Malcev}.
The hardware of a two-tape Minsky machine consists of two tapes and a head. The tapes are infinite to the right and are divided into infinitely many cells numbered from the left to the right, starting with zero. The external alphabet consists of $0$ and $1$. The first cells on both tapes always contain $1$ and all other cells have $0$. The head may acquire one of several internal states: $q_0,q_1,\ldots,q_n$; $q_0$ is the {\em terminal} state. At every moment the head looks at one cell of the first tape and at one cell of the second tape.
The program of a Minsky machine consists of a set of commands of the form
\bee\label{f1}
q_i \varepsilon \sigma \rightarrow q_j T_{\a}T_{\b},
\eee
where $1\leq i\leq n$, $0\leq j\leq n$, $\varepsilon,\sigma\in \{0,1\}$, $\a,\b\in \{-1,0,1\}$, and $\a\geq 0$ if $\varepsilon=1$ and $\b\geq 0$ if $\sigma=1$. This means that if the head is in the state $q_i$ observing a cell containing $\varepsilon$ on the first tape and a cell containing $\sigma$ on the second tape, then it acquires the state $q_j$ and the first (the second) tape is shifted $\a$ (resp. $\b$) cells to the left relative to the head. If $\a=-1$, for example, then the first tape is shifted one cell to the right.
A configuration of a Minsky machine can be described by a triple $[i,m,n]$, where $m$ and $n$ are the numbers of the cells observed by the head in the first and the second tapes, respectively, and $q_i$ is the internal state of the head. We write
\bes
[i,m,n]\rightarrow [j,p,q],
\ees
if a Minsky machine at the configuration $[i,m,n]$ gets the configuration [j,p,q] in one step, i.e., as a result of execution of one (a unique!) command of the type (\ref{f1}).
Recall that in algorithmic theory the set of natural numbers includes $0$, i.e., $\mathbb{N}=\{0,1,2,\ldots\}$.
Minsky \cite{Malcev} proved that for every partial recursive function $f: \mathbb{N}\rightarrow \mathbb{N}$ there exists a Minsky machine that calculates $f(x)$, i.e., for every natural $x$ it passes from the configuration $[1,2^x,0]$ to the configuration $[0,2^{f(x)},0]$ if $f(x)$ is defined, and operates infinitely, never reaching the terminal state $q_0$, if $f(x)$ is not defined.
We say that a Minsky machine has a cycle if there exists a configuration $[i,m,n]$ such that the machine starting work at this configuration returns to the same configuration in a finite number of positive steps. A Minsky machine without cycles is called {\em acyclic}.
We need the next lemma.
\begin{lm}\label{l1} \cite{Um7}
Let $S$ be a recursively enumerable subset of natural numbers $\mathbb{N}$. Then there exists a two-tape acyclic Minsky machine that for every $x\in \mathbb{N}$ starting work at the configuration $[1,2^{2^x},0]$ reaches $[0,1,0]$ in finitely many steps if $x\in S$ and operates infinitely if $x\notin S$.
\end{lm}
First of all we assume that the basic set of derivations $\Delta = \{\delta_1,\ldots, \delta_m\}$ contains at least two elements. Moreover, we may assume that $\Delta = \{\delta_1,\delta_2\}$ since the other derivations do not hurt our proofs.
We also fix a recursively enumerable subset $S$ of the set of natural numbers $\mathbb{N}$ and fix an acyclic Minsky machine $M$ that calculates the characteristic function of $S$ as in Lemma \ref{l1}. Assume that (\ref{f1}) is the set of all commands of $M$.
Let $\Phi$ be an arbitrary ring. We consider all our algebras over $\Phi$. In the case of positive solutions of algorithmic problems we have to assume that $\Phi$ is constructive (or computable). But it is not mandatory for negative solutions. Of course, we assume that $\Phi$ contains a nonzero unity.
We consider $\Phi$ as a differential ring with the trivial action of the derivation operators. Let $A=\Phi\{x_1,x_2,q_0,q_1,\ldots,q_n\}$ be the free differential algebra over $\Phi$ in free differential variables $x_1,x_2,q_0,q_1,\ldots,q_n$.
With each command of $M$ of the type (\ref{f1}) we associate the element
\bes
f(i,\varepsilon,\sigma)=x_1^{\varepsilon}x_2^{\sigma}\delta_1^{1-\varepsilon}
\delta_2^{1-\sigma}(q_i)
-x_1^{\varepsilon}x_2^{\sigma}\delta_1^{1-\varepsilon+\a}
\delta_2^{1-\sigma+\b}(q_j)
\ees
of the algebra $A$, where $1\leq i\leq n$ and $\varepsilon, \sigma= 0,1$. Denote by $I$ the differential ideal of $A$ generated by all elements $f(i,\varepsilon,\sigma)$.
Denote by $J$ the differential ideal of $A$ generated by the elements
\bes
\delta_1(x_2), \delta_2(x_1).
\ees
Put also
\bes
f_m=x_1x_2\delta_1^{2^{2^m}}(q_1)-x_1x_2\delta_1(q_0)
\ees
for all $m\in \mathbb{N}$.
\begin{pr}\label{p1}
Element $f_m$ of $A$ belongs to the differential ideal $I+J$ if and only if $m\in S$.
\end{pr}
The rest of this section is devoted to the proof of this proposition.
Denote by $B$ the quotient algebra $A/J$.
\begin{lm}\label{l2} The algebra $B$ is a polynomial algebra over $\Phi$ in the polynomial variables
\bee\label{f2}
\delta_1^i(x_1),
\delta_2^i(x_2), \delta_1^i\delta_2^j(q_0),\ldots, \delta_1^i\delta_2^j(q_n),
\eee
where $i,j\geq 0$.
\end{lm}
\Proof Let $R$ be a polynomial algebra over $\Phi$ in the set of variables (\ref{f2}). We can turn $R$ to a differential algebra by
\bes
\delta_1(\delta_1^i(x_1))=\delta_1^{i+1}(x_1), \delta_2(\delta_1^i(x_1))=0,
\delta_2(\delta_2^i(x_2))=\delta_1^{i+1}(x_2),\\ \delta_1(\delta_2^i(x_2))=0,
\delta_1(\delta_1^i\delta_2^j(y))=\delta_1^{i+1}\delta_2^j(y),
\delta_2(\delta_1^i\delta_2^j(y))=\delta_1^i\delta_2^{j+1}(y),
\ees
for all $i,j\geq 0$ and $y\in \{q_0,\ldots,q_n\}$.
Consider the differential homomorphism $\varphi: A\rightarrow R$ defined by $\varphi(x)=x$ for all $x=x_i,q_i$. Obviously, $\varphi(J)=0$ and it easy to check that the induced homomorphism $A/J\rightarrow R$ is an isomorphism. $\Box$
We continue to work with the algebra $B$. The images of elements of $A$ in $B$ will be written in the same way as in the algebra $A$. The images of $f(i,\varepsilon,\sigma), f_m$, and $I$ will be denoted by $g(i,\varepsilon,\sigma), g_m$, and $\widetilde{I}$, respectively.
Notice that $B$ is homogeneous with respect to each of its polynomial generators (\ref{f2}). Moreover, the elements $g(i,\varepsilon,\sigma)$ and $g_m$ are homogeneous with respect to
each of the polynomial variables
\bee\label{f3}
\delta_1^i(x_1), \delta_2^i(x_2), \ \ i\geq 0,
\eee
and with respect to the group of variables
\bee\label{f5}
\delta_1^i\delta_2^j(q_0),\ldots, \delta_1^i\delta_2^j(q_n), \ \ i,j\geq 0.
\eee
Denote by $V$ the set of all monomials in the set of commuting variables (\ref{f2}). Every element of the universal enveloping algebra $B^e$ can be uniquely represented as a linear
combination of elements of the form
\bee\label{f6}
v\delta_1^i\delta_2^j, \ \ v\in V, i,j\geq 0.
\eee
Let $\deg$ be the standard polynomial degree function on $B$, i.e., $\deg(y)=1$ for all elements from (\ref{f2}). All elements $g(i,\varepsilon,\sigma)$ and $g_m$ are homogeneous with respect to $\deg$ and
\bes
\deg(g(i,\varepsilon,\sigma))=1+\varepsilon+\sigma, \ \ \deg(g_m)=3.
\ees
We also define polynomial degree functions $\deg_1$ and $\deg_2$ on $B$ as follows:
$\deg_1(\delta_1^i(x_1))=i+1$ for all $i\geq 0$ and $\deg(y)=0$ for all other variables from (\ref{f2}); $\deg_2(\delta_2^j(x_2))=j+1$ for all $j\geq 0$ and $\deg(y)=0$ for all other variables from (\ref{f2}). For any $v\in V$ put $\mathrm{Deg}(v)=(\deg_1(v),\deg_2(v))$. Let $\leq$ be the lexicographic order on $\mathbb{N}^2$ (recall that $\mathbb{N}$ includes $0$).
For any $v\in V$ denote by $\overline{v}$ its highest homogeneous part with respect to $\mathrm{Deg}$.
The elements $g(i,\varepsilon,\sigma)$ and $g_m$ are also homogeneous with respect to $\mathrm{Deg}$.
\begin{lm}\label{l3}
\bes
\overline{\delta_1^s\delta_2^tg(i,\varepsilon,\sigma)}
=(\delta_1^s(x_1))^{\varepsilon}(\delta_2^t(x_2))^{\sigma}\delta_1^{(s+1)(1-\varepsilon)}
\delta_2^{(t+1)(1-\sigma)}(q_i)\\
-(\delta_1^s(x_1))^{\varepsilon}(\delta_2^t(x_2))^{\sigma}\delta_1^{(s+1)(1-\varepsilon)+\a}
\delta_2^{(t+1)(1-\sigma)+\b}(q_j).
\ees
\end{lm}
\Proof We consider only the case $\varepsilon=1$ and $\sigma=0$ since the other cases can be treated similarly. We have
\bes
g(i,1,0)=x_1\delta_2^1(q_i)
-x_1\delta_1^{\a}\delta_2^{1+\b}(q_j).
\ees
Consequently,
\bes
\delta_1^s\delta_2^tg(i,1,0)=\delta_1^s\delta_2^t(x_1\delta_2^1(q_i)
-x_1\delta_1^{\a}\delta_2^{1+\b}(q_j))\\
=\delta_1^s(x_1\delta_2^{t+1}(q_i)
-x_1\delta_1^{\a}\delta_2^{t+1+\b}(q_j))\\
=\sum_{r=0}^s(\delta_1^r(x_1)\delta_1^{s-r}\delta_2^{t+1}(q_i)
-\delta_1^r(x_1)\delta_1^{s-r+\a}\delta_2^{t+1+\b}(q_j)).
\ees
Consequently,
\bes
\overline{\delta_1^s\delta_2^tg(i,1,0)}=\delta_1^s(x_1)\delta_2^{t+1}(q_i)
-\delta_1^s(x_1)\delta_1^{\a}\delta_2^{t+1+\b}(q_j).
\ees
This proves the statement of the lemma for $\varepsilon=1$ and $\sigma=0$. $\Box$
With each element of $B$ of the form
\bee\label{f7}
u= \delta_1^a(x_1)\delta_2^b(x_2)\delta_1^s\delta_2^t(q_i), \ \ a,b\geq 1, s,t\geq 0,
\eee
we associate the configuration $[i,s,t]$ of the Minsky machine $M$.
Denote by $V_{\varepsilon \sigma}$ the set of all elements of $B^e$ of the form
\bes
w=(\delta_1^a(x_1))^{1-\varepsilon}(\delta_2^b(x_2))^{1-\sigma}\delta_1^s\delta_2^t,
\ees
where $a,b\geq 1$ and $s,t\geq 0$
Every $v\in V$ can be uniquely represented as $v=v_1v_2$, where $v_1$ is a monomial in the variables (\ref{f3}) and $v_2$ is a monomial in the variables (\ref{f5}). We have $\mathrm{Deg}(v)=\mathrm{Deg}(v_1)$ and $\mathrm{Deg}(v_2)=(0,0)$. We denote $v_1$ by $\{v\}$.
\begin{lm}\label{l4}
Let $u$ and $v$ be two elements of the form (\ref{f7}). Then
\bee\label{f8}
u-v=\overline{w g(i,\varepsilon,\sigma)}
\eee
for some $w\in V_{\varepsilon \sigma}$ if and only if $\{u\}=\{v\}$ and $[u]\rightarrow [v]$ as a result of the execution of the command (\ref{f1}) (or $[v]\rightarrow [u]$ if $1+1=0$ in $\Phi$).
\end{lm}
\Proof We consider only the case $\varepsilon=1$ and $\sigma=0$. Then
$w\in V_{1 0}$ has the form
\bes
w=\delta_2^r(x_2)\delta_1^s\delta_2^t.
\ees
By Lemma \ref{l3},
\bes
\overline{w(g(i,1,0))}=\delta_2^r(x_2)\overline{\delta_1^s\delta_2^t(g(i,1,0))}\\
=\delta_1^s(x_1)\delta_2^r(x_2)
\delta_2^{t+1}(q_i)
-\delta_1^s(x_1)\delta_2^r(x_2)\delta_1^{\a}
\delta_2^{t+1+\b}(q_j)
\ees
Assume that $1+1\neq 0$ in $\Phi$. Then (\ref{f8}) holds if and only if
\bes
u=\delta_1^s(x_1)\delta_2^r(x_2)
\delta_2^{t+1}(q_i), \ \ v=\delta_1^s(x_1)\delta_2^r(x_2)\delta_1^{\a}
\delta_2^{t+1+\b}(q_j).
\ees
Notice that $u,v$ has the form (\ref{f7}), $\{u\}=\{v\}$, and $[u]=[i,0,t+1]$ and $[v]=[j,\a,t+1+\b]$. We get $[u]\rightarrow [v]$ as a result of the execution of the command (\ref{f1}).
If $1+1= 0$ in $\Phi$, then
\bes
v=\delta_1^s(x_1)\delta_2^r(x_2)
\delta_2^{t+1}(q_i), \ \ u=\delta_1^s(x_1)\delta_2^r(x_2)\delta_1^{\a}
\delta_2^{t+1+\b}(q_j)
\ees
is possible. In this case we get $[v]\rightarrow [u]$.
$\Box$
For each $\varepsilon, \sigma= 0,1$, denote by $W_{\varepsilon \sigma}$ the set of all elements of the form
\bes
x_1^{1-\varepsilon}x_2^{1-\sigma}\delta_1^i\delta_2^j, \ \ \ i,j\geq 0,
\ees
such that $i=0$ if $\varepsilon=1$ and $j=0$ if $\sigma=1$. In particular, we have $W_{1 1}=\{1\}$.
\begin{co}\label{c1}
Let $u$ and $v$ be two elements of the form (\ref{f7}) such that $\{u\}=\{v\}=x_1x_2$. Then the equality (\ref{f8}) holds only if $w\in W_{\varepsilon \sigma}$ and in this case
\bes
u-v=w g(i,\varepsilon,\sigma).
\ees
\end{co}
\Proof Again consider only the case $\varepsilon=1$ and $\sigma=0$. If $\{u\}=\{v\}=x_1x_2$, then using the proof of Lemma \ref{l4}, we get
\bes
s=0, r=0,
\ees
and consequently,
\bes
u=x_1x_2
\delta_2^{t+1}(q_i), v=x_1x_2\delta_1^{\a}
\delta_2^{t+1+\b}(q_j), w=x_2\delta_2^t\in W_{1 0}.
\ees
Then
\bes
w(g(i,1,0))=u-v. \ \ \Box
\ees
\begin{co}\label{c2}
If $m\in S$, then $g_m\in \widetilde{I}$.
\end{co}
\Proof
If $m\in S$, then there exists a sequence of configurations
\bes
[1,2^{2^m},0]=c_0\rightarrow c_1\rightarrow\ldots c_r=[0,1,0].
\ees
of the Minsky machine $M$. For each configuration $c_i$ there exists a unique element $u_i$ of the form (\ref{f7}) such that
$[u_i]=c_i$ and $\{u_i\}=x_1x_2$. Notice that $g_m=u_0-u_r$. By Lemma \ref{l3} and Corollary \ref{c1}, we have $u_i- u_{i+1}\in \widetilde{I}$ for all $0\leq i<r$. Consequently,
\bes
g_m=u_0-u_r=(u_0-u_1)+(u_1-u_2)+\ldots+(u_{r-1}-u_r)\in \widetilde{I}. \ \ \ \Box
\ees
\begin{lm}\label{l5}
If $g_m$ is a linear combination of elements of the form
\bes
w g(i,\varepsilon,\sigma),
\ees
where $w\in W_{\varepsilon \sigma}$, then $m\in S$.
\end{lm}
\Proof Put $u=x_1x_2\delta_1^{2^{2^m}}(q_1)$ and $v=x_1x_2\delta_1(q_0)$. Then $g_m=u-v$. By Lemma \ref{l4}, we may assume that
\bee\label{f9}
u-v=\lambda_1(u_1-v_1)+\lambda_2(u_2-v_2)+\ldots+\lambda_r(u_r-v_r),
\eee
where all $u_i,v_i$ are elements of the form (\ref{f7}) and $[u_i]\rightarrow [v_i]$ for all $1\leq i\leq r$. Assume that $r$ is the minimal number satisfying (\ref{f9}). This condition immediately implies that $u_i\neq v_i$.
In order to prove that $m\in S$, it is sufficient to show the existence of a sequence of configurations of the form
\bes
[u]\rightarrow\ldots\rightarrow [v].
\ees
If $u_i=u_j$, then $v_i=v_j$ since $[u_i]\rightarrow [v_i]$ for all $i$. Consequently, we may assume that all $u_1,u_2,\ldots,u_r$ are different. The machine $M$ at the configuration $[v]=[0,1,0]$ immediately stops its work since it is in the internal state $q_0$. For the same reason, the machine at the configurations $[u_1],[u_2],\ldots,[u_r]$ is not in the internal state $q_0$.
This means that $v$ contains $q_0$ and $u_1,u_2,\ldots,u_r$ do not contain it.
All elements $u,v,u_1,v_1,\ldots,u_r,v_r$ belong to a linearly independent set of elements (\ref{f7}). Then the equality (\ref{f9}) implies that
$v$ coincides with one of $v_1,v_2,\ldots,v_r$. Without loss of generality, we may assume that $v=v_1$. If $u=u_1$, then $[u]=[u_1]\rightarrow [v_1]=[v]$. Otherwise (\ref{f9}) implies that
$u_1$ coincides with one of $v_2,\ldots,v_r$. Without loss of generality, we may assume that $u_1=v_2$.
Continuing this discussion, we may assume
that $v=v_1, u_1=v_2,\ldots,u_s=v_{s+1}$, $u\neq u_1,u_2,\ldots,u_s$, and $s$ is the maximal number with this property. If $u=u_{s+1}$, then
\bes
[u]=[u_{s+1}]\rightarrow [v_{s+1}]=[u_s]\rightarrow\ldots \rightarrow [v_1]=[v].
\ees
If $u\neq u_{s+1}$, then (\ref{f9}) implies that
$u_{s+1}$ coincides with one of $v_1,v_2,\ldots,v_r$.
If $u_{s+1}=v_j$ for some $1\leq j\leq s+1$, then we get
\bes
(u_j-v_j)+(u_{j+1}-v_{j+1})+\ldots+(u_{s+1}-v_{s+1})=0.
\ees
This allows to reduce the number $r$ in (\ref{f9}).
Consequently, $u_{s+1}$ coincides with one of $v_{s+2},\ldots,v_r$.
We may assume that $u_{s+1}=v_{s+2}$ and this contradicts the maximality of $s$. $\Box$
We intentionally avoided to use the acyclicity of $M$ in the proof of Lemma \ref{l6}. The next lemma is not true for machines without cycles.
\begin{lm}\label{l6} The set of elements of the form
\bee\label{f10}
\overline{w g(i,\varepsilon,\sigma)},
\eee
where $w\in V_{\varepsilon \sigma}$, $1\leq i\leq n$, and $\varepsilon,\sigma\in \{0,1\}$,
are linearly independent over $\Phi$.
\end{lm}
\Proof By Lemma \ref{l4}, every element of the form (\ref{f10}) can be represented as $u-v$, where $u$ and $v$ are elements of the form (\ref{f7}) such that $\{u\}=\{v\}$ and $[u]\rightarrow [v]$.
First of all, notice that $u-v\neq 0$. If $u=v$, then $[u]\rightarrow [v]=[u]$ becomes a nontrivial cycle of the machine $M$. Recall that $M$ is acyclic.
A nontrivial linear dependence of elements of the form
(\ref{f10}) can be written in the form
\bee\label{f11}
\lambda_1(u_1-v_1)+\lambda_2(u_2-v_2)+\ldots+\lambda_r(u_r-v_r)=0,
\eee
where $0\neq \lambda_1,\lambda_2,\ldots,\lambda_r\in \Phi$ and $u_i$ and $v_i$ are elements of the form (\ref{f7}) such that $\{u_i\}=\{v_i\}$ and $[u_i]\rightarrow [v_i]$ for all $i$.
We noticed that $u_i=u_j$ implies $v_i=v_j$. Therefore, we may assume that all $u_1,u_2,\ldots,u_r$ are different. Start with $v_1$ as in the proof of Lemma \ref{l5}. We have $[u_1]\rightarrow [v_1]$.
Then (\ref{f11}) implies that $u_1$ coincides with one of $v_2,v_3,\ldots,v_r$. We may assume that $u_1=v_2$. If $u_2=v_1$, then we get a cycle $[u_2]\rightarrow [v_2]=[u_1]\rightarrow [v_1]=[u_2]$ of the machine $M$. We also know that $u_2\neq v_2$. Consequently, $u_2$ coincides with one of $v_3,\ldots,v_r$.
Assume
that $u_1=v_2,\ldots,u_s=v_{s+1}$ and $s$ is the maximal number with this property. If $u_{s+1}=v_j$ for some $1\leq j\leq s+1$, the we get a cycle
\bes
[u_{s+1}]\rightarrow [v_{s+1}]=[u_s]\rightarrow[v_s]\rightarrow\ldots \rightarrow [v_j]=[u_{s+1}]
\ees
of $M$. Consequently, $u_{s+1}\neq v_1,v_2,\ldots,v_{s+1}$. Then (\ref{f11}) implies that
$u_{s+1}$ coincides with one of $v_{s+2},\ldots,v_r$. We may assume that $u_{s+1}=v_{s+2}$ and this contradicts to the maximality of $s$. $\Box$
\begin{lm}\label{l7}
If $g_m\in \widetilde{I}$, then $m\in S$.
\end{lm}
\Proof Let $g_m\in \widetilde{I}$. Then
\bee\label{f12}
g_m=\sum_{s,i,\varepsilon,\sigma} \lambda_{s,i,\varepsilon,\sigma} w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma),
\eee
where $\lambda_{s,i,\varepsilon,\sigma}\in \Phi$ and $w_s(\varepsilon,\sigma)\in B^e$ are elements of the form (\ref{f6}).
Notice that all elements $w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma)$ and $g_m$ are homogeneous with respect to each set of variables
(\ref{f3})-(\ref{f5}) and with respect to degree function $\deg$. Recall that
\bes
\deg(g_m)=3, \deg(g(i,\varepsilon,\sigma))=1+\varepsilon +\sigma.
\ees
Consequently, we may assume that $w_s(\varepsilon,\sigma)\in V_{\varepsilon,\sigma}$ in (\ref{f12}).
Suppose that there exists at least one $w_s(\varepsilon,\sigma)$ that does not belong to
$W_{\varepsilon,\sigma}$. In this case we get $\mathrm{Deg}(w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma))>(1,1)$ by Corollary \ref{c1}. Let $(c,d)$ be the highest degree of all elements $w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma)$ participating in (\ref{f12}) with respect to $\mathrm{Deg}$. We have $(c,d)>(1,1)=\mathrm{Deg}(g_m)$. Then the equality (\ref{f12}) implies a nontrivial linear dependence of the highest homogeneous parts $\overline{w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma)}$ of elements $w_s(\varepsilon,\sigma) g(i,\varepsilon,\sigma)$ with the degree $(c,d)$. It is impossible by Lemma \ref{l6}.
Therefore, we may assume that all $w_s(\varepsilon,\sigma)$ in (\ref{f12}) belong to $W_{\varepsilon,\sigma}$. Then Lemma \ref{l5} implies that $m\in S$. $\Box$
{\bf Proof of Proposition 1}. Notice that $f_m\in I+J$ in the algebra $A$ if and only if $g_m\in \widetilde{I}$ in the algebra $B$. By Corollary \ref{c2} and Lemma \ref{l7}, $g_m\in \widetilde{I}$ if and only if $m\in S$. $\Box$
Proposition \ref{p1} immediately implies the next result.
\begin{theor}\label{t1}
The ideal membership problem for differential polynomial algebras with at least two basic derivations is algorithmically undecidable.
\end{theor}
\Proof Let $S$ be a recursively enumerable but nonrecursive set \cite{Malcev}. This means that there is no algorithm which determines for every natural $m$ whether $m\in S$ or not. Assume that the algebra $A$ and its ideal $I+J$ are constructed by the commands of an acyclic machine $M$ calculating the characteristic function of $S$. By Proposition \ref{p1}, $m\in S$ if and only if $f_m\in I+J$. Consequently, there is no algorithm which determines for all $m$ whether $f_m\in I+J$ or not. $\Box$
\section{The subalgebra membership problem}
\hspace*{\parindent}
Let $\Delta = \{\delta_1,\ldots, \delta_m\}$ be a basic set of derivation operators and $\Theta$ is the free commutative monoid on $\Delta$.
Let $A=\Phi\{x_1,x_2,\ldots,x_n\}$ be a differential polynomial algebra over $\Phi$ and let
$I=[f_1,f_2,\ldots,f_r]$ be a differential ideal of $A$ generated by $f_1,f_2,\ldots,f_r$.
Let $B=\Phi\{x_1,x_2,\ldots,x_n,t\}$ be a differential polynomial algebra with one more free differential variable $t$. We identify $A$ with the corresponding subalgebra of $B$. Denote by $S_I$ the differential subalgebra of $B$ generated by
\bes
x_1,x_2,\ldots,x_n,\delta_1(t),\delta_2(t),\ldots,\delta_m(t),tf_1,tf_2,\ldots,tf_r.
\ees
\begin{pr}\label{p2}
Let $f\in A$. Then $f\in I$ if and only if $tf\in S_I$.
\end{pr}
\Proof Let $M$ be the free commutative monoid generated by all elements
$x_i^{\theta}$, where $1\leq i\leq n$ and $\theta\in \Theta$. Then every element of $A^e$ is a linear combination of elements of the form
\bes
m\theta, \ \ m\in M, \theta\in \Theta.
\ees
If $f\in I$, then
\bes
f=\sum_{m,\theta,i} \lambda_{m,\theta,i}m\theta f_i
\ees
for some $\lambda_{m,\theta,i}\in \Phi$, $m\in M$, $\theta\in \Theta$, and $1\leq i\leq r$.
Let $T$ be the subalgebra of $S_I$ generated by the elements
\bes
x_1,x_2,\ldots,x_n,\delta_1(t),\delta_2(t),\ldots,\delta_m(t).
\ees
Notice that $M\subset A\subset T$. It is easy to check that
\bee\label{f13}
\theta(tf_i)=t\theta(f_i)+g, g\in T.
\eee
Consequently,
\bes
tf=\sum_{m,\theta,i} \lambda_{m,\theta,i}mt\theta(f_i)=\sum_{m,\theta,i} \lambda_{m,\theta,i}m\theta (tf_i)+g, \ \ g\in T.
\ees
Therefore, $tf\in S_I$.
Denote by $\deg_t$ the polynomial degree function on $B$ such that $\deg_t(t^\theta)=1$ and $\deg_t(x_i^\theta)=0$ for all $i$ and $\theta$. All generates of the subalgebra $S_I$ are homogeneous with respect to $\deg_t$. Denote by $H$ the subspace of all homogeneous elements of degree $1$ of $S_I$ with respect to $\deg_t$. Then every element of $H$ is a linear combination of elements of the form
\bes
mt^{\theta_1}, m \theta(tf_i),
\ees
where $\theta_1,\theta\in \Theta$, $|\theta_1|\geq 1$, $m\in M$, and $1\leq i\leq r$.
Taking into account (\ref{f13}), we may assume that every element of $H$ is a linear combination of elements
\bes
mt^{\theta_1}, mt \theta(f_i).
\ees
Moreover, every element of $H$ divisible by $t$ is a linear combination of elements
\bes
mt \theta(f_i), \ \ m\in M, \theta\in \Theta, 1\leq i\leq r.
\ees
Assume that $tf\in S_I$ for some $f\in A$. We have $tf\in H$ and $tf$ divisible by $t$. Then
\bes
tf=\sum_{m,\theta,i} \lambda_{m,\theta,i} mt \theta(f_i).
\ees
Consequently,
\bes
f=\sum_{m,\theta,i} \lambda_{m,\theta,i} m \theta(f_i)\in I. \ \ \Box
\ees
\begin{theor}\label{t2}
The subalgebra membership problem for differential polynomial algebras with at least two basic derivations is algorithmically undecidable.
\end{theor}
\Proof By Theorem \ref{t1}, the ideal membership problem is undecidable. By Proposition \ref{p2}, if the ideal membership problem for $A$ is undecidable, then the subalgebra membership problem for $B$ is also undecidable. $\Box$
The method of this section gives the undecidability of the subalgebra membership problem for free algebras of many varieties of algebras \cite{U95A}. But this method does not work for subfields of fields. Recall that the subfield membership problem for fields of rational functions $k(x_1,x_2,\ldots,x_n)$ over a constructive field $k$ is also positively solved by means of Gr\"obner bases \cite{Sweedler93}. The subfield membership problem for differential fields of rational functions remains open.
| 185,660
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TITLE: Are morphisms of parametrized spectra themselves parametrized morphisms of spectra?
QUESTION [2 upvotes]: Let $X$ be a fixed parametrizing space. Let $E$ and $E'$ be two spectra and let $E_X$ and $E'_X$ be their trivial parametrized versions. Intuitively I imagine that the morphisms of parametrized spectra $E_X\to E'_X$ should correspond to maps $X\to \operatorname{Map}(E, E')$ where the mapping space on the right should be the underlining infinite loop space of the function spectrum, if you wish $\operatorname{Map}(E, E')=\Omega^\infty F(E, E').$
Is this, or at least anything similar to it, actually true?
REPLY [5 votes]: This is true.
To prove it I will use the fact that maps of parametrized spectra can be computed as natural transformations of functors from $X$ into the $\infty$-category of spectra (cfr. this paper) and the formula for computing the space of natural transformation (e.g. see here, proposition 2.3):
$$\mathrm{Map}(E_X,E'_X) = \int_{x\in X} \mathrm{Map}(E_x,E'_x) = \int_{x\in X} \mathrm{Map}(E,E') = \mathrm{Map}(\tilde O(X), \mathrm{Map}(E,E')) = \mathrm{Map}(X,\mathrm{Map}(E,E'))$$
Here $\tilde O(X)$ is the twisted arrow category of $X$, seen as an $\infty$-groupoid, which is just equivalent to $X$ (one way to see this is that the canonical projection $\tilde O(X)\to X$ is a left fibration with contractible fibers).
| 146,699
|
\begin{document}
\maketitle
\begin{abstract}
In this article, we study complete Type I ancient Ricci flows with positive sectional curvature. Our main results are as follows: in the complete and noncompact case, all such ancient solutions must be noncollapsed on all scales; in the closed case, if the dimension is even, then all such ancient solutions must be noncollapsed on all scales. This furthermore gives a complete classification for three-dimensional noncompact Type I ancient solutions without assuming the noncollapsing condition.
\end{abstract}
\section{Introduction to the Main Results}
The study of ancient solutions to the Ricci flow, ever since Hamilton had published his program \cite{Ha1}, has been an important topic in the field of Ricci flow. Ancient solutions are Ricci flows whose existing intervals extend to negative infinity. They are of great importance because they usually arise as blow-up limits at finite-time singularities of the Ricci flow, and to this kind of blow-up limits, a term not improper, ``singularity models'', is assigned.
Perelman \cite{P} proved that a Ricci flow on a closed manifold cannot become locally collapsed within finite time. Subsequently one may conclude that every singularity model must be $\kappa$-noncollapsed on all scales. This precisely means the following.
\begin{Definition}[$\kappa$-noncollapsing]
A Ricci flow $(M^n,g(t))$ is called $\kappa$-noncollapsed on all scales, where $\kappa$ is a positive number, if for any point $(x,t)$ in space-time and any positive scale $r$, it holds that $\displaystyle\operatorname{Vol}_{g(t)}\big(B_{g(t)}(x,r)\big)\geq\kappa r^n$ whenever $R\leq r^{-2}$ on $B_{g(t)}(x,r)$. Here $R$ stands for the scalar curvature
\end{Definition}
The noncollapsing notion defined above is sometimes called the \emph{strong} noncollapsing in the literature. The \emph{weak} noncollapsing notion is defined similarly with only the ``whenever $R\leq r^{-2}$ on $B_{g(t)}(x,r)$'' statement replaced by ``whenever $|Rm|\leq r^{-2}$ on $B_{g(t)}(x,r)\times [t-r^2,t]$''. It is known that for an ancient solution with bounded and nonnegative curvature operator, the weak noncollapsing is equivalent to the strong noncollapsing condition, with possibly a different $\kappa$. The noncollapsing condition which we use throughout this paper is the strong one. We remark that these two notions are not equivalent in general. For instance, a closed nonflat and Ricci-flat (static) Ricci flow is weakly noncollapsed but not strongly noncollapsed.
Since, according to Hamilton \cite{Ha1}, ancient solutions are critical to the understanding of the singularity formation in the Ricci flow (see, for instance, Perelman's proof of the canonical neighborhood theorem \cite{P}), it makes sense to assume the noncollapsing condition when studying ancient solutions. With this assumption, many groundbreaking works are done, and the most outstanding one is of Perelman \cite{P}. See also \cite{B1}, \cite{B2}, \cite{B}, and \cite{L}, etc., to list but a few.
It is well-known that not all ancient solutions are noncollpased. But what if some further conditions are added? Concerning this a question is proposed in \cite{CLN}:
\begin{quotation}
\noindent\textbf{Problem 9.41.} \emph{Are nonflat Type I ancient solutions with nonnegative curvature operator $\kappa$-solutions?}
\end{quotation}
Recall that an ancient solution $(M,g(t))_{t\in(-\infty,w)}$ is called \emph{Type I} if
\begin{eqnarray*}
\limsup_{t\rightarrow-\infty}|t|\big|Rm_{g(t)}\big|<\infty.
\end{eqnarray*}
Without any further qualification, the answer to the above question is obviously ``no'', either in the closed case or in the complete and noncompact case. One may immediately think of $\mathbb{S}^{n-1}\times\mathbb{S}^1$ or $\mathbb{S}^{n-2}\times\mathbb{R}\times\mathbb{S}^1$ as counterexamples. There are also much more sophisticated counterexamples constructed. For instance, Fateev \cite{F} constructed an ancient solution on the Hopf fiber bundle, and Bakas-Kong-Ni \cite{BKN} generalized this construction to all odd-dimensional spheres; all these ancient solutions are Type I, collapsed, and with positive curvature operator.
In this article, we give a relatively satisfactory answer to Problem 9.41 in \cite{CLN} as quoted above. First of all, to rule out the possibility of a compact flat factor, with which the ancient solution is always collapsed, we would like to assume that the sectional curvature is strictly positive. This condition also largely simplifies the geometry in the complete and noncompact case, since the underlying manifold must be diffeomorphic to the Euclidean space by the Gromoll-Meyer theorem.
\begin{Theorem} \label{MainTheorem1}
Let $(M^n,g(t))_{t\in(-\infty,w)}$, where $0<w\leq\infty$, be a complete and noncompact Type I ancient solution with positive sectional curvature. Then $(M^n,g(t))_{t\in(-\infty,0]}$ is $\kappa$-noncollapsed on all scales for some $\kappa>0$.
\end{Theorem}
The Type I, closed, and collapsed examples constructed in \cite{BKN} are only in odd dimensions, while the cases of even dimensions are yet open. The following theorem shows that there are no such collapsed examples in even dimensions.
\begin{Theorem}\label{MainTheorem2}
Let $(M^n,g(t))_{t\in(-\infty,w)}$, where $n$ is an even number and $0<w\leq\infty$, be a closed Type I ancient solution with positive sectional curvature. Then $(M^n,g(t))_{t\in(-\infty,0]}$ is $\kappa$-noncollapsed on all scales for some $\kappa>0$.
\end{Theorem}
The critical observations applied to the proofs of Theorem \ref{MainTheorem1} and Theorem \ref{MainTheorem2} are some injectivity radius estimates resulted from the Gromoll-Meyer theorem and the Klingenberg theorem. These classical theorems imply that an ancient solution as described in Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2} has a Type I injectivity radius lower bound. This is sufficient to conclude the existence of an asymptotic shrinker, which in turn implies noncollapsedness; the authors used a similar idea in \cite{ChZ} to prove the noncollapsedness of a more general type of ancient Ricci flows---the locally uniformly Type I ancient solutions. Here we emphasize that the Type I injectivity radius lower bound itself does not directly imply the noncollapsedness; see section 2 for more details concerning this point.
An immediate application of Theorem \ref{MainTheorem1} is the following classification of three-dimensional Type I ancient solutions without assuming the noncollapsing condition. This is a generalization of \cite{Z1} (or \cite{H}). In consequence, all noncompact three-dimensional Type I ancient solutions must be noncollapsed.
\begin{Corollary} \label{3d}
A three-dimensional noncompact Type I ancient solution must be the standard cylinder $\mathbb{S}^2\times\mathbb{R}$, $(\mathbb{S}^2\times\mathbb{R})/\mathbb{Z}_2$, or $\operatorname{\mathbb{R}P}^2\times\mathbb{R}$. Hence it must also be $\kappa$-noncollapsed on all scales for some $\kappa>0$.
\end{Corollary}
In \cite{CLN}, it is asked in Problem 9.40 whether a Type I ancient solution with positive curvature operator is closed. While we are not yet able to give an answer to this question, the following Corollary rules out one possibility of its asymptotic shrinker---the standard cylinder. This also means that if such an ancient solution did exist, then its geometry could not be very simple (though topologically it is the Euclidean space).
\begin{Corollary} \label{noncompact}
A Type I complete and noncompact ancient Ricci flow with nonnegative curvature operator and positive sectional curvature cannot have the standard cylinder $\mathbb{S}^{n-1}\times\mathbb{R}$ as its asymptotic shrinker.
\end{Corollary}
Ni \cite{Ni2} classified all closed Type I ancient and noncollapsed Ricci flows with nonnegative curvature operator. It is noted in \cite{BKN} that, because of the examples therein, the noncollapsing condition in the classification of \cite{Ni2} cannot be dropped. Nonetheless, because the examples in \cite{BKN} are only in odd dimensions, this conclusion is not decisive in even dimensions. Indeed, the noncollapsing condition can be dropped in even dimension, at least for the strictly positive curvature operator case.
\begin{Corollary} \label{compact}
An even-dimensional closed Type I ancient Ricci flow satisfying the strict $\operatorname{PIC}-2$ curvature condition (and in particular, with positive curvature operator) must be a standard shrinking round space form.
\end{Corollary}
Here we remark that, unlike \cite{Ni2}, we are not able to deal with the case when the curvature operator admits a zero eigenvalue, since in this case we can no longer prove the noncollapsedness. For instance, one may think of $\mathbb{S}^1\times\mathbb{S}^{2m+1}$, where the $\mathbb{S}^1$ factor is static and the $\mathbb{S}^{2m+1}$ factor is the standard shrinking sphere.
\\
This paper is organized as follows. In section 2 we use the Gromoll-Meyer theorem and the Klingenberg theorem to derive the Type I injectivity radius lower bound for the ancient solutions in question. In section 3 we review the fact that Perelman's entropy and the Nash entropy converge to the entropy of the asymptotic shrinker. In section 4 we show that the existence of asymptotic shrinker implies noncollapsedness. In section 5 we prove all the corollaries.
\emph{Acknowledgment.} The second author would like to thank Professor Jiaping Wang, Professor Lei Ni, and Professor Bennett Chow for many helpful discussions.
\section{The injectivity radius}
The injectivity radius estimates are provided by the following classical theorems of Gromoll-Meyer and Klingenberg.
\begin{Proposition}[Gromoll-Meyer; c.f. Theorem 1.168 in \cite{CLN}]\label{Gromoll-Meyer}
Let $(M^n,g)$ be a complete and noncompact Riemannian manifold satisfying $0<\operatorname{sec}\leq K$, where $K$ is a positive number. Then the injectivity radius of $(M^n,g)$ satisfies
\begin{eqnarray*}
\operatorname{inj}(g)\geq\frac{\pi}{\sqrt{K}}.
\end{eqnarray*}
Moreover, $(M^n,g)$ is diffeomorphic to the standard Euclidean space.
\end{Proposition}
\begin{Proposition}[Klingenberg; c.f. Theorem 1.115 in \cite{CLN}]\label{Klingenberg}
Let $(M^n,g)$ be an even-dimensional closed orientable manifold satisfying $0<\operatorname{sec}\leq K$, where $K$ is a positive number. Then the injectivity radius of $(M^n,g)$ satisfies
\begin{eqnarray*}
\operatorname{inj}(g)\geq\frac{\pi}{\sqrt{K}}.
\end{eqnarray*}
\end{Proposition}
The results above imply that the ancient solutions in question have Type I injectivity radii lower bounds.
\begin{Lemma}\label{inj-radius}
Let $(M^n,g(t))_{t\in(-\infty,w)}$ be an ancient solution as described in either Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2}. Then there exists a constant $c>0$, such that the injectivity radius of $g(t)$ satisfies
\begin{eqnarray*}
\operatorname{inj}(g(t))\geq c\sqrt{|t|},
\end{eqnarray*}
for all $t\in(-\infty,0)$.
\end{Lemma}
\begin{proof}
By the Type I and the positive sectional curvature conditions, we have that there exists $C>0$, such that
\begin{eqnarray*}
0<\operatorname{sec}(g(t))\leq\frac{C}{|t|},
\end{eqnarray*}
for all $t\in(-\infty,0)$. The lemma then follows from Proposition \ref{Gromoll-Meyer} and Proposition \ref{Klingenberg}. Note that though Proposition \ref{Klingenberg} requires orientability, which is not assumed in the statement of Theorem \ref{MainTheorem2}, yet one may always consider the double cover if necessary.
\end{proof}
Here we remark again that this Type I injectivity radius lower bound does not imply $\kappa$-noncollapsedness on all scales, since at some point on the manifold, the curvature could decay faster than Type I. Nevertheless, this estimate is sufficient for the existence of an asymptotic shrinker, which in turn implies $\kappa$-noncollapsedness on all scales.
\\
\section{The Asmptotic Shrinker}
Perelman \cite{P} and Naber \cite{Na} proved the existence of the asymptotic shrinker for ancient solutions under the assumptions of nonnegative curvature operator and of Type I curvature bound, respectively. They both also assumed the noncollapsing condition. However, it turns out that the only place where they applied this condition was to obtain an injectivity radius lower bound for a blow-down sequence, and they need this injectivity radius lower bound only at base points to conclude the convergence. This, of course, can be covered by Lemma \ref{inj-radius}. A more important fact is that Perelman's entropy and the Nash entropy on the ancient solution converge to the entropy of the asymptotic shrinker; see Proposition \ref{Asymptotic_Shrinker} below. Let us first review these notions of entropy.
Let $(M,g(t))_{t\in(-\infty,w)}$ be an ancient solution, where $0<w\leq\infty$. Let $(x_0,t_0)\in M\times(-\infty,w)$ be a fixed point in space-time and $u: M\times (-\infty,t_0)\rightarrow\mathbb{R}_+$ the fundamental solution to the conjugate heat equation $\displaystyle -\partial_t u-\Delta u+Ru=0$ based at $(x_0,t_0)$. We write $u$ as
\begin{eqnarray*}
u:=(4\pi\tau)^{-\frac{n}{2}}e^{-f},
\end{eqnarray*}
where $\tau=t_0-t\in(0,\infty)$. Then Perelman's entropy and the Nash entropy based at $(x_0,t_0)$ are respectively defined as
\begin{eqnarray*}
\mathcal{W}_{x_0,t_0}(\tau)&:=&\int_M\Big(\tau\big(|\nabla f|^2+R\big)+f-n\Big)udg_t,
\\
\mathcal{N}_{x_0,t_0}(\tau)&:=&\int_M fudg_t-\frac{n}{2}.
\end{eqnarray*}
It is a well known fact that both $\mathcal{W}_{x_0,t_0}(\tau)$ and $\mathcal{N}_{x_0,t_0}(\tau)$ are increasing in $t$ (or decreasing in $\tau$). Furthermore, we have
\begin{eqnarray*}
\lim_{\tau\rightarrow 0+}\mathcal{W}_{x_0,t_0}(\tau)=\lim_{\tau\rightarrow 0+}\mathcal{N}_{x_0,t_0}(\tau)=0.
\end{eqnarray*}
The following result is already well-established in literature.
\begin{Proposition}\label{Asymptotic_Shrinker}
Let $(M,g(t))_{t\in(-\infty,w)}$, where $0<w\leq\infty$, be an ancient solution as described in Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2}. Let $(x_0,t_0)\in M\times(-\infty,w)$ be an arbitrarily fixed point and $\displaystyle u:=(4\pi\tau)^{-\frac{n}{2}}e^{-f}$ the conjugate heat kernel based at $(x_0,t_0)$. Let $\tau_i\nearrow\infty$ be an increasing sequence of positive numbers. Then, the following sequence of tuples
\begin{eqnarray*}
\Big\{\big(M,g_i(t),(x_0,-1),f_i\big)_{t\in(-\infty,0)}\Big\}_{i=1}^\infty
\end{eqnarray*}
converges, possibly after passing to a subsequence, to the canonical form of a shrinking gradient Ricci soliton, called the \emph{asymptotic shrinker}
\begin{eqnarray*}
\big(M_\infty,g_\infty(t),(x_\infty,-1),f_\infty\big)_{t\in(-\infty,0)},
\end{eqnarray*}
where $f_\infty$ is the potential function, satisfying
\begin{eqnarray*}
Ric_{g_\infty}+\nabla^2f_\infty=\frac{1}{-2t}g_\infty.
\end{eqnarray*}
The convergence $g_i\rightarrow g_\infty$ is in the pointed Cheeger-Gromov-Hamilton \cite{Ha3} sense, and the convergence $f_i\rightarrow f_\infty$ is in the locally smooth sense. Here $g_i$ and $f_i$ are obtained by time-shifting and parabolic scaling as follows
\begin{eqnarray*}
g_i(t)&:=&\tau_i^{-1}g(\tau_it+t_0),
\\
f_i(\cdot,t)&:=&f(\cdot,\tau_it+t_0).
\end{eqnarray*}
Furthermore, Perelman's entropy and the Nash entropy converge to the entropy of the asymptotic shrinker. By this we mean
\begin{eqnarray}\label{eq1}
\int_M(4\pi|t|)^{-\frac{n}{2}}e^{-f_\infty}dg_\infty=1,
\\\nonumber
\lim_{\tau\rightarrow \infty}\mathcal{W}_{x_0,t_0}(\tau)=\lim_{\tau\rightarrow \infty}\mathcal{N}_{x_0,t_0}(\tau)=\mu_\infty,
\end{eqnarray}
where
\begin{eqnarray}\label{eq2}
\mu_\infty&=&\int_M \Big(|t|(|\nabla f_\infty|^2+R_{g_\infty})+f_\infty-n\Big)(4\pi|t|)^{-\frac{n}{2}}e^{-f_\infty}dg_\infty
\\\nonumber
&=&\int_M f_\infty(4\pi|t|)^{-\frac{n}{2}}e^{-f_\infty}dg_\infty-\frac{n}{2}
\end{eqnarray}
is a negative constant independent of time $t$, which we call \emph{the entropy of the asymptotic shrinker}.
\end{Proposition}
\begin{proof}
The proof of this proposition can be modified from, for instance, \cite{CZ} or \cite{X}. First of all, the Type I condition implies that there exists a positive number $C$, such that
\begin{eqnarray}\label{tp1}
|Rm_{g(t)}|\leq\frac{C}{|t|}
\end{eqnarray}
for all $t\in(-\infty,0)$. We then observe that for all $(x,t)\in M\times (-\infty,0)$, it holds that
\begin{eqnarray}\label{vol}
\operatorname{Vol}_{g(t)}\big(B_{g(t)}(x,\sqrt{|t|})\big)\geq c|t|^{\frac{n}{2}},
\end{eqnarray}
where $c>0$ is a constant. Suppose this is not true, then one may find a sequence of counterexamples $\{(x_i,t_i)\}_{i=1}^\infty\subset M\times(-\infty,0)$, such that the scaled Ricci flows $(M,g_i(t))_{t\in(-\infty,0)}$, where $g_i(t):=|t_i|^{-1}g_i(t|t_i|)$, all satisfy (\ref{tp1}), but
\begin{eqnarray}\label{vol2}
\operatorname{Vol}_{g_i(-1)}\big(B_{g_i(-1)}(x_i,1)\big)\rightarrow 0.
\end{eqnarray}
By Lemma \ref{inj-radius}, we have
\begin{eqnarray*}
\operatorname{inj}(g_i(-1),x_i)\geq c>0.
\end{eqnarray*}
Hence, by \cite{Ha3}, the sequence of Ricci flows $\displaystyle \big\{(M,g_i(t),(x_i,-1))_{t\in(-\infty,0)}\big\}_{i=1}^\infty$ converges, possibly after passing to a subsequence, to a smooth ancient solution $(M_\infty,g_\infty(t),(x_\infty,-1))_{t\in(-\infty,0)}$. In particular, we have
\begin{eqnarray*}
\operatorname{Vol}_{g_\infty(-1)}\big(B_{g_\infty(-1)}(x_\infty,1)\big)>0,
\end{eqnarray*}
and this contradicts (\ref{vol2}).
One may then follow the proofs in \cite{CZ} or \cite{X} to conclude this Proposition. Obviously, the noncollapsing condition in these proofs can be replaced by (\ref{vol}).
\end{proof}
\section{The Nash Entropy and Noncollapsing}
It turns out that from Proposition \ref{Asymptotic_Shrinker} it is sufficient to conclude that the ancient solution is noncollapsed on all scales everywhere. This follows from an observation made in \cite{MZ}. Let $(M,g(t))_{t\in(-\infty,w)}$, where $0<w\leq\infty$, be an ancient solution as describe in Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2}. One may generally regard $(-\infty,w)$ as the maximum existing interval of $g(t)$, in which case $t=w$ is the singular time (whether it is infinity or not). The following lemma says $g(t)$ has bounded geometry as long as it is regular.
\begin{Lemma}
For all $t\in(-\infty,w)$, it holds that $$\sup_M\big|Rm_{g(t)}\big|<\infty\text{ and }\displaystyle\inf_{x\in M}\operatorname{Vol}_{g(t)}\big(B_{g(t)}(x,1)\big)>0.$$
\end{Lemma}
\begin{proof}
If $w<\infty$, then by the definition of finite singular time, this is the first instance at which $g(t)$ has unbounded curvature. If $w=\infty$, then this means that $g(t)$ has bounded curvature for all $t$. The volume lower bound for unit balls follows from a straightforward volume distortion estimate.
\end{proof}
From this time-wise geometry bound, we may conclude the following proposition; this is a combination of Proposition \ref{Asymptotic_Shrinker} above and Proposition 4.6 in \cite{MZ}, where the second author together with Zilu Ma proved (as a consequence of Corollary 5.11 in \cite{Ba}) that on an ancient solution with bounded geometry on each time-slice, Perelman's entropies and the Nash entropies based at all points converge to the same number as the time approaches negative infinity.
\begin{Proposition}\label{globalentropy}
Let $(M,g(t))_{t\in(-\infty,w)}$, where $0<w\leq\infty$, be an ancient solution as describe in Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2}. Then for all $(x,t)\in M\times(-\infty,w)$, the following holds
\begin{eqnarray*}
\lim_{\tau\rightarrow\infty}\mathcal{W}_{x,t}(\tau)=\lim_{\tau\rightarrow\infty}\mathcal{N}_{x,t}(\tau)=\mu_\infty,
\end{eqnarray*}
where $\mu_\infty$ is the entropy of any one of asymptotic shrinkers based at any point (for their entropies are all equal). In particular, we have
\begin{eqnarray}
\mathcal{N}_{x,t}(\tau)\geq \mu_\infty
\end{eqnarray}
for all $(x,t)\in M\times(-\infty,w)$ and for all $\tau>0$.
\end{Proposition}
Though in the original proof of Perelman \cite{P}, he uses the bound of the $\mu$ functional to show the no local noncollapsing theorem, yet the second author showed that the boundedness of the Nash entropy could also be used to prove the noncollapsedness at its base point; this is the following Proposition.
\begin{Proposition}[Theorem 6.1 in \cite{Ba}]\label{noncollapsing}
Let $(M,g(t))$ be a Ricci flow and $(x,t)$ a point in the space time. Let $r$ be a positive scale such that $[t-r^2,t]$ is in the existing interval and $R\leq r^{-2}$ on $\displaystyle B_{g(t)}(x,r)$. Then, it holds that
\begin{eqnarray*}
\operatorname{Vol}_{g(t)}\big(B_{g(t)}(x,r)\big)\geq c\exp\big(\mathcal{N}_{x,t}(r^2)\big)r^n.
\end{eqnarray*}
Here $c$ is a positive constant depending only on the dimension.
\end{Proposition}
\emph{Remark:} Though Bamler \cite{Ba} proved the above result for Ricci flows on closed manifolds, yet one may check the proof of Theorem 6.1 in \cite{Ba} and easily verify its validity for Ricci flows with bounded geometry on each time-slice; one may need to apply Theorem 4.4 of \cite{MZ} in this verification. Fortunately, all the Ricci flows we work with in this paper satisfy this condition. On the other hand, the second author proved that bounded Nash entropy implies weak noncollapsing, and this proof does not need bounded geometry on each time-slice; see Proposition 3.3 in \cite{Z1}.
\bigskip
\begin{proof}[Proof of Theorem \ref{MainTheorem1} and Theorem \ref{MainTheorem2} ]
Let $(M,g(t))_{t\in(-\infty,w)}$ be an ancient solution as described in either Theorem \ref{MainTheorem1} or Theorem \ref{MainTheorem2}. Let $(x,t)\in M\times(-\infty,w)$ be an arbitrary space-time point, and $r$ any scale that satisfies
\begin{eqnarray*}
R\leq r^{-2} \text{ on } B_{g(t)}(x,r).
\end{eqnarray*}
Since, by Proposition \ref{globalentropy}, we have
\begin{eqnarray*}
\mathcal{N}_{x,t}(r^2)\geq \mu_\infty\in(-\infty,0),
\end{eqnarray*}
where $\mu_\infty$ is the entropy of one of the asymptotic shrinkers, it then follows from Proposition \ref{noncollapsing} that
\begin{eqnarray*}
\operatorname{Vol}_{g(t)}\big(B_{g(t)}(x,r)\big)\geq ce^{\mu_\infty} r^n;
\end{eqnarray*}
this finishes the proof.
\end{proof}
\section{Applications}
In this section, we prove all the corollaries proposed in the introduction section.
\begin{proof}[Proof of Corollary \ref{3d}]
Let $(M^3,g(t))$ be a three-dimensional noncompact Type I ancient solution. By Chen \cite{C}, $g(t)$ has nonnegative sectional curvature everywhere. If its sectional curvature is strictly positive, then, by Theorem \ref{MainTheorem1}, it is also noncollapsed. It follows from \cite{Z2} (or \cite{H}) that such ancient solution does not exist.
If $g(t)$ ever attains zero sectional curvature somewhere, then by the strong maximum principle of Hamilton \cite{Ha2}, $(M^3,g(t))$ splits locally and hence its universal cover must be the standard shrinking cylinder $\mathbb{S}^2\times\mathbb{R}$. Furthermore, the only noncompact quotients of $\mathbb{S}^2\times\mathbb{R}$ are the $\mathbb{Z}_2$ quotients. The reason is that the projection of a group action on the $\mathbb{R}$ factor can only be either the reflection or the identity---if it is ever a translation, then this action will generate an infinity group action on $\mathbb{S}^2\times\mathbb{R}$, and the quotient space must be compact. This finishes the proof of the corollary.
\end{proof}
\begin{proof}[Proof of Corollary \ref{noncompact}]
We argue by contradiction. Assume one of the asymptotic shrinkers is the standard cylinder $\mathbb{S}^{n-1}\times\mathbb{R}$, then all the results obtained in section 3 of \cite{LZ} can be applied to this ancient solution. In particular, it satisfies a canonical neighborhood theorem and hence always has a non-neck-like region at each time (c.f. Theorem 1.3 in \cite{LZ}), it always splits as $\mathbb{S}^{n-1}\times\mathbb{R}$ at space infinity (c.f. Proposition 3.9 in \cite{LZ}), it satisfies the neck stability theorem of Kleiner-Lott \cite{KL} (c.f. Theorem 3.11 in \cite{LZ}), and all such ancient solutions form a compact space (c.f. Theorem 1.2 in \cite{LZ}). Knowing all these facts, one may follow the arguments in \cite{Z2} line by line to conclude that such an ancient solution does not exist.
\end{proof}
\begin{proof}[Proof of Corollary \ref{compact}]
The $\operatorname{PIC}-2$ condition implies the positive sectional curvature condition. Hence, according to Theorem \ref{MainTheorem2}, such ancient solution must be $\kappa$-noncollapsed on all scales for some $\kappa>0$. The conclusion then follows from Corollary 0.4 in \cite{Ni2}.
\end{proof}
| 33,684
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The songs therapy also decreased the dad and mom' strain, says Joanne Loewy, the study's direct author, director of your Armstrong center and co-editor with the journal Music and Medication
This self confidence interval tells us that, at the least many of the time, the ratio of results involving the groups is just one. Equally, if an absolute measure of impact, like a big difference in signifies in between groups, has a 95% CI that includes zero in its vary, we will conclude there is absolutely no difference between the teams.
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Have you ever read our post on case scientific studies as well as flipped classroom from the Journal of Faculty Science Teaching? It led to an NSF grant which includes authorized us to further more acquire the use of case reports to teach STEM topics by like videos to aid a "flipped classroom" style. Click the website link down below to read the short article.
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TITLE: Number of monomials in the complement of $\langle LT(I) \rangle$.
QUESTION [0 upvotes]: I'm trying to find the number of monomials in the complement of $\langle LT(I) \rangle$, with both the lexicographic ($>_{lex}$) and graduated lexicographic ($>_{grlex}$) orders for the ideal $I = \langle x^{4}y - z^{6}, x^{2} - y^{3}z, x^{3}z^{2} - y^{3} \rangle \subset \mathbb{K}[x,y,z]$.
In the first case, the lexicographic ($>_{lex}$), what I get is the following:
$$\langle LT(I) \rangle = \langle x^2, xz^6, y^3, yz^6, z^{67} \rangle,$$
whilst in the other, ($>_{grlex}$), what I get is
$$\langle LT(I) \rangle = \langle x^9, x^2y^7, y^9, x^7y, x^4y^4, x^5z, z^6, x^3z^2, y^3z \rangle.$$
My question is if there has to be the same number of monomials in the complement of $\langle LT(I) \rangle$ regardless of the order taken to compute the Groebner basis, or it can be a different number. Because when I compute both of them, what I get is 80 monomials and 52 respectively, and I'm not sure if it is well done.
Is there a result which could help me?
REPLY [0 votes]: The monomials in the complement of $\operatorname{LT}(I)$ form the normal basis $\operatorname{NB}(I)$ of the quotient ring $\mathbb{K}[x,y,z]/I$ with respect to the given monomial ordering, i.e. a $\mathbb{K}$-vector space basis of the quotient ring $\mathbb{K}[x,y,z]/I$. In particular, the cardinality of the complement of $\operatorname{LT}(I)$ is the cardinality of $\operatorname{NB}(I)$, which is the dimension of the $\mathbb{K}$-vector space $\mathbb{K}[x,y,z]/I$. This dimension is independent of any choice of monomial ordering, so the cardinality of the complement of $\operatorname{LT}(I)$ is independent of the choice of monomial ordering, even though the complement of $\operatorname{LT}(I)$ may be different for different monomial orderings.
In the example the cardinality of the complement of $\operatorname{LT}(I)$ is $97$:
sage: R.<x,y,z> = PolynomialRing(QQ, order='lex')
sage: I = R.ideal(x^4*y - z^6, x^2 - y^3*z, x^3*z^2 - y^3)
sage: len(I.normal_basis())
97
| 93,688
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Da$h drops off a new visual for "22 Tabs", from his recent "V.I.C.E.S." project.Can you appreciate the bars here? Hear more of Da$h on Intoxicated Scarfaces, featuring Remy Banks.
Quotable Lyrics"...sinister literature Read my riddles it'll get rid of ya quicker than R italin Red bones leave me with a ridden dick, how real is this?
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Quot eit
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TITLE: Can we always span a decomposable form via constant coefficients?
QUESTION [1 upvotes]: Let $M$ be a smooth manifold of dimension $d$. Let $1 < k <d$ be an integer. Let $\omega^i$ be a local frame of the exterior power bundle $\Lambda_{k}(T^*M)$.
Does there exists numbers $a_i \in \mathbb{R}$ , such that $a_i \omega^i$ is decomposable?
(I am using the summation convention here).
Of course, there always exist functions $a_i$ satisfying the above, since by definition, if $\omega^i$ is a frame, we can span every smooth element with it.
My guess is that the answer might be negative, since being decomposable element is $\Lambda_k(V)$ a closed condition (rather than an open one, see Plucker relations). Here $V$ is a fixed real vector space .
I am OK with shrinking the local neighbourhood on where the $\omega_i$ are defined.
REPLY [2 votes]: Of course $a_1 = \cdots = a_{{}_dC_k} = 0$ is always a solution. Excluding this case, however, the answer in general is no.
Here's how we can construct a counterexample: The lowest values $d, k$ for which there are indecomposable (local) $k$-forms on a $d$-manifold are $d = 4$, $k = 2$, and for $k = 2$ we have a lemma useful here:
A $2$-form $\alpha$ is locally decomposable iff $\alpha \wedge \alpha = 0$.
For any local coframe $(\theta^i)$ of $TM$, define $\theta^{ab} := \theta^a \wedge \theta^b$, and for a coordinate function $\tau$, set
\begin{alignat*}{3}
\omega^1 &:= \cos \phantom{2} \tau \,\theta^{12} - \sin \phantom{2} \tau \,\theta^{34},& \qquad
\omega^2 &:= \sin \phantom{2} \tau \,\theta^{12} + \cos \phantom{2} \tau \,\theta^{34} \\
\omega^3 &:= \cos 2 \tau \,\theta^{13} - \sin 2 \tau \,\theta^{24},& \qquad
\omega^4 &:= \sin 2 \tau \,\theta^{13} + \cos 2 \tau \,\theta^{24} \\
\omega^5 &:= \cos 3 \tau \,\theta^{14} - \sin 3 \tau \,\theta^{23},& \qquad
\omega^6 &:= \sin 3 \tau \,\theta^{14} + \cos 3 \tau \,\theta^{23} . \\
\end{alignat*}
It's straightforward to check that this is a frame of $\bigwedge^2 T^*M$.
By the lemma, if $\Omega := \sum a_i \omega^i$ (with constant $a_i$) is decomposable then $\Omega \wedge \Omega = 0$; so, to show that this frame gives a counterexample, it's enough to show that this equation forces $a_1 = \cdots = a_6 = 0$.
Substituting gives
$$f(\theta) \, \theta^1 \wedge \theta^2 \wedge \theta^3 \wedge \theta^4 = 0,$$ and applying angle sum identities to expand $f$ in the linearly independent functions $\cos j \tau, \sin j \tau$, $j \in \{0, 1, 2, \ldots\}$ gives
$$
f(\tau) = 2 a_5 a_6 \cos(6 \tau) + (a_6^2 - a_5^2) \sin (6 \tau) + \cdots
= 0 $$
where $\cdots$ denotes terms with $j \neq 6$. By linear independence these coefficients are both zero, so $a_5 = a_6 = 0$. Then successively considering $j = 4$ and $j = 2$ gives that the remaining $a_i$ are zero, too, and we're done.
| 26,923
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TITLE: Must a proper curve minus a point be affine?
QUESTION [11 upvotes]: Let $C$ be a proper smooth geometrically connected curve over a field $K$, and let $P\in C(K)$ be a point. Must $C - P$ be affine?
EDIT: By Riemann-Roch, you can definitely find functions $f_1,\ldots,f_r : C-P\longrightarrow\mathbb{A}^n_K$, but how do you guarantee that for some $n$, you can find enough such $f_i$'s such that this gives you an embedding?
EDIT: Is the same true with $C$ not smooth?
REPLY [23 votes]: Theorem
If a nonzero finite number of points $p_1,\dots, p_r $ are deleted from $C$ the resulting curve will be affine.
Indeed consider the divisor $D=p_1+\dots+ p_r $ on $C$.
Since it has positive degree some positive multiple $nD$ of it will be very ample.
Thus we get an embedding of $j:C\to \mathbb P^N$ (for some huge $N$) and a hyperplane section divisor $\Delta =H\cap j(C)$ on $j(C)$ such that $j^*\Delta=nD$.
But then $C\setminus \{p_1,\dots, p_r\}$ is isomorphic to $j(C)\cap (\mathbb P^N\setminus H)\cong j(C)\cap \mathbb A^N$ (the complement of a hyperplane in projective space is affine space) and since this last variety $j(C)\cap \mathbb A^N$ is clearly affine, so is $C\setminus \{p_1,\dots, p_r\}$.
Edit
The theorem is valid even if $C$ is singular.
To see that, consider the finite normalization morphism $n:\tilde C\to C$ and delete the inverse image of $\{p_1,\dots, p_r\}$, obtaining the smooth curve $C'=\tilde C\setminus n^{-1}(\{p_1,\dots, p_r\})$ which is affine by the result already proved for smooth curves.
Now consider the restricted finite morphism $n':C'\to C\setminus \{p_1,\dots, p_r\}$.
Since $C'$ is affine and the finite morphism $n'$ is surjective the curve $C\setminus \{p_1,\dots, p_r\}$ will also be affine by Chevalley's Theorem (EGA $_{II}$, Théorème (6.7.1), page 136), and we are done.
| 87,564
|
India and the US are unlikely to sign a key foundational defence agreement for mutual access to high accuracy geospatial maps at an upcoming meeting between the defence and foreign leaders next month as technical issues on sharing of data have still not been worked out.
However, an industrial security pact has been finalised at a recent meeting of the Defence Policy Group (DPG) in Washington and the agreement is likely to be formally inked – enabling US military manufacturers to share high end technology with their Indian partners.
Sources told ET that the Basic Exchange and Cooperation Agreement (BECA) was discussed at the DPG meet led by Defence secretary Sanjay Mitra from the Indian side but `too many issues’ still remain unresolved.
The agreement, which would give India access to a database of global maps that is critical for precise targeting of weapon systems and operational planning, has been in discussions for several years with sources saying that all queries by both sides on how it will be operationalized have not yet been answered.
Two other of the so called foundational agreements have already been inked – one for sharing of military logistics and another that enables transfer of secure communication equipment to enhance interoperability.
The Communications Compatibility and Security Agreement (COMCASA) is currently being operationalized with the equipping of new equipment on US origin platforms in service with Indian forces and the activation of systems that had come as an integral part of others.
Two sides have worked out all formalities for the Industrial Security Annex (ISA) pact will add to existing agreements on protection of classified military information, sources said, adding that this is set to be inked at the 2+2 meeting next month between the Defence and Foreign ministers of the two nations.
The pact is critical for any transfer of technology by a US firm to its Indian partners. The pact is expected to enable the India US Defence Technology and Trade Initiative (DTTI) that was signed in 2012 but has not resulted in any major project on the ground yet.
The ISA will involve Indian government assurances on the safety and security of technology against transfer and access to third parties. Once operational, US companies wishing to transfer technology to India will be able to do so through the government route. For critical technology, US companies will submit documentation to the US government that will share it with their Indian counterparts through diplomatic channels.
This technology piece will then be sent to the Indian industry by the government which will first satisfy itself that adequate safeguard mechanisms are in place for its protection. The ISA will be vital for US companies to participate in all upcoming Make in India projects.
| 308,239
|
TITLE: Possible values of a solution to a differential equation
QUESTION [1 upvotes]: The equation looks like this:
$$x'={-tx\over (t+1)},$$ $x(0)=2$
The solution is
$$x(t)={2(t+1)\over e^t},$$ $t\in\mathbb R,$
and I can't figure out why $t$ isn't limited to $(-1,+\infty)$, since in the separable function $g(t) = t/(t+1)$, $t$ is obviously limited to $(-\infty,-1)\cup (-1,+\infty)$.
Thanks for any kind of advice.
REPLY [0 votes]: A solution $y$ to differential equation $x'(t)=\dfrac{-tx(t)}{t+1}$ is, by definition, a function $y$ defined on a non trivial interval $I$ such that $\forall t\in I\left(y'(t)=\dfrac{-ty(t)}{t+1}\right)$.
The function $x\colon \mathbb R\to \mathbb R, t\mapsto 2(t+1)e^{-t}$ isn't a solution to the given ODE simply because it isn't true that $x'(-1)=\dfrac{-(-1)x(-1)}{-1+1}$ (nor false, for that matter).
Thus a maximal solution to the starting IVP is $y\colon (-1,+\infty)\to \mathbb R, t\mapsto 2(t+1)e^{-t}$.
| 53,053
|
\section{Quiver varieties}
We assume that \field is an algebraically closed field. Let
$\xwt\in \cN^I$ be fixed and let $W$ be an $I$-graded vector space
of dimension \xwt. Let $\Ga_*$ be the quiver defined in
Introduction; that is, we adjoin to $\Ga$ a new vertex $*$ and
$\xwt_i$ arrows from $*$ to $i$ for each $i\in I$. We identify
$W_i$ with $\bigoplus_{h:*\ar i}\field\cdot h$. Let as before $V$
be an $I$-graded vector space of dimension \xrt. As in
Introduction, a pair $(\xrt,n)$ with $\xrt\in\cN^I$ and $n\in\cN$
will be considered as an element from $\cN^{I_*}$.
There is an obvious identification
$$M(\xrt,\xwt):=\Rep(\ub \Ga_*,(\xrt,1))=\Rep(\ub \Ga,\xrt)\oplus\Hom_I(W,V)\oplus\Hom_I(V,W).$$
The elements of this space will be represented as triples
$(x,p,q)$. Note that $G_{(\xrt,1)}=(\prod_{i\in
I}\GL_{\xrt_i}\xx\Gm)/\Gm\iso\GL_\xrt$. Therefore the moment map
may be considered as a map $\mu_*:M(\xrt,\xwt)\ar\gl_\xrt^*$. It
is given by the formula
$$\mu_*(x,p,q)=\mu(x)+pq.$$
We fix once and for all $\te\in \cZ^{I_*}$,
$\te=(-1,\dots,-1,\sum\xrt_i)$ and consider stability and
semistability conditions in $M(\xrt,\xwt)$ with respect to \te.
\begin{lemma} Stability and semistability conditions in
$M(\xrt,\xwt)$ are equivalent. An element $(x,p,q)\in
M(\xrt,\xwt)$ is stable if and only if any $I$-graded,
$x$-invariant subspace $V'\sb V$ s.t. $q(V')=0$ is zero.
\end{lemma}
\begin{proof} Assume that $(x,p,q)$ is semistable and there
exists an $I$-graded, $x$-invariant subspace $V'\sb V$ s.t.
$q(V')=0$. Then $V'$ may be considered as an $I_*$-graded,
$(x,p,q)$-invariant subspace of $V\oplus\field$ of dimension
$(\dim V',0)$. From semistability condition we get $-\dim V'\ge0$,
thus $V'=0$. It follows that the last condition of the lemma
holds.
Conversely, assume that the last condition of the lemma holds. Let
$V'\oplus V_*$ be some proper, $I_*$-graded, $(x,p,q)$-invariant
subspace of $V\oplus\field$. If $V_*=\field$, then $\dim V'<\xrt$
and therefore $\te\cdot(\dim V',1)=\sum_i\xrt_i-\sum_i\dim
V'_i>0$. If $V_*=0$, then $q(V')=0$ and by our assumption $V'=0$.
This implies that $(x,p,q)$ is stable.
\end{proof}
\begin{remark} For any subscheme $X\sb M(\xrt,\xwt)$ we write
respectively $X^n$, $X^s$, $X^{ns}$ to denote the subschemes of
$X$ consisting respectively of nilpotent, stable, nilpotent and
stable elements.
\end{remark}
\begin{definition} We define the quiver variety
$\lM=\lM(\xrt,\xwt)$ to be the quotient $\mu_*\inv(0)^s\GIT
\GL_\xrt$. Define $\lL=\lL(\xrt,\xwt):=\mu_*\inv(0)^{ns}\GIT
\GL_\xrt$.
\end{definition}
\begin{remark} It is easy to see that $\lL(\xrt,\xwt)$ is the
preimage of zero under the projective morphism $\mu_*\inv(0)^s\GIT
\GL_\xrt\ar \mu_*\inv(0)\GIT \GL_\xrt$. It is known that an
element $(x,p,q)\in M(\xrt,\xwt)^s$ is nilpotent if and only if
$x$ is nilpotent and $p=0$, see e.g.\ \cite[Lemma 5.9]{Nak1} or
\cite[Lemma 2.22]{Lus1}.
\end{remark}
Let $T$ denotes the Tits form of the quiver \Ga and $T_*$ denotes
the Tits form of the quiver $\Ga_*$. As in Introduction, we define
$d=d(\xrt,\xwt):=1-T_*(\xrt,1)=\xrt\cdot\xwt-T(\xrt)$.
\begin{theorem}[{Nakajima \cite[Section 3]{Nak2}}] Variety $\lM$
is smooth and variety $\lL$ is projective. The complex manifold
$\lM(\cC)$ is symplectic and its subvariety $\lL(\cC)$ is a
Lagrangian subvariety homotopic to $\lM(\cC)$. The dimension of
$\lM$ equals $2d(\xrt,\xwt)$.
\end{theorem}
For a scheme $X$ of finite type over $\cZ_f$, we define
$$h^i(X):=\dim H^i(X(\cC),\cQ),\qquad h^i_c(X):=\dim H^i_c(X(\cC),\cQ).$$
Note that $h^i_c(X)$ can also be defined as $\dim
H^i_c(X_{\ub\cF_p},\cQ_l)$ (for large enough prime $p$) by the
base change theorem \cite[Thorem 1.8.7]{FK1} and comparison
theorem \cite[Thorem 1.11.6]{FK1}.
\begin{lemma}\label{lem mult of modules} $\dim
L(\ub\xwt)_{\ub\xwt-\xrt}=h^{2d}_c(\lM)$.
\end{lemma}
\begin{proof} It is well-known (see e.g.\ \cite{Nak2} or
\cite{Saito1}) that $\dim L(\ub\xwt)_{\ub\xwt-\xrt}$ equals the
number of irreducible components of \lL i.e., $h^{2d}_c(\lL)$. We
note that
$$h^{2d}_c(\lL)=h^{2d}(\lL)=h^{2d}(\lM)=h^{2d}_c(\lM),$$
where the last equality follows from the Poincar\'e duality.
\end{proof}
| 172,399
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TITLE: I can't induce a contradiction by assuming negation, what does that imply?
QUESTION [0 upvotes]: Suppose I want to prove that $A \Rightarrow B$. So I assume $A$ and $\neg B$ and hope to arrive at a contradiction. But suppose $\neg B$ does not induce any contradiction - I don't mean "suppose I fail to spot any contradictions", I mean let's pretend I have proven that no contradictions arise (if that's even formally possible to do). Does this mean that I have proven $A \not \Rightarrow B$?
I mean, I have shown that $\neg B$ is consistent with $A$, or that $\neg B \not \Rightarrow \neg A$. But I'm no formal logic expert so I'm not convinced that this is equivalent to $A \not \Rightarrow B$. Help?
REPLY [1 votes]: $(A \Rightarrow B) \Leftrightarrow (\neg B \Rightarrow \neg A)$
The right statement is called the contrapositive.
What they are saying is that the satisfaction of $A$ results in a satisfaction of $B$.
If you happen to show that there are cases where $\neg B$ is satisfied but $\neg (\neg A)$ isn't then you have proved that $\neg B \not \Rightarrow \neg A$. At this point, this is telling us that there are cases where $\neg B $ and $A$ are simultaneously verified. What does this mean? it means that we can have $A$ satisfied and not be satisfied by $B$. Thus it does imply $A \not \Rightarrow B$ and your assumption is correct.
| 185,213
|
1864-04-05 [Letter, 1864 Apr. 5]
Dublin Core
Title
1864-04-05 [Letter, 1864 Apr. 5]?- a member of the 22nd Iowa Volunteer Infantry, to his father, Pospisil, Joseph. In Czech. Dated 5 Apr. 1864. 2 pages.
Creator
Pospisil, John (Jan), 1839?-
Date
04/05/1864
1860; 1861; 1862; 1863; 1864; 1865; 1866; 1867; 1868; 1869 National Czech & Slovak Museum & Library.
National Czech & Slovak Museum & Library
Coverage
United States--History--Civil War, 1862-1865
| 197,661
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BLACKCAPS coach John Bracewell today announced the BLACKCAPS squad for the Twenty20 match and first two one-day international matches of The National Bank Series against the West Indies.
Chris Cairns is added to the squad for the Twenty20 match only, which will be his last match for the BLACKCAPS before he retires from international cricket.
“With a return to form of many players at domestic level there was fierce competition for top order batsmen. It was very much a case of deciding who to leave out rather than who to select,” Bracewell said.
“With injuries to key players and the retirement of Chris Cairns there are some issues around selecting all rounders of international quality.
“Michael Mason is currently the best line and length bowler in the country. With the gap left by injuries to Kyle Mills, Andre Adams and Jacob Oram, he has a real opportunity to prove himself at international level.
“Mills is recovering from his groin strain but does not have enough recent cricket to be considered for section. He will be returning to club cricket this weekend and will play a State Championship match before the team for the remainder of the one-day series is announced.
“Adams is recovering from his fractured hand and should be returning to cricket soon. We do not know how long Oram will be out of the frame as he recovers from his bruised heel.”
BLACKCAPS squad to face West Indies in The National Bank Series: Stephen Fleming (captain), Daniel Vettori (vice captain), Nathan Astle, Shane Bond, James Franklin, Peter Fulton, Jamie How, Hamish Marshall, Michael Mason, Brendon McCullum, Jeetan Patel, Scott Styris, Lou Vincent
| 138,166
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\begin{document}
\bibliographystyle{spmpsci}
\title{Elementary Integral Series for Heun Functions}
\subtitle{With an Application to Black-Hole Perturbation Theory}
\author{Pierre-Louis Giscard \and
Aditya Tamar
}
\institute{Pierre-Louis Giscard \at
Univ. Littoral C\^ote d’Opale, UR 2597, LMPA, Laboratoire de Mathématiques Pures et Appliqu\'ees Joseph Liouville, F-62100 Calais, France. \\
\email{giscard@univ-littoral.fr}
\and
Aditya Tamar \at
Independent Researcher, Delhi, India. \\
\email{adityatamar@gmail.com}
}
\date{Received: date / Accepted: date}
\maketitle
\begin{abstract}
Heun differential equations are the most general second order Fuchsian equations with four regular singularities.
An explicit integral series representation of Heun functions involving only elementary integrands has hitherto been unknown and noted as an important open problem in a recent review.
We provide explicit integral representations of the solutions of all equations of the Heun class: general, confluent, bi-confluent, doubly-confluent and triconfluent, with integrals involving only rational functions and exponential integrands. All the series are illustrated with concrete examples of use. These results stem from the technique of path-sums, which we use to evaluate the path-ordered exponential of a variable matrix chosen specifically to yield Heun functions.
We demonstrate the utility of the integral series by providing the first representation of the solution to the Teukolsky radial equation governing the metric perturbations of rotating black holes that is convergent everywhere from the black hole horizon up to spatial infinity.
\keywords{Heun Equations \and Integral Representation \and Path Sums \and Volterra equation \and Neumann Series \and Teukolsky Equation}
\PACS{02.30.Hq \and 02.30.Rz \and 04.70.Bw \and 04.70.-s}
\end{abstract}
\section{Introduction}
\label{intro}
The study of Heun equations has generated significant interest in both mathematics and physics lately. From a mathematical standpoint, recent results have uncovered relation between Heun equations other equations of paramount importance for physics.
For example, it was found by means of antiquantisation procedures \cite{Slav1} and monodromy preserving transformations \cite{Takem1} that the Heun equations share a bijective relationship with Painlev{\'e} equations \cite{Slav1,Slav2,Slav4}. This permitted in-depth studies on the integral symmetry properties of equations of the Heun class \cite{Slav3} and to determine generating polynomial solutions of the Heun equation by formulating a Riemann-Hilbert problem for the Heun function \cite{RieHibH}. The reduction of certain Heun equations under non-trivial substitutions to hypergeometric equations has also been possible by means of pull-back transformations based on Belyi coverings \cite{BelyiHeun} and polynomial transformations \cite{Maier1,Maier2}.
In contrast, in spite of the increasing use of Heun functions in physics (in quantum optics \cite{Xie2010,Moham}, condensed matter physics \cite{Crampe,Dorey}, quantum computing \cite{QuantComp}, two-state problems \cite{QTS1,QTS2} and more \cite{Hort1}), few studies \cite{Hort1,Hort2} have specifically focused on determining their properties most relevant to physical applications. For example, the lack of integral expansions of these functions involving only elementary integrands has been clearly identified as a major obstacle when extracting physical meaning from the mathematical treatment of black holes quasinormal modes \cite{Hort1,Hort2}, yet remains unaddressed in the mathematical literature.
The present works tackles this issue by determining a novel integral representation of the Heun equations involving elementary functions that is tailored to physical applications. In particular, we demonstrate the applicability of the novel integral representation to the Teukolsky equation \cite{TeukEqn} that governs the metric perturbations of rotating black holes and further explore which physical observables pertinent to black hole perturbation theory can be obtained from the integral form. The present progress in integral representation is enabled by the method of path-sum \cite{Giscard2015}, which generates the linear Volterra integral equation of the second kind satisfied by any function involved a system of coupled linear differential equations with variable coefficients.
\begin{comment}
The study of Heun equations has generated significant interest in both mathematics and physics lately. From a mathematical standpoint, recent results have uncovered relation between Heun equations other equations of paramount importance for physics.
For example, it was found by means of antiquantisation procedures \cite{Slav1} and monodromy preserving transformations \cite{Takem1} that the Heun equations share a bijective relationship with Painlev{\'e} equations \cite{Slav1,Slav2,Slav4}. This permitted in-depth studies on the integral symmetry properties of equations of the Heun class \cite{Slav3} and to determine generating polynomial solutions of the Heun equation by formulating a Riemann-Hilbert problem for the Heun function \cite{RieHibH}. The reduction of certain Heun equations under non-trivial substitutions to hypergeometric equations has also been possible by means of pull-back transformations based on Belyi coverings \cite{BelyiHeun} and polynomial transformations \cite{Maier1,Maier2}.
In contrast, in spite of the increasing use of Heun functions in physics (in quantum optics \cite{Xie2010,Moham}, condensed matter physics \cite{Crampe,Dorey}, quantum computing \cite{QuantComp} and two-state problems \cite{QTS1,QTS2} and more \cite{Hort1}), few studies \cite{Hort1,Hort2} have specifically focused on determining their properties most relevant to physical applications. For example, the lack of integral expansions of these functions involving only elementary integrands has been clearly identified as a major obstacle when extracting physical meaning from the mathematical treatment of black holes quasinormal modes \cite{Hort1,Hort2}, yet remains unaddressed in the mathematical literature.
The present works tackles this issue by determining a novel integral representation of the Heun equations involving elementary functions that is tailored to physical applications. In particular, we demonstrate the applicability of the novel integral representation to the Teukolsky equation \cite{TeukEqn} that governs the metric perturbations of rotating black holes and further explore which physical observables pertinent to black hole perturbation theory can be obtained from the integral form. The present progress in integral representation is enabled by the method of path-sum \cite{Giscard2015}, which generates the linear Volterra integral equation of the second kind satisfied by any function involved a system of coupled linear differential equations with variable coefficients.
\end{comment}
This paper is organised as follows. In Section \ref{sec:2} we give the minimal necessary background on Heun equations. This section concludes in \S\ref{sec:IntRep} with a review of existing integral representations of Heun functions and their major drawback as noted in the recent mathematical-physics literature.
Section \ref{sec:ExplicitRes} is a self-contained presentation of the novel, elementary integral representations of all functions of Heun class, illustrated with concrete examples. This section contains none of the proofs, all of which are deferred to Appendix~\ref{AppendixProofs}. Then, in Section~\ref{sec:black hole} we give the elementary integral series representation of the solution to the Teukolsky radial equation. This representation is the first one to be convergent from the black hole horizon up to spatial infinity. This stands in contrast to the state-of-the-art MST formalism \cite{Mano1997}, that uses \textit{two} hypergeometric series (one convergent at the horizon and the other at infinity) that must then be matched after an analytic continuation procedure. This last step requires the introduction of an auxiliary parameter lacking physical correspondence, at the very least obscuring the physical picture. The convergence of the integral series over the entire domain from the black hole horizon up to spatial infinity therefore alleviates the need for such parameters lacking physical correspondence when calculating solutions of the Teukolsky radial equation. These solutions are of primary importance for computing quantities of physical interest such as gravitational wave fluxes \cite{Fujita2004} and quasinormal modes \cite{Zhang2013}.
We conclude in \S\ref{sec:conclusion} with a brief discussion of the novel integral series and future prospects of the method of path-sum from which they stem for solving the coupled system of Teukolsky angular and radial equations.
\section{Heun Differential Equations}
\label{sec:2}
\subsection{Mathematical Context}
The most general linear, homogenous, second order differential equation with polynomial coefficients is given by the Fuchsian equation \cite{SpecFunc} which has the following form
\begin{equation*}
P(z)\frac{d^2y(z)}{dz^2} + Q(z)\frac{dy(z)}{dz} + R(z)y(z) = 0,~ z \in \mathbb{CP}^1,
\end{equation*}
where $\mathbb{CP}^1$ is the Riemann sphere. In the above equation, if the function $K_{QP} = Q(z)/P(z)$ has a pole of at most first order and $K_{RP} = R(z)/P(z)$ has a pole of at most second order at some singularity $z=z_0$, then $z_0$ is called a \textit{Fuchsian} singularity, otherwise it is an \textit{irregular} singularity.
The above equation is a \textit{Fuchsian equation} if all its singularities are Fuchsian singularities. Now, any Fuchsian equation with exactly four singular points can be mapped onto a Heun equation \cite{HeunOrig} by transformation in dependent or independent variables. These transformations are called s-homotopic and M{\"o}bius transformations respectively. The Heun equation is a straightforward generalisation of the hypergeometric equation, a Fuchsian equation with exactly three singular points \cite{SpecFunc}.
\subsection{General Heun Equation}
As mentioned in the Introduction, the Heun differential equation is the most general Fuchsian equation with four regular singularities. The canonical form of the equation, also known as the General Heun Equation (GHE) is given by the following equation and conditions:
\begin{equation}
\frac{d^2y(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \frac{\delta}{z - 1} + \frac{\epsilon}{z - t} \bigg]\frac{dy(z)}{dz} + \frac{\alpha\beta z - q}{z(z-1)(z-t)}y(z) = 0, \label{eq:Heun}
\end{equation}
where $q \in \mathbb{C}$ is called the \textit{accessory parameter}. The corresponding Riemann-P symbol is as follows:
\begin{equation*}
\begin{pmatrix}
0 & 1 & a & \infty \\
0 & 0 & 0 & \alpha&;z \\
1-\gamma & 1-\delta & 1 - \epsilon & \beta
\end{pmatrix}
\end{equation*}
where the parameters satisfy the Fuch's condition:
\begin{equation*}
1 + \alpha + \beta = \gamma + \delta + \epsilon
\end{equation*}
The GHE has four singular points at $z = 0,1,t,\infty$. Concerning its solutions, Maier, completing a task initiated by Heun \cite{Heun1888} himself has shown that solutions of the GHE have Coxeter group $D_4$ as their automorphism group \cite{Maier192}. This means that 192 solutions can be generated using the symmetries of $D_4$, much more than the 24 solutions of the Gauss Hypergeometric equation determined by Kummer \cite{Whittaker}. We refer the reader to \cite{Maier192} for the complete list of solutions and their relations as well as to \cite{SpecFunc} for a further discussion of their properties.
For specific parameter values the Heun equation reduces to other well-known equations of importance: e.g. setting $\epsilon = 0, \gamma = \delta = 1/2$ yields the Mathieu equation, which has found widespread applicability in the theoretical and experimental study of vibration phenomenon \cite{Mathieu,Mathieu5}, electromagnetic scattering from elliptic waveguides \cite{Mathieu1,Mathieu2,Mathieu3}, ion traps in mass spectrometry \cite{Mathieu4}, stability of floating ships \cite{Mathieu6}. Furthermore, the confluent form of the Heun equation has found wide ranging applications in quantum particle confinement and interaction potentials \cite{ConfPot,IntPot} and in the Stark effect \cite{Slav5,SpecFunc}.
\subsection{Confluent Heun Equations}
\label{sec:2.1}
The GHE contains 4 regular singularities. If we apply a \textit{confluence} procedure to two of its singularities such that we get an irregular singularity, we call the resultant equation a confluent Heun equation (CHE). The CHE contains at least one irregular singular point besides the regular singular points. We can construct local solutions in the vicinity of this irregular singular points by the means of (generally divergent) Thom{\'e} series \cite{SpecFunc}. The number of parameters in the CHE are reduced by one. Thus by applying the confluence procedure laid out in \cite{SpecFunc} to the singularities at $z = t$ and $z = \infty$ in equation 2, we get the CHE:
\begin{equation}\label{eq:HeunConfluent}
\frac{d^2y(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \frac{\delta}{z - 1} + \epsilon \bigg]\frac{dy(z)}{dz} + \frac{\alpha z - q}{z(z-1)}y(z) = 0.
\end{equation}
By continuing application of the confluence procedure, we obtain the bi-confluent Heun equation
\begin{equation}\label{eq:HeunBiConfluent}
\frac{d^2y(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \delta + \epsilon z \bigg]\frac{dy(z)}{dz} + \frac{\alpha z - q}{z}y(z) = 0,
\end{equation}
and related doubly-confluent Heun equation
\begin{equation}\label{eq:HeunDoublyConfluent}
\frac{d^2y(z)}{dz^2} + \bigg[ \frac{\delta}{z^2}+\frac{\gamma}{z} + 1 \bigg]\frac{dy(z)}{dz} + \frac{\alpha z - q}{z^2}y(z) = 0,
\end{equation}
as well as the triconfluent Heun equation
\begin{equation}\label{eq:HeunTriConfluent}
\frac{d^2y(z)}{dz^2} + \bigg[ \gamma + \delta z + \epsilon z^2 \bigg]\frac{dy(z)}{dz} + (\alpha z - q)y(z) = 0.
\end{equation}
We refer the reader to \cite{SpecFunc} for further general informations on these functions.
\begin{comment} THIS IS EITHER INCORPORATED IN THE INTRO OR LEFT OUT. TOO MUCH TEXT
\section{Heun Equations in Mathematical Physics}
\label{sec:3}
Heun's differential equations have found a wide range of applicability in mathematical physics. While an ad-hoc list of the various applications is given in \cite{Hort1}, in this section we wish to focus on specific applications where the authors feel that an integral representation might provide novel physical insight. However, we also recognise that at the expense of focusing on certain applications in detail, we have to forego the discussion on applicability of Heun equations to a number of physical problems such as in quantum optics \cite{Xie2010,Moham}, certain applications in condensed matter physics \cite{Crampe,Dorey}, quantum computing \cite{QuantComp}, quantum two-state problem \cite{QTS1,QTS2} and other applications already laid out in \cite{Hort1}. In particular, we postpone the discussion on application to black hole physics to Section \hyperlink{sec:5}{V}.
\subsection{Exactly solvable quantum mechanical potentials}
The Schrodinger equation governing the state of a wavefunction is one of the most ubiquitous mathematical object in quantum physics. However, despite its wide applicability, exact analytical solutions for specific forms of the potential are extremely rare. In almost all standard quantum mechanical textbooks like \cite{Liboff}, the only cases that are discussed are those of the potential well $(V_{pw})$, harmonic oscillator $(V_{ho})$ and hydrogen atom $(V_{hyd})$ that have an extremely simple mathematical form:
\begin{gather}
V_{pw}(x) = \text{constant},\\
V_{ho}(x) = \frac{kx^2}{2}, \\
V_{hyd}(x) = -\frac{e^2}{4\pi\epsilon x},
\end{gather}
where, $k$ is the spring constant, $e$ is the charge of the electron $(9.1\times 10^{-31} C)$ and $\epsilon$ is the permittivity of the medium. Even for these simple cases, the solutions to the Schr{\"o}dinger equation are described by special mathematical functions: Hermite polynomials for the eigenfunction of the harmonic oscillator and Laguerre polynomials for the radial part of the hydrogen wavefunction. The important mathematical properties of these special functions are given in \cite{Arfken}.
Notwithstanding the challenges of finding such solvable potentials, it is a remarkable fact that Ishkhanyan and collaborators have been able to find a number of potentials whose solutions can be exactly found using hypergeometric and Heun functions. By termination of series expansion of Heun's functions that reduces them to the hypergeometric case, Ishkhanyan found the solutions to the following potentials: inverse square root potential \cite{Ishk1}, Lambert-W potential \cite{Ishk2} and independent Gauss hypergeometric equation \cite{Ishk3}. The entire class of Schr{\"o}dinger potentials that can be solved using the general Heun functions was given in a remarkable paper by him \cite{Ishk4}.
The commonality in the two aforementioned scenarios, analysis of the Schr{\"o}dinger's equation and solutions in terms of Heun's functions, is the fact that the treatment in both cases is done in differential form. When we switch over to analysing the behaviour of potentials in integral form, a host of mathematical subtleties come into the picture. For example in the case of the non-smooth Kurasov-type potentials \cite{Kurasov}, the seminal results of \cite{Kurasov} in distribution theory, which were obtained in differential form, were also obtained in integral form by \cite{Lange} .This is despite the fact that the equivalence between the differential and integral approaches are assumed to be valid only for smooth potentials. Two important insights were obtained from studying the distribution theory in terms of integral equations: it demonstrates the Schr{\"o}dinger's integral equation as a viable tool for analysis of singular potentials as well as hints at deeper equivalence between the integral and differential approaches.
Therefore, once we have the machinery of the integral form of the Heun equations at hand, a holistic study of exactly solvable potentials (such as the one's found by Ishkhanyan) can be pursued by utilising the Schr{\"o}dinger's integral equation. It is to be noted that a seemingly straightforward analysis in the differential form is by no means a guarantee of simplicity in the integral treatment. This has become apparent in the treatment of the seemingly simple Dirac delta prime potential, that governs point interactions in quantum mechanics has been a scope of intense debate, with special reference to the suitable boundary conditions (see references in \cite{Lange})
Lastly, we would also like to state that just like there are limited number of exactly solvable quantum mechanical potentials for the Schr{\"o}dinger's equation in differential form, there are a limited number of time-dependent Schr{\"o}dinger's propagators \footnote[1]{While discussing time-dependent Schr{\"o}dinger propagators, it would be remiss to not mention the path-integral formulation of quantum mechanics \cite{RPF} and while it has no direct relevance to the integral form of Heun's function, it most certainly can be studied for utilising the path-sum to provide a rigorous mathematical definition for the path-integral \cite{PathInt}. These issues are being currently investigated by the authors.} for various potentials \cite{Gaveau}. By understanding the differential and integral behaviour of the potentials whose solutions are governed by Heun's equations, obtaining the propagators for the physical systems being described for these equations could provide a much more general and robust analytical framework.
\subsection{Bethe Ansatz}
While it is practically impossible to match the breadth of applicability of studying various aspects related to the Schr{\"o}dinger's equation mentioned in the previous section, in this section we would like to discuss the potential applications of an integral representation of the Heun function \textit{specifically in the path sum formulation} due to the analytical richness of the path-sum integral form at the first order itself.
In the study of integrable models, the thermodynamic Bethe ansatz (TBA) has proven to be extremely useful for computing the free-energy of 1+1 dimensional integrable system \cite{Yang}. There was an unexpected result found that provided a correspondence between the limits of these integral equations and the Toda hierarchy in nonlinear differential equations. Now it has been shown that a non-trivial solution of the TBA equations can be found in terms of Airy functions \cite{Fendley}. The results of the paper are based on utilising intricate results involving the Painlev{\'e} III equations. As noted by us earlier, it is known that equations of the Heun class share a bijective relationship with Painlev{\'e} equations. Therefore, under the path sum formulation, it may be possible to obtain mathematical insight into further obtaining closed form solutions of the TBA.
\end{comment}
\subsection{Integral representations of Heun functions}
\label{sec:IntRep}
Erd{\'e}lyi was the first to give an integral equation relating the values taken at two points by a general Heun function \cite{Erdelyi}. His equation, a Fredholm integral equation, involves an hypergeometric kernel and can be used to obtain a series representation of Heun functions as sums of hypergeometric functions with coefficients determined via recurrence relations. Applications of this result in the special cases of Mathieu and Lam{\'e} equations were discussed by Sleeman \cite{Sleeman}. Naturally, since Erd{\'e}lyi's breakthrough many mathematical works on Heun equations were concerned with integral transformations involving Heun functions. In particular, based on the work of Carlitz\cite{Carlitz}, Valent found an integral transform for the Heun equation in terms of Jacobi polynomials \cite{Valent}; Ishkhanyan gave expansions of the confluent Heun functions involving incomplete beta functions \cite{Ishkhanyan_2005}; El Jaick and coworkers \cite{El_Jaick_2011} provided novel transformations and classified expansions for Heun functions involving hypergeometric kernels; and Takemura found an elliptic transformation relating Heun's functions for different parameters based on the Weierstrass sigma function \cite{Takem2}. This brief list of contributions is far from exhaustive, we refer to the recent review \cite{Hort1} for more details.
The common feature of all of these integral transforms is that they contain higher transcendental functions which makes them physically opaque and of limited use for practical calculations. In addition, the resulting series representations for the Heun functions have insufficient radiuses of convergence \cite{Cook2016} causing difficulties for black hole perturbation theory (see Section~\ref{sec:5}).
These issues were noted in the recent review \cite{Hort1}
on Heun's functions, the current state of research on this being described as follows :\\[-1.1em]
\textit{``No example has been given of a solution of Heun's equation expressed in the form of a definite integral or contour integral involving only functions which are, in some sense, simpler.[...] This statement does not exclude the possibility of having an infinite series of integrals with `simpler' integrands''.}\\[-1.1em]
In this work, we constructively prove the existence of such a representation for all types of Heun's functions and for all parameters, in the form of infinite series of integrals whose integrands involve only rational functions and exponentials of polynomials. Furthermore, we show that the series converges everywhere except at the singular points of the Heun function. We show that any Heun function, general or (bi-, doubly-, tri-)confluent, is a sum of exactly two functions each of which satisfy a linear Volterra equation of the second kind with explicitely identified elementary kernels. In particular, any Heun function $H(z)$ itself satisfies a linear integral Volterra equation of the second kind with such an elementary kernel if either there is at least one non-singular point $z_0\in\mathbb{R}$ where $H(z_0)=H'(z_0)$ or there is a point where $H(z_0)=0$.
\section{Elementary integral series for all types of Heun functions}\label{sec:ExplicitRes}
Owing to the emphasis of the present work on concrete results and a physical application, all the technical mathematical proofs are deferred to Appendix~\ref{AppendixProofs}.
\subsection{Notation}
The $\ast$ notation is useful to denote iterated integrals. Let $K(z,z_0)$ be a function of two variables that is continuous over $]z_0,z[$. We denote $K(z,z_0)=K^{\ast1}(z,z_0)$ and, for any integer $n>1$,
$$
K^{\ast n}(z,z_0)=\int_{z_0}^{z} K^{\ast (n-1)}(z,\zeta_1)K(\zeta_1,z_0)d\zeta.
$$
In other terms $K^{\ast n}$ is the Volterra composition \cite{Volterra1924} of $K$ with itself $n$-times. The only type of integral series that is required to present all results of this section is the following
\begin{align*}
G(z,z_0)&:= \sum_{n=1}^\infty K^{\ast n}(z,z_0),\\
&=K(z,z_0)+\int_{z_0}^z K(z,\zeta_1)K(\zeta_1,z_0)d\zeta_1\nonumber\\
&\hspace{5mm}+\int_{z_0}^z \int_{\zeta_1}^z K(z,\zeta_2)K(\zeta_2,\zeta_1)K(\zeta_1,z_0)d\zeta_2d\zeta_1\,+\cdots, \nonumber
\end{align*}
see also Eq.~(\ref{Gexplicitform}). In the appendix, we show that once $z_0$ is fixed, the above series converges over any subinterval of $\mathbb{R}$ which does not contain a singularity of $K(z,z_0)$.
A bound on the convergence speed of the series is also provided.
The function $G(z,z_0)$ defined above, is solution to the linear Volterra integral equation of the second kind
\begin{equation}\label{VolterraG}
G(z,z_0)=K(z,z_0)+\int_{z_0}^z K(z,\zeta)G(\zeta,z_0)d\zeta,
\end{equation}
or, in $\ast$ notation, $G=K+K\ast G$. Thus, the function $G$ can either be evaluated from the integral series or by solving the above Volterra equation.
\subsection{Results}
We emphasize that all results stated remain valid for complex parameter values. This is crucial notably when forming solutions of the Teukolsky equation in the study of quasinormal modes, for which the frequency parameter takes complex values (see Eq.~\ref{eq:18}).
\begin{corollary}[General Heun Equation]\label{GenHeunCorrolary}
Let $H_G(z)$ be solution of the General Heun Equation,
\begin{equation*}
\frac{d^2H_G(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \frac{\delta}{z - 1} + \frac{\epsilon}{z - t} \bigg]\frac{dH_G(z)}{dz} + \frac{\alpha\beta z - q}{z(z-1)(z-t)}H_G(z) = 0,
\end{equation*}
with initial conditions $H_G(z_0)=H_0$ and $\dot{H}_G(z_0)=H'_0$, assuming that $z_0\in\mathbb{R}$ is not a singular point of $H_G$. Denote $I$ the largest real interval that contains $z_0$ and does not contain any singular point of $H_G$. Then, for any $z\in I$,
\begin{align*}
H_G(z)&=H_0+H_0\int_{z_0}^z \!G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\left(\!e^{z-z_0}-1+\!\int_{z_0}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right),
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$ and
\begin{align*}
K_{1}(z,z_0)&=\\
&\hspace{-15mm}1+ e^{-z}\int_{z_0}^z \Big\{\frac{\zeta_1^{\gamma } (\zeta_1-1)^{\delta }(t-\zeta_1)^{\epsilon }}{z^{\gamma }
(z-1)^{\delta } (t-z)^{\epsilon }}e^{\zeta_1} \left(\frac{q-\alpha \beta
\zeta _1}{\left(\zeta
_1-1\right) \zeta _1
\left(\zeta
_1-t\right)}-\frac{\epsilon
}{t-\zeta _1}-\frac{\gamma }{\zeta
_1}-\frac{\delta }{\zeta
_1-1}-1\right)\Big\}d\zeta_1,\\
K_{2}(z,z_0)&=\left(\frac{q-\alpha \beta z}{(z-1) z
(z-t)}-\frac{\epsilon
}{t-z}-\frac{\gamma
}{z}-\frac{\delta }{z-1}-1\right)e^{z-z_0}-\frac{q-\alpha \beta z}{(z-1) z
(z-t)}.
\end{align*}
\end{corollary}
~\\[-1em]
\begin{example}[Elementary integral series converging to a general Heun function]
In order to illustrate concretely the above corollary, consider the following General Heun equation (here with arbitrary parameters),
\begin{equation}\label{ex:Hg1}
\frac{d^2H_G(z)}{dz^2} + \bigg[ \frac{2}{z} + \frac{7}{z - 1} + \frac{(-1)}{z - 4} \bigg]\frac{dH_G(z)}{dz} + \frac{(3/2)z - 1}{z(z-1)(z-4)}H_G(z) = 0,
\end{equation}
with initial conditions $H_G(6)=H'_T(6)=1$. Here, the largest real interval containing $6$ and none of the singular points $0$, $1$ and $t=4$ is $I=]4,+\infty[$.
Thus Corollary~\ref{GenHeunCorrolary} indicates that for any $z\in]4,+\infty[$,
\begin{align*}
H_G(z) &= 1+\int_{z_0}^z G_{1}(\zeta,z_0)d\zeta,\\
&=1+\sum_{n=1}^\infty \int_{6}^z K_{1}^{\ast n}(\zeta,6)d\zeta,\\
&=1+\int_{6}^z K_{1}(\zeta,6)d\zeta+\int_{6}^z\int_{6}^{\zeta} K_{1}(\zeta,\zeta_1)K_{1}(\zeta_1,6)d\zeta_1d\zeta+\cdots,
\end{align*}
with the kernel $K_1$ given by
\begin{align*}
K_{1}(z,z_0)
&=1- e^{-z}\frac{(z-4)}{z^2(z-1)^7}\int_{z_0}^z\!e^{\zeta_1}\frac{\zeta _1\left(\zeta _1-1\right)^6}{2 \left(\zeta
_1-4\right)^2}
\left(2 \zeta _1^3+10 \zeta _1^2-67 \zeta
_1+14\right)d\zeta_1.
\end{align*}
In Fig.~(\ref{fig:Ex1}), we show a purely numerical evaluation of $H_G(z)$ together with analytical estimates based on the first few orders of the above series, i.e. we give $H^{(m)}_G(z) := 1+\sum_{n=1}^m \int_{6}^z K_{1}^{\ast n}(\zeta,6)d\zeta$, with $m=1,2,3$ and $m=6$. This exhibits the convergence of the Neumann series representation of the path-sum formulation of a general Heun function, as predicted by the theory.
\begin{figure}[!h]
\centering
\includegraphics[width=1\textwidth]{HeunFigGen.pdf}
\caption{\label{fig:Ex1}\textbf{Convergence to a general Heun function with elementary integrals.} Numerical evaluation of the general Heun function solution of Eq.~(\ref{ex:Hg1}) (solid black line), together with the first integral approximands of it: $H^{(1)}_G(z)$ (dotted magenta line), $H^{(2)}_G(z)$ (dot-dashed red line), $H^{(3)}_G(z)$ (dashed blue line) and $H^{(6)}_G(z)$ (solid blue line, very close to the numerical solution). For orders $m\geq 9$, we reach the numerical solution to within machine precision. Note that the integral series given here is convergent on $z\in ]4,+\infty[$ but we show only the interval $z\in[6,26]$ for illustration purposes.}
\end{figure}
~\\[-1em]
The results above continue to hold should e.g. $z_0=3$, in which case $I=]1,4[$; $z_0=1/2$ implying $I=]0,1[$; or $z_0=-20$ giving $I=]-\infty,0[$. In other terms, the integral representation given for the General Heun function is valid everywhere on $z\in\mathbb{R}\backslash\{0,1,t=4\}$ but can only be used in an interval $I$ where initial conditions for $H_G$ are available.
\end{example}
\newpage
\begin{corollary}[Confluent Heun Equation]\label{corr:Conf}
Let $H_C(z)$ be solution of the Confluent Heun Equation,
\begin{equation*}
\frac{d^2H_C(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \frac{\delta}{z - 1} + \epsilon \bigg]\frac{dH_C(z)}{dz} + \frac{\alpha z - q}{z(z-1)}H_C(z) = 0,
\end{equation*}
with initial conditions $H_C(z_0)=H_0$ and $\dot{H}_C(z_0)=H'_0$, assuming that $z_0\neq0$ and $z_0\neq 1$. If $z_0<0$, let $I=]-\infty,0[$, if $0<z_0<1$ let $I=]0,1[$, and else for $z_0>1$ let $I=]1,+\infty[$. Then, for any $z\in I$,
\begin{align*}
H_C(z)&=H_0+H_0\int_{z_0}^z \!G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\left(\!e^{z-z_0}-1+\!\int_{z_0}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right),
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$, and
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-z}\int_{z_0}^z \Big\{
\frac{e^{\zeta\epsilon}\zeta^{\gamma } \left(\zeta-1\right)^{\delta}}{e^{z \epsilon} z^{\gamma } (z-1)^{\delta }} e^{\zeta}
\left(\frac{q-\alpha \zeta}{\left(\zeta-1\right) \zeta}-\frac{\gamma }{\zeta}-\frac{\delta }{\zeta-1}-\epsilon -1\right)\Big\}d\zeta,\\
K_{2}(z,z_0)&=\left(\frac{q-\alpha z}{\left(z-1\right) z}-\frac{\gamma }{z}-\frac{\delta }{z-1}-\epsilon -1\right)e^{z-z_0}-\frac{q-\alpha z}{(z-1) z}.
\end{align*}
\end{corollary}
~\\[-1em]
\begin{example}[Convergence to a Confluent Heun function]
Let us now consider the following Confluent Heun function $H_C(z)$ satisfying
\begin{equation}\label{ex:HC}
\frac{d^2H_C(z)}{dz^2} + \bigg[ \frac{3}{z} + \frac{(2/3)}{z - 1} + 4 \bigg]\frac{dH_C(z)}{dz} + \frac{5 z - 1}{z(z-1)}H_C(z) = 0
\end{equation}
with initial conditions $H_C(-5)=0$ and $H'_C(-5)=1$. Suppose that we wish to evaluate $H_C$ on the interval $z\in]-\infty,0[$, i.e. on both sides $z<z_0$ and $z>z_0$ of the conditions at $z_0=-5$.
Then Corollary~\ref{corr:Conf} indicates that, for any $z\in]-\infty,0[$, we have
\begin{equation*}
H_C(z) = e^{z+5}-1+\int_{-5}^z(e^{z-\zeta}-1)G_{2}(\zeta,-5)d\zeta,
\end{equation*}
with $G_2=\sum_{n=1}^\infty K_2^{\ast n}$ and
\begin{align*}
K_{2}(z,z_0)=\frac{3 (5 z-1)-e^{z-z_0} (3 z+4) (5 z-3)}{3 (z-1) z}.
\end{align*}
We emphasize that these results hold for all $z\in]-\infty,0[$ since this interval is divergence free, more precisely $K_2$ is bounded continuous on any compact subinterval of $]-\infty, 0[$ and the integral series for $G_2$ is thus guaranteed to converge on this entire domain (this is shown in the appendix). Note that when considering $z<z_0$, all integrals remain the same as for $z>z_0$.
In Fig.~(\ref{fig:ExConf}) below, we show a purely numerical evaluation of $H_C(z)$ together with the truncated integral series approximations
\begin{align}
H_C^{(m)}(z) &:= \int_{-5}^z(e^{z-\zeta}-1)\Big(1+\sum_{n=1}^m K_{2}^{\ast n}(\zeta,-5)d\zeta\Big),\label{PSsol}\\
&=\int_{-5}^z(e^{z-\zeta}-1)\Big(1+K_{2}(\zeta,-5)+\int_{-5}^\zeta K_2(\zeta,\zeta_1)K_2(\zeta_1,-5)d\zeta_1 +\cdots\Big).\nonumber
\end{align}
Since kernel $K_2$ is singular at $z=0$ just as $H_C$ is, we expect the convergence speed of the integral series to slow down when approaching the singular point, as predicted by the bound of Eq.(\ref{Gbound}) presented in the appendix. This does not preclude analytically obtaining the correct asymptotic behavior for $H_C(z)$ as $z\to 0^-$. Indeed this follows from the behavior of $K_2$ under the same limit. We demonstrate such a procedure in \S\ref{TRESol}.
\begin{figure}[!h]
\centering
\includegraphics[width=1\textwidth]{HeunFigConf.pdf}
\caption{\label{fig:ExConf}\textbf{Convergence to a Confluent Heun function with elementary integrals over the interval $]-\infty,0[$.} Numerical solution of the Eq.~(\ref{ex:HC}) (solid black line) with conditions $H_C(-5)=0$, $H_C'(-5)=1$, together with its integral approximands as per Eq.~(\ref{PSsol}), $H^{(2)}_C(z)$ (dotted magenta line), $H^{(20)}_C(z)$ (dot-dashed red line) and $H^{(40)}_C(z)$ (dashed blue line, very close to the numerical solution). Convergence near $z=0$ is slowed down due to $K_{2}$ being singular at $z=0$ just as $H_C$ is. Still, the integral series is convergent over the entire domain $z\in]-\infty,0[$, a crucial property for perturbative black hole theory that is \emph{unique} to the present approach. Here as in subsequent examples we plot the various functions over smaller intervals for $z$, for illustration purposes.}
\end{figure}
\end{example}
~\\
\FloatBarrier
\newpage
\begin{corollary}[Biconfluent Heun Equation]\label{corr:BiConf}
Let $H_B(z)$ be solution of the Biconfluent Heun Equation,
\begin{equation*}
\frac{d^2H_B(z)}{dz^2} + \bigg[ \frac{\gamma}{z} + \delta + \epsilon z \bigg]\frac{dH_B(z)}{dz} + \frac{\alpha z - q}{z}H_B(z) = 0,
\end{equation*}
with initial conditions $H_B(z_0)=H_0$ and $\dot{H}_B(z_0)=H'_0$, assuming that $z_0\neq0$. If $z_0>0$, denote $I=]0,+\infty[$ otherwise let $I=]-\infty,0[$. Then, for any $z\in I$,
\begin{align*}
H_B(z)&=H_0+H_0\int_{z_0}^z \!G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\left(\!e^{z-z_0}-1+\!\int_{z_0}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right),
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$, and
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-z}\int_{z_0}^z \Big\{\frac{\zeta _1^{\gamma }}{z^{\gamma}} e^{\zeta _1-\frac{1}{2} \left(z-\zeta _1\right) \left(2 \delta +\epsilon \left(\zeta
_1+z\right)\right)} \Big(\frac{q-\alpha \zeta _1}{\zeta _1}-\frac{\gamma }{\zeta _1}-\delta -\zeta _1 \epsilon -1\Big)\Big\}d\zeta_1,\\
K_{2}(z,z_0)&=\Big(\frac{q-\alpha z}{z}-\frac{\gamma }{z}-\delta -z \epsilon -1\Big)e^{z-z_0}-\frac{q-\alpha z}{z}.
\end{align*}
\end{corollary}
\begin{example}[Evaluating a Biconfluent Heun function via Volterra equations]
Let us now consider the following Biconfluent Heun function $H_B(z)$ satisfying
\begin{equation}\label{ex:HB}
\frac{d^2H_B(z)}{dz^2} + \bigg[ \frac{(1/10)}{z} + 1 + 6 z \bigg]\frac{dH_B(z)}{dz} + \frac{(-1) z - 2}{z}H_B(z) = 0,
\end{equation}
with initial conditions $H_B(2/3)=0$ and $H'_B(2/3)=-4$. Then Corollary~\ref{corr:BiConf} indicates that for $z>0$,
\begin{align}\label{VolterraBiConf}
H_B(z)&=2+2 \int_{2/3}^z \!\!G_{1}(\zeta,2/3)d\zeta-6\left(\!e^{z-2/3}-1+\!\int_{2/3}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,2/3)\!\right) d\zeta,
\end{align}
with $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$ for $i=1,2$, and
\begin{align*}
K_{1}(z,z_0)&=1+\int_{z_0}^z\frac{(19-10 \zeta (6 \zeta +1)) e^{-(z-\zeta ) (3 \zeta +3 z+2)}}{10 \zeta ^{9/10} z^{1/10}}d\zeta,\\
K_{2}(z,z_0)&=\frac{\big(19-10 z (6 z+1)\big) e^{z-z_0}-10 (z+2)}{10 z}.
\end{align*}
Instead of evaluating functions $G_1$ and $G_2$ as the integral series, we may directly solve the linear integral Volterra equations that they satisfy, see Eq.~(\ref{VolterraG}). Such equations are very well behaved and numerically easy to solve, so that we can evaluate $H_B$ thanks to Eq.~(\ref{VolterraBiConf}) with high numerical accuracy.
In Fig.~(\ref{fig:ExBiConf}) below, we show the numerical evaluation of $H_B(z)$ obtained using a standard differential equations numerical solver versus the procedure described above.
\begin{figure}[!h]
\centering
\includegraphics[width=.92\textwidth]{HeunFigBiGen.pdf}
\caption{\label{fig:ExBiConf}\textbf{Biconfluent Heun function from Volterra equations.} Numerical solution of the Eq.~(\ref{ex:HC}) (solid black line), together with the function predicted by Eq.(\ref{VolterraBiConf}) (dashed blue line). The two are indistinguishable. The integral representation of Eq.~(\ref{VolterraBiConf}) is valid for $z\in]0,+\infty[$, we here show only $z\in]0,10[$ for illustration purposes.}
\end{figure}
\end{example}
~\\
\FloatBarrier
\begin{corollary}[Doubly-confluent Heun Equation]\label{DoublyConfluentCorr}
Let $H_D(z)$ be solution of the Doubly-confluent Heun Equation,
\begin{equation*}
\frac{d^2H_D(z)}{dz^2} + \bigg[ \frac{\delta}{z^2}+\frac{\gamma}{z} + 1 \bigg]\frac{dH_D(z)}{dz} + \frac{\alpha z - q}{z^2}H_D(z) = 0
\end{equation*}
with initial conditions $H_D(z_0)=H_0$ and $\dot{H}_D(z_0)=H'_0$, assuming that $z_0\neq0$. If $z_0>0$, denote $I=]0,+\infty[$ otherwise let $I=]-\infty,0[$. Then, for any $z\in I$,
\begin{align*}
H_D(z)&=H_0+H_0\int_{z_0}^z \!G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\left(\!e^{z-z_0}-1+\!\int_{z_0}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right),\,
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$, and
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-z}\int_{z_0}^z \Big\{\frac{\zeta _1^{\gamma }}{z^{\gamma }} e^{-\frac{\delta }{\zeta _1}+2 \zeta _1+\frac{\delta }{z}-z} \Big(\frac{q-\alpha \zeta _1}{\zeta _1^2}-\frac{\gamma }{\zeta _1}-\frac{\delta }{\zeta _1^2}-2\Big)\Big\}d\zeta_1,\\
K_{2}(z,z_0)&=\Big(\frac{q-\alpha z}{z^2}-\frac{\delta }{z^2}-\frac{\gamma }{z}-2\Big)e^{z-z_0}-\frac{q-\alpha z}{z^2}.
\end{align*}
\end{corollary}
~\\[-2em]
\begin{example}[Convergence to a Doubly-Confluent Heun function]
Let us now consider the following Doubly-Confluent Heun equation, once again with arbitrarily chosen parameters for the example,
\begin{equation}\label{ex:Hd2}
\frac{d^2H_D(z)}{dz^2} + \bigg[ \frac{(-2)}{z^2}+\frac{1}{z} + 1 \bigg]\frac{dH_D(z)}{dz} + \frac{10 z - (-1)}{z^2}H_D(z) = 0
\end{equation}
with initial conditions $H_D(1)=H'_D(1)=1/2$.
Then Corollary~\ref{DoublyConfluentCorr} indicates that for $z\in]0,+\infty[$,
\begin{align*}
H_D(z) &= \frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty \int_{1}^z K_{1}^{\ast n}(\zeta,1)d\zeta,
\end{align*}
with
\begin{align*}
K_{1}(z,z_0)=1+\frac{e^{-z-\frac{1}{z}}}{z}\int_{z_0}^ze^{2\zeta_1+\frac{2}{\zeta_1}}\frac{1}{\zeta_1 }(1-2\zeta_1^2+11\zeta_1)d\zeta_1
\end{align*}
In Fig.~(\ref{fig:Ex1}) below, we show a purely numerical evaluation of $H_D(z)$ together with analytical approximations based on the first few orders of the above series, i.e. we give $H_D^{(m)}(z) := 1+\sum_{n=1}^m \int_{1}^z K_{1}^{\ast n}(\zeta,1)d\zeta$, with $m=3,5,8$. This demonstrates again the convergence of the Neumann series representation of the path-sum formulation of a general Heun function, as predicted by the theory. Here the exact $H_D(z)$ and $H_D^{(m)}(z)$ become indistinguishable for $m\geq 9$.
\begin{figure}[h!]
\centering
\includegraphics[width=1\textwidth]{HeunFigDoublyConfluent.pdf}
\caption{\label{fig:Ex1}\textbf{Convergence to Doubly-Confluent Heun function with elementary integrals.} Numerical solution of the Eq.~(\ref{ex:Hd2}) (solid black line), together with its integral approximands $H^{(3)}_D(z)$ (dotted magenta line), $H^{(5)}_D(z)$ (dot-dashed red line) and $H^{(8)}_D(z)$ (dashed blue line). Note the integral series provided here is convergent for $z\in]0,+\infty[$ and show only the interval $z\in[1,9]$ for illustration purposes.}
\end{figure}
\end{example}
\newpage
\begin{corollary}[Triconfluent Heun Equation]\label{TriconfluentCoro}
Let $H_T(z)$ be solution of the Triconfluent Heun Equation,
\begin{equation*}
\frac{d^2H_T(z)}{dz^2} + \bigg[ \gamma + \delta z + \epsilon z^2 \bigg]\frac{dH_T(z)}{dz} + (\alpha z - q)H_T(z) = 0,
\end{equation*}
with initial conditions $H_T(z_0)=H_0$ and $\dot{H}_T(z_0)=H'_0$. Then, for any $z\in \mathbb{R}$,
\begin{align*}
y(z)&=H_0+H_0\int_{z_0}^z \!G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\left(\!e^{z-z_0}-1+\!\int_{z_0}^z\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right),
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$ and
\begin{align*}
K_{1}(z,z_0)&=1-e^{-\frac{1}{6} z \left(6 \gamma +2 z^2 \epsilon +3 \delta z+6\right)}\!\!\int_{z_0}^z\!\! e^{\frac{1}{6} \zeta \left(6 \gamma +3 \delta \zeta +2 \zeta ^2 \epsilon +6\right)} \big(\zeta (\alpha +\delta +\zeta
\epsilon )+\gamma -q+1\big) d\zeta,\\
K_{2}(z,z_0)&=-\big(\epsilon z^2+(\alpha+\delta)z -q+\gamma+1\big)e^{z-z_0}-(q-z\alpha).
\end{align*}
\end{corollary}
~\\[-1em]
\begin{example}[Convergence to a complex-valued Triconfluent Heun function]
Consider the Triconfluent Heun function defined as the solution to
\begin{equation}\label{ex:HT}
\frac{d^2H_T(z)}{dz^2} + \bigg[ 2 - z + 7 z^2 \bigg]\frac{dH_T(z)}{dz} + (z - (2+i))H_T(z) = 0,
\end{equation}
where $i^2=-1$, and with initial conditions $H_T(-5)=H'_T(-5)=2$. Corollary~\ref{TriconfluentCoro} indicates that for $z\in\mathbb{R}$,
\begin{align*}
H_D(z) &= 2+2 \int_{z_0}^z G_{1}(\zeta,-10)d\zeta,
\end{align*}
with $G_1=\sum_{n=1}^\infty K_1^{\ast n}$ and
\begin{align*}
K_1(z,z_0)=1-e^{\frac{1}{6} (3-14 z) z^2-3z}\int_{z_0}^ze^{-\frac{1}{6} (3-14 \zeta) \zeta^2+3\zeta} (7\zeta^2+1-i)\,d\zeta.
\end{align*}
We show in Fig.~(\ref{fig:HeunT}) convergence to the complex-valued triconfluent Heun function by the integral series
\begin{align*}
H_T^{(m)}(z) &:= 2+2 \int_{z_0}^z \sum_{n=1}^m K^{\ast n}_{1}(\zeta,-10)d\zeta,\\
&=2+2\int_{z_0}^z K_{1}(\zeta,-10)d\zeta+\int_{z_0}^z \int_{z_0}^z K_{1}(z,\zeta_1)K_1(\zeta_1,-10)d\zeta_1d\zeta+\cdots
\end{align*}
With this example, we emphasize that all the integrals representations obtained here remain valid for complex-valued Heun functions.
\begin{figure}[h!]
\centering
\includegraphics[width=1\textwidth]{HeunFigTriRe.pdf}\\
\includegraphics[width=1\textwidth]{HeunFigTriIm.pdf}
\caption{\label{fig:HeunT}\textbf{Convergence to a complex-valued Triconfluent Heun function with elementary integrals.} Numerical solution of the Eq.~(\ref{ex:HT}) (solid black line), together with its integral approximands $H^{(1)}_T(z)$ (dotted magenta line), $H^{(3)}_T(z)$ (dot-dashed red line) and $H^{(6)}_T(z)$ (dashed blue line). Top figure: real parts of these quantities. Bottom figure: imaginary parts of these quantities. Note that the integral series provided is convergent over the entire real line, we here show only the interval $z\in [-5,7]$ for illustration purposes.}
\end{figure}
\end{example}
Having focused on concrete evaluations of various Heun functions in the illustrative examples, we now turn to using the elementary integral series in the field of black hole physics.
\newpage
\section{Application to Black-Hole Perturbation Theory}
\label{sec:black hole}
\subsection{Motivations}
The theory of metric perturbations of Kerr black holes is governed by the Teukolsky equation \cite{TeukEqn}. This equation provides the basic mathematical framework to study
the stability of Schwarzschild \cite{Finster2009} and Kerr black holes \cite{Costa2020} and yields physical insights in the broader field of gravitational wave astrophysics \cite{Sasaki2003}. With the advent of event detections by LIGO \cite{LIGO2016,LIGO2020}, obtaining a better analytical grasp over the solutions of the Teukolsky equation is paramount in modeling the ringdown stage \cite{London2014} of a binary black hole merger using accurate waveform templates \cite{McWilliams2019}.
\\
In the frequency domain, the Teukolsky equation can be decoupled into radial and angular components \cite{TeukEqn}. Determining the analytical solutions of the radial equation has been an active area of research since the first formulation of the equations \cite{Mano1996,Sasaki2003}. To this end, state-of-the-art approaches all rely on the same strategy: i) obtain two series expansions of the solution, one convergent near the black hole horizon the other at spatial infinity; and ii) match both expansions at some intermediate radial point.
The standard implementation of this strategy, due to Mano, Suzuki and Takasugi (MST) \cite{Mano1996,Mano1997}, relies on a series of hypergeometric functions at the black hole horizon and of Coulomb wave functions at spatial infinity. Matching both expansions requires the introduction of an auxiliary parameter $\nu$.
We stress that this parameter is \textit{not} part of the original parameters of the Teukolsky equation. Rather $\nu$ is a mathematical checkpost introduced to establish the convergence and matching of the hypergeometric and Coulomb series \cite{Fujita2004}. The MST strategy successfully yields accurate numerical data for studying gravitational wave radiation from Kerr black holes \cite{Sasaki2003,Fujita2004}. It is \textit{``the only existing method that can be used to calculate the gravitational waves emitted to infinity to an arbitrarily high post-Newtonian order in principle.''} \cite{Sasaki2003}. At the same time, it has been explicitly recognised that the mathematical complexity of the formalism obscures physical insights into the problem \cite{Sasaki2003}.
In particular, the auxiliary parameter $\nu$, which has been called "renormalised angular momentum" to make it more palatable, has limited correspondence to physical phenomenon, if any.
More recently, explicit, analytic solutions to the Teukolsky equation have been established in terms of Heun functions \cite{Fiziev2}.
Yet, Cook and Zalutskiy \cite{Cook1} note that in order to extract physical quantities of interest out of this approach, one is forced to
revert to Leaver's formalism \cite{Leaver1985} because \textit{``the series solution around z = 1 has a radius of convergence no larger than 1, far short of infinity''}. Thus, just as for the MST formalism the problem is, in essence, that we are lacking a single representation of the solution to the Teukolsky radial equation that is convergent from the black hole horizon up to spatial infinity.
The integral series provided in this work addresses this issue completely since it converges on this entire domain, thereby retaining the crucial features of the MST formalism that lead to its widespread applicability, while also not requiring any auxiliary, unphysical parameter.
In a similar vein, we can assert that our formalism is suited for practical numerical and even analytical, calculations since the integral series are rapidly convergent, and their asymptotic behavior is analytically available. We may therefore also hope that the integral series representation will help solve the well-recognised computational difficulties that emerge from the MST formalism when applied to gravitational wave physics, in particular for the two body problem \cite{Bini2013}, and in the gravitational self force program \cite{Sago2003,Hikida2004,Kavanagh2016}.
\\
For completeness, we begin with a brief discussion of the theory of the Teukolsky equation and its reduction to Heun form. We then give the series representation of its solution. Finally, we establish its asymptotics at both the black hole horizon ($z\to 1^+$) and spatial infinity ($z\to +\infty$).
\subsection{The Teukolsky Equation : background}
The Teukolsky Equation \cite{TeukEqn} is a gauge invariant equation \cite{GaugeInv} that governs the curvature perturbations of the Kerr black hole \cite{MWT}. By making use of the Newman-Penrose formalism \cite{NewmanPenrose}, the single master equation for the spin $(s)$ weighted scalar wave function $_{s}\psi$ in Boyer-Lindquist co-ordinates $\{t,r,\theta,\phi \}$ \cite{BoyerLindquist} and the Kinnersley tetrad \cite{Kintetrad} is written as:
\begin{gather}
\bigg[ \frac{(r^2 + a^2)^2}{ \Delta} - a^2 \sin^2 \theta \bigg] \frac{ \partial^2 _{s}\psi}{\partial t^2} +
\bigg(\frac{4Mar}{\Delta}\bigg) \frac{\partial^2 _{s}\psi}{\partial t \partial \phi} + \bigg[ \frac{a^2}{\Delta} - \frac{1}{sin^2 \theta} \bigg] \frac{\partial^2 _{s}\psi}{\partial \phi^2} \notag \\
- \Delta^{-s} \frac{\partial}{\partial r} \bigg( \Delta^{s+1} \frac{\partial _{s}\psi}{\partial r} \bigg) - \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \bigg( \sin \theta \frac{ \partial _{s}\psi}{\partial \theta} \bigg) -2s \bigg[ \frac{a(r-M)}{\Delta} + \frac{i \cos \theta}{\sin^2 \theta} \bigg] \frac{\partial _{s}\psi}{\partial \phi} \notag \\ -2s \bigg[ \frac{M(r^2 - a^2)}{\Delta} - r - ia\cos \theta \bigg] \frac{\partial}{\partial t} + (s^2 \cot^2 \theta - s)_{s}\psi = 4 \pi \Sigma T \label {eq:TeukolskyEqn}
\end{gather}
where the auxiliary variables are given by:
\begin{gather}
\Sigma \equiv r^2 + a^2\cos^2 \theta,
\Delta \equiv r^2 - 2Mr + a^2
\end{gather}
Here, $M$ is the mass of the black hole, $a$ is its angular momentum (per unit mass), $T$ is the source term built from the energy-momentum tensor \cite{TeukEqn} and the spin parameter $s = 0, \pm 1/2, \pm 1, \pm 2 \pm 3/2$ for scalar, neutrino, electromagnetic, gravitational and Rarita-Schwinger \cite{RaritaSchwinger} fields respectively. It reduces to the Bardeen-Press equation \cite{BardeenPress} in the non-rotating $(a = 0)$ case.
The equation \ref{eq:TeukolskyEqn} can be separated in time \cite{Krivan1997} and frequency domain
\cite{TeukEqn}. The latter can be performed for the vacuum case $(T = 0)$ by the following separation ansatz:
\begin{gather}
_{s}\psi(t,r,\theta, \phi) = e^{-i \omega t}e^{i m \phi}S(\theta)R(r).
\end{gather}
For the radial function $R(r)$ we obtain the Teukolsky Radial Equation (TRE):
\begin{gather} \label{eq:TRE}
\Delta^{-s} \frac{d}{dr} \bigg[ \Delta^{s+1} \frac{d R(r)}{dr} \bigg] + \bigg[ \frac{K^2 -2is(r-M)K}{\Delta} + 4is\omega r - \lambda \bigg]R(r) = 0,
\end{gather}
where,
\begin{gather}
K \equiv (r^2 + a^2)\omega -am, \quad
\lambda \equiv \, _{s}A_{lm}(a\omega) + a^2\omega^2 -2am\omega.
\end{gather}
For the angular equation, we make $x \equiv \cos \theta$. Now the function $S(\theta) = \, _{s}S_{lm}(x;a\omega)$ is the spin weighted spheroidal function \cite{Breuer1977} which gives the
solution for the Teukolsky Angular Equation (TAE):
\begin{align} \label{eq:TAE}
&\partial_{x} \bigg[(1-x^2)\partial_{x}[_{s}S_{lm}(x;c)] \bigg] + \bigg[ (cx)^2 - 2csx + s \\&\hspace{35mm} +\, _{s}A_{lm}(c) - \frac{(m+sc)^2}{1-x^2} \bigg] \,_{s}S_{lm}(x;c) = 0, \nonumber
\end{align}
where $c = a\omega$ is the oblateness parameter, $m$ is the azimuthal separation constant and $_{s}A_{lm}(c)$ is the angular separation constant. The equations \ref{eq:TRE} and \ref{eq:TAE} are coupled equations which require simultaneous evaluation of the parameters $\omega$ and $_{s}A_{lm}(c)$. Given a value for $_{s}A_{lm}(c)$, we can solve \ref{eq:TRE} for the complex frequency $\omega$ and given the latter, we can solve \ref{eq:TAE} as an eigenvalue problem for $_{s}A_{lm}(c)$.
\subsection{Teukolsky Radial Equation in Heun Form}
We now reduce the Teukolsky Radial Equation to the non-symmetrical Heun form, which allows us to represent its solution with the results of Section.~\ref{sec:ExplicitRes}. There is one small consideration to be noted: depending on the sign of the spin $s$ we wish to operate in, certain parameters of the CHE form of the TAE and TRE flip their signs as given in \cite{Fiziev2}. However this is not of relevance for our purposes since our main aim is to work with the CHE \textit{form} of the equations that obviously remains irrespective of the sign of the spin parameter.
The radial function $R(r)$ solution to Eq.~\ref{eq:TRE} has three singularities: an irregular singular point at $r = \infty$ and two regular singular points corresponding to the roots of $\Delta = 0$, which are
\begin{equation*}
r_{\pm} = M \pm \sqrt{M^2 - a^2}
\end{equation*}
The values $r_{\pm}$ correspond to the event and Cauchy horizon respectively (for an in-depth introduction to the notation and terminology on back-hole mathematics, we refer the reader to \cite{MWT}). Having identified these, we may now map the Teukolsky Radial Equation into an Heun equation. We close following the standard treatment \cite{Cook1}. We begin by letting the radial function $R$ be of the form
\begin{equation}
R(r) = (r-r_{+})^{\xi}(r-r_{-})^{\eta}e^{\zeta r}H(r), \label{eq:RtoH}
\end{equation}
where the parameters $\zeta, \xi, \eta$ are given by
\begin{align}
&\zeta = \pm i\omega \equiv \zeta_{\pm},\qquad
\xi = \frac{-2 \pm (s + 2i\sigma_{+})}{2} \equiv \xi_{\pm},\label{eq:18}\\
&\eta = \frac{ -s \pm (s - 2i\sigma_{-})}{2} \equiv \eta_{\pm},\qquad
\sigma_{\pm} = \frac{2\omega M r_{\pm} - ma}{r_{+} - r_{-}}.\nonumber
\end{align}
With the dimensionless variables
\begin{align*}
\bar{r} \equiv \frac{r}{M}, \quad
\bar{a} \equiv \frac{a}{M}, \quad
\bar{\omega} \equiv M\omega, \quad
\bar{\zeta} \equiv M\zeta,
\end{align*}
we transform the radial coordinate $r$ into the dimensionless variable $z$ defined by
\begin{equation*}
z = \frac{r - r_{-}}{r_{+} - r_{-}} = \frac{ \bar{r} - \bar{r}_{-}}{\bar{r}_{+} - \bar{r}_{-}}
\end{equation*}
Now, any of the eight possible combinations of the parameters $\{ \zeta, \xi, \eta \}$ given in Eqs.~(\ref{eq:18}) will reduce the Teukolsky Radial Equation (\ref{eq:TRE}) into the following
equation for the auxiliary function $H$,
\begin{equation}
\frac{d^2 H(z)}{dz^2} + \Bigg(\frac{\gamma}{z} + \frac{\delta}{z-1} +4p\Bigg)\frac{d H(z)}{dz} + \frac{4\alpha pz - \sigma}{z(z-1)}H(z) = 0 \label{eq:HeunH}
\end{equation}
which is a Confluent Heun equation. Here, the following variables have been introduced to clarify the equation,
\begin{align}
&p = (\bar{r}_{+} - \bar{r}_{-})\frac{\bar{\zeta}}{2}, \qquad \hspace{-2mm}
\alpha = 1 + s + \xi + \eta -2\bar{\zeta} + s\frac{i\bar{\omega}}{\bar{\zeta}}, \label{Param}\\
&\gamma = 1 + s + 2\eta, \qquad
\delta = 1 + s + 2\xi, \nonumber\\
&\sigma = \,_{s}A_{lm}(\bar{a}\bar{\omega}) + \bar{a}^2\bar{\omega}^2 -8\bar{\omega}^2 + p(2\alpha + \gamma - \delta) + \bigg(1 + s - \frac{\gamma + \delta}{2} \bigg) \bigg( s + \frac{\gamma + \delta}{2} \bigg).\nonumber
\end{align}
Furthermore, the local solutions at the singularities have the exact same form for all eight combinations of the parameters $\{ \zeta, \xi, \eta \}$ given in Eqs.~(\ref{eq:18}). More precisely, we get
\begin{subequations}
\begin{align}
&\lim_{z \to 0} R(z) \sim z^{-s+i\sigma_{-}} \quad \text{or} \quad z^{-i\sigma_{-}}, \label{eq:locsol1}\\
&\lim_{z \to 1} R(z) \sim (z-1)^{-s-i\sigma_{+}} \quad \text{or} \quad (z-1)^{i\sigma_{+}},\label{eq:locsol2}\\
&\lim_{z \to \infty} R(z) \sim z^{-1-2s+2i\bar{\omega}}e^{i(\bar{r}_{+} - \bar{r}_{-})\bar{\omega}z} \quad \text{or} \quad z^{-1-2i\bar{\omega}}e^{-i(\bar{r}_{+} - \bar{r}_{-})\bar{\omega}z} \label{eq:locsol3}
\end{align}
\end{subequations}
Now, the above forms correspond to behaviour of the perturbations at the boundary conditions of the event and Cauchy horizon and spatial infinity. By suitable choice of the signs in \ref{eq:locsol1},\, \ref{eq:locsol2} and \ref{eq:locsol3}, we can obtain expressions for quantities of physical interest such as Quasinormal Modes and Totally Transmitting Modes \cite{Cook1}. Also, see \cite{Navaes2019,Suzuki1998,Suzuki1999,Yoshida2010} for applications of the Heun form of the Teukolsky equations.
The equation can be solved by various methods such as Frobenius series about the singular points \cite{Fiziev2} and continued fractions \cite{Leaver1985}.
\subsection{Representation of the Teukolsky radial function convergent on $]1,+\infty[$}\label{TRESol}
\subsubsection{Elementary integral series}
The solution of the Confluent Heun equation~(\ref{eq:HeunH}) satisfied by the auxiliary function $H(z)$ is described by Corollary~(\ref{corr:Conf}). Since the singular points are located at $z=0,1,+\infty$, given any initial conditions for $H(z_0)$ and $\dot{H}(z_0)$ at $z_0\in]1,+\infty[$, the integral series representation of $H(z)$ is guaranteed to converge on the entire domain $]1,+\infty[$. This crucial property stands in stark contrast with the hypergeometric and Coulomb series, which converge close to 1 and to $+\infty$, respectively. Because of this, we do not need to introduce the unphysical parameter $\nu$.
Recall that the Teukolsky radial function $R$ and auxiliary function $H$ are related by Eq.~(\ref{eq:RtoH}). The auxiliary function is a confluent Heun function given by the following integral series representation, convergent for any $z\in]1,+\infty[$,
\begin{align*}
H(z)&=H_0+H_0\int_{z_0}^z G_{1}(\zeta,z_0)d\zeta
+(H'_0-H_0)\!\left(\!e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right)\!,
\end{align*}
where $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$, and
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-(1+4p) z} z^{-\gamma } (z-1)^{-\delta }\times\\
&\hspace{5mm}\int_{z_0}^z \Big\{
e^{(1+4p)\zeta}\zeta^{\gamma } \left(\zeta-1\right)^{\delta}
\left(\frac{\sigma-4\alpha p \zeta}{\left(\zeta-1\right) \zeta}-\frac{\gamma }{\zeta}-\frac{\delta }{\zeta-1}-4p -1\right)\Big\}d\zeta,\\
K_{2}(z,z_0)&=\left(\frac{\sigma-4\alpha p z}{\left(z-1\right) z}-\frac{\gamma }{z}-\frac{\delta }{z-1}-4p -1\right)e^{z-z_0}-\frac{\sigma-4\alpha p z}{(z-1) z}.
\end{align*}
Here we assumed $z_0\in]1,+\infty[$ then $H_0:=H(z_0)$, $H'_0:=\dot{H}(z_0)$ and all parameters are given by Eq.~(\ref{Param}).\\
Witnessing to the fact that the above representation is convergent for all $z\in]1,+\infty[$, we here recover the asymptotic behavior of $H(z)$ in both limits $z\to 1^+$ and $z\to +\infty$. We emphasize that this is not possible with any single series representation of $H(z)$, which converges either in the vicinity of $1^+$ or of $+\infty$.
\subsubsection{Asymptotic behavior for $z\to +\infty$}
From now on, we write $F(z)\sim_{a.e.}$ to present the leading term of the asymptotic expansion of the function $F(z)$, disregarding constant factors. For example, we would write $1+2/z\sim_{a.e} z^{-1}$ as $z\to 0$.\\
We begin by determining the asymptotic behavior of $K_1(z,z_0)$ for $z\gg 1$. This depends on two cases: $p=0$ and $p\neq 0$. We suppose first that $p=0$ and assume that $\delta+\gamma>0$. In this situation, the confluent Heun function becomes a well understood hypergeometric function \cite{Erdelyi1955,Motygin2018OnEO} for which we will nonetheless show that we recover the correct asymptotic behavior. Setting $p=0$ we get, as $z\to +\infty$,
\begin{align*}
K_1(z,z_0)&\sim_{a.e}1+ e^{-(1+4p) z} z^{-\gamma } (z-1)^{-\delta }\left(-e^{z } z ^{\gamma +\delta }+e^{z_0} z_0^{\gamma +\delta }\right),\\
&\sim_{a.e}e^{-z}z^{-\delta-\gamma}e^{z_0}z_0^{\gamma +\delta }.
\end{align*}
Then $K_1(z,z_0)$ is asymptotically the product of a function depending only on $z$ and of a function depending only on $z_0$. This property is sufficient to determine the asymptotic behavior of $G_1$ in closed-form \footnote{This is because the solution of a linear Volterra integral equation of the second kind with kernel $K_1(z,z_0)=k(z)l(z_0)$ is known exactly \cite{giscardvolterra}.}
$$
G_1(z,z_0)\sim_{a.e}e^{-z}z^{-\delta-\gamma}e^{z_0}z_0^{\gamma +\delta }e^{\int_{z_0}^ze^{-\zeta}\zeta^{-\delta-\gamma}e^{\zeta}\zeta^{\gamma +\delta }d\zeta}=(z_0/z)^{\delta+\gamma}
$$
implying that $\int_{z_0}^z G_1(\zeta,z_0)d\zeta\sim_{a.e.} z^{1-\delta-\gamma}$ for $z\to+\infty$. Analyzing $K_2$ and $G_2$ yields the same results. Indeed, with $p=0$, we have
$$
K_2(z,z_0)\sim_{a.e.}\left(\frac{-1}{z}(\gamma+\delta)-1\right)e^{z-z_0},
$$
which is the product of a function of $z$ and a function $z_0$ so we determine
$$
G_2(z,z_0)\sim_{a.e.}\left(\frac{-1}{z}(\gamma+\delta)-1\right)e^{z-z_0}e^{-(z-z_0)}(z_0/z)^{\gamma+\delta},
$$
that is $G_2\sim_{a.e.}z^{-\gamma-\delta}$. From there $e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\sim_{a.e.}z^{1-\gamma-\delta}$. Thus for $p=0$ and $\delta+\gamma>0$, we get $H(z)\sim_{a.e.}z^{1-\delta-\gamma}$ regardless of the conditions at $z_0$ and provided $\delta+\gamma>0$, as expected \cite{Erdelyi1955}. Further cases arise for $\delta+\gamma\leq 0$ but we do not discuss these here as
they correspond to well known hypergeometric results.\\
Let us now suppose that $p\neq 0$. Then, since
$$
e^{4p\zeta}\zeta^{\gamma } \left(\zeta-1\right)^{\delta} f(\zeta)=-e^{4p\zeta}\zeta^{\gamma +\delta}\left(\frac{4\alpha p+\gamma+\delta}{\zeta}+4p+O(1/\zeta^2)\right),$$
we have, asymptotically for $z\to +\infty$,
$$
K_1(z,z_0)\sim_{a.e.}1-e^{-4p z}z^{-\gamma}z^{-\delta}\times z^{\gamma +\delta } \left(e^{4 p z}-4 \alpha p \,E_{-\gamma -\delta +1}(-4 p z)\right).
$$
where $E_{n}(z)$ is the exponential integral function, with asymptotic expansion $E_{n}(x)\sim_{a.e.}e^{-x}/x$ as $x\to+\infty$. This result greatly simplifies $K_1$, reducing it to
$$
K_1(z,z_0)\sim_{a.e}-\frac{\alpha}{z},\text{ as }z\to+\infty.
$$
This allows us to determine the asymptotic behavior of $G_1$ straightforwardly as
$$
G_1(z,z_0)\sim_{a.e}\frac{-\alpha}{z}e^{\int_{z_0}^z -\alpha /\zeta\, d\zeta}= -\alpha z^{-1-\alpha},
$$
and therefore $\int_{z_0}^z G_1(\zeta,z_0)d\zeta\sim_{a.e.} z^{-\alpha}$ for $z\to+\infty$.\\
\begin{comment}
We proceed similarly for $K_2$ and $G_2$. Since
$
K_2(z,z_0)=f(z)e^{z-z_0}+O(1/z),
$
we have asymptotically $K_2(z,z_0)\sim_{a.e.} f(z) e^z e^{-z_0}$ for $z\to+\infty$. Then $K_2(z,z_0)$ is asymptotically the product of a function depending only on $z$ and of a function depending only on $z_0$. As earlier, this means that $G_2$ is available asymptotically
$$
G_2(z,z_0)\sim_{a.e.} e^{z-z_0}f(z) e^{\int_{z_0}^zf(\zeta)d\zeta},\text{ as } z\to+\infty.
$$
The above yields, asymptotically for $z\to +\infty$,
$$
G_2(z,z_0)\sim_{a.e.} e^{z-z_0}f(z)e^{-z(1+4p)}(1-z)^{\sigma-\alpha-\delta}z^{-\sigma-\delta}.
$$
Then, asymptotically for $z\to +\infty$,
$$
\left(\!e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right)\sim_{a.e.}(-1)^{\sigma+\alpha+\delta }e^{-4pz} z^{-\alpha -\delta}z^{-\gamma}.
$$
Gathering our results, we conclude that
\begin{align*}
&H(z)\sim_{a.e.} -H_0\frac{e^{4p z_0}}{4p}z_0^{\gamma+\delta}\,e^{-4p z}z^{-\gamma-\delta}\\&\hspace{20mm}+(-1)^{q+\alpha+\delta }(H_0'-H_0)\,e^{-4pz} z^{-\alpha -\delta}z^{-\gamma},\text{ as }z\to+\infty,
\end{align*}
which gives the same asymptotic behavior as obtained from the analysis of either the series representation of $H(z)$ converging for $z\to+\infty$ \cite{Motygin2018OnEO,Cook1,Ronveaux1995}.
\end{comment}
We proceed similarly for $K_2$ and $G_2$. We have
$
K_2(z,z_0)=f(z)e^{z-z_0}+O(1/z),
$
so that asymptotically $K_2(z,z_0)\sim_{a.e.} f(z) e^z e^{-z_0}$ for $z\to+\infty$. Then $K_2(z,z_0)$ is asymptotically the product of a function depending only on $z$ and of a function depending only on $z_0$. We therefore obtain
$$
G_2(z,z_0)\sim_{a.e.} e^{z-z_0}f(z) e^{\int_{z_0}^zf(\zeta)d\zeta},\text{ as } z\to+\infty.
$$
The right hand-side is
\begin{align*}
e^{z-z_0}f(z) e^{\int_{z_0}^zf(\zeta)d\zeta}&=e^{-4 p (z-z_0)}\left(\frac{z_0-1}{z-1}\right)^{\delta} \left(\frac{z_0}{z}\right)^{\gamma
+\sigma} \left(\frac{1-z_0}{1-z}\right)^{4 \alpha p-\sigma } \\&\hspace{-5mm}\times\frac{1}{z(z-1)}(\gamma -z (\gamma +\delta +4 p (\alpha
+z-1)+z-1)+\sigma )
\end{align*}
which yields the asymptotic result,
$$
G_2(z,z_0)\sim_{a.e.} e^{-4p z}z^{-\delta-\gamma-4\alpha p},\text{ as }z\to+\infty.
$$
This implies that
$$
\left(\!e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right)\sim_{a.e.}
e^{-4pz} z^{-4\alpha p -\delta-\gamma}.
$$
Gathering our results, we conclude that when $p\neq 0$,
\begin{align*}
&H(z)\sim_{a.e.} z^{-\alpha}~~~\text{or}~~~H(z)\sim_{a.e.} e^{-4pz} z^{-4\alpha p -\delta-\gamma},~\text{ as }z\to+\infty,
\end{align*}
which gives the same asymptotic behavior as obtained from series designed to converge when $z\to+\infty$ \cite{Motygin2018OnEO,Cook1,Ronveaux1995}. The result for $p=0$ yields the correct asymptotics of the hypergeometric function obtained in this case.
\subsubsection{Asymptotic behavior for $z\to 1^+$}
In this situation, we begin with
\begin{align*}
K_{1}(z,z_0)&\sim_{a.e}1+ e^{-(1+4p) z} (z-1)^{-\delta }\int_{z_0}^z \Big\{
e^{(1+4p)\zeta} \left(\zeta-1\right)^{\delta-1}
\left(c+c'(\zeta-1)\right)\Big\}d\zeta,
\end{align*}
where $c=\sigma-4\alpha p - \delta$ and $c'=\gamma+4p+1$.
In order to progress without presenting cumbersome equations, denote $F(\zeta)$ the following indefinite integral
\begin{align*}
F_\delta(\zeta):&=\int e^{(1+4p)\zeta} \left(\zeta-1\right)^{\delta}d\zeta,\\
&=-e^{4 p+1} (\zeta-1)^{\delta +1} E_{-\delta }\big(-(1+4 p) (\zeta-1)\big),
\end{align*}
where $E_{n}(x)$ is the exponential integral function. In particular $E_{n}(x)\sim_{a.e.}x^{n-1}c_1+c_2$ as $x\to 0^+$ and where $c_1$ and $c_2$ are non-zero real constants that are irrelevant here. This implies $F_\delta(\zeta)\sim_{a.e.}(\zeta-1)^{1+\delta}$.
Now given that
$$
K_1(z,z_0)=1+ e^{-(1+4p) z} (z-1)^{-\delta }\big(c F_{\delta-1}(z)-cF_{\delta-1}(z_0)+c'F_{\delta}(z)-c'F_{\delta}(z_0)\big).
$$
then
$$
K_1(z,z_0)\sim_{a.e.}1,\text{ as }z\to1^+.
$$
This implies that $G_1(z,z_0)\sim_{a.e} e^{z-z_0}$ and therefore $\int_{z_0}^z G_1(\zeta,z_0)d\zeta \sim_{a.e.} 1$ as $z\to 1^+$.
\begin{comment}
We progress on noting that
$$
F_\delta(z)=e^{1+4p}(z-1)^\delta\,E_{-\delta}\big(-(1+4p)(z-1)\big)\xrightarrow[z\to 1^+]{} -e^{4 p+1} \Gamma (\delta +1) (-4 p-1)^{-\delta -1}=O(1)
$$
here assuming that $\delta$ is not a negative odd number.
where $c_2=(-1)^{\delta+1}e^{1+4p}\Gamma(1+\delta)(1+4p)^{-1-\delta}$ is a constant. Then
$$
K_1(z,z_0)\sim_{a.e.}
$$
\begin{align*}
\int_{z_0}^z \Big\{
e^{(1+4p)\zeta} \left(\zeta-1\right)^{\delta-1}
\left(c+c'(\zeta-1)\right)\Big\}d\zeta&=-ce^{1+4p}(z-1)^\delta E_{-1-\delta}\big(-(1+4p)(z-1)\big)+c'e^{1+4p}(z-1)^{\delta+1}E_{-\delta}\big(-(1+4p)(z-1)\big)
\end{align*}
\end{comment}
For $K_2$ and $G_2$ we begin by noting that for $z$ close to 1,
$$
K_2(z,z_0)\sim_{a.e.}\frac{4\alpha p}{z-1}(1-e^{z-z_0})-\frac{\delta}{z-1},\text{ as } z\to 1^+.
$$
from which it follows that
$
G_2(z,z_0)\sim_{a.e.}(z-1)^{-\delta}
$ for $z\to 1^+$, and therefore
$$
\left(\!e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right)\sim_{a.e.}(z-1)^{1-\delta},\text{ as }z\to 1^+.
$$
Note that this assumes that $\delta>0$. If this is not the case, then the asymptotics is $O(1)$.
\begin{comment}
Now either $z_0$ is close to 1, or it is not.
Assuming first $z_0$ close to 1 with $z_0>z>1$, we observe that $K_2(z,z_0)\sim_{a.e.} -\delta/(z-1)$, which implies the asymptotic result
$$
G_2(z,z_0)\sim_{a.e.} -\frac{\delta}{z-1}e^{-\int_{z_0}^z \delta/(z-1)}\sim_{a.e.}(z-1)^{-\delta},\text{ as }z\to 1^+.
$$
This then yields
$$
\left(\!e^{z-z_0}-1\!+\!\int_{z_0}^z\!\!(e^{z-\zeta}-1)G_{2}(\zeta,z_0)d\zeta\right)\sim_{a.e.}(z-1)^{1-\delta}+C,\text{ as }z\to 1^+.
$$
Here $C$ is a constant that depends on all parameters. In particular, if $\delta<1$ then the above becomes $o(1)$. If instead $z_0$ is not close to 1, we have
$$
K_2(z,z_0)\sim_{a.e.}\frac{4\alpha p-\sigma-\delta}{z-1}
$$
as $z\to 1^+$. Then, defining $c:=4\alpha p-\sigma-\delta$, we get
$$
G_2(z,z_0)\sim_{a.e.}\frac{c}{z-1}\left(\frac{1-z_0}{1-z}\right)^c
$$
implying that
$$
e^{z-z_0}-1+\int_{z_0}^z (e^{z-\zeta}-1)G_2(\zeta,z_0)d\zeta=e^{z-z_0}-1-\left(\frac{1-z}{1-z_0}\right)^c \left(c e^{z-1} E_{1-c}(z-1)+1\right)+c'
$$
with $c'$ a constant. Under the limit $z\to1^+$, $E_{1-c}(z-1)\sim_{a.e.}(z-1)^{-c} \Gamma(c)-\frac{1}{c}$, entails that the above is $o(1)$.\\[-.5em]
\end{comment}
Gathering our results, we get that
$$
H(z)\sim_{a.e} 1~\text{ or }~(z-1)^{1-\delta},\text{ as }z\to 1^+.
$$
which gives the same asymptotic behavior as obtained from series representations of $H(z)$ for $z$ close to $1$ \cite{Motygin2018OnEO,Cook1,Ronveaux1995}.
\subsection{Remarks on the Teukolsky Angular Equation}
The Teukolsky angular equation \ref{eq:TAE} has two regular singular points at $x = \pm 1$ and an irregular singular point at infinity. Just like the radial equation, we can transform it to either the Bocher symmetrical form \cite{Cook1} or the non-symmetric canonical form of the confluent Heun equation \cite{Fiziev2}. It follows that any solution to the angular equation has an integral series representation as described in this work.
The radial and angular Teukolsky equations are coupled equations, as shown e.g. by the presence of the frequency parameter $\omega$ and of the angular eigenvalue $_{s}A_{lm}$ in both the angular and radial equations. Therefore, when it comes to determining physical quantities of interest, such as quasinormal modes, the two equations must be solved \textit{simultaneously} (we refer the reader to \cite{Berti2006,Fiziev2,Leaver1985,Fackrell1977,Hughes2000} for methods to that end). While using integral series to solve both the radial and angular equations separately and then match the solutions is feasible, a truly ambitious alternative approach would be solve the coupled system directly with the path-sum formalism. Indeed, natively this formalism was designed to solve systems of coupled (differential) equations with variable coefficients. So much so that in order to solve the Heun equations and get an integral series representation from path-sum, the first step (see Appendix~\ref{AppendixProofs}) is to map any Heun equation back onto a \emph{system} of coupled differential equations. We believe such an approach to be feasible not only for the system comprising the angular and radial Teukolsky equations, but also for the underlying pair of coupled equations in the Penrose-Newmann formalism from which Teukolsky obtained his equation \cite{TeukEqn}. This is beyond the scope of this work.
\begin{comment}Witnessing to the fact that the above representation is convergent both when $z\to 1^+$ and $z\to+\infty$, we recover the asymptotic behavior of $R(z)$ in both limits:
\begin{itemize}
\item Assuming $\delta>1$, we have the asymptotic expansiosn for $z\to 1^+$, $K_1(z,z_0)\sim (z-1)^{-\delta}$ and similarly for $G_1$, while $K_2,G_2\sim(z-1)^{-1}$.
After integration this yields $H(z)\sim(z-1)^{1-\delta}$ and thus $R(z)\sim(z-1 )^{-s-i\sigma_+}$. If $\delta\leq 1$, all divergences disappear from both $G_1$ and $G_2$ and $H$ and $R$ are convergent in the limit $z\to 1+$.\\
\item In the limit $z\to+\infty$, $K_2(z,z_0)\sim(-4p+1) e^{z-z_0}$ implying $G_2(z,z_0)\sim -(1+4p) e^{-4p z}$ and from there $(H_0'-H_0)(e^z-(1/4p)e^{-4p z})$
$$
K_1\sim -e^{(1 + 4 p) z} z^{\gamma+\delta}
$$
\end{itemize}
\end{comment}
\section{Conclusion}\label{sec:conclusion}
In this work, we present novel integral series representations for all functions of Heun class. The major advantage of these representations is that 1) they involve only elementary integrands (rational and exponential functions); 2) they are unconditionally convergent everywhere except at the singular points of the Heun function being studied; and 3) they demonstrate that all functions of Heun class can be obtained from one or at most two Volterra equations of the second kind. Points 1) and 2) above are crucial in order to obtain physically well-behaved solutions of the homogenous Teukolsky radial
equation by means of Heun functions, as this necessitates a series representation that is convergent from the black hole horizon up to spatial infinity.
This is not feasible with state-of-the-art techniques involving hypergeometric and Coulomb series representations of confluent Heun function. The former is convergent \textit{only} near the horizon while the later is convergent \textit{only} at spatial infinity. In order to match both representations of the solutions, a book-keeping unphysical parameter $\nu$ has to be introduced which, at the very least, obscures the physical picture.
Unlike the above MST strategy, the integral series proposed here converge over the entire spatial domain from the horizon up to infinity, thus bypassing the need for parameters that are not already present in the Teukolsky equation.
While this work is devoted to establishing the well-posedness of the integral series formalism, the next obvious step is to use it to actually compute quantities of physical interest for the rapidly growing field of gravitational wave astrophysics. These include gravitational wave fluxes \cite{Fujita2004}, quasinormal modes \cite{London2014,Cook1} and totally transmitting modes \cite{Cook1}, all of which should now be accessible without Leaver's method (which suffers from numerical stability issues) nor the MST strategy. We hope that the formalism can also help resolve mathematical difficulties that arise in implementing the MST formalism in the various aspects of the two body problem in general relativity \cite{Bini2013,Kavanagh2016}.
Finally, we stress that our novel mathematical results were obtained by applying the method of path-sum to Heun's equation. This method, relying on the algebraic combinatorics of walks on graphs, was originally designed to solve systems of coupled differential equations and compute matrix functions. While it already proved successful in the fields of quantum dynamics, matrix theory and combinatorics, we think that this work opens new venues for its use in ordinary differential equations and general relativity. In particular, path-sum is natively adapted to solve directly the system of coupled equations which, in the Penrose-Newman formalism, underlies the Teukolsky equation.
\begin{acknowledgements}
P.-L. G. is supported by the Agence Nationale de la Recherche
young researcher grant No. ANR-19-CE40-0006.
\end{acknowledgements}
\newpage
\setcounter{section}{0}
\renewcommand{\thesection}{\Alph{section}}
\section{Appendix: Proof of the results}\label{AppendixProofs}
The method of proof is as follows: we map the Heun equation onto a system of two coupled linear first order differential equations with variable coefficients. The solution of such systems is given by a formal object called a path-ordered exponential, which we present below. Then we use the path-sum method to evaluate this path-ordered exponential. Finally we extract the desired Heun function from the path-sum solution.
\subsection{Path-ordered exponentials}
All the results are corollaries of the general purpose method of path-sum, which permits the exact calculation of path-ordered exponentials of finite variable matrices. The path-ordered exponential $\mathsf{U}(z)$ of a variable matrix $\mathsf{M}(z)$ is the unique matrix solution to the system of coupled first order ordinary linear differential equations with variable coefficients encoded by $\mathsf{M}(z)$, i.e.
\begin{equation}\label{OrdExpSys}
\frac{d}{dz}\mathsf{U}(z,z_0)=\mathsf{M}(z).\mathsf{U}(z,z_0),
\end{equation}
and such that for all $z_0$, $\mathsf{U}(z_0,z_0)=\mathsf{Id}$ is the identity matrix of relevant dimension. The solution of Eq.~(\ref{OrdExpSys}) is the path-ordered exponential $\mathsf{U}(z,z_0)$ of $\mathsf{M}$, denoted
\begin{align*}
\mathsf{U}(z,z_0)&=\mathcal{P}e^{\int_{z_0}^{z} \mathsf{M}(\zeta) d\zeta},\\
&=\mathsf{Id}+\int_{z_0}^z \mathsf{M}(\zeta_1)d\zeta_1+\frac{1}{2}\int_{z_0}^z\int_{z_0}^z \mathcal{P}\{\mathsf{M}(\zeta_2)\mathsf{M}(\zeta_1)\}d\zeta_2d\zeta_1+\cdots
\end{align*}
where $\mathcal{P}$ the path-ordering operator,
$$
\mathcal{P}\big\{\mathsf{M}(\zeta_2)\mathsf{M}(\zeta_1)\big\}=\begin{cases}
\mathsf{M}(\zeta_2)\mathsf{M}(\zeta_1),&\text{if }\zeta_2\geq \zeta_1\\
\mathsf{M}(\zeta_1)\mathsf{M}(\zeta_2),&\text{otherwise.}
\end{cases}
$$
We refer the reader to \cite{Dyson1952} for the origins of this notation.\\[-1em]
Although used primarily to gain analytical understanding into the dynamics of quantum systems driven by time-dependent forces, path-sum relies solely on the algebraic combinatorics of walks on graphs that is valid
irrespectively of the nature or size of the matrix $\mathsf{M}$. It is also only distantly related to the famous Feynman's path-integrals.
The interest here is that when calculating path-ordered exponentials, the method natively generates integral representations of the solutions. The strategy thus consists in calculating the ordered exponential of a matrix $\mathsf{M}(z)$ designed so that the solution of Eq.~(\ref{OrdExpSys}) should involve the desired Heun's function.
In order to recover an integral representation for all of Heun's functions, remark that Eqs.~(\ref{eq:Heun})--\ref{eq:HeunTriConfluent}) all take the form
\begin{equation}\label{HeunEq}
y''(z)-B_1(z)y'(z)-B_2(z)y(z)=0,
\end{equation}
We thus focus on obtaining the integral representation of the solution of Eq.~(\ref{HeunEq}) in terms of integrals involving $B_1$ and $B_2$, irrespectively of what these functions are. To this end, we begin by exihibiting a matrix $\mathsf{M}(z)$ whose path-ordered exponential involves a function solution to Eq.~(\ref{HeunEq}).
\begin{proposition}\label{MForm}
Let $y(z)$ be a solution of Eq.~(\ref{HeunEq}) with initial conditions $y(z_0)=y_0$ and $y'(z_0)=y'_0$. Let
\begin{equation}\label{Mz}
\mathsf{M}(z)=\begin{pmatrix}1&1\\
B_1(z)+B_2(z)-1&B_1(z)-1\end{pmatrix},
\end{equation}
and let $\mathsf{U}(z,z_0):=\mathcal{P}e^{\int_{z_0}^{z} \mathsf{M}(\zeta) d\zeta}$ be the path-ordered exponential of $\mathsf{M}$.
Then
$$
y(z) = y_0\mathsf{U}_{11}(z,z_0) +(y'_0-y_0)\mathsf{U}_{12}(z,z_0).
$$
\end{proposition}
\begin{proof}
By direct differentiation. Let $\psi(z)=(\psi_1(z),\psi_2(z)\big)^\mathrm{T}$ such that $\dot{\psi}(z)=\mathsf{M}(z).\psi(z)$. This implies
$$
\dot{\psi_1}=\psi_1+\psi_2,\qquad \dot{\psi_2}=(B_1+B_2-1)\psi_1+(B_1-1)\psi_2,
$$
where we omitted the $(z)$ arguments to alleviate the notation. Then
$$
\ddot{\psi_1}=\dot{\psi_1}+(B_1+B_2-1)\psi_1+(B_1-1)( \dot{\psi_1}-\psi_1),
$$
which is
$$
\ddot{\psi_1}-(B_1-1+1)\dot{\psi_1}-(B_1+B_2-1-B_1+1)\psi_1=0,
$$
i.e. $\ddot{\psi_1}-B_1\dot{\psi_1}-B_2\psi_1=0$. This is precisely Eq.~(\ref{HeunEq}). Now, since $\psi_1(z_0)=y_0$ is the desired initial condition, and since $\dot{\psi}(z_0)=\mathsf{M}(z_0).\psi(z_0)$, then to get $\dot{\psi}_1(z_0)=y'_0$ we must have $\psi_2(z_0)=y'_0-y_0$. From there and given that $\psi(z)=\mathsf{U}(z,z_0).\psi(z_0)$, we obtain
$$
\psi_1(z)=y_0\mathsf{U}_{11}(z,z_0) +(y'_0-y_0)\mathsf{U}_{12}(z,z_0),
$$
which completes the proof.
\end{proof}
\subsection{Path-sum formulation}
We may now use the method of path-sum to calculate the path-ordered exponential of $\mathsf{M}$ to recover the desired integral representations. We first state and prove the general result concerning Eq.~(\ref{HeunEq}) before giving its corollaries in the specific cases of the general Heun, confluent, biconfluent, doubly-confluent and triconfluent Heun functions.
\begin{theorem}\label{GenTheorem}
Let $\mathsf{M}(z)$ be given as in Eq.~(\ref{Mz}), let $\mathsf{U}(z,z_0)$ be its path-ordered exponential.
Then
$$
\mathsf{U}_{11}(z,z_0)=1+\int_{z_0}^z G_{1}(\zeta,z_0)d\zeta,
$$
where $G_{1}(z,z_0)$ satisfies the linear integral Volterra equation of the second kind
$$
G_{1}(z,z_0)=K_{1}(z,z_0)+\int_{z_0}^z K_{1}(z,\zeta) G_{1}(\zeta,z_0) d\zeta,
$$
with kernel
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-z}\int_{z_0}^z \Big\{e^{\zeta}\,e^{\int_{\zeta}^z B_1(\zeta')d\zeta'} (B_1(\zeta)+B_2(\zeta)-1)\Big\}d\zeta,
\end{align*}
Furthermore,
$$
\mathsf{U}_{12}(z,z_0)=\int_{z_0}^z(e^{z-\zeta}-1)\big(1+G_{2}(\zeta,z_0)\big) \,d\zeta,
$$
where $G_{2}(z,z_0)$ satisfies the linear integral Volterra equation of the second kind
$$
G_{2}(z,z_0)=K_{2}(z,z_0)+\int_{z_0}^z K_{2}(z,\zeta) G_{2}(\zeta,z_0) d\zeta,
$$
with kernel
\begin{align*}
K_{2}(z,z_0)&=(B_1(z)+B_2(z)-1)e^{z-z_0}-B_2(z).
\end{align*}
\end{theorem}
Given that a linear Volterra integral equations of the second kind always has an explicit solution in the form of a Neumann series of the kernel obtained from Picard iteration, we present below the ensuing elementary integral series representations for the solution $y_0 \mathsf{U}_{11}(z,z_0)+(y'_0-y_0)\mathsf{U}_{12}(z,z_0)$ of Eq.~(\ref{HeunEq}). This will be greatly facilitated by Volterra compositions, presented in the proof of the Theorem.
\begin{remark}
If the initial conditions are such that $y_0=y'_0$, then by Proposition~\ref{MForm}, the solution $y(z)$ is directly proportional to $\mathsf{U}_{11}$. By Theorem~\ref{GenTheorem} this implies that the derivative $\dot{y}(z)$ of the solution of Eq.~(\ref{HeunEq}) satisfies a linear Volterra integral equation of the second kind with kernel $K_{1}$ given above. In other terms, any solution of any Heun equation which has at least one point $z_0$ for which $y(z_0)=y'(z_0)$ satisfies such a Volterra integral equation with kernel $K_{1}$. This is the first known integral equation satisfied by Heun functions in terms of elementary functions.
Similarly if $y(z_0)=0$, then the solution $y(z)$ is proportional to $\mathsf{U}_{12}$, itself an integral of $G_{2}$ which satisfies a linear Volterra integral equation of the second kind.
\end{remark}
\begin{proof}
The central mathematical concept enabling the path-sum formulation of path-ordered exponentials is the $\ast$-product. This product is defined on a large class of distributions \cite{GiscardPozza2020}, however for the present work only its definition on smooth functions of two variables is required. For such functions the $\ast$-product reduces to the Volterra composition, a product between functions first expounded by Volterra and P\'er\`es in the 1920s \cite{Volterra1924} and which had largely fallen out of use by the early 1950s for a reason that appears, restrospectively, to be the lack of a mathematical theory of distributions.
The Volterra composition of two smooth functions of two variables $f(z,z_0)$ and $g(z,z_0)$ is
$$
\big(f\ast g\big)(z,z_0)=\int_{z_0}^{z} f(z,\zeta)g(\zeta,z_0) d\zeta\,\Theta(z-z_0),
$$
with $\Theta(.)$ the Heaviside theta function under the convention that $\Theta(0)=1$.
This extends to functions of less than two variables, for example if $h(z)$ is a smooth function of one variable, then
\begin{align*}
\big(h\ast g\big)(z,z_0)&=h(t')\int_{z_0}^{z} g(\zeta,z_0) d\zeta\,\Theta(z-z_0),\\
\big(g\ast h\big)(z,z_0)&=\int_{z_0}^{z} g(z,\zeta)h(\zeta) d\zeta\,\Theta(z-z_0).
\end{align*}
That is, the variable of $h(z)$ is always treated as the left variable of a function of two variables.
The identity element for the $\ast$-product is the Dirac distribution, denoted $1_\ast\equiv \delta(z-z_0)$, an observation which we here accept without proof as it would require presenting the full theory of the $\ast$-product \cite{GiscardPozza2020}. Similarly we accept without proof that for any bounded function $f(z,z_0)$ of two variables, $f^{\ast 0}=1_\ast$, while $f^{\ast 1}=f $ and $f^{\ast n+1}=f\ast f^{\ast n}=f^{\ast n}\ast f$ \cite{Volterra1924}.
Furthermore, if $f$ is bounded the Neumann series $\sum_{n=0}^\infty \big(f^{\ast n}\big)(z,z_0)$ converges superexponentially and thus unconditionally \cite{Linz1985} to an object, called the $\ast$-resolvent $R_f$ of $f$, given by
\begin{align*}
R_f(z,z_0)&=\sum_{n=0}^\infty \big(f^{\ast n}\big)(z,z_0),\\
&=\delta(z-z_0)+f(z,z_0)\Theta(z-z_0)+\int_{z_0}^z f(z,\zeta_1)f(\zeta_1,z_0)d\zeta_1\Theta(z-z_0)\\
&\hspace{10mm}+\int_{z_0}^z\int_{\zeta_1}^z f(z,\zeta_2)f(\zeta_2,\zeta_1)f(\zeta_1,z_0)d\zeta_2d\zeta_1 \Theta(z-z_0)+\cdots.
\end{align*}
Seeing this as steming from a Picard iteration entails an additional property of $\ast$-resolvents, namely that they solve the Volterra equation of the second kind with kernel $f$,
\begin{equation}\label{VolterraR}
R_f=1_\ast + f \ast R_f,
\end{equation}
or, in explicit integral notation, and showing all distributions
$$
R_f(z,z_0)=\delta(z-z_0)+\!\!\int_{z_0}^{z}\!\!\!f(z,\zeta) R_f(\zeta,z_0)d\zeta\, \Theta(z-z_0).
$$
Thus we have $R_f\ast (1_\ast - f)= 1_\ast$ and are therefore justified in writing $R_f=\big(1_\ast-f\big)^{\ast-1}$. In order to avoid distributions altogether, it is more convenient to define $G_f:=R_f-1_\ast$ and rewite Eq.~(\ref{VolterraR}) as
\begin{equation*}
G_f=f+f\ast G_f,
\end{equation*}
which is an ordinary linear integral Volterra equation of the second kind.
The Neumann integral series obtained from Picard iterations for $G_f$ as above is now
$$
G_f=\sum_{n=1}^\infty f^{\ast n},
$$
and it is a well-established result \cite{Linz1985} that the convergence of this series is guaranteed provided $f$ is continuous and bounded. In this case, truncating the series at order $m$, yields a relative error of at most
$$
\left|G_f(z,z_0)-\sum_{n=1}^m f^{\ast n}(z,z_0)\right|\leq \frac{\kappa_f^m}{m!}
$$
with $\kappa_f:=\sup_{\zeta,\zeta'\in]z_0,z[:\,\zeta\geq z'}|f(\zeta,\zeta')|$.\\[-0em]
Path-sum expresses the path-ordered exponential of any finite variable matrix in terms of a finite number of Volterra compositions and $\ast$-resolvents. The path-sum formulation of the path-ordered exponential of the $2\times 2$ matrix $\mathsf{M}(z)$ is
\begin{align*}
\mathsf{U}_{11}(z)&=1\ast R_1,\\
R_1&=\left(1_\ast-\mathsf{M}_{11}-\mathsf{M}_{12}\ast\big(1_\ast-\mathsf{M}_{22}\big)^{\ast-1}\ast\mathsf{M}_{21}\right)^{\ast-1},
\end{align*}
where the $\ast$-multiplication by $1$ on the left is a short-hand notation for an integral with respect to the left variable, since for any $f$ smooth, $(1\ast f)(z,z_0)=\int_{z_0}^z f(\zeta,z_0)d\zeta\Theta(z-z_0)$. Furthermore, since $\mathsf{M}_{22}$ depends on a single variable, its $\ast$-resolvent can be shown to be
$$
\big(1_\ast-\mathsf{M}_{22}\big)^{\ast-1}=1_\ast+ \mathsf{M}_{22}e^{1\ast \mathsf{M}_{22}},
$$
or equivalently
$$
\big(1_\ast-\mathsf{M}_{22}\big)^{\ast-1}(z,z_0)=\delta(z-z_0)+\mathsf{M}_{22}(z)e^{\int_{z_0}^z\mathsf{M}_{22}(\zeta)d\zeta}\Theta(z-z_0).
$$
Now the form of $\mathsf{U}_{11}$ as claimed in the theorem follows upon writing the $\ast$-products as explicit integrals with $\mathsf{M}$ given by Proposition.~(\ref{MForm}).
For $\mathsf{U}_{12}$, the path-sum formulation reads
$$
\mathsf{U}_{12}=1\ast \big(1_\ast-\mathsf{M}_{11}\big)^{\ast-1}\ast \mathsf{M}_{12}\ast R_{2},
$$
where
$$
R_{2}=\left(1_\ast-\mathsf{M}_{22}-\mathsf{M}_{21}\ast\big(1_\ast-\mathsf{M}_{11}\big)^{\ast-1}\ast\mathsf{M}_{12}\right)^{\ast-1},
$$
and the theorem result for $\mathsf{U}_{12}$ follows upon writing the $\ast$-products as explicit integrals with $\mathsf{M}$ given by Proposition.~(\ref{MForm})
\end{proof}
Since $R_{1}$ and $R_{2}$ are $\ast$-resolvents, we may express them as the unconditionally convergent Neumann series involving the corresponding kernels $K_{1}$ and $K_{2}$, i.e. $R_{i}=1_\ast+\sum_{n=1}^\infty K_{i}^{\ast n}$ or equivalently $G_{i}=\sum_{n=1}^\infty K_{i}^{\ast n}$, $i=1,2$. This yields an explicit representation for the solution of Eq.~(\ref{HeunEq}) as series of elementary integrals:
\begin{theorem}\label{TheoExplicit}
Let $y(z)$ be the unique solution of
$$
y''(z)-B_1(z)y'(z)-B_2(z)y(z)=0,
$$
such that $y(z_0)=y_0$ and $y'(z_0)=y'_0$. Then
\begin{align*}
y(z)&=y_0+y_0\int_{z_0}^z G_{1}(\zeta,z_0)d\zeta
+(y'_0-y_0)\int_{z_0}^z(e^{z-\zeta}-1)\big(1+G_{2}(\zeta,z_0)\big)d\zeta,
\end{align*}
where $G_{1}$ and $G_{2}$ satisfy linear Volterra integral equations of the second kind with kernels respectively given by
\begin{align*}
K_{1}(z,z_0)&=1+ e^{-z}\int_{z_0}^z \Big\{e^{\int_{\zeta}^z B_1(\zeta')d\zeta'}e^{\zeta} \big(B_1(\zeta)+B_2(\zeta)-1\big)\Big\}d\zeta,\\
K_{2}(z,z_0)&=(B_1(z)+B_2(z)-1)e^{z-z_0}-B_2(z).
\end{align*}
In consequence, $G_{1}$ and $G_{2}$ have the following representation as integral series inbvolving elementary integrands,
\begin{align}
G_{i}(z,z_0)&=\sum_{n=1}^\infty K_{i}^{\ast n}(z,z_0),\label{Gexplicitform}\\
&=K_{i}(z,z_0)+\int_{z_0}^z K_{i}(z,\zeta_1)K_{i}(\zeta_1,z_0)d\zeta_1\nonumber\\
&\hspace{3mm}+\int_{z_0}^z \int_{\zeta_1}^zK_{i}(z,\zeta_2)K_{i}(\zeta_2,\zeta_1)K_{i}(\zeta_1,z_0)d\zeta_2d\zeta_1+\nonumber\\
&\hspace{5mm}+\int_{z_0}^z \int_{\zeta_1}^z\int_{\zeta_2}^z K_{i}(z,\zeta_3)K_{i}(\zeta_3,\zeta_2)K_{i}(\zeta_2,\zeta_1)K_{i}(\zeta_1,z_0)d\zeta_3 d\zeta_2d\zeta_1+\cdots,\nonumber
\end{align}
for $i=1,2$.
The series representation is guaranteed to converge to $G_i$ everywhere except at the singular points of $K_i$. More precisely, let $]z_0,z_1[$ be an open interval over which $K_i$ is divergent free and let $\kappa_i:=\sup_{\zeta,\zeta'\in ]z_0,z_1[:\, \zeta\geq \zeta'}|K_i(\zeta,\zeta')|$. Then
\begin{equation}\label{Gbound}
\left|G_i(z,z_0)-\sum_{n=1}^m K_i^{\ast n}(z,z_0)\right| \leq \frac{\kappa_i^m}{m!}.
\end{equation}
\end{theorem}
This immediately provides the Corollaries of the main text for the general Heun, confluent, biconfluent, doubly-confluent and triconfluent Heun's functions upon replacing $B_1$ and $B_2$ appearing in Eq.~(\ref{HeunEq}) and Theorem~\ref{TheoExplicit} with their values as dictated by Eqs.~(\ref{eq:Heun}--\ref{eq:HeunTriConfluent}).
\begin{comment}
\section{Appendix: behaviour of Heun functions} \label{Appendix}
The theory of Heun functions, in general form \cite{Batic} as well for the confluent case, relevant to the analysis of the Teukolsky Radial and Angular Equations (TRE and TAE) has been discussed in various papers \cite{Fiziev1,Fiziev2,Fiziev3,Cook1,Suzuki1999,Suzuki1998,Yoshida2010}. Here, we discuss the nature and local behaviour of solutions for the Confluent Heun Equation (CHE) since that is most relevant to our paper's analysis. We follow the notation of \cite{Cook1}.
In addition to the form mentioned in Section 2.3, a useful form of the CHE that will be used to study the TRE is the non-symmetrical canonical form:
\begin{equation}
\frac{d^2 H(z)}{dx^2} + \Bigg( 4p + \frac{\gamma}{z} + \frac{\delta}{z-1} \Bigg)\frac{d H(z)}{dz} + \frac{4\alpha pz - \sigma}{z(z-1)}H(z) = 0 \label {eq:A1}
\end{equation}
where, $H(z)$ is the Heun function and $p, \alpha, \gamma, \delta$ and $\sigma$ are the parameters that define the equation. The equation has two regular singular points at $z=0, z=1$ with characteristic exponents $\{0,1-\gamma \}$ and $\{0,1-\delta\}$ respectively. The local behaviour of the solutions at these singular points is given by:
\begin{gather}
\lim_{z\to 0} H(z) \rightarrow 1 \quad \textrm{or} \quad z^{1-\gamma} \label{eq:localzero}
\end{gather}
and,
\begin{gather}
\lim_{z \to 1} H(z) \rightarrow 1 \quad \textrm{or} \quad (z-1)^{1-\delta} \label{eq:localone}
\end{gather}
It also has an irregular singular point at $z=\infty$ with the behaviour of solutions as:
\begin{equation}
\lim_{z \to \infty} H(z) \rightarrow z^{-\alpha} \quad \textrm{or} \quad e^{-4pz}z^{\alpha - \gamma - \delta} \label{eq:localinf}
\end{equation}
Another form of the CHE which is commonly used for study of the TAE is the \textit{B{\^o}cher symmetrical form}. It is obtained by making the following substitution in the non-symmetrical form:
\begin{equation}
H^{(B)} \equiv (z-1)^{\frac{\mu+\nu}{2}}(z+1)^{\frac{\mu-\nu}{2}}e^{-pz}H((1-z)/2),
\end{equation}
which gives the following form:
\begin{equation}
\frac{d}{dz} \Bigg( (z^2 -1)\frac{d H^{(B)}(z)}{dz} \Bigg) + \Bigg( -p^2(z^2 - 1) + 2p\beta z - \lambda -\frac{\mu^2 + \nu^2 +2\mu\nu z}{z^2-1} \Bigg) H^{(B)}(z) = 0
\end{equation}
The regular singular points are now at $z = \pm1$ while the irregular point at $z = \infty$ remains unchanged. The new parameters map to the ones of the normal form as follows:
\begin{gather}
\gamma = \mu + \nu +1, \quad \delta = \mu - \nu +1, \quad \alpha = -\beta + \mu + 1, \notag \\
\sigma = \lambda - 2p(\beta - \mu - \nu -1) - \mu(\mu + 1),
\end{gather}
\subsection{Local solutions}
The local solutions in terms of power series for the singularities of the CHE can be obtained from two functions. For the regular singularities, we have $Hc^{(r)}(p,\alpha, \gamma, \delta, \sigma;z)$ which is defined as:
\begin{gather}
Hc^{(r)}(p,\alpha, \gamma, \delta, \sigma;0) = 1,\\
Hc^{(r)}(p,\alpha, \gamma, \delta, \sigma;0) = \sum_{k=0}^{\infty}c_{k}^{(r)}z^k,
\end{gather}
while for the irregular singularity we have $Hc^{(ir)}(p,\alpha, \gamma, \delta, \sigma;z)$ which is defined as:
\begin{gather}
\lim_{z \to \infty} z^{\alpha}Hc^{(ir)}(p,\alpha,\gamma,\delta,\sigma;z) = 1, \\
Hc^{(ir)}(p,\alpha, \gamma, \delta, \sigma;z) = \sum_{k=0}^{\infty}c_{k}^{(ir)}z^{\alpha-k}.
\end{gather}
The coefficients $c_k^{(r)}$ and $c_k^{(ir)}$ are defined by three term recurrence relations:
\begin{gather}
0 = f_{k}^{(r)}c_{k+1}^{(r)} + g_{k}^{(r)}c_{k}^{(r)} + h_{k}^{(r)}c_{k-1}^{(r)} \quad \text{with}, \\
c_{-1}^{(r)} = 0, c_{0}^{(r)} = 1,\notag \\
g_{k}^{(r)} = k(k - 4p + \gamma + \delta -1) - \sigma, \notag\\
f_{k}^{(r)} = - (k + 1)(k + \gamma) \notag \\
h_{k}^{(r)} = 4p(k + \alpha -1)
\end{gather}
and,
\begin{gather}
0 = f_{k}^{(ir)}c_{k+1}^{(ir)} + g_{k}^{(ir)}c_{k}^{(ir)} + h_{k}^{(ir)}c_{k-1}^{(ir)} \quad \text{with}, \\
c_{-1}^{(ir)} = 0, c_{0}^{(ir)} = 1,\notag \\
g_{k}^{(ir)} = (k + \alpha)(k - 4p + \alpha - \gamma - \delta +1) - \sigma, \notag\\
f_{k}^{(ir)} = - 4p(k + 1) \notag \\
h_{k}^{(ir)} = -(k + \alpha -1)(k + \alpha - \gamma)
\end{gather}
As mentioned earlier, we can write the solutions local to regular singular points $z=0,z=1$ in terms of $Hc^{(r)}$ and for irregular singular point $z = \infty$ in terms of $Hc^{(ir)}$. Thus, keeping in mind the characteristic exponents for each of the solutions, we have the local solutions as follows.\\
For $z=0$,
\begin{gather}
Hc^{(r)}(p,\alpha, \gamma, \delta, \sigma;z), \\
z^{1-\gamma} Hc^{(r)}(p, \alpha + 1 -\gamma, 2- \gamma, \delta, \sigma + (1- \gamma)(4p - \delta);z).
\end{gather}
For $z=1$,
\begin{gather}
Hc^{(r)}(-p,\alpha, \gamma, \delta, \sigma - 4p\alpha; 1- z), \\
(z-1)^{1-\delta} Hc^{(r)}(-p, \alpha + 1 -\delta, 2- \delta, \gamma, \sigma - (1- \delta) \gamma - 4p(\alpha +1 - \delta); 1- z).
\end{gather}
For $z=\infty$
\begin{gather}
Hc^{(ir)}(p,\alpha, \gamma, \delta, \sigma; z), \\
e^{-4pz} Hc^{(ir)}(-p, -\alpha + \gamma +\delta, \gamma, \delta, \sigma - 4p\gamma; z).
\end{gather}
\subsection{Confluent Heun Functions}
The confluent Heun functions are those solutions that are simultaneously Frobenius solutions for two adjacent singular points. These functions arise in solution of eigenvalue problems in the TRE and TAE wherein the suitable imposition of the boundary conditions at the two singular points leads to a particular value of a parameter of the equation which satisfies both the boundary conditions. This particular parameter corresponds to the frequency in TRE and the angular separation constant in the TAE. These considerations are discussed in more detail in the main body of the paper.
\end{comment}
\providecommand{\noopsort}[1]{}\providecommand{\singleletter}[1]{#1}
| 176,310
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TITLE: A problem on conditional probability
QUESTION [0 upvotes]: In the above problem, $P(A)$ is the probability that one of the tubes drawn is good ,so $\frac{(6C_1+6C_2)}{10C_2} =\frac{7}{15}$ . So I'm getting a final answer of $\frac{5}{7}$.
Which one is correct?
Please explain
REPLY [1 votes]: The phrases "one of the tubes is tested and is found to be good" is different from "at least one of the tubes is good." The difference is that in the former, we only know about the first tube. You seem to be attempting to calculate the odds, based on at least one being good.
As a shorthand, if you did the test before drawing the second tube, you would have: "One tube is drawn and tested and shown to be good. Another tube is drawn. What is the probability that it is good?" Clearly, in this case, there are 9 tubes left, of which 5 are good, so the probability is $5/9$.
Whether you draw the second tube before or after the test has no effect on which tubes are drawn, so the correct answer must be the same.
The explanation is using the formula $P(B|A) = {P(A \cap B) \over P(A)}$ to calculate the conditional probability.
| 9,549
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Master Higgins wrote:DC's animation schedule pisses me the hell off.
LiQuid wrote:Batman is boring. Batman is the least interesting thing about this show by about a mile and a half. I said before that I like lighthearted takes on him. Adam West's Batman is great cause it's unapologetically goofy. The star of this show is the gold/silver age sensibilities and the crazy supporting cast. Batman almost plays the straight man to whatever his team up of the week is.
This show doesn't retroactively make me like modern Batman comics or Batman TAS, or any of the other junk he's been in. If there was a Batman comic on the shelf that was anything like this show (yes I know there is/was a comic adaptation of the show, I'm looking into it), I'd definitely read that, but none of the current 4 mainline comics are.
Users browsing this forum: No registered users and 3 guests
| 245,942
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Coronavirus Dow Stocks Bear Market Trend Forecast Update - Part1Stock-Markets / Stock Markets 2020 Mar 09, 2020 - 09:43 AM GMT
By: Nadeem_Walayat
This analysis continues.
This rest of this analysis has first been made available to Patrons who support my work. Coronavirus Dow Stocks Bear Market - March and April 2020 Trend Forecast
- Stock Market Trend Implications
- Coronavirus Global Recession 2020
- Short-term Trend Analysis
- Long-term Trend Analysis
- ELLIOTT WAVES
- Formulating a Stock Market Trend Forecast
- Dow Stock Market Forecast Conclusion
- Investing in AI to Kill the Coronavirus
- AI Stocks Q1 Buying Levels Current State.
- KIlling the Coronavirus!.
| 231,344
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Capital.com is a reasonably new trading platform that has been making waves on the market. To begin with, this app supplies over 3,000 instruments, consisting of a fantastic range of stock CFDs. This includes over 3118 shares and 85 cryptocurrencies, which is a great amount, so you can trade firms from all over the globe. It also supplies ETFs, indices, assets, as well as forex.
This online trading broker is likewise very cheap, without any compensations and low spreads. One more plus is that Capital.com provides spread betting (just offered for UK) in addition to CFDs, indicating any type of revenues you make from spread wagering are tax-free.
Maybe the most effective feature of this stock trading system is its outstanding trading devices and also attributes. It has an innovative AI-powered trading system with customised trading insights, technology signs, analytic tools, as well as progressed charts, as well as it likewise has superb academic resources, consisting of webinars.
Capital.com is certified by the FCA, so itís an extremely secure trading platform. Although it doesnít offer PayPal, it does offer some wonderful different payment approaches,.
Doesn’t accept PayPal.
VantageFX: Global Trading Platform with Zero Payment.
VantageFX LogoVantageFX is a worldwide CFD broker that offers whatever from stocks as well as forex to commodities, futures, as well as indices. The system provides utilize approximately 500:1 for significant money sets, making it among the very best trading platforms for large settings. The minimum contract dimension for a typical account is additionally just 0.01.
With a conventional account, CFD trading at VantageFX is 100% commission-free. Spreads begin at 1.4 pips, which is slightly above standard. Nevertheless, there are no down payment, withdrawal, or inactivity charges to worry about on your account.
VantageFX also supplies ECN accounts with a minimum deposit of just $500. You can trade with spreads from 0.0 pips and also pay a payment of $3 per lot.
VantageFX supplies its own trading system for the internet and also iOS as well as Android smart phones. Itís fairly thorough, with complete display graphes, dozens of technical indicators, and also seamless order access.
Conversely, this broker likewise gives you accessibility to MetaTrader 4 as well as MetaTrader 5. You can also make the most of social trading with ZuluTrade, Myfxbook, as well as DupliTrade. All 3 of these platforms allow copy trading.
VantageFX is managed platforms.
Supports MetaTrader 4 as well as 5.
Regulated in the UK and Australia.
What we don’t such as:.
Just supplies CFDs.
Above-average forex spreads.
Ally Invest.
PROS.
Deals detailed economic services, including banking, lendings, and also its own robo-advisor.
Reduced pricing for both options trading as well as mutual funds.
DISADVANTAGES.
Charges a commission of $4.95 + $0.01 per share on low-priced stocks.
Minimum initial investment: $0.
Financial items: Stocks, bonds, ETFs, mutual funds, options, and FOREX.
Costs: $0 for stocks, ETFs, and also choices (+ $0.50 per agreement); $1 for bonds; $4.95 for mutual funds; $4.95 + $0.01 per share for stocks valued under $2.
Bonus: $100 to $3,000 for brand-new accounts with down payments varying from $10,000 to $2 million plus.
Ally Invest supplies self-directed investing that not just offers commission-free sell stocks, ETFs, as well as choices, however likewise lower prices than the competitors for the majority of various other financial investments.
However what truly makes Ally Invest stand apart is detailed monetary solutions. They supply banking and borrowing through Ally Financial institution, however additionally use their very own robo-advisor, in case you desire assistance handling a minimum of some of your portfolio.
Zacks Profession.
PROS.
Offers a vast array of study registrations, both free and also premium.
DISADVANTAGES.
They do bill a little commission on stock trades, not a standard method.
Needs a minimum preliminary investment of $2,500.
Minimum initial financial investment: $2,500.
Monetary products: Stocks, ETFs, alternatives, mutual funds, bonds, crypto.
Costs: $0.01 per share ($ 1 minimum per trade) for stocks and ETFs more than $1 per share; or else, 1% of the value of the profession; for choices, $1 for the first contract, after that $0.75 per added agreement; $27.50 for mutual funds.
Perk: None.
Zacks Profession is created with active investors in mind. That’s due to the fact that the company provides several of the best study in the market. This consists of Zacks Financial investment Research, which has three various costs membership plans.
Though you will spend for these plans, Zacks Profession supplies some of the most effective investing suggestions in the market. For instance, individual stocks are ranked as Solid Buy, Buy, Hold, Market, or Solid Sell, which will certainly offer experienced investors a substantial trading advantage. This is a platform made for energetic, serious financiers.
Crypto.com.
If you’re trying to find the very best crypto trading system for 2022 ñ you might want to take a look at Crypto.com. The platform provides over 250 digital currencies from a selection of blockchain networks and also jobs. You can conveniently open up an account with Crypto.com in under 10 minutes.
And also the platform permits you to deposit funds immediately with ACH on a fee-free basis. Debit and bank card are additionally supported yet you will be charged 2.99% of the purchase quantity. What we additionally like regarding Crypto.com is that you can get cryptocurrency and also down payment the tokens into an interest-bearing account.
crypto.com review.
You’ll have access to APRs of approximately 14.5% ñ yet specific interest rates do vary relying on the cryptocurrency. Furthermore, to maximize your passion, you will need to stake CRO symbols and also opt for a 3-month term. Alternatively, flexible crypto passion accounts are likewise used, albeit, at lower rates.
When it pertains to fees, Crypto.com charges a variable commission. This begins at 0.4% per slide however does decrease as your trading quantities go beyond particular turning points. Staking CRO tokens also offers you access to lower compensations. If you’re also aiming to access non-fungible tokens, Crypto.com provides among the most effective NFT marketplaces for this objective.
AssetsCrypto.
CommissionsUp to 0.40% perslide.
Minimum Down payment$ 20.
Minimum TradeNo minimal mentioned.
Pros.
Over 250+ cryptocurrency markets.
Reduced trading charges.
Make passion on your idle cryptocurrencies.
NFT industry without purchaser fees.
Fantastic mobile app.
Immediate as well as cost-free deposits using ACH.
Disadvantages.
Does not use various other assets in crypto just.
Webull: Popular System for Trading US Stocks and Crypto.
In a comparable nature to eToro, Webull also allows you to trade stocks and crypto using one risk-free hub. Concerning the last, the system sustains 37 cryptocurrencies ñ a lot of which are large-cap symbols. You can trade crypto here from just $1 and also there are no commissions charged. Spreads begin at 1% when trading Bitcoin and extra on various other electronic currencies.
If you’re a lot more thinking about stocks, Webull supplies thousands of US-listed markets ñ all on a commission-fee basis. The minimum stock trade demand is just $5. Remember, Webull does not provide global shares apart from a very moderate selection of ADRs. Furthermore, you can additionally trade options as well as ETFs.
One more disadvantage with Webull is that domestic bank cables are billed at $8 per deposit. ACH, on the other hand, can be utilized fee-free. We likewise located that Webull is doing not have when it involves easy investment tools. Seasoned investors are catered for with a selection of charting functions as well as technological signs. Extra experienced and advanced traders might intend to check out the most effective MT5 brokers for advanced charting tools. There are likewise instructional overviews for newbies.
AssetsStocks, crypto, ETFs, options.
Commissions0% payment plus spreads.
Minimum Down payment$ 0.
Minimum TradeNo minimal mentioned.
Pros.
No minimum down payment.
Minimum stock investment is simply $5.
Disadvantages.
Global stocks supplied via ADRs just.
No copy trading devices.
Does not support debit/credit cards or e-wallets.
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We have been waiting for the day where Sunil is going to be turned as a Comedian again after failing as hero for lot of times. Now none other than Director Trivikram Srinivas very close friend of Sunil is making him a comeback with the movie of NTR.
Jr NTR is having the prestigious project with Trivikram soon and in the same movie Sunil is going to impress the audience again with his comedy skills. The project is stated to be started soon where in Sunil’s role not only as Comedy but also will be interesting character in the film. NTR and Trivkram already read the script with Sunil role in it and NTR felt excited for this role.
Also Read – Babai Balayya vs Abbai NTR for Dussehra!!
Might be the bigger twist here is that Sunil doing with wonderful comedy. Might be just killed by the goons which triggers NTR to take the revenge. Sources said Trivikram not only gave Sunil a bigger role as comedian but also as character artist.
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Full Name
Tiaki Lance Ereuti Phillips
Background
Background
NZ Herald story here
Hawkes Bay Today story here
Hawkes Bay Today story 2 May 2008
"Dad say do nw, pigs r al at da national." "That was one of the text messages a mother sent to her partner, giving the green light to rob a Hastings bank last year. In the Napier District Court on Wednesday, the Crown stated Natasha Janelle Ross, 21, sent the text at 1.51pm, on April 27, to her boyfriend, Tiaki Phillips, informing him police were attending to matters at the National Bank in Queen St, so it would be a "gud time" to rob the other National Bank, at Stortford Lodge.
She was aware of police whereabouts as she had a scanner in her car. Earlier that day, Ross, of Hastings, had sent a text message stating the Lowe Walker helicopter used by police was locked in its shed. Phillips responded to her texts and robbed the targeted bank. He pointed a pistol at bank staff, then jumped a teller's counter and took money from the cashier's draw. Phillips then went to a room at the back of the bank and stole coins and notes from a desk.
Judge Bridget Mackintosh believed Ross felt emotionally pressured to assist with the crime due to her relationship with Phillips. "It's one of those situations when you end up in a situation without strength to disassociate (yourself) with what was involved. "However, "you walk a tightrope between going to jail, or not". She sentenced Ross to nine months' home detention and 250 hours' community work. Phillips was sentenced to six years' imprisonment in October last year.
0
Alias
.
Date of Birth
Born 1987
Offences
Attempted murder of another inmate at Hawke's Bay Regional prison in October 2009
Dangerous driving, wounding with intent to cause grievous body harm, unlawful possession of a firearm, giving false details and perverting the course of justice in Hastings in early 2007
Current Location
Prison
Victims
**Suppressed**
Offender Affiliations
Harley Collier
Gang Affiliations
Mongrel Mob
Parole
Sentenced to 8 years 9 months in October 2007
A cumulative 8 year term was imposed in November 2010
Making a total sentence of 16 years 9 months
Additional Photos & Files
Associated Media Links
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synonyms for picked on
synonyms for picked on
- badger
- torment
- bully
- get at
- tease
- bait
- blame
- carp
- cavil
- criticize
- foment
- goad
- hector
- incite
- instigate
- provoke
- quibble
- start
- find fault
- get to
- pick at
antonyms for picked on
MOST RELEVANT
Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group.
TRY USING picked on
See how your sentence looks with different synonyms.
QUIZ
No Drama (Except K-Dramas) With This K- Trends QuizSTART THE QUIZ
How to use picked on in a sentence
When alone she sometimes picked it up and kissed the cold glass passionately.THE AWAKENING AND SELECTED SHORT STORIESKATE CHOPIN
All this while Squinty was chewing on the apple which he had picked up from the ground after he had jumped over the rope.SQUINTY THE COMICAL PIGRICHARD BARNUM
Her eyes were not nearly as soft as they had been, while she picked up the hanging folds of pink cloth, and went on.ROSEMARY IN SEARCH OF A FATHERC. N. WILLIAMSON
It was the Town Crier, with whom, as with a brother artist, he had picked acquaintance the day before.THE JOYOUS ADVENTURES OF ARISTIDE PUJOLWILLIAM J. LOCKE
Aristide picked it up and began to dance and shake his fist at the invisible police.THE JOYOUS ADVENTURES OF ARISTIDE PUJOLWILLIAM J. LOCKE
He picked out that simple little study of Cramer in D major in the first book—you know it well—and asked me to play it.MUSIC-STUDY IN GERMANYAMY FAY
Jean Baptiste laughed when he had completed the letter, picked up one of his books and looking through it, found the place.THE HOMESTEADEROSCAR MICHEAUX
SYNONYM OF THE DAY
OCTOBER 26, 1985
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