text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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"""
qlqw dumps your queue into ${XDG_RUNTIME_DIR} on every song change.
"""
import os
import threading
import urllib.parse
from gi.repository import Gtk, GLib
from quodlibet import _
from quodlibet import app
from quodlibet.plugins.events import EventPlugin
from quodlibet.qltk import Icons
from quodlibet.util.dprint import print_d
from quodlibet.errorreport import errorhook
from quodlibet import commands
LOGGING_IS_ENABLED = False
class QlqwError(RuntimeError):
"""Do-nothing exception specific to qlqw."""
class QlqwBackend:
"""Provides threaded I/O for queue writing."""
FILE_PREFIX = "file://"
TARGET_PATH = os.path.join(
os.getenv("XDG_RUNTIME_DIR"),
"qlqw.txt"
)
def __init__(self):
self.called_event = threading.Event()
self.last_queue_hash = None
def parse_queue(self, dumped_queue_string):
parsed = list()
for line in urllib.parse.unquote(dumped_queue_string).splitlines():
if not line.startswith(self.FILE_PREFIX):
raise QlqwError("unexpected queue entry: ``{}''".format(line))
path = line[len(self.FILE_PREFIX):]
if not os.path.exists(path):
raise QlqwError("nonexistent queue entry: ``{}''".format(path))
parsed.append(path)
return parsed
def get_queue(self):
dumped_string = commands.registry.run(app, "dump-queue")
queue_hash = hash(dumped_string)
if not dumped_string or queue_hash == self.last_queue_hash:
return None
self.last_queue_hash = queue_hash
return self.parse_queue(dumped_string)
def commit_queue(self, queue_list):
with open(self.TARGET_PATH, "w") as qfp:
for line in queue_list:
qfp.write(line)
qfp.write("\n")
def get_and_commit_queue(self):
queue_list = self.get_queue()
if queue_list is not None:
self.commit_queue(queue_list)
def fire(self):
"""
Arms this backend to write the queue. Called from the main
plugin context.
"""
self.called_event.set()
def _run_loop(self):
self.called_event.wait()
self.called_event.clear()
self.get_and_commit_queue()
def run_loop(self):
"""
Main entry point of this class. Waits for the main plugin
context to request a queue write. Operates from the I/O context.
"""
while True:
try:
self._run_loop()
except Exception:
errorhook()
class Qlqw(EventPlugin):
"""
This class does nothing but deliver queue write requests to the
backend (residing in a separate I/O thread). It does not fail
meaningfully, deferring error handling to said backend.
"""
PLUGIN_ID = "qlqw"
PLUGIN_NAME = _("Write Queue Aggressively")
PLUGIN_DESC = _("Commits queue contents to disk on every song "
"change.")
PLUGIN_ICON = Icons.DIALOG_ERROR
def __init__(self):
self.__enabled = False
self.backend = QlqwBackend()
backend_thread = threading.Thread(None, self.backend.run_loop)
backend_thread.setDaemon(True)
backend_thread.start()
def plugin_on_song_started(self, song):
if not self.__enabled:
return
# TODO(j39m): rate-limit the present method.
self.backend.fire()
def enabled(self):
self.__enabled = True
def disabled(self):
self.__enabled = False
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,145 |
Q: Why can electricity flow only in one direction through a diode? A few days ago I was soldering a small thing which contained a diode, a battery and some other useless things.
Unfortunately, I soldered the diode reversed and it didn't work. When I reversed it again it started to light.
I was trying to find an answer in Google, but the only thing I could find was the fact that electricity in diode can flow only in one direction, but my question is why? How does it work? Is this fact connected with the construction of a diode?
A: Let me try yet one more type of explanation, which I will confine to the PN junction diode (covers virtually all diodes used in modern circuits).
The diode consists of a p-doped region (p-type) slapped up against an n-doped region (n-type). In the p-type, the electron (e-) flow is largely accomplished by electrons moving from hole to hole. This is, electrically, exactly analogous (and is often visualized) as holes moving in a direction opposite to e- flow (although there is no physical movement of postivie charge) In the n-type, there are loosely bound e- which can be donated (moved).
At the PN junction of the diode, loosely bound e- in the n-type fall into the holes of the adjacent p-type. What you then have is an abundance of e- in a thin layer of the p-type layer at the junction, and a depletion of them (creating a net positive charge) in a thin layer of the n-type. This sets up a voltage field of positive in the n-type relative to negative in the p-type. This pushes any free e- in the n-type further away from the junction. The result is a thin PN layer which has no free holes and no free e-. The layer becomes an insulator.
Now, if you apply a positive voltage to the p-type and a negative at the n-type, e- in the p-type are removed, making free holes. Simultaneously, the positive voltage is conteracting the reverse voltage which had been set up in the PN junction, and e- in the n-type are force closer to the p-type, where they can cross over and fill up the new holes. Current flows.
If, however, you apply positive voltage to the n-type, and negative to the p-type ("reverse-biasing" the diode) you simply reinforce the voltage gradient which was already naturally set up in the PN junction. The e- are forced even farther away from the PN junction, and the insulative boundary (depletion region) thickens. No current flows.
To get more in-depth than that might take a good portion of a graduate course in materials science. I hope what I have written suffices.
A: A diode consists of two materials known as p-type and n-type semiconductors, connected in series which allows current to flow through them differently. In the n-type semiconductor, electrons travel with enough energy such that they're not attached to an atom and are said to be in the conduction energy band. For the p-type semiconductor, electrons "hop" from atom to atom, but lacking the energy to free them, are said to be in the valency energy band.
At the interface between the n-type and p-type materials, a travelling electron has to move either from the n-type to the p-type in one direction, the p-type to the n-type in the other, to continue moving. Is there a difference between the two directions?
Well, an electron moving from the n-type to the p-type material can occur spontaneously because the free electron's energy is released as radiation and it can move to a lower energy state, attached to an atom in the p-type semiconductor. But to move from the p-type to the n-type it has to gain energy from somewhere, and this isn't spontaneous because there is no guarantee of some other process providing this energy.
Think of a ball at the top of a hill: It can move from the top to the bottom while spontaneously releasing energy, but some other process must provide the energy to take it from the bottom to the top. And this analogy provides a basic explanation for why a diode conducts in one direction, but not in the other.
A: This really depends on the type of diode you are talking about but for the most common types there is something called a PN-Junction that allows current to flow in only one direction. This page:
http://en.wikipedia.org/wiki/PN_junction
on wikipedia explains how a PN-junction works. The short version is the voltage has to be above a certain threshold defined by the materials in the junction before the juice can flow. This shows up in the I-V characteristic curve as a point in the plot where the derivative is discontinuous. The details on other features of the I-V curve are up for grabs depending on the materials used but it's the junction and the application of a voltage beyond that threshold that means you have a diode on your hands.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,751 |
Aarne Ihamo Salovaara (* 25. Februar 1887 in Kotka als Arne Nylenius; † 11. September 1945 ebenda) war ein finnischer Leichtathlet und Turner.
Bei den Olympischen Spielen 1908 in London wurde er Vierter im Speerwurf. Im Speerwurf (freier Stil) kam er nicht unter die ersten neun, im Diskuswurf nicht unter die ersten elf.
Beim Turnen der Spiele in London gewann er im Mannschaftsmehrkampf mit der finnischen Riege die Bronzemedaille. 1912 in Stockholm folgte im freien Turnen die Mannschaftssilbermedaille.
Weblinks
Speerwerfer (Finnland)
Diskuswerfer (Finnland)
Turner (Finnland)
Olympiateilnehmer (Finnland)
Teilnehmer der Olympischen Sommerspiele 1908
Teilnehmer der Olympischen Sommerspiele 1912
Finne
Geboren 1887
Gestorben 1945
Mann | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,888 |
We are experts in site-to-site and client VPN access from several vendors including Cisco, Barracuda and Sonicwall. We can create your entire WAN for you, or help support your existing infrastructure. Take the guesswork out of setup and hire an expert!
We have designed and implemented many different solutions for clients all over the globe. Contact us today to discuss your needs! | {
"redpajama_set_name": "RedPajamaC4"
} | 1,862 |
Andy Rotherham | Tedeschi Trucks Band, Drive By Truckers, and Marcus King Band
The Wolf Trap Performing Arts Center outside of Washington, D.C. is a national treasure. The lovely amphitheater with cozy lawn seating and plenty of covered reserved seating is owned by the National Park Service. This means they're good with patrons bringing in coolers and picnic blankets to enjoy dinner and drinks before (and during) the shows because they are not most concerned with selling you $15 beers.
Over the years the venue has evolved into hosting great rock and roll while still retaining a wide array of programming. I remember some of their early forays into edgier music, a 1989 show with the Pogues, Mojo Nixon, and the Violent Femmes ended up somewhat chaotic as ushers used to the National Symphony and orderly seating confronted a crowd that really wanted to get close to the bands.
These days it's a well oiled machine with a great summer line up. On July 11thWolf Trap again hosted the Tedeschi Trucks Band summer tour, Wheels of Soul. The day gave way to a lovely evening with pleasant temperatures and one of those classic Virginia sunsets that eases into darkness like a gentle stain. This year's TTB roster featured Drive By Truckers and the Marcus King Band in support. Marcus King opened the night with a rocking set. His dad is bluesman Marvin King so he has a leg up, but he's unbelievably only 22 and has assembled a fantastic band fusing Southern rock and the blues with a forward looking sound. A prodigious talent, he read the crowd well telling a story about a party at Virginia Tech before landing his set with a rocking "Virginia." They might have been the third band on a three-band bill but earned a standing ovation from an appreciative audience.
Drive By Truckers followed up with a strong set. Whatever the configuration of the band, Mike Cooley and Patterson Hood know how to get it done and the band worked through an earnest set. What can't you say about a band that's produced great Southern-infused music for so long, is socially aware, and still parties on stage? They drew a lot from "American Band," an album that seems more resonant all the time and Hood reminded the audience as he is wont to do in the words of Patti Smith, to just "love each other motherf**ers." Hard to argue.
Finally, TTB took the stage. I like their hard charging sound more in smaller theaters, the kind they play during the colder months rather than the summer amphitheaters. And you get more songs. But they packed a lot into a thirteen-song set. I continue to argue they're the not-to-miss live act on the road these days. Alecia Chakour is a backup singer, c'mon! Highlights included a hot "Get What You Deserve" encore and Marcus King joining on a cover of "Key to the Highway." Backup singer Mike Mattison sang out in front and Mark Rivers did as well.
Midway through the set a stage hand brought out sheet music and Susan Tedeschi sang "Lord Protect My Child," one of three Dylan covers in their set. It's a song the band had not played in a few years, only twice since 2012, and just ten times as of this writing. Beautifully delivered, it needed little explanation for a crowd looking for a few hours of escape but not amnesia from our contemporary troubles.
Music-Loving Education Champion: Andy Rotherham
Venue: Wolf Trap Performing Arts Center
City: Vienna, Virginia
Date of Show: July 11, 2018
Photo by Andy Kahn for JamBase.com, which features his take on the show | Members Of Drive-By Truckers & Marcus King Band Join Tedeschi Trucks Band In Virginia
As The Crow Flies | Tim Taylor
Tim Taylor | Amos Lee at Denver Botanic Gardens | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 965 |
\section{Introduction}
\noindent
The year 1887 can be considered as the starting year of the relativistic revolution in physics. In this year Voigt published the article \cite{1}: ``On Doppler's Principle,'' which marked the beginning of the relativistic era. In the first part of this article, Voigt demanded covariance to the wave equation in inertial frames, assumed the invariance of the speed of light in these frames, and obtained a set of spacetime transformations now known as Voigt's transformations. In modern notation these transformations can be written as \cite{2}:
\begin{equation}
x'=x -vt,\;\; t'=t-vx/c^2,\;\; y'=y/\gamma,\;\; z'=z/\gamma,
\end{equation}
where $\gamma=1/\sqrt{1-v^2/c^2}$ is the Lorentz factor. Voigt's transformations are similar to the well-known Lorentz transformations of special relativity:
\begin{equation}
x'=\gamma(x -vt),\;\; t'=\gamma\big(t-\!vx/c^2\big), \;\; y'=y, \;\; z'=z.
\end{equation}
If the right-hand side of Voigt's transformations in equation (1) is multiplied by the Lorentz factor $\gamma$ then the Lorentz transformations in
equation (2) are obtained. Despite the similarity between Voigt and Lorentz transformations, the former are rarely mentioned in standard textbooks \cite{3,4,5,6}. Voigt's transformations have been discussed in old textbooks \cite{7, 8, 9, 10}, old papers \cite{11, 12}, recent articles \cite{13,14,15,16,17,18,19,20,21} and specialized books \cite{22, 23}.
Some initial comments enlighten the conceptual and historical importance of Voigt's 1887 paper \cite{1}: (i) Voigt derived his transformations by demanding covariance to the homogeneous wave equation under inertial frames, which implied the form invariance of this equation, and this is one application of what would be later known as the first postulate of special relativity. We must say, however, that Voigt didn't explicitly mention the terms ``covariance'' and ``inertial frames.'' He used these concepts in practice; (ii) Covariance of the wave equation carried the invariance of the speed of light, and this would be later known as the second postulate of special relativity. Remarkably, Voigt applied in practice the postulates of special relativity to the wave equation 18 years before Einstein explicitly enunciated these postulates \cite{24};
(iii) Voigt followed a formal procedure that allowed him to derive a first example of the now known conformal symmetry of spacetime. A general discussion of this symmetry was presented in 1909 by Bateman \cite{25} and Cunningham \cite{26}; (iv) The well-established Newtonian absolute time: $t'=t$ was questioned by Voigt's non-absolute time: $t'=t-vx/c^2$. According to Ives \cite{12} this was the first suggestion that: ``...a `natural' clock would alter its rate on motion.'' In the same sense, Simonyi \cite{23} has noted that when discarding $t'=t$, Voigt was ``...opening the possibility for the first time in the history of physics to call into question the concept of the absolute time.'' Voigt's non-absolute time was re-introduced in 1895 by H. A. Lorentz \cite{27} who called it ``the local time.''
Interestingly, in a paper devoted to the Doppler effect, Voigt was inadvertently laying the fundamentals of special relativity nearly two decades before Einstein. As pointed out by Ernst and Hsu \cite{18}: ``He was very close to suggesting a conceptual framework for special relativity.'' Unfortunately, Voigt's transformations are not usually mentioned in standard textbooks \cite{4,5,6} despite they imply the same transformation law for velocities of special relativity \cite{13}. We believe that two basic reasons are key to understanding why Voigt's transformations have aroused scant interest among authors of standard textbooks: Firstly, these transformations do not form a group \cite{14} which makes them little attractive from a physical point of view. Secondly, the original derivation of these transformations presented by Voigt \cite{1} is certainly not pedagogical and not easy-to-follow.
In this paper we hope to call attention to Voigt's transformations: (a) by briefly reviewing Voigt's original derivation of these transformations and stressing the scant impact produced by them among Voigt's contemporaries; (b) by discussing the relation of these transformations to the Lorentz transformations and (c) by presenting an alternative derivation of Voigt's transformations from the conformal invariance of the d'Alembert operator.
\section {Voigt's 1887 paper}
For a modern reader, Voigt's 1887 paper \cite{1}: ``On Doppler's Principle'' looks like a note, something like a technical report, rather than a research paper. It does not contain an abstract nor a first paragraph explaining the idea and purpose of the paper. Also it does not contain references.
Many years later, Voigt would explain what he considered to be the basic idea developed in his seminal paper \cite{28}:
\begin{quote} ``...{\bf it is about the applications of Doppler's principle}, which occur in special parts, though not on the basis of the electromagnetic theory, but on the basis of the elastic theory of light. However, already then some of the same consequences were given, which were later gained from the electromagnetic theory.''
\end{quote}
(Hereafter, some statements in quotations will be emphasized by writing them in bold). Clearly, Voigt mainly focussed his attention on the Doppler effect. The derivation of the transformations in equation~(1) was not his main objective. More specifically, in his 1887 paper, Voigt studied the propagation of oscillating disturbances through an elastic uniform incompressible medium. His basic equation was the homogeneous vector wave equation of the elastic theory of light. As usual in those times, he considered the scalar components of this vector equation. According to Voigt \cite{1}: ``It is known that the differential equations for the oscillations of an elastic incompressible medium read
\begin{equation}
\frac{\partial^2 u}{\partial t^2}=\omega^2\Bigg(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}
+\frac{\partial^2 u}{\partial z^2}\Bigg),\;\; \frac{\partial^2 {\rm v}}{\partial t^2}=\omega^2\Delta {\rm v},\;\; \frac{\partial^2 w}{\partial t^2}=\omega^2\Delta w,
\end{equation}
where $\omega$ is the propagation velocity of the oscillations, or more precisely, the propagation velocity of plane waves with constant amplitude.''
Voigt wrote $\partial^2 u/\partial t^2=\omega^2\Delta u$ for the first equation in (3), where $\Delta$ is the d'Alembert operator. The wave of the elastic incompressible medium is represented by $(u, {\rm v}, w)$. With small changes, we are following the original notation of Voigt's paper. He also assumed the continuity equation:
$\partial u/\partial x+ \partial {\rm v}/\partial y+ \partial w/\partial z=0.$ Voigt's central idea was to transform equation~(3) from the rest frame to another frame moving with a constant velocity, and thus finding the formula for the Doppler effect. Without providing any explanation, he wrote a set of linear transformations relating the space coordinates $(x,y,z)$ and the time $t$ measured in the rest frame with the space coordinates $(\xi,\eta,\zeta)$
and the time $\tau$ measured in the moving frame:
\begin{eqnarray}
\xi&=&xm_1+yn_1+zp_1-\alpha t,\\
\eta&=&xm_2+yn_2+zp_2-\beta t,\\
\zeta&=&xm_3+yn_3+zp_3-\gamma t,\\
\tau&=&t-(ax+by+cz).
\end{eqnarray}
Voigt introduced 15 unknown constants ($m_1,...,a,..., \alpha,...$) in equations~(4)-(7). Notice that $\gamma$ is not the gamma factor of special relativity. After transforming the first wave equation displayed in (3), in which $u=u(x,y,z,t$), and
using equations~(4)-(7), Voigt obtained a large expression for the transformed wave equation:
\begin{eqnarray}
\frac{\partial^2 (U)}{\partial \tau^2}\big(1-\omega^2(a^2+b^2+c^2)\big)\nonumber\\
\qquad\quad=\omega^2\Bigg\{\frac{\partial^2 (U)}{\partial\xi^2}\Bigg(m_1^2\!+\!n_1^2\!+\!p_1^2\!-\!\frac{\alpha^2}{\omega^2}\Bigg)\!+\!\frac{\partial^2 (U)}{\partial \eta^2}\Bigg(m_2^2\!+\!n_2^2\!+\!p_2^2
\!-\!\frac{\beta^2}{\omega^2}\Bigg)\!+...\nonumber\\
\qquad\quad\;\;\;+2\frac{\partial^2(U)}{\partial\eta\partial\zeta}\Bigg(m_2m_3+n_2n_3+p_2p_3-\frac{\beta\gamma}{\omega^2}\Bigg)+ ...\Bigg\}.
\end{eqnarray}
Here $(U)=(U)(\xi,\eta,\zeta,\tau$) is the transformed function corresponding to $u=u(x,y,z,t$). Thus he compared the transformed wave equation in equation~(8) with the equation
\begin{equation}
\frac{\partial^2 (U)}{\partial \tau^2}=\omega^2\Bigg(\frac{\partial^2 (U)}{\partial \xi^2}+\frac{\partial^2 (U)}{\partial \eta^2}
+\frac{\partial^2 (U)}{\partial \varsigma^2}\Bigg),
\end{equation}
which exhibits the same form as the first equation appearing in (3). Voigt justified the validity of equation~(9) with only four words: {\bf ``as it must be''} (or ``da ja sein muss'' in the original German version). Voigt'comparison of equations~(8) and (9) allows him to obtain an algebraic system of 9 equations containing 15 unknown constants ($m_1,...,a,..., \alpha,...$). Some of these equations are: $1-\omega^2(a^2+b^2+c^2) = m_1^2+n_1^2+p_1^2-\alpha^2/\omega^2,
1-\omega^2(a^2+b^2+c^2)= m_2^2+n_2^2+p_2^2-\beta^2/\omega^2,...,
m_2m_3+n_2n_3+p_2p_3=\beta\gamma/\omega^2,...$
Voigt's comparison among equations~(8), (9) and the first equation displayed in equation~(3) can be better understood by considering the implied expression
\begin{eqnarray}
\frac{\partial^2 u}{\partial t^2}-\omega^2\Bigg(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}
+\frac{\partial^2 u}{\partial z^2}\Bigg)\!\nonumber\\
\qquad\qquad\qquad= {\rm K}\;\Bigg[\frac{\partial^2 (U)}{\partial \tau^2}-\omega^2\Bigg(\frac{\partial^2 (U)}{\partial \xi^2}+\frac{\partial^2 (U)}{\partial \eta^2}+\frac{\partial^2 (U)}{\partial \varsigma^2}\Bigg)\Bigg].
\end{eqnarray}
The parameter ${\rm K}$ in equation~(10) is defined by ${\rm K}\!=\!m_1^2\!+n_1^2\!+\!p_1^2\!-\!\alpha^2/\omega^2$. Equation (10) expresses the covariance of the d'Alembertian of a scalar function defined in both (rest and moving) frames. The typical argument of covariance can be now applied to equation~(10): If the wave equation $\partial^2 (U)/\partial \tau^2\!-\!\omega^2[\partial^2 (U)/\partial \xi^2\!+\!\partial^2 (U)/\partial \eta^2\!+\!\partial^2 (U)/\partial \varsigma^2]\!=\!0$ is valid in the moving frame then it is valid in the rest frame: $\partial^2 u/\partial t^2\!-\!\omega^2[\partial^2 u/\partial x^2\!+\!\partial^2 u/\partial y^2\!+\!\partial^2 u/\partial z^2]\!=\!0$ since ${\rm K}\not\!=\!0$. Therefore the covariance of the homogeneous wave equation introduced by Voigt implied the form invariance of this equation under inertial frames, and this is an application of what would be later called by Einstein the first postulate of special relativity. When demanding covariance to the homogeneous wave equation, Voigt assumed (without explicitly specifying it) the constancy of the speed of light $\omega$ in both (rest and moving) frames. This constancy would be later called by Einstein the second postulate of special relativity.
In order to solve his troublesome algebraic system of 9 equations with 15 unknowns, Voigt made an assumption \cite{1}: ``supposed that $\alpha, \beta$ and $\gamma$ [the components of the relative velocity between the rest and moving frames] are given, then we have 12 available constants, so we can arbitrarily use three of them.'' From this point the reading of Voigt's paper becomes somewhat difficult to follow, due in part to an abuse of notation. He re-defined variables and made a number of additional assumptions with the purpose of finding the solution for his troublesome algebraic system. In essence, Voigt restricted his calculations to work in the standard configuration in which the space coordinates $(x_1,y_1,z_1)$ and the time $t$ are associated with the rest frame and the space coordinates $(\xi_1,\eta_1,\zeta_1)$ and the time $\tau$ are associated with the moving frame. He also wrote the $x-$component of the inter-frame velocity as $\alpha=\chi.$ At the end of his calculations, Voigt obtained the following set of transformations
\begin{equation}
\xi_1=x_1 - \chi t,\;\; \eta_1=y_1q,\;\; \zeta_1=z_1q,\;\; \tau=t-\chi x_1/\omega^2,
\end{equation}
where $q\!=\!\sqrt{1-\chi^2/\omega^2}$. If we identify $(\xi_1,\eta_1,\zeta_1)$ with $(x',y',z')$; the time $\tau$ with the time $t'$; $(x_1,y_1,z_1)$ with $(x,y,z)$ and $q$ with $1/\gamma$ (where $\gamma$ is the Lorentz factor) then we can see that the transformations in equation~(11) are the same as those given in equation~(1). Voigt then proceeded to generalize equation~(11) to be valid for an inter-frame velocity in three dimensions. The remainder of the paper was devoted to a study of the Doppler effect. Because our main concern here are the transformations in equation~(11), or equivalently, those in equation~(1), and not the discussion of the Doppler effect, let us stop here our review of Voigt's 1887 paper.
Let us re-construct the original derivation of Voigt's transformations following a modern and simplified approach. We will see that what Voigt really did in his 1887 paper was to discover one first example of the conformal symmetry of spacetime. Consider the scalar field $F(x,y,z,t)$ in the frame $S$ satisfying the homogeneous wave equation:
\begin{equation}
\frac{\partial^2 F}{\partial x^2}+ \frac{\partial^2F}{\partial y^2} + \frac{\partial^2F}{\partial z^2} -\frac{1}{c^2}\frac{\partial^2F}{\partial t^2}=0.
\end{equation}
Consider the following set of transformations connecting the space coordinates $(x',y',z')$ and the time $t'$ in the frame $S'$ with the space coordinates $(x,y,z)$ and the time $t$ in the frame $S$:
\begin{equation}
x'=k_1x -vt,\;\; t'=t-k_2x,\;\; y'=k_3y,\;\; z'=k_4z,
\end{equation}
where $k_1,k_2, k_3$ and $k_4$ are constants to be determined and $v$ the relative speed between the frames $S$ and $S'$. From equation~(13) we can derive the transformation laws:
\begin{eqnarray}
\frac{\partial}{\partial y}=k_3\frac{\partial}{\partial y'},\\
\frac{\partial}{\partial z}=k_4\frac{\partial}{\partial z'},\\
\frac{\partial }{\partial x}=k_1\frac{\partial }{\partial x'}-\!k_2\frac{\partial }{\partial t'},\\
\frac{\partial }{\partial t}= \frac{\partial }{\partial t'}-\!v\frac{\partial }{\partial x'}.
\end{eqnarray}
It follows that
\begin{eqnarray}
\frac{\partial^2}{\partial y^2}=k_3^2\frac{\partial^2}{\partial y'^2},\\
\frac{\partial^2}{\partial z^2}=k_4^2\frac{\partial^2}{\partial z'^2},\\
\frac{\partial^2}{\partial x^2}=k_1^2\frac{\partial^2}{\partial x'^2}-2k_1k_2\frac{\partial^2}{\partial x'\partial t'} +k_2^2\frac{\partial^2}{\partial t'^2},\\
\frac{\partial^2}{\partial t^2}=\frac{\partial^2}{\partial t'^2}-2v\frac{\partial^2}{\partial t'\partial x'} +v^2\frac{\partial^2}{\partial x'^2}.
\end{eqnarray}
Using these transformation laws, equation~(12) is transformed as follows
\begin{eqnarray}
\Bigg(k_1^2-\frac{v^2}{c^2}\Bigg)\frac{\partial^2 F}{\partial x'^2}+ k_3^2\frac{\partial^2F}{\partial y'^2} + k_4^2\frac{\partial^2F}{\partial z'^2}\nonumber\\
\qquad\qquad -(1-k_2^2c^2)\frac{1}{c^2}\frac{\partial^2F}{\partial t'^2}+ \bigg(\frac{2v}{c^2}-2k_1k_2\Bigg)\frac{\partial F}{\partial x'\partial t'} =0,
\end{eqnarray}
where now $F(x',y',z',t')$ is defined in the frame $S'.$ Conformal covariance demands
\begin{equation}
\bigg(k_1^2-\frac{v^2}{c^2}\bigg)= k_3^2=k_4^2= (1-k_2^2c^2),\;\; \frac{v}{c^2}-k_1k_2=0.
\end{equation}
By making use of these relations, equation~(22) becomes
\begin{equation}
\bigg(k_1^2-\frac{v^2}{c^2}\bigg)\Bigg[\frac{\partial^2 F}{\partial x'^2}+\frac{\partial^2F}{\partial y'^2} + \frac{\partial^2F}{\partial z'^2} -\frac{1}{c^2}\frac{\partial^2F}{\partial t'^2}\Bigg] =0.
\end{equation}
From equations~(12) and (24) we obtain the relation
\begin{eqnarray}
\frac{\partial^2 F}{\partial x^2}+ \frac{\partial^2F}{\partial y^2} + \frac{\partial^2F}{\partial z^2} -\frac{1}{c^2}\frac{\partial^2F}{\partial t^2}\nonumber\\
\qquad\qquad=\Bigg(k_1^2-\frac{v^2}{c^2}\Bigg)\Bigg[\frac{\partial^2 F}{\partial x'^2}+\frac{\partial^2F}{\partial y'^2} + \frac{\partial^2F}{\partial z'^2} -\frac{1}{c^2}\frac{\partial^2F}{\partial t'^2}\Bigg].
\end{eqnarray}
A solution for the system of algebraic equations displayed in equation~(23) is given by
\begin{equation}
k_1=1,\;\; k_2=v/c^2,\;\; k_3=1/\gamma,\;\; k_4=1/\gamma,
\end{equation}
where $\gamma\!=\!1/\sqrt{1\!-\!v^2/c^2}$. Using equations~(13) and (26), we get Voigt's transformations given in equation~(1). Equation~(25) with $k_1\!=\!1$ can compactly be written as
\begin{equation}
\Box^{2}=\frac{1}{\gamma^{2}}\Box'^{2},
\end{equation}
where $\Box^{2}\!\equiv\!\nabla^2\!-\!(1/c^2)\partial^2/\partial t^2$ is the d'Alembert operator in $S$ and $\Box'^{2}\!\equiv\!\nabla'^2-(1/c^2)\partial^2/\partial t'^2$ denotes this operator in $S'$. In section 5 we will present an alternative derivation of Voigt's transformations based on the conformal covariance expressed in equation~(27).
It is pertinent to mention a common and recurrent misunderstanding, which can be found, for example, in a paper by Rott (See reply in Ref.~\cite{18}):
``It is undisputed that Voigt discovered in 1887 the invariance of the wave equation with respect to the transformation that is named today after Hendrik Lorentz.'' Strictly speaking, Voigt's transformations [equation~(1)] are not equivalent to the Lorentz transformations [equation~(2)]. Doyle has correctly pointed out \cite{29}: ``What Voigt actually did was show that the wave equation is covariant under his transformation ---certainly a physical result.'' Authors of books have also incorrectly identified Voigt's transformations with the Lorentz transformations.
Here are some examples. When referring to the Lorentz transformations, Whittaker \cite{8} claimed: ``It should be added that the transformation
in question had been applied to the equation of vibratory motions many years
before by Voigt,...'' Regarding the origin of the Lorentz transformation, Sears and Brehme \cite{4} wrote: ``In point of fact W. Voigt (1850-1919) first published the transformation in 1887.'' Similarly, French \cite{5} wrote: ``...the Lorentz transformations had, in essence, been discovered in 1887 by W. Voigt, who in that year published a theoretical paper about the Doppler effect (which can be regarded as the problem of observing a wave motion from different inertial frames).'' O'Rahilly \cite{6} has also incorrectly identified Voigt's transformations with the Lorentz transformations. He pointed out: ``The transformation is usually called by the name of Lorentz. But we wish to point out ,..., that the formula was first explicitly given and fully employed by W. Voigt in 1887.''
In the next section we will discuss more deeply the misunderstanding of considering equivalent the Voigt and Lorentz transformations.
\section {The scant impact of Voigt's 1887 paper}
At the risk of being somewhat reiterative, we wish to highlight three important aspects of the first part of Voigt's 1887 paper, which could explain its scant impact among physicists of that time: (i) The main purpose of Voigt in his 1887 paper was not to propose a new set of spacetime transformations, which should replace the well-established Galilean transformations, a very ambitious idea from a conceptual point of view, but simply to study the transformation of oscillating disturbances through an elastic incompressible medium and deduce the formula for the Doppler effect; (ii) The formal process by which Voigt derived a set of transformations that maintained covariance of the wave equation was not discussed. As mentioned above, the only four words used by Voigt to justify this covariance were {\bf ``as it must be''}; and (iii) He did not provide any physical interpretation of his non-absolute time: $t'=t-vx/c^2$ nor did he say anything about the invariance of the speed of light $c$ in inertial frames. Apparently, Voigt did not realize the great conceptual importance of his transformations. For a modern reader it is hard to understand why Voigt did not discuss other important and immediate predictions of his transformations, like some restricted forms of the time dilation and length contraction. Only a few lines of elementary algebra would have been sufficient. He had all the ingredients at the hand!
Regarding Voigt's ideas developed in his 1887 paper,
Hsu has pointed \cite{22}: ``If the physicists of the time had been imaginative enough, they might have recognized the potential of these ideas to open up a whole new view of physics.'' We believe that a considerable number of physicists of that time were imaginative enough and that they did not appreciate Voigt's 1887 paper because the first part of this paper (that introducing his transformations) was poorly written. This seems to have been ``the extraordinary failure'' of Voigt.
Despite the great initial acceptance of special relativity in the beginnings of the 1900's, Voigt did not seem to have been interested in claiming the paternity of the non-absolute time: $t'=t-vx/c^2$. According to the standard account, Lorentz \cite{27} was the first in introducing this time in 1895 but with the name ``local time.'' The replacement of the absolute time by a non-absolute time was a crucial idea in the construction of special relativity. Poincar\'e \cite{30} recognized the great conceptual importance of the local time by claiming that it was Lorentz's ``most ingenious idea'' (in French: ``L'id\'ee la plus ing\'enieuse a \'et\'e celle du temps local'').
Moreover, circumstantial evidence points out that Voigt was not interested enough in promoting his
transformations among his contemporaries. For example, Voigt corresponded with Lorentz since 1883, but he did not seem to have mentioned his 1887 paper to Lorentz, at least during the period 1887-1907. It was not until 1908 that Voigt sent his 1887 paper to Lorentz. In a response letter, Lorentz wrote \cite{31}:
\begin{quote}
``Of course I will not miss the first opportunity to mention, that {\bf the concerned transformation
and the introduction of a local time has been your idea.}''
\end{quote}
In a letter addressed to Wiechert in the year of 1911, Lorentz \cite {32} mentioned again the priority of Voigt in the discovery of the Lorentz transformations. It is also interesting to note that Lorentz explicitly pointed out in his book \cite{9} that his space-time transformation were first introduced by Voigt. Lorentz wrote:
\begin{quote}
\noindent ``In a paper ``\"{O}ber das Doppler�sche Princip,'' published in 1887 (G\"{o}tt. Nachr., p. 41) and wich to my regret has escaped my notice all these years, Voigt has applied to equations of the form (6)
(\S\,3 of this book)[$\triangle\!\psi\!-\!(1/c^2)\partial^2\psi/\partial t^2\!=\!0$] a transformation equivalent to the formulae (287) and
(288) [$x'\!=\!kl(x\! -\!wt),t'\!=\!kl(t\!-\!wx/c^2), y'\!=\!ly, z'\!=\!lz $]. {\bf The idea of the transformations used above (and in \S\,44) might therefore have been borrowed from Voigt and the proof that it does not alter the form of the equations for the
free ether is contained in his paper.}''
\end{quote}
In this paragraph Lorentz pointed out two things: (i) that his transformations were equivalent to those of Voigt and (ii) that Voigt's transformations do not alter the form of the wave equation. The transformations introduced by Lorentz: $x'\!=\!kl(x\! -\!wt),t'\!=\!kl(t\!-\!wx/c^2), y'\!=\!ly, z'\!=\!lz $ involve both the ``standard'' Lorentz transformations ($w\!=\!v, l\!=\!1$ and $k\!=\!\gamma$) and the Voigt transformations ($w\!=\!v, l\!=\!1/\gamma$ and $k\!=\!\gamma$). The standard Lorentz transformations transform the scalar wave equation $\Box^2 F\!=\!0$ into $\Box'^2 F\!=\!0$ (a perfect invariance!) and Voigt's transformations transform $\Box^2 F\!=\!0$ into $\Box'^2 F/\gamma^2\!=\!0$ (a form of covariance). In other words: the homogeneous scalar wave equation is invariant under Lorentz transformations and covariant under Voigt's transformations. This subtle difference between invariance and covariance of the wave equation didn't seem to have been relevant for Lorentz. Strikingly, this difference is not usually noted by most current authors!
Because Voigt and Lorentz transformations ultimately yield the form invariance of the wave equation [in the case of Voigt's transformations $\Box'^2 F/\gamma^2\!=0$ implies $\Box'^2 F=0$ because $1/\gamma^2\not=0]$, some authors
have incorrectly identified the Lorentz transformations with Voigt's transformations. These authors seem to minimize the difference between covariance and invariance of the wave equation. For example, when discussing the evolution and interpretation of the Lorentz transformations, Pais \cite {33} does not hesitate to say that Voigt was ``the first to write down Lorentz transformations [equations~(2)].'' He points out that wave equations of the type $\Box^2 \phi=0$ ``retains their form if one goes over the new space-time variables [equations (1)].'' He states that ``These [equations~(1)] are the Lorentz transformations [equations~(2)] up to scale factor.'' Pais's statement is incorrect because equations~(1) transform $\Box^2 \phi \!=\!0$ into $\Box'^2 \phi/\gamma^2\!=\!0$ and not into $\Box'^2 \phi\!=\!0$.
It is fair to say that Lorentz always recognized Voigt's 1887 paper. In 1914 he commented on a paper by Poncar\'e and wrote \cite{34}:
\begin{quote}
``These considerations published by myself in 1904,
have stimulated Poincar\'e to write his article on the
dynamics of electron where he has given my name to
the just mentioned transformation. I have to note as
regards this that {\bf a similar transformation has been
already given in an article by Voigt published in 1887}
and I have not taken all possible benefit from it.''
\end{quote}
Minkowski didn't seem to have noted some difference between equations~(1) and equations~(2). In a physics meeting of 1908 he claimed \cite{35}:
\begin{quote}
\noindent ``I want to add that the transformations, which play the main role in the relativity principle, {\bf were first mathematically discussed by Voigt in the year 1887.}''
\end{quote}
Voigt's 1887 paper was cited by E. Kohl in 1903 in Annalen der Physik \cite{11}. However, other authors that constructed the special theory of relativity like Einstein and Poincar\'e did not mention Voigt's 1887 paper.
The fact that Voigt's paper remained unnoticed during the period 1887-1892, was clearly
pointed out by Pauli in his 1921 book on relativity \cite{7}. He wrote :
\begin{quote}
``As long ago as 1887, in a paper still written from the point of view of the elastic solid theory of light, Voigt mentioned that it is mathematically convenient to introduce a local time $t'$ into a moving reference frames.... In this way the wave equation $\triangle\phi-(1/c^2)\partial^2\phi/\partial t^2=0$
could be made to remain valid in the moving reference frame, too. {\bf These remarks, however, remained completely unnoticed}, and a similar transformation was not again suggested until 1892 and 1895, when H. A. Lorentz published his fundamental papers on the subject.''
\end{quote}
Another 1921 book that mentioned the work of Voigt was that of Kopff \cite{10}.
As a recognition to Voigt, his 1887 paper was reprinted in 1915 in occasion of the tenth anniversary of the principle of relativity (See Ref. \cite {36}). In connection with this recognition, Doyle writes \cite{29}: ``However, Voigt did live to see his [1887] paper chosen
to be reprinted in its entirety in the
Physikalische Zeitschrift (a German
Physics Today of his time) on the
occasion of what the editors called
simply the tenth `birthday celebration of the principle of relativity.'''
In the reprinted version of his 1887 paper (See Ref. \cite{36}), Voigt included some additional comments, the second of them is particularly disconcerting. When referring to his transformations [equation~(11) in the present paper], he wrote:
\begin{quote}
``This is, except for the factor $q$ which is irrelevant for the application, {\bf exactly the Lorentz transformation of the year 1904.}''
\end{quote}
We know now that the factor $q$ (our $1/\gamma$ in modern notation) is generally significant for applications \cite {37}. Apparently, Voigt himself committed the same interpretative mistake as that of Lorentz and Minkowski: he identified his transformations with the Lorentz transformations. Notice that Voigt's comment was made in 1915 when special relativity was already a well-established theory.
\section {The connection between the Voigt and Lorentz transformations}
The relation between the Lorentz and Voigt transformations is transparently established using four-dimensional spacetime notation. Greek indices $\alpha, \beta, \ldots$ run from 0 to 3; Latin indices $i,j,\ldots$ run from 1 to 3. Coordinates are labeled as $x^{\alpha}\!=\!(x^0, x^1,x^2, x^3)\!=\!(ct,x,y,z)$ in the frame $S$ and $x'^{\alpha}\!=\!(x'^0, x'^1,x'^2, x'^3 )\!=\!(ct',x',y',z')$ in the frame $S'$. Summation convention on repeated indices is adopted. As is well-known, the Lorentz transformations in equation~(2) can be written as
\begin{equation}
x'^\alpha={\rm \Lambda}_\beta^\alpha x^\beta,
\end{equation}
where
\begin{equation}
{\rm \Lambda}^\alpha_\beta= \left(
\begin{array}{cccc}
\gamma & -v\gamma/c & 0 & \quad 0 \\
-v\gamma/c & \gamma & 0 &\quad 0 \\
0 & 0& 1&\quad 0\\
0& 0 & 0 & \quad 1
\end{array}
\right),
\end{equation}
is the Lorentz matrix. Voigt's transformations in equation~(1) can be written as
\begin{equation}
x'^\alpha={\rm V}_\beta^\alpha x^\beta,
\end{equation}
where the Voigt matrix is defined by
\begin{equation}
{\rm V}^\alpha_\beta=\left(
\begin{array}{cccc}
1 & -v/c & 0 & 0\\
-v/c & 1 & 0 & 0\\
0 & 0& 1/\gamma & 0\\
0& 0 & 0 & 1/\gamma\\
\end{array}
\right).
\end{equation}
Clearly, the relation between the Lorentz and Voigt matrices is given by
\begin{equation}
{\rm \Lambda}^\alpha_\beta=\gamma{\rm V}_\beta^\alpha,
\end{equation}
i.e., the Lorentz matrix is proportional to the Voigt matrix. Using this proportionality we can infer the properties of the latter from those of the former. The Lorentz matrices satisfy the relation ${\rm \Lambda}^\alpha_\theta{\rm \Lambda}^\theta_\beta=\delta^\alpha_\beta,$ where $\delta^\alpha_\beta$ is the Kronecker delta (See Refs. \cite{38} and \cite{39}).
This means that ${\rm \Lambda}^\theta_\beta$ is the inverse of ${\rm \Lambda}^\alpha_\theta$. This inverse can also be denoted as $({\rm \Lambda}^{-1})^\theta_\beta$. From ${\rm \Lambda}^\alpha_\theta{\rm \Lambda}^\theta_\beta=\delta^\alpha_\beta$ and equation~(32) it follows that
\begin{equation}
{\rm V}^\alpha_\theta [\gamma^2{\rm V}^\theta_\beta]=\delta^\alpha_\beta.
\end{equation}
Therefore $\gamma^2{\rm V}^\theta_\beta$ can be interpreted as the inverse of ${\rm V}^\alpha_\theta$. This inverse can also be denoted as ${\rm (V^{-1}})^\theta_\beta$. In its explicit form, this inverse transformation reads
\begin{equation}
({\rm V^{-1}})^\theta_\beta=\left(
\begin{array}{cccc}
\gamma^2 & -v\gamma^2/c & 0 &\quad 0\;\\
-v\gamma^2/c & \gamma^2 & 0 &\quad 0\;\\
0 & 0& \gamma&\quad 0\;\\
0& 0 & 0 & \quad \gamma\;\\
\end{array}
\right).
\end{equation}
The Lorentz matrices are defined to be those satisfying ${\rm \Lambda}^\mu_\alpha\eta_{\mu\nu}{\rm \Lambda}^\nu_\beta=\eta_{\alpha\beta}$, where
\begin{equation}
\eta_{\alpha\beta}=\left(
\begin{array}{cccc}
-1 & \;0 & \;0 & \; 0 \;\\
\;0 & 1 & \;0 & 0\\
\; 0 & 0& 1 & 0\\
\;0 & 0 & 0 & 1\\
\end{array}
\right).
\end{equation}
It follows that the Voigt matrices can be defined as the set of matrices satisfying
\begin{equation}
{\rm V}^\mu_\alpha\gamma^2\eta_{\mu\nu}{\rm V}^\nu_\beta=\eta_{\alpha\beta}.
\end{equation}
We can write equation~(36) in the compact form: ${\rm V^T}\gamma^2\eta{\rm V}=\eta$, where ${\rm V^T}$ is the transpose matrix of ${\rm V}$ [notice that ${\rm V^{-1}}=\gamma^2{\rm V}^T]$. Despite the close relation between the Lorentz and Voigt matrices, the latter do not form a group \cite{14}.
To show this we consider two Voigt matrices ${\rm V_1}$ and ${\rm V_2}$. We will investigate if their product ${\rm V_1}{\rm V_2}$ is also another Voigt matrix (closure property). We have ${\rm V_1^T}\gamma_1^2\eta{\rm V_1}\!=\!\eta$ and ${\rm V_2^T}\gamma_2^2\eta{\rm V_2}=\eta$. Let ${\rm V_3}\!=\!{\rm V_1}{\rm V_2}.$ Thus ${\rm V_3^T}\gamma_3^2\eta{\rm V_3}\!=\![{\rm V_1^T}{\rm V_2^T}]\gamma_3^2\eta[{\rm V_1}{\rm V_2}].$ If $\gamma_3\!=\!\gamma_1\gamma_2$ then
\begin{equation}
{\rm V_3^T}\gamma_3^2\eta{\rm V_3}={\rm V_1^T}\gamma_1^2[{\rm V_2^T}\gamma_2^2\eta{\rm V_2}]{\rm V_1}={\rm V_1^T}\gamma_1^2\eta{\rm V_1}=\eta.
\end{equation}
From this equation it appears to be that the Voigt matrices satisfy the closure property. But this is not so because the assumption $\gamma_3=\gamma_1\gamma_2$ is incorrect. It can be shown that \cite{40}: $\gamma_3=\gamma_1\gamma_2[1+v_1v_2/c^2]$. The Voigt matrices do not satisfy the closure property and
therefore the Voigt transformations do not form a group since two successive Voigt transformations do not yield another Voigt transformation. This makes them unattractive from a physical point of view because the physical equivalence of the inertial frames is broken.
As pointed out by Levi-Leblond \cite{41}: ``The physical equivalence of the inertial frames implies a group structure for the set of all inertial transformations.''
Remarkably, in the low-velocity limit $v \!<\!<\!c$ we have $\gamma\approx 1$ and therefore
\begin{equation}
{\rm \Lambda}^\alpha_\beta\approx{\rm V}_\beta^\alpha.
\end{equation}
The Lorentz and Voigt matrices coincide in the weakly relativistic regime of special relativity. Consequently, the Lorentz and Voigt transformations coincide in this limit. The low-velocity limit of Voigt's transformations take the form
\begin{eqnarray}
x'= x -vt,\;\; t'=t-vx/c^2,\;\; y'\approx y,\;\; z'\approx z.
\end{eqnarray}
Notice that the time transformation in equation~(37) reduces to the Galilean transformation: $t'=t$ only if the additional condition $ct \!>\!>\!x$ is imposed \cite{42}. In the low-velocity limit we have $\gamma_3\approx\gamma_1\gamma_2$ and then two sets of successive Voigt transformations yield another set of Voigt's transformations, and this satisfies the closure property. We then conclude that Voigt's transformation approximately form a group in the low-velocity limit $v \!<\!<\!c$. The main objection against Voigt's transformations is seen to disappear when considering low-velocities compared with the speed of light.
\section {Alternative derivation of Voigt's transformations}
In this section we will demand the conformal covariance of the d'Alembert operator expressed in equation~(27) to obtain Voigt's transformations.
Consider the standard configuration and equation~(27) expressed as
\begin{eqnarray}
\bigg[\frac{\partial}{\partial x}-\frac{1}{c}\frac{\partial}{\partial t}\!\bigg]\bigg[\frac{\partial}{\partial x}+\frac{1}{c}\frac{\partial}{\partial t}\bigg]+\!\frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}\nonumber\\
\qquad\qquad=\frac{1}{\gamma^2}\Bigg(\bigg[\!\frac{\partial}{\partial x'}-\frac{1}{c}\frac{\partial}{\partial t'}\bigg]\bigg[\!\frac{\partial}{\partial x'}+\frac{1}{c}\frac{\partial}{\partial t'}\!\bigg]+ \frac{\partial^2}{\partial y'^2} + \frac{\partial^2}{\partial z'^2}\Bigg).
\end{eqnarray}
By assuming linearity for the transformations of derivative operators, we can write
\begin{eqnarray}
\bigg[\frac{\partial}{\partial x}\!-\!\frac{1}{c}\frac{\partial}{\partial t}\bigg] =\frac{A}{\gamma}\bigg[\frac{\partial}{\partial x'}-\frac{1}{c}\frac{\partial}{\partial t'}\bigg],\\
\bigg[\frac{\partial}{\partial x}\!+\!\frac{1}{c}\frac{\partial}{\partial t}\bigg]=\frac{A^{-1}}{\gamma} \bigg[\frac{\partial}{\partial x'}+\frac{1}{c}\frac{\partial}{\partial t'}\bigg],\\
\frac{\partial}{\partial y}=\frac{1}{\gamma}\frac{\partial}{\partial y'},\\
\frac{\partial}{\partial z}=\frac{1}{\gamma}\frac{\partial}{\partial z'}.
\end{eqnarray}
Insertion of these quantities into the left-hand side of equation~(40) leads to an identity. The factor $A$ is independent of the derivative operators but can depend on the velocity $v$ and $A^{-1}=1/A$. In order to determine $A$, we demand that the expected linear transformation relating primed and unprimed time-derivative operators should appropriately reduce to the corresponding Galilean transformation \cite{43}: $\partial/\partial t= \partial/\partial t'-v\partial/\partial x'.$ Our demand is consistent with a linear transformation of the general form: $\partial/\partial t= F(v)[\partial/\partial t' -v\partial/\partial x'],$ where $F(v)$ depends on the velocity $v$ so that $F(v)\!\to\! 1$ when $v<<c$. From this general transformation it follows that if
$\partial/\partial t\!=\!0$ then $\partial/\partial t'\!=\!v\partial/\partial x'$ because $F(v)\neq 0$. Using this result in equations (41) and (42) we get
\begin{eqnarray}
\frac{\partial}{\partial x}=\frac{A}{\gamma}\bigg[\frac{\partial}{\partial x'}\!-\!\frac{v}{c}\frac{\partial}{\partial x'}\bigg],\quad
\frac{\partial}{\partial x}=\frac{A^{-1}}{\gamma}\bigg[\frac{\partial}{\partial x'}\!+\!\frac{v}{c}\frac{\partial}{\partial x'}\bigg].
\end{eqnarray}
By combining these equations we can derive expressions for $A$ and $A^{-1}$,
\begin{equation}
A=\frac {[1+v/c]^{1/2}}{[1-v/c]^{1/2}},\quad
A^{-1}=\frac {[1-v/c]^{1/2}}{[1+v/c]^{1/2}},
\end{equation}
which can conveniently be written as
\begin{equation}
A=\gamma\bigg[1+\frac{v}{c}\bigg],\quad A^{-1}=\gamma\bigg[1-\frac{v}{c}\bigg].
\end{equation}
Using these relations in equation~(35) we obtain
\begin{eqnarray}
\bigg[\frac{\partial}{\partial x}-\frac{1}{c}\frac{\partial}{\partial t}\bigg]=\bigg[1+\frac{v}{c}\bigg] \bigg[\frac{\partial}{\partial x'}-\frac{1}{c}\frac{\partial}{\partial t'}\bigg],\\
\bigg[\frac{\partial}{\partial x}+\frac{1}{c}\frac{\partial}{\partial t}\bigg]=\bigg[1-\frac{v}{c}\bigg] \bigg[\frac{\partial}{\partial x'}+\frac{1}{c}\frac{\partial}{\partial t'}\bigg].
\end{eqnarray}
By adding and subtracting equations (48) and (49) we can obtain the corresponding transformation laws connecting unprimed and primed operators, which are added to the transformation laws for $\partial/\partial y$ and $\partial/\partial z$, obtaining
\begin{equation}
\frac{\partial}{\partial x}\!=\!\frac{\partial}{\partial x'}-\!\frac{v}{c^2}\frac{\partial}{\partial t'},\;
\frac{\partial}{\partial t}\!=\! \frac{\partial}{\partial t'}-\!v\frac{\partial}{\partial x'},\; \frac{\partial}{\partial y}\!=\!\frac{1}{\gamma}\frac{\partial}{\partial y'},\;
\frac{\partial}{\partial z}\!=\!\frac{1}{\gamma}\frac{\partial}{\partial z'}.
\end{equation}
These relations are the Voigt transformations for derivative operators of the standard configuration. The first two relations in equation~(50) imply coordinate transformations of the form: $x'=x'(x,t)$ and $t'=t'(x,t).$ To find the explicit form of these transformations we can use the first two relations displayed in equation (50) to obtain
\begin{equation}
\frac{\partial x'}{\partial x}= 1,\;
\frac{\partial x'}{\partial t}= -v,\; \frac{\partial t'}{\partial t}= 1,\;
\frac{\partial t'}{\partial x}=-\frac{v}{c^2}.
\end{equation}
From the first relation in equation (51) it follows the equation (A): $x'\!=\! x \!+ \!g_1(t),$ where $g_1(t)$ can be determined (up to a constant) by deriving (A) with respect to the time $t$ and using the second relation in (51): $\partial x'/\partial t\!=\!\partial g_1(t)/\partial t\!=\!-v.$ This last equality implies (B): $g_1(t)\!=\!-vt\!+\! x_0,$ where $x_0$ is a constant. From (A) and (B) we obtain
\begin{equation}
x'=x -vt + x_0.
\end{equation}
The third relation in equation~(51) implies (C): $t'\!=\!t\! +\! g_2(x),$ where $g_2(x)$ can be obtained (up to a constant) from deriving (C) with respect to $x$ and using the last relation in equation~(41):
$\partial t'/\partial x\!=\!\partial g_2(x)/\partial x\!=\!-v/c^2$. This last equality implies (D):
$g_2(x)\!=\! -v x/c^2\!+\! t_0,$
where $t_0$ is a constant. From (C) and (D) we conclude
\begin{equation}
t'=t\!-\!vx/c^2 + t_0.
\end{equation}
The origins of the frames $S$ and $S'$ coincide at $t\!=\!t'\!=\!0$. It follows that $x_0\!=\!0$ and $t_0\!=\!0$. In this way we obtain the Voigt transformations for the $x$ and $t$ coordinates of the standard configuration:
\begin{equation}
x'=x -vt,\quad t'\!=t\!-\!vx/c^2.
\end{equation}
The transformations for the $y$ and $z$ coordinates are easily derived. From the last two relations in equation~(50) we get $\partial y'/\partial y\!=\!1/\gamma$ and $\partial z'/\partial z\!=\!1/\gamma$. They imply $y'\!=\!y/\gamma\! + \!y_0$ and $z'\!=\!z/\gamma + z_0$, where $y_0$ and $z_0$ are constants, which vanish because the origins of the frames $S$ and $S'$ coincide at the time $t\!=t'=0$. Thus
\begin{equation}
y'=y/\gamma, \quad z'=z/\gamma.
\end{equation}
Equations~(54) and (55) are the Voigt transformations of the standard configuration. A direct manipulation of equations~(54) and (55) yields the corresponding inverse transformations \cite{14}:
\begin{equation}
x=\gamma^2\big(x'+vt'\big),\;\; t=\gamma^2\big(t'+vx'/c^2\big), \;\; y=\gamma y',\;\; z=\gamma z'.
\end{equation}
\section {Conclusion}
In the creation of special relativity, we traditionally find the names of Lorentz, Larmor, Poincar\'e and Einstein. They appear to be the main actors. Voigt is relegated to being a minor player, in the best of cases. But this tradition is not faithful to the history of physics. Although Voigt did not derive the Lorentz transformations, he was the first in applying the postulates of special relativity to a physical law: he demanded covariance to the homogeneous wave equation with respect to inertial frames and assumed the invariance of the speed of light in these frames, obtaining a set of transformations which introduced the non-absolute time $t'=t-vx/c^2$ and thereafter the time was never the same! We can say that Voigt signed the death certificate of the Newtonian absolute time and that this marked the beginning of the relativistic era. In retrospect, we can criticize the extraordinary failure of Voigt of having lost the opportunity of explicitly expressing the postulates of special relativity before Einstein, and also of having written his paper in a very unattractive form. But we cannot dispute him on his idea of having extracted a spacetime symmetry from the wave equation, thus opening the gate to relativity. This is the main legacy of Voigt's 1887 paper.
To finish, we would like to point out that we are in agreement
with Rott \cite{16}, who on the centennial of Voigt's 1887 paper, claimed that W. Voigt was a ``Relativity's forgotten figure.'' He wrote \cite{16}: ``...this year is also the centennial of a theoretical paper [Voigt's 1887 paper] that is largely forgotten, but has a certain role in the history of the theory of relativity even though it has no documented impact on the actual historical development.''\\
\vskip 50pt
\noindent {\bf References}
\vskip 10pt
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,990 |
Q: What is the difference between scheduler's standby() and pauseAll()? I'm using Quartz Scheduler v.1.8.0.
What's the difference between scheduler.standby() and scheduler.pauseAll()?
standby() -
Temporarily halts the Scheduler's firing of Triggers.
pauseAll() -
Pause all triggers - similar to calling
pauseTriggerGroup(group) on every
group, however, after using this
method resumeAll() must be called to
clear the scheduler's state of
'remembering' that all new triggers
will be paused as they are added.
Based on what I've understood from the API documentation, I'm not able to easily/clearly differentiate/distinguish from each one of them. I'm seeing both of them serving the same purpose - temporarily pause/halt all the triggers in the scheduler, and subsequently followed by a start() (for standby) or resumeAll() (for pauseAll) to clear the scheduler's state. Is there any other difference?
Hope experts can help me in understanding any subtle difference.
A: There is difference when scheduler is resumed after standby and pauseAll.
I have made difference in bold in following description from API docs.
standby :
void standby()
throws SchedulerException Temporarily halts the Scheduler's
firing of Triggers.
When start() is called (to bring the
scheduler out of stand-by mode),
trigger misfire instructions will NOT
be applied during the execution of the
start() method - any misfires will be
detected immediately afterward (by the
JobStore's normal process).
The scheduler is not destroyed, and
can be re-started at any time.
pauseAll :
void pauseAll()
throws SchedulerException Pause all triggers
- similar to calling pauseTriggerGroup(group) on every
group, however, after using this
method resumeAll() must be called to
clear the scheduler's state of
'remembering' that all new triggers
will be paused as they are added.
When resumeAll() is called (to
un-pause), trigger misfire
instructions WILL be applied.
A: Here are what I got from source code v1.8.6:
standby() simply freezes the scheduler thread, which means no more trigger will be fired from now on, even those new triggers added later.
start() just resume the scheduler thread, and will not apply misfire policies immediately. But all misfires will be applied later naturally.
pauseAll() is similar to call pauseTriggerGroup() on every now existing trigger groups, which means those new trigger groups added later will be fired normally. And please notice that there is nothing to do with pauseJob() or pauseJobGroup(), it's just about what happening on triggers and their groups.
resumeAll() is similar to call resumeTriggerGroup() on every now existing trigger groups. In addition, misfires will be applied during execution of resumeAll().
A: The difference is in trigger misfire instructions applying behavior.
When you call start() after standby(), any misfires, which appear while standby, will be ignored.
When you call resumeAll() after pauseAll(), all misfires, which appear while scheduler was paused, will be applyed.
A: start() and standby() are per-instance methods. Other instances running in clustered mode will continue to trigger jobs.
resumeAll() and pauseAll() is applicable to the whole cluster.
A: While pauseAll() pauses all schedules (already created by that moment), standby() pauses a scheduler itself. Thus when you create a new schedule, after pauseAll() it will be scheduled and run appropriately, but in case of standby() it runs not earlier than a 'start()' method will be called on a scheduler.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,692 |
Q: Viewing a set of images and asking for input from pims with opencv, matplotlib, or PIL, or?-- quickly and efficiently I need a way to iterate quickly through image files and ask for user input at each image. This is for the development of a binary classification training set. Hopefully I will import the images as pims frames such as
frames = pims.ImageSequence("*.jpg",as_grey=True)
Then maybe I can load all of the image files in some kind of sequence with maybe some kind of structure along the lines of (obviously this won't work):
images = [plt.imshow(f) for f in frames]
So that I can do something along the lines of:
classification = []
for i in images:
magic_function_which_plots_image_without_a_delay(i)
value = input()
if value==1:
classification.append(True)
else:
classification.append(False)
So I can have an array of True and False values, one associated with each image. If I do this some thousands of times I hope to have a training set for my binary classification problem. Does anyone know how to view images quickly? Any help is very much appreciated !!
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,798 |
<?php
namespace Podlove;
class File_Type_List_Table extends \Podlove\List_Table {
function __construct(){
global $status, $page;
// Set parent defaults
parent::__construct( array(
'singular' => 'file_type', // singular name of the listed records
'plural' => 'file_types', // plural name of the listed records
'ajax' => false // does this table support ajax?
) );
}
function column_name( $file_type ) {
$actions = array(
'edit' => sprintf(
'<a href="?page=%s&podlove_tab=%s&action=%s&file_type=%s">' . __( 'Edit', 'podlove-podcasting-plugin-for-wordpress' ) . '</a>',
filter_input(INPUT_GET, 'page', FILTER_SANITIZE_STRING),
filter_input(INPUT_GET, 'podlove_tab', FILTER_SANITIZE_STRING),
'edit',
$file_type->id
),
'delete' => sprintf(
'<a href="?page=%s&podlove_tab=%s&action=%s&file_type=%s">' . __( 'Delete', 'podlove-podcasting-plugin-for-wordpress' ) . '</a>',
filter_input(INPUT_GET, 'page', FILTER_SANITIZE_STRING),
filter_input(INPUT_GET, 'podlove_tab', FILTER_SANITIZE_STRING),
'delete',
$file_type->id
)
);
return sprintf('%1$s %2$s',
/*$1%s*/ $file_type->name,
/*$3%s*/ $this->row_actions( $actions )
);
}
function column_id( $file_type ) {
return $file_type->id;
}
function column_file_type( $file_type ) {
return $file_type->type;
}
function column_mime( $file_type ) {
return $file_type->mime_type;
}
function column_extension( $file_type ) {
return $file_type->extension;
}
function get_columns(){
return array(
'id' => __( 'ID', 'podlove-podcasting-plugin-for-wordpress' ),
'name' => __( 'Name', 'podlove-podcasting-plugin-for-wordpress' ),
'file_type' => __( 'File Type', 'podlove-podcasting-plugin-for-wordpress' ),
'mime' => __( 'MIME Type', 'podlove-podcasting-plugin-for-wordpress' ),
'extension' => __( 'Extension', 'podlove-podcasting-plugin-for-wordpress' )
);
}
function prepare_items() {
// number of items per page
$per_page = 1000;
// define column headers
$columns = $this->get_columns();
$hidden = array();
$sortable = $this->get_sortable_columns();
$this->_column_headers = array( $columns, $hidden, $sortable );
// retrieve data
// TODO select data for current page only
$data = \Podlove\Model\FileType::all();
// get current page
$current_page = $this->get_pagenum();
// get total items
$total_items = count( $data );
// extrage page for current page only
$data = array_slice( $data, ( ( $current_page - 1 ) * $per_page ) , $per_page );
// add items to table
$this->items = $data;
// register pagination options & calculations
$this->set_pagination_args( array(
'total_items' => $total_items,
'per_page' => $per_page,
'total_pages' => ceil( $total_items / $per_page )
) );
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,372 |
{"url":"http:\/\/mathematica.stackexchange.com\/questions?page=64&sort=faq","text":"# All Questions\n\n4k views\n\n### What is the cleanest way to prevent divide-by-zero warnings?\n\nIf I evaluate {1, 2, 3, 4}\/{5, 6, 0, 8}, obviously I get one warning: ...\n467 views\n\n### How can I make a cell that contains an image which can be shifted?\n\nWhen you're looking at some help pages, they may have cells containing images which can be shifted within the cell by left-clicking and dragging the mouse. Here is a sample: tutorial\/...\n395 views\n\n### Using MapIndexed only at certain elements of a list\n\nMapIndexed is a very handy built-in function. Suppose that I have the following list, called list: ...\n353 views\n\n### Does NRoots own an abstract counterpart? If not, can we write one?\n\nWe know when solving linear algebra equations, despite its abstract syntax, LinearSolve is much faster compared to Solve: ...\n417 views\n\n### Giving JLink Huge Memory by Default\n\nI have noticed that Mathematica makes use of Java by way of JLink, for example, in the extensions such as DatabaseLink and (I think) RLink. It also seems like Java is a first-class citizen for writing ...\n281 views\n\n### Boundary effect of Graphics3D object Tube[]\n\nI draw a Graphics3D object of two tubes: Graphics3D[{Tube[{{-0.2, -1, 0}, {-0.2, 1, 0}}, 0.05], Tube[{{0, 0, -1}, {0, 0, 1}}, 0.05]}, Boxed -> False] And ...\n354 views\n\n### Using Except with strings\n\nSuppose that I have the following sample list, called list: list = {\"bBb\", \"aAa\", \"aaa\"}; Now suppose I want to select the ...\n399 views\n\n### Selectively clear DownValues\n\nHow can I clear a subset of a symbol's DownValues ? 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I tried to incorporate ...\n1k views\n\n### How to estimate the matrix condition number in the 2-Norm?\n\nThe Mathematica documentation says it is possible to estimate the matrix condition number in norms 1, 2, and \u221e. But the 2-norm raises a message. This is an extract from reference documentation \"...\n6k views\n\n### Notation of partial derivative\n\nI want to write partial derivatives of functions with many arguments. Why is it that when I type f[x,y] ctrl+6 (0,1) it turns out to be bad syntax? The output of ...\n381 views\n\n### Implementation of tensor product formula\n\nFor the B\u00e9zier surface, which owns the following matrix definition: \\begin{align*} \\mathbf S(u,v)&=\\sum_{i=0}^m \\sum_{j=0}^n \\mathbf P_{i,j} B_{m,i}(u) B_{n,j}(v)\\\\ &=\\small \\begin{pmatrix}...\n416 views\n\n### How to provide your own example data for a custom package?\n\nA nice feature of the mathematica documentation is the copious use of ExampleData in illustrating functionality. 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I would need to specify the rotation of the ellipsoid ...\n249 views\n\n### 10.0.2 breaks a recursive \u201ctrie\u201d query\n\nThe 10.0.2 Dataset type system complication breaks a recursive trie constructor that worked in 0.1: ...\n536 views\n\n### Using NSolve in the complex plane\n\nI am trying to find numerical values for $\\Im\\ e^{\\frac{1}{2} +i\\ y} = \\Im\\ \\zeta(\\frac{1}{2} +i\\ y)$ for a given range. I have tried: ...\n800 views\n\n### Export list elements each to it's own page in a multi-paged PDF document\n\nI have a list, myList, of let's say 20 (different) objects myList = Table[\"object\", {20}] I know would like to export every ...\n159 views\n\n### Tracking Initialization\n\nHow to track Initialization step by step? Any tricky solution is ok. One condition, x should be ...\n895 views\n\n### Customize your FrameTicks in a MATLAB way\n\nThe figure below is made by MATLAB. However, the scientific form of numbers in FrameTicks that how MATLAB deals with is quite different from that of Mathematica. ...\n730 views\n\n### How can I find the vertexes of a Polygon?\n\nI made a polygon with the following vertexes : ...\n6k views\n\n### Mathematica sporadically crashes on open\n\nSometimes, when I open Mathematica, while the opening splash screen still says \"Initializing Kernel Connections\", Windows will tell me that \"Wolfram Mathematica 9 has stopped working\". This will ...\n197 views\n\n### Problem constructing Association tables with some operators\n\nfixed in 10.0.2 First, set up the example Dataset from the documentation: titanic = ExampleData[{\"Dataset\", \"Titanic\"}]; ...\n301 views\n\n### Dataset: cannot reproduce a basic example. A bug? [duplicate]\n\nJust to report: I try to reproduce the very first example for the Dataset from the Documentation Center: ...\n749 views\n\n### How can I display a multigraph with different colored edges?\n\nI'm new to Mathematica(version 9) and I'm trying to display a multigraph with different colored edges. Here's my code so far: ...\n480 views\n\n### Changing the atom color for plotting xyz-files\n\nMathematica takes the standard color code for atoms when plotting an *.xyz file, but I want to customize the color for the specific atoms. I really become desperate as I did not get it so far. ...\n435 views\n\n### Sorting function for non commuting bosons\n\nI am trying to write a sorting function which will sort expressions involving products of bosonic objects which do not commute. For example, I can have objects like $a,\\ a^\\dagger,\\ b,\\ b^\\dagger$ ...\n170 views\n\n### Passing an unevaluated part of an association to a function\n\nI'm making a UI widget that controls sound volume on a speaker using a slider, e.g.: ...\n316 views\n\nIs there a possibility to suppress the Mathematica Kernel taskbar tab when accessing it through .NETLink on Windows?\n2k views\n\n### Simplifying nested If statements\n\nConsider : list = {100,102,103,99,106,107,104,112} I want to consider, extract only the \"big moves\" of that time series. In this case the moves bigger then 5 ...\n208 views\n\n### Does PrivateFontOptions -> {\u201cOperatorSubstitution\u201d ->False} fail in version 10?\n\nPlease consider the following, version 10.0.1, OS X Mavericks: ...\n111 views\n\n### SparseArray output triggers \u201cunsafe dynamic content\u201d warning\n\nThe first basic example in the documentation of SparseArray: ...\n670 views\n\n### Finding the coefficient of a certain power in a generating function\n\nI want to compute $(t^1+\\dots +t^5)(t^2+\\dots+ t^6)(t^3+\\dots +t^9)$ and find the coefficient of $t^{15}$ for example. $t$ is an indeterminate Now in Maple this is simply as ...","date":"2016-07-26 10:20:31","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5559527277946472, \"perplexity\": 2759.5537426875}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-30\/segments\/1469257824757.8\/warc\/CC-MAIN-20160723071024-00202-ip-10-185-27-174.ec2.internal.warc.gz\"}"} | null | null |
Sabrina Bartlett, née le à Hammersmith (Londres), est une actrice britannique. Elle est connue pour son rôle de Sophie dans Da Vinci's Demons.
Biographie
Après plusieurs années dans la danse, Sabrina Bartlett entre à la dont elle sort diplômée en 2013.
En 2016, elle apparait dans l'épisode final de la sixième saison de la série Game of Thrones en tant que déguisement d'Arya Stark.
En 2017, elle tient le rôle de la Princesse Isabella dans la première saison de la série Knightfall diffusée sur History.
Pas chanteuse à l'origine, elle apprend le chant d'opéra pour le rôle de Siena Rosso dans La Chronique des Bridgerton en 2020. Malgré cela, elle est doublée dans la série par la chanteuse Rowan Pierce.
Elle est choisie pour incarner le rôle de Mariette Larkin dans le reboot de la série d'ITV, The Darling Buds of May en 2021.
Filmographie
Courts-métrages
2014 : The Crossing
2015 : Respite
Télévision
Séries télévisées
2014 : Doctor Who : Quayle's Ward
2014 : Holby City : Gabriella 'Gabi' Mendoza
2014 : Suspects : Hannah Stevenson
2014 : The Passing Bells : Katie
2015 : Da Vinci's Demons : Sophia
2015 : Inspecteur Barnaby : Tina Tyler
2015 : Poldark : Keren Smith
2016 : Game of Thrones : Handmaid
2017 : Versailles : Mathilde
2017 : Knightfall : Isabella / Princesse Isabella
2020 : La Chronique des Bridgerton : Siena Rosso
Téléfilms
2015 : Virtuoso : Birgit
Liens externes
Notes et références
Actrice anglaise de télévision
Naissance en septembre 1991
Naissance à Hammersmith | {
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} | 9,538 |
{"url":"https:\/\/cs.stackexchange.com\/questions\/84432\/number-of-2-3-trees-of-depth-4","text":"# Number of 2-3 trees of depth 4\n\nI have a task to find a number of 2-3 trees... I don't quite know what it means, I'm asked to: find the number of 2-3 trees of depth $4$? Is there a specific way to find number of 2-3 trees of depth $n$?\n\nThanks!\n\n\u2022 I think I would be best to ask for clarification from the one who set this question for you. Nov 25 '17 at 2:26\n\u2022 Draw them all. Count them.\n\u2013\u00a0Raphael\nNov 25 '17 at 11:33\n\u2022 I'm guessing the 2-3 trees mean B-trees with the parameters set to 2 and 3 here?\n\u2013\u00a0Raphael\nNov 25 '17 at 11:35\n\nDenote by $T_d$ the number of 2-3 trees of depth at most $d$, so $T_0 = 1$ (the tree consisting of a single node) and $T_1 = 3$ (the tree consisting of a a single node; a root with two children; a root with three children). More generally, a 2-3 tree of depth at most $d$ is either just a single node or consists of a root with two or three children, each of which is a 2-3 tree of depth at most $d-1$. Therefore $T_d = 1 + T_{d-1}^2 + T_{d-1}^3$.\nFinally, the number of 2-3 trees of depth exactly $d$ is $T_d - T_{d-1}$. For example, the number of 2-3 trees of depth exactly 1 is $T_1 - T_0 = 3-1 = 2$.\n\u2022 So, in case of B-trees, where all leaves should be at the same level, it seems your analysis would lead to the recursion $B_0=1$, $B_d = B_{d-1}^2 + B_{d-1}^3$. (?) Nov 25 '17 at 14:17","date":"2022-01-27 15:37:23","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.590320348739624, \"perplexity\": 226.21449796262053}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320305266.34\/warc\/CC-MAIN-20220127133107-20220127163107-00652.warc.gz\"}"} | null | null |
Fitness Classes Nottingham - Welcome to the directory of Nottingham fitness classes and recommended exercise classes in Nottingham. It features fitness classes in Nottingham and Sherwood, and includes maps and photos of Nottingham exercise classes who offer fitness training and cardiovascular fitness. Find contact details and reviews of your nearest exercise class or fitness class in Nottingham and add your own review. Do you want to advertise a exercise class in Nottingham? Advertise your fitness training business on the Nottingham Fitness Classes Directory – IT'S FREE! | {
"redpajama_set_name": "RedPajamaC4"
} | 6,587 |
{"url":"https:\/\/en.m.wikipedia.org\/wiki\/Semicolon","text":"# Semicolon\n\nThe semicolon or semi-colon[1] ; is a symbol commonly used as orthographic punctuation. In the English language, a semicolon is most commonly used to link (in a single sentence) two independent clauses that are closely related in thought. When a semicolon joins two or more ideas in one sentence, those ideas are then given equal rank.[2] Semicolons can also be used in place of commas to separate the items in a list, particularly when the elements of that list contain commas.[3]\n\n;\nSemicolon\nIn\u00a0UnicodeU+003B ; SEMICOLON (;)\n\u061b\nArabic semi colon Ethiopic semicolon Bamum semicolon\n\nThe semicolon is one of the least understood of the standard marks, and so it is not as frequently used by many English speakers.[4]\n\nIn the QWERTY keyboard layout, the semicolon resides in the unshifted homerow beneath the little finger of the right hand and has become widely used in programming languages as a statement separator or terminator.[5]\n\n## History\n\nDe \u00c6tna. 1496 print by Aldine Press.\n\nIn 1496,[a] the semicolon ; is attested in Pietro Bembo's book De Aetna\u00a0[it] printed by Aldo Manuzio.[9][6] The punctuation also appears in later writings of Bembo. Moreover, it is used in 1507 by Bartolomeo Sanvito, who was close to Manuzio's circle.[10]:\u200a49\n\nIn 1561, Manuzio's grandson, also called Aldo Manuzio, explains the semicolon's use with several examples in Orthographiae ratio. In particular, Manuzio motivates the need for punctuation (interpung\u014d) to divide (distingu\u014d) sentences, and thereby make them understandable. The comma, semicolon, colon, and period are seen as steps, ascending from low to high; the semicolon thereby being an intermediate value between the comma , and colon :. Here are four examples used in the book to illustrate this:[11][10]:\u200a49\n\nPublica, privata; sacra, profana; tua, aliena.[b]\nPublic, private; sacred, profane; yours, another's.\n\nRatio docet, si adversa fortuna sit, nimium dolendum non esse; si secunda, moderate laetandum.[c]\nReason teaches, if fortune is adverse, not to complain too much; if favorable, to rejoice in moderation.\n\nTu, quid divitiae valeant, libenter spectas; quid virtus, non item.[d]\nYou, what riches are worth, gladly consider; what virtue (is worth), not so much.\n\nEtsi ea perturbatio est omnium rerum, ut suae quemque fortunae maxime paeniteat; nemoque sit, quin ubivis, quam ibi, ubi est, esse malit: tamen mihi dubium non est, quin hoc tempore bono viro, Romae esse, miserrimum sit.[e]\nAlthough it is a universal confusion of affairs(,) such that everyone regrets their own fate above all others; and there is no one, who would not rather anywhere else in the world, than there, where he is, prefer to be: yet I have no doubt, at the present time for an honest man, to be in Rome, is the worst form of misery.\n\nAround 1580, Henry Denham starts using the semicolon \"with propriety\" for English texts and more widespread usage picks up in the next decades.[10]:\u200a52\u200a[12]\n\nAround 1640,[f] in Ben Jonson's book The English Grammar, the character ; is described as \"somewhat a longer breath\" compared to the comma. The aim of the breathing, according to Jonson, is to aid understanding.[14][g]\n\nIn 1644, in Richard Hodges' The English Primrose, it is written:[16]\n\nAt a comma, stop a little; [...] At a semi-colon, somewhat more; [...] At a colon, a little more than the former; [...] At a period, make a full stop; [...]\n\nIn 1762, in Robert Lowth's A Short Introduction to English Grammar, a parallel is drawn between punctuation marks and rest in music:[17]\n\nThe Period is a pause in quantity or duration double of the Colon; the Colon is double of the Semicolon; and the Semicolon is double of the Comma. So that they are in the same proportion to one another as the Sembrief, the Minim, the Crotchet, and the Quaver, in Music.\n\nIn 1798, in Lindley Murray's English Grammar, the semicolon is introduced as follows:[18]\n\nThe Semicolon is used for dividing a compound sentence into two or more parts, not so closely connected as those which are separated by a comma, nor yet so little dependent on each other, as those which are distinguished by a colon.\n\nThe semicolon is sometimes used, when the preceding member of the sentence does not of itself give a complete sense, but depends on the following clause; and sometimes when the sense member would be complete without the concluding one; [...]\n\n## Natural languages\n\n### English\n\nAlthough terminal marks (i.e. full stops, exclamation marks, and question marks) indicate the end of a sentence, the comma, semicolon, and colon are normally sentence-internal, making them secondary boundary marks. The semicolon falls between terminal marks and the comma; its strength is equal to that of the colon.[19]\n\nThe plural of semicolon in English is semicola or semicolons.\n\nThe most common use of the semicolon is to join two independent clauses without using a conjunction like \"and\".[20] Semicolons are followed by a lower case letter, unless that letter would ordinarily be capitalised mid-sentence (e.g., the word \"I\", acronyms\/initialisms, or proper nouns). In older English printed texts, colons and semicolons are offset from the preceding word by a non-breaking space, a convention still current in present-day continental French texts. Ideally, the space is less wide than the inter-word spaces. Some guides recommend separation by a hair space.[21] Modern style guides recommend no space before them and one space after. They also typically recommend placing semicolons outside ending quotation marks, although this was not always the case. For example, the first edition of The Chicago Manual of Style (1906) recommended placing the semicolon inside ending quotation marks.[22]\n\nApplications of the semicolon in English include:\n\n\u2022 Between items in a series or listing when the items contain internal punctuation, especially parenthetic commas, where the semicolons function as the serial commas for the entire series or listing. The semicolon divides the items on the list from each other, to avoid having a jumble of commas with differing functions which could cause confusion for the reader. This is sometimes called the \"super comma\" function of the semicolon:\n\u2022 The people present were Jamie, a man from New Zealand; John, the milkman's son; and George, a gaunt kind of man with no friends.\n\u2022 Several fast food restaurants can be found within the following cities: London, England; Paris, France; Dublin, Ireland; and Madrid, Spain.\n\u2022 Here are three examples of familiar sequences: one, two, and three; a, b, and c; first, second, and third.\n\u2022 (Fig. 8; see also plates in Harley 1941, 1950; Schwab 1947).\n\u2022 Between closely related independent clauses not conjoined with a coordinating conjunction, when the two clauses are balanced, opposed or contradictory:[23]\n\u2022 My wife would like tea; I would prefer coffee.\n\u2022 I went to the basketball court; I was told it was closed for cleaning.\n\u2022 I told Kate she's running for the hills; I wonder if she knew I was joking.\n\u2022 In rare instances, when a comma replaces a period (full stop) in a quotation, or when a quotation otherwise links two independent sentences:\n\u2022 \"I have no use for this,\" he said; \"you are welcome to it.\"\n\u2022 \"Is this your book?\" she asked; \"I found it on the floor.\"\n\nIn a list or sequence, if even one item needs its own internal comma, use of the semicolon as the separator throughout that list is justified, as shown by this example from the California Penal Code:\n\nA crime or public offense is an act committed or omitted in violation of a law forbidding or commanding it, and to which is annexed, upon conviction, either of the following punishments:\n\n1. Death;\n2. Imprisonment;\n3. Fine;\n4. Removal from office; or,\n5. Disqualification to hold and enjoy any office of honor, trust, or profit in this State.[24]\n\n### Arabic\n\nIn Arabic, the semicolon is called fasila manqoota (Arabic: \u0641\u0627\u0635\u0644\u0629 \u0645\u0646\u0642\u0648\u0637\u0629) which means literally \"a dotted comma\", and is written inverted \u061b. In Arabic, the semicolon has several uses:\n\n\u2022 It can be used between two phrases, in which the first phrase causes the second.\n\u2022 Example: \"He played a lot; so, his clothes became dirty\". (Arabic: \u0644\u064e\u0639\u0650\u0628\u064e \u0643\u064e\u062b\u0650\u064a\u0631\u064b\u0627\u061b \u0641\u064e\u0671\u062a\u064e\u0651\u0633\u064e\u062e\u064e\u062a\u0652 \u0645\u064e\u0644\u064e\u0627\u0628\u0650\u0633\u064f\u0647\u064f.)\n\u2022 It can be used between two phrases, where the second is a reason for the first.\n\u2022 Example: \"Your sister did not get high marks; she didn't study\". (Arabic: \u0644\u0645 \u062a\u062d\u0642\u0642 \u0623\u062e\u062a\u0643 \u062f\u0631\u062c\u0627\u062a \u0639\u0627\u0644\u064a\u0629\u061b \u0644\u0623\u0646\u0647\u0627 \u0644\u0645 \u062a\u062f\u0631\u0633 .)\n\n### Greek, Church Slavonic\n\nIn Greek and Church Slavonic, the question mark looks exactly the way a semicolon looks in English, similar to the question mark used in Latin.[25][26] To indicate a long pause or to separate sections that already contain commas (the semicolon's purposes in English), Greek uses, but extremely rarely, the Greek: \u03ac\u03bd\u03c9 \u03c4\u03b5\u03bb\u03b5\u03af\u03b1, romanized:\u00a0\u00e1no tele\u00eda, lit.'up dot', an Interpunct \u00b7.\n\nChurch Slavonic with a question mark: \u0433\u0434\u0463 \u0454\u0441\u0442\u044c \u0440\u043e\u0436\u0434e\u0439\u0441\u0467 \u0446\u0430\u0440\u044c \u0457\u0443\u0434e\u0439\u0441\u043a\u0456\u0439; (Where is the one who is born king of the Jews? \u2013 Matthew 2:1)\n\nGreek with a question mark: \u03a4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03b4\u03b9\u03b1\u03c3\u03cd\u03bd\u03b4\u03b5\u03c3\u03b7; (What is an interpunct?)\n\n### French\n\nIn French, a semicolon (point-virgule, literally \"dot-comma\") is a separation between two full sentences, used where neither a colon nor a comma would be appropriate. The phrase following a semicolon has to be an independent clause, related to the previous one but not explaining it. (When the second clause explains the first one, French consistently uses a colon.)\n\nThe dash character is used in French writing too, but not as widely as the semicolon. Usage of these devices (semicolon and dash) varies from author to author.\n\n## Literature\n\nJust as there are writers who worship the semicolon, there are other high stylists who dismiss it \u2014 who label it, if you please, middle-class.\n\nLynne Truss, Eats, Shoots, and Leaves[27]\n\nSome authors have avoided and rejected the usage of the semicolon throughout their works. Lynne Truss stated:\n\nSamuel Beckett spliced his way merrily through such novels as Molloy and Malone Dies, thumbing his nose at the semicolon all the way. James Joyce preferred the colon, as he thought it was more authentically classical. P. G. Wodehouse did an effortlessly marvelous job without it, George Orwell tried to avoid the semicolon completely in Coming Up for Air (1939), Martin Amis included just one semicolon in Money (1984), and Umberto Eco was congratulated by an academic reader for using zero semicolons in The Name of the Rose (1983).[28]\n\nIn response to Truss, Ben Macintyre, a columnist in The Times, wrote:\n\nAmericans have long regarded the semi-colon with suspicion, as a genteel, self-conscious, neither-one-thing-nor-the other sort of punctuation mark, with neither the butchness of a full colon nor the flighty promiscuity of the comma. Hemingway, Chandler and Stephen King wouldn't be seen dead in a ditch with a semi-colon (though Truman Capote might). Real men, goes the unwritten rule of American punctuation, don't use semi-colons.[29]\n\nSemicolon use in British fiction has declined by 25% from 1991 to 2021.[30]\n\n## Character encoding\n\nThe semicolon has an assigned value in computer character encoding standards. In ASCII it is encoded as 0x3B, in EBCDIC it is encoded as 0x5E, and in Unicode it is encoded as U+003B.\n\nUnicode contains encoding for several semicolon characters:\n\n\u2022 U+003B ; SEMICOLON \u2013 inherited from ASCII\n\u2022 U+037E \u037e GREEK QUESTION MARK\n\u2022 U+061B \u061b ARABIC SEMICOLONArabic script\n\u2022 U+1364 ETHIOPIC SEMICOLONGe\u02bdez script\n\u2022 U+204F REVERSED SEMICOLON \u2013 used in old writing systems, such as Hungarian Runic[31] and Sindhi language\n\u2022 U+236E APL FUNCTIONAL SYMBOL SEMICOLON UNDERBAR \u2013 used in the APL programming language\n\u2022 U+2E35 TURNED SEMICOLON \u2013 \"indicates sudden glottal closure\"\n\u2022 U+A6F6 BAMUM SEMICOLONBamum script\n\u2022 U+FE14 PRESENTATION FORM FOR VERTICAL SEMICOLON \u2013 determines orientation when wide-character scripts are written vertically instead of horizontally\n\u2022 U+FE54 SMALL SEMICOLONSmall Form Variants are for compatibility with Chinese National Standard CNS 11643\n\u2022 U+FF1B FULLWIDTH SEMICOLON \u2013 for use in wide-character scripts such as kanji\n\u2022 U+E003B TAG SEMICOLON \u2013 deprecated tags block\n\n## Computing\n\n### Programming\n\nIn computer programming, the semicolon is often used to separate multiple statements (for example, in Perl,[32] Pascal,[33] and SQL;[34] see Pascal: Semicolons as statement separators). In other languages, semicolons are called terminators[5] and are required after every statement (such as in PL\/I,[35] Java,[36] and the C[37] family). Today semicolons as terminators has largely won out, but this was a divisive issue in programming languages from the 1960s into the 1980s.[38] An influential and frequently cited study in this debate was Gannon & Horning (1975), which concluded strongly in favor of semicolon as a terminator: \"The most important [result] was that having a semicolon as a statement terminator was better than having a semicolon as a statement separator.\"[39] The study has been criticized as flawed by proponents of semicolon as a separator,[40] due to participants being familiar with a semicolon-as-terminator language and unrealistically strict grammar. Nevertheless, the debate ended in favor of semicolon as terminator. Therefore, semicolon provides structure to the programming language.\n\nSemicolons are optional in a number of languages, including BCPL,[41] Python,[42] R,[43] Eiffel,[44] and Go,[45] meaning that they are part of the formal grammar for the language, but can be inferred in many or all contexts (e.g. by end of line that ends a statement, as in Go and R). As languages can be designed without them, semicolons are considered an unnecessary nuisance by some.[40]\n\nThe use of semicolons in control-flow structures and blocks of code is varied\u00a0\u2013 semicolons are generally omitted after a closing brace, but included for a single statement branch of a control structure (the \"then\" clause), except in Pascal, where a semicolon terminates the entire if...then...else clause (to avoid dangling else) and thus is not allowed between a \"then\" and the corresponding \"else\", as this causes unnesting.\n\nThis use originates with ALGOL 60[46] and falls between the comma ,\u00a0\u2013 used as a list separator\u00a0\u2013 and the period\/full stop .\u00a0\u2013 used to mark the end of the program. The semicolon, as a mark separating statements, corresponds to the ordinary English usage of separating independent clauses and gives the entire program the gross syntax of a single ordinary sentence. Of these other characters, whereas commas have continued to be widely used in programming for lists (and rare other uses, such as the comma operator that separates expressions in C), they are rarely used otherwise, and the period as the end of the program has fallen out of use. The last major use of the comma, semicolon, and period hierarchy is in Erlang (1986), where commas separate expressions; semicolons separate clauses, both for control flow and for function clauses; and periods terminate statements, such as function definitions or module attributes, not the entire program. Drawbacks of having multiple different separators or terminators (compared to a single terminator and single grouping, as in semicolon-and-braces) include mental overhead in selecting punctuation, and overhead in rearranging code, as this requires not only moving lines around, but also updating the punctuation.[47]\n\nIn some cases the distinction between a separator and a terminator is strong, such as early versions of Pascal, where a final semicolon yields a syntax error. In other cases a final semicolon is treated either as optional syntax or as being followed by a null statement, which is either ignored or treated as a NOP (no operation or null command); compare trailing commas in lists. In some cases a blank statement is allowed, allowing a sequence of semicolons or the use of a semicolon by itself as the body of a control-flow structure. For example, a blank statement (a semicolon by itself) stands for a NOP in C\/C++, which is useful in busy waiting synchronization loops.\n\nAPL uses semicolons to separate declarations of local variables[48]:\u200a61\u200a and to separate axes when indexing multidimensional arrays, for example, `matrix[2;3]`.[48]:\u200a220,\u200a275\n\nOther languages (for instance, some assembly languages[49] and LISP[50] dialects, CONFIG.SYS and INI files) use semicolons to mark the beginning of comments.\n\nExample C code:\n\n```int main() {\nint x, y;\nx = 1; y = 2;\nprintf(\"X + Y = %d\", x + y);\nreturn 0;\n}\n```\n\nOr in JavaScript:\n\n```var x = 1; var y = 2;\nalert(\"X + Y = \" + (x + y));\n```\n\nConventionally, in many languages, each statement is written on a separate line, but this is not typically a requirement of the language. In the above examples, two statements are placed on the same line; this is legal, because the semicolon separates the two statements. Thus programming languages like Java, the C family, Javascript etc. use semicolons to obtain a proper structure in the respective languages.\n\n### Data\n\nThe semicolon is often used to separate elements of a string of text. For example, multiple e-mail addresses in the \"To\" field in some e-mail clients have to be delimited by a semicolon.\n\nIn Microsoft Excel, the semicolon is used as a list separator, especially in cases where the decimal separator is a comma, such as ` 0,32; 3,14; 4,50`, instead of `0.32, 3.14, 4.50`.\n\nIn Lua,[51] semicolons or commas can be used to separate table elements.\n\nIn MATLAB and GNU Octave,[52] the semicolon can be used as a row separator when defining a vector or matrix (whereas a comma separates the columns within a row of a vector or matrix) or to execute a command silently, without displaying the resulting output value in the console.\n\nIn HTML, a semicolon is used to terminate a character entity reference, either named or numeric.[53] The declarations of a style attribute in Cascading Style Sheets (CSS) are separated and terminated with semicolons.[54]\n\nThe file system of RSX-11 and OpenVMS, Files-11, uses semicolons to indicate a file's version number.[55] The semicolon is permitted in long filenames in the Microsoft Windows file systems NTFS and VFAT, but not in its short names.[56]\n\nIn some delimiter-separated values file formats, the semicolon is used as the separator character, as an alternative to comma-separated values.\n\n## Mathematics\n\nIn the argument list of a mathematical function ${\\displaystyle f(x_{1},x_{2},\\dots ;a_{1},a_{2},\\dots )}$\u00a0, a semicolon may be used to separate variables from fixed parameters.[citation needed]\n\nIn differential geometry, a semicolon preceding an index is used to indicate the covariant derivative of a function with respect to the coordinate associated with that index.\n\nIn the calculus of relations, the semicolon is used in infix notation for the composition of relations: ${\\displaystyle A;B\\ =\\ \\{(x,z):\\exists y\\ \\ xAy\\ \\land \\ yBz\\}.}$\n\nThe ; Humphrey point is sometimes used as the \"decimal point\" in duodecimal numbers: 54;612 equals 64.510.\n\n## Other uses\n\nThe semicolon is commonly used as parts of emoticons, in order to indicate winking or crying, as in `;)` and `;_;`.\n\nProject Semicolon is the name of a faith-based anti-suicide initiative (since the semicolon continues a sentence rather than ending it) which has led to the punctuation mark becoming a highly symbolic and popular tattoo,[57] which is most commonly done on the wrist.\n\n## Notes\n\n1. ^ Note: this is February 1495 by the Venetian calendar, or MVD in Roman numerals. Several texts mention the earliest printing year as 1494 without any attested source. This may be due to a misunderstanding: the text De Aetna itself recalls a conversation from September 1494 (Gregorian calendar); the earliest attested printing occurred in February 1496 (Gregorian calendar).[6][7] See also the colophon at the end of the book: impressum venetiis in aedibus aldi romani mense februario anno M.V.D. (\"printed in Venice at the house of Aldo in February 1495 [m.v.]\").[8]\n2. ^ Manuzio notes: using just commas here is not sufficient, and using a colon instead of a semicolon would slow down the sentence too much.[11]\n3. ^ Manuzio notes: if there were a comma after non esse, then the sentence would be as if rushed forward head-first; however, since the sentence is in two parts, it must stop for a little bit before continuing. Moreover, he says, we can't use a colon, since Ratio docet governs both parts of the sentence equally, not just the first.[11]\n4. ^ Manuzio shows another example of where we need a semicolon instead of a colon; Tu and spectas govern both parts of the sentence equally.[11]\n5. ^ Manuzio notes, this passage is taken from Cicero's letter to Torquatus (see Epistulae ad Familiares). He uses this example, he says, to show situations where both a semicolon and a colon is needed for division.[11]\n6. ^ According to the British Library, the book was \"written in 1623 and lost in a fire, but rewritten and published after Jonson's death\".[13]\n7. ^ The 1640 version of the text calls the character ; a subdistinction,[15] the 1692 version names it a semicolon.[14] Moreover, the order of the comma and semicolon seem to have been reversed by mistake in the 1640 version. Thus, the comma is mistakenly described as having a longer breathe than the semicolon.[15]\n\n## References\n\n1. ^ \"Learning English \u2013 BBC World Service\". bbc.co.uk.\n2. ^ \"Using Semicolons\". The Writing Center. Retrieved 2020-11-08. A semicolon is most commonly used to link (in a single sentence) two independent clauses that are closely related in thought. When a semicolon is used to join two or more ideas (parts) in a sentence, those ideas are then given equal position or rank.\n3. ^ Scribendi.com. \"Learning English Grammar: How to Correctly Use a Semicolon\". Archived from the original on 8 August 2014. Retrieved 30 July 2014.\n4. ^ \"For Love of the Semicolon\". Insights to English. Retrieved 2020-11-09.\n5. ^ a b M\u00f6ssenb\u00f6ck, H. \"Introduction to C# \u2013 The new language for Microsoft .NET\" (PDF) (subtitle: Statements). Linz: University of Linz. p.\u00a034. Retrieved 2011-07-29. Empty statement:\u00a0; \/\/\u00a0; is a terminator, not a separator.\n6. ^ a b Kidwell, Carol (2004). Pietro Bembo: Lover, Linguist, Cardinal. Montreal: McGill-Queen's University Press. p.\u00a012. ISBN\u00a0978-0-7735-7192-1.\n7. ^ Nuvoloni, Laura (2012-08-09). \"Pietro Bembo and the University Library copy of the De Aetna of 1496\". Cambridge University Library. Retrieved 2022-10-26.\n8. ^ Mariano, Bianca Maria (1991). \"Il \"de Aetna\" Di P. Bembo E Le Varianti Dell'edizione 1530\". Aevum (in Italian). 65 (3): 441\u2013452. JSTOR\u00a020858679.\n9. ^ Bembo, Pietro (1496). De Aetna (in Latin). Venice: Aldus Manutius.\n10. ^ a b c Parkes, Malcolm Beckwith (1993). Pause and Effect: An Introduction to the History of Punctuation in the West. University of California Press. ISBN\u00a00-520-07941-8.\n11. Aldo, Manuzio (1561). Orthographiae ratio (in Latin). Venice. p.\u00a052.\n12. ^\n13. ^ \"Ben Jonson's The English Grammar\". British Library. Retrieved 2022-10-22.\n14. ^ a b Jonson, Ben (1692). The Works of Ben Jonson, which were Formerly Printed in Two Volumes, are Now Reprinted in One. (3rd folio). London: Printed by Thomas Hodgkin, for H[enry] Herringman, E. Brewster, T. Bassett, R[ichard] Chiswell, M. Wotton, G. Conyers. p.\u00a0690.\n15. ^ a b Jonson, Ben (1640). The English Grammar. p.\u00a083.\n16. ^ Hodges, Richard (1969). The English Primrose, 1644. Menston, (Yorks.) Scolar P. p.\u00a0N3. ISBN\u00a0978-0-85417-116-3.\n17. ^ Lowth, Robert (1762). A Short Introduction to English Grammar: with Critical Notes ... London: J. Hughs. p.\u00a0158.\n18. ^ Murray, Lindley (1798). \"Part IV: Of Punctuation. Chapter II: Of the Semicolon\". English Grammar, Adapted to Different Classes of Learners. With an Appendix, Containing Rules and Observations, For Assisting The More Advanced Students to Write With Perspicuity and Accuracy (4th\u00a0ed.). London: Wilson, Spence, and Mawman. p.\u00a0226.\n19. ^ The Cambridge Grammar of the English Language, Chapter 19, \u00a7\u00a07.\n20. ^ \"Semicolon\". 7 January 2016.\n21. ^ One book\/five ways\u00a0: the publishing procedures of five university presses. USA: The Association of American University Presses. 1978. p.\u00a0276 \u2013 via Internet Archive. COLONS AND SEMI-COLONS Should be preceded by a hair space only.\n22. ^ Spencer, Dave (15 February 2011). \"Review\". glyphic.design (review of The Chicago Manual of Style, 16th edition). Retrieved 28 August 2019.\n23. ^ \"Purdue OWL: Commas\". purdue.edu.\n24. ^ \"California Penal Code \u00a7 15\". California Office of Legislative Counsel. 1872. Retrieved 23 September 2020.\n25. ^ Paine, Stephen W. (1961). Beginning Greek a Functional Approach. Oxford University Press. p.\u00a013.\n26. ^ Truss 2004, p.\u00a0111\n27. ^ Truss 2004, p.\u00a0107.\n28. ^ Truss 2004, pp.\u00a088, 108\u2013109.\n29. ^ MacIntyre, Ben (19 November 2005). \"Is it worth busting your (Lynne) truss over a comma with a hat on?\". The Times. London.\n30. ^ Will Lloyd (24 November 2021). \"The melancholy decline of the semicolon\". UnHerd.\n31. ^ Everson, Michael; Szelp, Andr\u00e9 Szabolcs (2009-07-22). \"Proposal for encoding generic punctuation used with the Hungarian Runic script\" (PDF).\n32. ^ \"perlsyn - Perl syntax\". Perldoc Browser (subtitle: Simple Statements). 28 May 2022. Archived from the original on 12 November 2022.\n33. ^ \"Free Pascal wiki\". Free Pascal and Lazarus Wiki. 4 July 2021. Archived from the original on 22 October 2021.\n34. ^ \"Documentation: 15: 4.1. Lexical Structure\". PostgreSQL. Archived from the original on 21 October 2022. Retrieved 28 November 2022.\n35. ^ IBM System\/360 Operating System PL\/I Language Specifications (PDF). IBM. July 1966. p.\u00a014. C28-6571-3. Retrieved 2021-01-04. A statement, which is a string of characters, is always terminated by the special character, semicolon.\n36. ^ \"The Java\u00ae Language Specification - Chapter 14. Blocks, Statements, and Patterns\". Oracle Help Center. Oracle. 23 February 2022. Archived from the original on 8 June 2022.\n37. ^ ISO\/IEC JTC1\/SC22\/WG14 (7 September 2007). \"ISO\/IEC 9899:TC3 \u2014 Programming languages \u2014 C\" (PDF). p.\u00a0132. Archived from the original on 16 April 2020.\n38. ^ Meyer, Bertrand (1997). Object-Oriented Software Construction (Second\u00a0ed.). p.\u00a0897. ISBN\u00a0978-0-13-629155-8.\n39. ^ Perlis, Alan J; Sayward, Frederick; Shaw, Mary (1981). Software Metrics: An Analysis and Evaluation. MIT Press. p.\u00a0324. ISBN\u00a0978-0-26-216083-4.\n40. ^ a b Meyer, Bertrand. Principles of language design and evolution (PDF) (Technical report). Archived (PDF) from the original on 9 January 2012.\n41. ^ Richards, Martin (21 July 1967). \"The BCPL Reference Manual\" (PDF). Dennis M Ritchie's personal page. Archived (PDF) from the original on 11 June 2015.\n42. ^ \"The Python Language Reference - 7. Simple statements\". Python Software Foundation. 22 January 2016. Archived from the original on 7 October 2017.\n43. ^ \"R Language Definition\". The Comprehensive R Archive Network (subtitle: 3.2 Control structures). 31 October 2022. Archived from the original on 21 November 2022.\n44. ^ \"Eiffel: The Syntax\". Gobo Eiffel Project. 2 September 2019. Archived from the original on 8 August 2022.\n45. ^ \"The Go Programming Language Specification\". The Go Programming Language (subtitle: Tokens). 29 June 2022. Archived from the original on 27 November 2022.\n46. ^ \"Revised Report on the Algorithmic Language Algol 60\". mass:werk (subtitle: 4. Statements). 1 January 1963. Archived from the original on 25 June 2007.\n47. ^ Katz, Damien (9 March 2008). \"What Sucks About Erlang\". Archived from the original on 14 June 2013.\n48. ^ a b \"Dyalog APL Language Reference Manual\" (PDF). Dyalog Limited. April 2011. Archived (PDF) from the original on 4 March 2016. Retrieved 14 February 2012.\n49. ^ \"Chapter 3: The NASM Language\". NASM - The Netwide Assembler. 28 August 2020. Archived from the original on 25 October 2022.\n50. ^ \"Common Lisp HyperSpec\" (subtitle: Section 2.4.4). LispWorks. Archived from the original on 18 June 2013.\n51. ^ \"Lua 5.4 Reference Manual\" (subtitle: 3.4.9 \u2013 Table Constructors). 13 January 2022. Archived from the original on 1 October 2022.\n52. ^ \"GNU Octave (version 7.3.0)\" (subtitle: 1.2.2 Creating a Matrix). GNU Octave. 2 November 2022. Archived from the original on 28 November 2022.\n53. ^ \"Entity\". MDN. 21 September 2022. Archived from the original on 25 November 2022.\n54. ^ \"CSS Syntax\". W3Schools. Archived from the original on 25 February 2021.\n55. ^ \"Ext File Specs, Using File Specification Differences\". VMS Help. Consiglio Nazionale delle Ricerche. Archived from the original on 7 August 2020.\n56. ^ \"Filename - Win32 Apps\". Microsoft Learn. Microsoft. 31 May 2018. Archived from the original on 28 August 2019.\n57. ^ Steyer, Carly (7 July 2016). \"RELIGION Global Semicolon Tattoo Trend Is A Sign Of Strength Among Faithful Individuals Dealing With Mental Health Problems\". The Huffington Post. Retrieved 2016-10-02.\n\nSources\n\n\u2022 Gannon, J. D.; Horning, J. J. (1975). The impact of language design on the production of reliable software. Proceedings of the international conference on Reliable software. pp.\u00a010\u201322. doi:10.1145\/800027.808420.\n\u2022 Truss, Lynn (2004). Eats, Shoots & Leaves: The Zero Tolerance Approach to Punctuation. New York: Gotham Books. ISBN\u00a01-59240-087-6.","date":"2022-12-09 03:56:14","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 2, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6771780252456665, \"perplexity\": 8313.143775197543}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-49\/segments\/1669446711376.47\/warc\/CC-MAIN-20221209011720-20221209041720-00570.warc.gz\"}"} | null | null |
Heike Brehmer (born 5 November 1962) is a German politician of the Christian Democratic Union (CDU) who has been serving as member of the Bundestag and former district counselor of the district of Aschersleben-Staßfurt.
Early life and career
Heike Brehmer was born in Staßfurt. She completed an apprenticeship in gastronomy and afterwards an apprenticeship as a merchant for agriculture and food industry (Land- und Nahrungsgüterwirtschaft, LNG). She later qualified to a business administration graduate at a Fachhochschule (FH).
From 1990 to 1994, she was a municipality director in Tarthun and from 1994 to 2003 head of the joint administrative office Bördeaue.
From 2008 to 2009, she was managing director of the Studentenwerk Magdeburg.
Political career
Career in local politics
From 1990 to 2003, Brehmer was a member of the Aschersleben-Staßfurt district council, which she chaired between 1999 and 2003. On 27 April 2003, she was elected in the run-off election 61.4% as the successor of Thomas Leimbach district administrator of the district of Aschersleben-Staßfurt. On 22 April 2007, Heike Brehmer joined as CDU candidate for the district council post of the new Salzlandkreis district. In the run-off election on 6 May 2007, she lost against the candidate of the SPD, Ulrich Gerstner.
Member of Parliament, 2009–present
In June 2009, Brehmer was nominated by the CDU as the direct candidate for the constituency of Harz (constituency 69). In the 2009 federal election, she was able to move as directly elected representative for the first time in the German Bundestag, receiving 33% of the first votes. In 2013 federal election she was re-elected with 46%, in 2017 federal election with 36.4%.
In parliament, Brehmer has served on the Committee on Labor and Social Affairs (2009–2013), the Committee on Tourism (2014–2017, since 2021), and the Committee on Petitions (2018–2021). She chaired the Committee on Tourism (2014–2017) and the Committee on Petitions (2018–2021). In 2014, she joined the parliament's Council of Elders, which – among other duties – determines daily legislative agenda items and assigning committee chairpersons based on party representation.
Within her parliamentary group, Brehmer has been the deputy chairman of the workers' group and led the group of CDU parliamentarians from Saxony-Anhalt from 2013 until 2021.
Since 2009, Brehmer has been a member of the Federal Council of the CDU Association for Municipal Policies (Kommunalpolitische Vereinigung, KPV), since 2010 also deputy chairman of the CDU Saxony-Anhalt under the leadership of chairman Holger Stahlknecht.
Other activities
Konrad Adenauer Foundation, Member (since 2022)
Federal Agency for Civic Education (BPB), Member of the Board of Trustees (since 2022)
Political positions
In June 2017, Brehmer voted against Germany's introduction of same-sex marriage.
Personal life
Brehmer is of Roman Catholic denomination, divorced and has two children.
References
1962 births
Living people
Members of the Bundestag for Saxony-Anhalt
People from Staßfurt
Members of the Bundestag 2021–2025
Members of the Bundestag 2017–2021
Members of the Bundestag 2013–2017
Members of the Bundestag 2009–2013
Members of the Bundestag for the Christian Democratic Union of Germany
Recipients of the Cross of the Order of Merit of the Federal Republic of Germany | {
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ComicListNews
X-Statix and The X-Cellent return this March
by Charles LePage December 17, 2022
written by Charles LePage December 17, 2022
Peter Milligan, Michael Allred, and Laura Allred are returning with more mutant madness in THE X-CELLENT #1.
The press release follows:
Media Release — This last year saw the long-awaited return of Peter Milligan, Michael Allred, and Laura Allred's X-STATIX saga in X-CELLENT, and Marvel Comics is thrilled to announce that there's more to come this March! The trio of comic superstars will reunite for even more mutant celebrity exploits in the pages of a five-issue sequel series, THE X-CELLENT. The series will continue the offbeat, thrilling adventures of X-STATIX and further explore their newly-introduced supervillain counterparts known as the X-CELLENT.
A breakout hit of the 2000s, X-STATIX stunned fans with its dark wit and unique take on Marvel super heroics. More timely and relevant than ever, fans can once again visit this fascinating and strange corner of the Marvel Universe and all its fan-favorite characters including U-Go Girl, Zeitgeist, Doop, and more!
Your favorite celebrity super villains are back! Zeitgeist is still on a mission to achieve social media godhood, no matter who he has to kill! But the spotlight won't be big enough when the next generation of the X-Statix drop in!
"There are few finer feelings in comics than teaming up again with Mike Allred and creating THE X-CELLENT, the latest chapter of the strange mutant journey that began with X-STATIX!" Milligan said.
"Playing with Peter Milligan and our beloved Marvel Mutant creations is always pure comic book bliss!" Allred added. "And this next arc of THE X-CELLENT really kicks everything up to ELEVEN!"
Be there for the next chapter of Peter Milligan, Michael Allred, and Laura Allred's mutant celebrity saga this March! For more information, visit Marvel.com.
THE X-CELLENT #1 (OF 5)
Written by PETER MILLIGAN
Art and Cover by MICHAEL ALLRED & LAURA ALLRED
On Sale 3/15
To find a comic shop near you, visit www.comicshoplocator.com
About Marvel Entertainment
Marvel Entertainment, LLC, a wholly-owned subsidiary of The Walt Disney Company, is one of the world's most prominent character-based entertainment companies, built on a proven library of more than 8,000 characters featured in a variety of media for over eighty years. Marvel utilizes its character franchises in entertainment, licensing, publishing, games, and digital media.
For more information visit marvel.com.
Two new variant covers for MOSELY #1 revealed
Teenage Mutant Ninja Turtles and Usagi Yojimbo reunite in WHEREWHEN | {
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Q: From Google JWT to Access and Refresh tokens using React and Node.js I have a MERN application and I want to obtain the following behaviour in my app:
*
*A user logs in with with google using the GoogleLogin component in the @react-oauth/google package
*After the user logs in and gives permission to my app to access some Google API (e.g. Google Calendar), I want to send the JWT generated from the login process to the Node.js backend
*I want to "extract" access and refresh tokens from this JWT (decoding it in some way?)
*I want to store the tokens in a persistent way so that I can access the user's calendar whenever I want
I managed to get this done with the GoogleLogin from react-google-login package, which gives you a code that can be exchanged for tokens in the backend. But this library has been recently deprecated (when I try to log in with this button I get this error: idpiframe_initialization_failed). I don't know anymore how to use this library.
This is the code that I use to login:
<GoogleOAuthProvider clientId={"MY_CLIENT_ID"}>
<GoogleLogin
onSuccess={this.onSuccess}
onFailure={this.onFailure}
/>
</GoogleOAuthProvider>;
And this is the object I get when the log in is successful:
{
clientId: "MY_CLIENT_ID"
credential: "JWT_string"
select_by: "btn"
}
I'm not sure this is the right approach to use in order to get access and refresh tokens to use in the backend, or if there's a better way to do so with the new Google Identity Services SDK.
A: What a coincidence!! I'm currently working on a blog to explain OIDC, OAuth, and the new Google Identity Services, and somehow I saw this question.
let me give a TLDR version to explain different response_type first.
*
*access_token and refresh_token are part of Implicit Flow with response_type=token and Authorization code flow with response_type=code as defined in OIDC 1.0 Core spec.
*The JWT (included in "credentials" field of the returned JSON object from Google Server), is Implicit Flow with response_type=id_token.
Here is a small demo link in case you want to play with different auth methods.
I want to "extract" access and refresh tokens from this JWT (decoding it in some way?)
I don't think so. Since id_token was never meant for that purpose.
I want to store the tokens persistently so that I can access the user's calendar whenever I want
I would suggest not using id_token since you need access_token and refresh_token. You can consider the following flow:
*
*Use Authorization code flow with response_type=code to get the code.
*Using the code get the refresh_token by making a POST request to https://oauth2.googleapis.com/token. (The response will also contain access_token. Just ignore it.) Or you can also use google-api-nodejs-client library.
*Store the refresh token safely.
*Make a POST request to https://www.googleapis.com/oauth2/v4/token with the following BODY to get the access_token.
{
"grant_type": "refresh_token",
"refresh_token": "YOUR_REFRESH_TOKEN",
"client_id": "YOUR_VALUE.apps.googleusercontent.com",
"client_secret": "YOUT_SECRET_VALUE"
}
Do let me know if that helps. Cheers.
| {
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Old pertussis vaccine could help combat food allergy
Research has linked the phasing out of whole-cell whooping cough vaccines with a spike in childhood food allergies.
Researchers found that children who had received one or more doses of whole-cell vaccine in the late 1990s were 23% less likely to be diagnosed with a food allergy.
According to a new Australian study, using two different types of the whooping cough vaccine could have the added benefit of boosting protection against life-threatening allergies to foods like eggs, milk, soy, peanuts, tree nuts, wheat, fish and shellfish.
'Acellular' whooping cough vaccines are currently used in Australia, but vaccine researcher Professor Tom Snelling from the Wesfarmers Centre of Vaccines and Infectious Diseases said children could benefit from a return to older 'whole-cell' vaccines that were replaced in the late 1990s.
'Since use of the whole-cell vaccine was phased out, researchers noticed an increase in both the number of cases of food allergies and their severity,' Professor Snelling said.
'Researchers reviewed the cases of 500 children diagnosed with food allergy by specialist allergists over the past 20 years, discovering that children who had received one or more doses of whole-cell vaccine in the late 1990s were 23% less likely to be diagnosed with a food allergy than those who didn't.'
Professor Snelling pointed out that allergic and other serious reactions caused by vaccines are extremely rare, which supports the theory that some vaccines could reduce the risk of serious allergies.
'There are currently 250,000 young Australians living with severe food allergy, and up to three in every 10 babies born each year will develop either a food-related allergy or eczema,' he said.
'These allergies occur when the immune system reacts to everyday substances such as different types of food. We believe that by harmlessly mimicking infections, some vaccines such as the whole-cell whooping cough vaccine have the potential to help steer the immune system away from developing allergic reactions.
'This study adds evidence that a single initial dose of the whole-cell vaccine might have the additional benefit of partially protecting young babies against developing life-threatening food allergies.'
A $3.9 million National Health and Medical Research Council grant will be used to further investigate the findings and conduct a controlled study involving 3000 Australian babies throughout 2020.
'Babies participating in the study will be randomly assigned to receive either one dose of whole-cell whooping cough vaccine followed by two doses of acellular vaccine, or to just have the usual schedule of three doses of the acellular whooping cough vaccine,' Professor Snelling said.
'Participants will be followed until they are 12 months old to confirm whether the whole-cell vaccine truly helps to protect against food allergies in infancy and, if successful, a new vaccine schedule could form part of an effective strategy to combat the rise in food allergies.'
However, even if the vaccines are found to protect against food allergies a new whooping cough vaccine may soon be needed, with researchers warning the virus is evolving into a potential superbug.
Log in below to join the conversation.
food allergy pertussis vaccination whooping cough
GPs 'critical' to world-first allergy campaign
Imported food found to contain unlabelled allergens
Taking allergy seriously | {
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\section{WZW model}
\ \ \ Wess-Zumino-Witten (WZW) model [1] occupies an important place in
the conformal field theory: it is a model of two-dimensional
quantum massless fields exactly soluble for energy spectrum and
correlation functions; it is also a generating theory for
a rich family of other conformal field theories which may be
obtained from the WZW one by various versions of the so called
coset construction [2][3]. It is a $1+1$-dimensional analogue of
$0+1$-dimensional particle on the group $G$ (for simplicity,
we shall assume $G$ to be a compact matrix group). The motion
of the particle on $G$ is described by the classical equation
$$\partial_t(g\partial_tg^{-1})=0.\eqno(1)$$
On the quantum level, the system may be solved by harmonic analysis
on $G$. The quantum space of states is $L^2(G)$ and
carries the representation of $G\times G$.
$$L^2(G)=\bigoplus\limits_\lambda V_\lambda
\otimes{\overline V_\lambda}\eqno(2)$$
where $V_\lambda$ denotes the irreducible highest weight (HW) $\lambda$
representation of $G$. The Hamiltonian is proportional to the
quadratic Casimir operator. The multiplication
by matrix elements of $g\in G$ may be expressed in the realization
(2) of $L^2(G)$ as a bilinear combination of Clebsch-Gordan
coefficients for the tensor product with the fundamental
representation, i.e. as a bilinear expression in intertwiners of
representations of $G$.
\vskip 0.2cm
The WZW model may be viewed as describing the particle on
the loop group $LG$. The classical equation of motion is
$$\partial_-(g\partial_+g^{-1})=0\eqno(3)$$
where $\partial_\pm\equiv\partial/\partial(x^1\pm x^0)$
and we shall consider the cylindrical geometry with $x^1$ taken
mod $2\pi$.
The quantum theory is solved by harmonic analysis on the
Kac-Moody group $\widehat{LG}$, the central extension
of the loop group. The space of states is
$$\bigoplus\limits_{\lambda\ {\rm int.}}
V_{k,\lambda}\otimes{\overline V_{k,\lambda}}
\eqno(4)$$
where the level $k$, a fixed positive integer, is the value of the
central charge of $\widehat{LG}$ and $\lambda$ runs through the
finite ($k$-dependent) set of the so called integrable highest weights.
The symmetry of the model is $\widehat{LG}\times\widehat{LG}$ where the
factors act on, respectively, left- and
right-moving degrees of freedom. The spaces $V_{k,\lambda}$
and $\overline V_{k,\lambda}$ combine diagonally in eq. (4)
due ot the coupling between the left- and right-movers which share
a finite number of degrees of freedom. The field operators of the
WZW theory are bilinear combinations of intertwiners of
representations of $\widehat{LG}$.
\vskip 0.2cm
There are several ways to see manifestations of the hidden
quantum group symmetry in the WZW model\footnote{See also [4]-[6]
for the related work concerning the minimal or Liouville theories
and [7]-[9] for an attempt at a more general, less model dependent
approach relating the quantum-group-like symmetries to the theory of
superselection sectors and field statistics in the
spirit of [10][11]}. The labeling of the
representations $V_{k,\lambda}$ is similar to that of the
``good'' representations [12] of the Drinfeld-Jimbo quantum
group [13][14] $\hspace{0.05cm}{\cal U}_q(\cal G)$ for $q={\rm e}^{\pi
i/(k+h{\check{}
\hspace{0.05cm}})}$
($h{\check{}\hspace{0.05cm}}$ is the dual Coxeter number of $\cal G$,
the Lie algebra of $G$).
The exchange algebra of the intertwining operators is given
by the $6j$-symbols of ${\cal U}_q(\cal G)$ [15]-[17]. In the free
field description of the WZW theory [18][19], the quantum group
describes the homology of the contours of screening charge integrals
[20]-[23]. Let $j(x)$ be the current giving the infinitesimal
version of a representation $\pi$ of $\widehat{LG}$ by
$$\pi(X)=\int\limits_0^{2\pi}{\rm tr}\hspace{0.05cm}j(s)\hspace{0.05cm}
X(s)\hspace{0.05cm}ds\eqno(5)$$
for $X\in L{\cal G}$. In representation of central charge $k$,
current $j$ satisfies the commutation relations
$$[j(x)_1,j(y)_2]=(j(x)_1C-Cj(y)_2)\hspace{0.05cm}\delta(x-y)
-(ik/2\pi)\hspace{0.05cm}C\hspace{0.05cm}\delta'(x-y)\eqno(6)$$
where $\hspace{0.05cm}j_1\equiv j\otimes Id$\hspace{0.05cm},
$\hspace{0.05cm}j_2\equiv Id\otimes j\hspace{0.05cm}$ and
$\hspace{0.05cm}C=\sum t^a\otimes t^a\in{\cal G}\otimes{\cal G}
\hspace{0.05cm}$, $\hspace{0.08cm}{\rm tr}\hspace{0.08cm}
t^at^b=\delta^{ab}\hspace{0.05cm}$.
Eq. (6) may be viewed as the defining relation of the Kac-Moody
algebra $\widehat{L{\cal G}}$. Various aspects of the
$\hspace{0.05cm}{\cal U}_q(\cal G)$ symmetry in the
WZW model may be formally explained by postulating that the monodromy
operators
$$M(x)=P\hspace{0.07cm}{\rm e}^{\int\limits_x^{x+2\pi}j(s)
\hspace{0.07cm}ds}\eqno(7)$$
satisfy the $\hspace{0.05cm}{\cal U}_q(\cal G)$ relations
$$M_1R^-M_2(R^+)^{-1}=R^-M_2(R^+)^{-1}M_1\eqno(8)$$
where $R^\pm$ is a pair of solutions of the quantum Yang-Baxter
equation
$$R^\pm_{12}R^\pm_{13}R^\pm_{23}=R^\pm_{23}R^\pm_{13}R^\pm_{12}.\eqno(9)$$
Operators (7) appeared (implicitly) in [24] and were explicitly
discussed as generators of $\hspace{0.05cm}{\cal U}_q(\cal G)$
in [25]. The problem is that, as written above, they are very singular
objects due to the short distance singularity of currents $j(x)$.
\section{Lattice Kac-Moody algebra}
\ \ \ A possible way out from the above difficulty has been proposed
in a series of papers [26]-[29] which introduced the lattice version
of the Kac-Moody algebra\footnote{See [30]-[33] for related proposals
concerning the Liouville and Toda theories}. This allows
to build the regularized version
of the WZW model preserving essentially all the symmetries of the
continuum WZW theory and at the same time making explicit the quantum
group symmetry hidden in the continuum version of the theory.
\vskip 0.2cm
View ${\bf Z}_N$ as an $N$-point lattice in $S^1$.
The lattice Kac-Moody algebra ${\cal K}_N$ proposed in [26][27]
is given by matrix generators $J(n)$, $n\in {\bf Z}_N$, satisfying
the quadratic relations
\begin{eqnarray*}
\hbox to 3cm{$J(n)_1J(n)_2$\hfill}&=&R^+J(n)_2J(n)_1R^-,\\
\hbox to 3cm{$J(n)_1R^-J(n-1)_2$\hfill}&=&J(n-1)_2J(n)_1,\\
\hbox to 3cm{$J(n)_1J(m)_2$\hfill}&=&J(m)_2J(n)_1\ \ \
{\rm for}\ \ |n-m|>1.
\end{eqnarray*}
The other relations are the deformations of the $det=1$ conditions
for $SL(n)$ etc. For $SL(2)$ they read
$J_{11}(n)J_{22}(n)-q^{-1}J_{21}(n)J_{12}(n)=q^{1/2}$
(here the subscripts ``ij'' refer to the matrix element of $J(n)$).
The $SL(2)$ $R$-matrices are
$$R^+ = q^{1/2}\left(\matrix{q^{-1}&0&0&0\cr 0&1&q^{-1}-q&0\cr
0&0&1&0\cr 0&0&0&q^{-1}}\right) ,\hspace{0.1cm}R^- =
q^{-1/2}\left(\matrix{q&0&0&0\cr
0&1&0&0\cr 0&q-q^{-1}&1&0\cr 0&0&0&q}\right).$$
Operators $J(n)$ should be thought of as regularized versions
of parallel transport operators $\ P\hspace{0.07cm}\exp[{(k+
h\check{}\hspace{0.05cm})^{-1}\int_{2\pi n/N}
^{2\pi(n+1)/N}j(s)\hspace{0.07cm}ds}]$\hspace{0.2cm}[26].
\vskip 0.2cm
Algebra ${\cal K}_N$ contains quantum group $\hspace{0.05cm}
{\cal U}_q({\cal G})\hspace{0.05cm}$ as the monodromy
which is now, unlike in eq. (7), defined by a regular expression:
$$M(n)=J(n+N-1)J(n+N-2)\cdots J(n).\eqno(10)$$
$M(n)$ satisfy relations (8). For the $SL(2)$ case, we may write
$$M=q^{3/2}\left(\matrix{q^{-S^3}&(q^{-1}-q)S^+\cr 0&q^{S^3}}\right)
\left(\matrix{q^{S^3}&0\cr (q-q^{-1})S^-&q^{-S^3}}\right)\eqno(11)$$
where
\begin{eqnarray*}
[S^3 ,\hspace{0.05cm}S^\pm] &=& \pm S^\pm ,\cr
[S^+ , S^{-} ] &=& (q^{2S^3}-q^{-2S^3})/(q-q^{-1}).
\end{eqnarray*}
Algebras ${\cal K}_N$ may be also viewed as interpolating
between quantum group ${\cal U}_q({\cal G})$ and the
enveloping algebra ${\cal U}(\widehat{L{\cal G}})$ when
$N$ goes from $1$ to $\infty$.
\section{Conformal invariance on lattice}
\ \ \ The algebraic manifestation of the conformal invariance
underlying the continuum Kac-Moody algebras is the action
of the orientation-preserving diffeomorphisms
of the circle $D\in{\rm Diff}_+S^1$ on the currents
$$j(x)\mapsto {_{dD(x)}\over{^{dx}}}\hspace{0.05cm}j(D(x))$$
which induces automorphisms of $\widehat{L{\cal G}}$. The latter
are unitarily implementable in the HW representations:
$${_{dD(x)}\over{^{dx}}}\hspace{0.05cm}j(D(x))=U_D\hspace{0.08cm}
j(x)\hspace{0.08cm}{U_D}^{-1}.$$
$D\mapsto U_D$ is a projective representation of ${\rm Diff}_+S^1$
giving, on the infinitesimal level, a representation of the Virasoro
algebra.
\vskip 0.2cm
This structure essentially descends to the lattice. Let
$$D:{\bf Z}\rightarrow{\bf Z}$$
be an increasing map such that $D(n+N')=D(x)+N$ ($N'\leq N$).
Thus $D$ describes blocking of intervales of lattice ${\bf Z}_N$
into those of ${\bf Z}_{N'}$. $D$ induces a ``block spin''
homomorphism of the lattice Kac-Moody algebras
${\cal D}:{\cal K_{N'}}\rightarrow{\cal K_N}$
$${\cal D}(J(n))=J(D(n+1)-1)\cdots J(D(n)+1)J(D(n)).\eqno(12)$$
{\bf On the lattice, the conformal transformations are represented
by a (local) renormalization group!}
\vskip 0.2cm
If we relabel algebras ${\cal K}_N$ by arbitrary triangulations $T$
of $S^1$ (i.e. splittings of $S^1$ into intervals $l_n$)
renaming the generators as $J(l_n)$ then the block spin homomorphisms
give rise to homomorphisms $\iota_{T',T}:{\cal K}_T\rightarrow
{\cal K}_{T'}$ for $T'$ finer than $T$. One may define
the continuum limit algebra\footnote{Its generators are
$\ P\hspace{0.07cm}\exp[\int_x^y(k+h\hspace{0.04cm}\check{})^{-1}
j(s)\hspace{0.05cm}ds]$\ \ which
are singular in the standard continuum Kac-Moody algebra}
${\cal K}_{\infty}$ as the inductive
limit of algebras ${\cal K}_T$. Any $D\in{\rm Diff}_+S^1$ defines
an isomorphism ${\cal D}:{\cal K}_T\rightarrow{\cal K}_{D(T)}$
which descends to ${\cal K}_\infty$. As a result, ${\rm Diff}_+S^1$
acts by automorphisms on ${\cal K}_\infty$ showing that the block
spin homomorphisms indeed encode the conformal invariance.
\vskip 0.2cm
In continuum, the generators of the Virasoro
algebra implementing conformal invariance in the HW representations
of the Kac-Moody algebra are given in terms of current $j(x)$
by the Sugawara construction. On the lattice, the block spin
homomorphisms $\cal D$ may be also implemented in the class of
representations that we shall study below. If $D$ is a rigid
rotation of ${\bf Z}_N$ then we may expect that the implementing
maps are expressible in terms of generators of ${\cal K}_N$.
Such lattice Sugawara expressions for the (Minkowski time) transfer
matrix are not known yet except for the $U(1)$ case, see below.
\section{Free field representations of the Kac-Moody algebras}
\ \ \ In continuum, there are various ways to approach the construction
of the HW representations $V_{k,\lambda}$ of the Kac-Moody algebras.
\begin{enumerate}
\item In the algebraic approach [34] one starts from the concept
of Verma modules of $\widehat{L{\cal G}}$ and analyzes
their reducibility studying the structure
of singular vectors with the Kac-Kazhdan determinant formula,
constructing the BGG resolution etc.
\item In the geometric Borel-Weil type approach [35] one constructs
$V_{k,\lambda}$ as the space of holomorphic sections of a line bundle
over $LG/T$ ($T$ is the Cartan subgroup of $G$).
\item Finally, more recently, another algebraic approach to
the HW representations of $\widehat{L{\cal G}}$ has been obtained
[18][19][36][37] by representing
current $j(x)$ by free fields and constructing
$V_{k,\lambda}$ as a cohomology of a complex of Fock space (Wakimoto)
modules of the Kac-Moody algebra.
\end{enumerate}
\vskip 0.2cm
{}From those three approaches at least the last one carries over
to the lattice so let us sketch some of its essential points.
We shall stick to the $SL(2)$ case. The free fields one uses
to represent $j(x)$ are chiral scalar field $\phi(x)$
satisfying
$$[\phi(x),\phi(y)]={_{\pi i}\over^{2(k+2)}}\hspace{0.07cm}{\rm sgn}
(x-y)\eqno(13)$$
or the corresponding
$u(1)$ current $\partial\phi$ \hspace{0.05cm}(\hspace{0.03cm}
$\phi(x+2\pi)=\phi(x)+\pi r/(k+2)$
where $r$ is the momentum) and the $\beta\gamma$ system
$\beta(x)$, $\gamma(x)$ (periodic in $x$) s.t.
$$[\beta(x),\gamma(y)]={_{2\pi i}\over^{k+2}}\hspace{0.07cm}\delta(x-y)
\eqno(14)$$
(all other commutators vanish). These fields act in the standard Fock spaces
${\cal F}_r$ (labeled by the eigenvalues of $r$).
$$j={_{k+2}\over^{2\pi}}\left(\matrix{i\partial\phi+:\beta\gamma:&
\partial\beta-2i\beta\partial\phi-:\beta^2\gamma:\cr
\gamma&-i\partial\phi-:\beta\gamma:}\right)\eqno(15)$$
gives the $\widehat{L\hspace{0.04cm}sl(2)}$ currents and turns
the Fock spaces into the (Wakimoto)
$\widehat{L\hspace{0.04cm}sl(2)}$-modules. For $r=1,\dots,k+1$,
using the (regularized) powers of the screening charge integral
$$Q(x)=\int\limits_x^{x+2\pi}\gamma(s)\hspace{0.05cm}:{\rm e}^{2i\phi(s)}:
\hspace{0.05cm}ds\hspace{0.05cm},\eqno(16)$$
one obtains the following ($x$-independent) complex of
Fock space modules ($p\equiv k+2$)
$$\cdots\ \smash{\mathop{\rightarrow}\limits^{Q^r}}\ {\cal F}_{-r+2p}\
\smash{\mathop{\rightarrow}\limits^{Q^{p-r}}}\ \ {\cal F}_r\
\ \smash{\mathop{\rightarrow}\limits^{Q^r}}\ {\cal F}_{-r}\
\smash{\mathop{\rightarrow}\limits^{Q^{p-r}}}\ {\cal F}_{r-2p}\
\smash{\mathop{\rightarrow}\limits^{Q^r}}\ \cdots\eqno(17)$$
whose cohomology is concentrated at ${\cal F}_r$ and provides
the HW representation of $\widehat{L\hspace{0.04cm}sl(2)}$
of level $k$ and spin ${_1\over^2}(r-1)$ [19]. The
intertwining chiral fields are
$$h=\left(\matrix{\beta\hspace{0.05cm}:{\rm e}^{-i\phi}:\hspace{0.05cm}
Q&\ \beta\hspace{0.05cm}:{\rm e}^{-i\phi}:\cr
:{\rm e}^{-i\phi}:\hspace{0.05cm}Q&\ :{\rm e}^{-i\phi}:}
\right).\eqno(18)$$
Each raw of the matrix defines an operator between the complexes
(17) which projected onto their cohomology gives the component
with magnetic number $\pm 1/2$ of the spin $1/2$ primary field
of the WZW model [38]. The real fields of the WZW theory are
bilinear combinations of fields $h$ and their right-moving partners.
\section{Lattice free fields}
\ \ \ One may build the representations of the lattice Kac-Moody
algebra ${\cal K}_N$ using (deformed) free fields. Below,
$\ q^{1/2}\equiv{\rm e}^{\pi i/(2(k+2))}\ $ so that $q$ is a primitive
$2p$-th root of unity. We shall need lattice versions of the
$u(1)$ current and of the $\beta\gamma$ system.
\subsection{$u(1)$ current}
The lattice $u(1)$ algebra ${\cal U}_N$
is generated by invertible elements $\Theta(n)$,
$n\in{\bf Z}_N$, with the relations
$$\Theta(n)\hspace{0.05cm}\Theta(n+1)
=q^{1/2}\hspace{0.05cm}\Theta(n+1)
\hspace{0.05cm}\Theta(n)\eqno(19)$$
(all other generators commute). Think about $\Theta(n)$ as of a regularized
version of $\ \exp[{i\int_{2\pi n/N}^{2\pi(n+1)/N}\partial
\phi(s)\hspace{0.05cm}ds}]\hspace{0.07cm}$. $\ {\cal U}_N$ becomes
a $*$-algebra if we put $\Theta(n)^*=\Theta(n)^{-1}$.
\vskip 0.2cm
Consider, for $N$ odd, the representations of
$\hspace{0.05cm}{\cal U}_N$ acting in the space spanned by
orthonormal vectors $|\underline\alpha\rangle$ where
${\underline\alpha}\equiv(\alpha_0,\alpha_2,\dots,\alpha_{N-1})\hspace{0.05cm}$,
$\hspace{0.05cm}\alpha_{2n}\in{\bf Z}_{4p}\hspace{0.05cm}$,
$\hspace{0.05cm}\sum\alpha_{2n}=0$:
\begin{eqnarray*}
\Theta(2n)\hspace{0.08cm}
|{\underline\alpha}\rangle\ \ \ \ \ &=&z_{2n}\hspace{0.05cm}
q^{\alpha_{2n}/2}\hspace{0.05cm}
|{\underline\alpha}\rangle\ \ {\rm for}\ 2n<N-1,\\
\Theta(2n+1)\hspace{0.05cm}
|{\underline\alpha}\rangle&=&z_{2n+1}\hspace{0.05cm}|(\alpha_0,\dots,
\alpha_{2n}+1,\alpha_{2n+2}-1,\dots\alpha_{N-1})\rangle\hspace{0.05cm},\\
\Theta(N-1)\hspace{0.11cm}|{\underline\alpha}\rangle &=&
z_{N-1}\hspace{0.05cm}q^{\alpha_{N-1}/2+(N+1)/4}\hspace{0.05cm}
|(\alpha_0-1,\alpha_2,..\\
&&\hspace{4.8cm}\dots,\alpha_{N-3},\alpha_{N-1}+1)\rangle
\end{eqnarray*}
\vskip -0.2cm
\noindent where $|z_n|=1$.
\vskip 0.3cm
\noindent{\bf Proposition.} [39] {\it The above formulae
give irreducible $*$-representations
of $\hspace{0.05cm}{\cal U}_N$. They are
equivalent iff they correspond to the same
eigenvalues of the central elements $\Theta(n)^{4p}=z(n)^{4p}$ and
$\Pi\equiv q^{-1/2}\Theta\nobreak(\nobreak N\nobreak-\nobreak 1\nobreak)
\cdots\Theta(1)\Theta(0)=\prod z_n$. Every irreducible
$*$-representation of $\hspace{0.05cm}
{\cal U}_N$ is equivalent to one of the above.}
\vskip 0.3cm
Below, we shall study uniquely representations with $\Theta(n)^{4p}=1$
(we could have included this condition into the defining relations
of $\hspace{0.05cm}{\cal U}_N$). In those representations,
the rigid rotation of ${\bf Z}_N$ by two units may be implemented
by elements $U\in{\cal U}_N$:
$$U\hspace{0.05cm}\Theta(n+1)=\Theta(n-1)\hspace{0.05cm}U$$
where
$$U\hspace{0.05cm}=\hspace{0.05cm}(4p)^{-N/2}
\sum\limits_{\alpha_n\in{\bf Z}_{4p}}q^{-{_1\over^2}\alpha_0\alpha_{N-1}}
\hspace{0.05cm}\Theta(N-1)^{\alpha_{N-1}}\hspace{0.05cm}\cdots
\hspace{0.05cm}\Theta(1)^{\alpha_1}\hspace{0.05cm}\Theta(0)^{\alpha_0}.
\eqno(20)$$
Notice that, except for the boundary term, $U$ is a product
of local expressions. Eq. (20) gives the lattice version
of the Sugawara construction of the WZW Hamiltonian for the
abelian group.
\vskip 0.2cm
The case of $N$ even is similar.
\subsection{$\beta\gamma$ system}
\ \ \ Consider algebra ${\cal B}$ with generators ${\rm B}$,
$\Gamma$ and relation
$$q\hspace{0.05cm}{\rm B}\hspace{0.05cm}\Gamma-q^{-1}\hspace{0.05cm}
\Gamma\hspace{0.05cm}{\rm B}=q-q^{-1}.\eqno(21)$$
${\rm B}^p$ and $\Gamma^p$
generate the center of $\cal B$ and we have different classes
of representations depending on the eigenvalues of those elements.
\vskip 0.2cm
\begin{enumerate}
\item If in an irreducible representation
$\Gamma^p\not=0$ then $\Gamma$ is invertible and we may
introduce\footnote{I owe this observation to L. Faddeev and A. Volkov}
${\rm D}=\Gamma^{-1}-{\rm B}$ satisfying
$$\Gamma\hspace{0.05cm}{\rm D}=q^2\hspace{0.05cm}{\rm D}\hspace{0.05cm}
\Gamma.$$ If ${\rm D}=0$ we get the 1-dimensional representation\footnote{This
possibility was pointed to me by R. Kashaev}
of $\cal B$. If $\rm D\not=0$, the representations of $\cal B$ are
periodic: we may find a basis $|s\rangle$,
$s\in{\bf Z}_p$\hspace{0.05cm}, s.t.
$$\Gamma\hspace{0.05cm}|s\rangle=\zeta_1\hspace{0.05cm}|s+1\rangle,
\ \ \ {\rm D}\hspace{0.05cm}|s\rangle=\zeta_2\hspace{0.05cm}q^{-2s}
\hspace{0.05cm}|s\rangle.$$ The periodic representations are characterized
by the eigenvalues of $\hspace{0.05cm}\Gamma^p=\zeta_1^p$ and
of $\hspace{0.05cm}{\rm B}^p=\zeta_1^{-p}
+\zeta_2^p$.
\vskip 0.2cm
\item The case $\Gamma^p=0$ but ${\rm B}^p\not=0$ may be
treated similarly.
\vskip 0.2cm
\item Finally, if $\Gamma^p=0$ and ${\rm B}^p=0$, we either have
a trivial representation or a HW representation in $p$-dimensional
space spanned by states $|s\rangle$, $s=0,1,\dots,p-1,$ with the action
$$\Gamma\hspace{0.07cm}|s\rangle=
\cases{|s+1\rangle\ \ {\rm if}\ s<p-1,\hspace{1.06cm}\cr
0\ \ \ {\rm if}\ \ s=p-1,\hspace{1.06cm}}$$
$${\rm B}\hspace{0.05cm}|s\rangle=\cases{(1-q^{-2s})\hspace{0.05cm}
|s-1\rangle\ \ {\rm if}\ s>0,\cr \hspace{0.05cm}0\ \ \ {\rm if}\ \ s=0.}$$
\end{enumerate}
\vskip 0.2cm
The lattice $\beta\gamma$ system is obtained by taking a copy of algebra
$\cal B$ for each lattice site:
$${\cal B}_N=\bigotimes\limits_{n\in{\bf Z}_N}{\cal B}\hspace{0.05cm}.$$
We shall denote its generators as ${\rm B}(n)$ and $\Gamma(n)$.
${\cal B}_N$ may be represented in the space spanned by vectors
$\hspace{0.05cm}|{\underline s}\rangle\hspace{0.05cm}$, $\hspace{0.05cm}{\rm B}
(n)\hspace{0.05cm},\ \Gamma(n)\hspace{0.05cm}$ acting on the $n$-th
component of ${\underline s}\equiv(s_0,s_1,\dots,s_{N-1})\hspace{0.05cm}$.
Below, we shall only consider the HW representations of
${\cal B}_N$ so that we could include ${\rm B(n)}^p=\Gamma(n)^p=0$
into the defining relations of ${\cal B}_N$. The periodic
representations might be also of interest but we shall not
study them here.
\vskip 0.2cm
In the HW representation, the rigid rotation of ${\bf Z}_N$
by one unit is implemented by element $U\in{\cal B}_N$,
\begin{eqnarray*}
\hspace*{-0.39cm}U\hspace{0.05cm}=&&\hspace{-0.05cm}p^{-N}\hspace{-0.18cm}
\sum\limits_{\alpha_n,\hspace{0.03cm}\beta_n\in{\bf Z}_p}\
\prod\limits_{n=0}^{N-1}(1-q^{-2\beta_n})^{-1}
(1-q^{-2(\beta_n-1)})^{-1}\cdots(1-q^{-2})^{-1}\\
&&\cdot\hspace{0.05cm}\Gamma(N-1)^{\beta_0}
\hspace{0.05cm}(1-\Gamma(N-1){\rm B}(N-1))^{\alpha_{N-1}}
\hspace{0.05cm}{\rm B}(N-1)^{\beta_{N-1}}\cdots\cdots\cdot\\
&&\cdots\hspace{0.05cm}\Gamma(1)^{\beta_2}\hspace{0.05cm}
(1-\Gamma(1){\rm B}(1))^{\alpha_1}\hspace{0.05cm}{\rm B}(1)^{\beta_1}
\hspace{0.05cm}\Gamma(0)^{\beta_1}\hspace{0.05cm}
(1-\Gamma(0){\rm B}(0))^{\alpha_0}\hspace{0.05cm}{\rm B}(0)^{\beta_0}
\hspace{0.05cm},
\end{eqnarray*}
such that
\begin{eqnarray*}
U\hspace{0.05cm}{\rm B}(n+1)&=&{\rm B}(n)\hspace{0.05cm}U\hspace{0.05cm},\\
U\hspace{0.05cm}\Gamma(n+1)&=&\Gamma(n)\hspace{0.05cm}U\hspace{0.05cm}.
\end{eqnarray*}
\subsection{Lattice Wakimoto representation}
\ \ \ It will be convenient to put together the lattice $u(1)$ fields
and the $\beta\gamma$ system in a somewhat twisted way defining their
action on vectors $\hspace{0.05cm}|{\underline\alpha},{\underline s}\rangle
\hspace{0.05cm}\equiv|{\underline\alpha}\rangle\otimes|{\underline s}\rangle
\hspace{0.05cm}$ by
\begin{eqnarray*}
\Theta(n)\hspace{0.05cm}|{\underline\alpha},{\underline s}\rangle
&=&q^{s_{n+1}-s_n}\hspace{0.05cm}(\Theta(n)\hspace{0.05cm}|
{\underline\alpha}\rangle)\otimes|{\underline s}\rangle\hspace{0.05cm},\\
{\rm B}(n)\hspace{0.05cm}|{\underline\alpha},{\underline s}\rangle
&=&|{\underline\alpha}\rangle\otimes({\rm B}(n)\hspace{0.05cm}
|{\underline s}\rangle)\hspace{0.05cm},\\
\Gamma(n)\hspace{0.05cm}|{\underline\alpha},{\underline s}\rangle
&=&|{\underline\alpha}\rangle\otimes(\Gamma(n)\hspace{0.05cm}
|{\underline s}\rangle)\hspace{0.05cm}.
\end{eqnarray*}
Taking different representations of $\hspace{0.05cm}{\cal U}_N
\hspace{0.05cm}$ and the HW representation of
$\hspace{0.05cm}{\cal B}_N$, we obtain this way irreducible representations
(labelled by the eigenvalue $z$ of $\Pi$) of algebra $\hspace{0.05cm}{\cal U}_N
\tilde{\otimes}{\cal B}_N\hspace{0.05cm}$, \hspace{0.05cm}a twisted
tensor product of $\hspace{0.1cm}{\cal U}_N\hspace{0.05cm}$
and $\hspace{0.05cm}{\cal B}_N\hspace{0.05cm}$. We shall denote
by ${\cal H}_z$ the corresponding representation space.
\vskip 0.2cm
\addtocounter{equation}{21}
\renewcommand{\arabic{equation}}{\arabic{equation}}
The lattice version of the Wakimoto realization (15) of the Kac-Moody
currents by free fields is given by the formulae
\begin{eqnarray}
J_{11}(n)&=&\Theta(n)+q^{-1/2}\Theta(n)^{-1}{\rm B}(n+1)
\Gamma(n)\hspace{0.05cm},\cr
J_{12}(n)&=&-\Theta(n){\rm B}(n)+q^{-1/2}\Theta(n)^{-1}{\rm B}(n+1)
(1-\Gamma(n){\rm B}(n))\hspace{0.05 cm},\\
J_{21}(n)&=&q^{1/2}\Theta(n)^{-1}\Gamma(n)\hspace{0.05cm},\cr
J_{22}(n)&=&q^{1/2}\Theta(n)^{-1}(1-\Gamma(n){\rm B}(n))
\nonumber
\end{eqnarray}
which define a homomorphism from the lattice Kac-Moody algebra
$\hspace{0.05 cm}{\cal K}_N\hspace{0.05 cm}$ to
$\hspace{0.05 cm}{\cal U}_N\tilde{\otimes}{\cal B}_N\hspace{0.05 cm}$
and turn each representation space ${\cal H}_z$ of the latter into a
(Wakimoto) ${\cal K}_N$-module. These modules are irreducible
if $z^{2p}\not=1$. For $z=q^r$, $r=-p+1,\dots,p-1,p$, their reducibility
may be studied by adapting to the lattice the cohomological constructions
of [19].
\section{Bernard-Felder cohomology}
We shall have to adjoin to the $u(1)$ algebra $\hspace{0.05cm}{\cal U}_N$
the zero mode $\Psi(0)$ s.t.
\begin{eqnarray*}
\hbox to 4.5cm{\hspace{2cm}$\Theta(0)\hspace{0.05cm}\Psi(0)$\hfill}&=&
q^{-1/2}\hspace{0.05cm}\Psi(0)\hspace{0.05cm}\Theta(0)\hspace{0.05cm},\cr
\hbox to 4.5cm{\hspace{2cm}$\Theta(N-1)\hspace{0.05cm}\Psi(0)$\hfill}&=&
q^{-1/2}\hspace{0.05cm}\Psi(0)\hspace{0.05cm}\Theta(N-1)
\end{eqnarray*}
and all other commutators are trivial. $\Psi(0)$, which divides the
eigenvalue of $\Pi$ by $q$, may be implemented
in the sum of representation spaces of $\hspace{0.05cm}{\cal U}_N$
with $\Pi=q^r$. We shall also let it act in $\oplus {\cal H}_{q^r}$ by
$$\Psi(0)\hspace{0.05cm}|{\underline\alpha},{\underline s}\rangle
=q^{s_0}\hspace{0.05cm}(\Psi(0)\hspace{0.05cm}|{\underline\alpha}\rangle)
\otimes|{\underline s}\rangle\hspace{0.05cm}.$$
More symmetrically, we may construct the lattice $u(1)$ vertex
operator ($\sim$ ${\rm e}^{i\phi(2\pi n/N)}$)
\begin{eqnarray}
\Psi(n)&=&\Theta(n-1)\hspace{0.05cm}...\hspace{0.05cm}
\Theta(1)\hspace{0.05cm}\Theta(0)\hspace{0.05cm}\Psi(0)\ \ \ \ \ \ \ \ \
\ \ \ {\rm for}\ n>0\hspace{0.05cm},\cr
\Psi(n)&=&\Theta(n)^{-1}...\hspace{0.05cm}\Theta(-2)^{-1}
\Theta(-1)^{-1}\Psi(0)\ \ \ \ {\rm for}\ n<0\ .
\end{eqnarray}
The screening charge integral is now defined by
$$Q(n)=\Pi^{-1}\sum\limits_{m=n}^{n+N-1}\Gamma(m)\Psi(m)^2
\hspace{0.05cm},\eqno(24)$$
compare eq.\hspace{-0.05cm} (16). It is related to the lower left
matrix element of the monodromy matrix (10) in the Wakimoto
realization (22):
$$Q(n)=q\hspace{0.05cm}M_{21}(n)\hspace{0.05cm}\Psi(n)^2\hspace{0.05cm}$$
and thus (see eq.\hspace{-0.05cm} (11))
to the $\hspace{0.05cm}{\cal U}_q(sl(2)$ lowering operator. The relation of
the screening charge integrals to quantum groups has been observed
in [20] and was developed in [21]-[23] into a theory of topological
realizations of quantum groups.
\vskip 0.2cm
$Q(n)$ acts as an operator from ${\cal H}_{z}$ to ${\cal H}_{z'}$,
$z'=q^{-2}z$,
in a nilpotent
way: $Q(n)^p=0$. Besides, for $r=0,1,\dots,p$\hspace{0.05cm},
powers of $Q(n)$ define complexes of ${\cal K}_N$-modules
\begin{eqnarray*}
&\hspace{1.87cm}0\ \rightarrow\ {\cal H}_{q^{-r}}\
\smash{\mathop{\rightarrow}\limits^{Q^{p-r}}}
\ {\cal H}_{q^{r}}\ \smash{\mathop{\rightarrow}
\limits^{Q^{r}}}\ {\cal H}_{q^{-r}}\ \rightarrow\ 0&\ ,
\end{eqnarray*}
\vskip -0.4cm
\begin{eqnarray*}
&\hspace{2cm}0\ \rightarrow\ {\cal H}_{q^r}\
\smash{\mathop{\rightarrow}\limits^{Q^{r}}}
\ {\cal H}_{q^{-r}}\ \smash{\mathop{\rightarrow}\limits^{Q^{p-r}}}\ {\cal
H}_{q^r}\
\rightarrow\ 0&\ .
\end{eqnarray*}
In other words, the powers of $Q(n)$ above (independent, in fact, of $n$)
commute with the action of ${\cal K}_N$ as given by eqs.\hspace{-0.05cm}
(22). One may show that the above complexes are exact in the middle.
We conjecture that the remaining cohomology
\addtocounter{equation}{1}
\renewcommand{\arabic{equation}}{\arabic{equation}}
\begin{eqnarray}
\hspace{1cm}\ \ \ {\cal H}_{q^r}\supset
{\rm ker}\hspace{0.05cm}\hspace{0.05cm} Q^r\hspace{0.05cm}\equiv
{\cal H}'_{q^r}\hspace{0.05cm}
\ \cong\ \hspace{0.05cm}{\cal H}_{q^{-r}}\hspace{-0.05cm}/\hspace{0.05cm}
\hspace{0.05cm}{\rm im}\hspace{0.05cm}\hspace{0.05cm} Q^{r}\hspace{0.05cm}
\equiv\hspace{0.05cm}{\cal H}'_{q^{-r}}
\end{eqnarray}
\vskip -0.2cm
\noindent and
\vskip -0.7 cm
\begin{eqnarray}
\hspace{1cm}{\cal H}_{q^{-r}}\supset
{\rm ker}\hspace{0.05cm}\hspace{0.05cm} Q^{p-r}\hspace{0.05cm}
\equiv\hspace{0.05cm}{\cal H}''_{q^{-r}}\hspace{0.05cm}
\ \cong\ \hspace{0.05cm}{\cal H}_{q^{r}}\hspace{-0.05cm}/\hspace{0.05cm}
\hspace{0.05cm}{\rm im}\hspace{0.05cm}\hspace{0.05cm} Q^{p-r}
\hspace{0.05cm}\equiv\hspace{0.05cm}{\cal H}''_{q^r}
\end{eqnarray}
gives irreducible representations of ${\cal K}_N$.
This is true, for example, for $N=1$ when ${\cal K}_N$ reduces to
$\hspace{0.05cm}{\cal U}_q(sl(2))$\hspace{0.05cm}.
\vskip 0.2cm
\section{WZW theory on lattice}
\ \ \ The chiral intertwining operators on the lattice are
$$h(n)=\left(\matrix{(\hspace{0.05cm}\Pi-\Pi^{-1})\hspace{0.05cm}\Psi(n)+
{\rm B}(n)\hspace{0.05cm}\Psi(n)^{-1}\hspace{0.05cm} Q(n)&
\ {\rm B}(n)\hspace{0.05cm}\Psi(n)^{-1}\cr
\Psi(n)^{-1}\hspace{0.05cm} Q(n)&\ \Psi(n)^{-1}}\right).\eqno(27)$$
In comparison with the expression (18), notice additional
term in the uper left corner of (27) killed by Wick ordering renormalization
in the continuum.
Intertwiners (27) map space $\oplus {\cal H}_{q^r}$ into itself and descend
to cohomology (25),\hspace{0.05cm}(26).
\vskip 0.2cm
The above constructions have been adapted to the left-moving sector
of the WZW theory. In the ones for the right-moving sector (distinguished
below by the bar), we should
replace $q$ by $\bar q=q^{-1}$. The space of states of the lattice WZW theory
may be taken as
$${\cal H}'\ =\ \bigoplus\limits_{r=1}^{p-1}{\cal H}'_{q^r}
\otimes\bar{\cal H}'_{q^r}$$
or as ${\cal H}''$ using the doubly-primed spaces. The real fields
of the lattice WZW model are then
$$g(n,m)=h(n)\hspace{0.05cm}\bar h(m)^{-1}\eqno(28)$$
acting in spaces ${\cal H}'\ {\rm or}\ {\cal H}''$, $\hspace{0.05cm}
g(n,m)=g(n+N,m+M)$.
\vskip 0.2cm
It should be stressed that the resulting theory lives
on Minkowski lattice ($n,\hspace{0.05cm}m$ in eq.\hspace{-0.05cm}
(28) are integer valued light-cone variables). The model has
essentially all the symmetries of the continuum WZW theory (in deformed form),
including the conformal covariance. Its relation to quantum groups
is explicit. Let us conclude by listing some
open problems.
\vskip 0.2cm
\begin{enumerate}
\item As mentioned above, the lattice counterpart of the Sugawara
construction of energy-momentum is not known apart from the abelian
case.
\vskip 0.1cm
\item The continuum limit is not easy to understand even for the
$u(1)$ case. One should expect it to take place in a weak form,
for vacuum expectation values or traces of products of operators.
Studying the first ones would require a choice of vacuum on
the lattice (among many states invariant under lattice translations).
The traces (e.g. the characters of the lattice Kac-Moody algebra)
seem more accessible.
\vskip 0.1cm
\item In the above constructions we did not study the unitarity
properties of representations of ${\cal K}_N$ or of the
intertwining fields. One knows that in the continuum this
is difficult within the free field approach, even
when the cohomology of the Fock space modules gives
unitary HW representations of the Kac-Moody algebra.
It remains to be seen if there exists a (deformed?) version
of unitarity properties on the lattice.
\vskip 0.1cm
\item There is a close relation between the two-dimensional
(chiral) WZW model and the three dimensional Chern-Simons theory [40].
For example, the space of the fixed-time Chern-Simons states on the disc
is the basic representation of the Kac-Moody algebra [41].
The Chern-Simons theory seems to possess lattice
versions given by constructions [42][43] of 3-manifold invariants
from quantum groups. The lattice WZW model described above should
be related to the latter. It seems that the clue to understanding
this relation should be the fusion of lattice Kac-Moody algebras
generalizing the coproduct for quantum groups and allowing to glue
spaces of Chern-Simons states for more complicated topologies.
\end{enumerate}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,953 |
From 10th April 2017 the scheme of charges for Building Regulations applications has been revised.
This is paid after you have completed a full plans application, and covers our costs in administering the application and checking your plans and specifications.
This is payable after the first inspection has been made on a full plans application. The cost covers visiting the site, inspecting the work and issuing a completion certificate.
This is payable when the building notice is given to us. This charge covers site inspections and any design checking.
This must be paid when you have completed a regularisation application for unauthorised work. The charge covers all administration, site checks and inspections.
Fees are not payable when work is carried out to buildings which have specific facilities added to support disabled people.
Pay at Wigan Life Centre (card payments only).
VAT is payable on all charges except if you have already started work (regularisation charges). | {
"redpajama_set_name": "RedPajamaC4"
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Q: Can't run geoproject from d3-geo-projection I'm trying to follow Mike Bostok tutorial. Every time I want to run geoproject I obtain the following error:
/usr/local/lib/node_modules/d3-geo-projection/bin/geoproject:5
d3 = Object.assign({}, require("d3-geo"), require("../")),
^
TypeError: undefined is not a function
at Object.<anonymous> (/usr/local/lib/node_modules/d3-geo-projection/bin/geoproject:5:17)
at Module._compile (module.js:460:26)
at Object.Module._extensions..js (module.js:478:10)
at Module.load (module.js:355:32)
at Function.Module._load (module.js:310:12)
at Function.Module.runMain (module.js:501:10)
at startup (node.js:129:16)
at node.js:814:3
| {
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Q: вывести простые числа от 1 до n. Пожалуйста подскажите в чем ошибка, не могу разобраться Пожалуйста подскажите в чем ошибка, не могу разобраться
let n = 50,
count = 0;
let b = false;
for (let i = 2; i <= n; i++) {
for (let j = 1; j <= i; j++) {
if (i % j == 0) {
count++;
}
}
if (count == 2) {
console.log(i);
}
}
A: Первый вопрос, а что за count? Просто для вывода она не нужна (может, для того чтобы посчитать количество?..). Ну окей, давай подумаем над алгоритмом:
1) Нам нужно перебрать все числа от 1 до n, с проверкой на простоту. Окей, цикл for для перебора всех чисел в студию! (метка 'outer')
2) Когда a число называется простым? Когда есть такое число 1<b<a, a%b=0. Собственно эта проверка у вас и была, но опять же, зачем count?? Если срабатывает условие, то число явно не простое, и нам нужно продолжить поиск для других a
3) Если таки весь вложенный цикл отработал, и условие не сработало - поздравляю, у нас простое число, давайте напишем об этом!
let n = 50;
let b = false;
outer: for (let i = 2; i <= n; i++) {
for (let j = 2; j*j <= i; j++) {
if (i % j == 0) {
continue outer;
}
}
console.log(i);
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 532 |
{"url":"https:\/\/codereview.stackexchange.com\/questions\/145445\/pojo-like-class-in-python","text":"# POJO-like class in Python\n\nI simply want to organize some data (class= MarketOnCloseSecurity) to be used in another class MarketOnClosePortfolio. The former class would not have any methods but just a constructor that includes a ticker for a company symbol and two Pandas dataframes bars and signals.\n\nclass MarketOnCloseSecurity():\n\"\"\"Encapsulates the notion of a portfolio of positions based\non a set of signals as provided by a Strategy.\n\nRequires:\nsymbol - A stock symbol which forms the basis of the portfolio.\nbars - A DataFrame of bars for a symbol set.\nsignals - A pandas DataFrame of signals (1, 0, -1) for each symbol.\n\"\"\"\n\ndef __init__(self, symbol, bars, signals):\nself.symbol = symbol\nself.bars = bars\nself.signals = signals\n\n\nNow the MarketOnClosePortfolio would have all the methods to calculate a portfolio.\n\nIs creating a class with essentially no methods (besides constructor) in python a good practice?\n\nThe idea is to pass MarketOnClosePortfolio a list of MarketOnCloseSecurity objects.\n\nAn alternative (non-OOP) solution would do something like list_of_MOCS = [{'symbol_1':(bars_1,signals_1)},...,{'symbol_N':(bars_N,signals_N)}] so basically a list containing a map with the key being the ticker and value being a tuple of the two dataframes.\n\nMarketOnClosePortfolio will identify the positions (buy\/sell) and how based on each MarketOnCloseSecurity and form a Portfolio Dataframe. It will then calculate metrics, such as holdings, cash, and returns based on the summation of each MarketOnCloseSecurity. The code is not complete, but I have included it for reference.\n\nclass MarketOnClosePortfolio(Portfolio):\n\"\"\"Encapsulates the notion of a portfolio of positions based\non a set of signals as provided by a Strategy.\n\nRequires:\nmarket_on_close_securities\ninitial_capital - The amount in cash at the start of the portfolio.\"\"\"\n\ndef __init__(self, market_on_close_securities, initial_capital=100000.0):\nself.market_on_close_securities = market_on_close_securities\nself.initial_capital = float(initial_capital)\nself.positions = self.generate_positions()\n\ndef generate_positions(self):\n#loop through each MOCS (market_on_close_security) to calculate positions\n\n# Initialize index for positions Dataframe. Picking index 0 was arbitary they all have same index length\npositions = pd.DataFrame(index=self.market_on_close_securities[0].signals.index).fillna(0.0)\nfor security in self.market_on_close_securities:\npositions[security.symbol] = 100 * security.signals['signal'] # This strategy buys 100 shares\n\nreturn positions\n\ndef backtest_portfolio(self):\n\nportfolio_series = self.positions[self.symbol] * self.bars[0]['close_price'].astype(float)\n\npos_diff = self.positions.diff()\n\n# Will need to use this for multiple columns (Eventually)\nsymbol_columns = [self.symbol]\n\nportfolio = pd.DataFrame(index=self.bars[0].index, columns=symbol_columns)\nportfolio[self.symbol] = portfolio_series.values\n\n# Sum holdings from each company\nportfolio['holdings'] = portfolio.sum(axis=1)\nportfolio['cash'] = self.initial_capital - (pos_diff[self.symbol] * self.bars[0]['close_price'].astype(float)).cumsum()\nportfolio['cash'][0] = self.initial_capital\nportfolio['total'] = portfolio['cash'] + portfolio['holdings']\nportfolio['returns'] = portfolio['total'].pct_change()\n\nreturn portfolio\n\n\nThe backtest_portfolio() will determine the total returns\/holdings based on each MarketOnCloseSecurity. This implementation is not complete, and is currently the method used when MarketOnClosePortfolio was adapted for only one security.\n\n\u2022 I think it's ok. But, most of time, i use NamedTuple for POJO like objects. Not sure that it is possible with Dataframes. \u2013\u00a0Arnial Oct 27 '16 at 17:49\n\u2022 namedtuples are awesome. You can inherit from them too in cases where you don't need to alter them just a tiny bit - like this \u2013\u00a0Daerdemandt Oct 27 '16 at 18:01\n\u2022 Knowing how you would use this data into MarketOnClosePortfolio will help us give more taillored advices. \u2013\u00a0301_Moved_Permanently Oct 27 '16 at 19:32\n\u2022 Updated with what I have so far. Let me know what you think! Also thanks for the input so far from everyone. \u2013\u00a0user3547551 Oct 27 '16 at 19:42\n\nAn alternative (non-OOP) solution would do something like list_of_MOCS = [{'symbol_1':(bars_1,signals_1)},...,{'symbol_N':(bars_N,signals_N)}] so basically a list containing a map with the key being the ticker and value being a tuple of the two dataframes.\n\nI thought, that a better alternative would be:\n\nMOCS = {\n'symbol_1': (bars_1, signals_1),\n'symbol_2': (bars_2, signals_2),\n...\n'symbol_N': (bars_N, signals_N),\n}\n\n\nEspecially given the fact that pd.DataFrame(MOCS) would yield a valid structure.\n\nHowever, reading at your usage in generate_positions and expected usage in backtest_portfolio; and especially the fact that you will need to explode the data-structure anyway, I would keep the class approach and use collections.namedtuple as largely suggested in the comments. It provides both the advantages of a class (attributes access) and a tuple (immutable + index access).\n\nI would however change generate_positions to create a dictionnary that will hold the structure of the resulting dataframe before building said dataframe. And incorporate that method directly into the constructor as it does not provide much added value on its own:\n\nfrom collections import namedtuple\n\nMarketOnCloseSecurity = namedtuple('MarketOnCloseSecurity', 'symbol bars signals')\n\nclass MarketOnClosePortfolio(Portfolio):\n\"\"\"Encapsulates the notion of a portfolio of positions based\non a set of signals as provided by a Strategy.\n\nRequires:\nmarket_on_close_securities\ninitial_capital - The amount in cash at the start of the portfolio.\"\"\"\n\ndef __init__(self, market_on_close_securities, initial_capital=100000.0):\nself.market_on_close_securities = market_on_close_securities\nself.initial_capital = float(initial_capital)\n\n#loop through each MOCS (market_on_close_security) to calculate positions\nsecurities = {\nsecurity.symbol: 100 * security.signals['signal'] # This strategy buys 100 shares\nfor security in market_on_close_securities\n}\nself.positions = pd.DataFrame(securities)\n\ndef backtest_portfolio(self):\n...\n\n\nI would also extend the namedtuple to provide extra utilities methods:\n\nclass MarketOnCloseSecurity(namedtuple('MarketOnCloseSecurity', 'symbol bars signals')):\nreturn amount * self.signals['signal']\n\ndef close_price(self):\nreturn self.bars['close_price'].astype(float)\n\n\nSo that you can have more meaningful actions in MarketOnClosePortfolio:\n\nclass MarketOnClosePortfolio(Portfolio):\n\"\"\"Encapsulates the notion of a portfolio of positions based\non a set of signals as provided by a Strategy.\n\nRequires:\nmarket_on_close_securities\ninitial_capital - The amount in cash at the start of the portfolio.\"\"\"\n\ndef __init__(self, market_on_close_securities, initial_capital=100000.0):\nself.market_on_close_securities = market_on_close_securities\nself.initial_capital = float(initial_capital)\n\nsecurities = {","date":"2020-01-18 01:53:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.2981865108013153, \"perplexity\": 3938.3931678685894}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-05\/segments\/1579250591431.4\/warc\/CC-MAIN-20200117234621-20200118022621-00450.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.physicsforums.com\/threads\/hypermatrices-and-hyperdeterminant.181210\/","text":"# Hypermatrices and Hyperdeterminant\n\n1. Aug 21, 2007\n\n### Sangoku\n\ncould someone provide a link to this subject ??\n\ni have taken a look to 'Mathworld' and understood basic idea but i would need to know if there is a method to develop the Hyperdeterminant of Hypermatrix (in more than 2 dimension) of\n\n$$A_{jklm}$$ as product of its 'Eigenvalues' (whatever this means) just in the similar case of 2-dimension.\n\nFor example we could define the 'eigenvalues' as the numbers satisfying\n\n$$Det|A_{ijkl}-\\lambda I_{ijkl}|=0$$\n\nwhere I here is the Identity Hypermatrix in more than 2 dimension\n\nthanks.. if possible could someone provide a link to a .ps or .pdf file about the subject ??\n\nLast edited: Aug 21, 2007\n2. Aug 21, 2007\n\n### Chris Hillman\n\nYou might be confusing two distinct notions\n\nhttp:\/\/mathworld.wolfram.com\/Hyperdeterminant.html\n\nOK, but in that case you should have given the link yourself instead of asking us to search for it, eh?\n\nWell, the \"hyperdeterminant\" described in the Mathworld article is a kind of generalized resultant which tells you whether six \"sparse\" quadratic forms (in three variables) have a common zero. (The classical resultant of Sylvester tells you whether two univariate polynomials have a common root.) Note these forms are not independent of one another. In any case, this must be the second definition offered by Cayley, which evidently belongs to the domain of commutative algebra.\n\nYou seem to be talking about the first definition, some kind of generalization of the determinant to hypermatrices, whatever Cayley took that to mean.\n\nI don't think it is clear what \"identity\", \"eigenvalue\", or \"eigenvector\" might mean for \"hypermatrices\", but it sounds like you expect that the \"hyperdeterminant\" should have formal properties similar to familiar and crucially important properties of the determinant. So if you think you have concocted a possible definition, you should carefully explore its formal properties and compare with those of the usual determinant. In particular, is your notion in some sense \"multiplicative\"? Does it reverse sign under interchange of something analogous to \"rows\"?\n\nIn some important circumstances it is possible interpret a fourth rank tensor ${A_{jk}}^{mn}$ as a linear operator taking the bivector $$X_{mn}[\/itex] to the bivector ${A_{jk}}^{mn} \\, X_{mn}$. Then one can refer to the eigenvalues of the fourth-rank tensor, meaning the eigenvalues of this operator on bivectors, and the \"eigenbivectors\", without trying to invent any truly novel. Can you obtain the paper by Gelfand et al.? Last edited: Aug 22, 2007 3. Aug 22, 2007 ### Sangoku Yes, HIllman.. but unfortunately Algebra is not my forte, i can vaguely understand the concept , a NOn-linear system (Polyomial) solution can be obtained if its Hyperdeterminant is non-vanishing or similar however i would like to know if * An Hypermatrix can be Diagonalized \/ Diagonal by 'Hyperblocks' * There exist a Charasteristic Polynomial of an Hypermatrix * As i pointed above perhaps we can define the 'Eigenvalues' (in any case) as the numbers that make the Hyperdeterminant of [tex] A_{\u00edjkl}-\\lambda I_{ijkl}$$\nso 'I? is the identity matrix with diagonal terms equal to 1 and 0 elsewhere\n* There is an analogue to 'Fredholm alternative theorem' with Non-linear equations of the\nform $$g(x)+f(x)=\\lambda\\int_{a}^{b}dyK(y,x)(f^{2}(y)+f(y))$$\n\n4. Aug 22, 2007\n\n### Chris Hillman\n\nDon't shoot the messenger\n\nI have no idea what you mean by \"nonlinear system (polynomial) solution\". The point of my post was to urge you to rethink what you are looking for, if only so that you can restate your question so that we can (perhaps) help you answer it.\n\nI sense you are frustrated that I am throwing questions back at you; maybe you didn't understand what I meant, so let me try again. Your questions don't make sense unless you define your terms. Indeed, your real question seems to be \"what are the definitions of my terms?\" I don't know either, and in fact I was trying to get you to discover why no creature with all the properties you demand may exist at all. To find out, I suggest that if you think hard about questions like the following you may be able to start answering your own questions:\n\nAfter reviewing the definitions and formal properties of matrix, block, diagonalization of a matrix, ask yourself: what do you mean by hypermatrix? Hyperblock? Diagonalization of a hypermatrix? What formal properties if any do you demand that these things possess?\n\nAfter reviewing the definition and properties of characteristic polynomial of a matrix, ask yourself: what do you mean by characteristic polynomial of a hypermatrix? What properties if any do you demand that these things possess?\n\nAfter reviewing the definition and formal properties of eigenvalues, eigenvectors of a matrix, identity matrix, ask yourself: what do you mean by eigenvalues, eigenvectors of a hypermatrix, identity hypermatrix? What properties if any do you demand that these things possess?\n\nAfter reviewing the statement, proof, and some applications of the version of the Fredholm alternative theorem which you have in mind, and after noting down how the other concepts you have mentioned enter in, formulate and try to answer questions similar to those I suggested above.\n\nDid you notice that you used the word \"hyperdeterminant\" again, but at this point you do not know how you want to define this term? We can't possibly answer your question without knowing the definition you have in mind, can we? I've already pointed out that the definition you saw at mathworld is probably not the one you want, and I am suggesting that there simply may not be anything with all the properties you appear to demand, except perhaps if you come up with a much more sophisticated definition of hypermatrix than what you found at mathworld.\n\nJust noticed something: you appear to be confusing dimension of a vector space with rank of a tensor. I have to ask: are you really sure you \"understand the basic idea\"?\n\nAnd yet you are throwing around terms like \"characteristic polynomial\", \"hyperdeterminant\", \"eigenvalue\", \"Fredholm alternative\"?\n\nI am not sure what you are objecting to. The suggestion that you look up a paper by Gelfand et al? The suggestion that (if you are unable or unwilling to find what you need in the literature) you must figure out for yourself what you are looking for, and be prepared to prove for yourself that it doesn't exist (if in fact it does not exist)? Only you can decide what properties you are willing to give up if neccessary in order to concoct a \"hyperdeterminant\" possessing the properties you really need (if this is even possible).\n\nLast edited: Aug 22, 2007\n5. Aug 22, 2007\n\n### Sangoku\n\nWell, from you told .. my idea is if to define the entries of a Matrix you need a basis of vector then perhaps to define a 3-d Hypermatrix you would need a 'bivector basis $$v_{i} \/\\ v_{j}$$ (whatever a bivector means) and similar for higuer-order Matrix is that correct?\n\nAlso perhaps you can 'diagonalize' or 'triangularize a n-dimensional matrix as the Product\n\n$$PAP^{-1} =D$$ where P,D and A are 'Hypercube' hypermatrices\n\nI'm taking basis ideas from usual algebra generalized to Hyperdeterminants.. however i can be wrong, the basis idea i had was to find an interpretation of\n\n$$\\int_{V} d\\mathcal V e^{-aQ_{ij}x^{i}x^{j}-bA_{ijkl}x^{i}x^{j}x^{k}x^{l}}$$\n\nWhere 'Q' is a Quadratic form and 'A' is a 4-linear form, so we could give an approximation to this in term of the Hypedeterminant of\n\n$$B_{ijkl}=A_{ijkl}+Q_{ij00}$$\n\nSo as an approximation the integral above is $$Constant x (Hyperdet[B_{ijkl}])^{-1\/2}$$\n\nI belive that to define an Hypermatrix you will need a basis made of multi-vectors (generalization of vector using exterior product), i can understand this,and perhaps also the property that the Hyperderminant of an n-dimensional Hypermatrix can be described (minors ??) as the determinant of some (n-1)-dimensional Matrices , this would be a key concept to develop an hyperdeterminant by recursion\n\n6. Aug 22, 2007\n\n### mathwonk\n\nhow about hyperdrive? star wars is a good reference. or hyper BS, for which there are many sources.\n\n7. Aug 22, 2007\n\n### Chris Hillman\n\nSad to say, Sangoku, I'm with mathwonk. I tried hard to Assume Good Faith, but I can't keep that up indefinitely as evidence accumulates which points in a different direction.\n\nDespite the good advice which I took the time to offer you in two posts, I see that once again:\n\n1. you throw out standard terms (dimension, basis, bivector, exterior product, recursion) in a manner which suggests that (i) you have no idea what they mean (ii) you have no interest in learning a subject (algebra) for which you say you have no aptitude,\n\n2. you completely ignore my advice to to recognize that you haven't defined other terms (hypermatrix, hyperdeterminant) and can make no progress until you do so,\n\n3. overall, you seem to be generating nonsense faster than anyone here is likely to tolerate.","date":"2017-03-24 18:01:50","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7746595144271851, \"perplexity\": 694.0931990754005}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-13\/segments\/1490218188550.58\/warc\/CC-MAIN-20170322212948-00296-ip-10-233-31-227.ec2.internal.warc.gz\"}"} | null | null |
Introductory 'bonus' rates are designed to attract new customers. They are temporarily higher interest rates, which act as a 'minimum rate guarantee' for the introductory period, however long that might be. This means that you will always get at least some interest on your savings for the period outlined. If you are especially crafty, you can move your money to a different type of account as soon as the introductory rate ends, but check the terms to make sure you have the option before attempting that trick.
Look at what access you will have to your money. The terminology here can give you clues. "Easy access" refers to both instant access accounts (i.e. you can make an immediate withdrawal of cash from a branch) and no-notice accounts (which can take a little longer to get the cash, often a few days). With some accounts you must give notice to withdraw money or pay a penalty. The notice period can be as long as 180 days. However, the easier it is to get to your money the more likely it is that the interest rate will be lower.
Some accounts will only allow you a limited number of withdrawals per year. With others you will not earn any interest for months where you make a withdrawal. The impact of these limits and penalties will vary depending on how often you wish to withdraw from your account.
Sometimes you will be quoted two different interest rates. There is the Gross rate, which is a set percentage paid monthly; and there is the Annual Equivalent Rate (AER), which is a theoretical annual interest rate that factors in compounded interest over the year. When you are comparing two interest rates, be sure you compare the same type of rate.
Check the exact name of the account. Some banks will hook you with a high interest rate account, then they'll try to sell you on a different account with a similar name. Always check if you are changing accounts.
Basic rate taxpayers are normally charged 20% tax on their interest, and this is often not factored into the account interest rate advertised by the bank. So be sure and take 20% off of your interest calculation. For 'additional rate' taxpayers there can be up to 40% or 50% tax. | {
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\section{Introduction}
The dark energy has become one of the longstanding important central problems in modern cosmology and theoretical physics
ever since the discovery of the accelerating expansion of the Universe in 1998 \cite{riess-1998-116, Perlmutter-1999-517}.
Evidence exists from Type Ia supernova (SN Ia) observations~\cite{riess-1998-116} that
the universe is expanding at an accelerating rate, which implies the existence of dark energy having negative pressure $p_\Lambda$ and density $\rho_\Lambda$ satisfying
the equation of state $\omega_\Lambda\equiv p_\Lambda/\rho_\Lambda<-1/3$.
Many dark energy models, such as the quintessence~\cite{PhysRevLett.80.1582,ZlatevWangSteinhardt:PRL1999,SteinhardtWangZlatev:PRD1999},
$k$-essence~\cite{ChibaOkabeYamaguchi:PRD2000,ArmendarizPiconMukhanovSteinhardt:PRL2000,ArmendarizPiconMukhanovSteinhardt:PRD2001,ArmendarizPiconDamourMukhanov:PLB1999,GarrigaMukhanov:PLB1999},
phantom~\cite{phantom}, Chaplygin gas~\cite{Chaplygin} and
tachyon field~\cite{Padmanabhan:PRD2002,BaglaJassalPadmanabhan2003,AbramoFinelli2003,AguirregabiriaLazkoz2004,GuoZhang2004} models have been suggested; however,
these models usually require a fine-tuning of parameters or unnatural properties of matter to explain the observed data~\cite{CopelandGarousiSamiTsujikawa2005}.
The simplest candidate for dark energy is the cosmological constant with the equation of state $\omega_\Lambda=-1$, which has a theoretical value that is too large to be the observed dark energy~\cite{CC}.
The current observation favor the $\Lambda CDM$ model as the cosmological model. The $\Lambda CDM$ model consists of a cosmological constant $\Lambda$
as the origin of dark energy and a cold dark matter giving rise to galaxies and the large scale structure of the universe.
Although the $\Lambda CDM$ model is favored by observation, it suffers from two cosmological constant problems~\cite{CC,Weinberg1989}: (1) Why is the observed value so small?
(2) Why is its value now on the order of the present matter density?
To solve these problems, numerous models of dark energy have been proposed over the last two decades~\cite{CC,Padmanabhan2003,CopelandSamiTsujikawa2006,FriemanTurnerHuterer2008,CaldwellKamionkowski2009,SilvestriTrodden2009,LiLiWangWang2011,BambaCapozzielloNojiriOdintsov2012}.
However, the nature of dark energy still remains a deep mystery.
We believe that the dark energy problem is essentially an issue of quantum gravity.
Only a full understanding of quantum gravity through a well-established quantum theory can solve the dark energy problem.
The most fundamental principle of quantum gravity is thought to be the holographic principle, which may play an essential role in solving the dark energy problems.
A. Cohen {\it et al.}~\cite{CohenKaplanNelson:PhysRevLett.82.4971} suggested that in the effective quantum field theory, a long-distance infrared (IR) cut-off is related to a short-distance
ultraviolet (UV) cut-off due to the bound of energy in a region of size $L$ by the formation of a black hole of the same size $L$,
which is inspired by the existence of the limited bound for the total entropy $S$ of the system of the same size $L$ in blackhole thermodynamics~\cite{Bekenstein1973,Hawking1975}.
These interrelations between the UV cut-off and the IR cut-off are called UV/IR mixing,
which may be demonstrated explicitly in noncommutative field theory~\cite{MinwallaRaamsdonkSeiberg1999} and, in particular, in emergent quantum gravity~\cite{HSYang,LeeYang2014}.
In 2004, by applying the holographic principle to dark energy, based on the idea of Cohen {\it et al.}~\cite{CohenKaplanNelson:PhysRevLett.82.4971}, Li~\cite{Li2004-603}
suggested the holographic dark energy model in which the IR cut-off is chosen as the size of the future event horizon of the Universe.
The holographic dark energy can drive the accelerating expansion of the Universe and is in good agreement with the present cosmological observational data~\cite{LiLiWangZhang2009}.
Thus, the holographic dark energy has now become one of the most competitive candidates for dark energy~\cite{WangWangLi:2017}.
However, the above approach brings about some criticisms due to the use of the future event horizon as the present IR cut-off. These criticisms are summarized as the causality problem, circular logic problem and non-locality problem.
The causality problem occurs due to the fact that the evolution of the Universe depends on the future information for the Universe because the future even horizon is defined on the future information of the Universe.
The circular logic problem states that the future event horizon exists only in accelerating universe.
In addition, the equations of motion become non-local because the future event horizon is defined globally.
Fortunately, these problems are resolved with the action principle ~\cite{LiMiao:2012} and the equation of motion including the initial value~\cite{KimLeeLee2013}.
In Ref.~\citealp{LiMiao:2012}, by introducing two new fields in the action, holographic dark energy is shown to obey the causality
because the future event horizon as a present cut-off is not an input but is automatically determined by the present cut-off from the equations of motion.
However, there unsatisfactory points still remain because the action is not written in a general covariant form. Complete resolutions of these problems might be expected for a complete quantum gravity theory.
In the recent decade, the holographic paradigm of dark energy has been widely studied, and many theoretical models including action principle~\cite{LiMiao:2012},
such as entanglement entropy~\cite{MullerLousto1995}, holographic gas~\cite{LiLiLinWang2009}, Casimir energy~\cite{CasimirPolder1948,FischettiHartleHu1979,HartleHu1979,HartleHu1980,LiMiaoPang2010-55,LiMiaoPang2010-9026},
and entropic force~\cite{Verlinde2011,LiWang2010}, have been proposed to theoretically explain the origin of holographic dark energy. In particular, that links exist between gravity and quantum information (quantum entanglement, especially)~\cite{LeeLeeKim2015}
or, more specifically, between dark energy and the quantum entanglement of the vacuum of the Universe is very interesting~\cite{LeeLeeKim2007}.
Quantum entanglement~\cite{nielsen} is a physical resource for the quantum key distribution and quantum computing and is an order parameter for certain condensed matter~\cite{PhysRevA.66.032110}.
Recently, interest in quantum entanglement in the theoretical physics community has been growing~\cite{VanRaamsdonk:2016exw,Brustein:2000,Takayanagi:2012}.
For example, entanglement was proposed to be the source of gravity \cite{Lee:2010bg,Faulkner:2013ica,Oh:2017pkr} and spacetime~\cite{VanRaamsdonk:2010pw}.
Before these works, studies on the role of entanglement in cosmology had been published. (See Refs.~\cite{MartinMartinez:2012sg,MartinMartinez:2014sg} for a review.)
Those works usually focused on entanglement generation during inflation ~\cite{Nambu:2008my}.
In this paper, we review briefly the dynamics of dark energy and describe the properties of the holographic dark energy model.
In particular, we explain the idea linking holographic dark energy with vacuum entanglement~\cite{LeeLeeKim2007}.
The holographic dark energy model is based on the holographic principle proposed by 't Hooft and Susskind~\cite{hooft-1993,Susskind:1995} and may solve the cosmological constant problem.
A fundamental connection between gravity and quantum information (quantum entanglement, especially) has been proposed in a series of works~\cite{LeeLeeKim2007,Kim:2007vx,Kim:2008re,Lee:2010bg,kias}.
For example, the Einstein equation for gravity was suggested to be equal to the entanglement first law
by considering quantum entanglement entropy of local Rindler horizons based on Jacobson's idea linking the Einstein equation to the first law of thermodynamics~\cite{Lee:2010bg,kias}.
In Section II, we briefly introduce the basic knowledge of modern cosmology, such as the Friedmann-Lemaitre-Robertson-Walker (FLRW) cosmology, and the dynamics of dark energy.
Section III is devoted to a discussion of the properties of and the problems in holographic dark energy.
In Section IV, we analyze the causality problems in holographic dark energy in detail.
In Section V, we describe the idea linking quantum entanglement to holographic dark energy. Section V contains conclusions and discussions.
Throughout the review, we use natural units $c=\hbar = 1$ and a metric signature $(-,+,+,+)$.
\section{Modern Cosmology and Dark Energy}
In this section, we will give a brief introduction to the theoretical base of modern cosmology and dark energy.
\subsection{FLRW cosmology}
Modern cosmology is based on two fundamentals. The first is Einstein's equations which describe the dynamics of the universe:
\beq \label{EinsteinEquation}
G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi G T_{\mu \nu},
\eeq
where $G_{\mu\nu}$, $R_{\mu\nu}$, $g_{\mu\nu}$ and $R$ represent the Einstein tensor, the Ricci tensor, the metric tensor, and the Ricci scalar, respectively, $T_{\mu\nu}$ is the energy-momentum tensor, and $G$ denotes the universal gravitational constant.
Let us consider an ideal perfect fluid as the source for the energy-momentum tensor $T_{\mu\nu}$. In this case, the energy-momentum tensor is given by
\beq \label{EnergyMomentum}
T_{\mu \nu} = (\rho + p)u_{\mu} u_{\nu} + g_{\mu \nu} p.
\eeq
Here, $\rho$ is the total energy density, $p$ is the pressure density of the fluid, and $u_{\mu}$ is the fluid four velocity.
The second is the cosmological principle, which is based on the assumption of isotropy and homogeneity of the Universe, which is true approximately on large scales.
Based on the presence of generic symmetries, the Universe can be described by using FLRW metric~\cite{LiddleLyth2000, Weinberg1972,KolbTurner1990,Padmanabhan2000}:
\beq \label{FLRWmetric}
ds^2 = - dt^2 + a^2 (t) \left[ \frac{dr^2}{1-Kr^2} + r^2 (d\theta^2 + \sin^2 \theta d\phi^2 )\right].
\eeq
Here, $a(t)$ is the scale factor with the cosmic time $t$, and the coordinates $r$, $\theta$ and $\phi$ are the comoving coordinates. A freely moving body comes to rest under these coordinates.
The constant $K$ characterizes the geometry of 3-dimensional space, where closed, flat and open universe correspond to $K=+1, 0, -1$, respectively.
By inserting Eq. (\ref{FLRWmetric}) into Eq. (\ref{EinsteinEquation}), we can get the two Friedmann equations:
\beq \label{FriedmannEq1}
3 M^2_p H^2 =\rho -\frac{3 M^2_p K}{a^2},
\eeq
\beq \label{FriedmannEq2}
\frac{\ddot{a}}{a} = - \frac{1}{6 M^2_p} (\rho + 3 p).
\eeq
Here, we have defined the Hubble parameter $H\equiv \dot{a}/a $, which determines the expansion rate of the universe, and the dot represents the derivative with respect to the cosmic time t.
For the late analysis, instead of $G$, the reduced Planck mass $M_p = \sqrt{1/8 \pi G}$ is used .
From the two Friedmann equations, Eqs. (\ref{FriedmannEq1}) and (\ref{FriedmannEq2}), we can get the continuity equation, which describes the conservation of the energy-momentum tensor:
\beq \label{continuityEq}
\dot{\rho} + 3 H ( \rho + p)=0.
\eeq
The expansion rate of the universe may be, in principle, positive or negative; these correspond to expansion or contraction of the universe, respectively.
In 1929, Hubble's astronomical observation showed that the universe is expanding~\cite{Hubble:1929}, and in 1998, Riess {\it et al.} and Perlmutter {\it et al.} discovered that the universe is expanding at an increasing rate~\cite{riess-1998-116, Perlmutter-1999-517}.
In a theoretical aspect, from the second Friedmann equation, we can see that the pressure $p$ will affect the cosmic acceleration expansion. With the definition of the equation of state $w$ in $p = w \rho$, if $w < - 1/3$, the universe will accelerate.
\subsection{Dark energy}
The energy density on the right-hand side of the two Friedmann equations includes all components present in the Universe, namely, radiation, baryon matter, dark matter, and dark energy:
\beq
\label{energyDensity}
\rho = \sum_i \rho^{0}_i (1+z)^{3(1+w_i)},
\eeq
where $\rho^{(0)}_i$ and $w_i$ correspond to the present energy density and the equation of state of each component, respectively.
By introducing the dimensionless density parameter $\Omega(t) = \rho (t) / \rho_c (t) $ with the critical density $\rho_c (t)=3 M^2_p H^2 (t)$, We can rewrite Eq. (\ref{FriedmannEq1}) as
\beq
\label{EqDensityParameter1}
H^2 = H^2_0\sum_i \Omega^{0}_i (1+z)^{3(1+w_i)},
\eeq
where $H_0$ and $\Omega^{0}_i =\rho^0_i / \rho^0_c $ are the Hubble parameter and the density parameter for each component at the present epoch, respectively.
Here, $z=-1+ a_0 / a = -1 + a^{-1}$ is the red-shift parameter with the present scale factor $a_0$ taken to be $ 1$.
Obviously, $\sum_i \Omega^0_i = 1$.
The explicit form of Eq. (\ref{EqDensityParameter1}) for each component of the universe is
\beq
\label{EqDensityParameter2}
H = H_0 \sqrt {\Omega^{0}_r (1+z)^4 + \Omega^{0}_b (1+z)^3 + \Omega^{0}_{DM} (1+z)^3 + \Omega^{0}_K (1+z)^2 + \Omega^{0}_{DE} X(z)},
\eeq
where $\Omega^0_r$, $\Omega^0_b$, $\Omega^0_{DM}$, $\Omega^0_K$ and $\Omega^0_{DE}$, denote the present density parameter for each component: radiation, baryon matter, dark matter, spatial curvature and dark energy, respectively.
Here the dark energy density function is defined from the continuity equation, Eq. (\ref{continuityEq}), as
\beq
\label{DEfunction}
X(z) \equiv \frac{\rho_{DE} (z)}{\rho^0_{DE}}= exp[3 \int^z_0 dz \frac{1+ w(z)}{1+z}],
\eeq
where $w \equiv p_{DE} / \rho_{DE}$ is the equation of state of dark energy.
With the total fraction of matter density $\Omega_M = \Omega_b + \Omega_{DM}$ and the effective energy density of spatial curvature $\rho_K = - 3 M^2_p K / a^2$,
the most important quantity characterizing the properties of dark energy is the dark energy function $X(z)$~\cite{HutererStarkman2003, HutererCooray2005,HuangLiLiWang2009,WangLiLi2011,LiLiWangZhangHuangLi2011,WangHuLiLi2016}.
In fitting the current observational data so far, the $\Lambda CDM$ model, which is the simplest model with the equation of state $w=-1$, still gives the best performance,
so this model is viewed as the standard model of physical cosmology.
However, the standard model of cosmology has two cosmological constant problems~\cite{Padmanabhan2002,Padmanabhan2005}: (1) Why is $\rho_\Lambda \approx 10^{-120} M^4_p$ so tiny?
This poses the most severe naturalness problem in theoretical physics, which is the disagreement between the theoretical large value of the zero-point energy suggested by quantum field theory and the observed small values of the vacuum energy density.
(2) Why now is $\rho_\Lambda \approx \rho_M$? While a cosmological constant is by definition time-independent, the matter energy density is diluted as $a^{-3}$ as the Universe expands.
This is called a coincidence problem. Despite many challenges of the past two decades to solve the problem of dark energy, unfortunately, the nature of dark energy still remains unresolved.
In fact, because the dark energy problem is essentially a matter of quantum gravity, only a full understanding of quantum gravity can solve this problem.
Therefore, the holographic principle as the most fundamental principle of quantum gravity will surely play an important role in solving the dark energy problem.
The holographic dark energy model based on the holograpic principle is likely to present a fundamental answer to this problem because this model is in good agreement with the current observational data.
\section{Holographic dark energy}
Many dark energy models based on quantum field theory (QFT), such as the quintessence~\cite{PhysRevLett.80.1582},
$k$-essence~\cite{ArmendarizPiconMukhanovSteinhardt:PRL2000}, and phantom~~\cite{phantom} models, exist.
However, in these models, the zero-point energy of quantum fields given by
\beq
\rho_\Lambda\sim \int^\Lambda_0 k^2 dk \sqrt{k^2+m^2}\sim {\Lambda^4}
\eeq
contributes to the cosmological constant about $O(10^{70}) GeV^4$ for the UV cutoff $\Lambda\sim M_P$,
where $M_P$ is the Planck mass.
This value is $O(10^{120})$ times larger than the observational data. In short, the number of degrees of freedom of QFT
is too large for dark energy to explain the observational data.
This difficulty can be overcome when we consider the holographic principle by saying that the actual number of degrees of freedom of a region is proportional to its area not to its volume.
From this viewpoint, $\rho_\Lambda$ is a function of the square of the IR length scale $L$ and the UV cutoff scale $\Lambda$.
From dimensional analysis, we expect
\beq
\rho_\Lambda=\sum_{i=0}C_i M_P^{4-2i}L^{-2i}.
\eeq
The $C_0$ term is the problematic vacuum energy and the terms with $C_i$ $(i\ge 2)$ are negligible
compared to the $C_1$ term.
Hence, dark energy with $\rho_\Lambda=C_1 M_P^{2}L^{-2}$ is a reasonable guess, which
is further justified by Cohen {\it et al.}~\cite{CohenKaplanNelson:PhysRevLett.82.4971}
who considered the fact that the quantum vacuum energy in a region of size $L$, $O(L^3\Lambda^4)$, cannot be larger than the black-hole mass of the same size, $O(LM_P^2)$; thus, we have
\beq
\label{energybound}
L^3\rho_\Lambda \leq LM_P^2.
\eeq
Saturating this bound, Eq. (\ref{energybound}) again gives $\rho_\Lambda=C_1 M_P^{2}L^{-2}$.
Therefore, imagining a holographic dark energy model with
\beq
\label{holodark}
\rho_\Lambda=\frac{3 d^2 M_P^2}{ L^2 },
\eeq
where $d$ is an $O(1)$ parameter characterizing the equation of state, is natural.
One can easily derive a useful relation from the Friedmann equation $\rho=3M^2_P H^2$:
\beq
HL=d,
\eeq
which holds for a flat and dark energy dominated universe.
A natural length scale for cosmology is the Hubble radius $L=H^{-1}$, and
$\rho_\Lambda=3 d^2 M_P^2 H^2$ is a simple guess for holographic dark energy.
Interestingly, the present Hubble parameter $H=H_0\sim 10^{-33}~eV$ gives a holographic dark energy
comparable to the observed value $\rho\sim
10^{-10} eV^4$. The success of this simple estimate is quite remarkable. Compared to other models, holographic dark energy models do not need an $ad~hoc$ canceling mechanism
or fine-tuning to solve the cosmological constant problem.
However, Hsu~\cite{hsu} pointed out that the Friedmann equation $\rho=3M^2_P H^2$ for Hubble radius requires holographic dark energy to behave like ordinary
matter and does not give an expansion of the universe.
Huang and Li~\cite{1475-7516-2004-08-013} showed that if the future event horizon ($R_h$) is used for $L$, holographic dark energy of the form
\beq \label{holodark2}
\rho_\Lambda=\frac{3 d^2 M_P^2}{ R_h^2 },
\eeq
can give an accelerating universe.
Here,
\beq
R_h=a\int^\infty_t \frac{dt'}{a}=a\int^\infty_a \frac{da'}{Ha'^{2}}=\frac{d}{H}.
\label{Rh0}
\eeq
Differentiating $R_h/a$ with respect to $a$ gives the differential equation
\beq
-\frac{1}{Ha^2}=d\frac{d}{da}\left(\frac{1}{Ha} \right),
\eeq
which has a power-law solution $H=\alpha a^{-1+1/d}$.
Therefore, $R_h=d/H\propto a^{1-1/d}$, which yields the dark energy density
\beq
\rho_\Lambda=3 \alpha^2 M_P^2 a^{-2(1-1/d) },
\eeq
which has an equation of state $\omega=-1/3-2/(3d)$.
Thus, for $d>0$, holographic dark energy gives an accelerating expansion of the Universe.
If we include other matter, the equation of state at the present for
holographic dark energy becomes ~\cite{Li2004-603,1475-7516-2004-08-013}
\beq
\label{omega}
w_0 =-\frac{1}{3} \left(1+\frac{2\sqrt{\Omega^0_\Lambda} }{d}\right),
\eeq
and
its change rate is ~\cite{Li2004-603,HuangGong2004}
\beq
\label{omega3}
w_1
=
\frac{\sqrt{\Omega^0_\Lambda} \left( 1 - \Omega^0_\Lambda \right)}{3d} \left(1+\frac{2\sqrt{\Omega^0_\Lambda} }{d}\right),
\eeq
where $\Omega_\Lambda^0$ is the density parameter of holographic dark energy at the present.
Observational data for dark energy are usually given with a parametrization like
$w_\Lambda (z) \simeq w_0+w_1 (1-R)$, where $R$ is the scale factor of the Universe at the redshift $z$.
For the fiducial value $\Omega^0_\Lambda= 0.688$ and $d=1$, Eqs. (\ref{omega}) and (\ref{omega3}) give $w_0=-0.886$ and $w_1=0.229$.
If $d = 1$, $w_\Lambda (z)$ asymptotically approaches $-1$, and the Universe becomes a de Sitter-like universe.
If $d < 1$, $w_\Lambda (z)$ will cross the $w_\Lambda (z)=-1$ boundary, and holographic dark energy acts as a phantom dark energy with a big rip.
If $d > 1$, $w_\Lambda (z)>1$ and holographic dark energy acts as quintessence.
We can determine the value of $d$ by using the current observational data, which favor $d$ smaller than $1$.
For example, a combination of the Planck data, the baryon acoustic oscillation (BAO), and Type Ia supernovae (SN)
yields a matter density parameter $\Omega_{m0}=0.288^{+0.015}_{-0.013}$ and $d=0.768^{+0.112}_{-0.068}$~\cite{Wang:2013zca}.
Although the holographic dark energy model is in good agreement with current observational data, this model causes fundamental criticisms because it uses the future event horizon as an IR cut-off.
These problems are summarized in three major problems. The first is the causality problem: the evolution of the Universe depends on the future information of the universe.
The second is the circular logic problem: the future event horizon exists only in the accelerating Universe. The third is the non-locality problem: the equation of motion is non-local because the future event horizon is defined globally.
It has been shown that these problems can be resolved by the action principle and an analysis of equations of motion~\cite{LiMiao:2012,KimLeeLee2013}.
\section{Causality of the Holographic Dark Energy Model}
In this section, the attempts to resolve these problems based on holographic dark energy itself rather than introducing other interactions are summarized in some detail.
The distance to the future event horizon in a flat Robertson-Walker spacetime is given by
\begin{equation} \label{Rh}
R_h(t) \equiv a(t) \int_t^\infty \frac{dt'}{a(t')} .
\end{equation}
The equation of motion of the Universe in the presence of holographic dark energy is given by
\begin{equation} \label{HDE}
3M_p^2 H^2 = \rho_\Lambda + \rho_{nh},
\end{equation}
where $\rho_\Lambda$ represents the energy density of the holographic dark energy in Eq.~\eqref{holodark2} and $\rho_{nh}$ represents the energy density of all other forms of matters satisfying local equations of motion.
A major objection to the holographic dark energy model comes from the form of the holographic dark energy in Eq.~\eqref{holodark2} because it depends on the future evolution of the Universe as in Eq.~\eqref{Rh}.
Because of the future dependence, doubt arises that it may violate causality or has a circular logic problem because the future event horizon is defined globally~\cite{Cai2007} as in Eq.~\eqref{Rh}.
The problems can be summarized as follows:
\begin{itemize}
\item {\it Causality problem:} Assume that a creature who can modify the future event horizon exists.
Because the equation of motion depends on the size to the future event horizon, if the horizon is modified, that may affect the present motion of the universe.
This raises the problem of causality: "How can we expect the next state of the universe from our current data when we have no future knowledge?"
\item {\it Circular logic:} Given the current data on the Universe, we cannot know whether the Universe will eventually be in the phase of accelerating expansion or not.
The future event horizon is not determined until the Universe finishes its final evolution.
Once holographic dark energy is introduced, on the other hand, the universe is destined to expand at an accelerating rate.
If we do not know the present existence of the future event horizon, how can we use it?
This constitutes the essence of the circular logic problem. How can we use an assumption (future event horizon) based on the accelerating expansion to explain the accelerating expansion.
When we point out that holographic dark energy based on horizons other than the future event horizon does not give the accelerating expansion of the Universe, the problem becomes worse.
\end{itemize}
Various attempts have been made to overcome these shortcomings.
In the Brans-Dicke theory of gravity, by adopting the Hubble scale as the IR cut-off instead of the future event horizon, Gong {\it et al.}~\cite{gong:064029,GongLiu:2008} developed the extended holographic dark energy model.
A no-go theorem exists that the Hubble scale cannot be chosen as an IR cut-off for a Universe governed by the Brans-Dicke theory of gravity~\cite{gong:064029,GongLiu:2008}.
However, given a potential term for the scalar field in the Brans-Dicke gravity, Liu {\it et al.} succeed to prove that the holographic dark energy can be generated with the Hubble horizon~\cite{Liu:2010}.
Many authors considered different IR cut-off scales, such as the Ricci curvature radius~\cite{Nojiri:2005pu} or the age of the Universe~\cite{Cai2007,Wei2008}.
Other attempts~\cite{horvat:087301,0295-5075-71-5-712,gong:064029,pavon-2005-628} use gravity action~\cite{LiMiao:2012}, non-minimal coupling
~\cite{0295-5075-71-5-712,gong:064029} or the interaction between dark energy and dark matter ~\cite{pavon-2005-628}.
\subsection{Properties of the future event horizon}
Let us briefly summarize the properties of the future event horizon.
In order to understand causality, even though co-moving time is intrinsic to a co-moving observer, co-moving time may not be the best parameter to describe time.
Based on a conformal time measured from infinity,
\begin{equation} \label{eta}
\eta = \int_\infty ^t \frac{dt'}{a(t')},
\end{equation}
the event horizon appears to be causal, as discussed by Li in Ref.~\citealp{Li2004-603}.
Adopting that time, the flat Robertson-Walker metric takes the form
\begin{equation} \label{metric:2}
ds^2 = a^2(\eta) ( - d\eta^2 + d{\bf x}^2) .
\end{equation}
Now, the range of conformal time has a finite upper bound, for instance $\eta \in (-\infty,0)$.
Due to this finite upper limit, a light ray starting from the origin at time $\eta$ cannot reach a certain distance arbitrarily, but has a horizon at $ r = -\eta$.
Now, the formula for the distance of the future event horizon, $R_h = a(\eta)|\eta|$, appears to be causal.
This justification, however, resorts to the existence of a peculiar conformal time, defined by the integral from the future infinity.
Therefore, the circular reasoning problem remains unsolved.
Second, the future event horizon is always located at outside the Hubble horizon if $\dot R_h > 0$, as can be seen from the equation satisfied by the distance to the horizon,
\begin{equation} \label{dRh}
H^{-1}\dot{R}_h = R_h -H^{-1},
\end{equation}
as given in Ref.~\citealp{Kim:2007kw}.
All observers in the universe will be surrounded by future event horizons, and access to remote information in the Universe will be limited to the horizon.
The absence of this information is represented by a kind of entropy given by the horizon area in Planck unit; which is similar to the black hole entropy.
Davies~\cite{Davies:1988} and Pollock and Singh~\cite{Pollock:1989} proved the generalized second law of thermodynamics for the Robertson-Walker space time, which states that the total entropy of gravity and matter never increases through physical process.
In a sense, Eq.~\eqref{dRh} provides a clue to the resolution of the causality problem because this equation implies that the horizon satisfies a causal evolution equation even though horizon itself is defined globally.
\subsection{Resolution of the causality problem based on the equation of motion}
The circular logic problem is rooted in the assumption that the future accelerating expansion is a natural consequence because the holographic dark energy comes from the existence of the future event horizon.
However, the universe may not be destined to expand at a rate that accelerates in the future, as shown in Ref.~\citealp{KimLeeLee2013}, even though the energy density of the Universe depends on the future event horizon.
This implies that the future accelerating expansion is not a natural consequence of the existence of a future event horizon.
Therefore, a breakthrough must exist in the problem. The discussions in this subsection follows the work in Ref.~\citealp{KimLeeLee2013}.
Let us specify the origin of the causality problem in the evolution equation, Eq.~\eqref{HDE}.
Then, we isolate it as a boundary condition from a well-posed differential equation, which can be determined from initial data.
Suppose that we have a creature that can modify the future event horizon, and someday in the future, he changes the horizon.
Now, we can ask the following two questions:
\begin{enumerate}
\item Does this behavior modify the current evolution of the Universe in which the creature lives?
\item Does this behavior imply a violation of causality for us?
\end{enumerate}
The answer to the first question must be ``yes".
The evolution equation justifies the statement that the modification of the horizon area actually changes the current evolution.
However, this does not make the answer to the second question also ``yes".
To answer the second question, we review the Eq.~\eqref{HDE} carefully.
We separate the future-dependent part from the true dynamics described by a well-posed second-order differential equation.
By inserting the future event horizon into the holographic dark energy, we can rewrite the equation of motion, Eq.~\eqref{HDE}, as
\begin{equation} \label{HDE:2}
\int_t^\infty \frac{dt'}{a(t')}
= \frac{d}{a(t) \sqrt{H^2 - \rho_{nh}/3M_p^2 }} ,
\end{equation}
where we assume $d> 0$.
Note that the future-dependent part can be localized on the left-hand side of Eq.~\eqref{HDE:2}.
We can separate that part from the others by setting $\int_t^\infty dt'/a(t') = r_\infty - \int_0^t dt'/a(t')$, where $r_\infty = \int_0^\infty dt/a(t)$ bears the future dependence of the equation of motion.
When Eq.~\eqref{HDE:2} is differentiated with respect to $t$, the term $r_\infty$ disappears, and well-posed second-order differential equation is obtained:
\begin{equation} \label{HDE:3}
\dot H = H^2- \frac{\rho_{nh} - \dot{\rho}_{nh}/2H }{3 M_p^2} +
\frac{[H^2 - \rho_{nh}/(3M_p^2)]^{3/2}}{d H} .
\end{equation}
Finally, we divide the evolution equation, Eq.~\eqref{HDE}, into two pieces; one is an evolution equation~\eqref{HDE:3}
and the other is the term that appears to contain future information, $r_\infty$.
Note that the value $r_\infty$ does not affect the evolution equation, Eq.~\eqref{HDE:3}.
Therefore, all evolution can be resolved from some initial information without any future information from Eq.~\eqref{HDE:3}.
Because the evolution of the Universe can be determined in the absence of $r_\infty$,
solving the equations of motion, Eq.~\eqref{HDE:3}, reveals that the horizon can be determined from current information.
In this sense, we do not need information about $r_\infty$ to determine the distance to the event horizon.
In fact, $r_\infty$ is not always well defined.
It is finite only when the scale factor behaves as $a(t) \sim t^m ~(m< 1)$ for small $a$ and as $a(t) \sim t^n~(n> 1)$ in the future.
However, even if $r_\infty$ is ill-defined, the evolution equation, Eq.~\eqref{HDE:3}, works well and reproduces the Einstein equation.
If someone says that causality has been violated, he means that from the past data, something unexpected happens at present (due to the future action).
However, as seen in Eq.~\eqref{HDE:3}, all future evolutions are predictable from the past data.
If `causality' is defined in this sense, then the answer to the second question is ''No''; 'causality' is not violated for us.
Now, let us discuss the role of $r_\infty$.
Even though $r_\infty$ is defined by the integral over time, we may also obtain its value by using the limit
\begin{equation} \label{rinfty}
r_\infty = \lim_{\epsilon \to 0} \int_\epsilon^\infty \frac{dt'}{a(t')}
= \lim_{\epsilon \to 0} \frac{d}
{a(\epsilon)\sqrt{H(\epsilon)^2 - \rho_{nh}(\epsilon)/3M_p^2}} .
\end{equation}
Therefore, rather than integrating over the whole evolution, the limit to obtain $r_\infty$ may simply be taken.
From this point of view, the value $r_\infty$ plays the role of an initial boundary condition rather than a constant dependent on the future.
Summarizing, if a creature modifies the future event horizon, that simply means that he manages to change the initial condition given as $r_\infty$ or the boundary condition.
In that case, the creature may move from one universe to another universe (multi-universe theory).
On the other hand, an observer in a universe may not recognize the change in the initial condition because he/she has been living in the modified universe.
Rather, the observer will see a universe that follows the evolution equation, Eq.~\eqref{HDE:3}, with a modified initial condition from the beginning.
Thus, in the point of view of the observer, causality is not violated.
The circular logic problem arises from ignorance about the future event horizon.
However, the cosmic horizon has a topological form and cannot be created or removed by any classical means.
The absence or existence of the future event horizon can be determined from the beginning of the Universe.
Thus, the cosmological solutions are divided into two classes: the universe with/without a future event horizon.
How can we find out where we are living in the two classes?
In the previous calculations, the evolution of the Universe was determined from the sum of a second-order differential equation, Eq.~\eqref{HDE:3}, and a boundary condition, Eq.~\eqref{rinfty}.
Given the present data on the Universe, one may trace the Universe backward to identify the initial value of $r_\infty$.
In doing so, future information need not be known.
Even if $r_\infty$ is not known, the future can be predicted from the evolution equation, Eq.~\eqref{HDE:3}, because it is a well-posed differential equation.
Therefore, in principle, we may be able to recognize where we live of the two classes, if the present data on the Universe are known in detail.
The logic so far only holds when the assumptions about the homogeneity of the universe are valid in the Universe because we predict the future evolution on the basis of that logic.
Until we have more accurate data on the present universe, including dark matter and geometry, we cannot tell whether we are living in a universe with a future event horizon or without one.
However, the use of Eq.~\eqref{HDE} for the evolution at the next moment can be justified because the whole evolution of the Universe is governed by a well-posed differential equation, Eq.~\eqref{HDE:3},
and the future evolution of the Universe is only related to the boundary condition $r_\infty$, which can be given at the beginning of the Universe.
\subsection{Action approach to the causality problem}
In the previous section, we dealt with the causality problem by analyzing the equation of motion.
However, if one wants to explain holographic dark energy based on fundamental physics, it should be described by an action, which we do not know as of yet.
In Ref.~\citealp{LiMiao:2012}, by considering the Robertson-Walker metric,
\beqa
\label{Metric:Li}
ds^2= -N^2 (t) dt^2 + a^2 (t) \left[ \frac{dr^2}{1-Kr^2} + r^2 d\Omega^2\right],
\eeqa
Li and Miao proposed an action for the holographic dark energy with the following form:
\beq
\label{S:Li}
S = \frac{1}{16\pi} \int dt \left[ \sqrt{-g} \Big(R - \frac{12d^2}{a^2 L^2} \Big) -\lambda \Big(\dot L + \frac{N}{a} \Big) \right] + S_M,
\eeq
where $R$ is the Ricci scalar, $\sqrt{-g} = N a^3$, $N^2= -g_{tt}$ is the lapse function of the metric, and $S_M$ denotes the action of all matter fields.
Note also that $\lambda(t)$ plays the role of a Lagrange multiplier constraining the dynamics of $L(t)$.
With the redefintion of $N dt$ as dt, the equation of motion derived from the action becomes
\begin{equation} \label{EOM:Li}
\frac{\ddot a}{a} = - \frac{\lambda}{6a ^4} - \frac{4\pi}{3} (\rho_M + 3p_M) .
\end{equation}
The auxiliary fields behave as
\beqa
L(t) &=& \int_t^\infty \frac{dt'}{a(t')} + L(\infty), \no
\lambda(t) &=& - \int_0^t dt' \frac{24 a(t') d^2}{L^3(t')} ,
\eeqa
where we assume $\lambda(0)=0$ and $L(\infty) =0$ for the solution in the work.
Note that $\lambda < 0$ for large enough time $(d> 0)$.
When the matter effect is ignorable, Eq.~\eqref{EOM:Li} predicts an accelerating expansion.
Specifically for a holographic-dark-energy-dominated universe with $d<1$, the explicit solution for the Universe predicts a big rip universe:
\begin{equation} \label{BigRip}
a \sim (- t)^{(1-\sqrt{1+ 8d^2})/(3- \sqrt{1+ 8d^2})} ,
\end{equation}
where the time ends at $t=0$.
The model was analyzed again in the literature~\cite{Li:2012fj} and the results were compared to the observational data in Ref.~\citealp{Cui:2014sma}.
However, the model needs to be revised because it is given as an effective action only on the comoving time $t$, and the metric component $a(t)$ appears explicitly on the action.
To be well-behaving, the action should keep symmetries such as scale invariance and invariance under general coordinate transformations.
The action in Eq.~\eqref{S:Li}, is unsatisfactory in this respect.
An action, including a tachyonic scalar field, was considered in Ref.~\citealp{Rozas-Fernandez:2014vna} to overcome this problem.
\subsection{Other approaches}
Other approaches have tried to avoid the causality problem of holographic dark energy.
Here, we summarize them in their simplest forms.
For example, Nojiri and Odintsov~\cite{Nojiri:2005pu} introduced a $1/R$-like term to the action as an IR behavior at small curvature.
The correction term will be non-negligible when the curvature is small, such as in the present Universe.
In a similar spirit, Gao {\it et al.}~\cite{Gao2009} introduced an energy density proportional to the Ricci scalar curvature, $\rho_\Lambda \propto R$.
An interesting model for the agegraphic dark energy model was also developed by Cai~\cite{Cai2007}.
The model applies the Heisenberg uncertainty principle to the age of the Universe to determine the limit of the distance accuracy, $\delta t = \lambda t_p^{2/3} t^{1/3}$,
where $\lambda $ is a dimensionless constant on the order of unity and $t_p$ is the Planck time.
The quantum uncertainty principle between time and energy present an energy density of metric fluctuations of Minkowski spacetime of the form $\rho_q \propto m_p^2/t^2$.
Choosing the time scale $t$ as the age of the Universe, we may find an energy density, which is similar to that of holographic dark energy.
On the whole, the causality problem of holographic dark energy is resolved at the equation-of-motion level.
However, at the fundamental level of action, the resolution was done only partly, so additional work is necessary.
Therefore, explaining the microscopic origin of dark energy is desirable.
One of the explanations is based on entanglement, which is to be reviewed in Section V.
\section{Entanglement and dark energy}
In this section, we review the idea that holographic dark energy is from a quantum entanglement of the quantum field vacuum of the Universe.
In this model, dark energy is identified as a thermal energy (entanglement energy) associated with
the entanglement entropy $S_{ent}$ of the Universe, and its microscopic origin is suggested.
Quantum entanglement is a kind of quantum nonlocal correlation that quantum states can have.
A good measure of entanglement for pure states is $S_{ent}$.
Let us consider a bipartite system with subsystems $A$ and $B$, which has a density
matrix $\rho_{AB}$. Then, $S_{ent}$ is given by
$S_{ent}=-Tr(\rho_A ln \rho_A)$
for a reduced density matrix $\rho_A\equiv Tr_B
\rho_{AB}$.
An interesting fact is that the quantum vacuum of fields in the Universe has natural entanglement due to
the Reeh-Schlieder theorem~\cite{reeh,werner}.
For example, we consider a Hamiltonian for the massless scalar field $\phi$ in the Minkowski spacetime
~\cite{Srednicki1993}
\beq
H=\int d^3 x (|\nabla \phi(x)|^2+ |\pi(x)|^2),
\eeq
where $\pi(x)$ is the momentum of the field. We can divide the space into two parts: one inside (A) and the other outside (B) a spherical region like a cosmic horizon. Discretization of a radial coordinate with a UV cut-off gives
an effective Hamiltonian for discretized field oscillators~\cite{Srednicki1993}.
Then, one finds that the leading term in entanglement entropy between $A$ and $B$ is proportional
to the area of the boundary between the two subsystems.
We expect this area law to hold for a more general space-time.
Therefore, $S_{ent}$ for a spherical region with a radius $r$ can be expressed as
\beq
\label{Sent}
S_{ent}=\frac{\beta r^2}{b^2},
\eeq
where $\beta$ is a constant depending on the fields, and $b$ is the UV cutoff.
$S_{ent}$ in this form is consistent with the holographic principle.
Numerical calculations gives a value $\beta\simeq 0.3$ for a massless field ~\cite{Srednicki1993,PhysRevD.52.4512}.
We need to add up the contributions from other $j$ fields with constants $\beta=\beta_j$ and spin degrees of freedom $N_j$ ~\cite{PhysRevD.52.4512}.
Therefore, we expect
\beq
S_{ent}=\sum_j \beta_j N_j\frac{ r^2}{b^2}\equiv \frac{\alpha r^2}{l_P^2},
\eeq
where the Planck length is $l_P=\sqrt{G}$ in $c=\hbar =1$ unit. Choosing the UV cut-off, $b=1/M_P$, gives
\beq
\label{alpha}
\alpha=\frac{1}{8\pi}\sum_j \beta_j N_j.
\eeq
Bianchi showed~\cite{Bianchi:2012br} that a variation of $S_{ent}$ is equal to a variation of the Bekenstein-Hawking entropy
\beq
S_{BH}=\frac{A}{4G}.
\eeq
If we identify $S_{ent}$ with $S_{BH}$, then $\alpha=\pi$.
\begin{figure}[hbtp]
\includegraphics[width=0.32\textwidth]{horizon2.eps}
\caption{ For an observer at $O$ the space inside the cosmic event horizon with radius $R_h$ can be divided into two subspaces, one inside (A) and the other outside (B) the spherical horizon.
The horizon surface $\Sigma$ has the entanglement entropy $S_{ent}\propto {R_h}^2$ and entanglement energy $E_{ent}\equiv \int_\Sigma T_{ent} dS_{ent}$, which can be dark energy.}
\label{RindlerHorizon}
\end{figure}
What is the relation between $S_{ent}$ and holographic dark energy? In Ref.~\citealp{Lee:2010bg}, the entanglement first law
\beq
\label{firstlaw}
dE_{ent}=k_B T_{ent} dS_{ent}
\eeq
was first derived and was shown to be equivalent to the Einstein equation,
where $dE_{ent}$ is the variation of entanglement energy associated with the local Rindler horizon as shown in Fig.~\ref{RindlerHorizon},
where $k_B$ is the Boltzmann constant and $T_{ent}$ is the horizon temperature.
We assume that the first law holds at a cosmic horizon.
In Ref.~\citealp{LeeLeeKim2007}, we suggested the entanglement energy $E_{ent}$ associated with the future cosmic event horizon is holographic dark energy.
(There are similar suggestions based on the Verlinde's idea ~\cite{Li:2010cj,Zhang:2011,Wei:2011,Easson:2011}.)
Integrating $dE_{ent}$ on the spherical surface ${\Sigma}$ of the event horizon we obtain
\beq
\label{eent}
E_{ent}=\int_{\Sigma} dE_{ent}= k_B T_{ent} \int_{\Sigma} dS_{ent}= \frac{\alpha R_h}{2 \pi l_P^2},
\eeq
where $T_{ent}= 1 /2\pi k_B R_h $ is the Gibbons-Hawking temperature of the horizon.
The entanglement energy density within $R_h$ is given by
\beq
\label{rho}
\rho_{\Lambda}=\frac{3 E_{ent}}{4 \pi R_h^3}
=\frac{3 \alpha M_P^2}{\pi R_h^2} ,
\eeq
From the holographic dark energy $\rho_{\Lambda}={3 d^2 M_P^2}/{ R_h^2} $ in Eq.~\eqref{holodark2}, we can obtain the holographic dark energy parameter
\beq
\label{d1}
d=\sqrt{\frac{\alpha}{\pi}}.
\eeq
That this model makes the holographic dark energy parameter $d$, in principle, calculable, is worth emphasizing.
Before Ref.~\citealp{LeeLeeKim2007}, the parameter $d$ was obtained only by observations.
\begin{figure}[hbtp]
\includegraphics[width=0.32\textwidth]{wmap2018.eps}
\caption{ Observational constraints on the dark energy equation of state $w_0$ and $w_1$ from WMAP+BAO+$H_0$+SN. (The data are extracted from
Ref. ~\citealp{Komatsu:2010fb}). The star represents our theoretical prediction with $d=1$ while the circle is for the standard model $(d=0.67)$.}
\label{ObservationalConstraints}
\end{figure}
If we use approximate values for the parameters for field $j$, $\beta_j \simeq 0.3$ and $\sum_j N_{j}=118$, in the standard model of particle physics,
we obtain $d=0.67$, which is similar to the observed data $d=0.768^{+0.112}_{-0.068}$~\cite{Wang:2013zca}.
In Fig.~\ref{ObservationalConstraints}, the star and the circle indicate the theoretical predictions with $d=1$ and $d=0.67$, respectively.
Our model provides a way to derive $d$ and $\omega_\Lambda$ in an $ab~initio$ manner for the first time.
However, the above estimate is based on a flat space-time calculation.
In a future work, we need to study quantum entanglement in an expanding universe.
What is the modern version of the entanglement first law?
The reduced density matrix can be represented by a modular Hamiltonian $H_A$
as $\rho_A=e^{-H_A}/Z$, where $Z$ is the partition function.
Then, the variation of entanglement across the boundary is
\beqa
\delta {{S}_{ent}}&=&-tr(\delta {{\rho }_{A}}\log ({{\rho }_{A}}))-tr({{\rho }_{A}}{{\rho }_{A}}^{-1}\delta {{\rho }_{A}}) \no
&=&tr(\delta {{\rho }_{A}}{{H}_{A}}) \no
&=&\delta \left\langle {{H}_{A}} \right\rangle.
\eeqa
Usually, $\rho_A$ is not a thermal state, but in locally flat space-time, $\rho_A$ looks like a thermal state.
We expect this first law to hold for a cosmic horizon with Hawking radiation.
\section{Conclusions}
The problem of identifying the dark energy that explains the cosmic acceleration expansion found in 1998 has become the biggest issue of modern cosmology and theoretical physics. In essence,
the problem of dark energy is a matter of quantum gravity. Therefore, holographic principles as a fundamental principle of quantum gravity will play a very essential role in solving the problem of dark energy.
In this review paper, a holographic dark energy model based on the holographic principle was introduced to analyze the dark energy dynamics.
In this model, the IR cut-off of the UV/IR mixing in the holographic principle suitable for the accelerating expansion of the universe is the size of the horizon of the future event,
which enters into the formula of the holographic dark energy density.
The analysis of the equation of state in this model shows that the case of the constant parameter with $d\lesssim 1$ is favored for the current observational data.
Causality problems and circular logic problems about the holographic dark energy model can be resolved with the action principle and the initial condition of the equations of motion.
Two auxiliary fields, which are introduced from the action principle, make all equations of motion local.
The use of the event horizon as the current cut-off is automatically determined from the equation of motion; therefore, the evolution of the universe is determined by the present initial condition.
However, because the action introduced in Ref.~\citealp{LiMiao:2012} does not have the covariant form with full symmetry, it still remains unsatisfactory.
Jacobson's idea linking gravity with thermodynamics requires a microscopic degree of freedom.
Quantum entanglement of the quantum field vacuum in curved space-time is a natural candidate for the microscopic degrees of freedom for the thermodynamic version of gravity.
If we extend this idea to the dark energy problem, we can easily arrive at the holographic dark energy model associated with an entanglement of cosmic horizons.
Interestingly, a rough estimate in this model gives a holographic dark energy comparable to the current observational data.
Additionally, that the holographic dark energy model associated with an entanglement of cosmic horizons provides a theoretical basis for the calculation of dark energy parameter $d$ is worth emphasizing.
However, to get the exact equation of state we still need more precise calculations for the entanglement in curved space-time.
\section*{ACKNOWLEDGEMENT}
This work was supported by the Daejin University Research Grants in 2018.
\section*{REFERENCES}
| {
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Rangers Chose Girardi, Not Boyle, Over Stralman
May 2, 2016 by Glen Miller
Two summers ago the New York Rangers made the decision to sign unrestricted free agent defenseman Dan Boyle to a two-year, $9MM deal. The same day, Anton Stralman left Manhattan to join the Tampa Bay Lightning with a five-year contract worth $22.5MM. Most who follow the team felt the two signings were linked, that the Rangers essentially chose Boyle over Stralman; a decision, which in hindsight didn't work out well. It's understandable to look at those choices and draw that same conclusion. Yet that conclusion is erroneous; the Rangers effectively moved on from Stralman the moment the club agreed to a long-term extension with Dan Girardi.
First, Stralman and Boyle were totally different in terms of what they did well. The Rangers coveted Boyle because they felt he would fill their need for a PP QB. The team had been looking for someone to take charge from the point on the man advantage for years. The Blue Shirts finished the 2013-2014 regular season 15th in PP efficiency. The year before they placed 23rd. In the five seasons before joining the Blue Shirts, Boyle placed 16th, 10th, 14th, 7th and 4th in power play points among defensemen. No Ranger blue liner placed in the top 16 since Michael Del Zotto finished 12th during the 2009-2010 campaign. Signing Boyle to boost the man-advantage attack did make sense.
Dan Boyle (Amy Irvin / The Hockey Writers)
On the other hand, Stralman wasn't viewed as someone who could help the power play. During the 2013-2014 season, the Rangers used Stralman on the man advantage for just 22:32. That's a total accumulated in 81 games played that season. Granted, it's possible or even likely the Rangers missed the boat on their evaluation of Stralman's abilities as he has tallied 23 PP points in two years since joining Tampa. While not the production of an elite performer it's still solid.
Regardless of whether the team underestimated Stralman's capabilities, the team desired to improve their man-advantage and overall offensive attack and decided to pursue Boyle to address those areas. Stralman was someone the team utilized in their top-four, as a right-side defenseman, and who played a majority of his minutes at even strength or killing penalties. After signing Girardi to his six-year, $33MM extension, and with fellow right-handed defenseman Kevin Klein also on the roster, the Blue Shirts had room for only one more right defenseman. With Girardi and Klein around to kill penalties and capable of playing shutdown roles against top competition, the team decided they wanted a more offensively inclined blue liner to augment the lineup and signed Boyle.
Dan Girardi doesn't just score points, he contributes in a plethora of stat categories. (Timothy T. Ludwig-USA TODAY Sports)
The free agent market that summer also dictated the Rangers' maneuverings. Boyle, because of his age, was limited to short-term offers. Stralman, at 10 years Boyle's junior was in line for a much longer term, as evidenced by the five-year contract he would ultimately receive. The club had a coming roster crunch with young defenders Dylan McIlrath and Brady Skjei in the pipeline and lots of key contributors approaching free agency in the near future. Since July 1st, 2014 when both Boyle and Stralman inked their free agent deals, the Blue Shirts have had to re-sign centers Derick Brassard and Derek Stepan, winger Mats Zuccarello, while also extending Marc Staal. Those moves accounted for a total of $21.7MM in cap charges. That doesn't include the need to re-sign Chris Kreider, Kevin Hayes and JT Miller, all of whom are scheduled to be restricted free agents this summer. Boyle's two-year commitment fit much better in the team's long range budget.
Signing Boyle made sense on paper for a variety of reasons. And while in hindsight, that signing certainly didn't work out as hoped, it wasn't what precluded the club from re-singing Stralman. Dan Girardi's massive extension is what essentially sealed Stralman's fate. The moment Girardi inked his new deal, it signaled that unless Stralman would come back for far less than what he was worth on the open market, his days in New York were all but done. Remember, there were reports the club had offered Stralman's camp a three-year contract in advance of free agency so there was interest in keeping the defender. But the Rangers had no intention keeping both Girardi and Stralman together for half a decade. From an on-ice perspective they filled the same role. For long-range budgetary reasons, it didn't make sense. Boyle and Stralman together? Perhaps. Boyle and Girardi? Yes. Stralman and Girardi? No, and that is why the Rangers are likely stuck with a severely declining Girardi while Stralman continues to perform well in Florida.
Glen Miller
Native Southern Californian who somehow adopted the New York Rangers as my favorite team. I've written for quite a few websites during my days as a blogger including: Inside Hockey, Ranger Nation and SNY.tv. By day, I work as an operations manager.
Categories New York Rangers Tags Anton Stralman, Dan Boyle, Dan Girardi | {
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{"url":"https:\/\/www.semanticscholar.org\/paper\/On-the-homology-of-locally-compact-spaces-with-ends-Diestel-Sprussel\/de4edd497592ad35db947c86ec5178ee62e9328d","text":"# On the homology of locally compact spaces with ends\n\n@article{Diestel2009OnTH,\ntitle={On the homology of locally compact spaces with ends},\nauthor={Reinhard Diestel and Philipp Sprussel},\njournal={Topology and its Applications},\nyear={2009},\nvolume={158},\npages={1626-1639}\n}\n\u2022 Published 29 October 2009\n\u2022 Mathematics, Computer Science\n\u2022 Topology and its Applications\n8 Citations\n\n## Figures from this paper\n\nThe fundamental group of a locally finite graph with ends: a hyperfinite approach\n\u2022 Mathematics\n\u2022 2012\nThe end compactification |\\Gamma| of the locally finite graph \\Gamma is the union of the graph and its ends, endowed with a suitable topology. We show that \\pi_1(|\\Gamma|) embeds into a nonstandard\nOn the coincidence of zeroth Milnor-Thurston homology with singular homology\n\u2022 Mathematics\n\u2022 2014\nIn this paper we prove that the zeroth Milnor-Thurston homology group coincides with singular homology for Peano Continua. More- over, we show that the canonical homomorphism between these ho- mology\nLabeled Trees Generating Complete, Compact, and Discrete Ultrametric Spaces\n\u2022 Mathematics\nAnnals of Combinatorics\n\u2022 2022\nWe investigate the interrelations between labeled trees and ultrametric spaces generated by these trees. The labeled trees, which generate complete ultrametrics, totally bounded ultrametrics, and\n\n## References\n\nSHOWING 1-10 OF 25 REFERENCES\nThe homology of locally finite graphs with ends\n\u2022 Mathematics\n\u2022 2008\nWe show that the topological cycle space of a locally finite graph is a canonical quotient of the first singular homology group of its Freudenthal compactification, and we characterize the graphs for\nThe homology of a locally finite graph with ends\n\u2022 Mathematics\nComb.\n\u2022 2010\nA new singular-type homology for non-compact spaces with ends is constructed, which in dimension 1 captures precisely the topological cycle space of graphs but works in any dimension.\nThe Cycle Space of an Infinite Graph\n\u2022 R. Diestel\n\u2022 Mathematics\nCombinatorics, Probability and Computing\n\u2022 2005\nA new \u2018singular\u2019 approach is presented that builds the cycle space of a graph not on its finite cycles but on its topological circles, the homeomorphic images of $S^1$ in the space formed by the graph together with its ends.\nDuality in Infinite Graphs\n\u2022 Mathematics\nCombinatorics, Probability and Computing\n\u2022 2006\nThis work shows that obstructions fall away when duality is reinterpreted on the basis of a \u2018singular\u2019 approach to graph homology, whose cycles are defined topologically in a space formed by the graph together with its ends and can be infinite.\nOn Infinite Cycles I\n\u2022 Mathematics\nComb.\n\u2022 2004\nWe adapt the cycle space of a finite graph to locally finite infinite graphs, using as infinite cycles the homeomorphic images of the unit circle S1 in the graph compactified by its ends. We prove\nEulerian edge sets in locally finite graphs\n\u2022 Mathematics\nComb.\n\u2022 2011\nIn a finite graph, an edge set Z is an element of the cycle space if and only if every vertex has even degree in Z. We extend this basic result to the topological cycle space, which allows infinite\nCycle\u2010cocycle partitions and faithful cycle covers for locally finite graphs\n\u2022 Mathematics\nJ. Graph Theory\n\u2022 2005\nIt is shown that if Seymour\u2019s faithful cycle cover conjecture is true for finite graphs then it also holds for locally finite graphs when infinite cyles are allowed in the cover, but not otherwise.\nGeodetic Topological Cycles in Locally Finite Graphs\n\u2022 Mathematics\nElectron. J. Comb.\n\u2022 2009\nWe prove that the topological cycle space C(G) of a locally finite graph G is generated by its geodetic topological circles. We further show that, although the finite cycles of G generate C(G), its","date":"2022-07-04 02:25:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8121301531791687, \"perplexity\": 1688.1713317316169}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656104293758.72\/warc\/CC-MAIN-20220704015700-20220704045700-00334.warc.gz\"}"} | null | null |
Biologia:
Fura: mamífer carnívor amb pelatge marronós i de cos allargat.
Fura (Pimelia elevata): insecte escarabat, subfamília de la Pimelia pròpia de les Illes Pitiüses.
Automoció:
Montesa Fura (escúter): prototipus d'escúter que presentà Montesa el 1958
Montesa Fura (ciclomotor): ciclomotor de fora d'asfalt que comercialitzà Montesa el 1970
SEAT Fura: automòbil produït per SEAT entre 1981 i 1986
Altres:
La Fura: setmanari del Penedès
La Fura dels Baus: companyia catalana de teatre | {
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Рашид () — нефтегазоконденсатное месторождение на шельфе ОАЭ. Открыто в 1973 году. Плотность нефти составляет 0,8398-0,8498 г/см3 или 38° API. Начальные запасы нефти составляют 200 млн тонн, а газа — 100 млрд м³.
Оператором месторождении является дубайская нефтяная компания Dubai Petroleum Establishment.
Нефтяные месторождения ОАЭ
Газовые месторождения ОАЭ
Морские месторождения нефти и газа | {
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A Brief Introduction to Libertarian Legal Theory
Libertarianism is in essence a philosophy which deals with the proper role of law in society, although it is not often considered as such. Indeed, Frédéric Bastiat in his seminal work The Law discusses this role of law, and concludes that the law should merely...
Libertarianism is in essence a philosophy which deals with the proper role of law in society, although it is not often considered as such. Indeed, Frédéric Bastiat in his seminal work The Law discusses this role of law, and concludes that the law should merely be utilized in the protection of person and property. He built on the classical liberal tradition started by the likes of John Locke and the American founding fathers, which believed that governments are created for the purpose of safeguarding the lives, liberty, and the justly acquired estates of citizens.
As a law student, I appreciate the need for Libertarian Legal Theory if the cause of liberty is to gain widespread intellectual favor outside of the United States. In this article I want to provide a brief introduction to the legalistic realm of libertarianism, in the South African context. This will most likely turn out to only be a translation of the commonly accepted principles underlying the philosophy into legal theory. While I did come to the phrase 'Libertarian Legal Theory' by myself, I must recognize Stephan Kinsella as apparently the only notable libertarian to use the term prominently. He has done a considerable amount of work on the matter within the American legal context. I intend to publish a much longer, more academically-inclined text on Libertarian Legal Theory, which I have already started working on, in the future. The concepts which will be mentioned here, such as 'positivism', 'statocentrism', 'common law' etc. will therefore not be elaborated upon fully.
Legal positivist John Austin described the law according to command theory. This means that law is a command handed down by the more powerful entity (the State or sovereign) to the weaker entity (the individual or citizen), and is to be obeyed by such weaker entity if he hopes to escape the sanction, or punishment, that will follow on his disobedience. Sir Frederick Pollock disagreed with the command theory and said law simply needs a 'political community' of persons who regard the set rules as binding upon them. I however regard these two 'theories' as saying the exact same thing. A binding rule is a rule with a consequence. The law can aptly be described as a system of rules which carry force – i.e. if it is not complied with, the State will initiate force against the disobedient subject.
One can say that Libertarian Legal Theory confines itself only to matters of force. Feminist Legal Theory may concern itself with the fact that historically only three women have been elected to the International Court of Justice. Critical Legal Studies may take issue with the fact that the 'logic' underlying law has been almost exclusively determined by white men. Legal positivists concern themselves with the integrity of the legal system as an essentially self-contained ecosystem. Libertarian Legal Theory will find overlap with many other schools of jurisprudence, but concerns mostly State initiatory force. Many other concerns do flow from this, for example, when is property justly acquired, or when is legislation legitimately binding? On the other hand, Libertarian Legal Theory would most likely not have a definitive opinion on the distinction between international and national law, as the principles it contains apply across boundaries.
The principles upon which Libertarian Legal Theory rests, from the South African common law perspective, is the doctrine of self-ownership, informed consent, and the nemo plus juris rule. All three of these lead to one another as a matter of course, and in a way, say the exact same thing in practice. However there are some theoretical differences which distinguish them. I will examine each principle briefly.
Self-ownership
'Self-ownership' is simply one way to say that individual human beings are not slaves or computers. By virtue of our capacity to rationalize – i.e. make decisions which we can justify through reason, and accept the consequences and responsibility for those decisions – as individuals and not a collective hivemind, are we said to 'own' ourselves. This does not mean we are islands unto ourselves and completely isolated from the rest of humanity. By our nature, humans are social beings and interact, individually as well as in groups (with the most relevant groups being the family and the community). But we retain our individuality in perpetuity. A person in a given group, can at any stage consciously diverge from the group's opinion, culture and interests. It is naturally impossible to alienate or lose one's own individuality. Self-ownership is thus not a highly conflated philosophical concept but a logical result of our existence.
In our common law, self-ownership finds expression in the doctrine of legal subjectivity. Every individual is a legal subject; however under the erstwhile Roman law, slaves, and even prisoners of war, were not legal subjects. Juristic entities are also regarded as legal subjects, although it is preferred to refer to their legal subjectivity as legal personality. From legal subjectivity flows what is known in South African law as subjective rights. This is, however, a misnomer which came about as a result of a direct translation of 'subjektiewe reg' from Afrikaans to English. In our context of Libertarian Legal Theory, it is preferable to refer to individual rights. Libertarian and classical liberal theory describe the doctrine of individual rights as the ability of every individual to pursue his own interests unhindered, while not violating the same ability of other individuals. The common way to describe this is that every person has a right to life, liberty, and property.
It is because we 'own' ourselves that only we, as individuals, have the entitlement to alienate or modify our legal and moral abilities (i.e. our rights) and the product they yield. The main ability concerned here is our labor – physical or mental – and all that follows from it. Within the context of Libertarian Legal Theory, the main question to be asked in a given legal issue, is whether or not consent is present. Any damage or apparently harmful consequence that results from a consensual situation is no matter of law, as provided by a maxim in our common law known as volenti non fit injuria (a willing person is not wronged: he who consents cannot be injured). The consent must be informed, i.e. the person must have the mental capacity to provide consent. This is not always naturally possible, and thus, consent is not an absolute requirement.
Allow me to illustrate by using two examples:
Person A wants to try an exotic new drink on the market. The producer of the drink, company B, warns its customers that the drink contains a type of poisonous chemical which gives it its exotic flavor, and that they have no way of judging how the poison affects a given individual customer. Person A nonetheless decides to try the drink, and subsequently dies. Person A provided informed consent and therefore B cannot be held criminally liable.
Person A is being held hostage in his home. His neighbor, B, a private security contractor and former police officer, sees what is happening inside. B realizes that A is in grave danger and breaks down the door to A's home, proceeds inside and disarms the hostage takers. The door – A's property – has been broken without his consent. However, he could not provide consent, and B will still not be liable.
I will refer to the general principles underlying libertarianism to explain this second example above.
The 'non-aggression principle' (or axiom) is the overarching principle in libertarian thought which can be said to encompass all the principles I am discussing here: self-ownership, informed consent and nemo plus juris. According to this principle, no person may initiate force against another's person or property. However, persons may use defensive force to guard their own person and property against assault, as well as the person or property of another. This second, highlighted leg of the 'NAP' is essentially what justifies the existence of government within libertarian theory. Bastiat also describes the law, through the State, as a collective tool for defense.
Clearly, in our example, B is acting within the ambit of the NAP since he is using defensive force to save A from the hostage takers. South African common law adequately makes provision for such a situation, as well. Under our law of delict (tort), a defense in court that can be raised to escape liability for damaging the property of another, is necessity (self-defense is also such a defense). According to the defense of necessity, "[damage causing] conduct directed against an innocent person for the purpose of protecting an interest of the actor or a third party (including the innocent person) against a dangerous situation" (Neethling, Potgieter and Visser) is lawful, and will exclude delictual wrongfulness on the part of the actor.
Therefore, the common law provides an answer. The NAP, being the overarching principle in libertarian theory, is assisted by rules of law which have developed over centuries. The NAP, or the principle of informed consent, is not being 'limited' by the common law, but is merely being applied within the context of the matter at hand. Principles of common law can also assist in justifying why a doctor may amputate a leg while the patient is unconscious and cannot consent, and why a parent can kill a neighbor's large dog running toward his baby child.
The common law, from most accounts, is accepted by libertarian theorists. Both Stephan Kinsella, as well as Bruno Leoni (author of Freedom and the Law) regard the common law as a more justifiable law, over statutes. I am inclined to agree to an extent. The problem with law, thus, arises in the case of legislation and statutory law.
Nemo plus juris
Legislation is often based on the fallacy of circular reasoning. Take this example as an illustration:
Libertarian A asks positivist B why the Equality Act of 2000 is binding on his private business. B tells him that the Equality Act is authorized by Section 9 of the Constitution of 1996. But what authorizes the Constitution? The people, answers B, via the social contract. How is the social contract manifested in South Africa, asks A. By the Constitution of 1996, says B.
Often times, it does not even get as far as the social contract. Even so-called 'legal naturalists' fall into the trap of using the Constitution to justify itself, by referring to Section 1(c). The Constitution itself makes no reference to a prior 'social contract' and instead bases its own legitimacy on the common will, as can be seen in the Preamble.
The 'common will' is problematic as a justification for the existence of the Constitution (and by extension, all other legislation), with simple reference to something called the nemo plus juris – a common law maxim. Translated from Latin, the rule basically says that no person can delegate or transfer a right which he does not have himself, to someone else. To put it more simply: you cannot give someone else something if it's not even yours to begin with. This is not only a rule in law, but ordinary common sense. Needless to say, the nemo plus juris rule in the South African context forms a foundational element in Libertarian Legal Theory.
In light of this, what is Libertarian Legal Theory's position on democracy? In my opinion, it has none. 'Democracy' has become a nice buzzword in legal-political speak over the past century, and almost by default implies 'a good system'. However, as I said above, Libertarian Legal Theory, and arguably libertarianism itself, concerns itself with initiatory State force. How the State is elected or composed is irrelevant. Where democracy does become relevant is when the citizenry attempt to transfer rights, entitlements and obligations which they themselves do not possess, to others (usually the State). The nemo plus juris rule logically precludes citizens from empowering the State to initiate force against fellow citizens or foreigners.
My endorsement of some common law principles above should not mislead you to believe that the South African common law is libertarian in nature. While our 'Roman-Dutch' common law does have distinctively libertarian features, as Professor John Dugard has written, and while it is certainly more libertarian than our Constitution and statutory law, it is certainly not by itself libertarian. This is why I believe Libertarian Legal Theory should be a legal approach in and of itself, with common sense influences, where compatible, from the common law.
In summary, Libertarian Legal Theory is an evolution from the (classical) liberal theories of law which were observed in the United States, and to an extent in the United Kingdom, before and in the early parts of the twentieth century. South Africa, unfortunately, has no such history and has arguably always had a very collectivistic and statocentric approach to law. The principles underlying Libertarian Legal Theory are self-ownership, informed consent and nemo plus juris. The non-aggression principle is the foundational principle of libertarianism as a whole and overarches Libertarian Legal Theory. These principles, when there is uncertainty, can find application by relying on common law methods and rules which have developed over the centuries.
I invite any lawyers, law students or others knowledgeable in the law to contribute to my project to write a comprehensive analysis (or formulation?) of Libertarian Legal Theory from the South African legal perspective. Libertarian Legal Theory has only to an extent been developed in the Anglo-American common law, and I believe there is intellectual value in translating it into the Roman-Dutch tradition.
Libertarian Legal Theory
Corporate Responsibility and Shifting the Blame
South African Is Losing Its Liberties, New Report Shows | {
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Cape Pallarenda Conservation Park
Townsville, Townsville Area
Cape Pallarenda was a quarantine station in the early 1900s and a strategic defence location in World War II.
Nestled in a scenic coastal location amongst open woodland and vine thickets, the historic quarantine station, established in 1915, was initially used to quarantine passengers on incoming ships. During World War II the area became a strategic defence location. Concrete structures were built on the headland in 1943 to protect Townsville and the harbour from raiding enemy ships. American and Australian armies set up camps on nearby beaches and used the Quarantine Station as a hospital.
Spend time in the station's historic display centre to find out more about the quarantine days. Then set off on foot or by mountain bike to explore the shared Cape Pallarenda Trails to enjoy scenic coastal views and discover the historic World War II structures on Cape Pallarenda headland. Choose from short strolls to longer hikes or rides around the slopes of Many Peak Range. Explore picturesque beaches and forested slopes of Many Peak Range. Enjoy a picnic on the foreshore. Look for wallabies, lizards and many kinds of birds in the woodland.
Cape Pallarenda Road
qpws@npsr.qld.gov.au
https://www.npsr.qld.gov.au/parks/cape-pallarenda | {
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HomeHistorical CollectionsThe International Union for Land Value Taxation and Free Trade, 1926-1991Series Three
Series Three
First IU Conference in Copenhagen, July 1926
W.R. Lester, "Natural Law in Social Life"
Fourth IU Conference Edinburgh, July 29-August 4, 1929
No. X – Henry George, "Is our Civilization Just to Working Men?"
No. 1 – F. Folke and K.J. Kristensen, "The Work for LVT and Free Trade in Denmark, 1926 to 1929"
No. 2 – Jakob E. Lange, "Danish Agrarian and Social Evolution…"
No. 3 – Sophus Berthelesen, "Social Democrats and the Henry George Policy"
No. 4 – E.J. Craigie, "LVT in Australia for Federal, State and Local Purposes"
No. 7 – P.J. O'Regan, "The Progress of the Henry George Movement in New Zealand"
No. 9 – J.L. Björner, "The Banner of Free Trade"
No. 12 – K.J. Kristensen, "Land Valuation in Denmark"
No. 14 – C.H. Nightingale, "Government Valuation of Land in New Zealand"
No. 15 – A. Daudé Bancel and Sam Meyer, "International Peace and Economic Peace"
No. 16 – Carl Marfels, "The Riddle of Modern Society"
No. 19 – A.W. Dakers, "Land Tenure and Taxation—History of English System"
No. 22 – Axel Fraenckel, "The Physiocrats and Henry George"
No. 23 – Engineer D. de Clercq., "The Reclamation of the Zuider Zee"
No. 25 – Charles H. Smithson, "The Beneficence of Natural Law"
No. 29 – Alex Paletta, "The Valuation and Taxation of Land in Germany"
No. 24 – F.G.H. Anderson, "Some Facts, Fallacies and Reflections…"
No. 31 – F.R. Jones, "International Trade, Tariffs and Land-Value Taxation"
No. 33 – Anders Vedel, "War and the Land Question"
No. 38 – Byron W. Holt, "Evils of Unstable Money"
No. 47 – Alan C. Thompson & Arthur W. Roebuck, "The Progress of Henry George Thought in Canada"
Fifth IU Conference in London, September 1-5, 1936
No. 3 – Bue Björner, "Reactions and Results of Trade Restrictions"
No. 6 – S. Vere Pearson, "Causes of Fear"
No. 8 – K.J. Kristensen, "Land Valuation in Denmark"
No. 9 – F.C.R. Douglas, "Official Testimony on Land Value Rating"
No. 10 – William N. McNair, "The Exemption of Improvements and LVT in Pittsburgh"
No. 13 – Abel Brink, "Ten Years of Land Value Taxation in Denmark"
No. 15 – Arthur R. McDougal, "Agriculture and the Economic Depression"
No. 18 – Councillor H. Kolthek, "State Interference in Industry"
No. 19 – W.R. Lester, "True Free Trade and Laissez-Faire"
No. 21 – Pastor Chr. Norlev, "Problems of Population"
No. 22 – A. Daudé-Bancel, "The Land Question in Russia"
No. 25 – Jakob E. Lange, "Henry George. America—Europe"
No. 26 – E.J. McManus, "Remedies for Raising Wages: True and False"
Seventh IU Conference in Swanwick, England, August 14-August 21, 1949
No. 4 – K.J. Kristensen, "Land-Value Taxation in Denmark"
No. 17 – Viggo Starcke, "The History of Land-Taxation in Denmark"
Eighth IU Conference in Odense, Denmark, July 28-August 4, 1952
Ashley Mitchell, "International Trade (Without Statistics)"
No. 2 – Dan Björner, "The Henry George Movement in Denmark"
Tenth IU Conference in West Germany, July 25-August 1, 1959
J. Rupert Mason, "Irrigation and Land Values in California"
David B. Ascher,"The Building: Land-Ratio"
Glenn E. Hoover, "The First Step in Land-Value Taxation"
Eleventh IU Conference in New York, August 30-September 5, 1964
Rolland O'Regan, "Report from New Zealand"
Robert Tideman, "Land Tax Exemptions" and Philip Knab, "Report from Austria"
J.H. Kristensen, "The Georgist Situation in Denmark"
Thirteenth IU Conference in Isle of Man, September 8-15, 1973
Geoffrey W. Etsy, "The Vietnam War—An American Tragedy of Ignorance"
Joseph Zashin, "Land Reform and the Tax Map Plan"
Sixteenth IU Conference in Cambridge, UK, August 4-11, 1984
G.R.A. Dunkley, "Unemployed Marginal Labor and Man's Right to Work"
Anne Haila, "Effects of Land Value Tax on Land Use"
Nineteenth IU Conference in London, March 21-27, 1991
Geoffrey Dunkley, "Land Tenure: A Time Bomb Ticking South Africa" | {
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Q: Enable Button only when 2 TextInput fields are not empty - code and screenshot attached In a Flex mobile app (or any Flex 4 application) - how can you enable/disable a button depending on the contents of 2 other fields?
I am pasting my very simple test code below and the problem with it is the syntax error in Flash Builder 4.7: The entity name must immediately follow the '&' in the entity reference. - which probably means that ampersand is a special character, but how to solve this (probably frequent) problem?
TestApp.mxml:
<?xml version="1.0" encoding="utf-8"?>
<s:ViewNavigatorApplication
xmlns:fx="http://ns.adobe.com/mxml/2009"
xmlns:s="library://ns.adobe.com/flex/spark"
firstView="views.Home"
applicationDPI="160">
</s:ViewNavigatorApplication>
views/Home.mxml:
<?xml version="1.0" encoding="utf-8"?>
<s:View xmlns:fx="http://ns.adobe.com/mxml/2009"
xmlns:s="library://ns.adobe.com/flex/spark"
title="How to enable Login button?">
<s:layout>
<s:VerticalLayout paddingTop="8"
horizontalAlign="center" gap="8" />
</s:layout>
<s:Label text="Username:" />
<s:TextInput id="_username" />
<s:Label text="Password:" />
<s:TextInput id="_password" />
<s:Button id="_login"
label="Login"
enabled="{_username.text.length > 0 && _password.text.length > 0}" />
</s:View>
A: You shall replace && with && when you write it outside of CDATA in mxml.
It is also better to use _username.text!='' instead of _username.text.length > 0 because it will cause warnings at runtime as text is not an eventdispatcher and it can't report length changes. However, it will update button's availablity because text will be changed itself and TextInput will report it causing binding update.
A: when using binding expressions. mxml does not consider && rather write your expression as
& amp ; & amp ;
....also mxml does not read escape characters e.g. \t for tab...rather write your expression in actionscript then use data binding .cheers ;-)
| {
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{"url":"https:\/\/ngtnews.com\/doe-to-provide-7m-for-fuel-engine-co-optimization-tech-advancements","text":"# DOE to Provide \\$7M for Fuel, Engine Co-Optimization Tech\n\nThe U.S. Department of Energy (DOE) says that it will provide up to \\$7 million in project funding to accelerate the introduction of affordable, scalable and sustainable high-performance fuels for use in high-efficiency, low-emission engines as part of the Co-Optimization of Fuels and Engines (Co-Optima) initiative.\n\nAccording to the DOE, this initiative is a collaboration between its Bioenergy Technologies Office (BETO) and Vehicle Technologies Office (VTO) and brings together DOE national laboratories and industry stakeholders to conduct tandem fuel and engine research, development, and deployment assessments. This funding research aims to help maximize energy savings and on-road vehicle performance while dramatically reducing transportation-related petroleum consumption and greenhouse-gas (GHG) emissions.\n\nEligibility for this funding opportunity is restricted to U.S. Institutions of Higher Education and nonprofit research institutions that operate as a division under the U.S. Institutions of Higher Education. This restricted eligibility applies to both prime recipients and sub-recipients, says the DOE.\n\nThis year, BETO and VTO jointly funded a consortium of nine DOE national laboratories to begin a multiyear project in support of the Co-Optima initiative. Projects selected under this funding opportunity will complement the ongoing DOE national laboratory project and support the broader Co-Optima initiative. The DOE says that the recipients selected for funding are expected to interface with the national laboratory consortium throughout the performance of their projects.\n\nAccording to the\u00a0energy department, the national laboratory project includes two parallel research thrusts: 1) improvement of near-term conventional spark-ignition engine efficiency; and 2) enabling the full operability of advanced compression ignition engines. The research cycle for each thrust includes identifying fuel candidates, understanding their characteristics and combustion performance, and determining market-transformation requirements \u2013 such as cost, GHG reduction, feedstock requirements, scalability and infrastructure compatibility \u2013 while actively engaging with stakeholders and future collaborators.\n\nThe DOE is seeking proposals that address one or more of the following sub-topics:\n\n\u2022 Fuel characterization and fuel property prediction;\n\u2022 Kinetic measurement and mechanism development;\n\u2022 Emissions and environmental impact analysis;\n\u2022 Impact of fuel chemistry and fuel properties on particulate emissions;\n\u2022 Small-volume, high-throughput fuel testing; and","date":"2017-07-27 00:55:35","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8197957277297974, \"perplexity\": 10023.217836170013}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-30\/segments\/1500549426693.21\/warc\/CC-MAIN-20170727002123-20170727022123-00153.warc.gz\"}"} | null | null |
Myotonic Dystrophy Type 1
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We're engaging targets previously considered inaccessible and undruggable
Using our proprietary Endosomal Escape Vehicle (EEV™) platform, we aim to develop intracellular therapeutics with the potential to fundamentally improve the standard of care across a wide range of diseases.
Intracellular therapeutics have the potential to alter the treatment landscape for patients suffering from devastating diseases by targeting and engaging the underlying drivers of disease. Despite significant advances in understanding disease drivers, obstacles to effective treatment remain, in part because approximately 75% of all disease-causing targets are located inside of cells. Small molecules can permeate cell membranes but tend to be rapidly cleared by the body before they reach the intended tissue and can be associated with off-target effects.
On the other hand, biological therapeutics are highly targeted and potent but are limited in their ability to reach intracellular targets of interest. These limitations often necessitate high therapeutic doses and can be associated with less-than-optimal therapeutic activity.
We believe our EEV™ platform will enable specific and potent therapeutics to enter cells for treatment using intracellular target engagement, broad biodistribution, deep tissue penetration and a broad potential therapeutic index. Our intracellular therapeutics pass through the phospholipid bilayer, allowing the cell to fold around the therapy and internalize it. They also have the ability to access multiple cell and tissue types including previously difficult to access tissues. By accessing these tissues, we have observed that therapies can be effectively delivered to the cytosol and multiple organelles within cells.
Entrada has developed a patented library of EEVs that enable the intracellular engagement of therapeutics against previously inaccessible and undruggable disease-causing targets.
Endosomal Escape Vehicle
Even when a therapeutic is successful in penetrating a cell, only less than 2% of the drug will escape the early endosome to reach its intended intracellular target. As a result, high doses of drug product are often needed to produce a therapeutic effect, which could potentially cause systemic dose-related toxicity. To effectively capitalize on both known biology and future discoveries, a better way of targeted intracellular delivery of therapeutics is needed.
We believe we have discovered a potential solution.
Entrada's EEV™ platform harnesses the inherent endocytic mechanism of cells. We have observed that our approach enables higher intracellular target engagement, lower drug concentrations and reduced or minimal toxicity compared to alternative approaches. Importantly, our EEV™ platform may solve a fundamental problem related to intracellular target engagement which is independent of cellular uptake – the efficient escape from the early endosome.
In our preclinical studies, we have observed that greater than 90% of EEV-conjugated material is taken up by the tissues of the body. Once inside the cell, these studies indicate that the EEV-conjugated material rapidly and efficiently escapes from the early endosome. Because of the low-pH conditions in the early endosome, the binding affinity of the EEV to the inner endosome wall increases, resulting in the successful formation and budding of unstable vesicles which then collapse and release therapeutic contents into the cell cytosol. In our preclinical studies, we observed that approximately 50% of the EEV-conjugated material escaped the endosome to reach the intracellular disease target, indicating a potentially significant improvement over the less than 2% observed in current biologics.
Given the flexibility and modular nature of the EEV™ platform, we are able to design and develop therapeutics with adequate exposure to intracellular compartments of target cells and tissues at reasonable doses.
Our EEV™ platform is designed to enable the development of intracellular therapeutics that either modulate, inhibit, degrade or replace an intracellular target to correct the underlying disease pathophysiology. It is a simple and scalable construct designed to translate from preclinical to clinical development across our therapeutic programs.
Ultimately, we believe that the significant increase in intracellular target exposure enabled by EEV conjugation has the potential to translate into substantial improvements to the efficacy, safety, tolerability, manufacturability and cost of future medicines.
A Versatile Platform
There's more to explore
Discover our pipeline of programs
Read the latest press releases | {
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<?php
DrawHeader('Balance Report');
if($_REQUEST[modfunc]=='')
{
echo '<BR>';
PopTable('header','Search');
echo "<FORM action=Modules.php?modname=$_REQUEST[modname]&modfunc=list method=POST>";
echo '<TABLE>';
Warehouse('searchstu');
echo '<TR><TD>Balance Between</TD><TD><INPUT type=text name=balance_low> & <INPUT type=text name=balance_high></TD></TR>';
echo '<TR><TD>Balance Not Zero</TD><TD><INPUT type=checkbox name=not_zero value=Y></TD></TR>';
echo '<TR><TD>Balance</TD><TD>';
echo '<SELECT name=editor>
<OPTION value="Student Billing">Student Billing</OPTION>
<OPTION value=Lunch>Lunch</OPTION>
</SELECT>';
echo '</TD></TR>';
PrepareSchool(SessionSchool(),'',SessionCurSchool());
Warehouse('searchgrade');
Warehouse('searchyear');
echo '<TR><TD colspan=2 align=center>';
Buttons('Find','Reset');
echo '</TD></TR>';
echo '</TABLE>';
PopTable('footer');
}
if($_REQUEST[modfunc]=='list')
{
if($_REQUEST[balance_low]>$_REQUEST[balance_high])
{
$tmp = $_REQUEST[balance_high];
$_REQUEST[balance_high] = $_REQUEST[balance_low];
$_REQUEST[balance_low] = $tmp;
}
DrawHeader($_REQUEST[editor].' Balances - School Year: '.DispYear($_REQUEST[year]).' as of '.ProperDate(DBDate()));
if($_REQUEST[editor]=='Student Billing')
{
$sql = "SELECT ssm.STUDENT_ID,ssm.SCHOOL,ssm.GRADE,s.FIRST_NAME,s.LAST_NAME,s.MIDDLE_NAME ";
$sql .= ",(SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_FEES WHERE STUDENT_ID=ssm.STUDENT_ID),0) -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID AND (LUNCH_PAYMENT!='Y' OR LUNCH_PAYMENT IS NULL) ),0)
".FROM_DUAL.") as BALANCE ";
$sql .= "FROM STU_SCHOOL_MEETS ssm,STUDENTS s
WHERE ssm.SYEAR='$_REQUEST[year]' AND s.STUDENT_ID=ssm.STUDENT_ID ";
if(($_REQUEST[balance_low] || $_REQUEST[balance_low]==='0') && ($_REQUEST[balance_high] || $_REQUEST[balance_high]==='0'))
$sql .= "AND (SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID),0) -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_FEES WHERE STUDENT_ID=ssm.STUDENT_ID),0) ".FROM_DUAL.")
BETWEEN '$_REQUEST[balance_low]' AND '$_REQUEST[balance_high]' ";
if($_REQUEST[not_zero]=='Y')
$sql .="AND (SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID),0) -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_FEES WHERE STUDENT_ID=ssm.STUDENT_ID),0) ".FROM_DUAL.")
!= '0' ";
if($_REQUEST[sch])
$sql .= "AND ssm.SCHOOL='$_REQUEST[sch]' ";
if($_REQUEST[grade])
$sql .= "AND ssm.GRADE='$_REQUEST[grade]' ";
if($_REQUEST[stuid])
$sql .= "AND ssm.STUDENT_ID='$_REQUEST[stuid]' ";
if($_REQUEST[first])
$sql .= "AND s.FIRST_NAME LIKE '".strtoupper($_REQUEST[first])."%' ";
if($_REQUEST[last])
$sql .= "AND s.LAST_NAME LIKE '".strtoupper($_REQUEST[last])."%' ";
$sql .= "ORDER BY s.LAST_NAME,s.FIRST_NAME,s.MIDDLE_NAME ";
}
else
{
$sql = "SELECT ssm.STUDENT_ID,ssm.SCHOOL,ssm.GRADE,s.FIRST_NAME,s.LAST_NAME,s.MIDDLE_NAME ";
$sql .= ",(SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT_LUNCH WHERE STUDENT_ID=ssm.STUDENT_ID),0) ".FROM_DUAL.") -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID AND LUNCH_PAYMENT='Y' ),0) as BALANCE ";
$sql .= "FROM STU_SCHOOL_MEETS ssm,STUDENTS s
WHERE ssm.SYEAR='$_REQUEST[year]' AND s.STUDENT_ID=ssm.STUDENT_ID ";
if(($_REQUEST[balance_low] || $_REQUEST[balance_low]==='0') && ($_REQUEST[balance_high] || $_REQUEST[balance_high]==='0'))
$sql .= "AND (SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID AND LUNCH_PAYMENT='Y' ),0) -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT_LUNCH WHERE STUDENT_ID=ssm.STUDENT_ID),0) ".FROM_DUAL.")
BETWEEN '$_REQUEST[balance_low]' AND '$_REQUEST[balance_high]' ";
if($_REQUEST[not_zero]=='Y')
$sql .="AND (SELECT COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_ACT WHERE STUDENT_ID=ssm.STUDENT_ID),0) -
COALESCE((SELECT sum(AMOUNT) FROM STU_BILLING_FEES WHERE STUDENT_ID=ssm.STUDENT_ID),0) ".FROM_DUAL.")
!= '0' ";
if($_REQUEST[sch])
$sql .= "AND ssm.SCHOOL='$_REQUEST[sch]' ";
if($_REQUEST[grade])
$sql .= "AND ssm.GRADE='$_REQUEST[grade]' ";
if($_REQUEST[stuid])
$sql .= "AND ssm.STUDENT_ID='$_REQUEST[stuid]' ";
if($_REQUEST[first])
$sql .= "AND s.FIRST_NAME LIKE '".strtoupper($_REQUEST[first])."%' ";
if($_REQUEST[last])
$sql .= "AND s.LAST_NAME LIKE '".strtoupper($_REQUEST[last])."%' ";
$sql .= "ORDER BY s.LAST_NAME,s.FIRST_NAME,s.MIDDLE_NAME ";
}
$QI = DBQuery($sql);
$RET = DBGet($QI,array('LAST_NAME'=>'GetCapWords','FIRST_NAME'=>'GetCapWords','SCHOOL'=>'GetSchool','GRADE'=>'GetGrade','BALANCE'=>'Currency'));
$columns = array('LAST_NAME'=>'Last Name','FIRST_NAME'=>'First Name','STUDENT_ID'=>'Student ID','SCHOOL'=>'School','GRADE'=>'Grade','BALANCE'=>'Balance');
ListOutput($RET,$columns,'Student','Students');
}
?> | {
"redpajama_set_name": "RedPajamaGithub"
} | 6,878 |
Site Updated April 15, 2019. Pricing, terms and conditions subject to change without prior notice. This is not an offer to sell or lease any real property. All photos shown may not be from specific communities and no representation is warranted in any manner whatsoever. Copyright 2019 Blandford Homes, LLC. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,432 |
Q: Django Haystack - are facets limited to 10? I'm indexing a list of products and each product has a list of categories. Categories are about 1000.
In the search page I would like to list all the categories which have more than 1 result.
I'm collecting facets with:
results.facet_counts()
Unfortunately this just returns the top 10 categories!
I can't seem to find a way to have it return more.
I'm using the latest version of Haystack and ElasticSearch.
Hope someone can help! :)
A: You can specify in your json query the size of your facet, which is the number of entries you will get back:
{
"query" : {
"match_all" : { }
},
"facets" : {
"tag" : {
"terms" : {
"field" : "tag",
"size" : 10
}
}
}
}
You need to configure a high value if you want to have back all the entries.
Not sure how you can add this parameter with the library you're using. Hope the answer will be useful anyway.
A: You can override the get_queryset() method of your custom FacetedSearchView as:
def get_queryset(self):
options = {"size": 0} #unlimited results, use with caution
qs = super().get_queryset()
for field in self.facet_fields:
qs = qs.facet(field, **options)
return qs
You can also define a global setting ELASTIC_FACET_SIZE and use that instead.
from settings import ELASTIC_FACET_SIZE
def get_queryset(self):
options = {"size": ELASTIC_FACET_SIZE} #unlimited results, use with caution
qs = super().get_queryset()
for field in self.facet_fields:
qs = qs.facet(field, **options)
return qs
This way you won't have to mess with the haystack package source code
A: Now they are limited to 100, see https://github.com/toastdriven/django-haystack/issues/658
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,408 |
Pterolophia apicefusca is a species of beetle in the family Cerambycidae. It was described by Stephan von Breuning in 1938.
References
apicefusca
Beetles described in 1938 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,691 |
{"url":"https:\/\/csns.cysun.org\/department\/cs\/forum\/topic\/view?id=5318082","text":"Author Message\najoshi6\nPosts: 46\nPosted 23:41 Feb 13, 2016 |\n\nhello professor,\n\nIn homework 4 , for Point-3 we have to add department, program and additional details.\n\nwe are adding them but only department is populated not additional details nor program.\n\nhow this can be done?\n\nthank you\n\ncysun\nPosts: 2935\nPosted 08:46 Feb 14, 2016 |\n\nChances are you either need to add @Transactional to the DAO methods that update the database, or add {CascadeType.MERGE, CascadeType.PERSIST} or CascadeType.ALL to the collections to be saved.","date":"2023-03-25 14:28:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.25237029790878296, \"perplexity\": 11510.072145744989}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296945333.53\/warc\/CC-MAIN-20230325130029-20230325160029-00289.warc.gz\"}"} | null | null |
package com.axt.jvmcallgraph;
import java.util.Collection;
import java.util.HashSet;
import java.util.Set;
import org.objectweb.asm.ClassVisitor;
import org.objectweb.asm.MethodVisitor;
import org.objectweb.asm.Opcodes;
import com.google.common.collect.HashMultimap;
import com.google.common.collect.Multimap;
class MethodCallCollector extends ClassVisitor {
private String className;
private Set<MethodInfo> calledMethods;
private Multimap<MethodInfo, MethodInfo> calleeMethods = HashMultimap.create();
public MethodCallCollector(Collection<MethodInfo> calledMethods) {
super(Opcodes.ASM5);
this.calledMethods = new HashSet<MethodInfo>(calledMethods);
}
@Override
public MethodVisitor visitMethod(int access, final String outerName, final String outerDesc, String signature, String[] exceptions) {
return new MethodVisitor(Opcodes.ASM5) {
@Override
public void visitMethodInsn(int opcode, String owner, String name, String desc, boolean itf) {
MethodInfo mi = new MethodInfo(owner, name, desc);
if (calledMethods.contains(mi)) {
MethodInfo outerMethodInfo = new MethodInfo(className, outerName, outerDesc);
outerMethodInfo.setAccess(access);
calleeMethods.put(mi, outerMethodInfo);
// int INVOKEVIRTUAL = 182;
// int INVOKESPECIAL = 183;
// int INVOKESTATIC = 184;
// int INVOKEINTERFACE = 185;
//TODO
// int INVOKEDYNAMIC = 186; // visitInvokeDynamicInsn
// System.out.println("CALL " + outerMethodInfo + " -> " + mi + " " + opcode);
}
}
};
}
public void setClassName(String className) {
this.className = className;
}
Collection<MethodInfo> getCallees(MethodInfo method) {
return this.calleeMethods.get(method);
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 8,689 |
Lepidiota caudata är en skalbaggsart som beskrevs av Blackburn 1890. Lepidiota caudata ingår i släktet Lepidiota och familjen Melolonthidae. Inga underarter finns listade i Catalogue of Life.
Källor
Skalbaggar
caudata | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,773 |
Sebastian de Souza est un acteur britannique né le à Oxford. Il est connu pour son rôle dans la troisième génération de la série Skins, interprétant le personnage de Matthew « Matty » Levan.
Biographie
Sebastian Denis de Souza à Oxford au Royaume-Uni, le . Il est le second fils de Elinor Kelly et , un producteur d'opéra portugais. Il a un frère aîné nommé Tristan. Il est issu d'une famille d'origine portugaise et indienne.
Élève à la Brockhurst et Marlston House Preparatory School à Boxford, Sebastian développe une grande fascination pour le théâtre, la musique et l'écriture. À l'âge de 11 ans, il joue dans la pièce de théâtre de son école, dans une grange, l'histoire de The Willows at Christmas, une adaptation de William Horwood. En 2006, alors qu'il n'avait que 13 ans, il a co-écrit avec le club cinéma de son école et joue dans un court métrage nommé Hangman, qui sera présenté en première à la . Cette même année, il remporte un prix de musique de la pour ses prestations en tant que pianiste, clarinettiste et chanteur.
Plus tard, il interprétera de nombreux rôles dans de diverses pièces de théâtre tels que Le Songe d'une nuit d'été, Les Grandes Espérances ou encore Henri V, à la . Il remportera par ailleurs un prix pour ses interprétations en 2010.
Il entame par la suite ses études au pensionnat de dans le comté de Berkshire et à la à Oxford, où il a obtenu un baccalauréat en anglais, histoire et philosophie. Durant cette période, il a également suivi les ateliers de théâtre au , près de Newbury. Depuis 2010, il est membre du National Youth Theatre.
Filmographie
Acteur
Cinéma
2014 : Plastic de Julian Gilbey : Rafa
2016 : Kids in Love de Chris Foggin : Milo
2018 : Ophélie (Ophelia) de Claire McCarthy : Edmund
Télévision
Séries télévisées
2011-2012 : Skins : Matthew « Matty » Levan (14 épisodes)
2012-2013 : The Borgias : Alphonse d'Aragon (11 épisodes)
2015 : Crossing Lines : Matteo (saison 3, épisode 4)
2016 : Recovery Road : Wes Stewart (10 épisodes)
2018 : Les Médicis : Maîtres de Florence : Sandro Botticelli (saison 3, récurrent saison 2 - 16 épisodes)
2020 : Normal People : Gareth (2 épisodes)
2020 : The Great : Leo (8 épisodes)
Téléfilms
2017 : Maigret au Picratt's : Philippe Martinot
Scénariste
Cinéma
2016 : Kids in Love
2018 : Ophélia: Edmund
Notes et références
Liens externes
interview « Meet The 'Skins': Sebastian De Souza » digitalspy.co.uk, 4 min 17 s
Acteur britannique
Naissance en avril 1993
Naissance à Oxford | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,789 |
Gustav Müller may refer to:
Gustav Müller (astronomer) (1851–1925), German astronomer
Gustav Müller (serial killer) (1865–?), German bigamist, murderer and serial killer
Gustav Wilhelm Müller (1857–1940), German zoologist
Gustl Müller (1903–1989), German Nordic combined and cross-country skier
See also
Gustave Mueller (disambiguation) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,275 |
Many people are aware of the need of gutters in their respective homes but most don't take the necessary measures to maintain and keep the gutter in working condition. Homeowners need to cultivate the habit of cleaning their gutters frequently and not waiting until something goes wrong. To avoid disaster in your homes, you need to be more proactive in maintaining the sanity of your gutter.
It is very easy to forget about it especially when you have many things to take care of, but failure to clean your gutters can have repercussions. So, how often do you need to clean your gutter? Why do you need to clean your gutter? What are the consequences of not cleaning your gutter? All these questions will be answered in this article.
Firstly, gutters mainly prevent water damage and convey water droplets from the roof to other locations where they are linked with other water bodies.
How often should you clean your gutter?
Strive to get your gutter cleaned at least twice in a year, preferably between the spring and the autumn. This is considered as the ideal time because rainfall is at its peak during this season and leaves tend to wither and fall at the same time.
If there is a need for you to clean your gutter more than twice, you should ensure it is done. This is common among homeowners that have trees close to their houses which lead to leaves clogging in gutters. You should also inspect your gutter once in 3 months if possible depending on the foliage of the resident trees and the amount of rainfall.
Why do you need to clean your gutter?
Cleaning your gutter lowers the risk of any damage to your house as clean gutters avoid water damage due to erosion and reduces the process as water is effectively drained to the gutters.
In addition to rusting, houses that have many trees around it can be prone to fire outbreak which threatens that vicinity. Flammable debris and dry leaves can lead to fire disaster which won't be restricted to the gutter and can easily spread resulting into a serious home fire.
What are the consequences of not cleaning your gutter?
Inadequate cleaning of your gutter attracts various pests such as mosquitoes (and other such insects) mice, squirrels. Damaged gutters will cause cracks and holes in your buildings from which other pests like rodents and insects can gain access to your house.
Stagnant water in the gutter can also lead to various bacteria and fungi growing in that area thereby increasing the pollen counts and making life difficult for people who are allergic to such or prone to sickness.
Clogged gutters can also be responsible for erosion, altering the foundation and destroying your gardens. This can lead to water flowing into your windows and wall, leading to severe damage that can be expensive to fix.
It is therefore compulsory to clean your gutter frequently and effectively to prevent these consequences and stay healthy. You wouldn't want to be spending your cash on hospital bills and repairs of you house. To avoid these, ensure you maintain your gutter. If you feel you are not up to the task, you can hire professionals who clean and repair gutters.
Watch the video and you will see the most powerful vac of its type in the world.
Using the worlds most powerful gutter cleaning vac I can access most gutters, including over conservatories etc without so much as leaving the ground.
Gutter Cleaning…But not as you know it!
When was the last time you had your gutters cleaned out?
Have you not been able to have them cleaned due to obstructions such as conservatories or other such buildings? With the Worlds most powerful gutter clearing vac that's no longer an problem!
The worlds highest powered gutter cleaning machine.
Poor access or conservatories not an issue.
No damage to gutters from ladders.
Also as most gutter waste is ideal for composting I would require that I am permitted to empty the extracted gutter contents either in your garden waste bin, your composter or on your garden.
Whilst this is fantastic for cleaning out gutters in what would normally be hard to access locations such as above conservatories etc the SkyVac equipment still needs to be able to move around your property. So Access to all areas is required and requires a min gap equivalent to a normal door gap i required.
Safety is paramount and if high winds or an electrical thunderstorm is imminent I will have to reschedule. Also parked cars which sit under, or close to the gutter line will need to be moved.
For business premises please contact me for a quote.
As you can see height and obstructions are rarely and issue.
Check out my online calendar for bookings.
Some limited Saturday AM slots available upon request. Please email me direct for Sat availability. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,684 |
{"url":"https:\/\/www.groundai.com\/project\/central-exclusive-diffractive-production-of-k-k-k-k-via-the-intermediate-state-in-proton-proton-collisions\/","text":"Central exclusive diffractive production of K^{+}K^{-}K^{+}K^{-}via the intermediate \\phi\\phi state in proton-proton collisions\n\n# Central exclusive diffractive production of K+k\u2212K+k\u2212via the intermediate \u03d5\u03d5 state in proton-proton collisions\n\nPiotr Lebiedowicz Institute of Nuclear Physics Polish Academy of Sciences, Radzikowskiego 152, PL-31342 Krak\u00f3w, Poland \u2003\u2003 Otto Nachtmann Institut f\u00fcr Theoretische Physik, Universit\u00e4t Heidelberg, Philosophenweg 16, D-69120 Heidelberg, Germany \u2003\u2003 Antoni Szczurek 111Also at Faculty of Mathematics and Natural Sciences, University of Rzesz\u00f3w, Pigonia 1, PL-35310 Rzesz\u00f3w, Poland. Institute of Nuclear Physics Polish Academy of Sciences, Radzikowskiego 152, PL-31342 Krak\u00f3w, Poland\n###### Abstract\n\nWe present a study of the exclusive reaction at high energies. We consider diffractive mechanisms with the intermediate state with its decay into the system. We include the -channel exchanges and the -channel exchange mechanism. This state is a candidate for a tensor glueball. We discuss the possibility to use the process in identifying the odderon exchange. An upper limit for the coupling is extracted from the WA102 experimental data. The amplitudes for the processes are formulated within the tensor-pomeron and vector-odderon approach. We adjust parameters of our model to the WA102 data and present several predictions for the ALICE, ATLAS, CMS and LHCb experiments. The distributions in four-kaon invariant mass, in rapidity distance between the two mesons, in a special \u201cglueball filter variable\u201d, and in proton-proton relative azimuthal angle are presented. A measurable cross section of order of a few nb is obtained including the experimental cuts relevant for the LHC experiments. The distribution in rapidity difference of both -mesons could shed light on the coupling, not known at present. We discuss the possible role of the , , and resonances observed in the channel in radiative decays of .\n\n## I Introduction\n\nDiffractive studies are one of the important parts of the physics programme for the RHIC and LHC experiments. A particularly interesting class are central-exclusive-production (CEP) processes, where all centrally produced particles are detected; see Sect.\u00a05 of N.Cartiglia:2015gve (). In the last years there was a renewed interest in exclusive production of pairs at high energies related to successful experiments by the CDF Aaltonen:2015uva () and the CMS Khachatryan:2017xsi () Collaborations. These measurements are important in the context of resonance production, in particular, in searches for glueballs. The experimental data on central exclusive production measured at Fermilab and CERN all show visible structures in the invariant mass. As we discussed in Ref.\u00a0Lebiedowicz:2016ioh () the pattern of these structures has mainly resonant origin and is very sensitive to the cuts used in a particular experiment (usually these cuts are different for different experiments). In the CDF and CMS experiments only large rapidity gaps around the centrally produced dimeson system are checked but the forward and backward going (anti)protons are not detected. Preliminary results of similar CEP studies have been presented by the ALICE Schicker:2012nn () and LHCb McNulty:2016sor () Collaborations at the LHC. Although such results will have diffractive nature, further efforts are needed to ensure their exclusivity. Ongoing and planned experiments at RHIC (see e.g. Sikora:2018cyk ()) and future experiments at the LHC will be able to detect all particles produced in central exclusive processes, including the forward and backward going protons. Feasibility studies for the process with tagging of the scattered protons as carried out for the ATLAS and ALFA detectors are shown in Staszewski:2011bg (). Similar possibilities exist using the CMS and TOTEM detectors; see, e.g., Albrow:2014lrm ().\n\nIt was known for a long time that the frequently used vector-pomeron model has problems from the point of view of field theory. Taken literally it gives opposite signs for and total cross sections. A way how to solve these problems was discussed in Nachtmann:1991ua () where the pomeron was described as a coherent superposition of exchanges with spin 2 + 4 + 6 + \u2026 . The same idea is realised in the tensor-pomeron model formulated in Ewerz:2013kda (). In this model pomeron exchange can effectively be treated as the exchange of a rank-2 symmetric tensor. In Ewerz:2016onn () it was shown that the tensor-pomeron model is consistent with the experimental data on the helicity structure of proton-proton elastic scattering at \u00a0GeV and small from the STAR experiment\u00a0Adamczyk:2012kn (). In\u00a0Ref.\u00a0Lebiedowicz:2013ika () the tensor-pomeron model was applied to the diffractive production of several scalar and pseudoscalar mesons in the reaction . In Bolz:2014mya () an extensive study of the photoproduction reaction in the framework of the tensor-pomeron model was presented. The resonant () and non-resonant (Drell-S\u00f6ding) photon-pomeron\/reggeon production in collisions was studied in Lebiedowicz:2014bea (). The\u00a0central exclusive diffractive production of continuum together with the dominant scalar , , and tensor resonances was studied by us in Lebiedowicz:2016ioh (). The meson production associated with a very forward\/backward system in the and processes was discussed in Lebiedowicz:2016ryp (). Also the central exclusive production via the intermediate and states in collisions was considered in Lebiedowicz:2016zka (). In Lebiedowicz:2018sdt () the reaction was studied. Recently, in Lebiedowicz:2018eui (), the exclusive diffractive production of the in the continuum and via the dominant scalar , , , and tensor , resonances, as well as the photoproduction contributions, were discussed in detail. In Lebiedowicz:2019por () a possibility to extract the couplings from the analysis of pion angular distributions in the Collins-Soper system of reference was studied.\n\nThe identification of glueballs in the reaction, being analysed by the STAR, ALICE, ATLAS, CMS and LHCb Collaborations, can be rather difficult as the dipion spectrum is dominated by the states and mixing of the pure glueball states with nearby mesons is possible. The partial wave analyses of future experimental data could be used in this context. Studies of different decay channels in central exclusive production would be very valuable. One of the promising reactions is with both mesons decaying into the channel.\n\nThe advantage of this process for experimental studies is the following. The is a narrow resonance and it can be easily identified in the spectra. On the other hand, non- backgrounds in these spectra should have a broad distribution. However, identification of possible glueball-like states in this channel requires calculation\/estimation both of resonant and continuum processes. It is known from the WA102 analysis of various channels that the so-called \u201cglueball-filter variable\u201d () Close:1997pj (), defined by the difference of the transverse momentum vectors of the outgoing protons, can be used to select out known states from non- candidates. It was observed by the WA102 Collaboration (see, e.g., Barberis:1996iq (); Barberis:1997ve (); Barberis:1998ax (); Barberis:1999cq (); Barberis:2000em (), Kirk:2000ws (); Kirk:2014nwa ()) that all the undisputed states are suppressed at small in contrast to glueball candidates. It is therefore interesting to make a similar study of the dependence for the system decaying into in central collisions at the LHC.\n\nStructures in the invariant-mass spectrum were observed by several experiments. Broad structures around 2.3\u00a0GeV were reported in the inclusive reaction Booth:1985kv (); Booth:1985kr (), in the exclusive Etkin:1985se (); Etkin:1987rj () and Aston:1989gx (); Aston:1990wf () reactions, in central production Armstrong:1986ky (); Armstrong:1989hz (); Barberis:1998bq (), and in annihilations Evangelista:1998zg (). In the radiative decay an enhancement near \u00a0GeV with preferred was observed Bisello:1986pt (); Bai:1990hk (); Ablikim:2008ac (); Ablikim:2016hlu (). The last partial wave analysis Ablikim:2016hlu () shows that the state is significant, but a large contribution from the direct decay of , modeled by a phase space distribution of the system, was also found there. Also the scalar state , and two additional pseudoscalar states, and the were observed. Three tensor states, , and , observed previously in Etkin:1985se (); Etkin:1987rj (), were also observed in . It was concluded there that the tensor spectrum is dominated by the . The nature of these resonances is not understood at present and a tensor glueball has still not been clearly identified. According to lattice-QCD simulations, the lightest tensor glueball has a mass between 2.2\u00a0GeV and 2.4\u00a0GeV, see e.g. Morningstar:1997ff (); Morningstar:1999rf (); Hart:2001fp (); Loan:2005ff (); Gregory:2012hu (); Chen:2005mg (); Sun:2017ipk (). The and states are good candidates to be tensor glueballs. For an experimental work indicating a possible tensor glueball see Longacre:2004jn (). Also lattice-QCD predictions for the production rate of the pure gauge tensor glueball in radiative decays Yang:2013xba () are consistent with the large production rate of the in the Ablikim:2013hq (), Ablikim:2016hlu () and Ablikim:2018izx () channels.\n\nIn the present paper we wish to concentrate on CEP of four charged kaons via the intermediate state. Here we shall give explicit expressions for the amplitudes involving the pomeron-pomeron fusion to () through the continuum processes, due to the - and -channel reggeized -meson, photon, and odderon exchange, as well as through the -channel resonance reaction (). The pseudoscalar mesons having and can also be produced in pomeron-pomeron fusion and may contribute to our reaction if they decay to . Possible candidates are e.g. and which were observed in radiative decays of \u00a0Ablikim:2016hlu (). The same holds for scalar states with and , for example, the scalar meson. We will comment on the possible influence of these contributions for CEP of pairs. Some model parameters will be determined from the comparison to the WA102 experimental data Barberis:1998bq (); Barberis:2000em (). In order to give realistic predictions we shall include absorption effects calculated at the amplitude level and related to the nonperturbative interactions.\n\n## Ii Exclusive diffractive production of four kaons\n\nIn the present paper we consider the process, CEP of four mesons, with the intermediate resonance pair,\n\n pp\u2192pp\u03d5\u03d5\u2192ppK+K\u2212K+K\u2212. (1)\n\nIn Fig.\u00a01 we show diagrams for this process which are expected to be the most important ones at high energies since they involve pomeron exchanges. Fig.\u00a01\u00a0(a) shows the continuum process. In Fig.\u00a01\u00a0(b) we have the process with intermediate production of an resonance\n\n pp\u2192pp(f2\u2192\u03d5\u03d5)\u2192ppK+K\u2212K+K\u2212. (2)\n\nIn the place of the we can also have an - and an -type resonance. That is, we treat effectively the processes (1) - (2) as arising from the process, the central diffractive production of two vector mesons in proton-proton collisions.\n\nThe production of can also occur through diagrams of the type of Fig.\u00a01 but with reggeons in the place of the pomerons. For example, in Fig.\u00a01\u00a0(a) we can replace the pomerons by reggeons and the intermediate by a pomeron. In Fig.\u00a01\u00a0(b) we can replace one or two pomerons by one or two reggeons. For high energies and central production such reggeon contributions are expected to be small and we shall not consider them in our present paper. We shall treat in detail the diagrams with pomeron exchange (Fig.\u00a01) and diagrams involving odderon and also photon exchange; see Figs.\u00a02 and 3 below.\n\nA resonance produced in pomeron-pomeron fusion must have and charge conjugation , but it may have various spin and parity quantum numbers. See e.g. the discussion in Appendix\u00a0A of Lebiedowicz:2013ika ().\n\nIn Table\u00a01 we have listed intermediate resonances that can contribute to the reaction (2) and to other processes with two vector mesons in the final state. It must be noted that the scalar state , and three pseudoscalar states, , and , which were observed in the process Ablikim:2016hlu (), are only listed in PDG\u00a0Tanabashi:2018oca () and are not included in the summary tables. Clearly these states need confirmation.\n\nTo calculate the total cross section for the reactions one has to calculate the 8-dimensional phase-space integral 222In the integration over four-body phase space the transverse momenta of the produced particles (, , , ), the azimuthal angles of the outgoing protons (, ) and the rapidities of the produced mesons (, ) were chosen as integration variables over the phase space. numerically Lebiedowicz:2009pj (). Some modifications of the reaction are needed to simulate the reaction with in the final state. For example, since the is an unstable particle one has to include a smearing of the masses due to their resonance distribution. Then, the general cross-section formula can be written approximately as\n\n \u03c32\u21926 = [B(\u03d5(1020)\u2192K+K\u2212)]2 (3) \u00d7\u222bmax{mX3}2mK\u222bmax{mX4}2mK\u03c32\u21924(...,mX3,mX4)f\u03d5(mX3)f\u03d5(mX4)dmX3dmX4\n\nwith the branching fraction Tanabashi:2018oca (). We use for the calculation of the decay process the spectral function\n\n f\u03d5(mXi)=C\u03d5(1\u22124m2Km2Xi)3\/22\u03c0m2\u03d5\u0393\u03d5(m2Xi\u2212m2\u03d5)2+m2\u03d5\u03932\u03d5, (4)\n\nwhere , is the total width of the resonance, its mass, and is found from the condition\n\n \u222b\u221e2mKf\u03d5(mXi)dmXi=1. (5)\n\nThe quantity smoothly decreases the spectral function when approaching the threshold, , and takes into account the angular momentum of the state.\n\nTo include experimental cuts on charged kaons we perform the decays of mesons isotropically 333This is true for unpolarised \u2019s. In principle our model also makes predictions for the polarisation of the \u2019s and the anisotropies of the resulting decay distributions. Once a good event generator for our reaction is available all these effects should be included. in the rest frames and then use relativistic transformations to the overall center-of-mass frame.\n\nIn principle, there are other processes contributing to the final state, for example, direct continuum production and processes with resonances:\n\n pp\u2192ppK+K\u2212K+K\u2212, (6) pp\u2192ppf0,2K+K\u2212\u2192ppK+K\u2212K+K\u2212, (7) pp\u2192ppf0,2f0,2\u2192ppK+K\u2212K+K\u2212, (8) pp\u2192pp(f2\u2192f0f0)\u2192ppK+K\u2212K+K\u2212. (9)\n\nHere stands for one of the scalar or tensor mesons decaying to . It should be noted that a complete theoretical model of the process should include interference effects of the processes (1), (2), (6) to (9). However, such a detailed study of the reaction will only be necessary once high-energy experimental data for the purely exclusive measurements will be available. We leave this interesting problem for future studies. The GenEx Monte Carlo generator Kycia:2014hea (); Kycia:2017ota () could be used in this context. We refer the reader to Ref.\u00a0Kycia:2017iij () where a first calculation of four-pion continuum production in the reaction with the help of the GenEx code was performed.\n\n## Iii pp\u2192pp\u03d5\u03d5\n\nHere we discuss the exclusive production of in proton-proton collisions,\n\n p(pa,\u03bba)+p(pb,\u03bbb)\u2192p(p1,\u03bb1)+\u03d5(p3,\u03bb3)+\u03d5(p4,\u03bb4)+p(p2,\u03bb2), (10)\n\nwhere , and denote the four-momenta and helicities of the protons and and denote the four-momenta and helicities of the mesons, respectively.\n\nThe amplitude for the reaction (10) can be written as\n\n M\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=(\u03f5(\u03d5)\u03c13(\u03bb3))\u2217(\u03f5(\u03d5)\u03c14(\u03bb4))\u2217M\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5, (11)\n\nwhere are the polarisation vectors of the meson.\n\nWe consider here unpolarised protons in the initial state and no observation of polarisations in the final state. Therefore, we have to insert in (3) the cross section summed over the meson polarisations. The spin sum for a meson of momentum and squared mass is\n\n \u2211\u03bb=0,\u00b11\u03f5(\u03d5)\u03bc(\u03bb)(\u03f5(\u03d5)\u03bd(\u03bb))\u2217=\u2212g\u03bc\u03bd+k\u03bck\u03bdm2X. (12)\n\nBut in our model the terms do not contribute to the cross section since we have the relations\n\n p3\u03c13M\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=0,p4\u03c14M\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=0, (13)\n\nwhich will be shown below in Secs.\u00a0III.1 and III.2.\n\nTaking also into account the statistical factor due to the identity of the two mesons we get for the amplitudes squared (to be inserted in in (3))\n\n (14)\n\nTo give the full physical amplitude for the reaction we should include absorptive corrections to the Born amplitudes discussed below. For the details how to include the -rescattering corrections in the eikonal approximation for the four-body reaction see Sect.\u00a03.3 of Lebiedowicz:2014bea ().\n\n### iii.1 \u03d5-meson exchange mechanism\n\nThe diagram for the production with an intermediate -meson exchange is shown in Fig.\u00a01\u00a0(a). The Born-level amplitude can be written as the sum\n\n M(\u03d5\u2212exchange)\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=M(^t)\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5+M(^u)\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5 (15)\n\nwith the - and -channel amplitudes:\n\n M(^t)\u03c13\u03c14=(\u2212i)\u00afu(p1,\u03bb1)i\u0393(Ppp)\u03bc1\u03bd1(p1,pa)u(pa,\u03bba)i\u0394(P)\u03bc1\u03bd1,\u03b11\u03b21(s13,t1)\u00d7i\u0393(P\u03d5\u03d5)\u03c11\u03c13\u03b11\u03b21(^pt,\u2212p3)i\u0394(\u03d5)\u03c11\u03c12(^pt)i\u0393(P\u03d5\u03d5)\u03c14\u03c12\u03b12\u03b22(p4,^pt)\u00d7i\u0394(P)\u03b12\u03b22,\u03bc2\u03bd2(s24,t2)\u00afu(p2,\u03bb2)i\u0393(Ppp)\u03bc2\u03bd2(p2,pb)u(pb,\u03bbb), (16)\n M(^u)\u03c13\u03c14=(\u2212i)\u00afu(p1,\u03bb1)i\u0393(Ppp)\u03bc1\u03bd1(p1,pa)u(pa,\u03bba)i\u0394(P)\u03bc1\u03bd1,\u03b11\u03b21(s14,t1)\u00d7i\u0393(P\u03d5\u03d5)\u03c14\u03c11\u03b11\u03b21(p4,^pu)i\u0394(\u03d5)\u03c11\u03c12(^pu)i\u0393(P\u03d5\u03d5)\u03c12\u03c13\u03b12\u03b22(^pu,\u2212p3)\u00d7i\u0394(P)\u03b12\u03b22,\u03bc2\u03bd2(s23,t2)\u00afu(p2,\u03bb2)i\u0393(Ppp)\u03bc2\u03bd2(p2,pb)u(pb,\u03bbb), (17)\n\nwhere , , , , . Here and denote the effective propagator and proton vertex function, respectively, for the tensorial pomeron. The corresponding expressions, as given in Sect.\u00a03 of Ewerz:2013kda (), are as follows\n\n i\u0394(P)\u03bc\u03bd,\u03ba\u03bb(s,t) = 14s(g\u03bc\u03bag\u03bd\u03bb+g\u03bc\u03bbg\u03bd\u03ba\u221212g\u03bc\u03bdg\u03ba\u03bb)(\u2212is\u03b1\u2032P)\u03b1P(t)\u22121, (18) i\u0393(Ppp)\u03bc\u03bd(p\u2032,p) = \u2212i3\u03b2PNNF1(t){12[\u03b3\u03bc(p\u2032+p)\u03bd+\u03b3\u03bd(p\u2032+p)\u03bc]\u221214g\u03bc\u03bd(p\/\u2032+p\/)}, (19)\n\nwhere \u00a0GeV. For extensive discussions of the properties of these terms we refer to Ewerz:2013kda (). Here the pomeron trajectory is assumed to be of standard linear form, see e.g. Donnachie:1992ny (); Donnachie:2002en (),\n\n \u03b1P(t)=\u03b1P(0)+\u03b1\u2032Pt,\u03b1P(0)=1.0808,\u03b1\u2032P=0.25GeV\u22122. (20)\n\nOur ansatz for the vertex follows the one for the in (3.47) of Ewerz:2013kda () with the replacements and . This was already used in Sect.\u00a0IV\u00a0B of Lebiedowicz:2018eui ().\n\nIn the hadronic vertices we should take into account form factors since the hadrons are extended objects. The form factors in (19) and in the vertex are chosen here as the electromagnetic form factors only for simplicity,\n\n F1(t)=4m2p\u22122.79t(4m2p\u2212t)(1\u2212t\/m2D)2, (21) FM(t)=11\u2212t\/\u039b20; (22)\n\nsee Eqs.\u00a0(3.29) and (3.34) of Ewerz:2013kda (), respectively. In (21) is the proton mass and \u00a0GeV is the dipole mass squared. As we discussed in Fig.\u00a06 of Lebiedowicz:2018eui () we should take in (22) \u00a0GeV instead of \u00a0GeV used for the vertex in Ewerz:2013kda ().\n\nThen, with the expressions for the propagators, vertices, and form factors, from Ewerz:2013kda () can be written in the high-energy approximation as\n\n M(\u03d5\u2212exchange)\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=2(p1+pa)\u03bc1(p1+pa)\u03bd1\u03b4\u03bb1\u03bbaF1(t1)FM(t1)\u00d7{V\u03c13\u03c11\u03bc1\u03bd1(s13,t1,^pt,p3)\u0394(\u03d5)\u03c11\u03c12(^pt)V\u03c14\u03c12\u03bc2\u03bd2(s24,t2,\u2212^pt,p4)[^F\u03d5(^p2t)]2+V\u03c14\u03c11\u03bc1\u03bd1(s14,t1,\u2212^pu,p4)\u0394(\u03d5)\u03c11\u03c12(^pu)V\u03c13\u03c12\u03bc2\u03bd2(s23,t2,^pu,p3)[^F\u03d5(^p2u)]2}\u00d72(p2+pb)\u03bc2(p2+pb)\u03bd2\u03b4\u03bb2\u03bbbF1(t2)FM(t2), (23)\n\n (24)\n\nwith two rank-four tensor functions,\n\n \u0393(0)\u03bc\u03bd\u03ba\u03bb(k1,k2)=[(k1\u22c5k2)g\u03bc\u03bd\u2212k2\u03bck1\u03bd][k1\u03bak2\u03bb+k2\u03bak1\u03bb\u221212(k1\u22c5k2)g\u03ba\u03bb], (25) \u0393(2)\u03bc\u03bd\u03ba\u03bb(k1,k2)=(k1\u22c5k2)(g\u03bc\u03bag\u03bd\u03bb+g\u03bc\u03bbg\u03bd\u03ba)+g\u03bc\u03bd(k1\u03bak2\u03bb+k2\u03bak1\u03bb) \u2212k1\u03bdk2\u03bbg\u03bc\u03ba\u2212k1\u03bdk2\u03bag\u03bc\u03bb\u2212k2\u03bck1\u03bbg\u03bd\u03ba\u2212k2\u03bck1\u03bag\u03bd\u03bb \u2212[(k1\u22c5k2)g\u03bc\u03bd\u2212k2\u03bck1\u03bd]g\u03ba\u03bb; (26)\n\nsee Eqs.\u00a0(3.18) and (3.19) of Ewerz:2013kda (). In Lebiedowicz:2018eui () we have fixed the coupling parameters of the tensor pomeron to the meson based on the HERA experimental data for the reaction Derrick:1996af (); Breitweg:1999jy (). We take the coupling constants \u00a0GeV and \u00a0GeV from Table\u00a0II of Lebiedowicz:2018eui () (see also Sect. IV\u00a0B there). The relations (13) are now easily checked from (23) - (III.1) using the properties of the above tensorial functions; see (3.21) of Ewerz:2013kda (). We can then make in (23) the following replacement for the -meson propagator\n\n \u0394(\u03d5)\u03c11\u03c12(^p)\u2192\u2212g\u03c11\u03c12\u0394(\u03d5)T(^p2), (27)\n\nwhere we take the simple Breit-Wigner expression for .\n\nThe amplitude (23) contains a form factor taking into account the off-shell dependences of the intermediate -mesons. The form factor is normalized to unity at the on-shell point and parametrised here in the exponential form\n\n ^F\u03d5(^p2)=exp\u239b\u239d^p2\u2212m2\u03d5\u039b2off,E\u239e\u23a0, (28)\n\nwhere the cut-off parameter could be adjusted to experimental data.\n\nWe should take into account the fact that the exchanged intermediate object is not a simple spin-1 particle ( meson) but may correspond to a Regge exchange, that is, the reggeization of the intermediate meson is necessary, see e.g. Lebiedowicz:2016zka (). The \u201creggeization\u201d of the amplitude given in Eq.\u00a0(23) is included here approximately, by replacing the -meson propagator both in the - and -channel amplitudes by Lebiedowicz:2016zka ()\n\n \u0394(\u03d5)\u03c11\u03c12(^p)\u2192\u0394(\u03d5)\u03c11\u03c12(^p)(s34s0)\u03b1\u03d5(^p2)\u22121, (29)\n\nwhere we take and with from Collins:1977 () and \u00a0GeV.\n\n### iii.2 f2 resonance production\n\nNow we consider the amplitude for the reaction (10) through the -channel -meson exchange as shown in Fig.\u00a01 (b). The , and mesons could be considered as potential candidates; see Table\u00a01.\n\nThe Born amplitude for the fusion is given by\n\n M(PP\u2192f2\u2192\u03d5\u03d5)\u03c13\u03c14\u03bba\u03bbb\u2192\u03bb1\u03bb2\u03d5\u03d5=(\u2212i)\u00afu(p1,\u03bb1)i\u0393(Ppp)\u03bc1\u03bd1(p1,pa)u(pa,\u03bba)i\u0394(P)\u03bc1\u03bd1,\u03b11\u03b21(s1,t1)\u00d7i\u0393(PPf2)\u03b11\u03b21,\u03b12\u03b22,\u03c1\u03c3(q1,q2)i\u0394(f2)\u03c1\u03c3,\u03b1\u03b2(p34)i\u0393(f2\u03d5\u03d5)\u03b1\u03b2\u03c13\u03c14(p3,p4)\u00d7i\u0394(P)\u03b12\u03b22,\u03bc2\u03bd2(s2,t2)\u00afu(p2,\u03bb2)i\u0393(Ppp)\u03bc2\u03bd2(p2,pb)u(pb,\u03bbb), (30)\n\nwhere , , , , , , and .\n\nThe vertex can be written as\n\nA possible choice for the coupling terms is given in Appendix\u00a0A of Lebiedowicz:2016ioh (). The corresponding coupling constants are not known and should be fitted to existing and future experimental data. In the following we shall, for the purpose of orientation, assume that only is unequal to zero.\n\nIn practical calculations we take the factorized form for the form factor\n\n ~F(PPf2)(q21,q22,p234)=~FM(q21)~FM(q22)F(PPf2)(p234) (32)\n\nnormalised to . We will further set\n\n ~FM(t)=11\u2212t\/~\u039b20,~\u039b20=1GeV2; (33) F(PPf2)(p234)=exp\u239b\u239d\u2212(p234\u2212m2f2)2\u039b4f2\u239e\u23a0,\u039bf2=1GeV. (34)\n\nFor the vertex we take the following ansatz:\n\n i\u0393(f2\u03d5\u03d5)\u03bc\u03bd\u03ba\u03bb(p3,p4) = i2M30g\u2032f2\u03d5\u03d5\u0393(0)\u03bc\u03bd\u03ba\u03bb(p3,p4)F\u2032(f2\u03d5\u03d5)(p234) (35) \u2212i1M","date":"2021-01-23 22:59:59","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8694485425949097, \"perplexity\": 1943.8469665088323}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-04\/segments\/1610703538741.56\/warc\/CC-MAIN-20210123222657-20210124012657-00084.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.sigterritoires.fr\/index.php\/en\/2018\/10\/12\/","text":"## GIS and Decision Support (5): Theoretical Fundamentals (Part 1)\n\nIn the previous series of articles we have discussed examples of\napplication for the Boolean and fuzzy logic, as well as two commands for ArcGis\nallowing the application of fuzzy logic to geographic information. We will discuss\nnow the theory that underlies this type of treatment.\n\nIntroduction\n\nWe have a set \u03a9 of objects to classify according to a set C\nof criteria. The number of objects is finite. The partial evaluations of the objects\naccording to each criterion take values \u200b\u200bin easily identifiable sets.\nA partial objective will be seen as a fuzzy set restricting the acceptable\nvalues \u200b\u200bof the associated criterion. Therefore, we accept the implicit hypothesis\nthat each objective defines a total order for \u03a9 .\nWe will use as an example the case of a set \u03a9 representing the\npixels of a study area that we wish to classify according to their ability to\nreceive aquaculture breeding sites. The criteria set C is the dataset\nlayers available: bathymetry, slope, substrate, productivity, etc. Each of\nthese info layers adopts easily identifiable values : favourable, somehow favourable,\nunfavourable, and so on.\nFor each layer of information we will set a goal, for example, for bathymetry\nthat is at least favourable, for productivity that is at least unfavourable,\nand so on. The goal is none other than the subset of the acceptable values \u200b\u200bof\nthe info layer.\nFinally, we accept the hypothesis that each layer of information can be\nclassified in its entirety by the set goal, that is to say that we are able for\neach pixel to determine the corresponding value of the layer.","date":"2018-11-14 13:16:29","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8303017616271973, \"perplexity\": 1618.08024617052}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-47\/segments\/1542039742020.26\/warc\/CC-MAIN-20181114125234-20181114151234-00506.warc.gz\"}"} | null | null |
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Sonic boom – Tame Impala's Kevin Parker
PsychRock Staff — November 30, 2015
Tame Impala frontman Kevin Parker at home in Fremantle. Picture: Iain Gillespie/The West Australian
Kevin Parker's hair flows majestically over his shoulders, the Tame Impala frontman depicted as a Norse god or surfer himbo, as psychedelic rainbow riffs explode from his guitar.
The impressive piece of fan art — possibly a gift from a recent tour of South America, where fans are particularly passionate about the Perth psych-rock band — sits among other colourful tributes inside the 29-year-old's South Fremantle home.
A framed Time magazine with rapper Kanye West on the cover hangs nearby. Parker is a fan of the controversial hip-hop superstar — he recently took proud ownership of a signed set-list — but not as fervent as his girlfriend Sophie Lawrence, who works for a Perth advertising firm.
The celebrated musician continues the tour past magazines stacked neatly on a coffee table surrounded by a few empty wine bottles and bespoke ashtrays, and out to the dishevelled backyard.
Where one might assume a successful rock star would sink their gold-plated swimming pool, a ramshackle enclosure for rabbits or chickens lies empty, aside from tangled wire and long grass.
Parker has no time for gardening. Most of his time at home is spent in the studio, consisting of a room full of guitars, another bursting with synthesisers and recording gear, plus a makeshift storeroom for stage lights, so he can work on Tame Impala's live show as well as the music that has made the band one of the hottest rock acts on the planet.
He did manage to give an outside wall one coat of pink paint before embarking on their recent five-week US tour, which culminated in a breathtaking late-afternoon performance to about 50,000 fans at the Governors Ball in New York.
Tame Impala's third album Currents, released on Friday, could well be the biggest album ever released by a WA act in terms of sales and pure unadulterated buzz. The album, which UK music bible NME described as "astronomically anticipated", is expected to easily eclipse the band's previous outing, the ARIA Award-winning and Grammy-nominated Lonerism, which has sold more than 200,000 copies in the US.
Parker wrote, performed, recorded, produced and mixed the new album in his modest home studio. The appropriately titled Lonerism and equally inward-looking 2010 debut Innerspeaker, which features the apt Solitude is Bliss, were also almost entirely solo creations.
"It's not like I'm brooding or miserable or withdrawn, " Parker once told British music magazine Uncut. "I'm just happier on my own."
Parker was born in Sydney but raised in WA. His parents split when he was three and, after initially living with his mother Rosalind, he moved in with his father Jerry, elder brother Stephen and stepmother Rhonda, who lived in Cottesloe.
His dad, an accountant and "closet musician", owned plenty of instruments as well as an impressive record collection dominated by the Beatles, the Beach Boys, the Shadows and Supertramp.
While Parker learnt guitar and the drums from a young age, the big bang for the future globetrotting rocker came in the late 90s, when he worked out how to capture his sonic experiments.
He taped himself bashing away on his dad's drum kit on the home stereo, then cajoled a melody out of a Casio keyboard as the drum loop played, recording both on a second tape deck. And so it went — more layers, more tapes. The rudimentary results left the junior George Martin's "mind blown".
"My world opened up in terms of music, " Parker says in a South Fremantle cafe, where no one gives the world-famous muso a second glance. "I didn't need other people."
Initially, he kept these early songs to himself but eventually shared a few with his best mate and now Tame Impala guitarist, Dominic Simper.
The pair became friends during Year 9 music lessons at John XXIII College in Mt Claremont. "It became instantly obvious that we were the only two people in the class who could play their instruments and cared about music, " Simper says from Leuven, in the Flemish part of Belgium, where he moved recently with his Dutch girlfriend.
Simper and Parker would thrash Nirvana and Rage Against the Machine covers, before Parker began to reveal his homemade recordings. "He'd do it all on his own at home and not even give me a glimpse until he was satisfied that it was finished, " Simper says. "It was much like it is now."
Later they played in the Dee Dee Dums, runners-up in the 2005 Ampfest battle of the bands to fellow travellers Mink Mussel Creek. The two bands merged and Tame Impala evolved out of "a big collection of Perth freaks", as Simper puts it, in late 2007.
Tame Impala in concert and on record are two completely different projects. Simper says Parker's touring buddies don't mind that the recordings are a one-man show. "All the guys in the band are completely free to pursue any creative projects they want to, " he says. "There's no reason we'd feel creatively stifled or anything."
The members of Tame Impala are not only involved with other bands, they're spread out across the globe. London-based guitarist/keyboardist Jay Watson has his Gum solo project, and plays in Pond with former Tame Impala member Nick Allbrook. Bassist Cam Avery, who is living in Los Angeles, leads blues-rockers the Growl and occasionally plays in Pond. (Parker has also played drums with Pond and produces their albums.) Originally from Toulouse, France, drummer Julien Barbagallo now lives in Melbourne.
None of the other projects can match Tame Impala for popularity or acclaim, however. The figurehead of this music-obsessed collective, Parker is frequently described as a genius for his futuristic yet reverential take on trippy guitar-based pop.
When Tame Impala trumped much-fancied, electronic boy wonder Flume at the 2013 ARIA Awards to win album of the year, plus best group and best rock album for Lonerism, Avery paid tribute to Parker on national TV.
"The album Kevin wrote himself, it not only impressed everybody else, it impressed us a whole bunch as well, " the bassist said of the album ranked the best of 2012 by NME, Rolling Stone and Triple J. "He's a prolific, amazing musician and we all appreciate being part of his life."
Simper says labelling Parker as a musical genius doesn't give his close friend enough credit for the sheer hard work put into these seemingly spontaneous sonic creations.
"He's a very rational, scientific guy, " the guitarist says. "He's freakishly talented but it's also hours and hours of work."
Simper concedes Parker has the X factor. "There has to be something else because his music makes a connection with people that you can't quantify."
Parker's longest-serving cohort continues to act as a sounding board. Simper says he was left "literally speechless" last year when he first heard the sprawling eight minutes of Let it Happen, the first single from Currents.
"I got to listen to (the songs) as if I was a fan, " he says. "They're on a different level of songwriting and craft that he hasn't shown before."
Parker may work in much the same way as he has done since childhood but the results are in a different universe.
Yes, I'm Changing — a highlight of the new album — acknowledges that the touring, awards and adulation from fans, critics and fellow musicians have to have some impact. "They say people never change but that's bulls…, they do, " Parker croons in the electronic ballad.
"Generally, the right thing to say is, like 'Oh no, this isn't changing me, I'm still the same old bloke', and try to deny that the excessive amount of travelling or success aren't changing you, " he says, sipping on a long mac. "In the end, it does. It makes you see things differently. It completely rearranges your view on life."
Parker believes the evolution from bedroom recordings to bands formed in Daglish and Shenton Park share-houses to international rock stardom has had a positive influence, citing his new-found self-respect as the biggest personal shift.
"I never used to think I had anything to say or my output amounted to anything, " he says. "I always assumed that whatever was going on in my head was completely removed from any relevance to anyone else.
"I always assumed it was a total fluke, " he laughs, "that I made any music that resonated with anybody."
That's despite Tame Impala almost immediately attracting famous friends, including Flaming Lips leader Wayne Coyne and superstar producer Mark Ronson. The latter enlisted Parker to sing on three tracks from his recent album, Uptown Special. After joining him on stage at last month's Glastonbury festival in the UK, Parker will tour Australia with Ronson this month, including a Perth gig on July 22.
Ronson, who scored a global smash hit with Uptown Funk last year, is a massive fan of Parker. The London-born producer told The West Australian in February that the Tame Impala sound was already hugely influential and "his biggest records are yet to come. The groundswell of love and admiration for the band is pretty unparalleled, I think."
"That was happening from day dot, " Parker says of the big-name support. "One of the first gigs we ever played in London, Noel Gallagher and Noel Fielding came and a bunch of other high-profile people."
Dreamy single Feels Like We Only Go Backwards won APRA Song of the Year in 2013 but not even that peer-voted honour means as much to Parker as the growing numbers of fans turning up at shows.
A new self-respect and confidence came from "going to countries on the other side of the world and seeing people devoted to what you're doing. It's playing a song live and seeing people … literally in tears. That's the most profound effect, not some famous person thinks you're good."
That poise led Parker to "expose himself" on Currents. After submerging his vocals on previous albums, the third outing boasts his most direct and audible lyrics. "Honesty is a surprisingly elusive quality in music, " he says.
Parker's first outing as a musician came mid-2008, when he was offered a record deal by Sydney-based label Modular. He immediately dropped out of an astronomy course at university, releasing a self-titled EP featuring indie hit Half Full Glass of Wine later the same year.
Tame Impala's relationship with Modular, and its founder Stephen Pavlovic in particular, has since dissolved. Tame Impala publisher BMG had taken Pavlovic, the label and partner Universal Music to court for almost $600,000 in unpaid royalties from US sales.
Tame Impala manager Jodie Regan said from London on the eve of Parker's Glastonbury appearance that "the Modular situation" had been brewing for at least two years. She is hell-bent on retrieving the money from Pavlovic, who is no longer employed by the label. (Like the royalties, the former Modular man is missing — Regan assumes he's with family in Croatia.)
"Basically, the band has to win in this situation, " says Regan, who has been based in Los Angeles since March last year. The major labels involved, German giant BMG and the band's current home, Universal imprint Interscope, both stood to "make a lot more money out of Tame Impala. So can someone please take responsibility."
However, she stresses the importance of not letting the ongoing legal battle detract from what Parker and his bandmates have achieved or could achieve in the future.
For his part, Parker says the Modular situation was "the storm on the horizon" but, after initially investigating for himself, he decided to leave the lawyers to sort it out.
"It really screwed with my creativity, " he says. "All of a sudden they started talking about this album being worth this much and this album being worth this much … That whole world is so foreign to me."
Parker remains remarkably sanguine about the alleged exploitation by the label that helped launch Tame Impala into the stratosphere. His glass is half full, not half empty.
"That's the thing, " he says. "I wouldn't be where I am today if they hadn't done that … if that's the trade-off for me being able to make music for a living, release it and travel around the world, then that's a f…ing fair trade-off. I'm happy with that."
Parker relishes being able to play top festivals around the globe, including Australia's own Splendour in the Grass later this month. Tame Impala knocked back Glastonbury this year, preferring a more prominent slot at the iconic festival in 2016. (The touring schedule is already pencilled in until August next year, according to Regan.)
In the end, however, the gigs and record deals are a sideshow to Parker's solo explorations in the studio. While famed US producer Dave Fridmann helped him polish previous albums, Currents was done alone.
"Doing it on your own, you just go insane, " Parker says at the cafe. "Your perspective goes out the window." He admits to doing more than 1000 vocal takes on the R&B-influenced single Cause I'm a Man.
How do you know when a song is done? "You don't, " Parker laughs ruefully. "You just run out of time. You have someone pull the plug on it."
Currents was due in January but the deadline was pushed back several times. Parker worked around the clock to complete the album before jetting out for Tame Impala's performance at California's Coachella Festival in mid-April.
"In the few months leading up to finishing, the album was all day, every day, " Parker says. "There wasn't even a consideration that I would do anything else. It wasn't even like 'It's Friday night, I'm going out to have a drink with friends'. That didn't even exist.
"All the time I was working as though the whole album was going to be done in a few days. Even when I wasn't working on it, even when I was eating, I was thinking about the next thing or I'd fall asleep thinking about it and I'd wake up thinking about it."
Parker isn't quite like Beach Boys genius Brian Wilson, sitting at his grand piano in the sandpit trying to coax heavenly melodies out of a tortured psyche, but he says music is no longer simply for fun.
"It's not really a pleasure these days, " he admits. "It's more instinctual. I wouldn't know how to switch off if I tried. As soon as you conquer something, as soon as you satisfy yourself with one section of music another one opens up. It's an infinite world of discovery and fulfilment."
Perth's musical mastermind will continue to push himself to the limit for the same simple reason he first pressed play and record simultaneously on his father's tape deck. "I could be just around the corner from writing the best song I've ever written."
Source: yahoo.com
PsychRock Staff
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\section*{Appendix}
\renewcommand{\thesubsection}{\Alph{subsection}}
\subsection{Details of the Training Setup}
\label{app-subsec-training-setup}
\subsubsection{Data Augmentation}
\label{app-subsubsec-data-augmentation}
To increase the number of training samples and their diversity, data augmentation was used. The augmentation includes geometric transformation such as flipping (left-right, up-down), rotation of multiple of 90 degrees, random affine transformation (translation -10 to 10 px, scaling 0.9 to 1.1, rotation -5 to 5 degrees, shear -5 to 5 px), random elastic deformation (deformation range -4 to 4 px, grid size of control points 64 px). A training sample has 0.5 probability of being processed with one of the transformations. The probability for applying each transformation is: flipping left-right (1/24), flipping up-down (1/24), rotation of multiple of 90 degrees (1/12), affine transformation (1/6), elastic deformation (1/6), no transformation (1/2). To make the trained model robust to noise, in addition to the geometric transformations, we also augmented data by adding Gaussian noise to the pixel value with a zero mean and a standard deviation between 0.01 and 0.03. The probability of adding the noise is 0.5.
\subsubsection{Training Settings}
\label{app-subsubsec-training-settings}
The $\lambda$ value in the training loss Eq. (\ref{method-eq-total-loss}) was set to 10 to make the scale of the two terms similar. Adam optimizer was used to minimize the loss function with a learning rate 0.0001. The number of training samples in a batch is 4. The network was trained with 100 epochs to ensure convergence.
\subsection{Parameter Tuning for Catheter Tip Tracking}
\label{app-subsec-parameter-tuning}
This section describes the details of tuning the parameters of optical flow and particle filter for catheter tip tracking.
\subsubsection{Tuning Optical Flow Parameters}
\label{exp-subsubsec-of}
The approach of \cite{farneback2003two} was used as the optical flow implementation in Algorithm \ref{method-alg-dl-bayes}. A grid search to find the optimal parameter setting was done on the following parameters of the method: (1) the image scale to build the pyramids, (2) the number pyramid levels, (3) the averaging window size, (4) the number of iterations, (5) the size of the pixel neighborhood used to find polynomial expansion in each pixel, and finally (6) the standard deviation of the Gaussian that is used to smooth derivatives used as a basis for the polynomial expansion.
The above parameters were tuned independently of the deep neural network, as optical flow directly estimates the catheter tip motion between two frames. To tune the parameters, we tracked the catheter tip in X-ray fluoroscopy starting from the ground truth tip position in the first frame using the motion field between two adjacent frames estimated with optical flow. The average and median distance between the tracked tip position and the ground truth were used as the evaluation criteria for the tuning.
The method of \cite{farneback2003two} was implemented by using the OpenCV function \texttt{calcOpticalFlowFarneback}. With consideration of the suggested parameter values from the documentation, the parameter setting chosen for optical flow from the grid search is $\texttt{pyr\_scale}=0.5$, $\texttt{levels}=3$, $\texttt{winsize}=10$, $\texttt{iterations}=30$, $\texttt{poly\_n}=5$, $\texttt{poly\_sigma}=1.1$. Details of the parameters can be found on the function documentation page\footnote{\url{https://docs.opencv.org/2.4/modules/video/doc/motion_analysis_and_object_tracking.html?}}.
\subsubsection{Tuning Particle Filter Parameters}
\label{exp-subsubsec-pf}
The parameters to tune for the particle filter are the number of samples $N_s$ and the variance of process noise $\sigma_v^2$. When tuning them, we fixed the parameters of the trained network and the optical flow method, and used their optimal parameter settings during this experiment. Following Algorithm \ref{method-alg-dl-bayes}, we tracked the catheter tip from the ground truth position (probability map) in the first frame, and used the mean and median distance between the tracked and the true position as the validation metric.
The tracking results on the validation (tracking) set are shown in Table \ref{exp-tab-tune-particle}. The table shows that 100 samples are suboptimal, while 1000 samples seem sufficient, as 10000 samples result in tracking accuracies similar to 1000 samples. It also shows that the optimal choices of the standard deviation of the process noise are 4 or 5 px for the downsampled images. One possible reason for such choices may be that they are similar to the size of guiding catheters. In general, good choices for $N_s$ are 1000 and 10000, for $\sigma_v$ are 4 and 5. By considering the mean, the standard deviation and the median of tracking errors, the parameter setting $\sigma_v=5$, $N_s=1000$ was chosen.
\begin{table*}[th!]
\centering
\caption{Catheter tip tracking errors (mm) on the validation (tracking) dataset of different parameter settings for particle filter. The tracked tip point was rounded to the pixel center. The error of all images (mean $\pm$ std / median) are presented. Red cells show the good choices of parameters; \textit{bold} number indicates the chosen setting.}
\begin{tabular}{c c c c}
\toprule
$\sigma_v$ & \multicolumn{3}{c}{$N_s$} \\
\cmidrule(lr){2-4}
(px) & 100 & 1000 & 10000 \\
\midrule
3 & 1.52 $\pm$ 2.19 / 0.79 & 1.49 $\pm$ 2.18 / 0.79 & 1.48 $\pm$ 2.18 / 0.79 \\
4 & 1.50 $\pm$ 2.17 / 0.79 & \cellcolor{red!30}1.46 $\pm$ 2.17 / 0.79 & \cellcolor{red!30}1.47 $\pm$ 2.18 / 0.79 \\
5 & 1.52 $\pm$ 2.21 / 0.79 & \cellcolor{red!30}\textbf{1.47 $\pm$ 2.17 / 0.74} & \cellcolor{red!30}1.47 $\pm$ 2.19 / 0.74 \\
6 & 1.53 $\pm$ 2.39 / 0.79 & 1.49 $\pm$ 2.33 / 0.79 & 1.48 $\pm$ 2.29 / 0.74 \\
7 & 1.56 $\pm$ 2.42 / 0.79 & 1.50 $\pm$ 2.29 / 0.74 & 1.50 $\pm$ 2.39 / 0.74 \\
8 & 1.58 $\pm$ 2.41 / 0.79 & 1.51 $\pm$ 2.40 / 0.74 & 1.51 $\pm$ 2.42 / 0.74 \\
9 & 1.56 $\pm$ 2.22 / 0.79 & 1.53 $\pm$ 2.43 / 0.79 & 1.52 $\pm$ 2.45 / 0.61 \\
10 & 2.25 $\pm$ 6.18 / 0.79 & 1.54 $\pm$ 2.46 / 0.79 & 1.53 $\pm$ 2.47 / 0.61 \\
\bottomrule
\end{tabular}
\label{exp-tab-tune-particle}
\end{table*}
\subsection{Detection, tracking and roadmapping without catheter segmentation}
\label{app-subsec-detection-only}
Training of the network in Figure \ref{method-figure-network} requires catheter labels for detection and segmentation. As the segmentation labels are often more expensive to acquire than the detection label in practice, we also investigated the performance of catheter tip detection, tracking and dynamic coronary roadmapping without segmenting the catheter. To this end, we used a similar encoder-decoder network architecture as Figure \ref{method-figure-network} which computes only the detection output directly after the last \textit{up} block of the decoder with a $1\times1$ convolution followed by a spatial softmax layer, instead of having two outputs. We then followed the same way as the approach using the network with segmentation to search for (hyper-)parameters for the approach without segmentation. The following parameter setting was chosen for the experiments in this section: for deep learning, the basic channel number is 64, the depth is 5, the dropout rate is zero; for particle filtering, $\sigma_v=3$, $N_s=10000$. With this setup, we compared the performance of the approach without catheter segmentation to the proposed approach with segmentation on catheter tip detection and tracking (Appendix \ref{app-subsubsec-detection-tracking}) and dynamic coronary roadmapping (Appendix \ref{app-subsubsec-roadmapping}) on the test set from Table \ref{setup-tab-tip-experiment}.
\subsubsection{Catheter tip detection and tracking}
\label{app-subsubsec-detection-tracking}
The same metrics in Table \ref{exp-tab-tracking-test} are used to report the accuracy of catheter tip detection and tracking without catheter segmentation. Table \ref{app-tab-tracking-test} and Figure \ref{app-figure-boxplot-all_dist} both manifest that the segmentation sub-task improves the accuracy of catheter tip detection and tracking. Although the improvement on the tracking task is marginal and not statistically significant ($p=0.06$), the segmentation helps to reduce the magnitude and amount of outliers (large errors).
\begin{table*}[h]
\centering
\caption{Catheter tip tracking errors (mm) with and without catheter segmentation on the test (tracking) dataset. $\dagger$ indicates that the difference between the detection with and without segmentation is statistically highly significant with the two-sided Wilcoxon signed-rank test ($p < 0.001$). No statistically significantly difference is observed between the tracking with and without segmentation using the two-sided Wilcoxon signed-rank test ($p = 0.06$).}
\begin{tabular}{l c c c c}
\toprule
\multirow{2}{*}{Evaluation Metrics} & \multicolumn{2}{c}{With Segmentation} & \multicolumn{2}{c}{Without Segmentation} \\
\cmidrule(lr){2-3}
\cmidrule(lr){4-5}
& Detection$\dagger$ & Tracking & Detection & Tracking \\
\midrule
Maximal error of all images & 108.20 & \textbf{17.72} & 133.94 & 23.2 \\
Median error of all images & \textbf{0.96} & \textbf{0.96} & \textbf{0.96} & \textbf{0.96} \\
Mean error of all images & 5.62 $\pm$ 15.91 & \textbf{1.29 $\pm$ 1.76} & 9.32 $\pm$ 19.73 & 1.75 $\pm$ 3.17 \\
\midrule
Average of sequence median error & 6.26 $\pm$ 17.11 & \textbf{1.03 $\pm$ 0.49} & 9.34 $\pm$ 18.64 & 1.42 $\pm$ 2.14 \\
Average of sequence mean error & 6.83 $\pm$ 13.88 & \textbf{1.29 $\pm$ 0.94} & 10.41 $\pm$ 15.94 & 1.69 $\pm$ 1.97 \\
\bottomrule
\end{tabular}
\label{app-tab-tracking-test}
\end{table*}
\begin{figure}[t]
\centering
\subfloat[Overall view of tracking errors]{
\includegraphics[width=0.23\textwidth]{sections/appendix/figures/tracking/boxplot_all_dists.pdf}
\label{app-subfigure-boxplot-all-dist}
}
\subfloat[A zoom-in view of (a)]{
\includegraphics[width=0.23\textwidth]{sections/appendix/figures/tracking/boxplot_all_dists_zoom_in.pdf}
\label{app-subfigure-boxplot-all-dist-zoom-in}
}
\caption{Tracking errors on all test images with and without catheter segmentation.}
\label{app-figure-boxplot-all_dist}
\end{figure}
\subsubsection{Dynamic coronary roadmapping}
\label{app-subsubsec-roadmapping}
In this experiment, the same setup in Section \ref{exp-subsec-roadmapping} was used to assess the accuracy of DCR using catheter tip tracking without segmenting the catheter. Table \ref{app-tab-roadmapping} indicate that tracking the catheter tip with catheter segmentation improves the DCR accuracy compared to that without catheter segmentation. Although the improvement is not statistically significant ($p=0.43$), the approach with segmentation is more robust by making less large roadmapping errors (Figure \ref{app-figure-roadmapping-boxplot}).
\begin{table*}[h]
\centering
\caption{The statistics of DCR accuracy (mm) via catheter tip tracking with and without catheter segmentation. With the two-sided Wilcoxon signed-rank test, no statistically significantly difference is observed between the DCR with and without segmentation ($p = 0.43$).}
\begin{tabular}{l c c}
\toprule
& With Segmentation & Without Segmentation \\
\midrule
\textbf{All point pairs} & & \\
Maximal distance & 20.24 & 25.20 \\
Median distance & 1.43 & 1.43 \\
Mean distance & 2.07 $\pm$ 2.08 & 2.44 $\pm$ 3.15 \\
\midrule
\textbf{Frame mean distance} & & \\
Median distance & 1.42 & 1.40 \\
Average distance & 1.91 $\pm$ 1.52 & 2.23 $\pm$ 2.59 \\
\bottomrule
\end{tabular}
\label{app-tab-roadmapping}
\end{table*}
\begin{figure}
\centering
\includegraphics[width=0.48\textwidth]{sections/appendix/figures/roadmapping/boxplot.png}
\caption{Accuracy (mm) of DCR via catheter tip tracking with and without catheter segmentation.}
\label{app-figure-roadmapping-boxplot}
\end{figure}
\section{Scenario Setup and Method Overview}
\label{method-sec-overview}
The proposed method assumes that the scenario of performing dynamic coronary roadmapping to guide a PCI procedure consists of an offline phase and an online phase. An overview of the method is shown in Figure \ref{method-figure-overview}.
\begin{figure*}[t]
\centering
\includegraphics[width=0.99\textwidth]{sections/methods/figures/overview7.pdf}
\caption{The overview of the proposed dynamic coronary roadmapping method. The colored blocks with a dash line border denote objects acquired in the online phase; the colored blocks with a solid line border are objects originated from the offline phase.}
\label{method-figure-overview}
\end{figure*}
\subsection{Offline Phase}
\label{method-subsubsec-roadmap-preparation}
This phase (Step 0 in Figure \ref{method-figure-overview}) is performed off-line before the actual roadmapping is conducted. In this stage, roadmaps of coronary arteries containing multiple cardiac phases are created from an X-ray angiography sequence acquired with injection of contrast agent. A roadmap can be a vessel model in the form of centerlines, contours, masks, etc. It may also contain information of clinical interest, e.g. stenosis. Since the main focus of this paper is on accurate overlay of a roadmap, we do not investigate how to create the most suitable roadmaps, but use the images containing only vessels and catheters that are created using the layer separation method by \cite{ma2015layer} as the roadmaps to show the concept of dynamic coronary roadmapping. Along with the XA sequence, ECG signals are also acquired and stored for later selecting a roadmap that has similar cardiac phase to a given X-ray fluoroscopic frame in the online phase (see details in Section \ref{method-sec-roadmap}). Once the image sequence and ECG signals are acquired, the catheter tip location in every frame is obtained to serve as a reference point for roadmap transformation. In this work we manually annotated the catheter tip in the offline XA sequence. In real clinical scenarios, the annotation work can be done by the clinician or a person who assists the intervention, such as a technician or a nurse.
\subsection{Online Phase}
\label{method-subsubsec-roadmap-transformation}
This is when the dynamic roadmapping is actually performed. In this phase, non-contrast X-ray fluoroscopic images with the same view angles as the roadmaps created during the offline phase are acquired sequentially. At the same time, ECG signals along with the roadmapping frames are also obtained and are compared with the stored ECG to select the most matched roadmap (Step 1 in Figure \ref{method-figure-overview}; see details in Section \ref{method-sec-roadmap}). This is to compensate the change of vessel shape and position between frames due to cardiac motion. Simultaneously, the catheter tip location in the acquired X-ray fluoroscopic images is tracked online using the proposed deep learning based Bayesian filtering method in Section \ref{method-sec-bayesian-filter} (Step 2 in Figure \ref{method-figure-overview}). The displacement of catheter tip between the current image and the selected roadmap image is then obtained and are applied to transform the roadmap. Finally, the transformed roadmap is overlaid on the current non-contrast frame to guide the procedure (Step 3 in Figure \ref{method-figure-overview}).
\section{ECG matching for Roadmap Selection}
\label{method-sec-roadmap}
Roadmap selection in this work is achieved by comparing the ECG signal associated with the fluoroscopic image and the ECG of the angiographic sequence, such that the most suitable candidate roadmap is selected where the best match of the ECG signals is found. The selected roadmap has the same (or very similar) cardiac phase with the X-ray fluoroscopic image, which compensates the difference of vessel shape and pose induced by cardiac motoin. An approach similar to the ECG matching method by \cite{kim2018registration} is used to accomplish this task.
To select roadmaps images based on ECG, a temporal mapping between X-ray images and ECG signal points needs to be built first. We assume that ECG signals and X-ray images are well synchronized during acqusition. In the offline phase, the beginning and the end of the image sequence are aligned with the start and end ECG signal points; the XA frames in between are then evenly distributed on the timeline of ECG. This way, a mapping between the stored sequence images and its ECG signal can be set up: for each image, the closest ECG signal point to the location of the image on the timeline can be found; for each ECG point, an image that is closest to this point on the timeline can be similarly located. Once the mapping is available, all images with good vessel contrast filling and the ECG points that are associated to these images are selected from the XA sequence for the pool of roadmaps. In this process, at least one heartbeat of frames should be acquired, which is generally the case in our data. In the online phase, similar to the approach of \cite{kim2018registration}, for acquisition of each image, a block of $N_{ECG}$ latest ECG signal points is constantly stored and updated in the history buffer. This is considered as the ECG signal corresponding to the fluoroscopic frame.
To compare the ECG signals associated with the angiographic sequence and the online fluoroscopic image, a temporal registration of the two signals using cross-correlation is applied (\cite{kim2018registration}). The two ECG signals are first cross-correlated for every possible position on the signals, resulting in a 1D vector of correlation scores. The candidate frame for dynamic overlay is then selected as the one associated with the point on the ECG of the angiographic sequence that is corresponding to the highest correlation score.
\section{Bayesian Filtering for Catheter Tip Tracking}
\label{method-sec-bayesian-filter}
Bayesian filtering is a state-space approach aiming at estimating the true state of a system that changes over time from a sequence of noisy measurement made on the system (\cite{arulampalam2002tutorial}). One popular application area of this approach is tracking objects in a series of images.
\subsection{Theory of Bayesian Filtering}
\label{method-subsec-bayesian-filter}
Bayesian filtering typically includes the following components: hidden system states, a state transition model, observations and a observation model. Let $\textbf{x}_k \in \mathbb{R}^2$ $(k=\{0, 1, 2, ...\})$ denote the state, the location of guiding catheter tip in the $k$-th frame, a 2D vector representing the coordinates in the X-ray image space. The transition of the system from one state to the next state is given by the state transition model $\textbf{x}_k=f_k(\textbf{x}_{k-1}, \textbf{v}_{k-1})$, where $\textbf{v}_{k-1} \in \mathbb{R}^2$ is an independent and identically distributed (i.i.d.) process noise, $f_k: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a possibly nonlinear function that maps the previous state $\textbf{x}_{k-1}$ to the current state $\textbf{x}_k$ with noise $\textbf{v}_{k-1}$. The observation $\textbf{z}_k$ in this work is defined as the $k$-th X-ray image of a sequence, so that $\textbf{z}_k \in \mathbb{R}^{w \times h}$, where $w$ and $h$ are the width and height of an X-ray image. We further define the observation model as $\textbf{z}_k=h_k(\textbf{x}_k, \textbf{n}_k)$, where $\textbf{n}_k \in \mathbb{R}^{n_k}$ is an i.i.d~measurement noise ($n_k$ is the dimension of $\textbf{n}_k$), $h_k: \mathbb{R}^2 \times \mathbb{R}^{n_k} \rightarrow \mathbb{R}^{w \times h}$ is a highly nonlinear function that generates the observation $\textbf{z}_k$ from the state $\textbf{x}_k$ with noise $\textbf{n}_k$. The state transition model $f_k$ and the observation model $h_k$, respectively, can also be equivalently represented using probabilistic forms, i.e. the state transition prior $p(\textbf{x}_k|\textbf{x}_{k-1})$ and the likelihood $p(\textbf{z}_k|\textbf{x}_k)$ from which $\textbf{x}_k$ and $\textbf{z}_k$ can be obtained by sampling.
With these definitions and $p(\textbf{x}_0)$, the inital belief of $\textbf{x}_0$, Bayesian filtering seeks an estimation of $\textbf{x}_k\ (k\geq1)$ based on the set of all available observations $\textbf{z}_{0:k}=\{\textbf{z}_i, i=0,...,k\}$ up to time $k$ via recursively computing the posterior probability $p(\textbf{x}_k|\textbf{z}_{0:k})$ as Eq.(\ref{method-eq-bayes-filter}) (\cite{arulampalam2002tutorial}):
\begin{equation}
\label{method-eq-bayes-filter}
p(\textbf{x}_k|\textbf{z}_{0:k}) \propto p(\textbf{z}_k|\textbf{x}_k)\underbrace{\int p(\textbf{x}_k|\textbf{x}_{k-1})p(\textbf{x}_{k-1}|\textbf{z}_{0:k-1}) \mathrm{d}\textbf{x}_{k-1}}_\text{$p(\textbf{x}_k|\textbf{z}_{0:k-1})$} .
\end{equation}
\noindent Assuming the initial probability $p(\textbf{x}_0|\textbf{z}_0)=p(\textbf{x}_0)$ is known, based on Eq.(\ref{method-eq-bayes-filter}), Bayesian filtering runs in cycles of two steps: prediction and update. In the prediction step, the prior probability $p(\textbf{x}_k|\textbf{z}_{0:k-1})$, the initial belief of $\textbf{x}_k$ given previous observations, is estimated by computing the integral in Eq.(\ref{method-eq-bayes-filter}). In the update step, the prior probability is corrected by the current likelihood $p(\textbf{z}_k|\textbf{x}_k)$ to obtain the posterior $p(\textbf{x}_k|\textbf{z}_{0:k})$.
In Section \ref{method-subsec-likelihood}, we will firstly introduce how to model the likelihood. Then in Section \ref{method-subsec-posterior}, a way to represent and efficiently approximate the posterior will be discussed. Finally in Section \ref{method-subsec-summary}, a summary of the complete catheter tip tracking method will be given.
\subsection{A Deep Learning based Likelihood}
\label{method-subsec-likelihood}
Directly modeling the likelihood $p(\textbf{z}_k|\textbf{x}_k)$ is challenging due to (1) the complexity of the generation process $h_k$ and (2) the computational complexity of $p(\textbf{z}_k|\textbf{x}_k)$ for every value $\textbf{x}_k \in \mathbb{R}^2$. In this work, we simplify the problem by only computing the likelihood $p(\textbf{z}_k|\textbf{x}_k)$ in the image pixel space, i.e. the integer pixel coordinate. For a subpixel $\textbf{x}_k$, the value of $p(\textbf{z}_k|\textbf{x}_k)$ can possibly be approximated by interpolation. To this end, we propose to use a deep neural network $\mathcal{D}$ to approximate $p(\textbf{z}_k|\textbf{x}_k)$ for integer pixel locations. The network takes an image $\textbf{z}_k$ as input and outputs a probability of observing the input $\textbf{z}_k$ for every pixel location $\textbf{x}_k$. Therefore, the approximated likelihood is a function of $\textbf{x}_k$, denoted as $\mathcal{D}_{\textbf{z}_k}(\textbf{x}_k)$. Since $\textbf{x}_k$ is defined within the scope of the image pixel space, $\mathcal{D}_{\textbf{z}_k}(\textbf{x}_k)$ is essentially a probability map having the same dimension and size with the input image $\textbf{z}_k$, in which the entry at each location $\textbf{x}_k^{j}$ ($j=1, 2, \dots, wh$) in the map represents the probability of observing $\textbf{z}_k$ given $\textbf{x}_k^{j}$. It is worth mentioning that the deep neural network is used for approximation of $p(\textbf{z}_k|\textbf{x}_k)$, which should be clearly distinguished from the generation model $h_k$ that maps an $\textbf{x}_k$ to $\textbf{z}_k$. The existence of $h_k$ is merely for the convenience of definition, its explicit form, however, is not required in the context of this work.
To obtain the training labels, we assume that there exists a mapping $h_k$, such that the training label can be defined as a distance-based probability map, i.e. the farther away $\textbf{x}_k$ is from the ground truth tip location in the image $\textbf{z}_k$, the less possible it is to observe $\textbf{z}_k$ given $\textbf{x}_k$ through the process $h_k$. This definition matches the intuition that from a location $\textbf{x}_k$ that is far from the ground truth tip location, the probability of observing a $\textbf{z}_k$ with the catheter tip being located at the ground truth position should be low. For simplicity, a 2D Gaussian probability density function (PDF) $\mathcal{N}(\textbf{x}_k;\textbf{x}_k^{\prime}, \sigma^2\mathbf{\mathit{I}})$ centered at the ground truth tip location $\textbf{x}_k^{\prime}$ with variance $\sigma^2\mathbf{\mathit{I}}$ in the image space is used as the label to train the network (Figure \ref{method-subfigure-detection-heatmap}). Note that this training label makes the estimation of $p(\textbf{z}_k|\textbf{x}_k)$ equivalent to a catheter tip detection problem such that the deep neural network learns features of catheter tip and outputs high probability at locations where the features are present. Due to this reason, we also call $p(\textbf{z}_k|\textbf{x}_k)$ ``detection output'' or ``detection probability'' and call the estimation of $p(\textbf{z}_k|\textbf{x}_k)$ ``catheter tip detection'' in the context of this paper.
\begin{figure}[t]
\centering
\subfloat[]{
\includegraphics[width=0.14\textwidth]{sections/methods/figures/XA.png}
\label{method-subfigure-xa}
}
\subfloat[]{
\includegraphics[width=0.14\textwidth]{sections/methods/figures/catheter_mask.png}
\label{method-subfigure-catheter-mask}
}
\subfloat[]{
\includegraphics[width=0.14\textwidth]{sections/methods/figures/detection_heatmap.png}
\label{method-subfigure-detection-heatmap}
}
\caption{Input and ground truth labels for the deep neural network: (a) an input X-ray fluoroscopic image, (b) the binary catheter mask of (a) for catheter segmentation, (c) a 2D Gaussian PDF ($\sigma=4$ px) for likelihood estimation for (a).}
\label{method-figure-labels}
\end{figure}
\begin{figure*}[t]
\centering
\includegraphics[width=0.98\textwidth]{sections/methods/figures/network2.pdf}
\caption{A joint segmentation and detection network for catheter tip detection. This figure shows an example network with 4 levels of depth (the number of down or up blocks). Meaning of abbreviations: \textit{Conv}, 2D convolution; \textit{Bn}, batch normalization; \textit{Relu}, ReLU activation; \textit{Dp}, dropout; \textit{Concat}, concatenation; \textit{Ch}, number of channels; \textit{S}, segmentation output; \textit{D}, detection output. The number above an image or feature maps indicates the number of channels; the number of channels in the residual network in a block is shown above the block; \textit{c} is the basic number of channels, the channel number in the first down block. The number next to a rectangle denotes the size of the image or feature maps. Red arrows indicate a change of number of channels.}
\label{method-figure-network}
\end{figure*}
\indent The network that we use follows a encoder-decoder architecture with skip connections similar to U-net (\cite{ronneberger2015u}). Additionally, similar to the work by \cite{milletari2016v}, residual blocks (\cite{he2016deep}) are adopted at each resolution level in the encoder and decoder to ease gradient propagation in a deep network. The encoder consists of 4 \textit{down} blocks in which a residual block followed by a stride-2 convolution is used for extraction and down-scaling of feature maps. The number of feature maps is doubled in each downsampling step. The decoder has 4 \textit{up} blocks where a transposed convolution of stride-2 is used for upsampling of the input feature maps. Dropout is used in the residual unit of the \textit{up} block for regularization of the network. Between the encoder and the decoder, another residual block is used to process the feature maps extracted by the encoder. The detailed network architecture is shown in Figure \ref{method-figure-network}.
\indent Due to similar appearance between a guiding catheter tip and corners of a background structure, such as vertebral bones, lung tissue, stitches or guidewires, ambiguity may exist when the network is expected to output only one blob in the probability map. To alleviate the issue, we adopt a similar strategy as \cite{laina2017concurrent}, using a catheter mask (Figure \ref{method-subfigure-catheter-mask}) as an additional label to jointly train the network to output both the catheter segmentation heatmap and the likelihood probability map. The segmentation heatmap is obtained by applying a $1\times1$ convolution with ReLU activation on the feature maps of the last \textit{up} block. To compute the likelihood probability map, a residual block is firstly applied on the feature maps of the last \textit{up} block. The output feature maps are then concatenated with the segmentation heatmap as one additional channel, followed by a $1\times1$ convolution. Finally, to ensure the network detection output fits the definition of a probability map on image locations, following the $1\times1$ convolution, a spatial softmax layer is computed as Eq.(\ref{method-eq-spatial-softmax}):
\begin{equation}
\label{method-eq-spatial-softmax}
D_{k,l} = \frac{e^{A_{k,l}}}{\sum_{i,j}^{}e^{A_{i,j}}} ,
\end{equation}
\noindent where $A$ is the output feature map of the $1\times1$ convolution, $A_{i,j}$ denotes the value of $A$ at location $(i,j)$, $D$ is the final output of the detection network, a 2D probability map representing $p(\textbf{z}_k|\textbf{x}_k)$. The details are shown in Figure \ref{method-figure-network}.
\indent The training loss is defined as a combination of the segmentation loss and the detection loss. The segmentation loss $L_s$ in this work is a Dice loss defined by Eq.(\ref{method-eq-dice-loss}):
\begin{equation}
\label{method-eq-dice-loss}
L_s = 1 - \frac{2\sum_{i,j}M_{i,j}S_{i,j}}{\sum_{i,j}M_{i,j}^2+\sum_{i,j}S_{i,j}^2}
\end{equation}
\noindent where $M$ denotes the ground truth binary catheter masks, $S$ is the segmentation heatmap. The loss function for detection $L_d$ is mean square error (MSE) given by Eq.(\ref{method-eq-mse-loss}):
\begin{equation}
\label{method-eq-mse-loss}
L_d = \frac{1}{w \times h}\sum_{i \leq w,j \leq h}|T_{i,j}-D_{i,j}|^2
\end{equation}
\noindent where $T$ denotes the ground truth PDF, $w$ and $h$ are the width and height of an image\iffalse$n_s$ is the number of training samples in a batch\fi. The total training loss $L$ is defined as Eq.(\ref{method-eq-total-loss}):
\begin{equation}
\label{method-eq-total-loss}
L = L_s + \lambda L_d
\end{equation}
\noindent where $\lambda$ is a weight to balance $L_s$ and $L_d$.
\subsection{Approximation of the Posterior with Particle Filter}
\label{method-subsec-posterior}
Once the deep neural network in Section \ref{method-subsec-likelihood} is trained, its weights are fixed during inference for computing the posterior $p(\textbf{x}_k|\textbf{z}_{0:k})$ for new data. Idealy, the network detection output $p(\textbf{z}_k|\textbf{x}_k)$ should be a Gaussian PDF during inference, as it is trained with labels of Gaussian PDFs. However, due to similar appearance of background structures or contrast residual, the detection output is unlikely to be a perfect Gaussian (possibly non-Gaussian or having multiple modes), which prevents the posterior $p(\textbf{x}_k|\textbf{z}_{0:k})$ in Eq.(\ref{method-eq-bayes-filter}) being solved with an analytical method. In practice, the posterior can be approximated using a particle filter method (\cite{arulampalam2002tutorial}).
Particle filter methods approximate the posterior PDF by a set of $N_s$ random samples with associated weights $\{\textbf{x}_k^i, w_k^i\}_{i=1}^{N_s}$ (\cite{arulampalam2002tutorial}). As $N_s$ becomes very large, this discrete representation approaches the true posterior. According to \cite{arulampalam2002tutorial}, the approximation of the posterior $p(\textbf{x}_k|\textbf{z}_{0:k})$ is given by Eq.(\ref{method-eq-approximation}):
\begin{equation}
\label{method-eq-approximation}
p(\textbf{x}_k|\textbf{z}_{0:k}) \approx \sum_{i=1}^{N_s}w_k^i\delta(\textbf{x}_k-\textbf{x}_k^i)
\end{equation}
\noindent where $\delta(\cdot)$ is the Dirac delta function. The weight $w_k^i$ can be computed in a recursive manner as Eq.(\ref{method-eq-weight-update}) once $w_{k-1}^i$ is known (\cite{arulampalam2002tutorial}):
\begin{equation}
\label{method-eq-weight-update}
w_k^i \propto w_{k-1}^i\frac{p(\textbf{z}_k|\textbf{x}_k^i)p(\textbf{x}_k^i|\textbf{x}_{k-1}^i)}{q(\textbf{x}_k^i|\textbf{x}_{k-1}^i, \textbf{z}_k)}
\end{equation}
\noindent where $q(\textbf{x}_k|\textbf{x}_{k-1}^i, \textbf{z}_k)$ is an importance density from which it should be possible to sample $\textbf{x}_k^i$ easily. For simplicity, a good and convenient choice of the importance density is the prior $p(\textbf{x}_k|\textbf{x}_{k-1}^i)$ (\cite{arulampalam2002tutorial}), so that the weight update rule (\ref{method-eq-weight-update}) becomes $w_k^i \propto w_{k-1}^i p(\textbf{z}_k|\textbf{x}_k^i)$.
A sample can be drawn from $p(\textbf{x}_k|\textbf{x}_{k-1}^i)$ in the following way. First, a process noise sample $\textbf{v}_{k-1}^i$ is sampled from $p_v(\textbf{v}_{k-1})$, the PDF of $\textbf{v}_{k-1}$; then $\textbf{x}_k^i$ is generated from $\textbf{x}_{k-1}^i$ via the state transition model $\textbf{x}_k^i=f_k(\textbf{x}_{k-1}^i, \textbf{v}_{k-1}^i)$. In this work, $p_v(\textbf{v}_{k-1})$ is set to be a Gaussian $\mathcal{N}(\textbf{0},\, \sigma_v^2\mathbf{\mathit{I}})$. The choice of motion model for $f_k$ is important for an accurate representation of the true state transition prior $p(\textbf{x}_k|\textbf{x}_{k-1})$. A random motion cannot characterize well the motion of catheter tip in XA frames. In this work, we estimated the motion from adjacent frames using an optical flow method, as this approach 1) takes into account of the observation $\textbf{z}_k$, which results in a better guess of the catheter tip motion, and 2) enables estimation of a dense motion field where the motion of a sample $\textbf{x}_k^i$ can be efficiently obtained. Therefore, $f_k$ is defined as Eq.(\ref{method-eq-state-transition-flow}):
\begin{equation}
\label{method-eq-state-transition-flow}
\textbf{x}_k=\textbf{x}_{k-1}+\textbf{u}_{k-1}(\textbf{x}_{k-1})+\textbf{v}_{k-1}
\end{equation}
\noindent where $\textbf{u}_{k-1}(\cdot)$ is the motion from frame $k\!-\!1$ to frame $k$ estimated with optical flow using the method by \cite{farneback2003two}, $\textbf{u}_{k-1}(\textbf{x}_{k-1})$ is the motion from state $\textbf{x}_{k-1}$.
Once samples are drawn and their weights are updated, the so-called ``resampling'' of the samples should be performed to prevent the degenaracy problem, where all but one sample will have negligible weight after a few iterations (\cite{arulampalam2002tutorial}). The resampling step resamples the existing samples according to their updated weights and then resets all sample weights to be $1/N_s$, so the number of effective samples which have actual contribution to approximate $p(\textbf{x}_k|\textbf{z}_{0:k})$ is maximized (\cite{arulampalam2002tutorial}). If the resampling is applied at every time step, the particle filter becomes a sampling importance resampling (SIR) filter, and the weight update rule follows Eq.(\ref{method-eq-weight-update-simplest}).
\begin{equation}
\label{method-eq-weight-update-simplest}
w_k^i \propto p(\textbf{z}_k|\textbf{x}_k^i)
\end{equation}
\noindent The final decision on catheter tip location in frame $k$ can then be computed as the expectation of $\textbf{x}_k$, $\hat{\textbf{x}}_k =\int\textbf{x}_kp(\textbf{x}_k|\textbf{z}_{0:k})d\textbf{x}_k$, which is in this case, the weighted sum of all samples:
\begin{equation}
\label{method-eq-final-tip-location}
\hat{\textbf{x}}_k = \sum_{i=1}^{N_s}w_k^i\textbf{x}_k^i.
\end{equation}
\subsection{Summary}
\label{method-subsec-summary}
The overall catheter tip tracking using a deep learning based Bayesian filtering method is summarized in Algorithm \ref{method-alg-dl-bayes}.
\begin{algorithm*}
\caption{Deep learning based Bayesian filtering for online tracking of catheter tip in X-ray fluoroscopy}
\label{method-alg-dl-bayes}
\begin{algorithmic}[1]
\REQUIRE $\{\textbf{z}_0, \dots, \textbf{z}_T\}$ (sequentially observed frames), $\mathcal{D}$ (A trained network from Section \ref{method-subsec-likelihood}), $p(\textbf{x}_0)$ (the initial PDF), $\sigma_v^2$ (the variance of $\textbf{v}_{k-1},\, k=1, \dots, T$), $T$ (number of frames for tracking), $N_s$ (number of samples)
\STATE{Draw $\textbf{x}_0^i \sim p(\textbf{x}_0)$, set $w_0^i=1/N_s$, $\forall\, i=1, \dots, N_s$}
\FOR{$k$ = 1 \TO $T$}
\STATE{Compute $\textbf{u}_{k-1}$ from $\textbf{z}_{k-1}$ to $\textbf{z}_k$ using the optical flow method of \cite{farneback2003two}}
\FOR{$i$ = 1 \TO $N_s$}
\STATE{Draw $\textbf{v}_{k-1}^i \sim \mathcal{N}(\textbf{0},\, \sigma_v^2\mathbf{\mathit{I}})$}
\STATE{Compute the motion of $\textbf{x}_{k-1}^i$: $\textbf{u}_{k-1}^i = \textbf{u}_{k-1}(\textbf{x}_{k-1}^i)$}
\STATE{Draw $\textbf{x}_k^i \sim p(\textbf{x}_k|\textbf{x}_{k-1}^i)$: $\textbf{x}_k^i = \textbf{x}_{k-1}^i+\textbf{u}_{k-1}^i+\textbf{v}_{k-1}^i$}
\STATE{Update weight $w_k^i = p(\textbf{z}_k|\textbf{x}_k^i) = \mathcal{D}_{\textbf{z}_k}(\textbf{x}_k^i)$}
\ENDFOR
\STATE{Normalize $w_k^i \leftarrow w_k^i/\sum_{i=1}^{N_s}w_k^i$, $\forall\, i=1, \dots, N_s$}
\STATE{Prediciton in frame $k$: $\hat{\textbf{x}}_k = \sum_{i=1}^{N_s}w_k^i\textbf{x}_k^i$ }
\STATE{Resample $\{\textbf{x}_k^i, w_k^i\}_{i=1}^{N_s}$ using the method of \cite{arulampalam2002tutorial}} (so all $w_k^i$ are set to $1/N_s$ again)
\ENDFOR
\end{algorithmic}
\end{algorithm*}
\section*{References}
\input{sections/references/references.bbl}
\section{Introduction}
\subsection{Clinical Background}
\label{intro-subsec-clinical}
Percutaneous coronary intervention (PCI) is a minimally invasive procedure for treating patients with coronary artery disease. During these procedures, medical instruments inserted through a guiding catheter are advanced to treat coronary stenoses. A guiding catheter is firstly positioned into the ostium of the coronary artery. Through the guiding catheter, a balloon catheter carrying a stent is introduced over a guidewire to the stenosed location. The balloon is then inflated and the stent is deployed to prevent the vessel from collapsing and restenosing.
PCI is typically performed with image-guidance using X-ray angiography (XA). Coronary arteries are visualized with X-ray opaque contrast agent. During the procedure, interventional cardiologists may repeatedly inject contrast agent to visualize the vessels, as the opacification of coronary arteries only lasts for a short period. The amount of periprocedural contrast use has been correlated to operator experience, procedural complexity, renal function and imaging setup (\cite{piayda2018dynamic}). Furthermore, the risk for contrast induced nephropathy has been associated to contrast volume (\cite{tehrani2013contrast}). Manoeuvring guidewires and material, however, typically occurs without continuous contrast injections. In these situations, the navigation of devices is guided with "vessel-free" fluoroscopic images. Cardiologists have to mentally reconstruct the position of vessels and stenosis based on previous angiograms.
\subsection{Dynamic Coronary Roadmapping}
\label{intro-subsec-dcr}
Dynamic coronary roadmapping (DCR) is a promising solution towards improving visual feedback and reducing usage of contrast medium during PCI (\cite{elion1989dynamic, zhu2010image, manhart2011self, kim2018registration}). This approach dynamically superimposes images or models of coronary arteries onto live X-ray fluoroscopic sequences. The dynamic overlay serves as a roadmap that provides immediate feedback to cardiologists during the intervention, so as to assist in navigating a guidewire into the appropriate coronary branch and proper placement of a stent at the stenotic site with reduced application of contrast agent. Studies with a phantom setup using research software (\cite{kim2018registration}) or on first cases of clinical interventions using commercially available systems (\cite{dannenberg2016road, yabe2018impact, takimura2018usefulness}) have investigated the usefulness of DCR in assisting PCI, reporting that DCR helps to reduce procedure time, radiation dose and contrast volume.
To develop a DCR system, it is important but yet a challenge to accurately overlay a roadmap of coronary arteries onto an X-ray fluoroscopic image, as limited information of vessels is present in the target fluoroscopic image for inferring the compensation of the vessel motion resulting from patient respiration and heartbeat. The methods that have been proposed on motion compensation for DCR can be generally grouped into two categories: direct roadmapping and model-based approaches.
Direct roadmapping methods use information from X-ray images and ECG signals to directly correct the motion caused by respiration and heartbeat. The first DCR system (\cite{elion1989dynamic}) used digital subtraction of a contrast sequence and a mask sequence to create a full cardiac cycle of coronary roadmaps. The roadmaps were stored and later synchronized with the live fluoroscopic sequence by aligning the R waves of their corresponding ECG signals. This system compensates the cardiac motion of vessels, yet does not correct the respiratory motion during interventions. \iffalse In addition, it ignored the potential motion variance between different cardiac cycles.\fi Two later studies by \cite{zhu2010image} and \cite{manhart2011self} introduced image-based respiratory motion compensation methods for DCR. Their methods assumed an affine respiratory motion model in ECG-gated fluoroscopic frames and recovered the respiratory motion from soft tissues with special handling of static structures. The limitation of these approaches is that they require relevant tissue to be sufficiently visible in the field of view for reliable motion compensation which is not always the case. In addition, they require to be run on cardiac-gated frames. \iffalse reliable motion compensation is not always possible especially when relevant tissue is not or only weakly visible.\fi In a more recent work by \cite{kim2018registration}, binary vessel masks were created as the roadmaps from at least one cardiac cycle of angiographic images. Temporal alignment of the roadmaps and the fluoroscopic sequence, which compensated the cardiac motion of vessels, was performed by registering ECG signals using cross-correlation. Additionally, the respiratory motion was corrected by aligning the guidewire centerline in the fluoroscopy to the contour of vessels in the angiogram where the roadmaps were created. The system has been shown useful in a phantom-based study, nevertheless no accuracy evaluation of the spatiotemporal alignment was presented. Furthermore, the spatial registration relies on robust extraction of vessels and guidewires which is often challenging for X-ray images.
Unlike direct roadmapping, the model-based approaches build a model to predict motion in fluoroscopic frames. The motion models are often functions that relate the motion of roadmaps to surrogate signals derived from images or ECG, so that once the surrogates for fluoroscopic frames are obtained, the motion can be computed by the model. For cardiac interventions including PCI, the organ motion is mainly affected by respiratory and cardiac motion. Many previous works often built a motion model parameterized by a cardiac signal derived from ECG and a respiratory signal obtained from diaphragm tracking (\cite{shechter2005prospective, timinger2005motion, faranesh2013integration}) or automatic PCA-based surrogate (\cite{fischer2018mr}). Some other works model only the respiratory motion in cardiac-gated images (\cite{schneider2010model, king2009subject, peressutti2013novel}). For a complete review on respiratory motion modeling, we refer readers to the survey article by \cite{mcclelland2013respiratory}. One limitation of the model-based approaches is that the motion models are often patient-specific, which requires training the model every time for a new subject. Additionally, once the surrogate values during inference are out of the surrogate range for building the model, e.g. for abnormal motion, extrapolation is needed, which may hamper accurate motion compensation.
\subsection{Interventional / Surgical Tool Tracking}
\label{intro-subsec-tracking}
Tracking interventional tools is relevant for motion compensation (\cite{schneider2010model, brost2010respiratory, ma2012clinical, baka2015respiratory, ambrosini2017hidden}). In particular for PCI, the guiding catheter tip typically remains within the coronary ostium which is in the field of view during the largest part of the intervention, making it a suitable landmark for tracking. \cite{baka2015respiratory} have shown that catheter tip motion during PCI can be modeled as a combination of cardiac and respiratory motion. As using catheter tip displacement can only correct translational motion, \cite{baka2015respiratory} further showed that, compared to a rigid motion model for the respiratory motion, modeling only the translational part of the respiratory motion deteriorated the accuracy marginally, which confirms the observations by \cite{shechter2004respiratory} that the rotational part of respiratory motion is small. These findings motivate motion compensation for DCR through tracking the catheter tip in X-ray fluoroscopic sequences.
Many works have been proposed to address the problem of tracking interventional or surgical tools in medical images for various applications. \iffalse For example, tracking different types of catheters in X-ray fluoroscopic images for electrophysiology (EP) procedure \cite{brost2010respiratory, ma2012clinical, yatziv2012toward, wu2012fast, ma2013real, wu2015fast, baur2016cathnets}, for transcatheter aortic valve implantation (TAVI) \cite{chang2016robust}, for transcatheter arterial chemoembolization (TACE) \cite{ambrosini2017hidden, ambrosini2017fully}; tracking guidewire in X-ray fluoroscopy \cite{wang2009robust, pauly2010machine, petkovic2010guidewire, honnorat2011graph, heibel2013interventional, chang2016robust, vandini2017robust}; tracking surgical tools for minimally invasive surgery (MIS) \cite{speidel2006tracking, speidel2014visual, rieke2016real, garcia2016real, laina2017concurrent, bouget2017vision, du2018articulated}.\fi The tracking methods from these works can be generally categorized into two kinds of approaches: tracking by detection, and temporal tracking.
The tracking by detection approaches treat tracking as a detection problem, which rely on features only from the current image without using information from previous frames. For example in electrophysiology procedure, as the catheters present specific features in shape or intensity, \textit{ad hoc} methods were proposed based on, e.g. blob detection, shape constrained searching and model-/template- based detection (\cite{ma2012clinical, ma2013real}). \cite{chang2016robust} modeled the catheter tracking problem by optimizing the posterior in a Bayesian framework, in which the catheter was represented as a B-spline tube model and was tracked by fitting the B-spline to measurements based on gray intensity and vesselness image. \cite{baur2016cathnets} proposed a convolutional neural network (CNN) to detect catheter electrodes in X-ray images, which treated catheter detection as a segmentation problem. The method used a weighted cross-entropy loss to cope with the class imbalancing problem due to the small size of the target. \cite{laina2017concurrent} and \cite{du2018articulated} tracked surgical instruments using a deep network having an encoder-decoder architecture. Their approaches combined instrument segmentation and detection in a multi-task learning problem to make the tool detection in a cluttered background more robust.
Different from tracking by detection, which relies solely on the current image, temporal tracking also uses information from previous frames. The temporal information can reduce the search space for detection, or put additional constraints in the model, making the tracking more robust.
Temporal information has been used in various ways. Some methods mainly relied on a detection model, but incorporate temporal information in the preprocessing (\cite{brost2010respiratory}) or post-processing (\cite{garcia2016real}) step or in the input (\cite{rieke2016real, ambrosini2017fully}). Approaches based on background estimation have been used for catheter (\cite{yatziv2012toward}) or guidewire (\cite{petkovic2010guidewire}) tracking. In these approaches, the background was recursively updated for every frame, and was used for enhancing the foreground containing instruments. Apart from those, many works adopted a Bayesian framework for tracking instruments via a \textit{maximum a posteriori} (MAP) formulation. Representations based on key points (\cite{wu2015fast}), B-splines (\cite{wang2009robust, pauly2010machine, honnorat2011graph, heibel2013interventional}), or segment-like features (\cite{vandini2017robust}) have been used to model catheters or guidewires. Markov random field (MRF) was used to model the position or deformation of the control points in the B-spline (\cite{pauly2010machine, honnorat2011graph, heibel2013interventional, wu2015fast}). \iffalse discrete optimization methods were employed to find the optimal deformation in a \textit{maximum a posteriori} (MAP) formulation.\fi In the work by \cite{vandini2017robust}, temporal information was incorporated in the prior term using Kalman filter. Particularly, learning-based approaches were used in several works to obtain the likelihood for a more robust measurement using probabilistic boosting tree (\cite{wang2009robust, wu2012fast}) or support vector regression (\cite{pauly2010machine}). In addition, temporal tracking models based on Bayesian filtering were also a popular approach for instrument tracking. \cite{ambrosini2017hidden} used a hidden Markov model (HMM) to track catheter tip in a 3D vessel tree, for which the likelihood was obtained based on the 3D-2D registration outcome. \cite{speidel2006tracking} used particle filters to track surgical tools in medical images. They used a likelihood based on the segmentation of instruments, and a dynamic model that incorporates samples from two previous time steps. In a later work, \cite{speidel2014visual} used a multi-object particle filter to track multiple instrument regions simultaneously, in which a particle is the concatenation of the states of several objects.
Despite of many existing works on inverventional or surgical tool tracking in medical images, an automatic approach for tracking the tip of guiding catheter in X-ray fluoroscopy for PCI has not been investigated yet. The challenges of this task are: (1) different from the catheters for EP that can be viewed as blobs or a circle, the guiding catheter for PCI presents a dark tubular appearance which shows no prominent features; (2) the shape of the guiding catheter tip segment varies depending on the orientation of the C-arm, making feature-/model- based detection challenging; (3) the background may contain structures that have similar appearance to a catheter tip, such as vertebral structures or residual contrast agent, which makes robust tracking difficult.
\subsection{Contributions}
\label{intro-subsec-contribution}
We propose and evaluate a novel approach for dynamic coronary roadmapping. The approach compensates changes in vessel shapes and cardiac motion by selecting the roadmap of the same cardiac phase through ECG alignment, and corrects the respiratory induced motion via tracking the tip of the guiding catheter. Our contributions are:
\begin{enumerate}
\item We develop a new way to perform dynamic coronary roadmapping on free breathing, non-cardiac-gated X-ray fluoroscopic sequences. Particularly, the respiratory-induced vessel motion is robustly compensated via the displacement of catheter tip.
\item We proposed a deep learning based method within a Bayesian filtering framework for online detection and tracking of guiding catheter tip in X-ray fluoroscopic images. The method models the likelihood term of Bayesian filtering with a convolutional neural network, and integrates it with particle filtering in a comprehensive manner, leading to more robust tracking.
\item We evaluate the proposed approach visually and quantitatively on clinical X-ray sequences, achieving low errors on both tracking and roadmapping tasks.
\item The proposed DCR method runs in real-time with a modern GPU, thus can potentially be used during PCI in real clinical settings.
\end{enumerate}
\section{Experiments and Results}
\label{exp-sec}
The following experiments are performed to assess the methods. First, In Section \ref{exp-subsec-detection}, the training of the deep neural network is described. Then in Section \ref{exp-subsec-tracking}, the accuracy of catheter tip tracking using the optimized trained network and the tuned particle filter is presented. Section \ref{exp-subsec-roadmapping} describes the accuracy evaluation of dynamic coronary roadmapping via the proposed catheter tip tracking method. Finally, in Section \ref{exp-subsec-time}, we measure the processing time of the proposed DCR approach.
\subsection{Training the Deep Neural Network}
\label{exp-subsec-detection}
The purpose of this experiment is to train the deep neural network to output reasonable likelihood probability map. The network hyperparameters were tuned to optimize the detection performance.
The training and validation data for detection mentioned in Section \ref{setup-subsec-tracking} were used for training the deep neural network. The evaluation metric mentioned in Section \ref{setup-subsec-trainingnetwork}, the mean Euclidean distance between the ground truth and the predicted tip location averaged over all validation frames, was used as the validation criteria for selecting the optimal training epoch and the network hyperparameters. When we evaluated hyperparameter settings, we firstly selected the training epoch with the lowest mean validation error for each setting, then the settings were compared using the model weights (trainable network parameters) of their chosen epochs.
The network hyperparameters we investigated in the experiments include (1) the basic channel number, i.e. the number of channels or feature maps in the first down block, (2) the network depth level, the number of down or up blocks, and (3) the dropout probability.
The validation errors for different hyperparameter settings using the experimental settings in Section \ref{setup-subsec-trainingnetwork} are shown in Table \ref{exp-tab-training-network}. The table shows that the hyperparameter setting with the lowest mean error, which has 4 level in depth and 64 channels in the first down block, achieves a validation error of about 2 mm. The table also shows other good choices of network architecture that have a small validation error (shown in red in Table \ref{exp-tab-training-network}): 32 channels in the first down block with 4 or 5 levels in depth, or 64 channels with 3 or 4 depth levels. The dropout regularization improves the accuracy of the model, compared to the ones without dropout.
The learning curves of the training process with the chosen hyperparameter setting are illustrated in Figure \ref{exp-figure-learning-curves}. The curves indicate that both segmentation and detection reach convergence after training 100 epochs.
We did not investigate a model with more than 64 channels or 5 depth levels, because (1) it will further increase the processing time which makes online applications less feasible; (2) the results in Table \ref{exp-tab-training-network} show that such a setting (64 channels, 5 depth levels) starts increasing the validation error compared to those less complex models.
The subsequent experiments will be based on the network trained with the chosen hyperparameter setting indicated in Table \ref{exp-tab-training-network} (64 channels, 4 depth levels, dropout 0.2, also see Table \ref{exp-tab-all-hyperparameters}).
\begin{table*}[h]
\centering
\caption{Validation errors (mm) for different hyperparameter settings. Red cells show the settings with the 10 smallest validation errors. \textit{bold} number indicates the setting with the lowest error.}
\begin{tabular}{c c c c c c c c}
\toprule
Basic Number & Depth & \multicolumn{6}{c}{Dropout} \\
\cmidrule(lr){3-8}
of Channels & Level & none & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 \\
\midrule
8 & 3 & 5.43 & 4.99 & 5.02 & 5.37 & 4.38 & 4.24 \\
& 4 & 4.17 & 4.45 & 4.25 & 5.04 & 4.75 & 4.36 \\
& 5 & 3 & 4.14 & 3.53 & 4.28 & 3.95 & 4.11 \\
16 & 3 & 3.74 & 4.29 & 3.57 & 4.11 & 3.74 & 3.4 \\
& 4 & 3.36 & 3.11 & 3.63 & 3.33 & 3.36 & 3.78 \\
& 5 & 3.38 & 2.89 & 3.16 & 2.52 & 2.71 & 2.74\\
32 & 3 & 2.99 & 3.02 & 3.26 & 2.82 & 3.26 & 2.56 \\
& 4 & 2.87 & 2.34 & 2.46 & 2.6 & 2.65 & \cellcolor{red!30}2.27 \\
& 5 & 3.04 & 2.51 & \cellcolor{red!30}2.21 & \cellcolor{red!30}2.29 & 2.3 & \cellcolor{red!30}2.25 \\
64 & 3 & \cellcolor{red!30}2.19 & 2.54 & 2.34 & \cellcolor{red!30}2.27 & \cellcolor{red!30}2.26 & 2.49 \\
& 4 & 2.55 & 2.31 & \cellcolor{red!30}\textbf{2.04} & 2.44 & \cellcolor{red!30}2.22 & \cellcolor{red!30}2.27 \\
& 5 & 2.42 & \cellcolor{red!30}2.29 & 2.73 & 2.77 & 2.61 & 2.85 \\
\bottomrule
\end{tabular}
\label{exp-tab-training-network}
\end{table*}
\begin{figure}[t]
\centering
\subfloat[Total loss]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/training/learning_curves/total_loss.pdf}
\label{exp-subfigure-lineplot-total-loss}
}
\subfloat[Detection error (mm)]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/training/learning_curves/detection_error.pdf}
\label{exp-subfigure-lineplot-detection-error}
}
\\[-1ex]
\subfloat[Segmentation loss]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/training/learning_curves/seg_loss.pdf}
\label{exp-subfigure-lineplot-seg-loss}
}
\subfloat[Detection loss]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/training/learning_curves/det_loss.pdf}
\label{exp-subfigure-lineplot-det-loss}
}
\caption{Learning curves for the chosen hyperparameter setting.}
\label{exp-figure-learning-curves}
\end{figure}
\subsection{Catheter Tip Tracking}
\label{exp-subsec-tracking}
The purpose of this experiment is to assess the accuracy of catheter tip tracking with the proposed method in Section \ref{method-sec-bayesian-filter}. Guiding catheter tip is tracked in X-ray fluoroscopy using Algorithm \ref{method-alg-dl-bayes} based on a trained network with the optimal hyperparameter setting from Section \ref{exp-subsec-detection}. First, the parameters of the optical flow method used in Algorithm \ref{method-alg-dl-bayes} and particle filter were tuned on the validation data for tracking in Section \ref{setup-subsec-tracking} (see Appendix \ref{app-subsec-parameter-tuning} for details). Then in Section \ref{exp-subsubsec-tracking}, we evaluated the tracking accuracy with the tuned optimal parameter setting (see Table \ref{exp-tab-all-hyperparameters}) on the test dataset, and compared the proposed tracking method with alternative approaches using only the detection network in Section \ref{method-subsec-likelihood} or using only optical flow. Finally, in Section \ref{exp-subsubsec-initialization}, we investigated tracking accuracy with different ways of tip initialization in the first frame.
\begin{table}[h]
\centering
\caption{The chosen (hyper-)parameters for different building blocks of the catheter tip tracking algorithm. The parameters of the optical flow method can be found in Appendix \ref{exp-subsubsec-of}.}
\begin{tabular}{l l l}
\toprule
Building block & (hyper-)parameters & value \\
\midrule
Deep learning & Basic channel number & 64 \\
& Depth & 4\\
& Dropout & 0.2 \\
\midrule
Particle filter & $\sigma_v$ (px) & 5 \\
& $N_s$ & 1000 \\
\bottomrule
\end{tabular}
\label{exp-tab-all-hyperparameters}
\end{table}
\subsubsection{Tracking Methods Evaluation}
\label{exp-subsubsec-tracking}
In this experiment, the proposed tracking method in Algorithm \ref{method-alg-dl-bayes} uses the ground truth tip probability map of the first frame as the initial PDF $p(\textbf{x}_0)$ to draw samples. This method is referred to as ``Tracking''. In addition, we compared the proposed method with three alternatives. The first one tracks catheter tip using only the detection network in Section \ref{method-subsec-likelihood} with the chosen network architecture and trained parameters in Section \ref{exp-subsec-detection}, therefore, no temporal information is used. This method is referred to as ``Detection (Net)''. The other two methods in this experiment use only optical flow to track catheter tip starting from the ground truth tip position in the first frame. The motion field towards the current frame, estimated by the two methods, was based on the deformation from the previous frame or the first frame in the sequence, respectively. The same implementation setting as in Appendix \ref{exp-subsubsec-of} was used for these two methods. They are called ``Optical Flow (previous)'' and ``Optical Flow (first)'', or in short form, ``OF (pre)'' and ``OF (1st)''. Additionally, we refer the interested readers to Appendix \ref{app-subsubsec-detection-tracking} where the influence of catheter segmentation on the detection and tracking approaches is reported.
The tracking accuracies of all methods reported in this section were obtained on the test set from Table \ref{setup-tab-tip-experiment}. The mean, the median and the maximal tracking error between the predicted and the ground truth tip position of all test images are reported in Table \ref{exp-tab-tracking-test}. In addition, as the sequences in the test set have different lengths, we also computed the mean and the median error per sequence, and report the the average of the sequence mean and median errors, so that each sequence contributes equally in these metrics. Table \ref{exp-tab-tracking-test} shows that the results from the detection network have large average errors which are caused by some completely failed cases. The proposed tracking method has median errors of about 1 mm and mean errors of about 1.3 mm. It achieves the lowest errors compared to the other 3 methods on all listed evaluation criteria.
\begin{table*}[h]
\centering
\caption{Catheter tip tracking errors (mm) of the 4 methods on the test (tracking) dataset. $\dagger$ indicates that the difference between that method and the ``Tracking'' method are statistically highly significant with the two-sided Wilcoxon signed-rank test ($p < 0.001$).}
\begin{tabular}{l c c c c}
\toprule
\multirow{2}{*}{Evaluation Metrics} & Optical Flow$\dagger$ & Optical Flow$\dagger$ & Detection Net$\dagger$ & Tracking \\
& (previous) & (first) & (Section \ref{method-subsec-likelihood}) & \\
\midrule
Maximal error of all images & 29.16 & 20.83 & 108.20 & \textbf{17.72} \\
Median error of all images & 1.78 & 1.22 & \textbf{0.96} & \textbf{0.96} \\
Mean error of all images & 3.74 $\pm$ 4.93 & 3.05 $\pm$ 4.05 & 5.62 $\pm$ 15.91 & \textbf{1.29 $\pm$ 1.76} \\
\midrule
Average of sequence median error & 2.35 $\pm$ 2.52 & 2.64 $\pm$ 3.52 & 6.26 $\pm$ 17.11 & \textbf{1.03 $\pm$ 0.49} \\
Average of sequence mean error & 2.59 $\pm$ 2.69 & 3.31 $\pm$ 2.81 & 6.83 $\pm$ 13.88 & \textbf{1.29 $\pm$ 0.94} \\
\bottomrule
\end{tabular}
\label{exp-tab-tracking-test}
\end{table*}
Figure \ref{exp-figure-boxplot-all_dist} illustrates the boxplots of tracking errors made by the 4 methods on all test images. It shows that the proposed tracking approach outperforms the detection method by avoiding making extremely large errors (Figure \ref{exp-subfigure-boxplot-all-dist}); meanwhile, it maintains as accurate as the detection method for cases with small errors, and is more accurate than the methods based solely on optical flow (Figure \ref{exp-subfigure-boxplot-all-dist-zoom-in}).
\begin{figure}[h]
\centering
\subfloat[Overall view of tracking errors]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/boxplot_all_dists.pdf}
\label{exp-subfigure-boxplot-all-dist}
}
\subfloat[A zoom-in view of (a)]{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/boxplot_all_dists_zoom_in.pdf}
\label{exp-subfigure-boxplot-all-dist-zoom-in}
}
\caption{Tracking errors for the 4 methods on all test images.}
\label{exp-figure-boxplot-all_dist}
\end{figure}
Figure \ref{exp-figure-line-plot} shows longitudinal views of tracking errors of the 4 methods on 4 example sequences. Although the optical flow methods show high accuracy when the target is on the track (row 4), they present periodic error patterns in two sequences due to large cardiac motion. The detection method shows peaks of large errors, this is because temporal relation between frames is not modeled by the approach, thus the detection on different frames is independent of each other. The proposed tracking method overcomes the problems that other methods have and presents accurate detection on these 4 sequences. The tracking results of the 4 methods on example frames from the 4 sequences are illustrated in Figure \ref{exp-figure-tracking-compare-examples-small-2}.
\begin{figure*}[t]
\centering
\subfloat{
\parbox[t]{0.23\textwidth}{
\centering
{\normalsize Optical Flow (previous)} \\[1ex]
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/ofpre_0.pdf}
}
}
\subfloat{
\parbox[t]{0.23\textwidth}{
\centering
{\normalsize Optical Flow (first)} \\[1ex]
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/offirst_0.pdf}
}
}
\subfloat{
\parbox[t]{0.23\textwidth}{
\centering
{\normalsize Detection} \\[1ex]
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/detect_0.pdf}
}
}
\subfloat{
\parbox[t]{0.23\textwidth}{
\centering
{\normalsize Tracking} \\[1ex]
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/tracking_0.pdf}
}
}
\\[-2ex]
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/ofpre_1.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/offirst_1.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/detect_1.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/tracking_1.pdf}
}
\\[-2ex]
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/ofpre_25.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/offirst_25.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/detect_25.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/tracking_25.pdf}
}
\\[-2ex]
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/ofpre_28.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/offirst_28.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/detect_28.pdf}
}
\subfloat{
\includegraphics[width=0.23\textwidth]{sections/experiments/figures/tracking/lineplot/examples/tracking_28.pdf}
}
\caption{Longitudinal view of tracking errors made by the 4 methods on 4 test sequences (one sequence per row). The x-axis denotes the time steps of a sequence, the y-axis is the tracking error (mm).}
\label{exp-figure-line-plot}
\end{figure*}
\begin{figure*}[h]
\centering
\includegraphics[width=0.9\textwidth]{sections/experiments/figures/tracking/examples/compare_examples3.pdf}
\caption{Tracking results on example frames from the same 4 sequences in Figure \ref{exp-figure-line-plot}. The blue point indicates the predicted catheter tip location; the red point shows the ground truth location. (Best viewed in color)}
\label{exp-figure-tracking-compare-examples-small-2}
\end{figure*}
Figure \ref{exp-figure-tracking-examples-small-2} illustrates how the proposed tracking method works on the same 4 frames in Figure \ref{exp-figure-tracking-compare-examples-small-2}. It shows that the prior hypotheses (samples) assists to focus on the correct target location and results in reliable posterior estimation, especially when the detection produces ambiguity in cases of multiple catheters or contrast residual presented in images.
\begin{figure*}[h]
\centering
\includegraphics[width=0.99\textwidth]{sections/experiments/figures/tracking/examples/cmap_hot/tracking_examples4_hot.pdf}
\caption{Workflow of the proposed tracking method on the same 4 frames in Figure \ref{exp-figure-tracking-compare-examples-small-2}. The high probability is shown with bright color in the detection map. Samples or particles are presented as green dots. The blue point indicates the predicted catheter tip location; the red point shows the ground truth location. (Best viewed in color)}
\label{exp-figure-tracking-examples-small-2}
\end{figure*}
\subsubsection{Catheter Tip Initialization}
\label{exp-subsubsec-initialization}
In this experiment, the initial PDF $p(\textbf{x}_0)$ from which samples are drawn in the proposed tracking is investigated (Algorithm \ref{method-alg-dl-bayes}). In particular, we explored and evaluated the tracking accuracy with an automatic initialization using the probability map obtained from the trained detection network in Section \ref{method-subsec-likelihood} with the chosen setting in Section \ref{exp-subsec-detection}.
Figure \ref{exp-figure-tip-init} shows the boxplot of tracking errors on all test images with automatic initialization (Auto) and manual initialization (Manual) for which the ground truth tip probability map of the first frame was used. The tracking with automatic initialization presents an accuracy similar to the one with manual initialization for small tracking errors, but has more large tracking errors which influence the mean error over all test images (Table \ref{exp-tab-auto-init}). We, therefore, defined the tracking errors on the right side of the gap in the boxplot ($>$ 40 mm) as outliers, and explored the statistics without those outliers.
Table \ref{exp-tab-auto-init} indicates that, the mean and median error of the tracking with automatic initialization excluding the outliers are only slightly higher than the tracking with manual initialization and the detection method. While the tracking with automatic initialization has 100 outliers in total from 6 sequences, the detection method that has 10 sequences containing 106 outliers.
Unlike the detection method for which the outliers are mainly presented as the peaks in the longitudinal views (Figure \ref{exp-figure-line-plot}), the outliers for the tracking with automatic initialization are more consistent over time. Figure \ref{exp-figure-auto-init} shows the temporal change of tracking errors for the 6 sequences with outliers using the tracking with automatic initialization. For the 3 sequences on the top row, the tracking with automatic initialization makes large errors at the beginning, but becomes accurate very fast in a few frames; for the 3 sequences on the bottom row, however, the tracking errors remain large till the end of the sequences.
Figure \ref{exp-figure-auto-init-examples} shows example frames to give an insight of the tracking with automatic initialization on the 6 sequences in Figure \ref{exp-figure-auto-init}. For the 3 sequences on the top row (Figure \ref{exp-subfigure-auto-init-examples-1}), although the initialization on the first frame (frame 0) is overall not correct, the true tip positions are still covered by some samples; once the detection in subsequent frames is correct, the tracker can still converge to the right target. For the 3 sequence on the bottom row (Figure \ref{exp-subfigure-auto-init-examples-2}), the initializations of samples are ambiguous in frame 0; the detection in subsequent frames focuses on a wrong area also given by the initial samples due to residual of contrast agent or multiple catheters, the tracker then tends to find the wrong target.
\begin{figure}[t]
\centering
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/tracking/init/boxplot.pdf}
\caption{Catheter tip tracking errors (mm) with manual and automatic initialization.}
\label{exp-figure-tip-init}
\end{figure}
\begin{table*}[h]
\centering
\caption{Catheter tip tracking errors (mm) of detection and tracking with manual and automatic initialization}
\begin{tabular}{l c c c}
\toprule
& \multirow{2}{*}{Detection} & \multicolumn{2}{c}{Tracking} \\
\cmidrule(lr){3-4}
& & Manual init. & Automatic init. \\
\midrule
Maximal error & 108.20 & 17.23 & 98.58 \\
Median error & 0.96 & 0.96 & 0.96 \\
Mean error & 5.62 $\pm$ 15.91 & 1.29 $\pm$ 1.76 & 5.16 $\pm$ 13.91 \\
\midrule
No. of outliers ($>$ 40 mm) & 106 & 0 & 100 \\
No. of sequences with outliers & 10 & 0 & 6 \\
\midrule
Maximal error of inliers & 31.06 & 17.23 & 28.28 \\
Median error of inliers & 0.96 & 0.96 & 0.96 \\
Mean error of inliers & 1.17 $\pm$ 1.78 & 1.29 $\pm$ 1.76 & 1.34 $\pm$ 2.15 \\
\bottomrule
\end{tabular}
\label{exp-tab-auto-init}
\end{table*}
\begin{figure}[t]
\centering
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_1.pdf}
}
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_25.pdf}
}
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_29.pdf}
}
\\[-2ex]
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_23.pdf}
}
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_30.pdf}
}
\subfloat{
\includegraphics[width=0.15\textwidth]{sections/experiments/figures/tracking/init/examples/lineplot/auto_init_33.pdf}
}
\caption{Longitudinal views of tracking errors (mm) for the 6 sequences with outliers using automatic initialization.}
\label{exp-figure-auto-init}
\end{figure}
\begin{figure}
\centering
\subfloat[Sequence 1-3 on the top row in Figure \ref{exp-figure-auto-init}]{
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/tracking/init/examples/cmap_hot/autoinit1_hot.pdf}
\label{exp-subfigure-auto-init-examples-1}
}
\\%[-0.1ex]
\subfloat[Sequence 4-6 on the bottom row in Figure \ref{exp-figure-auto-init}]{
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/tracking/init/examples/cmap_hot/autoinit2_hot.pdf}
\label{exp-subfigure-auto-init-examples-2}
}
\caption{Examples frames from the 6 sequences in Figure \ref{exp-figure-auto-init}. The high probability in the detection heatmap is highlighted as bright color. Particles are presented as green dots. The red dots in the last column indicate the ground truth tip location. (Best viewed in color)}
\label{exp-figure-auto-init-examples}
\end{figure}
\subsection{Dynamic Coronary Roadmapping}
\label{exp-subsec-roadmapping}
In this experiment, the accuracy of dynamic coronary roadmapping using the proposed method with manual tip initialization was evaluated. For roadmap selection with ECG matching (Section \ref{method-sec-roadmap}), the number of online ECG signal points $N_{ECG}$ was manually determined so that the ECG signal stored in the buffer corresponding to 12 X-ray frames (0.8 second in acquisition time). Following the setup in Section \ref{setup-subsec-roadmapping}, we used the distance between the two points in each point pair as the evaluation metric for DCR (the length of a yellow line segment in Figure \ref{setup-figure-distance-vector}). As each frame may have different numbers of point pairs, depending on the length of the target guidewire, the average point pair distance per frame was also computed for evaluation. These distances were evaluated on 409 selected frames with manual annotation of guidewires and vessel centerlines (Section \ref{setup-subsec-roadmapping}).
In the experiment, we compared the DCR with the proposed tracking method to those with manual tip tracking and without tracking. All three approaches were based on the same ECG matching method (Section \ref{method-sec-roadmap}) for selecting roadmaps. The accuracy of the DCR without tracking in Table \ref{exp-tab-roadmapping} shows that the mean distances are reduced to less than 3 mm by compensating only cardiac motion via roadmap selection with ECG matching. Table \ref{exp-tab-roadmapping} also shows that the DCR with the proposed method achieves median distances of about 1.4 mm and mean distances of about 2 mm. The boxplots of the distances of all point pairs and the frame mean point distances of all 409 evaluation frames are illustrated in Figure \ref{exp-figure-roadmapping-boxplot}. The comparison of the three DCR approaches from Table \ref{exp-tab-roadmapping} and Figure \ref{exp-figure-roadmapping-boxplot} indicates that the accuracy of the proposed DCR method has shown improvement over the DCR without tracking, and is only slightly less than the DCR with manual tip tracking (although the difference is statistically significant). Additionally, interested readers are referred to Appendix \ref{app-subsubsec-roadmapping} where the influence of catheter segmentation on the accuracy of DCR is investigated.
Table \ref{exp-tab-roadmapping-error-distribution} shows how the frame mean point distances of the 409 evaluation frames are distributed. The DCR with the proposed method has similar error distribution as the one with manual tip tracking: they both have about 1/3 of the distances less than 1 mm and 1/3 of the distances between 1 and 2 mm. The proposed method has slightly more distances larger than 5 mm than manual tip tracking. Both methods are more accurate than the DCR without tracking on intervals of small errors ($<$ 2 mm).
Figure \ref{exp-figure-roadmapping-examples} shows overlays of selected roadmaps on example frames of 4 sequences with the three DCR approaches. The DCR without tracking presents mismatch of catheters, guidewires or residual of contrast agent in the images, whereas the other methods improve the alignment and show good match between the structures in the original X-ray image and the roadmaps. Compared to the DCR with manual tip tracking, the proposed method show similar visual alignment of the roadmaps to the original X-ray images. For a dynamic view of a roadmapping example, we refer readers to the supplemental material.
\begin{table*}[h]
\centering
\caption{The statistics of DCR accuracy (mm) with three different tracking scenarios. With the two-sided Wilcoxon signed-rank test: $\dagger$ denotes that the difference between the DCR without tracking and that with the proposed tracking method is statistically highly significant ($p < 0.001$); * indicates a statistically significantly difference between the DCR using manual tip tracking and that with the proposed tracking approach ($p < 0.05$).}
\begin{tabular}{l c c c}
\toprule
& Without Tracking$\dagger$ & Proposed Tracking Method & Manual Tip Tracking* \\
\midrule
\textbf{All point pairs} & & & \\
Maximal distance & 27.19 & 20.24 & 13.12 \\
Median distance & 1.97 & 1.43 & 1.35 \\
Mean distance & 2.94 $\pm$ 2.83 & 2.07 $\pm$ 2.08 & 1.85 $\pm$ 1.72 \\
\midrule
\textbf{Frame mean distance} & & & \\
Median distance & 2.11 & 1.42 & 1.38 \\
Average distance & 2.76 $\pm$ 2.08 & 1.91 $\pm$ 1.52 & 1.75 $\pm$ 1.30 \\
\bottomrule
\end{tabular}
\label{exp-tab-roadmapping}
\end{table*}
\begin{figure}
\centering
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/roadmapping/boxplot4.png}
\caption{Accuracy (mm) of DCR with three different tracking scenarios.}
\label{exp-figure-roadmapping-boxplot}
\end{figure}
\begin{table*}[h]
\centering
\caption{Distribution of frame mean point distances of the 409 evaluation frames.}
\begin{tabular}{l c c c c c c}
\toprule
\multirow{2}{*}{Tracking Methods of DCR} & \multicolumn{6}{c}{Error Intervals (mm)} \\
\cmidrule(lr){2-7}
& $<$ 1 & 1-2 & 2-3 & 3-4 & 4-5 & $\geq$ 5 \\
\midrule
Without tracking & 81 & 115 & 69 & 47 & 31 & 66 \\
Proposed tracking method & 131 & 145 & 61 & 32 & 17 & 23 \\
Manual tip tracking & 139 & 144 & 61 & 35 & 20 & 10 \\
\bottomrule
\end{tabular}
\label{exp-tab-roadmapping-error-distribution}
\end{table*}
\begin{figure}
\centering
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/roadmapping/examples/roadmapping_examples3.pdf}
\caption{Examples of superimposition of selected roadmaps (red) on X-ray fluoroscopic frames. (Best viewed in color)}
\label{exp-figure-roadmapping-examples}
\end{figure}
\subsection{Processing Time}
\label{exp-subsec-time}
The processing time of all steps in the proposed DCR method was measured with the hardware and software setup in Section \ref{setup-setup-implementation}. The ECG matching method for roadmap selection was running in Python on the CPU of the Linux machine; the deep neural network and the optical flow component of the tracking method were running on the GPU.
In the experiments, the runtimes for roadmap selection (step 1) and roadmap transformation (step 3) in Figure \ref{method-figure-overview} were negligible ($<$ 1 ms / frame). The runtime of the proposed catheter tip tracking method is shown in Table \ref{exp-tab-tracking-time} and Figure \ref{exp-figure-tracking-time}. The average time to compute the likelihood with the deep learning setup (DL) is 31.5 ms / frame. The particle filtering (PF) step, which consists of the optical flow estimation, sample propagation, sample weight update and normalization, prediction and resampling, takes on average 23 ms / frame. Therefore, the average tracking time in total is 54.5 ms / frame. The total average time of the proposed DCR including roadmap selection, catheter tip tracking and roadmap transformation is still less than the acquisition time of our data (66.7 ms / frame, 15 fps), indicating that the proposed DCR method would run in real-time with our setup.
\begin{table*}[h]
\centering
\caption{Statistics of the runtime of catheter tip tracking (ms / frame) on the test (tracking) dataset.}
\begin{tabular}{c c c c}
\toprule
& Deep Learning & Particle Filtering & Total Tracking Time \\
\midrule
Mean & 31.5 $\pm$ 10.3 & 23.0 $\pm$ 8.7 & 54.5 $\pm$ 12.3 \\
Median & 35.1 & 22.8 & 57.7 \\
\bottomrule
\end{tabular}
\label{exp-tab-tracking-time}
\end{table*}
\begin{figure}[t]
\centering
\includegraphics[width=0.48\textwidth]{sections/experiments/figures/time/boxplot.pdf}
\caption{Runtime of catheter tip tracking (ms / frame) on all test frames.}
\label{exp-figure-tracking-time}
\end{figure}
\section{Experimental Setup}
\subsection{Data}
\label{setup-subsec-data}
Anonymized clinical imaging data were used for our experiments. The data were acquired with standard clinical protocol using Siemens AXIOM-Artis system, and are from 55 patients who underwent a PCI procedure at the Department of Cardiology at Erasmus MC in Rotterdam, Netherlands. Out of these data, we selected data from 37 patients which were acquired since the year 2014 to develop our method, and used the data from the other 18 patients acquired before the year 2013 for evaluation. The detailed information about the data is listed in Table \ref{setup-tab-data}.
\begin{table}
\centering
\caption{Basic information of the acquired X-ray image data for our experiments. The number in the parenthesis next to the pixel size indicates the possible image size.}
\begin{tabular}{l l l}
\toprule
Data & Development & Evaluation \\
\midrule
No. patients & 37 & 18 \\
No. sequences & 354 & 34 \\
Frame rate (fps) & 15 & 15 \\
Image size (px) & 512 $\times$ 512 & 512 $\times$ 512 \\
& 600 $\times$ 600 & 600 $\times$ 600 \\
& 776 $\times$ 776 & 776 $\times$ 776 \\
& 960 $\times$ 960 & 1024 $\times$ 1024 \\
& 1024 $\times$ 1024 & \\
Pixel size (mm) & 0.108 (1024) & 0.139 (1024) \\
& 0.139 (1024) & 0.184 (600) \\
& 0.184 (600) & 0.184 (776) \\
& 0.184 (776) & 0.184 (1024) \\
& 0.184 (960) & 0.216 (512) \\
& 0.184 (1024) & 0.279 (512) \\
& 0.216 (512) & \\
\bottomrule
\end{tabular}
\label{setup-tab-data}
\end{table}
In order to evaluate the proposed roadmapping method, for which an off-line angiographic sequence is required for roadmap preparation and an online fluoroscopic sequence taken from the same C-arm position is needed for performing the actual roadmapping (see Section \ref{method-sec-overview}), we selected the contrast frames from a real clinical sequence to simulate the off-line sequence, and chose the non-contrast frames from the same clinical sequence to simulate the online sequence. The selected contrast sequence were ensured sufficiently long to cover at least one complete cardiac cycle.
\subsection{Data Split for Catheter Tip Detection and Tracking}
\label{setup-subsec-tracking}
To develop the catheter tip tracking method, 1086 X-ray fluoroscopic images selected from 260 non-contrast sequences of 25 patients from the development set were used for training the network from Figure \ref{method-figure-network}; 404 images from 94 non-contrast sequences of another 12 patients from the development set were used as validation set for the network model and hyperparameter selection. In the training and validation sets, 4-5 frames were randomly selected from each sequence, which are not necessarily continuous. To tune the parameters for tracking, 1583 images from 88 sequences out of the 94 from the same 12 patients of the validation set were used (6 sequences were not selected for this task due to very short sequence length not more than 5 frames). Finally, to evaluate catheter tip tracking accuracy, 1355 images from 34 non-contrast sequences of 18 patients from the evaluation set were used for testing. The frames selected for tracking from each sequence must be continuous; the number of selected frames for tracking might vary, depending on the number of the non-contrast frames in the sequences. Details of the datasets for training, validation and test are listed in Table \ref{setup-tab-tip-experiment}.
\begin{table*}
\centering
\caption{Dataset of training, validation and test for detection and tracking of catheter tip in X-ray fluoroscopic frames.}
\begin{tabular}{l l l l l}
\toprule
& Training & Validation & Validation & Test \\
& (detection) & (detection) & (tracking) & (tracking) \\
\midrule
No. patients & 25 & 12 & 12 & 18 \\
No. sequences & 260 & 94 & 88 & 34 \\
No. frames & 1086 & 404 & 1583 & 1355 \\
Continous frames? & No & No & Yes & Yes \\
\bottomrule
\end{tabular}
\label{setup-tab-tip-experiment}
\end{table*}
\subsection{Experimental Settings for Training the Deep Network}
\label{setup-subsec-trainingnetwork}
This section describes the basic experimental settings for training the deep neural network. Details of the training setup can be found in Appendix \ref{app-subsec-training-setup}.
\subsubsection{Preprocessing}
As the image data have different size ranging from $512\times512$ to $1024\times1024$, all images were resampled to a grid of $256\times256$ before being processed by the neural network. In addition, the image intensities were rescaled to the range from 0 to 1.
\subsubsection{Training label}
The standard deviation $\sigma$ of the Gaussian PDF for the training label of the detection network was set to 4 pixels in the resampled image space ($256\times256$). This choice corresponds to the estimation of the maximal possible catheter tip radius. An example of the Gaussian PDF is shown in Figure \ref{method-subfigure-detection-heatmap}.
\subsubsection{Evaluation Metric}
To select hyperparameters and model weights in training, an evaluation metric is required. As the deep network is essentially a catheter tip detector, accurate detection of the tip location is desired. Therefore, we chose the location with the highest value in the detection output, and computed the Euclidean distance between the chosen location and the ground truth tip coordinate as the evaluation metric to tune the deep network.
\subsection{Setup for Evaluating Dynamic Coronary Roadmapping}
\label{setup-subsec-roadmapping}
It is in general a challenge to evaluate the roadmapping accuracy, as the structure of interest, e.g. coronary arteries in our case, is not directly visible in the target image. One possible choice introduced by \cite{zhu2010image} is to use the guidewire as a surrogate of the target vessel centerline in non-contrast images, as guidewire is always inside vessels and commonly present in image sequences during interventions. In this work, we follow a similar strategy to evaluate the accuracy of dynamic coronary roadmapping.
The first step is to select frames for roadmapping evaluation. From each non-contrast sequence in the test set for tracking in Section \ref{setup-subsec-tracking}, we uniformly select 8-20 frames to annotate guidewire. The number of the selected frames from each sequence depends on the sequence length, the frame interval size and guidewire visibility. For some rare cases in our data where no guidewire is present in the image, we discarded that non-contrast frame, and chose those frames with little vessel contrast from the same sequence and annotated the vessel centerline. The selection results in 409 frames from 34 sequences in total. Once the target non-contrast frames for evaluating roadmapping are chosen, their corresponding angiographic frames were found using the ECG matching method in Section \ref{method-sec-roadmap}. We then annotated the centerline of the vessel corresponding to the guidewire in the non-contrast frames.
The next step is performing the transformation of the labelled vessel centerline from the angiographic frame to its corresponding target non-contrast frame via displacement of catheter tip in the two frames. This step simulates the roadmapping transformation in the last step in Figure \ref{method-figure-overview}.
Finally, the distance between the guidewire annotation in the target frame and the transformed vessel centerline is reported as the roadmapping accuracy. In order to compute the distance between two point sets of annotations (e.g. Figure \ref{setup-subfigure-without-vector}), point-point correspondence between the two sets is required (Figure \ref{setup-subfigure-with-vector}). The point sets were firstly resampled with the point interval being 1 mm. We then followed the approach of \cite{van2008averaging} to find such correspondences which minimizes the sum of the Euclidean distance of all valid point-point correspondence paths. This way guarantees no cross-over connection and each point in one set is connected to at least one point in the other set. As the annotated point sets may have different size, the point correspondences to endpoints are excluded such that we only focused on the distance between corresponding sections, not the entire centerlines (Figure \ref{setup-subfigure-with-vector-zoomin}). Once the point-point correspondence is available, the distance between the two points in a pair can be used for evaluating the accuracy of DCR.
\begin{figure}[h]
\centering
\subfloat[]{
\includegraphics[width=0.15\textwidth]{sections/setup/figures/WithoutVector0.png}
\label{setup-subfigure-without-vector}
}
\subfloat[]{
\includegraphics[width=0.15\textwidth]{sections/setup/figures/WithVector1.png}
\label{setup-subfigure-with-vector}
}
\subfloat[]{
\includegraphics[width=0.15\textwidth]{sections/setup/figures/WithVectorZoomIn2.png}
\label{setup-subfigure-with-vector-zoomin}
}
\caption{Correspondence between the labelled guidewire (green) and the transformed vessel centerline (red). The yellow lines connecting the two point sets illustrate the correspondence between red and green points.}
\label{setup-figure-distance-vector}
\end{figure}
\subsection{Implementation}
\label{setup-setup-implementation}
The proposed method was developed in Python. The framework used for developing the deep learning approach for likelihood approximation is PyTorch. The major experiments of dynamic coronary roadmapping were performed on a computer with an Intel Xeon E5-2620 v3 2.40 GHz CPU and 16 GB RAM running Ubuntu 16.04. The deep neural network and the optical flow method were running on an NVIDIA GeForce GTX 1080 GPU. The approach for evaluating dynamic coronary roadmapping was developed and running in MeVisLab on a computer with an Intel Core i7-4800MQ 2.70 GHz CPU and 16 GB RAM running Windows 7.
\section*{Acknowledgment}
This work was supported by NWO-domain TTW (The Division of Applied and Engineering Sciences in The Netherlands Organisation for Scientific Research), IMAGIC project under the iMIT program (grant number 12703). Ries and Simon van Walsum are acknowledged for their contribution in the manual annotations.
\section{Discussion}
We have presented a new approach to perform online dynamic coronary roadmapping on X-ray fluoroscopic sequences for PCI procedures. The approach compensates the cardiac-induced vessel motion via selecting offline-stored roadmaps with appropriate cardiac phase using ECG matching, and corrects the respiratory motion of vessels by online tracking of guiding catheter tip in X-ray fluoroscopy using a proposed deep learning based Bayesian filtering. The proposed tracking method represents and tracks the posterior of catheter tip via a particle filter, for which a likelihood probability map is computed for updating the particle weights using a convolutional neural network. In the experiments, the proposed DCR approach has been trained and evaluated on clinical X-ray sequences for both tracking and roadmapping tasks.
One prerequisite of accurate tracking with the proposed approach is to obtain a reasonably good likelihood estimation, which requires to train the deep neural network to detect catheter tip well. In this work, we have investigated the influence of three network hyperparameters on the performance of the detection network (Section \ref{exp-subsec-detection}): the basic channel number and network depth level are model capacity parameters, the dropout adds regularization to the model. The experiment showed that the detection accuracy improves when the basic channel number and the network depth level increase (Table \ref{exp-tab-training-network}). This observation matches the expectation that a more complex model has higher capacity to model the variation in the data, hence results in better accuracy. When the complexity reaches a certain level, e.g. 64 basic channels and 5 level of depth, the network performance does not increase much compared to those with simpler settings, implying that the model starts overfitting on our dataset. \iffalse The experiment also shows that the dropout layers can further boost the model performance compared with the networks without dropout.\fi
In addition to the deep neural network, the other important component of the proposed tracking approach is the sampling in the particle filter that yields the samples for representing the prior and the posterior of catheter tip position. First, a sufficient number of samples in the whole sample space are required to well characterize the probability distributions (see Appendix \ref{exp-subsubsec-pf}). Second, the sample dynamics plays an important role in tracking, in particular, as indicated by Eq.(\ref{method-eq-state-transition-flow}), the process noise and the sample motion. The process noise influences the tracking accuracy, according to Table \ref{exp-tab-tune-particle} in Appendix \ref{exp-subsubsec-pf}. Additionally, sample motion is another key aspect of sample dynamics. Motion estimation has previouly been incorporated in a motion-based particle filter, such as adaptive block matching (\cite{bouaynaya2009optimality}). In our work, optical flow was chosen for motion estimation, as its non-parametric nature allows to characterize the complexity of motion in X-ray fluoroscopy well. In addition, the advantage of such approach from a theoretical point of view is that it takes into account of the current observation, leading to a more optimal importance density (\cite{arulampalam2002tutorial}) compared to random motion.
The tracking results in Section \ref{exp-subsubsec-tracking} show that the proposed tracking approach is able to track the catheter tip in X-ray fluoroscopy accurately with an average tracking error of about 1.3 mm. It also shows advantages over methods based only on optical flow or the detection network. The OF (pre) method relies heavily on tracking in the previous frame, hence the error could accumulate. The OF (first) method may suffer from large motion from the first frame to the current frame. The detection method uses information only from the current frame, no temporal relation between frames is utilized; therefore, it results in spikes in the longitudinal view, as shown in Figure \ref{exp-figure-line-plot}. The proposed tracking method has a CNN to provide an accurate observation on the current frame which improves the accuracy of optical flow tracking within the framework of Bayesian filtering. In the meantime, the optical flow based particle filter maintains and propagates the prior knowledge from the initial tip position to provide a constraint on searching for the potentially correct positions, which is useful especially when the CNN detector fails to find the correct target area. The association of knowledge from two sources together improves the tracking accuracy compared to each single source.
The initial state is a also key component of tracking approaches. In the context of Bayesian filtering, the initial state provides the prior knowledge of the tracking target. Most tracking algorithms assume a known initial state from which the target is tracked, e.g. our proposed method with manual initialization in Section \ref{exp-subsubsec-initialization}. In this case, the prior knowledge is provided by human. In Section \ref{exp-subsubsec-initialization}, we also investigated a scenario where the initial state is given by the detection network, so that the complete tracking process is fully automated. The results indicate that, the proposed tracking method with automatic initialization works reasonably well on most sequences even when the initialization is sometimes incorrect (Figure \ref{exp-subfigure-auto-init-examples-1}). This is because (1) the true position is covered by a few samples, and (2) the correct detection in later frames corrects the initial mistake in the first frame. The automatic initialization fails when (1) a wrong position is covered by a few samples and (2) the wrong detection in subsequent frames confirms the mistake in the initial frame (Figure \ref{exp-subfigure-auto-init-examples-2}). This happens when there is contrast agent remaining in the image or there are multiple catheters, which are the major sources causing ambiguity in detection. In practice, the automatic initialization would work well when contrast agent is washed out and only one catheter is present in the field of view, otherwise manual initialization would be needed which requires only one click to initiate tracking.
Dynamic coronary roadmapping is the direct application of the catheter tip tracking results. In our experiments, the DCR was performed with manual tip initialization to show the potential of the proposed tracking method, and was compared with the DCR without tracking and with manual tracking. The results indicate that using catheter tip tracking can improve DCR accuracy, as the respiratory-induced vessel motion is corrected by the displacement of catheter tip in addtion to cardiac motion correction. The results also show that the proposed DCR reaches a good accuracy (mean error is about 2 mm) and performs only slightly worse than its best case, the DCR with manual tip tracking which is not applicable for intraoperative use. Additionally, according to a previous study by \cite{dodge1992lumen}, the average lumen diameters of human coronary arteries are between 1.9 mm (distal left anterior descending artery) and 4.5 mm (left main artery). This means that the accuracy achieved with the proposed approach is comparable with the size of coronary arteries.
Apart from catheter tip tracking, several other possible factors in different steps of the experiments may influence the final DCR accuracy. First, in the offline phase, the signal of contrast agent may become too strong and completely cover the catheter tip, complicating the tip visibility in some cases. In this situation, the uncertainty in the manual tip annotation may result in errors in roadmap transformation. Second, in the roadmap selection step, the offline-stored roadmaps are only discrete samples of complete cardiac cycles which might not fully characterize every possible change in the cardiac motion. This problem could possibly be addressed in the future by interpolating frames between the existing frames in the data. Additionally, variation exists between different cardiac cycles (\cite{mcclelland2013respiratory}), therefore, choosing a roadmap from another cycle may cause inaccuracy for cardiac motion compensation. Finally, the way of DCR evaluation in Section \ref{setup-subsec-roadmapping} might also introduce inaccuracies in the error measurement. Since guidewires often attach to the inner curves of a vessel to take the shortest path, the small difference between the annotation of guidewire and vessel centerlines was ignored in the evaluation.
In addition to accuracy, processing speed is also critical for intraoperative applications. The results in Section \ref{exp-subsec-time} indicate that the total processing time of the proposed DCR approach is less than the image acquisition time, meaning that it runs in real-time on our setup. To build a real-time system for PCI in practice, the overall latency of the complete system needs to be considered. It is also worth noticing that the DL and PF steps of the proposed tracking method are independent from each other. In practice, in case more than one GPU are available, the proposed DCR approach can be further accelerated by paralleling the DL and PF steps, making them running on different GPUs.
Compared to the previous works on DCR, the proposed approach in this paper shows advancement in several aspects. First, our systems works on non-cardiac-gated sequences which does not require additional setups for cardiac motion gating that were needed for some methods (\cite{zhu2010image, manhart2011self}). Second, our approach compensates both respiratory- and cardiac-induced vessel motion, which is more accurate than systems that correct only cardiac motion (\cite{elion1989dynamic}). In addition, the proposed DCR approach follows a data-driven paradigm that learns target feature from sequences acquired from different patients and various view angles, making it more robust than the method that relies on traditional vesselness filtering (\cite{kim2018registration}) or methods that require specific tissue being present (\cite{zhu2010image, manhart2011self}). These are the major advantages of the proposed DCR over the existing direct roadmapping systems. Compared to model-based motion compensation, our approach does not require the extraction of motion surrogate signals and train a motion model for each new patient, but can be directly run with a trained model.
The proposed deep learning based Bayesian filtering method has several advantages over the existing instrument tracking methods. First, the deep learning component enables a more general framework to detect instruments in medical images than methods tailored for specific tools (\cite{ma2012clinical, ma2013real}). Compared to the existing detection methods based on deep learning (\cite{baur2016cathnets, laina2017concurrent, du2018articulated}), our approach takes into account of the information between frames; the Bayesian filtering framework allows interaction between temporal information and the detection of a convolutional neural network, making the tracking more robust. Bayesian frameworks have been used in many previous temporal instrument tracking methods. Particularly, the likelihood term in some works was designed based on registration or segmentation outcomes (\cite{ambrosini2017hidden, speidel2006tracking}) or traditional machine learning approaches with handcrafted features (\cite{wang2009robust, wu2012fast, pauly2010machine}). In our method, we approximated the likelihood with a deep neural network learned from the clinical data which exempts the need of feature engineering but yet possesses more discriminative power; the network directly outputs the probability map, making it more straightforward to use. Finally, compared to the existing instrument tracking approaches based on Bayesian filtering (\cite{ambrosini2017hidden, speidel2006tracking, speidel2014visual}), the state transition in our method was based on the motion estimated from two adjacent frames, which is more reliable than totally random motion or artificially-designed state transition models.
From a practical point of view, the proposed DCR approach could potentially fit into the clinical workflow of PCI. The offline phase of the method can be done efficiently by a person who assists the procedures: selecting and creating roadmaps from an angiography acquisition, annotating the catheter tip (one point) in the images. This phase is typically done before a fluoroscopy acquisition during which the guidewire advancement and stent placement are performed. In the online phase, when a fluoroscopic image is acquired, the proposed system selects the most suitable roadmap, tracks the catheter tip and transforms the roadmap to prospectively show a vessel overlay on the fluorosocpic image. The online updated coronary roadmap can provide real-time visual guidance to cardiologists to manipulate interventional tools during the procedure without the need of administering extra contrast agent.
In the future, it may be worth investigating the following directions related to this work. As the data used in this study was acquired from one hospital using a machine from a single vendor, it would be interesting to evaluate the proposed approach on multi-center data acquired with machines from different vendors. Next, since the ECG signals of our data appear to be regular, it may be necessary in a future study to acquire data with irregular ECG that could be obtained in practice, and validate the proposed approach on those data. Besides, it would be also interesting to validate our approach during PCI procedures in an environment simulating the real clinical settings. Additionally, from a methodological point of view, although the proposed tracking method is invariant under different view angles, the whole DCR approach works only when the offline and online phase have the same view angle, i.e. it is a 2D roadmapping system. Therefore, one future direction would be to develop a 3D DCR system that would work with various view angles in the online phase.
\section{Conclusion}
We have developed and validated a novel approach to perform dynamic coronary roadmapping for PCI image guidance. The approach compensates cardiac motion through ECG alignment and respiratory motion by guiding catheter tip tracking during fluoroscopy with a deep learning based Bayesian filtering method. The proposed tracking and roadmapping approaches were trained and evaluated on clinical X-ray image datasets and were proved to perform accurately on both catheter tip tracking and dynamic coronary roadmapping tasks. Our approach also runs in real-time on a setup with a modern GPU and thus has the potential to be integrated into routine PCI procedures, assisting the operator with real-time visual image guidance.
| {
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Logan City er befolkningsmæssigt den tredje største by i den australske delstat Queensland, med sine ca. 177.000 (2006) indbyggere.
Det første nybyggere kom til Logan i 1842. Logan City fik bystatus i 1981.
Navnet Logan er efter kaptajn Patrick Logan, noget som fik en lokal politiker i 2006 til at foreslå at omdøbe byen til Silver City. Bagrunden var at Patrick Logan var kendt for at være hensynsløs tyran, noget han mente kunne forbinde byen med kriminalitet.
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handedly performs massive renovation projects. He is a writer, photographer, cave explorer, and world traveler. Ben has been to 36 world countries and all 7 continents. In 2007 he appeared on the Rachel Ray Show. In 2010 he appeared on HGTV's, All American Handyman. Most recently, Ben auditioned with 30,000 others for Fox's hit series Masterchef with Gordon Ramsay. He was cast on the show and placed 5th of the top 100 contestants, appearing on all 22 episodes of the show. Gordon Ramsay's production company has signed Ben to a talent contract and is currently working with him to develop a TV show of his own.
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Contents
Title Page
Introduction
01/ Easy
02/ Fast
03/ Healthy
04/ Classic
05/ Teatime
06/ Dessert
07/ Children's
Decorating ideas
08/ Celebration
09/ Festive
Cake essentials
List of Recipes
Thank yous
Copyright
About the Publisher
Introduction
I love every type of cake, whether light and buttery or dense and fudgy. In fact, I have adored baking ever since I was a child, when I began making cakes with my mother and sister. I was transfixed by the whole process – from stirring the ingredients in a big bowl to popping the filled tin in the oven and waiting for the magic to happen. Then, hey presto, the mixture had turned into a proper cake, beautifully risen and golden and with mouth-watering smells filling the kitchen. Our finished creation was then carefully removed and decorated before being placed on a plate and proudly presented to the rest of the family or to friends.
I know I'm not alone in my passion for cakes. Indeed, I get asked for my cake recipes more than any other type of dish! This book is aimed at anyone who has ever enjoyed a slice of cake – whether baking is your favourite pastime or if you've never tried it before. My previous book, _Bake_ , was dedicated to all types of baking, from pies and casseroles to puddings and biscuits. This volume focuses exclusively on cake baking, showing you how to quickly master the basics, add effortless decoration and expand your repertoire of great-looking, great-tasting cakes to give, share and enjoy.
There is always something in my diary that offers the perfect excuse to bake a cake. With this in mind, I've divided the book into sections, each with a selection of cakes for different occasions, be it a simple picnic or coffee with a friend, or an elaborate event like a wedding, birthday or special celebration. You'll find chapters devoted to fast cakes, festive cakes, classic and teatime cakes, novelty cakes for children's parties and recipes catering specifically for people with allergies or food intolerances.
The recipes in this book are all easy to follow, with no particular level of skill or experience required. Most are quick and simple to make, while a few will take a little more time and effort. The sections on cake decoration include recipes for honeycomb, chocolate squares and fondant-icing shapes, each of which will help you add that special finishing touch or real 'wow' factor. Icing and decoration are usually tailored to specific cakes, but you can experiment with mixing ideas from different recipes and be as creative as you like.
There is something in _Cake_ for everyone, from the casual cook to the committed baker, and I hope you enjoy using this book as much as I've enjoyed creating the recipes for it. I believe that divine recipes are made to be shared and cake tastes best when enjoyed with friends or family. It's true that cakes are an indulgence and not something to have every day, but eaten every so often they are a genuine treat. Even a tiny sliver can really brighten your day or transform an otherwise ordinary get-together into a memorable occasion – magic not only in the making but in the eating!
You don't always need a special reason to bake a cake and this chapter is full of cakes that are really easy to put together. The recipes are not elaborate; there are no grand tiers or complicated icings here. The cakes are straightforward without involving too much technique or hours of labour. These cakes are the ones that I keep in a tin in the kitchen, ready for when a friend calls round or invites me over for coffee. They are also good for popping into a lunchbox or at the end of a simple supper, when you don't need a fancy dessert, just a little slice of something sweet. Easily achievable and simply delicious, these are cake recipes that you'll return to again and again.
01/ Easy
Lemon poppy-seed cake
Lemon poppy-seed muffins
Apricot, pistachio and saffron muffins
Pear and ginger muffins
Coconut and lime cake
Apple and walnut cake
Bakewell cake
Marbled chocolate crumble cake
Lemon crumble cupcakes
Lemon curd
Tropical cake
Toscatårta or Swedish almond cake
Lemon and rosemary polenta cake
Orange sour-cream cake
Lemon poppy-seed cake
Poppy seeds are used in cakes and many other dishes across the world, their fragrant nutty taste going especially well with the citrus flavour of lemon or orange. There is a lot of milk in this recipe, which makes the crumb very tender and soft. It's baked in a bundt tin, which gives more of a crusty outer layer, though if you don't have one, a standard round cake tin would do just as well, or you could use the mixture to make muffins. If you like, you could decorate with candied orange slices, placing these on the iced cake instead of sprinkling over the poppy seeds.
**Prep time:** 15 minutes
**Baking time:** 50–70 minutes
**Ready in:** 1 hour 40 minutes
**Serves:** 10–14
325g (11½oz) plain flour, plus extra for dusting
1½ tsp baking powder
1½ tsp salt
400g (14oz) caster sugar
25g (1oz) poppy seeds, plus 1–2 tsp to decorate
225ml (8fl oz) sunflower or vegetable oil, plus extra for greasing
4 eggs
1 tsp vanilla extract
325ml (11½fl oz) milk
Finely grated zest of 1 lemon
Candied orange slices, to decorate (optional)
For the icing
200g (7oz) icing sugar, sifted
3 tbsp lemon juice
2.5 litre (4⅓ pint) bundt tin (about 23cm/9in in diameter) or 25cm (10in) diameter cake tin
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the bundt tin with sunflower or vegetable oil and dust with flour. If you're using a standard type of cake tin, grease the sides and line the base with a disc of baking parchment.
Sift the flour, baking powder and salt into a large bowl, add the sugar and poppy seeds and mix together.
In a separate bowl, whisk together the remaining ingredients until combined. Tip this mixture into the dry ingredients and mix together using a wooden spoon to make a smooth batter. You may need to use a whisk briefly to get rid of any lumps of flour.
Tip into the prepared tin and bake for 50–60 minutes (60–70 minutes if using a standard cake tin). When cooked, the cake should feel springy to the touch and a skewer inserted into the centre should come out clean.
Remove the cake from the oven and allow to sit for just 2 minutes, then loosen the edges with a small, sharp knife, place a wire rack upside down on top of the cake and carefully turn it over. Gently remove the tin and allow to cool completely.
While the cake is baking, or while it's cooling, you can make the icing. Beat the icing sugar and lemon juice together until smooth. Carefully transfer the cake to a cake stand or serving plate, then drizzle the icing backwards and forwards from the centre to the outside of the cake in a zigzig pattern. Sprinkle over the poppy seeds straight away and the candied orange slices, if you're using them, before the icing has a chance to dry.
Orange poppy-seed cake
Make the cake as in the recipe, replacing the lemon zest in the sponge with the finely grated zest of 1 orange and using orange juice instead of lemon juice in the icing.
Lemon poppy-seed muffins
**Prep time:** 10 minutes
**Baking time:** 20–25 minutes
**Ready in:** 1 hour
**Makes:** 12 muffins
215g (7½oz) plain flour
1 tsp baking powder
Pinch of salt
265g (9½oz) caster sugar
15g (½oz) poppy seeds, plus 2 tsp for sprinkling
150ml (5fl oz) sunflower or vegetable oil
3 eggs
½ tsp vanilla extract
215ml (7½fl oz) milk
Finely grated zest of ½ large lemon
For the icing
130g (4½oz) icing sugar, sifted
1–2 tbsp lemon juice
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180ºC (350ºF), Gas mark 4, and line the muffin tray with the paper cases.
Mix the ingredients for the sponge following the main recipe, then divide the batter between the paper cases, filling each case three-quarters full.
Bake for 20–25 minutes or until lightly golden on top and springy to the touch. Allow to cool in the muffin tray for 5 minutes, then transfer to a wire rack to cool down fully.
Make the icing, then drizzle it over the cooled muffins and sprinkle with the poppy seeds to finish.
Apricot, pistachio and saffron muffins
I adore the combination of the sweet juicy apricots with the exotically perfumed saffron. Use peaches (fresh or tinned) in place of the apricots if you wish. These muffins are delicious served with a spoonful of crème fraîche and a piping-hot cup of coffee or tea.
**Prep time:** 10 minutes
**Baking time:** 35 minutes
**Ready in:** 1 hour 15 minutes
**Makes:** 12 muffins
225ml (8fl oz) buttermilk
2 eggs
Pinch of saffron
300g (11oz) plain flour
2 tsp baking powder
½ tsp bicarbonate of soda
½ tsp salt
200g (7oz) caster sugar
125g (4½oz) butter, cut into 1cm (½in) cubes
300g (11oz) fresh or tinned apricots (about 7 fresh apricots, stones removed), roughly chopped
75g (3oz) unsalted shelled pistachios, roughly chopped
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180ºC (350ºF), Gas mark 4, and line the muffin tray with the paper cases.
Place the buttermilk, eggs and saffron in a bowl and whisk together until smooth.
In a separate large bowl, sift together the flour, baking powder, bicarbonate of soda and salt and mix in the sugar. Add the cubes of butter and rub in with your fingertips until the mixture resembles coarse breadcrumbs.
Add the wet ingredients and mix together, then fold in the apricots and pistachios. Spoon the batter into the muffin cases, filling each case about three-quarters full.
Bake for about 35 minutes or until golden brown on top and springy to the touch. Allow the muffins to cool for 5 minutes, then remove from the tin and place on a wire rack to finish cooling.
Pear and ginger muffins
These soft, moist muffins are flavoured with chunks of pear and the gentle heat of ginger. The riper the pears, the sweeter the muffins will be. They would be perfect for a picnic or a quick snack on the run.
**Prep time** : 10 minutes
**Baking time** : 20 minutes
**Ready in** : 45 minutes
**Makes** : 12 muffins
275g (10oz) plain flour, sifted
200g (7oz) caster sugar
2 tsp baking powder
1 tbsp ground ginger
1 tsp salt
4 eggs, beaten
150ml (5fl oz) vegetable oil
300g (11oz) pears, peeled, cored and cut into 1cm (½in) dice
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Place all the ingredients, except the pears, in a large bowl and whisk together until smooth, then fold in the pears and divide between the muffin cases, filling each case three-quarters full. Bake for about 20 minutes or until well risen, golden and feeling springy to the touch.
Remove from the oven and allow to cool for 5 minutes, then take from the muffin tray and place on a wire rack to cool down fully.
Apple and ginger (or cinnamon) muffins
Follow the recipe above, replacing the pears with the same quantity of chopped eating apples and either keeping the ginger or replacing it with 2 teaspoons of ground cinnamon.
Coconut and lime cake
Ingredients grown together seem to go together. Just as apples and blackberries are a perfect match, so too are coconut and lime. The coconut flavour in this recipe comes from both desiccated coconut and coconut milk. The coconut milk is an unusual addition, but it is nonetheless an excellent baking ingredient that helps make the cake lovely and soft due to its high oil content. If at all possible, serve this cake outside on a summer's evening, ideally accompanied by piña coladas or margaritas!
**Prep time** : 20 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 8–12
150ml (5fl oz) coconut milk (see the tip)
50g (2oz) desiccated coconut
25ml (1fl oz) milk
250g (9oz) butter, softened, plus extra for greasing
250g (9oz) caster sugar
3 eggs, beaten
Juice and finely grated zest of 2 limes
275g (10oz) self-raising flour
50g (2oz) coconut flakes, toasted (see the tip), to decorate
For the icing
15ml (½fl oz) coconut milk
250g (9oz) icing sugar, sifted
Juice of ½–1 lime
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a square of baking parchment. Pour the coconut milk into a saucepan and place over a high heat. As soon as it boils, remove from the heat and stir in the desiccated coconut and the milk.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together with the lime juice and zest in a small bowl, then gradually add the eggs to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently to mix.
Tip the batter into the prepared tin and smooth the top with a spatula or palette knife. Place in the oven and bake for 30–35 minutes. When the cake is ready, a skewer inserted into the centre will come out very slightly sticky.
Leave to cool in the tin for 10 minutes, then use a small, sharp knife to loosen the edges and carefully remove the cake from the tin before leaving on a wire rack to finish cooling.
As the cake cools, make the icing. In a bowl, whisk together the coconut milk, icing sugar and the juice of half a lime, adding more if the mixture seems too stiff. Carefully pour the icing over the cake, to cover it in an even layer, then sprinkle over the toasted coconut flakes.
Tip This recipe uses a small (165ml) tin of coconut milk. If you can't get a small tin, then use a larger one and store any leftover milk in the fridge, where it will keep (in an airtight container) for up to four days, or it can be frozen. Coconut flakes can be toasted in a dry, hot frying pan for a few minutes.
Apple and walnut cake
Some spices have their favourite accompaniments: cinnamon, for example, seems to prefer apple to almost anything else. The two are combined here with crunchy walnuts and muscovado (soft brown) sugar for its unrefined, caramel flavour. I've used some wholemeal flour in this recipe as I like the difference it makes, adding a little weightiness in both texture and taste. The cake makes a delicious dessert, eaten warm with whipped cream, or you could serve it in the afternoon with tea.
**Prep time** : 20 minutes
**Baking time** : 55 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 10–12
300g (11oz) plain flour
1 tsp ground cinnamon
2 tsp baking powder
½ tsp bicarbonate of soda
1 tsp salt
175g (6oz) wholemeal flour
225g (8oz) soft light brown sugar
175ml (6fl oz) sunflower oil, plus extra for greasing
3 eggs
150ml (5fl oz) milk
3 eating apples, peeled, cored and chopped into 1cm (½in) dice
75g (3oz) walnuts, roughly chopped
For the glaze
150g (5oz) soft light brown sugar
75g (3oz) butter
25cm (10in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, then grease the sides of the cake tin with sunflower oil and line the base with a disc of baking parchment.
Sift the plain flour, cinnamon, baking powder, bicarbonate of soda and salt into a large bowl, then add the wholemeal flour and sugar and mix together.
In a separate bowl, whisk together the sunflower oil, eggs and milk, then tip these into the flour and whisk until you have a smooth batter, and stir in the apples and walnuts. Pour the mixture into the prepared tin and bake for 55 minutes or until a skewer inserted into the centre of the cake comes out clean.
About 5 minutes before the cake is due to come out, make the glaze. Put the sugar, butter and 1 tablespoon of water into a saucepan and place over a medium heat. Stir until the sugar has dissolved, then remove from the heat.
When the cake is cooked, take it out of the oven and let it sit in the tin for 10 minutes. Using a small, sharp knife, loosen around the edges and carefully remove the cake from the tin before transferring to a serving plate.
Reheat the glaze and brush all over the cake, then allow to cool before serving. The glaze means this cake will keep for up to a week in an airtight container.
Bakewell cake
The Derbyshire town of Bakewell is responsible for the invention of the classic tart to which the town gave its name. The divine combination of raspberries and almonds makes for a gorgeous recipe – its inventor must have been someone who baked very well indeed! This cake is an evolution of the idea, using fresh or frozen raspberries rather than raspberry jam. It's a simple recipe but a perfect example of how a simple combination of flavours can be deliciously effective.
**Prep time** : 15 minutes
**Baking time** : 50–55 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 6–8
150g (5oz) butter, softened, plus extra for greasing
150g (5oz) caster sugar
2 eggs
A few drops (not more than ⅛ tsp) of almond essence or extract
50ml (2fl oz) milk
150g (5oz) self-raising flour, sifted
150g (5oz) ground almonds
150g (5oz) fresh or frozen (and defrosted) raspberries
25g (1oz) flaked almonds
Icing sugar, for dusting
Softly whipped cream and fresh raspberries, to decorate
20cm (8in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs and the almond essence or extract together in a small bowl for a few seconds or just until combined, then gradually add the eggs to the creamed butter mixture, beating all the time. Next beat in the milk, then add the flour and ground almonds, carefully folding these in just until they are mixed.
Next add the raspberries and fold in gently so as not to break them up too much. Tip the batter into the prepared tin, then scatter over the flaked almonds. Bake for 50–55 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes. Use a small, sharp knife to loosen the edges, then carefully remove the cake from the tin and leave on a wire rack to cool down completely before transferring to a serving plate.
Dust with icing sugar. Serve with softly whipped cream if you wish, or decorate with fresh raspberries.
Marbled chocolate crumble cake
I adore marbled cakes – they look so special with the different colours of sponge swirling into each other. In this recipe the crumble topping adds a contrasting crunch as well as an extra hit of chocolate. Serve on its own or after dinner with some softly whipped cream or crème fraîche.
**Prep time** : 25 minutes
**Baking time** : 50–60 minutes
**Ready in** : 2 hours
**Serves** : 6–8
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs
1 tsp vanilla extract
225g (8oz) plain flour
2 tsp baking powder
50ml (2fl oz) milk
25g (1oz) cocoa powder
Icing sugar, for dusting
For the crumble topping
125g (4½oz) plain flour, sifted
75g (3oz) caster sugar
75g (3oz) unsalted butter, chilled and cut into cubes
75g (3oz) dark or milk chocolate, in chips or roughly chopped into pieces
23cm (9in) diameter spring-form or loose-bottomed cake tin with 6cm (2½in) sides
First make the crumble topping. Using your fingertips, rub together the flour, sugar and butter in a large bowl until it resembles thick breadcrumbs, then mix in the chocolate pieces. Set aside in the fridge while you make the sponge.
Preheat the oven to 180°C (350°F), Gas mark 4, and butter the sides and the base of the cake tin. If you're using a spring-form tin, make sure the base is upside down so there's no lip and the cake can slide off easily when cooked. Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs and vanilla extract together in a small bowl for a few seconds or just until combined, then gradually add the eggs to the creamed butter and sugar mixture, beating all the time. Sift in the flour and baking powder and fold in carefully, then add the milk and mix gently to combine. Tip half of the cake mixture into another large bowl, then sift the cocoa powder into this bowl and fold it in.
Place the two different cake mixtures in the prepared tin by alternating heaped tablespoonfuls of the vanilla batter with the chocolate one, then, with a skewer or similar implement, gently draw swirls through the cake mixture to 'marbleise' it. Try not to over-mix or you won't get that wonderful marbled effect.
Scatter the crumble mixture evenly over the top of the cake mixture and bake for 50–60 minutes or until the crumble is golden and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges using a small, sharp knife and remove the sides of the tin. Place the cake (sitting on the base of the tin) on a wire rack and allow to cool completely.
Use a palette knife or metal fish slice to loosen the bottom of the cake from the base of the tin, then slide the palette knife or fish slice under the cake and carefully ease it onto a plate. Dust with icing sugar to serve.
Lemon crumble cupcakes
This recipe takes lemon cupcakes to the next level. The lemon curd is easy to make – though you could use a bought variety, of course – and adds a wonderfully sweet-sharp citrus softness. The crumble topping is quick to put together, meanwhile, and adds a great contrasting crunch.
**Prep time** : 30 minutes (excluding the lemon curd)
**Baking time** : 30 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 12 cupcakes
200g (7oz) butter, softened
200g (7oz) caster sugar
4 eggs
200g (7oz) self-raising flour
Juice of 1 lemon
150g (5oz) lemon curd
For the crumble topping
75g (3oz) plain flour
75g (3oz) caster sugar
Finely grated zest of 1 lemon
100g (3½oz) butter, chilled and cut into 1cm (½in) cubes
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
First make the crumble topping. Sift the flour into a bowl, then add the sugar and lemon zest and mix together to combine. Use your fingertips to rub the butter into the flour until the mixture forms coarse flakes and crumbs, then set aside in the fridge until needed.
To make the cupcake batter, cream the butter in a large bowl or in an electric food mixer until soft. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and add the lemon juice, then fold in gently to incorporate.
Divide half of the batter between the muffin cases, filling each about one-third full. Add roughly ½ tablespoon of lemon curd to each paper case, so that it forms a small dollop in the middle. Then divide the other half of the batter between the cases, spooning it over the lemon curd. Finally divide the lemon crumble mixture between the cupcakes.
Bake for about 30 minutes or until nice and golden on top and lightly springy to the touch in the middle. Don't worry if some of the curd bubbles over the top of the cupcakes while they cook.
Allow to cool for 10 minutes before removing from the muffin tray and placing on a wire rack to cool down completely.
Lemon curd
**Prep time** : 10 minutes
**Cooking time** : 10 minutes
**Ready in** : 30 minutes
**Makes** : 300g (11oz)
75g (3oz) butter
150g (5oz) caster sugar
Juice and finely grated zest of 3 lemons
2 eggs
1 egg yolk
In a saucepan over a very low heat, melt the butter with the sugar, lemon zest and juice. Place the eggs and egg yolk in a bowl and beat together well, then pour into the pan.
Stir carefully over a low heat until the mixture has thickened enough to coat the back of a spoon. Take care not to let the mixture get too hot, or the eggs may scramble.
Remove from the heat and pour into a bowl or a sterilised jar (tip). The lemon curd will thicken further as it cools. Stored in the fridge, it will keep for a couple of weeks.
Orange curd
Make in the same way, substituting the lemons with two oranges and reducing the amount of caster sugar to 125g (4½oz).
Tip To sterilise jars for jams and preserves, wash them in hot soapy water, then rinse and dry. Place the jars upturned on a baking tray in the oven (preheated to at least 130°C/250°F/Gas mark ½) for approximately 15 minutes or until completely dry. Leave them upturned on a clean tea towel until ready to use. Alternatively, you can put them through a hot cycle in the dishwasher.
Tropical cake
Fresh pineapple is almost reason enough to head to sunnier climes – tangy slices dripping with juice can be one of the great culinary pleasures of a holiday in the tropics. The pineapples we get here can be a little dry by comparison. For this recipe it's fine to use tinned pineapple, however, as it is more predictable in the amount of moisture and sweetness it contains. With the coconut and mango, this cake is a real celebration of tropical flavours.
**Prep time** : 25 minutes
**Baking time** : 45 minutes
**Ready in** : 2 hours
**Serves** : 8–12
250g (9oz) tinned pineapple (drained weight of a 432g tin)
200g (7oz) soft light brown sugar
225g (8oz) butter, softened, plus extra for greasing
3 eggs
125g (4½oz) self-raising flour, plus extra for dusting
100g (3½oz) desiccated coconut
1 large mango, peeled and sliced, to decorate
For the icing
250g (9oz) mascarpone
50g (2oz) icing sugar, sifted
23cm (9in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and dust with flour and line the base with a square of baking parchment.
Place the pineapple in a food processor and whiz for a minute or two until puréed, then put into a saucepan with the sugar. Set over a low heat and cook, stirring occasionally, for about 5 minutes or until the mixture turns a light brown colour. Then set aside and allow to cool.
Beat the cooled pineapple purée with the butter in a large bowl or in an electric food mixer. Whisk the eggs together in a small bowl for just a few seconds until mixed, then gradually add the eggs to the pineapple and butter mixture, beating continuously. Sift in the flour, add the coconut and fold in gently to combine.
Tip the batter into the prepared tin, then bake in the oven for about 45 minutes or until a skewer inserted into the centre of the cake comes out clean. Allow the cake to sit in the tin for about 20 minutes, then use a small, sharp knife to loosen the edges and carefully remove the cake from the tin before leaving on a wire rack to cool down completely.
To make the icing, whisk together the mascarpone and icing sugar, then spread over the cooled cake using a palette knife and finish by decorating with the sliced mango.
Toscatårta or Swedish almond cake
There is a banquet of baked Swedish treats, from _kanelbullar_ (cinnamon buns) to _kladdkaka_ (sticky chocolate cake), but one of the most well known is this relatively simple but absolutely divine almond cake – a light sponge topped with sweetened buttery almonds. Traditionally eaten at Christmas, it goes down just as well in the summer. Great on its own or with sweet, ripe strawberries.
**Prep time** : 15 minutes
**Baking time** : 45–50 minutes
**Ready in** : 1 hour 20 minutes
**Serves** : 8–10
3 eggs
150g (5oz) caster sugar
150g (5oz) plain flour, plus extra for dusting
1½ tsp baking powder
2 tsp vanilla extract
3 tbsp milk
75g (3oz) butter, melted, plus extra for greasing
For the topping
50g (2oz) butter
100g (3½oz) flaked almonds
50g (2oz) caster sugar
2 tsp plain flour
3 tbsp double or regular cream
1 tsp vanilla extract
23cm (9in) diameter spring-form or loose-bottomed cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the base and sides of the cake tin with butter and dust with flour. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Using a hand-held electric beater or an electric food mixer, whisk together the eggs and the sugar for 5–7 minutes or until thick and mousse-like.
Sift in the flour and baking powder and pour in the vanilla extract, milk and melted butter, then fold everything in until combined. Tip the mixture into the prepared tin and bake for 30–35 minutes or until a skewer inserted into the centre of the cake comes out moist but not totally clean, as the mixture will still need another 10 minutes of cooking. Increase the heat to 200°C (400°F), Gas mark 6.
Just before the 30–35 minutes are up, make the topping. Place the butter in a saucepan over a medium heat. When it has melted, add the remaining ingredients and bring to the boil, allowing the mixture to bubble away for 1 minute.
After the cake has been cooking for 30–35 minutes, remove it from the oven and spoon the almond mixture evenly over the top. Place it back in the oven and bake for about 10–15 minutes or until the topping is golden.
Leave to cool in the tin for 10 minutes. Then, using a small, sharp knife, loosen around the edges and carefully remove the sides of the tin before placing the cake (still on the base of the tin) on a wire rack to cool down fully.
To transfer to a plate, use a palette knife to loosen the bottom of the cake from the tin, then slide the knife under the cake and ease it onto the plate to serve.
Lemon and rosemary polenta cake
The polenta and ground almonds in this cake mean that it's already moist even before it's smothered in a rosemary-lemon syrup as it emerges from the oven. The polenta gives a slight crunch too, which contrasts with the moist crumb. If you use a gluten-free baking powder, the cake will be totally gluten free. Stored in an airtight container, it will keep for a week or so.
**Prep time** : 30 minutes
**Baking time** : 1 hour 15 minutes–1 hour 25 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 10–12
225g (8oz) fine polenta
1 tsp baking powder
450g (1lb) butter, softened, plus extra for greasing
450g (1lb) caster sugar
450g (1lb) ground almonds
6 eggs
Finely grated zest of 2 lemons
3 tsp finely chopped rosemary leaves, plus sprigs of rosemary to decorate
For the syrup
Juice of 2 lemons
2 large sprigs of rosemary
100g (3½oz) caster sugar
25cm (10in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Mix the polenta and baking powder together in a bowl. In a separate large bowl or in an electric food mixer, cream the butter until soft. Add the sugar and beat until the mixture is light and fluffy, then beat in the ground almonds.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Add the lemon zest and chopped rosemary, then gently fold in the polenta and baking powder to combine.
Tip the batter into the prepared tin and bake on the lowest shelf of the oven for between 1 hour 15 minutes and 1 hour 25 minutes or until a skewer inserted into the centre of the cake comes out clean. The cake will cook to a deep golden-brown colour and may dip a little in the middle.
While the cake is cooking, make the syrup. Place all the ingredients in a saucepan, along with 50ml (2fl oz) of water. Place the saucepan on the hob, and bring to the boil, stirring to dissolve the sugar. Boil for 3–4 minutes until ever so slightly thickened, then remove from the heat and discard the rosemary sprigs.
When the cake is cooked, take it out of the oven and let it sit in the tin for 10 minutes. Loosen around the edges using a small, sharp knife, then carefully remove the cake from the tin and transfer to a serving plate. Reheat the syrup and pour over the cake, then leave the cake to cool down completely before serving. I like to decorate the centre with a few flowering sprigs of rosemary.
Orange sour-cream cake
Cakes made with sour cream are especially moist and have a very slight tang to them, which here is complemented by the bittersweet marmalade glaze. That tang is nicely accentuated by serving with spoonfuls of rich, thick crème fraîche. Kept covered, this cake will keep for 3–4 days.
**Prep time** : 10 minutes
**Baking time** : 40–50 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
**200g (7oz) butter, softened, plus extra for greasing**
**200g (7oz) caster sugar**
**2 large eggs, beaten**
**Finely grated zest of 1 orange**
**200ml (7fl oz) sour cream**
**300g (11oz) plain flour**
**2 tsp baking powder**
**For the glaze**
**Juice of 1 orange**
**100g (3½oz) marmalade**
**20cm (8in) diameter cake tin with 6cm (2½ in) sides**
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Gradually add the eggs to the creamed butter mixture, beating all the time. Next beat in the orange zest and sour cream, then sift in the flour and baking powder and fold in to combine.
Tip the mixture into the prepared cake tin, then bake for 40–50 minutes or until a skewer inserted into the centre of the cake comes out clean.
While the cake is cooking, make the glaze. Place the orange juice and marmalade in a saucepan. About 5 minutes before the cake has finished cooking, place the pan on the hob and bring to the boil, stirring to dissolve the marmalade. Then remove from the heat.
When the cake is cooked, take it out of the oven and let it sit in the tin for just 5 minutes. Loosening around the edges using a small, sharp knife, carefully remove the cake from the tin, peeling away the baking parchment, then transfer to a serving plate.
Straight away pour the marmalade glaze over the cake (after reheating it if it has had a chance to cool down), then allow the cake to cool down fully while soaking up the syrup.
Sometimes we need a cake quickly. It could be a last-minute panic or when there is simply too much else to fit into the day to dedicate two hours to cake making. These fast recipes are for times like that. People often assume that if you have made a cake it has taken hours of work and you must be a genius, but that isn't necessarily the case. Making a cake can take minutes rather than hours. Some of these faster cakes are made in the food processor, some are quite simple and some are not iced, but none of them take long to make. These cakes are not about delicate sugar craft or laborious preparation, they are smart recipes that make great-tasting cakes without cutting any corners.
02/ Fast
Muscovado Madeira cake
Winter breakfast muffins
Macadamia nut and lemon cakes
Lemongrass coconut cake
Fast cinnamon yoghurt cake
Raspberry and blueberry friands
Marzipan cake
Marzipan
Apple, oat and pecan bars
Raspberry and coconut squares
Crunchy peanut butter banana muffins
Upside-down peach and saffron cake
Muscovado Madeira cake
This classic English cake got its name from the sweet Madeira wine that it was traditionally served with back in the eighteenth century. Nowadays it's more often eaten with tea, although if you do have a bottle of Madeira or dessert wine, it would go perfectly with either of those. This version is flavoured with the deep molasses sweetness of muscovado (soft brown) sugar. It's a versatile recipe and you could include a handful of glace cherries, the finely grated zest of an orange or even a teaspoon of ground cinnamon if you like, adding these at the same time as the flour. (See also the additional variations.)
**Prep time** : 10 minutes
**Baking time** : 40–45 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 6–8
175g (6oz) butter, softened
175g (6oz) dark soft brown sugar
5 eggs
275g (10oz) self-raising flour
900g (2lb) loaf tin
Preheat the oven to 180°C (350°F), Gas mark 4, then line the base and sides of the loaf tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for just a few seconds until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently to combine. Tip the batter into the prepared tin and bake for 40–45 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool for 5 minutes. Carefully lift the cake out of the tin using the baking parchment, then peel away the paper and leave on a wire rack to finish cooling down.
Lemon Madeira cake
Make the cake as above, substituting caster sugar for the brown sugar and adding the finely grated zest of 1 lemon to the mixture with the eggs.
Coffee Madeira cake
Replace the brown sugar with caster sugar and mix in 3 tablespoons of coffee essence (Camp or Irel) with the eggs before adding to the batter.
Tip If any of this cake is left over, you could use it for making Cake pops.
Winter breakfast muffins
A hint of spice and a little ginger is sometimes all I need to perk me up on a frosty winter's morning. Well, that and ten minutes standing by the Aga with a big cup of coffee! I like the plump juiciness of sultanas, but if you'd prefer you could use currants or raisins instead.
**Prep time** : 15 minutes
**Baking time** : 30 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 12 muffins
300g (11oz) plain flour
½ tsp bicarbonate of soda
½ tsp salt
2 tsp mixed spice
100g (3½oz) caster sugar
125g (4½oz) butter, cut into 1cm (½in) cubes
2 eggs
175ml (6fl oz) buttermilk
100g (3½oz) golden syrup
25g (1oz) stem ginger in syrup (drained weight), finely chopped
50g (2oz) sultanas
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Sift the flour, bicarbonate of soda, salt and spice into a large bowl, then add the sugar and mix together. Add the butter and rub together with your fingertips until the mixture resembles breadcrumbs.
In another bowl, whisk together the eggs, buttermilk and golden syrup. Tip this into the flour mixture along with the chopped stem ginger and the sultanas, and beat until well mixed.
Divide the batter between the muffin cases, filling each about three-quarters full.
Bake the muffins for about 30 minutes or until well risen and springy to the touch. Allow the muffins to cool for about 5 minutes before removing them from the tin and placing on a wire rack to finish cooling.
Tip If you like, you could use some of the syrup from the stem-ginger jar to brush over the top of the muffins after they come out of the oven.
Macadamia nut and lemon cakes
This quick recipe uses a food processor to whiz up macadamia nuts before incorporating them into a lemon sponge mixture. The sharpness of the lemon contrasts so well with the rich and buttery nuts. If you can't get hold of macadamias, you can replace them with ready-ground almonds for an even speedier cake, as the almonds won't need whizzing in the food processor.
**Prep time** : 20 minutes
**Baking time** : 15–18 minutes
**Ready in** : 50 minutes
**Makes** : 12 cakes
50g (2oz) macadamia nuts
150g (5oz) self-raising flour, sifted
150g (5oz) caster sugar
3 eggs
Finely grated zest of 2 lemons
100g (3½oz) butter, melted
For the icing
200g (7oz) lemon curd (to make it yourself,
25g (1oz) macadamia nuts (or almonds), lightly toasted (see the tip) and roughly chopped
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Place the macadamia nuts in a food processor and whiz for a minute or two until fairly fine. Add the rest of the ingredients and pulse just until combined. Divide the batter between the muffin cases, filling each up to three-quarters full.
Bake for 15–18 minutes or until the cakes spring back lightly to the touch. Take out of the oven and allow to cool for 5 minutes, then remove from the tin and place on a wire rack to cool down fully.
When the cupcakes are cool, spread generously with lemon curd, then scatter over the toasted macadamia nuts.
Tip To toast nuts, either scatter them over a baking tray in a single layer and toast them in the oven (preheated to 180°C/350°F/Gas mark 4) for 4–5 minutes or until golden brown, or place them in a frying pan and toast over a medium–low heat for a similar length of time. In either case, the nuts need to be shaken every so often to prevent them burning on one side.
Lemongrass coconut cake
Coconut and lemongrass, two quintessentially Southeast Asian ingredients, are combined here in this deliciously moist cake. The lemongrass is added to a syrup that infuses the sponge with its aromatic flavour. Found in supermarkets as well as in Asian food shops, the taste of lemongrass is certainly reminiscent of lemons but has a unique floral flavour all of its own.
**Prep time** : 15 minutes
**Baking time** : 40–45 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
4 stalks of lemongrass, base and tops trimmed, outer leaves removed but reserved for the syrup
250g (9oz) caster sugar
4 eggs
200g (7oz) butter, softened, plus extra for greasing
125g (4½oz) desiccated coconut
125g (4½oz) plain flour, plus extra for dusting
2 tsp baking powder
Greek yoghurt or crème fraîche, to serve
For the syrup
Reserved trimmings and outer leaves of the lemongrass
75g (3oz) caster sugar
23cm (9in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3. Butter the sides of the cake tin and dust with flour, then line the base with a disc of baking parchment.
Slice the lemongrass stalks quite thinly into rounds about 3mm (⅛in) thick, then place in a food processor with the caster sugar and whiz for 1–2 minutes or until the lemongrass is finely pureed and very aromatic. Add the eggs, butter and coconut and whiz again until combined, then sift the flour and baking powder together and add to the machine, whizzing very briefly just until the ingredients come together.
Tip the mixture into the prepared tin and bake for 40–45 minutes or until a skewer inserted into the centre of the cake comes out clean.
While the cake is cooking, make the syrup. Roughly chop the lemongrass trimmings, place in a saucepan with the sugar and 75ml (3fl oz) of water and set over a high heat. Stir the mixture until the sugar is dissolved, then bring to the boil and boil for 2 minutes before removing from the heat and leaving to infuse.
When the cake has finished baking, take it out of the oven and let it sit in the tin for 10 minutes. Loosen around the edges using a small, sharp knife and carefully remove the cake from the tin before transferring to a serving plate.
Reheat the syrup, then pierce holes all over the cake with a skewer and pour the hot syrup through a sieve onto the cake, moving the pan and sieve around as you pour so that the syrup covers the top of the cake. Allow the cake to cool down completely.
Serve with a dollop of natural Greek yoghurt or crème fraîche.
Fast cinnamon yoghurt cake
The yoghurt in this cake gives it a gentle tang and ensures it is wonderfully light. The sweetness comes mostly from honey, and being made with sunflower or vegetable oil, this cake has a deliciously soft crumb. It will keep in an airtight box for up to a week.
**Prep time** : 10 minutes
**Baking time** : 35 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 6–8
225g (8oz) self-raising flour
1 tsp baking powder
1 tbsp ground cinnamon
75g (3oz) ground almonds
100g (3½oz) caster sugar
2 eggs
50g (2oz) honey
250ml (9fl oz) natural yoghurt
150ml (5fl oz) sunflower or vegetable oil, plus extra for greasing
To decorate
1 tbsp icing sugar
½ tsp ground cinnamon
23cm (9in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides of the cake tin with vegetable oil and line the base with a disc of baking parchment.
Sift the flour, baking powder and cinnamon into a large bowl, add the ground almonds and sugar and mix everything together.
In a separate bowl, whisk together the eggs, honey, yoghurt and vegetable oil. Add this mixture to the dry ingredients, then carefully fold in to combine.
Tip the batter into the prepared tin and bake for about 35 minutes or until golden on top and a skewer inserted into the centre of the cake comes out clean. Remove from the oven and allow to cool in the tin for 10 minutes. Use a small, sharp knife to loosen the edges, then carefully remove the cake from the tin and leave to cool down completely on a wire rack before transferring to a serving plate.
Mix together the icing sugar and cinnamon and dust over the cake to serve.
Raspberry and blueberry friands
A friand is a type of cake that is very popular down under in both New Zealand and Australia. The sponge is extremely light, made with an egg-white foam and ground almonds. These will stay wonderfully moist for up to three days if kept in an airtight box.
**Prep time** : 10 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 cakes
7 egg whites (about 250ml/9fl oz)
150g (5oz) icing sugar, plus extra for dusting
50g (2oz) plain flour
100g (3½oz) ground almonds
100g (3½oz) butter, melted
24 fresh or frozen (and defrosted) raspberries
24 fresh or frozen (and defrosted) blueberries
Icing sugar, to dust
12-cup muffin tray and 12 muffin cases
Preheat the oven to 190°C (375°F), Gas mark 5, and line the muffin tray with the paper cases.
Whisk the egg whites in a bowl using a hand-held electric whisk or in an electric food mixer, beating until the egg whites are foamy but not holding stiff peaks. Sift in the icing sugar and flour, then add the ground almonds and melted butter and stir just to combine.
Divide the mixture between the muffin cases, filling each three-quarters full, then add two raspberries and two blueberries to the centre of each cake, pressing the fruit in lightly. Bake for 20–25 minutes or until a light golden colour and slightly springy to the touch.
Remove the friands from the oven and leave in the tin for 5 minutes, then remove from the tray and place on a wire rack to cool completely.
Arrange on a plate and dust with icing sugar to serve.
Marzipan cake
This cake is incredibly quick to make, with the food processor doing all of the work. The marzipan gives the cake a wonderfully moist texture and an intense almondy flavour. You can make your own marzipan, and I've included a recipe opposite, or, for a super-fast cake, just use shop-bought marzipan if you prefer.
**Prep time** : 5 minutes (excluding the marzipan)
**Baking time** : 50 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 8–12
125g (4½oz) caster sugar
125g (4½oz) marzipan (to make it yourself)
100g (3½oz) butter, softened, plus extra for greasing
4 eggs
A few drops (not more than ⅛ tsp) of almond essence or extract
75g (3oz) plain flour, plus extra for dusting
¾ tsp baking powder
¼ tsp salt
20cm (8in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 170°C (325°F), Gas mark 3. Butter the sides of the cake tin and dust with flour, then line the base with a disc of baking parchment.
Place the sugar and marzipan in a food processor and whiz together until the marzipan is finely ground and the mixture resembles grains of sand.
Next add the butter, eggs and almond essence or extract and whiz until smooth and fluffy. Sift the flour, baking powder and salt into the machine and pulse a few times, just until all the ingredients are incorporated. (Try not to over-mix or your cake will be a little heavy.)
Scrape the batter into the prepared tin and bake the cake on a lower shelf in the oven for about 50 minutes or until the top is golden brown and a skewer inserted into the centre comes out clean.
Remove from the oven and allow to sit for 5 minutes, then use a small, sharp knife to loosen the edges of the cake tin. Allow to cool completely in the tin before carefully removing the cake and transferring to a serving plate.
Marzipan
**Prep time** : 5 minutes
**Cooking time** : 10 minutes
**Ready in** : 30 minutes
**Makes** : 450g (1lb)
225g (8oz) caster or granulated sugar
175g (6oz) ground almonds
A few drops (not more than ⅛ tsp) of almond essence or extract
1 egg white, lightly beaten
Place the sugar in a heavy-based saucepan and pour in 75ml (3fl oz) of water, then continue to heat for 5–10 minutes or until a sugar thermometer reads 110–115°C (230–235°F). If you don't have a thermometer, this is the 'thread stage'. The mixture will be thick and syrupy and the last couple of drops that fall from a spoon will form a thread.
Remove the pan from the heat and stir until the syrup is cloudy. Add the ground almonds, almond essence or extract and egg white and mix well. Transfer to a bowl and allow the marzipan to cool and become pliable.
Tip Marzipan is incredibly useful for so many cakes, from icing to decoration and even as a key ingredient in the actual sponge (as in the recipe opposite). It will keep in the fridge, covered, for up to three months, and it can be frozen too.
Apple, oat and pecan bars
Eating apples are best in this recipe as they keep their shape while baking, whereas cooking apples tend to disintegrate as they cook. These bars are full of texture from the apples, and have the slight chewiness of oats and the crunch of pecans. They're good candidates for lunchboxes and will keep for 4–5 days in an airtight container.
**Prep time** : 10 minutes
**Baking time** : 25–30 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 12 bars
2 eating apples (unpeeled), cored, cut into quarters and roughly chopped
150g (5oz) butter, softened, plus extra for greasing
150g (5oz) light soft brown sugar
50g (2oz) porridge oats
2 eggs
200g (7oz) self-raising flour, sifted
50g (2oz) pecans, plus 12 to decorate
23 × 30cm (9 × 12in) Swiss roll tin
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides of the Swiss roll tin with butter and line the base with baking parchment.
Place the prepared apples in a food processor and pulse a few times until they're in small pieces. Add the butter and sugar and cream together for 20 seconds or so, then add all the remaining ingredients and pulse just until mixed.
Tip the mixture into the prepared tin and place the pecans on top, spaced apart to form an even grid (4 × 3). Bake for 25–30 minutes or until risen and golden, then remove from the oven and leave in the tin to cool down completely.
When cool, cut into 12 bars, each with one pecan on top, and remove from the tin.
Raspberry and coconut squares
Raspberry jam and coconut are one of those combinations that seem meant for each other and they've found their way into quite a few of my recipes. This cake is simple and so fast to make – perfect as an after-school treat or a last-minute addition to finish off lunch.
**Prep time** : 20 minutes
**Baking time** : 15–20 minutes
**Ready in** : 1 hour
**Makes** : 24 squares
225g (8oz) butter, plus extra for greasing
225g (8oz) caster sugar
4 eggs
225g (8oz) self-raising flour
250g (9oz) raspberry jam
50g (2oz) desiccated coconut, to decorate
23 × 30cm (9 × 12in) Swiss roll tin
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides of the Swiss roll tin with butter and line the base with baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently to incorporate.
Tip the batter into the prepared tin and smooth the top with a palette knife or spatula. Bake for 15–20 minutes or until golden on top and springy to the touch.
Remove from the oven and allow to cool for 5 minutes, then loosen the edges with a small, sharp knife. Place a wire rack upside down on top of the cake and carefully turn over. Remove the tin and peel off the baking parchment, then turn over again onto another wire rack so the cake is upright, and leave it to cool down completely.
Once the cake has cooled, cover with raspberry jam, then sprinkle the coconut evenly over the cake and cut into 24 squares to serve.
Crunchy peanut butter banana muffins
The combination of peanut butter and banana works so well in these muffins, which are really easy to make. They'll keep for 4–5 days in an airtight container, though when their amazing smell draws everyone into the kitchen, I can't promise they'll last more than 4–5 minutes!
**Prep time** : 10 minutes
**Baking time** : 20–25 minutes
**Ready in** : 50 minutes
**Makes** : 12 muffins
75g (3oz) butter, softened
200g (7oz) caster sugar
150g (5oz) crunchy peanut butter
2 very ripe bananas (about 200g/7oz when peeled), peeled and mashed
4 eggs, beaten
250g (9oz) self-raising flour
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar, peanut butter and the mashed bananas, then beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating continuously. Sift in the flour and fold in gently to combine.
Divide the batter between the muffin cases, filling each up to three-quarters full, and bake for 20–25 minutes or until risen and golden. Place on a wire rack to cool for 5 minutes, then remove from the tin and leave on the rack to finish cooling.
Upside-down peach and saffron cake
An upside-down cake takes so little time to bring together. Here I've combined lightly caramelised peaches with that most magical of spices, saffron. If you can get good fresh peaches, then do use them, but they're quite hard to find, even in late summer. Tinned peaches work perfectly well otherwise. Delicious as a dessert, this cake would also be lovely to serve guests for afternoon tea.
**Prep time** : 10 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour
**Serves** : 6–8
50g (2oz) butter
175g (6oz) caster sugar
4 fresh peaches, quartered and stones removed, or 1 × 410g tin of sliced peaches
200g (7oz) plain flour
1 tsp baking powder
½ tsp bicarbonate of soda
¼ tsp salt
2 eggs
200ml (7fl oz) buttermilk
75ml (3fl oz) sunflower oil
Large pinch of saffron
Greek yoghurt or softly whipped cream, to serve
26cm (10½ in) diameter ovenproof frying pan (measured across the top)
Preheat the oven to 180°C (350°F), Gas mark 4.
Place the butter in the frying pan and melt over a medium heat, then sprinkle over 50g (2oz) of the sugar and cook for 3 minutes or until light golden, stirring regularly. Remove from the heat and add the peaches in a single layer, cut side down.
Sift the flour, baking powder, bicarbonate of soda and salt into a large bowl, add the remaining caster sugar and mix together. In a separate bowl, whisk together the eggs, buttermilk, sunflower oil and saffron. Make a well in the centre of the dry ingredients, then pour in the buttermilk mixture and whisk together to form a very soft batter.
Pour the batter evenly over the peaches, taking care not to disturb them in the pan, then place in the oven and bake for 30–35 minutes or until slightly springy in the middle and a skewer inserted into the centre of the cake comes out clean.
Allow to cool for just 5 minutes before turning out by placing an inverted serving plate over the top of the pan and turning the pan and plate over together in one quick movement.
Serve warm or at room temperature with Greek yoghurt or softly whipped cream.
It seems that more and more people have food intolerances or are avoiding animal products or gluten through necessity or choice. The cakes in this chapter are geared towards people with particular allergies or intolerances and are also for people who simply would like to bake cakes that are a little better for you. For children in particular, there is just no substitute for a great cake and I am so often asked for cake recipes that are egg or gluten free. While it is tricky to make a cake with no eggs, I've included a few delicious recipes that have no eggs, but still work wonderfully. Having an intolerance or allergy should not mean having to forgo cakes completely. These recipes have been carefully crafted and don't compromise on flavour or texture, so they can be enjoyed by all.
03/ Healthy
Chocolate, rum and almond cake
Sugar-free banana bread
30-day bran and pumpkin seed muffins
Tahini honey cake
Gluten- and dairy-free pecan brownies
Vegan frosted lemon cake
Dairy-free blueberry crumble cake
Date and almond honey cake
Dairy- and egg-free vanilla cupcakes
Wholemeal chocolate cake
Vegan raspberry muffins
Sweet potato muffins
Vegan chocolate cake
Apple and blackberry oat muffins
Chocolate, rum and almond cake
Alcohol and chocolate have long been combined in desserts, the alcohol accentuating and offsetting the rich, fruity flavour of dark chocolate. It also shows how versatile chocolate can be: often associated with sweet cakes and dishes for children, here it is elegant and sophisticated, perfect for a dinner party. This will keep for 4–5 days and, as it contains no flour, has the advantage of being gluten free. For an extra impressive finishing touch why not decorate with chocolate squares instead of the blanched almonds?
**Prep time** : 15 minutes
**Baking time** : 38–45 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 8–10
225g (8oz) dark chocolate, in drops or broken into pieces
225g (8oz) ground almonds
250g (9oz) caster sugar
150g (5oz) butter, softened, plus extra for greasing
6 eggs, beaten
4 tbsp rum
8 blanched almonds, to decorate
Softly whipped cream or crème fraîche, to serve
For the glaze
75ml (3fl oz) double or regular cream
125g (4½oz) dark chocolate, in drops or broken into pieces
2 tbsp rum
20cm (8in) diameter spring-form cake tin with 6cm (2½ in) side
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment. Make sure the base of the tin is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the chocolate in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat and set aside.
Place the ground almonds, sugar and butter in a food processor and whiz until well mixed. Add the eggs, rum and melted chocolate and whiz until blended.
Tip the cake mixture into the prepared tin and cook for 38–45 minutes or until puffed up and slightly cracked on the surface but still quite moist in the centre – a skewer inserted into the centre of the cake will still be slightly sticky.
Remove from the oven and allow to cool in the tin for 20 minutes, then loosen the sides using a small, sharp knife and carefully remove the cake from the tin before transferring to a wire rack to cool down fully.
While the cake is baking, or while it's cooling, make the glaze. Pour the cream into a saucepan and bring to the boil, then take off the heat, add the chocolate, stirring to melt, and mix in the rum.
Transfer the cooled cake to a serving plate and pour over the chocolate glaze (reheating if necessary), then decorate with the blanched almonds by placing these in an evenly spaced circle, with the nuts pointing inwards, around the edge of the cake. Serve in slices with softly whipped cream or crème fraîche.
Variations
I've used rum in this recipe, but you could just as easily use brandy, whiskey or even orange liqueur. The flavours will all be slightly different, but the effect will be much the same.
Sugar-free banana bread
Agave syrup is made from the agave plant, grown primarily in Mexico. The syrup is extracted from the plant and produced in concentrated form. Sticky and slightly sweeter than sugar, it's considered healthier as it has a lower Gl (glycaemic index), and releases energy more steadily than pure sugar. In this recipe the agave combines with the natural sweetness of bananas for a gorgeously healthy take on banana bread. I love to eat this straight from the oven, but it keeps well for a good few days. A slice or two would be perfect for a picnic.
**Prep time** : 10 minutes
**Baking time** : 55–65 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 6–10
125g (4½oz) plain flour
¾ tsp bicarbonate of soda
½ tsp salt
150g (5oz) wholemeal flour
350g (12oz) mashed bananas (about 4 bananas)
3 eggs
100ml (3½fl oz) agave syrup
75ml (3fl oz) sunflower or extra-virgin olive oil
50ml (2fl oz) natural yoghurt
Finely grated zest of 1 lemon
900g (2lb) loaf tin
Preheat the oven to 180°C (350°F), Gas mark 4, and line the base and sides of the loaf tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
Sift the plain flour, bicarbonate of soda and salt into a large bowl, then mix in the wholemeal flour. In a separate bowl, whisk together the remaining ingredients.
Add this mixture to the dry ingredients, folding together to combine. Tip the batter into the prepared loaf tin, then bake for 55–65 minutes or until the top is just turning golden brown and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to sit for 10 minutes. Carefully lift the cake out of the tin using the baking parchment, then peel away the paper and leave on a wire rack to finish cooling.
30-day bran and pumpkin seed muffins
These are called '30-day muffins' because the batter will keep happily in the fridge for a month! They're sugar free and you can easily make them dairy free too by using rice or soya milk. The pumpkin seeds are healthy, packed with nutrition, but they're also delicious – lightly toasted in the oven for a characteristic crunch. I like to eat these muffins still warm for breakfast, but they make a handy and healthy lunchbox treat too.
**Prep time** : 10 minutes
**Baking time** : 15–22 minutes
**Ready in** : 45 minutes
**Makes** : 12 muffins
200g (7oz) plain flour
1 ½ tsp bicarbonate of soda
¾ tsp salt
200g (7oz) wholemeal flour
50g (2oz) bran
50g (2oz) pumpkin seeds, plus 25g (1oz) for sprinkling
3 eggs
200ml (7fl oz) light agave syrup
100ml (3½fl oz) sunflower oil
50ml (2fl oz) milk or rice or soya milk
12-cup muffin tray and up to 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases (or fewer if you prefer – see the tip below).
Sift the plain flour, bicarbonate of soda and salt into a large bowl and mix in the wholemeal flour, bran and pumpkin seeds.
In a separate bowl, whisk together the remaining ingredients. Make a well in the centre of the flour and pumpkin seed mixture and pour in the wet ingredients, whisking together to incorporate fully.
Fill each of the muffin cases three-quarters full, then scatter with the additional pumpkin seeds and cook for 15–22 minutes or until springy to the touch. Allow to cool in the tin for 5 minutes, then remove from the tin and place on a wire rack to finish cooling.
Fruity bran and pumpkin seed muffins
I often add a small handful or so of dried fruit to the batter, such as raisins, chopped dried apricots or dates. Make up the recipe as above, adding the fruit to the fully mixed batter before filling the muffin cases.
Tip If you'd like to make fewer than 12 muffins, cover the remaining batter and place in the fridge, where it will keep for up to a month. It will thicken over time, so thin slightly by whisking in a little more milk or water – the mixture should be quite sloppy.
Tahini honey cake
I love tahini, the peanut butter-like paste made from sesame seeds. Tahini is ubiquitous in Middle Eastern cooking, used in all sorts of salads and sauces. It also finds its way into many sweet dishes and desserts, usually sweetened with honey. This recipe combines these flavours with the citrus tang of oranges and a touch of cinnamon to make a cake that would be ideal served after a big, generous Middle Eastern meal or perfect with a cup of coffee. The cake also has the advantage of being vegan (for those vegans who choose to eat honey).
**Prep time** : 10 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour
**Serves** : 6–8
Sunflower oil, for greasing
75g (3oz) wholemeal flour
1 tsp bicarbonate of soda, sifted
1 tsp ground cinnamon
Pinch of salt
100g (3½oz) fresh or frozen white breadcrumbs
125g (4½oz) honey, plus
1 tsp for drizzling
100ml (3½fl oz) light tahini
Juice and finely grated zest of 2 oranges
20cm (8in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides of the cake tin with sunflower oil and line the base with a disc of baking parchment.
Place the flour in a food processor with the bicarbonate of soda, cinnamon, salt and breadcrumbs and whiz for about 30 seconds or until fine.
In a bowl, mix together the 125g (4½oz) of honey, along with the remaining ingredients. Add to the food processor with the dry ingredients and whiz together for about 20 seconds or until it just comes together. Tip the batter into the prepared tin and bake for 30–35 minutes or until a skewer inserted into the centre of the cake comes out clean.
When the cake is cooked, take it out of the oven and let it sit in the tin for 5 minutes. Loosen around the edges using a small, sharp knife, then carefully remove the cake from the tin and transfer to a serving plate.
Drizzle over the remaining honey and allow to cool down completely before serving.
Gluten- and dairy-free pecan brownies
You'll be amazed at how gorgeously fudgy these brownies are despite not containing a hint of dairy or gluten. The pecans aren't essential but I love the crunchy contrast they provide with the softness of the brownies.
**Prep time** : 10 minutes
**Baking time** : 45 minutes
**Ready in** : 2 hours
**Makes** : 12 brownies
125g (4½oz) dark chocolate, in drops or broken into pieces
225g (8oz) soft dark brown sugar
125ml (4½fl oz) sunflower oil, plus extra for greasing
½ tsp vanilla extract
2 eggs
75g (3oz) ground almonds
25g (1oz) rice flour
½ tsp salt
½ tsp baking powder or gluten-free baking powder
75g (3oz) pecans, chopped, plus 12 whole (shelled) pecans to decorate
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, grease the sides of the tin with sunflower oil and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the chocolate in a heatproof bowl and set over a saucepan of gently simmering water. Leave just until melted, stirring occasionally, then remove from the heat and add the sugar, sunflower oil, vanilla extract and eggs and whisk together well. Next add the ground almonds, rice flour, salt, baking powder and chopped pecans and fold in until well mixed.
Tip into the prepared tin, then place the 12 whole pecans in an even grid on top. Bake for about 45 minutes or until the middle of the cake still wobbles slightly when you gently shake the tin – a skewer inserted into the centre should come out with a little moisture on it.
Remove from the oven, place on a wire rack and allow to cool in the tin for at least an hour before carefully removing the cake and cutting into 12 squares (each with a pecan in the middle) to serve.
Vegan frosted lemon cake
Whether you're vegan by choice or intolerant to dairy foods and eggs, finding cakes that both work and taste good can be tricky. It's certainly not impossible, though, as this cake proves with its intense lemony flavour and a soft, light crumb that is just as good as anything made with butter or eggs. For the icing I've used a soya spread, which works wonderfully when sweetened and thickened with the icing sugar. You could use any other citrus fruits to flavour this cake as well (see the variations).
**Prep time** : 20 minutes
**Baking time** : 45–50 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 8–12
450g (1lb) plain flour, plus extra for dusting
2 tsp baking powder
1 tsp bicarbonate of soda
300g (11oz) caster sugar
125ml (4½fl oz) sunflower oil, plus extra for greasing
Finely grated zest of 3 lemons
Juice of 3 lemons plus enough water to make 300ml (½ pint)
For the frosting
175g (6oz) soya spread
500g (1lb 2oz) icing sugar, sifted
Pinch of salt
Finely grated zest of 1 lemon
1 tsp lemon juice
23cm (9in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3. Grease the sides of the cake tin with sunflower oil and dust with flour, then line the base with a disc of baking parchment.
Sift the flour, baking powder and bicarbonate of soda into a large bowl and mix in the sugar. In a separate bowl, mix together all the remaining ingredients. Add these to the dry ingredients and stir well until the mixture comes together.
Pour the batter into the prepared tin and bake for 45–50 minutes or until springy to the touch and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes. Using a small, sharp knife, loosen around the edges, then carefully remove the cake from the tin and leave on a wire rack to finish cooling.
As the cake cools, make the frosting. Place all the ingredients in a large bowl or in an electric food mixer. Using either a hand-held electric beater or the food mixer, whisk on full speed for 2–3 minutes or until light and fluffy.
Once the cake is cool, use a bread knife to slice it horizontally in half. Place the bottom half of the cake, cut side up, on a cake plate or stand, then spread over some of the frosting, to about 5mm (¼in) thick. Place the top half of the cake on top, cut side down, then spread the remaining frosting over the top and sides of the cake.
Vegan frosted orange cake
Make up the cake in the same way, substituting the lemons with the zest and juice of three oranges (with the juice topped up to 300ml/½ pint with water, if necessary) and reducing the sugar to 275g (10oz).
Vegan frosted lime cake
Follow the recipe as above, substituting the lemons with the zest and juice of four limes (with the juice topped up to 300ml/½ pint with water, if necessary).
Dairy-free blueberry crumble cake
This dairy-free cake has a divinely thick, sticky crumble topping. Naturally sweet blueberries are dotted throughout the cake and scattered on top where they burst during baking and ooze their syrupy juices into the topping. If you're intolerant to eggs, this cake works well with an egg replacer, which you can buy from health-food shops. Of course, you can use real eggs otherwise.
**Prep time** : 20 minutes
**Baking time** : 50–60 minutes
**Ready in** : 2 hours
**Serves** : 8–12
225g (8oz) plain flour
2 tsp baking powder
1 tsp bicarbonate of soda
½ tsp salt
50g (2oz) caster sugar
50g (2oz) wholemeal flour
150ml (5fl oz) soya or rice milk
50ml (2fl oz) vegetable oil
1 tbsp cider vinegar
2 tsp vanilla extract
3 eggs, beaten (or 3 tsp egg replacer mixed with 6 tsp water)
250g (9oz) fresh or frozen (and defrosted) blueberries
For the crumble topping
50g (2oz) plain flour
½ tsp ground cinnamon
150g (5oz) soft dark brown sugar
2 tbsp soya spread
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then line the base and sides of the cake tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
First make the topping. Sift the flour and cinnamon into a large bowl, add the sugar and mix together. Using your fingers, rub in the soya spread until the mixture forms a crumbly topping.
Sift the plain flour, baking powder, bicarbonate of soda and salt into a separate bowl and mix in the sugar and wholemeal flour. In another bowl, whisk together the soya or rice milk, vegetable oil, vinegar and vanilla extract until well combined.
Add the beaten eggs, or egg replacer mixture, to the wet ingredients, then gradually pour them into the flour mixture and mix just until combined.
Fold in about half of the blueberries, then pour the batter into the prepared tin. Scatter the remaining blueberries over the top of the cake and then sprinkle the crumble topping evenly over the top of the batter.
Bake for 50–60 minutes or until a skewer inserted into the centre of the cake comes out clean. The top will be golden brown and look uneven, but don't worry as that's part of its charm.
Allow the cake to sit in the tin for 20 minutes before lifting out using the baking parchment, then carefully peel away the paper and allow to cool down fully on a wire rack.
Cut into squares and arrange on a plate to serve.
Date and almond honey cake
This fantastically dense, moist cake has echoes of sticky toffee pudding. It contains no refined sugar, all the sweetness coming from the honey and dates, while the wholemeal flour imparts its lovely nutty flavour.
**Prep time** : 15 minutes
**Baking time** : 45–50 minutes
**Ready in** : 2 hours
**Serves** : 6–8
100g (3½oz) chopped dates
200g (7oz) butter, softened, plus extra for greasing
200g (7oz) honey, plus
2 tbsp for drizzling
3 eggs
100g (3½oz) ground almonds
125g (4½oz) wholemeal flour
1½ tsp baking powder
25g (1oz) flaked almonds
20cm (8in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment. Place the tin on a baking sheet, as some butter may seep out during cooking if you are using a spring-form cake tin.
Place the dates in a saucepan and pour in 50ml (2fl oz) of water. Set over a high heat and cook for 2–3 minutes or until soft, then remove from the heat and set aside.
Cream the butter and the 200g (7oz) of honey until soft in a large bowl or in an electric food mixer. Whisk the eggs together in a small bowl for a few seconds until just mixed, then gradually add them to the creamed butter and honey mixture, beating all the time.
Stir in the cooked dates, along with any remaining cooking liquid, followed by the ground almonds, then add the flour and baking powder and fold in gently to incorporate. Tip the mixture into the prepared tin, smoothing the surface gently with a palette knife, then scatter the flaked almonds evenly over the top.
Place in the oven and bake for 45–50 minutes or until a skewer inserted into the centre of the cake comes out clean. It will be quite dark-looking, but don't worry – the cake will be perfectly moist inside.
Remove from the oven and allow to cool in the tin for 10 minutes. Loosen the sides using a small, sharp knife and carefully remove the cake from the tin before transferring to a cake stand or plate.
Use a skewer to pierce a few holes in the top of the cake, then drizzle over the 2 tablespoons of honey and allow to cool before cutting into slices to serve.
Dairy- and egg-free vanilla cupcakes
Even if you can eat dairy products or eggs, you'll love these simple vanilla cupcakes. The extra raising agents and vinegar really lift these, making them especially light and airy. The vanilla icing uses soya spread, which works just as butter does to form a rich sweet icing. If you don't have a dairy intolerance, however, you could simply replace the soya spread with softened butter.
**Prep time** : 15 minutes
**Baking time** : 15–20 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
1 tbsp white wine vinegar or cider vinegar
275ml (9½fl oz) soya or rice milk
75ml (3fl oz) sunflower or vegetable oil
1 tsp vanilla extract
225g (8oz) self-raising flour
¾ tsp baking powder
½ tsp bicarbonate of soda
¼ tsp salt
200g (7oz) caster sugar
For the icing
125g (4½oz) soya spread
350g (12oz) icing sugar, sifted
Pinch of salt
2 tsp vanilla extract or seeds scraped from ½ vanilla pod
12-cup bun tray and 12 bun cases
Piping bag with a 4 or 5mm (¼in) star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Preheat the oven to 180°C (350°F), Gas mark 4, and line the bun tray with the paper cases.
In a bowl, mix together the vinegar, soya or rice milk, sunflower or vegetable oil and vanilla extract. In a separate bowl, sift together the flour, baking powder, bicarbonate of soda and salt, then mix in the sugar. Add the wet ingredients to the flour mixture and whisk to combine completely – this may take a minute or so.
Divide the batter between the paper cases, filling each case three-quarters full. Bake in the oven for 15–20 minutes or until golden brown and springy to the touch. Allow to cool in the tin for 5 minutes, then remove from the tray and place on a wire rack to cool down completely.
As the cupcakes cool, make the icing. Whisk all the ingredients together (this is easiest using an electric food mixer or a hand-held electric beater) until soft and light.
Using a palette knife, the back of a spoon or a piping bag, spread the icing over the top of each cupcake before placing on a plate to serve.
Wholemeal chocolate cake
A distinctive cake that's really worth trying. Wholemeal flour contains all of the wheat grain and is high in fibre, proteins and minerals. It's not just that it's healthier, but it tastes fantastic too, with the nutty flavour complementing the dark chocolate and brown sugar.
**Prep time** : 15 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Serves** : 6–10
125g (4½oz) wholemeal flour
1 tsp bicarbonate of soda
¼ tsp salt
50g (2oz) dark chocolate, in drops or broken into pieces
50g (2oz) butter, softened, plus extra for greasing
200g (7oz) soft light or dark brown sugar
2 eggs
1 tsp vanilla extract
150ml (5fl oz) sour cream
Cocoa powder, for dusting
23cm (9in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the tin and line the base with a disc of baking parchment.
Sift the flour, bicarbonate of soda and salt into a bowl and set aside. Some of the bran will be left in the sieve – you can return it to the flour bag.
Place the chocolate in a heatproof bowl and set over a saucepan of just simmering water. Leave just until the chocolate has melted, stirring from time to time, then remove from the heat and set aside.
Cream the butter until soft in a large bowl or electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time.
Beat in the vanilla extract and the melted chocolate, then gradually stir in the flour mixture along with the sour cream. Mix just until combined, then pour the batter into the prepared tin.
Bake for 20–25 minutes or until a skewer inserted into the centre of the cake comes out clean. Remove from the oven and allow to cool in the tin for 10 minutes. Loosen around the edges with a small, sharp knife, then carefully remove the cake from the tin and leave on a wire rack to finish cooling.
Once cool, transfer to a cake plate or stand and dust with cocoa powder to serve.
Vegan raspberry muffins
These don't have a great deal of sugar in them, as much of the sweetness comes from the raspberries. For that reason, if you'd like to replace the raspberries with blackberries – which would make a great substitute – do taste the blackberries first and if they're quite sour then increase the sugar in the recipe.
**Prep time** : 10 minutes
**Baking time** : 15–20 minutes
**Ready in** : 1 hour
**Makes** : 10 muffins
225g (8oz) plain flour
3 tsp baking powder
1 tsp bicarbonate of soda
¼ tsp salt
75g (3oz) caster sugar
Finely grated zest of 1 orange
150g (5oz) fresh or frozen (and defrosted) raspberries
225ml (8fl oz) soya or rice milk
50ml (2fl oz) sunflower oil
1 tbsp cider vinegar
1 tsp vanilla extract
12-cup muffin tray and 10 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Sift the flour, baking powder, bicarbonate of soda and salt into a large bowl, then mix in the sugar and orange zest. Carefully fold in the raspberries and set aside.
In a separate bowl, whisk together the remaining ingredients, then add these to the flour mixture, folding in to combine.
Divide the batter between the muffin cases, filling each three-quarters full, then bake for 15–20 minutes or until golden on top and springy to the touch.
Remove from the oven and allow to cool in the tray for 5 minutes, then take the muffins out of the tray and place on a wire rack to cool down completely.
Sweet potato muffins
The aptly named sweet potato provides these muffins with natural sweetness and gives a divinely moist crumb. The sweetness is given a little boost with agave syrup, making these delicious treats surprisingly healthy.
**Prep time** : 10 minutes (excluding cooking the sweet potato)
**Baking time** : 30–35 minutes
**Ready in** : 1 hour
**Makes** : 12 muffins
2 sweet potatoes (unpeeled) (about 400g/14oz)
50g (2oz) butter, softened
75ml (3fl oz) buttermilk
175ml (6fl oz) natural yoghurt
100ml (3½fl oz) light agave syrup
1 egg
125g (4½oz) plain flour
1 tsp baking powder
½ tsp bicarbonate of soda
½ tsp ground or grated nutmeg
1 tbsp ground cinnamon
1 tsp ground ginger
100g (3½oz) wholemeal flour
12-cup muffin tray and 12 muffin cases
Preheat the oven to 200°C (400°F), Gas mark 6.
To cook the sweet potatoes, place on a baking tray and bake for about 1 hour or until tender when pierced with a fork. Cut open and scoop out the flesh and discard the skins. Place the flesh in a bowl to cool – you'll need 300g (11oz) of the cooked flesh in total. Alternatively, cook the sweet potatoes in a microwave: pierce each potato a few times with a fork, then cook on a high heat for 10–15 minutes, turning over halfway through. When cooked the sweet potatoes will be tender to the touch and soft all the way through.
Turn down the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
In a bowl, beat the butter with the cooked flesh of the sweet potato. In a separate bowl, whisk together the buttermilk, yoghurt, agave syrup and egg. Then add to the sweet potato mixture and beat in until well mixed.
Sift the plain flour, baking powder, bicarbonate of soda and spices into a separate bowl, then mix in the wholemeal flour. Add the dry ingredients to the sweet potato mixture and fold everything together to combine.
Divide the batter between the muffin cases, filling each three-quarters full. Bake for 30–35 minutes or until lightly springy to the touch. Allow to cool for 5 minutes before removing from the tin and placing on a wire rack to cool down fully.
Vegan chocolate cake
This cake is inspired by a recipe by Giorgio Locatelli that is a favourite of his daughter Dita, who is allergic to eggs. I have added a chocolate frosting to it to make it even more indulgent. Everyone, whether allergic to eggs or not, will love this cake.
**Prep time** : 45 minutes
**Baking time** : 45–50 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 12–16
450g (1lb) plain flour, plus extra for dusting
50g (2oz) cocoa powder
2 tsp baking powder
2 tsp bicarbonate of soda
300g (11oz) caster sugar
125ml (4½fl oz) sunflower oil, plus extra for greasing
2 tbsp white wine vinegar or cider vinegar
2 tsp vanilla extract
For the frosting
125g (4½oz) soya spread
300g (1loz) icing sugar, sifted
50g (2oz) cocoa powder, sifted
1 tbsp vanilla extract
Pinch of salt
23cm (9in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 170°C (325°F), Gas mark 3. Grease the sides of the cake tin with sunflower oil and dust them with flour, then line the base of the tin with a disc of baking parchment.
Sift the flour into a large bowl with the cocoa powder, baking powder and bicarbonate of soda and mix in the sugar. In a separate bowl, mix together the remaining ingredients with 300ml (½ pint) of water.
Add the wet ingredients to the flour mixture, stirring well to combine, then tip the batter into the prepared tin. Bake for 45–50 minutes or until springy to the touch and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes. Then loosen around the edges using a small, sharp knife and carefully remove the cake from the tin before leaving on a wire rack to finish cooling.
While the cake is cooling, make the frosting. Using a hand-held electric beater or an electric food mixer, whisk all the ingredients together on full speed for 4–5 minutes or until light and thick.
Once the cake is cold to the touch, use a bread knife to cut it horizontally in two. Place the bottom half on a cake plate or stand, then spread over some of the frosting. Sandwich together with the top half, spreading the remaining frosting over the top and sides of the cake. Place in the fridge for about 20 minutes to allow it to set before serving.
Apple and blackberry oat muffins
This has to be my favourite partnership of fruit flavours, and it's serendipitous that they come into season at exactly the same time. This recipe uses eating apples rather than cookers because they're sweeter, though they will need puréeing first as they don't break down as easily when cooked.
**Prep time** : 15 minutes
**Baking time** : 25–30 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 12 muffins
425g (15oz) eating apples, peeled, cored and roughly chopped (400g (14oz) once peeled and cored)
200g (7oz) soft light brown sugar
100ml (3½fl oz) sunflower oil
75g (3oz) porridge oats
225g (8oz) plain flour, sifted
3 tsp baking powder
36 fresh or frozen (and defrosted) blackberries (about 125g/4½oz)
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Place the apples and sugar in a saucepan and add 3 tablespoons of water, then cover with a lid and bring to a simmer. Continue to simmer for 10 minutes, stirring occasionally, then tip into a food processor or blender and whiz for a minute or until smooth.
Pour the apple purée into a bowl and stir in the sunflower oil and oats. Sift in the flour and baking powder and stir in just until mixed. Divide the batter between the muffin cases, filling each case three-quarters full, then lightly press three or four blackberries into each muffin.
Bake for 25–30 minutes or until golden brown and lightly springy to the touch in the middle. Remove from the oven and leave in the tin to cool for 5 minutes, then remove the muffins from the tin and place on a wire rack to cool down completely.
Classic cakes have withstood the test of time. They remain firm favourites not only because they taste so good, but because we hold a special affection for them. Classic cake recipes often bring back childhood memories, making them one of the most comforting foods. As these cakes are so well loved, for the most part I haven't strayed too far from the original recipes. Sometimes, though, I've added a different flavour or changed a classic shape, to put a new spin on an old favourite. Each cake is a classic from a particular place. Lamingtons are an institution in Australia and the French Brittany butter cake is the perfect showcase for Brittany's famous butter. If you are new to some of the recipes, I urge you to try them, as these cakes are classics for a reason.
04/ Classic
Baked Alaska with hot chocolate sauce
Hot chocolate sauce
Carrot cake
Ginger carrot cupcakes
Caramel carrot cake
Classic NYC crumb cake
St Clement's drizzle cake
Dark treacly gingerbread squares
Angel food cake
Rose-water icing with pistachios
Lavender icing
Lamingtons
Battenberg cake
Dundee cake
Madeleines
Chocolate-dipped madeleines
Lemon-striped madeleines
Victoria sponge
Brittany butter cake
Chocolate hazelnut cheesecake
Baked Alaska with hot chocolate sauce
A warm crisp meringue surrounding frozen ice cream atop a delicate chocolate sponge – a temperature contrast that is always impressive. The key to this culinary feat lies in the meringue, which acts as an effective insulator so the ice cream remains frozen even in the heat of the oven. A really fun dessert for a special occasion, either served on its own or topped with this extremely moreish hot chocolate sauce.
**Prep time** : 45 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 8–10
1 litre (1¾ pints) vanilla, coffee or chocolate ice cream
For the sponge base
50g (2oz) dark chocolate, in drops or broken into pieces
125g (4½oz) butter, softened, plus extra for greasing
125g (4½oz) caster sugar
2 eggs, beaten
125g (4½oz) self-raising flour, sifted
For the meringue
3 egg whites
200g (7oz) caster sugar
Pinch of cream of tartar
1 litre (1¾ pint) pudding basin (20cm/8in in diameter)
20cm (8in) diameter cake tin
Line the pudding basin with a double layer of cling film, leaving enough hanging over the edge to cover the sides of the basin. Place the ice cream in the basin (you may need to allow it to soften for a few minutes, but don't let it melt). Press down to get a smooth surface (giving it a few gentle bangs to help remove any air holes), then cover the top with the cling film and return it to the freezer.
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Place the chocolate in a heatproof bowl and set over a pan of gently simmering water. Leave until just melted, stirring occasionally, then remove from the heat and set aside.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Gradually add the eggs to the creamed butter mixture, beating all the time. Beat in the melted chocolate, then gently fold in the flour to combine.
Tip into the prepared cake tin and bake for 25–30 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove the cake from the oven and turn the temperature up to 220°C (425°F), Gas mark 7, if cooking the meringue straight away.
Allow the cake to cool down in the tin for 10 minutes, then loosen around the edges with a small, sharp knife and carefully remove the cake before leaving on a wire rack to cool down fully.
Meanwhile, make the meringue. In a large, spotlessly clean bowl, whisk the egg whites until they form soft peaks. Add half the caster sugar and the cream of tartar and whisk until stiff peaks form and the meringue is satiny and glossy. Fold in the remaining sugar with a metal spoon or spatula.
Once it has cooled, place the cake base in the middle of a baking sheet (one that is completely flat, with no 'lip', so that the baked Alaska can be removed easily). Take the ice cream out of the freezer. With the help of the cling film, remove the ice cream from the pudding bowl and place it upside down onto the cake, removing all the cling film.
Quickly spoon over the meringue, spreading it thickly and in peaks over the ice cream and the sides of the cake, down to the baking sheet. You can cook the baked Alaska immediately or you can return it to the freezer for up to 1 hour. It will take an extra 3–4 minutes in the oven if cooking from frozen. To cook, place in the centre of the hot oven for 3–4 minutes or until the meringue is set on the outside and golden in colour.
Carefully transfer the baked Alaska to a serving plate, using a palette knife dipped in hot water to help you slide and push it onto the plate. Serve straight away on its own or with chocolate sauce.
Variation
If you prefer, you can make this dish using a plain version of the sponge. Follow the recipe as above but simply omit the chocolate from the mixture.
Hot chocolate sauce
**Prep time** : 5 minutes
**Cooking time** : 5 minutes
**Ready in** : 10 minutes
**Makes** : about 400ml (14fl oz)
200ml (7fl oz) double or regular cream
200g (7oz) dark chocolate, in drops or chopped into pieces
1–2 tbsp rum or brandy (optional)
To make the hot chocolate sauce, place the cream in a saucepan and bring to the boil. Add the chocolate and stir until just melted, then add the rum or brandy (if using). Reheat the sauce gently when needed.
Chocolate orange sauce
Add 1–2 tablespoons of orange liqueur (such as Cointreau or Grand Marnier) and 1 teaspoon of finely grated orange zest to the basic chocolate sauce.
Carrot cake
Carrot cake is a classic. Carrots were originally added to cakes to sweeten them when sugar was rare and expensive. Despite sugar being readily available today, carrot cake remains universally popular, the carrots adding not just sweetness but moisture and a lightly resistant texture. Here are three different types of carrot cake. This first one is cooked in a loaf tin where the cream-cheese icing has just a hint of orange. Then there are some cupcakes, which are infused with ginger – both ground and crystallised. These are followed by a layer cake, where the icing is laden with the caramel flavour of dulce de leche.
**Prep time** : 20 minutes
**Baking time** :1 hour–1 hour 15 minutes
**Ready in** : 2 hours
**Serves** : 8–10
2 eggs
150ml (5fl oz) sunflower or vegetable oil
200g (7oz) soft light brown sugar
300g (11oz) peeled and grated carrots (weight when grated)
75g (3oz) pecans or walnuts, chopped (optional)
175g (6oz) self-raising flour
½ tsp bicarbonate of soda
1 tsp ground cinnamon
1 tsp mixed spice
Pinch of salt
For the icing
100g (3½oz) cream cheese
Finely grated zest of 1 orange
200g (7oz) icing sugar, sifted
900g (2lb) loaf tin
Preheat the oven to 150°C (300°F), Gas mark 2, and line the base and sides of the loaf tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
Whisk the eggs in a large bowl to break them up, then whisk in the oil, sugar, grated carrots and chopped nuts. Sift in the remaining ingredients and bring the mixture together using a wooden or large metal spoon.
Pour the mixture into the prepared loaf tin, smoothing the surface with a palette knife, and bake in the oven for between 1 hour and 1 hour 15 minutes or until a skewer inserted into the centre of the cake comes out clean.
Allow to cool in the tin for about 5 minutes, then carefully lift the cake out of the tin with the lining paper. Peel away the baking parchment and place on a wire rack to cool down fully before you ice it.
While the cake is cooling, make the icing. Beat all the ingredients together, then use a palette knife to spread evenly over the cake.
Tip Try to use a serrated knife to cut this cake into slices, as it is dense but quite crumbly.
Ginger carrot cupcakes
**Prep time** : 20 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
For adding to the cake batter
50g (2oz) crystallised ginger, finely chopped
For the icing
200g (7oz) cream cheese
200g (7oz) icing sugar, sifted
½ tsp ground ginger
To decorate
75g (3oz) crystallised ginger, finely chopped
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Mix together the ingredients for the cake batter following the second paragraph of the Carrot cake method, omitting the nuts and adding the crystallised ginger instead, then divide the mixture between the muffin cases, filling each case three-quarters full.
Bake for 20–25 minutes or until well risen and springy to the touch, then remove from the oven and allow to cool for 5 minutes before removing from the tin and placing on a wire rack to cool down fully.
Meanwhile, make the icing. Whisk all the ingredients together in a bowl, then use a palette knife to spread evenly over each cupcake.
As a final touch, sprinkle over the chopped crystallised ginger to decorate.
Caramel carrot cake
**Prep time** : 20 minutes
**Baking time** : 30–35 minutes
**Ready in** : 2 hours
**Serves** : 8–10
For the icing
200g (7oz) icing sugar, sifted
200g (7oz) cream cheese
3 tbsp dulce de leche (tip)
To decorate
Walnut or pecan halves (optional)
Two 20cm (8in) diameter cake tins
Preheat the oven to 160°C (325°F), Gas mark 3, then grease the cake tins with oil and line each base with baking parchment.
Mix the ingredients for the sponge following the second paragraph of the Carrot cake method, then divide the batter between the two prepared tins and bake for 30–35 minutes or until lightly browned on top and springy to the touch.
Remove the cakes from the oven and allow to sit in the tins for 10 minutes, then use a small, sharp knife to loosen the edges and carefully remove each cake from its tin before transferring to a wire rack to cool down completely.
While the cakes are baking, or cooling, make the icing. Place all the ingredients in a mixing bowl and beat together until well mixed. To assemble the cake, place one layer upside down on a serving plate and spread half the icing on the top. Add the second layer, placing it right side up, then spread the rest of the icing over the top. You might like to decorate the top of the cake with walnut or pecan halves.
Classic NYC crumb cake
New York's melting pot of cultures has produced a city of amazing culinary diversity. Central to this have been the bakeries, with the Polish, Jewish and German ones being especially prominent. The classic New York City crumb cake derives from the German _Krümelkuchen_. The cake here is adapted from a recipe in _Baked in America_ by David Lesniak and David Muniz, two wonderful American bakers who run a fabulous café and bakery in London called Outsider Tart, where they make some of the most delicious baked treats this side of the Atlantic. Delicious eaten as a snack, or served to friends and family, the cake will keep for 2–3 days.
**Prep time** : 20 minutes
**Baking time** : 45–55 minutes
**Ready in** : 1 hour 45 minutes
**Makes** : about 24 squares
225g (8oz) butter, softened
350g (12oz) caster sugar
2 tsp vanilla extract
4 eggs, beaten
350ml (12fl oz) sour cream
525g (1lb 2½oz) plain flour
1 tsp baking powder
½ tsp bicarbonate of soda
1 tsp salt
For the crumb topping
150g (5oz) butter, melted
300g (11oz) plain flour
1 tbsp ground cinnamon
225g (8oz) soft dark brown sugar
¼ tsp salt
25 × 35cm (10 × 14in) roasting tin with 5cm (2in) sides (approximate size)
Preheat the oven to 180°C (350°F), Gas mark 4, then line the base and sides of the roasting tin with baking parchment, with the paper coming just above the sides of the tin to enable the cake to be lifted out easily.
First make the crumb. Pour the melted butter into a large bowl or into an electric food mixer. Sift in the flour and cinnamon and add the remaining ingredients, then mix briefly – for just 30–60 seconds if using a food mixer – until coarse and crumbly.
To make the cake batter, first cream the butter until soft in a separate large bowl or in the food mixer. Pour in the sugar and beat until the mixture is light and fluffy. Add the vanilla extract and gradually mix in the eggs. Add the sour cream, then (with the mixer on a low speed, if you're using it) sift in the remaining ingredients, and fold in to combine.
Tip the batter into the prepared tin and smooth the surface with a spatula or the back of a spoon. Sprinkle over the crumb mixture, making sure it covers the batter completely.
Bake on the lower shelf in the oven for 45–55 minutes or until golden brown on top and a skewer inserted into the centre of the cake comes out clean. Check the cake after 30 minutes and if the top is already a rich golden brown, cover the tin with foil (to prevent the cake from burning) and continue to cook for the remaining time.
Remove the cake from the oven and allow to cool in the tin completely. Then lift the cake out using the baking parchment, peel away the paper and cut into squares to serve.
St Clement's drizzle cake
The St Clement's combination of oranges and lemons is a real classic, named of course after the bells in the famous nursery rhyme. This drizzle cake is different from other syrup cakes: rather than dissolving the sugar in a hot syrup first, it's simply mixed with the citrus juices before being drizzled over the cake. The undissolved sugar crystals are essential in this recipe for their slight crunch.
**Prep time** : 10 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–10
175g (6oz) butter, softened
175g (6oz) caster sugar
2 eggs
Juice and finely grated zest of 1 lemon
Juice and finely grated zest of 1 orange
50ml (2fl oz) milk
175g (6oz) self-raising flour, sifted
100g (3½oz) granulated sugar
900g (2lb) loaf tin
Preheat the oven to 180°C (350°F), Gas mark 4, and line the base and sides of the loaf tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the caster sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Beat in the lemon and orange zest, followed by the milk.
Fold in the flour until combined, then tip the mixture into the prepared loaf tin. Bake for about 45 minutes or until nice and golden on top and a skewer inserted into the centre of the cake comes out clean.
While the cake is cooking, pour the lemon and orange juice into a bowl and mix in the granulated sugar.
When the cake comes out of the oven, prick it with a skewer a few times, then drizzle the sugary juices all over the cake, allowing them to soak in. Leave the cake to cool completely in the tin, then carefully lift it out with the lining paper. Peel away the baking parchment and leave on a wire rack to finish cooling down.
Place on a plate or board and cut into slices to serve.
Dark treacly gingerbread squares
The strong, almost bitter, flavour of treacle goes so well with the equally strong taste of ginger. The bitterness is mellowed with the brown sugar, while the treacle, along with the milk, makes this cake lovely and moist. The finished squares will keep for up to 10 days if stored in an airtight container.
**Prep time** : 15 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 16 large or 25 smaller squares
75g (3oz) butter, plus extra for greasing
100g (3½oz) soft dark brown sugar
175g (6oz) black treacle
150ml (5fl oz) milk
1 egg
225g (8oz) plain flour
2 tsp ground ginger
1 tsp baking powder
¼ tsp bicarbonate of soda
½ tsp salt
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the butter, brown sugar and treacle in a saucepan over a medium heat. Stir together until melted and combined, then remove from the heat and whisk in the milk, followed by the egg. Set aside while you prepare the other ingredients.
Sift the flour, ground ginger, baking powder, bicarbonate of soda and salt into a large bowl. Make a well in the centre, then pour in the wet ingredients and mix together thoroughly, beating the mixture with a wooden spoon just until it comes together, and making sure there are no lumps of flour.
Tip the batter into the tin and bake for 20–25 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then carefully remove the cake and leave on a wire rack to cool down fully before placing on a chopping board and cutting into either 16 large squares or 25 smaller ones.
Dark treacly gingerbread loaf
Line the base and sides of a 900g (2lb) loaf tin with baking parchment before tipping in the cake mixture. Baked like this, the cake will take about an hour to cook. Once cooled, cut into slices and spread with butter to serve.
Angel food cake
Angel food cake is a classic American dessert, especially popular in the South. Its name derives from it being almost magically light, making it seem like the 'food of the angels'. Once the cake is made, you can cover it with rose-water or lavender icing, if you wish. Iced or plain, this cake is delicious served with summer berries and whipped cream, to which you could add a dash of rose or orange-blossom water or a few drops of lavender essence for a refreshing, floral note.
**Prep time** : 20 minutes (excluding the icing)
**Baking time** : 20–28 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 6–8
Butter, for greasing
125g (4½oz) plain flour, plus extra for dusting
Pinch of salt
175g (6oz) caster sugar
½ tsp cream of tartar, sifted
8 egg whites (250ml/9fl oz)
2 tsp vanilla extract
23cm (9in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tin and dust with flour, then line the base with a disc of baking parchment.
Sift the flour and salt into a bowl and mix in 150g (5oz) of the caster sugar. In a separate bowl, mix together the remaining sugar with the cream of tartar.
Tip the egg whites into a large, spotlessly clean bowl or in an electric food mixer. Using a hand-held electric beater or the food mixer, whisk the egg whites for about 20 seconds or until they begin to turn cloudy and frothy. With the beater or mixer still running, add the mixed sugar and cream of tartar, then continue to whisk until the mixture is glossy and forms stiff peaks.
Carefully fold in the vanilla extract, followed by the dry ingredients, adding these a quarter at a time and folding in very lightly, so as not to knock any air out of the mixture.
Tip the mixture into the prepared tin and bake for 20–28 minutes or until golden on top and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 15 minutes. Using a small, sharp knife to loosen the edges, carefully remove the cake from the tin and leave on a wire rack to finish cooling before transferring to a plate to serve.
While the cake is cooling, make the icing, if using, and decorate the cake following the instructions given in the individual recipes.
Variation
Instead of icing the cake, you could serve it with softly whipped cream and berries such as raspberries, blueberries, strawberries or blackberries, or a mixture. You could either top the cake with about 450ml (16fl oz) of softly whipped cream and a generous scattering of berries, or alternatively you could simply serve each slice with cream and berries.
Tip Being a delicate, fat-free sponge, it needs to be cut very carefully with a serrated knife.
Rose-water icing with pistachios
**Prep time** : 5 minutes
**Makes** : enough for 1 large cake
1–2 tbsp rose water
250g (9oz) icing sugar, sifted
To decorate
75g (3oz) unsalted shelled pistachios, roughly chopped
Fresh or crystallised rose petals (optional)
Stir together 1 tablespoon of the rose water with the icing sugar until well mixed and runny enough for drizzling – adding a few more drops of rose water if it seems too stiff. Use a spoon to drizzle all over the cake in zigzags, then, before the icing has a chance to dry, quickly sprinkle the pistachios all over the cake and scatter with the fresh or crystallised rose petals, if using.
Lavender icing
**Prep time** : 5 minutes
**Makes** : enough for 1 large cake
¼ tsp lavender essence
1–2 tbsp boiling water
250g (9oz) icing sugar, sifted
To decorate
About 8 lavender flowers (optional)
In a small bowl, mix together the lavender essence with 1 tablespoon of boiling water. Sift the icing sugar into a separate bowl, then add the lavender mixture and stir together until well mixed and runny enough for drizzling – add a few more drops of hot water if it seems too stiff. Use a spoon to drizzle all over the cake in a zigzag pattern, and scatter with the lavender flowers, if using.
Lamingtons
Lamingtons are an Aussie favourite, eaten all over the country. While their exact origins are disputed, it's generally agreed that they were named after Lord Lamington, Governor of Queensland in the late nineteenth century. Some claim the cakes are so called because they resemble the homburg hats Lamington liked to wear. Others suggest that the name came about after the governor's cook accidentally dropped some sponge cake in melted chocolate and found it was delicious when rolled in coconut. Whatever the case, these cakes are loved down under for a reason – they're not difficult to make and they look fabulous. Perfect for picnics or as an after-school treat.
**Prep time** : 30 minutes
**Baking time** : 30 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 16 squares
6 eggs
150g (5oz) caster sugar
¼ tsp salt
175g (6oz) self-raising flour
1 tsp vanilla extract
75g (3oz) butter, melted
200g (7oz) desiccated coconut, to decorate
For the icing
175g (6oz) milk chocolate, in drops or broken into pieces
50g (2oz) butter
175ml (6fl oz) milk
200g (7oz) icing sugar
50g (2oz) cocoa powder
20cm (8in) square cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the eggs, sugar and salt in a large bowl or in an electric food mixer. Using a hand-held electric whisk or the food mixer, whisk for 6–8 minutes or until the mixture is light and thick. To test if it's thick enough, lift the whisk out and draw a figure of eight in the bowl – the '8' should remain visible for a couple of seconds.
Sift in the flour and fold it into the batter, along with the vanilla extract and melted butter. Tip the mixture into the prepared cake tin and bake for 30 minutes or until the cake begins to pull away from the sides of the tin and a skewer inserted into the centre comes out clean.
Remove from the oven and allow to cool in the tin for 5 minutes before carefully removing the cake and leaving on a wire rack to cool down fully.
While the cake is cooking or cooling, make the icing. Place the chocolate, butter and milk in a heatproof bowl and set over a saucepan of just simmering water. Allow the butter and chocolate to melt, stirring from time to time, then remove from the heat. Sift together the icing sugar and cocoa powder and whisk into the melted chocolate, then transfer to a large bowl and set aside to cool.
When the cake has finished cooling, place it on a chopping board and cut into 16 squares.
Place the coconut in a shallow bowl and put it next to the bowl of chocolate icing. Dip each sponge square into the chocolate icing, ensuring each side is just covered in the mixture and allowing any excess to drip off, then gently roll in the coconut. Transfer to a wire rack to allow the icing to set, then repeat with the rest of the squares.
Orange lamingtons or lemingtons
Replace the chocolate icing with 400g (14oz) of orange or lemon curd and instead of the desiccated coconut use the same quantity of toasted chopped almonds.
Jamingtons
Replace the chocolate icing with 400g (14oz) of raspberry or strawberry jam.
Battenberg cake
I remember eating Battenberg cake after school one day at my friend's house when I was about eight. I had never seen such a beautiful cake in all my life: the pink-and-white chequerboard sponge wrapped in rich almondy marzipan was to me the height of sophistication. Even now, I can't eat it without being transported back to that time. The recipe may look long, but it really isn't complicated – just follow it step by step.
**Prep time** : 30 minutes
**Baking time** : 25–30 minutes
**Ready in** : 2 hours
**Serves** : 8–10
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
3 eggs, beaten
2 tbsp milk
175g (6oz) self-raising flour
Pink food colouring
150g (5oz) apricot jam
350g (12oz) marzipan (to make it yourself)
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a square of baking parchment.
Cut out another square of baking parchment the same size as the base of the tin (using the tin as a template to draw around). Fold in one side by 3cm (1¼ in) to form a flap, then fold the opposite side by the same amount to make a second flap. Fold the paper in half parallel to the folds for the flaps, folding over the middle section so that there is a flap on each side and, viewed from the side, the folded shape looks like an upside-down letter 'T'. This will be used as a partition to separate the two different coloured sponges.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Gradually add the eggs, beating all the time, then beat in the milk and sift in the flour, folding it in to combine.
Divide the mixture between two bowls, adding a few drops of the pink food colouring to one of the bowls and mixing in.
Place the paper shape in the centre of the tin so that it stands upright with the central fold uppermost and the flaps facing down, flat against the base of the tin. You'll see that the paper now forms a partition with a compartment on either side. Supporting the paper with one hand, pour one of the batters into one compartment. Continuing to hold the paper so that it doesn't cave in under the weight of the first batter, pour the second bowlful of batter into the other compartment. You can then take your hand off the paper as it should now be held upright by the batter pressing in on either side. (You might want to ask someone to hold the paper 'wall' steady while you pour in the batter.)
Place the filled cake tin in the oven (with the partition in place) and bake for 25–30 minutes or until well risen and a skewer inserted into the centre of each section of cake comes out clean.
Remove from the oven and leave on a wire rack to cool for 10 minutes, then, loosening round the edges using a small, sharp knife, carefully remove the cake from the tin and place on the wire rack to cool down completely.
When the cake is cool, divide it in half so you have a rectangle in each colour. Then cut each section in half lengthways so you have two rectangles in each colour. Trim all the edges so each rectangle has the same dimensions. The cross sections should be a square, with the width the same as the depth. Push the jam through a sieve and, using a palette knife, spread the jam over two adjoining sides of each rectangle of cake, then stick them together along the jammed sides to form a chequerboard pattern (when viewed from the end of the cake).
Next roll out the marzipan on a work surface lightly dusted with icing sugar. It should be rolled out into a rectangle the same length as the cake and four times the width, with any excess trimmed away. Spread the remaining jam over the marzipan, then place the cake along one edge so that cake and marzipan align lengthways, and roll up the cake in the marzipan. Cut off each end of the cake to neaten, then cut into slices to serve.
Dundee cake
Dundee cake is a traditional Scottish fruitcake. It looks gorgeous with its characteristic rings of blanched almonds on top. Unlike many fruitcakes, it doesn't have any spices, the flavour coming from the wonderful array of dried fruit it contains. It will keep for weeks, and it's worth waiting a few days before cutting into it, in order to allow the flavours to develop properly.
**Prep time** : 20 minutes
**Baking time:** 2 hours–2 hours 30 minutes
**Ready in:** 4 hours
**Serves:** 8–12
150g (5oz) butter, softened
150g (5oz) caster sugar
4 eggs
225g (8oz) plain flour, sifted
1 tsp baking powder
100g (3½oz) currants
100g (3½oz) sultanas
50g (2oz) glacé cherries, cut in half
50g (2oz) whole mixed candied peel, finely chopped
50g (2oz) ground almonds
Finely grated zest of 1 orange
Finely grated zest of 1 lemon
50g (2oz) whole blanched almonds
20cm (8in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 150°C (300°F), Gas mark 2. Line the base and sides of the cake tin with baking parchment and wrap a collar of parchment or foil around the outside of the tin to prevent the cake from drying out during cooking.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or until just mixed, then gradually add them to the creamed butter mixture, beating all the time.
In a separate bowl, mix together all the remaining ingredients apart from the whole almonds, then stir into the cake batter. Mix well, then tip into the prepared tin. Carefully arrange the whole almonds in concentric circles on top of the mixture, laying them lightly on top without pressing in.
Bake on the lowest shelf in the oven for between 2 hours and 2 hours 30 minutes or until springy to the touch and a skewer inserted into the centre of the cake comes out clean. It's best to check the cake halfway through cooking: if it's a deep golden brown already, place a sheet of baking parchment or foil over the top to stop it browning any further.
Remove from the oven and place on a wire rack to cool for about 30 minutes. Remove the collar of paper or foil and, using a small, sharp knife to loosen the edges, carefully remove the cake from the tin and leave on a wire rack to finish cooling before transferring to a serving plate.
Madeleines
Madeleines are the quintessential delicate treat. The airy batter is baked in the traditional shell-shaped moulds to make a cake that is just crisp on the outside and elegantly light in the middle. This recipe is quick and easy to make, but there are many twists you can give to the basic cake, which are all delicious variations on a classic theme.
**Prep time** : 20 minutes (excluding any icing)
**Baking time** : 12–15 minutes
**Ready in** : 50 minutes
**Makes:** 12 madeleines
1 egg
50g (2oz) caster sugar
50g (2oz) plain flour, plus extra for dusting
¼ tsp baking powder
50g (2oz) butter, melted, plus extra for greasing
½ tsp vanilla extract
Icing sugar, for dusting
12-hole madeleine tray
Preheat the oven to 180°C (350°F), Gas mark 4. Brush a little melted butter over the madeleine moulds (making sure to coat every ridge) and dust a little flour into each one, tapping out any excess.
Place the egg and sugar in a large bowl or in an electric food mixer. Using a hand-held electric beater or the food mixer with its whisk attachment, whisk on a high speed for about 5 minutes or until the mixture is pale, thick and mousse-like and has grown almost three times in volume.
Sift the flour and baking powder into the whisked egg and sugar and carefully fold in, then fold in the melted butter and vanilla extract, taking care not to over-mix. Either pouring the batter directly from the bowl or using a tablespoon to spoon it in, divide the batter between the madeleine moulds, filling each almost to the top.
Bake for 12–15 minutes or until golden and lightly springy to the touch. (Try not to overcook them or they will be dry.) Remove from the oven and carefully remove each madeleine from its mould using a palette knife, then place on a wire rack to cool, if you must, as there are few things more delicious than warm madeleines served straight from the oven with nothing more than a light dusting of icing sugar.
Chocolate-dipped madeleines
**Prep time** : 30 minutes
**Baking time** : 12–15 minutes
**Ready in** :1 hour 15 minutes
**Makes** : enough for 12 cakes
For the chocolate coating
100g (3½oz) dark or white chocolate, in drops or broken into pieces
Prepare and bake the madeleines as and while they are cooling, prepare the chocolate. Place the chocolate in a heatproof bowl and set over a pan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Holding one of the madeleines by one end, dip it into the chocolate so that it comes about a third of the way up. Remove from the chocolate, shaking off any excess into the bowl, then place on a sheet of baking parchment to set. Repeat with the rest of the madeleines until all of them are dipped.
Double chocolate madeleines
Make up the madeleines as in the main recipe, adding 1 tablespoon of cocoa powder at the same stage as the flour, then dip in the melted chocolate and leave to set.
Chocolate almond madeleines
Follow the basic madeleine recipe, but rather than vanilla extract, add a few drops (not more than ⅛ tsp) of almond essence. Dip in the melted chocolate and then roll in 125g (4½oz) of toasted nibbed (chopped) almonds (tip) before leaving to set.
Madelamingtons
Follow the basic madeleine recipe, adding 25g (1oz) of desiccated coconut at the same stage as the flour. Dip in the melted chocolate and then roll in 100g (3½oz) of desiccated coconut before leaving to set.
Lemon-striped madeleines
**Prep time** : 30 minutes
**Baking time** : 12–15 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : enough for 12 cakes
For adding to the sponge
Finely grated zest of 1 lemon
For the icing
l00g (3½oz) icing, sifted
1–2 tbsp lemon juice
To the basic recipe, add the lemon zest along with the flour. Bake the madeleines as in the main recipe, then remove from the oven and allow to cool.
When the cakes are cool, make the icing. In a bowl, mix together the icing sugar with 1 tablespoon of lemon juice. You want the icing to be thick enough to set but just liquid enough to drizzle. If it seems too thick, you may need to add a few more drops of lemon juice to get the right consistency.
Place the madeleines, ridged side up, on a plate. Set the bowl of icing close to the madeleines, then dip a spoon into the icing and lift it out again, allowing some of the icing to drip off so there isn't too much on the spoon. Holding the spoon over one of the madeleines, carefully drizzle over the icing in stripes or zigzags. Repeat with all of the madeleines, dipping the spoon back into the icing as needed.
Victoria sponge
The classic sponge cake was named after Queen Victoria, who enjoyed a slice of it with her tea. In its traditional form, it consists of two layers of cake with a raspberry jam and cream filling and dusted with icing or caster sugar. I've provided a version of it here as it's a classic, but it's a seriously versatile recipe with lots of variations that you can try – see the five different filling ideas.
**Prep time** : 10 minutes (excluding any filling)
**Baking time** : 18–25 minutes
**Ready in** : 1 hour
**Serves:** 6–8
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
3 eggs
175g (6oz) plain flour, plus extra for dusting
1 tsp baking powder
1 tbsp milk
Icing or caster sugar, for sprinkling
Two 18cm (7in) diameter sandwich tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter and flour the sides of each tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds until just mixed, then gradually add them to the butter mixture, beating all the time. Sift in the flour and baking powder, then add the milk and fold in gently to incorporate.
Divide the mixture between the two tins and make a slight hollow in the centre of each cake so that when it rises it doesn't peak too much in the centre, making it difficult to sandwich together with the other half.
Place in the centre of the oven and bake for 18–25 minutes or until golden on top and springy to the touch.
Remove from the oven and allow to cool in the tins for 10 minutes, then loosen around the edges of each cake using a small, sharp knife and carefully remove from the tins before leaving on a wire rack to cool down completely.
Once cool you can sandwich the cakes together using one of the fillings detailed. To assemble the Victoria sponge, place one cake upside down on a plate and spread over your chosen filling. Place the second cake on top, right side up, then sprinkle over the icing or caster sugar to finish.
Chocolate Victoria sponge
Make up the cake batter as in the recipe above, adding 25g (1oz) of cocoa powder with the flour for a chocolate sponge and sandwiching together with chocolate buttercream – using a quarter of the quantities specified).
Orange or lemon Victoria sponge
Add the finely grated zest of a lemon or orange with the flour. For an orange-flavoured sponge, you could sandwich the sponge layers together with the orange and mascarpone filling. For a lemon-flavoured version of this filling – to go with a lemon sponge – you could substitute the orange curd with lemon curd instead.
Fillings
Jam and cream or jam only
Spread 3–4 rounded tablespoons of jam, such as raspberry, strawberry or blackberry, over the first cake, followed by a layer of 125ml (4½fl oz) of double or regular cream, whipped until almost stiff. This is best eaten on the day it's made. If you want the cake to last for a few days (without putting it in the fridge), omit the cream and use just jam in the centre; this will keep in an airtight box for 2–3 days.
Fresh fruit and cream
Substitute the jam with 200g (7oz) of fresh soft fruit, such as raspberries, blueberries (slightly mashed first) or sliced strawberries, adding these to the top of the whipped cream.
Rhubarb and cream
Place 100g (3½oz) of sliced rhubarb in a saucepan with 50g (2oz) of caster sugar and a splash of water, cover with a lid and cook over a gentle heat for 10 minutes or until the rhubarb is soft. Bring to the boil, then take off the lid and boil, stirring continuously, until thickened. Pour into a bowl and allow to cool, then fold 125ml (4½fl oz) of double or regular cream, whipped until almost stiff, into the cooked rhubarb and spread over the first cake.
Elderflower and strawberry
Fold 2 tablespoons of elderflower cordial into 125ml (4½fl oz) of double or regular cream, whipped until almost stiff, and spread over the first cake, then add 175g (6oz) of sliced strawberries in an even layer on top of the cream. Sprinkle the finished cake with icing or caster sugar or, if you prefer, use elderflower sugar.
Orange and mascarpone
Whisk 100g (3½oz) of orange curd into 150g (5oz) of mascarpone and spread over the first cake. Once the cake is assembled, decorate the top with candied orange slices and sprinkle with caster sugar.
Brittany butter cake
This cake is a pure expression of the butter for which Brittany is famous. With so few ingredients, the flavour of the butter has nowhere to hide; the sugar is only there to sweeten the cake and the flour to give structure. Delicious eaten on its own, it also goes perfectly with strawberries and cream.
**Prep time** : 20 minutes
**Baking time** : 30 minutes
**Ready in** : 1 hour 10 minutes
**Serves:** 8–12
1 egg yolk, for the glaze
225g (8oz) plain flour, sifted
225g (8oz) caster sugar
225g (8oz) butter, softened, plus extra for greasing
6 egg yolks
25cm (10in) diameter cake tin
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tin and line the base with a disc of baking parchment.
Whisk together the egg yolk with 2 teaspoons of water for the glaze and set aside.
Either in a large bowl using a wooden spoon or in an electric food mixer using the paddle beater, mix together the flour and sugar, then add the butter and egg yolks and beat together until the mixture resembles a stiff dough.
Press into the prepared tin, and flatten with a spatula. Brush with the glaze, then decorate by drawing a fork across the cake in a criss-cross pattern of lines, each set of lines roughly 5cm (2in) apart in a sort of chequerboard design, following the traditional style for the cake.
Bake for 30 minutes or until it is a deep golden colour and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen the edges using a small, sharp knife and carefully remove the cake from the tin before placing on a wire rack to cool down completely.
Variation
If you like, you can add a tablespoon of rum or brandy to the dough, at the same time as the egg yolks, for another note of flavour.
Chocolate hazelnut cheesecake
Light this is not, but if you're craving something rich and indulgent, this American-style baked cheesecake, with its fabulous creamy topping and crisp buttery base, will certainly fit the bill. The cream cheese topping is loaded with chocolate and melted butter, offset by the delicate tang of sour cream. The toasted hazelnuts, roughly chopped, add another layer of texture to this divine dessert.
**Prep time** : 30 minutes
**Baking time** : 1 hour 30 minutes
**Ready in** : 4 hours
**Serves:** 10–12
100g (3½oz) butter, plus extra for greasing
350g (12oz) dark chocolate, in drops or broken into pieces
450g (1lb) cream cheese
100ml (3½fl oz) sour cream
3 eggs
175g (6oz) caster sugar
150g (5oz) hazelnuts, toasted, skinned and roughly chopped (see the tip)
For the biscuit base
375g (13oz) dark chocolate digestive biscuits
3 tbsp double or regular cream
75g (3oz) butter, melted
25cm (10in) diameter spring-form cake tin with 6cm (2½ in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, and grease the sides of the cake tin with butter. Make sure the base of the cake tin is upside down, so there's no lip and the cake can slide off easily when cooked.
First make the biscuit base. Place the biscuits in a food processor and whiz until they form coarse crumbs. Alternatively, place the biscuits in a plastic bag and crush them using a rolling pin until finely broken down with only a few coarse pieces remaining. Tip the biscuit crumbs into a large bowl and mix with the cream and melted butter, then place in the cake tin, pushing the mixture down into the base to cover it as evenly as possible. Chill in the fridge while you make the topping.
Place the butter and chocolate in a large heatproof bowl and set over a saucepan of just simmering water, stirring occasionally until melted and combined. When melted, remove from the heat and whisk in the cream cheese until smooth, then whisk in the sour cream.
In a separate bowl using a hand-held electric beater or in an electric food mixer, whisk together the eggs and sugar for 2–3 minutes or until mousse-like in consistency, then turn off the machine and fold in the chocolate mixture along with the chopped hazelnuts.
Add the mixture to the chilled biscuit base, spreading it in an even layer, then bake for 1 hour 30 minutes or until almost set. It should wobble a little if you gently shake the tin, while the top will be slightly cracked.
Allow to cool in the tin on a rack for an hour, then chill, still in the tin, in the fridge for at least an hour before serving. Loosen around the edges using a small, sharp knife and unclip and remove the sides of the tin. Then use a long, sharp knife to loosen the bottom of the cake from the base and a palette knife or metal fish slice to slide the cake onto a plate. (If you don't feel brave enough to slide it off the base, just leave it on and place it like this on the plate.)
Tip You can buy toasted hazelnuts, but for the best flavour you should toast your own. To toast them, spread them out on a baking tray and cook in the oven (preheated to 200°C/400°F/Gas mark 6) for 7–10 minutes or until their skins have darkened (rub the skin off one or two to check that the nuts are golden). Wrap the nuts in a clean tea towel (this slightly stains the tea towel, so don't use your favourite one!) and rub them for a few seconds to remove the skins, which should come off easily. Pour the skinned nuts back onto the baking tray, then either place the tray outside and let the skins just blow away (you can speed up the process by blowing gently on the hazelnuts) or just pick the hazelnuts out from the skins. Next use a knife to roughly chop the nuts, leaving some larger pieces.
'There are few hours in life more agreeable than the hour dedicated to the ceremony known as afternoon tea.'
– _The Portrait of a Lady_ , Henry James
Few and far between are the clays in which I don't drink tea. There is no drink that I find more comforting or calming. I will happily drink tea in the afternoon by itself, but even a small piece of cake transforms a five-minute break from work or chores into a real treat. There is no rule as to what makes a cake good with tea (or indeed with coffee). Sometimes I like a spiced fruitcake, a rich chocolate cake or a fluffy, nutty muffin. But nothing too advanced or complicated, and usually without too much icing. Here are my favourite cakes that are the perfect partners to a soothing cup of comfort.
05/ Teatime
Cheesecake brownies
Banoffee blondies
Cappuccino squares
Ras el hanout spiced fruitcake
Chocolate and pecan muffins with maple glaze
Coffee mascarpone cake
Ginger golden syrup loaf
Banana, almond and honey muffins
Boiled fruitcake
Nutty toffee cake
Cardamom yoghurt cake
Ginger peach muffins
Pear crumble cake
Double chocolate peanut butter brownies
Cheesecake brownies
I adore the effect of pale cheesecake swirled through dark sponge in these brownies. They look fabulous and the rich, strong flavour of the chocolate is nicely complemented by the slight tang of the cream cheese. These are perfect for those times when a standard brownie just doesn't seem indulgent enough.
**Prep time** : 20 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 16 brownies
100g (3½oz) butter, cut into cubes
100g (3½oz) dark chocolate, in drops or broken into pieces
125g (4½oz) caster sugar
Pinch of salt
3 eggs, beaten
75g (3oz) self-raising flour
1 tbsp cocoa powder
100g (3½oz) dark chocolate, in chips or chopped into pieces
For the cheesecake mixture
200g (7oz) cream cheese
2 egg yolks
75g (3oz) caster sugar
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the butter and chocolate in a saucepan and melt over a low heat. Stir just until smooth, then remove from the heat and beat in the sugar and salt, followed by the eggs. Sift in the flour and cocoa powder, add the chocolate chips and stir in. Tip the batter into the prepared tin and spread out using a spatula.
In a clean bowl, beat all the ingredients for the cheesecake mixture together until smooth. Dot heaped teaspoonfuls of this mixture across the top of the brownie batter in the tin, then use a skewer or knife to swirl the two mixtures together, right across the tin and up to the edges.
Bake for 30–35 minutes or until the middle of the cake still wobbles slightly when you gently shake the tin – a skewer inserted into the centre should come out with a little moisture on it. Remove from the oven, place on a wire rack and allow to cool completely in the tin before carefully removing the cake and cutting into squares to serve.
Banoffee blondies
Blondies, with their white chocolate vanilla flavour, are a close relation of the brownie. I like to mix additional ingredients into blondies and I love the classic 'banoffee' combination of banana and toffee when mixed with white chocolate. These blondies take a little extra effort, but it's well worth it. The almond praline adds a lovely toffee-almond crunch to contrast with the soft sweetness of the bananas. Eat either with vanilla ice cream, while the cakes are still warm, or as a picnic treat.
**Prep time** : 20 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 16 blondies
100g (3½oz) butter
225g (8oz) white chocolate, in drops or broken into pieces
175g (6oz) soft light brown sugar
1 egg, beaten
200g (7oz) peeled very ripe bananas (about 2), mashed
2 tsp vanilla extract
225g (8oz) plain flour
1 tsp baking powder
¼ tsp salt
For the almond praline
100g (3½oz) caster sugar
100g (3½oz) whole almonds (skin still on)
20cm (8in) square cake tin with 5cm (2in) sides
First make the almond praline. Place the caster sugar in a frying pan (preferably non-stick if you have one), spreading it over the base of the pan, then scatter over the almonds. Cook on a medium heat until the sugar turns a deep golden caramel. (You may swirl the pan once the sugar starts caramelising, but do not stir it.) Tip out onto a baking tray lined with baking parchment and allow to cool and set, then chop until the consistency of very coarse breadcrumbs, or pulse in a food processor.
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the butter and white chocolate in a saucepan and set over a low heat. When the butter has melted, remove from the heat and stir to melt the chocolate. Pour into a large bowl and beat in the sugar, egg, mashed bananas and vanilla extract, then stir in the almond praline.
Sift in the remaining ingredients and fold in thoroughly to mix, then spoon the mixture into the prepared tin and bake for 30–35 minutes or until just set and golden on top.
Either allow to cool completely in the tin or allow to cool in the tin for 5–10 minutes and serve while still warm. Carefully remove the cake from the tin and cut into squares for serving.
Cappuccino squares
A light and fluffy buttercream icing mimics the layer of frothy milk in these playful, cappuccino-style cakes. Unsurprisingly, these squares go especially well with coffee, or indeed cappuccino.
**Prep time** : 20 minutes
**Baking time** : 40 minutes
**Ready in** : 1 hour 45 minutes
**Makes** : 16 squares
200g (7oz) butter, softened, plus extra for greasing
200g (7oz) caster sugar
3 eggs
325g (11½oz) plain flour, sifted
3 tsp baking powder
150ml (5fl oz) milk
2 tbsp coffee essence (ideally Camp or Irel)
For the vanilla buttercream icing
200g (7oz) butter, softened
150g (5oz) icing sugar, sifted
2 tsp vanilla extract
3 tbsp milk
To decorate
1 tbsp cocoa powder, for dusting
16 chocolate-covered coffee beans (optional)
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place all the ingredients for the sponge in a food processor and whiz for 10–20 seconds or just until combined. Alternatively, cream the butter until soft using a hand-held electric beater, then beat in the sugar, whisk in the eggs one at a time and fold in the remaining ingredients.
Tip the mixture into the prepared tin, smooth over the top with a palette knife and bake in the oven for 40 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes before carefully removing the cake from the tin and leaving on a wire rack to finish cooling.
While the cake is in the oven, or while it's cooling, make the vanilla buttercream icing. Place all the ingredients in the food processor, having cleaned the bowl and mixer blade, and whiz until light and fluffy. Alternatively, cream the butter until soft using the electric beater, then beat in the remaining ingredients.
Transfer the cake to a plate and spread the icing over the top using a palette knife, dust with cocoa powder then cut into squares to serve. As a finishing touch, you could decorate the tops with chocolate-covered coffee beans, adding several beans to the centre of each square.
Ras el hanout spiced fruitcake
Ras el hanout refers to a Moroccan blend of spices. It doesn't refer to a specific blend, however: ras el hanout just means 'top of the shop' in Arabic, indicating a mix of the very best spices a market stall or shop may have to offer. This recipe uses a blend of spices that may be a little unusual for a cake but convey all the exotic aromas of a Marrakech market. Interestingly, this cake has no fat, the large amount of dried fruit doing the job instead, keeping it lovely and moist. Stored in an airtight container, it will keep for a few weeks.
**Prep time** : 15 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
Butter, for greasing
75g (3oz) raisins
75g (3oz) dried apricots, chopped
75g (3oz) dried blueberries or dried cranberries (or use raisins)
75g (3oz) sultanas
75g (3oz) currants
250ml (9fl oz) hot black tea
½ tsp ground cinnamon
Pinch of ground cloves
Pinch of chilli powder
½ tsp ground cardamom
¼ tsp ground coriander
¼ tsp ground cumin
¼ tsp turmeric
½ tsp ground nutmeg
225g (8oz) self-raising flour
150g (5oz) soft light brown sugar
1 egg, lightly beaten
20cm (8in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Put all the dried fruit in a heatproof bowl and pour over the hot tea. Stir thoroughly, then add the ground spices and leave to cool for 10 minutes.
Sift the flour into a large bowl and mix in the sugar, then add the soaked fruit mixture and the beaten egg and stir until all the ingredients are completely incorporated.
Tip the mixture into the prepared cake tin and bake for about 45 minutes or until well risen and springy to the touch – a skewer inserted into the centre should come out clean.
Remove from the oven and allow to cool for 10 minutes. Using a small, sharp knife to loosen the edges, carefully remove the cake from the tin and leave on a wire rack to cool down fully before transferring to a serving plate.
Chocolate and pecan muffins with maple glaze
Half muffin, half cupcake, these aren't quite as indulgent as a fully fledged cupcake but a bit of a treat nonetheless. I like the speckled appearance created by the chocolate chips as well as the contrast in texture when combined with the crunchy pecans.
**Prep time** : 10 minutes
**Baking time** : 25 minutes
**Ready in** : 1 hour
**Makes** : 12 muffins
200g (7oz) butter
200g (7oz) caster sugar
4 eggs
100g (3½oz) milk or dark chocolate, in drops or broken into pieces
75g (3oz) pecans, roughly chopped
200g (7oz) self-raising flour
25ml (1fl oz) maple syrup, for the glaze
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Beat in the eggs one at a time, then tip in the chocolate and pecans and fold in with the flour to combine.
Divide the batter between the muffin cases, filling each case three-quarters full. Bake for about 25 minutes or until golden on top and lightly springy to the touch. Transfer the tin to a wire rack and brush each muffin with some of the maple syrup. Allow to cool for 5 minutes, then remove from the tin and place on the wire rack to finish cooling.
Coffee mascarpone cake
The rich creamy mascarpone provides the perfect foil to this simple coffee sponge. The coffee essence is very useful because it provides coffee flavour without adding too much liquid to your cake mixture. If you can't get hold of it, you can always substitute the same quantity of very strong coffee for the essence. The cocoa isn't just for decoration: it's essential to this recipe, with its magic mocha twist.
**Prep time** : 20 minutes
**Baking time** : 20 minutes
**Ready in** : 1 hour
**Serves** : 8–12
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
3 eggs, beaten
2 tbsp coffee essence (ideally Camp or Irel)
175g (6oz) self-raising flour, plus extra for dusting
Cocoa powder, for dusting
For the filling and icing
250g (9oz) mascarpone
3 tbsp icing sugar, sifted
1 tbsp coffee essence (ideally Camp or Irel)
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins, dusting them lightly with flour, and line the base of each tin with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Gradually add the eggs and coffee essence to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently to mix.
Divide the batter between the two prepared tins, making a slight hollow in the centre so they will rise with a flat top. Place in the oven and bake for about 20 minutes or until well risen, golden brown and springy to the touch.
Remove from the oven and allow to sit in the tins for 10 minutes. Loosen round the edges of each cake using a small, sharp knife, then carefully remove from the tins and leave on a wire rack to cool down completely.
Meanwhile, make the icing. Place the mascarpone in a bowl, and mix in the icing sugar along with the coffee essence.
Place one of the cakes upside down on a plate, then spread over half the icing. Place the other cake, right side up, on top of the filling, then spread with the rest of the icing and dust with the cocoa powder.
Ginger golden syrup loaf
I love stem ginger – little pieces of peeled young ginger preserved in a jar of sweet sugary syrup. It keeps its flavour and spiciness well, making it ideal for chopping up for use in a cake, as in this recipe, or spooned over vanilla ice cream. The sweet syrup is just divine – I like to use it in cocktails or, as here, drizzled over the cake while it's still hot so it really soaks up that gorgeous ginger taste.
**Prep time** : 10 minutes
**Baking time** : 50–60 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–10
225g (8oz) golden syrup
100g (3½oz) butter
100g (3½oz) caster sugar
75g (3oz) stem ginger in syrup (drained), chopped, reserving 3 tbsp of syrup for the glaze
200ml (7fl oz) milk
2 eggs, beaten
225g (8oz) self-raising flour
1 tbsp ground ginger
1 tsp bicarbonate of soda
1 tsp salt
900g (2lb) loaf tin
Preheat the oven to 180°C (350°F), Gas mark 4, then line the base and sides of the loaf tin with baking parchment, with the paper coming above the sides of the tin to enable the cake to be lifted out easily.
Place the golden syrup, butter and sugar in a saucepan and stir over a medium-low heat until the butter has melted and the mixture is smooth. Set aside and allow to cool, then mix in the ginger, milk and eggs.
Meanwhile, sift the remaining ingredients into a large bowl. Make a well in the centre, then pour in the wet ingredients and mix thoroughly.
Tip the mixture – which will be quite runny – into the prepared loaf tin and bake in the centre of the oven for 50–60 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and drizzle or brush the reserved stem ginger syrup over the loaf while it's still hot.
Allow to cool for 10 minutes in the tin, then carefully lift the cake out of the tin with the lining paper and place on a wire rack. Peel away the baking parchment and leave to cool down fully.
Banana, almond and honey muffins
The honey in these muffins brings out the natural sweetness of the ripe bananas. When baking it's best to try and use the ripest bananas you can – the darker the better. As a banana darkens it converts more of its starch to sugar and will make these muffins moist and even sweeter.
**Prep time** : 10 minutes
**Baking time** : 25 minutes
**Ready in** : 1 hour
**Makes** : 12 muffins
125g (4½oz) plain flour
2 tsp baking powder
½ tsp salt
175g (6oz) wholemeal flour
350g (12oz) peeled very ripe bananas (3–4 in total), mashed
2 eggs
75ml (3fl oz) sunflower oil
150g (5oz) honey
175g (6oz) almonds (skin still on), chopped
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Sift the plain flour into a large bowl with the baking powder and salt and mix in the wholemeal flour. Place the bananas in a separate bowl with the eggs, sunflower oil and honey and whisk to combine. Tip into the dry ingredients and beat until smooth, adding the almonds just as the mixture is coming together.
Divide the batter between the muffin cases – they should each be about three-quarters full – then bake for about 25 minutes or until golden on top and springy to the touch.
Remove the tray from the oven and allow the muffins to cool for 5 minutes, then take them from the tray and place on a wire rack to finish cooling.
Boiled fruitcake
In researching this book I found so many places in which the traditional cake was really just another variation of a fruitcake. The fruits and spices change from place to place, but the basic formula is similar. My recipe for fruitcake isn't tied to any specific place; it's a combination of ideas from different regions. The dates when boiled and cooked add their divine toffee taste. The golden syrup is the only sugar here, which makes the cake especially moist.
**Prep time** : 15 minutes
**Baking time** : 1 hour 30 minutes–2 hours
**Ready in** : 3 hours
**Serves** : 6–8
150g (5oz) butter, plus extra for greasing
200g (7oz) golden syrup
225g (8oz) raisins
225g (8oz) currants
100g (3½oz) sultanas
100g (3½oz) pitted dates, chopped
100g (3½oz) whole mixed candied peel, chopped
150ml (5fl oz) milk
225g (8oz) plain flour
1 tsp mixed spice
½ tsp grated nutmeg
½ tsp bicarbonate of soda
Pinch of salt
2 eggs, beaten
20cm (8in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 150°C (300°F), Gas mark 2. Butter the sides of the cake tin and line the base with a disc of baking parchment. Double wrap the outside of the cake tin with extra baking parchment or foil, which will help prevent the cake from drying out.
Place the butter in a saucepan with the golden syrup, dried fruit, mixed peel and milk. Set over a low heat and stir until the butter has melted, then increase the heat to a simmer. Simmer for 5 minutes, stirring occasionally, then remove from the heat and set aside.
In a large bowl, sift together the flour, mixed spice, nutmeg, bicarbonate of soda and salt. Add the cooled fruit mixture to the dry ingredients, along with the beaten eggs, and mix together thoroughly.
Tip the mixture into the prepared tin and bake on the lowest shelf in the oven for between 1 hour 30 minutes and 2 hours or until a skewer inserted into the centre of the cake comes out clean. Check the cake after an hour (without taking it out of the oven) and if it is browning too much on top, cover with baking parchment or a sheet of foil for the remainder of the cooking time. When cooked, remove from the oven and place on a wire rack to cool completely in the tin.
When completely cool, loosen around the edges with a small, sharp knife and carefully remove the cake from the tin. You can cut the cake into slices now, if you like, or leave for a few days in an airtight container to let the flavours develop. Stored like this, it will keep for a few weeks.
Nutty toffee cake
Here, the sweetness of dates and maple syrup combines with the glorious crunchiness and rich flavour of three different types of nut – almonds, walnuts and brazils. You could use just one type of nut if you'd prefer. Arranging them in a ring around the outside of the cake is best: not only does it look good, but it also prevents the centre of the cake from sinking a little under their weight.
**Prep time** : 10 minutes
**Baking time** : 40 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 8–10
225g (8oz) mixed whole almonds, brazil nuts and walnuts (see the tip), or 225g (8oz) one type of nut (almonds, brazil nuts or walnuts)
200g (7oz) butter, softened, plus extra for greasing
200g (7oz) soft brown sugar
4 eggs
150g (5oz) pitted dates, chopped
150g (5oz) self-raising flour, plus extra for dusting
50ml (2fl oz) maple syrup
23cm (9in) spring-form or loose-bottomed cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides and base of the cake tin and dust with flour. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Tip 100g (3½oz) of the nuts into a food processor and whiz for a couple of minutes until fine. Add the butter and sugar, then whiz briefly until the mixture is soft and fluffy. Add the eggs one at time, pulsing just to combine, then add the dates and pulse a few times to combine.
Tip the mixture into a large bowl, then sift in the flour, folding in just until combined. Pour the batter into the prepared tin, then carefully arrange the remaining nuts in a circle or band about 5cm (2in) wide around the edge of the cake, leaving the centre free of nuts otherwise they'd weigh down the centre of the cake.
Bake for about 40 minutes or until golden brown on top and well risen – a skewer inserted into the centre of the cake should come out clean. Remove from the oven and immediately brush the nuts and top of the cake with the maple syrup.
Allow to cool for 20 minutes, then loosen around the edges using a small, sharp knife and carefully remove the sides of the cake tin. Place on a wire rack to cool down fully, then use a palette knife or metal fish slice to loosen the bottom of the cake from the base of the tin and, with the help of the knife or fish slice, ease the cake onto a plate to serve.
Tip When using walnuts, make sure to taste them before you add them to a cake, as if they are rancid or bitter they will affect the cake's flavour.
Cardamom yoghurt cake
Cardamom is one of my favourite spices and it surprisingly works just as well in cakes as in curries. Its floral fragrance is enhanced here with just a little orange zest. You can buy ready-ground cardamom, which will work well in this cake. For the best flavour, use whole cardamom pods, which you can peel and grind yourself (see the tip).
**Prep time** : 25 minutes
**Baking time** : 50–55 minutes
**Ready in** : 2 hours
**Serves** : 10–12
250g (9oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
Juice and finely grated zest of ½ large or 1 small orange
1 tsp freshly ground cardamom seeds (from about 12–14 pods) (see the tip) or 1 tsp ready-ground cardamom
4 eggs
225ml (8fl oz) natural yoghurt
350g (12oz) plain flour
1½ tsp baking powder
¼ tsp salt
75g (3oz) fresh raspberries, to decorate
For the icing
225g (8oz) icing sugar, sifted
¼ tsp ready-ground cardamom or the ground seeds of 5 cardamom pods (see the tip)
25–30ml (1–1¼fl oz) natural yoghurt
2.5 litre (4⅓ pint) bundt tin (about 23cm/9in in diameter) or 25cm (10in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, and butter the bundt tin. If you're using a standard type of cake tin, butter the sides and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Next beat in the orange juice and zest and the ground cardamom.
Add the eggs one at a time, beating well between each addition, then mix in the yoghurt. Next sift in the flour, baking powder and salt, folding in just until combined.
Tip the batter into the prepared tin and smooth the top with a spatula or palette knife. Bake the cake for 50–55 minutes or until a skewer inserted into the centre of the cake comes out clean. Allow to cool for just 5 minutes, then, if using a bundt tin, place a wire rack on the top and invert the cake so it is upside down, then remove the tin, turn upright again and leave to cool on the wire rack. For removing a standard type of cake tin, follow the instructions before leaving to cool.
As the cake cools, make the icing. Beat together the icing sugar, cardamom and 25ml (1fl oz) of the yoghurt, adding a tiny bit more yoghurt if the mixture seems too stiff. (It should be a thick drizzling consistency: too thin and the icing will slide off the cake – too thick and you won't be able to drizzle it.) Place the cake on a cake stand or serving plate, then drizzle the icing backwards and forwards from the centre to the outside of the cake in a zigzag pattern (or in zigzags across the top if it's been made in a standard tin). Decorate with the raspberries immediately while the icing is still wet so they stick to the icing.
Tip To remove the seeds from cardamom pods, lightly crush the pods, then pick out the black seeds and discard the husks. Crush the black seeds using a pestle and mortar or place in a plastic bag and use a rolling pin.
Ginger peach muffins
Sweet peaches are given a boost here with some slightly spicy ginger, both fresh and ground. The wholemeal flour provides these muffins with a hint of nuttiness, and a general feeling that they must be healthy for you, as well as tasting good. They would be welcome at any time of the day, but especially as a mid-morning or mid-afternoon snack.
**Prep time** : 10 minutes
**Baking time** : 25–30 minutes
**Ready in** : 1 hour
**Makes** : 12 muffins
200g (7oz) plain flour
1 tsp baking powder
1 tsp bicarbonate of soda
1 tsp ground ginger
½ tsp salt
100g (3½oz) wholemeal flour
100g (3½oz) soft light brown sugar
75g (3oz) butter, melted
150ml (5fl oz) buttermilk
1 egg
50g (2oz) peeled root ginger, finely grated
200g (7oz) fresh or tinned sliced peaches (drained), cut into 1–2cm (½–¾in) chunks
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Sift the plain flour into a large bowl with the baking powder, bicarbonate of soda, ground ginger and salt, and mix in the wholemeal flour and brown sugar. In a separate bowl, whisk together the melted butter, buttermilk, egg and grated ginger. Pour the wet ingredients into the flour mixture and mix together until well combined, then fold in the peaches.
Divide the batter between the muffin cases, filling each about three-quarters full, then bake for 25–30 minutes or until well risen and golden. Remove from the oven and allow to cool in the tin for 5 minutes before transferring from the tin onto a wire rack to cool down fully.
Pear crumble cake
Each of the layers in this cake is essential to the finished dish. The bottom layer consists of soft sponge. Next comes a layer of moist pears flavoured with a little cinnamon and lightly caramelised sugar, while on top is a layer of crumble enhanced by the texture and flavour of hazelnuts and brown sugar. It's absolutely worth the little extra effort it takes to make.
**Prep time** : 25 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour 40 minutes
**Serves** : 6–10
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs
225g (8oz) self-raising flour
Double or regular cream, whipped, to serve
For the cooked pears
400g (14oz) peeled and cored pears, chopped
1 tsp cinnamon
50g (2oz) butter
50g (2oz) caster sugar
For the crumble topping
100g (3½oz) butter
150g (5oz) plain flour, sifted
100g (3½oz) demerara or light soft brown sugar
100g (3½oz) hazelnuts, toasted, skinned and roughly chopped (see the tip)
23cm (9in) diameter spring-form or loose-bottomed cake tin with 6cm (2½ in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, and butter the sides and base of the cake tin. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
First cook the pears. Place the chopped fruit in a saucepan with the cinnamon, butter and sugar and set over a medium heat, then cook, uncovered and stirring every now and again, for 5–10 minutes or until most of the liquid has evaporated. Remove from the heat and set aside.
To make the crumble topping, melt the butter gently in a saucepan, then add the flour, sugar and chopped, toasted hazelnuts. Stir to mix and then set aside while you make the sponge.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds until just mixed, then gradually add the eggs to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently until mixed, then tip the batter into the prepared tin.
Add the cooked pears to the tin, spreading them out in an even layer over the mixture. Sprinkle the crumble evenly over the pears, then bake for about 45 minutes or until well risen and deep golden in colour.
Remove from the oven and allow to cool in the tin for 10 minutes. Loosening around the edges using a small, sharp knife, carefully remove the sides of the tin and place the cake on a wire rack to cool down, unless you'd prefer to serve it warm.
When you're ready to serve, use a palette knife or fish slice to loosen the bottom of the cake from the base of the tin, then slide the knife under the cake and carefully ease it onto a plate. Cut the cake into slices and serve warm or cold with softly whipped cream, to which you could add a little icing sugar and a splash of brandy or pear liqueur.
Double chocolate peanut butter brownies
Fudgy, sticky and unapologetically rich, these brownies bring together two quintessentially American treats – brownies and peanut butter. For me, crunchy peanut butter is essential, as the chunks of peanut break up the softness of the brownie. I like to serve these with coffee, but they're so good I'll happily eat them regardless of whether I've a cup to hand!
**Prep time** : 15 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour
**Makes** : 16 brownies
225g (8oz) dark chocolate, in drops or broken into pieces
225g (8oz) butter
300g (11oz) caster sugar
3 eggs, beaten
75g (3oz) plain flour
1 tsp baking powder
100g (3½oz) white chocolate, in chips or chopped into pieces
200g (7oz) crunchy peanut butter
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven 180°C (350°F), Gas mark 4, then line the base and sides of the baking tin with parchment paper. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the chocolate and butter in a heatproof bowl and set over a saucepan of simmering water. Stir occasionally just until melted, then set aside.
In a separate large bowl using a hand-held electric beater or using an electric food mixer, whisk together the sugar and eggs for a few minutes or until pale and light. Pour in the melted chocolate and continue to whisk until thickened, then sift in the flour and the baking powder, folding in just to combine.
Pour the mixture into the prepared tin, then scatter over the white chocolate chips or pieces. Using a teaspoon, dot the peanut butter in heaped teaspoonfuls across the batter, swirling them in with the back of your spoon.
Bake in the oven for 30–35 minutes or until softly set in the centre – the cake should wobble slightly in the middle when you gently shake the tin. Remove from the oven and set on a wire rack to cool completely in the tin before carefully removing and cutting into squares to serve.
Tip Do let the cooked brownie mixture cool completely in the tin before cutting up and serving, as when warm the brownies have a tendency to fall apart.
Cake may be my favourite dessert. There are few more satisfying ways to finish a meal than a forkful (or five) of a moist tender cake accompanied perhaps by its own special sauce or some whipped cream. One of the advantages of serving cake for dessert is that cake can be made ahead, either hours or a day beforehand, leaving you free to enjoy spending time with your friends without having to worry about making dessert. Rather than an icing topping, many of these cakes have a sauce that can be served warm and gently drizzled over individual slices as a sophisticated end to an evening meal. These cakes work wonderfully well with a liqueur or dessert wine. And if there are any leftovers they are lovely the next day with a cup of coffee. Whichever recipe you choose to make, these cakes are all worth saving room for!
06/ Dessert
Marmalade steamed pudding
Banana cake with butterscotch sauce
Butterscotch sauce
Walnut and orange cake
Orange crème anglaise
Chocolate fudge cake
Irish coffee cups
Chocolate and orange torte with caramelised oranges
Caramelised oranges
Cherry marzipan cake
After dinner mint cake
Swedish apple cake
Sticky toffee date cakes
Dulce de leche cake
Raspberry frangipane cake
Caramelised orange upside-down cake
Rhubarb and custard Swiss roll
Blueberry and coconut cake with lemon cream
Flourless fig, pine nut and ricotta cake
Caramelised pear upside-down cake
Petits gâteaux
Praline buttercream icing
Peanut buttercream icing
Chocolate buttercream and white chocolate drizzle icing
Berry buttercream icing
Intensely chocolatey beetroot brownies
Raspberry and white chocolate cheesecake
Marmalade steamed pudding
A steamed pudding is essentially a cake that has been steamed rather than baked. As you'd expect, the result is moist and dense rather than light and airy. It's a type of dish that I find supremely comforting – there's just something about dipping a spoon into a thick soft pudding soaked with syrup. Marmalade works particularly well, with a citrus tang that offsets the rich sponge to perfection.
**Prep time** : 15 minutes
**Cooking time** : 1 hour
**Ready in** : 1 hour 15 minutes
**Serves** : 8–10
200g (7oz) butter, softened, plus extra for greasing
200g (7oz) caster sugar
3 eggs, lightly beaten
250g (9oz) plain flour
1 tsp baking powder
½ tsp bicarbonate of soda
50ml (2fl oz) milk
125ml (4½fl oz) marmalade
Double or regular cream, whipped, or orange crème anglaise, to serve
1.2 litre (2 pint) heatproof pudding basin
Lightly butter the pudding basin. Cream the butter until soft in a large bowl or an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Sift in the flour, baking powder and bicarbonate of soda and fold into the batter until just incorporated, then mix in the milk. Spoon the marmalade into the base of the prepared pudding basin and tip in the sponge batter.
To prepare the pudding for steaming, first cut out a piece of foil and a disc of baking parchment at least 6cm (2½in) wider in circumference than the top of the basin, and secure the disc over the lip of the basin by tying a piece of string around it.
Select a saucepan not much larger than the basin. Place the pudding in the saucepan and carefully pour in enough hot water to come no higher than three-quarters of the way up the basin.
Cover the pan with a lid and simmer for approximately 1 hour or until it feels springy to the touch and a skewer inserted into the centre of the pudding comes out clean. Keep the water topped up in the saucepan during cooking, otherwise it may boil dry, burning the pudding.
Carefully remove the basin from the pan and turn out onto a warmed serving plate, allowing the marmalade to ooze down the sides. Serve with softly whipped cream or orange crème anglaise.
Jam steamed pudding
Replace the marmalade with the same quantity of jam, such as strawberry, raspberry or plum.
Banana cake with butterscotch sauce
I adore the way bananas work in cakes. Soft and sweet, their flavour permeates the sponge, adding moisture to make a lovely soft crumb. This cake is ideal for adding to lunchboxes but if you're eating it for dessert, try it with a spoonful of vanilla ice cream and this delicious butterscotch sauce drizzled over the top.
**Prep time** : 15 minutes
**Baking time** : 40–45 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
250g (9oz) plain flour
2 tsp baking powder
½ tsp bicarbonate of soda
250g (9oz) caster sugar
150ml (5fl oz) sunflower oil, plus extra for greasing
250g (9oz) peeled ripe bananas (2–3 in total), mashed
2 eggs
50ml (2fl oz) natural yoghurt
1 tsp vanilla extract
Vanilla ice cream or butterscotch sauce, to serve
23cm (9in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the cake tin with sunflower oil and line the base with a disc of baking parchment.
In a large bowl, sift together the flour, baking powder and bicarbonate of soda, then mix in the sugar. Using a separate bowl, whisk together the remaining ingredients until they are fully combined.
Pour the wet ingredients into the flour mixture and stir to mix together. Tip the batter into the prepared tin and cook for 40–45 minutes or until a skewer inserted into the centre of the cake comes out with just a small amount of moisture left on it.
Remove from the oven and allow to cool in the tin for 10 minutes, then use a small, sharp knife to loosen the edges and carefully remove the cake from the tin before leaving on a wire rack to cool down completely, unless you'd prefer to eat it warm.
Serve in slices on its own or with vanilla ice cream and butterscotch sauce.
Butterscotch sauce
**Prep time** : 5 minutes
**Cooking time** : 3 minutes
**Ready in** : 10 minutes
**Makes** : 500ml (18fl oz)
100g (3½oz) butter
250g (9oz) golden syrup
100g (3½oz) soft dark brown sugar
200ml (7fl oz) double or regular cream
1 tsp vanilla extract
¼ tsp salt
Place all the ingredients in a saucepan and set over a medium heat. Bring to the boil and continue to boil, stirring occasionally, for 3 minutes, then remove from the heat. This sauce can be reheated and it will keep very well in an airtight container in the fridge for up to six months.
Walnut and orange cake
Walnut and orange go surprisingly well together, the slight crunch of the nuts contrasting with the soft, citrusy sponge. Walnuts are better the fresher they are – left too long they can go a little bitter. If you can't get good ones, use pecans instead. The orange crème anglaise isn't essential to this recipe, but it goes so well with this cake, making the perfect end to a meal.
**Prep time** : 10 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour 20 minutes
**Serves** : 8–10
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs
Juice and finely grated zest of 1 orange
225g (8oz) plain flour, sifted
2 tsp baking powder
200g (7oz) walnuts, chopped
Orange crème anglais, to serve
23cm (9in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter and sugar mixture, beating all the time. Next beat in the orange juice and zest, then gently fold in the flour, baking powder and walnuts until fully incorporated.
Tip the batter into the prepared tin and bake for about 45 minutes or until well risen and golden brown on top, and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges using a small sharp knife and carefully remove the cake from the tin before leaving on a wire rack to finish cooling down, unless you're serving it warm.
Place on a cake plate and serve either warm with the orange crème anglaise, if you wish, or cooled down as a snack with a cup of tea or coffee.
Orange crème anglaise
**Prep time** : 15 minutes
**Cooking time** : 5–8 minutes
**Ready in** : 25 minutes
**Makes** : 600ml (1 pint)
500ml (18fl oz) milk
5 egg yolks
100g (3½oz) caster sugar
Pinch of salt
1 tsp finely grated orange zest
Pour the milk into a saucepan and bring to the boil. Meanwhile, whisk the egg yolks, sugar and salt together in a large bowl for a few minutes until pale and thick. Gradually whisk the hot milk into the beaten eggs and sugar and pour the whole mixture back into the saucepan, adding the orange zest.
Return to a low heat and cook gently for 5–8 minutes, stirring all the time, until the custard thickens slightly (enough to just coat the back of the spoon). If it boils it will scramble, so just keep it at a bare simmer. Once it's cooked, remove from the heat and pour into a warm jug to serve. If reheating, do so very gently on a low heat so that the eggs don't scramble.
Chocolate fudge cake
There are times when I don't want chocolate to be diluted or compromised by any other ingredient – I want a full-on onslaught of chocolate! At times like these, I turn to this cake. Here, every ingredient lives to serve the chocolate, giving it texture, moisture and fudginess. This cake makes an ideal dessert, especially when served with a spoonful of lightly whipped cream.
**Prep time** : 20 minutes
**Baking time** : 1 hour
**Ready in** : 2 hours 15 minutes
**Serves** :10–12
225g (8oz) dark chocolate, in drops or broken into pieces
225g (8oz) butter, plus extra for greasing
325g (11½oz) caster sugar
6 eggs, separated
200g (7oz) ground almonds
75g (3oz) plain flour, sifted
1 tsp baking powder
2 tsp vanilla extract
Pinch of salt
For the icing
275g (10oz) icing sugar
100g (3½oz) cocoa powder
125g (4½oz) butter
175g (6oz) caster sugar
25cm (10in) diameter spring-form or loose-bottomed cake tin with 6cm (2½in) sides
Preheat the oven to 160°C (325°F), Gas mark 3, then butter the sides of the cake tin and line the base with a disc of baking parchment. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the chocolate in a heatproof bowl set over a pan of simmering water. Leave just until melted, stirring occasionally, then set aside.
Either in a large bowl using a wooden spoon or in the bowl of an electric mixer using the paddle beater, cream the butter until soft. Add the sugar and beat until the mixture is light and fluffy. Add the egg yolks one at a time, beating well between each addition. Mix in the ground almonds and the melted chocolate, followed by the flour, baking powder and vanilla extract.
Tip the egg whites into a separate bowl, add a pinch of salt, and beat until stiff but still smooth in appearance – do not over-beat. Fold the egg whites into the cake mixture and then pour the mixture into the prepared tin.
Bake for about 1 hour or until a skewer inserted into the centre of the cake comes out clean but moist. Leave in the tin for 20 minutes, then loosen the edges of the cake using a small, sharp knife and remove the sides of the tin before carefully transferring to a serving plate to cool down fully.
While the cake is cooling, make the icing. Sift the icing sugar and cocoa powder into a mixing bowl. Place the butter, caster sugar and 100ml (3½fl oz) of water in a saucepan and set over a medium heat. Stir all the ingredients together until the butter is melted and the sugar has dissolved, then pour into the dry ingredients and mix together well.
When the cake has cooled, pour over the icing, allowing it to drizzle down the sides.
Irish coffee cups
Warm coffee cakes served in teacups, with a whiskey sauce and whipped cream, make a smart and playful dessert dish. The perfect pudding to serve at a dinner party, in fact, as both the cakes and sauce can be made ahead of time. The cakes are better if made on the same day, but the sauce can be made up to a couple of days ahead. Reheat the sauce in a saucepan and reheat the cakes in the oven (preheated to 130°C/250°F/Gas mark½) for 10 minutes or just until warmed through.
**Prep time** : 30 minutes
**Baking time** : 25–30 minutes
**Ready in** : 1 hour
**Makes** : 4 cakes
125g (4½oz) butter, softened, plus extra for greasing (optional)
125g (4½oz) soft light brown sugar
2 eggs
1 tbsp coffee essence (ideally Camp or Irel)
125g (4½oz) plain flour
1 tsp baking powder
150ml (5fl oz) double or regular cream, softly whipped, to serve
For the sauce
125g (4½oz) caster sugar
150ml (5fl oz) double or regular cream
50ml (2fl oz) whiskey (ideally Irish)
Four small ovenproof teacups and their saucers or four 6cm (2½ in) diameter ramekins
Preheat the oven to 180°C (350°F), Gas mark 4, and butter the insides of the cups if you intend to tip the cakes out of them to serve. Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together with the coffee essence in a small bowl for just a few seconds until mixed, then gradually add the eggs to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently to combine.
Divide the mixture between the teacups or ramekins and place these in a deep-sided ovenproof dish or roasting tin. Pour boiling water into the dish so it comes to halfway up the sides of the cups, then bake for 25–30 minutes or until the centre of each cake is springy to the touch.
While the cakes are cooking, make the sauce. Place the sugar in a saucepan with 75ml (3fl oz) of water and bring to the boil, stirring to dissolve the sugar as the mixture heats up. Turn the heat up to high and continue to boil – but without stirring – until the syrup goes a deep golden caramel colour. Cook the sauce just to the point when it starts smoking. (If it's not cooked enough you won't get a strong enough caramel flavour, but overcooked it will taste slightly bitter.) You may need to swirl the pan, rather than stirring the mixture, towards the end of caramelising to ensure it cooks evenly.
Turn down the heat and immediately stir in the cream and whiskey, taking care as it may bubble and spit, then set aside. You may need to whisk it for a few seconds on a low heat to help the caramel dissolve.
When the cakes are ready, remove from the oven and allow to cool for a couple of minutes. Place on saucers (if using cups), or tip them out onto warm plates. Serve with the warm whiskey sauce and a spoonful of whipped cream.
Chocolate and orange torte with caramelised oranges
This rich thick torte has a smooth texture, similar to a baked chocolate mousse, as it contains no ground almonds or flour. Chocolate and orange are a favourite combination of mine, the orange zest in the torte complemented by the caramelised oranges, which are optional but make a wonderful accompaniment to this dish.
**Prep time** : 25 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
150g (5oz) butter, plus extra for greasing
300g (11oz) dark chocolate, in drops or broken into pieces
5 eggs, separated
50g (2oz) golden caster sugar
Finely grated zest of 2 small oranges
Double or regular cream, whipped, fresh orange slices or caramelised oranges, to serve
20cm (8in) diameter spring-form or loose-bottomed cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the base and sides of the cake tin. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the butter and chocolate in a heatproof bowl and place over a pan of simmering water, Leave just until melted, stirring occasionally, then remove from the heat.
Using an electric hand-held beater or an electric food mixer, whisk together the egg yolks and half the sugar until pale and thick. Fold in the melted chocolate and the orange zest and mix together to combine.
Using the hand-held beater or food mixer (and a spotlessly clean bowl in either case), whisk the egg whites together with the remaining sugar until they are stiff and glossy. Gently fold this into the chocolate mixture, then tip into the prepared cake tin and bake for 20–25 minutes or until puffed up and slightly cracked around the edges.
Allow to sit in the tin for 20 minutes, then use a small, sharp knife to loosen the edges and remove the sides of the tin before carefully transferring to a serving plate.
Serve with whipped cream or fresh orange slices or with caramelised oranges.
Spicy chocolate torte
For a different, spicy twist to this dish, you could replace the orange zest with half a teaspoon each of ground cinnamon and ground cardamom, serving the torte with crème fraîche or softly whipped cream instead of the caramelised oranges.
Caramelised oranges
**Prep time** : 10 minutes
**Cooking time** : 5–8 minutes
**Ready in** : 45 minutes
4 oranges
125g (4½oz) caster sugar
Using a small, sharp knife and working over a bowl to catch the juices, peel all the oranges. First cut the ends off each orange, then carefully cut away the peel and pith in a spiral until you have a fully peeled orange with only flesh and no pith.
Squeeze all the juice from the orange trimmings into the bowl, then cut the peeled oranges crossways into slices 5mm (¼in) thick and add these to the orange juice.
Place the sugar and 75ml (3fl oz) of water in a saucepan and set over a medium heat. Stir the mixture to dissolve the sugar, then increase the heat and bring to the boil. Without stirring the mixture, continue to boil, uncovered, for about 5–8 minutes or until you have a deep whiskey-coloured caramel. As the mixture caramelises you can swirl the pan to ensure it browns evenly.
Pour in a further 75ml (3fl oz) of water and swirl the pan over a low heat to dissolve the caramel. Remove from the heat, then tip the orange slices and any juice into the caramel and allow to cool.
Transfer to a bowl and serve with the Chocolate and orange torte, with or without some lightly whipped cream.
Cherry marzipan cake
Cherries and marzipan are a divine combination, the almonds complementing the fruit in much the same way as raspberries in a Bakewell tart. When fresh cherries are in season, replace half the dried cherries with pitted fresh ones. (If you use all fresh, the mixture will be too wet.)
**Prep time** : 15 minutes
**Baking time** : 1 hour 10 minutes–1 hour 20 minutes
**Ready in** : 2 hours 15 minutes
**Serves** : 4–6
100ml (3½fl oz) sunflower oil, plus extra for greasing
4 eggs
150g (5oz) caster sugar
75ml (3fl oz) milk
1 tsp vanilla extract
225g (8oz) plain flour
2 tsp baking powder
150g (5oz) dried cherries
Icing sugar, for dusting
For the topping
225g (8oz) marzipan (to make it yourself)
1 tsp ground cinnamon
1 egg, beaten
25g (1oz) caster sugar
1 tbsp plain flour
20cm (8in) diameter spring-form or loose-bottomed cake tin
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides and base of the cake tin with a little sunflower oil. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the sunflower oil in a large bowl with the eggs, sugar, milk and vanilla extract and whisk together until smooth. Sift in the flour and baking powder and add the cherries, folding these in to combine. Tip the batter into the prepared tin, using a spatula to smooth the top.
Next make the topping. Crumble the marzipan into another bowl and add the remaining ingredients. Beat the mixture together just until blended, then spoon on top of the cake batter, spreading it out to cover the cake in an even layer. In cooking, some of the marzipan will sink to the bottom. Don't worry about this, however, as it creates a lovely effect.
Bake for between 1 hour and 10 minutes and 1 hour and 20 minutes or until a skewer inserted into the centre of the cake comes out clean. Allow to cool for 10 minutes, then use a small, sharp knife to loosen the edges before removing the sides of the tin and carefully transferring the cake to a plate.
Dust with icing sugar and cut into slices to serve.
After dinner mint cake
Using after dinner mints in a cake is a smart and fast way to infuse it with that cooling combination of mint and chocolate. It does make a nice evening's treat, but really you could eat this at any time of the day. The icing uses fresh mint, which gives a lovely green colour as well as a deliciously fresh flavour. You could add chocolate-covered mint leaves for an impressive, finishing touch, either using them to decorate this cake or serving them on their own with coffee after supper as your own version of after dinner mints.
**Prep time** : 25 minutes
**Baking time** : 40–45 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 6–9
125g (4½oz) dark chocolate, in drops or broken into pieces
50ml (2fl oz) milk
150g (5oz) butter, softened, plus extra for greasing
150g (5oz) caster sugar
3 eggs
200g (7oz) plain flour, sifted
2 tsp baking powder
100g (3½oz) after dinner mints (such as After Eight Mints), cut or broken into quarters
Chocolate mint leaves, to decorate (optional)
For the icing
15g (½oz) mint leaves, finely chopped
250g (9oz) icing sugar, sifted
A few drops of green food colouring (optional)
100g (3½oz) butter, softened
1 tbsp lemon juice
A few drops of mint essence (optional)
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the tin has a removable base, butter the sides of the cake tin and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the chocolate and milk in a heatproof bowl and set over a pan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat and set aside.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Continuing to beat, pour in the melted chocolate, then add the eggs, one at time, beating continuously. Add the flour, baking powder and mints and gently fold in just until mixed. (You may find that some of the mints sink to the bottom of the cake during cooking, but this won't matter.)
Tip the mixture into the prepared tin and bake for 40–45 minutes or until a skewer inserted into the centre of the cake comes out with no batter sticking to it (although there may still be traces of mint or chocolate).
Remove from the oven and allow to cool for 10 minutes in the tin before carefully removing the cake and leaving on a wire rack to cool down completely.
Meanwhile, make the icing. Place the chopped mint in a food processor with 2 tablespoons of the icing sugar and the food colouring and whiz until combined. Alternatively, you could beat these ingredients together in a bowl with a wooden spoon.
Add the butter, followed by the rest of the icing sugar and the lemon juice, and including some mint essence if you'd like to enhance the mint flavour, then whiz again until all the ingredients are combined and the mixture is smooth. Use a palette knife to spread the icing over the top of the cake, then decorate with the chocolate mint leaves, if using.
Swedish apple cake
Sweden is famous for its baked goods (see also the Swedish almond cake) and variations of this basic cake are to be found all over the country – buttery apples mixed with the fragrant cinnamon they love so much. Serve this cake on its own with a cup of coffee or as a dessert with softly whipped cream.
**Prep time** : 15 minutes
**Baking time** : 50 minutes
**Ready in** : 1 hour 45 minutes
**Serves** :8–10
3 cooking apples (such as Bramleys)
250g (9oz) plain flour
2 tsp baking powder
1 tsp ground cinnamon
150g (5oz) caster sugar
150g (5oz) butter, melted, plus extra for greasing
3 eggs, beaten
For the glaze
25g (1oz) butter
3 tbsp caster sugar
1 tsp ground cinnamon
25cm (10in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Peel the apples, then cut two into 1cm (½in) dice and one into thin slices/wedges about 5mm (¼in) thick.
Sift the flour, baking powder and cinnamon into a large bowl or into the bowl of an electric mixer, then add the sugar and mix together. Add the two diced apples, then the melted butter and the eggs, and mix all the ingredients together until combined.
Tip the batter into the prepared tin, then lay the apple slices on top, so that they fan out around the edge of the cake. Bake for about 50 minutes or until golden brown and a skewer inserted into the centre of the cake comes out clean.
While the cake is in the oven, make the glaze. Melt the butter in a pan over a medium heat, then add the sugar, stirring until it dissolves. Stir in the cinnamon, then remove from the heat and set aside.
When the cake has finished cooking, remove from the oven and allow to cool for 10 minutes. Loosen around the edges using a small, sharp knife and carefully remove the cake from the tin before transferring to a serving plate.
Reheat the glaze, then pour over the cake and allow to cool completely while soaking up the spicy syrup.
Sticky toffee date cakes
Similar to sticky toffee pudding, these cakes are so much faster to make. The rich toffee sauce is absolutely not optional – its thick buttery sweetness is what transforms the cakes into a fabulous dessert. It would be perfect for a dinner party, as both the cakes and the sauce can be made ahead and just reheated.
**Prep time** : 15 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 cakes
150g (5oz) pitted dates, roughly chopped
1 tsp bicarbonate of soda, sifted
75g (3oz) butter, softened
100g (3½oz) caster sugar
2 eggs
150g (5oz) self-raising flour
For the sauce
125g (4½oz) butter
125g (4½oz) soft dark brown sugar
75ml (3fl oz) double or regular cream
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Place the pitted dates in a saucepan with 200ml (7fl oz) of water and bring to the boil, then remove from the heat and mix in the bicarbonate of soda.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the creamed butter and sugar, beating all the time. Sift in the flour, then add the cooked dates and any water in the pan and carefully fold in until combined.
Divide the cake mixture between the muffin cases, filling each three-quarters full, then bake for 20–25 minutes or until well risen and deep golden in colour.
While the cakes are cooking, mix together the sauce ingredients in a saucepan and bring to the boil, stirring to dissolve the sugar. Boil for 2 minutes until thickened, then remove from the heat.
To serve, remove the paper cases from the muffins, place one (or two) on warmed plates and spoon over the warm sauce.
Tip Any of the sauce that you don't use can be stored in the fridge in an airtight container for up to six months.
Dulce de leche cake
This is a divine cake. It uses one of my favourite ingredients, dulce de leche, a sweet, thick caramel that I often crave just by itself. Here, ground almonds are added to a light sponge which is filled and iced with a gloriously rich, buttery icing made with the dulce de leche. Served as a dessert, this cake needs no accompaniment.
**Prep time** : 30 minutes
**Baking time** : 25 minutes
**Ready in** : 1 hour 30 minutes
**Serves** :8–12
175g (6oz) butter, softened, plus extra for greasing
100g (3½oz) soft light brown sugar
1 tsp vanilla extract
200g (7oz) dulce de leche (see the tip)
2 eggs
100g (3½oz) ground almonds
175g (6oz) self-raising flour
25g (1oz) flaked almonds, toasted (see the tip), to decorate (optional)
For the filling and icing
50g (2oz) butter, softened
175g (6oz) dulce de leche (see the tip)
1 tsp vanilla extract
150g (5oz) icing sugar, sifted
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of each cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar, vanilla extract and dulce de leche and beat the mixture for a good few minutes until the mixture is smooth and light.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the creamed butter mixture, beating all the time. Tip in the ground almonds, then sift in the flour, gently folding in until combined.
Divide the mixture between the tins and bake for about 25 minutes or until a skewer inserted into the centre of each cake comes out almost clean. Remove from the oven and allow to cool in the tins for 10 minutes, then loosen round the edges of each tin using a small, sharp knife and carefully remove the cake before leaving on a wire rack to cool down completely.
To make the filling and icing for the cake, beat the butter with the dulce de leche and vanilla extract until very soft, then gradually add the icing sugar and beat until smooth.
Place one cake upside down on a serving plate, then spread over a third of the icing. Place the second cake on top, right side up, then spread the rest of the icing over the top. Decorate with toasted flaked almonds, if you wish.
Tip Dulce de leche is boiled condensed milk. You can buy ready-boiled condensed milk, which is sold as dulce de leche or caramel. If you want to make your own, boil unopened tins of condensed milk for 2 hours and let the tins cool _completely_ before opening. I like to prepare a few tins at a time and store them in the cupboard, where they'll keep for months.
Raspberry frangipane cake
A frangipane is a rich buttery almond sponge that goes superbly well with raspberries. All this cake needs is a dollop of lightly whipped cream to make a delicious dessert. While not exactly virtuous, this cake has the advantage of being gluten free.
**Prep time** : 10 minutes
**Baking time** : 35–40 minutes
**Ready in** : 1 hour 10 minutes
**Serves** : 6–10
200g (7oz) butter, plus extra for greasing
200g (7oz) caster sugar
4 large eggs
Finely grated zest of 1 lemon
200g (7oz) ground almonds
25g (1oz) cornflour
200g (7oz) fresh raspberries
Icing sugar, for dusting
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add the eggs to the creamed butter mixture, beating all the time. Beat in the lemon zest, then add the ground almonds and cornflour and mix to combine.
Tip the batter into the prepared tin, smoothing the surface with a palette knife or spatula. Scatter half the raspberries evenly over the top and lightly push them into the batter. Bake for around 35–40 minutes or until the top is golden and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 15 minutes before carefully removing the cake and leaving on a wire rack to cool down fully.
Scatter the remaining raspberries over the top and dust with icing sugar, then cut into squares to serve.
Caramelised orange upside-down cake
Peeled slices of orange look lovely atop this upside-down cake, infusing it with their tangy citrus flavour. The batter combines the zest of the oranges and olive oil for a distinctive-tasting sponge. I like to serve this for dessert with a little custard or orange crème anglaise.
**Prep time** : 15 minutes
**Baking time** : 30–35 minutes
**Ready in** : 1 hour
**Serves** :6–8
25g (1oz) butter, softened
75g (3oz) soft light brown sugar
3–4 oranges
175g (6oz) plain flour
1½ tsp baking powder
½ tsp salt
125g (4½oz) caster sugar
100ml (3½fl oz) milk
100ml (3½fl oz) extra-virgin olive oil
2 large eggs
Custard or orange crème anglaise, to serve
26cm (10½ in) diameter ovenproof frying pan (measured across the top)
Preheat the oven to 180°C (350°F), Gas mark 4.
Spread the butter out in the frying pan and sprinkle evenly with the brown sugar. Finely grate the zest of 2 oranges and set aside.
Using a small, sharp knife and working over a bowl to catch the juices, peel all the oranges. First cut the ends off each orange, then carefully cut away the peel and pith in a spiral until you have a fully peeled orange with only flesh and no pith. Cut each peeled orange crossways into slices about 1cm (½in) thick. Arrange these in a single layer over the sugar in the pan.
Sift the flour, baking powder and salt into a large bowl and stir in the caster sugar. In a separate bowl, whisk together the milk, olive oil and eggs with the orange zest and the reserved juices. Tip this mixture into the dry ingredients and whisk together until well combined.
Pour the batter over the orange slices, taking care not to disturb them, then bake for 30–35 minutes or until a skewer inserted into the centre of the cake comes out clean. Remove from the oven and allow to sit for just 5 minutes before placing a plate over the top and then carefully (but quickly!) flipping the cake over and turning it out onto the plate.
Delicious served on its own or with custard or orange crème anglaise.
Rhubarb and custard Swiss roll
Rhubarb is such a great spring ingredient, coming into the shops at a time when there's little else growing locally and hardly any fresh fruit. In this dish, the natural sharpness of the rhubarb is tempered with sugar. It combines extraordinarily well with custard and when rolled up in a light, vanilla-scented sponge makes for a truly special cake.
**Prep time** : 25 minutes
**Baking time** : 12–15 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 6
Butter, melted, for greasing
4 eggs
125g (4½oz) caster sugar, plus 3 tbsp for sprinkling
2 tbsp warm water
1 tsp vanilla extract
125g (4½oz) plain flour, plus extra for dusting
For the filling
200g (7oz) rhubarb (about 2 stalks, trimmed), cut into 5mm (¼in) slices
125g (4½oz) caster sugar
200ml (7fl oz) milk
1 vanilla pod, split lengthways, or 2 tsp vanilla extract
3 egg yolks
15g (½oz) cornflour
100ml (3½fl oz) whipped double or regular cream (measured when whipped)
25 × 38cm (10 × 15in) Swiss roll tin
Preheat the oven to 190°C (375°F), Gas mark 5. Line the base of the Swiss roll tin with baking parchment, brush the base and sides of the tin with melted butter and dust with flour.
Using a hand-held electric beater or an electric food mixer, whisk together the eggs, caster sugar, water and vanilla extract until light and fluffy. Sift in the flour, about one-third at a time, and fold it into the mixture.
Carefully pour the mixture into the prepared Swiss roll tin and bake in the oven for 12–15 minutes or until the centre of the sponge is slightly springy to the touch and the edges have shrunk a little from the sides of the tin.
Take a piece of baking parchment slightly larger in size than the tin and spread out on a work surface. Sprinkle the paper evenly with caster sugar (this is to stop the cake sticking to the paper). Quickly flip the Swiss roll tin over onto the sugared paper, then carefully remove the tin and baking parchment from the bottom of the cake.
Place a clean, slightly damp tea towel over the cake while it cools – this will prevent it drying out and cracking when you roll it.
Meanwhile, make the filling. Place the rhubarb in a saucepan with 75g (3oz) of the caster sugar and 25ml (1fl oz) of water and place on a medium heat, stirring to dissolve the sugar. Bring to the boil and allow to boil, uncovered and stirring regularly, for 10–15 minutes or until the rhubarb is completely soft and the mixture is quite thick. Tip out onto a plate and allow to cool.
Next make the custard. Place the milk and the split vanilla pod (if using) in another pan and bring to the boil. Whisk the egg yolks (and vanilla extract, if using) with the remaining sugar (I like to use a hand-held electric beater for this) for a few minutes or until pale and light. Then briefly whisk in the cornflour.
Pour the hot milk and vanilla pod (if using) onto the egg mixture, whisking as you pour, then tip it all back into the saucepan and cook, stirring all the time, over a low heat for a few minutes or until it forms a thick custard.
Pour it into a bowl and allow to cool, remove the vanilla pod, then fold in the cooled rhubarb and the whipped cream – you can leave it slightly marbled (not fully mixed) if you prefer.
When the sponge is completely cold, spread over the rhubarb and custard mixture, then, with one of the short sides facing you, roll up the Swiss roll away from you and carefully transfer to a serving plate.
Sprinkle with a little extra caster sugar to finish, and cut into slices about 2cm (¾in) thick to serve.
Blueberry and coconut cake with lemon cream
Full of antioxidants and seriously healthy, blueberries are often hailed as a superfood. They are certainly good for you, but let's not overlook the fact that they taste pretty good too! Cooked in this cake, their natural sweetness is brought out, as each berry is a burst of fresh flavour within the soft sponge. The lemon zest and lemon curd cut through the sweet berries and the rich cream, providing a contrasting sharpness. This cake is divine as a dessert or served at teatime, eaten outside or at least with the sun shining through the windows.
**Prep time** : 30 minutes
**Baking time** : 20 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs, beaten
Finely grated zest of 1 lemon
200g (7oz) plain flour, sifted
2 tsp baking powder
25g (1oz) desiccated coconut
150g (5oz) fresh or frozen (and defrosted) blueberries
For the lemon cream
200ml (7fl oz) double or regular cream
100g (3½oz) lemon curd (to make it yourself)
75g (3oz) fresh blueberries
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of each cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Gradually add the eggs to the creamed butter mixture, beating all the time. Add the lemon zest, flour, baking powder and coconut, carefully folding them into the batter until fully incorporated, then fold in the blueberries.
Divide the mixture between the prepared cake tins, then bake for 20 minutes or until well risen and golden or until a skewer inserted into the centre of each cake comes out clean. Remove from the oven and place on a wire rack for 5 minutes, then loosen round the edges of each tin using a small, sharp knife and carefully remove each cake before leaving on a wire rack to cool down fully.
As the cakes cool, prepare the lemon cream. In a bowl, whip the cream until it forms firm peaks, then whip in the lemon curd until just stiff.
Place one cake, upside down, on a serving plate, then spread over half the lemon cream mixture. Add the other cake, right side up, on top, then spread over the remaining lemon cream and scatter with the blueberries.
Flourless fig, pine nut and ricotta cake
This festive, flourless concoction is quite a sophisticated dish, the pine nuts providing a savoury contrast to the sweetness of the dried fruit. This cake is moist enough to serve without the icing, but the fresh, tangy taste of the ricotta complements it really well.
**Prep time** : 30 minutes
**Baking time** : 45 minutes
**Ready in** : 2 hours
**Serves** : 10–12
250g (9oz) butter, plus extra for greasing
250g (9oz) caster sugar
150g (5oz) ricotta cheese
Finely grated zest of 2 lemons
4 eggs
150g (5oz) pine nuts, lightly toasted (see the tip)
100g (3½oz) pitted dates, chopped
200g (7oz) dried figs, chopped
75g (3oz) ground almonds, plus 1 tbsp for sprinkling
For the icing
100g (3½oz) ricotta cheese
200g (7oz) icing sugar, sifted
Juice of ½ lemon
23cm (9in) diameter cake tin
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides and line the base of the cake tin with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar, ricotta and lemon zest and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for just a few seconds until mixed, then gradually add them to the creamed butter mixture, beating all the time. Add 50g (2oz) of the toasted pine nuts, together with the remaining ingredients, and fold into the batter until well mixed in.
Tip the mixture into the prepared cake tin and bake for about 45 minutes or until a skewer inserted into the centre of the cake comes out moist but clean. Remove from the oven and allow to sit for 10 minutes, then loosen around the edges using a small, sharp knife. Scatter a tablespoon of ground almonds over the top of the cake, then place a plate on top and carefully turn the cake upside down and out onto the plate (the almonds will help stop the cake sticking to the plate you're turning with). Then transfer the cake to a serving plate and leave to cool down completely.
While the cake is cooling, make the icing by beating together all the ingredients until smooth. Once the cake is cool, drizzle over the icing and sprinkle with the remaining pine nuts.
Tip To toast pine nuts, scatter them over a baking tray in a single layer and toast them in the oven (preheated to 180°C/350°F/Gas mark 4) for 4–5 minutes or until golden brown, or place them in a frying pan and toast over a medium-low heat for a similar length of time. In either case, shake the nuts every so often to prevent them burning on one side.
Caramelised pear upside-down cake
Upside-down cakes are such a comforting dessert and they're supremely simple to put together. This dish includes whole pieces of pear, which caramelise on the bottom of the tin during baking, as well as the grated flesh of the fruit, which is stirred into the cake batter. The effect is to infuse the sponge with moisture and that perfect pear flavour. A little whipped cream is the only accompaniment this cake needs.
**Prep time** : 10 minutes
**Baking time** : 40 minutes
**Ready in** : 1 hour
**Serves** : 8
75g (3oz) butter
125g (4½oz) brown sugar
4 pears (about 375g/13oz)
150g (5oz) plain flour
2 tsp ground cinnamon
1 tsp bicarbonate of soda
½ tsp salt
175g (6oz) caster sugar
4 eggs
100ml (3½fl oz) sunflower oil
1 tsp finely grated orange zest
Double or regular cream, whipped, to serve (optional)
26cm (10½in) diameter ovenproof frying pan (measured across the top)
Preheat the oven to 180°C (350°F), Gas mark 4.
Put the butter into the frying pan and melt over a low heat, then remove from the hob and sprinkle over the brown sugar.
Grate one pear, including the skin (but avoiding the pips and core), and set aside. Peel the remaining three pears, slice each into quarters and cut away the core. Arrange the pear quarters, cut side down, in a spiral pattern in the bottom of the pan.
Sift the flour, cinnamon, bicarbonate of soda and salt into a bowl and mix in the caster sugar. In a separate bowl, whisk together the eggs, sunflower oil, orange zest and grated pear, then mix this into the dry ingredients, whisking to combine.
Pour the batter into the frying pan, taking care not to disturb the spiral pattern, then bake for about 40 minutes or until a skewer inserted into the centre of the cake comes out clean. Allow to cool on a wire rack for just 5 minutes, then place a plate on top and carefully turn the cake upside down (the right way up, in other words) and out onto the plate.
Serve while warm or at room temperature, with a little softly whipped cream if you wish.
Petits gâteaux
I adore these neat little cakes. Each square looks so pretty: covered in icing with four tidy corners, they're quite Parisian in appearance. I've given a few variations – just choose whichever flavours or colours you prefer. The colourful berry buttercream cakes would be lovely to hand round at a particularly special teatime such as a baby shower, while the hazelnut praline variation would be perfect served as an after-dinner treat.
**Prep time** : 20 minutes (excluding the icing)
**Baking time** : 25–30 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 16 squares
175g (6oz) butter, softened
175g (6oz) caster sugar
3 eggs
175g (6oz) plain flour
2 tsp baking powder
20cm (8in) square cake tin with 5cm (2in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently to combine.
Tip the batter into the prepared cake tin and bake for 25–30 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes before carefully removing the cake and transferring to a wire rack to finish cooling.
Meanwhile, make one of the buttercream icings of your choice.
When the cake has cooled down, transfer to a chopping board and use a sharp, serrated knife to cut it into 16 equal-sized squares.
Use a palette knife to ice the top and sides of each of the 16 squares. If you'd like an especially neat look, fill a jug or bowl with boiling water and keep dipping the palette knife into the hot water before you apply it to the buttercream so that you can carefully level off the edges.
Chocolate petits gâteaux
For a chocolate version of these squares, add 25g (1oz) of cocoa powder to the sponge mixture along with the flour and ice with chocolate buttercream icing, drizzling with white chocolate icing to decorate.
Peanut butter petits gâteaux
Add 50g (2oz) of smooth or crunchy peanut butter to the cake mixture along with the sugar and then ice with peanut buttercream icing.
Praline buttercream icing
**Prep time** : 20 minutes
**Cooking time** : 6–8 minutes
**Ready in** : 45 minutes
**Makes** : enough for 16 sponge squares
450g (1lb) butter, softened
2 tsp vanilla extract
600g (1lb 5oz) icing sugar
6 tbsp milk
For the hazelnut praline
200g (7oz) caster sugar
200g (7oz) hazelnuts
First make the praline. Line a baking tray with baking parchment and set aside, then place the sugar in a frying pan and scatter the hazelnuts over the sugar.
Place the pan over a medium heat to allow the sugar to caramelise, not stirring but swirling the pan every so often to ensure it caramelises evenly. Cook for about 6–8 minutes or until the sugar has completely melted and is a deep golden colour, swirling the pan again so that the hazelnuts are coated in the caramel.
Transfer the coated nuts to the prepared baking tray and spread apart with a fork. Once cool, break up the praline using your hands, then place in a food processor and whiz until it resembles slightly coarse breadcrumbs. Alternatively, place in a plastic bag and crush using a rolling pin.
Next make the buttercream icing. Beat together the butter and vanilla extract until very soft in a large bowl using a hand-held electric beater or in the bowl of an electric mixer. Gradually sift in the icing sugar, beating all the time, until it is fully incorporated. Pour the milk in, a little at a time, and continue to beat. Add just enough so the mixture is softly spreadable, then whisk until it is light and fluffy.
Spread the buttercream over the sponge squares (see the main recipe), then sprinkle the praline all over the iced cakes.
Peanut buttercream icing
**Prep time** : 10 minutes
**Makes** : enough for 16 sponge squares
250g (9oz) smooth or crunchy peanut butter
200g (7oz) butter, softened
250g (9oz) cream cheese
2 tsp vanilla extract
750g (1lb 10oz) icing sugar
2–3 tbsp milk
Cream together the peanut butter, butter, cream cheese and vanilla extract in a large bowl using a wooden spoon, hand-held electric beater or in an electric food mixer. Gradually sift in the icing sugar and continue beating until fully incorporated, then add the milk and mix in to combine. If you'd like it especially fluffy and you are using a food mixer, continue to beat for a few minutes, then spread over the sponge squares (see the main recipe).
Chocolate buttercream and white chocolate drizzle icing
**Prep time** : 15 minutes
**Makes** : enough for 16 sponge squares
250g (9oz) milk chocolate, in drops or broken into pieces.
450g (1lb) butter, softened
900g (2lb) icing sugar
50g (2oz) cocoa powder, sifted
For the drizzle icing
75g (3oz) white chocolate, in drops or broken into pieces
75g (3oz) butter, softened
100g (3½oz) icing sugar, sifted
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
To make the buttercream, place the milk chocolate in a heatproof bowl and set over a saucepan of gently simmering water. Leave just until melted, stirring occasionally, then remove from the heat and set aside.
Cream the butter until very soft in a large bowl using a hand-held electric beater or in the bowl of an electric food mixer. Sift in the icing sugar and cocoa powder and continue to mix. Then add the melted chocolate and beat in until fully incorporated.
Spread the buttercream over the sponge squares (see the main recipe), then make the white chocolate drizzle icing.
Place the white chocolate in a clean heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Add the butter and icing sugar to the bowl and beat together until well mixed. Add the icing to the piping bag or freezer bag and carefully pipe a simple design, such as a zigzag pattern or spirals, onto each square (see Decorating ideas).
Berry buttercream icing
**Prep time** : 10 minutes
**Makes** : enough for 16 sponge squares
225g (8oz) hulled strawberries or raspberries, plus 16 strawberry halves or 16 raspberries, to decorate
1½ tbsp lemon juice
275g (10oz) butter, softened
900g (2lb) icing sugar
Place the strawberries or raspberries and lemon juice in a food processor and whiz for about a minute to purée. Push through a sieve into a large bowl then set aside.
Cream the butter until very soft in a large bowl using a hand-held electric beater or in the bowl of an electric mixer. Gradually sift in the icing sugar, beating all the time, until it is fully incorporated, then pour in the fruit purée, a little at a time and beating continuously.
Spread the buttercream over the sponge squares (see the main recipe), then place a strawberry half or a raspberry on top of each cake to decorate.
Variation
For a pale purple icing instead of a pink one, make the buttercream as above, replacing the pureed strawberries or raspberries with the same quantity of blackberries or blueberries.
Intensely chocolatey beetroot brownies
It might seem strange adding beetroot to a brownie, but surely no stranger than adding carrot to a cake? The beetroot gives sweetness and moisture while the flavour only enhances the intense chocolate flavour. You can cook the beetroot from scratch for this recipe, or use leftover cooked beetroot, if you have any. Do make sure to let these cakes cool before serving as they are too soft when warm, though if you have to try one straight from the oven, who can blame you? If you're serving them for dessert, they are heavenly with a dollop of softly whipped cream.
**Prep time** : 15 minutes (excluding cooking the beetroot)
**Baking time** : 30–35 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 16 brownies
275g (10oz) beetroot, leaves removed
250g (9oz) dark chocolate, in drops or broken into pieces
250g (9oz) butter, softened, plus extra for greasing
300g (11oz) caster sugar
3 eggs
75g (3oz) plain flour
50g (2oz) cocoa powder
½ tsp baking powder
Pinch of salt
20cm (8in) square cake tin with 5cm (2in) sides
Prepare the beetroot by trimming the stalks to about 2cm (¾in) from the end, then wash carefully under a cold tap. Do not scrub them – simply rub off any dirt with your fingers. You don't want to damage the skin or tails otherwise the beetroot will 'bleed' while cooking.
Place the beetroot in a saucepan and cover with cold water. Bring to a simmer, then cover with a lid and continue to simmer for between 30 minutes and 1 hour, depending on the size and age of the beetroot. They are cooked when their skins rub off easily and a knife can easily be inserted into the centre of each beetroot.
When they are cooked, rub off the skins and discard, then cut into chunks. Place in a food processor and whiz for a few minutes to form a smooth purée, then set aside.
Preheat the oven to 180°C (350°F), Gas mark 4. If the cake tin has a removable base, butter the sides and line the base with a square of baking parchment, otherwise line the base and sides of the tin.
Place the chocolate in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat and set aside.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for just a few seconds until mixed, then gradually add these to the creamed butter mixture, beating all the time. Beat in the puréed beetroot and melted chocolate, then sift in the remaining ingredients, folding these in until fully combined.
Tip the batter into the prepared cake tin, smoothing the top with a palette knife or spatula. Bake in the oven for 30–35 minutes or until the centre of the cake is almost set but still wobbles when you gently shake the tin. Remove from the oven and place on a wire rack to cool. Allow to cool completely in the tin before carefully removing the cake and cutting into squares to serve. If you can resist them for that long, these will keep in an airtight container for a few days.
Raspberry and white chocolate cheesecake
With nearly a kilogram of cream cheese and a lot of Oreo cookies, this cheesecake is uncompromisingly full on. This is not a cake for a weekday lunchtime – it demands an occasion or celebration as a valid excuse for such outrageous decadence. But should you have any of this left over after the celebration, it is sure to brighten up a more ordinary day!
**Prep time** : 30 minutes
**Baking time** : 1 hour 30 minutes
**Ready in** : 4 hours
**Serves** : 10–14
175g (6oz) white chocolate, finely chopped
350g (12oz) fresh or frozen (and defrosted) raspberries
900g (2lb) cream cheese
300g (11oz) caster sugar
25g (1oz) plain flour, sifted
4 eggs
2 tbsp single or regular cream
2 tsp vanilla extract
Icing sugar, for dusting
For the biscuit base
250g (9oz) Oreo cookies, coarsely broken
75g (3oz) butter, melted, plus extra for greasing
23cm (9in) diameter spring-form cake tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, and butter the base and sides of the cake tin. Make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
First make the biscuit base. Place the cookies in a food processor and whiz until they form coarse crumbs. Add the melted butter and whiz together until mixed. Alternatively, place the cookies in a plastic bag and crush them using a rolling pin, then add to the pan with the melted butter and mix together to combine.
Tip the crumb mixture into the prepared cake tin, spreading it out evenly to cover the bottom of the tin and pressing it down firmly.
Next make the cheesecake topping. Place the white chocolate in a heatproof bowl and set this over a pan of just simmering water, stirring occasionally until melted.
While the chocolate is melting, purée the raspberries in the food processor (having first cleaned the bowl), or simply mash them with a fork, then push them through a sieve, discarding the seeds.
Using a hand-held electric beater or an electric food mixer, mix the cream cheese and sugar together until smooth and fluffy. Beat in the flour, followed by the eggs, one at a time, then beat in the cream, vanilla extract, puréed raspberries and melted white chocolate.
Pour the mixture into the prepared tin, spreading it evenly over the biscuit base, then bake on the lowest shelf in the oven for about 1 hour 30 minutes or until the cheesecake is softly set in the middle. It should wobble a little when you gently shake the tin and the top will be slightly cracked.
Allow to cool down fully in the tin, then place in the fridge and leave to set for 2 hours.
To serve, loosen around the edges using a small, sharp knife, then unclip and remove the sides of the tin. Use a palette knife or metal fish slice to loosen the bottom of the cake from the base and ease the cake onto a plate. (If you don't feel brave enough to slide the cheesecake off the base of the tin, just leave it on and place it like this on the serving plate.) Dust with icing sugar to serve.
To a child, the birthday cake can be the most important part of their party. Yet to the baker, it can cause the most amount of stress. It really doesn't need to be stressful though, and I've devoted this chapter to showing you how to make children's party cakes that are simple and don't take too much time. There are also some more elaborate recipes for novelty cakes, and as long as you carefully follow each step these cakes are not difficult to make. I have specially created the recipes so that you don't need specific shaped tins, but can use cake tins that you probably have in your cupboard anyway. Children's cakes are such fun to decorate so you can let your imagination take over. Minimalism is certainly not the aim here, and you can adapt the cakes using your own ideas. Whether it is because of their inventive decoration or their layers of sponge and ice cream, these cakes are sure to impress your children and their friends.
07/ Children's
Arctic roll
Hot chocolate fairy cakes
Chocolate and toffee ice-cream cake
Layered chocolate and caramel cake
Chocolate hedgehog cake
Almond hedgehog cake
Strawberry cupcakes
Strawberry ice-cream cake with strawberry coulis
Banana ice-cream cake with butterscotch sauce
Chocolate ice-cream cake with chocolate sauce
Teddy bear cake
Chocolate teddy bear cake
Snowman cake
American frosting
Princess cake
Mini mango cupcakes
Cake pops
Coloured glacé icing
Sprinkles
Peanut butter pops
Butterfly cake
Coloured buttercream icing
Berry buttercream icing
Coloured glacé icing
Letter or number cake
Coloured buttercream icing
Berry buttercream icing
Coloured glacé icing
Coloured sugar
Chocolate letter or number cake
Peanut butter letter or number cake
Castle cake
Coloured buttercream icing
Berry buttercream icing
Chocolate castle cake
Decorating ideas
Fondant icing
Fast fondant flowers
Fondant roses
Fondant ghosts
Fondant football shirts
Fondant butterflies
Fondant birds
Fondant snowmen
Fondant dinosaurs
Fondant penguins
Arctic roll
This nostalgia-drenched dessert will always be a retro classic, but it has seen a recent revival. My amazing assistant Josh Heller, to whom this recipe is dedicated, has particularly fond memories of it. As he says, 'To an eight-year-old boy it's probably the pinnacle of all desserts.' It's very simple to make and will be appreciated by both the adults who remember it from their youth and the youth who are eating it for the first time! It may miss the point a little, but it's delicious served with some fresh raspberries. Once made, it will keep in a container in the freezer for up to a month, though it's best eaten on the day when the sponge is still fresh.
**Prep time** : 30 minutes
**Baking time** : 15 minutes
**Ready in** : 1 hour
**Serves** : 6–8
1 litre (1¾ pints) vanilla ice cream
Butter, for greasing
100g (3½oz) caster sugar, plus extra for dusting
4 eggs
100g (3½oz) plain flour
1 tsp baking powder
200g (7oz) raspberry jam
23 × 30cm (9 × 12in) Swiss roll tin
Remove the ice cream from the freezer and allow to soften for 10 minutes until mouldable but not melting. If your kitchen is very warm, put the ice cream in the fridge and leave for about half an hour instead. Scoop the ice cream out in blobs to form a line on a large sheet of baking parchment and use the paper to roll it into a sausage 23cm (9in) long and 5cm (2in) in diameter. Roll up the ice cream in the paper, twist the ends of the baking parchment to close and quickly return to the freezer, leaving it there to freeze until solid (30 minutes at least).
Meanwhile, preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the Swiss roll tin and line the base with baking parchment.
Place the sugar and eggs in a large bowl or an electric food mixer and whisk together for several minutes until light, thickened and tripled in size. Sift the flour and baking powder over the mixture, then fold in just until combined.
Tip into the prepared tin and smooth the top using a spatula or palette knife, then bake for 10–14 minutes or until it is just springy to the touch in the centre. Remove from the oven and leave the tin to cool down on a wire rack for 5 minutes.
Dust the sponge with caster sugar, then place a sheet of baking parchment slightly larger than the tin on top of the sponge. Quickly, but carefully, flip the sponge over onto the paper, remove the tin and baking parchment from the base of the sponge and place on the wire rack to finish cooling.
To assemble the Arctic roll, lift the cooled sponge together with the baking parchment on to your work surface and spread the raspberry jam all over the sponge.
Take the ice-cream cylinder from the freezer and unwrap the paper from around it, then place on the sponge, aligning with the short edge 2cm from the end. Carefully roll up the ice cream in the sponge. Dust with more caster sugar and cut into slices to serve. If you're not eating the Arctic roll immediately, it can be kept in the freezer and slightly defrosted in your kitchen for about 20 minutes before eating.
Hot chocolate fairy cakes
My children love these little chocolate cakes. A hot chocolate in a bun complete with creamy icing and mini marshmallows. You could scatter with chocolate sprinkles rather than dusting with cocoa powder, if you'd prefer.
**Prep time** : 15 minutes
**Baking time** : 15–20 minutes
**Ready in** : 50 minutes
**Makes** : 12 fairy cakes
125g (4½oz) butter, softened
125g (4½oz) caster sugar
2 eggs
125g (4½oz) plain flour
1½ tsp baking powder
2 tbsp cocoa powder
2 tbsp milk
For the icing
75g (3oz) butter, softened
300g (11oz) icing sugar, sifted
1 tbsp boiling water
To decorate
36 mini marshmallows
1 tsp cocoa powder, sifted, for dusting
12-cup bun tray and 12 bun cases
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Preheat the oven to 180°C (350°F), Gas mark 4, and line the bun tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the creamed butter mixture, beating all the time. Sift in the flour, baking powder and cocoa powder, pour in the milk and fold in gently to combine.
Divide the mixture between the paper cases, filling each about two-thirds full, then bake for 15–20 minutes or until well risen and lightly springy to the touch. Remove from the oven and allow to cool for 5 minutes before removing from the tin and transferring to a wire rack to finish cooling.
As the cakes cook or cool, make the icing. Beat together all the ingredients for a couple of minutes or until light and fluffy.
Place a heaped tablespoon of icing on each cake or use the piping bag or freezer bag to a pipe a rosette of icing onto the cake (see Cake essentials). Add three mini marshmallows to the centre of each cake, then dust with cocoa powder.
Chocolate and toffee ice-cream cake
Two layers of mocha sponge sandwich a layer of vanilla ice cream with swirls of caramel. The effect of the ice-cream layer inside the cake is impressive to party guests, be they eight or eighty. This cake is definitely best served on the day of eating, though it will keep in the freezer for up to a month. If you're serving it from the freezer, make sure to take it out about 15 minutes before eating just so the cake softens slightly. If you have time, chocolate curls (see Decorating ideas) make a gorgeous decoration for this cake.
**Prep time** : 20 minutes
**Baking time** : 25–30 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–8
100g (3½oz) dark chocolate, in drops or chopped into pieces
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
3 eggs
1 tsp coffee essence (ideally Camp or Irel)
175g (6oz) plain flour
2 tsp baking powder
400ml (14fl oz) good-quality vanilla ice cream
200g (7oz) dulce de leche (see the tip)
Cocoa powder, sifted, for dusting (optional) or Chocolate curls, to decorate
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line the base of each tin with a disc of baking parchment.
Place the chocolate in a large heatproof bowl and set over a saucepan of gently simmering water. Stir occasionally until melted, then set aside.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the creamed butter mixture, beating all the time. Next beat in the coffee essence and melted chocolate, then sift in the flour and baking powder, folding it in just to combine.
Divide the mixture between the prepared tins, then bake for 25–30 minutes or until springy to the touch and a skewer inserted into the centre of each cake comes out clean. Remove from the oven and allow to cool on a wire rack for 10 minutes, then loosen around the edges with a small, sharp knife and carefully remove each cake from its tin before leaving on a wire rack to cool down completely.
Wash one of the tins, then line completely with a double layer of cling film. Fill with the ice cream and use a spoon to swirl in the dulce de leche, then cover the top with another layer of cling film and place in the freezer for about 30 minutes or until completely frozen.
To serve, place one cake upside down on a stand or cake plate. Carefully remove the ice-cream cake from the tin, peeling away all the cling film. Place on top as the next layer, then place the remaining cake over this. Dust with cocoa powder or top with chocolate curls (see Decorating ideas) and serve before the ice cream has a chance to melt!
Layered chocolate and caramel cake
One of my guilty pleasures is a tin of boiled condensed milk, or actually condensed milk in any shape or form. Once boiled, the sugars in the milk caramelise to make a thick toffee-like sauce. It's perfect for cooking or eating straight from the tin (with or without a spoon!). It's also possible to buy ready-boiled condensed milk, which is sold in tins or jars as dulce de leche – or you can make your own (see the tip). Here the dulce de leche is combined with chocolate in the sponge and the icing for three layers of undiluted pleasure!
**Prep time** : 45 minutes
**Baking time** : 25–30 minutes
**Ready in** : 2 hours
**Serves** : 8–10
300g (11oz) butter, plus extra for greasing
300g (11oz) caster sugar
6 eggs
1½ tsp vanilla extract
300g (11oz) self-raising flour
2 tsp baking powder
1 × 397g tin of dulce de leche (to make it yourself, see the tip)
2 tbsp cocoa powder
For the icing
350g (12oz) milk chocolate, in drops or broken into pieces
150ml (5fl oz) double or regular cream
Pinch of salt
Three 20cm (8in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line each base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs and vanilla extract together in a separate bowl for a few seconds or just until mixed, then gradually add the eggs to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently to combine.
Tip a third of the batter into one of the prepared cake tins and divide the other two-thirds evenly between two separate bowls. Add 4 tablespoons of the dulce de leche to one of the bowls and stir it in, then tip into one of the two empty cake tins. Sift the cocoa powder into the remaining bowl of cake mixture and stir it in, then put the batter into the remaining tin.
Place all three cakes in the preheated oven and bake for 25–30 minutes or until springy to the touch and a skewer inserted into the centre of each sponge comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges of each cake using a small, sharp knife and carefully remove from the tin before leaving on a wire rack to cool down fully.
While the sponges are cooking, or while they are cooling down, make the icing. Place the chocolate in a heatproof bowl and set over a saucepan of barely simmering water. Leave just until melted, stirring occasionally, then remove from the heat and stir in the cream and salt. Place in the fridge to chill just until spreadable.
When the sponges have cooled down, place the dulce de leche cake upside down on a cake plate. Spread over half of the dulce de leche left in the tin, then place the chocolate cake on top, right side up. Spread over the rest of the dulce de leche from the tin and add the vanilla cake, right side up, as the final layer. Finally, spread the chocolate icing all over the cake using a palette knife dipped into boiling-hot water for a smooth finish.
Chocolate hedgehog cake
Like the Teddy bear cake, this is another creative cake that looks really cute. I've used chocolate fingers for the spikes, though if you prefer you can use chocolate flakes. I've also provided a vanilla and almond variation for those children who aren't so keen on chocolate (do they exist?).
**Prep time** : 1 hour
**Baking time** : 50–55 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 8–12
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
3 eggs
2 tsp vanilla extract
175g (6oz) plain flour, plus extra for dusting
2 tsp baking powder
25g (1oz) cocoa powder
2 tbsp milk
For the icing
125g (4½oz) milk chocolate, in drops or broken into pieces
225g (8oz) butter, softened
450g (1lb) icing sugar
25g (1oz) cocoa powder
To decorate
2 × 125g packets of chocolate fingers
1 × glacé cherry or 1 red Smartie
2.4 litre (4 pint), 23cm (9in) diameter or similar capacity ovenproof pudding basin/Pyrex bowl (measured across the top)
Large round or oval plate
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the inside of the pudding basin bowl and dust with flour.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs and vanilla extract together in a small bowl, then gradually add the eggs to the creamed butter mixture, beating all the time. Sift in the flour, baking powder and cocoa powder and fold in gently, then add the milk and mix gently to combine.
Tip the batter into the prepared pudding basin/bowl and bake for 50–55 minutes or until springy to the touch and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool for 5 minutes in the basin bowl. Use a palette knife to loosen the edges of the cake, then place a wire rack upside down over the basin bowl and carefully flip over. Lift off the basin bowl and leave the cake to cool completely on the wire rack.
As the cake is cooking, or while it's cooling down, make the icing. Place the chocolate in a heatproof bowl and set over a saucepan of gently simmering water. Leave just until melted, stirring occasionally, then remove from the heat and set aside.
Cream the butter until soft in a large bowl or in an electric food mixer, then sift in the icing sugar and cocoa powder and continue to mix. Pour in the melted chocolate and beat in until well combined.
Place the cake domed side down on a work surface. If the base of the cake is uneven, use a bread knife to level it off. Spread a little of the icing 2–3mm (⅛in) thick over the base of the cake, then cut the cake in half so that you have two half domes.
Sandwich together the iced sides to form a roughly oval shape, then spread a little icing on the large plate and place the cake on top so it sticks down.
Spread the remaining icing all over the cake, and use a palette knife to shape the face and the snout of the hedgehog (see photograph). Use a fork to add texture to the snout and the back of the hedgehog, then decorate the iced cake.
Cut the chocolate fingers in half crossways and arrange pointing backwards as the spines of the hedgehog. To make the eyes, cut a chocolate finger in half and insert into the cake with the uncut surfaces pointing outwards. Place the glace cherry or Smartie at the tip of the snout to make a nose.
Almond hedgehog cake
**Prep time** : 1 hour
**Baking time** : 50–55 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 8–12
For the icing
150g (5oz) butter, softened
2 tbsp milk
2 tsp vanilla extract
200g (7oz) icing sugar
To decorate
100g (3½oz) flaked almonds, toasted (see the tip)
3 × brown Smarties
Mix the cake as in the main recipe, omitting the cocoa powder and using only 1 tablespoon of milk, then bake.
Meanwhile, make the icing. Using a hand-held electric beater or an electric food mixer, beat together the butter, milk and the vanilla extract until very soft. Then gradually sift in the icing sugar, beating all the time, until it is fully incorporated and light and fluffy.
Assemble and ice the cake as above, then, rather than using chocolate fingers, insert toasted flaked almonds at an angle all over the cake for the spines. Place two of the Smarties on the cake for the eyes and one at the tip of the snout to make a nose.
Strawberry cupcakes
These cupcakes are deceptively simple to make and they taste divine. They are so pretty with the naturally pink icing full of sweet strawberry flavour. If you'd like to make these for a celebration, then you could try decorating them with the crystallised rose petals (see Decorating ideas).
**Prep time** : 30 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour 15 minutes
**Makes** : 12 cupcakes
125g (4½oz) butter, softened
200g (7oz) caster sugar
3 eggs
150ml (5fl oz) milk
1 tsp vanilla extract
300g (11oz) plain flour
3 tsp baking powder
½ tsp salt
For the icing
150g (5oz) fresh or frozen (and defrosted) strawberries
1 tbsp lemon juice
175g (6oz) butter, softened
600g (1lb 5oz) icing sugar, sifted
Strawberries, halved, to decorate
12-cup muffin tray and 12 muffin cases
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Preheat the oven to 160°C (325°F), Gas mark 3, and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Beat in the milk, followed by the vanilla extract, then sift in the flour, baking powder and salt and fold in gently to combine.
Divide the mixture between the muffin cases, filling each case about two-thirds full. Bake for 20–25 minutes or until well risen, lightly golden in colour and springy to the touch. Allow to cool for 5 minutes, then remove from the tin and place on a wire rack to cool completely.
While the cupcakes are cooking, or while they are cooling down, make the icing. Place the strawberries and lemon juice in a food processor and whiz for about a minute to purée them. Push through a sieve into a bowl, then set aside.
In a separate bowl, cream together the butter and icing sugar until combined, then gradually add the fruit purée, beating continuously.
Spread the icing onto the cooled cupcakes using a palette knife or the back of a spoon, or pipe the icing onto the cakes (see Cake essentials) and decorate each cake with half a strawberry.
Variation
Try replacing the strawberries for the icing with the same quantity of raspberries, blackberries or blueberries – which can be either fresh or frozen (and then defrosted). If using blackberries or blueberries the colour will no longer be pink but a lovely purple shade. Decorate each cake with a blackberry, raspberry or blueberry.
Strawberry ice-cream cake with strawberry coulis
This is the recipe to go for when your children (or you!) don't just want ice cream with cake, they want ice-cream cake! With its layers of ice cream and sponge all frozen together, it looks very impressive – especially when served with the strawberry coulis – but is actually quite simple to put together. I've included two further variations on the theme, but you could create your own versions. As long as you stick to the same basic quantities for the cake and the ice cream, it's really up to you what flavours you choose. While this cake can be made up to a month ahead and kept in the freezer, it is best eaten on the day when the sponge is fresh.
**Prep time** : 20 minutes
**Baking time** : 20 minutes
**Ready in** : 4 hours
**Serves** : 10–12
Butter, for greasing
4 eggs
125g (4½oz) caster sugar
125g (4½oz) self-raising flour
400ml (14fl oz) vanilla ice cream
400ml (14fl oz) strawberry ice cream
For the strawberry coulis
250g (9oz) fresh or frozen (and defrosted) strawberries, plus sliced fresh strawberries to decorate
1–2 tbsp caster sugar
Juice of ½ large lemon
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line the base of each tin with a disc of baking parchment.
Place the eggs and sugar in a large bowl or an electric food mixer. Using a hand-held electric beater or the food mixer, whisk on a high speed for about 3 minutes or until the mixture has roughly tripled in volume, then sift in the flour and fold in gently just until combined.
Divide the mixture between the prepared cake tins, then bake for about 20 minutes or until golden on top and a skewer inserted into the centre of each cake comes out clean.
Remove from the oven and allow to cool in the tins for 10 minutes, then loosen around the edges of each cake using a small, sharp knife and carefully remove from the tin before leaving on a wire rack to cool down fully.
Wash and dry both tins, then line completely with a double layer of cling film. Remove the ice cream from the freezer and allow to soften until it is easy to scoop. Fill one of the tins with vanilla ice cream and the other with strawberry. Cover the top with cling film then leave in the freezer for about 30 minutes or until completely frozen.
Meanwhile, make the strawberry coulis. Whiz all the ingredients together in a food processor or blender.
Taste the mixture, adding more sugar or lemon juice if necessary, then push through a sieve. Pour into a serving jug and keep in the fridge (for up to 3–4 days) until you're ready to serve.
To assemble the cake, place one sponge upside down on a work surface or serving plate. Take the vanilla ice cream out of the freezer and remove from the tin, peeling away all the cling film. Place this ice cream layer on top of the sponge, then place the second half of the cake on top, right side up. Next take the strawberry ice-cream layer out of the freezer and remove it from the tin, peeling away the cling film as before. Place on top of the cake and either serve immediately or wrap in cling film and place in the freezer where it will keep for up to a month.
To serve, decorate the top of the cake with sliced strawberries and serve in slices accompanied by the strawberry coulis.
Banana ice-cream cake with butterscotch sauce
**Prep time** : 20 minutes
**Baking time** : 25 minutes
**Ready in** : 4 hours
**Serves** : 10–12
For the butterscotch sauce
100g (3½oz) butter
250g (9oz) golden syrup
100g (3½oz) soft dark brown sugar
200ml (7fl oz) single or regular cream
1 tsp vanilla extract
¼ tsp salt
To decorate
1 banana, peeled and sliced
Make the cake as in the main recipe opposite, replacing the strawberry ice cream with banana ice cream and decorating the top with slices of banana instead of strawberries.
While the cake tins of ice cream are setting in the freezer, make the butterscotch sauce (to replace the strawberry coulis). Place all the ingredients in a saucepan and set over a medium heat, allowing the butter to melt and the sugar to dissolve. Bring to the boil and continue to boil, stirring regularly, for 3 minutes, then remove from the heat and set aside. The sauce will keep for up to 3 months in the fridge.
Reheat the sauce when ready to serve and pour it warm over slices of the finished ice-cream cake.
Chocolate ice-cream cake with chocolate sauce
**Prep time** : 20 minutes
**Baking time** : 25 minutes
**Ready in** : 4 hours
**Serves** : 10–12
For the chocolate sauce
150ml (5fl oz) double or regular cream
150g (5oz) dark or milk chocolate, or 75g (3oz) of each, in drops or broken into pieces, plus extra, grated, or chocolate curls, to decorate
½ tsp vanilla extract (optional)
Make the cake as in the main recipe, replacing the strawberry ice cream with chocolate ice cream and decorating the top with grated chocolate or chocolate curls instead of sliced strawberries.
While the cake tins of ice cream are setting in the freezer, make the chocolate sauce (to replace the strawberry coulis). Pour the cream into a saucepan and bring to the boil, then remove from the heat. Add the chocolate and stir until it melts, then stir in the vanilla extract, if using. Keep the sauce in the fridge (for up to a week) until you're ready to serve. Reheat gently to serve, pouring it over slices of the finished ice-cream cake.
Teddy bear cake
I love this cake as it looks so impressive and yet doesn't require any special tins, just a few rounds of sponge and a bit of creativity in putting things together. You can be as playful as you'd like with the decorations. You'll see that I've given a couple of alternative versions as well – a chocolate teddy and a snowman, which would be perfect for a winter party. The cake can be prepared up to two days in advance and iced either while fresh or on the day of the party. You'll need to store it somewhere cool but not in the fridge.
**Prep time** : 1 hour
**Baking time** : 1 hour 10 minutes
**Ready in** : 2 hours 45 minutes
**Serves** : 18–22
400g (14oz) butter, softened, plus extra for greasing
400g (14oz) sugar
8 eggs
400g (14oz) plain flour
4 tsp baking powder
For the icing
450g (1lb) butter, softened
900g (2lb) icing sugar, sifted
A few drops of food colouring, such as pink, red, blue or yellow
To decorate
6–7 small sweets (such as chocolate buttons, Smarties or Jelly Tots) for the eyes, nose and buttons
1 liquorice lace for the mouth
25cm (10in) diameter cake tin and 18cm (7in) diameter cake tin
6-cup muffin tray and 5 muffin cases
Large plate or 40 × 50cm (16 × 20in) cake board
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tins and line the base of each tin with baking parchment, then line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a separate bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently to combine.
Fill the muffin cases and smaller tin three-quarters full with the batter, then put the rest of the mixture into the larger cake tin (it should be about half full).
Bake the muffins and the smaller cake for 20–25 minutes or until they have risen and are golden brown on top. The larger cake will take about 35 minutes.
Remove each cake from the oven when it is ready and allow to cool in the tin for 10 minutes. Then loosen around the edges using a small, sharp knife and carefully remove each cake from its tin before leaving on a wire rack to cool down completely. Allow the muffins to cool in the tin for 5 minutes, then remove from the tray and place on a wire rack to cool.
Place the large cake right side up on the lower part of the large plate or cake board to make the teddy's body, then place the smaller cake directly above for the head (like a snowman). Arrange four muffins around the body for the arms and legs, then cut the remaining muffin in half vertically and place on top of the head for ears.
To make the icing, beat together all the ingredients for a couple of minutes or until light and fluffy, adding more food colouring if you want a deeper shade.
Using a palette knife or spatula, spread the icing all over the teddy bear. If you want it to look smooth, dip the palette knife regularly into a jug or bowl of boiling-hot water.
Next decorate the cake. Add two sweets (chocolate buttons, Smarties or Jelly Tots) for the eyes and one for the nose. Using the liquorice lace, make a rounded 'W' right under the nose, with the centre of the 'W' elongated to resemble the mouth. Place 3–4 sweets down the front of the body for the teddy's buttons, or cover the whole body in sweets if you like.
Chocolate teddy bear cake
**Prep time** : 1 hour
**Baking time** : 1 hour 10 minutes
**Ready in** : 2 hours 45 minutes
**Serves** : 18–22
For adding to the sponge
2 tbsp cocoa powder
For the icing
250g (9oz) milk (or dark) chocolate, in drops or broken into pieces.
450g (1lb) butter, softened
900g (2lb) icing sugar
50g (2oz) cocoa powder
Sift in the cocoa powder to the cake mixture as you add the flour, then make and assemble the uniced cakes into the teddy's body, with the muffins for arms, legs and ears, as in the main recipe above. Then make the chocolate icing.
Place the chocolate in a heatproof bowl and set over a saucepan of gently simmering water. Stir occasionally just until melted and set aside to cool until no warmer than tepid.
Cream the butter until soft in a large bowl or in an electric food mixer. Sift in the icing sugar and cocoa powder and continue to mix. Then add the melted chocolate and beat in until fully incorporated.
Using a palette knife or spatula, spread the icing all over the teddy bear, dipping the knife or spatula into boiling-hot water for a smooth finish.
Decorate in the same way as in the main recipe, though for greater colour contrast you might want to use white chocolate buttons or Smarties for the eyes and nose.
Snowman cake
The icing I've suggested here, cream cheese whipped with icing sugar, is very easy to make, but if you're feeling a little more adventurous you could try the American frosting instead. If making this cake in advance, it's best to store it in the fridge (because of the cream-cheese icing), but make sure to take it out of the fridge 1–2 hours before serving.
**Prep time** : 1 hour
**Baking time** : 1 hour 10 minutes
**Ready in** : 2 hours 45 minutes
**Serves** : 18–22
350g (12oz) butter
350g (12oz) caster sugar
7 eggs
350g (12oz) plain flour
3½ tsp baking powder
For the cream-cheese icing
450g (1lb) cream cheese
450g (1lb) icing sugar, sifted
To decorate
10–12 sweets resembling coal (such as chocolate buttons or chocolate-covered toffees) for the eyes and buttons
Baby carrot for the nose (just for decoration!)
Chocolate fingers for the arms
Follow the Teddy bear cake method but using these sponge quantities. Omit the muffins and just make two round cakes, with one larger as in the main recipe. Place the cooled cakes on the plate or cake board to form the snowman, then make the icing.
In a bowl, beat together the cream cheese and icing sugar until well mixed. Then spread all over the snowman using a palette knife, making little peaks in the icing to resemble snow (see the last step in the American frosting). Alternatively, make up the American frosting and ice the snowman with this.
To decorate the snowman, add the sweets for the eyes and buttons. Insert the baby carrot for the nose and add chocolate fingers for the arms.
American frosting
This delicious soft icing makes a beautiful 'snow' that can be smoothed or peaked, but you'll need to follow the instructions carefully. Quick and accurate decisions are necessary in judging when the icing is ready and it must then be applied to the cake immediately. If the icing is not cooked enough, it will still taste good, but will not dry out properly on the outside. If cooked too much, it will be difficult to spread.
**Prep time** : 7 minutes
**Ready in** : 7 minutes
**Makes** : enough for one large cake
4 large egg whites
250g (9oz) caster sugar
Pinch of salt
Make sure the cakes are ready before you start, as this icing begins to set very quickly. Place all the ingredients in a heatproof bowl, add 2 tablespoons of cold water and set over a saucepan of simmering water. (The bowl should sit snugly over the pan with its base high enough above the water that it does not come into contact with it.)
Whisk slowly using a balloon whisk until the sugar has completely dissolved and the mixture is foamy. Continue to heat and whisk until the mixture reaches 60°C (140°F) when measured with a sugar thermometer – this will take about 4 minutes. If you don't have a thermometer, you can gauge whether the mixture is ready by how it feels and looks: it should be hot to the touch, glossy white in appearance and starting to thicken.
Quickly remove the bowl from the pan and pour the mixture into the bowl of an electric food mixer fitted with the whisk attachment. Alternatively, whisk in the original bowl using a hand-held electric beater. Whisk on a high speed for about 3–5 minutes or until the frosting is very thick, glossy and has cooled.
Spread quickly over the cakes with a palette knife, regularly dipping into a jug of boiling-hot water. The icing sets very quickly at this stage, so speed is essential. If you'd like a 'peaked' snowy look, dip the flat of your palette knife into the icing and quickly lift it away, creating a peak. Continue in this fashion across the cake, if you like.
Princess cake
This is a novelty cake and a half. Every bit of effort you put into making this cake will be so worth it when you see the birthday girl's face light up with delight. The cake can be prepared up to two days in advance and iced either while fresh or on the day of the party. You'll need to store it somewhere cool but not in the fridge.
**Prep time** : 45 minutes
**Baking time** : 25–35 minutes
**Ready in** : 2 hours
**Serves** : 18–20
400g (14oz) butter, softened, plus extra for greasing
400g (14oz) caster sugar
8 eggs, beaten
400g (14oz) plain flour, plus extra for dusting
4 tsp baking powder
For the icing (to cover the 'skirt' of the doll)
450g (1lb) butter, softened
3 tsp vanilla extract
¼ tsp red or pink food colouring (or a colour of your choice)
750g (1lb 10oz) icing sugar, sifted
2 tbsp milk
To decorate
Sprinkles, coloured sugar crystals (to make them yourself,), edible glitter, icing flowers or other decorations of your choice
Three 18cm (7in) diameter cake tins
12cm or 16cm (5in or 6½in) diameter ovenproof pudding basin/Pyrex bowl (measured across the top)
One doll, clothed from the waist up
One 50cm (20in) long ribbon
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tins and dust with flour, then line the base of each tin with a disc of baking parchment. Next butter and flour the inside of the pudding basin.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is pale and fluffy, then gradually add the eggs to the creamed butter mixture, beating all the time.
Sift together the flour and baking powder and add to the wet ingredients, folding in just until the mixture comes together. Half fill the pudding basin or bowl with the batter, then divide the rest between the three tins, making a slight hollow in the centre of the cake in the tins so that the cake does not have a peak when cooked, making it easier to assemble the pile of cakes.
Bake in the oven for 25–35 minutes (the cooking time will depend on how evenly you've divided your mixture between the bowl and tins) or until golden on top and a skewer inserted into the centre of each cake comes out clean.
Remove from the oven and leave to cool for 5 minutes, then loosen around the edges of each cake using a small, sharp knife. Place a plate upside down on top of the cake and carefully flip over. Remove the basin/bowl or tin and any baking parchment, then transfer to a wire rack and leave to cool down completely.
As the cakes cool, wrap the doll from just above her waist down to her toes in cling film (so you can re-use her later).
Make the icing by beating the butter until very soft, then add the vanilla extract, food colouring and half the icing sugar. Continue to beat until light and fluffy, then add the rest of the sugar, followed by the milk, and beat again.
Pile the three cake-tin sponges on top of each other and place the basin/bowl cake (upside down) on top. Then, using a long-bladed knife, cut out a hole about 2cm (¾in) wide through the centre of the cakes, going from the top of the basin/bowl cake down through to the bottom of the third cake-tin sponge. (You could use the discarded cake for the Cake pops, if you wish.) This will be the hole that the doll stands in.
Separate the cakes again and place one of the cake-tin sponges on a cake stand or plate. Spread with some of the icing (as if very generously buttering toast), then top with the second cake. Spread with the same amount of icing and add the third cake, then spread with more icing and top with the upturned basin/bowl cake.
Insert the doll into the hole running through the cakes (they should come up to her hips). Fill a jug with boiling water and place a palette knife in it. Now you can start icing all around the cakes and up to the doll's waist to make a rather voluminous skirt! Spreading it with a palette knife (dipped in the hot water) will help to give a smooth finish.
Decorate with icing flowers, edible glitter, hundreds and thousands, Smarties, or whatever you like (see Children's decorating ideas), but do it now before the outer layer of the icing begins to dry a little. Tie a ribbon around the doll's waist (where the icing meets her clothes) and serve.
Mini mango cupcakes
These cupcakes are a taste of the tropics with the luscious flavour of mango and the rich crunch of toasted coconut. They are a cute little size, which children will love and parents can eat with coffee after the party.
**Prep time** : 20 minutes
**Baking time** : 10–12 minutes
**Ready in** : 40 minutes
**Makes** : 24 cupcakes
100g (3½oz) butter, softened
100g (3½oz) caster sugar
2 eggs
100g (3½oz) plain flour
1 tsp baking powder
25g (1oz) desiccated coconut
25g (1oz) coconut flakes, toasted (see the tip), to decorate
For the icing
50g (2oz) mango (about ¼ mango), peeled
200g (7oz) icing sugar, sifted
50g (2oz) butter, softened
24-cup mini bun tray with 24 mini bun cases
Piping bag with a 4 or 5mm (¼in) nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Preheat the oven to 180°C (350°F), Gas mark 4, and line the bun tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and add the desiccated coconut, then fold in gently to incorporate.
Divide the batter between the paper cases, using 2 teaspoons – one to scoop the mixture off the other. Bake for 10–12 minutes or until golden brown and springy to the touch, then allow to cool in the tray for 5 minutes before placing on a wire rack to cool down fully.
As the cakes cool, make the icing. First roughly chop the mango and liquidise using a hand blender or food processor, or simply chop it finely, and push through a sieve to make a thick purée. In a large bowl, beat together the icing sugar, butter and mango purée until fluffy.
Add to the piping bag or freezer bag, if using, then pipe the icing onto the cakes (see Cake essentials), or apply the icing with a teaspoon if you prefer, and sprinkle over the toasted coconut flakes.
Cake pops
Cake pops are such a great treat for children's parties. They're not hard to make but take a little time between each step. They're absolutely worth it, though, and the young guests will definitely think so! They're also a great way of using any leftover cake – any type of sponge cake would do, in fact. Just having balls of cake on sticks can be fun, but the decoration ideas are really what count here. I've suggested a few variations for icing and decorating your cake pops, but be creative and use whatever decorations spring to mind.
**Prep time** : 45 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 12 cake pops
175g (6oz) milk or white chocolate, in drops or broken into pieces
350g (12oz) Madeira cake or any basic sponge cake, chocolate or plain
For the coloured chocolate coating
400g (14oz) white chocolate, in drops or chopped into pieces
A few drops of your choice of food colouring (optional)
12 lollipop sticks
Polystyrene block into which to insert the cake pops for coating and decorating (optional)
First line a baking sheet with baking parchment.
Place the chocolate in a heatproof bowl and set over a saucepan of simmering water, stirring occasionally just until melted. Crumble the cake into the melted chocolate, and stir until well mixed.
Use your hands to roll the mixture into balls each about as large as a golf ball. Insert a lollipop stick into each ball, place on the lined baking sheet and leave in the fridge for about 30 minutes to firm up.
While the cake pops are in the fridge, make the chocolate coating. Place the chocolate in a heatproof bowl and set over a pan of simmering water, stirring occasionally just until melted. Remove from the heat and stir in the food colouring (if you'd like to colour the white chocolate), adding more or less to obtain the exact shade you'd like.
Remove the cake pops from the fridge, then dip each one into the chocolate coating and either insert upright into the polystyrene block or place back on the lined baking sheet. (Inserting the cake pops in the polystyrene block will ensure they have a perfectly round shape rather than a flat side from lying on the baking parchment.) Place somewhere cool, but not the fridge, and leave for 20–30 minutes or just until the chocolate has set.
Before the coating sets on the cake pops, you can roll them in one of the sprinkle options . Alternatively, once the coating has set, you could decorate with the coloured glacé icing.
Coloured glacé icing
**Prep time** : 10 minutes
**Makes** : 100g (3½oz)
100g (3½oz) icing sugar, sifted
A few drops of your choice of food colouring (in a contrasting colour to the chocolate icing)
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
When the coloured chocolate coating has set on the cake pops (see the main recipe), make the glacé icing. In a bowl, mix together the icing sugar and food colouring with just enough water (½–1 tablespoon) to make an icing the consistency of thick double cream.
Spoon the icing into the piping bag or freezer bag, then carefully pipe shapes onto the cake pops – such as spirals, zigzags, faces, or a name or number – before inserting in the polystyrene block, or placing on the lined baking sheet, to set.
Tip For more colours, once you've mixed the icing sugar and water, divide the icing into two or three separate bowls and add different types of food colouring to each, then pipe or drizzle the different icings onto the cake pops for a lovely multi-coloured effect.
Sprinkles
Sugar crystals
100g (3½oz) coloured (or multi-coloured) sugar crystals (or hundreds and thousands)
Popping candy
100g (3½oz) popping candy
Crushed honeycomb
100g (3½oz) honeycomb (see Decorating ideas), crushed into a powder but with a few larger chunks remaining
After dipping the cake pops in the coloured chocolate coating, you can roll them in one of these toppings. Sprinkle them onto a plate, then roll each of the cake pops in the mixture before leaving to set.
Peanut butter pops
**Prep time** : 45 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : 12 cake pops
100g (3½oz) milk or dark chocolate, broken into pieces
100g (3½oz) smooth or crunchy peanut butter
350g (12oz) Madeira cake or any basic sponge cake
To decorate
400g (14oz) dark, milk or white chocolate, in drops or broken into pieces
150g (5oz) salted peanuts, chopped
Put the chocolate and peanut butter into a heatproof bowl and place over a saucepan of simmering water, stirring occasionally just until melted. Crumble the cake into the melted chocolate and peanut butter mixture, and stir until well mixed.
Roll the mixture into balls and insert a lollipop stick into each ball, as in the main recipe, before chilling in the fridge for 30 minutes.
Meanwhile, melt the chocolate for decorating in a heatproof bowl over a pan of simmering water, stirring occasionally just until melted. Place the chopped peanuts on a plate next to the bowl of chocolate.
Remove the balls from the fridge, then dip each one in the chocolate and roll in the chopped peanuts. Insert either into the polystyrene block or place on the lined baking sheet (see main recipe) and leave somewhere cool, but not the fridge, for 20–30 minutes or just until the chocolate has set.
Tip Here are some recipes that leave leftover cake – perfect for making into cake pops:
Muscovado Madeira cake
Barbie cake
Butterfly cake
Letter or number cake
Castle cake
Butterfly cake
This pretty cake is so easy to make: it uses just one round cake tin and there's no difficult trimming. You can decorate it however you like, too. I've provided a few ideas; but you can get as creative as you like, playing with different colours and decorations.
**Prep time** : 45 minutes
**Baking time** : 30–35 minutes
**Ready in** : 2 hours
**Serves** : 10–14
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs
225g (8oz) plain flour
2½ tsp baking powder
To decorate
3 chocolate fingers (or 4 chocolate matchsticks) for the body of the butterfly
Edible glitter, hundreds and thousands or coloured sugar (to make it yourself, and 2 liquorice wheels for the antennae
23cm (9in) diameter cake tin
Large plate or 40 × 50cm (16 × 20in) cake board
Preheat the oven, 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently to combine. Tip the mixture into the prepared tin and bake for 30–35 minutes or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges using a small, sharp knife and carefully remove the cake before leaving on a wire rack to cool down fully. To make the butterfly, cut the cake in half down the middle, then make two cuts at an angle from the centre line, slicing through the cake. This will give you two large and two small wedges. Cut the edges off the circle and turn these around so the edges are facing each other, and angle the four wedges so they look like wings.
Next ice the cake using either the coloured buttercream icing, dyed whatever colour you like, or the berry buttercream icing for a lovely shade of pale pink or purple. First place the cake on the plate or cake board, sticking it down with a small blob of icing on each corner, then use a palette knife to ice the cake all over. (Have a bowl or jug of boiling-hot water to hand for dipping your knife into to make it easier to spread.)
Once the cake is iced, place a few chocolate fingers or matchsticks aligned in the centre of the cake to make up the body. You can then use the coloured glacé icing to pipe shapes onto each wing (see Cake essentials) before carefully shaking over edible glitter, coloured sugar or hundreds and thousands. Place two liquorice wheels at the top of the cake for the butterfly's antennae.
Coloured buttercream icing
**Prep time** : 10 minutes
**Makes** : 475g (1lb 1oz)
200g (7oz) butter, softened
1–2 tbsp milk
1 tsp vanilla extract
275g (10oz) icing sugar
A good few drops of a food colouring of your choice
Using a hand-held electric beater or an electric food mixer, beat together the butter, milk and vanilla extract until very soft. Then gradually sift in the icing sugar, beating all the time, until it is fully incorporated. Continue to beat until very soft – if beating in the food mixer, continue whisking the mixture until it is light and fluffy – then whisk in just enough food colouring to give your desired shade for the icing.
Berry buttercream icing
**Prep time** : 10 minutes
**Makes** : 900g (2lb)
150g (5oz) fresh or frozen (and defrosted) strawberries or raspberries
1 tbsp lemon juice
175g (6oz) butter, softened
600g (1lb 5oz) icing sugar, sifted
Place the strawberries or raspberries and lemon juice in a food processor and whiz for about a minute to puree. Push through a sieve into a large bowl, then set aside. In a separate bowl, cream together the butter and icing sugar until combined, then gradually add the fruit puree, beating continuously.
Coloured glacé icing
**Prep time** : 5 minutes
**Makes** : 250g (9oz)
250g (9oz) icing sugar, sifted
A drop or two of food colouring (in a contrasting colour to the main icing)
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
In a bowl, mix together the icing sugar and food colouring with just enough water (½–1 tablespoon) to make an icing that you can use to drizzle over the cake (the consistency of thick double cream).
Use this to fill the piping bag or freezer bag and decorate the cake with swirls or stripes (see Decorating ideas). It looks lovely if you use the icing to draw the outline of the wings to accentuate them. If you'd rather not use a piping bag, you could use a teaspoon instead. Place the bowl of icing next to the cake and dip your spoon into it, then lift the spoon out and drizzle on stripes or swirls. For a lovely multi-coloured effect, see the tip.
Letter or number cake
There's something so lovely about having a cake created just for you. Whether it gives the age of the child or the first initial of their name, it is a unique creation that will make them feel really special. Of course, this cake isn't just for children, it's perfect for adults celebrating a big birthday, or even retirement. The basic cake is a simple sponge baked in a roasting tin, which gives a nice large rectangle from which you can cut out any number or letter you like. This does result in quite a lot of trimmings, but these can be put to good to use in Cake pops!
**Prep time** : 20 minutes
**Baking time** : 35–45 minutes
**Ready in** : 2 hours
**Serves** : 8–14 (depending on the number/letter)
350g (12oz) butter, softened, plus extra for greasing
350g (12oz) caster sugar
6 eggs
350g (12oz) plain flour
3½ tsp baking powder
To decorate
Sprinkles, hundreds and thousands, edible glitter (optional)
25 × 35cm (10 × 14in) roasting tin with 5cm (2in) sides (approximate size)
Large plate or 40 × 50cm (16 × 20in) cake board
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the roasting tin and line the base and sides with baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Whisk the eggs together in a bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently until mixed.
Tip the batter into the prepared roasting tin and bake on the lower shelf in the oven for 35–45 minutes (depending on the size of the tin) or until a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges using a small, sharp knife. Place a large wire rack upside down on top of the cake and carefully turn over. Remove the tin and baking parchment and leave to finish cooling on the wire rack.
When the cake has cooled, transfer to a chopping board. Draw a template of your letter or number on a piece of paper and cut this out. Place on top of the cake and use a sharp knife to cut around the template.
Now you can ice the cake. The amount of icing you need may vary slightly depending on the letter or number (a number '8' will need more icing than the figure '1', for instance). Select one of the coloured icings or, if you prefer, use the icing from the chocolate or peanut butter versions of this cake. First add a little of the icing to the base of the cake – a small blob on each corner to hold it in place – before placing it on the plate or board. Spread the icing all over the cake using a palette knife, dipping regularly into a jug or bowl of boiling-hot water for a smooth effect.
Once your cake is covered in icing, you can decorate it, playing with colours and designs. You could scatter over sprinkles, for instance, or hundreds and thousands. Another option would be to sift over some coloured sugar – ideally in a contrasting colour to the icing). I like to cut out dots, stars or stripes with paper, lay them on the cake, then sprinkle over the coloured sugar or edible glitter. This looks great when you take the paper away. Alternatively, use coloured glacé icing to pipe designs onto the iced cake (see Cake essentials). Or for additional ideas.
Coloured buttercream icing
**Prep time** : 10 minutes
**Makes** : 700g (1½lb)
200g (7oz) butter, softened
450g (1lb) icing sugar, sifted
50ml (2fl oz) milk
1 tsp vanilla extract
A few drops of your choice of food colouring
Cream the butter and sugar together in a large bowl or in an electric mixer until fluffy and light. Continuing to beat, gradually add the milk with the vanilla extract and food colouring – adding just enough to achieve the desired shade.
Berry buttercream icing
**Prep time** : 10 minutes
**Makes** : enough for one large cake
150g (5oz) fresh or frozen (and defrosted) strawberries, raspberries, blueberries or blackberries
1 tbsp lemon juice
175g (6oz) butter, softened
600g (1lb 5oz) icing sugar, sifted
This is an alternative coloured icing which uses fruit for colour and flavour. Choose strawberries or raspberries for a gorgeous pink colour and blueberries or blackberries for a light purple shade. Place the fruit and lemon juice in a food processor and whiz for about a minute to purée. Push through a sieve into a large bowl and then set aside. In a separate bowl or using an electric food mixer, cream together the butter and icing sugar until combined, then gradually add the fruit purée, beating continuously.
Coloured glacé icing
**Prep time** : 5 minutes
**Makes** : 100g (3½oz)
100g (3½oz) icing sugar, sifted
A few drops of your choice of food colouring
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
In a bowl, mix together the icing sugar and food colouring with just enough water (½–1 tablespoon) to make an icing the consistency of thick double cream.
Use this to fill the piping bag or freezer bag and carefully pipe stripes, swirls or even writing onto the iced cake (see Cake essentials). If you'd rather not use a piping bag, you could use a teaspoon instead. Place the bowl of icing close to the cake and dip your spoon into it, then lift the spoon out and let some icing drip off. Hold the spoon close to the iced surface of the cake as you carefully drizzle on stripes, zigzags or swirls. For a lovely multi-coloured effect, see the tip.
Coloured sugar
**Prep time** : 10 minutes
**Makes** : 125g (4½oz)
125g (4½oz) icing sugar
3–4 drops of food colouring (ideally in a contrasting colour to your choice of icing)
Place the icing sugar and food colouring in a sieve set over a bowl. Push the icing sugar through the sieve into the bowl, then sift it again. Keep doing this maybe three or four times until the food colouring has tinted the sugar. Use this for sifting over the iced cake as suggested in the main recipe.
Chocolate letter or number cake
**Prep time** : 20 minutes
**Baking time** : 35–45 minutes
**Ready in** : 2 hours
**Serves** : 8–14 (depending on the number/letter)
For adding to the sponge
2 tbsp cocoa powder
For the icing
125g (4½oz) milk chocolate, in drops or broken into pieces
225g (8oz) butter, softened
450g (1lb) icing sugar
25g (1oz) cocoa powder
To decorate
100g (3½oz) white chocolate, in drops or broken into pieces
Make the cake as in the main recipe, adding the cocoa powder with the flour.
To make the icing, first place the chocolate in a heatproof bowl and set over a saucepan of gently simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Cream the butter until very soft in a large bowl using a hand-held electric beater or in an electric food mixer, then sift in the icing sugar and cocoa powder and continue to mix. Pour in the melted chocolate and beat in until well mixed. Use a palette knife to spread the icing all over the cake.
Place the white chocolate in a heatproof bowl and set over a pan of simmering water. Leave just until melted, stirring from time to time, then remove from the heat and place next to the cake. Dip a teaspoon or dessertspoon into the chocolate, then drizzle on stripes, zigzags or swirls.
Peanut butter letter or number cake
**Prep time** : 20 minutes
**Baking time** : 35–45 minutes
**Ready in** : 2 hours
**Serves** : 8–14
For adding to the sponge
75g (3oz) smooth or crunchy peanut butter
For the icing
125g (4½oz) smooth or crunchy peanut butter
100g (3½oz) butter, softened
125g (4½oz) cream cheese
1 tsp vanilla extract
375g (13oz) icing sugar
1–2 tbsp milk
Salted peanuts (optional), to decorate
Make the cake as in the main recipe, adding the peanut butter with the sugar to the creamed butter.
To make the icing, cream together the peanut butter, butter, cream cheese and vanilla extract in a large bowl using a hand-held electric beater or in an electric food mixer. Gradually sift in the icing sugar and continue to beat until all the ingredients are well mixed, then add the milk and beat in to combine. If you'd like it especially fluffy and you are using a food mixer, then continue to beat for a few minutes.
Use a palette knife to spread all over the cake, then decorate with pieces of peanut brittle or chocolate-covered peanuts (see Decorating ideas), or just salted peanuts.
Castle cake
This might just be the dream cake for more than a few children I know. It takes a bit of time and a lot of ingredients, but if you have the time it's definitely worth the effort. I realise this cake uses a lot of sugar and butter, but then it could feed about 40 children. I've provided suggestions for how to decorate it, but let your imagination run wild. You could cover the whole thing with different-coloured sweets, for instance, or even place knights or princesses standing outside! You can make this cake in four batches if you've only one tin, or you can make it in two batches if you have enough tins. These are the instructions for making the cake in two batches; if you're making it in four batches, you need to use a quarter of the quantities listed for the ingredients each time. The photograph opposite is for the chocolate version of this cake.
**Prep time** : 1 hour 30 minutes (depending on the number of tins)
**Baking time** : 30 minutes–1 hour
**Ready in** : 3 hours
**Serves** : about 40
700g (1½lb) butter, softened, plus extra for greasing
700g (1½lb) caster sugar
12 eggs
700g (1½lb) self-raising flour
To decorate
Coloured sugar crystals or hundreds and thousands
5 ice-cream cones for the turrets
Mini marshmallows, sugar cubes or square Liquorice Allsorts for the battlements
Wafers, rectangular biscuits, chocolates (such as Curly Wurlys), thin matchsticks, for a door and windows and for edging the sides and liquorice lace for a drawbridge and ropes
Two 20cm (8in) square cake tins with 5cm (2in) sides
Large plate or a 35cm (14in) square cake board
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line the base of each tin with baking parchment.
Cream half the butter until soft in a large bowl or in an electric food mixer. Add half the sugar and beat until the mixture is light and fluffy.
Whisk six eggs together in a bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Sift in half the flour and fold in gently until mixed.
Tip the batter into the prepared cake tins and bake for 25–30 minutes or until a skewer inserted into the centre of each cake comes out clean.
Remove from the oven and allow to cool in the tins for 10 minutes. Then loosen around the edges of each cake using a small, sharp knife and carefully remove from the tin before leaving on a wire rack to cool down completely.
Repeat the process for the next two cakes until you have four square cakes.
Next make the icing, using either the coloured buttercream icing or the berry buttercream for a gorgeous pink or purple icing made using fresh berries. Or if you prefer, use the icing from the chocolate version of this cake.
If your cakes are domed or a little uneven on top, use a serrated knife to neaten the tops so they are level. Add a little blob of icing onto the four corners of one of the cakes, then place upside down on the serving plate or cake board (the icing helps hold the cake in place).
Next spread the icing onto the cake to about 5mm (¼in) thick. Add another cake on top and spread that cake with icing, then put a third cake on top.
Take the fourth cake and cut in half then half again to get four equal-sized squares. Spread icing onto one of these smaller squares and place this, icing side down, in the centre of your main cake. Repeat with another two squares so you have another three-layered tier on top of the main cake.
Cut the remaining square of cake into four small squares. Place two of these squares on top of each other in the centre of the main cake, adding icing between each layer so they stick together and on to the main cake. You'll be left with two small squares – feel free to eat these as I think it makes this cake just too tall (and you deserve a treat at this stage!). Or you could otherwise save them for making Cake pops.
It's now time to ice the cake. Spread the icing all over the cake using a palette knife, regularly dipping in a bowl of boiling-hot water to make it easier to spread. You won't need all the icing; some should be saved for the ice-cream cones.
Now the cake is iced all over, you can decorate it, starting with the ice-cream cones for the turrets.
Scatter either coloured sugar crystals, hundreds and thousands or sprinkles onto a plate, spreading them out. Spread the icing, not too thickly, on the outside of one ice-cream cone. This is easy to do if you place the cone on two fingers (your forefinger and middle finger) so that they support the cone while you ice it with your other hand. Then, with your two fingers still inside the cone and the forefinger of your other hand supporting the top, roll the cone in the sugar or sprinkles. After the cone is well coated, use the same technique to carefully place on one corner of the bottom tier of the castle. Repeat with the remaining four cones, placing the final cone on the very top of the cake.
Next make the battlements. Using square-shaped sweets such as mini marshmallows or Liquorice Allsorts, space these evenly apart around the middle and top tier of the cake to look like turrets.
After that you can play around with whatever decorations you like. A door is a great touch: I like to use pink wafers if it's a pink cake, or any rectangular biscuits or chocolates otherwise. You could also cut these in half and stick them onto the cake to make windows. You could also 'edge' the sides of the cake with chocolate matchsticks for a neat look.
If you like, you can make a drawbridge by laying Curly Wurlys (cut to size) or white chocolate fingers side by side at the castle entrance, using a little icing to stick them down. If you're feeling adventurous, you can complete the effect by making two ropes for the drawbridge out of thin chocolate matchsticks or liquorice lace, with one end of the 'rope' stuck to the end of the drawbridge and the other to the castle, just above and to one side of the door.
Coloured buttercream icing
**Prep time** : 15 minutes
**Makes** : 3kg (6½lb)
1.2kg (2lb 10oz) butter, softened
1 tbsp vanilla extract
1.6kg (3½lb) icing sugar
150–200ml (5–7fl oz) milk
A good few drops of your choice of food colouring
If using a hand-held electric beater, you can make this in one batch in a large mixing bowl. If using an electric food mixer, you will need to do it in 2–3 batches, unless you have a very large mixer.
Using a hand-held electric beater or an electric food mixer, beat together the butter and vanilla extract until very soft. Then gradually sift in the icing sugar, beating all the time, until it is fully incorporated. Gradually pour in the milk while continuing to beat, adding just enough so the mixture is softly spreadable, then whisk until it is light and fluffy. Whisk in just enough food colouring to give your desired shade.
Berry buttercream icing
**Prep time** : 15 minutes
**Makes** : 3.75kg (8¼lb)
600g (1lb 5oz) fresh or frozen (and defrosted) strawberries, raspberries, blueberries or blackberries
50ml (2fl oz) lemon juice
700g (1½lb) butter, softened
2.4kg (5lb 5oz) icing sugar, sifted
If using a hand-held electric beater, you can make this in one batch in a large mixing bowl. If using an electric food mixer, you will need to do it in 2–3 batches, unless you have a very large mixer.
Place the fruit and lemon juice in a food processor and whiz for about a minute to purée. Push through a sieve into a large bowl, then set aside. Using a hand-held electric beater or an electric food mixer, cream together the butter and icing sugar until combined, then gradually add the fruit purée, beating continuously.
Chocolate castle cake
**Prep time** : 1 hour 30 minutes (depending on the number of tins)
**Baking time** : 30 minutes–1 hour
**Ready in** : 3 hours
**Serves** : about 40
For adding to the sponge
100g (3½oz) cocoa powder, sifted
For the icing
625g (1lb 6oz) milk chocolate, in drops or broken into pieces
1.13kg (2½lb) butter, softened
2.25kg (5lb) icing sugar
125g (4½oz) cocoa powder
To decorate
Chocolate sprinkles for the ice-cream cones
Chocolate-covered toffees for the battlements
Bourbon biscuits or chocolate fingers for the door and windows
Thin matchsticks for edging the sides
Chocolate fingers or Curly Wurlys for a drawbridge and a liquorice lace for the ropes (optional)
Make the cake as in the main recipe, decreasing the quantity of self-raising flour to 650g (1lb 7oz) and adding the cocoa powder to the sponge mixture with the flour.
To make the icing (which you may need to do in 2–3 batches, though in the main recipe it can be made in one batch in a large mixing bowl using a hand-held electric beater), first place the chocolate in a heatproof bowl and over a saucepan of gently simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Cream the butter until soft in a large bowl using a hand-held electric beater or in an electric food mixer, then sift in the icing sugar and cocoa powder and continue to mix. Pour in the melted chocolate and beat until well combined.
Ice the cake as in the main recipe, and decorate in the same way, covering the iced ice-cream cones in chocolate sprinkles instead of coloured ones, and using chocolate-covered toffees for the battlements, Bourbon biscuits or chocolate fingers for a door and windows and thin matchsticks for edging the sides. If making a drawbridge, use chocolate fingers or Curly Wurlys for the bridge itself and a cut-up liquorice lace for the ropes.
Tip Should you have any icing left over, store it in the fridge, where it will last for up to a week in an airtight container. The berry buttercream will keep for up to three days in the fridge. You can then use it for buns or sponge cakes – making sure to take it out of the fridge about an hour beforehand to allow it to soften.
Decorating ideas
If you'd like to spend a little more time decorating and want to get really creative, then fondant decorations offer a hugely versatile way of taking your cake to the next level. Sugarcraft, as it's also known, in which intricate structures are created from sugar and coloured with edible dyes, can take hours or even days to create. The ideas included below are much more do-able and simple to achieve, but they look impressive even so.
You can make up the fondant icing recipe I've provided and add food colouring. Alternatively, you can buy pre-coloured fondant icing from supermarkets and specialist shops or online.
Fondant icing
For coloured fondant icing, you can either mix a few drops of food colouring in with the egg white and glucose (as the recipe below), or you can add it to the icing when it's made, kneading it in until the colour is even and the icing is smooth. Any fondant shapes made from this icing should keep for up to two weeks if stored in an airtight box.
**Prep time** : 10 minutes
**Makes** : 450g (1lb)
1 egg white, whisked
50g (2oz) liquid glucose
¼ tsp vanilla extract
A few drops of your chosen food colouring (optional)
400g (14oz) icing sugar, sifted, plus extra for dusting
In a bowl, mix together the egg white, glucose, vanilla extract and food colouring, if using. Place the icing sugar in a separate bowl or an electric food mixer and gradually add the egg-white mixture, beating continuously until all the ingredients come together.
Place the icing on a spotlessly clean worktop that has been generously dusted with icing sugar and knead the icing for a minute or two or until it is completely smooth on the surface.
Fast fondant flowers
**Prep time** : 10 minutes
**Makes** : 20–30 flowers
150g (5oz) white fondant icing
Icing sugar, for dusting
A few drops each of 2 different types of food colouring
Flower-shaped cookie cutter or a flower template
Divide the fondant icing into two pieces, one about a quarter the size of the other. If you would like to use the flower template, trace the shape using baking parchment and cut it out.
Dust a work surface with icing sugar, then add a few drops of your chosen colour to the larger piece (this will be for the main part of the flowers) and knead until the colour is even and the icing smooth. Use a rolling pin to roll out to a thickness of about 5mm (¼in), then cut out flowers using a flower-shaped cookie cutter or place the template on the rolled-out icing and cut around it with a sharp knife to make the flowers.
Add a few drops of a different food colouring to the smaller piece of fondant icing (this will be for the centre of your flowers) and knead it in as before.
Next, break off small pieces of this icing and to roll them into little balls (about the size of a lentil) with your hands. Use a little water to moisten the flowers, then press one ball each into the centre of the flowers to stick them on.
Once the flowers are assembled, if you're placing them on top of a cake covered in fondant icing, moisten the bottom first with a little water and then press them down. Alternatively, place on baking parchment and leave in a cool, dry place for a few hours. They can then be placed on top of any iced cake or cupcakes.
Fondant roses
For each rose, you'll need five balls of icing each roughly the size of a hazelnut. The roses in the step-by-step photographs are all the same colour, but you can mix them up, making petals of different colours. I love to have a variety – some white, some pale pink and some a bit darker pink – adding different amounts of pink food colouring to batches of plain fondant icing.
**Prep time** : 45 minutes
**Makes** : about 25–30 roses
100g (3½oz) white fondant icing
A few drops of pink food colouring (optional)
Icing sugar, for dusting
You can either leave the fondant icing white or, for pink roses, mix a few drops of food colouring into the icing, or into batches of it for petals in varying shades of pink. Dust your worktop with icing sugar and knead a few drops of food colouring into the icing until the colour is even and the icing smooth.
To make a rose, pinch off five little pieces of the icing, each about the size of a hazelnut, and roll each one into a ball. Place the five small balls, spaced apart, between two sheets of cling film. Press each down with your thumb, flattening it so that it forms an oval shape, slightly thinner on one long side. When you have flattened all five balls like this, remove the top layer of cling film.
To make the centre of a rose, roll one of the pieces into a spiral, with the thin side up. Take another piece and roll it, thin side up, around the centre of the rose, covering the overlap. Take the third piece, thin side up again, and wrap it around the other two pieces, covering the overlap once again. Continue with the last two pieces in the same way, then slightly curve the edges of the petals outwards with your fingertips to make a rose. The rose will probably have quite a long base so pinch this off and remove it. The discarded pieces can be reused to make more roses.
Stick them immediately to the top of your cake by moistening the bottom first with a little water and then pressing them down. Alternatively, you can store them in an airtight box, where they will keep for up to two weeks.
Fondant ghosts
**Prep time** : 30 minutes
**Makes** : 6–10 ghosts
200g (7oz) white fondant icing
Icing sugar, for dusting
A couple of drops of black or brown food colouring
Plain-edge 5cm (2in) diameter cookie cutter (optional)
First take six small pieces of fondant icing, each about the size of a walnut in its shell, and use your hands to shape them into balls.
Dust you worktop with icing sugar and roll out the remaining icing to a thickness of about 5mm (¼in), then using the cookie cutter, or a straight-sided glass about 5cm (2in) in diameter, cut out six circles. Drape each circle over one of the balls, then use your fingers to create a few folds.
To make eyes for the ghosts, roll just a tiny amount of fondant icing and add a drop or so of black or brown food colouring. Dust your work surface with icing sugar and knead until the colour is even and the icing smooth. Break off tiny pieces and roll into balls for the eyes. Moisten the balls with a little water and stick a couple onto each ghost.
Stick the ghosts to your cake by moistening the bottom of the ghosts first with a little water and then pressing them down. Alternatively, you can store them in an airtight box, where they will keep for up to two weeks.
Tip These spooky shapes would make a great alternative decoration to the Halloween pumpkin squares.
Fondant football shirts
**Prep time** : 20 minutes
**Makes** : 20–30 shirts
150g (5oz) white fondant icing
A few drops of you chosen food colouring (optional)
Icing sugar, for dusting
For the icing for piping
100g (3½oz) icing sugar, sifted
A few drops of your chosen food colouring
Shirt template
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
First trace around the shirt template using baking parchment and cut out the shape. To make a white football shirt, just leave the icing uncoloured, or to colour it add a few drops of food colouring, then dust your work surface with icing sugar and knead until the colour is even and the icing smooth.
Dust the worktop with more icing sugar and roll out the icing to about 5mm (¼in) thick. Place the shirt template on the rolled-out icing and use a knife to cut out shirt shapes, then make up the following icing to draw different colours or patterns for you team on each shirt.
Mix together the icing sugar and a few drops of food colouring in a bowl, adding just enough water (½–1 tablespoon) to make an icing the consistency of thick double cream. Fill the piping bag or freezer bag with the icing and carefully draw stripes or team numbers on each shirt. The shirts will stick on to a freshly iced cake or can be moistened with a little water and pressed lightly down onto the cake.
Tip A crowd-pleasing way of showing off these shirts is to put them on individual iced sponge squares (or cupcakes) and then arrange them on a plate all in one block, like a cake 'team', and then separate the individual 'players' to be eaten!
Fondant butterflies
**Prep time** : 45 minutes
**Makes** : 20–30 butterflies
150g (5oz) white fondant icing
A few drops of your chosen food colouring
Icing sugar, for dusting
Cornflour, for dusting
For the icing for piping
100g (3½oz) icing sugar, sifted
A few drops of your chosen food colouring
Butterfly-shaped cookie cutter or a butterfly template and cardboard for making a few drying 'stands'
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
You can make white butterflies, but if you'd prefer them to be coloured, add a few drops of food colouring to the fondant icing, dust your worktop with icing sugar and knead until the colour is even and the icing smooth. If you would like to use the butterfly template, trace the shape using baking parchment and cut it out.
On a work surface dusted with icing sugar, roll out the icing to a thickness of about 5mm (¼in). Cut out the butterflies using a butterfly cutter or alternatively place the paper template on the rolled-out icing and use a sharp knife to cut around it to make the butterflies.
Next, take a small square of cardboard and fold in half, then fold each half over again so you have a 'M' shape if you look at it side on. Dust the cardboard with cornflour, then use a palette knife to transfer each butterfly to a cardboard stand so the middle of its body is in the fold of the 'M' and the wings are angled upwards. Repeat until all the butterflies are on cardboard, with their wings angled upwards, then leave in a cool dry room so they will dry off and stiffen slightly.
To make the icing for the piping, mix together the icing sugar and a few drops of food colouring in a bowl, adding just enough water (½–1 tablespoon) to make an icing the consistency of thick double cream. Fill the piping bag or freezer bag with the icing and carefully draw patterns on your butterflies. The butterflies will stick on to a freshly iced cake or moisten the body of the butterflies with a little water first and then gently press them down on to the cake.
Fondant birds
**Prep time** : 30 minutes
**Makes** : about 20 birds
150g (5oz) white fondant icing
A few drops each of 3 different types of food colouring
Icing sugar, for dusting
Bird template
First trace the inner and outer bird templates using baking parchment, then cut out the shapes.
Divide the fondant icing into two, then break a small amount off one of the pieces. The largest piece will be for making the bodies of your birds. Add a few drops of your chosen food colouring to this, dust a worktop with icing sugar and knead until you have the colour you desire and the icing is smooth.
The larger of the two remaining pieces of fondant icing will be for the wings. Add a few drops of a different colour and knead again until the colour is even and the icing is smooth. To the final small piece of icing, add, a drop or two of another colour (this will for the beaks), then knead until smooth.
Dust your work surface with a little icing sugar and roll out each piece of icing until about 5mm (¼in) thick. Using a sharp knife and with the body template as guide, cut out the bodies of the birds from the largest piece of rolled-out icing. Next cut out the same number of wings from the larger of the remaining two pieces of icing, and finally cut out the beaks from the third piece of icing.
Moisten a wing piece using your fingertips and a little water, then press this onto a body piece, moist side down, gently rubbing around the icing so the wing sticks. Next moisten the edge of the beak and carefully press it onto the edge of the body so it sticks.
To make an eye, roll just a tiny amount of fondant icing (it could be any colour) and make a tiny ball. Moisten the ball a little and stick down on the body of the bird. Repeat until all the birds have been completed.
Once made, you can either add them straight away to iced cupcakes, or, if you prefer, leave them to dry overnight, then carefully place them flat on to a cake or cakes covered in a thick layer of buttercream.
Fondant snowmen
**Prep time** : 45 minutes
**Makes** : 6–10 snowmen
150g (5oz) white fondant icing
A few drops each of orange and black food colouring
Icing sugar, for dusting
Fine paintbrush (optional)
Toothpick (optional)
Snowman template (optional)
To make a snowman, break off a piece of fondant icing about the size of a walnut in its shell, then roll into a ball with your hands – this will be the body. Next break off a slightly smaller piece from the main block of fondant icing and roll again into a ball – this will be the head of the snowman. Moisten the top of the body with a little water and press on the head. (To ensure they stick together, you can push a toothpick through the body and the head, breaking the stick off to make it the correct size if necessary.)
To make a carrot-shaped nose, break off a small piece of fondant icing and add a good few drops of orange food colouring. Dust your work surface with a little icing sugar and knead until the colour is even and the icing smooth. Then break off a tiny piece and roll into a small ball, pinch the ends to form a teardrop shape and roll a little to form a cone. Moisten the end with a little water, then press onto the head to form a nose.
Break off another piece of fondant icing and add a good few drops of black food colouring. Dust your work surface with a little icing sugar and knead the fondant until the colour is even and the icing smooth. Break off several tiny dots (they don't need to be rolled) and moisten these a little before attaching to the snowman as eyes and buttons. Alternatively, dip the very end of your paintbrush into the black food colouring and carefully paint dots on the head for eyes and down the body for buttons.
Repeat the above steps to make the remaining snowmen. Stick the snowmen to your freshly iced cake or you can moisten the bottom of the snowmen first with a little water and then press them gently down on to the cake. You can store them in an airtight box, where they will keep for up to two weeks. Alternatively you can use the snowman template to cut out fondant icing into fun shapes.
Tip These would also work on a Christmas cake.
Fondant dinosaurs
**Prep time** : 40 minutes
**Makes** : about 12–15 dinosaurs
A few drops of your chosen food colouring
150g (5oz) white fondant icing
Icing sugar, for dusting
For the icing for piping
100g (3½oz) icing sugar, sifted
A few drops of your chosen food colouring
Dinosaur-shaped cookie cutter or a dinosaur template
Piping bag with a 1 or 2mm (1/16in) nozzle or a freezer bag with 1 or 2mm (1/16in) cut from one corner
Add a few drops of food colouring to the fondant icing, then dust your worktop with icing sugar and knead until the colour is even and the icing smooth. If you would like to use the dinosaur template, trace the shape opposite using baking parchment and cut it out.
On a work surface dusted with icing sugar, roll out the icing to about 5mm (¼in) thick. Cut out dinosaur shapes using a cookie cutter or place the paper template on the rolled-out icing and use a sharp knife to cut out the dinosaurs.
To make the icing for piping, mix together the icing sugar and a few drops of food colouring in a bowl, adding just enough water (½–1 tablespoon) to make an icing the consistency of thick double cream. Fill the piping bag or freezer bag with the icing and carefully draw stripes, dots or other patterns onto your dinosaurs. They will stick on to a freshly iced cake or you can moisten the bottom of the dinosaurs first and press them gently on to the cake.
Fondant penguins
**Prep time** : 45 minutes
**Makes** : 6–10 penguins
200g (7oz) white fondant icing
A few drops each of black and yellow food colouring
Icing sugar, for dusting
Plain-edge approx 2cm (¾in) diameter cookie cutter and a very small heart-shaped cutter (optional)
Toothpick (optional)
Break off about a quarter of the fondant icing and from that piece break off another quarter, leaving you with three pieces of icing, decreasing in size.
To the largest piece, add a good few drops (or more) of black food colouring, dust your work surface with a little icing sugar and knead until the colour is even and the icing smooth.
To make a penguin, break off a piece of the black fondant icing about the size of a walnut in its shell, then roll into an oval with your hands – this will be the body. Next break off a smaller piece and roll again into a ball for the penguin's head.
Dust your work surface with a little icing sugar, take the larger piece of white fondant icing and roll out until very thin. Using a cookie cutter or a straight-edged glass roughly the same size in diameter, cut out a circle. Moisten one side of the circle and the body of the penguin with a little water, then press the circle onto the body, rubbing around the icing to firmly attach it. This forms the white 'breast' of the penguin.
To make the penguin's face, use either a very small heart-shaped cutter or a knife to cut out a small heart shape in the icing. Moisten one side of the heart with a little water and press onto the front of the head, rubbing the icing and pressing in with your hands to make sure it sticks on firmly.
Break off another two pieces of black fondant icing for the feet – they should be quite small, each about the size of a cherry stone.
Break off two more pieces of black icing to form the wings – they should be larger than the foot pieces. Use your fingers to form then into flat oval shapes.
Next stick the feet onto the bottom of the penguin's body, moistening with a little water. Then stick the wings to the body, moistening again. Finally stick the head onto the top of the body. (To ensure they stick together, you can push a toothpick through the body and the head, breaking the stick off to make it the correct size if necessary.)
To make the eyes, break off two small pieces of black fondant icing the same size and roll into tiny balls.
To the smaller piece of white fondant, add a few drops of yellow food colouring. Knead until the colour is even and the fondant is smooth. Break off a small piece and roll into a ball the same size as the eyes, then pinch the end a little to form a beak shape. Stick the eyes and beak to the head, moistening with a little water.
Repeat the above steps to make the remaining penguins. Stick the penguin to your freshly iced cake or you can moisten the bottom of the penguin first with a little water and then press them gently down on to the cake. Alternatively, you can store them in an airtight box, where they will keep for up to two weeks.
Tip These cheeky chaps would make an alternative festive decoration to the top of the Christmas cake.
A celebration can be transformed by cake. Having a cake as a centrepiece to your celebration can make an occasion extra special, whether it is a grand and extravagantly decorated cake, a simple but stunning classic, or a collection of beautiful cupcakes. This chapter has a range of cakes, some of which are relatively quick to put together and some that ask a little more from you in terms of time and effort. While many of these cakes are tailored to certain celebrations such as a wedding or an anniversary, you do not have to be tied down to specifics as you can adapt your cake's decoration to suit your particular celebration, be that a baby shower or a retirement party. Be as inventive as you like with decorations, and I hope you will enjoy making these cakes whatever the event, even if it is just to celebrate a Tuesday afternoon!
08/ Celebration
Hazelnut meringue cake
Anniversary chocolate mousse layer cake
Strawberry white chocolate cake
Chocolate cupcakes
Chocolate meringue frosting
Chocolate honeycomb topping
Peanut butter topping
Easter nests
Rose-water pistachio cupcakes
Pistachio and rose-water buttercream icing
Orange and almond syrup cake
Pistachio cake
Hazelnut praline triple-layered cake
White chocolate macadamia nut cake
Mini wedding fruitcakes
Vanilla wedding cake
White chocolate anniversary cake
Red velvet cupcakes
Perfect Peanut butter cake
Hazelnut meringue cake
Half meringue and half cake, this is an elegant dish that uses a few ingredients to impressive effect. It contains no flour but uses ground hazelnuts and egg yolks to give structure and wonderful flavour to a meringue base. Topped, pavlova-style, with softly whipped cream and strawberries or raspberries, this would make a super end to a summer meal.
**Prep time** : 20 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour and 15 minutes
**Serves** : 6–10
Butter, for greasing
225g (8oz) hazelnuts (skin still on), toasted (see the tip)
6 eggs, separated
200g (7oz) caster sugar
For the topping
350ml (12fl oz) double or regular cream, softly whipped
250g (9oz) fresh raspberries or sliced hulled strawberries or a mixture of the two
A few fresh mint leaves (optional)
Icing sugar, for dusting
23cm (9in) diameter cake spring-form or loose-bottomed tin with 6cm (2½in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, then butter the sides of the cake tin and line the base with a disc of baking parchment. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the toasted hazelnuts in a food processor and whiz for a minute or so until they form a coarse powder with a few larger chunks for texture.
Using a hand-held electric beater or an electric food mixer, whisk the egg yolks and sugar together for 5–7 minutes or until pale, thick and mousse-like, then fold in the hazelnuts.
In a separate, spotlessly clean bowl (and having cleaned the electric beater, if using), whisk the egg whites until they form stiff peaks, then gently fold them into the yolk mixture until incorporated.
Tip the batter into the prepared tin and bake for about 45 minutes or until a skewer inserted into the centre comes out clean and the cake begins to come away slightly from the edges of the tin. Don't worry if it dips slightly in the middle.
Remove from the oven and allow to cool in the tin for 10 minutes. Loosen the sides using a small, sharp knife, then carefully remove the sides of the tin and leave the cake on a wire rack to cool down completely before transferring to a serving plate.
Spoon the whipped cream all over the top of the cake, leaving a gap about 1cm (½in) wide around the edge, and scatter the raspberries or strawberries over the cream. Decorate with a few mint leaves, if using, then dust with icing sugar.
Anniversary chocolate mousse layer cake
Four layers of chocolate sponge sandwiched together with a rich chocolate mousse, this is an elegant cake that would be ideal for a wedding anniversary celebration. A couple of tablespoons of rum, brandy or other spirits add to the intensity of the mousse, but if you'd rather not include any alcohol, vanilla extract makes an excellent substitute. The recipe below is based upon one for a classic Genoese cake that calls for browned or clarified butter and requires the eggs and sugar to be whisked over a flame. This version of the recipe omits those steps as I don't feel they're strictly necessary: the resulting cake is still very light and it tastes fabulous.
**Prep time** : 45 minutes
**Baking time** : 30–35 minutes
**Ready in** : 2 hours
**Serves** : 8–12
100g (3½oz) butter, plus extra for greasing
8 eggs
300g (11oz) caster sugar
200g (7oz) plain flour, plus extra for dusting
75g (3oz) cocoa powder
3 tsp baking powder
¼ tsp salt
1 tsp vanilla extract
For the mousse
275g (10oz) dark chocolate, in drops or broken into pieces
6 eggs, separated
175g (6oz) butter, softened
2 tbsp rum, brandy, whiskey or orange liqueur, or 1 tsp vanilla extract
To decorate
Chocolate curls, made using either dark, milk or white chocolate or a mixture
Cocoa powder or icing sugar, for dusting (optional)
Two 23cm (9in) diameter cake tins with 6cm (2½in) (minimum) sides
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tins and dust with flour, then line the base of each tin with a disc of baking parchment.
To make the sponge, first melt the butter and set aside for 5–10 minutes to cool until tepid.
Place the eggs and sugar in a large bowl or an electric food mixer. Using an electric hand-held beater or the food mixer, whisk for 6–8 minutes or until the mixture is light and thick. To test if it's thick enough, lift the whisk out of the mixture and draw a figure of eight in the bowl with the batter that's left on the whisk: the '8' should remain visible on the surface for a couple of seconds.
Sift in the flour, cocoa powder, baking powder and salt and fold in gently but thoroughly, then pour in the vanilla extract and the melted butter and fold in again.
Quickly divide the batter between the prepared cake tins before it has a chance to lose any volume, then bake for 30–35 minutes or until the mixture springs back when lightly pressed with a finger and a skewer inserted into the centre of each cake comes out clean.
Remove from the oven and allow to cool in the tins for 5 minutes, then loosen around the edges of each tin using a small, sharp knife and carefully remove the cake before transferring to a wire rack to cool down completely.
While the cakes are cooling, make the chocolate mousse. Place the chocolate in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Beat in the egg yolks one at a time, then beat in the butter, spirits or orange liqueur or vanilla extract and keep beating until smooth. Set aside to cool while you whisk the egg whites.
In a spotlessly clean bowl or using the food mixer (having cleaned the bowl and whisk), whisk the egg whites until they form stiff peaks. Fold in a quarter of the chocolate mixture, then carefully fold in the rest just until mixed. Set aside and leave to cool at room temperature – not in the fridge, otherwise the mousse will harden and become difficult to spread.
To assemble the cake, first use a bread knife to carefully slice each cake in half horizontally. Place the bottom half of one of the cakes, cut side up, on a plate or cake stand. Spread over some of the mousse in a layer about 5mm (¼in) thick, then sandwich with the other half of the cake, placing it cut side down. Spread over more of the mousse, just as thickly, then sandwich with the bottom half of the second cake, cut side up. Repeat for the next layer until you have four layers of cake sandwiched together with three layers of mousse. Cover the top and sides of the assembled cake with the remaining mousse, smoothing it over with a palette knife.
Decorate the top with chocolate curls, if using, or dust with cocoa powder or icing sugar.
Strawberry white chocolate cake
This recipe uses a lot of fresh strawberries. They're sliced and mixed into the cake, making it extremely moist and really bursting with that summery strawberry flavour. The cake has a slightly rustic and uneven top, but that's part of its charm. Moist as it is, this cake needs no accompaniment (other than a blob of softly whipped cream, if you wish).
**Prep time** : 10 minutes
**Baking time** : 50–60 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 6–10
175g (6oz) butter, softened, plus extra for greasing
175g (6oz) caster sugar
2 eggs
175g (6oz) self-raising flour, sifted
100g (3½oz) ground almonds
75g (3oz) white chocolate, in chips or chopped into pieces
450g (1lb) fresh strawberries, hulled and sliced
23cm (9in) diameter spring-form or loose-bottomed cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the base and sides of the cake tin. If you're using a spring-form tin make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed. Gradually add the eggs to the creamed butter mixture, beating all the time. Add the flour and ground almonds and fold in to combine, then fold in the chocolate chips and the strawberries.
Tip the mixture into the prepared tin and bake for 50–60 minutes or until well risen and golden on top and a skewer inserted into the centre of the cake comes out clean.
Remove from the oven and allow to cool in the tin for 20 minutes. Loosen around the edges using a small, sharp knife and remove the sides of the tin before carefully transferring to a serving plate and leaving to cool down completely.
Chocolate cupcakes
Cupcakes have become incredibly popular in recent years. One of the reasons for this is their versatility. With a huge range of flavours, styles and toppings, the only limit is your imagination. Here are four different ways of topping a delicious basic cupcake recipe.
**Prep time** : 10 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
For the basic cupcakes
100g (3½oz) butter, softened
150g (5oz) caster sugar
2 eggs
125ml (4½fl oz) milk
175g (6oz) plain flour
25g (1oz) cocoa powder
2 tsp baking powder
Pinch of salt
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Pour in the milk and beat until mixed. Sift in the flour, cocoa powder, baking powder and salt and mix gently to combine.
Divide the mixture between the muffin cases, filling each case two-thirds full. Bake for 20–25 minutes or until well risen and lightly springy to the touch. Allow to cool for 5 minutes, then remove from the tray and place on a wire rack to finish cooling down.
While the cupcakes are cooling, you could select one of the options for topping them, depending on what feels right for the occasion.
Chocolate meringue frosting
**Prep time** : 7 minutes
**Makes** : enough to cover 12 cupcakes
4 large egg whites
250g (9oz) caster sugar
Pinch of salt
25g (1oz) cocoa powder
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Make sure the cupcakes are ready before you start, as this icing begins to set very quickly. Place all the ingredients, except the cocoa powder, in a heatproof bowl, add 2 tablespoons of cold water and set over a saucepan of simmering water. (The bowl should sit snugly over the pan with its base high enough above the water that it does not come into contact with it.)
Whisk slowly, using a balloon whisk, until the sugar has completely dissolved and the mixture is foamy. Continue to heat and whisk until the mixture reaches 60°C (140°F) when measured with a sugar thermometer – this will take about 4 minutes. If you don't have a thermometer, you can gauge whether the mixture is ready by how it feels and looks: it should be hot to the touch, glossy white in appearance and starting to thicken.
Quickly remove the bowl from the pan and pour the mixture into the bowl of an electric food mixer fitted with the whisk attachment. Alternatively, whisk in the original bowl using a hand-held electric beater. Whisk on a high speed for about 3–5 minutes or until the frosting is very thick, glossy and has cooled. Sift in the cocoa powder and continue to whisk just until mixed.
Spread quickly over the cupcakes with a palette knife and use the back of a teaspoon to shape them, regularly dipping the spoon into a jug of boiling-hot water. Alternatively, the icing could be quickly placed in the piping bag or freezer bag and piped over the cupcakes (see Cake essentials).
Chocolate honeycomb topping
**Prep time** : 10 minutes
**Makes** : enough to cover 12 cupcakes
150g (5oz) butter, softened
2 tbsp double or regular cream
2 tsp vanilla extract
350g (12oz) icing sugar
2 tbsp cocoa powder
About 12 tbsp honeycomb pieces to decorate
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Using a hand-held electric beater or an electric food mixer, beat together the butter, cream and vanilla extract until very soft. Then gradually sift in the icing sugar and the cocoa powder, beating all the time, until fully incorporated. Continue to beat until very soft, then (if using a food mixer) whisk the mixture until it is light and fluffy.
Pipe (see Cake essentials) or spread the icing (using a palette knife or the back of a spoon) onto the cooled cupcakes. Then sprinkle over the honeycomb pieces (see Decorating ideas), allowing about a tablespoon per cupcake. I like to try and include a mixture of small and large chunks.
Peanut butter topping
**Prep time** : 15 minutes
**Makes** : enough to cover 12 cupcakes
125g (4½oz) smooth or crunchy peanut butter
100g (3½oz) butter, softened
125g (4½oz) cream cheese
1 tsp vanilla extract
375g (13oz) icing sugar
1–2 tbsp milk
125g (4½oz) salted peanuts, roughly chopped, to decorate
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Place the peanut butter, butter, cream cheese and vanilla extract in a large bowl or in an electric food mixer and cream together until well mixed. Sift in the icing sugar a little at a time and beat in to combine, then add the milk and mix in well. If you're using a food mixer, you could continue to beat for a few minutes until the mixture is very fluffy.
When the cupcakes are cool, either pipe the icing onto them (see Cake essentials) or spread it using a palette knife or the back of a teaspoon. Sprinkle a few chopped peanuts over each cupcake.
Easter nests
**Prep time** : 20 minutes
**Makes** : enough for 12 cupcakes
For the icing
100g (3½oz) butter, softened
225g (8oz) icing sugar, sifted
25ml (1fl oz) milk
1 tsp vanilla extract
A few drops of green food colouring (optional)
For the nests
100g (3½oz) dark or milk chocolate, in drops or broken into pieces
4 Shredded Wheat
36 sugar-coated mini chocolate eggs
To make the nests, first line a baking tray with baking parchment. Then place the chocolate in a heatproof bowl over a pan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Crumble the Shredded Wheat and add these to the melted chocolate, mixing in until well coated. Working quickly, add about 2 teaspoons of the mixture to form a small mound on the baking parchment. Use your fingers to press down the centre to form a rough nest shape, then add 3 mini eggs, pressing them down into the nest. Repeat with the rest of the mixture and eggs until you have 12 nests, then leave on the tray to cool and set.
Meanwhile, make the icing. Place the butter and sugar in a large bowl or in an electric food mixer and cream together until fluffy and light. Gradually add the milk with the vanilla extract and food colouring (if using) and continue beating the mixture until smooth. When the cupcakes are cool, ice them using a palette knife or the back of a teaspoon. Place a nest on top of each cake, pressing carefully but firmly into the icing.
Rose-water pistachio cupcakes
Rose water makes a fantastic addition to cakes. Its fragrance is quite intense so you don't need to use too much to get a really lovely effect. Pistachios complement the flavour really well, while the green colour of the nuts looks gorgeous against the pale icing covering the cupcakes in this recipe. I've used a simple rose-water icing here, but you could spread the cakes with the pistachio and rose-water buttercream icing if you prefer. Decorated with crystallised pink rose petals (see Decorating ideas), these look truly pretty and celebratory. The cakes can be made a day ahead and would be fabulous at a wedding as an
**Prep time** : 15 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
150g (5oz) butter, softened
150g (5oz) caster sugar
2 eggs
1 tbsp rose water
200ml (7fl oz) buttermilk
225g (8oz) self-raising flour
25g (1oz) unsalted shelled pistachios, chopped, to decorate
Crystallised rose petals, to decorate
For the icing
250g (9oz) icing sugar, sifted
1 tbsp rose water
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together with the rose water in a small bowl for a few seconds or just until mixed, then gradually add this to the creamed butter mixture, beating all the time. Next beat in the buttermilk, then sift in the flour and fold in gently to combine.
Divide the mixture between the muffin cases, filling each three-quarters full, then bake for 20–25 minutes or until golden on top and lightly springy to the touch. Allow to cool for 5 minutes before removing from the muffin tin and placing on a wire rack to cool down completely.
Meanwhile make the icing. Mix together the icing sugar and rose water with 1 tablespoon of water until you have a thick icing, adding a tiny extra splash of rose water or water if it seems too thick to spread. Spread the icing using the back of a teaspoon or with a palette knife dipped into hot water – or drizzle over the icing – and sprinkle with the chopped pistachios. Alternatively, you could ice the cakes with the pistachio and rose-water buttercream icing.
Transfer to a serving plate and scatter with crystallised rose petals (see Decorating ideas), if using, to finish.
Pistachio and rose-water buttercream icing
**Prep time** : 10 minutes
**Makes** : enough to cover 12 cupcakes
50g (2oz) unsalted shelled pistachios
250g (9oz) icing sugar, sifted
125g (4½oz) butter, softened
1 tbsp rose water
Place the pistachios in a food processor and whiz for a good few minutes to form a fine powder, then add the icing sugar and pulse just to mix.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the pistachio and sugar mixture a little at a time until combined, then add the rose water and continue to beat until the mixture is light and fluffy.
Orange and almond syrup cake
Dense, moist and richly orange-flavoured, this is an impressive cake even before it's covered in the chocolate icing. You can omit the icing if you like – the cake is delicious without it – but I love the extra intensity that the dark chocolate brings, especially if it's flavoured with a little orange liqueur. Adding crystallised orange peel (see Decorating ideas) would give a beautiful and really impressive finish to this cake, making it perfect for a proper celebration.
**Prep time** : 25 minutes
**Baking time** : 45–50 minutes
**Ready in** : 2 hours
**Serves** : 6–8
200g (7oz) butter, softened, plus extra for greasing
300g (11oz) caster sugar
Finely grated zest of 2 oranges
200g (7oz) ground almonds
4 eggs, beaten
100g (3½oz) plain flour, sifted
1 tsp baking powder
Pinch of salt
Crystallised orange peel, to decorate
For the syrup
Juice of 2 oranges
100g (3½oz) caster sugar
For the icing
75g (3oz) butter, softened
100g (3½oz) dark chocolate, in drops or broken into pieces
1 tbsp orange liqueur or orange juice
20cm (8in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 160°C (325°F), Gas mark 3, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Place the butter, sugar and orange zest in a large bowl and beat together until soft and creamy. Add half the ground almonds and continue to mix, then add the eggs, a little at a time, making sure they're well mixed in. Sift in the flour, baking powder and salt and add the rest of the almonds, folding these in until combined.
Tip the cake batter into the prepared tin, spreading it out and levelling the top with a spatula or palette knife. Bake for 45–50 minutes or until a skewer inserted into the centre comes out with just a small amount of moisture sticking to it.
While the cake is cooking, make the syrup. Place the orange juice and sugar in a saucepan. Stir until mixed and bring to the boil. Allow to boil for just 1 minute, then remove from the heat and set aside.
When the cake is cooked, take it out of the oven and let it sit in the tin for just 5 minutes. Loosening around the edges using a small, sharp knife, carefully remove the cake from the tin and transfer to a serving plate.
Straight away pour the hot syrup over the cake (reheating the syrup if it has had a chance to cool down) and leave to cool completely while soaking up the syrup.
To make the icing, place the butter in a heatproof bowl with the chocolate and orange liqueur or orange juice and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then pour the icing over the cake. (This is a wet icing so don't worry if it drips down the sides of the cake.)
Let the icing set and then decorate with pieces of crystallised orange peel (see Decorating ideas), if you wish.
Pistachio cake
Pistachios and oranges are both redolent of sunnier climes, where pistachios arrive at the market in giant sacks, not tiny plastic packets, and oranges are sold by the boxload. This recipe combines both those flavours, as the sponge is made using ground pistachios and orange zest and the finished cake is drenched in a tangy orange glaze.
**Prep time** : 15 minutes
**Baking time** : 30–40 minutes
**Ready in** : 1 hour 30 minutes
**Serves** : 6–10
150g (5oz) unsalted shelled pistachios
225g (8oz) plain flour
2 tsp baking powder
100g (3½oz) butter, softened, plus extra for greasing
225g (8oz) caster sugar
4 eggs
100ml (3½fl oz) milk
Finely grated zest of 2 oranges
Double or regular cream, whipped, or Greek yoghurt, to serve
For the glaze
Juice of 1 orange
100g (3½oz) soft light brown sugar
23cm (9in) diameter cake tin
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Place 100g (3½oz) of the pistachios in a food processor and pulse a good few times until coarsely ground. Tip into a large bowl and sift in the flour and the baking powder, mixing together to combine.
Cream the butter until soft in a separate bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the creamed butter mixture, beating all the time. Next beat in the milk and orange zest, then tip in the ground pistachios and flour and fold gently to combine.
Tip the mixture into the prepared tin and smooth the surface with a spatula. Bake for 30–40 minutes or until a skewer inserted into the centre of the cake comes out clean.
While the cake is in the oven, make the glaze. Place the orange juice and sugar in a saucepan and set over a medium heat. Stir until the sugar is dissolved, then bring to the boil and immediately remove from the heat.
When the cake is cooked, take it out of the oven and let it sit in the tin for just 5 minutes. Loosening around the edges using a small, sharp knife, carefully remove the cake from the tin and transfer to a serving plate.
Straight away pour the hot glaze over the cake (reheating it if it has had a chance to cool down), chop the remaining pistachios and scatter over the cake to decorate, then allow to cool completely while soaking up the glaze.
Serve with a spoonful of softly whipped cream or natural Greek yoghurt.
Hazelnut praline triple-layered cake
A triple-layered praline cake makes a fabulous birthday treat. The three layers of sponge are lightened with a good amount of whisked egg whites. For the filling, praline crumbs are mixed into a divinely rich custard cream. The cake is topped in a thick, snowy-white American frosting, crisp on the outside and fluffy and marshmallow-like beneath. As it's covered in icing, the cake will keep for 3–4 days in an airtight container. If you don't have an airtight box big enough, you can use a large mixing bowl upturned over the cake.
**Prep time** : 1 hour 30 minutes
**Baking time** : 30 minutes
**Ready in** : 3 hours
**Serves** : 14–18
375g (13oz) plain flour
4 tsp baking powder
1 tsp salt
225g (8oz) butter, softened, plus extra for greasing
675g (1½lb) caster sugar
325ml (11½fl oz) milk
2 tsp vanilla extract
9 egg whites (about 250ml/9fl oz)
For the praline
100g (3½oz) caster sugar
100g (3½oz) hazelnuts (skin still on)
For the custard cream
25g (1oz) caster sugar
3 egg yolks
175ml (6fl oz) milk
15g (½oz) cornflour
1 tsp vanilla extract
100ml (3½fl oz) double or regular cream
For the frosting
4 large egg whites
250g (9oz) caster sugar
Pinch of salt
Three 20cm (8in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line the bases with a disc of baking parchment.
First make the sponge. Sift the flour, baking powder and salt into a bowl. Cream the butter until soft in a large bowl or in an electric food mixer. Add 450g (1lb) of the sugar and beat until the mixture is light and fluffy. Add about a third of the sifted flour along with about a third of the milk and continue to mix gently, in thirds, until all of the flour and milk is well mixed in, then stir in the vanilla extract.
In the bowl of an electric mixer, whisk together the egg whites until foamy, then add the remaining sugar and whisk until the meringue holds stiff glossy peaks. Mix in a quarter of the meringue to the cake mixture, then carefully fold in the rest until fully incorporated.
Tip the mixture into the prepared cake tins and bake for 30 minutes or until a skewer inserted into the centre of each cake comes out clean. Remove from the oven and allow to cool for 10 minutes, then loosen the sides of each tin with a small, sharp knife and carefully remove the cakes from the tins before placing on a wire rack to cool completely. (The sponge can be made up to a day in advance and kept in an airtight container.)
To make the praline, first line a baking tray with baking parchment and set aside. Place the sugar in a frying pan and scatter the hazelnuts over the sugar. Place the pan over a medium heat to allow the sugar to caramelise, swirling the pan every so often to ensure it caramelises evenly. Cook until the sugar has completely melted and is a deep golden colour and the hazelnuts are coated evenly.
Transfer the coated nuts to the prepared baking tray. Before the caramel has a chance to harden, set apart about 10 hazelnut clusters (with 4–5 hazelnuts in each cluster) for decorating (see photograph). Using two forks, spread apart the remaining hazelnuts and leave the praline to cool completely. Once cool break up the praline using your hands, then place the pieces (but not the reserved clusters) in a food processor and whiz until it resembles slightly coarse breadcrumbs.
To make the custard cream, place the sugar in a saucepan with the egg yolks, milk, cornflour and vanilla extract and whisking all the time bring just to the boil, then reduce the heat to low. Then cook, continuing to whisk, until thickened. Immediately remove from the heat before transferring to a bowl to cool completely. In a separate bowl, whip the cream just until it holds stiff peaks. Add the praline to the cooled custard and mix in, then carefully fold in the whipped cream. Cover the praline custard cream and place in the fridge until you are ready to use it.
You can now assemble the cake. Place one of the cakes on a cake stand or plate. Spread with half of the praline custard, then cover with a second cake. Spread the other half of the custard cream over the cake, then top with the third cake. Use a pastry brush to brush off any excess crumbs from the cake.
Next make the frosting. First place a palette knife in a jug or bowl and put the kettle on. It makes it really easy to frost this cake if you can use a palette knife that has been dipped in hot water. Place all the frosting ingredients in a heatproof bowl, add 2 tablespoons of cold water and set over a saucepan of simmering water. (The bowl should sit snugly over the pan, with its base high enough above the water that it does not come into contact with it.)
Whisk slowly by hand until the sugar has completely dissolved and the mixture is foamy. Continue to heat and whisk until the mixture reaches 60°C (140°F) when measured with a sugar thermometer – this will take about 4 minutes. If you don't have a thermometer, you can gauge whether the mixture is ready by how it feels and looks: it should be hot to the touch, glossy white in appearance and starting to thicken.
Quickly remove the bowl from the pan and pour the mixture into the bowl of an electric food mixer fitted with the whisk attachment. Alternatively, whisk in the original bowl using a hand-held electric beater. Whisk on a high speed for about 3–5 minutes or until the frosting is very thick, glossy and has cooled.
Pour boiling water into the jug or bowl holding the palette knife. Before the frosting has a chance to cool and therefore set, spread it with the hot, wet palette knife over the top and all around the sides of the cake, covering it as evenly as possible. You can go for a smooth appearance or a slightly peaked look by tapping the flat side of the palette knife over the frosting. As you pull it up, it should create little peaks (see Cake essentials). Do this all over the cake.
Decorate around the top edge of the iced cake with the reserved hazelnut praline clusters.
White chocolate macadamia nut cake
Macadamia nuts hail all the way from Australia. With a higher oil content than almost any other nut, they are perhaps the most decadent, their rich buttery flavour combining wonderfully with the sweet vanilla flavour of white chocolate. Being so rich means that a few macadamias go a long way, which is fortunate as they can be expensive. If you can't find them or would prefer not to use them, walnuts or pecans would do very well instead.
**Prep time** : 15 minutes
**Baking time** : 45–50 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 6–8
200g (7oz) butter, softened, plus extra for greasing
200g (7oz) caster sugar
4 eggs
150ml (5fl oz) sour cream
200g (7oz) plain flour, sifted
2 tsp baking powder
100g (3½oz) white chocolate, as chips or finely chopped
150g (5oz) macadamia nuts (or walnuts or pecans), toasted (see the tip) and roughly chopped
For the icing
150g (5oz) white chocolate, in drops or broken into pieces
50ml (2fl oz) sour cream
20cm (8in) diameter cake tin with 6cm (2½in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tin and line the base with a disc of baking parchment.
Cream the butter until soft in a large bowl or in an electric food mixer until soft. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Next beat in the sour cream and then fold in the flour, baking powder and chocolate chips, along with 100g (3½oz) of the toasted nuts.
Tip the cake batter into the prepared tin and bake for 45–50 minutes or until a skewer inserted into the centre comes out clean.
Remove from the oven and allow to cool for 10 minutes. Loosening round the edges using a small, sharp knife, carefully remove the cake from the tin and leave on a wire rack to finish cooling before transferring to a serving plate or cake stand.
While the cake is cooking, or while it's cooling down, make the icing. Place the chocolate in a heatproof bowl and set over a pan of simmering water. Leave just until melted, then remove from the heat and whisk in the sour cream until smooth.
Drizzle the icing over the cake and sprinkle over the remaining nuts to finish.
Mini wedding fruitcakes
If you'd like an alternative to one large wedding cake, this recipe is worth considering. I love the idea of each guest being given their own mini cake rather than just a slice of a big cake. You could decorate these with roses, bows or something to fit your wedding theme. These mini cakes will keep for a week or two in an airtight container, so you can make them well ahead of time. The recipe can be doubled if you have a large enough bowl or electric food mixer, or alternatively you can make them in as many batches as you need.
**Prep time** : 1 hour (including the fondant roses)
**Baking time** : 40–45 minutes
**Ready in** : 2 hours
**Makes** : 12 cakes
125g (4½oz) butter, softened, plus extra for greasing
125g (4½oz) soft dark brown sugar
3 eggs
300g (11oz) mixed fruit
50g (2oz) chopped mixed candied peel
25g (1oz) ground almonds
125g (4½oz) plain flour, plus extra for dusting
1 tsp mixed spice
For the fondant icing
1 egg white, whisked
50g (2oz) liquid glucose
½ tsp vanilla extract
400g (14oz) icing sugar, sifted, plus extra for dusting
To decorate
250g (9oz) marzipan (to make it yourself)
25g (1oz) apricot jam
12-cup muffin tray
Preheat the oven to 150°C (300°F), Gas mark 2, then butter the cups of the muffin tray and dust them with flour.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds just until mixed, then gradually add these to the creamed butter mixture, beating all the time. Add the dried fruit and peel and ground almonds and sift in the flour and mixed spice, then fold in to incorporate.
Divide between the 12 muffin cups and bake for 40–45 minutes or until a skewer comes out clean and the cakes feel springy to the touch. Remove from the oven and allow to cool for 5 minutes, then use a small, sharp knife to help release the cakes from the tray and place on a wire rack to cool down fully while you make the fondant icing.
In a bowl, mix together the egg white, glucose and vanilla extract. Place the icing sugar in a separate bowl or an electric food mixer and gradually add the egg-white mixture, beating continuously until all the ingredients come together. Place the icing on a spotlessly clean worktop that has been generously dusted with icing sugar and knead for a minute or two until it is completely smooth on the surface, then cover with an upturned bowl or cling film and set aside.
Roll out the marzipan to 5mm (¼in) thick and cut out 12 circles, each 6cm (2½in) in diameter. Then do the same with 250g (9oz) of the fondant icing. Brush the top of each cooled cake with apricot jam, place a circle of marzipan on top of each, then brush the top of that with apricot jam. Add a circle of icing on top of each.
Use the remaining fondant icing to make fondant roses (see Children's decorating ideas) for the top of each cake.
Use a small amount of apricot jam to stick the roses to the centre of the cakes, then repeat for each cake.
Variation
If you like, you could use coloured fondant icing for decorating the cakes (adding a few drops of food colouring to the egg white and glucose mixture before combining with the icing sugar), rolling it out and cutting out shapes using a shaped cutter or even with a template cut out of paper. You can make whatever you like to fit with your theme – bows, shoes or hearts, for example.
Vanilla wedding cake
The grand effect of this cake belies its relative simplicity to make. When it's cut into slices, guests may be pleasantly surprised to discover that each beautifully iced tier contains not a heavy fruitcake but a soft, moist sponge. The two things this recipe needs are time and equipment – a number of different tins (listed). It's well worth the effort, though – a beautiful centrepiece to that special day and a delicious cake that everyone will adore. Unless your oven is very large (or you have a double oven) and your mixing bowl is more than about 35cm (14in) in diameter, you will need to make the sponge for the cake in two batches. The instructions given below are based upon making the cake in this way.
**Prep time** : 40 minutes
**Baking time** : 1 hour 30 minutes
**Ready in** : 3 hours
**Serves** : 32–40
800g (1¾lb) butter, softened, plus extra for greasing
800g (1¾lb) caster sugar
800g (1¾lb) plain flour
8 tsp baking powder
16 eggs, beaten
For the icing
1kg (2lb 3oz) caster sugar
20 egg yolks
1kg (2lb 3oz) butter, softened
1 tbsp vanilla extract
To decorate
Fresh flowers, such as rosebuds, lilies or lavender; edible flowers, such as primroses, violets, lavender, borage flowers, mint leaves, lemon verbena leaves or sweet cicely; crystallised flowers or petals
Four round cake tins: one 30cm (12in) in diameter; one 25cm (10in) diameter; one 18cm (7in) in diameter; one 12cm (5in) in diameter
Large plate or 40cm (16in) square cake board
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides and line the base of each cake tin with baking parchment.
Cream half the butter until soft in a large bowl or in an electric food mixer. Add half the sugar and beat until the mixture is light and fluffy. Meanwhile sift together half the flour with 4 teaspoons of the baking powder.
Crack eight of the eggs into a separate bowl and whisk together just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Add the sifted flour and baking powder, folding in to combine. Tip the batter into the 30cm (12in) cake tin and bake for 50 minutes or until a skewer inserted into the centre comes out clean.
As the large cake bakes, repeat the last two steps above for the remaining ingredients and divide between the prepared tins, filling each to a depth of 2–3cm (¾–1¼in). Wait until the large cake has finished cooking and you've removed it from the oven before you mix the flour and baking powder into the wet ingredients to complete the second batch of batter.
Bake the cakes for 25–35 minutes or until well risen and golden and a skewer inserted into the centre of each cake comes out clean. They will all probably be ready at different times, so you'll need to keep a close eye on them. When each cake is ready, remove it from the oven and allow to cool in the tin for 10 minutes, then loosen around the edges using a small, sharp knife and carefully remove from the tin before transferring to a wire rack to cool down completely.
To make the buttercream icing, first make a sugar syrup. Place the sugar in a saucepan with 400ml (14fl oz) of water and set over a medium heat, stirring very regularly to dissolve the sugar. Once the sugar is dissolved, turn up the heat and bring to the boil. Continue to boil for 8–12 minutes or until a sugar thermometer reads 110°–115°C (230°-235°F), then remove from the heat. If you don't have a thermometer, this is the 'thread stage'. The mixture will be thick and syrupy and the last couple of drops that fall from a spoon will form a thread.
While the syrup is boiling, whisk together the egg yolks for 2–3 minutes in the cleaned bowl of the food mixer until pale and fluffy, then turn the speed to slow and add the hot syrup in a steady stream. Once the syrup has been incorporated, turn the speed back to high and continue to whisk until the mixture has cooled down. This will take about 15 minutes: you may want to cover the food mixer with a tea towel to avoid too much mess, as the contents of the bowl may splash during mixing. Continue whisking until the mixture is stiff and mousse-like. To test if it's thick enough, pull the whisk out of the mixture and draw a figure of 8 into the bowl with the mixture that's left on the whisk; the '8' should remain visible for a couple of seconds.
In a separate bowl (you may want to use the food mixer for this too, having first decanted the mousse mixture from the bowl), beat the butter until very soft. Then add the vanilla extract and the mousse mixture, adding a large spoonful of this at a time, and keep mixing until fully combined.
To assemble the cake, first slice each cake horizontally in half with a bread knife. Place one half of the 30cm (12in) cake on the cake plate or board, securing it to the plate or board with a few blobs of icing. Spread the top with a layer of icing about 3mm (⅛in) thick, then sandwich with the second half. Spread the top with more icing, then add one half of the 25cm (10in) cake. Continue with the rest of the layers, from the largest to the smallest, until you have a tiered cake, then carefully spread the rest of the icing all over the cake using a palette knife. To ensure that each 'step' of the cake looks well defined, it's best have a jug or bowl of boiling-hot water to hand in which to dip your palette knife before using it to smooth over the surface of the icing.
Decorate the finished cake with fresh or crystallised flowers or petals (see Decorating ideas). (Photographs.)
Lemon or orange wedding cake
Add the finely grated zest of four lemons or oranges and 50ml (2fl oz) of lemon or orange juice to the cake with the eggs and bake and assemble as in the main recipe. Replace the 1 tablespoon of vanilla extract in the icing with 2 tablespoons of lemon or orange juice and decorate the iced cake with either Candied or Crisp lemon or orange slices (see Decorating ideas).
Coffee wedding cake
Add 100ml (3½fl oz) coffee essence (ideally Camp or Irel) to the cake mixture along with the eggs, then cook and assemble the cake as. To the icing, add 2 tablespoons of coffee essence to replace the vanilla extract. When the cake is iced, carefully place about 225g (8oz) toasted skinned hazelnuts (for how to toast them) in a band at the base of each tier. In the centre of the top tier you could add a large flower or perhaps miniature models of a bride and groom.
Tip The cake can be made up to three days in advance if kept covered. Probably, you won't have an airtight box big enough to store this cake. In which case, the best thing to do is to keep the cake on a surface and cover it with a large upturned mixing bowl or a large plastic storage box upturned. If you don't have a box or bowl big enough, place wine bottles around the cake (making sure the bottles are taller than the cake), and drape cling film, tin foil or even a table cloth over the cake to cover it.
White chocolate anniversary cake
The elegance of this cake makes it ideal for an anniversary celebration. If it's a silver or diamond anniversary, you could decorate the cake with edible silver or diamonds! The cake itself consists of two delicate layers of sponge encased in a luscious white chocolate ganache. The glucose syrup is essential to the ganache, transforming the chocolate and cream into a thick buttery icing that won't set hard.
**Prep time** : 45 minutes
**Baking time** : 25 minutes
**Ready in** : 3 hours 30 minutes
**Serves** : 8–10
Butter, softened, for greasing
4 eggs
125g (4½oz) caster sugar
125g (4½oz) plain flour, sifted, plus extra for dusting
1½ tsp baking powder
50g (2oz) white chocolate, as chips or chopped into pieces Chocolate curls,
made using either dark, milk or white chocolate or a mixture, to decorate
For the ganache
300ml (11fl oz) double or regular cream
45g (1½oz) glucose syrup
400g (14oz) white chocolate, in drops or broken into pieces
75g (3oz) butter, softened and cut into 2cm (¾in) cubes
Two 18cm (7in) diameter cake tins
First make the ganache. Place the cream and glucose syrup in a large saucepan and bring to the boil, stirring occasionally, then immediately remove from the heat and stir in the chocolate. Stir to melt the chocolate – you may need to place it back over a low heat, just so all the chocolate is melted. Remove from the heat and allow to cool until tepid.
Add the butter to the melted chocolate, a few cubes at a time, beating the mixture until all the butter has been incorporated. Leave in the fridge for about 2 hours, stirring a couple of times as it cools. It should stiffen but still be spreadable. Make sure to take the ganache out of the fridge at least 30 minutes before icing the cake. It may be necessary to beat the mixture with a wooden spoon to make it spread more easily.
While the ganache is cooling, you can make the cakes. Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides of the cake tins and dust with flour, then line each base with a disc of baking parchment.
Using a hand-held electric beater or an electric food mixer, whisk the eggs and sugar together on a high speed for several minutes or until light and mousse-like.
In a separate bowl, mix together the flour, baking powder and chocolate pieces, then fold into the egg and sugar mixture. Divide the batter between the prepared tins and bake for about 25 minutes or until golden on top and lightly springy to the touch.
Remove from the oven and allow to cool in the tins for 10 minutes, then loosen around the edges using a small, sharp knife and carefully remove each cake before transferring to a wire rack to cool down fully.
When the cakes are cool, place one upside down on a cake plate or stand, then spread over some of the ganache to about 4–5mm (¼in) thick. Place the second cake on top the right way up and use the rest of the ganache to cover the cake completely. Using a palette knife or the back of a spoon, lightly flick the ganache to make peaks all over the cake. If you prefer, you can simply smooth the cake with a palette knife, dipping it from time to time into a jug or bowl of boiling-hot water.
Now you can decorate your cake, selecting one of the options below, if you like. You could otherwise leave the cake undecorated, or scatter with chocolate curls (see Decorating ideas), either in white or contrasting milk or dark chocolate.
Chocolate anniversary cake
Make the cake as above, using milk or dark chocolate instead of white chocolate in both the ganache and sponge. You could then sprinkle over dark chocolate curls (see Decorating ideas).
Silver/golden wedding anniversary
Use edible silver or gold glitter to scatter over the cake, or make patterns or spell out words using edible silver or gold balls. Another option is to cut out paper stencils for the first initial of each person in the couple. Place the stencils on the cake and sprinkle the glitter on and around them, then carefully lift the paper off the surface of the cake to reveal the two letters marked out in the ganache.
Ruby wedding anniversary
Scatter with crystallised rose petals (see Decorating ideas), or arrange hulled strawberry halves in a ring around the top of the cake and use red coloured sugar (see Letter or number cake) for dusting over stencils of the couple's initials.
Diamond wedding anniversary
Use edible diamonds (available from specialist food shops or online) to spell out the initials of the couple or, if you prefer, arrange them in a sparkling 'necklace' around the base of the cake.
Red velvet cupcakes
These red velvet cupcakes are perfect for Valentine's Day. Red velvet is a classic American cake that has become popular on this side of the Atlantic in recent years. The red tint was originally obtained by mixing cocoa powder with an acidic ingredient such as buttermilk or vinegar. It's worth giving it a helping hand with some red food colouring to make the red even more vivid. The velvet refers to the delicate and light crumb, which comes from folding in whisked egg whites. When decorated with the brilliant white American frosting, the striking colour contrast should make them irresistible to your Valentine.
**Prep time** : 15 minutes
**Baking time** : 15–20 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
100g (3½oz) butter, softened
200g (7oz) caster sugar
1 tsp vanilla extract
2 eggs, separated
175ml (6fl oz) buttermilk
½ tsp red food colouring
175g (6oz) plain flour
25g (1oz) cornflour
¾ tsp baking powder
1 tbsp cocoa powder
Pinch of salt
1 tsp white vinegar
1 tsp bicarbonate of soda, sifted
For the frosting
4 large egg whites
250g (9oz) caster sugar
Pinch of salt
To decorate
Crystallised rose petals made using pink or red roses, or red coloured sugar
12-cup muffin tray and 12 muffin cases
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
Preheat the oven to 180°C (350°F), Gas mark 4 and line the muffin tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is pale and fluffy, then add the vanilla extract and the egg yolks, and beat again.
Mix the buttermilk together with the food colouring – it should be very red, so add more if you need to. Sift the flour, cornflour, baking powder and cocoa powder into the butter and egg mixture and add the red buttermilk. Fold everything together gently but thoroughly.
Whisk the egg whites and salt in a large, spotlessly clean bowl until stiff peaks form. Add one-quarter to the batter and mix in. Add the remaining egg whites in two stages, folding them in gently with a spatula until just incorporated and retaining as much air in the mixture as possible.
In a small bowl, mix together the vinegar and bicarbonate of soda until the mixture bubbles up, then gently fold this into the batter. Quickly divide between the muffin cases, filling each case two-thirds full, then bake in the oven for 15–20 minutes or until the cupcakes are well risen and springy to the touch.
Leave in the tin for 5 minutes, then remove the cupcakes and place on a wire rack to cool down while you make the frosting.
Make sure the cupcakes are ready before you start, as this icing begins to set very quickly.
Place all the frosting ingredients in a heatproof bowl, add 2 tablespoons of cold water and set over a saucepan of simmering water. (The bowl should sit snugly over the pan with its base high enough above the water that it does not come into contact with it.)
Whisk slowly, using a balloon whisk, until the sugar has completely dissolved and the mixture is foamy. Continue to heat and whisk until the mixture reaches 60°C (140°F) when measured with a sugar thermometer – this will take about 4 minutes. If you don't have a thermometer, you can gauge whether the mixture is ready by how it feels and looks: it should be hot to the touch, glossy white in appearance and starting to thicken.
Quickly remove the bowl from the pan and pour the mixture into the bowl of an electric food mixer fitted with the whisk attachment. Alternatively, whisk in the original bowl using a hand-held electric beater. Whisk on a high speed for about 3–5 minutes or until the frosting is very thick, glossy and has cooled.
Use immediately to spread quickly over the cupcakes with a palette knife and use the back of a teaspoon to shape them, regularly dipping the spoon into a bowl of boiling-hot water. Or the icing could be quickly placed in the piping bag or freezer bag, if you prefer, and piped over the cupcakes (see Cake essentials). The icing sets very quickly at this stage, so speed is essential.
Once the icing has set, you can then decorate with the crystallised rose petals, using either one large petal or three small petals per cake, or sift a little red sugar over each cake.
Cream-cheese frosting
For a much easier and much faster icing, beat together 300g (11oz) of cream cheese and 300g (11oz) of icing sugar until smooth and then pipe or spoon this onto the cupcakes, using the back of a teaspoon to shape the icing on each cake.
Perfect Peanut butter cake
Peanut butter is simply sublime in cakes and puddings. A jar of it can always be found in our house, as the whole family adores its creamy, crunchy goodness. This cake uses dulce de leche (a type of caramel – see the note) and chocolate for a three-pronged attack of deliciousness! I make this cake for my children but can never resist cutting myself the biggest slice.
**Prep time** : 15 minutes
**Baking time** : 30 minutes
**Ready in** : 1 hour
**Serves** : 8–12
200g (7oz) butter, softened, plus extra for greasing
75g (3oz) crunchy or smooth peanut butter
200g (7oz) caster sugar
4 eggs
200g (7oz) plain flour
2 tsp baking powder
For the filling and icing
100g (3½oz) milk chocolate, in drops or broken into pieces
3 tbsp double or regular cream
50g (2oz) crunchy or smooth peanut butter
100g (3½oz) dulce de leche
To decorate
50g (2oz) salted peanuts, roughly chopped
1 tbsp icing sugar, for dusting
Two 20cm (8in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line with a disc of baking parchment.
Cream together the butter and the peanut butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Sift in the flour and baking powder and fold in gently.
Divide the mixture between the prepared tins and bake in the oven for about 30 minutes until golden on top and a skewer inserted into the centre comes out clean. Remove from the oven and allow to cool in the tins for 5 minutes, then loosen around the edges using a small, sharp knife and carefully remove each cake from its tin and leave on a wire rack to cool down completely.
To make the icing, place the chocolate in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat and beat in the cream, peanut butter and dulce de leche.
Place one cake upside down on a cake plate or stand, then spread over half the icing. Sandwich the second cake on top, sitting the right way up, and spread the rest of the icing over the top.
To decorate the iced cake, scatter over the peanuts and dust with icing sugar. Alternatively, you could scatter over peanut brittle or a handful of chocolate-covered peanuts (see Decorating ideas).
Across the world cake plays a fundamental part in cultural and religious festivities. Food is woven into the fabric of every important festival and cake will almost always make an appearance at some point in the festivities. This chapter includes seasonal classics, such as the festive cakes made to celebrate Christmas, but there are also less familiar recipes from other countries, such as Vínarterta from Iceland and Italy's panforte, both of which are traditionally made in the winter. I have also included some of my own recipes for marking other festivals such as Halloween and Bonfire Night. There is a great range of divine recipes here, from spicy and fruity to delicate and light, and any of them will make cake the most delicious part of your festivities.
09/ Festive
Halloween pumpkin squares
Barmbrack or Báirín breac
Bonfire Night cupcakes
Maple pecan cake
Ballymaloe mincemeat crumble cake
Mincemeat
Chocolate chestnut torte
Spiced pear and ginger cheesecake
Vínarterta
Christmas cake
Royal icing
Fondant icing
Miriam's light Christmas pudding
Pecan Yule log with caramel sauce
Clementine cake
Mincemeat muffins with eggnog icing
Fondant holly leaves
Panforte
Gaff's porter cake
Halloween pumpkin squares
These pumpkin squares are perfect fare for Halloween. The pumpkin makes the sponge very moist, while the gentle spiciness of the cake is offset by the zesty cream-cheese icing. I like to use black icing to draw spiders and webs on the squares, but they taste just as good unadorned, of course – in which case you could make and eat them at any time of the year!
**Prep time** : 30 minutes (excluding cooking the pumpkin)
**Baking time** : 1 hour
**Ready in** : 2 hours
**Makes** : 20 squares
450g (1lb) peeled butternut squash or pumpkin, chopped into 1cm (½in) cubes, or 1 × 425g tin of good-quality cooked pumpkin (such as Libby's)
200g (7oz) butter, softened
300g (11oz) soft brown sugar
4 eggs
300g (11oz) self-raising flour
2 tsp bicarbonate of soda
1 tsp ground cinnamon
½ tsp ground nutmeg
1 tsp ground ginger
1 tsp salt
150g (5oz) sultanas
Juice and finely grated zest of 1 orange
For the cream-cheese icing
400g (14oz) cream cheese
100g (3½oz) butter, softened
200g (7oz) icing sugar, sifted
Juice and finely grated zest of 1 orange
For the black icing
100g (3½oz) icing sugar, sifted
½-1 tbsp orange juice or water
A few drops of black food colouring
20 × 28cm (8 × 11in) roasting tin
Piping bag with a 4 or 5mm (¼in) plain or star-shaped nozzle or a freezer bag with 4 or 5mm (¼in) cut from one corner (optional)
To cook the fresh butternut squash or pumpkin, if using, place in a saucepan and add 500ml (18fl oz) of water. Bring to the boil and then reduce the heat and simmer, uncovered, for 20–25 minutes or until completely tender. Drain the water and leave to cool, then set aside 425g (15oz) of the cooked squash or pumpkin to use in the cake batter.
Preheat the oven to 180°C (350°F), Gas mark 4, then line the sides and base of the tin with baking parchment, with the paper coming just above the sides of the tin to enable the cake to be lifted out easily.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Mash the cooked squash or pumpkin with a fork, then add this or the tinned pumpkin to the cake mixture, beating all the time.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add these to the mixture, beating all the time. (It may look slightly curdled at this point, but don't worry.) Sift in the flour, bicarbonate of soda, spices and salt, add the remaining ingredients and fold in gently to combine.
Tip the batter into the prepared roasting tin and bake on the lowest shelf in the oven for about 1 hour or until a skewer inserted into the centre of the cake comes out clean. Remove from the oven and leave to cool for 10 minutes in the tin, then lift it out of the tin with the lining paper, peel away the baking parchment and leave on a wire rack to cool down completely.
As the cake cools, make the cream-cheese icing, beating together all the ingredients until well mixed and fluffy. When the cake is cool, spread the top with the icing, using a palette knife to smooth the surface. If you like, you can tap the icing all over using the flat side of the palette knife to create little peaks. Once the cake is iced, you can cut it into squares before making the black icing for decorating each cake.
Put the icing sugar and food colouring into a bowl and mix with just enough orange juice or water to make an icing the consistency of thick double cream. Fill the piping bag or freezer bag with the icing and carefully pipe a spider or web onto each of the iced squares.
Jack o' lantern cake
Make the cake mixture as in the main recipe, but bake it in a 25cm (10in) diameter round cake tin instead. Prepare one and a half times the quantity of the cream-cheese icing, adding a few drops of orange food colouring to the mixture, and cover the whole cake using a palette knife dipped in hot water for a smooth finish. Use a knife to mark the ridges of the pumpkin and, with the black icing, pipe on three triangles, two for the eyes and one for the nose, and an evil toothy grin (see photograph) and fill in the piped shapes with sugar sprinkles.
Barmbrack or Báirín breac
Barmbrack is a traditional Irish sweetened bread not dissimilar to the Welsh bara brith. In Gaelic it's known as báirín breac or 'speckled loaf', due to the way the dough is dotted with raisins or sultanas. When barmbrack was baked for Halloween, the tradition was to add to the cake mixture a pea, a stick, a piece of cloth, a coin and a ring. Each item had a special significance for the person who discovered it in their slice of cake. The person who received the pea wouldn't marry that year; the stick meant an unhappy marriage, the cloth indicated poverty and the coin riches, while the person who found the ring would wed within the year. Nowadays it's usually just a ring that's added to the mixture.
**Prep time** : 30 minutes
**Baking time** : 45 minutes
**Ready in** : 1 hour 45 minutes (if using dried yeast); 2 hours 45 minutes (if using fresh yeast)
**Serves** : 8
225g (8oz) strong white flour, plus extra for dusting
2 tbsp mixed spice
¼ tsp salt
25g (1oz) butter, plus extra for greasing
1 × 7g (¼oz) sachet fast-action dried yeast (or 14g/½oz fresh yeast – for method)
50g (2oz) caster sugar
150ml (5fl oz) milk
1 egg, beaten
200g (7oz) mixed dried fruit (either ready-mixed or your own mixture of sultanas, raisins and currants)
25g (1oz) chopped mixed peel
900g (2lb) loaf tin
First butter the sides and base of the loaf tin.
Sift the flour, spice and salt into a large bowl or into an electric food mixer and add the butter, yeast (if using fresh yeast, for the method) and sugar. Beat together either in the bowl or with the food mixer using the dough hook attachment.
Warm the milk just until tepid, then add to the flour mixture along with the egg. Mix until the dough comes together and then knead well – for 8 minutes by hand or for 5 minutes in the food mixer. Add the dried fruit and mixed peel and knead for a further 2 minutes to mix them in.
Tip the dough out onto a floured work surface (the mixture will be quite wet, but don't worry as this is fine) and mould into a rough loaf shape. Place in the prepared loaf tin, then cover with a light tea towel or napkin and leave to rise in a warm part of your kitchen (by a radiator, for instance, or a sunny window) for 1 hour or until doubled in size.
Meanwhile, preheat the oven to 180°C (350°F), Gas mark 4.
Remove the tea towel or napkin and bake for 45 minutes or until deep golden brown on top. When you think the loaf is ready, gently loosen the sides with a palette knife and tip it out of the tin. If it's fully cooked, it should sound slightly hollow when you tap it on the base and feel springy when you lightly squeeze the sides. Place it on a wire rack to cool.
Slice up the loaf and serve fresh – or it is delicious toasted and buttered.
Barmbrack made with fresh yeast
Mix 14g (½oz) of fresh yeast in a bowl with the tepid milk and leave for 2 minutes or until creamy. Mix with the beaten egg, then pour into the dry ingredients and knead as. Place the dough in a large bowl, cover with cling film and leave in a warm part of your kitchen to rise for 1–2 hours or until doubled in size. It's ready when you press your finger into the dough and it does not spring back (the indentation stays put). Remove the cling film, tip onto a floured work surface and knock back the dough – by punching it in and then kneading all the air out of it – for about 2 minutes. Now form the dough into a loaf shape and pop it into the tin before leaving it to rise as in the main recipe on the previous page.
Bonfire Night cupcakes
These cute little cupcakes with their faux fires make a delicious Bonfire Night treat. And they are somewhat kinder to your teeth than a toffee apple!
**Prep time** : 25 minutes
**Baking time** : 15 minutes
**Ready in** : 1 hour
**Makes** : 12 cupcakes
125g (4½oz) butter, softened
125g (4½oz) caster sugar
2 eggs, beaten
125g (4½oz) self-raising flour
1 tbsp cocoa powder
For the icing
75g (3oz) white chocolate, in drops or broken into pieces
75g (3oz) butter, softened
100g (3½oz) icing sugar, sifted
4 chocolate Flake bars
A few drops of orange food colouring
12-cup bun tray and 12 bun cases
Piping bag with a 2 or 3mm (⅛in) nozzle or a freezer bag with 2 or 3mm (⅛in) cut from one corner
Preheat the oven to 180°C (350°F), Gas mark 4, and line the bun tray with the paper cases.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy. Gradually add the eggs to the butter mixture, beating all the time. Sift in the flour and cocoa powder and fold in to combine.
Divide the mixture between the bun cases, filling each case three-quarters full, and bake for about 15 minutes or until well risen and lightly springy when you gently press one or two of the cakes in the middle. Allow to cool for 5 minutes in the tin, then transfer to a wire rack and allow to cool down completely.
Meanwhile, make the icing. Place the white chocolate in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat and allow to cool until barely tepid.
Add the butter and icing sugar and beat together until smooth, then spread most of the icing on the cupcakes, reserving 2 tablespoonfuls.
Break the Flake bar into shards (these will be the 'logs') and arrange on the top of each iced cupcake in the shape (roughly!) of a bonfire.
Add a few drops of orange food colouring to the remaining icing, then place in the piping bag or freezer bag and pipe small blobs of orange icing on top of the chocolate logs for flames.
Maple pecan cake
Thanksgiving is a festival about food and this maple pecan cake is a traditional dessert that brings together two of my favourite American ingredients. Maple syrup and pecan nuts are the very best of friends. That they both hail from the same continent may be a coincidence, but there is no better sweetener for pecans and no better nut to combine with maple syrup. The sponge here is quite light, with the maple-pecan flavour dispersed throughout. The icing tastes more strongly of the syrup, but this is counterbalanced by a little sharpness from the cream cheese. The pecans make a wonderfully crunchy addition, though you don't need to caramelise them if you'd prefer not to; just toast them instead.
**Prep time** : 20 minutes
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Serves** : 6–10
200g (7oz) self-raising flour
¼ tsp salt
200g (7oz) soft light brown sugar
125ml (4½fl oz) buttermilk
200g (7oz) butter, melted, plus extra for greasing
50ml (2fl oz) maple syrup
2 eggs, beaten
75g (3oz) pecans, finely chopped
For the pecan praline
50g (2oz) caster sugar
50g (2oz) pecans
For the icing
300g (11oz) cream cheese
25ml (1fl oz) maple syrup
200g (7oz) icing sugar, sifted
Two 18cm (7in) diameter cake tins
Preheat the oven to 180°C (350°F), Gas mark 4, then butter the sides of the cake tins and line the base of each tin with a disc of baking parchment.
Sift the flour and salt into a large bowl and mix in the sugar. In a separate bowl, whisk the buttermilk with the butter, maple syrup and eggs. Add the wet ingredients to the flour and sugar, mixing together to combine, then fold in the chopped pecans.
Divide the batter between the prepared tins and bake for 20–25 minutes or until golden on top and springy to the touch.
Remove from the oven and allow to cool in the tin for 10 minutes. Loosen around the edges using a small, sharp knife and carefully remove each cake from its tin before transferring to a wire rack to finish cooling down.
Meanwhile, make the pecan praline. First line a baking tray with a sheet of baking parchment, then place the sugar in a non-stick frying pan or saucepan and scatter the nuts over the top. Cook over a medium heat until the sugar turns a caramel colour. Do not stir the pan, but carefully 'swirl' it to allow the sugar to caramelise evenly and to coat the nuts in caramel. Alternatively, you could just toast the pecans instead (following the tip).
Pour the mixture into the paper-lined baking tray. Using two forks, separate the nuts to make sure they don't stick together, then set aside to cool and harden.
To make the icing, whisk together all the ingredients until smooth. Place one cake upside down on a plate and spread over some of the icing, so that it's about 5mm (¼ in) thick. Place the other sponge on top, right side up, then cover the whole cake with the icing, using a palette knife dipped in hot water to smooth the surface.
Decorate with the caramelised or toasted pecans to finish.
Ballymaloe mincemeat crumble cake
This is a lovely Christmas dessert that we make at the Cookery School. A very satisfying cake, full of flavour and texture: the mincemeat keeps it firmly festive, while the layer of sponge is delicate and light and the crumble topping good and crunchy. I've included a recipe for mincemeat, but feel free to use your own favourite variety. At other times of the year I like to replace the mincemeat with 300g (11oz) of raspberry jam. This cake is best served warm as a pudding, though if you have any left over it makes an excellent snack with a cup of tea or coffee.
**Prep time** : 25 minutes (excluding the mincemeat)
**Baking time** : 45–50 minutes
**Ready in** : 1 hour 15 minutes
**Serves** : 8–10
100g (3½oz) butter, softened, plus extra for greasing
100g (3½oz) soft light brown sugar
2 eggs
½ tsp vanilla extract
2 tbsp milk
175g (6oz) self-raising flour
550g (1lb 3oz) mincemeat (to make it yourself,
Icing sugar, for dusting
Double or regular cream, whipped, to serve
For the crumble topping
100g (3½oz) self-raising flour, sifted
75g (3oz) caster sugar
75g (3oz) butter, chilled and cut into 1cm (½ in) cubes
25g (1oz) flaked almonds
22cm (8½in) diameter spring-form or loose-bottomed cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4, and butter the sides and base of the cake tin. If you're using a spring-form tin, make sure the base of the tin is upside down, so there's no lip and the cake can slide off easily when cooked.
First make the crumble topping. Place the flour and caster sugar in a bowl, then add the butter and, using your fingertips, rub it in until the mixture resembles coarse breadcrumbs. Stir in the almonds and set aside.
To make the sponge, first cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl with the vanilla extract and milk for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and fold in gently to combine.
Tip the mixture into the prepared tin, then spoon in the mincemeat, spreading it evenly over the batter, before sprinkling over the crumble topping.
Place on the lowest shelf in the oven and bake for 45–50 minutes or until golden brown on top and a skewer inserted into the centre of the cake comes out just clean. Remove from the oven and allow to cool in the tin for 20 minutes, then loosen around the edges using a small, sharp knife and remove the sides of the tin before carefully transferring the cake to a serving plate.
Dust the cake with lots of icing sugar, then serve warm with softly whipped cream.
Mincemeat
This delicious mincemeat will keep happily in a cool dark place for at least a year. If you are preparing your own suet (the fat that surrounds the beef kidney), make sure that every trace of blood has been removed before you whiz it in the food processor, otherwise it will cause the mincemeat to go off.
**Prep time** : 20 minutes
**Cooking time** : 8–10 minutes
**Ready in** : 50 minutes
**Makes** : 2.7kg (6lb)
2 large cooking apples, peeled, cored and cut into large chunks
Juice and finely grated zest of 2 oranges and 2 lemons
250g (9oz) shredded suet, or butter, chilled and grated
275g (10oz) raisins
275g (10oz) sultanas
275g (10oz) currants
125g (4½oz) chopped mixed peel
650g (1lb 7oz) soft dark brown sugar
50g (2oz) nibbed (chopped) almonds, or chopped pecans
2 tsp mixed spice
75ml (3fl oz) Irish whiskey or brandy
Place the apple chunks in a small saucepan with 1 teaspoon of water, cover with a lid and cook over a low heat for 8–10 minutes or until the apples are cooked down to a pulp. Remove from the heat and set aside to cool.
Mix with all the remaining ingredients in a large bowl and place in sterilised jars (for sterilising jars, see the tip). Leave to mature for at least two weeks, if possible, before using. If properly stored (somewhere cool in sterilised jars), this mincemeat should last for 1–2 years.
Tip To prepare suet properly, first peel off all the papery membrane that surrounds the suet. Next, use a knife to cut away any traces of kidney, blood or any gristle. Then roughly chop the suet and put it into a food processor. Pulse a few times until it is the consistency of rough breadcrumbs. Suet, once prepared, can be frozen.
Chocolate chestnut torte
Chestnuts are a truly festive ingredient. Their beguiling sweetness makes them ideal for adding to stuffing for the Christmas turkey, as well as making into truffles, tarts and tortes. This flourless cake is almost mousse-like, the only solid ingredient being the chestnuts themselves. It must be allowed to cool completely before you remove it from the tin. I like to decorate this torte with marrons glacés – chestnuts candied in sugar syrup – if I can get hold of them. Hugely popular in southern France and northern Italy, they are available either whole or as a purée. (You can buy them in specialist food shops and delis.) The whole ones would look, and taste, fabulous with this cake.
**Prep time** : 20 minutes
**Baking time** : 30 minutes
**Ready in** : 1 hour
**Serves** : 8–10
200g (7oz) dark chocolate, in drops or broken into pieces
200g (7oz) butter, plus extra for greasing
200g (7oz) cooked and peeled chestnuts (available vacuum-packed; to roast them yourself, see the tip)
200ml (7fl oz) milk
3 eggs, separated
125g (4½oz) caster sugar
Pinch of salt
To decorate
Cocoa powder, for dusting (optional)
8–10 marrons glacés (optional)
23cm (9in) diameter spring-form or loose-bottomed cake tin with 6cm (2½ in) sides
Preheat the oven to 170°C (325°F), Gas mark 3, then butter the base and sides of the cake tin. If you're using a spring-form tin, make sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
Place the chocolate and butter in a heatproof bowl and set over a saucepan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Place the chestnuts in a separate saucepan with the milk and bring to the boil, then boil for 3 minutes before removing from the heat. Tip into a food processor and whiz for a few minutes to form a coarse purée.
In a large bowl, whisk together the egg yolks and 100g (3½oz) of the caster sugar for a few minutes or until light-textured and pale in colour. Stir in the melted chocolate along with the chestnut purée and mix in thoroughly.
In a separate, spotlessly clean bowl, whisk together the egg whites and salt until frothy, then add the remaining sugar and continue to whisk until the mixture forms stiff peaks. Fold carefully into the chocolate and chestnut mixture, then tip into the prepared tin.
Bake for about 30 minutes or until just set – a skewer inserted into the centre of the cake will still be slightly sticky. Remove from the oven and place, in the tin, on a wire rack to cool down completely. Loosen around the edges using a small, sharp knife, then remove the sides of the tin before carefully transferring the cake to a serving plate.
Dust the top of the cake with cocoa powder or decorate with marrons glacés, if using. You can serve the torte warm, if you prefer, leaving it in the oven (preheated to 130°C/250°F/Gas mark½) for about 10 minutes to warm through.
Tip If you like, you can roast your own chestnuts over an open fire – the smell and the resulting sweet treats are divine, though it's a little easier to roast them in the oven. For the recipe above, you'll need 300g (11oz) of whole chestnuts. Use a sharp knife to cut a small cross in each nut, then place in a roasting tin and cook in the oven (preheated to 200°C/400°F/Gas mark 6) for about 30 minutes. Allow the nuts to cool before peeling them.
Spiced pear and ginger cheesecake
Cheesecake makes a fabulous alternative for the festive season. This recipe is very Christmassy in fact, as the pears are poached with mulled wine spices before being added to the cheesecake topping. The base contributes even more to the festive mix, with the ginger nuts, ground ginger and mixed spice.
**Prep time** : 30 minutes
**Baking time** : 40–45 minutes
**Ready in** : 3 hours
**Serves** : 8–10
Butter, for greasing
450g (1lb) cream cheese
150g (5oz) caster sugar
4 eggs, beaten
For the poached pears
50g (2oz) caster sugar
1 cinnamon stick
1 star anise
3 whole cloves
4 pears, peeled, quartered and cored
For the biscuit base
175g (6oz) ginger nut biscuits
1 tsp ground ginger
1 tsp mixed spice
75g (3oz) butter, melted
23cm (9in) diameter spring-form cake tin with 6cm (2½ in) sides
Preheat the oven to 180°C (350°F), Gas mark 4. Butter the sides and base of the cake tin, making sure the base is upside down, so there's no lip and the cake can slide off easily when cooked.
First poach the pears. Place the sugar and spices in a saucepan with 200ml (7fl oz) of water and set over a medium heat. Heat the syrup, stirring until the sugar has dissolved, then raise the heat and bring to the boil. Reduce the heat to a simmer, then add the pears and cover with a disc of baking parchment.
Simmer for 10 minutes or until the pears have softened, then remove from the syrup and place on a plate to cool. Reserve the poaching syrup.
Next make the biscuit base. Place the biscuits and spices in a food processor and whiz for a few minutes or until the mixture resembles coarse breadcrumbs. Add the melted butter and whiz again just until mixed. Alternatively, place the biscuits in a plastic bag and crush them using a rolling pin, then tip into the pan with the melted butter, add the spices and mix together to combine.
Tip the mixture into the prepared cake tin, spreading it out evenly to cover the bottom of the tin and pressing it down firmly.
Next make the cheesecake topping. Pour the pear-poaching syrup into a bowl, using a sieve to catch the whole spices – which you can keep for decorating the cheesecake later. Add the cream cheese, sugar and eggs and whisk together until well mixed and smooth.
Arrange the pears in a single layer on the biscuit base, then pour over the cream-cheese mixture. Bake for about 40–45 minutes or just until golden on top – it will be slightly cracked on top and should wobble a little in the middle when you gently shake the tin. Remove from the oven and allow to cool in the tin, then place in the fridge and leave to set for 2 hours.
To serve, loosen around the edges using a small, sharp knife and then unclip and remove the sides of the tin. Use a palette knife or metal fish slice to loosen the bottom of the cake from the base and, with the help of the palette knife or fish slice, carefully ease the cake onto a plate. (If you are worried about sliding the cheesecake off the base of the tin, then fear not: you can always serve it from the base. Simply transfer it, base and all, to your serving plate.)
If you like, you can place the reserved whole spices on top of the cheesecake to serve.
Vínarterta
My mum is Icelandic and this is the cake that she remembers most from her childhood; it's a real favourite of both hers and mine. Traditionally made at the start of winter, it's then stored and eaten over the Christmas period as it keeps well. With seven layers, the cake does take some time to make, but it isn't tricky if you take it step by step. After making, the cake should be covered and stored for a week to let it soften and 'mature'. It will keep for a few months and steadily improves, as the flavours are then given the chance to really get to know each other. You will need to cook this recipe in batches, as it's unlikely you'll have enough baking sheets, oven space or wire racks! For that reason, it's worth putting the remaining dough in the fridge while you cook the first or second batch, as it shouldn't really be out of the fridge for more than about 40 minutes before baking because it becomes difficult to roll.
**Prep time** : 1 hour 15 minutes
**Baking time** : about 1 hour (depending on the number of baking sheets and amount of oven space)
**Ready in** : a week!
**Serves** : 12–14
450g (1lb) plain flour, plus extra for dusting
2 tsp baking powder
½ tsp salt
1 tsp ground cardamom (to grind it yourself, see the tip)
225g (8oz) butter, softened
225g (8oz) caster sugar
2 eggs
1 tbsp milk
1 tsp vanilla extract
Icing sugar, for dusting
For the filling
975g (2lb 2½oz) pitted prunes
200g (7oz) caster sugar
1 tsp ground cardamom (to grind it yourself, see the tip)
1 tsp vanilla extract
First make the cake dough. Sift the flour, baking powder and salt into a bowl and mix in the ground cardamom. Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Mix in the milk and vanilla extract, followed by the dry ingredients, and stir together until combined.
Flatten out slightly to form a thick disc – if it's in a ball, the middle won't chill sufficiently – then wrap in cling film and leave in the fridge to chill for 30 minutes while you make the filling.
Place the prunes and sugar in a saucepan, add 400ml (14fl oz) of water and set over a medium heat. Cover and cook for 15 minutes, stirring regularly, until the prunes are tender, then remove the lid and cook uncovered for another 5 minutes, stirring regularly, until the mixture has thickened.
Remove from the heat and allow to cool for 5 minutes before blending until smooth with the ground cardamom and vanilla extract in a food processor or using a hand-held blender. Set aside and leave to cool down completely.
Preheat the oven to 180°C (350°F), Gas mark 4.
Remove the dough from the fridge and divide it into seven equal pieces, each about 135g (4¾oz) in weight. Shape into balls, then press down with the palm of your hands and place on to floured sheets of baking parchment before rolling out into circles 23cm (9in) in diameter using the base of a cake tin or a plate as a guide. (It's best to roll out the number of circles that will fit into the oven in one batch, keeping the rest of the dough covered in the fridge until you're able to use it.) When rolling out, try and keep the dough in a circle and use the rolling pin to enlarge it, tucking it in with your hands to shape the dough into a disc if it is losing its circular shape. Make sure, too, that it isn't sticking to the parchment by dusting frequently with flour and lifting the dough using a palette knife or metal fish slice. Trim each circle to the right size, if necessary, using a sharp knife and the cake-tin base or plate as a guide.
Transfer each circle to a baking sheet (no need to grease or line the sheet) and place in the oven. Bake for 12–16 minutes or until golden brown around the edges, then allow each cake to cool on its sheet for 2–3 minutes before running a palette knife or metal fish slice underneath and lifting out onto a wire rack to finish cooling. Repeat as necessary until you have seven quite thin, biscuit-like cakes.
To assemble the cakes, place one on your serving plate, then use a knife (a palette knife is ideal) to evenly spread over the filling. It should be spread quite thickly (to about 5mm/¼in), but if you're concerned that you might be using too much, simply weigh the filling – you need six layers' worth in total, allowing 200g (7oz) for each layer. Cover with another round of cake, then spread over more filling. Keep going until all seven layers of cake are sandwiched together with six layers of prune filling.
Place in an airtight container or cover with cling film and store for a minimum of a few days, ideally a week. After a week, dust generously with icing sugar and cut into slices to serve. This cake will keep for at least a month if stored in an airtight container.
Christmas cake
Christmas cake is a classic for a reason: the strong fruit flavours go so well with the almond paste and sweet white icing. The cake also keeps for months, so you can serve slices whenever people drop by during the Christmas holidays. This recipe is a play on the classic, with the slightly spicy addition of crystallised ginger. There are three icing paths you can take, too. The first, outlined in the main recipe, is to cook the cake again once it's been covered in almond paste – toasting the paste really accentuates the almond flavour, and my mother-in-law, Darina, has been doing it like this for years. The second option is to ice over the almond paste with royal icing. Made from whisked egg whites and sugar, this icing will dry to a crisp shell, and can be 'peaked' all over to resemble snow. The third option is to use fondant icing, also known as sugar paste, which is rolled out and laid over the almond paste and can be made perfectly smooth. Choose whichever option you prefer.
**Prep time** : 1 hour 30 minutes
**Baking time** : 2 hours 30 minutes-3 hours 15 minutes
**Ready in** : 7 hours (or a few days)
**Serves** : 15–20
275g (10oz) butter, softened, plus extra for greasing
275g (10oz) soft light brown sugar
5 eggs
1 tsp finely grated orange zest
75g (3oz) ground almonds
275g (10oz) plain flour
2 tsp mixed spice
For the fruit
225g (8oz) sultanas
225g (8oz) raisins
125g (4½oz) pitted dates, chopped
50g (2oz) currants
125g (4½oz) chopped mixed peel
50g (2oz) crystallised ginger, finely chopped
125ml (4½fl oz) brandy or whiskey
For the almond paste
450g (1lb) ground almonds
450g (1lb) caster sugar
1 egg
2 tbsp brandy or whiskey
A couple of drops (not more than ⅛ tsp) of almond essence or extract
Icing sugar, for dusting
For brushing the cake
1 small egg white, lightly beaten
2 egg yolks, lightly beaten
23cm (9in) diameter cake tin with 6cm (2½ in) sides or 20cm (8in) square tin with 5cm (2in) sides
Sugar paste cutters in festive shapes (optional)
Large plate or a 28cm (11 in) square cake board
First prepare the fruit for the cake. Place the dried fruit, mixed peel and crystallised ginger in a bowl, pour over the brandy or whiskey and allow to soak for at least 2 hours.
Preheat the oven to 150°C (300°F), Gas mark 2. Line the cake tin with baking parchment and wrap a collar of brown paper or a double layer of baking parchment around the outside to help prevent the cake from drying out as it bakes.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy, then add the eggs, one at a time and beating well between each addition.
Stir in the orange zest and ground almonds, then sift in the flour and mixed spice and fold in gently. Fold in the soaked dried fruit, together with any brandy or whiskey left in the bowl, and transfer the mixture to the prepared cake tin.
Bake in the oven for 2 hours 30 minutes to 3 hours 15 minutes or until a skewer inserted into the centre of the cake comes out clean. Cover the cake, still in the tin, with foil and allow to cool. Once the cake has cooled, carefully tip it out of the tin and wrap in foil until you are ready to cover it with almond paste. (The cake will keep like this for a couple of months.)
Preheat the oven to 220°C (425°F), Gas mark 7.
To make the almond paste, first mix the ground almonds and sugar together in a bowl. In another bowl, beat the egg, add the brandy or whiskey and the almond essence or extract, then add to the dry ingredients and mix to a stiff paste. (You may not need to add all the egg mixture.) Sprinkle a work surface with icing sugar, turn the almond paste out of the bowl and gently knead until smooth.
Remove the foil and baking parchment from the cake. Take about half of the almond paste and place it on a work surface dusted with icing sugar. Roll out until it is slightly bigger than the cake itself and about 1cm (½in) thick. Brush the top of the cake with the lightly beaten egg white and turn it upside down onto the almond paste. Cut around the edge of the cake, then carefully turn the cake the right side up with the lid of almond paste attached to the top. (Alternatively, you could cut the almond paste out using the cake tin as a template, and place it on top of the cake.)
Next measure the circumference of the cake with a piece of string. Roll out one long strip of almond paste (or two shorter strips joined together) the same length as the string, and trim both short edges to the same height as the cake. Brush the cake and the almond paste lightly with egg white and press the strip against the sides of the cake, but without overlapping or there will be a bulge. Trim away any overlapping pieces of almond paste, then use a straight-sided tumbler to even the edges and smooth the join, and rub the cake well with your hand to ensure a nice flat surface. (If you would like to ice the cake at this stage, omit the next two steps and follow the recipes for making and applying either royal icing or fondant icing – .)
Now carefully place the cake on a large, greased baking sheet. Roll out the remainder of the almond paste to about 5mm (¼in) thick and cut out shapes in the paste, such as Santa Claus, using sugar paste cutters, if you like. Brush the whole surface of the cake with the beaten egg yolks and stick the shapes on top and around the sides, if you wish. Brush these with egg yolk as well.
Bake the cake for 10–20 minutes (not too near the top of the oven or it may burn) until it is golden and toasted. Remove from the oven, allow to cool, then carefully – using a palette knife or metal fish knife to ease it off the baking sheet – transfer to a serving plate or cake board.
It can be nice to tie a ribbon around the finished cake (whichever icing option you choose), then decorate the top with a sprig of holly leaves – either fresh or fondant – or whatever festive decorations you like, such as some adorable fondant penguins.
Royal icing
**Prep time** : 10 minutes
**Makes** :725g (1lb 9½oz)
4 egg whites
675g (1½lb) icing sugar, sifted
1⅓ tsp glycerine
Place the egg whites and icing sugar in a large bowl or in an electric food mixer. Using either a hand-held electric beater or the food mixer, whisk for several minutes or until the icing stands up in stiff peaks, then whisk in the glycerine just for a second or two to combine.
Use a palette knife to spread the icing all over the cake, covering the almond paste, then use the flat of the palette knife to gently lift up the icing in small peaks all over for a 'snow scene' effect.
Fondant icing
**Prep time** : 10 minutes
**Makes** : 675g (1½lb)
1½ egg whites, whisked
100g (3½oz) liquid glucose
½ tsp vanilla extract
600g (1lb 5oz) icing sugar, sifted, plus extra for dusting
In a bowl, mix together the egg white, glucose and vanilla extract. Place the icing sugar in a separate bowl or an electric food mixer and gradually add the egg-white mixture, beating continuously until all the ingredients come together.
Place the icing on a spotlessly clean worktop that has been generously dusted with icing sugar and knead it for a minute or two or until it is completely smooth on the surface.
Dust your worktop again with icing sugar and roll out the icing into a round about 30cm (12in) in diameter. Make sure that the worktop doesn't get sticky by regularly lifting up the icing with a palette knife or metal fish slice and dusting the work surface beneath it with icing sugar to stop the icing sticking to it.
Brush the almond paste with boiling-hot water so that the fondant icing will stick to it, then carefully lift the round of icing and place over the cake. Press and smooth the icing all over the cake with your hands. Use a straight-sided tumbler to roll all over the cake, then, with a sharp knife, trim away any excess from the 'skirt'. Keep rolling over the cake with the tumbler for a very smooth and neat finish. If making a square cake, press something flat like a hardback book against each side to neaten and flatten the sides.
Miriam's light Christmas pudding
This recipe comes from my friend Miriam Murphy, using a recipe that was handed down from her grandmother, to her mother, to her. Miriam's grandfather could no longer eat Christmas pudding as he found it too heavy, so his wife created this light version of the traditional dish using cider, to enhance the fruitiness of the pudding, and no suet. There are no spices either, keeping things simple, but if you'd like it spicier you could try adding a teaspoon of cinnamon or mixed spice. The brandy cream makes a lovely light alternative to brandy butter, although you can use plain whipped cream if you prefer.
**Prep time** : 30 minutes
**Baking time** : 2 hours-2 hours 30 minutes
**Ready in** : 6 hours (but do make at least a week ahead)
**Serves** : 10–14
100g (3½oz) butter, softened
175g (6oz) soft light brown sugar
3 eggs
100g (3½oz) plain flour
75g (3oz) ground almonds
75g (3oz) fresh white breadcrumbs
50ml (2fl oz) brandy
For the fruit
225g (8oz) mixed dried fruit (either ready-mixed or your own mixture of sultanas, raisins, currants and/or cranberries)
75g (3oz) chopped mixed peel
75g (3oz) glace cherries, halved
1 × 300ml can of cider
For the brandy cream (optional)
200ml (7fl oz) double or regular cream
1–2 tbsp icing sugar, sifted
3 tbsp brandy or rum, or 3 tbsp Cointreau with 1 tsp finely grated orange zest or double or regular cream, whipped, to serve
1.8 litre (3 pint) ovenproof pudding basin
To prepare the fruit for the pudding, place the mixed dried fruit in a bowl with the chopped peel and glacé cherries and pour over the cider. Mix together and leave to soak for at least a few hours or preferably overnight
When you're ready to make the pudding, preheat the oven to 140°C (275°F), Gas mark 1.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Sift in the flour and tip in the ground almonds and breadcrumbs, followed by the soaked fruit and any remaining cider, then stir everything together until well mixed.
Spoon the mixture into the pudding basin and level the top. Gently bang the bowl on the work surface to release any air bubbles.
Next cut a disc of baking parchment about 6cm (2 ½in) larger in diameter than the rim of the basin. Place it over the pudding basin and tie with a long piece of string under the lip of the bowl. Place in an ovenproof saucepan a bit bigger than the basin and carefully pour hot water around the bowl to come three-quarters of the way up the sides.
Cover with the saucepan lid and bring to the boil, then immediately transfer to the oven to cook for between 2 hours and 2 hours 30 minutes, checking once or twice during cooking that the water in the saucepan has not evaporated and topping up with water as necessary.
The pudding can also be cooked on the hob, rather than in the oven, for the same amount of time on a low heat. The pudding is ready when a skewer inserted into the centre comes out moist, but with no mixture sticking to it.
Remove the bowl from the water and allow to cool just slightly, then remove the paper lid and pour over the brandy. Either serve immediately or place the baking parchment lid over the top again, then cover with foil and store in a cool place, where it will keep for up to two months.
To serve, reheat (if necessary) by steaming for 1 hour in the same way as above, and make the brandy cream, if using. Whip the cream until almost stiff and fold in the remaining ingredients. When the pudding is ready, turn it out onto a plate and serve with the brandy cream or a dollop of plain, softly whipped cream.
Pecan Yule log with caramel sauce
This retro-style festive treat makes the perfect centrepiece for a Christmas party or dinner. The cake itself is given a nutty flavour and slight crunch by the toasted pecans, which are ground before being mixed into the batter for the sponge. The bark effect for the log is easily achieved by running a fork along the icing. Served with a rich caramel sauce, this cake not only looks fantastic, but tastes as good as it looks!
**Prep time** : 30 minutes
**Baking time** : 15–20 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 8–12
250g (9oz) pecans, toasted (see the tip)
6 eggs, separated
½ tsp salt
125g (4½oz) caster sugar
75g (3oz) butter, melted
25g (1oz) plain flour
For the icing and sauce
275g (10oz) caster sugar
275ml (9½fl oz) double or regular cream
100g (3½oz) butter, softened and cut into cubes
½ tsp salt
350g (12oz) dark chocolate, in drops or broken into pieces
To decorate
Sprig of holly leaves
Icing sugar, for dusting
23 × 30cm (9 × 12in) Swiss roll tin Large plate or large cake board
Preheat the oven to 180°C (350°F), Gas mark 4, then line the sides and base of the Swiss roll tin with baking parchment, with the paper coming just above the sides of the tin to enable the cake to be lifted out easily.
Place the toasted pecans in a food processor and pulse a good few times until the nuts are finely chopped, but not fully ground.
Place the egg whites and salt in a large, spotlessly clean bowl or in an electric food mixer. Using a hand-held electric beater or the food mixer, whisk until foamy. Tip in 50g (2oz) of the sugar and continue to whisk until stiff and glossy peaks form, then set aside.
In a separate bowl, whisk the egg yolks with the rest of the sugar for a few minutes or until thickened. Add the egg-white mixture and the melted butter and fold in carefully, then sprinkle over the finely chopped pecans and sift in the flour, folding it in to combine.
Tip into the prepared tin, smoothing the batter with a spatula to form an even layer. Bake for 15–20 minutes until it feels lightly springy in the centre, then cover with a clean, slightly damp tea towel and place the cake, still in the tin, on a wire rack to cool.
Next make the caramel. Place the sugar in a saucepan with 50ml (2fl oz) of water and set over a medium heat. Cook, stirring regularly, until the sugar dissolves, then increase the heat and bring to the boil. Once the mixture has come to the boil, remove the spoon and continue to boil – without stirring, though you can swirl the pan every so often to help it cook evenly – for about 10 minutes. The caramel is ready when the mixture has turned a deep golden whiskey colour.
Remove the pan from the heat and whisk in the cream followed by the butter a few bits at a time, followed by the salt – the mixture will bubble vigorously. Place the pan over the heat again and stir until all the caramel bits dissolve, then remove from the hob and measure out 225ml (8fl oz) of this sauce, pouring it into a jug for serving later.
To make the icing add the chocolate to the remaining caramel in the saucepan and whisk until smooth. Transfer to a bowl and allow to stand, stirring occasionally, until spreadable. This should take about an hour if left in a cool place, but not the fridge as it will solidify.
When the cake is cool and you're ready to roll it up (though it will sit very well like this for 24 hours), remove the tea towel covering the cake, lift the cake out with the baking parchment still attached and place on a work surface next to your cake board or plate.
Spread some of the icing over the cake as if you were thickly buttering a slice of bread, then, beginning at one short side, and using the baking parchment to help, carefully roll up the cake. It may crack a little, but don't worry about this. Transfer the rolled-up cake from the baking parchment to the serving plate or board, removing the paper. Position the cake so the seam is at the bottom, then carefully tuck in the edges of the roll to neaten it.
Roughly 5cm (2½ in) from one end of the log, make a cut running diagonally across the cake, with the other end about 12cm (5in) from the end of the log (see photograph). Spread a little of the icing on the cut side of the smaller section of cake and then place it about a third of the way down the log to look like a branch, pressing it in on its iced surface.
Now spread the rest of the icing all over the cake to completely cover it. To decorate, use a fork to make lines all along the log and branch to look like bark, making concentric circles at each end. This may get a little messy, so if you like you can neaten the serving plate or board by cleaning around the log' with kitchen paper.
Decorate with the holly leaves and lightly dust with icing sugar (to resemble snow). The cake will keep somewhere dry and cool (but not the fridge) for 2–3 days, covered with a large upturned bowl.
When you're ready to serve, cut into slices and pour over the reserved caramel sauce (reheating this if you like).
Clementine cake
Intensely moist and citrusy, this cake will be an expression of whichever fruit you'd prefer to use, with the flavour of its juice and zest. You could use Clementines, tangerines, satsumas or even mandarins if you can get hold of them. The syrup in this cake means it will last for up to 10 days if kept in an airtight container.
**Prep time** : 20 minutes
**Baking time** : 40–50 minutes
**Ready in** : 1 hour 45 minutes
**Serves** : 8–12
200g (7oz) caster sugar
100g (3½oz) ground almonds
50g (2oz) fresh white breadcrumbs
2 tsp baking powder
4 eggs, beaten
175ml (6fl oz) sunflower oil, plus extra for greasing
Finely grated zest of 5 Clementines
For the syrup
75g (3oz) caster sugar
Juice of 3 Clementines
20cm (8in) diameter cake tin
Preheat the oven to 180°C (350°F), Gas mark 4, then grease the sides of the cake tin with sunflower oil and line the base with a disc of baking parchment. (It's a good idea to place a baking sheet under the tin before cooking this cake, or line the floor of the oven with a sheet of foil, in case there are any leaks.)
Place the sugar in a bowl and mix together with the ground almonds, breadcrumbs and baking powder. In a separate bowl, whisk together the eggs, sunflower oil and clementine zest. Pour the wet ingredients into the breadcrumb mixture and stir everything together until they are thoroughly mixed, then tip the batter into the prepared tin.
Bake on a lower shelf in the oven for 40–50 minutes or until lightly springy in the centre, golden brown on top and a skewer inserted into the centre of the cake comes out clean.
As the cake cooks, peel two of the Clementines and slice crossways into rounds about 5mm (¼in) thick.
Next make the syrup. Place the caster sugar and clementine juice in a saucepan and set over a medium heat, stirring to dissolve the sugar. Bring to the boil and allow the mixture to boil, without stirring it, for 3 minutes or until slightly syrupy. Remove from the heat and set aside.
When the cake is cooked, take it out of the oven and let it sit in the tin for 10 minutes. Loosening around the edges using a small, sharp knife, carefully remove the cake from the tin and transfer to a serving plate.
Arrange the clementine slices over the top, then reheat the syrup and pour it over the cake. Allow the cake to cool completely before serving.
Mincemeat muffins with eggnog icing
A fun, festive treat for anyone who's looking for an alternative to mince pies. The eggnog icing isn't strictly necessary but offers another layer of Christmas cheer. Freshly grating the nutmeg yourself will give an even better flavour than using ready ground. The mincemeat recipe here is very quick to make and can be used immediately, but you can of course use ready-made mincemeat if you prefer.
**Prep time** : 20 minutes (including the mincemeat)
**Baking time** : 20–25 minutes
**Ready in** : 1 hour
**Makes** : 12 muffins
200g (7oz) butter
150g (5oz) caster sugar
4 eggs
200g (7oz) plain flour
2 tsp baking powder
For the quick mincemeat
50g (2oz) mixed dried fruit (either ready-mixed or your own mixture of raisins, sultanas and currants)
50g (2oz) soft dark brown sugar
½ tsp mixed spice
1 cooking apple, peeled, cored and grated
Juice and finely grated zest of 1 orange
1 tbsp brandy or whiskey
_or_ 175g (6oz) ready-made mincemeat
For the icing
4 egg yolks
1 tsp cornflour
50g (2oz) caster sugar
25ml (1fl oz) brandy or whiskey
100ml (3½fl oz) double or regular cream
½ tsp finely grated nutmeg, plus extra for dusting
12-cup muffin tray and 12 muffin cases
Preheat the oven to 180°C (350°F), Gas mark 4, and line the muffin tray with the paper cases.
Place all the mincemeat ingredients in a saucepan and set over a medium heat, allowing it to simmer, uncovered, for 10 minutes or until nearly all the liquid has gone. Remove from the heat and set aside to cool.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the creamed butter mixture, beating all the time. Beat in the cooled mincemeat, then sift in the flour and baking powder and fold in gently to combine.
Divide between the muffin cases, filling each three-quarters full, then bake for 20–25 minutes or until well risen, deep golden on top and lightly springy to the touch. Remove from the oven and allow to cool for 5 minutes before removing from the tin and placing on a wire rack to cool down completely.
As the muffins cool, make the icing. Whisk together all the icing ingredients and place in a saucepan. Bring to the boil, stirring continuously, then remove from the heat and transfer to a large bowl or an electric food mixer. Using a hand-held electric beater or the food mixer, whisk continuously for a few minutes until the mixture has cooled and become very thick.
Spoon the icing onto the cooled muffins, and sprinkle with extra grated nutmeg for decoration. For an even more festive touch, you could decorate the tops of the muffins with fondant holly leaves, adding one or two to each cake, depending on the size of the leaves.
Fondant holly leaves
**Prep time** : 30 minutes
**Makes** : about 20–30 holly leaves
½ egg white, whisked
35g (1¼oz) liquid glucose
¼ tsp vanilla extract
A few drops of green food colouring
200g (7oz) icing sugar, sifted, plus extra for dusting
Holly leaf sugar paste cutter
In a bowl, mix together the egg white, glucose, vanilla extract and food colouring. Place the icing sugar in a separate bowl and gradually beat in the egg-white mixture.
Place the icing on a spotlessly clean work surface that has been dusted with icing sugar and knead for a minute or two or until it is completely smooth on the surface.
Dust your worktop again with icing sugar and roll out the icing to about 5mm (¼in) thick, then use a sugar paste cutter to cut out the holly leaves. Using a sharp knife, score one line along the length of each leaf and add lines radiating out to the spikes to form a leaf shape. Then place the leaves on the cakes, adding one or two per cake, depending on the size.
Tip This recipe uses half an egg white. The easiest way of halving an egg white is to first separate the egg, then whisk the white lightly in a bowl and pour off half.
Tip For additional shapes made out of fondant icing.
Panforte
Probably the oldest type of cake in this book, panforte dates back to the twelfth century. It hails from the Italian city of Siena, in which you'll find it in numerous guises. Panforte is similar to fruitcake but is made in a very different way: a mixture of honey, butter and sugar is boiled to make a thick syrup, then mixed with the other ingredients before being baked. The result is distinctively chewy and quite crunchy due to all the nuts in the cake. Traditionally eaten at Christmas, panforte is now available all year round. It can be served with tea or coffee, or as a pudding with a glass of sweet dessert wine. Once made, it will last, stored in an airtight container, for a good three months.
**Prep time** : 25 minutes
**Baking time** : 50–55 minutes
**Ready in** : 2 hours
**Serves** : 10–12
150g (5oz) hazelnuts, toasted, skinned and coarsely chopped (see the tip)
150g (5oz) pecan halves, toasted (see the tip) and coarsely chopped
75g (3oz) pitted dates, coarsely chopped
50g (2oz) candied orange peel, coarsely chopped
75g (3oz) dried cherries or dried cranberries
50g (2oz) dried figs, coarsely chopped
50g (2oz) plain flour, sifted
150g (5oz) honey
150g (5oz) caster sugar
25g (1oz) butter
23cm (9in) diameter cake tin
Preheat the oven to 150°C (300°F), Gas mark 2, and line the sides and base of the cake tin with baking parchment, with the paper coming just above the sides of the tin to enable the cake to be lifted out easily.
In a large bowl, mix the nuts and dried fruit with the flour and set aside.
Place the honey in a saucepan with the sugar and butter and set over a medium-low heat. Stir together, allowing the butter to melt, and continue to heat for 5–10 minutes or until a sugar thermometer reads 110°-115°C (230–235°F). If you don't have a thermometer, this is the 'thread stage'. The mixture will be thick and syrupy and the last couple of drops that fall from a spoon will form a thread. At this stage, immediately remove from the heat and add to the fruit and nuts.
Beat until well mixed: this is quite hard work and may take a while – it is easiest using an electric food mixer. Press the mixture into the prepared tin and bake for 50–55 minutes or until the cake feels firm in the middle.
Remove from the oven and place on a wire rack to cool down completely. When cool, remove from the tin with the lining paper, peel off the baking parchment and place on a serving plate.
Gaff's porter cake
Traditionally made for St Patrick's Day, this cake is flavoured with stout, for which 'porter' is another name. While you can serve it straight from the oven, it is best eaten a few days or even a week later, to give all the flavours a proper chance to get to know each other. This recipe comes from a friend of mine, Gaff, who swears this is the very best porter cake you'll find. You can make it with the stout of your choice, though Gaff prefers to use Murphy's.
**Prep time** : 15 minutes
**Baking time** : 1 hour 30 minutes
**Ready in** : 2 hours 30 minutes
**Serves** : 8–12
225g (8oz) butter, softened, plus extra for greasing
225g (8oz) soft light brown sugar
3 eggs
225ml (8fl oz) Irish stout (such as Beamish, Murphy's or Guinness)
450g (1lb) plain flour
½ tsp bicarbonate of soda
1 tsp mixed spice
1 tsp ground cinnamon
225g (8oz) sultanas
100g (3½oz) chopped mixed peel
450g (1lb) mixed dried fruit (either ready-mixed or your own mixture of sultanas, raisins and currants)
50g (2oz) almonds (skin still on), chopped, or nibbed (chopped) almonds
23cm (9in) diameter cake tin with 6cm (2½ in) sides
Preheat the oven to 150°C (300°F), Gas mark 2. Butter the sides of the cake tin and line the base with a disc of baking parchment, then double wrap the outside of the cake tin with extra baking parchment or with foil.
Cream the butter until soft in a large bowl or in an electric food mixer. Add the sugar and beat until the mixture is light and fluffy.
Whisk the eggs together in a small bowl for a few seconds or just until mixed, then gradually add them to the butter mixture, beating all the time. Pour in the stout and beat it in, then sift in the flour, bicarbonate of soda and spices and fold in with the remaining ingredients until fully incorporated.
Tip the mixture into the prepared tin and bake for 1 hour 30 minutes or until a skewer inserted into the centre of the cake comes out clean. Check the cake halfway through baking and if it is looking a deep golden brown already, cover with baking parchment or foil for the remainder of the cooking time.
Remove from the oven and allow to sit in the tin for 10 minutes, then loosen around the edges using a small, sharp knife and carefully remove the cake from the tin before leaving on a wire rack to cool completely.
Cut a slice now, if you like, but for a better flavour, store in an airtight container for a few days before eating.
Cake essentials
Cake making has a real magic to it – a few ingredients mixed together and then transformed behind the oven door – but rather than magic, much of it is pure science. The recipes are formulas that should be followed exactly: ingredients need to be carefully weighed and cake tins measured. There is no secret to cake making however, and you won't go wrong if you follow the steps in the recipe and keep in mind these tips and suggestions:
– Weigh all your ingredients carefully.
– Measure your cake tins, as the baking times in each recipe are specific to the size of the tin called for.
– Don't open the oven door too early or your cake may collapse.
– Make sure your oven is preheated to the correct temperature before placing the cake inside.
– When creaming make sure your butter is softened and cream butter and sugar together thoroughly; when I say light and fluffy, I really mean it.
Mixing the cake
When making the batter for a sponge, sugar is mixed with butter in the creaming process. It's important to do this step really thoroughly. If the mixture isn't light and fluffy, it won't have enough air bubbles and it won't combine properly with the eggs when you add them.
When you do add the eggs, doing so gradually will prevent the mixture from curdling. If it does curdle, add a tablespoon of the flour weighed out for adding after the eggs, and mix together before you add the remaining eggs. If the mixture does curdle a little at this point, don't worry as the cake will be fine when it's cooked.
When adding the flour to your sponge, it's generally folded or mixed in gently. Extended beating will work the gluten in the flour and toughen the batter, resulting in a heavier cake.
Another method for making sponges involves simply putting all the ingredients into a food processor or electric food mixer and whizzing just until combined. This produces a less delicate sponge but it is convenient, if you're short of time.
Some sponges are fat free. Eggs or egg yolks are whisked with sugar to make a light mousse; plain flour is then folded in with baking powder and the resulting sponge is very light.
When is the cake ready?
There are two good ways of checking when a cake is ready:
– A skewer inserted into the centre should come out clean with no raw mixture sticking to it; this shows that the middle of the cake – which takes the longest time to cook – is fully cooked.
– The centre of the cake should feel slightly springy to the touch and not at all wet.
Cake tins
How you remove a cake from its tin depends on the type of tin you're using. Below I've given instructions on the best way of doing this for each type of tin. In every case, the cake is left to cool the right way up, following my own preference. You can leave your cake to cool upside down, of course, but allowing it to cool upright means that there will be no grid marks left in the top of the cake from the wire rack, which can spoil its appearance.
Standard These are cake tins in one piece – they can be round or square – without a removable base. They are easy to use, but to remove a cake from them they must be tipped upside down. Some cakes may be too fragile for this, however, or may have a topping that needs to remain the right way up, in which case a loose-bottomed or spring-form tin would be more appropriate.
To remove a cake from a standard tin, first loosen around the edges using a small, sharp knife. Then place a spare plate upside down on top of the cake and carefully turn the cake over, still in its tin. Remove the tin and peel away any baking parchment, then place a wire rack upside down on top of the cake and carefully turn upright again, leaving the cake on the wire rack to cool down completely.
Loaf Suitable for making loaf-shaped cakes and fruit loaves, these also come in one piece. To remove the cake, follow the same procedure as for a standard tin. Where the tin is fully lined, as specified in some recipes, the cake can alternatively be lifted out with the baking parchment.
Spring-form This is a type of round cake tin with a removable base and sides that are kept together by a clasp, which is ideal if you can't turn the cake upside down and would like to slide it off the base onto a serving plate. Most spring-form bases have a 'lip', however, and if this is uppermost, you won't be able to slide the cake off easily. In each of the recipes in this book for which a spring-form tin is specified, I have included the instruction to turn the base of the tin over before you fill the tin.
When the cake is cooked, loosen around the edges using a small, sharp knife, then carefully unclasp and remove the sides of the tin. If the cake can be turned over, follow the same procedure as for a standard tin. If it can't be turned over, use a palette knife or metal fish slice to loosen the bottom of the cake from the base of the tin, then slide the palette knife or fish slice under the cake and ease it onto a serving plate.
Loose-bottomed This type of tin – either round or square – consists of two sections: a loose base and a ring or square-sided frame for the sides. The base doesn't have a 'lip', hence the cake can be slid off onto a serving plate or flipped over if the cake isn't too delicate for this.
When the cake is cooked, loosen around the edges using a small, sharp knife. Then place a heavy jar, sturdy glass or small bowl on your work surface and set the cake on top of it, still in its tin. Carefully push down the sides of the tin, then lift away the cake and base from the jar, glass or bowl. If the cake can be turned over, follow the same procedure as for a standard tin. If it can't be turned over, use a palette knife or metal fish slice to slide the cake from the base of the tin, as you would for a spring-form tin.
Sandwich This is a round tin generally with a diameter of 18–20cm (7–8in) and relatively shallow sides. Sandwich tins usually come in pairs, as they are used in recipes where two cakes are sandwiched together. They tend not to have removable bases, so cakes baked in them should be removed as for a standard type of cake tin.
Bundt This tin is used for making a ring-shaped cake, which has a ridge design that can be more or less elaborate depending on the tin. The resulting cake looks fantastic and as it's ring-shaped, with a larger surface area than one made in a standard tin, it has a crisper crust. It's especially important to make sure the inside of each ridge is greased with butter or oil before adding the cake mixture so that the cake can be tipped out easily. To remove the cake, first loosen around the edges using a small, sharp knife, then place a wire rack on top of the cake and flip it over. Remove the tin and leave on the rack to cool down fully.
Muffin trays These come in various sizes, but for this book a standard 12-cup tray has been used in every instance, where each hole is 7.5cm (2¾ in) wide, 3cm (1¼in) deep and has a volume of 100ml (3½fl oz).
Bun trays Bun trays also come in different sizes, but in this book a 12-cup tray is specified, smaller than the one used for muffins, where each hole is 6cm (2½in) wide, 2cm (¾in) deep and has a volume of 55ml (2fl oz).
Size of tin
All the recipes in this book call for a specific size of tin, hence it's important to measure the tin before you begin, to ensure that you have the correct one for the timings in a particular recipe. If you don't have a tin the right size but you have something similar, you can still bake the cake; you'll just need to adjust the cooking times slightly. If your tin is wider, the cooking time will be less as the cake will be shallower (the same amount of mixture but spread over a larger area). Likewise, if the width of your tin is slightly smaller, the cake will be deeper and will take a little longer to cook.
Lining cake tins
Lining a tin with paper helps to prevent the cake from sticking. I prefer baking parchment to greaseproof paper, as I find it works better. I also use non-stick silicone coated baking parchment as it can be used again and again. I keep discs of silicone baking parchment that I have cut out for each of my cake tins and I simply store them in the tins ready to bake, washing and drying them each time.
To line the base of a round or square cake tin, stand the tin on a piece of baking parchment or silicone baking parchment and draw around the base with a pencil, then cut out the circle or square and place in the base of the tin. To line the base and sides of a square tin or loaf tin, place the tin on a piece of baking parchment and draw around the base. Draw another square or oblong around the tin, larger than the tin itself this time, to mark out the paper that will line the sides of the tin. Cut around this square/oblong and then cut from each corner just to the corner of the smaller square or oblong. Fit the paper inside the tin and fold up the sides, allowing the paper to overlap at the corners and come above the sides of the tin so that the cake can be lifted out easily once baked.
Greasing tins
The sides of cake tins are almost always greased to prevent them from sticking to the cake. I think it's worth greasing even non-stick tins as they often do stick! To grease a tin, apply a little softened butter to the sides using a butter wrapper or a piece of baking parchment, making sure to cover the whole surface. If the recipe uses oil rather than butter, you can grease the tin using the same type of oil specified in the recipe.
When you've finished using a cake or loaf tin and have washed and dried it, place it in the oven while it's still warm from baking the cake to enable it to dry completely. This will help prevent it from rusting, keeping it in good condition so that the next cake you bake doesn't stick.
Timings
The following times appear at the top of each recipe to help you assess how long it will take to prepare and bake the cake in question.
**Prep time** This is the amount of time – after you've weighed all your ingredients – that you'll spend actually making the cake, whether that's creaming, whisking, mixing or icing.
**Baking time** This shows the amount of time the cake will be in the oven, allowing you to spend time doing something else.
**Ready in** This includes the prep. time, cooking time and any cooling time.
Ovens
Fan versus conventional
Conventional ovens radiate heat from the top and bottom. The oven temperature varies based on which shelf you use: the bottom shelf is the coolest part of the oven and the top shelf is the hottest. Fan or convection ovens circulate the heated air, so the temperature is roughly the same throughout. The great advantage of this is that you can bake your cake in any part of the oven, or use all of the oven's shelves at the same time if you're cooking more than one cake, without having to move the tins around for even baking.
All the recipes in this book were tested using a conventional oven and baking the cake on the middle shelf (unless otherwise stated), and all oven temperatures are for conventional rather than fan ovens. If you have a fan oven, reduce the temperature by about ten per cent and the cooking time by a few minutes (the latter varying, depending on your oven).
Cooking times
Never take the cooking time as gospel! Oven temperatures will vary and your oven may run hotter or cooler than the thermostat states. For the best results, buy an oven thermometer for checking the temperature.
Preheating the oven is essential when baking. It should be allowed to come up to the correct temperature before you put in your cake. Never put a cake into an oven when it's cold and then switch it on, as the cake won't cook properly.
Similarly, it's important not to open the oven too soon: if you do, there's a risk the cake will collapse. A good general rule is not to open the oven before three-quarters of the cooking time specified in the recipe has elapsed.
Other equipment
You really don't need anything elaborate or specialist to make a cake. A mixing bowl, a wooden spoon and a cake tin are pretty much all you need. Cakes are easily achievable with just those three things, although other equipment can come in handy, as you'll see from the list below:
Mixing bowls It's worth having a few different bowls of varying sizes. Pyrex bowls are especially useful. They are heatproof and are good for gently melting chocolate or heating eggs over simmering water. I also have a large stainless-steel bowl and a plastic one.
Spatula A large spatula is useful for scraping the last of the mixture from the bowl as well as for folding mixtures together.
Palette knife A long, rounded tool, this is useful for spreading icing and, being flat and flexible, for sliding underneath cakes to help manoeuvre them.
Whisk I like to have two different-sized whisks, a small one for whisking just a couple of eggs and a bigger one for whisking larger quantities of eggs or cream.
Rolling pin This isn't just for rolling out dough; it can also be used as a pestle to grind toasted nuts or spices, or for crushing biscuits to make a cheesecake base.
Grater Having a good-quality fine grater to hand makes grating lemon or orange zest so much easier.
Wire rack Metal cooling racks are important as they enable air to circulate round your cake, helping it to cool down quickly.
Pastry brush Used to brush on glazes or syrups or to brush away excess crumbs from your cake before icing it.
Piping bag A piping bag with a range or nozzles in different shapes and sizes is very useful if you like to decorate your cakes quite precisely. (For more on these,.)
Sugar paste cutters These come in all different shapes and sizes and they're great for quickly cutting out shapes from fondant icing to use as decoration. You can also use cookie cutters to make fondant shapes.
Icing turntable This is like a cake stand but with a plate that revolves. It makes icing easier but is really not essential. Instead, I tend to place a cake on an upturned plate when icing it, so that the 'lip' of the plate doesn't get in the way, making for a neater finish.
Cake board Large cake boards are useful for particularly big cakes and are available at most kitchen shops, but you could of course use a chopping board, large plate or even a tray instead.
Electric food mixer Almost all the cakes in this book can be made without using a free-standing electric food mixer, being achievable with just a large bowl, a wooden spoon and perhaps a whisk. However, a food mixer does take a lot of the work out of cake making, and if you're an avid baker I would really recommend getting one. If you don't have a food mixer then a hand-held electric beater is a good substitute, especially for making meringues – making them using just a hand whisk is a mighty task indeed!
Ingredients
Making self-raising flour
Self-raising flour is essentially plain flour that has been mixed with baking powder. If you don't have any self-raising flour, you can easily make your own by adding 2 level teaspoons of baking powder for every 150g (5oz) of plain flour and mixing or sifting together well before using. Or in case a smaller quantity of flour is needed, you can add 1 level teaspoon of baking powder for every 75g (3oz) of plain flour.
Using eggs
All the eggs used in this book are medium-sized.
When you're whisking egg whites, make sure to use a spotlessly clean, dry bowl and whisk or the egg whites won't properly whisk up into peaks.
To separate an egg, crack the centre of the egg on the side of a bowl, then hold it over the bowl and carefully break the shell into two halves, holding each half upright in each hand, with one half containing the yolk. Pass the yolk back and forth a few times between each half shell, allowing the white to drop down into the bowl until you are left with just the yolk. It is important not to let any of the yolk get into the white and it is a good idea to separate each egg into a smaller bowl before tipping it into a larger one containing the other egg whites, just in case you have an accident and some egg yolk escapes into your egg whites.
Egg yolks can be kept covered for use within two hours (although they are best used immediately). Egg whites can be stored for much longer; kept covered in the fridge, they will remain fresh for a few weeks provided they're used within the 'use by' date. Egg whites can also be frozen – they'll keep like this for up to two months.
Melting chocolate
Chocolate can go from melted to burnt quite quickly, so it should never be melted by being placed straight in a saucepan and set over the heat. Instead, it should be placed, broken into pieces, into a heatproof bowl and then set over a saucepan of simmering water. (The bowl should fit snugly in the pan, its base not touching the water.) This ensures a regular and relatively low heat. If the water is left at a rolling boil for too long, the chocolate will burn. Indeed, if you are unable to keep an eye on the chocolate to ensure it doesn't burn, it's a good idea to take it off the heat once the water comes to the boil. Left like this, it will just melt slowly, with no risk of overheating. Don't place melted chocolate in the fridge to cool, as this causes the fat to 'bloom' on the surface, creating an ugly white-grey mark.
Icing
Glacé icing Glacé icing is made simply by mixing a liquid (usually lemon or orange juice or water) with icing sugar. It can be thick or thin, depending on how much liquid is used.
Fondant icing Also called sugar paste, fondant icing is made from icing sugar, egg white and glucose syrup. It is mouldable and can be rolled out to cover cakes or shaped into anything you like, from roses to bows.
American frosting American frosting is made from egg whites and sugar that are whisked together and heated to make a meringue-like mixture which is then spread over the cake. American frosting will dry out on the surface but remain light and almost marshmallowy on the inside.
Buttercream icing There are two types of buttercream icing. The first and most simple kind – frequently used in this book – consists of butter and icing sugar beaten together so they are light and fluffy. Flavourings are also added, such as vanilla extract or lemon zest. The second type of buttercream icing, also known as French buttercream, is more complicated, consisting of a sugar syrup that is boiled and added slowly to whisked egg yolks. Although not as quick to make, it results in an icing that is lighter but still rich and creamy.
Cream-cheese icing A mixture of icing sugar, cream cheese and some additional flavouring, this icing is rich but offset by a slight tang from the cream cheese.
Royal icing Royal icing is a white icing made with egg whites and sugar. It should set quite hard but with the inclusion of a little glycerine, it won't set too hard.
If you'd like a smooth flat icing, keep a bowl of hot water close to the cake and periodically dip your palette knife into it. For a 'snowy' effect with royal icing, use your palette knife (not dipped in hot water) to gently lift small peaks all over the cake.
Piping bags
Piping bags are useful for precise cake decorating. The control you get with one means you can easily create exact shapes such as swirls or zigzags or write letters and spell out names. Different nozzles also allow you to make specific shapes. For example, using a star-shaped nozzle you create rosettes. It's amazing how effective a simple cake can look with a few piped rosettes of whipped cream or buttercream icing dotted around the top. They also look quite pretty as a base for candles to sit in on top of a birthday cake.
Making a piping bag
On those occasions where delicate decoration is needed or something quite precise, then there's no substitute for a piping bag or bottle with a small nozzle attached. Yet often a homemade paper piping bag, or even just a simple plastic sandwich or freezer bag with the corner snipped off, will suffice. A paper piping bag is not as sturdy as one made from plastic, but it still works well for piping something quite soft such as glacé icing, melted chocolate or American frosting.
To make a paper piping bag, cut out a piece of baking parchment, about 25cm (10in) square, then fold it in half to make a triangle. Cut along the fold to give you two triangles, enough for two piping bags. Hold one triangle in front of you with the long end on top (so that the triangle is upside down). If you are right-handed, bring (but do not fold) the top corner on the right down to the bottom, right-angled corner. Twist your hand to turn the paper so that it forms a cone, making sure the two corners are aligned. Using your left thumb and index finger, hold the two aligned corners in place so that the cone is upright with the held-together corners on the far side of the cone. Then with your right hand bring the so far unused corner down and wrap it around so that it is at the back of the cone, aligned with the other two corners. Make sure the point of the cone doesn't have a big hole: if you look down through the cone, you should not be able to see through it. You can secure the edges by either folding them in to hold everything in place, or stapling them together. Now you can fill the paper piping bag with your chosen icing, folding down the top to enclose your filling and using scissors to snip a tiny piece off the end – not too much or the filling will fall out too quickly.
Alternatively, using a plastic sandwich or freezer bag with 1–2mm (1/16in) cut out of one corner works very well.
How to fill a piping bag
Hold the bag with one hand upright and fold down the top of the bag with your other hand to form a 'cuff' over the hand holding the bag. Half fill the bag with icing – do not let it get too full or the icing may be forced out of the wrong end.
To close the bag, unfold your 'cuff' and twist the top of the bag closed, forcing the icing down the bag towards the nozzle. Hold the bag with your thumb and forefinger, keeping the twist closed. Place your other hand nearer the nozzle and use your top hand to gently squeeze the bag to push out the icing.
Using a piping bag
To write letters or precise shapes Select a plain nozzle for your bag, then fill with your chosen icing. Holding the bag at a 45° angle from the cake and using gentle but consistent pressure, carefully guide the nozzle to spell out whatever letter or shape you like. This is the method to use to draw on stripes, zigzags or spirals.
To pipe dots Select a plain nozzle for your bag, then fill with your chosen icing. Holding the bag as upright as you can, with the nozzle just above the cake, carefully squeeze out just enough icing to form a dot, then lift off and repeat as desired.
To pipe icing beads Select a plain nozzle for your bag, then fill with your chosen icing. Holding the bag so it is at a 45° angle, place the nozzle just above the cake and gently squeeze until a small dot is formed. Continue squeezing and pull the top down slightly to form a teardrop shape, then stop squeezing, lift off and repeat as desired.
To pipe rosettes Choose a star-shaped nozzle for your bag, then fill with your chosen icing. A liquid type of icing such as melted chocolate or glacé icing, or something that's very stiff like fondant icing, won't be suitable for making rosettes, whereas buttercream icing, American frosting, royal icing or whipped cream are ideal. Holding the bag upright over the cake, about 2cm (¾in) from the surface, squeeze it to make a rosette the size of your choice, pushing more icing through for a larger rosette. Lift off your piping bag and repeat as desired. If you'd rather not use a piping bag, you can use a spoon to drizzle patterns on the cake using glace icing. Place the bowl of icing close to the cake and dip your spoon into it, then lift the spoon out and let some icing drip off. Hold the spoon close to the cake as you drizzle on stripes, zigzags or swirls.
Decorating ideas
Candied lemon or orange slices
**Prep time** : 20 minutes
**Cooking time** : 30 minutes
**Ready in** : 1 hour 30 minutes
**Makes** : enough to cover two 20–25cm (8–10in) diameter cakes
200g (7oz) caster sugar, plus 4 tbsp for sprinkling 4 lemons or oranges (unpeeled), cut crossways into slices about 5mm (¼in) thick and any pips removed
Line a baking tray with baking parchment and sprinkle over 2 tablespoons of the caster sugar. Place the 200g (7oz) sugar and 500ml (18fl oz) of water in a saucepan or large frying pan and set over a medium heat, then bring to the boil, stirring occasionally. Add the lemon or orange slices and continue to boil for 5 minutes, turning the slices over halfway through.
Reduce the heat to low and keep the syrup simmering for about 30 minutes, turning over the slices occasionally. The syrup will reduce during cooking, but keep an eye on it and don't let it reduce too much. If it does it will caramelise and start to brown and you want to avoid this, otherwise the lemon or orange will caramelise too and harden, whereas you want them to remain soft.
After 30 minutes the lemon or orange slices will have softened and become translucent. Carefully remove them from the pan and place in the baking tray on the sugared baking parchment, spreading the slices out so they're not too close to each other. Allow to cool completely, then sprinkle with another 2 tablespoons of caster sugar. Either decorate your cake with the candied slices or place in an airtight container, where they'll keep for up to two months.
Crisp lemon or orange slices
**Prep time** : 5 minutes
**Cooking time** : 6–8 hours
**Ready in** : 6–8 hours
**Makes** : enough to cover two 20–25cm (8–10in) diameter cakes
4 oranges or lemons (unpeeled), cut into slices as thin as you can get them (2–3mm/⅛in if possible)
Icing sugar, for dusting
Preheat the oven to its lowest setting.
Place the citrus slices on a rack in a baking tray and leave in the oven for 6–8 hours or until completely crisp. When they are ready, remove from the oven and allow to cool. Dust with icing sugar and use to decorate your cake. Stored in an airtight box, these will keep for up to a week undusted, but once dusted in icing sugar they'll keep for only a day or two.
Chocolate-covered coffee beans
Sweet-tasting with a coffee kick, these work well scattered over or neatly arranged on top of a coffee or chocolate cake. They can be made up to a couple of weeks in advance and stored in a box or jar somewhere cool but not the fridge.
**Prep time** : 10 minutes
**Makes** : 100g (3½oz)
75g (3oz) dark, milk or white chocolate, in drops or broken into pieces
25g (1oz) coffee beans
First line a baking tray with baking parchment. Place the chocolate in a heatproof bowl set over a pan of simmering water. Stir occasionally just until melted, then remove from the heat.
When the chocolate has melted, stir in the coffee beans until they are well coated, then using two forks lift out a few coffee beans at a time and place them on the baking parchment. Separate out the beans using the forks so they aren't touching, then leave somewhere cool to set.
Chocolate-covered mint leaves
**Prep time** : 15 minutes
**Makes** : roughly 20 leaves
75g (3oz) dark chocolate, in drops or broken into pieces
A handful of good-quality, large fresh mint leaves
Line a baking tray with baking parchment.
Place the chocolate in a heatproof bowl and set over a pan of simmering water. Leave just until melted, stirring occasionally, then remove from the heat.
Holding a mint leaf by the base, or by the stalk if it's still attached, carefully dip into the melted chocolate. Lift it out and hold over the bowl to allow any excess to drip off, then place on the baking parchment to set. (You may find it easier to hold the mint leaves using tweezers.) Repeat with the rest of the mint leaves and place somewhere cool to set.
Chocolate-covered peanuts
**Prep time** : 10 minutes
**Makes** : 175g (6oz)
75g (3oz) dark or milk chocolate, in drops or broken into pieces
100g (3½oz) roasted salted peanuts
Line a baking tray with a sheet of baking parchment.
Place the chocolate in a heatproof bowl and set over a pan of just simmering water. Leave just until the chocolate has melted, stirring occasionally, then remove from the heat and stir in the peanuts.
When the nuts are well coated, tip onto the lined baking tray and spread out well so they don't cluster or stick together too much. Allow to cool until set, then add to your cake to decorate.
Chocolate squares
Wafer thin, these chocolate shapes make an impressive but easy decoration that can be made in advance (up to a couple of weeks) and stored until needed. They can be stuck around the top or sides of a cake, or inserted into piped rosettes of buttercream or whipped cream.
**Prep time** : 10–15 minutes
**Ready in** : 30–45 minutes
**Makes** : about 30 squares
100g (3½oz) dark, milk or white chocolate, in drops or broken into pieces, plus an additional 25g (1oz) (optional, using a contrasting type of chocolate to the main variety)
Line a baking sheet with baking parchment (alternatively, you could use the shiny side of a cereal packet, which works just as well).
Place the 100g (3½oz) of chocolate in a heatproof bowl, then set over a saucepan of simmering water and stir occasionally just until melted. Remove from the heat and pour onto the baking sheet or cereal packet, spreading the melted chocolate with a palette knife and carefully evening it out to just 2–3mm (⅛ in) thick.
If you like, you can also melt an additional 25g (1oz) of chocolate, using a contrasting variety to the 100g (3½oz), such as white or milk chocolate if you're using dark chocolate for the squares. When this has melted, drizzle over the thin layer of chocolate and leave as it is or swirl the different types of chocolate together using a small, sharp knife or a cocktail stick.
Leave somewhere cool, but not in the fridge, for 20–30 minutes or until the chocolate has set and is matt in appearance. Transfer to a chopping board, still with the paper attached, and trim with a large knife to straighten the edges.
Next cut the chocolate carefully to make squares or diamonds. You can cut in a grid pattern or, for diamonds, cut in parallel straight lines, then cut another set of parallel lines at an angle to the first ones (as in the photograph). Alternatively, use a cookie cutter to cut hearts, stars or whatever shapes you'd like.
Carefully lift the chocolate pieces off the baking parchment or cereal box, and store in an airtight box, somewhere cool but not in the fridge.
Chocolate curls
**Prep time** : 15 minutes
**Makes** : enough to cover one 18cm (7in) cake
300g (11oz) dark, milk or white chocolate, in drops or chopped into pieces
Place the chocolate in a heatproof bowl over a saucepan of water on a medium heat and bring to a rolling boil, then turn off the heat. Allow the chocolate to melt slowly, stirring occasionally.
Place a baking tray or roasting tin upside down on your work surface. Once the chocolate has completely melted, pour it over the upside-down tray or tin and, using a palette knife, spread it out so that it's 3–4mm (⅛in) thick and about the size of an A4 sheet of paper.
Place the chocolate somewhere cool (but not the fridge as this will be too cold – see tip) and allow it to slowly set. The chocolate is set when it is no longer shiny – it should become matt in appearance.
Using a swivel-bladed vegetable peeler or a cheese slicer, run along the top of the chocolate to shave off curls. Either place the curls directly onto a cake or, if you'd like to make them ahead of time, transfer them onto a plate or into an airtight box and place somewhere cool (but not the fridge). Stored in an airtight container, they will keep for up to two weeks.
Tip It's important that the chocolate is completely set before using it, yet not too cold. If it's too cold, it will not 'curl', but if it's not sufficiently set the curls will collapse.
Crystallised orange peel
**Prep time** : 10 minutes
**Cooking time** : 20 minutes
**Ready in** : 1 hour
**Makes** : enough to cover one 20–25cm (8–10in) diameter cake
2 oranges
200g (7oz) caster sugar
First line a baking tray with a sheet of baking parchment.
Peel the oranges very thinly with a peeler, being careful not to include the white pith. Cut the peel into thin strips (each as thin as a matchstick), then place in a small saucepan and cover with cold water.
Bring to a simmer and allow to cook for 5 minutes. Drain the peel and return to the saucepan. Cover again with fresh water and simmer for another 5 minutes, then drain once more and set aside.
Next make the syrup. Place 150g (5oz) of the sugar and 150ml (5fl oz) of water in the same saucepan that you used for boiling the zest, and set over a medium heat, stirring to dissolve the sugar. Bring to the boil and allow the syrup to bubble for 2 minutes, without stirring the mixture, then return the zest to the saucepan and simmer gently for 6–8 minutes – covering the pan with a lid if the syrup is thickening – until the strips of peel look glossy and translucent.
Remove from the syrup using a small sieve or slotted spoon and spread the zest out to cool on the paper-lined baking tray, reserving the orange-scented syrup for homemade lemonade or cocktails, or to make more crystallised peel another time.
Once the zest is cool, pour the remaining sugar over it and use your fingertips to coat each piece of orange peel in the sugar grains. Allow the crystallised peel to sit in a dry atmosphere for about 30 minutes or until it has dried slightly, then use straight away or store in an airtight box or jar where it will keep for up to a month.
Crystallised flowers
**Prep time** : 30 minutes
**Ready in** : about 3 hours
**Makes** : 50–60 petals or about 30–40 flowers (depending on size)
1 egg white, lightly whisked
About 75g (3oz) caster sugar
Edible flowers, such as primroses, violets or violas, or rose petals
Dip a small paintbrush into the egg white and carefully brush the flowers, using just enough egg white to cover the surface of each flower. Sprinkle the sugar over the petals, being careful to cover every petal.
Gently shake off any excess sugar and arrange carefully on a tray lined with baking parchment. Place somewhere dry (and even slightly warm, such as near a radiator or in an airing cupboard) and leave to dry for a few hours or overnight. Remove from the tray and use to decorate your cake. Stored in an airtight box, these will remain at their best for up to 1–2 weeks, after which time their colour begins to fade.
Tip for how to make fondant roses.
Honeycomb
Honeycomb looks seriously impressive but is actually not too tricky to make. It requires glucose syrup, which you can get from some supermarkets, as well as chemists and specialist food shops. It's easier to make if you have a sugar thermometer, but it's not essential as you can check for the 'hard crack' stage instead. The honeycomb can be stored in an airtight container, where it will keep for a couple of weeks. Try crumbling over ice cream or adding to melted chocolate.
**Prep time** : 15 minutes
**Cooking time** : 5–10 minutes
**Ready in** : 45 minutes
**Makes** : 500g (1lb 2oz)
325g (1½lb) caster sugar
50g (2oz) honey
125g (4½oz) glucose syrup
1 tbsp bicarbonate of soda
Line a large baking tray with baking parchment.
Place the sugar, honey and glucose syrup in a large saucepan and add 50ml (2fl oz) of water. Bring to the boil, stirring until the sugar dissolves, then reduce the heat to medium and simmer without stirring for 5–10 minutes or until the syrup turns a light golden colour and a sugar thermometer dipped into the mixture reads 149°C (300°F). Alternatively, check whether it has reached the 'hard crack' stage. To test for this, use a spoon to drop a small amount of syrup into a bowl of very cold water. If it's reached the right temperature in the pan, the syrup will solidify into hard, brittle threads that immediately break when you bend them. (Bear in mind that the syrup is now very hot and can easily burn; allow it to cool in the water for a few moments before touching it.)
Immediately remove the pan from the heat and quickly whisk in the bicarbonate of soda. The mixture will treble in size very quickly, so be careful it doesn't spill over. Pour onto the baking parchment, swirling the tray to spread the mixture evenly. Leave to cool down fully and harden before breaking into chunks. Cut with a knife or break with your hands into small pieces about the size of rough gravel. Use pieces to decorate your cake, then place the rest in an airtight container where it will keep for a couple of weeks. It is delicious stirred into ice cream or dipped into melted chocolate and allowed to set.
Peanut brittle
**Prep time** : 5 minutes
**Cooking time** : 20–25 minutes
**Ready in** : 30–40 minutes
**Makes** : 400g (14oz)
225g (8oz) butter, diced
225g (8oz) caster sugar
½ tsp salt
150g (5oz) salted peanuts, roughly chopped
Line a baking tray with baking parchment.
Place all the ingredients, except the peanuts, in a heavy-based saucepan with 50ml (2fl oz) of water and bring slowly to the boil, stirring until the sugar has dissolved. Reduce the heat to low and simmer for 15–20 minutes without stirring until a sugar thermometer dipped into the mixture reads 150°C (302°F). It will be deep golden brown in colour. Below this temperature and the brittle will not set; above and it will taste burnt. If you don't have a sugar thermometer, check that the 'hard crack' stage has been reached: if you drop a teaspoonful of the syrup into a bowl of iced water it will solidify into hard, brittle threads that immediately break when you bend them. (Bear in mind that the syrup is now very hot and can easily burn; allow it to cool in the water for a few moments before touching it.)
Immediately (and taking great care as it is very hot) add the peanuts to the syrup, without stirring the mixture, then pour onto the baking tray and swirl the tray around to spread evenly. It will begin to set almost straight away, so you need to work quickly.
Once the brittle has completely hardened, after 5–10 minutes, slam the baking tray on the counter to break into small pieces. Chop or break up the pieces if they are not small enough.
Scatter some of the peanut brittle over the cake to decorate, placing any remaining pieces in an airtight container where they will keep for 3–4 days.
Tip See the recipe for Hazelnut praline triple-layered cake if you would like to make hazelnut praline as a decorative topping for your cake.
Salted peanut praline
**Prep time** : 5 minutes
**Cooking time** : 10 minutes
**Ready in** : 45 minutes
**Makes** : 200g (7oz)
100g (3½oz) caster sugar
100g (3½oz) salted peanuts
First line a baking tray with a sheet of baking parchment.
Place the sugar in a frying pan or saucepan and scatter the nuts over the top. Set over a medium heat until the sugar turns a caramel colour. Do not stir, but instead carefully 'swirl' the pan to allow the sugar to caramelise evenly and to coat the nuts in caramel.
Pour the mixture onto the prepared baking tray and, using two forks, separate out the nuts to make sure they don't stick together, then allow to cool and harden. Roughly chop the salted peanut praline and scatter over your cake to decorate.
Stored in a jar or airtight container, this will keep for a couple of weeks.
Elderflower sugar
**Prep time** : 10 minutes
**Ready in** : 2–3 hours
**Makes** : 450g (1lb)
For sprinkling over 1 large cake
2 elderflower heads
Finely grated zest of ½ lemon
50g (2oz) caster sugar
For the store cupboard
About 20 elderflower heads
Finely grated zest of 2 lemons
400g (14oz) caster sugar
Carefully knock or pick just the flowers from the elderflower heads and mix in a bowl along with the lemon zest and sugar. Add half of the mixture to a food processor and whiz for a few minutes, then return to the bowl with the rest of the mixture.
Spread out on a baking tray lined with baking parchment and leave to dry somewhere warm – such as the side of a stove, perched over a radiator or in a sunny window – for a few hours or until the sugar has dried. Stored in a sealed container, the sugar will last for months if you want to make the larger quantity. Try sprinkling it over pancakes or porridge.
Tip for how to make decorative coloured sugar.
List of Recipes
30-day bran and pumpkin seed muffins
After dinner mint cake
Almond hedgehog cake
American frosting
Angel food cake
Anniversary chocolate mousse layer cake
Apple and blackberry oat muffins
Apple and walnut cake
Apple, oat and pecan bars
Apricot, pistachio and saffron muffins
Arctic roll
Baked Alaska with hot chocolate sauce
Bakewell cake
Ballymaloe mincemeat crumble cake
Banana cake with butterscotch sauce
Banana ice-cream cake with butterscotch sauce
Banana, almond and honey muffins
Banoffee blondies
Barmbrack or Báirín breac
Battenberg cake
Berry buttercream icing
Berry buttercream icing
Berry buttercream icing
Berry buttercream icing
Blueberry and coconut cake with lemon cream
Boiled fruitcake
Bonfire Night cupcakes
Brittany butter cake
Butterfly cake
Butterscotch sauce
Cake pops
Cappuccino squares
Caramel carrot cake
Caramelised orange upside-down cake
Caramelised oranges
Caramelised pear upside-down cake
Cardamom yoghurt cake
Carrot cake
Castle cake
Cheesecake brownies
Cherry marzipan cake
Chocolate and orange torte with caramelised oranges
Chocolate and pecan muffins with maple glaze
Chocolate and toffee ice-cream cake
Chocolate buttercream and white chocolate drizzle icing
Chocolate castle cake
Chocolate chestnut torte
Chocolate cupcakes
Chocolate fudge cake
Chocolate hazelnut cheesecake
Chocolate hedgehog cake
Chocolate honeycomb topping
Chocolate ice-cream cake with chocolate sauce
Chocolate letter or number cake
Chocolate meringue frosting
Chocolate teddy bear cake
Chocolate, rum and almond cake
Chocolate-clipped madeleines
Christmas cake
Classic NYC crumb cake
Clementine cake
Coconut and lime cake
Coffee mascarpone cake
Coloured buttercream icing
Coloured buttercream icing
Coloured buttercream icing
Coloured glacé icing
Coloured glacé icing
Coloured glacé icing
Coloured sugar
Crunchy peanut butter banana muffins
Dairy- and egg-free vanilla cupcakes
Dairy-free blueberry crumble cake
Dark treacly gingerbread squares
Date and almond honey cake
Double chocolate peanut butter brownies
Dulce de leche cake
Dundee cake
Easter nests
Fast cinnamon yoghurt cake
Fast fondant flowers
Flourless fig, pine nut and ricotta cake
Fondant birds
Fondant butterflies
Fondant dinosaurs
Fondant football shirts
Fondant ghosts
Fondant holly leaves
Fondant icing
Fondant icing
Fondant penguins
Fondant roses
Fondant snowmen
Gaff's porter cake
Ginger carrot cupcakes
Ginger golden syrup loaf
Ginger peach muffins
Gluten- and dairy-free pecan brownies
Halloween pumpkin squares
Hazelnut meringue cake
Hazelnut praline triple-layered cake
Hot chocolate fairy cakes
Hot chocolate sauce
Intensely chocolatey beetroot brownies
Irish coffee cups
Lamingtons
Lavender icing
Layered chocolate and caramel cake
Lemon and rosemary polenta cake
Lemon crumble cupcakes
Lemon curd
Lemon poppy-seed cake
Lemon poppy-seed muffins
Lemongrass coconut cake
Lemon-striped madeleines
Letter or number cake
Macadamia nut and lemon cakes
Madeleines
Maple pecan cake
Marbled chocolate crumble cake
Marmalade steamed pudding
Marzipan cake
Marzipan
Mincemeat muffins with eggnog icing
Mincemeat
Mini mango cupcakes
Mini wedding fruitcakes
Miriam's light Christmas pudding
Muscovado Madeira cake
Nutty toffee cake
Orange and almond syrup cake
Orange crème anglaise
Orange sour-cream cake
Panforte
Peanut butter letter or number cake
Peanut butter pops
Peanut butter topping
Peanut buttercream icing
Pear and ginger muffins
Pear crumble cake
Pecan Yule log with caramel sauce
Perfect Peanut butter cake
Petits gâteaux
Pistachio and rose-water buttercream icing
Pistachio cake
Praline buttercream icing
Princess cake
Ras el hanout spiced fruitcake
Raspberry and blueberry friands
Raspberry and coconut squares
Raspberry and white chocolate cheesecake
Raspberry frangipane cake
Red velvet cupcakes
Rhubarb and custard Swiss roll
Rose-water icing with pistachios
Rose-water pistachio cupcakes
Royal icing
Snowman cake
Spiced pear and ginger cheesecake
Sprinkles
St Clement's drizzle cake
Sticky toffee date cakes
Strawberry cupcakes
Strawberry ice-cream cake with strawberry coulis
Strawberry white chocolate cake
Sugar-free banana bread
Swedish apple cake
Sweet potato muffins
Tahini honey cake
Teddy bear cake
Toscatårta or Swedish almond cake
Tropical cake
Upside-down peach and saffron cake
Vínarterta
Vanilla wedding cake
Vegan chocolate cake
Vegan frosted lemon cake
Vegan raspberry muffins
Victoria sponge
Walnut and orange cake
White chocolate anniversary cake
White chocolate macadamia nut cake
Wholemeal chocolate cake
Winter breakfast muffins
Thank yous
My huge, heartfelt thanks to Lizzy Gray, Elizabeth Woabank and Hannah MacDonald and to the amazing team at Collins who brought this book together. Also thanks to: Martin Topping, Lucy Sykes-Thompson, Heike Schussler, Elen Jones, Annie Rigg, Polly Webb-Wilson, Phil Webb, Annabel Hornsby, Sam Head, Helen Cuddigan, Fiona Lindsay and everyone at Limelight, and Darina and Tim Allen and all at Ballymaloe.
Thanks to Irish designer Lucy Downes of Sphere One (www.sphereone.ie) whose gorgeous cashmere I wore throughout the photography and filming. An unbelievably ginormous thank you to Josh Heller.
Lastly, a great big thank you to my family to whom I dedicate this book: Isaac, Joshua, Lucca and Scarlett Lily.
Copyright
First published in 2012 by Collins
An imprint of HarperCollinsPublishers
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London W6 8JB
www.harpercollins.co.uk
FIRST EDITION
Text © Rachel Allen 2012
Photography © Philip Webb 2012
Front Cover Photography © Wendy Carrig. Back Cover Photography © Philip Webb
Rachel Allen asserts the moral right to be identified as the author of this work.
All rights reserved under International and Pan-American Copyright Conventions. By payment of the required fees, you have been granted the non-exclusive, non-transferable right to access and read the text of this ebook on-screen. No part of this text may be reproduced, transmitted, down-loaded, decompiled, reverse engineered, or stored in or introduced into any information storage and retrieval system, in any form or by any means, whether electronic or mechanical, now known or hereinafter invented, without the express written permission of HarperCollins.
A catalogue record for this book is available from the British Library.
The author and the publishers would like to thank the following for permission to include these cakes:
NYC crumb cake from _Baked in America_ by David Lesniak and David Muniz, Ebury Press, 2011. Vegan chocolate cake adapted by the kind permission of Giorgio Locatelli.
Coconut photograph © Image Source/Getty Images.
Source ISBN: 9780007309054
Ebook edition © August 2012 ISBN: 9780007490868
Version 1.0
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| {
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Matt Cooper, the host of Today FM's popular radio show The Last Word will interview the motor racing legend Tommy Byrne in a post-screening Q&A at the IFI's special opening night screening of Crash and Burn next Friday 2nd December. The film's award-winning director Seán Ó Cualáin will also take part in the live interview where audience members will have the opportunity to ask Tommy and Sean questions too.
The five-star reviewed Crash and Burn has been going down a storm with audiences at the Galway Film Fleadh and the Cork Film Festival in Ireland as well as the prestigious Sheffield DocFest and this week's screening in Picturehouse Central London in the UK.
The film will be released in Irish cinemas on Friday 2nd December. The special opening night screening will take place in the Irish Film Institute on Friday 2nd December at 6.30pm. Tickets are now on sale on the IFI website. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,221 |
The United Food and Commercial Workers has stepped up its fight against Bush administration use of "no match" Social Security letters against workers whose on-the-job identification doesn't match what's in government files. The government uses the program to pressure companies to fire employees and to force workers to prove "legal" status or face deportation.
Companies have pulled out the no match letters just weeks before union representation elections, Lopez said, and the organizing campaigns fell apart.
The UFCW and its allies note that, as late as 2006, before the no match program began, the Social Security database contained 18 million discrepancies. These ranged from transposed Social Security numbers to mix-ups with married and maiden names. Some 70 percent of the "no matches" involved records of U.S. citizens.
In one example, Fernando Tinoco, a 30 year legal resident of the U.S., tried to get a job with Tyson Foods on Chicago's South Side. The company got a "no match" letter for him and, despite written statements from Social Security that he had been a citizen for 18 years, the human resources director refused to believe him, threatening to call the police and have him arrested when he showed up with the letter. He never got the job and eventually became a custodian at a Catholic high school. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,543 |
\section{Introduction}
The continuous interest in inverse spectral theory \cite{band2009, kac1966} has reached many important results connected with cavities, billiards and their inherent quantum chaology \cite{berry1989, stoeckmann1999}. Apart from the purest interest of Kac's question and the shape of drums, fresh applications of quantum mechanics seem to draw the attention to similar problems in different areas such as propagation in crystals. Recent examples in experiments connected with the quantum behaviour of matter \cite{bloch2005, oberthaler1996} illustrate the important role of tight-binding lattices and their shapes. This is also true for other emulations of solids in mesoscopic physics, which sacrifice quantumness in favour of geometrical flexibility, with the aim of exploring more and more effects inexpensively \cite{laurent2007, sadurni2010-1, sadurni2013, franco2013}. Along with these examples comes the necessity of dealing with inverse spectral problems defined on lattices of any dimensionality, in particular tight-binding chains. Thus, inverse lattice design can be conceived as an answer to the problem of a discrete shape producing a specific spectral feature.
The incarnations of tight-binding models are of a great variety, but the focus of this paper shall be centered in waveguides, since they can be regarded as elongated variants of the more traditional drum. See figure \ref{fig:I.0}. Waveguides can also be open at their ends, which makes them accessible through scattering, with slight perturbative effects in their resonating frequencies when compared with their closed counterparts. The question is, can we hear the shape of a pipe?
In this paper we provide a general answer to the following two practical questions: Is it possible to produce an arbitrary finite spectrum by the mere use of a one-dimensional chain {\it without\ }local potentials? Are these results applicable to the transport properties of a scattering device? The answers shall be delivered by characterizing the space of all tight-binding hamiltonians that produce a suitably defined spectrum, together with their plausible realization in bent waveguides operating with a well-established binding mechanism at corners \cite{exner1989, schult1989, sadurni2010-2, bittner2013}.
Similar problems have been proposed in previous works; notably, the system of interest has a close resemblance with quantum graphs, for which there are extensive studies \cite{kuchment2004, kuchment2005, kurasov2005}. However, there are also sharp differences between our waveguides and one-dimensional wires conforming a graph. For instance, evanescent transport is possible only in the presence of a non-vanishing thickness, and it is through this mechanism that we shall establish finite difference equations. In this respect, it is important to mention that significant results in the field of graphs can be found in \cite{pankrashkin2006}, relating the spectrum of a continuous one-dimensional Laplacian with that of a finite difference operator defined on graph vertices. Nevertheless, the design of a specific graph producing an arbitrary spectrum is rarely given in the literature.
We present our discussion as follows. In section \ref{sec:2} we characterize the spaces of hamiltonians for a given spectrum and we generate a set of algebraic equations that relate hopping amplitudes (couplings) in terms of energies as parameters. This part is of a mathematical nature and it is carefully presented.
As an application, in section \ref{sec:3} we study bent waveguide realizations of tight-binding arrays.
We demonstrate that exactly solvable systems of the inverse problem can be emulated by explicit design of the lattice: the Dirac oscillator, the finite oscillator and partially isospectral chains are presented. We conclude in section \ref{sec:4}.
\begin{figure}[h!]
\begin{center} \includegraphics[width=12cm]{bentwaveguides.eps} \end{center}
\caption{\label{fig:I.0} Systems described in this paper: bent waveguides with multiple corners and tight-binding arrays of resonators.}
\end{figure}
\section{The problem of inverse design in nearest neighbour models \label{sec:2}}
We shall work with nearest neighbour tight-binding systems of identical sites, which belong to the wider class of Jacobi matrices. We must mention that previous work such as \cite{gesztezy1997} and references therein, deals with the inverse spectral theory of such operators. Although the theory has more general hypotheses related to all posible tridiagonal matrices, the claimed solution in such references rests on the assumption of two known spectra for two Hamiltonians belonging to the same sequence -- see Corollary 2.5 in \cite{gesztezy1997}. The intertwining of levels and the specific form of the spectral function allow to determine univocally the elements of one of the Hamiltonians (either the first or second in the sequence). However, our inverse problem is restricted to the knowledge of only one spectrum. We shall see that this explains the existence of infinite solutions. The determination of such a space of solutions can be seen as a complementary result to the known literature.
Let us establish our mathematical notation: We choose units such that $\hbar=1$, we denote by $N+1$ the number of sites in the array and we assume that any state $|\phi\>$ can be written as a superposition of a complete set of localized (atomic) states $|n\>$:
\begin{eqnarray}
|\phi \> = \sum_{n=1}^{N+1} \phi_n |n\>.
\label{1.1}
\end{eqnarray}
We consider the wavefunction $\< x|n \>$ to be concentrated around the site located at $x_n$ (e.g. a Wannier function \cite{wannier1937, marzari2012}, but without band index for simplicity). Since we deal with arrays of identical sites, we must have identical localized functions related by translations $\<x | n-1\> = \<x-x_n+x_{n-1} | n\>$. The atomic states themselves $|n\>$ can be translated by means of a unitary operator $T$
\begin{eqnarray}
T |n\> = |n-1\>, \qquad T^{\dagger} |n\> = |n+1\>.
\label{1.2}
\end{eqnarray}
The matrix elements of $T$ are clearly given by $\<n|T|m\> =\delta_{n,m-1}$. We can define an observable $\hat N$ associated to the site label $n$ with the property
\begin{eqnarray}
\hat N |n\>=n |n\>, \qquad \<n|\hat N|m \>=n \delta_{n,m}.
\label{1.3}
\end{eqnarray}
For any function $F$ of the operator $\hat N$ alone, we must have
\begin{eqnarray}
F(\hat N) |n\>=F(n) |n\> \equiv F_n |n\>.
\label{1.4}
\end{eqnarray}
The stationary Schr\"odinger equation corresponding to nearest neighbour chains with variable couplings but without local potentials is thus
\begin{eqnarray}
H |\phi_k \> = E_k |\phi_k \> = \left[ F(\hat N) T + T^{\dagger} F(\hat N) \right] |\phi_k \>, \quad 1 \leq k \leq N+1,
\label{1.5}
\end{eqnarray}
where now $F(\hat N)$ encodes the information on the variable couplings, according to their location in the chain. In fact, (\ref{1.5}) can be written as a recurrence if we set
\begin{eqnarray}
|\phi_k\> = \sum_{n=1}^{N+1} \phi_n^{k} |n\>,
\label{1.6}
\end{eqnarray}
from which the following relation holds:
\begin{eqnarray}
E_k \phi_n^k = F_n \phi_{n+1}^k + F_{n-1}^* \phi_{n-1}^k, \quad 1 \leq k \leq N+1, \quad 1 \leq n \leq N+1.
\label{1.7}
\end{eqnarray}
Since the system is finite, the previous relation is supplemented with fictitious boundary conditions $\phi_0=\phi_{N+2}=0$. Similarly, the couplings to fictitious sites must vanish $F_0 = F_{N+1}=0$.
With these conditions, the physical block of the system is determined by the non-vanishing elements of the operator $F(\hat N)$. Since we are interested in completely connected chains, we impose $F_n \neq 0$ for $1 \leq n \leq N$, but it is important to stress that $N$ can be arbitrarily large.
A few results can be established already: The spectrum $\sigma(H)$ -- with $H$ as in (\ref{1.5})-- is not quite arbitrary.
\begin{theorem} If $E_k \in \sigma(H)$, then $-E_k \in \sigma(H)$, i.e. $\sigma(H)$ is symmetrical around the origin. \end{theorem} To prove this assertion, we multiply by $(-1)^{n+1}$ both sides of (\ref{1.7}) and define $\psi_{n}^k =(-1)^n \phi_n^k$, leading to
\begin{eqnarray}
- E_k \psi_n^k = F_n \psi_{n+1}^k + F_{n-1}^* \psi_{n-1}^k,
\label{1.8}
\end{eqnarray}
which implies that $|\psi_k\> = \sum_n \psi_n^k|n\>$ is an eigenstate of $H$ with eigenvalue $- E_k$ $\square$.
A simple corollary of this result is that for $N$ even, $0 \in \sigma(H)$, i.e. at least one $E_k$ should vanish.
Another result of interest here is the nature of wavefunctions in connection with couplings. \begin{theorem} Let $\tilde H = |F(\hat N)| T + T^{\dagger} |F(\hat N)|$ with $|F(\hat N)| |n\> \equiv |F_n| |n\>$. Then $\sigma(H) = \sigma(\tilde H)$. \end{theorem} The proof is given by a gauge transformation when $F_n$ is allowed to be complex. If we set $F_n = e^{i \delta_n}|F_n|$, $\Delta_n \equiv \sum_{j=1}^{n-1} \delta_j$ and multiply both sides of (\ref{1.7}) by $e^{i \Delta_n}$, we find
\begin{eqnarray}
E_k ( e^{i \Delta_n}\phi_n^k) = F_n (e^{i \Delta_{n+1}} \phi_{n+1}^k) + F_{n-1}^* (e^{i \Delta_{n-1}}\phi_{n-1}^k).
\label{1.9}
\end{eqnarray}
From this relation it follows that the new state
\begin{eqnarray}
|\psi_k\> \equiv \sum_{n=1}^{N+1} e^{i \Delta_{n}} \phi_{n}^k | n \>
\label{1.10}
\end{eqnarray}
has energy $E_k$ with hamiltonian $\tilde H$ $\square$.
Therefore, without loss of generality, we can work with real and positive couplings $F_n$ in one-dimensional systems without loops. Although we are not concerned now with a non-trivial topology of the chain, it is worth mentioning that the influence of magnetic fields would destroy (\ref{1.9}).
In passing, we note that in certain applications related to transmission of information along chains (e.g. bosonic or fermionic matter in optical lattices \cite{bloch2005, oberthaler1996}) one needs a second-quantization scheme. The formalism previously developed can be generalized easily to such situations: The local-field replacements $\phi_n \mapsto \hat a_n$, $\phi_n^* \mapsto \hat a^{\dagger}_n$, with canonical relations $\left[ \hat a_n , \hat a^{\dagger}_m \right]=\delta_{n,m}$ or $\left\{ \hat a_n , \hat a^{\dagger}_m \right\}=\delta_{n,m}$ lead to spectral problems similar to (\ref{1.7}).
Returning to our problem as stated in (\ref{1.5}), we recognize that $H$ can be identified with the position operator $X$ of an $f$- or $q$- deformed oscillator (e.g. deformations of Heisenberg algebras {\it \`a la\ }Man'ko \cite{biedenharn1989, macfarlane1989, manko1993}). This type of quantum optical application requires an operator $\hat N -1 $ representing the occupation number of photons at a given frequency, i.e. a normal mode in a cavity. What we note from this connection is that indeed, position operators have symmetric spectra about the origin and that eigenvalues can be distributed either continuously, discretely, in bounded regions, in closed intervals, or in the whole real line.
Finally, the inverse spectral problems that we tackle in this paper must be fed \footnote{An exception may arise when one of the eigenfunctions of $H$ is known from the outset. We touch upon this case in other sections.}with the following information:
\begin{itemize}
\item[ i)] A symmetric, but otherwise arbitrary set $\sigma(H)$, \item[ii)] A set of real coupling constants $F_n$, \item[iii)] An arbitrary but fixed size $N+1$.
\end{itemize}
With this in mind, we characterize the spaces of solutions in what follows.
\subsection{The many solutions of the inverse spectral problem for variable couplings \label{sec:2.1}}
In this section we determine the space $\mbox{$\cal S\,$}$ of all possible nearest-neighbour hamiltonians that lead to the same spectrum - complying, of course, with the requirements of the previous section. We shall see that such spaces can be characterized by orbits generated by linear operators, which however do not correspond to the whole space of $(N+1) \times (N+1)$ unitary matrices due to the nearest-neighbour restriction. We shall see as well that the gauge transformations defined by (\ref{1.10}) are but a trivial part of $\mbox{$\cal S\,$}$. For a positive integer $n$, let us denote by $\mbox{$\cal U\,$}(n)$ the group of unitary $n\times n$ matrices and let us define $U_H(n)=\left\{ UHU^{\dagger} | U \in \mbox{$\cal U\,$}(n)\right\}$. Then, for a fixed $\sigma(H)$ we have an equivalence class $\mbox{$\cal S\,$}_H$ containing $H$ such that
\begin{eqnarray}
\left[ U_H(1) \right]^{N+1} \subset \mbox{$\cal S\,$}_H \subset U_H (N+1)
\label{1.11}
\end{eqnarray}
where the contentions are proper except for $N=0$ or vanishing couplings. The process by which we shall work consists in the introduction of two auxiliary operators $A$ and $X$; $A$ encodes the information of the couplings and $X$ connects all the hamiltonians isospectral to the homogeneous chain. This is shown diagramatically in figure \ref{fig:1} and we proceed to explain it.
\begin{figure}
\begin{center}
\includegraphics[width=15cm]{map.eps}
\end{center}
\caption{Diagram showing the maps between isospectral tridiagonal hamiltonians (lower row) and their characterization in terms of $X$ (upper row).}
\label{fig:1}
\end{figure}
First, we determine the conditions for a set of couplings that lead to a certain $\sigma(H)$. Suppose that $\tilde H$ satisfies $\sigma(H)=\sigma(\tilde H)$; then there must be a unitary matrix $U$ such that:
\begin{eqnarray}
U \left\{ F(\hat N) T + T^{\dagger} F^{\dagger}(\hat N) \right\} U^{\dagger} = \tilde F(\hat N) T + T^{\dagger} \tilde F^{\dagger}(\hat N).
\label{1.12}
\end{eqnarray}
Now, we can prove that each hamiltonian is susceptible of a factorization
\begin{eqnarray}
H= A(\hat N) H_0 A^{\dagger}(\hat N), \qquad \tilde H= \tilde A(\hat N) H_0 \tilde A^{\dagger}(\hat N), \qquad H_0= T+T^{\dagger},
\label{1.13}
\end{eqnarray}
reflecting the fact that couplings can be obtained by lateral transformations acting on the homogenous chain hamiltonian $H_0$ (if the chain is finite, the system is sometimes referred to as a locally periodic structure \cite{griffiths2000}). The operator $A(\hat N)$ is constructed by observing that
\begin{eqnarray}
A(\hat N) H_0 A^{\dagger}(\hat N) = A(\hat N) A^{\dagger}(\hat N + 1) T + T^{\dagger} A(\hat N + 1) A^{\dagger}(\hat N),
\label{1.14}
\end{eqnarray}
leading to $F(\hat N) = A(\hat N) A^{\dagger}(\hat N +1)$, and from here we can find the elements of $A$ using the recurrence
\begin{eqnarray}
A_n A^*_{n+1} = F_n
\label{1.15}
\end{eqnarray}
or in polar form (i.e. $A_n \equiv |A_n| e^{i a_n}, F_n \equiv |F_n| e^{i f_n}$)
\begin{eqnarray}
|F_n|= |A_n| |A_{n+1}|, \qquad f_n = a_n - a_{n+1}.
\label{1.16}
\end{eqnarray}
The solutions are
\begin{eqnarray}
a_n = \sum_{j=0}^{n-1} f_j, \qquad f_0 \in \Re
\label{1.17}
\end{eqnarray}
and
\begin{eqnarray}
|A_{n+1}| = \frac{\prod_{k=0}^{\left[ | n/2 |\right]} |F_{n-2k}| }{ \prod_{k=0}^{\left[ | n/2 |\right]} |F_{n-2k-1}| } \times A_1^{(-)^n}, \qquad 1 \leq n \leq N, \quad A_1 \in \Re_+
\label{1.18}
\end{eqnarray}
Furthermore, since $F_n$ can be chosen real and positive, we may work with
\begin{eqnarray}
A_{n+1} = \frac{\prod_{k=0}^{\left[ | n/2 |\right]} F_{n-2k} }{ \prod_{k=0}^{\left[ | n/2 |\right]} F_{n-2k-1} } \times A_1^{(-)^n}, \qquad 1 \leq n \leq N, \quad A_1 \in \Re \backslash \{ 0 \}
\label{1.19}
\end{eqnarray}
and $a_n \equiv 0$. Now that we have constructed $A$ from any given $F$, we can also define its {\it inverse\ }restricting ourselves to the physical block, i.e. $A^{-1}A = A A^{-1} = \v 1_{N+1}$, with $\v 1_{N+1}$ the $ (N+1)\times (N+1) $ identity matrix. With this, we can transform (\ref{1.12}) into
\begin{eqnarray}
H_0 = X H_0 X^{\dagger}, \qquad X \equiv \tilde A^{-1}(\hat N) U A(\hat N).
\label{1.20}
\end{eqnarray}
Our task is now simplified to the determination of all matrices $X$ subject to (\ref{1.20}). In other words, it is enough to analyze the locally periodic chain.
\subsubsection{The matrix $X$} The solutions of (\ref{1.20}) are multifold. We must note that $X$ could be non-unitary, which renders a solution space beyond the set of all symmetries of the periodic chain. We also note that the validity of (\ref{1.20}) is restricted to the physical $N+1$ dimensional block, which is more conveniently described by $N+1$ dimensional projectors $P$ as follows:
\begin{eqnarray}
X P (T+T^{\dagger}) P^{\dagger} X^{\dagger} = P (T+T^{\dagger}) P^{\dagger}.
\label{1.21}
\end{eqnarray}
In order to find $X$, we compute the matrix elements of our operators in the eigenbasis of $P (T+T^{\dagger}) P^{\dagger}$. For infinite systems, the states in question are Bloch waves. For finite systems, both wavefunctions $ \< m | k \>$ and eigenvalues $E_k$ of $H_0$ are known (in fact, they are related to the Dirichlet kernel). We have
\begin{eqnarray}
\sum_{n,m} \< k | X | m\> \< m | P (T+T^{\dagger}) P^{\dagger} | n \> \< n | X^{\dagger} | k' \> = E_k \delta_{k,k'}
\label{1.22}
\end{eqnarray}
which means that the matrix elements of $X$ in mixed bases $X_{k; m} = \< k | X |m \>$ constitute an operator that diagonalizes $H_0$. Therefore, we must have
\begin{eqnarray}
X_{k;m} = e^{-i \alpha(k)} \< k | m \>, \qquad X^*_{m; k} = e^{i \alpha(k)} \< m|k \>.
\label{1.23}
\end{eqnarray}
Here we have a freedom of choice, since the phase encoded by $\alpha(k)$ is so far arbitrary \footnote{In the limit $N \rightarrow \infty$, these phases are linked to the problem of the most localized Wannier functions. In linear arrays, this problem has been solved \cite{marzari2012}.}. This possibility renders $X$ unitary, but more general choices are attainable such as $\left[X, P(T+T^{\dagger})P^{\dagger} \right]=0$ and $(XX^{\dagger}-1)P(T+T^{\dagger})P^{\dagger}=0$, leading to
\begin{eqnarray}
XX^{\dagger} = 1 + P_0, \qquad P_0 (T+T^{\dagger}) = (T+T^{\dagger})P_0 = 0,
\label{1.24}
\end{eqnarray}
which is indeed more general than a unitarity condition. In fact, the projector $P_0$ onto the null space of $H_0$ exists whenever $N$ is even, making $N+1$ odd and at least one eigenvalue would vanish. For our purposes (general $N$), the choice (\ref{1.23}) is a reasonable one, as it contains a sufficient number of real parameters $\alpha(1), \alpha(2), ..., \alpha(N+1)$.
In addition to (\ref{1.21}), we have the condition $U=\tilde A X A^{-1}$ with $U$ unitary. This implies
\begin{eqnarray}
(A^{\dagger})^{-1} X \tilde A^{\dagger} \tilde A X^{\dagger} A^{-1} = \v 1_{N+1}
\label{1.25}
\end{eqnarray}
or
\begin{eqnarray}
X \tilde A^{\dagger} \tilde A X^{\dagger} = A^{\dagger} A.
\label{1.26}
\end{eqnarray}
From here we see that given a matrix $\tilde A^{\dagger} \tilde A$, we can build all possible $A^{\dagger} A$ by choosing a specific $X$, which in turn amounts to a choice of the phases $\alpha(k)$. This expression can be given also in terms of matrix elements:
\begin{eqnarray}
|A_n|^2 \delta_{n,n'} &=& \sum_{k,k',m,m'} \<n|X|k\> \< k | m \> \< m | \tilde A^{\dagger} \tilde A | m'\> \< m' | k' \> \<k' | X^{\dagger} | n' \> \nonumber \\
&=& \sum_{k,k',m} e^{i(\alpha(k')-\alpha(k))} \< n | k \> \< k | m \> \< m | k' \> \< k' | n' \> |\tilde A_m|^2 .
\label{1.27}
\end{eqnarray}
The diagonal part of (\ref{1.27}) represents indeed the orbits of a vector with components $|\tilde A_m|^2, m=1,...,N+1$ under the action of a linear operator that depends on the parameters $\alpha(k)$. On the other hand, the off-diagonal part restricts the possible $\alpha$'s as functions of $\tilde A$. This establishes that only some pairs $A, \tilde A$ can be related according to restrictions such as
\begin{eqnarray}
\sum_{k,k',m} \sin \left(\alpha(k')-\alpha(k) \right) \< n | k \> \< k | m \> \< m | k' \> \< k' | n' \> |\tilde A_m|^2 = 0,
\label{1.28}
\end{eqnarray}
\begin{eqnarray}
|\sum_n |A_n|^2 \< k| n \> \< n| k' \>|^2 = | \sum_m |\tilde A_m|^2 \< k| m \> \< m| k' \>|^2,
\label{1.30}
\end{eqnarray}
\begin{eqnarray}
\prod_{n=1}^{N+1} |\tilde A_n|^2 = \prod_{n=1}^{N+1} | A_n|^2,
\label{1.31}
\end{eqnarray}
which can be easily deduced.
To summarize this part, we underscore that the general isospectral problem for tight-binding hamiltonians of equal sites can be reduced to the isospectral problem of the (locally) periodic chain by means of a family of transformations represented by the matrix $X$. Then by varying $X$ we obtain all the possible orbits of a given matrix $A^{\dagger} A$ using the relation $ X A^{\dagger} A X^{\dagger} = \tilde A^{\dagger} \tilde A \equiv A^{\dagger} A \left[ \alpha \right]$, i.e. a functional of $\alpha$. The final step consists in a reconstruction of $F$ using once more $F_n = |A_n| |A_{n+1}|$, where additional phases are irrelevant due to (\ref{1.9}). The parameter $A_1$ is lost in the expression for $F_n$ (see the power $(-1)^n$ in (\ref{1.18})), and the generation of new couplings $\tilde F_n$ comes exclusively from the variation of all possible $\alpha$'s. We refer once more to figure \ref{fig:1}.
\subsection{Algebraic equations for couplings and energies \label{sec:2.2}}
Here we obtain a set of algebraic equations whose solutions determine the systems that lead to a given (target) spectrum. Our approach in this part is a rather direct one: we shall find equations in the variables $\{ F_n \}_{n=1}^{N+1}$ and parameters $\sigma(H)=\{ E_k \}_{k=1}^{N+1}$.
We start by defining $\lambda = -E$, $\Phi_N(\lambda) = \mbox{det}\left( H+ \lambda \right)$, leaving us with a secular equation $\Phi_N(\lambda)=0$. An expansion of $\Phi_{N}(\lambda)$ by the method of minors leads to a recurrence relation in the size of the system $N+1$:
\begin{eqnarray}
\Phi_N(\lambda) = \lambda \Phi_{N-1}(\lambda) - F_N^2 \Phi_{N-2}(\lambda).
\label{2.1}
\end{eqnarray}
We want to determine the coefficients $\Lambda_{N-1}^{j}$ in the following power expansion
\begin{eqnarray}
\Phi_N(\lambda)= \sum_{j=0}^{N+1} \Lambda_N^j \lambda^j.
\label{2.2}
\end{eqnarray}
To this end, we replace (\ref{2.2}) in (\ref{2.1}) and collect terms in $\lambda^j$ to obtain
\begin{eqnarray}
\Lambda_{N}^{j} = \Lambda_{N-1}^{j-1} - F_N^2 \Lambda_{N-2}^{j}.
\label{2.3}
\end{eqnarray}
This is a double recurrence in size $N$ and order $j$. In order to solve it, we need to specify boundary conditions on the lattice $(N,j)$ (see figure \ref{fig:2}). The index $j$ is limited by $N+1$, so we have $\Lambda_{N}^{N+1}=1$ (monic polynomial). For $j=0$, we have a lower boundary condition provided by (\ref{2.3}) if we set $\Lambda_{N-1}^{-1}=0$, i.e.
\begin{eqnarray}
\Lambda_N^0 = - F_N^2 \Lambda_{N-2}^0
\label{2.4}
\end{eqnarray}
which is solved by
\begin{eqnarray}
\Lambda_{N}^0 = \cases{0&for $N$ even \\ (-1)^{(N-1)/2} \prod_{k=1}^{(N-1)/2} F_{N-2k}^2 & for $N$ odd \\}
\label{2.5}
\end{eqnarray}
\begin{figure}[h!]
\begin{center} \includegraphics[width=7cm, height=7cm]{Diagram} \end{center}
\caption{\label{fig:2} Lattice representation of the recurrence relation (\ref{2.3}) in red and its boundary conditions in blue.}
\end{figure}
It is important to recognize that $N=N_{\mbox{max}}$ in figure \ref{fig:2} is not a boundary condition, but the mere statement that the recurrence shall be solved for a fixed (arbitrary) $N$.
Now we take advantage of the fact that (\ref{2.3}) is of order $1$ in $j$, for the method to solve (\ref{2.3}) consists in iterations of $\Lambda_{N-2}^j$ in terms $\Lambda_{N-3}^j$, $\Lambda_{N-4}^j$ in terms $\Lambda_{N-5}^j$, and so forth. This leads to a relation
\begin{eqnarray}
\Lambda_N^j = \sum_{n=0}^{(N-j+1)/2} (-1)^n \Lambda_{N-2n-1}^{j-1} \prod_{l=0}^{n-1} F_{N-2l}^{2}
\label{2.6}
\end{eqnarray}
which puts $\Lambda_{N}^j$ in terms of all the coefficients of a preceding order in $j$, as expected. Now we continue this iterative process by substituting $\Lambda_{N-2n-1}^{j-1}$ in (\ref{2.6}) using (\ref{2.6}) again, but with properly shifted indices. The process ends when we reach $\Lambda_{M}^0$, specified by (\ref{2.5}). The formula thus obtained is
\begin{eqnarray}
\Lambda_{N}^{j} = \cases{ 0 & if $N-j$ even \\ \sum_{n_0 = 1, ...., n_{j-1}=1}^{N_{0,j}, ..., N_{j-1,j}} (-1)^{\sum_{i=0}^{j-1} n_i } \prod_{m=0}^j \left\{ \prod_{l_m=0}^{n_m-1} F_{M_m}^2 \right\} & if $N-j$ odd \\}
\label{2.7}
\end{eqnarray}
with the shorthands
\begin{eqnarray}
N_{k,j} \equiv \frac{N-j+1}{2} - \sum_{i=0}^{k-1} n_i, \qquad 0\leq k \leq j-1,
\label{2.8}
\end{eqnarray}
\begin{eqnarray}
M_{m} \equiv N - 2 l_m - 2 \sum_{i=0}^{m-1} n_i - m, \qquad 0\leq m \leq j,
\label{2.8}
\end{eqnarray}
and with $n_m, l_m$ dummy indices. Summations with negative upper limits are simply zero. This general expression for the coefficients of the characteristic polynomial indicates that we are dealing with polynomials of $2N$-th order in the couplings. Also, the coefficients are homogeneous functions of degree $N+1-j$, and setting $\Lambda_{N}^j = \mbox{constant}$ defines a number of $\left[| N/2 |\right]$ surfaces that are shape-invariant under global scaling.
What we have achieved so far is an explicit calculation of the characteristic polynomial for tight-binding hamiltonians of variable couplings. Since we are interested in the solutions of the inverse spectral problem, we proceed now to the determination of $\Phi_N(\lambda)$ in terms of $\{ E_k \}_{k=1}^{N+1}$. We have
\begin{eqnarray}
\Phi_N(\lambda) = \prod_{k=1}^{N+1} (\lambda + E_k)
\label{2.9}
\end{eqnarray}
from which the coefficients are obtained by computing the corresponding products:
\begin{eqnarray}
\Lambda_N^j = \sum_{\{ k_i \}} \prod_{r=1}^{N-j} E_{k_r}, \qquad \{ k_i \} \equiv \{ k_i =1, ..., N+1 \forall i; k_i \neq k_j \forall i \neq j\}.
\label{2.10}
\end{eqnarray}
This formula specifies uniquely the characteristic polynomial for a given spectrum; therefore, if we want to find the couplings $F_n$ we must equate (\ref{2.10}) and (\ref{2.7}). The resulting system reads
\begin{eqnarray}
\left( \frac{1 - (-)^{N+j}}{2} \right)\sum_{ \{n_k\}}^{\{N_{k,j}\}} (-1)^{\sum_{i=0}^{j-1} n_i } \prod_{m=0}^j \left\{ \prod_{l_m=0}^{n_m-1} F_{M_m}^2 \right\} = \sum_{\{ k_i \}} \prod_{r=1}^{N-j} E_{k_r}.
\label{2.11}
\end{eqnarray}
This is a set of $\left[| (N+1)/2 |\right]$ equations, giving rise to a number of $N-\left[| (N+1)/2 |\right]=\left[| N/2 |\right]$ free variables and consequently a $\left[| N/2 |\right]$-dimensional surface representing $\mbox{$\cal S\,$}_H$. We shall see that these relations can be solved uniquely by providing a set of constraints in some of the $F$'s. The numerical method of choice is iterative, but we ellaborate more on this in section \ref{sec:2.5}.
\subsubsection{Minimal conditions for a unique solution} We have seen that specifying the energies leads to a surface $\mbox{$\cal S\,$}$ of models. It is possible to fix the values of $\left[| N/2 |\right]$ couplings in order to obtain a single point in $\mbox{$\cal S\,$}$. However, there is another way of defining one and only one model by means of the wavefunction; we show this by considering two cases. If $N$ is even, the size $N+1$ is odd and one of the energies vanishes. Suppose that the corresponding eigenvector has components $\phi_n^{k}$, with $k$ fixed. The relation (\ref{1.7}) implies
\begin{eqnarray}
F_n = -\frac{\phi_{n-1}^k}{\phi_{n+1}^k} F_{n-1} = F_{1} (-1)^{n-1} \prod_{k} \frac{\phi_{n-1-k}^k}{\phi_{n+1-k}^k}
\label{2.12}
\end{eqnarray}
fixing all $F$'s up to an overall scale factor $F_1$. If $N$ is odd, we may use our knowledge of any energy $E_k \neq 0$ to write a similar recurrence
\begin{eqnarray}
F_n = \frac{\phi_{n}^k}{\phi_{n+1}^k} E_{k} -\frac{\phi_{n-1}^k}{\phi_{n+1}^k} F_{n-1}
\label{2.13}
\end{eqnarray}
which can be solved by iteration. Once the couplings are fixed, the information of other wavefunctions and energies is also fixed.
\subsection{Problems with analytic solutions \label{sec:2.3}}
Quite often one finds that the problem of couplings for a given spectrum is accompanied by additional restrictions. In fact, the latter can make the inverse problem easier to solve, as shown in recent works \cite{sadurni2010-1, sadurni2013, franco2013}.
The aforementioned restrictions may come in the form of Lie algebraic relations satisfied by the hamiltonian or by the operators contained in it. Such relations affect directly the couplings $F_n$ and impose their functional form up to gauge transformations. In table \ref{tab:table1} we give some examples related to Lie algebraic structures imposed as restrictions. The family presented here is related to $\mbox{sl}(2, \mathbb{C})$, which has three generators. This is in fact the most general structure that one may propose, since the hamiltonian $H = FT + (FT)^{\dagger}$ is composed of only two operators $FT, T^{\dagger}F^{\dagger}$ belonging to a Cartan basis. We note however, that the possibilities provided by this well- studied object are sufficiently rich: compact and non-compact subsets of the special linear group give rise to finite or infinite spectra.
\begin{table}[t
\caption{\label{tab:table1}%
A class of solvable inverse problems and their algebraic structures
}
\begin{tabular}{lcccc}
\br
\textrm{ -- } &
\textrm{Dirac oscillator}&
\textrm{Finite oscillator}&
\textrm{Position operator}&
\textrm{Infinite chain}
\\
\mr
Restrictions & $\begin{array}{c} \{ FT,T^{\dagger}F^{\dagger} \}=H^2, \\ (FT)^2=0 \end{array}$ & $\begin{array}{c} \left[ FT,T^{\dagger}F^{\dagger} \right]= 2 J_z, \\ FT= J_+ \end{array}$ & $\begin{array}{c} \left[ FT,T^{\dagger}F^{\dagger} \right]= \hat N, \\ FT = a \end{array}$ & $F=1$ \\ &&&&\\
Hamiltonian & $H = \sigma_+ a + \sigma_- a^{\dagger}$ & $H = J_x $ & $H = x$ & $H = T + T^{\dagger}$ \\
Group & S$(2)$ & SU$(2)$ & Heisenberg & $(\mathbb{Z}, +)$\\
Spectrum & $ \{ \pm \sqrt{n} \}_{n \in \mathbb{N}} $ & $ \{ -\frac{N}{2}, ..., \frac{N}{2} \}$ & $\Re$ & $\left[ -2,2 \right] \subset \Re$ \\
Coupling $F_n$ & $\sqrt{n}$ or $1$ & $\sqrt{(n-1)(N+1-n)}$ & $\sqrt{n}$ & $1$ \\
\end{tabular}
\end{table}
\subsection{Exact solutions for short chains \label{sec:2.4}}
A depiction of the spaces $\mbox{$\cal S\,$}$ is provided for some examples. The low dimensional cases illustrate in a simple manner the multiplicity of solutions for our inverse problems, which are presented for $N=2,3,4$, as well as the explicit constraints (\ref{2.11}). This renders $\mbox{$\cal S\,$}$ as a surface or as an intersection of surfaces in the space of couplings.
\begin{figure}[h!]
\begin{center} \includegraphics[width=15cm]{rivera_2D_3D_4D} \end{center}
\caption{\label{fig:1.2} Curves and surfaces in the space of couplings $F_1, F_2, ...$, representing isospectral chains. (a) $N=2$, a circle; (b) $N=3$, four loops at the intersections of a sphere with a hyperbolic cylinder; (c) $N=4$, a surface generated by successive intersections of surfaces, each represented by a value of $F_4$.}
\end{figure}
\begin{itemize}
\item[$N=2$:]
$\sigma(H)= \{ - \sqrt{F_1^2 + F_2^2} , 0 , + \sqrt{F_1^2 + F_2^2} \}$ invariant if $F_1^2 + F_2^2=$constant.
\item[$N=3$:]
$\sigma(H)= \{ -E_1 , -E_0 ,+E_0 , +E_1 \}$, with the constraints
\begin{eqnarray}
F_1^2 +F_2^2+F_3^2= E_0^2 + E_1^2, \nonumber \\
F_1^2 F_3^2 = E_0^2 E_1^2.
\label{2.14}
\end{eqnarray}
\item[$N=4$:]
$\sigma(H)= \{ -E_2 , -E_1 ,0, +E_1 , +E_2 \}$, with the constraints
\begin{eqnarray}
F_1^2 +F_2^2+F_3^2 + F_4^2= E_1^2 + E_2^2, \nonumber \\
F_1^2 F_3^2 + F_1^2 F_4^2 + F_2^2 F_4^2 = E_1^2 E_2^2.
\label{2.14}
\end{eqnarray}
\end{itemize}
As shown before, in general, adjacent values of $N$ (i.e. with an equal $\left[| (N+1)/2 |\right]$) give rise to sets $\mbox{$\cal S\,$}$ with the same dimensionality. It is found that $N=1$ yields a circle and $N=2$ yields a loop defined by the intersection of a sphere and a quartic hyperbolic cylinder. The connected components of these surfaces have the same topology. This is displayed in figures \ref{fig:1.2}(a) and (b). When $N=4$, the geometric representation is generated by the successive intersections of two families of surfaces in the space $F_1, F_2, F_3$, giving rise to a familiy of curves parameterized by $F_4$; as a final result a 2d surface in the space $F_1, F_2, F_3$ is obtained. This is shown in figure \ref{fig:1.2}(c).
\begin{figure}[h!]
\begin{center} \includegraphics[width=12cm]{rivera-distributions} \end{center}
\caption{\label{fig:1.3} Energy and coupling distributions. Upper row: (a) Cosine distribution of energies and (b) peaked distribution of couplings around $F=1/2$ with a nonvanishing width. The peak represents a locally periodic system. Lower row: (c) Gaussian distribution for energies and (d) a distribution of couplings with a smaller variance and non-vanishing average.}
\end{figure}
\subsection{Coupling distributions from random spectra: numerical examples \label{sec:2.5}}
When $N$ is large, the solutions of (\ref{2.11}) can be obtained numerically. As a demonstration, we generate a target spectrum from random distributions and find the family of couplings that produces it. This procedure gives more importance to overall trends rather than specific details. Our goal is to obtain the most probable couplings for two types of spectra: i) a dispersion relation given by a cosine law, emulating a periodic chain and ii) energies of a large system that are located predominantly in a gaussian band with zero average and adjustable width. The exploration of $\mbox{$\cal S\,$}$ is achieved by choosing random initial conditions in the iterative method (e.g. Newton-Raphson).
Example i) is relevant to our general theory, given the role of $X$ in (\ref{1.20}). This offers interesting results: If the system is periodic, the spectrum must follow a cosine distribution law, but other possibilities for the couplings may give rise to the same dispersion relation, despite the lack of local periodicity. This is shown in figures \ref{fig:1.3}(a) and (b); in (b) we find indeed a peak at $F=1/2$ as the most problable value (i.e. the periodic case), accompanied by a nonvanishing width that reveals the existence of other systems with the same spectrum (note however that the eigenfunctions do not coincide). The average and deviation are $\< F \> = 0.45$ and $\<\Delta F\>=0.20$ respectively.
The second example gives a distribution of couplings (see histograms in figures \ref{fig:1.3}(c) and (d)) with a decay faster than exponential. This is natural, given the fact that the spectral span is roughly the width $\sigma$ of the gaussian in (c), and the maximum coupling of the system must be constrained also by a gaussian band, i.e. the coupling distribution and the energy distribution have a similar asymptotic behaviour. The average of the coupling distribution is $\< F \>=0.17$ . We also find that the most probable value of the coupling is $F^*=0.11$ and it is considerably greater than zero (here $0 \leq F \leq 0.5$); this is explained by noting that nonzero couplings may give rise to zero or close to zero energy eigenvalues. Thus, if the spectral distribution is dominated by the centre of a band, we obtain matrix elements of $H$ that are predominantly {\it displaced\ }from the centre, as shown in figure \ref{fig:1.3}(d).
\section{Bent waveguide realizations \label{sec:3}}
We test our tight-binding models with a physical system that provides the required flexibility. Adjustable couplings, as well as the existence of evanescent modes draw our attention to systems of waveguides with several bendings. Such realizations are indeed susceptible of manipulations that produce partial isospectrality: we should bear in mind that the Dirichlet boundary value problem posed by bent waveguides leads to nontrivial behavior at high energies or frequencies, connecting the problem with wave-like manifestations of chaos \cite{richter1999}. However, at very low energies we can take advantage of trapped states at corners \cite{bittner2013, exner2015, carini1992, londergan1999, carini1997-1} (in fact, only one bound state at a right bending angle), providing an exponentially evanescent coupling of two corners as a function of the distance between them. The idea then is to accomodate as many levels as necessary below the propagation threshold of the guide; this is done by adjusting the distances between the corners contained in the array. See figures \ref{fig:I.0}, \ref{fig:1.4} and \ref{fig:1.4.1}. In this way we engineer the necessary couplings for a given spectrum and finally tame the beast.
It is important to mention that, from the physical point of view, some realizations put restrictions on the width of waveguides \cite{exner1989-2}. Our realizations are of a different nature, and the existence of bound states in corners is independent of the scale. In fact, they have been confirmed experimentally even with microwaves \cite{bittner2013}. This can be easily explained by noting that the classical theorems found by Exner and Seba rely on the smoothness of the waveguide -- in contrast to our case-- as well as the independent variation of width and curvature. Quoting their result, ``a bound state with energy below the first transversal mode exists for all sufficiently small $d$'', which ensures the existence of a trapped mode for a certain range of the width and ``the proof... establishes existence of the critical width $d_0$...''. In addition, Goldstone and Jaffe (see fig. 1 in \cite{goldstone1992}) proved by a variational method the existence of bound states for all widths in twisted tubes of {\it constant\ } section.
Right-angle bends, on the other hand, have infinite curvature at the corner and do not have constant section. For a single corner, one is convinced that increasing the separation distance between boundaries amounts to a global scaling of the system -- see the explanation below (7) in \cite{sadurni2010-2}. In connection with scaling properties in the limit of vanishing width, some works have shown the persistence of evanescent modes in effectively one-dimensional wires \cite{cacciapuoti2007}, but here we provide an explicit dependence of the evanescence length in terms of the geometry.
\subsection{The validity of tight-binding approximations \label{sec:3.1}}
The equation to solve here is of the Helmholtz type (a quantum particle trapped in the cavity or TE modes in a good conductor):
\begin{eqnarray}
\left[ \nabla^2 + k^2 \right] \phi(x,y) = 0, \quad
\phi(x,y) = 0 \quad \mbox{if} \, (x,y) \in \partial \Omega.
\label{3.1}
\end{eqnarray}
The associated Hamiltonian for this equation is simply $H= -\nabla^2$. For two adjacent corners 1 and 2, we find the coupling $F$ by means of an overlap between trapped modes
\begin{eqnarray}
F = \int \int_{\Omega} \phi_1(x,y) H \phi_2(x,y) dxdy,
\label{3.2}
\end{eqnarray}
giving rise to the following block representation of a two-corner system at very low energies:
\begin{eqnarray}
H_{\mbox{\scriptsize Low} E} = \left( \begin{array}{cc} E_b & F \\ F & E_b \end{array} \right).
\label{3.3}
\end{eqnarray}
Here, $k^2 = E_b$ is the energy of a trapped mode in an isolated corner. Numerical calculations show that for an infinite guide $E_b = 0.93 E_t$ \cite{schult1989}, that is, a fraction of the propagation threshold $E_t = \pi^2 / L^2$ if the straight segment has a width $L$. It has been shown theoretically and experimentally \cite{bittner2013} that such trapped modes have exponentially vanishing tails as functions of the distance $d$ from the corner, measured along a straight segment. We write $\phi(x,y) \sim \exp \left( - d / 2 \lambda \right)$ with $\lambda \sim 1 / \sqrt{E_b}$. Therefore, the function $F$ can be written to a very good approximation as
\begin{eqnarray}
F \equiv \Delta(d) = \Delta(0) \exp \left( -\frac{d}{\lambda} \right).
\label{3.4}
\end{eqnarray}
We have verified this law numerically for a system of two corners in different configurations, the results are shown in figures \ref{fig:1.4} and \ref{fig:1.4.1}. This is also in agreement with the classic results by Carini and co workers \cite{carini1997-1}.
With these considerations we justify the applicability of tight-binding models for the description of spectra and wave functions of more intricate systems of waveguides. We also perform numerical computations of the 2d problem for each realization, in order to give a fair comparison with our predictions.\\
\begin{figure}[h!]
\begin{center} \includegraphics[width=7cm]{Decay-Law_Delta} \end{center}
\caption{\label{fig:1.4} Numerical results for the decay of energy splittings as a function of distance $d$ in units of width $L$. The close agreement with the red curve is a clear indicative of exponentially vanishing couplings; the parameters are $\lambda = 1.3 L$, $\Delta(0)=0.43 E_{b}$. This has been computed for U and S configurations shown in figure \ref{fig:1.4.1}.}
\end{figure}
\begin{figure}[h!]
\begin{center} \includegraphics[width=12cm]{UandS-sys-l3d} \end{center}
\caption{\label{fig:1.4.1} Symmetric and antisymmetric states for S and U waveguides. The finite separation distance between corners produces a level splitting. }
\end{figure}
\begin{figure}[t!]
\begin{center} \includegraphics[width=14cm]{OscDirac-11-dimeros} \end{center}
\caption{\label{fig:1.6} A bent waveguide representing a Dirac oscillator. The logarithmically deformed array comprises 11 dimers. Its spectrum corresponds to panel (b) of figure \ref{fig:1.5}.}
\end{figure}
\begin{figure}[h!]
\begin{center} \includegraphics[width=12cm]{erivera_OscDirac_3syst} \end{center}
\caption{\label{fig:1.5} Spectra of waveguides representing Dirac oscillators of three different sizes: (a) $N=10$ (b) $N=22$ (c) $N=32$. A better agreement with the theoretical prediction is achieved as the size of the array is increased. More energy levels (blue dots) can be fitted by a square root law (solid black curve) centered at the bound state energy (red dashed line).}
\end{figure}
\subsection{The realization of exactly solvable configurations \label{sec:3.2}}
Here we provide waveguide designs that produce the models mentioned in table \ref{tab:table1}, as well as the examples discussed in section \ref{sec:2.4}.
\subsubsection{The Dirac oscillator} This system was recently produced experimentally in \cite{franco2013}, with suggestions provided in \cite{sadurni2010-1}. The incarnation using microwave resonators provided a natural way to adjust couplings by varying distances between dielectric disks. Here we resort to the variation of segment lengths and their orientation to build a sequence of dimers; the intra dimer distance has a constant value in compliance with a constant coupling, while the inter distance of consecutive dimers is varied with a law of the type $\log \sqrt{n}$. For the explicit 1d hamiltonian of this system, see table \ref{tab:table1}.
The resulting waveguide and the numerically obtained spectrum are shown in figures \ref{fig:1.5}, \ref{fig:1.6}. We observe an excellent agreement with our predictions: a square root law for the spectrum is obtained for half the energies (around the centre of the band). A parabolic curve is fitted to the lowest energies applying the method of least squares to the lower half of the spectrum\footnote{Finite size effects can alter the upper half of the eigenvalues. For this reason we do not include them in the parabolic fit.}.
\begin{figure}[t!]
\begin{center} \includegraphics[width=13cm]{equiespaciado} \end{center}
\caption{\label{fig:1.7} Numerical spectra for waveguides obeying the coupling prescriptions of finite oscillators. (a) and (b) show results for various spectral spacings $\varepsilon$ in terms of $\zeta \equiv (E_t-E_b)/ l$ (see inset) and sizes $N=2l=6$ and $8$. (c) and (d) show especial examples where the agreement with equispaced spectra is outstanding.}
\end{figure}
\begin{figure}[h!]
\begin{center} \includegraphics[width=13cm]{WaveFunc-Eql4d10-sys4} \end{center}
\caption{\label{fig:1.8} Examples of wavefunctions for a finite oscillator with 9 sites, $l=4$. (a) $m=1$, (b) $m=4$, (c) $m=6$, (d) $m=9$. The spectrum is shown in figure \ref{fig:1.7}(d).}
\end{figure}
\subsubsection{Equispaced spectrum or finite oscillator}
Finite oscillators \cite{atakishiyev1991, atakishiyev2001} and relativistic rotors \cite{sadurni2013} have been proposed in the past, with the aim of producing systems that evolve coherently due to their equispaced frequencies. In one of such applications it was even possible to produce a schematic spectrum for baryons. These constructions do not suffer the effects of finite size, as in the previous case. In its simplest form (see table \ref{tab:table1}), the spectrum corresponds to that of $J_x$, i.e. a projection of angular momentum orthogonal to its quantization axis. Spectra and wavefunctions are presented in figures \ref{fig:1.7} and \ref{fig:1.8}.
\begin{figure}[h!]
\begin{center} \includegraphics[width=10cm]{all-sys-4} \end{center}
\caption{\label{fig:1.9} The spectra for triads of partially isospectral systems, two sizes: (a) $N=3$ (four sites), (b) $N=4$ (five sites). Abscissa: three systems and a randomly generated target spectrum. Ordinate: energy eigenvalues. The arrays were obtained by finding the couplings and solving for the lengths of the waveguide segments. The boundary value problem was solved numerically and the agreement is quite reasonable. The uncertainties due to discretization were also considered and the corresponding spectrum was calculated when lengths were rounded up (dashed red lines) and rounded down (dashed blue lines). The isospectrality is better achieved numerically in case (c). The arrays are shown in fig. \ref{fig:1.10}}
\end{figure}
\begin{figure}[h!]
\begin{center} \includegraphics[width=14cm]{waveFunc-isos-3sys-1-m} \end{center}
\caption{\label{fig:1.10} Wavefunctions for partially isospectral waveguides. (a), (b) and (c) correspond to systems 1, 2 and 3 in figure \ref{fig:1.9}(b), first excited state.}
\end{figure}
\subsubsection{Partially isospectral systems for low $N$}
As a final application, we build a family of bent waveguide arrays that possess the same spectrum below propagation threshold. This is most easily done for low values of $N$, already studied in section \ref{sec:2.4}. Three types of waveguides with $N=4, 5$ are designed such that their spectra below threshold coincide. See figures \ref{fig:1.9} and \ref{fig:1.10}. This example is of the utmost importance, since it shows that at least $N+1$ levels of a waveguide (or billiard, if closed at the ends) can be produced by an infinite number of systems. Although the shape of the pipe cannot be determined uniquely using the first $N+1$ levels, we have shown that the set of all pipes can be analytically determined for low $N$.
\section{Conclusion \label{sec:4}}
Inverse lattice design in one dimension shows that arbitrary symmetric spectra can be emulated in various types of tight binding systems. Arbitrary but finite spectra can be also produced by embedding it in half of the energy band obtained from a deformed chain; in this case a bent waveguide operating below threshold was used. The technical goals that we have achieved in this paper are the characterization of the space of all isospectral nearest-neighbour models, the explicit form of the algebraic equations that relate couplings and energies, as well as their analytical (short chains) and numerical (long chains) solutions. Moreover, we designed bent waveguides of variable segments that emulate a number of paradigmatic cases, such as relativistic hamiltonians (a Dirac oscillator) an equispaced spectrum without finite size effects (a finite oscillator) and a few partially isospectral 2d systems (isospectrality only below threshold). In our view, these results are valuable both practically and fundamentally.
They can be readily used in order to engineer the resonances of many experiments, particularly with microwaves. From the mathematical point of view, we have shown the multivaluedness of solutions and the algorithm to obtain matrix elements from eigenvalues.
\ack
We are grateful to CONACyT for financial support under project CB2012-180585. E.R.-M. also wishes to thank CONACyT for {\it beca-cr\'edito\ } 245104.
\newpage
\section*{References}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,875 |
Conus harasewychi est une espèce de mollusques gastéropodes marins de la famille des Conidae.
Comme toutes les espèces du genre Conus, ces escargots sont prédateurs et venimeux. Ils sont capables de « piquer » les humains et doivent donc être manipulés avec précaution, voire pas du tout.
Description
Description originale : « Coquille trapue, solide, large à travers l'épaule ; épaule quelque peu arrondie, nettement coronée avec de grands boutons surélevés ; verticille du corps avec de nombreux cordons fins, perlés, en spirale ; les cordons en spirale deviennent plus forts sur la moitié antérieure de la coquille ; verticille du corps uniformément jaune moutarde foncé avec une fine bande blanche amorphe autour du milieu du corps ; coronations de l'épaule et verticilles de la flèche de couleur blanche ; pointe antérieure de la coquille et région siphonnée brun foncé ; intérieur de l'ouverture lavande pâle ».
La taille de la coquille atteint .
Distribution
Cette espèce marine est présente au large de la Floride et des Bahamas à une profondeur de .
Taxinomie
Première description
L'espèce Conus harasewychi a été décrite pour la première fois en 1987 par le malacologiste américain Edward James Petuch dans la publication intitulée « Charlottesville, Virginia: The Coastal Education and Research Foundation ».
Synonymes
Conus (Dauciconus) harasewychi Petuch, 1987 · appellation alternative
Purpuriconus harasewychi (Petuch, 1987) · non accepté
Notes et références
Voir aussi
Liens externes
Identifiants taxinomiques : CoL : XXJR - GBIF : 5193177 - IRMNG : 10862279 - SeaLifeBase : 567396 - TAXREF : 1561 - WoRMS : 429441
Espèce de gastéropodes (nom scientifique)
Conidae
Gastéropode décrit en 1987 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,791 |
You've Got to Have the Guts
You've Got to Have the Guts
Life after Childhood Cancer
Esther Radstaak
isbn 9789461851192 paperback
isbn 9789461851208 ebook
Original Dutch title: Je moet het durven
1st print november 2015
Design: Eric Jan van Dorp
Translation: Aviva Dassen, Karin Wouters, Jacob Gibbons
Cover photo: Irene Hoekstra
Published by Village, an imprint of VanDorp Publishers
Postbox 42
3956 ZR Leersum
The Netherlands
www.vandorp.net / info@vandorp.net
Copyright©2015 Village / VanDorp Publishers
Copyright©2015 Esther Radstaak
No part of this publication may be reproduced in any form without the express written permission of the publisher.
ESTHER RADSTAAK
YOU'VE GOT TO HAVE THE GUTS
Life after Childhood Cancer
VILLAGE
All revenues from this book will be donated to the
VOKK, the Dutch Association for Childhood Cancer
Foreword
As a television news reporter, I rarely experience a personal story that actually touches me. I speak to victims – people who were duped or discriminated against – on a regular basis, but normally I can shake it off quite easily. From a reporter's perspective, it's always nice to find a person who is actually involved in the story. That is how I met Esther; I had to report on the consequences of childhood cancer and a large national scientific study that was launched during that time. I visited Esther one afternoon in her house in Ede, and was greeted by a cute dog, delicious home- made cake, and a big garden.
Esther is a childhood cancer survivor, and she was introduced to me by her treatment specialist. I noticed straight away how tense she was, but I also saw her willpower to show herself to the outside world in both her vulnerability and strength. That is what truly struck me that afternoon. She was a person with a past, a very clear mission, and an unbelievable passion, who dared and wanted to share her story. Go ahead and try to do the same, exposing yourself in front of millions of people.
That afternoon, we both felt that we got along quite well. That is a big advantage when wanting to make a good documentary, but such chemistry is by no means a guarantee that everything would work out. After we were done, we both felt positive about the interview, and that is how we parted our ways. I phoned her the day after to express my admiration, and to ask for her response after seeing the broadcast. She was surprised, and immediately indicated that she wanted to write a book about it. In all honesty,
I had my reservations at the time. She kept in touch through text and told me about her progress. And here we are: that book is in your hands right now.
Beautifully written from the point of view of a twelve-year- old, and later when transformed into the perspective of an adult, this book is accessible and remarkably detailed. I don't doubt for a second that her many years of therapy have contributed to the fact that she remembers so much from before, and that she is able to commit her story to paper.
In all honesty, I think everyone who has been through a traumatic experience will be able to relate to this book. It provides an insight in the experiences of a teenager who suffers from a horrible disease. Additionally, the role of the parents during the disease process is described poignantly and can be of great help to others.
Esther, many congratulations on finishing this accessible book. In your own words, you have indeed come full circle.
\- Dutch TV reporter Theo Verbruggen
Contents
Foreword
That Was Then
Cold Hands
The Flags Flow High
New Life
The Door Cracks
Ed and Es
Crying for No Reason
Epilogue
That Was Then
It seemed so idyllic. My husband Kees, our daughter Marloes, and I – we lived a good life together. We had a beautiful home, I liked my job, and we were spared any financial concern. An outsider would look at our family with envy, but what they could not see is that all of those jealousy-inspiring living conditions were lost on me. On the inside I felt like a zombie, and a deeply unhappy one. Every day I was faced with the sensation of looking at myself from a distance, and feeling like I had no say in what happened to me.
This also came with its physical repercussions. It became more and more difficult to eat, and, without choosing to, I started to lose weight. 'You're overworked,' my mother told me, but I denied it violently. I knew what burnouts felt like. This was different, but I couldn't quite put a finger on what exactly it was.
This estrangement of myself and the desire to escape it became so great that I wanted to die. After all I'd been through, after all those years of fighting, I knew I should be happy to be alive, but the truth was that I didn't feel that way. At the most unexpected moments, bits and pieces of my oncological past continued to surface. It was as if the sick little girl of times past wanted to let me hear her voice sometimes.
When my scale indicated a meagre fifty kilos, I knew what I had to do. I sent an email to my previous oncologist, with whom I kept in touch, and told her that I wanted to move up my checkup. I knew I wasn't doing well, but I couldn't put into words what I was going through, let alone explain it to others.
I walked into the hospital with legs like jelly. This was the hospital where I used to be treated, and where I had a job at the time. I hoped not to run into people I knew, so I didn't have to explain why I wasn't at work. The door of my oncologist's practice room was open, and when she saw me she motioned me to come inside. She seemed startled when she saw what was left of me and how skinny I was. I couldn't do much more than stutter.
'I'm at my wits' end... I should be grateful to be alive, right? Then why am I bursting with guilt for still being here when others are not? I stole two years of my parents' and sister's lives, because everything revolved around me. Am I going insane?'
I hardly dared to look at her and nearly drowned in my own sorrow. The specialist's response was unexpectedly fierce.
'You're not crazy at all. You really have got to stop trying to please others at your own expense. Your parents decided to have children, so they should be there for you, especially when you're seriously ill. That is what parenthood is, you know. How could you possibly think that you are to blame for being sick for two years?'
I wasn't used to her forcefulness, but it did me some good.
'There's another thing,' she added. 'I hope you've got the time today, because the psychologist of the children's clinic is waiting on you. I forwarded your email to him straight away. I'll skip the physical tests for now, since they're not too important.'
That's where our fixed ritual ended. When I was young, every time after sampling blood, the oncologist would draw a smiley face on the white plaster and made hairs out of the cotton ball under it. I was too old for that now.
In the waiting room, I seated myself amongst the young cancer patients. I used to be able to deal with this quite well, but on that day I couldn't. Their sadness floored me, and the parents' powerlessness was unbearable. The assistant at the desk saw me growing more upset, and guided me to an empty consultation room.
'I'll ask the psychologist if she can come in a bit sooner,' she promised me, leaving me very thankful for her empathy.
It was Friday, June 11, 2004. The psychologist found the notes she took fourteen years ago. They were notes of a couple of conversations we'd had after my granddad passed away. She wasn't surprised that I was back. One of the possible causes of my problems, she thought, was post-traumatic stress disorder.
At that time the extent of the term 'PTSD' was unfamiliar to me. The thing that mainly got through to me was that this woman, a professional, ensured me that the way I felt was perfectly normal, and that I'm not the only one to feel this way. Filled with hope, I wanted to make an appointment for a follow-up session, but I had to deal with a setback straight away. The psychologist is not a specialist in the field of EMDR (Eye Movement Desensitisation and Reprocessing, an effective treatment to process traumatic experiences), so she would not be able to give me the treatment I needed. She promised to find someone who would, but that meant that I had to completely bare my soul to a new psychologist. I hated that.
The psychologist was true to her word, though. A few weeks later, she sent me the phone number of a colleague of hers, who saw my post-traumatic stress issues as 'a challenge'.
When I finally mustered the courage to pick up the phone, I found out that it wasn't that easy to make an appointment. First of all, I had to have a referral from my GP. Then I had to wait until there was a place available.
'But I received this phone number to make an appointment,' I argued. 'The psychologist ought to be aware of this.'
One night a couple of weeks later, when my husband and daughter dropped me off at Ed van der Leeuw's house, the path to the front door seemed like a mile. Once inside and seated, my new psychologist first told me about himself, and then it was my turn.
My own behaviour surprised me, being able to put my worries and concerns into words at the speed of a waterfall. My life had lost its glory. I didn't function, and I didn't want to deal with it anymore. I couldn't focus, had many headaches, and suffered from vertigo. I slept badly, and when I finally managed to fall asleep, nightmares would be waiting for me – nightmares filled with horrible details from the period in which I was hospitalised. I'd rather be invisible. Why should I have the right to live?
I even had the nerve to make a joke when I told him that sadness overwhelmed me, and that I was afraid to drown – even though I knew how to swim just fine.
Even though my days seemed like eternities, this hour flew by. The psychologist thought he could help me, and inquired whether I felt safe enough to try therapy with him. His closing question hit the bull's eye: 'Would you say you are currently surviving?'
I couldn't do anything but to answer with a straightforward 'yes'. I couldn't guarantee, though, that I'd be able to keep on. Sometimes I just wished I'd get run over and get the rest I so longed for without being able to do anything about it.
My husband and daughter didn't drive me to my second appointment with the therapist. Suddenly I was afraid to not be able to find the way on my own. Yet, I found it. As I told the therapist about the way I experienced my disease process – in an unstructured manner – the psychologist surprised me with a confronting question.
'Did people shout at home when you were growing up?'
'Shout?!' I echo jokingly. 'No, not at all. We never talked about our feelings. This stopped entirely when I got so sick, and you could see the worry on my parents' faces. It didn't even cross my mind to burden them with what was going on with me. I was willing to do anything for my parents so they wouldn't feel sad. I didn't even want to upset or worry my doctors.'
Not long after this conversation, I got a phone call that illustrated perfectly how communication between me and my parents worked.
'Yes, hi. It's your father. Don't startle, but—'
I felt the adrenaline surging through my body and I cramped up.
'—We just got out of the hospital. I have rectal cancer. No metastases. They're treating me with chemo tablets, and they will eventually start radiation.'
I knew what my dad was going through and tried to make him feel better.
'That's reasonably good news. If there aren't any metastases and if there's a possible treatment...'
'Yeah.'
Silence on my end.
'I'm hanging up now – I still have to call your sister.'
That was it. While I was still letting the news sink in, I thought to myself, 'Could this be an opportunity to talk to my parents about the past?'
That hope vanished quickly, though, when my dad had to be admitted to the hospital to undergo heavy surgery. When I came to see him, he already had a catheter in, and there was another one in the packaging on the cabinet next to his bed.
'Es,' my mother called me by my nickname, 'Have you seen a catheter before?' She didn't await my response, took the thing, and wanted to explain it to me.
I responded, surprised, 'Do you really think that after all my abdominal surgeries I've never seen a catheter?'
'Yes, but they're adding a bit of water, and then the balloon fills itself. That's how it keeps itself up in your bladder.'
'I know how it works, Mum, and I also know how it feels when they're trying to pull it out and forget to take out the liquid first.'
'Oh, right,' my mother said, and continued to make small talk.
A couple of months later, when my dad was readmitted to the hospital, it once again became obvious how parents and children can experience one and the same situation in completely different ways. As we were walking through the hallway, my mother and I were talking.
'Do you understand, Es, that it isn't all that great at home when someone is in the hospital? You were very jealous in the past, because you thought that we were having fun while you weren't around. It really isn't that way.'
'I was a child, Mum!' I retorted. 'I was a very sick child.'
My mother raised me and my sister with the best intentions and with a lot of love. It must have been terribly sad not being able to do anything, standing next to the bed of your dying child. Yet too much concern can also be suffocating. My dad couldn't be of much more comfort either.
'Dad, how does it feel for you, a hospital bed and being hooked up to parenteral nutrition?'
He looked at me, but didn't say anything. I kept trying, though.
'I thought it was awful to be fed that way. That dependency on others...'
Then I did get a response.
'Oh, Es, that was then. My disease is happening now.'
That's all I got out of him. I tried on multiple accounts to strike up conversation about my childhood emotions, but the reactions were always the same: 'We don't have to talk about that anymore. That was then, this is now.'
I still dream about before, especially about the angiography I had. I shared this dream with my psychotherapist.
It was holiday season in the autumn of 1980. I woke up and looked around me. I wasn't at home, nor in the hospital, yet I still felt a shrinking sensation. I repressed it and went back to sleep. Then I heard someone calling my name, and I actually awoke. My mother was standing next to me and told me to get dressed.
'Why?' I asked.
'Listen carefully, Esther,' my mother said solemnly. 'We are going to the hospital for a blood vessel examination. We didn't tell you because we didn't want to get you worried, but we have to. I promise we won't hide anything from you anymore. Do you hear me, Es? Never again.'
We were at the radiology department of the hospital within the hour. On our way there, we barely talked. Mum waited in the waiting room while the nurse came with me to the treatment room. I had to lie down, because the doctor was going to insert a cannula in my groin to make the contrast medium show up clearly in my veins. That's how they were going to see if there were blood vessels in or around the tumour. I got local anaesthesia, but to me it felt as if it wasn't working at all. I was in tremendous pain.
The nurse sat at my bed and let me hold her hand. She told me that I could squeeze it if I wanted to. I think I completely pulverised her hand, and I screamed when they took another attempt at stabbing at my artery. The nurse went to the toilet so often that I suspected something was wrong with her bladder. The doctors grew more and more frustrated and angry. They were wiping the blood from my groin and tried it for the umpteenth time. They told my mum that I was supposed to stay in the operation room for forty-five minutes, but it took hours. I wanted to walk away from the pain and from the angry faces. Eventually the doctors gave up, and the examination failed.
'You have to stay here tonight,' one of them told me. 'We'll try again tomorrow, this time through your feet.'
At the ward, my mum told me that the nurse was sent home in distress. She didn't have to go to the bathroom at all; she just couldn't bear to see me in so much pain.
The next day, I was put on a hard slab, which stood almost completely upright. The torture started again. In the waiting room, I could hear the people say, 'Do you hear those screams? Sounds like they got somebody pretty bad!'
Every time I wake up from this dream, I feel physical pain, exactly the way I felt it during the examination I underwent as a child. It's still weird to me, because I find it easy to talk about the entire period of sickness without getting emotional. When I tell that story, it's about a different girl, not about me. But in my dream, I always experience the pain myself.
My mum still can't talk about it. In order to help me tell my story, I asked her to send me her memories via email, but she stopped doing that halfway through the process.
One day my husband Kees asked me to get him some oatmeal to have for breakfast. The next morning, he heated the milk in the microwave. Upon hearing the beep, he poured the oats into the bowl, but when he started stirring something happened to me. Kees hadn't even brought the spoon to his mouth before I ran out of the room, retching. I re-experienced the mornings when I was still sleep-drunk, and my mother gave me yoghurt loaded with Fantomalt and Profitar so I would get all my nutrients. When I eventually came back from the toilet, I turned on the TV until the kitchen was safe terrain once more.
When my husband, my daughter Marloes, or I get sick, even with a simple stomach flu, I start to feel stressed out and unfit to do anything.
'I can't see, hear, or smell nausea or vomiting,' I confessed to Ed van der Leeuw. 'If I have to vomit myself, I panic and the tears start rolling down my cheeks. I'd prefer to lock myself up, so nobody can see me.' To my surprise, my psychotherapist told me that my extreme, irrational fear of vomiting has a name:
'emetophobia'. I'm not alone in this one, either. There is an association for people who once became nauseated from fear and developed a phobia of nausea and vomiting as a result.
Naturally, I realised that the goal of psychotherapy sessions is to process repressed emotions. What I didn't know, however, was that it can make you incredibly tired and cold.
After the session, I decided to heat up a little by taking a shower. I turned the knobs and undressed. The water was not quite warm enough, so I turned up the hot water more and more. Slowly but surely, my body temperature was getting back to normal.
'Discover your own body,' Ed told me, because I found intimacy very difficult. I felt immensely guilty towards my husband and daughter. What a waste of a wife I was to Kees, and how could I possibly be a good mother for Marloes? If I were gone, if I didn't exist at all, they would be able to grieve and move on. That would be much better than to have me as an extra burden. Over and over again, I cursed that damned cancer for still having such a grasp on me after twenty years. My tears washed away with the water. I didn't want to 'discover my own body' at all. The doctors did that for me, more than enough, and what they found was a malignant tumour. There were so many times where I had to take off my underwear for those intrusive internal examinations. I remained on the shower floor, sitting until my skin was all wrinkly. After I towel dried myself, I checked the mirror to make my hair look decent. Stupid hair – you should have seen it before my chemotherapy.
Cold Hands
The GP rubbed his hands together to make them a little less cold, but when he put them on my tummy, a pang of pain shot right through me.
Mum brought him in, because I was the only one in the family who kept throwing up after we had all caught the flu. I couldn't eat anything – I couldn't even stand up straight, because my body hurt that much.
The doctor knocked and pushed, and his face showed concern. I was allowed to get dressed again, and he walked downstairs with my mother to deliberate.
It was almost Easter. Accordingly, there was a chocolate hen with Easter eggs on my bedside table. I wished I felt like eating them.
When mum returned to me, she told me she had to take me to the hospital, and that it was possible that I'd have to stay there for a while. We purposely didn't bring any clothes or toiletries. If they wanted to hospitalise me, I could at least go home to pick up my pyjamas and toothbrush.
We went to the Martini Hospital in the city of Groningen, where I was given a double room – just for me. But there was no way that I was allowed to leave even for the briefest moment. My room was close to the nursery, and the crying of those babies was harrowing. I pitied those little kids, being without their parents and all.
While Mum drove home to fill everyone in and get my things, I stared out the window from my bed. There was a cat chasing a seagull. Only moments later, I was laughing: the tables had turned, and the seagull started chasing the cat.
When Mum returned, we decorated the room a little to make it cosier. She stayed during some further tests, but then she really had to go. At the door, she turned to look at me and waved.
Evening turned to night, but I couldn't sleep. I felt very sad, lonely, and my tummy hurt. They gave me a button to press if I needed anything. I had taken it in my hands a couple of times, but put it back each time. Then I decided to press anyway. Nervously, I looked towards the door, expecting a nurse to come in soon. Instead, the intercom creaked.
'What's the matter?'
Startled by the noise and a bit let down, I didn't dare utter a word and instead dove under the sheets. Tears of disappointment started to roll down my cheeks. Even though I was twelve, it was my first night in a hospital. They could at least come to check up on me, couldn't they?
Like our GP, the doctors in the hospital all had cold hands. They thought my tummy hurt because of an abscess, but just to be sure, they sent me to the Groningen University Medical Centre – a larger hospital in the city. My mum was allowed to ride with me in the ambulance, and Dad drove behind us. I had promised to wave to him while we were driving there, but the windows were so high that Mum had to wave for me. The ambulance seemed old and was rather creaky. I was very happy when we finally arrived. Meanwhile I felt weak as a kitten, not having been able to keep any food or drink inside. The nurses at my ward – 'Older Children' – kept encouraging me to eat, but I felt nauseated and full the whole time. Even pills for stomach aches and fever were difficult for me to swallow.
The nurses had to check up on what I consumed and what came out. They didn't allow me to go to the toilet: I had to bedpan.
I did have to take it to the bathroom – but the thing was so heavy, and I was only twelve years old! I contemplated drinking even less, so I wouldn't have to go to the toilet anymore. But no, I understood that it could be dangerous not to drink enough liquid. Thankfully, my mum was willing to carry the chamber pot for me during visiting hours, so I could conserve my strength.
'I want to go home!' I told my mother. 'I want everything to be normal again.'
Her honest response upset me.
'Es, they're not talking about admittance for a couple of days, but more likely weeks. They're going to perform surgery on you, so the doctor can determine what exactly is wrong with you. You'll probably be staying here for about six weeks, so you'd be home halfway through May.'
Forthright as it may have been, the message was nevertheless destructive. I collapsed into her arms, crying. I had to stay in this room for six whole weeks!
A couple of days before my surgery, the weather outside was beautiful, and one of my nurses asked me if I wanted to go for a stroll. I refused. When my parents came to visit, they immediately asked me why I didn't want to go outside. 'I'd like to go outside with you,' I responded, 'but not with a nurse.' A little later, the three of us walked past some surprised faces at the nurses' desk.
On the day of the surgery itself, I was lucky to be the first person to be operated on. Afterwards I woke up in a different room, waiting for my mum and dad to arrive. When they eventually came, they told me that they'd been waiting for the hospital to call them, but they never called. Mum had already sat down on the sofa with a photo album of mine, crying, because she thought I wasn't with them anymore. In the afternoon, they called the hospital themselves. It appeared that the doctor who was supposed to phone them had fallen ill, and that nobody had taken over his task.
An oncologist came in to explain what was wrong with me. His face – as well as my parents' faces – betrayed his lack of good news. I thought I probably wasn't allowed to go home.
I heard the word 'cancer', but the rest of it passed me by without really sinking in.
Later, after the first chemotherapy session, Professor Jan Aalders told me that the X-ray showed a massive lump in my belly. The lump had teeth, so it seemed. Also, my blood had an excess of some substance with a strange name, which had the doctors very worried.
When I was in surgery, the specialist discovered that the lump had grown out to reach quite a few organs. Such a lump is called a tumour. He had the tissue sample examined quickly in the lab, where they found that the tumour was riddled with malignant cells. I had cancer. The official word for it is 'teratoma', a tumour belonging to the group of germ cell tumours. It had attached itself to one of my ovaries.
Of course, I knew the word cancer. When it was pronounced at home, it was in a whisper and with an undertone of concern.
The doctor said he was unable to remove the cancer during surgery. It was the size of a big melon, which explained why my tummy was so big and painful. It would have to shrink before they could operate on me. To that end, I was to undergo chemotherapy. This would make me feel even worse than I already felt. He wanted to start treatment as soon as possible.
Resigned, I listened to the oncologist's story. Would a man like him lie? He had full eyebrows behind thick, black glasses. The only thing I wanted was not to feel sick, and he told me he was going to make me even sicker. My mum and dad's eyes glistened, and mum's head nodded to the beat of the doctor's words. I thought of a melon with little teeth.
Three days after my surgery, a nurse brought me to the treatment room. I lied there, shivering, until an attending physician had the time to install an IV drip. The cold remained, though, so they had to put my hands in a tub of hot water to open up my arteries a bit more.
'Here it goes,' the doctor said, and just when I started to wonder what was going, he injected the needle with a plastic casing under my skin. The nurse held my hand tightly so I wouldn't be able to pull away as a reflex. A little bit of blood trickled through the needle, which meant that it was put in correctly. They pulled the needle out, so the plastic remained. It was taped to my hand with plasters.
Mum was there the next day, when I was getting hooked up to the chemo. Her face was washed out completely. As soon as the poison meant to shrink the melon trickled into my body, I started to throw up. Vomit left my body in big waves, and after every wave I collapsed back into my pillows. Would this really continue for five days? Nurses came walking to and fro with spittoons, feeling my wrist, taking my temperature, spraying with salt and chlorhexidine solutions to keep sores from developing in my mouth.
Visiting hours brought the necessary distraction. Long before the clock above my room's door struck three, I followed its big hand with every minute that passed. And yes, it finally moved – Mum would be with me soon. I pricked up my ears to catch the sound of her heels in the hallway, or her nervous cough. Where could she be?
She finally arrived around half past four. She'd been driving circles on the parking terrain before she could find a spot. The mail she brought along every time showed me what was happening outside of hospital walls. The notice board above my bed became filled up more and more with cards. Neighbourhood kids bought six cards together, and sent me one every day – except for Sunday, when the post didn't run.
My classmates wondered why the teacher told them I was doing fine when I wasn't allowed to go home. Every day, one of them brought me notes from the school day in a notebook. On the weekend, Mum and Dad brought the notebook to the hospital, keeping me posted on their day-in day-out as well. Mum suggested that I could write about a day in the hospital as a surprise for them, but I didn't share her enthusiasm. How could I possibly tell them about fun things in the hospital?
Mum set me up straight in bed, turned the bedside table so the table part was in front of me, opened the notebook on an empty page, and put a pen in my hand.
Well, great.
I stared straight ahead and avoided her gaze. Eventually I scribbled some stuff about an ordinary day for me at the hospital. I couldn't bring myself to describe what actually happened to me, or how I felt about it all.
Every night, my dad phoned the ward to ask how I was doing. He often heard that I was homesick and that I wanted to go home very much. In the morning on the weekends, he was allowed to be with me for an extra hour, and I liked that. The nurses liked it too, because on weekend shifts they can always do with a couple of extra helping hands.
What actually happened to me was much less rosy. I lay in bed five days a week and felt deadly ill. That's nothing to write about, right? After the five-day treatment I was allowed a short week to recover a bit, and then the second part of my chemo started, this time a series of injections. They made me less sick and I'd be able to focus on my surroundings.
Since a few days before, I'd had a roommate, one who loves good food. She wasn't on chemo because she had heart problems. Seeing her beaming expression every time people came to visit and her parents and sisters unwrapped delicious treats they'd brought her was wonderful. Apparently that was also possible.
We were let into the central bathroom together. She had a shower, and I sat down in the walk-in bath, where my view consisted of stacks of nappies on top of the cupboards with towels. What did they mean, calling the ward 'older children'? We laughed, splashed the water, and chatted a lot. Suddenly I found myself saying, 'I think I'd like a bowl of custard'. My roommate didn't hesitate for a second: she stepped out of the shower, wrapped a towel around her, and disappeared. A bit later, she returned holding a bowl of custard. 'Eat!' She dropped her towel and got back into the shower, as if nothing had happened.
The second chemo course required another IV drip. I knew what was going to happen that time, and I dreaded the days of nausea and hours of retching. It may have been because I was so tense, but the doctor had to poke around for some time before he found a vein, the sweat glistening on his forehead.
When the valves to the IV bottle opened and the stuff started to drip, it didn't take long before I started to throw up. Because of the retching, I felt my feeding tube – which went through my nose into my stomach – come up. I pulled off the plaster on my nose, which was supposed to keep the tube in place, and threw it in a tray while I vomited. When the waves of nausea subsided, I pressed the button.
The attending nurse considered the entire situation, cleaned up the mess, and put three cups of yoghurt with syrup on my bedside table. Her face spoke volumes: 'When I return, those cups will be empty.'
I felt very small and sad. How could anyone demand such a thing of me? Reluctantly, I attempt to finish a cup. But here we go, my stomach started to jolt and emptied itself almost immediately.
I didn't push the button that time, but the nurse walked in and saw what happened. She disappeared without saying a word, and I took a sigh of relief.
The feeling of relief lasted only a couple of minutes. She returned with another filled cup in her hand, which she set down next to me demonstratively. This time, she did grace me with a couple of words: 'Every time you throw up the contents, I'll put down a new one.'
My parents found a timid little daughter when they came to visit.
'What are those cups doing there?' my mum asked me immediately.
'I have to drink all three of them,' I whispered.
'Then just do it!'
'I feel so sick...'
'What does that matter? That's what you have the spittoons for, right? Just hurry, so you get it over with soon.'
'But Muuum, you don't get it. The nurse will bring another cup, and she'll keep doing it until I keep it in.'
In moments like these, my mum could be very tenacious.
'Here, dad, take it and drink.' My parents both gulped down an entire cup, and wiped their white yoghurt-moustaches from their upper lips with the backs of their hands. The third cup took me the whole day to finish, sip after sip.
I woke up feeling cheerful the next morning. It was the period between the five-day chemotherapy and the injections, and I felt reasonably okay during that period. Also, my grandma and grandpa came to visit on that day. I was looking forward to it, even though I worried about my granddad. He hated hospitals tremendously, and avoided them when he could. I counted down the hours impatiently. After lunch, we had an obligatory rest period, and I hated that. I didn't want to go to sleep – I didn't have that resting period at home either. Also, I wouldn't be able to sleep anyway. I'd only stay awake, brooding. There was no escape, though. The curtains were all drawn closed, and the ward was silent for an hour.
After the rest hour I rose from my bed and glanced behind me at my pillow. It was covered with hair – my hair. First I let out a piercing scream, after which I burst out in tears. A nurse tried to console me, but that proved difficult. When I tried to smooth my hair, I found my hands filled with strands of my own hair. How would I have to face my grandparents? Before I was able to answer that question for myself, they were at my bed already. Their visible sadness about my despair cut straight through me.
'Mum, those loose hairs start to knot on my head. The itch is driving me insane. Please, take scissors and cut it all off.'
'First you have to finish your crackers with cheese. Else there'd be a lot of hair on them, and then you won't eat them anymore.'
I knew my mother well enough to know that she was iron- fisted in these situations, so I reluctantly shoved the crackers in my mouth. There.
'Now bring on the scissors.'
It was clear that my mother had immense difficulties cutting her daughter's hair to the point of baldness. I pretended that there was a different girl who mourned the loss of her thick, shiny hair that had made her classmates so jealous.
My mother left a little tuft on the front of the girl's head. Those hairs were still holding on.
The Flags Flew High
It was in the hospital that my fascination with the medical world started. In every ward, there were loads of people saving lives all day long. Sure, every once in a while there may be a shrew who forcefully compels a child to empty her cup, but they were probably convinced that it was for the child's own good.
Yet, most doctors and nurses weren't used to explaining difficult diagnoses and treatments to their child patients. Usually they talked to the parents, and asked them to answer their children's questions.
In the University Medical Centre Groningen, I learned for the first time that the substance my blood carried too much of was called 'alpha-1-fetoprotein' – a protein normally produced in the liver and intestines of children before birth.
The pathological examination during my surgery showed that the tumour tissue they removed contained malignant germ cells. Those are the cells that become the gametes, for girls in the ovaries, and for boys in the testicles. In the Netherlands, gametes are found in approximately fifteen children per year, and I was one of them.
There was some unrest in the ward 'Older Children'. Men in fancy suits were swarming through the building. The nursing staff was nervous and responded curtly to requests and comments.
'What are all of those bigwigs doing here?' my mum asked the nurse who came to take my temperature and check on my IV.
'Budget cutbacks. They're here to see if staff can be fired.'
The indignation was visible on her face, and she hurried along. Later that afternoon, the nurse freed up some time to explain to us that the hospital board ran an examination to see if the same work can be executed by fewer people. Healthcare costs were rising, and the simplest solution is to save money by cutting salaries.
As she told me this, her colleague coughed dramatically to warn that the 'dull suits' were about to show up. The nurse startled and dove behind and nearly under my bed. When the coast was clear, we laughed about the incident. But it left a strong impression on me – a grown woman had to hide to make sure it wasn't discovered that she had time to talk to a patient and her visitors.
It was 14 May, 1980. The day after, exactly six weeks after my admittance, I was allowed to go home. I still had to go through chemo to shrink the tumour; to do this I had to be admitted for a few days at a time in the middle of the week. The injections I'd be receiving were administered by the polyclinic. I also had to gain some strength before undergoing another surgery later in the year. In anticipation of the operation, I was allowed to try get back to normal life – as far as my lack of energy and nearly bald head would let me.
I was bouncing excitedly on my hospital bed and found myself swaying between ecstasy and nervousness. Would my bed still be in my room? And would the hen with Easter eggs still be on my bedside table? Suddenly I found it all-important that everything still looked the same as when I'd had to evacuate my familiar environment so abruptly. Surely they weren't mistaken – surely it wasn't another person that was allowed to go home while I had to stay?
I was frightened when the nurse came into my room, followed by my parents. None of them bore the cheerful looks I had expected given my pending homecoming. Why weren't they as happy as I was?
'Esther,' the nurse said, 'I am going to take you off the parenteral nutrition now, but then you'll have to try to put it back in. That is the condition for you to go home again.'
I looked at my mum and dad in despair. They shrugged at me apologetically – clearly, it wasn't their idea. The nurse loosened the plaster on my nose and her look implied the question. I nodded. In one swift movement, she pulled the tube up out of my stomach, through my esophagus and throat, and out through my nose. I started to retch and tried to suppress the urge to vomit by coughing it away. Next it was time for me to try. I tried three or four times, but each time I couldn't get the tube further than my nose, and I already started to retch. My parents tried it, but were also unable to do it. Eventually we grew hopeless, and Mum ended the torment.
'Nurse, if Es spits out the tube, we'll return to the hospital. We live so close by, it won't be any trouble.' The nurse agreed to the proposition.
When home, I entered the living room first. It was small and the ceiling was low compared to the hospital room. I was as surprised during the drive home – I didn't remember the meadows were that green.
My sister Marion, my junior by almost four years, looked at me expectantly. We would be able to take baths on Sunday evenings again, but we wouldn't be able to play hairdresser like we did before. She was glad I was home again. Marion wasn't allowed to come to the hospital, because visitors for the children's ward had to be older than fourteen.
My grandparents lived in Helpman, a district in Groningen, and came to visit after a short trip to Haren, where we lived at the time. My mother made her world-famous apple pie the day before, but that was for the others, not for me. Chemo had affected the mucous membranes in my entire digestive system, starting with my mouth, in such a way that I could hardly get any solid food in.
A couple of days after my homecoming, the parenteral nutrition made me so nauseated that I had to throw up. While doing so, I noticed that the tube also started to come up. That meant that I had to go back to the hospital... I already grew nervous just thinking about having the tube inserted another time.
'Mum,' I proposed when my stomach had settled a little, 'from this point on I will drink that foul concoction from a glass. I don't want a tube in my nose anymore. Didn't you see the looks on people's faces when we were buying groceries? And then I won't need to lie in bed at any specific times to be fed through that thing. I want to go to bed when I'm tired.'
'As long as you get your nutrients,' my mother agreed.
Mum took two glasses from the cupboard. She filled one with tube-food fluid and the other with 7Up. I pinched my nose hard and poured the contents of the first glass down my throat in one breath. Immediately afterwards I washed away the taste with the fizzy drink, crossing my fingers in the hope that it wouldn't come back up.
I had to return to the hospital all the same. We made an appointment at the polyclinic for another chemo session. Usually I had to throw up immediately from the stress, and that day was no different. To break the tension, Mum and I pretended to be drunk. We stumbled across the parking lot to the polyclinic.
'Where is your tube?' was the first thing the nurse asked. I explained how I ingested it all, and they put me on the scale straight away. A surprise awaited us all: I'd gained nine hundred grams!
Everybody rejoiced, and as a reward I got to go to the dietician, who made a schedule for me to get through the day, small portions of food at a time. I was all set from six in the morning till eleven at night, and so was my mother – who had to prepare my food. Food was plated before me in the tiniest portions. I had to have fifteen cocktail nuts at eleven, two pieces of cheese at twelve, et cetera, and that's how I got through the entire day.
The blood test results were fine, so the same chemo rubbish could be injected once more. Mum drove home as soon as she could – we managed to get home before the vomiting started. A couple of days later, my mouth was full of sores again.
When I was admitted to the hospital in the spring, I was in the sixth grade. I had two more months to pass before the school year ended and I was supposed to start secondary school. I managed to get the energy to go to school a couple of days a week. My classmates were very busy rehearsing for the recital they'd sing at the end-of-year musical. I stood there and watched them, unable to participate, and that saddened me. The whole day, everybody in the school could hear the girls whoop, 'Simon Tahamata, yeah, yeah, yeah.' I didn't know who he was, and I didn't dare ask either.
I spent a lot of time in the tub, because the warm water made the pain in my body much more bearable – the hotter the water, the better. I watched the way the water crawled up over my skin. First it washed up over my flat tummy, then my legs, and eventually the water covered my growing breasts. I turned the tap and cut off the water flow, because the contents of the boiler in the attic were limited. If I stopped now, I could go again a bit later. It took hours for the boiler to fill up again with hot water.
I kept my wig in the closet in my room upstairs. We'd already had it measured and fit before my hair started to fall out and my mum trimmed it. It was made of nice material that looked like my own hair, but so far I'd only worn it to try it on. I wore a bandana all throughout summer, so the little tuft of hair – impossible to get rid of – came out from under it as if it were fringe. It almost seemed as if I didn't have cancer.
One day I was in the bathroom, drying off my feet. I didn't notice that my mum snuck up on me from behind. When I stood upright, she pulled the wig over my head in one movement, and in the mirror I saw the new Esther. I quite liked what I saw. The hair tickled on my bare shoulders, and I rather enjoyed the feel of it. Let the surgery come quickly, I thought, so I could get to my new school and start a new phase in my life.
Back in the hospital – at the nurse's request – I visited someone my age, who would grow bald soon due to the chemo. I showed her how cool you could look with a wig on.
'Judging from the scans, it seems as if the tumour has shrunk,' the gynaecologist explained, 'but we think it may have fused with one of the most important blood vessels in your abdomen. I deliberated with the surgeon, and we agreed to administer some more chemo sessions in the hope of making the operation more likely to succeed.'
Postponed – again. This meant that I had to go to the hospital a couple of more times, where the scent of roasted coffee only enhanced my nausea.
Outside, the sun was up high, and the sky was clear blue. I wasn't interested at all. I didn't get out of my parents' double bed for days and didn't want the curtains opened. I barely drank and I stopped eating. Every gulp came out as three times as much. Where did it all come from?
My parents witnessed this and let it be for a couple of days, but then decided to ask the oncologists for advice.
'No!' I protested, 'They'll stick me back in the ward again!' Mum and Dad promised that they wouldn't let me be admitted, but they couldn't keep their promise. When dad carried me in his arms and we entered the hospital again, I was hooked up to an IV drip straight away. I felt betrayed by my parents and the doctors.
By the time I gained some strength, school was already three weeks in. Luckily I was in the same grade as some of my classmates from primary school, so I didn't have to explain anything to them. On my first day, I was ambushed with a written exam. I only barely failed the test, without ever having been taught English, so I was quite proud of that.
During the autumn holiday, Mum unexpectedly took me to the radiology departmentfor a blood test that would answer the question of when I could finally undergo surgery.
That particular examination became one of my life's worst nightmares. Every time they attempted to insert the cannula and failed, the radiologist became more and more disgruntled, wiping my groin clean more roughly. Suddenly a spurt of blood shot out: well, at least they'd found the artery! Before they could inject the contrast medium, the needle darted out and he had to start all over again. After three hours, the team gave up. They must have been angry with themselves, but they snapped at me.
The next day they would try to inject the fluid through my foot.
The slab on which they tied me looked like the operation table I had in my Playmobil set. I couldn't be tough, though. It hurt so much, it seemed as if the needle went straight into my bones. Just like when my mum had cut my hair, I imagined another girl being tortured – not me. I heard her scream, but thank God it wasn't me.
After the procedure, I stumbled towards my mum. My sandals couldn't close anymore, because my feet were too swollen. Mum heard me during the whole thing. She motioned me to pick up my card from the desk, because she didn't have the strength to do it herself. I felt deeply ashamed. What was I doing to her?
The surgery dream kept recurring. Both my sister and I had tumours in our bellies, and we were given the task of spitting it out. Marion succeeded, but regardless of how often I tried, it would just stay stuck. I shot up straight in my bed and couldn't fall asleep again.
Mum and Dad contacted the oncologists as they worried about me more and more. They got a phone call during Marion's birthday party. To our relief, we were given a date for the surgery:
20 November, 1980. I'd been preparing for this for months and almost longed for it by now. Yet, when the time had actually come, I was anaesthetised for the surgery, and the gynaecologist had to tell me what he'd done afterwards.
'The surgeon and I saw a tumour of about eight centimetres in size, Esther. This means that it had shrunk to almost half of its original size. It was stuck to your bladder, intestines, and womb, but we managed to separate it from those. We removed your right ovary with the rest of the tumour, as well as your appendix. Your left ovary and womb looked fine, which my colleague told your mother directly after the surgery. We were overjoyed that we were able to keep those organs intact. That was our intention from day one: we want to make you better without taking away the possibility for you to have children. Congratulations, we made it!'
I was visited by my history teacher the week following my operation. I thought it was very kind of him, but I told him in all honesty that I'm not really interested in his topic of expertise. During the period of my illness, I figured out what I wanted to be when I grew up: a nurse at a children's ward. The teacher crushed that plan right after I told him. His partner was studying to become a nurse, and she shared the pain of every disease she came across.
'Don't do it, Es,' he advised. 'It will make you very unhappy.' I actually believed him.
When I was allowed to go home, I could finally get excited for December. We'd celebrate Christmas and New Year's, which we usually spent with family. Plus, my birthday was at the end of the month.
I was expected in the hospital halfway through December to undergo one more chemo session. 'Just to be safe,' the oncologists said, but that security lasted only a short while. One afternoon, my mum picked up the phone and I saw her age ten years in only a second. When she put down the phone, she told me what she'd heard in a stifled tone.
'Es, according to the pathologist, there were live tumour cells in the tissue they removed. They decided that you'd need more than one treatment. In fact, they've determined that you need to undergo them for another year. You'll be admitted for three days every six weeks.'
That prospect loomed over me like a dark cloud during the holiday season. During Christmas I got my very first period, and asked my mother for a flag. The girls in school always talk about 'a red flag that needs to fly' or whatever.
My parents saw my menstruation as the proof that I remained a 'complete woman'. They wanted to celebrate, but I didn't see the fun of that.
On New Year's Eve, I found myself able to wish my mum, dad, and sister a happy 1981. Mum wanted to visit the neighbours for a bit, like we do every year, but I couldn't.
'I know what the new year has in store for me,' I explained,
'and that's not exactly a happy new year.' They understood.
Esther with parenteral nutrition and her own hair, during the royal inauguration in 1980
Esther in wig, 1980
Esther on the 40-year wedding anniversary of her parents, her mother on the left
New Life
About ten minutes into Physics class, the door flew open and a classmate came in without saying anything. He sat down and rummaged through his backpack.
'Why are you this late?' the teacher asked, and the boy answered by giving him a disparaging look. 'I don't need to answer that. You were late last week, and you didn't give us a reason either.' The teacher sent the boy out, and he disappeared with a big grin on his face.
See, I'd like to have the guts to do that.
I noticed that very little of my self-esteem was left after a year of illness. My wig gave me a normal appearance, even though I had a chemo-face, but since my admittance to the school I was terrified someone would pull at my hair.
In the schoolyard a classmate remarked to me in passing,
'You're lucky, you know. You have to be at school for a couple of weeks, and then you're free for a while.' I cringed with discomfort and didn't know what to say to that.
At home, I told my mum about the incident. She said I should ask the boy to come to the hospital at some point, so he could see what happened to me. When he congratulated me on my frequent holidays a couple of days later, I had the guts to propose what my mum had said. He never teased me again.
I often heard my friends' parents complain about their children's rebellious behaviour. Personally I wouldn't think of misbehaving myself, considering the pain I'd already caused my mum and dad, grandparents, and my sister... Invisible, that's what I wanted to be. That's why it really bothered me when a substitute history teacher gave me an insufficient grade, and told me that it counted towards the final grade.
This time I wouldn't sit and watch this happen to me. I had to protest. I raised my hand and said that my grade should not be included in the list.
"I don't see why not," the teacher replied.
"I have an agreement with the regular teacher," I responded, which started the humiliating discussion I was so scared of having. The class was struck silent.
"You can say whatever you like. Why don't you explain what your agreement is?'
"Can I tell you after class?" I asked, talking more softly than I already did.
"What was that?"
"I was wondering if I could explain it after class..." "Do it now, or you'll keep the grade!"
Eventually, in broken sentences, I explained the course of my illness. If the teacher were to feel as horrible as I felt throughout the rest of the day, I'd feel incredibly sorry for him.
Back at the hospital, post-surgery examination showed that the amount of alpha-1-fetoprotein in my blood was normal and that it would stay normal. In addition to the six-week chemotherapy, they administered hormones to calm the remaining ovary. I also got other hormones in order to cause as little disruption to my bodily development as possible.
The day of the last chemotherapy session commenced. Mum brought me to the hospital and promised to be there with me for the last treatment, just as she was with me during the first.
In the morning I sat up in bed, excited.
"They're not going to do it!" I cheered. "My blood isn't good enough!"
A nurse came to explain that my renal function had declined sharply. In a short while, they'd take another blood sample, after which they'd assess whether we'd end up postponing and postponing until we'd finally just stop trying, or put a halt to it right now.
The next day my parents came to pick me up. They were on the telephone all morning to discuss whether I did or did not need the chemo. Eventually they found that there was too great a risk of doing further damage to my kidneys. My things were packed. Mum even took the air freshener she used when I'd still been attached to a drip and couldn't go to the bathroom. She used it in vain, though, because she just couldn't get rid of the smell of my chamber pot. I skipped through the hospital on the way back, feeling as if my life would start again.
Since I only needed to go to the UMCG for routine inspection from this day on, it took me little effort to get through high school successfully. Mathematics was my favourite subject, which was mainly due to the teacher, who would make his lessons humorous and fun to watch.
In 1984, I took my final written exam. Despite earlier good results, I was very nervous. My guts were working overtime and I could hardly be found away from the toilet. I was nervous to the point of having my parents phone the school to arrange for a teacher to escort me when I had to go to the toilet during the exam.
On the day the exam results came out, the phone rang at ten to five. Had I failed?
It turned out to be the hairdresser who asked if the school had called us yet. If they hadn't, I could count it as a success. We knew that, right? I swear we could shoot him, but instead we found our stress relief in a laughing fit, not holding back on the tears.
Moments later, a taxi stopped at the door. My grandparents, who had a caravan in Ameland, an island just off the northern coast of the Netherlands, had come back especially for this occasion. We didn't drop the word 'cancer' once on this festive day, not once! We all knew, though, that the past made our joy even fuller.
So, what would I do next? My dream of becoming a nurse had been nipped in the bud by my history teacher, but I did still want to work in healthcare. It would have been perfect if I'd gotten to work at the hospital at which I'd been treated and declared healthy. I chose to become a pharmacy assistant.
During my second year at university, we had to arrange an internship ourselves. I asked if I could do my internship at the UMCG department to which I'd been admitted, where I had to return on occasion for my inspections. I could, but my request would have to be discussed with the staff. An internship with children who have cancer is trying for anyone, and all the more so for a trainee who has experienced the disease first-hand. But my passion and tenacity turned out to be the decisive factor.
During the first day at 'my' polyclinic, I sat down next to a doctor behind the desk. It made a world of difference, sitting on this side. It felt amazing. The first young patient had to have blood sampled from the neck. He lay on the examining table, on his back with his head over the edge. The puncture succeeded on the first try.
Next, a girl came in who needed a lumbar puncture. She had to lie down with her knees drawn all the way up to her chin, but she struggled with all her might and her crying was heart- breaking. Two nurses had to come and help, and the five of us kept the child under control.
'Are you coping?' a nurse asked me. I nodded. Of course it affected me, but I found the whole clinical procedure very interesting, to see how the cerebrospinal fluid dripped into the tube through a hollow needle.
In my university studies, as well as for high school, I obtained my diploma at first try. Afterwards, I could start working at a community pharmacy, but first I had to undergo a medical inspection.
'You were once very sick, huh?' The doctor came straight to the point after we introduced ourselves and shook hands. It scared me a little. How could he know?
I told him about my cancer, and that I was cured. He was satisfied with that statement and didn't do anything else. Later, I learned that one of my teachers had recommended a comprehensive medical examination, although I hadn't missed a single day due to illness. My oncologist was livid about this sneaky behaviour, but I didn't feel like following up on it.
On 28 December, 1985, I turned eighteen. I was still studying to become a pharmacy assistant at the time, and went to our church's club almost every Saturday night. We chatted and danced to UB40 and Elton John. And suddenly, there was Kees. It took a while for the spark to hit, but two weeks after my birthday we were officially dating.
It didn't come to mind to inform my boyfriend about my illness, and we didn't speak about it at home either. I told him in passing, when it was convenient. Kees took my announcement as simply informative, and clearly didn't need further details. A year later around Christmas, when we were chattering away with my in-laws, we talked about getting married, but only when his brother, his wife, and their children returned from Africa.
'I hope I'll be able to have kids myself,' I mentioned in an unguarded moment. 'That may be difficult, given my history of illness.'
Kees's mother went pale and didn't say anything. The rest of the family also went quiet. My sister-in-law regained her voice first and said, 'I do hope so for you two, because you're great with children.'
It took until I was twenty-three for my soon to be brother-in-law to return so we could actually get married. My mum and I bought me a light green wedding dress and white pumps for quite a sum. At the city hall, she warned me not to pull jokes like saying, 'Yeah, let's try,' when the official asked the famous question. Afterwards, we had the ceremony in the church and I discovered how hard the pews were that we had to kneel on.
At night, we held a big dinner and a reception, where my oncologist and her daughter came up to congratulate me. She brought the teddy bear my dad and I had bought for her when she had a baby. I was in the middle of my chemo sessions at the time. I remembered that she jumped out of her bed, barefoot, to put the teddy in her baby's crib. Barefoot! We weren't allowed to do that when I was still in the hospital.
Easter was around the corner. Kees and I had moved to Assen, a city twenty kilometres south of my home town. We lived in a newly built home where our dads had done a lot of work. We'd consulted with a doctor to ensure that my previous medical treatments wouldn't have consequences for a child. For as far as they could see, the risks were minimal, but that wasn't a guarantee for a pregnancy. My diminished renal function was a complicating factor. This meant that I would have to go to the UMCG for monitoring and guidance if and when the time came.
Shop windows were stocked with bunnies, chickens, and eggs. I tried to ignore the memory of the chocolate chick in its basket that had stood on my night stand when I was rushed to the hospital for the first time. Why didn't I want to see that image? Was it too painful to think about that child's sadness because I wanted to have a baby?
I was four days late. Was that even possible after quitting birth control the month before? I rushed to the drugstore to buy a pregnancy test and read and reread the instructions. The waiting began.
Two lines – that means pregnant. Kees and I embraced one another and I started to hyperventilate. The same night, we drove to our parents. It may have been rash, but we wanted to share the news and our happiness. Still I was nervous for my mother's response. Wouldn't she start worrying immediately? My fear appeared to be unfounded. Everybody was incredibly enthusiastic and happy. The next evening, our doorbell rang. I opened the door and I was met with a huge bouquet of flowers, behind which I saw my parents' smiling faces. They thought it was wonderful.
When I was about eight weeks pregnant, I had to go to the hospital for a check-up. The gynaecologist was still making notes in the file of the woman who came in before me when he asked how I was doing.
'Well,' I heard myself say. 'Actually, really well!'
Professor Aalders's pen stopped mid-air. 'Are you trying to tell me what I think you're going to tell me?'
Two can play that game, doc!
'If you thought I'm trying to say I'm pregnant, then yes.'
The gynaecologist got up and walked around his desk to put his hands on my shoulders. 'I'm going to call an obstetrician right away to guide you through this. I want you to be monitored here from now on.'
A while later, I saw him standing in the hallway talking to a colleague. They discovered me standing there, and I heard him say, 'There's the topic of our conversation.' Their eyes were glistening, and out of embarassment I didn't know what to say.
When the examinations with the obstetrician were through, I stayed in my seat as long as it took him to realise that I had another question.
'Can I get an ultrasound?' I finally blurted out.
'Why?' he says. 'You had one in the beginning. Everything is fine, both with you and the baby.'
'Because until now, ultrasounds were meant to look at a malignant tumour in my belly. Now it's our child we'll be looking at.'
Moments later I danced my way to Radiology with a form in my hand. The lab technician showed me the life growing inside me, and proposed that he wants to take more ultrasounds, because I didn't have a lot of amniotic fluid. Fantastic!
At eight months pregnant, I was admitted to the hospital. My blood pressure sky-rocketed and my kidney function was declining slightly. The doctors thought it may be the beginnings of preeclampsia. I was able to console my mother, thank heavens: both our child and I were fine otherwise. Yet, trying to fall asleep at night, I was back at the department of 'Older Children' and my childhood fears made their reappearance. In my dream, I heard my sister walk down the stairs in my parents' house. I woke up in a panic and spent part of the night in the wide window sill, with a pillow behind my back, staring out the window so as to not give my dreams a chance. I was happy when I saw the sun rise.
'Do you remember her?' my mother asked a neonatologist when we were having coffee the next day.
'Most definitely yes,' the specialist said in passing. 'Her belly is as big as it was back in her days in the children's ward.'
Mum shut up and I pretended that the comment didn't fall as harshly as it did. Luckily the doctor came to apologise later that night. When we ran into him, he was on his way to a couple whose child would be born far too early, and was mentally preparing for what he had to tell them. He stayed next to my bed for half an hour, and together we took a trip down memory lane. He also asked me to keep him updated on my developments and to apologise to my mother for him.
Kees was getting a bit tired of driving to the hospital every night. He shared his irritation with me. 'It's great to be bored together, but when you're alone it's quite the opposite.' He took out three cans of tonic from his bag. 'Drink up, Es. Quinine can bring on labour.'
That night, I felt contractions starting and the next morning it was on. My husband was in the hospital on time, and we spent large parts of the day doing breathing exercises. Every two to three minutes I felt them coming, and Kees helped me through every contraction by massaging my back to take some of the pain away. He wanted to take pictures, but the contractions came so close together that he didn't get the time to even press the button.
I'd already lost all of my amniotic fluid, but there was no full dilation yet. At night, they decided to perform a caesarean section. At sixteen past ten, a small human being was lifted above the green cloth for a couple of seconds to be shown to me. After inspection, Kees came to sit by the head of my bed with our child. Finally, I could make the call I've been waiting to make for nine months. I called my parents to tell them they had become grandparents, and my sister Marion an aunt. We had a beautiful, healthy daughter. We named her Marloes.
The end of this long, joyful day didn't come without another surprise: I got afterbirth pains. They were so heavy that the nurse was preparing an injection to stop the pain. Despite the torture, I started laughing when I saw her squinting behind her thick glasses to try to get the needle in an ampoule. She didn't miss my leg, though, and I dozed off peacefully.
The Door Cracks
I was over the moon to have Marloes, but her birth also triggered many difficult memories. I thought about how great it would have been for my granddad to be with us. During that time, I thought about the time of his illness. I had to go to the hospital only once a year for inspection. First I had to go to the gynaecologist, who was very satisfied. Then I had to see the children's oncologist.
'How are things?' was her standard opening question. I tried to look away, but she didn't fall for that. She kept staring at me and asked, 'Tell me, what's up?'
Suddenly I started to cry.
'My granddad has lung cancer,' I managed to say. 'It's incurable.'
The oncologist spoke some words of consolation and wanted to know how I felt. To buy some time, I took a while to wipe away my tears. I couldn't tell her that I'd rather be sad about someone I cared less about, could I? That I had to think about all of the children in her ward that died while I was getting better?
I mumbled that things were all right with me.
'You're free to come see me whenever you feel the need to,' she offered.
I was embarrassed for bothering her with my sadness.
On my way to my grandparents, there were two hearses that passed me. I took it as a bad omen, but I tried to push away that fear. Usually, granddad would be sitting in his chair by the window, smiling from ear to ear. My heart melted thinking of my granddad, who only recently postponed a chemo session, because I wanted to go out for dinner with the whole family once again... My granddad, who after his diagnosis said, 'Our Es survived, so I'll probably survive this, too.' On my way to their house, very dear and tender memories surfaced in my thoughts, like the afternoon my grandma opened the door and put a finger to her lips immediately.
'Is it that time again?' I whispered. She nodded conspiratorially.
When my granddad was baking, the kitchen door was closed demonstratively for everyone. It was an exceptional day, because I got to help. The oven was already being preheated and I cracked the eggs over a bowl. Granddad cut the butter up into pieces, and was so focused whilst doing it that he didn't see me wiping a piece of egg shell out of the bowl. He put the mixer on the highest setting and the flour whirled up out of the bowl. I greased the tin and he filled it with the batter and put it in the oven.
Satisfied, we went to sit down with grandma. After the kitchen timer went off, it took ages for the cake to cool off sufficiently for us to taste.
There are more images that my mind conjures up, such as our stays on Ameland. My grandmother and grandfather had a caravan on a campsite on one side of the island, and my parents had theirs set up on a campsite on the mainland. My sister and I stayed over at my grandparents' a lot. When my granddad was shaving in the morning – with a shaving brush and cream – he chased us until we both had white smudges on our cheeks, and when I'd had a shower he was the only one allowed to comb my wet hair.
When I visited him on his sickbed that day, I'd bought the special Belgian chocolates my granddad liked so much. We had to use a knife to cut them into pieces, because he didn't have the strength to bite down and his mouth hurt. There was no better way to push away the awareness that the end is near than to make jokes about it.
'Would you like a fork with that?' I asked lightly. 'Aren't you fancy, eating chocolates with knife and fork?'
I was working in the pharmacy when Kees picked me up to say goodbye. Granddad had his eyes closed when we walked in, but they opened when he heard us. He thanked my husband for the great years he'd spent with him, and told him to take good care of me and for us to have a great time together. Then he tried to pull himself up to tell me something, but it took too much effort. I took his hands and told him quietly, 'Just lie down, Gramps.' For the first time in his life, he obeyed me.
He squeezed my hand a couple of times and fell into a deep sleep. The next day, my father called me to say he had died.
My granddad wanted to be buried in private, which is why I wanted his obituary in the paper after his funeral. I also wanted to get an obituary published in the name of his four grandchildren. It took me an hour to get the text through. Every time I heard the dial tone I felt the tears well up and put the phone down again. I didn't want to pass it on while crying, what would they think? When I was finally able to do it, I was exhausted.
After my grandfather's death, there was a period in which my dreams were about children who died in the hospital while I was being treated for the tumour. When I went to the hospital for chemo, on occasion a patient's room's curtains were drawn. If I'd be at the ward six weeks later, and the curtains were opened, I knew enough.
I got fits of dizziness which led me to the neurologist's door once more. When he checked my knee reflexes with the little hammer, my leg didn't move at all. He examined my arms, completely supple. My legs, on the other hand, were flexed continuously. That could be the effect of years of magnesium deficiency, which is why I had to take pills. Yet, that wasn't the cause of my melancholy. My obsession with death became so extreme that my GP prescribed very heavy medication. I became unfit for work and couldn't stay at home alone anymore. My mum and dad took me in lovingly, but my stay in my parents' house wasn't entirely to my benefit. It also brought about a lot of sadness from the period in which I'd had cancer.
It took me about three months to get a bit better and got to reduce my medication. I found the detoxing terrible. I'd never want to take anti-depressants ever again.
It's a given: life and death are simply connected. While I still think about my granddad, who always gave me the feeling that I was special to him, my own daughter Marloes saw the light. To my joy, I saw her grow into a healthy child.
A few days after her birth, she let us sleep through the night. I took her to the hospital when I had to return for my examinations, as I promised the staff. I was proud to see my child enjoy the attention.
Sooner than I realised, she was about to start preschool. When I brought her for the first time and I had to leave, she cried. Luckily, I got a phone call from the daycare worker later that day. Within minutes after my leaving, she started playing with the other kids. When she was on the phone, Marloes was crafting a tie for her dad.
When Marloes fell ill and had to undergo minor surgery, I found out how vulnerable I still was when it came to doctors and hospitals. Our daughter had had a cold for a while, was feverish, and ate far too little. Her daycare teacher suggested that she could have problems with her tonsils, and the GP and ear, nose and throat doctor shared that opinion. When I walked into the hospital to have her tonsils removed, I suddenly became that twelve-year-old again.
The sombre concrete and light scent of Lysol filled me with dread. On top of that, my friend, who had joined us for support, was sent away because only one adult was allowed to be with the child. I was here once, crying during the first night at the hospital, which would be succeeded by so many more lonely nights. The sound was just as tinny as it was back then, and the doctor who was about to operate on my child shook my hand coldly.
Marloes had as much of a hard time giving in to the narcosis as I did, and struggled to fall asleep. I recognised that feeling of slowly losing grip and control over your life... It constricted my throat. Finally her face changed from red, to pink, to grey.
I didn't realise how vulnerable I was. A couple of days after the surgery, Marloes was as healthy as a horse and bursting with energy. It didn't come to mind for a minute that I had to worry about a simple operation on a healthy toddler, or about an innocent childhood disease, but it may be about the weight of my own fears. I approach them in the same way I deal with emotions. I wasn't the one who was nauseated by the smell of the hospital. It was another woman who tensed up when her child had the mask put over her mouth. Wouldn't the specialist have been bothered by my crying over my daughter?
During the turn of the millennium, computers didn't crash, nuclear plants didn't explode, and planes didn't fall out of the sky. Es Radstaak had a job interview at her hospital: the UMCG!
A friend told her boss that I was a meticulous worker and I really wanted the job.
I saw my interviewers look at me doubtfully when I told them the real reason for my application. 'So many people have spent months and months to give me a future,' I explained. 'Naturally you'd take the chance to give something back.'
First I received a phone call from human resources. They told me that I would be kept in the pool for a while. A short while after the first call, they rung again and asked me if I still wanted to work for them. My answer was brief: 'Yes, definitely!'
I was standing in line at the staff restaurant when I suddenly felt an arm around me. To my excitement, the arm belonged to my previous gynaecologist.
'Well, I see you're back! Did you finally manage to get what you want?' He added in a whisper, 'And do we have any additions to the family?'
In a couple of words I told him that up till then I hadn't been able to get pregnant again, which is something Kees and I would have wanted.
'We're back on the medical road, doctor, and we're trying our best.'
The last part was definitely true, even though making love became a little forced when the necessary planning and frequency became more important than lust and love.
The whole medical business gave me more trouble. For five years, we took every path we could. There were keyhole surgeries and treatments, but eventually we came to the conclusion that we had to decide whether or not we wanted to try IVF-treatment. I lost sleep over it. What if I had twins, triplets, quadruplets growing into my belly? My kidneys didn't work properly anymore after the chemo. Would my body be able to handle another pregnancy? Won't the age gap between a new child and our daughter be too big, since Marloes was eight already? Kees backed me entirely and unconditionally, whatever choice I would make.
Still very doubtful, we made an appointment at the hospital. I was terribly nervous. What would the specialist think of us? We'd already caused him so much work.
Turns out I'd worried with no reason. When I confessed we didn't want to continue treatment, the doctor responded laconically.
'Why do you find it so difficult to say that you don't want to try further? I don't have to raise the child, that's your responsibility. You're in your right to decide whatever you want, whether it means trying or quitting.'
Not long after, it appeared that I had made the right decision by not starting the exhausting IVF-treatment. The period between my menstruations became shorter and shorter, and the menstruation itself became much heavier. I had to build my social life around my periods, because my physical ailments became a bigger hindrance to my normal daily functioning. A new hormonal treatment couldn't guarantee success, and the alternative was to remove my womb and the fibroids that had come into existence. I didn't need a lot of time to think that one through. That organ had proved its worth: we had a beautiful daughter. Now it did me more harm than good.
I loved my job at the lab in the UMCG. We got an intern – Arnold – whose nose about reached my hip due to the growth disorders he'd dealt with in his childhood. We got on very well together, shared the same sense of humour, but more importantly we shared a lot of hospital experiences. Arnold became depressed when one of his previous caregivers became incurably ill and he didn't dare contact her. When he finally managed to get the courage to do so, it was too late.
I called my husband to tell him that I'd be meeting someone after work to offer a listening ear, a shoulder, and perhaps a box of Kleenex. Later that day, Arnold and I found a place in the grass by the hospital and talked for hours.
He shared with me his experiences of his disease, and the many footsteps he'd walked in this particular hospital. I told him about my cancer and all it did to me when I was a child.
For the first time, the door – which had hidden my emotions for more than two decades – started to open just a crack. I couldn't determine if my defence mechanism had weakened after my grandfather's death.
I don't know if it was just the conversation I had with Arnold, but suddenly I felt as if a tidal wave came over me and dislodged the anchor that kept my emotions in place. Perhaps it was the surgery with which I decided never to be able to become a mother again. Perhaps it was the recent visit to my parents, where I dodged conversations that made me feel like a visiting child. Perhaps it was remembering the deaths of so many little patients in the hospital, or losing my father-in-law to cancer, or losing a colleague whose daughter hadn't even reached the age of one yet.
During the first few months of 2003, I couldn't find any possible way to shut the door again. The worst was that through reasoning and argumentation, I could only find reasons to be happy. Then why would I feel like I didn't want to exist anymore? I didn't know who to ask for help. The oncologist who saved my life when I was a child was the one to come to the rescue once more. She had consoled me before, when I was incredibly sad about my grandfather's pending death, and had promised that I could call her any time. She set up an appointment with me on very short notice, listened to me intently, and helped me find a good psychotherapist.
A month later, I stopped working for the UMCG's lab, which had meant so much to me. I was exhausted and suffered from lack of concentration. That is inexcusable for a pharmacy assistant.
Kees worked in the province of Utrecht, halfway across the country. Driving 165 kilometres twice a day started to take its toll. On occasion he'd sleep in a hotel to have to travel a little less. With him gone and Marloes in school, I remained alone between four walls, fretting and awaiting a message from the psychologist in Groningen that she'd found a therapist for me. Every now and again, my spouse would send me a link to a house for sale somewhere in the middle of the country. I always just clicked the message away. Moving house was the furthest thing from my mind.
One night, I was changing channels in front of the TV, and clicked into the middle of a broadcast of 'No Going Back'. A married couple was stuck on a tropical island, but they had to return to the mainland as soon as possible because of the husband's asthma. When they got ready to return to their favourite sunny place, the screen cut to black. Was my TV broken?
Slowly a text crept across the screen in white lettering. They weren't going back. The protagonist had died suddenly of a heavy asthma-attack. End of broadcast.
I sat staring at my TV for quite a while without any new images really getting through to me. Slowly I realised how stupid it was of me to keep asking my husband to drive to and fro because I didn't feel like going away from the place where I lived.
I opened the next link Kees sent me. I saw a picture of less than two by two centimetres, fell in love with it, and ordered the brochure. Kees didn't know what was going on. In the best of moods, we made an appointment to look at the house and drove to Ede.
'Save your potential enthusiasm for when the realtor has left,' he advised. I couldn't help but smile at the comment.
When the three of us viewed the house together, we tied the knot.
The first meeting with the man who would be helping me learn how to live throughout the next four years had passed with mutual satisfaction. Ed van der Leeuw lived in Meppel; although much closer to Assen than to Ede, that wouldn't be a reason not to move to the new house. The hassle that came with moving distracted me from the horrors I would have to go through, the ones I didn't know existed yet. I packed my bags and threw some stuff away. If only that were as easy with the baggage from my childhood.
Ed and Es
July 2005. Marloes was twelve years old. Kees, me, my sister Marion, and her husband were sitting in the assembly hall of the Cavaljéschool in Ede, and I was looking at my daughter, shining on the stage in the end-of-year musical their class had been rehearsing for over the past few weeks. I should have been standing there when I was her age, yelling for Simon Tahamata.
I had very mixed feelings about the whole thing. I saw my daughter change from an elementary school kid into a true teenager who would soon start rebelling against me and my husband. I enjoyed it immensely. She talked to us as if we were her classmates. Sometimes, we enjoyed her adolescent behaviour a little too much to also be able to raise her responsibly. At the same time, I felt a sadness that cut very deep when I realised that the period she was going through was entirely alien to me. I had skipped it.
We moved to Ede on my birthday, 28 December. It snowed on that day. I worked a lot during the months before our move, both in our new home as well as in therapy. From then on, I had to make a weekly trip of 125 kilometres to Meppel and back, to plough throw my life with medical psychologist and psychotherapist Ed van der Leeuw.
From the first meeting it was Es(ther) and Ed. During our first session on 4 August, 2004, he detected a whole load of pain and sadness looking for a way out. Now he observed that it wasn't just my being ill as a twelve- and thirteen-year old that caused me to feel the way I felt, but that my relationship with my parents also had to do with it.
I find that really hard to write down. It's hard to characterise because our relationship is quite complicated, but emotionally it's also very difficult to get it out. I know how much they love me, how caring they are, and I'm very grateful for that. I am burdened with the guilt of knowing what I've done to them by getting ill and claiming such a large part of their lives for so long. I discovered what the past did to me, but also what it did to them.
Normally children want to please their parents. When they reach puberty, that need becomes less. I, however, didn't experience puberty at all. I kept trying to please them, and because my disease stopped me from doing that – or I felt it did – I developed feelings of guilt. Because I never figured out who and what I was, I let their smothering care rain down on me for all those years. I never wanted and haven't even been able to resist my parents or anyone. Now that my body and mind were letting me know that my boundaries had been crossed, I had to change. Could I? And if I could, would I have the guts to confront my loved ones with the new Esther? It could be quite a shock...
In the first meetings with my psychotherapist, these questions weren't at all near the surface. He made me come to realise that I was only surviving. It would take years to discover what from.
The therapy I was going through consisted of conversations intermittently complemented by Eye Movement Desensitisation and Reprocessing sessions, EMDR for short. Ed explained to me the purpose of the therapy and how it worked, and Google did the rest. On a website about EMDR, I read that this type of therapy is meant mainly for people who keep being haunted by a shocking experience, such as an accident, sexual abuse, or – as with me – a life-threatening illness. The therapy was first developed in 1989 by an American psychologist, Francine Shapiro. In the next few years, it was developed further. Just on time, I thought as I was browsing through the material.
In our conversations, I told my psychotherapist about a few heavy experiences in my life. He asked me to relive these experiences and to try to recall certain images, thoughts, and feelings. After that, I needed to recall the experience, the target, again, while he distracted me with a stimulus. This can be induced by headphones, where a patient hears sounds alternating from the left or right side. In my case, the stimulus was the hand of the therapist, which moved from left to right about thirty centimetres in front of my face. I had to follow the hand quickly and fully focused, whilst articulating what happened to me.
The experience I kept going through was split into different fragments, or sets. In every set, I relived a small part of my trauma.
On the EMDR website, it said that a series of eye movements would make the memory lose its strength and emotional charge. This meant that it would become easier for me to think about the traumatic experience. It would also be possible for me to start developing new thoughts of insights, spontaneously, which can give a different and less threatening meaning to the event. As such, the shocking experience would find its way into my life without bothering me too much.
That seemed pretty nice, but nevertheless I was quite tense when I subjected myself to the first EMDR session. I had to return to that dreadful afternoon, when a horrible nurse put down three full cups of yoghurt with syrup because I spit out my parenteral nutrition cable. I saw a terrified child, lying in a hospital bed, nauseated and shivering. I was so small and vulnerable, and wanted to feel big and strong. That experience unravelled itself in seventeen sets.
Set 1: I feel very lonely.
Set 2: The cups of yoghurt look like litres to me.
Set 3: The nurse is very angry and gruff.
Set 4: A strong nausea arises.
Set 5: I spit out the yoghurt and she fetches a new cup. Set 6: I feel empty and lonely.
Set 7: The nurse is standing next to my bed and is furious. I want to leave.
Set 8: I see a pool of yoghurt, my body tenses up.
Set 9: I feel small and worthless.
Set 10: The nurse orders me to drink. Why?
Set 11: I'm afraid the nurse will return.
Set 12: I'm completely empty.
Set 13: I see myself looking at the clock. Visiting hour? I feel light-headed.
Set 14: My parents don't understand. Helpless. I have to defend myself again.
Set 15: The despair I see in my parents gives me relief.
Set 16: Mum and Dad each drink a cup of yoghurt. I feel understood.
Set 17: I am less alone. Calmness returns.
The effects of EMDR can be explained by the combination of the strong recollection of the traumatic event and the attention to the distracting stimulus, which triggers the natural defence system. Since a traumatic memory is both lively and intense, this process asks a lot of the brain. Yet, following the therapist's fingers with concentration and at a fast pace also requires a lot of focus. As there exists a clash between the contradicting activities the brain experiences, there isn't a lot of capacity left over to deal with the horror and liveliness of the memory. This offers the patient the possibility to give the memory a different meaning.
EMDR sessions are so exhausting because not only do patients deal with the mental re-experiencing of fear, sadness, and pain, they especially deal with the physical repercussions. Almost twenty-five years after all of my operations and treatments, I once again shot in a panic when my hair started to fall out. I was nauseated and almost had to vomit when the tube slid through my nose and into my stomach. I heard the cursing of the doctors when the drip didn't go in right for the umpteenth time. I smelled the stench of the toilets and was paralysed by tension when the mask was pressed on my face to anaesthetise me. My body twitched with feverish chills and my muscles ached terribly.
The psychotherapist explained that the bodily sensations I was undergoing during the therapy sessions were much too strong or heavy to process and come to terms with when I first experienced them as a child. He then instructed me to mentally go to a place where I felt safe. That was Ameland.
The image of me and my sister on the back of our parents' bikes, driving to Het Oerd and De Hôn, and returning over the beach at low tide. When Mum and Dad miscalculated or if we stuck around for too long, they were forced to plough their way through the loose sand. I saw the image of children, building a castle that would definitely be washed away by the rising sea at the end of the day. I cycled through the dunes and heard the shells crack under my tires. On the dike by the mudflat, surrounded by woolly sheep, I sucked in the sea air greedily. This was to become a fixed pattern for every visit to the psychotherapist. An exhausting EMDR session followed by Ed's friendly command,
'And now back to the sheep on the dike.'
At home, I couldn't manage to eat anything and I decided on some custard. After half an hour, blisters started to form in my mouth, just as they did after chemo. I was shivering on the sofa all afternoon. I was overjoyed when Kees returned and he guided me to bed. Cold shivers. Fever.
The therapist was a little shocked when I told him that my mucous membranes and subsequently my stomach and bowels were inflamed after the EMDR session.
'Did you recover after three days?' he asked. 'Those sorts of reactions can't take longer than that.' His worry didn't hide his excitement, though, because he was on the edge of his seat and he couldn't hide his grin. 'Fan-tas-tic! I can't believe that the brain is capable of causing such a somatic re-experience after so many years...'
Secretly I thought that he wasn't just concerned about me, but also a little elated because he saw the intensity of my physical reactions as a publication for a journal.
Between 2004 and 2008, I had over a hundred sessions with Ed. The first few years we had weekly sessions, then the intervals became greater, and even now I go see him sometimes.
In his office in Meppel there was a desk with the usual office supplies and a little happy Buddha. Filled bookcases. There was a Discman with headphones to be used for EMDR. After a while, a stack of batteries was added to it, because they appeared to be empty during a particularly intense session. There was a small TV which I never saw turned on, but nevertheless had a note stuck to it reading 'Don't turn off'. I saw cigarette ashes behind the standing lamp. When the therapist took some time off for a couple of weeks, the ashes were gone, as well as the sad-looking plant. With the amount of tears I had shed in this office, the floor would have to have been made of some absorbent material...
We dedicated many sessions to the experiences that have their roots in my years in the hospital. We talked about how I never went through puberty and how my lack of sexual experiences affected me. All of those times that a group of doctors – mostly men around that time – were standing around my bed in a semi-circle and I had to take my underwear off. One by one they conducted physical examinations and I couldn't always imagine a different girl lying there. My tainted, violated, and hurt body was not exactly a thing to be proud of or to be enjoyed. Additionally, the one who lived in that body didn't have any control over it. In my childhood years I just let everything happen to me. That's the way it goes, or so I thought. Later this developed into an aversion to every form of intimacy, to the extent that I started to dread meetings – even with family – where I had to engage in customary cheek-kissing. Everybody had to stay away from me. Even in a busy store I felt disgust when someone accidentally ran into me. Nobody could touch me.
Many of the conversations Ed and I had revolved around my relationship with my parents, especially with my mother. Although she was bursting with good intentions, I felt that she still had a lot of influence over my life. One example followed another. I had to grow up even in that respect, and that was new for me. After my womb was removed, my mum came to our house for a couple of days to help in the house. I had to watch from the sideline as everything went her way. She decided when I had to rest and looked at me disapprovingly when I got up after an hour. A week later I got an inflammation in my urinary tract and she had to let me know that she was right.
'That's why you should've rested more, Es.'
The phone call in which she diagnosed me as being 'burnt out' – which I didn't agree with – she ended with the statement,
'You're still overworked, Es.'
When the half-past-nine ferry docked at Ameland, Mum expected us to be with her for coffee at half past ten. My sister went into town at her leisure to check out a new sweater. When my mother called to let us know that 'the sweater won't run away,' Marion replied, 'Neither will the coffee!' I admired her, because I'd never dare to do that. I disregarded the things I wanted to do, didn't even ask Kees and Marloes what their plans were, and arrived at my mother's caravan at exactly half past ten.
When Marloes would stay over at my parents, I got told what she should wear.
Mum commented on the paint we chose for the walls in our living room. 'Where did you even get the idea in your head to put such a dark color on a wall that's already in the shade?'
She called me to tell me when I had to have my picture taken for an interview, and after the publication she kept whining about the bra-strap that was showing.
After careful deliberation with the teachers, Kees and I decided to let our daughter get used to her new classmates for two days before we definitely moved to Ede. Mum riled up the entire family to stop us from executing this vile plan.
In conversations with the psychotherapist about my relationship with my mum, more and more frequently I felt as if something gripped my throat and I couldn't speak. I could barely swallow away the anger when I told him about the lack of space, of my own ideas, my own thoughts, my own convictions.
Her patronising originated in my youth, but happened even then, because I continued to be treated as a child even though I was an adult woman.
'You're only an adult in years,' Ed responded. 'Once you have to spend less time surviving, you can start to discover what is necessary to exist in the here and now, and find pleasure and satisfaction in that.'
This was not the first time Ed made a comment that came like a slap in the face. I had to realise what I thought, what I felt, what I needed, what I wanted, and when I determined all that, I had to start behaving according to what I found. Plus, I also had to start to realise what my responsibilities were and which were someone else's. Good luck with that.
Theoretically, I understood that he was right, but at the same time I was scared. If I started to move towards change, the process was irreversible. What if my loved ones didn't like me anymore? How would my husband, my daughter, my parents, my sister respond?
Marloes asked if we could go shopping. I loved such an invitation from my adolescent daughter, and I happily freed up some time for it. We strolled through the city centre in Ede, and stood still in front of a shopping window. I was struck with shock. I saw a horribly weathered face next to such a happy young girl. Quickly, I glanced at the clothing in the window, as long as I didn't have to look at myself.
In Meppel, subjects such as being 'someone you want to be seen with' and 'looking good' were brought up. For the first time, I dared to go through what the disease had done to my self-image. My hair, which the chemo took from me – that was me.
Ed immediately responded with an EMDR session. The target is the image I see the moment I ran my hand through my hair and found that I was holding a strand of hair.
Set 1: Intense sadness at the discovery that my hair is starting to fall out.
Set 2: It can't be stopped.
Set 3: Grandma and Granddad are witnessing it. The look in my grandfather's eyes hurts.
Set 4: I feel empty, powerless. I lost all control.
Set 5: I know my parents will be with me soon and I am not at all looking forward to it.
Set 6: Another thing I have to lose because of the chemo.
Set 7: Fear of going bald. Being unsatisfied with the hair I have now.
Set 8: I come to terms with it.
Set 9: I feel a little calmer.
Set 10: Calmness and acceptance, but not without protest.
The therapist's notes kept repeating the phrase, 'Cries uncontrollably'. I think you could easily fill up a pool with the tears shed in four years of therapy.
When I returned the next week, I told him proudly, 'I bought a white skirt, a white top, and even a crème-coloured jumper.'
'And when are you going to wear it?' he asked, because he knew I often found clothes in my wardrobe that still had the price tag on them.
I couldn't tell him a date. 'It will happen, though,' I promised.
Crying for No Reason
All of a sudden, my psychotherapist asked, 'Which plant or flower do you identify with most?'
How on Earth should I respond to that? I started to outgrow the stage of willow and pine tree, but I had no idea what to answer.
He stared at the ceiling for a while and then determined, 'You are a hibiscus!'
Ed saw from my look that I had no idea what he was talking about, and continued, 'That's a flowering plant or bush. When the first flower ceases blossoming, there are new ones that follow immediately. The plant gives flowers all the time, just like that.'
I muttered something in objection, but his hand gesture silenced me.
'This is how the world sees you, Es. Trust me. Make sure you get to fully bloom and enjoy it, before the buds start to wither and die.'
Still bewildered, I relayed to Kees over the phone what Ed had said. Sometimes when I had a session, I would stay over at a friend's house in Assen so I didn't have to drive all the way back to Ede after such a tiring EMDR session.
The next day, my husband said, 'Go look outside. There you are.'
Kees had bought a hibiscus, and he clarified that I'd bloom every year. Moved and grateful, I kissed him. When I went to take a better look at myself, I read on the attached card that I had until the next summer to make that prediction come true.
It was still autumn, though, and at night a girl came to visit me in my dreams. She was sweet and vulnerable, and at silent moments she also came during the day, in the bathroom, under the shower, whenever I found myself alone. I didn't fear her, even though she showed me things from my past. It was almost as if she wanted to melt the then and the now into one entity. I longed to get to know her better.
In the supermarket, I was taking a jar of jam from the shelf when suddenly I felt my cheeks getting wet. Stupidly I looked to see if the liquid came from the jar, before I came to realise that I was crying. What the...? Buying jam isn't exactly an emotional experience. I was upset to the point of forgetting about the rest of my groceries and cycling home as fast as I could, to the environment I knew and trusted.
On our Crown Prince Willem Alexander's birthday, I decided to buy myself an ice cream cup from the ice cream parlour Bernardo's. I got the so-called Coupe Caramello. As I sat there spooning away the ice cream, I saw a mother and her child enjoying their ice cream as much as I did.
Teenagers also settled down. Carelessly, they threw down their bikes left and right because they wanted to order their favourite ice cream as soon as possible. A cat wouldn't be able to manoeuvre its way between the bikes. Not too much later, they cycled away chattering, and the sidewalk was empty again. My eye caught an elderly couple, licking their ice creams like toddlers. He spilled a little and she made him presentable with a handkerchief. What was wrong with me?! I was crying again!
Ed was euphoric when I told him what was going on. I couldn't quite see why. 'Because the blending of the little girl and the grown-up woman is a necessary condition for the development of your identity, which has stagnated for such a long time.'
'So you think bawling your eyes out for no reason is normal?' I ask hesitantly. If there was something I really wanted to be, it was to be normal.
'You're not going to be crazier than you already were, and your little bit of weird suits you very well. You will probably start crying more without anything causing it, but the frequency will reduce quickly. Let the tears flow; they should've been flowing years ago.'
I decided to use this sort of revival in my existence by looking around for a new job. The psychologist assured me that it would be good for me to meet new people and learn to find and protect my own boundaries. A local newspaper had an ad out for a pharmacy assistant, eight to twelve hours a week, in the nursing homes of De Gelderse Vallei. I applied but got turned down, because the hospital was in a rush and they'd rather have someone who already knew how their computer system worked. A couple of weeks later, I got a phone call anyway. My predecessor had wanted more hours and she'd left. Did I feel like coming in for an interview?
The job came to me like a lottery ticket. One day I hoped to be so proud of myself that I felt they should feel like the lucky ones, but I wasn't that far yet. For the final interview I drove to the pharmacy in Wolfheze.
'Are you available on every weekday, Mrs Radstaak?'
This was the time to practice my assertiveness. I took a deep breath and then I dove in. 'I can make it on every day except Wednesday morning, because I have other obligations then.'
I rehearsed this sentence, because this way it became clear immediately that there wouldn't be an explanation about said obligations. My interviewers accepted my condition, and I'd found another job that fit me like a rubber glove.
From experience I knew that the ups in my treatment – which was approaching its third year – would always be followed by downs. Around me there were some people who died: our male neighbour opposite us and a colleague of mine. Her husband couldn't handle the loss and soon after ended his own life. I think that events like those shove mortality in anyone's face, but for me, all of a sudden, all pleasure in my own existence just left. The days became heavy and dark once more. I was afraid that I would lose my new job and with my psychotherapist I also found out that I had a phobia for nausea and vomiting. Ed thought my recovery wasn't fast enough, and gave me an ultimatum. If in two months I wouldn't be able to find a moment every day in which I was completely happy, I had to start my medication. Another slap in the face.
He was shocked by my intense response and asked me what was going on. The build-up of saliva in my mouth had made me unable to speak. The snot started to run out of my nose, I started to sweat tremendously, and before I knew it, I burst into tears in his office. I didn't know what to do... I wanted to be out of there, and Ed let me leave. Luckily I realised that I really shouldn't get in the car in this state, so I roamed the streets of Meppel for some time.
Pills? Never again. The getting onto and off of antidepressants, which I'd had to take while I was overworked and after my granddad had died, brought me to the gates of hell. I would never go there again voluntarily.
During the process of coming to terms with the traumas of my childhood, I put appeals out on different Dutch forums: the forum of the Association of Parents, Children, and Cancer (Vereniging Ouders, Kinderen en Kanker, VOKK) and the Dutch cancer research fund. I already thought of a title for my post,
'Ovarian cancer as a child, now an adult and...?' I hoped to get some responses of people who were also confronted with the fact that they were different than others, but at a later age. Every time I notice I'm not the only one battling with an unprocessed past, I experience that recognition as a blessing. I received a couple of comments on my story, but one was especially intriguing: a private message from someone by the name of Zwannie Sterenborg. 'Are you the Esther who, in the UMCG, showed me how good a wig can look?'
Suddenly the incident sprung to mind. A nurse asked me if I could show a girl that a wig can make you look pretty cute. Nothing to be scared of or to remind you of clowns, like they did to the girl. Zwannie turned out to live in East-Groningen, not too far from the German border, and we both wanted to meet each other very badly. The central point between our towns was – concidentally – Meppel! On Wednesday, 16 May, 2007, I was the first of the two of us to reach the Meppel train station. I saw Zwannie's train roll in and saw the doors open. I recognised her and felt a sudden rush. With arms wide open, we ran towards each other. We hugged and kissed.
At that moment, which I often replay in my mind – slowed down and with music playing – it was as if I got permission to live.
Our grip loosened a little, so we could look at each other, and then we nearly choked the life out of one another again. We are survivors. We were deadly ill when we met each other so briefly years ago. Now we were standing there as healthy young women. We recounted memories until our behinds hurt from sitting down on the cold, hard benches for so long. Then we strolled through the centre of Meppel, without seeing a store nor building, until our feet couldn't carry us any further.
Zwannie had loved my wig with long, black hair, but she'd gotten a copy of her own short, blonde haircut. 'If only I'd known Sophie van der Stap's book then, which was about a girl who had no less than nine wigs, one for every mood.'
That Wednesday, an intense friendship began.
We barely sat down before Ed blurted out, 'The pills are off the table. I can clearly tell that there's been a massive turn in your life. Tell me what happened.'
I had such difficulty communicating what that meeting had moved in me. I'd met with a girl who'd gotten cancer at pretty much the same age as I did, who I'd talked to for one day after so long a time. How can anyone capture that feeling in words?
'It's not important,' Ed dryly interrupted my ramblings. 'It's of importance to me that the change is visible. I see a lust for life in you now.' He paused. 'I do hope you don't hold it against me that I threatened to put you on medication. You needed a shock. It turned out right, and that is what matters.'
Bit by bit, my desire not to live anymore started to make way for a bit of contentment, a tiny spark of optimism, and a dash of confidence. This also meant that I had to learn that my body had a right to be as well. Ed proposed that haptonomy may be able to help me. I just had to become aware of the fact that I have a body, but also that it was allowed to take up space, that I was allowed to cherish it, and that I could tell people when they got a little too close. I didn't think the sessions with the haptotherapist were successful, although I probably did learn a thing or two. I told her that I always shivered when someone pressed their bare foot against mine, or worse, on top of it. Her response to my opening up was to put her bare feet on mine. I cringed, but didn't say anything.
Wrong! I should have gotten furious!
When I called her to explain why I wanted to stop the therapy, I had to come and do that in the hour she reserved for me. In case I didn't have a lot to say, she said she'd bring a ball of wool and some knitting needles for her to knit the rest of the hour away...
Intentions can be more than good, but people aren't always the same.
Even when a patient is physically cured of cancer, the oncologist often decides to give chemotherapy or radiation. Just to be on the safe side, because you never know. For that reason I still drive to my therapist in Meppel every now and again, to ensure that my psychological recovery keeps on.
On 28 May, 2008, I officially ended my psychotherapy, which was a very memorable afternoon. I wore my white skirt, white top, and the fashionable, crème-coloured jumper. My skin was tan and I had stood in front of the mirror for quite some time to carefully put on makeup.
When I took place on my trusted spot on the sofa in the office, Ed stared at me with open mouth. He was completely flabbergasted by my metamorphosis and I enjoyed seeing him so thrown off guard for a change.
We talked about the time we first met, when I was still desperate to survive and would see it as a relief if I couldn't. We got to my years with Ed as a therapist, in which I discovered that I was being lived instead of living myself. Then we arrived in the present, where I finally regained the reins over my life and took them into my own hands.
'You have to be open for it and to want it yourself.' That's how I summarised four years of pain and effort.
'And you've got to have the guts,' Ed added. I agreed with him on that. We looked each other straight in the eye. Four years ago, I would have pretended not to hear his appreciation.
Ed informed the psychologist of the children's clinic in writing about the process and eventual result of the therapy. I got a copy, from which I cite:
'After the intake, it became clear that there was a large diversity of ailments and an intense disruption of functioning in practically every area of life.
'I diagnosed chronic PTSD with traumatic fatigue syndrome, related to the medical situation after the diagnosis teratoma re-ovarium 1980, and proposed to the patient to treat her with EMDR, in combination with hypnotherapeutic techniques.
'During the course of treatment, there was also much use of psycho-education and cognitive behavioural therapy.
'The treatment started in August 2004 and successfully ended on 28 May, 2008. The appointments were practically every week, with the exception of holiday periods. My advice of suggesting that the GP prescribe psychiatric drugs was declined by her resolutely.
'During EMDR sessions, many aspects related to the trauma caused by medical treatment were taken as targets. These were often on the basis of typical complaints, sensations, or the way she experienced situations from the 'now'. They appeared to have their origin in the experiences she had had during her period of illness.
'It also appeared that this period and the way she was cared for and guided by her parents played a hindering role in the development of her identity and her self-image.
'At the moment of termination of the treatment, Esther was doing very well. She now has had a job as pharmacy assistant in the Gelderse Vallei for quite some time, and is full of energy, humour, and initiative. She can be happy with and proud of herself now. Basically, now she can finally "live".'
We're alive and living, and to celebrate that fact we went to Mallorca in the summer of 2010. As per usual, I put some random clothing in my suitcase. Luckily I changed my mind when the weather chart showed that our holiday destination hit thirty degrees Celsius, so it'd be useful to bring a bikini. I had one – like many other clothing items, still with the tag on – but I was afraid to wear it. I finally dared to reluctantly show myself in daily life, and now I had to go almost naked.
One day walking back to our apartment from the pool, I saw myself in the mirror in the lift; I saw it. Before I could think, I also said it: 'Girl, you have some gorgeous legs. And so tanned, too!'
I startled myself and my eyes met Kees's. 'Finally,' he sighed and smiled, and nothing else needed to be said.
Marloes was training to become a veterinarian, and together with a classmate of hers they arranged an internship. They would go to Spain for nine weeks, to a dog shelter in the middle of nowhere, where a Dutch owner had to feed over six hundred dogs by himself. Earlier in life, I'd have resisted such an initiative. Difficult! Dangerous! But now I could rejoice in the excitement of my bold daughter, who was developing an independence I was incredibly proud of.
In the middle of her internship, Kees and I came to visit her for a week. Though I'd been rejecting getting a pet for years, because I didn't want to be burdened caring for something else, I completely melted when I saw how Marloes had made friends with a dog of a nondescript breed. We took him out for a walk and called him Fred, after a Dutch TV programme's imaginary dog. My attempts not to fall in love with the animal failed miserably. When I got back and Marloes let us know that there were people who wanted to view and potentially buy Fred, I found the idea unbearable. Even before our daughter had returned from her internship, Kees and I – with a banner – were at the airport awaiting the coming of our new family member, Fred. I appeared to be able to put demands on my family members to care for Fred and determine my boundaries for what I did and did not find acceptable.
Above all, though, I was able to enjoy the daily walks I took with our dog.
I have a lust for life.
Epilogue
When I saw so many people in the hospital find the best in themselves to be of service to others – day in, day out – it made an immense impression on me as a child. That experience was at the foundation of my biggest wish to work in health care after I was cured. To give back.
That same mentality made me not hesitate for a moment when I was asked to be interviewed for a local Dutch newspaper, Dagblad van het Noorden, exactly on the day of my thirty-year anniversary of being admitted to the hospital.
Children's oncologist Liedeke Postma and general practitioner Ria Blaauwbroek of the UMCG developed a system to improve the care for cancer survivors. The women's report can be found on the Internet. Through this system, GPs or regional hospitals can execute certain tests and inspections, to make a trip to Groningen redundant. Moreover, survivors now get advice about the potential effects of their treatment which surface later in life, and how they can deal with them. My interview would be an important part of the background story and would take up two pages in the newspaper. Both my parents and I received dozens of texts and phone calls of people who recalled the horror of thirty years ago, and were overjoyed by the fact that I was doing so well.
A year later, I got another call from the same oncologist. This time it was about the Dutch news broadcasting programme, NOS Journaal, which focused on a symposium in Amsterdam about the effects of childhood cancer later in life. There would be research on a large scale on the medical conditions of people who managed to survive cancer in their childhood years. Due to the improved methods of treatment, the survival rates have risen enormously. Almost 80 per cent of children survive the illness, whereas forty years ago this was at 25 per cent. On the other hand, different studies have pointed out that three quarters of all survivors are more likely to endure health problems later in life. I had to take medicine for years to combat my worsening kidney function, have regular urinary tract infections, and ten years ago I hit menopause prematurely.
The research is called the SKION LATER Study, which is being conducted by seven academic hospitals in the Netherlands. Between 2012 and 2016, more than six thousand childhood cancer survivors will be examined. The cure rate of children with cancer is supposed to keep rising, while the effects of the treatment in the long term will be diminished. All information can be found on the Foundation's website: www.skionlaterstudie.nl/english.
On the website, when you look for people who explain why they take part in the study, you can find Zwannie and myself among the list of people. My (and Fred's) TV-debut also harboured dozens of responses. One of the nicest is that of the NOS-reporter, who called me the day after the recordings to tell me how impressed she was with our conversation.
Once your name ends up in a card-index, you're public property. I can attest to that. One day, I had an envelope on my doormat which requested that I co-operate in a scientific study, where women who were cured of cancer would be examined on the basis of their reproductive function, ovarian reserve, and premature menopause. Apart from filling out a questionnaire and having blood sampled, participants also had to have an internal gynaecological ultrasound done. As the results of the VEVO-study had to be compared to women who did not undergo treatment, there had to be a control group.
Would the ex-patient perhaps have a sister eligible for the study?
Marion was hesitant, because an internal ultrasound wasn't exactly pleasant, but she participated anyway. Together we went to the hospital and had a look at each other's insides on the echo screen.
'Yours looks nice and clean,' my sister determined.
'If you're jealous, you just have to remove your appendix, ovary, and womb,' I advised.
You're not in the most flattering position when you're in the gynaecologist's chair. Perhaps that's why we started to laugh uncontrollably.
Seven years after I got cancer, the Association of Parents, Children, and Cancer (Vereniging Ouders, Kinderen en Kanker, VOKK) was founded to support and guide families of children with cancer. Books such as Chemo-Caspar and Radio-Robbie didn't exist in my time. There weren't any camps or trips for children – whether alone or with families. No get-togethers for youth, no trainings for teachers, and no forum on which I'd meet Zwannie much later. I think a lot of trouble would have been avoided if both me and my family had been able to rely on the things the VOKK does nowadays.
I think that one of the most engaging initiatives of the VOKK is the CancerChain.
The CancerChain is a reward system and guide at the same time, which consists out of brightly coloured beads. For every special occasion, one bead can be added to the chain. These are for instance for chemo or radiation, injections, scans, a lumbar puncture, hair loss, an amazing day or a horrible one, a surgery, or a day at the Intensive Care Unit. The CancerBeads make visible what is incomprehensible to a child. The CancerChain make them aware of what's happening to them and gives them a physical grip on the situation. Parents can use it as a guideline to explain to their child and to prepare them for what'll be coming. With the CancerChain in their hands, children tell their grandparents or classmates what they're going through. That's how the CancerChain helps them through their treatment.
After the illness, the CancerChain is a colourful memory of an intense time. For families who lost a child to cancer, it is a precious keepsake to cherish.
The events and experiences I've written down in this book, I see as my own symbolic CancerChain. And if the history of my illness appears in print and people who want to have it buy the book, everything goes to the VOKK in the name of the CancerChain.
One of the things I did to write this book was to go to Ameland, to my sheep on the dike. I was all but looking forward to asking my parents about their experiences, because I knew that my visit would bring up a lot of sadness. I'm not only grateful because they put me on this earth, but also for their willingness to let me share their memories to whatever extent they were capable of doing so. Our relationship is balanced now, because I no longer feel like a little girl, but like a grown woman. That's how I behave towards them and that's why they can see me as a grown-up daughter instead of the small, sick girl of days past. I'd very much like to thank my husband Kees and our daughter Marloes for the space they've given me over the past years to process my oncological past. And my sister Marion, even though she jokingly demanded to have this book dedicated to her...
Now for a couple of beads more to end this, to mark some other people who helped me put my life story down.
Firstly, Dr Liedeke Postma, who sacrificed her free Saturday to join me in finding my way through my medical file.
Then there's Professor Jan Aalders, who called me from abroad to make me more enthusiastic and wish me luck and who wrote the following:
'Your left ovary has done its work wonderfully, but your strong personality and positive attitude were as important when conquering your illness and going through this incredibly heavy process! Of course, the unconditional support of your parents and loved ones around you were of indescribable value as well.
All in all, my dear Esther, this has been a remarkable experience, especially for you, of course. It is one which I look on with much respect, thankful that you were able to conquer this disease and thankful that I was able to contribute to the multidisciplinary team that treated you. And especially thankful that I met you as a person in my professional career – it has been truly enriching!
I wish you and your loved ones all the best!
Third, there's Ed van der Leeuw, who was my outlet for years and who encouraged me to write this book. Moreover, I'd like to thank Zwannie Sterenborg, with whom I experienced the day where all the little pieces of our lives fell together. Valerie Strategier, who inspired me with her book Een Kikker in mijn Buik ('A Frog in my Belly', translated). I also want to thank Jeroen Terlingen, who helped me shape my story. And last but not least, the crew of 'Grand Café Restaurant Paal 13', who offered me a listening ear and took care of keeping me well-fed during my stays at Ameland, where I found the peace and quiet to write this book.
If there's only one person who recognises themselves in my story, who finds solace or inspiration, then I have reached my goal.
I devote the following poem by E.E. Cummings to my loved ones.
I carry your heart with me (I carry it in my heart)
I am never without it (anywhere I go you go, my dear;
and whatever is done
by only me is your doing, my darling)
I fear
no fate (for you are my fate, my sweet)
I want no world (for beautiful you are my world, my true)
and it's you are whatever a moon has always meant
and whatever a sun will always sing is you
Here is the deepest secret nobody knows
(here is the root of the root and the bud of the bud
and the sky of the sky of a tree called life; which grows
higher than soul can hope or mind can hide)
and this is the wonder that's keeping the stars apart
I carry your heart (I carry it in my heart)
From the same publisher:
I'm Not Strange, I Have Autism
Living with an Autism Spectrum Disorder
by Ellen van Gelder
ISBN 9789461850645
Village Publishers
Ellen van Gelder was almost 40 when she heard she has Asperger's Syndrome, a disorder within the Autism Spectrum. Although the diagnosis brought along the necessary emotions, it also brought relief - the puzzle pieces of her life finally fell into place. Ellen sought and found other fellow sufferers, and decided to use their input to create a clear ABC about life with Autism Spectrum Disorder/Condition.
4.5 Stars out of 5 on Amazon.co.uk
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D-Fenders Can't Keep Up With Mad Ants
By Joey Ramirez | March 12, 2017
A team that hadn't lost back-to-back games all season has suddenly dropped three in a row. The D-Fenders struggled on the defensive end against the playoff-caliber Fort Wayne Mad Ants, allowing 131 points in their 22-point loss.
D-Fender of the Game
Josh Magette: The All-Star point guard stayed steady for his team, tallying 23 points and seven assists. He tried to come through for Los Angeles in the fourth quarter, scoring 11 of his points, but the team wasn't able to close the gap on Fort Wayne.
Nwaba News
Before the game, the Lakers signed David Nwaba to a second 10-day contract and then assigned him to the D-Fenders for the night. The Cal Poly product couldn't stay out of foul trouble, finishing with five in 21 minutes while scoring 13 points. However, he was more upset with his team's play on the other end, where he has been the leader all season.
Prospect Watch
Vander Blue (2):
Despite the final outcome, Blue contributed his usual scoring surge by putting up 25 points on 10-of-18 shooting. He could have added to that total, but instead sat the majority of the fourth quarter with the final outcome already decided.
Key Stat
30: Fort Wayne's biggest lead of the night. The D-Fenders were never able to corral the Mad Ants' red-hot offense, which shot 52.2 percent from the field while hitting 14 3-pointers and 23 free throws. Fort Wayne was led by a 34-point night from Jordan Lloyd, while NBA vet Tyler Hansbrough added 12 points and 17 rebounds.
The D-Fenders will hit the road for all three of their games next week, beginning with back-to-back contests in Iowa on Wednesday and Friday. The Energy (11-29) are at the bottom of the D-League standings, but nonetheless had a pair of recent call-ups in Troy Williams (Memphis) and Wayne Selden Jr. (New Orleans)
D-Fenders Eliminated In Game 3 Against Vipers
D-Fenders Outshoot Vipers, Force Winner-Take-All Game 3
D-Fenders Fall to Reno Despite Zubac's Double-Double
D-Fenders Fall In Meeting Between D-League's Top Two Teams
D-Fenders Capture Win In First Meeting With Windy City
D-Fenders' Comeback Halted By OKC | {
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SS_ClassLoader::instance()->pushManifest($manifest, false);
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This petition is to be signed in hopes of generating a stir in the community about the legalization of human euthanisia. In the recent case with Terri, starving people to death is inhumane but in cases like hers, when the family is agreed and the patient has been deemed a "vegetable", Euthanisia should be available to end the suffering of all parties involved. Please sign this petition in hopes of ending human suffering.
Please consider making a small donation today. This will allow us to keep Euthanatization alive and available to a wide audience. | {
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{"url":"https:\/\/math.stackexchange.com\/questions\/2192940\/how-do-you-do-this-vector-projection\/2192981","text":"How do you do this vector projection?\n\nFor $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$.\n\nI don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this?\n\n\u2022 You do it the same way. There are just more components to compute. \u2013\u00a0amd Mar 19 '17 at 1:01","date":"2019-10-23 22:06:33","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9334585666656494, \"perplexity\": 233.34734087568296}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570987836295.98\/warc\/CC-MAIN-20191023201520-20191023225020-00495.warc.gz\"}"} | null | null |
Janka ist eine deutsche Indie-Pop-Band aus Hamburg und Berlin, die 2003 gegründet wurde.
Geschichte
Bereits 1993, also noch zu Schulzeiten, begannen Kai Gabriel, Thomas Liman und Tom Unguraitys, in Celle gemeinsam Musik zu machen und gründeten 1998 die Band Blobkanal. Der Titel Peter Shaw erschien bald darauf auf einem Sampler der Musikzeitschrift Visions und die Gruppe spielte im Vorprogramm von Kettcar und Gary. Das angekündigte Debütalbum erschien jedoch nie und im Juni 2003 löste die Band sich auf.
Gabriel, Liman und Unguraitys gründeten daraufhin schließlich gemeinsam mit Niclas Breslein von junges glueck die Band Janka.
Bereits 2004 erschien das Lied Wir teilen Wunden auf verschiedenen Samplern von u. a. Tapete Records. Im selben Jahr erschien mit der Unterstützung von Kettcar-Bassist Reimer Bustorff die von Plemo produzierte EP Unter Palmen mit sechs Titeln. 2005 wurde Breslein durch Ingmar Rehberg am Bass ersetzt, der wie die restlichen Bandmitglieder aus Celle stammt. In der Folgezeit tourte Janka mit befreundeten Bands wie Schrottgrenze, Herrenmagazin und junges glueck; 2006 veröffentlichten die vier Bands auch eine gemeinsame Split-Single. Am 27. Oktober 2006 erschien schließlich das Debüt-Album In die Arme von. Es wurde von Chris von Rautenkranz gemastert und schließlich beim Label Decoder Records veröffentlicht. Das Album erhielt größtenteils wohlwollende Kritiken, so meint Katja Scherle von laut.de: "Einfach, natürlich, lebendig und intelligent: Hamburger Schule kumuliert mit gesetztem Berlin-Pop" und Daniel Erk schreibt auf jetzt.de, dem Jugendmagazin der SZ: "Janka sind einer dieser Bands, die zu Unrecht unbemerkt unter dem aufgeregt hupenden Tomte-Juli-Helden-Indiepop-Radar hindurchschlüpfen". Zu der Single Punkt wurde auch ein professionelles Video gedreht. Mittlerweile ist die Band kaum noch aktiv, einige der Bandmitglieder spielen aber noch gemeinsam unter dem Namen Lucky Steve & The Magic Hoffmanns.
Diskografie
Alben und EPs
2004: Unter Palmen
2006: In die Arme von
Split 7″ Schrottgrenze / junges glueck / Herrenmagazin / Janka (7″, 2007; Eigenvertrieb)
Weblinks
Künstlerseite auf motor.de
Einzelnachweise
Deutsche Band
Indie-Band | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,532 |
/***************
* NAME: PrimDlg.h
* PROJECT: Scenario
* VERSION: 0.1 13.05.1997
* DESCRIPTION: attribute dialogs for basic primitives
* AUTHORS: Mike Hesser
* HISTORY: DATE NAME COMMENT
* 13.05.1997 mh initial version
***************/
#ifndef PRIMDLG_H
#define PRIMDLG_H
#include "typedefs.h"
#include "resource.h"
#include "object.h"
class CPrimSetPage : public CPropertyPage
{
public:
CPrimSetPage();
enum { IDD = IDD_PRIMSET };
BOOL m_bClosedTop, m_bClosedBottom;
int m_nSORAccurate;
OBJECT *m_pObject;
protected:
virtual BOOL OnInitDialog();
virtual void DoDataExchange(CDataExchange*);
DECLARE_MESSAGE_MAP()
};
#endif
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,662 |
#include "tensorflow/lite/kernels/internal/reference/elu.h"
#include <algorithm>
#include <limits>
#include "tensorflow/lite/c/common.h"
#include "tensorflow/lite/kernels/internal/cppmath.h"
#include "tensorflow/lite/kernels/internal/quantization_util.h"
#include "tensorflow/lite/kernels/internal/reference/process_broadcast_shapes.h"
#include "tensorflow/lite/kernels/internal/types.h"
#include "tensorflow/lite/kernels/kernel_util.h"
#include "tensorflow/lite/micro/kernels/kernel_util.h"
namespace tflite {
namespace {
// Input/output tensor index.
constexpr int kInputTensor = 0;
constexpr int kOutputTensor = 0;
// OLD-TODO(b/142762739): We should figure out a multi-threading plan for most
// of the activation ops below.
struct OpData {
int8_t table[256];
};
using TransformFunc = float (*)(float);
template <typename T>
void PopulateLookupTable(const TfLiteTensor* input, const TfLiteTensor* output,
const TransformFunc transform, OpData* data) {
if (sizeof(T) != 1) TF_LITE_FATAL("Lookup table valid only for 8bit");
const float inverse_scale = 1 / output->params.scale;
int32_t maxval = std::numeric_limits<T>::max();
int32_t minval = std::numeric_limits<T>::min();
for (int32_t val = minval; val <= maxval; ++val) {
const float dequantized =
input->params.scale * (val - input->params.zero_point);
const float transformed = transform(dequantized);
const float rescaled = TfLiteRound(transformed * inverse_scale);
const int32_t quantized =
static_cast<int32_t>(rescaled + output->params.zero_point);
data->table[static_cast<uint8_t>(static_cast<T>(val))] =
static_cast<T>(std::max(std::min(maxval, quantized), minval));
}
}
// OLD-TODO(b/143696793): move this to optimized_ops.
void EvalUsingLookupTable(const OpData* data, const TfLiteEvalTensor* input,
TfLiteEvalTensor* output) {
const int size = MatchingFlatSize(tflite::micro::GetTensorShape(input),
tflite::micro::GetTensorShape(output));
int8_t* output_data = tflite::micro::GetTensorData<int8_t>(output);
const int8_t* input_data = tflite::micro::GetTensorData<int8_t>(input);
for (int i = 0; i < size; ++i) {
output_data[i] = data->table[static_cast<uint8_t>(input_data[i])];
}
}
TfLiteStatus CalculateOpData(TfLiteContext* context, TfLiteNode* node) {
TF_LITE_ENSURE_EQ(context, NumInputs(node), 1);
TF_LITE_ENSURE_EQ(context, NumOutputs(node), 1);
const TfLiteTensor* input;
TF_LITE_ENSURE_OK(context, GetInputSafe(context, node, kInputTensor, &input));
TfLiteTensor* output;
TF_LITE_ENSURE_OK(context,
GetOutputSafe(context, node, kOutputTensor, &output));
TF_LITE_ENSURE_TYPES_EQ(context, input->type, output->type);
// Use LUT to handle quantized elu path.
if (input->type == kTfLiteInt8) {
OpData* data = static_cast<OpData*>(node->user_data);
TransformFunc transform = [](float value) {
return value < 0.0f ? std::exp(value) - 1.0f : value;
};
PopulateLookupTable<int8_t>(input, output, transform, data);
}
return kTfLiteOk;
}
void* EluInit(TfLiteContext* context, const char* buffer, size_t length) {
// This is a builtin op, so we don't use the contents in 'buffer', if any.
// Instead, we allocate a new object to carry information from Prepare() to
// Eval().
TFLITE_DCHECK(context->AllocatePersistentBuffer != nullptr);
return context->AllocatePersistentBuffer(context, sizeof(OpData));
}
TfLiteStatus EluPrepare(TfLiteContext* context, TfLiteNode* node) {
return CalculateOpData(context, node);
}
TfLiteStatus EluEval(TfLiteContext* context, TfLiteNode* node) {
const TfLiteEvalTensor* input =
tflite::micro::GetEvalInput(context, node, kInputTensor);
TfLiteEvalTensor* output =
tflite::micro::GetEvalOutput(context, node, kOutputTensor);
switch (input->type) {
case kTfLiteFloat32: {
reference_ops::Elu(tflite::micro::GetTensorShape(input),
tflite::micro::GetTensorData<float>(input),
tflite::micro::GetTensorShape(output),
tflite::micro::GetTensorData<float>(output));
return kTfLiteOk;
}
case kTfLiteInt8: {
const OpData* data = static_cast<OpData*>(node->user_data);
EvalUsingLookupTable(data, input, output);
return kTfLiteOk;
}
default:
TF_LITE_KERNEL_LOG(
context, "ELU only supports float32 and int8 currently, got %s.",
TfLiteTypeGetName(input->type));
return kTfLiteError;
}
}
} // namespace
TfLiteRegistration Register_ELU() {
return {/*init=*/EluInit,
/*free=*/nullptr,
/*prepare=*/EluPrepare,
/*invoke=*/EluEval,
/*profiling_string=*/nullptr,
/*builtin_code=*/0,
/*custom_name=*/nullptr,
/*version=*/0};
}
} // namespace tflite
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,872 |
\section{Introduction}
The scientific method is based upon the meticulous comparison of theoretical hypotheses with observations.
A hypothesis can be promoted to a foundational theory once it has rigorously
satisfied a multitude of observational tests.
Such is the case with the concordance cosmological model $\Lambda$CDM, named after the two dominant components that contribute to the current energy density of the Universe: the cosmological constant ($\Lambda$) and cold dark matter (CDM).
Despite their dominant contribution to the stress-energy density of the Universe, the fundamental nature of both dark matter and dark energy remains a mystery.
Determining the physical nature of these two components is a central challenge in modern physics.
Combined with the task of furthering our understanding of dark energy and dark matter is the requirement to test Einstein's theory of General Relativity (GR) in the hitherto un-explored cosmological regime.
Potential modifications to gravitational dynamics at cosmological length scales may also play a part in providing an explanation for dark energy and dark matter.
A principal pursuit of contemporary cosmology is therefore to stringently compare both GR and $\Lambda$CDM against a considerable collection of alternative models \cite{Copeland:2006wr, Clifton:2011jh, Joyce:2014kja, Bull:2015stt, Koyama:2015vza, Joyce:2016vqv, Ishak:2018his}.
The Large-Scale Structure (LSS) of the Universe provides an ideal testing ground for competing hypotheses.
Assuming that CDM can be treated as a perfect fluid, it undergoes gravitational collapse into localised over-densities, generating gravitational wells which the galaxy distribution subsequently traces.
By correlating galaxy positions over a large volume, a statistical description of how the underlying dark matter clusters can be obtained.
This is largely characterised by the two-point correlation function or the power spectrum in Fourier space.
Although the distribution of dark matter is not directly observable, modern cosmological surveys use observables such as galaxy clustering \cite{Alam:2016hwk} or weak lensing \cite{Asgari:2020wuj} to probe the underlying dark matter distribution.
The next generation of galaxy surveys such as Euclid \cite{Laureijs:2011gra} and LSST \cite{Ivezic:2008fe} have the capability to measure the cosmological galaxy distribution with extremely high precision, especially at length scales where the cosmological background becomes subdominant to baryonic and nonlinear gravitational physics.
Analytic methods are impractical in this regime as the evolution equations do not possess closed-form solutions.
Cosmological N-body simulations can provide highly accurate numerical predictions, yet their computational cost renders them unsuitable for constraining model parameters in Markov Chain Monte Carlo (MCMC) analyses.
Motivated by this issue, Refs.~\cite{Knabenhans:2018cng, Takahashi:2012em, Mead:2015yca} constructed emulators and nonlinear models for the matter power spectrum, with extensions to include deviations from $\Lambda$CDM developed in Refs.~\cite{Winther:2019mus, Mead:2016zqy, Zhao:2013dza}, ~\cite{Mead:2016zqy}.
These are fast and accurate methods which compute predictions for the shape of the matter power spectrum but are limited to an underlying hypothesis.
Recently, Ref.~\cite{Cataneo:2018cic} provided a method to predict the shape of the nonlinear matter power spectrum for a wide range of models which was subsequently implemented into a code called ${\tt ReACT}$ in Ref.~\cite{Bose:2020wch}.
Using this framework it is possible to generate a large dataset of mock matter power spectra for a broader class of extensions to $\Lambda$CDM with varying values of the model parameters.
Such tools enable one to extract information from a large range of length scales, substantially improving the constraining power.
MCMC analyses are frequently employed to determine whether physics beyond $\Lambda$CDM is present in cosmological data.
To consistently constrain beyond $\Lambda$CDM physics in such analyses, one must choose a finite set of parameters quantifying the new physics.
It turns out that the number of parameters needed to do this including nonlinear scales while remaining agnostic to the underlying fundamental physics is immense (see Ref.~\cite{Kennedy:2019nie} for example), making such an analysis currently unfeasible.
Consequently, current analyses either restrict themselves to the linear regime of structure formation or perform a model-by-model analysis.
It is worth noting that even if computational expense was not an issue, the simple inclusion of such a large additional parameter space would strongly penalise the extended-modelling on the basis of the Bayesian evidence.
It is therefore of interest to examine alternative approaches which do not rely on picking an effective set of parameters and are less computationally expensive.
With the ability to produce a large dataset consisting of power spectra for a variety of models, it is natural to consider the capability of Deep Neural Networks (DNNs) to classify power spectra according to their underlying cosmological model.
However, the prediction given by a trained DNN can be subject to several sources of uncertainty.
Adding a slight perturbation to the input, passing the input to a network with a different architecture or training on a separate subset of the training set could all drastically alter the result \cite{2013arXiv1312.6199S, 2014arXiv1412.1897N, gal2016dropout}.
Taking these issues into account is therefore crucial to obtaining statistically robust predictions with neural networks.
Quantifying the potential variability of the prediction, and in turn the confidence, is extremely difficult with DNNs. %
Bayesian Neural Networks (BNNs) try to model the uncertainty by replacing each weight in the network by a distribution initialised to a prior \cite{MacKay, Neal, Blundell2015, gal2016dropout, Charnock2020, jospin2020handson}.
Rather than obtaining the same point-like output for an example with every pass through the network, the BNN's prediction varies as each pass draws a different sample from the weight distribution.
By repeatedly passing an example to the BNN, one obtains a distribution of predictions conditioned on the training data, the network architecture, the noise in the data along with other potentially unknown sources of uncertainty.
A quantitative estimate of the classification uncertainty can be obtained in minutes once the BNN has been trained.
As an additional advantage, BNNs naturally provide a regularisation procedure for preventing overfitting \cite{blundell2015weight}.
BNNs have recently been applied in many fields such as gravitational waves \cite{Shen:2019vep, Lin:2020aps, Killestein:2021oqb}, the Cosmic Microwave Background \cite{2020PhRvD.102j3509H, hortua2020accelerating}, autonomous driving \cite{2019arXiv190909884M}, cellular image classification \cite{Deodato824862} and the detection and classification of supernovae \cite{Ramanah:2021bpb, 2020MNRAS.491.4277M}.
In this paper we explore the potential of BNNs to classify non-$\Lambda$CDM models from the matter power spectrum.
In particular, BNNs can be trained on as many deviations from $\Lambda$CDM as can be implemented in numerical codes such as {\tt ReACT}.
Even if none of these theories turn out to be the correct model they are all representative of possible sources of new physics.
We will therefore investigate whether BNNs can identify general deviations from concordance cosmology based on the observational features of known models.
The goal is, at the very least, to develop a promising tool to inform standard and more rigorous MCMC analyses by providing a refinement of the theoretical parameter space that needs to be explored. On the other hand, the possibility of constructing a well-defined probability distribution which accounts for all sources of uncertainty in the prediction from a BNN is an open research question~\cite{Mackay1995, MacKay1998, Graves2011, lakshminarayanan2017simple, depeweg2018decomposition, Ritter2018, shridhar2019uncertainty, wu2019quantifying, ovadia2019trust, hobbhahn2020fast}. Should this become possible, this method could be promoted to a statistical tool competitive to MCMC.
Regardless of this possibility, DNNs can simply be used as a tool to compress the information from the power spectrum in a small set of numbers, that can be in turn combined with other ML-based methods in rigorous statistical frameworks (such as Approximate Bayesian Computation, ABC \cite{fan2018abc}, or Likelihood-Free Inference \cite{Alsing:2019xrx}, LFI) to perform model selection in a fully ML-based fashion.
In any case, it it worthwhile to assess their potential, and this work is a first step in this direction.
\begin{figure}
\centering
\includegraphics[width=.49\textwidth]{plots/BaCoN_workflow.pdf}
\caption{ Representation of the workflow presented in the paper to study the presence of deviations from $\Lambda$CDM in the matter power spectrum. The key elements of the method, namely the use of a Bayesian Neural Network and a novel way to quantify the confidence, are colored in red.
}
\label{fig:workflow}
\end{figure}
In Fig.~\ref{fig:workflow} we show a schematic representation of the method.
Using ${\tt ReACT}$ to generate a training set of thousands of example matter power spectra for both $\Lambda$CDM and selected extensions,
we train two BNNs to classify the spectra according to the underlying model.
A five-label BNN is trained to classify an example spectrum as either $\Lambda$CDM or one of four chosen extensions while a two-label network is trained simply to classify between $\Lambda$CDM and non-$\Lambda$CDM.
Following the introduction of a novel method to construct a well-defined probability distribution from the output of a BNN in order to take into account the effect of the uncertainty in the final classification (thus preventing the network from being overconfident), we evaluate the performance of each network on the training, validation and test sets.
In addition, we determine the minimal deviation in the model parameters for each chosen non-$\Lambda$CDM model such that the five-label BNN classifies them as non-$\Lambda$CDM with some specified probability before passing the same spectra through the two-label BNN to compare their performance.
After studying how effective each BNN is at recognising spectra which do not belong to any class in the training set, we compare their predicted classification probabilities averaged over noise realisations for different values of the model parameters.
Finally, we examine the potential benefits of training specialist networks on selected subsets of the original classes in the training set.
While we only consider three well studied dark energy and modified gravity models in this work, this method can be extended to general extensions to $\Lambda$CDM such as massive neutrinos and Horndeski scalar-tensor theory as long as rapid and accurate predictions for the shape of the matter power spectrum can be computed. The effect of neutrino masses and baryonic feedback can now be taken into account and will be the subject of future work, given that it has been recently implemented in {\tt ReACT} \cite{Bose:2021mkz}.
This paper is laid out as follows.
Sec.~\ref{sec:BNNs} presents a concise theoretical background to both DNNs and BNNs.
Sec.~\ref{sec:training} then describes the generation and preparation of the training, validation and test data before they can be passed to the BNN, followed by a discussion of the BNN's architecture.
Sec.~\ref{sec:results} discusses the overall performance of each BNN.
We determine the values of the model parameters in each non-$\Lambda$CDM model such that the five and two-label BNN's are confident that a spectrum deviates from $\Lambda$CDM.
After analysing how sensitive each classification was to the noise in the power spectrum we also discuss the notion of specialist networks.
Sec.~\ref{sec:outlook} lays out potential avenues that should be explored in future studies before we conclude in Sec.~\ref{sec:conclusions}.
\section{Bayesian Neural Networks}\label{sec:BNNs}
Neural networks are becoming ever more widely employed in physics. The interested reader can find a review of the core concepts surrounding the use of neural networks in supervised classification problems in App.~\ref{app:NNintro}, to which we refer for the basic concepts used in the rest of the paper. For a thorough treatment see Refs.~\cite{Goodfellow-et-al-2016,Mehta:2018dln,Carleo:2019ptp}. In this section we discuss the aspects of BNNs relevant to this work, in particular the quantification of the classification uncertainty.
\subsection{Uncertainty in BNN classifiers}
\label{sec:BNN_uncertainty}
Two principal sources of uncertainty can be identified in the prediction given by a trained network for a new example, namely \emph{aleatoric} and \emph{epistemic} uncertainty \cite{KIUREGHIAN2009105}.
The former encompasses any intrinsic, non-reducible, uncertainty due to the stochasticity of the observations, while the latter describes uncertainty in the model.
High epistemic uncertainty quantifies limitations in modelling accuracy and could be reduced by, for example, choosing a more appropriate model architecture, adjusting the hyperparameters, or training with more data.
Differently from traditional NNs, BNNs can give an estimate of both sources of uncertainty. Technical details, including a discussion on the difference with traditional NNs, are given in App.~\ref{app:BNNtrain}. The key concept is the replacement of the networks' weights $w$ with distributions, so that an approximate posterior distribution of the weights, $q_{\theta}(w)$, can be learned by variational inference, instead of learning a single value for each weight as in traditional NNs.
This means that after training the learned distribution can be used to obtain predictions marginalised over the weights, rather than point-like prediction. This in turn allows to take into account potential variability in the network's output and the relative uncertainty as we will now show.
In the case of a classification problem with $N$ classes, the final layer of the network outputs an $N$-dimensional vector with components that sum to one and can therefore be interpreted as probabilities. We denote these components by $p(y_{i}^{\star}=1 | X^{\star}, w, \mathcal{D})$ ($i\in \{1, ..., N\}$) for a new example with features $X^{\star}$ and one-hot encoded label $y^{\star}$, for a given realisation of the weights $w$ and conditioned on the training data $\mathcal{D}$. %
Marginalisation over the weights can be obtained via Monte Carlo sampling from $q_{\theta}(w)$, giving for each component of the one-hot encoded label vector
\begin{align}\label{probabilityBayes}
&\mu_{i} \equiv p(y_{i}^{\star}=1 | X^{\star}, \mathcal{D}) \approx \frac{1}{N_{S}} \sum_{\alpha=1}^{N_{S}} \, p_{\alpha} , \\ \nonumber
& p_{\alpha} \equiv p(y_{i}^{\star}=1 | X^{\star}, w_\alpha, \mathcal{D})\, , \quad w_{\alpha} \sim q_{\theta}(w | \mathcal{D}) \, ,
\end{align}
where $N_{S}$ is the number of samples and throughout this paper we use Greek indices to denote MC samples and Latin indices to denote vector components.
Eq.~\ref{probabilityBayes} is the Monte Carlo approximation of the exact expression in Eq.~\ref{probabilityBayesExact}.
A prediction for the label for a new example with features $X^{\star}$ is obtained by assigning the label to the maximum output probability
$y_{\text{pred}}^{\star} = {\arg\max}_i \, \mu_{i}$ $(i=1...N)$, if this exceeds a chosen threshold probability $p_{th}$.
Defining $\mu$ to be the vector with components $\mu_i$ from Eq.~\eqref{probabilityBayes}, the full covariance of the classification is given by \cite{KWON2020106816}
\begin{align}\label{fullCov}
\Sigma_{q_{\theta}} &=
{\mathbb{E}}_{q_{\theta}} \Big[ Cov_{ p(y_{i}^{\star}=1 | X^{\star}, w, \mathcal{D})}(y^{\star}) \Big] \nonumber \\ & +Cov_{q_{\theta}}\Big[ {\mathbb{E}}_{ p(y_{i}^{\star}=1 | X^{\star}, w, \mathcal{D})}(y^{\star}) \Big] \, \nonumber \\
&= \frac{1}{N_{S}}\sum_{\alpha=1}^{N_{S}} \, \left( \text{diag}(p_\alpha) - p_\alpha^{\otimes 2} \right) + \frac{1}{N_{S}}\sum_{\alpha=1}^{N_{S}} (p_\alpha-\mu)^{\otimes 2} \, \nonumber \\
& = \text{diag} ( {\mu} )- { {\mu} }^{\otimes 2}\, ,
\end{align}
where the first line follows from the definition of the covariance
and the second from the use of Eq.~\eqref{probabilityBayes} with the following property of a multinomial distribution (which is used as the likelihood of the optimisation problem as customary in classification tasks, see App.~\ref{app:BNNtrain}): ${\mathbb{E}}_{ p(y_{i}^{\star}=1 | X^{\star}, w, \mathcal{D})}(y^{\star}) = p(y_{i}^{\star}=1 | X^{\star}, w, \mathcal{D})$.
This shows that the covariance is simply the standard multinomial covariance over the distribution of MC averages $\mu$.
The second term in the sum is the standard mean-squared error coming from the fact that the weights have a distribution $q_{\theta}(w | \mathcal{D})$, hence it corresponds to the epistemic uncertainty. The first term encodes the contribution to the variance marginalising over $q_{\theta}(w | \mathcal{D})$, and as such it describes the aleatoric uncertainty.
In order not to yield over-confident estimates of whether a given power spectrum is classified as $\Lambda$CDM or not it is important to accommodate both sources of uncertainty into the analysis.
When training on data coming from real-world observations, one has no means to reduce the aleatoric uncertainty (this is why this is sometimes referred to as ``uncertainty in the data''). In this paper we train a network on simulated noisy data, as described in \ref{sec:datagen}. In principle, the knowledge of the model from which the noise is drawn could be incorporated in the loss. Here we rather make the choice of treating noise as an effective aleatoric uncertainty, and including its effect in the classification uncertainty. Of course, a dependence on the noise model will be inherited during training. We note however that any data analysis tool relies on a model of the noise.
In order to compute the uncertainty, it must be kept in mind that despite $\mu_{i}$ being a probability by construction, it still does not represent an inferred ``true probability'' for the resultant classification as occurs in a likelihood or MCMC analysis.
The quantity $\mu_{i}$ should rather be interpreted as a parameter in itself used to classify a given spectrum if the magnitude exceeds the chosen threshold probability $p_{th}$.
Constructing a confidence in the classification at test time requires a joint distribution on $\mu_{i}$ to compute the sub-volume where $\mu_{i}>p_{th}$.
We shall detail in the following subsection how we utilise the uncertainty in Eq.~\eqref{fullCov} to estimate the confidence in a particular classification.
We stress here that while it is tempting to view BNNs as being able to provide a clear and statistically rigorous definition of probability in the classification, we should keep in mind that the model of the error is still subject to approximations, such as the variational approach described in this section and the choice of the parametric distribution $q_{\theta}(w)$.
For these reasons, it is also important to point out that the use of the definition ``Bayesian'' Neural Networks in the formulation used in this work comes from the Machine Learning literature, and the estimated classification probabilities should not be confused with the result of a truly Bayesian model selection as resulting, for example, from the computation of Bayesian evidences with a nested sampling algorithm.
Rather, at the current state of the art BNNs should be viewed as tools that at least enable one to introduce a model of the uncertainty, preventing overly optimistic interpretations of the results as well as providing an effective regularisation procedure.
\subsection{Quantifying the classification confidence}
\label{sec:construction_conf}
Currently there is no well-established method of quantifying the confidence in a prediction from a BNN.
In general, obtaining a classification confidence requires the definition of a probability distribution over the softmax output of the network.
One possibility is the Dirichlet distribution which is both analytic and possesses a natural interpretation as a distribution over probabilities being defined on the $N$-simplex.
Possible approaches include mapping the variance of the pre-softmax network output to Dirchlet parameters \cite{hobbhahn2020fast}, directly including a Dirichlet distribution in the loss function definition \cite{wu2019quantifying}, or training ``prior networks'' that directly output the parameters of the Dirichlet distribution \cite{malinin2018predictive}.
Another approach is to empirically define a ``confidence score'' using $\mu$ and the covariance in Eq.~\eqref{fullCov} \cite{Deodato824862,Lin:2020aps}.
In this work we introduce a novel approach which also directly utilises the covariance in Eq.~\eqref{fullCov}.
We consider a random variable $x \in \mathbb{R}^N$ distributed as a multivariate Gaussian truncated to lie between 0 and 1 with mean $\mu$ and covariance $\Sigma_{q_{\theta}}$ and compute the volume where $x_i>p_{\text{th}}$ $\forall i=1...N$ to obtain the confidence.
In practice, the definition of such a distribution is complicated by the fact that the components $x_i$ are not independent as both they and the means $\mu_i$ must sum to one.
This interdependency of the components $x_{i}$ implies one cannot define a multivariate Gaussian directly with the covariance Eq.~\eqref{fullCov}.
The full derivation of the resulting probability distribution we denote as $\mathcal{F}(x; \mu, \Sigma_{q_{\theta}})$ is outlined in App.~\ref{app:prob} with the final result being
\begin{align}\label{probDist}
\mathcal{F}(x; \mu, \Sigma_{q_{\theta}}) & = \, \delta\Big( 1-\sum_{j=1}^N x_j \Big) \times \sqrt{N} \\ \nonumber
& \times \prod_{i=1}^{N-1} {\mathcal{\tilde{N}}} \Big( \left[{B}^{-1} (x-\mu)\right]_i; 0, \left[{B}^{-1} \Sigma_{q_{\theta}}{B}\right]_{ii} \Big) \, ,
\end{align}
where $B$ is the matrix which diagonalises $ \Sigma_{q_{\theta}}$ and $\mathcal{\tilde{N}}$ denotes a Gaussian truncated between 0 and 1.
The Dirac delta-function enforces the constraint that the components must sum to one with the remaining terms being the product of $N-1$ one-dimensional Gaussians each with a variance given by the non-null eigenvalues of $\Sigma_{q_{\theta}}$.
By using the threshold probability $p_{th}$ and marginalising over the remaining labels, the probability an example is assigned the label $I$ can then be defined as
\begin{equation}\label{pGauss}
P_I \equiv \int_{p_{\text{th}}}^1 dx_I\, \int_0^1 dx_1...\hat{dx_I}...dx_N\, \mathcal{F}(x; \mu, \Sigma_{q_{\theta}}) \, ,
\end{equation}
where $\hat{dx_I}$ denotes that the integration on the $I$-th variable is omitted.
In practice, to compute the integrals in Eq.~\eqref{pGauss} we sample Eq.~\eqref{probDist} as outlined in App.~\ref{app:prob} and determine the fraction of samples which satisfy $x_I>p_{\text{th}}$.
If no components of a sample exceed $p_{th}$ then it is not assigned a label and the total fraction of such samples gives the probability the example is un-classifiable.
The probability $P_{I}$ encodes an estimate of the uncertainty in the classification and can be used to construct a first approximation of the confidence in the following manner.
Denoting $P_{\text{gauss}}(n\sigma)$ to be the usual volume of a Gaussian distribution in the interval centered on a mean value with width $n\times \sigma$ we define there to be a $n\sigma$ detection of a deviation from $\Lambda$CDM if $P_{\Lambda CDM}=1-P_{\text{gauss}}(n\sigma)$.
For example, a $2\sigma$ detection corresponds to $P_{\Lambda CDM}=1-P_{\text{gauss}}(2\sigma)=1-0.9545=0.0455$.
Moreover, if an example is classified with the label $I$ at less than $1\sigma$ confidence such that $P_I<0.68$, we shall not consider this a detection even if $\mu_I>p_{\text{th}}$.
Note that $1-P_{\Lambda CDM}$ represents the probability of an example not being $\Lambda$CDM, including the probability of the example being un-classifiable.
It therefore does not strictly represent the probability of a non-$\Lambda$CDM detection but also includes the probability that the BNN isn't able to determine which class from the training set the example belongs to.
%
\begin{table*}
\centering
\begin{tabular}{| c | c | c | c | c | c || c | c | c | c | }
\cline{2-10}
\multicolumn{1}{c | }{} & \multicolumn{5}{c || }{$\Lambda$CDM} & \multicolumn{4}{c |}{Extensions} \\ \hline
Parameter & $H_0$ & $n_s$ & $\Omega_m$ & $\Omega_b$ & $\sigma_8 (z=0)$ & $w_0$ & $w_a$ & $|{f}_{\rm R0}|$ & $\Omega_{\rm rc} $ \\ \hline
mean ($\mu$) &67.3 & 0.966& 0.316& 0.0494 & 0.766& -1 & 0 & 0 & 0 \\ \hline
variance ($\sigma$) & 0.4 & 0.007& 0.009 & 0.032 & 0.004& 0.097& 0.32& $10^{-5.5}$ & 0.173 \\ \hline
\end{tabular}
\caption{Mean model parameter values ($\mu$) and standard deviations ($\sigma$) used in $\tt{ReACT}$ for generating $\Lambda$CDM, $w$CDM, $f(R)$ and DGP matter power spectra for the training, validation and test data.}
\label{tab:params}
\end{table*}
\begin{table}
\centering
\begin{tabular}{| c || c | c | c | c | }
\hline
$z$ & 0.1 & 0.478 & 0.783 & 1.5 \\ \hline
$V(z) [{\rm Gpc}^3/h^3]$ & 0.283 & 3.34 & 6.27 & 10.43 \\ \hline
$\bar{n}(z) [{h^3}/{\rm Mpc^3}]$ & 0.0013 & 0.0010 & $8.3 \times 10^{-4}$ & $3.6 \times 10^{-4}$ \\ \hline
\end{tabular}
\caption{Chosen redshift bins, cosmological volume and number density parameters used to construct the Gaussian errors for each redshift bin in Eq.~\eqref{eq:Euclid_error}.}
\label{tab:errorpars}
\end{table}
\section{Training the Network}
\label{sec:training}
In this section we discuss the procedure of preparing the training, validation and test data, designing the network architecture and the subsequent hyperparameter optimisation.
\subsection{Generating and preparing matter power spectra} \label{sec:datagen}
We consider three well studied modifications to $\Lambda$CDM: the $f(R)$ gravity model described in Ref.~\cite{Hu:2007nk}, the Dvali-Gabadadze-Porrati (DGP) brane-world model of Ref.~\cite{Dvali:2000hr} and an evolving dark energy model as parameterised in Refs.~\cite{Chevallier:2000qy,Linder:2002et} ($w$CDM).
We compute dark matter power spectra for these theories utilising the recently developed code {\tt ReACT} \cite{Bose:2020wch} which calculates modified power spectra using the halo-model reaction method developed in Ref.~\cite{Cataneo:2018cic}. %
We sample the parameter space defining each model and pass the values to {\tt ReACT}, which generates power spectra in four redshift bins $z\in\{1.5,0.785,0.478,0.1\}$ and one hundred $k$-bins in the range $0.01 \leq k \leq 2.5 ~ h/{\rm Mpc}$ at equal intervals in log-space, according to that expected from a Euclid-like survey \cite{Laureijs:2011gra,Blanchard:2019oqi}. Details about the choices of the parameter space, redshift and $k$ ranges, are given in App~\ref{app:matpk}.
In addition to the aforementioned well studied extensions to $\Lambda$CDM we also include an additional class to represent potential ``unknown'' models.
Such models would imprint various signatures in the power spectrum that would be correlated in both space and time.
Since a priori we have no way of knowing what these signals are, we produce a dataset of filters with randomly generated features correlated in $k$ and $z$ before applying these to randomly selected spectra from the set of $\Lambda$CDM, $w$CDM, $f(R)$ and DGP model spectra.
We describe the method to generate this dataset in App.~\ref{app:randpk}.
For each of the five models considered, $\Lambda$CDM, $w$CDM, $f(R)$, DGP and random, we use 18,475 examples resulting in a total training dataset size of 92,375 examples.
Every example is a matrix of dimension $100 \times 4$ with each entry given by the value of the power spectrum in the particular $k$ and $z$ bin.
Of the 92,375 generated power spectra we set aside 15\% for the validation set with the remainder used in training the BNN.
Furthermore we generate a test set composed of 2500 examples per class.
Gaussian noise is then added to each spectrum in accordance to what one would expect from a Euclid-like survey \cite{Feldman:1993ky,Seo:2007ns,Zhao:2013dza,Blanchard:2019oqi}
\begin{equation}
\sigma_p(k) = \sqrt{\frac{4 \pi^2}{k^2 \Delta k V(z)} \times \left( P(k) + \frac{1}{\bar{n}(z)} \right)^2 + \sigma_{sys}^2} \, .
\label{eq:Euclid_error}
\end{equation}
The redshift dependent survey volume $V(z)$ and the shot noise $\bar{n}(z)$ are presented in Table~\ref{tab:errorpars}.
In addition, a constant systematic error of $\sigma_{sys}^2 =25 ~{\rm Mpc^6}/h^6$ is included to represent potential modelling inaccuracies.
This value of $\sigma_{\rm sys}$ is chosen such that we are able to recover the fiducial Planck parameters with $2\sigma$ confidence when performing an MCMC analysis using the nonlinear halofit Planck spectrum as our data vector and Eq.~\eqref{eq:Euclid_error} as our errors, with a $\chi^2$ likelihood.
We leave a thorough analysis of how much this choice affects the results to future work.
For each of the original set of 92,375 examples we generate ten spectra each with a different realisation of the Gaussian noise, ensuring that on top of recognising deviations from particular models, the network is more robust to different noise realisations for the same model.
In total, the number of training and validation examples is given by 923,750.
Finally, to ensure the data passed to the BNN are of comparable orders of magnitude across all scales and redshift bins we normalise each training example to a reference $\Lambda$CDM power spectrum with a cosmology given by the mean values in Table~\ref{tab:params}.
In Fig.~\ref{fig:Data_comparison} we display the process of how spectra generated with {\tt ReACT} are transformed before being passed to the BNN, including the addition of Gaussian noise followed by normalisation by a fiducial Planck spectrum.
Therefore the network is trained to detect deviations from $\Lambda$CDM for different noise realisations and choice of standard cosmological parameters.
\begin{figure*}
\centering
\includegraphics[width=.49\textwidth]{plots/fR_spectra.png}
\includegraphics[width=.49\textwidth]{plots/BNN_input_fR_ex.png}
\caption{\textbf{Left}: A pair of example $f(R)$ spectra with $f_{R0}= 8.34 \times 10^{-8}$ at redshifts $z=1.5$ and $z=0.1$ generated using {\tt ReACT} with the additional Gaussian noise.
Note that at low redshift cosmic variance dominates at low $k$ and at high redshift the shot noise dominates at high $k$.
\textbf{Right}: After normalising the noisy spectrum $P_{n}(k)$ by a fiducial Planck spectrum $P_{pl}(k)$ and centering around zero the spectra are ready to be passed to the BNN.
Due to this normalisation choice the BNN is trained to detect deviations from this fiducial Planck spectrum.
Note that in practice all four redshift bins are passed to the BNN.
Despite the presence of such a small modification, the five-label BNN classifies this spectrum as $\Lambda$CDM with only 5\% confidence, favouring the presence of a modification (see Fig.~\ref{fig:five-label_net}).
}
\label{fig:Data_comparison}
\end{figure*}
\subsection{Training and optimisation}
\label{sec:optimisation}
\begin{table*}
\centering
\begin{tabular}{| c | c | c | c | c | c | }
Operation Layer& Number of filters & Size & Stride & Output size & Number of parameters \\ \hline \hline
Input & - & - & - & 100x1x4 & - \\ \hline\hline
Convolution 1D flipout & 8 & 10 & 2 & 46x1x8 & 648\\ \hline
Max Pooling 1D & - & 2 & 2 & 23x1x8 & - \\ \hline
Batch normalization & - & - & - & 23x1x8 & 32 \\ \hline\hline
Convolution 1D flipout & 16 & 5 & 2 & 10x1x16 & 1296\\ \hline
Max Pooling 1D & - & 1 & 2 & 9x1x16 & - \\ \hline
Batch normalization & - & - & - & 9x1x16 & 64 \\ \hline\hline
Convolution 1D flipout & 32 & 2 & 1 & 8x1x32 & 2080\\ \hline
Batch normalization & - & - & - & 8x1x32 & 128 \\ \hline\hline
Global Average Pooling & - & - & - & 32 & - \\ \hline\hline
Dense flipout & - & - & - & 32 & 2080\\ \hline
Batch normalization & - & - & - & 32 & 128 \\ \hline\hline
Dense flipout & - & - & - & 5/2 & 325/130 \\ \hline
\end{tabular}
\caption{Description of the network's architecture}
\label{tab:architecture}
\end{table*}
\begin{figure*}
\centering
\resizebox{0.8\textwidth}{!}{
\includegraphics{plots/Architecture.pdf}}
\caption{Depiction of the BNN architecture employed for both the five-label and two-label classification tasks.
%
The height of each block illustrates the dimension size for each layer while the number of blocks per layer corresponds to the number of filters.
%
Additionally the dense blocks embedded in the first three transparent layers indicate the kernels for the first three one-dimensional convolutional layers scaled by their respective size.
}
\label{fig:model_architecture}
\end{figure*}
\begin{figure*}
\centering
\includegraphics[width=.49\textwidth]{plots/Accuracy_5_label.png}
\includegraphics[width=.49\textwidth]{plots/Accuracy_2_label.png}
\caption{\textbf{Left}: Evolution of the training and validation accuracy for the five-label network.
%
The accuracy on both data sets stabilises at $\sim 94.4 \%$ indicating that the five-label network is relatively robust.
%
\textbf{Right}: Evolution of the training and validation accuracy for a network trained to classify spectra between $\Lambda$CDM or non-$\Lambda$CDM, reaching 95.3\% accuracy for both datasets.
%
Note that the accuracy for both networks is evaluated with the output from a single pass through the BNN for all the training and validation spectra and it therefore simply gives an estimate of the network's performance.
}
\label{fig:LC_5labels_1D}
\end{figure*}
In this work we are concerned with the capability BNNs possess in tackling two questions.
The first is how effective BNNs can be in recognising the distinct features in the power spectrum for a particular modification to $\Lambda$CDM, such as $f(R)$ or DGP.
The second is the ability of BNNs to detect a deviation from $\Lambda$CDM in the power spectrum irrespective of the particular modification.
In practice, we train two BNNs with the same architecture, the first for five labels divided between $\Lambda$CDM and the four extensions and the second trained to distinguish between the two labels $\Lambda$CDM and non-$\Lambda$CDM.
Due to the fact that there are only four redshift bins it is beneficial to treat the data as four separate time-series and use one-dimensional convolutional layers.
Treated this way, the spectra are passed to the network with dimension $100 \times 1 \times 4$, or in analogy with image classification tasks, as $100 \times 1$ pixel images with four channels.
The architecture of the network used to train both the five-label and two-label network is displayed in Fig.~\ref{fig:model_architecture}.
Initially the structure consists of three 1D convolutional flipout layers with $8, 16$ and $32$ filters, kernel sizes of $10, 5$ and $2$ with strides of $2,2$ and $1$ respectively.
Each of the first two 1D convolutional layers are followed by a max pooling layer with a pool size of 2 and a pooling stride of 2 for the first max pooling layer and a pooling stride of 1 for the second max pooling layer.
After both of these max pooling layers there is a batch normalisation layer.
Following the final convolutional layer there is a global average pooling layer to reduce the filter size to one in order to pass it to a dense layer with 32 nodes.
Finally, after a further batch normalisation there is a softmax layer consisting of five or two neurons for either the five or two-label networks respectively.
The network's architecture is summarized in Table \ref{tab:architecture}.
The five-label and two-label networks consist of 6605 and 6410 trainable parameters respectively.
We set the initial learning rate $\text{lr}_0$ to be 0.01 with a decay rate 0.95 such that with a training set size $M$ and at each epoch $e$ the learning rate is
\begin{equation}
\text{lr}(e) = \text{lr}_0 \times 0.95^{(e/M)} \, .
\end{equation}
The batch size was set to $500 \times 5$. Each batch is composed of an equal number of power spectra for each of the labels in the training data. During training, we adjust the training set size by dropping a random subset of the data in order to have an integer number of batches of the same size.
In Fig.~\ref{fig:LC_5labels_1D} we show the evolution of the accuracy for a network trained to classify between $\Lambda$CDM, $w$CDM, $f(R)$, DGP and random spectra (top panel) and a network trained to distinguish simply between $\Lambda$CDM from non-$\Lambda$CDM (bottom panel).
In each classification task for both the training and the validation sets the five-label network asymptotes to an accuracy of $94.4\%$ and the two-label network asymtotes to a training accuracy of $95.3\%$.
Bear in mind that the accuracy is evaluated by passing examples once through the BNN with a single draw from the weight distribution and therefore only approximates the BNN's overall performance.
Note also that despite the fact the overall accuracy of the two-label network is slightly greater, it does not necessarily imply that it is generally better at detecting deviations from $\Lambda$CDM in the power spectrum (see Sec.~\ref{sec:results}).
\section{Results}
\label{sec:results}
In this section we determine the ability of both the five and two-label networks to classify previously unseen matter power spectra and perform tests to determine the robustness of the method.
In Sec.~\ref{sec:test_set} we study the overall performance of the network on the test set, followed by a calibration check in Sec.~\ref{sec:calibration} as well as test the robustness of the five-label BNN against variations in the training set (Sec.~\ref{sec:varTrain}). We then evaluate the performance on individual spectra in Sec.~\ref{sec:example}, including a study of the impact of noise on the classification in Sec.~\ref{sec:noise} and a comparison of the two and five-label networks in Sec.~\ref{sec:fivevtwo}.
We examine the ability of each network to recognise out-of-distribution examples which were not included in the training set in Sec.~\ref{sec:out_of_dist} before studying the constraints each network is capable of placing on the model parameters in Sec.~\ref{sec:constraints}. We finally comment on the relevance for future experiments in Sec.~\ref{sec:experiments}.
\subsection{Performance on the test set}
\label{sec:test_set}
\begin{figure*}
\centering
\includegraphics[width=.5\textwidth]{plots/Conf_matrices.pdf}
\caption[CONVERGENCE]{
The confusion matrices for the five-label network (Left) and the two-label network (Right) display the percentage of examples per class that are assigned to each class by the network.
%
A classification is obtained by assigning the label to be the maximum multinomial mean $\mu_{i}$ as long as it exceeds the threshold $p_{th}$.
%
If no $\mu_{i}$ exceeds $p_{th}$, the example is considered to be ``Not Classified'' (N.C.).
}
\label{fig:test_results}
\end{figure*}
\begin{figure*}
\centering
\resizebox{0.3\textwidth}{!}{
\includegraphics[width=0.3\textwidth]{plots/calibrationSCE5.pdf}
}
\resizebox{0.3\textwidth}{!}{
\includegraphics[width=0.3\textwidth]{plots/calibrationSCE2.pdf}
}
\caption[CONVERGENCE]{
The reliability diagrams for both the five-label (Left) and the two-label (Right) BNNs display the predictions of $\mu$ for examples in the test set divided into bins containing at least $0.5\%$ of the total test examples plotting against the resultant test accuracy evaluated on each bin.
%
A perfectly calibrated network corresponds to the bisector with the deviation from ``perfect calibration'' measured by the SCE defined in Eq.~\ref{SCE} also reported.
}
\label{fig:rel_diagram}
\end{figure*}
Now that the network has been trained, the next step is to evaluate its performance on the test set in order to determine how capable it is in classifying previously unseen examples.
To this end, ten copies of every test example are made with different noise realisations added to the same underlying spectra.
By computing the average of the output after five hundred MC samples using Eq.~\eqref{probabilityBayes}, each example's label is assigned to be the maximum $\mu_{i}$ as long as it exceeds the threshold value $p_{\text{th}}=0.5$.
If no $\mu_{i}$ is greater than the threshold the example is assigned the label ``Not Classifed'' (N.C.).
The resulting overall test accuracies are $94.9 \%$ and $95.8 \%$ for the five and two-label networks respectively.
As these results are comparable to the training and validation accuracies in Fig.~\ref{fig:LC_5labels_1D} the network can be considered to be robust.
In Fig.~\ref{fig:test_results} we show the confusion matrices for each network which provide information on the percentage of examples from each class that are classified accurately and if not, what class they were erroneously classified into.
Theories that show a greater degree of degeneracy in their effects on the matter power spectrum are more likely to be classified incorrectly.
For the five-label network, the strongest degeneracy exists between $w$CDM and $\Lambda$CDM, likely because the signatures of $w$CDM occur at length scales where the noise can dominate.
Indeed, $w$CDM modifications appear at the level of the cosmological background. In contrast, the other theories considered here can affect the higher order perturbations which leave a direct imprint on the power spectrum.
Following $\Lambda$CDM, $w$CDM also possesses a slight degeneracy with DGP.
By contrast, only 1\% of $f(R)$ examples were mis-classified which correspond to spectra with small values of $f_{R0}$ that are noise-dominated.
The high efficacy the BNN has in detecting $f(R)$ models warrants a more detailed analysis
which we discuss in Sec.~\ref{sec:specialist}.
\subsection{Calibration}\label{sec:calibration}
Another important test is to ensure the probabilities given by the network represent the likelihood for a prediction to be correct.
If this holds, the network is said to be well-calibrated.
By definition, a model is perfectly calibrated if the accuracy on all examples classified with probability $p$ is $p\times100 \%$.
Ensuring that DNNs are well-calibrated is a key step in assessing their reliability \cite{DBLP:conf/icml/GuoPSW17,kuleshov2018accurate,Nixon_2019_CVPR_Workshops}.
In Fig.~\ref{fig:rel_diagram} we present reliability diagrams for both the two and five-label networks which are constructed as follows.
First, we divide predictions for $\mu$ into bins.
The number of bins for each class, or component $\mu_{i}$, is chosen such that each bin contains at least $0.5\%$ of the total number of examples in the test set in order to avoid a large variance.
We then compute the accuracy in every bin for each class.
Let $B_i$ denote the set of bins for the $i$-th class, $n_{bi}$ the number of predictions in bin $b$ for class $i$ and $\text{acc}(b,i)$ and $\hat{\mu}(b,i)$ the corresponding accuracy and average probability for each $b$ and $i$ respectively.
Shown in Fig.~\ref{fig:rel_diagram} are the reliability diagrams displaying how $\text{acc}(b,i)$ varies with $\hat{\mu}(b,i)$ for both the five-label network and the two-label network.
For the $f(R)$ and random examples in the test set we find the probability is always either very close to 1 or 0 resulting in only two bins.
By construction, the reliability diagram would result in a straight line for a perfectly calibrated network.
We can quantify the deviation from perfect calibration by computing the Static Calibration Error (SCE), defined as \cite{Nixon_2019_CVPR_Workshops}
\begin{equation}\label{SCE}
\text{SCE} =\frac{1}{N}\sum_{i=1}^{N} \sum_{b=1}^{B_i} \frac{n_{bi}}{N_{\text{tot}}} \left| \text{acc}(b,i)- \hat{\mu}(b, i)\right| \, ,
\end{equation}
where $N_{\text{tot}}$ is the total number of test examples and $N$ is the number of labels.
From their values displayed in Fig.~\ref{fig:rel_diagram} we find that both networks are well-calibrated with an SCE of $ 0.3\%$.
Furthermore, we verified this value remains stable under changes in the number of bins and never exceeds $ \lesssim 0.5\%$.
\subsection{Robustness against variations in the training set}\label{sec:varTrain}
In this subsection we evaluate the impact of using the confidence introduced in Sec.~\ref{sec:construction_conf} to detect extensions to $\Lambda$CDM in the matter power spectrum and its usefulness in taking into account uncertainty due to the presence of noise in the training set.
Recall that, even when marginalised over the weights, the network's output is still conditioned on the training data (see Eq.~\eqref{probabilityBayes}).
Despite the fact we include multiple realisations of the noise for each clean spectrum to ensure the BNN is more robust to variations in the training set, there is no guarantee this eliminates significant fluctuations in the result with slight variations in the training set.
However, if the resulting classification of a particular example is highly dependent on the specific realisation of the noise during training, the associated uncertainty will be large for a similar example at test time.
As the covariance matrix in Eq.~\eqref{fullCov} contains an estimate of the aleatoric uncertainty, the classification confidence in Eq.~\eqref{pGauss} should be lower for such noise-dependent examples.
To explore this issue we train a second five-label network with an alternate partitioning of the data into training and validation sets before evaluating the probability $\mu$ for every example in the test set for both networks.
Note that the same realisation of the noise was added to each test example to ensure any variation in the result cannot be accounted for by the variation in the noise at test time.
We find that for 243 test examples, or $2\%$ of the test set, each network gives different predictions.
When considering our estimated confidence however, in 217 of these cases, or $89\%$ of the discrepancies, both networks yield a classification confidence of $<1\sigma$.
Of the remaining 26 discrepancies, only 3 give inconsistent predictions and in 23 cases one of the two network predictions has a confidence of $<1\sigma$.
\begin{figure*}
\vspace{0.5cm}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/fR_5label.png}
}
\vspace{0.5cm}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/dgp_5label.png}
}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/wcdm_5label.png}
}
\caption{Examples of a $w$CDM, $f(R)$ and DGP model which the BNN can correctly identify as non-$\Lambda$CDM at our chosen confidence level.
We compute $\mu$ and $\Sigma_{q_{\theta}}$ as described in Sec.~\ref{sec:construction_conf}, use them to construct the probability $\mathcal{F}$ defined in Eq.~\eqref{probDist}, then sample from this distribution. The corresponding samples are shown in green. The probabilities denoted by $P_I$ ($I= \text{DGP}, f_R,\Lambda\text{CDM}, \text{Random}, w\text{CDM}$) correspond to the fraction of samples where the $I$-th component of the sample lies above $0.5$, i.e. Eq.~\eqref{pGauss}. If a sample has no component above $0.5$, it is considered as unclassified. The fraction of samples for which this happens gives $P_{\text{Unclassified}}$.
}
\label{fig:five-label_net}
\end{figure*}
Since each discrepancy involves spectra with very small deviations from $\Lambda$CDM,
we generate an additional dataset of 200 example spectra for each of the three extenions DGP, $f(R)$ and $w$CDM with narrower ranges for the model parameters in the regime where each network may give different predictions, namely $f_{R0} \in \left[3\times10^{-8}, 1\times 10^{-7} \right] $, $\Omega_{rc} \in \left[0.002, 0.06\right]$, $w_{0} \in\left[-1.025, -0.975 \right]$ and $w_{a} \in \left[-0.1,0.1 \right]$.
In this case, we find a discrepancy in $10\%$ of the dataset but in all these cases at least one network has a confidence of $<1\sigma$.
In $94\%$ of these discrepant examples both networks yield a classification confidence of $<1\sigma$ while only in a single case does one network incorrectly classify an example with a confidence $>1\sigma$.
This analysis suggests that the confidence in Eq.~\eqref{pGauss} is a more realistic indicator of a prediction's reliability with respect to $\mu$.
\subsection{ Illustration on explicit examples}\label{sec:example}
In this subsection we illustrate the new method to compute the confidence, as introduced in \ref{sec:construction_conf}, on explicit examples.
We choose three underlying noiseless spectra belonging to the $f(R)$, DGP and $w$CDM classes, and add a fixed realisation of the noise to each of them, thus mimicking an actual observational situation where the network is given some noisy spectrum to classify.
We choose the parameters and the noise realisation so that the probability of being non-$\Lambda$CDM is around $95\%$ for each example. We will investigate the role of the noise and dependence on the strength of the modifications more extensively in Sec.~\ref{sec:noise} and Sec.~\ref{sec:constraints} respectively.
Following the procedure outlined in Sec.~\ref{sec:construction_conf}, we compute $\mu$ and $\Sigma_{q_{\theta}}$ with the five-label network for each spectrum. These are used to construct the distribution $\mathcal{F}$ using Eq.~\eqref{probDist}. This represents the distribution of possible outcomes of the network, taking into account the epistemic and aleatoric uncertainties. Then, according to the algorithm described in App.~\ref{app:prob} to compute the probabilities in Eq.~\eqref{pGauss}, samples are drawn from $\mathcal{F}$, each sample being a vector of dimensions equal to the number of classes (5) with values between 0 and 1, and where the dimensions $\{0,..., 4\}$ correspond respectively to the classes DGP, $f(R)$, $\Lambda$CDM, Random, and $w$CDM. The fraction of samples where the $I$-th component (with $I\in \{0,..., 4\}$) lies above $p_{th}=0.5$ is determined $\forall I$, which gives the integral in Eq.~\eqref{pGauss}. If a sample has no component above $p_{th}=0.5$, it is considered as unclassified. The fraction of samples for which this happens gives $P_{\text{Unclassified}}$.
In Fig.~\ref{fig:five-label_net} we display the results.
Samples from $\mathcal{F}$ are shown in green.
The first spectrum we consider (top panel) is an $f(R)$ spectrum with $f_{R0} = 8.34\times 10^{-8}$.
We find it is correctly classified as $f(R)$ with a probability of 88\%, with the remaining probability falling into $\Lambda$CDM with 5\% and unclassified with 7\%.
This remains consistent with the evaluation of the confusion matrix on the test set which showed there were no $f(R)$ spectra classified as DGP or $w$CDM.
In contrast, a DGP spectrum with $\Omega_{rc} = 0.0072$ (mid panel) is classified as DGP with only 54\% probability with 11\% $w$CDM, 30\% unclassified and 4\% $\Lambda$CDM showing the stronger degeneracy between DGP and $w$CDM.
Classifying $w$CDM spectra with small deviations in $w_{0}$ and $w_{a}$ is particularly difficult for the BNN due to their high degree of degeneracy with $\Lambda$CDM and the fact its features appear in noise-dominated regions of the power spectrum.
In this case, a spectrum with a deviation of $(w_{0}, w_{a}) = (-1.03, -0.04)$ (lower panel) is classified as $w$CDM with 78\% probability, 17\% unclassified and 4\% $\Lambda$CDM.
These modifications represent the minimum deviations from $\Lambda$CDM in each of our chosen extensions before the modifications become noise-dominated and the five-label network determines the spectra to either be unclassifiable or $\Lambda$CDM.
\begin{figure*}
\centering
\resizebox{.6\textwidth}{!}{
\includegraphics{plots/Two-label_network.pdf}
}
\caption{We display the distributions constructed from the output of the two-label network for the same spectra and Gaussian noise which were passed to the five-label network in Fig.~\ref{fig:five-label_net} in order to compare their relative performance in detecting deviations from $\Lambda$CDM in marginal examples.
Both the $w$CDM and DGP examples are classified as non-$\Lambda$CDM with a higher confidence than the five-label network, however the $f(R)$ spectrum is not correctly classified at high confidence}
\label{fig:two-label_net}
\end{figure*}
We then repeat the procedure for the same noisy spectra with the two-label network.
The result is shown in Fig.~\ref{fig:two-label_net}.
Note that in the case of two labels the probability of ``Not Classified'' is always zero as it is not possible to have two samples which are simultaneously above $0.5$.
We find that for DGP and $w$CDM the two-label network classifies the examples correctly with a higher probability than the five-label network.
However the $f(R)$ spectrum is not correctly classified with a high probability.
\subsection{Dependence on the noise}\label{sec:noise}
\begin{figure*}
\vspace{0.5cm}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/fR_marginalised_example.png}
}
\vspace{0.5cm}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/dgp_marginalised_example.png}
}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/wcdm_marginalised_example.png}
}
\caption{Taking the same clean $w$CDM, $f(R)$ and DGP spectra as were passed through the network in Fig.~\ref{fig:five-label_net}, we now pass them through the network one thousand times with each pass drawing a new sample from the weight distribution and with a new realisation of Gaussian noise added to the clean spectra.
The resulting distributions therefore give a measure on how the network's output varies when marginalising over the observational noise and the weights.
}
\label{fig:noise_marginalisation}
\end{figure*}
It is important to emphasise the role noise plays in determining the eventual classification probability for each example in Fig.~\ref{fig:five-label_net}.
In particular, it is possible that a different draw from the Gaussian noise in Eq.~\eqref{eq:Euclid_error} on top of the same underlying clean spectrum could change the resulting classification.
In order to obtain a measure on how much the noise affects the resulting classification for a given underlying spectrum, we compute $\mu$ and $\Sigma_{q_{\theta}}$ again starting from the same underlying noiseless $f(R)$, DGP, and $w$CDM spectra used for Fig.~\ref{fig:five-label_net}, but this time we further average the result over \emph{different} noise realisations for each spectrum.
In Fig.~\ref{fig:noise_marginalisation} we display the distribution of outputs from the five-label BNN varying the noise realisation. We stress that, differently from Fig.~\ref{fig:five-label_net}, the histograms in Fig.~\ref{fig:noise_marginalisation} do not represent samples from the distribution $\mathcal{F}$ in Eq.~\eqref{probDist}, but are different realisations of $\mu_i$ defined in Eq.~\ref{probabilityBayes} corresponding to different noise realisations on top of a given clean spectrum. This is an illustration of the potential variability of the network's output with noise.
From the $\mu$ and $\Sigma_{q_{\theta}}$ obtained from averaging over the noise we compute $P_I$, which now becomes a noise-averaged classification probability. The corresponding values are shown in Fig.~\ref{fig:noise_marginalisation}.
This probability gives a measure on how likely it is the network will pick up a deviation from $\Lambda$CDM given the distribution of Gaussian noise in Eq.~\eqref{eq:Euclid_error}.
For example, the $f(R)$ spectrum with $f_{R0} =8.34\times 10^{-8}$ possesses an average detection probability of $\sim 72\%$.
In contrast, the DGP and $w$CDM examples both have noise-averaged detection probabilities of less than $50\%$.
This implies that, even though the BNN classified each individual example correctly in Fig.~\ref{fig:five-label_net}, for our chosen model parameter values a correct classification was more likely to occur for the $f(R)$ example than for the DGP and $w$CDM examples given another realisation of Gaussian noise.
Note that this noise-averaged detection probability can be considered to be an invariant measure of the network's performance in classifying spectra with particular values of the model parameters.
\subsection{Performance of five and two-label BNNs}\label{sec:fivevtwo}
To compare the performance of the two-label and five-label networks more robustly, we compute the $\Lambda$CDM classification probability on the test set in each network, as well as compare the $P_{\mathrm{non-}\Lambda \mathrm{CDM}}$ from the two-label network with $1-P_{\Lambda \mathrm{CDM}}$ of the five-label network.
We find that in $\sim98\%$ of the cases where the example is correctly predicted as $\mathrm{non-}\Lambda \mathrm{CDM}$, the five-label network can correctly classify spectra at a higher confidence than the two-label network.
This is likely a result of the fact that the five-label network, possessing more final classes, can tune its layers to pinpoint specific features of each subclass, resulting in a higher confidence. By contrast, the two-label network needs to compress the information from any deviation into a single class, which can result in lower confidence due to contamination from the classes that are more difficult to distinguish from $\Lambda$CDM.
Of the $2\%$ of spectra where the two-label network was more confident, the probability in the five-label network was either split principally between two non-$\Lambda$CDM classes, not classified or belonged to $w$CDM.
This indicates that the two-label network may classify non-$\Lambda$CDM spectra which do not belong to any of the classes in the training set more confidently.
Such spectra are more evenly split by the five-label network between separate classes or classified as random (see Sec.~\ref{sec:out_of_dist}).
However, further investigation is required to determine the necessary conditions for the two-label network to outperform the five-label network and vice versa.
\begin{figure*}
\vspace{0.5cm}
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/gamma_classification.pdf}
}
\caption{If an example spectrum generated with a growth-rate parameter of $\gamma=0.52$ with a fixed noise realisation is passed through the five-label (Top) BNN, this estimates this spectrum is not $\Lambda$CDM at the $2\sigma$ confidence level.
However, the remaining probability is distributed between the other labels with no overall favoured class, highlighting the utility BNNs possess in determining a spectrum does not belong to any of the classes in the training set.
If the same spectrum with the same noise realisation is passed to the two-label network (Bottom) it is classified as non-$\Lambda$CDM with a higher confidence than the five-label network.
}
\label{fig:gamma_5label}
\end{figure*}
\begin{figure*}
\centering
\resizebox{.9\textwidth}{!}{
\includegraphics{plots/horndeski.jpg}
}
\resizebox{.5\textwidth}{!}{
\includegraphics{plots/Painted_spectra.png}
}
\caption{If an example that does not belong to any of the pre-trained classes is passed to the BNN, in this case the painting ``\emph{Blustery mountain road on an Autumn day}'' by Gregory Horndeski (Top), we find that both the five-label and two-label network classifies it as non-$\Lambda$CDM with a confidence $\gg$ 2$\sigma$.
In addition, the five-label BNN classifies it as random with extremely high confidence. %
To pass the painting to the BNN, the pixels were re-binned into a $100\times 4$ pixel grey-scale image which was multiplied by the fiducial Planck spectrum.
Gaussian noise was then added and the resulting noisy spectra were normalised to the Planck spectrum at each redshift bin (Bottom).
The final normalised spectrum has significant deviations from zero, consistent with the fact that the paining imprints large random deviations from the $\Lambda$CDM spectrum.
}
\label{fig:Horndeski_painting}
\end{figure*}
\subsection{Classification of out-of-distribution examples}
\label{sec:out_of_dist}
To investigate how each BNN classifies examples that do not belong to either the training, validation or test distributions, known as out-of-distribution examples, in this section we examine how each BNN classifies spectra generated both from the growth-index parameter $\gamma$ \cite{PeeblesBook, Wang:1998gt, Linder:2005in} and from a painting by Gregory Horndeski.
The growth-index is a frequently used phenomenological parameterisation designed to pick up deviations in the growth rate of structure from its $\Lambda$CDM value of $0.55$ arising from extensions to $\Lambda$CDM.
The parameterisation is defined by $D'(a) = \Omega_m(a)^{\gamma}$, where $D(a)$ is the linear density perturbation growth factor, $\Omega_m$ is the cosmological total matter density fraction and the prime denotes a logarithmic scale factor derivative.
To generate nonlinear spectra with varying values of $\gamma$ we first modify the linear power spectrum by applying the following parameterised growth factor
\begin{equation}
D(\gamma;a_f) = \int_{a_i}^{a_f} \left[ \frac{\Omega_{m,0}}{H(a)^2 a^3} \right]^\gamma \frac{a_i}{a} da \, ,
\end{equation}
where $a$ is the scale factor, $H(a)$ is the $\Lambda$CDM Hubble rate and $a_i=0.0001$ is the initial scale factor.
The modified linear spectrum is then simply $P_L(k,\gamma;a) = D(\gamma;a)^2 P_0(k)$, where $P_0(k)$ is the primordial power spectrum.
The modified nonlinear spectrum is produced by supplying the modified linear spectrum to the same halofit formula \cite{Takahashi:2012em} used in producing the training data.
We find that, while a spectrum generated with a growth-index of $\gamma = 0.55$ is correctly classified as $\Lambda$CDM, the associated confidence lies between $1\sigma$ and $2\sigma$ reflecting the fact that this parameterisation is only an approximation of $\Lambda$CDM.
Passing spectra generated with $\gamma = 0.54$ or $\gamma = 0.56$ to the five-label BNN shifts the $\Lambda$CDM classification probability to below $0.5$.
In Fig.~\ref{fig:gamma_5label} we display the sampled classification probabilities and the multinomial mean for each class for a spectrum generated with $\gamma = 0.52$.
With the classification probability of $\Lambda$CDM being $\approx 5\%$, this value of $\gamma $ represents the smallest deviation from $0.55$ such that the network can confidently classify the spectrum as not being $\Lambda$CDM.
Nonetheless, as no probability exceeds $0.5$, no class is favoured.
This ability to determine that a spectrum does not belong to the training set distribution demonstrates a unique capability of BNNs.
Note also that the failure of the five-label BNN to classify a spectrum generated from the growth-index as either $w$CDM, DGP or $f(R)$ further highlights the limitations of the growth-index parameterisation.
Taking the same spectrum with $\gamma=0.52$ with the same noise and passing it through the two-label network, we find that it is classified as non-$\Lambda$CDM with a higher confidence than the five-label network.
Although this suggests the two-label network is better suited to placing constraints on the growth-index, given the five-label network did not confidently classify the spectrum into any of the five-labels, it is an open question how useful such constraints would be in constraining more physically motivated models.
As a further test that the five-label network can identify spectra that do not belong to any known class of physical models, we pass the painting ``\emph{Blustery mountain road on an Autumn day}'' by Gregory Horndeski (see Fig.~\ref{fig:Horndeski_painting}) to the BNN.
Firstly, we convert it into a grey-scale image with $100\times4$ pixels.
This then acts as a similar filter to those constructed in App.~\ref{app:randpk} which we then apply to the fiducial Planck spectrum before finally adding Gaussian noise (see Eq.~\eqref{eq:filtertopk}).
The resulting matrix is then equivalent to a normalised input for the network.
These deviations are large enough such that the network can accurately determine that the painting is not a $\Lambda$CDM power spectrum.
However, it is also not ``Not Classifed''.
Rather, it is classified into the random class with 100\% probability, indicating that the random class is capable of picking up examples that contain deviations which are not comparable to any model included in the training set.
\subsection{Dependence on the strength of the modification }
\label{sec:constraints}
We have seen in Sec.~\ref{sec:noise} that a more reliable estimator of a BNN's ability to classify a non-$\Lambda$CDM spectrum with a particular modification strength is to pass the spectrum through the BNN multiple times with different realisations of the noise.
The resultant probability distribution quantifies not only whether a detection is possible, but also how probable it is the noise will alter the classification.
In this section we repeat this procedure for multiple $f(R)$, DGP and $w$CDM power spectra in the parameter range defined by the region where the five-label network transitions from classifying spectra as non-$\Lambda$CDM at low confidence to high confidence.
Specifically, we build a batch of power spectra composed of different noise realisations on top of the same underlying spectrum and predict the average classification likelihood $\mu$ marginalized over the weights for all the elements in the batch.
By further averaging the result over the batch, we obtain a noise-averaged classification likelihood for every example.
Using this to construct the probability distribution in Eq.~\eqref{probDist}, we then compute the corresponding average classification probability that the spectrum belongs to its true class and the average probability it is non-$\Lambda$CDM, defined as $1-P_{\Lambda \mathrm{CDM}}$ for both the five and two-label networks.
This process is then repeated for spectra with different modification strengths.
In order to obtain an estimate on how much the noise can shift the classification for particular values of the model parameters, we also construct a confidence band around the average classification probability as follows.
First, we remove the noise realisations such that any of the components of its predicted $\mu$ fall below the corresponding fifth or above the corresponding ninety-fifth percentile of the batch.
For each network the upper bound is then obtained by selecting the noise realisation among those remaining such that the probability $\mu_{i}$ (with $i$ being $f(R)$, DGP or $w$CDM for the five-label network, and non-$\Lambda$CDM for the two-label network) is maximised.
The minimum bound is obtained by taking the noise realisation that maximises the difference $ \mu_{\Lambda \mathrm{CDM}}-\mu_{i}$.
While for the two-label network this is equivalent to minimising $\mu_{ \mathrm{non}-\Lambda \mathrm{CDM}}$, in the case of the five-label network it ensures that the lower bound is the minimum of both $P_i$ and $P_{\mathrm{non}-\Lambda \mathrm{CDM}}$.
This would not be guaranteed by only taking the noise realisation that minimises $\mu_{i}$, due to the fact that we allow for an unclassified probability.
\begin{figure*}
\centering
\resizebox{0.49\textwidth}{!}{
\includegraphics{plots/constraints/fR_constraint.png}
}
\resizebox{0.49\textwidth}{!}{
\includegraphics{plots/constraints/dgp_constraint.png}
}
\caption{Noise-averaged non-$\Lambda$CDM classification probabilities and associated confidence bands for $f(R)$ (Top) and DGP (Bottom) spectra for the five and two-label networks as a function of $f_{R0}$ and $\Omega_{rc}$.
%
The five-label classification probabilities $P_{f(R)}$ and $P_{\mathrm{DGP}}$ are also shown.
%
One can see that the average non-$\Lambda$CDM classification probability for the five-label network provides the most robust indicator of the presence of a modification, confidently classifying spectra as non-$\Lambda$CDM for $f_{R0}\approx 9\times10^{-8}$ and $\Omega_{rc} \approx 0.008$ independently of the noise for $f(R)$ and DGP respectively.
}
\label{fig:fR_DGP_constraints}
\end{figure*}
\begin{figure*}
\centering
\resizebox{0.49\textwidth}{!}{
\includegraphics{plots/constraints/wcdm_constraint.png}
}
\resizebox{0.49\textwidth}{!}{
\includegraphics{plots/constraints/wa_constraint.png}
}
\caption{Noise-averaged non-$\Lambda$CDM and $w$CDM classification probabilities for deviations of $w_{0}$ (Top) and $w_{a}$ (Bottom) around their fiducial values.
%
On average, the five-label network is better at detecting deviations from $\Lambda$CDM in each case.
%
However the performance of each network remains highly sensitive to the noise in the ranges $-1.07\lesssim w_{0} \lesssim -0.94$ and $-0.25\lesssim w_{a} \lesssim 0.25$.
}
\label{fig:wcdm_constraints}
\end{figure*}
In Figs.~\ref{fig:fR_DGP_constraints} and \ref{fig:wcdm_constraints} we present the results for $f(R)$, DGP and $w$CDM.
One can see that in the case of $f(R)$ gravity the five-label network's non-$\Lambda$CDM classification probability is more capable of recognising small deviations in $f_{R0}$, on average classifying spectra as non-$\Lambda$CDM when $f_{R0}>8\times 10^{-8}$.
The same network becomes more confident that a spectrum specifically belongs to $f(R)$ for $f_{R0}>1\times 10^{-7}$.
Conversely, the two-label network's ability to confidently classify spectra as non-$\Lambda$CDM remains highly sensitive to the noise up to $f_{R0}> 1.4 \times 10^{-7}$.
In the case of DGP, while the five-label network's non-$\Lambda$CDM classification probability again provides the most reliable predictions, the two-label network's ability to classify spectra as non-$\Lambda$CDM outperforms the five-label network's ability to classify the spectra as DGP.
For values of $\Omega_{rc}>0.016$ both networks definitively determine all spectra are not $\Lambda$CDM independently of the noise.
Turning our attention to each network's ability to detect evolving dark energy, we show in Fig.~\ref{fig:wcdm_constraints} the noise-averaged classification probabilities for a range of $w$CDM power spectra.
In each case we vary either $w_{0}$ or $w_{a}$ fixing the non-varying parameter to their $\Lambda$CDM fiducial values of $\left(w_{0},w_{a}\right) = \left(-1,0 \right)$.
Again we find that for both $w$CDM parameters, the five-label non-$\Lambda$CDM classification probability is the most reliable indicator of a deviation from $\Lambda$CDM.
Despite both networks on average classifying $w$CDM as non-$\Lambda$CDM for deviations of $\Delta w_{0}\sim 0.05$ and $\Delta w_{a} \sim 0.2$, the five-label network is less sensitive to the noise.
We leave a detailed analysis of how the degeneracies between $w_{0}$ and $w_{a}$ affect the noise-averaged classification probability to future work.
\begin{figure*}
\centering
\resizebox{0.5\textwidth}{!}{
\includegraphics{plots/constraints/gamma_constraint.png}
}
\caption{Noise-averaged five-label and two-label classification probability for spectra generated using the growth-index $\gamma$.
%
For values of $\gamma>0.6$ the two-label network classifies the spectrum as non-$\Lambda$CDM with a much higher confidence than the five-label network.
%
However, we attribute this to the fact that the five-label network is better able to determine that this spectrum is an out-of-distribution example.
}
\label{fig:gamma_constraints}
\end{figure*}
\begin{table*}
\centering
\begin{tabular}{ | c | | c | c | c | c | c |}
\cline{2-6}
\multicolumn{1}{c | }{} & \multicolumn{5}{c |}{Extensions} \\ \hline
Parameter & $w_0$ & $w_a$ & ${f}_{\rm R0}$ & $\Omega_{\rm rc} $ & $\gamma$\\ \hline
Five-label & $\left(-1.05,-0.95 \right)$ & $\left( -0.20,0.17\right)$ & $8 \times 10^{-8}$ & 0.007 & $\left(0.52, 0.63\right)$\\ \hline
Two-label & $\left(-1.05, -0.95 \right)$ & $\left(-0.20, 0.19\right)$ & $10^{-7}$ & 0.008 & $\left(0.51, 0.59\right)$\\ \hline
\end{tabular}
\caption{Values of the minimum deviation in each of the model parameters such that both the five-label and two-label networks classify a spectrum as non-$\Lambda$CDM at $95\%$ confidence when averaged over Gaussian noise realisations.
In the case of $w$CDM, these bounds are obtained by fixing either $w_{0}$ or $w_{a}$ to their fiducial value and allowing the other to vary.
We refer the reader to Figs.~\ref{fig:fR_DGP_constraints},~\ref{fig:wcdm_constraints} and \ref{fig:gamma_constraints} for estimates of the variance in the classification probability for each of these values.
}
\label{tab:param_constraints}
\end{table*}
Having completed these tests for models belonging to the training set, we repeat the procedure for spectra generated with varying values of the growth-index $\gamma$ as outlined in Sec.~\ref{sec:out_of_dist}.
In Fig.~\ref{fig:gamma_constraints} we show the noise-averaged classification probability for deviations around the $\Lambda$CDM fiducial value of $\gamma=0.55$.
Due to the absence of a specific label for $\gamma$, in this case the lower and upper bounds for the confidence bands are constructed by selecting the noise realisations that maximise and minimize $\mu_{\Lambda \mathrm{CDM}}$ respectively.
Although it appears that the two-label network can pick up smaller deviations than the five-label network for $\gamma>0.55$, in this case it is because the five-label network recognises the spectrum does not belong to the models in the training set and thus assigns a lower classification probability.
For values of $\gamma < 0.55$ the spectra are more degenerate with DGP which may help each network recognise the spectrum as non-$\Lambda$CDM.
Even in this instance however, the probability is split between DGP and other non-$\Lambda$CDM labels as in Fig.~\ref{fig:gamma_5label}.
Note that the bounds on the growth index $\gamma$ were obtained from a network which was not trained on spectra generated from $\gamma$.
One would therefore expect these bounds to improve if such spectra were included in the training set.
\subsection{Impact on future experiments}
\label{sec:experiments}
In Table~\ref{tab:param_constraints} we summarise the values of the minimum magnitude of each model parameter for every $\Lambda$CDM extension such that the noise-averaged non-$\Lambda$CDM classification probability is approximately $95\%$ for both the five-label and the two-label network.
The uncertainty estimated in our approach to the BNN output is of similar order of magnitude of stage IV astronomical survey forecasts, despite the two methods not being directly comparable, nor do we use the observational probes of upcoming surveys, e.g.~weak lensing and galaxy clustering.
We note the $1\sigma$ cosmic shear forecasts of Ref.~\cite{Bose:2020wch} were $f_{\rm R0} \leq 10^{-7.25} $ and $\Omega_{\rm rc} \leq 0.08 $ which assumes an LSST-like survey and a multipole scale cut of $\ell_{\rm max} = 1500$.
The official Euclid Fisher forecast of Ref.~\cite{Blanchard:2019oqi} gives $w_0 = -1 \pm 0.097 (0.077) $ and $w_a =0 \pm 0.32 (0.24) $, which combines both galaxy clustering and weak lensing probes and pessimistic (optimistic) scale cuts.
The $2\sigma$ constraints estimated for Euclid on $\gamma$ are $\gamma = 0.55 \pm 0.036 (0.026)$ for the pessimistic (optimistic) analyses of Ref.~\cite{Blanchard:2019oqi} for WL+${\rm GC}_{\rm s}$.
This means that the method outlined here is able to pinpoint deviations from $\Lambda$CDM down to a level relevant for Euclid.
\section{Training specialist networks}
\label{sec:specialist}
Given the promising performance of the five-label network in detecting deviations in $f(R)$ models down to $f_{R0}\approx \mathcal{O}(10^{-8})$, in this section we discuss the potential gains that could be achieved by training additional networks on subsets of the original five classes in the training set.
Heuristically this follows the philosophy of an MCMC analysis in that in order to constrain a specific model it is beneficial to choose the most appropriate set of model parameters in the MCMC.
In the case of BNNs, if one is only interested in constraining a single model beyond $\Lambda$CDM then in order to maximise the performance of the BNN it is beneficial to only train the network on this model alongside $\Lambda$CDM.
Such a network would be ``specialised'' to pick up any deviation from the particular source of new physics one is interested in, at the expense of losing information on potential degeneracies between different models when trained on multiple theories.
In this subsection we discuss one such specialist network trained on 18,475 $f(R)$ and 18,475 $\Lambda$CDM power spectra, each of which is passed to the network during the training and validation process with ten different realisations of the noise for a total training and validation set of 369,500 power spectra.
We use the architecture displayed in Fig.~\ref{fig:model_architecture} where the final layer is now a binary classifier for the two new labels $f(R)$ and $\Lambda$CDM, finding that the training and validation accuracies reach approximately 99.5\%, exceeding that of the five-label network.
We now study how capable this specialist network is in constraining $f_{R0}$ in comparison with the five-label network.
In Fig.~\ref{fig:noise_avg_comparison} we display a plot of how the noise-averaged $f(R)$ classification probability varies for power spectra generated with values of $f_{R0} \in \left[3\times10^{-8},1\times10^{-7}\right]$ with associated confidence bands.
We find that the performance of the specialist $f(R)$ network exceeds that of the five-label network's $f(R)$ classification capability, retaining a noise-averaged detection confidence of $1\sigma$ for $f_{R0} \approx 5.5\times 10^{-8}$ where the equivalent noise-averaged detection probability for the five-label network is $<0.3$.
Furthermore, for values of $f_{R0}> 8\times 10^{-8}$ the classification probability for the specialist BNN asymptotes to one with only a few noise realisations decreasing this probability to $85\%$.
Spectra with values of $f_{R0}> 9\times 10^{-8}$ are classified at high confidence regardless of the noise realisation.
By contrast, the five-label network correctly classifies spectra independently of the noise when $f_{R0}> 1.2 \times 10^{-7} $.
Given the limited performance of the generic two-label $\Lambda$CDM vs non-$\Lambda$CDM network and the enhanced performance of the specialised $\Lambda$CDM vs $f(R)$ network we conclude that training a two-label network is principally beneficial when trained between well-defined physical models.
Tighter constraints on model parameters can also be attained with such a specialised network.
\begin{figure*}
\centering
\resizebox{0.5\textwidth}{!}{
\includegraphics{plots/constraints/fRspecialist_constraint.png}
}
\caption{We compare how the ability of both the five-label network and a specialist $f(R)$ network to correctly classify $f(R)$ spectra varies with the modification strength $f_{R0}$ when averaged over noise realisations.
%
It is clear that the specialist network outperforms the five-label network, with $f(R)$ spectra with $f_{R0} > 8 \times 10^{-8}$ being correctly classified largely independently of the noise realisation.
}
\label{fig:noise_avg_comparison}
\end{figure*}
\section{Outlook}\label{sec:outlook}
The potential of BNNs to provide insights into whether upcoming cosmological datasets contain signatures of physics beyond $\Lambda$CDM motivates further exploration.
For example, BNNs could be used to identify high priority areas in the theory space by selecting the most likely known theory or motivating the need for further model development, before performing parameter estimation and Bayesian model selection with standard techniques that require specific benchmarks from which deviations can be detected.
Indeed, our analysis of how the five-label BNN classified an example generated from the growth-index $\gamma$ in Fig.~\ref{fig:gamma_5label} demonstrates the advantages of using data from specific models over more generic parameterisations. %
With enough training examples from enough $\Lambda$CDM and non-$\Lambda$CDM power spectra generated from a larger set of model classes considered in this work, one may envision a sequence of pre-trained specialist networks such that the first is trained on as many deviations as possible from $\Lambda$CDM, with the latter networks trained on smaller subsets of the total number of classes.
When an unknown spectrum is passed to the first network it would determine which of the subsequently more specialised networks to pass the spectrum onto.
As the specialist networks are better at recognising the specific imprints of the models they are trained on, they would further classify the spectra until it falls into either a single class or is not confidently classified.
If a single model is indeed preferred following such an analysis, one could then proceed to constrain the model parameters, for example with a traditional MCMC analysis.
An additional advantage is that, once the network has been trained, one has a tool to rapidly indicate the presence of new physics in contrast to the many hours it would take to obtain constraints with MCMC which must be run on a theory-by-theory basis.
Importantly however, an MCMC possesses a well defined notion of confidence such that, for a given dataset and a given parameterisation it will converge to a unique set of confidence intervals for each parameter if allowed to run for a sufficient amount of time. Moreover, it allows the computation of quantities such as the Bayesian evidence, that have a well-defined interpretation in terms of model selection and a solid statistical ground, albeit being non-trivial to compute in practice \cite{Heavens:2017afc}.
As we have seen, defining a classification confidence from the output of a BNN can prove challenging as it is by no means trivial to account for the uncertainty arising in the training process, the noise in the data or the chosen network architecture. In particular, the notion of confidence introduced in this paper ensures that the resulting classification is not ``over-confident'' by encoding the uncertainty due to the noise.
However, we stress that this quantity is not directly comparable to the Bayesian evidence or other goodness-of-fit tests. If and how these notions are comparable remains an open question (see App.\ref{app:MCMC}).
In conclusion, performing a fair comparison between the two methods is not a straightforward endeavour and a more thorough analysis of their relative strengths and weaknesses in performing cosmological analyses will be a subject of future work.
For now, we see the use of our BNN as a supplementary and precursory tool to MCMC analyses that accelerates the search for new fundamental physics.
While we have restricted ourselves to the matter power spectrum in this work, it is not a directly observable quantity.
An additional study would therefore be to train BNNs on mock galaxy clustering and weak lensing data for a range of different theories to determine how capable BNNs are in detecting deviations from $\Lambda$CDM directly from observational data.
An interesting study in this context was performed in Ref.~\cite{Peel:2018aei}, using a convolutional neural network trained on simulated convergence maps.
Moreover, while we have restricted to a selected number of popular extensions to $\Lambda$CDM, this process is generically applicable to any non-standard theory for which it is possible to rapidly generate accurate power spectra.
Of particular interest would be to study the capability of BNNs to pick up signatures in the power spectrum from the presence of massive neutrinos, baryonic effects as well as modifications arising from Horndeski scalar-tensor theory.
In the case of massive neutrinos, training a specialist massive neutrino network in a similar manner to that performed for $f(R)$ gravity in Fig.~\ref{fig:noise_avg_comparison} could yield an estimate on how capable BNN's could be in indicating the presence of a non-vanishing neutrino mass. This could be achieved with the latest version of {\tt ReACT} \cite{Bose:2021mkz}.
With the ever-growing ability to model cosmological observables for a multitude of extensions to $\Lambda$CDM, in Sec.~\ref{sec:specialist} we discussed the possibility of training a hierarchy of increasingly more specialist networks to obtain a more confident classification for a spectrum belonging to an unknown class.
Note that one could also fine-tune an $N$-label BNN to distinguish between sub-classes by incorporating an additional layer on top of a previously trained $N$-label network and re-train on a smaller dataset containing the new labels.
Although training a two-label network from scratch takes a few hours on a GPU, we have included the option in $\tt{BaCoN}$ to fine-tune on new data in the event one wishes to adapt the BNN to classify new theories with a limited training set size.
With the capability of BNNs to extract the particular features in the data that were important in the resulting classification, it may also be possible to provide information on the length scales or redshift bins which should be probed to detect signatures of a particular theory.
One can then train a BNN on spectra from specific redshift or scale bins.
This may be especially useful for studying models such as $w$CDM whose signatures can be noise-dominated at large scales and low redshift.
It is also important to note that in this paper we have restricted to a $k$ range of $\left(0.01 - 2.5\right)h \mathrm{Mpc}^{-1}$.
We expect that the capability of BNNs to accurately classify models such as $f(R)$ will only increase with improvements in the ability to rapidly and accurately model power spectra at higher values of $k$.
Further avenues of exploration include studying the potential benefits of different choices of network architecture and hyperparameters.
It is also of interest to examine different methods of constructing probability distributions to account for various sources of uncertainty in the output of the BNN.
In this paper we have focused on the application of BNNs to a classification problem.
However, the question remains of how capable Neural Networks are in obtaining cosmological parameter constraints from unknown spectra in comparison with more traditional approaches such as MCMC.
Finally, although we have trained the network on data which lies within the bounds of Euclid errors, it is important to determine how effective BNNs could be in detecting new physics from other surveys such as LSST \cite{Ivezic:2008fe} or DES \cite{Troxel:2017xyo} as well as to investigate how the results vary with different choices of systematic error.
\section{Conclusions}
\label{sec:conclusions}
Over the coming years many new cosmological surveys will provide vast datasets which will determine whether $\Lambda$CDM remains concordance cosmology.
In this paper we have studied the ability of Bayesian Neural Networks to determine if a matter power spectrum is representative of $\Lambda$CDM or not.
By constructing a mapping from the output of a BNN to a well-defined probability distribution we were able to define a classification confidence for an individual spectrum that considers the uncertainty from the noise in the data, variations in the training procedure, the modelling uncertainty of the BNN and choice of hyperparameters.
We found that a five-label network trained to classify between $\Lambda$CDM, $f(R)$ gravity, DGP gravity, $w$CDM and a ``random'' class provided more reliable predictions than a two-label network trained to distinguish simply between $\Lambda$CDM and non-$\Lambda$CDM.
While generally being less sensitive to variations in the noise distribution, it can also determine whether a power spectrum does not belong to any class included in the training set.
Since the selection of the correct model is crucial when performing conventional statistical analyses such as with MCMCs, this ability could prove beneficial in indicating prospective models to consider.
However, the network used in this work is currently limited to classification tasks while the notion of model selection on firm statistical grounds in the context of BNNs remains an open problem.
Nevertheless, we found that when averaged over noise realisations the five-label BNN was able to recognise spectra as not being $\Lambda$CDM down to values of $f_{R0} \lesssim 10^{-7}$, $\Omega_{rc} \lesssim 10^{-2} $, $-1.05 \lesssim w_0 \lesssim 0.95 $, $-0.2 \lesssim w_a \lesssim 0.2 $, $0.52 \lesssim \gamma \lesssim 0.59 $, all of which are comparable with current forecasts, as discussed in Sec. \ref{sec:experiments}.
Specialist networks trained on specific subsets of the classes in the training set have the potential to improve such bounds even further.
We conclude that BNNs may provide a powerful new means to search for hints of new physics in cosmological datasets.
In particular, we anticipate they will serve as a powerful ``filter'', allowing us to narrow down the theory space before moving on to constrain model parameters with MCMCs while perhaps even signalling the presence of new physics that does not belong to any known model.
\section*{Acknowledgements}
We thank Tom Charnock for useful discussions. We thank Gregory Horndeski for suggesting and allowing the use of \emph{Blustery Mountain Road on an Autumn Day} in this work.
J.K., B.B., and L.L.~acknowledge the support by a Swiss National Science Foundation
(SNSF) Professorship grant (No.~170547).
The work of M.M.~is supported by the SwissMap National Center for Competence in Research.
BNN development was conducted on Google Colab.
Please contact the authors for access to research
materials.
\section*{Data and code availability}
Alongside this paper we publish the publicly available code Bayesian Cosmological Network ($\tt{BaCoN}$) which can be accessed at the github repository \url{https://github.com/Mik3M4n/BaCoN} with the training and test data available at \url{https://doi.org/10.5281/zenodo.4309918}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,170 |
\section{Introduction}
In the study of elliptic curves over number fields, the concept of modularity plays a central role. If $E$ is an elliptic curve over $\Q$ of conductor $N$, the modularity theorem \cite{bcdt,wil,tawil} states that there exists a non-trivial map of algebraic curves $X_0(N)\to E$, called a \emph{modular parametrization}, where $X_0(N)$ is the compact modular curve for $\Gamma_0(N)$. This fact has several important consequences, one of which is the fact that the $L$-function of $E$ coincides with the $L$-function of a weight $2$ newform of level $\Gamma_0(N)$. This in turn implies for example that the $L$-function has an analytic continuation to $\C$.
It is natural to ask for a generalization of this fact to elliptic curves over $\OQ$. Shimura proved \cite{shi2} that elliptic curves over $\OQ$ with complex multiplication (CM) admit a modular parametrization from the compact modular curve $X_1(N)$, for an appropriate positive integer~$N$. For curves without CM, the situation was more complicated. In 1992, Ribet \cite{rib1} proved that under Serre's modularity conjecture, an elliptic curve $E/\OQ$ without CM admits a modular parametrization from $X_1(N)$ if and only if $E$ is a $\Q$-curve, i.e.\ it is $\OQ$-isogenous to all of its Galois conjugates. The subsequent proof, in 2009, of the aforementioned conjecture by Khare and Wintenberger \cite{kwin} completed the characterization of this class of elliptic curves. Note that since there exists a natural map $X_1(N)\to X_0(N)$, the modularity theorem already shows that all elliptic curves over $\Q$ belong to this class. From now on, all $\Q$-curves will be implicitly assumed to be without CM.
Contrary to what happens for elliptic curves over $\Q$, the $L$-function $L(E/K,s)$ of a $\Q$-curve $E$ over a number field~$K\ne\Q$ is never the $L$-function of a newform. However, $L(E/K,s)$ may be a \emph{product} of $L$-functions of newforms. In fact, the proof of Ribet's theorem \cite{rib1} implies that abelian varieties of $\GL_2$-type are isogenous to products of abelian varieties attached to newforms of weight 2. For example, if $E$ is a $\Q$-curve over a quadratic field $K$ that is $K$-isogenous to its Galois conjugate, then its restriction of scalars to~$\Q$ is an abelian surface of $\GL_2$-type. Since $E$ and its restriction of scalars have the same $L$-function \cite{mil1}, it follows that $L(E/K,s)$ is a product of two $L$-functions of newforms of weight 2. This motivates the following definition \cite{guique}: a $\Q$-curve is said to be \emph{strongly modular} if its $L$-function is a product of $L$-functions of newforms of weight~2.
Guitart and Quer \cite{guique2,guique} gave necessary and sufficient conditions for a $\Q$-curve (and, more in general, for a building block) over a number field $K$ to be strongly modular: this happens if and only if $K/\Q$ is an abelian extension, $E$ is \emph{completely defined} over $K$ (i.e.\ all isogenies between conjugates of $E$ are defined over $K$) and the 2-cocycle attached to $E$ (cf.\ section \ref{strong_modularity}) is symmetric. It is easy to deduce from the results of \cite{gola}, \cite{gogui} and \cite{mil1} that every $\Q$-curve is geometrically isogenous to a strongly modular one. In this paper, we address the following question: what are necessary and sufficient conditions for a $\Q$-curve to admit a strongly modular model over some number field? Our main result is the following.
\begin{theorem}\label{main}
A $\Q$-curve admits a strongly modular model if and only if it has a model completely defined over an abelian number field $K$.
\end{theorem}
The paper is structured as follows. In section \ref{group_cohomology} we review some basic facts about group cohomology and group extensions that will be needed later. In section \ref{strong_modularity} we recall the construction of two invariants attached to a $\Q$-curve $E$ over a Galois extension $K/\Q$ with Galois group $G$. Both invariants depend only on the $K$-isogeny class of $E$. The first one is a 1-cocycle for $G$ with values in $K^{\times}/{(K^{\times})}^2$ yielding information about the smallest field over which the curve is completely defined, which we call the \emph{minimal field of complete definition} (Definition \ref{minimal_field_def}). The second one is a 2-cocycle for $G$ with values in $(\en(E)\otimes\Q)^{\times}\simeq \Q^{\times}$ carrying information about the strong modularity of $E$ and the field of definition of $E$ up to isogeny (Theorem \ref{descent} and Proposition \ref{inflation}). Section \ref{strong_mod_iso} is dedicated to the proof of our main theorem. The proof relies on two preliminary results. The first one is \cite[Lemma 6.1]{guique}, which characterizes strongly modular twists $E^{\gamma}$ of a $\Q$-curve $E$ over a number field~$K$ in terms of the arithmetic of $K(\sqrt{\lambda})$. The second result is that the 2-torsion of the Brauer group of $\Q$ consists of inflations of symmetric 2-cocycles for $\gal(\QA/\Q)$ with values in $\{\pm1\}$. This essentially allows us to ``twist'' the 2-cocycle of a $\Q$-curve over an abelian number field into a symmetric one. Finally, in section \ref{section_twist} we study in detail an instance of the problem above: Theorem \ref{main} is not effective in general, but what about $\Q$-curves $E$ over a quadratic field $K$? We prove that every such curve has a model completely defined over a biquadratic, and hence abelian, number field $L$ containing $K$; it therefore admits a strongly modular model. We construct explicitly all strongly modular models of~$E$ over~$L$ (Theorem~\ref{primQcurves}), the existence of which is regulated only by the arithmetic of~$L$. We explain how to determine those models that descend to subfields of~$L$, up to isogeny. As a corollary, we show how to construct all strongly modular twists of~$E$ over~$K$. We end the section with several examples exhibiting different behaviours.
\section*{Notation and conventions}
When $A$ is an abelian variety over a field $K$ and $F$ is an extension of $K$, the $\Q$-algebra of the $F$-endomorphisms of $A$ is denoted by $\en_F^0(A)$.
If $G$ and $A$ are abelian groups, we write $\ext^1(G,A)$ for the group of abelian extensions of $G$ by~$A$.
If $F/K$ is a Galois extension of fields and $A$ is a $\gal(F/K)$-module, we denote by $\H^i(F/K,A)$ the $i$-th cohomology group of $A$ with coefficients in $\gal(F/K)$. Analogously, we denote by $\rZ^i(F/K,A)$ the group of $i$-cocycles. If $c\in \rZ^i(F/K,A)$, its cohomology class is denoted by $[c]$. If $G$ is a group and $A$ is a $G$-module, we denote the action of~$G$ on~$A$ using left superscripts: for $\sigma \in G$ and $a\in A$, the image of~$a$ under~$\sigma$ is written as $\s a$. We denote by $A^G$ the submodule of $G$-invariants.
All profinite groups that we mention throughout the paper are endowed with their profinite topology; in particular, finite groups are discrete.
We fix an algebraic closure $\OQ$ of $\Q$, and write $\QA$ for the maximal abelian extension of $\Q$ inside $\OQ$. The absolute Galois group of $\Q$ is denoted by $\GQ$ and the Galois group of $\QA$ over $\Q$ is denoted by $\GQA$.
For a unitary ring $R$, we denote by $R^{\times}$ the group of units.
\section{Group cohomology, group extensions and embedding problems}\label{group_cohomology}
In this section we collect some standard facts and definitions from the theory of group cohomology that we will use later in the paper. For a complete treatment of profinite group cohomology, see for example \cite{neu} or \cite{serre}.
Let $G$ be a profinite group, let $N$ be a normal closed subgroup of~$G$, and let $A$ be a $G$-module. For every $i\geq 1$, the \emph{restriction map} $\res\colon \H^i(G,A)\to \H^i(N,A)$ is the natural map induced by the inclusion $N\subseteq G$, and the \emph{inflation map} $\infl\colon \H^i(G/N,A^N)\to \H^i(G,A)$ is the natural map induced by the projection $G\to G/N$.
The group $\H^1(N,A)$ is endowed with an action of $G/N$ defined in the following way: for $g\in G$ and $[c]\in \H^1(N,A)$, let ${}^g[c]$ be the cohomology class represented by the cocycle $h\mapsto {}^{g\!}c(g^{-1}hg)$ for all $h\in N$. It is easy to check that $N$ acts as the identity, so that the action factors through the quotient $G/N$.
\begin{theorem}[{{\cite[Proposition I.1.6.5]{neu}}}]\label{trans}
There is a natural map
$$\trg\colon \H^1(N,A)^{G/N}\longrightarrow \H^2(G/N,A^N)$$
fitting into an exact sequence
$$\begin{aligned}
0 & \longrightarrow \H^1(G/N,A^N) \stackrel{\infl}{\longrightarrow} \H^1(G,A) \stackrel{\res}{\longrightarrow} \H^1(N,A)^{G/N} \stackrel{\trg}{\longrightarrow} \\
& \stackrel{\trg}{\longrightarrow} \H^2(G/N,A^N) \stackrel{\infl}{\longrightarrow} \H^2(G,A).
\end{aligned}$$
\end{theorem}
The map $\trg$ in Theorem~\ref{trans} is called the \emph{transgression map}, and the exact sequence is called the \emph{inflation-restriction sequence}.
\begin{definition}
Let $G$ be a profinite abelian group and let $A$ be an abelian group regarded as a $G$-module with trivial action.
We say that a cocycle $c\in \rZ^2(G,A)$ is \emph{symmetric} if $c(\sigma,\tau)=c(\tau,\sigma)$ for all $\sigma,\tau\in G$.
\end{definition}
Note that the property of $c$ being symmetric only depends on the cohomology class of~$c$, because coboundaries are symmetric by the commutativity of $G$. Moreover, the product of two symmetric cocycles is clearly a symmetric cocycle. Thus the cohomology classes in $\H^2(G,A)$ represented by a symmetric cocycle form a subgroup of $\H^2(G,A)$, which we denote by $\H^2_\sym(G,A)$.
\begin{lemma}\label{extsym}
Let $G$ be a finite abelian group, and let $A$ be an abelian group equipped with the trivial $G$-action. There is a canonical isomorphism of abelian groups
$$\ext^1(G,A)\isom \H^2_\sym(G,A).$$
\end{lemma}
\begin{proof}
There is a well known isomorphism between $\H^2(G,A)$ and the group of central extensions of $G$ by~$A$; see for example \cite[Theorem 1.2.4]{neu}. Looking at the definition of this isomorphism, one sees that a cocycle is symmetric if and only if in the associated group extension $0\to A\to B\to G\to 1$ the group $B$ is abelian.
\end{proof}
\begin{lemma}\label{extpgr}
Let $G$ be a finite abelian group, let $A$ be an abelian group equipped with the trivial $G$-action, and let $p$ be a prime such that $pG=0$. Then there is a (non-canonical) isomorphism
$$\ext^1(G,A)\simeq\Hom(G,A/pA).$$
\end{lemma}
\begin{proof}
Using the properties of $\Hom$ and $\ext$ (see for example \cite{wei}), we reduce to the case $G=\Z/p\Z$. From the long exact sequence obtained by applying the $\ext$ functor to
$$
0 \longrightarrow \Z \stackrel{p}{\longrightarrow} \Z \longrightarrow \Z/p\Z \longrightarrow 0
$$
and the fact that $\ext^1(\Z,A)=0$, we get an isomorphism $\ext^1(\Z/p\Z,A)\simeq A/pA$. The claim follows using the canonical isomorphism $\Hom(\Z/p\Z,A/pA)\simeq A/pA$.
\end{proof}
\begin{theorem}[K\"unneth formula \cite{jan}]\label{prodcoh}
Let $G_1$, $G_2$ be profinite abelian groups acting trivially on a discrete abelian group $A$. Then there is a canonical isomorphism
$$\H^2(G_1\times G_2,A)\simeq \H^2(G_1,A)\oplus \H^2(G_2,A)\oplus \Hom(G_1\otimes G_2,A).$$
\end{theorem}
\begin{definition}\label{embedding}
Let $L/K$ be a Galois extension of fields with Galois group~$G$. Consider an extension
\begin{equation}\label{embprob}
1\longrightarrow H\longrightarrow \widetilde{G}\stackrel{\pi}{\longrightarrow} G\longrightarrow 1
\end{equation}
of profinite groups. A \emph{solution to the embedding problem} relative to the extension $L/K$ and the group extension \eqref{embprob} consists of a Galois extension $M/K$ with $L\subseteq M$ and an isomorphism $\iota\colon\gal(M/K)\isom \widetilde{G}$ such that $\pi\circ\iota$ equals the canonical map $\gal(M/K)\to G$.
\end{definition}
It is not hard to see that the solvability of the embedding problem given by the extension \eqref{embprob} depends only on the equivalence class of the extension.
To conclude this section, we recall a result from \cite{kim} for the case $p=2$ that will be useful later.
\begin{lemma}[{\cite[pp.\ 826--827]{kim}}]\label{trivialclass}
Let $L/K$ be a Galois extension of fields of characteristic different from~$2$ with Galois group~$G$. Let $1\to\{\pm1\}\to\widetilde{G}\to G\to 1$ be an extension of $G$ by $\{\pm1\}$, and let $c\in \H^2(G,\{\pm 1\})$ be the corresponding cohomology class. Then the embedding problem relative to $L/K$ and the above extension has a solution if and only if $c$ is in the kernel of the natural map
$$\varphi_L\colon \H^2(G,\{\pm1\})\to\H^2(G,L^{\times}).$$
\end{lemma}
\section{Strongly modular elliptic curves}\label{strong_modularity}
The goal of this section is to introduce the main objects of the paper, namely $\Q$-curves and strongly modular $\Q$-curves, and recall some of their basic properties. For more details, see \cite{golaque}, \cite{guique}, \cite{que} and \cite{quer2}.
\begin{definition}
Let $K$ be a Galois extension of $\Q$ inside $\OQ$. An elliptic curve $E/K$ is called a \emph{$\Q$-curve} if for every $\sigma \in\gal(K/\Q)$ there exists a $\OQ$-isogeny $\mu_{\sigma}\colon \s E\to E$. We say that $E$ is \emph{completely defined} over $K$ if in addition all $\OQ$-isogenies between the $\s E$ are defined over $K$.
\end{definition}
From now on, all our $\Q$-curves will be without CM.
Let $K$ be a Galois extension of $\Q$ and let $E$ be a $\Q$-curve over $K$. We recall the definition of two cohomology classes
$$
[\lambda]\in\H^1(K/\Q,K^{\times}/(K^{\times})^2),\quad
[\xi_K(E)]\in\H^2(K/\Q,\Q^{\times})
$$
attached to $E$ (the second under the assumption that $E$ is completely defined over~$K$) that encode arithmetic properties of the curve; see also \cite{pyl} or \cite{que}.
Let $G\coloneqq \gal(K/\Q)$. For every $\sigma\in G$ we choose a $\OQ$-isogeny $\mu_{\sigma}\colon \s E\to E$.
By the argument described in \cite[p.~4]{que}, if $E$ is given by an equation of the form $y^2=x^3+Ax+B$ with $A,B\in K$, then for all $\sigma\in G$ and all isogenies $\mu_{\sigma}\colon \s E\to E$, we can write
$$\mu_{\sigma}(x,y)=\biggl(F(x),\frac{1}{\lambda_{\sigma}}yF'(x)\biggr)$$
for some $F(x)\in K(x)$ and $\lambda_{\sigma}\in \OQ^{\times}$ such that $\lambda_{\sigma}^2\in K^{\times}$.
Because $E$ has no CM, for all $\sigma,\tau\in G$ there exists $m(\sigma,\tau)\in \Q^{\times}$ satisfying
\begin{equation}\label{lambda}
\lambda_{\sigma}\s\lambda_{\tau}=m(\sigma,\tau)\lambda_{\sigma\tau}
\end{equation}
This shows that the map
$$
\begin{aligned}
\lambda\colon G&\to K^{\times}/{(K^{\times})}^2\\
\sigma&\mapsto\lambda_{\sigma}^2
\end{aligned}
$$
is a 1-cocycle for the natural Galois action of $G$ on $K^{\times}/{(K^{\times})}^2$. A calculation shows that the cohomology class $[\lambda]$ of~$\lambda$ depends only on the $K$-isogeny class of $E$.
\begin{definition}\label{minimal_field_def}
We call the field $K(\lambda_{\sigma}\colon \sigma\in G)$ the \emph{minimal field of complete definition} for $E/K$.
\end{definition}
\begin{proposition}\label{minimal_field}
Let $L$ be the minimal field of complete definition for $E$. Then:
\begin{enumerate}[i)]
\item $L$ is Galois over $\Q$;
\item $E_L$ is completely defined over $L$;
\item $L/K$ is a multiquadratic extension;
\item if $M$ is a Galois extension of $\Q$ containing $K$ and $E_M$ is completely defined over $M$, then $L\subseteq M$.
\end{enumerate}
\end{proposition}
\begin{proof}
We observe that if $K/F$ is any Galois extension of subfields of $\OQ$ and $\lambda_1,\ldots,\lambda_n\in \OQ$, then the normal closure of $K(\lambda_1,\ldots,\lambda_n)$ over~$F$ equals $K(\s\lambda_i\colon i\in\{1,\ldots,n\},\sigma \in \gal(\overline{\Q}/F))$.
Point i) follows from this observation together with the identity \eqref{lambda}. Points ii), iii) and iv) are clear by construction.
\end{proof}
To construct the other cohomology class attached to $E$, suppose that $E$ is completely defined over $K$, so that every $\mu_{\sigma}$ is itself defined over $K$. The map
$$
\begin{aligned}
\xi_K(E)\colon G\times G &\longrightarrow (\en_K^0(E))^{\times}\simeq\Q^{\times}\\
(\sigma,\tau) &\longmapsto \mu_{\sigma}\s\mu_{\tau}\mu_{\sigma\tau}^{-1}
\end{aligned}
$$
is a 2-cocycle for the trivial action of $G$ on $\Q^{\times}$. With a small abuse of notation, we will often talk about ``the'' 2-cocycle attached to $E$, without mentioning explicitly the system of isogenies giving rise to it. A direct calculation shows that the cohomology class $[\xi_K(E)]$ of $\xi_K(E)$ depends only on the $K$-isogeny class of $E$. When $K=\OQ$, the 2-cocycle attached to $E$ will be denoted simply by $\xi(E)$.
Note that the image of $[\xi_K(E)]$ under the inflation map $\H^2(G,\Q^{\times})\to \H^2(\GQ,\Q^{\times})$ equals $[\xi(E)]$.
The decomposition $\Q^{\times}\simeq \Q^{\times}_+\times\{\pm 1\}$, where $\Q^{\times}_+$ is the multiplicative group of positive rational numbers, yields a decomposition of $\xi_K$ in a 2-cocycle $\xi_K^{\deg}\in \rZ^2(G,\Q^{\times}_+)$ and a 2-cocycle $\xi_K^{\pm}\in \rZ^2(G,\{\pm1\})$. Similarly, the isomorphism
$$\H^2(G,\Q^{\times})\simeq \H^2(G,\Q^{\times}_+)\times \H^2(G,\{\pm 1\})$$
yields a decomposition of $[\xi_K]$ into a \emph{degree component} $[\xi_K^{\deg}]\in \H^2(G,\Q^{\times}_+)$ and a \emph{sign component} $[\xi_K^{\pm}]\in \H^2(G,\{\pm 1\})$.
\begin{definition}
A $\Q$-curve $E$ over a number field $K$ is \emph{strongly modular} if its $L$-function $L(E/K,s)$ can be written as a product of $L$-functions attached to holomorphic newforms of weight~$2$ for congruence subgroups of the form $\Gamma_1(N)$.
\end{definition}
The newforms in such a product are unique up to ordering \cite[Proposition 3.4]{brufer}. The modularity theorem \cite{bcdt} implies that all elliptic curves over $\Q$ are strongly modular; this is not true for general $\Q$-curves.
\begin{theorem}[{{\cite[Theorem 2.3]{guique2}}}]\label{strongmod}
Let $E$ be an elliptic curve without CM over a number field $K$. Then $E$ is strongly modular over $K$ if and only if the following three conditions hold:
\begin{enumerate}[i)]
\item $K$ is abelian over $\Q$;
\item $E$ is completely defined over $K$;
\item the $2$-cocycle $\xi_K$ attached to $E$ is symmetric, i.e.\ $[\xi_K]\in \H^2_\sym(\gal(K/\Q),\Q^{\times})$.
\end{enumerate}
\end{theorem}
\begin{remark}\label{rem1}
A cohomology class in $\H^2(K/\Q,\Q^{\times})$ is symmetric if and only if both its sign component and its degree component are symmetric. The degree component of $[\xi_K]$ is always symmetric, since for $\sigma,\tau\in \gal(K/\Q)$, the fact that $\xi_K(\sigma,\tau)=\mu_{\sigma}\s\mu_{\tau}\mu_{\sigma\tau}^{-1}$ implies $\xi_K(\sigma,\tau)^2=\deg (\mu_{\sigma})\cdot \deg(\mu_{\tau})\cdot\deg(\mu_{\sigma\tau})^{-1}$, and since $K/\Q$ is an abelian extension, this implies $\xi_K(\sigma,\tau)^2=\xi_K(\tau,\sigma)^2$. Therefore condition~iii) above is equivalent to
\begin{enumerate}
\item[iii$'$)] $[\xi_K^{\pm}]\in \H^2_\sym(K/\Q,\{\pm 1\})$.
\end{enumerate}
\end{remark}
\section{Strongly modular \texorpdfstring{$\Q$}{}-curves up to isomorphism}\label{strong_mod_iso}
Every $\Q$-curve $E$ is geometrically isogenous to a strongly modular one. In fact, it is proved in \cite{gola} that there exists a newform $f$ such that the attached abelian variety $A_f$ admits a non-trivial morphism $(A_f)_{\OQ}\to E$. If $L$ is the splitting field of $f$, it is also proved in \cite{gola} that $A_f$ is isogenous over $L$ to $E'^n$ for some positive integer~$n$, where $E'$ is a $\Q$-curve over $L$. Note that $E$ is geometrically isogenous to $E'$ by the uniqueness of the decomposition up to isogeny. By \cite[Proposition~2]{gogui}, the restriction of scalars $\res_{L/\Q}(E')$ is isogenous to a product of abelian varieties of the form $A_g$ for some newform $g$. It follows that $E'/L$ is strongly modular, because the $L$-function of $E'/L$ coincides with the $L$-function of $\res_{L/\Q}(E')$; see \cite{mil1}.
Therefore, it is natural to ask how strong modularity behaves with respect to geometric \emph{isomorphisms}, rather than isogenies. It turns out that this is a more rigid property. In fact, we will prove the following theorem.
\begin{theorem}\label{isothm}
A $\Q$-curve $E/\OQ$ has a strongly modular model over a number field if and only if there exist an abelian number field $K$ and a model of $E$ completely defined over $K$.
\end{theorem}
Compared to Theorem \ref{strongmod}, the above result states that if one considers strong modularity up to geometric isomorphism, then the symmetry of the cocycle attached to the curve is redundant: if a $\Q$-curve is completely defined over an abelian number field, then there exists an appropriate model of the curve whose 2-cocycle is symmetric.
Let us rephrase the theorem above in cohomological terms. Let $E$ be a $\Q$-curve over a Galois number field $K$ with Galois group $G$, and let $\lambda\in \rZ^1(G,K^{\times}/{(K^{\times})}^2)$ be its associated 1-cocycle. A simple computation shows that for all $\gamma\in K^{\times}$, the cocycle attached to the twisted curve $E^{\gamma}$ is given by $\lambda^{\gamma}(\sigma)\coloneqq\lambda(\sigma)\cdot\frac{\s\gamma}{\gamma}$ for every $\sigma\in G$, and is thus cohomologous to $\lambda$. Now let $L$ be an extension of $K$ that is abelian over $\Q$, and let
$$\psi_L\colon \H^1(K/\Q,K^{\times}/{(K^{\times})}^2)\longrightarrow \H^1(L/\Q,L^{\times}/{(L^{\times})}^2)$$
be the composition of the inflation map with the map induced by the natural morphism $K^{\times}/{(K^{\times})}^2\to L^{\times}/{(L^{\times})}^2$. Proposition \ref{minimal_field} implies that $E$ has a model completely defined over $L$ if and only if $\psi_L([\lambda])$ is trivial. It follows that Theorem \ref{isothm} can be stated in the following way: if $E$ is a $\Q$-curve over $K=\Q(j(E))$, then $E$ has a strongly modular model if and only if $K$ is abelian and $[\lambda]$ belongs to the kernel of the map
$$\H^1(K/\Q,K^{\times}/{(K^{\times})}^2)\longrightarrow \H^1(\GQA,{\QA}^{\times}/{({\QA}^{\times})}^2).$$
\begin{example}
Consider the elliptic curve $E'\colon y^2=x^3+x+1$ over $\Q$. Let $K$ be the non-Galois number field $\Q(\alpha)$, where $\alpha$ is a root of $x^3+x+1$. The base-changed curve $E'_K$ has a non-trivial, rational 2-torsion point, namely $P=(\alpha,0)$. Now let $\varphi$ be the isogeny with kernel $\{O,P\}$ and let $E\coloneqq E'_K/\ker\varphi$. A Weierstrass equation for $E$ is
$$E\colon y^2=x^3 - (4+15\alpha^2)x + 22+14\alpha.$$
One can check that
$$j(E)=\frac{9580464+51659856\alpha+72060192\alpha^2}{961}\notin \O_K,$$
so $\Q(j(E))=K$ and $E$ has no CM. Moreover, $E$ is a $\Q$-curve: if $L$ is the Galois closure of $K$, then $E_L$ is $L$-isogenous to all its Galois conjugates, since by construction all of them are $L$-isogenous to $E'_L$. Thus, $E$ is a $\Q$-curve that does not have a strongly modular model.
\end{example}
In order to prove Theorem \ref{isothm}, we need two preliminary results.
The first one is proved in \cite{guique} and characterizes the twists of a $\Q$-curve defined over a Galois number field $K$ that are strongly modular.
\begin{lemma}[{{\cite[Lemma 6.1]{guique}}}]\label{quadtw}
Let $E$ be a $\Q$-curve completely defined over a Galois number field $K$. Let $\gamma\in K^{\times}$, and let $E^{\gamma}$ be the twisted curve. Let $\xi_K$ and $\xi_{K}^{\gamma}$ be the $2$-cocycles attached to $E$ and $E^\gamma$, respectively. Then $E^{\gamma}$ is completely defined over $K$ if and only if the field $K(\sqrt{\gamma})$ is Galois over $\Q$. In this case, the cohomology classes $[\xi_K]$ and $[\xi_K^{\gamma}]$ in $\H^2(K/\Q,\Q^{\times})$ differ by the cohomology class in $\H^2(K/\Q,\{\pm 1\})$ attached to the exact sequence
$$1 \longrightarrow \gal(K(\sqrt{\gamma})/K)\simeq \{\pm 1\} \longrightarrow \gal(K(\sqrt{\gamma})/\Q)\longrightarrow \gal(K/\Q)\longrightarrow 1.$$
In particular, twisting by $\gamma$ affects only the sign component of $[\xi_K]$ and not the degree component.
\end{lemma}
The second preliminary result shows that the 2-torsion of the Brauer group of $\Q$ is generated by cocycles inflated from certain symmetric ones.
\begin{proposition}\label{symmetric_cocycles}
The following hold:
\begin{enumerate}[i)]
\item There is a canonical isomorphism $\displaystyle \bigoplus_{p\;\text{prime}}\{\pm1\} \isom \H^2_\sym(\GQA,\{\pm1\})$.
\item The inflation map $\infl\colon \H^2_\sym(\GQA,\{\pm1\})\longrightarrow \H^2(\GQ,\{\pm1\})=\Br(\Q)[2]$ is an isomorphism.
\end{enumerate}
\end{proposition}
\begin{proof}
To prove i), recall that $\GQA$ is canonically isomorphic to $\widehat{\Z}^{\times}$. By Lemma~\ref{extsym}, we have a canonical isomorphism
$$
\varinjlim_n\ext^1((\Z/n\Z)^{\times},\{\pm1\}) \isom \varinjlim_n\H^2_\sym((\Z/n\Z)^{\times},\{\pm1\}) = \H^2_\sym(\widehat{\Z}^{\times},\{\pm1\}).
$$
Furthermore, we can rewrite the left-hand side using the canonical isomorphism
$$
\bigoplus_{p\;\text{prime}} \varinjlim_r \ext^1((\Z/p^r\Z)^{\times},\{\pm1\}) \isom \varinjlim_n\ext^1((\Z/n\Z)^{\times},\{\pm1\}).
$$
It therefore suffices to prove that for every prime number~$p$, there exists a (unique) isomorphism
$$
\{\pm1\} \isom \varinjlim_r \ext^1((\Z/p^r\Z)^{\times},\{\pm1\}).
$$
For odd $p$, all groups $\ext^1((\Z/p^r\Z)^{\times},\{\pm1\})$, and also their direct limit, are canonically isomorphic to $\ext^1((\Z/p\Z)^{\times},\{\pm1\})$, which has order~$2$ because $(\Z/p\Z)^{\times}$ is cyclic of even order. For $p=2$, we recall that there are isomorphisms
$$(\Z/2^r\Z)^{\times}\isom(\Z/4\Z)^{\times}\times\Z/2^{r-2}\Z \quad\text{for all } r\ge2$$
that are compatible in such a way that we obtain an isomorphism
$$
\ext^1((\Z/4\Z)^{\times},\{\pm1\}) \times \varinjlim_r\ext^1(\Z/2^r\Z,\{\pm1\}) \isom \varinjlim_r\ext^1((\Z/2^r\Z)^{\times},\{\pm1\}).$$
The group $\ext^1((\Z/4\Z)^{\times},\{\pm1\})$ has order~$2$, and it is well known that for all $r\ge1$, the groups $\ext^1(\Z/2^r\Z,\{\pm1\})$ have order~$2$ and the maps $\ext^1(\Z/2^r\Z,\{\pm1\}) \to \ext^1(\Z/2^{r+1}\Z,\{\pm1\})$ are trivial. This implies the claim.
To prove ii), consider the exact sequence
$$1\longrightarrow \gal(\OQ/\QA)\longrightarrow \GQ\longrightarrow \GQA\longrightarrow 1.$$
The inflation-restriction sequences obtained from the two $\GQ$-modules $\{\pm1\}$ and $\OQ^{\times}$ (with the natural $\GQ$-action) fit into a commutative diagram
$$
\begin{xymatrix}{
\H^1(\gal(\OQ/\QA),\{\pm 1\})^{\GQA}\ar[r]^-{\trg} \ar[d] & \H^2(\GQA,\{\pm 1\}) \ar[r]^-{\infl}\ar[d]^{\varphi_{\QA}} & \Br(\Q)[2]\ar[d]\\
\H^1(\gal(\OQ/\QA),\OQ^{\times})^{\GQA}\ar[r]^-{\trg} & \H^2(\GQA,{\QA}^{\times}) \ar[r]^-{\infl} & \Br(\Q).
}\end{xymatrix}$$
By Hilbert's theorem 90, $\H^1(\gal(\OQ/\QA),\OQ^{\times})$ is trivial, so the map
$$\infl\colon \H^2(\GQA,{\QA}^{\times}) \longrightarrow \Br(\Q)$$
is injective. On the other hand, also the map $\Br(\Q)[2]\to \Br(\Q)$ is injective, and therefore we have
\begin{equation}\label{kernels}
\ker\varphi_{\QA}=\ker(\infl\colon \H^2(\GQA,\{\pm 1\})\to \Br(\Q)[2]).
\end{equation}
By Lemma \ref{trivialclass}, we have $\H^2_\sym(\GQA,\{\pm1\})\cap \ker\varphi_{\QA}=\{1\}$, since a non-trivial element in this intersection would correspond to a non-trivial extension of $\QA$ that is Galois and abelian over $\Q$. This shows that the map
$$\infl\colon \H^2_\sym(\GQA,\{\pm1\})\to \Br(\Q)[2]$$
is injective.
To prove surjectivity, we recall that for every place $v$ of $\Q$, we have a canonical homomorphism $\inv_v\colon\Br(\Q_v)\to \Q/\Z$, and that we have a canonical exact sequence
$$
0 \longrightarrow \Br(\Q) \longrightarrow \bigoplus_v\Br(\Q_v) \longrightarrow \Q/\Z \longrightarrow 0,
$$
where $v$ runs over all places of $\Q$; the first non-trivial map is the product of the restriction maps $\res_v\colon \Br(\Q)\to\Br(\Q_v)$, and the second one is the sum of the maps $\inv_v$.
Let $\ell$ be a prime and consider the element $\varepsilon_{\ell}\in \H^2_\sym(\GQA,\{\pm1\})$ that under the isomorphism from i) corresponds to the element $(t_p)_p\in\bigoplus_{p\;\text{prime}}\{\pm1\}$ defined by
$$t_p=\begin{cases}-1 & \mbox{if } p=\ell\\1 & \mbox{otherwise}.\end{cases}$$
We will show that $\infl(\varepsilon_{\ell})\in \Br(\Q)[2]$ is ramified precisely at $\ell$ and $\infty$, by looking at the images of $\infl(\varepsilon_\ell)$ under the restriction maps $\res_v$.
We write $\Q(\zeta_{\ell^\infty})$ for the field obtained by adjoining all roots of unity of $\ell$-power order to~$\Q$. Let $v$ be a finite place of~$\Q$ different from~$\ell$, and let $\Q_v^\nr$ be the maximal unramified extension of $\Q_v$. Embedding $\Q_v(\zeta_{\ell^{\infty}})$ into $\Q_v^\nr$, we obtain a commutative diagram
$$
\begin{xymatrix}{
\H^2_\sym(\Q(\zeta_{\ell^\infty})/\Q,\{\pm1\}) \ar[rr]^-{\infl}\ar[d]^{\res_v}&& \Br(\Q)\ar[d]^{\res_v}\cr \H^2_\sym(\Q_v(\zeta_{\ell^\infty})/\Q_v,\{\pm1\})\ar[r]^-{\infl} & \Br(\Q_v^\nr/\Q_v)\ar[r]^-\sim \ar[d]^{\mathpzc v_*}& \Br(\Q_v)\cr
& \H^2(\Q_v^\nr/\Q_v,\Z).
}\end{xymatrix}
$$
Here $\mathpzc v_*$ is induced by the valuation map $\mathpzc v\colon {\Q_v^\nr}^{\times}\to\Z$, and is an isomorphism because the Brauer group of a finite field is trivial \cite[\S X.7 and \S XII.3]{ser3}. Now the composed map $\mathpzc v_*\circ\infl$ vanishes because units have valuation~$0$, so the map $\infl$ in the middle row vanishes as well. Since $\varepsilon_\ell$ is inflated from $\H^2_\sym(\Q(\zeta_{\ell^\infty})/\Q,\{\pm1\})$, it follows that for all finite places $v$ of $\Q$ different from~$\ell$ we have $\res_v(\infl(\varepsilon_{\ell}))=0$ in $\Br(\Q_v)$. Because $\varepsilon_{\ell}\in\H^2_\sym(\GQA,\{\pm1\})$ is non-trivial and the map $\infl\colon \H^2_\sym(\GQA,\{\pm1\})\to \Br(\Q)[2]$ is injective as shown above, it follows that $\infl(\varepsilon_{\ell})\in \Br(\Q)[2]$ is ramified precisely at $\ell$ and~$\infty$, which is what we wanted to show.
\end{proof}
\begin{proof}[Proof of Theorem \ref{isothm}]
The first implication is clear from Theorem \ref{strongmod}.
Let us now prove the converse. Assume that $E$ has a model $E_0$ completely defined over an abelian number field $K$. Let $\xi_K=\xi_K(E_0)\in \rZ^2(K/\Q,\Q^{\times})$ be the 2-cocycle attached to $E_0$. For every extension $L/K$ that is Galois over $\Q$, we denote by $[\xi_L]$ the inflation of $[\xi_K]$ to $\H^2(L/\Q,\Q^{\times})$. For every Galois extension $M/\Q$, we denote by
$$\varphi_M\colon \H^2(M/\Q,\{\pm1\})\to \H^2(M/\Q,M^{\times})$$
the canonical map induced by the inclusion $\{\pm1\}\subseteq M^{\times}$.
We will show that there exists an abelian extension $L/\Q$, containing $K$, such that ${(E_0)}_L$ has a strongly modular twist over $L$. By Theorem \ref{strongmod}, Lemma \ref{quadtw} and Remark \ref{rem1}, this happens if and only if there exist an abelian number field $L$ containing $K$ and an element $\gamma\in L^{\times}$ such that:
\begin{itemize}
\item $L(\sqrt{\gamma})$ is Galois over $\Q$;
\item if $c\in \H^2(L/\Q,\{\pm 1\})$ is the cohomology class attached to the exact sequence
$$1\longrightarrow \{\pm 1\}\longrightarrow \gal(L(\sqrt{\gamma})/\Q)\longrightarrow \gal(L/\Q)\longrightarrow 1,$$
then the class $[\xi_L^{\pm}]\cdot c$ in $\H^2(L/\Q,\{\pm 1\})$ is symmetric.
\end{itemize}
By Lemma \ref{trivialclass}, these conditions are satisfied for a given $L$ if and only if there exists $c\in\ker\varphi_L$ such that $[\xi_L^{\pm}]\cdot c$ is symmetric.
Writing $\H^2(\GQA,\{\pm1\})$ and $\H^2_\sym(\GQA,\{\pm1\})$ as direct limits indexed by finite abelian extensions of~$\Q$ containing $K$, we see that there exist $L$ and $c$ as above if and only if there exists $c'\in\ker\varphi_{\QA}$ such that $[\xi_{\QA}^{\pm}]\cdot c'\in \H^2(\GQA,\{\pm1\})$ is symmetric.
By \eqref{kernels}, the kernel of $\varphi_{\QA}$ equals the kernel of $\infl\colon \H^2(\GQA,\{\pm1\})\to \Br(\Q)[2]$. Therefore, it remains to prove that there exists $c\in \ker(\infl\colon \H^2(\GQA,\{\pm 1\})\to \Br(\Q)[2])$ such that $[\xi_{\QA}^{\pm}]\cdot c$ is symmetric. This amounts to saying that the image of $[\xi_{\QA}^{\pm}]$ in $\Br(\Q)[2]$ belongs to the image of $\H^2_\sym(\GQA,\{\pm1\})$. By Proposition~\ref{symmetric_cocycles}, this is always the case, and the proof is complete.
\end{proof}
\section{Twists of quadratic \texorpdfstring{$\Q$}{}-curves}\label{section_twist}
Theorem~\ref{isothm} is not effective: it only gives necessary and sufficient conditions for the existence of strongly modular twists, without providing an actual construction of them. In this section, for a $\Q$-curve over a quadratic field $K$, we will make the existence of strongly modular twists over the minimal field of complete definition $L$ effective in terms of the arithmetic of $L$. For certain purposes, such as studying $L$-functions of elliptic curves, it is useful to understand when the curve ``comes from a subfield'' up to isogeny. We will show how to distinguish twists coming from subfields of $L$; this distinction will allow us to characterize all strongly modular twists of $E$ over $K$ (Corollary \ref{corollary_twist}).
\begin{definition}
Let $E$ be a $\Q$-curve completely defined over a number field $L$. If there exist a subfield $K$ of~$L$ that is Galois over~$\Q$ and a $\Q$-curve $C$ completely defined over $K$ such that $E$ is $L$-isogenous to~$C_L$, we say that $E$ is \emph{inflated from $K$}. If no such $K$ and~$C$ exist, we say that $E$ is \emph{primitive}.
\end{definition}
The reason for the terminology ``inflated'' will be clarified by Proposition \ref{inflation}. We recall the following theorem.
\begin{theorem}[{{\cite[Theorem 8.2]{rib1}}}]\label{descent}
Let $L/K$ be a Galois extension of number fields, and let $E$ be a $\Q$-curve completely defined over $L$. Then the following are equivalent:
\begin{enumerate}[i)]
\item there exist isomorphisms $\mu_{\sigma}\colon \s E\to E$ of elliptic curves up to isogeny over $L$ such that
$$\mu_{\sigma}\s\mu_{\tau}=\mu_{\sigma\tau} \qquad \mbox{for all } \sigma,\tau\in \gal (L/K);$$
\item there exists an elliptic curve $C$ over~$K$ such that $E$ is $L$-isogenous to~$C_L$.
\end{enumerate}
\end{theorem}
Now let $K$ and~$L$ be Galois number fields with $K\subseteq L$, and consider the inflation map $\infl^K_L\colon \H^2(K/\Q,\Q^{\times})\to \H^2(L/\Q,\Q^{\times})$.
\begin{proposition}\label{inflation}
Let $E$ be a $\Q$-curve completely defined over $L$ and let $\xi_L(E)$ be its associated $2$-cocycle.
\begin{enumerate}[i)]
\item if $E$ is inflated from $K$, then $[\xi_L(E)]\in \im (\infl^K_L)$;
\item if $\gal(L/K)$ is contained in the center of $\gal(L/\Q)$ and $[\xi_L(E)]\in \im (\infl^K_L)$, then $E$ is inflated from $K$.
\end{enumerate}
\end{proposition}
\begin{proof}
First assume that $E$ is inflated from $K$. Let $C$ be a $\Q$-curve completely defined over $K$ such that $C_L$ is isogenous to~$E$. For all $\overline{\sigma}\in\gal(K/\Q)$, we fix a $K$-isogeny $\mu_{\overline{\sigma}}\colon {}^{\overline{\sigma}} C\to C$. Let $\xi_K(C)\in \rZ^2(K/\Q,\Q^{\times})$ be the 2-cocycle attached to~$C$ via the system of isogenies $\{\mu_{\overline{\sigma}}\}_{\overline{\sigma}}$, so that $\xi_K(C)(\overline{\sigma},\overline{\tau})=\mu_{\overline{\sigma}}{}^{\overline{\sigma}}\mu_{\overline{\tau}}\mu_{\overline{\sigma}\overline{\tau}}^{-1}$. Now for every $\sigma\in\gal(L/\Q)$, we set $\mu_{\sigma}=\mu_{\overline{\sigma}}\colon \s C_L\to C_L$, for $\overline{\sigma}$ the class of $\sigma$ in $\gal(K/\Q)$. Then the cocycle $\xi_L(C)$ corresponding to the system of isogenies $\{\mu_{\sigma}\}_{\sigma}$ represents the inflation of $[\xi_K(C)]$; since $C_L$ and $E$ are $L$-isogenous, we have $[\xi_L(C_L)]=[\xi_L(E)]$ and the claim follows.
To prove ii), note that since $\res^L_K\circ\infl^K_L=0$, we have $[\xi_L(E)]\in \ker(\res^L_K)$. Thus by Theorem \ref{descent} there exists a $\Q$-curve $C$ over $K$ such that $C_L$ is $L$-isogenous to $E$. Choose a system of isogenies $\{\mu_{\sigma}\colon \s C_L\to C_L\}_{\sigma\in\gal(L/\Q)}$ with the following properties:
\begin{itemize}
\item $\mu_{\sigma}=1$ whenever $\sigma\in \gal(L/K)$;
\item $\mu_{\sigma}=\mu_{\tau}$ whenever $\sigma\equiv\tau\bmod\gal(L/K)$.
\end{itemize}
Let $\xi_L(C_L)$ be the 2-cocycle attached to $C_L$ via the above system of isogenies. Now suppose that $C$ is not completely defined over $K$. Then there exist $\nu\in \gal(L/\Q)$, $\vartheta\in\gal(L/K)$ such that $\vt\mu_{\nu}=-\mu_{\nu}$; this implies
\begin{equation}\label{cocycles}
\xi_L(C_L)(\vartheta,\nu)=-1\mbox{ and }\xi_L(C_L)(\nu,\vartheta)=1.
\end{equation}
On the other hand, by hypothesis $[\xi_L(C_L)]=[\xi_L(E)]$ is inflated, so there exists a cocycle $c\in \rZ^2(K/\Q,\Q^{\times})$ such that $[\xi_L(C_L)]=\infl^K_L([c])$. Let $\widetilde{c}\in \rZ^2(L/\Q,\Q^{\times})$ be the cocycle defined by $\widetilde{c}(\sigma,\tau)=c(\overline{\sigma},\overline{\tau})$ for all $\sigma,\tau\in \gal(L/\Q)$, where $\overline{\cdot}$ denotes the equivalence class modulo $\gal(L/K)$. Let $\alpha\colon \gal(L/\Q)\to\Q^{\times}$ be a map such that
$$\widetilde{c}=\xi_L(C_L)\cdot\delta\alpha.$$
Note that the cocycle condition for $c$ implies $c(1,\nu)=c(\nu,1)$ for every $\sigma\in\gal(L/\Q)$. Thus, $\widetilde{c}(\nu,\vartheta)=\widetilde{c}(\vartheta,\nu)$ and by \eqref{cocycles}, this yields $\alpha(\nu\vartheta)=-\alpha(\vartheta\nu)$, a contradiction since $\nu\vartheta=\vartheta\nu$.
\end{proof}
From now on, $E$ is a $\Q$-curve without CM over a quadratic field $K=\Q(\sqrt{d})$, for $d\neq 0,1$ a square-free integer. The non-trivial automorphism of $K$ will be denoted by $\nu$. Moreover, we assume that $E/K$ is not strongly modular, i.e.\ that there exists an isogeny $\mu_{\nu}\colon \n E\to E$ of degree $m$ that cannot be defined over $K$.
\begin{lemma}\label{minimal}
The minimal field of complete definition $L$ of $E$ has Galois group $C_2\times C_2$ over\/~$\Q$.
\end{lemma}
\begin{proof}
It is clear from the construction of $L$ in the proof of Proposition \ref{minimal_field} that $L$ is a quadratic extension of $K$.
Suppose that $\gal(L/\Q)\simeq C_4$. Let $\gal(L/\Q)=\{1,\nu,\nu^2,\nu^3\}$, where by a slight abuse of notation, $\nu\in \gal(L/\Q)$ is a lift of $\nu\in \gal(K/\Q)$. We can set $\mu_{\nu^2}=\id$ and $\mu_{\nu^3}=\mu_{\nu}$. Let $\xi_L^{\pm}$ be the sign part of the 2-cocycle attached to $E_L$ via this system of isogenies. Since $\H^2(C_4,\{\pm1\})\simeq C_2\simeq \H^2_\sym(C_4,\{\pm1\})$, the cocycle $\xi_L^{\pm}$ must be symmetric. Then $\xi_L^{\pm}(\nu,\nu^2)=\mu_{\nu}\n\mu_{\nu^2}\mu_{\nu^3}^{-1}=\mu_{\nu}\mu_{\nu}^{-1}=1$. On the other hand, note that $\leftidx{^{\nu^2}\!}\mu_{\nu}$, which is an isogeny $\n E\to E$, cannot coincide with $\mu_{\nu}$, since this would imply that $\mu_{\nu}$ is defined over $K$. But $\mu_{\nu}$ and $\leftidx{^{\nu^2}\!}\mu_{\nu}$ have the same degree, and therefore $\leftidx{^{\nu^2}\!}\mu_{\nu}=-\mu_{\nu}$. Thus $\xi_L^{\pm}(\nu^2,\nu)=\mu_{\nu^2}\leftidx{^{\nu^2}\!}\mu_{\nu}\mu_{\nu^3}^{-1}=-\mu_{\nu}\mu_{\nu}^{-1}=-1$, which contradicts the symmetry of $\xi_L^{\pm}$.
\end{proof}
Let $e\neq 0,1$ be a squarefree integer such that $L=\Q(\sqrt{d},\sqrt{e})$ is the minimal field of complete definition for $E$. From now on, we set
$$K_e\coloneqq\Q(\sqrt{e}),\quad K_{de}\coloneqq\Q(\sqrt{de})\,\, \mbox{ and }\,\, G\coloneqq\gal(L/\Q)=\{1,\nu,\vartheta,\nu\vartheta\},$$
where $\vartheta$ is the generator of $\gal(L/K)$ and by a small abuse of notation the element $\nu\in G$ restricts to the non-trivial automorphism of $K$, which we also call $\nu$.
Let us compute the 2-cocycle $\xi_L\coloneqq\xi_L(E_L)\in\H^2(L/\Q,\Q^\times)$ attached to $E_L$. Let $\mu_{\vartheta}\colon \vt E_L=E_L\to E_L$ be the identity and let $\mu_{\nu\vartheta}=\mu_{\nu}\colon {}^{\nu\vartheta}E_L=\n E_L\to E_L$. Note that $\vt\mu_{\nu}\colon {}^{\nu\vartheta}E_L=\n E_L\to E_L=\vt E_L$ is an isogeny of the same degree as $\mu_{\nu}$; since $E_L$ has no CM, $\vt\mu_{\nu}$ coincides with $\mu_{\nu}$ up to sign. However, if it were $\vt\mu_{\nu}=\mu_{\nu}$, then $\mu_{\nu}$ would be defined over $K$, a contradiction; thus we have $\vt\mu_{\nu}=-\mu_{\nu}$, and hence ${}^{\nu\vartheta}\mu_{\nu}=-\n\mu_{\nu}$. The isogeny $\mu_{\nu}\n\mu_{\nu}$ equals multiplication by an integer $m\in \Z\setminus\{0,1\}$. By an easy computation we end up with the following table for the cocycle~$\xi_L$:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$\xi_L(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu$ & $1$ & $1$ & $m$ & $m$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $-m$ & $-m$\\ \hline
\end{tabular}
\end{center}
The sign component $[\xi_L^{\pm}]\in\H^2(L/\Q,\Q^\times)$ is represented by one of the following two non-cohomologous cocycles, depending on the sign of $m$.
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$\eta_1(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
\end{tabular}\qquad
\begin{tabular}{| c || c | c | c | c |}
\hline
$\eta_2(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
\end{tabular}
\end{center}
The table of $\xi_L$ shows that the curve $E_L$ is not strongly modular over $L$, because of Theorem \ref{strongmod}. The question we want to address is: which quadratic twists of $E_L$ are strongly modular? A first answer is provided by the following lemma.
\begin{lemma}\label{stronglymodulartwists}
Let $\gamma\in L^{\times}$. Then the twisted curve $E_L^{\gamma}$ is strongly modular over $L$ if and only if $L(\sqrt{\gamma})$ is Galois and non-abelian over\/~$\Q$.
\end{lemma}
\begin{proof}
Let $\xi_L^{\gamma}$ be the cocycle attached to $E_L^{\gamma}$. By Theorem \ref{strongmod}, Lemma \ref{quadtw} and Remark \ref{rem1}, the curve $E_L^{\gamma}$ is strongly modular if and only if $L(\sqrt{\gamma})$ is Galois over~$\Q$ and $[\xi_L^{\gamma,\pm}]\in \H^2_\sym(L/\Q,\{\pm 1\})$.
The group $\widetilde{G}\coloneqq\gal(L(\sqrt{\gamma})/\Q)$ is abelian if and only if the 2-cocycle attached to the exact sequence $1\to {\pm1} \to \widetilde{G}\to G\to 1$ is symmetric. Therefore when $\widetilde{G}$ is abelian, by Lemma \ref{quadtw} the symmetry of the cocycle~$\xi_L$ attached to $E_L$ does not change under twisting by $\gamma$, and this shows that $E_L^{\gamma}$ cannot be strongly modular. On the other hand, by Lemmas \ref{extsym} and~\ref{extpgr} we have $\H^2_\sym(G,\{\pm1\})\simeq C_2\times C_2$, while by Theorem \ref{prodcoh} we have $\H^2(G,\{\pm1\})\simeq C_2^3$. This shows that $\H^2(G,\{\pm1\})/\H^2_\sym(G,\{\pm1\})\simeq C_2$, which means that the product of two non-symmetric classes in $\H^2(G,\{\pm1\})$ is symmetric. Whenever $\widetilde{G}$ is non-abelian, we therefore have $[\xi_L^{\gamma}]\in \H^2_\sym(G,\{\pm1\})$.
\end{proof}
\begin{remark}\label{C2_cohomology}
Let us describe the structure of $\H^2(L/\Q,\{\pm 1\})$ more in detail. Recall that elements of this group correspond to equivalence classes of central extensions of the form $1\to \{\pm 1\}\to \widetilde{G}\to G\to 1$. There are four symmetric cohomology classes and four non-symmetric ones. The symmetric classes correspond to extensions with $\widetilde{G}\simeq C_2\times C_2\times C_2$ or $\widetilde{G}\simeq C_4\times C_2$. The non-symmetric classes correspond to extensions with $\widetilde{G}\simeq D_4$, the dihedral group of order $8$, or $\widetilde{G}\simeq H_8$, the group of quaternions. All extensions with $\widetilde{G}\simeq H_8$ are equivalent to each other, and the corresponding cohomology class is represented by the following cocycle:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$h_0(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $-1$ & $-1$ & $1$\\ \hline
$\nu$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu\vartheta$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
\end{tabular}
\end{center}
On the other hand, there are three non-equivalent extensions with $\widetilde{G}\simeq D_4$. These are uniquely determined by the image in $G$ of the cyclic subgroup of order 4 in $D_4$. If $\gamma\in L^{\times}$ is such that $\widetilde{G}=\gal(L(\sqrt{\gamma})/\Q)\simeq D_4$, $\sigma \in \widetilde{G}$ is an element of order 4 and $\overline{\sigma}$ is its image in $G$, then $L^{\overline{\sigma}}$ is the unique subextension such that $\gal(L(\sqrt{\gamma})/L^{\overline{\sigma}})\simeq C_4$. The following three cocycles represent these classes. The $C_4$-subextension is $L(\sqrt{\gamma})/K,L(\sqrt{\gamma})/K_e,L(\sqrt{\gamma})/K_{de}$, respectively.
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$h_d(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
$\nu$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
\end{tabular}\qquad\qquad
\begin{tabular}{| c || c | c | c | c |}
\hline
$h_e(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu$ & $1$ & $-1$ & $-1$ & $1$\\ \hline
$\nu\vartheta$ & $1$ & $-1$ & $-1$ & $1$\\ \hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$h_{de}(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
$\nu\vartheta$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
\end{tabular}
\end{center}
\end{remark}
\subsection{Distinguishing inflated and primitive twists}
Let $\gamma\in L^{\times}$ be such that $E_L^{\gamma}$ is strongly modular. By Lemma \ref{stronglymodulartwists}, $L(\sqrt{\gamma})$ is a non-abelian Galois extension of $\Q$, and $\widetilde{G}\coloneqq \gal(L(\sqrt{\gamma})/\Q)$ falls in precisely one of the following cases:
\begin{enumerate}[A.]
\item $\widetilde{G}\simeq H_8$;
\item $\widetilde{G}\simeq D_4$ and the unique $C_4$-subextension is $L(\sqrt{\gamma})/K$;
\item $\widetilde{G}\simeq D_4$ and the unique $C_4$-subextension is $L(\sqrt{\gamma})/K_e$;
\item $\widetilde{G}\simeq D_4$ and the unique $C_4$-subextension is $L(\sqrt{\gamma})/K_{de}$.
\end{enumerate}
Recall that the sign component $\xi_L^{\pm}$ of the 2-cocycle attached to $E_L$ equals one of the two classes $[\eta_1]$ and $[\eta_2]$ described below Lemma \ref{minimal}.
\begin{lemma}\label{inducedQcurves}
If $|m|\in (\Q^{\times})^2$, the curve $E_L^{\gamma}$ is inflated from the subfield $F\subseteq L$, according to the following table:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
\backslashbox{$[\xi_L^{\pm}]$}{case} & $A.$ & $B.$ & $C.$ & $D.$ \\ \hline\hline
$[\eta_1]$ & $K_{de}$ & $K_e$ & $K$ & $\Q$\\ \hline
$[\eta_2]$ & $K_e$ & $K_{de}$ & $\Q$ & $K$\\ \hline
\end{tabular}
\end{center}
If $|m|\notin (\Q^{\times})^2$, the curve $E_L^{\gamma}$ is inflated from $K$ if and only if C.\ or D.\ holds, and it is primitive if and only if A.\ or B.\ holds.
\end{lemma}
\begin{proof}
By Proposition \ref{inflation} and the fact that the inflation map preserves the degree and the sign components and the subgroup of symmetric classes, $E_L^{\gamma}$ is inflated from a Galois subfield $F\subseteq L$ if and only if $[\xi_L^{\gamma,\pm}]$ is the inflation of a cohomology class in $\H^2_\sym(F/\Q,\{\pm 1\})$ and $[\xi_L^{\gamma,\deg}]$ is the inflation of a cohomology class in $\H^2_\sym(F/\Q,\Q^{\times}_+)$.
By Lemma \ref{quadtw}, $[\xi_L^{\gamma}]$ is the product of $[\xi_L]$ with the class $[t]$ corresponding to the exact sequence
$$
1 \longrightarrow \gal(L(\sqrt{\gamma})/L) \longrightarrow \gal(L(\sqrt{\gamma})/\Q) \longrightarrow \gal(L/\Q) \longrightarrow 1.
$$
The elements $h_0$, $h_d$, $h_e$ and $h_{de}$ described in Remark \ref{C2_cohomology} represent cases A., B., C.\ and D.\, respectively.
The degree component of $\xi_L$ coincides with that of $\xi_L^{\gamma}$ and is represented by the following cocycle:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$\xi_L^{\deg}(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu$ & $1$ & $1$ & $|m|$ & $|m|$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $|m|$ & $|m|$\\ \hline
\end{tabular}
\end{center}
If $|m|\in (\Q^{\times})^2$, then $[\xi_L^{\deg}]$ is trivial. Thus $[\xi_L^{\gamma}]$ is inflated if and only if $[\xi_L^{\gamma,\pm}]$ is inflated. The non-trivial symmetric classes in $\H^2(L/\Q,\{\pm1\})$ are represented by the following cocycles $b_d$, $b_e$, $b_{de}$:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$b_d(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $-1$ & $-1$\\ \hline
\end{tabular}\qquad\qquad
\begin{tabular}{| c || c | c | c | c |}
\hline
$b_e(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
$\nu$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\nu\vartheta$ & $1$ & $-1$ & $1$ & $-1$\\ \hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$b_{de}(\cdot,\cdot)$ & $1$ & $\vartheta$ & $\nu$ & $\nu\vartheta$ \\ \hline\hline
$1$ & $1$ & $1$ & $1$ & $1$\\ \hline
$\vartheta$ & $1$ & $-1$ & $-1$ & $1$\\ \hline
$\nu$ & $1$ & $-1$ & $-1$ & $1$\\ \hline
$\nu\vartheta$ & $1$ & $1$ & $1$ & $1$\\ \hline
\end{tabular}
\end{center}
It is immediately clear that $[b_d]$ (resp.\ $[b_e]$, $[b_{de}]$) is the inflation of the unique non-trivial element in $\H^2(K/\Q,\{\pm1\})$ (resp.\ $\H^2(K_e/\Q,\{\pm1\})$, $\H^2(K_{de}/\Q,\{\pm1\})$). Thus our claim is equivalent to showing that the multiplication table of $[\eta_1],[\eta_2]$ by $[h_0],[h_d],[h_e],[h_{de}]$ is the following:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
$\cdot$ & $[h_0]$ & $[h_d]$ & $[h_e]$ & $[h_{de}]$ \\ \hline\hline
$[\eta_1]$ & $[b_{de}]$ & $[b_e]$ & $[b_d]$ & $1$\\ \hline
$[\eta_2]$ & $[b_e]$ & $[b_{de}]$ & $1$ & $[b_d]$\\ \hline
\end{tabular}
\end{center}
and it is easy to check that this is the case.
Assume now that $|m|\notin {(\Q^{\times})}^2$. The class $[\xi_L^{\deg}]$ is the inflation from $\H^2(K/\Q,\Q^{\times}_+)$ of the class $[c]$, where $c(1,1)=c(1,\nu)=c(\nu,1)=1$ and $c(\nu,\nu)=|m|$, while it is not the inflation of a class lying in $\H^2(F/\Q,\Q^{\times}_+)$ for any $F\in \{\Q,K_e,K_{de}\}$. Thus $E_L^{\gamma}$ is inflated if and only if it is inflated from $K$.
Therefore it is enough to check when $[\xi_L^{\gamma,\pm}]$ coincides with $[b_K]$. Since $[b_K]=[\eta_1]\cdot[\eta_2]$, the class $[\xi_L^{\gamma}]$ is inflated if and only if the class $[t]$ equals either $[\eta_1]$ or $[\eta_2]$. It is immediate to see that $[\eta_1]=[h_{de}]$ and $[\eta_2]=[h_e]$, and the proof is complete.
\end{proof}
The next step is to give necessary and sufficient conditions for the existence of primitive or inflated twists of $E_L$. Recall that $L=\Q(\sqrt{d},\sqrt{e})$ is a $C_2\times C_2$-extension of $\Q$ and $G=\gal(L/\Q)=\langle\nu,\vartheta\rangle$ where $\n\sqrt{d}=-\sqrt{d}$, $\vt\sqrt{e}=-\sqrt{e}$. For $a,b\in \Q$, we will denote by $(a,b)$ the quaternion algebra over $\Q$ with basis $\{1,i,j,ij\}$ such that $i^2=a$, $j^2=b$, $ij=-ji$. Recall that the \emph{reduced discriminant} of a quaternion algebra $B$ over $\Q$ is the product of the finite primes of $\Q$ where $B$ ramifies. A quaternion algebra is trivial in $\Br(\Q)$ if and only if it has reduced discriminant $1$.
\begin{theorem}[{{\cite[Theorems 4 and 5]{kim}}}]\label{embsol}
The following hold:
\begin{enumerate}[i)]
\item Let $H_8=\{\pm 1,\pm i,\pm j,\pm k\}$ be the group of quaternions. The embedding problem (cf.\ Definition \ref{embedding}) relative to $L/\Q$ and the group extension
$$1\longrightarrow C_2\longrightarrow H_8\stackrel{\pi}{\longrightarrow} G\longrightarrow 1$$
is solvable if and only if $(d,de)(e,de)(d,e)=1$ in $\Br(\Q)$, if and only if there exist $v_1,v_2,v_3,w_1,w_2,w_3\in \Q$ such that
$$\begin{cases}
d=v_1^2+v_2^2+v_3^2 & \\
e=w_1^2+w_2^2+w_3^2 & \\
v_1w_1+v_2w_2+v_3w_3=0.
\end{cases}$$
In this case, setting $t=1+\frac{v_1}{\sqrt{d}}+\frac{w_3}{\sqrt{e}}+\frac{v_1w_3-v_3w_1}{\sqrt{de}}$, the extensions solving the problem are exactly the ones of the form $L(\sqrt{qt})$, for $q\in \Q^{\times}$.
\item Let $D_4=\langle \sigma,\tau\colon \sigma^4=\tau^2=1,\,\, \sigma\tau=\tau\sigma^3\rangle$ be the dihedral group of order $8$. The embedding problem relative to $L/\Q$ and the group extension
$$1\longrightarrow C_2\longrightarrow D_4\stackrel{\pi}{\longrightarrow} G\longrightarrow 1$$
where $\pi(\sigma)=\vartheta$ and $\pi(\tau)=\nu$ is solvable if and only if $(-d,e)=1$ in $\Br(\Q)$.
In this case, if $x,y\in \Q$ are such that $d=ey^2-x^2$, the extensions solving this problem are exactly the ones of the form $L(\sqrt{q(ey+x\sqrt{e})})$ for $q\in \Q^{\times}$.
\end{enumerate}
\end{theorem}
Recall that the class $[\xi_L^{\gamma,\pm}]$ equals one of the classes $\eta_1,\eta_2$ described below Lemma~\ref{minimal}.
\begin{theorem}\label{primQcurves}
There exists $\gamma\in L^{\times}$ such that $E^{\gamma}$ is strongly modular if and only if at least one of the following conditions is satisfied:
\begin{enumerate}
\item[A.] the quaternion algebra $(-d,-e)$ has reduced discriminant 2;
\item[B.] the quaternion algebra $(-d,e)$ is trivial in $\Br(\Q)$;
\item[C.] the quaternion algebra $(d,-e)$ is trivial in $\Br(\Q)$;
\item[D.] the quaternion algebra $(d,-de)$ is trivial in $\Br(\Q)$.
\end{enumerate}
In particular, if $m\in {(\Q)^{\times}}^2$ then there exists $\gamma$ such that $E^{\gamma}$ is inflated from the subfield $F$, according to the following table:
\begin{center}
\begin{tabular}{| c || c | c | c | c |}
\hline
\backslashbox{$[\xi_L^{\pm}]$}{case} & $A.$ & $B.$ & $C.$ & $D.$ \\ \hline\hline
$[\eta_1]$ & $K_{de}$ & $K_e$ & $K$ & $\Q$\\ \hline
$[\eta_2]$ & $K_e$ & $K_{de}$ & $\Q$ & $K$\\ \hline
\end{tabular}
\end{center}
If $m\notin {(\Q)^{\times}}^2$, then there exists $\gamma$ such that $E^{\gamma}$ is primitive if and only if A. or B. holds, while there exists $\gamma$ such that $E^{\gamma}$ is inflated from $K$ if and only if C. or D. holds.
\end{theorem}
\begin{proof}
By Lemma \ref{stronglymodulartwists}, $E^{\gamma}$ is strongly modular if and only if $L(\sqrt{\gamma})/\Q$ is a non-abelian Galois extension. Thus, there exists $\gamma$ such that $E^{\gamma}$ is strongly modular if and only if the embedding problem
\begin{equation}\label{nonabelian}
1 \longrightarrow \gal(L(\sqrt{\gamma})/L) \longrightarrow \widetilde{G} \longrightarrow \gal(L/\Q) \longrightarrow 1
\end{equation}
has a solution with $\widetilde{G}$ non-abelian.
When $\widetilde{G}\simeq H_8$, by Theorem \ref{embsol} and the discussion at \cite[p.~239]{kim}, the embedding problem \eqref{nonabelian} is solvable if and only if the quadratic forms $S_{d,e}=\frac{1}{de}X^2+dY^2+eZ^2$ and $T=X^2+Y^2+Z^2$ are equivalent over $\Q$. This implies immediately that $d,e>0$ because $S_{d,e}$ and $T$ must have the same signature. Since the rank and the discriminant obviously coincide, it only remains to check that the Hasse-Witt invariants coincide. If $p$ is a prime, the Hasse-Witt invariant of $T$ at $p$ is $1$, while the Hasse-Witt invariant of $S_{d,e}$ at $p$ is
$$
\begin{aligned}
(de,d)_p(de,e)_p(d,e)_p&=(de,d)_p(de,e)_p(-d,-e)_p(d,-1)_p(-1,e)_p(-1,-1)_p=\\
&=(de,-de)_p(-d,-e)_p(-1,-1)_p=\\
&=(-d,-e)_p(-1,-1)_p,
\end{aligned}
$$
where we used bilinearity of the Hilbert symbol and the fact that $(a,-a)_p=1$ for every $a\in \Q^{\times}$ and every prime $p$. Since $(-1,-1)$ ramifies precisely at 2 and $\infty$, we see that this instance of the embedding problem is solvable if and only if A.\ holds.
When $\widetilde{G}\simeq D_4$, point ii) of Theorem \ref{embsol} shows that the embedding problem \eqref{nonabelian} is solvable if and only if B., C.\ or D.\ holds.
The other claims follow immediately from Lemma \ref{inducedQcurves}.
\end{proof}
\begin{corollary}\label{corollary_twist}
The curve $E$ has a strongly modular quadratic twist over $K$ if and only if the curve $E_L$ has a strongly modular twist that is inflated from $K$.
\end{corollary}
\begin{proof}
First recall that, by Theorem \ref{strongmod}, a $\Q$-curve over a quadratic field~$K$ is strongly modular if and only if it is completely defined over~$K$.
Let $\gamma\in K^{\times}$ be such that $E^{\gamma}$ is strongly modular over $K$. Then the base-changed curve $(E^{\gamma})_L$ is strongly modular over $L$ since its attached cocycle is the inflation of a symmetric one, and is therefore symmetric. On the other hand, $(E^{\gamma})_L$ is isomorphic to $(E_L)^{\gamma}$, and hence $E_L$ has a strongly modular twist inflated from~$K$.
Conversely, if $E_L$ has a strongly modular twist $(E_L)^{\gamma}$ inflated from $K$, then by Theorem \ref{primQcurves} at least one of $(d,-e)$ or $(d,-de)$ is trivial. Theorem \ref{embsol} shows that we can choose $\gamma\in K^{\times}$: it is enough to use point ii) of the theorem, replacing the map~$\pi$ by the one given by $\pi(\tau)=\vartheta$ and $\pi(\sigma)=\nu$ or $\pi(\sigma)=\nu\vartheta$. Then $d$ plays the role of~$e$ in the notation of the theorem, and it is clear that $\gamma\in K^{\times}$. Therefore $(E_L)^{\gamma}$ is $L$-isomorphic to $(E^{\gamma})_L$. Now $E^{\gamma}$ is completely defined over $K$, otherwise $E^{\gamma}_L$ would not be strongly modular since its attached cocycle would not be symmetric (cf.\ the discussion below Lemma \ref{minimal}).
\end{proof}
\subsection{Examples}
We will now give examples of $\Q$-curves with different behaviours with respect to the existence of primitive and inflated strongly modular quadratic twists.
\subsubsection*{Example 1}
The following example is borrowed from \cite{pyl}. Let $E$ be the following elliptic curve without CM over $K=\Q(\sqrt{-3})$:
$$E\colon y^2=x^3+2x^2+bx,$$
where $b\in \O_K$ is any element of trace $1$. There is an isogeny $\mu_{\nu}\colon \n E\to E$ such that $\mu_{\nu}\n\mu_{\nu}=-2$. The minimal field of definition of $E$ is $L=\Q(\sqrt{-3},\sqrt{-2})$. Since $(3,-2)$ is trivial in $\Br(\Q)$, by Theorem \ref{primQcurves} there are quadratic extensions of $L$ of type $D_4$ over $\Q$ with $C_4$-subextension $L(\sqrt{\gamma})/K$. Since $\alpha=1+\sqrt{-2}\in \Q(\sqrt{e})$ has norm $3=-d$, by Theorem \ref{embsol}, the set of all these extensions is $\left\{L\biggl(\sqrt{r+r/\sqrt{-2}}\biggr)\colon r\in \Q^{\times}\right\}$. The one found in \cite{pyl} corresponds to $r=2$. Let $\gamma=2-\sqrt{-2}$. An integral model for $E_L^{\gamma}$ is
$$E_L^{\gamma}\colon y^2=x^3+(4-2\sqrt{-2})x^2+b(2-4\sqrt{-2})x.$$
By Theorem \ref{primQcurves} there are no quadratic extensions of $L$ that are of type $H_8$ over~$\Q$, nor quadratic extension of type $D_4$ with $C_4$-subextension $\Q(\sqrt{-2})$ or $\Q(\sqrt{6})$. Thus, all strongly modular quadratic twists of $E$ are primitive over $L$. Note also that \cite[Proposition 6.2]{pyl}, which asserts that there are no quadratic twists of $E$ that are completely defined over $K$, follows immediately from Corollary \ref{corollary_twist}.
To construct other examples, consider the following family of $\Q$-curves given in~\cite{que}:
$$E_a\colon y^2=x^3-3\sqrt{a}(4+5\sqrt{a})x+2\sqrt{a}(2+14\sqrt{a}+11a),$$
where $a\in \Z$ is not a square. Then $E_a$ is defined over $K_a\coloneqq\Q(\sqrt{a})$, but its minimal field of complete definition is $L_a\coloneqq K_a(\sqrt{3})$.
\subsubsection*{Example 2}
Consider the curve $(E_6)_{L_6}$. With the notation of Theorem \ref{primQcurves}, we have $d=6$ and $e=3$. Since $j(E_6)=\frac{27625536+10768896\sqrt{6}}{125}$, it follows that $E_6$ has no CM.
The quaternion algebra $(-6,-3)=(-2,-3)$ has reduced discriminant 2, and therefore by Theorem \ref{primQcurves} it follows that $L_6$ has a quadratic extension of type~$H_8$ over~$\Q$. Following the notation of Theorem \ref{embsol}, we can pick $v_1=2$, $v_2=v_3=1$, $w_1$ and $w_2=w_3=-1$. Thus $t=1+\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}}$ and all extensions of type $H_8$ of $L_6$ are of the form $L_6(\sqrt{qt})$ with $q\in \Q^{\times}$. For example, letting $q=1$ and $\gamma=t$, an integral model for $(E_6)_{L_6}^{\gamma}$ is
$$(E_6)_{L_6}^{\gamma}\colon y^2 = x^3+Ax+B,$$
where
$$
\begin{aligned}
A&=4080384\alpha^3-13616640\alpha^2-412416\alpha+1375488,\\
B&=-25868537856\alpha^3+82215567360\alpha^2+2613252096\alpha-8305459200\\
\end{aligned}
$$
and $\alpha=\sqrt{2}+\sqrt{3}$. This is a primitive strongly modular curve over $L_6$.
Since $(-6,3)$ and $(6,-3)$ both have reduced discriminant $6$, there are no extensions of $L_6$ that are of type $D_4$ over $\Q$ with $C_4$-subextension $L_6/\Q(\sqrt{6})$ or $L_6/\Q(\sqrt{3})$. On the other hand, $(6,-18)$ is trivial in $\Br(\Q)$, so by Corollary \ref{corollary_twist} there exist strongly modular twists of $E_6$. To find them, note that by Theorem~\ref{embsol} it is enough to find $x,y\in \Q$ with $6y^2-x^2=18$. As $x=6$, $y=3$ solve this equation, letting $t=18+6\sqrt{6}$ we get that all extensions of $L_6$ of type $D_4$ over $\Q$ are of the form $L_6(\sqrt{qt})$ with $q\in \Q^{\times}$. Let us choose for example $q=1/6$ and $\gamma'=qt$. Then an integral model for $E_6^{(\gamma')}$ is
$$E_6^{(\gamma')}\colon y^2 = x^3 - (28512+11520 \sqrt{6}) x + 2594304+ 1059840 \sqrt{6}.$$
\subsubsection*{Example 3}
The curve $E_7$ is not strongly modular, but since $(7,-3)$ is trivial, by Corollary \ref{corollary_twist} it has strongly modular quadratic twists. For example setting $\gamma=7+2\sqrt{7}$, the curve $E^{\gamma}$ is strongly modular over $K_7$. An integral model is
$$E_7^{\gamma}\colon y^2 = x^{3} - (166992+61824 \sqrt{7}) x + 36452864+ 13804672 \sqrt{7}.$$
Since $(-7,-3)$ has reduced discriminant $3$ and $(-7,3)$ has reduced discriminant~$21$, Theorem \ref{primQcurves} shows that $(E_7)_{L_7}$ has no primitive strongly modular twists.
\subsubsection*{Example 4}
Finally, the curve $(E_5)_{L_5}$ has no strongly modular twists at all, since $(3,-5)$ has reduced discriminant~$10$, $(-3,-5)$ has reduced discriminant $5$ and both $(5,-3)$ and $(5,-15)$ have reduced discriminant~$15$.
\bibliographystyle{plain}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,933 |
Leaders in the Circular Economy Movement
Technology and the Circular Economy: Selkirk Technology Access Centre
Jacomien van Tonder
By Val Rossi and Kyla Girling of Writers Block Solutions
Whether it's reviewing a start-up idea sketched on the back of a napkin or working with an industry leader automating long-standing processes, the Selkirk Technology Access Centre (STAC) is lending its advanced manufacturing expertise to Lower Columbia professionals and helping take concepts from prototype to market.
The rural technology centre is one of 60 Technology Access Centres (TACs) across the country assisting Canadian businesses—particularly small-and-medium-sized enterprises—by providing applied research and innovation services. These centres help build new prototypes and solve unique business challenges, and provide specialized training on the latest equipment and emerging technologies. Their goal is to enhance productivity, competitiveness, and innovation for their industry partners.
There's no doubt that STAC in Trail is driving circular economy innovation, facilitating the more rapid development of cleantech, and changing the business landscape in the Lower Columbia and beyond. When it comes to training students for the jobs of tomorrow, fostering and retaining qualified local talent, and manufacturing goods this is especially apparent.
"One of the first questions we ask is: how can we help manufacture parts utilizing reclaimed or renewable resources?" explains STAC's Chair and Lead Researcher Jason Taylor. "We look at byproducts from companies' processes to see if there are opportunities. And we're very much a part of the local ecosystem, so sometimes those opportunities are finding ways to keep work local. We know who provides what services, and it cuts down on excessive transportation and outsourcing when we can provide certain services and then also act as a conduit for area referrals."
STAC boasts a suite of services designed to build competitiveness in research and development, specifically for companies exploring technical expertise and specialized equipment to advance product design and manufacturing or optimize a process or system. The centre offers 3D printing, 3D scanning, rapid prototyping, data specialist expertise, reverse engineering, metrology, customized advanced technology, and training. Another important facet of their operation is offering referrals to existing businesses in the region, as they are committed to supporting local companies rather than competing with them.
Rural Tech Centre Upskills Local Talent
Local manufacturing has become even more prevalent since the pandemic exposed how much Canada relies on foreign supply chains. Examining this has increased the desire to close the loop closer to home, supporting STAC's mission to strengthen the economy by giving local professionals the tools to do the work here. This open, collaborative approach is growing STAC's network, as the team continues to welcome qualified experts to the Metal Tech Alley region, which is known for technological and metallurgical advancement.
As a launch pad for innovation, the centre's presence in the region is helping attract more tech professionals, upping the area's already substantial reputation. Further highlighting the need to support the local manufacturing sector, Selkirk's new Digital Fabrication & Design program is now turning out qualified professionals, upskilling the local talent pool and supporting the established companies recruiting for advanced manufacturing and process operations.
"STAC is creating deep ties between business development and academic research, and fostering greater collaboration between local businesses," says Lower Columbia Initiative Corporation (LCIC)'s Economic Development Coordinator Rebecca Richards. "These factors promote innovation and the use of tech in industry, which will be important for regional economic growth and long-term stability."
Reverse engineering makes up about half of STAC's current workload and the future looks to be much the same, with industry giants like Kalesnikoff and Mercer Celgar changing their processes to look at more automation and just-in-time manufacturing and less warehousing of parts and machine equipment.
There's no project too small for STAC, which is more about customization than mass production and lends its expertise to individuals as well as companies; for example, a mechanic restoring a vintage car may turn to the centre for a discontinued part. The STAC team truly supports a broad range of projects that touch every industry or need imaginable.
STAC as a Metal Tech Alley Conduit
While acquiring and testing equipment to determine if it's really what a client may need normally can cost hundreds of thousands of dollars, STAC facilitates the process and provides individuals and businesses with significant cost savings. Their specialized staff also provide consultation and recommendations.
As awareness around its services grows, Taylor says the centre continues to gain further support, with close to $400,000 in new equipment secured this past year alone.Their efforts are supported by provincial and federal government funding, and those engaging STAC's services may either pursue possible research grant funding opportunities themselves or pay direct fees-for-service.
LCIC is working diligently on improving the circular economy framework in the region by establishing public policy and built-in business cases and investigating a possible industrial park. The organization is also hosting the Industrial Circular Economy Conference (ICE) online from June 8 – 10, 2021. The team behind Metal Tech Alley is proud that their movement towards a more sustainable economic structure aligns with STAC's mission. Working alongside industry leaders like Taylor and leaning on the natural resources and expertise already readily available in the region helps build a more robust future for all.
"STAC makes it possible for businesses to improve process efficiency and find solutions more easily by providing a designated facility for innovation and testing, as well as on-site support," explains Richards. "STAC is already helping businesses subvert their reliance on manufacturers for spare parts, which reduces the environmental impact and makes repairs cheaper and easier. Because companies are now better able to close loops in their supply chains locally, this is a positive step towards circularity."
For more information on STAC's projects click here
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This site presented by: | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,260 |
package cn.net.normcore.iorder.api.app;
import cn.net.normcore.iorder.common.SimpleResult;
import cn.net.normcore.iorder.filter.ValidateToken;
import cn.net.normcore.iorder.service.promotion.PromotionService;
import cn.net.normcore.iorder.vo.promotion.PromotionVo;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import java.util.Map;
/**
* 推广信息相关接口
* Created by 81062 on 2017/3/19.
*/
@Controller
@Path("/app/promotion")
@Produces(MediaType.APPLICATION_JSON)
@ValidateToken
public class PromotionApi {
@Autowired
private PromotionService promotionService;
/**
* 获取所有推广信息
*
* @return
*/
@GET
public Map<String, Object> promotion() {
Map<String, Object> result = SimpleResult.optimistic();
result.put("promotions", PromotionVo.listFromPromotions(promotionService.findAll()));
return result;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,288 |
Mario Kart races are a must and you have to be on your A-game at all times.
Every year since I was two, my cousins and I have gone to the Cape every summer.
Some of my favorite memories were made here with my favorite people. Now that the weather is cold and its dark outside, I figured that I'd share the rules of our Cape Cod vacations to make it seem less dark and more beach-y!
Rule 1: You laugh at the people driving the opposite way as you're going over the Bourne Bridge.
Rule 2: Then as you're leaving you realize that everyone is laughing at you.
Rule 3: Your Pinterest board is full of grey weathered sided cottages.
Rule 4: Your favorite ice cream shop is the one at the Cape.
Rule 5: It's sacrilegious for you to eat ice cream from anywhere else on the Cape.
Rule 6: You get ice cream from there every night.
Rule 7: You have a constant countdown for the next time you're going ... that starts from the day that you leave.
Rule 8: Your favorite flower is a hydrangea.
Rule 9: You're perfectly content sitting on the beach doing nothing for the whole day.
Rule 10: Mini-golf is all or nothing.
Rule 11: You've driven hours to Provincetown to see seals ... only to see none.
Rule 12: You've driven a duck boat.
Rule 13: You have a collection of duck quackers from each ride.
Rule 14: You're not really at the Cape until you get a picture of the bushes and send it to the group chat.
Rule 15: You're also a bowling pro.
Rule 16: You save your money for the penny candy store.
Rule 17: Hyannis mall is the place to go on a rainy day.
Rule 18: You spend the rest of your day playing board games.
Rule 19: You have a go-to rainy day movie every year, sometimes two.
Rule 20: The sunset is even more beautiful.
Rule 21: The fireworks are even brighter.
Rule 22: There is no whining because quite frankly, there is no reason.
Rule 23: Mario Kart races are a must and you have to be on your A-game at all times.
Rule 24: Freezing cold outdoor showers are just part of the experience.
Rule 25: The last night is capped off with a family talent show and surf and turf dinner. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,371 |
Being very new to blogging, I wasn't sure what people wanted to read about. I didn't think anyone would want to read about a mobile home family, but to my surprise there have been several hundred views! Exciting! So my new years resolution will have to be keeping up with this blog. That shouldn't be hard to do. We had a VERY BUSY summer of projects to transform this tiny mobile home into a fun place to live. I have some building plans to share. And we still have a "honey do" list a mile long!
So for this quick post, I'm giving ya'll a free printable that I made. ANGRY BIRDS Play Money in Boy Red Bird & Red Girl Bird. This one is perfect for a small gift, stocking stuffer, or as pinata filler or a party favor to complete your ANGRY BIRDS birthday.
Just print them out on letter sized (8.5" x 11") paper, front & back, and cut them out. The image is black & white like an ANGRY BIRDS coloring page. You can let your little ones color their money for a birthday craft. Or you can print it on colored paper.
Here is a picture of ours on colored paper.
Angry Birds Fan Art printable funny money in STAR WARS, Space, or Original Birds with Mighty Eagle, Mighty Dragon, King Pig, Red Girl Bird, & Yellow, Blue, Green Birds!
For "real" size bills, print groups of 4, front and back, on one sheet of paper or card-stock at full size.
Check out my other fun free holiday crafts!
Enjoy you free for home enjoyment printable play money!!! | {
"redpajama_set_name": "RedPajamaC4"
} | 3,399 |
Q: How can I redirect cards I created in Flutter to different pages? class EditNoteList extends StatelessWidget {
final String child;
EditNoteList({required this.child});
@override
Widget build(BuildContext context) {
return Center(
child: Padding(
padding: const EdgeInsets.only(
top: 36.0, left: 6.0, right: 6.0, bottom: 6.0),
child: Card(
child: InkWell(
splashColor: Colors.blue.withAlpha(30),
hoverColor: Colors.lightBlue[200],
onTap: () {
// if( Control for First Card)
// else if (Control for Second Card)
// else
Navigator.push(
context,
MaterialPageRoute(builder: (context) => const Card3()),
);
},
child: SizedBox(
width: 400,
height: 100,
child: Row(
mainAxisAlignment: MainAxisAlignment.spaceAround,
crossAxisAlignment: CrossAxisAlignment.start,
children: [
Center(
child: Text(
child,
style: TextStyle(
fontSize: 12,
fontFamily: 'Montserrat',
fontWeight: FontWeight.bold),
textAlign: TextAlign.center,
)),
],
),
),
),
),
),
);
I created my cards on this page and showed them on the page I want to show the cards with a different widget.
class _AddNotesState extends State<AddNotes> {
@override
Widget build(BuildContext context) {
return Scaffold(
body: Column(
children: const [Card1(), Card2(), Card3()],
),
);
}
}
class Card1 extends StatelessWidget {
const Card1({Key? key}) : super(key: key);
@override
Widget build(BuildContext context) {
return Container();
}
}
class Card2 extends StatelessWidget {
const Card2({Key? key}) : super(key: key);
@override
Widget build(BuildContext context) {
return Container();
}
}
class Card3 extends StatelessWidget {
const Card3({Key? key}) : super(key: key);
@override
Widget build(BuildContext context) {
return Container();
}
}
I want to show Card1, Card2 and Card3 widgets with the directions I will give in the codes on the first page. How can I redirect the page based on the card that the user clicked? I want the Card1 method to work when the user clicks the 1st card, the Card method when the 2nd card is clicked, and the Card3 method when he clicks the 3rd card.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,566 |
Q: Colored lines on monitor with freeze mostly on video playback My PC is showing colored lines and freezes and restarts moments after I starts playing some video. I have also seen this happening when I open an image. It will show the problem the instant I start video playback. I thought it was some problem with the driver but the lines are visible even on the bios boot screen when restart.
My hardware is as follows :
intel core i5 750 CPU,
ASUS ATI Radeon HD 5450 graphics card,
ASUS P7P55D motherboard,
2Gb Cosair DDR3 ram sticks x 2,
2 SATA hard drives,
It could be problem with Motherboard, video card, RAM or drivers. I am looking for some tips to identify the component that caused.
Thanks..
A: It sounds like your graphics card is overheating. This can cause lines and artifacts on the display. You should use a monitoring tool like GPUTemp. It could be the thermal paste on your GPU is not properly transferring the heat to your GPU fan.
A: It sounds to me like your computer's CPU heat sink is probably caked with dust. I would purchase a can of compressed air and blow the heat sink out.
A: This smells like a overheating and I would suggest that you test your hardware.
I've made a post here on how to stress-test your CPU, RAM and GPU, it should pinpoint where the problem lies.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,887 |
Reactor Core
The Red Lynx's Biplane
It's not over yet, Manx!
Characters, Male & Villains
Felony, Kooshmeister & MoDaD
Known as the most evil pilot in history, the Red Lynx dominated the skies during Megawar II, shooting down all that dared to oppose him. It wasn't until the debut of the Blue Manx did this pilot ever get shot down.
The Blue Manx, great grandfather to Mayor Manx, was made famous for his actions that day, and the Red Lynx was apparently killed in the crash. Many years later, Abner Doolittle, a worker as Megakat Harbor fished out the Red Lynx's bi-plane from the murky depths, and brought it ashore. The plane was transferred to the Megakat Museum of History, where it was put on display in the aviation section. Then, one moonlit night, the spirit of the Red Lynx manifested, and took off in the bi-plane, seeking revenge, and vowed to take it on the Blue Manx's only living descendent, Mayor Manx.
The only way to stop this avenging spirit would be for Manx to shoot down the ghost pilot. After much aerial dog fighting with the SWAT Kats, the Red Lynx was presumed shot down, but this was short lived, as he stole the new Enforcer jet prototype "The Blue Manx", and used it to try again to take out Manx. With the aid of the SWAT Kats, Mayor Manx was able to shoot him down on the cyclotron, while T-Bone distracted him in the Turbokat, silencing his ghost for eternity, or, until he resurfaces again…
He is named for Imperial German ace Manfred von Richtofen, whose nickname during World War I was the Red Baron. Unlike the Red Lynx, however, von Richtofen was not a war criminal and not remembered as an evil man.
Navigate entries
It's a deadly "close encounter" when T-Bone and Razor defend Megakat City from the awesome threat of Mutilor, the interstellar water pirate. More
Mr. Goodkat's in-depth overview of the Super Nintendo SWAT Kats: The Radical Squadron game including detailed background information, informative commentary about aspects of the gameplay, videos, tips, codes and more. More
Zed is the name assumed by a robot created by Dr. Lieter Greenbox. Zed began life as Greenbox's micro-brain repair unit. The Metallikats, badly damaged from a conflict with the SWAT Kats entered Greenbox's lab and used the device to repair themselves. Somehow the machine became sentient and taking the name "Zed" it followed the Metallik… More | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,993 |
Le Grand dauphin de l'océan Indien ou grand dauphin de l'Indo-Pacifique (Tursiops aduncus) est un dauphin du genre Tursiops.
Description
Cette espèce de Tursiops fut décrite par Ehrenberg en 1833.
T. aduncus est plus petit que le T. truncatus et possède un rostre proportionnellement plus long : le grand dauphin de l'Indo-Pacifique mesure en moyenne 2,60 m et pèse 230 kg maximum ; Il ne s'éloigne pas des côtes, forme de petits groupes sociables stables de 5 à 15 individus comme le grand dauphin, mais est en général moins joueur et plus farouche.
T. truncatus et T. adunctus existent dans les mêmes eaux. Les deux espèces ont longtemps été considérées comme une seule.
Les comparaisons d'un fragment de 386 points d'ébullition de la région mitochondrique de commande d'ADN (mtDNA) (n = 47) ont indiqué que les deux morphotypes étaient génétiquement distincts. Les analyses phylogénétiques ont prouvé que le type truncatus des eaux chinoises est plus étroitement lié à celui de l'océan Atlantique que le T. adunctus. Ces données moléculaires ont convenu complètement avec des classifications morphologiques des spécimens. Cette congruence est évidence forte que les morphotypes dans les eaux chinoises sont des reproducteurs isolés et comportent deux espèces distinctes, avec des implications importantes pour la conservation des dauphins Tursiops dans les eaux chinoises.
Les nouveaux résultats peuvent justifier d'autres subdivisions des espèces de Tursiops dans un avenir proche : Curry (1997) avait analysé 127 ordres mitochondriques de région de commande d'ADN pour étudier d'intra- et interspécifiques différences parmi des dauphins Tursiops.
Elle a identifié 73 haplotypes et les résultats, a combiné avec l'information sur la morphologie et l'écologie, soutenues par la suggestion qu'il y a des différences d'espèce-aux niveaux côtiers et des dauphins tursiops en pleine mer dans l'Atlantique nord occidental/golfe du Mexique.
L'évidence génétique récente suggère que T. adunctus soit plus étroitement lié aux espèces pélagiques de Stenella et de Delphinus, et en particulier aux Stenella frontalis, qu'au Tursiops truncatus. Si ces résultats sont confirmés, ils auraient plus que des implications taxonomiques et influencent considérablement des ségrégations basées sur la morphologie et le comportement social (Puits et Scott 2002).
Usage d'outils et automédication
Dans la Baie Shark, en Australie, des dauphins sauvages de l'espèce Tursiops aduncus utilisent des éponges marines comme outil. Ils s'en servent pour protéger leur rostre sensible lorsqu'ils fouillent les sédiments des fonds marins. Ce comportement, uniquement observé dans cette baie, a été observé pour la première fois en 1997. Seules certaines femelles utilisent les éponges comme outil ; il a été démontré grâce à une analyse de l'ADN mitochondrial de ces femelles qu'elles proviennent toutes de la même femelle ancestrale : ce phénomène semble donc être une transmission de savoir, de mère à fille exclusivement. Dans l'état actuel de nos connaissances, cette transmission culturelle matrilinéaire est unique chez les cétacés.
Des dauphins ont été observés faisant la queue dans des environnements naturels pour frotter certaines parties de leur corps contre des coraux (Rumphella aggregata, Sarcophyton sp.) et des éponges (Ircinia sp.) dans le nord de la mer Rouge. Il a été émis l'hypothèse que la présence de métabolites bioactifs explique ce comportement de frottement sélectif. Les trois invertébrés auxquels les dauphins accèdent préférentiellement, collectés et analysés contiennent effectivement dix-sept métabolites actifs, fournissant la preuve d'une automédication potentielle. Des frottements répétés permettent à ces métabolites actifs d'entrer en contact avec la peau des dauphins, ce qui pourrait les aider à atteindre l'homéostasie cutanée et être utile pour la prophylaxie ou le traitement auxiliaire contre les infections microbiennes.
Usage de jeux par les dauphins
Dans le lagon de Mayotte, dans l'archipel des Comores, des dauphins sauvages de l'espèce Tursiops aduncus utilisent des poissons-ballons comme jeu. Ils s'attaquent à un individu, qui gonfle pour se protéger. Les dauphins sauvages s'amusent alors à frapper le poisson-ballon latéralement avec leurs rostres, se faisant ainsi des passes.
Selon Rob Pilley, zoologue à la BBC, il pourrait même s'agir d'un comportement de prise de neurotoxine, mais cette interprétation est contestée.
Répartition
On trouve ce dauphin dans tout l'Océan Indien de la côte est-Africaine à l'Inde, et jusqu'au sud de la Chine et au nord de l'Australie.
Voir aussi
Articles connexes
Grand dauphin
Océan Indien
Références taxonomiques
Liens externes
Informations de CMS
Notes et références
Delphinidae
Cétacé (nom vernaculaire) | {
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{"url":"https:\/\/stats.stackexchange.com\/questions\/156603\/how-to-choose-between-fixed-effects-and-random-effects-model-in-meta-analysis","text":"# How to choose between fixed-effects and random-effects model in meta-analysis?\n\nIn meta-analysis packages, both fixed effects and random effects models are available. How do one choose between these two models? Since one is assessing different studies, should one not choose random effects model all the time?\n\n## 1 Answer\n\nYou use a fixed-effects model if you want to make a conditional inference about the average outcome of the $k$ studies included in your analysis. So, any statements you make about the average outcome only pertain to those $k$ studies and you cannot automatically generalize to other studies.\n\nYou use a random-effects model if you want to make an unconditional inference about the average outcome in a (typically hypothetical) population of studies from which the $k$ studies included in your analysis are assumed to have come. So, any statements you make about the average outcome in principle pertain to the that entire population of studies (assuming that the $k$ studies included in your meta-analysis are a random sample of the studies in the population or can in some sense be considered to be representative of all of those studies).\n\nA very common misconception is that the fixed-effects model is only appropriate when the true outcomes are homogeneous and that the random-effects model should be used when they are heterogeneous. However, both models are perfectly fine even under heterogeneity -- the crucial distinction is the type of inference you can make (conditional versus unconditional).\n\nIn fact, it is also perfectly fine to fit both models: Once to make a statement about the average outcome of those $k$ studies and once to try the more difficult task of making a statement about the average effect 'in general'.\n\n\u2022 Thanks for a clear answer. I believe power of any meta-analysis will be less for random-effects model. On the other hand, usually the idea is to find what is happening in the population rather than just in those studies. So I presume that random-effects model needs to be used most of the time. Are there any circumstances when fixed effects model is appropriate and random-effects model is not? \u2013\u00a0rnso Jun 12 '15 at 10:13\n\u2022 \"When you want to make a conditional inference about the average outcome of the $k$ studies included in your analysis.\" Yes, I am repeating my answer, but that's a circumstance when a fixed-effects model is appropriate and a random-effects model is not. \u2013\u00a0Wolfgang Jun 13 '15 at 12:29\n\u2022 Might you be able to cite a good paper or link which gives an example of the two types of inferences at a step-by-step level? \u2013\u00a0cerd Mar 3 '16 at 13:51","date":"2021-05-09 04:54:52","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7458136081695557, \"perplexity\": 558.3939001068165}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-21\/segments\/1620243988955.89\/warc\/CC-MAIN-20210509032519-20210509062519-00607.warc.gz\"}"} | null | null |
Dr. Angela Intili, MD is an obstetrics & gynecology specialist in Joliet, IL and has been practicing for 32 years. She graduated from Universita Degli Studi Di Pavia, Facolta Di Medicina E Chirurgia in 1987 and specializes in obstetrics & gynecology.
I was looking for a new OBGYN and heard about how great Dr. Intili is so I went to go see her. She is a very caring doctor who answered all of my questions and listened to my concerns. She made me feel very comfortable.
×Post a ResponseAre you Dr. Angela Intili, MD?
I have gone to her for all my pregnancies. She's very thorough and caring. Obviously be prepared to wait on occasion she's an OB who has to randomly delivery babies sometimes.
She is always extremely attentive and thorough. Her office staff is helpful and courteous. I have been going to her for years and have never had a bad experience. She delivered my son and it was a smooth and wonderful experience!
How was your experience with Dr. Intili?
Check the quality of care at hospitals where Dr. Intili treats patients. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,864 |
How Forest Burning Could Have Become Federal Policy
Huell Explores Iconic Ferndell in Griffith Park
Huell gets the quintessential urban experience when he visits a magical garden in Griffith Park and then lunches at a newly opened cafe amongst the lush green trees.
E2: Hecho en México
E4: Gustavo and Friends
Watch some of host Gustavo Dudamel's favorite performances from throughout the years.
Black Rock-High Rock: The Negotiation Behind Conservation
A proposal to protect Nevada's Black Rock Desert-High Rock Canyon received much criticism.
How SNPLMA Wove Urban Las Vegas Back Into the Environment
The legislation helped balance urban growth in Las Vegas with environmental conservation.
Sen. Cortez Masto's Pursuit of a Green Economy
Cortez Masto has a goal to grow a clean energy economy that includes jobs for the future.
california's gold
Huell Searches for the 'Crookedest' Street in the World
A small section of Vermont Street in the Potrero Hill section of San Francisco is just miles away from the more famous Lombard Street. Which street is crookeder?
20 Fascinating — and Walkable — Places to Discover DTLA's Broadway Commercial District
From department stores to hotels, restaurants, nightclubs, and even makeshift churches, here's a guide to the 20 most fascinating Broadway landmarks you can discover on foot.
Eat and Drink Your Way Through the Last Episode of 'In Concert at the Hollywood Bowl'
From Japanese katsu sandos to Tijuana-style tacos and Hong Kong buns, here are some purveyors from Smorgasburg's lineup that will help you relish the last days of summer.
How to Discover the Wonders of Elysian Park by Car
How many of these "hiding-in-plain-sight" gems in Elysian Park have caught your eye? Explore them via car.
How Forest Burning Could Have Become Federal Policy | KCET
Josh Garrett-Davis is the Gamble Assistant Curator of Western History, Popular Culture, and Firearms at the Autry Museum of the American West. He is a PhD candidate in U.S. history at Princeton University and the author of Ghost Dances: Proving Up on the Great Plains.
A century of forest growth at Bear Creek Fire Guard Station in Plumas National Forest | Image: Capital Public Radio, from USFS photographs
Forests in California and across the American West face an odd calamity. They are crowded with trees, crawling with beetles, and loaded to explode with wildfire at the next spark. While other environmental crises in history—for instance, destruction caused by the industrial revolution—may have been largely unavoidable, this one truly might not have happened.
In the first decades of the twentieth century, a range of powerful actors, virtually all white men, struggled to define how American foresters ought to deal with forests and fires. The fire historian Stephen Pyne has recounted their debate in multiple books, perhaps most compellingly in Year of the Fires: The Story of the Great Fires of 1910. A range of opinions and practices vied for prominence, but basically they fell into two camps.
One was a sort of libertarian, local-control model, in which foresters close to the land administered regular, small fires to protect themselves and the resources around them against large conflagrations. In California, high-profile voices like the poet Joaquin Miller and Sunset magazine (owned by the Southern Pacific Railroad, a huge timber landowner itself and a supporter of light burning), defined this as "the Indian way." A California forester wrote in 1904, "The white man has come to think that fire is a part of the forest, and a beneficial part at that." A 1910 Sunset article declared, "practical foresters contend and can demonstrate that from time immemorial fire has been the salvation and preservation of our California sugar and white pine forests."
The second model was a top-down, expert approach associated with President Theodore Roosevelt's administration and a new conservation movement.
On the right, a more open Sierra forest with a plume of smoke suggesting a Native fire. At left, the dark, dense forest that came to be seen as natural by some foresters. | Artwork: Edwin Deakin, "Donner Lake," 1869, courtesy of the California Historical Society Collection, Autry Museum of the American West
In 1905, Roosevelt established the U.S. Forest Service, and installed his friend Gifford Pinchot as chief forester. Pinchot had studied forestry in France and founded the Yale School of Forestry in a quest to rationalize America's timber management. He and his successors saw all fire as profligate waste of timber, an inefficiency incompatible with a modern nation.
1930s-era fire-suppression poster | Image: U.S. Forest Service
Or, for that matter, a modern empire. The United States had recently conquered a number of territories in the Caribbean and the Pacific, and it still held New Mexico and Arizona (not to mention Alaska and Hawai'i) as imperial territories. Like those imperial holdings, the vast tracts of federal land across the West could also be administered by centralized authorities. As Pyne writes, "They would keep fire out by keeping fire-setting people out, and regulating those who could enter. The premier agencies were forestry bureaus like those devised by Britain for India and France for Algeria. In the late nineteenth century they offered the boldest experiments in wholesale conservation."
The fire suppression model mostly won out, its partisans consistently denigrating light burning as "Paiute forestry." In 1910, a ferocious series of wildfires in the Northern Rockies — one called the Big Blowup — captured the nation's attention, burning more than 3 million acres and killing at least 85 people.
Reaction to the Big Blowup set the new Forest Service on a path of absolute fire suppression for generations. Twenty years later, when Big Blowup veteran Ferdinand Silcox became head of the Forest Service, he enacted what became known as the"10:00 a.m. policy": a Forest Service goal to extinguish all wildfires by 10:00 am on the day after the fire was reported.
But the dogma of what one historian called the "gospel of efficiency" in the first decades of the twentieth century did not need to wipe out the practice of controlled burning completely. As an analogy, consider the similar, efficiency-inspired movement toward "Simplified Spelling" of American English from the same era, promoted by Roosevelt and funded by Andrew Carnegie. This kindred folly would have removed and replaced every silent or counterintuitive letter from the language, using spellings like "hav," "ruf," and "abuv" on the grounds that it would reduce the size of, say, the Encyclopedia Britannica from 24 to 20 volumes.
A USFS bookmark | Image: courtesy Estes Park Museum
Fire suppression could have flamed out too. It could have been a footnote, an absurd overreach by Progressive Era experts. Some balanced approach might have emerged, the way Americans changed the spellings of "colour," "honour," and "civilised," but left "rough" as it was. Fire suppression did become more decentralized, undertaken cooperatively between state and federal governments. But the timings of the Big Blowup, partisan and regional political fights, and now-outmoded conservation theories allowed fire suppression to become gospel long enough for the forests to grow impossibly, dangerously thick. Today, there is no easy way to put fire back on the land in many parts of California and the rest of the West without the chance of spurring the destructive fires known as holocausts.
After Long Delay, LADWP Fixes Broken Pipeline on Big Pine Paiute Reservation
Klamath River Tribes in Crisis as Salmon Disappear
LADWP's Owens Valley Flood 'Emergency' is an Answered Prayer for the Paiute and Shoshone
To be clear, light burning was not the same as cultural burning, which Native Californians had practiced for thousands of years. Though it was described as the "Indian way," and sometimes even learned directly from Native people, the crucial aspects of spiritual and particular place-based knowledge were mostly lost in translation. It would be hard to square cultural burning with the capitalist markets for timber that fueled light-burners and fire-haters alike.
But keeping in mind that California Indian populations were at their nadir in the beginning of the twentieth century after genocidal violence and a near complete loss of their land, the "folk" tending of forests by newcomers likely would have left the landscape in much better shape as Native people regained their populations and political power over the next century.
It remains to be seen what compromises and collaborations can develop in coming decades. Can tribal, state, federal, and other stakeholders define—in a positive recasting of the term—a new "Paiute forestry," or Mono forestry, or Karuk forestry, or other variations based on resource needs, local knowledge, and tradition?
Banner image: The Meadow Fire in Little Yosemite Valley, Yosemite National Park, 2014 | Photo: National Park Service via NASA Earth Observatory
Co-produced by KCETLink and the Autry Museum of the American West, the Tending the Wild series is presented in association with the Autry's groundbreaking California Continued exhibition.
E1: Cultural Burning - How Native American Peoples Use Fire to Rejuvenate the Land
Suppressed for over a century, indigenous cultural burning is still practiced today and holds important lessons for managing the threat of destructive wildfires.
E2: Keeping the River - How the Klamath River's Native Peoples Maintain Their Relationship With Salmon
The Yurok, Karuk, and Hupa peoples have maintained a close relationship with the Klamath River. They have secured traditional fishing rights and mobilized against the threats of dams and agriculture, setting an example for Native environmental rights.
E3: Weaving Community - How Native Peoples are Rediscovering Their Basketry Traditions
Despite barriers to access, traditional gathering and basket weaving is still practiced across California as a new generation is rediscovering and preserving its cultural heritage.
E4: Decolonizing the Diet - How Native Peoples are Reclaiming Traditional Foods
The Chia Cafe Collective is working to revive Native food practices and raise awareness about the threats to native plants in Southern California.
How can we address California's increasingly devastating wildfires?
Learn More About This Discussion | {
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{"url":"http:\/\/www.ck12.org\/tebook\/Basic-Speller-Teacher-Materials\/r1\/section\/6.13\/","text":"<meta http-equiv=\"refresh\" content=\"1; url=\/nojavascript\/\"> Lesson Thirty-seven | CK-12 Foundation\nYou are reading an older version of this FlexBook\u00ae textbook: Basic Speller Teacher Materials Go to the latest version.\n\n# 6.13: Lesson Thirty-seven\n\nCreated by: CK-12\n0 \u00a00 \u00a00\n\n## Soft <g> and Hard <g>\n\n1. You've seen that a soft <c> spells the sound [s], as in acid, and that a hard <c> spells the sound [k], as in actor. You\u2019ve also seen that a soft <c> has to have either an <e>, $<\\mathrm{i}>$ , or <y> right after it.\n\nThe letter <g> sometimes spells the sound [j] as in gem, and it sometimes spells the sound [g] as in gum. When it spells the [j] sound, it is called soft <g>. When it spells the [g] sound, it is called hard <g>.\n\n2. Pronounce each of the following words. Pay special attention to the sounds being spelled by the <g> in each of them. Sort the words into the matrix:\n\n$&\\text{agent} && \\text{ignorance} &&\\text{agriculture} && \\text{college} && \\text{angel}\\\\&\\text{recognize} && \\text{grower} && \\text{gypped} &&\\text{digest} && \\text{angle}\\\\&\\text{argue} && \\text{genies} && \\text{intelligence} && \\text{disgusted} && \\text{regret}\\\\&\\text{sergeant} && \\text{discharge} && \\text{glimpse} && \\text{goddess} && \\text{legislator}\\\\&\\text{challenge} && \\text{gleamed} &&\\text{twig} && \\text{biology} && \\text{frog}$\n\nWords in which <g> spells ...\n[j]: [g]:\nWords with <e>, $\\mathrm{}$, or <y> right after the <g>:\n\nagent\n\ndigest\n\nsergeant\n\nbiology\n\nchallenge\n\nangel\n\ngenies\n\nlegislator\n\ndischarge\n\ngypped\n\nintelligence\n\ncollege\n\nWords with no <e>,$\\mathrm{}$, or <y> after the <g>:\n\nrecognize\n\ngoddess\n\nargue\n\nangle\n\nignorance\n\nregret\n\ngrower\n\nfrog\n\ngleamed\n\nagriculture\n\nglimpse\n\ntwig\n\ndisgusted\n\n3. You should have found that the letter <g> spells the [j] sound only when it has one of three letters right after it. The three letters are <e> , $\\underline{}$ , and <y>.\n\nThe letter <g> is called soft <g> when it spells the sound [j].\n\nA soft <g> always has one of three letters right after it: <e>, $\\underline{}$ , or <y>.\n\n4. Soft <g> always will have <e>, $<\\mathrm{i}>$, or <y> , after it. But not every <g> that has one of these three letters after it is a soft <g>! Look at these words, with hard <g>'s where we'd expect soft ones: get, together, hunger, give, and girl.\n\nSo we can't say that any <g> with <e>,$<\\mathrm{i}>$, or <y> after it will be soft. But we can say that any soft <g> will have <e>, $<\\mathrm{i}>$, or <y> after it.\n\n5. The letter <c> is soft when it has the letters <e> , $\\underline{}$ , or <y> after it. The soft <c> spells the sound [s].\n\n6. Soft <c> and <g> always have the letters <e> , $\\underline{}$ , or <y> after them.\n\n7. Combine these free stems and suffixes. Watch for cases of twinning and final <e> deletion:\n\nFree Stem + Suffix = Word\ngod + d + ess = goddess\nbiologist + s = biologists\ndisgust + ing = disgusting\ngold + en = golden\ngyp +p + ing = gypping\nintelligent + iy = intelligently\nlegislat$\\cancel{e}$ + or = legislator\nignor$\\cancel{e}$ + ance = ignorance\n\nTeaching Notes. The distinction between hard and soft <g> is a perfect historical parallel to that between hard and soft <c>. Notice that the two hard sounds, [k] and [g], are an unvoiced-voiced pair. That is, they are identical sounds except that [k] is unvoiced, [g] voiced. Both are pronounced well back in the mouth. Just as with hard and soft <c>, the distinction between hard and soft <g> arose from the influence of the following vowel on the pronunciation of the consonant sound being spelled by the <g>. Front vowels, usually spelled <e>, $\\mathrm{}$, or <y>, tended to urge the pronunciation of the preceding consonant more towards the front of the mouth, so that [g] developed into [j].\n\nThis explanation is particularly true of words that came to English from or through Latin and French (exs: gelatin, gender, general, genesis, genius, gentle, genuine, geography, germ, gesture, giant, gigantic, ginger, giraffe, gist, gymnasium, gypsum). In native English words (exs:geese, gild, girdle) and in words from German and Scandinavian (exs: get, geyser, gift, gill, girth, give, gear), hard <g> is common before <e>, $\\mathrm{}$, or <y>. The soft <g>, [j], by and large echoes developments in late Latin, when the consonant spelled <g> came to be pronounced [j] before front vowels, which were usually spelled with <e>, $\\mathrm{}$, or <y>.\n\nItem 2. The hard-soft distinction can help students keep straight the often-confused angle and angel. 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Q: In Ride4dApps, how does the deposit and withdraw dApp(the example in Waves IDE) handle the delay issue? Their seems to be a delay write key and transferring asset which means the deposit function could be manipulated what if someone deposits then immediately withdraws the money from the invoke account the dApp will still write a record with the amount specified?
A: ScriptResult(WriteSet(), TransferSet()) is a total inseparable result of transaction. All transactions are processed strictly one after another. So such problem will never happen.
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\section{Introduction}
A classical result due to Toponogov \cite{Toponogov} says that the length of any closed simple geodesic $\gamma$ on a closed Riemannian surface $M^2$ satisfies
\begin{eqnarray*}
L(\gamma)^2\inf_MK\le4\pi^2,
\end{eqnarray*}
where $K$ is the Gaussian curvature of $M$. Furthermore, if equality holds, then $M^2$ is isometric to the standard unit 2-sphere $\mathbb{S}^2\subset\mathbb{R}^3$ up to scaling (see \cite{HangWang} for a different proof).
A similar result could be imagined for minimal 2-spheres, instead of closed simple geodesics, in dimension 3. But, it turns out that there is no area bound for minimal 2-spheres in Riemannian 3-manifolds, as pointed out by Marques and Neves \cite{MarquesNeves}. Therefore, an extra hypothesis is needed.
It is well known that if $\Sigma^2$ is a stable minimal 2-sphere in a Riemannian 3-manifold $M^3$, then the area of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.Bray.Brendle.Neves}
A(\Sigma)\left(\frac{\inf_MR}{2}\right)\le4\pi,
\end{eqnarray}
where $R$ is the scalar curvature of $M$. Moreover, if equality holds, then $\Sigma$ is totally geodesic and $R$ is constant equal to $\inf_MR$ on $\Sigma$. If we further assume that $\Sigma$ is locally area-minimizing, then equality in \eqref{eq.Bray.Brendle.Neves} implies $M$ to be isometric to $(-\varepsilon,\varepsilon)\times\mathbb{S}^2$ up to scaling in a neighborhood of $\Sigma$, suposing that $\Sigma$ is embedded in $M$. This can be seen as a consequence of Bray, Brendle, and Neves' work \cite{BrayBrendleNeves} (see \cite{MicallefMoraru} for an alternative proof).
In dimension $n\ge3$, it is not difficult to construct manifolds $\Sigma^n$ with scalar curvature $R_\Sigma\ge\alpha_n>0$, for some constant $\alpha_n$ depending only on $n$, and arbitrarily large volume. For example, consider $\Sigma_r^n=\mathbb{S}^{n-1}\times S^1(r)$, where $\mathbb{S}^{n-1}\subset\mathbb{R}^n$ is the standard unit $(n-1)$-sphere and $S^1(r)\subset\mathbb{R}^2$ is the circle of radius $r>0$. Clearly, $R_{\Sigma_r}=(n-1)(n-2)$ and $\Vol(\Sigma_r)\longrightarrow\infty$ as $r\longrightarrow\infty$. However, these manifolds are not diffeomorphic to $\mathbb{S}^n$.
For the spherical case, Gromov and Lawson \cite{GromovLawson} developed a method which permits to construct metrics on $\Sigma^n=S^n$ with scalar curvature $R_\Sigma\ge n(n-1)$ and arbitrarily large volume if $n\ge3$.
These examples show that the analogous inequality to \eqref{eq.Bray.Brendle.Neves} is not true in general in dimension $n\ge3$ even for volume-minimizing hypersurfaces, as we can see taking $M^{n+1}=\mathbb{R}\times\Sigma^n$ with $\Sigma^n$ as above.
Bearing this in mind, Barros, Batista, Cruz, and Sousa \cite{BarrosBatistaCruzSousa} considered the case of Einstein 4-manifolds embedded in Riemannian 5-manifolds which minimize the volume in their homotopy classes. They proved:
\begin{theorem}[Barros-Batista-Cruz-Sousa]\label{theorem.BarrosBatistaCruzSousa}
Let $M^5$ be a complete Riemannian manifold with positive scalar curvature and nonnegative Ricci curvature. Suppose that $\Sigma^4$ is a two-sided closed Einstein manifold embedded in $M^5$ in such a way that $\Sigma$ minimizes the volume in its homotopy class. Then, the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.BarrosBatistaCruzSousa}
\Vol(\Sigma)^{1/2}\left(\frac{\inf_MR}{12}\right)\le\Vol(\mathbb{S}^4)^{1/2}.
\end{eqnarray}
Moreover, if equality holds, then $\Sigma$ is isometric to $\mathbb{S}^4$, $M$ is isometric to $(-\varepsilon,\varepsilon)\times\mathbb{S}^4$ in a neighborhood of $\Sigma$, and the Riemannian cover of $M$ is isometric to $\mathbb{R}\times\mathbb{S}^4$, up to scaling.
\end{theorem}
Our purpose in this work is to generalize Theorem \ref{theorem.BarrosBatistaCruzSousa} for manifolds that are not necessarily Einstein. To do so, from the above comments, it is necessary an extra term in \eqref{eq.BarrosBatistaCruzSousa}. Our first result is the following:
\begin{theorem}[Theorem \ref{main.theorem}]\label{main.theorem.introduction}
Let $M^5$ be a Riemannian manifold with scalar curvature $R$ satisfying $\inf_MR>0$ and nonnegative Ricci curvature. If $\Sigma^4$ is a two-sided closed hypersurface embedded in $M^5$ which is locally volume-minimizing, then the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.intro.main.theorem.1}
\Vol(\Sigma)\left(\frac{\inf_MR}{12}\right)^2\le\Vol(\mathbb{S}^4)+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma,
\end{eqnarray}
where $\mathring{\Ric_\Sigma}$ is the traceless Ricci tensor of $\Sigma$. Furthermore, if equality holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ and $M$ is isometric to $(-\varepsilon,\varepsilon)\times\mathbb{S}^4$ in a neighborhood of $\Sigma$, up to scaling.
\end{theorem}
Our second result is the following:
\begin{theorem}[Theorem \ref{main.theorem.2}]\label{main.theorem.introduction.2}
Let $M^5$ be a complete Riemannian manifold with scalar curvature $R$ satisfying $\inf_MR>0$ and nonnegative Ricci curvature. Suppose that $\Sigma^4$ is a two-sided closed manifold immersed in $M^5$ in such a way that $\Sigma$ minimizes the volume in its homotopy class. Then, the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.intro.main.theorem.2}
\Vol(\Sigma)\left(\frac{\inf_MR}{12}\right)^2\le\Vol(\mathbb{S}^4)+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma.
\end{eqnarray}
Moreover, if equality holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ and the Riemannian cover of $M$ is isometric to $\mathbb{R}\times\mathbb{S}^4$, up to scaling.
\end{theorem}
\begin{remark}
{\em The covering map of Theorem \ref{main.theorem.2} is explicit. In fact, it is given by $G(t,x)=\exp_x(tN(x))$, $(t,x)\in\mathbb{R}\times\Sigma$, where $\exp$ is the exponential map of $M$ and $N$ is a unit normal vector field defied on $\Sigma$.}
\end{remark}
\section{Preliminaries}
In this section, we are going to present some terminologies and useful results.
Let $\Sigma^n$ be a connected closed (compact without boundary) manifold of dimension $n\ge3$. Denote by $\mathcal{M}(\Sigma)$ the set of all Riemannian metrics on $\Sigma$. The {\em Einstein-Hilbert functional} $\mathcal{E}:\mathcal{M}(\Sigma)\to\mathbb{R}$ is defined by
\begin{eqnarray*}
\mathcal{E}(g)=\dfrac{\int_\Sigma R_gdv_g}{\Vol(\Sigma^n,g)^{\frac{n-2}{n}}},
\end{eqnarray*}
where $R_g$ is the scalar curvature of $(\Sigma,g)$. Denote by $[g]=\{e^{2f}g:f\in C^\infty(\Sigma)\}$ the conformal class of $g\in\mathcal{M}(\Sigma)$. The {\em Yamabe invariant} of $(\Sigma,[g])$ is defined as the following conformal invariant:
\begin{eqnarray*}
\mathcal{Y}(\Sigma,[g])=\inf_{\tilde g\in[g]}\mathcal{E}(\tilde g).
\end{eqnarray*}
The classical solution of the Yamabe problem by Yamabe \cite{Yamabe}, Trudinger \cite{Trudinger}, Aubin \cite{Aubin76} (se also \cite{Aubin98}), and Schoen \cite{Schoen} says that every conformal class $[g]$ contains metrics $\hat g$, called {\em Yamabe metrics}, which realize the minimum:
\begin{eqnarray*}
\mathcal{E}(\hat g)=\mathcal{Y}(\Sigma,[g]).
\end{eqnarray*}
Such metrics have constant scalar curvature given by
\begin{eqnarray*}
R_{\hat g}=\mathcal{Y}(\Sigma^n,[g])\Vol(\Sigma^n,\hat g)^{-\frac{2}{n}}.
\end{eqnarray*}
Furthermore,
\begin{eqnarray*}
\mathcal{Y}(\Sigma^n,[g])\le\mathcal{Y}(\mathbb{S}^n,[g_{\can}])
\end{eqnarray*}
and equality holds if and only if $(\Sigma^n,g)$ is conformally diffeomorphic to the standard unit $n$-sphere $\mathbb{S}^n\subset\mathbb{R}^{n+1}$ endued with the canonical metric $g_{\can}$. Therefore, as a consequence of Obata's theorem \cite[Proposition 6.1]{Obata}, if $\mathcal{Y}(\Sigma^n,[g])=\mathcal{Y}(\mathbb{S}^n,[g_{\can}])$ and $g$ has constant scalar curvature, then $(\Sigma^n,g)$ is isometric to $(\mathbb{S}^n,g_{\can})$ up to scaling.
When $n=4$, a very useful tool is the Gauss-Bonnet-Chern formula for the Euler characteristic $\chi(\Sigma)$ of a closed Riemannian manifold $(\Sigma^4,g)$, which reads as follows:
\begin{eqnarray*}
8\pi^2\chi(\Sigma)=\int_\Sigma\left(\frac{1}{4}|W_g|^2+\frac{1}{24}R_g^2-\frac{1}{2}|\mathring{\Ric_g}|^2\right)d\sigma,
\end{eqnarray*}
where $W_g$ and $\mathring{\Ric_g}=\Ric_g-(R_g/n)g$ are the Weyl and the traceless Ricci tensors of $(\Sigma,g)$, respectively.
Before finishing this section, we are going to state two important inequalities proved by Gursky \cite{Gursky}.
\begin{theorem}[Gursky]
Let $(\Sigma^4,g)$ be a closed Riemannian manifold. If $\Sigma$ has nonnegative scalar curvature, then
\begin{eqnarray}\label{gursky.aux.1.1}
\int_\Sigma|W_g|^2d\sigma\ge32\pi^2(\chi(\Sigma)-2)
\end{eqnarray}
and
\begin{eqnarray}\label{gursky.aux.1.2}
\mathcal{Y}(\Sigma,[g])^2\ge6\left(32\pi^2\chi(\Sigma)-\int_\Sigma|W_g|^2d\sigma\right).
\end{eqnarray}
\end{theorem}
\begin{remark}
{\em Clearly, \eqref{gursky.aux.1.1} and \eqref{gursky.aux.1.2} are trivial if $\chi(\Sigma)\le 2$ or $\chi(\Sigma)\le0$, respectively.}
\end{remark}
\section{The results}
Let $\Sigma^4$ be a closed hypersurface immersed in a Riemannian manifold $M^5$. Here, we suppose that $\Sigma$ is {\em two-sided}, that is, there exists a unit normal vector field $N$ defined on $\Sigma$.
\begin{proposition}\label{main.proposition}
Let $M^5$ be a Riemannian manifold with scalar curvature $R$ satisfying $\inf_MR>0$ and $\Sigma^4$ be a two-sided closed hypersurface immersed in $M^5$. If $\Sigma$ is stable minimal in $M$, then the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.main.prop}
\Vol(\Sigma)\left(\frac{\inf_MR}{12}\right)^2\le\Vol(\mathbb{S}^4)+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma,
\end{eqnarray}
where $\mathring{\Ric_\Sigma}$ is the traceless Ricci tensor of $\Sigma$. Furthermore, if equality holds, then
\begin{enumerate}
\item[\rm (i)] $\Sigma$ is isometric to $\mathbb{S}^4$ up to scaling,
\item[\rm (ii)] $\Sigma$ is totally geodesic in $M$,
\item[\rm (iii)] $\Ric(N,N)=0$ and $R=\inf_MR$ on $\Sigma$,
\end{enumerate}
where $\Ric$ is the Ricci tensor of $M$.
\end{proposition}
\begin{proof}
Since the left hand side of \eqref{eq.main.prop} and $\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma$ are scaling invariant and $\inf_MR>0$, without loss of generality, we may assume that $\inf_MR=12$. Since $\Sigma$ is stable minimal, the stability inequality says that
\begin{eqnarray}\label{main.prop.aux.1}
\int_\Sigma(|\nabla f|^2-(\Ric(N,N)+|A|^2)f^2)d\sigma\ge0
\end{eqnarray}
for all $f\in C^\infty(\Sigma)$, where $A$ is the second fundamental form of $\Sigma$ in $M$. Taking $f=1$ above and using the Gauss equation
\begin{eqnarray}\label{main.prop.aux.2}
\frac{1}{2}(R-R_\Sigma+|A|^2)=\Ric(N,N)+|A|^2,
\end{eqnarray}
we have
\begin{eqnarray}\label{main.prop.aux.3}
\int_\Sigma(R+|A|^2)d\sigma\le\int_\Sigma R_\Sigma d\sigma,
\end{eqnarray}
where $R_\Sigma$ is the scalar curvature of $\Sigma$. Therefore, observing that $R\ge12$ by hypothesis, it follows that
\begin{eqnarray}\label{main.prop.aux.4}
12\Vol(\Sigma)\le\int_\Sigma R_\Sigma d\sigma\le\Vol(\Sigma)^{1/2}\left(\int_\Sigma R_\Sigma^2d\sigma\right)^{1/2},
\end{eqnarray}
i.e.,
\begin{eqnarray}\label{main.prop.aux.5}
12^2\Vol(\Sigma)\le\int_\Sigma R_\Sigma^2d\sigma,
\end{eqnarray}
where above we have used the H\"older inequality.
Now, let $\phi\in C^\infty(\Sigma)$, $\phi>0$, be the first eigenfunction of the stability operator $L$ of $\Sigma$,
\begin{eqnarray}\label{main.prop.aux.6}
L=\Delta+(\Ric(N,N)+|A|^2),
\end{eqnarray}
associated to the first eigenvalue $\lambda=\lambda_1$, that is,
\begin{eqnarray}\label{main.prop.aux.7}
L\phi+\lambda\phi=0.
\end{eqnarray}
Because $\Sigma$ is stable, $\lambda\ge0$. Denote by $g_\Sigma$ the Riemannian metric on $\Sigma$ induced from $M$ and define a new metric $g=\phi^{2/3}g_\Sigma$. It is well known that the scalar curvatures of $g$ and $g_\Sigma$ are related according to the equations
\begin{eqnarray*}
R_g\phi&=&-6\Delta(\phi^{1/3})+R_\Sigma\phi^{1/3}\\
&=&-2\phi^{-2/3}\Delta\phi+\frac{4}{3}\phi^{-5/3}|\nabla\phi|^2+R_\Sigma\phi^{1/3},
\end{eqnarray*}
which imply
\begin{eqnarray}\label{main.prop.aux.9}
R_g\phi\ge-2\phi^{-2/3}\Delta\phi+R_\Sigma\phi^{1/3}.
\end{eqnarray}
Then, using \eqref{main.prop.aux.2}, \eqref{main.prop.aux.6} and \eqref{main.prop.aux.7} into \eqref{main.prop.aux.9}, we obtain
\begin{eqnarray*}
R_g\phi&\ge&\phi^{1/3}(R+|A|^2+2\lambda)\ge12\phi^{1/3}>0,
\end{eqnarray*}
thus $R_g>0$. In particular, $\mathcal{Y}(\Sigma,[g_\Sigma])>0$. Denoting by $W_\Sigma$ the Weyl tensor of $(\Sigma,g_\Sigma)$ and remembering that $\int_\Sigma|W_\Sigma|^2d\sigma$ is a conformal invariant of $(\Sigma,g_\Sigma)$ in dimension 4, it follows from Gursky's theorem that
\begin{eqnarray}\label{main.prop.aux.11}
\int_\Sigma|W_\Sigma|^2d\sigma\ge32\pi^2(\chi(\Sigma)-2).
\end{eqnarray}
Then, using \eqref{main.prop.aux.5}, \eqref{main.prop.aux.11} and the Gauss-Bonnet-Chern formula, we have
\begin{eqnarray*}
\Vol(\Sigma)&\le&\frac{1}{6}\left(\frac{1}{24}\int_\Sigma R_\Sigma^2d\sigma\right)\\
&=&\frac{1}{6}\left(8\pi^2\chi(\Sigma)-\frac{1}{4}\int_\Sigma|W_\Sigma|^2d\sigma+\frac{1}{2}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma\right)\\
&\le&\frac{8}{3}\pi^2+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma,
\end{eqnarray*}
which imply \eqref{eq.main.prop} because $\Vol(\mathbb{S}^4)=\frac{8}{3}\pi^2$.
If equality in \eqref{eq.main.prop} holds, then we have equality in \eqref{main.prop.aux.1} for $f=1$. Which means that $\lambda_1=0$ and $f=1$ is the first eigenfunction of $L$, i.e., $\Ric(N,N)+|A|^2=0$. On the other hand, equality in \eqref{eq.main.prop} also implies equality in \eqref{main.prop.aux.3} and \eqref{main.prop.aux.4}. Therefore,
\begin{eqnarray*}
0&\le&\int_\Sigma(R-12)d\sigma\le\int_\Sigma(R+|A|^2)d\sigma-12\Vol(\Sigma)\\
&=&\int_\Sigma R_\Sigma d\sigma-12\Vol(\Sigma)=0,
\end{eqnarray*}
thus $\Sigma$ is totally geodesic and $R=12$ on $\Sigma$. In particular, $\Ric(N,N)=0$ on $\Sigma$. Also, from \eqref{gursky.aux.1.2} we have $R_\Sigma=12$.
To finish, observe that equality in \eqref{eq.main.prop} implies equality in \eqref{main.prop.aux.11}. Therefore, since $\mathcal{Y}(\Sigma,[g_\Sigma])>0$, using Gursky's theorem we obtain
\begin{eqnarray*}
\displaystyle\mathcal{Y}(\mathbb{S}^4,[g_{\can}])^2&\ge&\mathcal{Y}(\Sigma,[g_\Sigma])^2\ge6\left(32\pi^2\chi(\Sigma)-\int_\Sigma|W_\Sigma|^2d\sigma\right)\\
&=&384\pi^2=\mathcal{Y}(\mathbb{S}^4,[g_{\can}])^2.
\end{eqnarray*}
Then, $\mathcal{Y}(\Sigma,[g_\Sigma])=\mathcal{Y}(\mathbb{S}^4,[g_{\can}])$ and $R_\Sigma=12$, which imply by the solution of the Yamabe problem and Obata's theorem that $\Sigma$ is isometric to $\mathbb{S}^4$.
\end{proof}
\begin{remark}
{\em It follows from the above proposition that if equality in \eqref{eq.intro.main.theorem.1} or \eqref{eq.intro.main.theorem.2} holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ up to scaling. In particular, $\Sigma$ is Einstein. In this case, we can use Barros-Batista-Cruz-Sousa's theorem to obtain Theorem \ref{main.theorem.introduction} and Theorem \ref{main.theorem.introduction.2}. But, for the sake of completeness, we are going to present the proofs of these theorems here.}
\end{remark}
Before proving our main results, we are going to state a very useful lemma due to Bray, Brendle, and Neves \cite{BrayBrendleNeves} (see \cite{Nunes} for a more detailed proof). The same technique has been used by many authors in the literature (e.g. \cite{AnderssonCaiGalloway,BarrosBatistaCruzSousa,Cai,MicallefMoraru}).
\begin{lemma}\label{lemma.foliation}
Let $M$ be a Riemannian $5$-manifold. If $\Sigma$ is a two-sided closed minimal hypersurface immersed in $M$ such that $\Ric(N,N)+|A|^2=0$ on $\Sigma$, then there exists a smooth function $w:(-\varepsilon,\varepsilon)\times\Sigma\to\mathbb{R}$, for some $\varepsilon>0$, satisfying
\begin{eqnarray*}
w(0,x)=0,\,\,\frac{\partial w}{\partial t}(0,x)=1\,\,\mbox{ and }\,\,\int_\Sigma(w(t,\cdot)-t)d\sigma=0
\end{eqnarray*}
for all $x\in\Sigma$ and $t\in(-\varepsilon,\varepsilon)$. Furthermore,
\begin{eqnarray*}
\Sigma_t=\{\exp_x(w(t,x)N(x))\in M:x\in\Sigma\}
\end{eqnarray*}
is a closed hypersurface immersed in $M$ with constant mean curvature for each $t\in(-\varepsilon,\varepsilon)$. Also, if $\Sigma$ is embedded in $M$, then $\{\Sigma_t\}_{t\in(-\varepsilon,\varepsilon)}$ is a foliation of a neighborhood of $\Sigma=\Sigma_0$.
\end{lemma}
All entities associated to $\Sigma_t$ will be denoted with a subscript $t$, except the mean curvature which will be denoted by $H(t)$. Furthermore, $\rho_t$ will denote the lapse function $\langle\frac{\partial}{\partial t},N_t\rangle$.
\begin{theorem}\label{main.theorem}
Let $M^5$ be a Riemannian manifold with scalar curvature $R$ satisfying $\inf_MR>0$ and nonnegative Ricci curvature. If $\Sigma^4$ is a two-sided closed hypersurface embedded in $M^5$ which is locally volume-minimizing, then the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.main.theorem}
\Vol(\Sigma)\left(\frac{\inf_MR}{12}\right)^2\le\Vol(\mathbb{S}^4)+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma.
\end{eqnarray}
Furthermore, if equality holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ and $M$ is isometric to $(-\varepsilon,\varepsilon)\times\mathbb{S}^4$ in a neighborhood of $\Sigma$, up to scaling.
\end{theorem}
\begin{proof}
Inequality \eqref{eq.main.theorem} follows immediately from Proposition \ref{main.proposition}, since all locally volume-minimizing hypersurfaces are stable minimal. Also, if equality in \eqref{eq.main.theorem} holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ up to scaling and $\Ric(N,N)=0=|A|^2$ on $\Sigma$. In particular, we can use Lemma \ref{lemma.foliation}. It is well known that
\begin{eqnarray*}
\frac{dH}{dt}(t)=-\Delta_t\rho_t-(\Ric(N_t,N_t)+|A_t|^2)\rho_t.
\end{eqnarray*}
Since $\rho_0=1$ and $\Sigma$ is compact, we may assume that $\rho_t>0$ for all $t\in(-\varepsilon,\varepsilon)$. Therefore, using that $\Ric\ge0$ and $H'(t)$ is constant on $\Sigma_t$, we have
\begin{eqnarray}
H'(t)\int_{\Sigma_t}\frac{1}{\rho_t}d\sigma_t&=&-\int_{\Sigma_t}\frac{\Delta_t\rho_t}{\rho_t}d\sigma_t-\int_{\Sigma_t}(\Ric(N_t,N_t)+|A_t|^2)d\sigma_t\label{main.theorem.aux.2}\\
&\le&-\int_{\Sigma_t}\frac{|\nabla_t\rho_t|^2}{\rho_t^2}d\sigma_t\le0,\label{main.theorem.aux.3}
\end{eqnarray}
which imply $H'(t)\le0$ for all $t\in(-\varepsilon,\varepsilon)$, and then
\begin{eqnarray}\label{main.theorem.aux.4}
H(t)\le0\le H(-t)\,\,\mbox{ for all }\,\,t\in[0,\varepsilon),
\end{eqnarray}
because $H(0)=0$. On the other hand, the first variation formula says that
\begin{eqnarray}\label{main.theorem.aux.5}
\frac{d}{dt}\Vol(\Sigma_t)=\int_{\Sigma_t}H(t)\rho_td\sigma_t.
\end{eqnarray}
Then, \eqref{main.theorem.aux.4} and \eqref{main.theorem.aux.5} imply
\begin{eqnarray*}
\Vol(\Sigma_t)\le\Vol(\Sigma_0)\,\,\mbox{ for all }\,\,t\in(-\varepsilon,\varepsilon).
\end{eqnarray*}
But, since $\Sigma=\Sigma_0$ is locally volume-minimizing, we have $\Vol(\Sigma_t)=\Vol(\Sigma)$ for all $t\in(-\varepsilon,\varepsilon)$, for a smaller $\varepsilon>0$ if necessary. Therefore,
\begin{eqnarray*}
0=\frac{d}{dt}\Vol(\Sigma_t)=\int_{\Sigma_t}H(t)\rho_td\sigma_t
\end{eqnarray*}
and \eqref{main.theorem.aux.4} imply $H(t)=0$ for all $t\in(-\varepsilon,\varepsilon)$. Using $H'(t)=0$ into \eqref{main.theorem.aux.2} and \eqref{main.theorem.aux.3}, we conclude that $\rho_t$ is constant on $\Sigma_t$ and $\Sigma_t$ is totally geodesic in $M$ for each $t\in(-\varepsilon,\varepsilon)$.
Now, we want to prove that $t\longmapsto N_t(x)$ is a parallel vector field along to the curve $t\longmapsto G(t,x)=\exp_x(w(t,x)N(x))$ for each $x\in\Sigma$. In fact, choosing a local coordinate system $x=(x_1,x_2,x_3,x_4)$ on $\Sigma$, we have
\begin{eqnarray*}
\left\langle\nabla_\frac{\partial G}{\partial t}N_t,\frac{\partial G}{\partial x_i}\right\rangle&=&-\left\langle N_t,
\frac{\partial^2 G}{\partial x_i\partial t}\right\rangle=-\frac{\partial}{\partial x_i}\left\langle N_t,\frac{\partial G}{\partial t}\right\rangle+
\left\langle\nabla_\frac{\partial G}{\partial x_i}N_t,\frac{\partial G}{\partial t}\right\rangle\\
&=&-\frac{\partial}{\partial x_i}\rho_t=0.
\end{eqnarray*}
Above we have used that $\nabla_{\frac{\partial G}{\partial x_i}}N_t=0$ since $\Sigma_t$ is totally geodesic.
Also,
\begin{eqnarray*}
\left\langle\nabla_\frac{\partial G}{\partial t}N_t,N_t\right\rangle=\frac{1}{2}\frac{\partial}{\partial t}\langle N_t,N_t\rangle=0.
\end{eqnarray*}
Thus, $N_t$ is parallel.
On the other hand, we know that $t\longmapsto d(\exp_x)_{w(t,x)N(x)}N(x)$ is also parallel along to $t\longmapsto G(t,x)$. Then, $N_t(x)=(d\exp_x)_{w(t,x)N(x)}N(x)$ by uniqueness of parallel vector fields, since $w(0,x)=0$ and $N_0(x)=N(x)$. In particular,
\begin{eqnarray*}
\rho_t=\left\langle\frac{\partial G}{\partial t},N_t\right\rangle=\frac{\partial w}{\partial t}.
\end{eqnarray*}
Now, because $\int_\Sigma(w(t,\cdot)-t)d\sigma=0$ and $\rho_t=\frac{\partial w}{\partial t}$ is constant on $\Sigma_t$, we obtain
\begin{eqnarray*}
0=\frac{d}{dt}\int_\Sigma(w(t,\cdot)-t)d\sigma=\int_\Sigma\left(\frac{\partial w}{\partial t}-1\right)d\sigma=\left(\frac{\partial w}{\partial t}
-1\right)\Vol(\Sigma),
\end{eqnarray*}
which imply $\frac{\partial w}{\partial t}(t,x)=1$ for all $(t,x)\in(-\varepsilon,\varepsilon)\times\Sigma$. Finally, because $\frac{\partial w}{\partial t}(0,x)=1$, we have $w(t,x)=t$ for all $(t,x)\in(-\varepsilon,\varepsilon)\times\Sigma$. Therefore, $G(t,x)=\exp_x(tN(x))$ and we can easily check that $G$ is an isometry from $(-\varepsilon,\varepsilon)\times\Sigma$ to a neighborhood of $\Sigma$ in $M$.
\end{proof}
\begin{remark}\label{remark}
{\em Supposing that $\Sigma$ is immersed instead of embedded into $M$ in the above theorem, it follows from the same proof that $\Sigma$ is isometric to $\mathbb{S}^4$ up to scaling and $G$ is a local isometry from $(-\varepsilon,\varepsilon)\times\Sigma$ to $M$, if equality in \eqref{eq.main.theorem} holds.}
\end{remark}
The proof presented below is essentially the same as in \cite{BarrosBatistaCruzSousa}, \cite{BrayBrendleNeves}, and \cite{Nunes}.
\begin{theorem}\label{main.theorem.2}
Let $M^5$ be a complete Riemannian manifold with scalar curvature $R$ satisfying $\inf_MR>0$ and nonnegative Ricci curvature. Suppose that $\Sigma^4$ is a two-sided closed manifold embedded in $M^5$ in such a way that $\Sigma$ minimizes the volume in its homotopy class. Then, the volume of $\Sigma$ satisfies
\begin{eqnarray}\label{eq.main.theorem.2}
\Vol(\Sigma)\left(\frac{\inf_MR}{12}\right)^2\le\Vol(\mathbb{S}^4)+\frac{1}{12}\int_\Sigma|\mathring{\Ric_\Sigma}|^2d\sigma,
\end{eqnarray}
Moreover, if equality holds, then $\Sigma$ is isometric to $\mathbb{S}^4$ and the Riemannian cover of $M$ is isometric to $\mathbb{R}\times\mathbb{S}^4$, up to scaling.
\end{theorem}
\begin{proof}
Inequality \eqref{eq.main.theorem.2} follows directly from Theorem \ref{main.theorem}. Suppose that equality in \eqref{eq.main.theorem.2} holds and define $G:\mathbb{R}\times\Sigma\to M$ by $G(t,x)=\exp_x(tN(x))$. We claim that $G$ is a local isometry. In fact, define $I=\{t>0:G|_{(0,t)\times\Sigma}\,\,\mbox{ is a local isometry}\}$ and observe that Remark \ref{remark} implies $\Sigma$ to be isometric to $\mathbb{S}^4$ up to scaling and $I\neq\emptyset$. In particular, $\mathring{\Ric_\Sigma}=0$. On the other hand, it is not difficult to see that $I$ is closed in $(0,\infty)$. In fact, suppose that $t_k\in I$ converges to $t\in(0,\infty)$. If $t\le t_k$ for some $k$ then $t\in I$ because $(0,t)\times\Sigma\subset(0,t_k)\times\Sigma$ and $G|_{(0,t_k)\times\Sigma}$ is a local isometry. Otherwise, if $t_k<t$ for all $k$ then $\bigcup_k(0,t_k)\times\Sigma=(0,t)\times\Sigma$ implies that $t\in I$ because $G|_{(0,t_k)\times\Sigma}$ is a local isometry (which is a local property) for each $k$. Let us prove that $I$ is also open. Given $t\in I$, we have that $\Sigma_t=G(t,\Sigma)$ is homotopic to $\Sigma$ in $M$, $\Vol(\Sigma_t)=\Vol(\Sigma)$, and $\mathring{\Ric_{\Sigma_t}}=0$, because $G:\{t\}\times\Sigma\to\Sigma_t$ is a local isometric. In particular, $\Sigma_t$ minimizes the volume in its homotopy class and attains the equality in \eqref{eq.main.theorem.2}. Therefore, it follows from Remark \ref{remark} that there exists $\varepsilon>0$ such that $G|_{(0,t+\varepsilon)\times\Sigma}$ is a local isometry. This proves that $I$ is open in $(0,\infty)$. Thus, $I=(0,\infty)$, i.e., $G|_{(0,\infty)\times\Sigma}$ is a local isometry. Analogously, we can prove that $G|_{(-\infty,0)\times\Sigma}$ is a local isometry. This, together with Remark \ref{remark}, implies that $G$ is a local isometry. In particular, $G$ is a covering map.
\end{proof}
\bibliographystyle{amsplain}
| {
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} | 624 |
{"url":"https:\/\/www.physicsforums.com\/threads\/prove-n-3-less-than-or-equal-to-3-n-for-n-1-2.345591\/","text":"# Prove n^3 less than or equal to 3^n for n=1,2,\n\n1. Oct 14, 2009\n\n### naele\n\n1. The problem statement, all variables and given\/known data\nProve that $$n^3 \\leq 3^n$$ for $n=1,2,...$\n\n2. Relevant equations\nBinomial theorem\n\n3. The attempt at a solution\nCheck for P(1)= $$1^3 \\leq 3^1$$. So the base case holds, now assume P(k) is true and show that it implies P(k+1).\n\nBy assumption $$3\\cdot k^3 \\leq 3\\cdot 3^n$$. And this is where I get stuck because obviously $$(k+1)^3 \\nleq 3k^3$$\n\n2. Oct 14, 2009\n\n### Office_Shredder\n\nStaff Emeritus\nIf that last part is obvious, you should be able to prove it. It'll be tough...\n\nConsider instead expanding (k+1) and finding an upper bound for that based on k3\n\n3. Oct 14, 2009\n\n### naele\n\nWell it's not \"obvious\" but plotting $$3k^3 - (k+1)^3$$ shows negative values for positive k. On the other hand, I'm not sure what you mean by finding an upper bound.\n\nExpanding gives $$k^3+3k^2 +3k+1$$. Comparing that with $$k^3 + k^3 +k^3$$. I could try a term by term comparison of $$3\\cdot 3^k=3^k + 3^k +3 ^k$$. The first terms of each side are ok because that's by assumption. But $$3k^2 > 3^k$$ so hmmm.\n\n4. Oct 14, 2009\n\n### Office_Shredder\n\nStaff Emeritus\nIf you can find a natural number k that breaks the inequality, you're done. You've proven the hypothesis to be false.\n\nBy inductive hypothesis\n$$3k^2<k^3<3^k$$ as long as $k \\geq 3$\n\nSo your base case is in fact going to be a couple of values of k, but that's not a real problem\n\n5. Oct 14, 2009\n\n### HallsofIvy\n\nStaff Emeritus\nDo you realize that you are saying that the very statement you are trying to prove is \"obviously\" false? $3k^3= k^{k+1}$ so if $(k+1)^3\\nleq 3k^3= 3^{k+1}$ then $n^2< n^3$ is not true for all n.","date":"2017-08-23 10:56:23","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7961588501930237, \"perplexity\": 494.68621620102874}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-34\/segments\/1502886118195.43\/warc\/CC-MAIN-20170823094122-20170823114122-00321.warc.gz\"}"} | null | null |
You all know how the Humble Bundle works. Pay whatever you want, get some games. Pay a little more, get more games. It's been a successful business model for years now, and the new Star Wars bundle stays true to the formula. A total of nine games are currently available, with more coming soon.
Pay whatever you want, and you'll nab Star Wars: Knights of the Old Republic, Jedi Knight: Jedi Academy, and the original Dark Forces.
Pay more than the average (which as of press time stands at $10.90) and you'll get three additional games; Knight of the Old Republic 2: The Sith Lords, Battlefront 2, and Republic Commando.
For a whopping $12, you'll receive all of the games already listed, plus Empire at War: Gold Park, The Force Unleashed: Ultimate Sith Edition, and The Force Unleashed 2.
All of the games can be purchased through Steam, and will be available until February 17th. So what are you waiting for? Stop bullseyeing womp rats in your T-16, and get yourself some classic Star Wars games. | {
"redpajama_set_name": "RedPajamaC4"
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Main Blog Get max value
Clouds Win Over Logistics
Maxim Berezin: "Many logistics companies turn to clouds to ensure, among other things, corporate information safekeeping and protection."
According to the National Research University Higher School of Economics, a driver is the most common occupation in Russia. It's not surprising: we need not only to make or grow a product, but also to deliver it to a buyer on time through the logistics chain.
In classic economic models, only 5% of costs go to production, while 95% cover storage, warehousing, loading and unloading, and other logistics operations, with transportation accounting for 30-50% of total logistics costs. Reducing transportation logistics expenses helps cut the total cost of a product, thus boosting competitiveness and market share of both the manufacturer and its transportation contractor. Indeed, cost reduction benefits everybody, and IT systems, in particular, cloud services definitely can help achieve it.
Cloud scenarios for logistics
Cloud-based data protection
Cheaper and faster
Audit and optimization before migration
Logistics companies leverage a range of business-specific applications, such as Yard Management System (YMS) to manage warehouse area and vehicle movement within it, Transportation Management System (TMS) to manage cargo movement from origin to destination, or Warehouse Management System (WMS) to control handling of goods and materials at the warehouse. As these systems are crucial for their business, transportation companies prefer to develop and manage them by themselves.
However, IT infrastructure for those systems is a different story. Indeed, server, telecom, and engineering infrastructure for a data center is a large non-core business cost item for them. Therefore many transportation and warehousing companies increasingly go for Infrastructure as a Service to maximize cost efficiency. A company just needs to enter into an agreement with one or several cloud providers and regularly pay a subscription fee to be able to remotely manage business applications and the entire logistics chain within a single cloud platform. In other words, logistics businesses can outsource all technical routine to a cloud service provider, thus freeing up their own personnel for strategic matters, such as customer base expansion or new service development and launching.
The logistics chain consists of a consignor, carrier, and customer. They are usually different companies or legal entities within one holding company, each having its own business processes and e-document management system. When it comes to cargo, the chain participants need to take account of many parameters, including weight, volume, packaging type, vehicle, delivery time and conditions, possible damage, insurance, loading and unloading times. By moving to a cloud and using a single communication environment, the participants can unify processes bringing them to a whole new level.
Moreover, a cloud can help foreign companies ensure business continuity and comply with uniform IT industry standards. Thus, SK Operations, a logistics and supply chain service provider, consolidated all of its data, previously hosted on various sites including foreign ones, within a single Russian cloud platform, thus securing itself against possible regulatory claims.
Furthermore, clouds are popular with companies that develop their own logistics management applications. For example, one logistics operator that provides transportation and safekeeping services, decided to host its proprietary warehouse management system in CROC Cloud. As a result, in addition to traditional benefits such as reduced IT CapEx, quick start of a new business, and no need for large initial investment, the customer also gained access to cloud-based test environments, allowing it to promptly introduce changes to software code, with system improvements not affecting its performance.
Many logistics companies turn to clouds to ensure, among other things, corporate information safekeeping and protection, and thus reduce risks and stop worrying about IT environment support. As a result, they gain access to a redundant infrastructure designed and built to global standards, with uninterrupted service guaranteed even during equipment failure thanks to the second and sometimes the third backup data center.
Large cloud providers employ information protections covering entire infrastructure perimeter, including communication channel protection and DDoS attack prevention tools, check their cloud platforms for vulnerabilities every year (sometimes several times a year) with the engagement of independent auditors, and build isolated segments compliant with regulatory requirements for customers seeking maximum security for personal data they process.
To avoid corporate data loss, service providers create disaster-tolerant architectures based on a data center or a cloud, such as a cloud replica of a customer's infrastructure to take over system operation in case of a force majeure on the customer's site. Such failure protection allows customers to pay only for the external resources actually used at the time of emergencies. In addition, many companies use cloud backup for the most critical data, which we believe to be a part of almost 80% of projects completed by CROC Cloud Services.
Software providers and developers also take into account the growing demand of logistics companies for outsourcing, and thus offer multiple systems on a SaaS model as a way to reduce license costs. For example, there are automated systems that show optimal routes accelerating delivery; supply management solutions minimizing human error and improving the efficiency; as well as systems that track vehicles and ensure cargo safety during transportation (by monitoring temperature mode, carriage conditions, etc.) for real-time delivery control subject to traffic and other factors. CROC Cloud Services also offers SaaS solutions for logistics, including its transportation and supply chain management system.
For large logistics market players, automation has long been a key pillar of business management. Thus, the adoption of a cutting-edge platform for transportation and warehousing management helped Delovye Linii cargo carrier reduce its transportation expenses (by 10%), storage OPEX, and the number of vehicles (by 15%). Moreover, the company achieved 25% better warehouse throughput and 35% faster cargo handling.
Before migrating applications to a cloud, it is important to analyze their architecture to make sure they will effectively operate in the new environment. Even though many developers now offer cloud-native systems, logistics companies still often stick to traditional systems that cannot be moved to a cloud as is because it may affect service performance and cause multiple errors. Audits have become an essential part of CROC Cloud Services' business, as project results directly depend on timely system analysis and further optimization. Moreover, we are always ready to contribute to a system improvement during its operation in the cloud, including by way of process or system fine-tuning, scaling, or replicating. Thus, we can create duplicate WMS for newly established warehouses and branches, ensure unified system sharing, move a customer to a more powerful DBMS without stopping production systems, or perform a virtually seamless migration to a system by another vendor.
By the way, logistics system functionality can be quickly expanded by using a micro service platform that helps create a standalone requirements automation system that is independent from the core logistics system (ERP, WMS, or TMS). This way, integration can be completed rather quickly with minimum changes to the codes of core systems. Thus, we can either add the above functional modules or move non-standard functionality to an external environment for fine-tuning, while returning it to the core system module when necessary. A great example here is the implementation of CRM or SRM systems that work together with existing logistics systems. Even though CRM or SRM modules can be deployed separately, a micro service approach brings added business value, for example, in the form of a practically unlimited functionality expansion with centralized management, which will subsequently reduce target system integration and infrastructure costs.
Author: Maxim Berezin
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Q: Flutter isolate not showing notifications when app killed Flutter isolates using the flutter isolate package do not show notificaitons when the app has been killed.
Tried using the isolate package with dart:isolate, that does not work.
import 'dart:async';
import 'dart:isolate';
import 'package:flutter/material.dart';
import 'package:flutter_isolate/flutter_isolate.dart';
'package:flutter_local_notifications/flutter_local_notifications.dart';
FlutterLocalNotificationsPlugin flutterLocalNotificationsPlugin = new FlutterLocalNotificationsPlugin();
Future onSelectNotification(String payload) async {
if (payload != null) {
debugPrint('notification payload: ' + payload);
}
}
void isolated(String args) async {
Timer.periodic(new Duration(seconds: 10), (Timer t) => print('hi!'));
Timer.periodic(new Duration(seconds: 10), (Timer y) async
{
var androidPlatformChannelSpecifics = AndroidNotificationDetails(
'your channel id', 'your channel name', 'your channel description',
importance: Importance.Max,
priority: Priority.High,
ticker: 'ticker');
var iOSPlatformChannelSpecifics = IOSNotificationDetails();
var platformChannelSpecifics = NotificationDetails(
androidPlatformChannelSpecifics, iOSPlatformChannelSpecifics);
await flutterLocalNotificationsPlugin.show(
0, 'plain title', 'plain body', platformChannelSpecifics,
payload: 'item x');
});
}
void main() async {
// initialise the plugin. app_icon needs to be a added as a drawable resource to the Android head project
// If you have skipped STEP 3 then change app_icon to @mipmap/ic_launcher
var initializationSettingsAndroid =
new AndroidInitializationSettings('appicon');
var initializationSettingsIOS = new IOSInitializationSettings();
var initializationSettings = new InitializationSettings(
initializationSettingsAndroid, initializationSettingsIOS);
flutterLocalNotificationsPlugin.initialize(initializationSettings,
onSelectNotification: onSelectNotification);
isolated("hi");
}
The code above is what the system sees when the app has been killed as the state classes have terminated, and no longer exist. If we can get the code above running then the isolate will still work when the app has been killed by the user in Android and iOS and the state classes are no longer there.
Thanks in advance.
Rob.
A: Yeah it can't be done in Flutter alone as the isolates are spawned by the parent process and they get killed when the app is terminated.
for Android use services which get spawned and persist after the app dies.
For iOS you can run a background thread which gets spawned after a minimum amount of time, but there's no guarantee as to when that will run. You can, at Apple's discretion, run apps as audio or location based, which will allow the background thread to run continuously.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,610 |
{"url":"https:\/\/solvedlib.com\/what-are-some-of-the-most-critical-components-of,118123","text":"# What are some of the most critical components of a good literature review? How will you...\n\n###### Question:\n\nWhat are some of the most critical components of a good literature review? How will you ensure that you write a successful literature review for your planned research proposal? What are some of the challenges you foresee when writing your literature review?\n\n#### Similar Solved Questions\n\n##### 9. Let C1 be the line segment connecting (0,0,0) to (1,1,0) and Cz be the line segment connecting (1,1,0) to (1,1,1).a) Parametrize both C1 and Cz and find |vl for each: (b) Show that Jcuce 1 - 3y2 + 2 ds (V2+3)\/2.\n9. Let C1 be the line segment connecting (0,0,0) to (1,1,0) and Cz be the line segment connecting (1,1,0) to (1,1,1). a) Parametrize both C1 and Cz and find |vl for each: (b) Show that Jcuce 1 - 3y2 + 2 ds (V2+3)\/2....\n##### Sing entries are posted? Exercise 3-22A Recording events in the general journal, posting to T-accounts, and...\nsing entries are posted? Exercise 3-22A Recording events in the general journal, posting to T-accounts, and LO preparing closing entries At the beginning of Year 2, Oak Consulting had the following normal balances in its accounts: Account Balance Cash Accounts Receivable Accounts Payable Common Stoc...\n##### 10x 2v Let f(x,v) and C be the path from (1,~ 3) to (2,0} along the 1+x2 parabola v=x?_ 4, Then Jvf < dr3\n10x 2v Let f(x,v) and C be the path from (1,~ 3) to (2,0} along the 1+x2 parabola v=x?_ 4, Then Jvf < dr 3...\n##### How do you factor by grouping ac-bc+ad-bd?\nHow do you factor by grouping ac-bc+ad-bd?...","date":"2022-08-12 05:46:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 2, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5216888785362244, \"perplexity\": 5045.236687832229}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-33\/segments\/1659882571584.72\/warc\/CC-MAIN-20220812045352-20220812075352-00555.warc.gz\"}"} | null | null |
Professional Begging
Begging has been around for thousands of years.
Wherever there are a large number of people living together – usually in a big town or a city – we will probably find beggars. People that are down on their luck and in need. With no regular income to support them, they have no choice but to live on the streets and beg for money.
But are all beggars real genuine cases? Or are some of them professional beggars? These are people who claim to be homeless and without a regular income but resort on the charity and kindness of others to help them.
Are professional beggars a real phenomenon – or just a modern-day myth?
In the following monologue, we meet Colin – a man who makes a living from begging on the streets.
Professional Beggar
It's not like I was really in dire need. I wasn't. But it was so easy.
Too easy, to be honest.
The money you can make from begging is just unbelievable.
I've been doing this for five years now — yes, I suppose you could say I am a professional beggar.
But the first year I did it, I was a real beggar. I was a genuine case. I had lost my job, then I lost my house — I had nowhere to live. Things were pretty desperate.
My life was at an all-time low. I was on the streets. I slept in shop doorways, underground car parks — anywhere I could get my head down for the night.
I was begging — but not just surviving.
But I found that begging is like a real job. If you present yourself in the right way and have the right attitude, then you can make a lot of money.
About a year later, I was back on my feet. I had somewhere to live, and I had money in the bank. But I still didn't have a job.
The job market in my city really sucks. There are no jobs for someone in my industry. I thought about doing other kinds of work, but the only vacancies were in places like fast-food outlets and coffee shop chains.
I couldn't do that. I thought I was too old for that kind of job.
So I ended up staying on the streets, and that's when I turned professional. I made begging a business.
For one thing, the money was just too good. I now make in just four hours what it would take me all day to make in my last job.
Plus, I would have to pay tax on whatever I earned. Begging is tax-free.
To do begging full-time was just a no-brainer to be honest.
If you want to be a professional beggar, I can give you some tips.
There are two things you really need to think about when begging — the first is, location is all-important. And the second is how you present yourself.
You can't go to any main tourist spot in the city. Those places are just teeming with beggars and other people trying to scam money out of the tourists.
Avoid those places at all costs.
And the second is to make sure you dress the part.
That means shabby-looking clothes, not too dirty, or people will just assume you are a drug addict. You need to look down on your luck but still trying to get out of the hole you are in.
Another thing you can definitely try is humour.
I have used a sense of humour to great effect in the last five years. A sign with a funny message or making a little joke as people pass by is all the difference between people giving you money or not.
Humour goes a long way. It brightens people's day.
People respect and admire someone that can smile through all adversity. They will think of you as a stronger person who is trying to make their life better.
And then they will reward you for it.
I know many people would really look down on me for doing what I am doing. They would say that there are people who are genuinely homeless and in real need.
These people rely on others' generosity and kindness.
But to be honest, I don't feel bad about it.
I was really homeless myself a few years ago. So I feel that I have earned the right to still beg on the streets.
It's the government's fault — they should do something about it. Give homes to the homeless, and make sure they have enough to eat.
But they don't do that.
And if people want to give me any of their spare change, then that is up to them.
dire need a genuine case pretty desperate
all-time low on the streets get my head down
the right attitude back on my feet job market
really sucks vacancies chains
tax earned tax-free
no-brainer tips tourist spot
teeming scam money at all costs
dress the part shabby-looking drug addict
down on your luck out of the hole sense of humour
to great effect goes a long way brightens someone's day
admire adversity reward you
look down on in real need rely on
generosity earned the right spare change
How long has Colin been begging?
Does Colin feel bad about begging? How does he describe it?
Describe Colin's life when he first started begging. What were his circumstances?
Where did he sleep?
How long did it take him to get back to a good position in his life?
Is it easy to get a job in Colin's town? Why/why not?
What jobs did Colin consider?
Why did he reject certain job offers?
How long does Colin work at begging to make enough for a full-day in his last job?
What are his two pieces of advice for begging?
Which place should people avoid if they are begging according to Colin?
What added extra tip does Colin suggest?
Does Colin feel bad about begging? Why/why not?
Who does he think should help beggars?
What do you think about Colin? Is he a good person or not? Why do you think this?
Do you think he should stop begging? Why/why not?
Could Colin have tried harder in getting a job? Explain your reasons.
Should he have accepted a job in the coffee shop? Why/why not?
How would you describe Colin's personality? Is he stupid? Clever? Go into detail and describe what kind of person you think he is.
Do you think it is possible to become a professional beggar?
Do professional beggars exist in your country? Give some details about this.
How do you know if a beggar is real or not?
Does your city have a homeless problem? What is being done about it?
Should people give money to beggars? Why/why not?
Have you ever given money to beggars? If so, why did you do so? If not, why did you not?
In your country, can beggars work in organised gangs? Talk about this if you can.
FOR TEACHERS: You can use this lesson as part of your English or ESL class.
FOR STUDENTS: You can use this lesson in your own time to help you improve your reading and speaking skills. It can also help you broaden your vocabulary.
Was this useful in your English/ESL class?
Let me know in the comments below!
Why not download this full and complete lesson plan for free? Click the link below!
A Beggar's Life
2 thoughts on "Professional Begging"
What a fascinating topic and one not often spoken about. I think this topic is good anywhere in the world for an ESL/EFL classes. There are always a few beggars around and it is interesting to get a look behind the scenes. Thank you for presenting the other side of the coin.
David Buckley
Thanks very much Leona! I hope it is useful to teachers and students. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,191 |
\section{Introduction}
\label{sect:intro}
The creation of stellar evolutional scheme requires a knowledge of
the fundamental parameters of stars in different stages of their
evolution. Eclipsing binary systems, especially W UMa binaries,
are the most important sources of such information. It is supposed
that they are result from the evolution of wide binaries by
angular momentum loss and mass-ratio reversals
(\citealt{Stepien+2006}, \citealt{Qian+2003}). Around 25 $\%$ of
main-sequence star binaries have separations small enough so when
their primaries ascend the giant branch, mass transfer via
Roche-lobe overflow puts the begining of a common envelope phase
(\citealt{Willems+Kolb+2004}). At this stage the two stars orbit
within a single envelope of material, quickly losing angular
momentum and spiraling towards each other (\citealt{Webbink+1984};
\citealt{Ivanova+etal+2013}). The common-envelope phase is
probably a short-lived stage that ends by envelope ejection and a
tighter binary or by a merger. But understanding of common
envelope stage remains one of the most important unsolved problems
in stellar evolution (\citealt{Ivanova+etal+2013}).
The components in a W UMa system have nearly equal surface
temperatures in spite of their often greatly different masses
(\citealt{Binnendijk+1965}). The model of \cite{Lucy+1968a},
\cite{Lucy+1968b} explained this effect by a common convective
photosphere which embedded two main sequence stars. As a result
one should expect the observable luminosities to have another
dependence on the mass ratio than would be the case of two main
sequence stars in detached configuration. The condition for equal
surface temperatures leads to specific period-luminosity-color
(PLC) relations of W UMa stars (\citealt{Rucinski+1994},
\citealt{Rucinski+Duerbeck+1997}) allowing currently to predict
their absolute magnitudes $M_V$ to about 0.25 mag
(\citealt{Rucinski+2004}). The PLC relations combined with the
ease detection make these binaries useful tracers of distance
(\citealt{Klagyivik+Csizmadia+2004}; \citealt{Gettel+etal+2006};
\citealt{Eker+etal+2009}). Moreover, the W UMa stars are important
targets for the modern astrophysics because they give information
for the processes of tidal interactions, mass loss and mass
transfer, angular momentum loss, merging or fusion of the stars
(\citealt{Martin+etal+2011}).
In this paper we present photometric observations and light curve
solutions of four W UMa binaries: V796 Cep (GSC 04502-00138,
TYC 4502-138-1), V797 Cep (GSC 04502-01040, 2MASS
J01424764+8007522), CSS J015341.9+381641, NSVS 3853195
(CSS J015404.1+382805). Table~1 presents the coordinates of our
targets and available information for their light variability.
\begin{table}
\begin{minipage}[t]{\textwidth}
\caption{Parameters of our targets from the VSX database}
\label{Tab1}
\centering
\begin{scriptsize}
\begin{tabular}{clcllcc}
\hline\hline
Target & RA & Dec & Period & V & Amplitude & Type \\
& & & [d] & [mag] & [mag] & \\
\hline
\noalign{\smallskip}
V796 Cep & 01 41 36.39 & +80 04 19.1 & 0.3929661 & 12.20 & 0.65 & EW \\
V797 Cep & 01 42 47.64 & +80 07 52.3 & 0.270416 & 14.60 & 0.40 & EW \\
CSS J015341.9+381641 & 01 53 41.95 & +38 16 41.1 & 0.347518 & 13.47 & 0.40 & EW \\
NSVS 3853195 & 01 54 04.05 & +38 28 05.2 & 0.29253 & 13.52 & 0.39 & EW \\
\hline\hline
\end{tabular}
\end{scriptsize}
\end{minipage}
\end{table}
\begin{table}
\begin{minipage}[t]{\textwidth}
\caption{Journal of our photometric observations}
\label{Tab2}
\centering
\begin{scriptsize}
\begin{tabular}{ccrrrr}
\hline\hline
Target& Date & Exposure $g'$ & Exposure $i'$ & Number $g'$ & Number $i'$ \\
\hline
V796 Cep, V797 Cep & 2015 Oct 25 & 90& 120 & 125& 125 \\
& 2015 Oct 26 & 90& 120 & 84& 82 \\
& 2015 Oct 27 & 90& 120 & 60& 59 \\
& 2015 Oct 28 & 90& 120 & 146& 146 \\
CSS J015341.9+381641, NSVS 3853195 & 2015 Nov 07 & 60 & 90 & 85 & 84 \\
& 2015 Nov 08 & 60& 90 & 67& 75 \\
& 2015 Nov 11 & 60& 90 & 84& 84 \\
& 2015 Nov 12 & 60& 90 & 43& 42 \\
& 2015 Nov 13 & 60& 90 & 84& 85 \\
\hline\hline
\end{tabular}
\end{scriptsize}
\end{minipage}
\end{table}
\begin{table}
\begin{minipage}[t]{\textwidth}
\caption{List of the standard stars}
\label{Tab3}
\centering
\begin{scriptsize}
\begin{tabular}{ccccll}
\hline\hline
Label & Star ID & RA & Dec & \emph{g'} & \emph{i' }\\
\hline
Target 1 & V0796 Cep & 01 41 36.39 & +80 04 19.10 & 12.320 & 11.789 \\
Target 2 & V0797 Cep & 01 42 47.64 & +80 07 52.30 & 14.966 & 14.025 \\
Chk & UCAC4 851-002007 & 01 41 16.52 & +80 04 21.76 & 13.755 (0.010) & 13.018 (0.010) \\
C1 & UCAC4 851-002085 & 01 45 07.01 & +80 10 45.03 & 13.238 (0.011) & 12.534 (0.011) \\
C2 & UCAC4 851-002011 & 01 41 28.03 & +80 11 18.42 & 13.870 (0.010) & 13.351 (0.012) \\
C3 & UCAC4 851-002002 & 01 40 56.97 & +80 04 14.51 & 13.448 (0.009) & 12.844 (0.010) \\
C4 & UCAC4 851-002062 & 01 43 56.73 & +80 02 08.49 & 14.205 (0.011) & 13.452 (0.013) \\
C5 & UCAC4 850-002063 & 01 41 51.38 & +79 56 58.55 & 13.257 (0.007) & 12.651 (0.009) \\
C6 & UCAC4 851-002028 & 01 42 25.41 & +80 01 00.68 & 13.898 (0.009) & 13.058 (0.009) \\
\hline
Target 3 & CSS J015341.9+381641 & 01 53 41.95 & +38 16 41.10 & 13.897 & 13.117 \\
Target 4 & NSVS 3853195 & 01 54 04.05 & +38 28 05.26 & 14.054 & 13.178 \\
Chk & UCAC4 643-007188 & 01 54 30.68 & +38 29 00.15 & 13.104 (0.014) & 11.754 (0.009) \\
C1 & UCAC4 644-007104 & 01 54 12.54 & +38 36 49.06 & 13.975 (0.016) & 13.902 (0.018) \\
C2 & UCAC4 643-007165 & 01 54 04.61 & +38 35 54.88 & 14.112 (0.011) & 13.419 (0.015) \\
C3 & UCAC4 643-007180 & 01 54 23.95 & +38 35 42.60 & 14.061 (0.014) & 13.369 (0.015) \\
C4 & UCAC4 643-007204 & 01 54 50.23 & +38 33 52.57 & 13.971 (0.019) & 13.147 (0.015) \\
C5 & UCAC4 643-007182 & 01 54 26.03 & +38 29 38.62 & 13.720 (0.009) & 13.012 (0.011) \\
C6 & UCAC4 643-007126 & 01 53 26.47 & +38 30 02.18 & 13.702 (0.016) & 12.978 (0.013) \\
C7 & UCAC4 642-006881 & 01 54 12.52 & +38 23 56.10 & 13.937 (0.012) & 12.907 (0.011) \\
C8 & UCAC4 642-006842 & 01 53 30.38 & +38 20 34.86 & 13.929 (0.024) & 13.071 (0.018) \\
C9 & UCAC4 642-006908 & 01 54 36.06 & +38 20 04.67 & 14.068 (0.013) & 11.491 (0.014) \\
C10 & UCAC4 642-006921 & 01 54 46.30 & +38 19 35.13 & 13.161 (0.021) & 11.764 (0.014) \\
C11 & UCAC4 643-007147 & 01 53 51.89 & +38 28 10.11 & 12.849 (0.018) & 12.191 (0.014) \\
\hline
\end{tabular}
\end{scriptsize}
\end{minipage}
\end{table}
\section{Observations}
Our CCD photometric observations of the targets in Sloan \emph{g',
i'} bands were carried out at Rozhen National Astronomical
Observatory with the 30-cm Ritchey-Chr\'{e}tien Astrograph
(located into the \emph{IRIDA South} dome) using CCD camera ATIK
4000M (2048 $\times$ 2048 pixels, 7.4 $\mu$m/pixel, field of view
35 $\times$ 35 arcmin). Information for our observations is
presented in Table~2. In fact, the pairs V796 Cep, V797 Cep and
CSS J015341.9+381641, NSVS 3853195 fall on two observed fields
(see coordinates in Table 1).
The data were obtained during photometric nights with seeing
within 1.1--1.9 arcsec and humidity below 70 $\%$. Twilight flat
fields were obtained for each filter, dark and bias frames were
also taken throughout the run. The frames were combined
respectively into a single master bias, dark and flat frames. The
standard procedure was used for reduction of the photometric
data (de-biasing, dark frame subtraction and flat-fielding) by
software \textsc{AIP4WIN2.0} (\citealt{Berry+Burnell+2006}).
We used aperture photometry with radius of 1.5 FWHM of the star
image, along with sky background measurements with annuli
enclosing a comparable area. The light variability of the targets
was estimated with respect to nearby comparison (constant) stars
in the observed field of each target, so called ensemble
photometry. A check star served to determine the observational
accuracy and to check constancy of the comparison stars. The CCD
ensemble photometry calculates the difference between the
instrumental magnitude of the target and a comparison magnitude
obtained from the mean of the intensities of the comparison stars.
The use of numerous comparison stars scattered over the CCD field
increases considerably the statistical accuracy of the comparison
magnitude (\citealt{Gilliland+Brown+1988},
\citealt{Honeycutt+1992}).
We performed the ensemble aperture photometry with the software
\textsc{VPHOT}. Table~3 presents the coordinates of the comparison
and check stars of our targets from the catalogue UCAC4
(\citealt{Zacharias+etal+2013}) and their magnitudes from the
catalogue APASS DR9 (\citealt{Henden+2016}). The values in
brackets correspond to the standard deviations of the standard
stars during the observational nights. The choice of comparison
and check stars in the same field of view of the targets means
practically equal extinctions for all stars. The transformation of
the obtained instrumental magnitudes to standard ones was made
manually. For this aim we used the mean color of the ensemble
comparison-star $\overline{(g'-i')}_{comp}$ and transformation
coefficients of our equipment (calculated earlier using standard
star field M67). The calculated corrections of the instrumental
magnitudes for our targets were from $-$0.0008 mag to 0.0003 mag
in $g'$ filter (within the observational precision) and from
$-$0.0258 mag to 0.0085 mag in $i'$ filter.
\begin{figure}
\centering
\includegraphics[width=8cm, angle=0]{0253fig1.eps}
\caption{Illustration of the \emph{q}-search analysis for V796 Cep:
the different isolines circumscribe the areas whose
normalized $\chi^2$ are smaller than the marked values; the empty
circle corresponds to the final value of the mass ratio and
orbital inclination given in Table 5.}
\label{Fig1}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=14cm,scale=1.00]{0253fig2.eps}
\caption{The folded light curves of the targets with their fits
and the corresponding residuals (shifted vertically by different amount to save space).}
\label{Fig2}
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=14cm,scale=1.00]{0253fig3.eps}
\caption[]{3D configurations of the targets}
\label{Fig3}
\end{center}
\end{figure}
We determined the times of the individual minima (Table 4)
by the method of \cite{Kwee+Woerden+1956}.
\begin{table}
\begin{center}
\caption[]{Times of minima of our targets \label{Tab4}}
\begin{scriptsize}
\begin{tabular}{cccc}
\hline\hline \noalign{\smallskip}
Target & Min. I & Min. II & IRIDA cycle \\
\hline
V0796 Cep & 2457321.43582(9) & - & 0.0 \\
& - & 22457322.41776(13) & 2.5 \\
& 2457323.40045(19) & - & 5.0 \\
& - & 2457324.38263(8) & 7.5 \\
& 2457324.57929(1) & - & 8.0 \\
\hline
V0797 Cep & 2457321.31715(74) & - & 0.0 \\
& - & 2457321.45201(25) & 0.5 \\
& 2457321.58629(22) & - & 1.0 \\
& - & 2457322.26135(33) & 3.5 \\
& 2457322.39815(24) & - & 4.0 \\
& - & 2457323.34511(22) & 7.5 \\
& 2457324.29034(18) & - & 11.0 \\
& - & 2457324.42626(3) & 11.5 \\
\hline
CSS J015341.9+381641 & - & 2457333.44320(11) & -0.5 \\
& 2457333.61496(31) & - & 0.0 \\
& - & 2457334.48599(14) & 2.5 \\
& 2457338.47982(18) & - & 14.0 \\
& - & 2457340.39396(31) & 19.5 \\
& 2457340.56578(15) & - & 20.0 \\
\hline
NSVS 3853195 & - & 2457333.44266(9) & -0.5 \\
& 2457333.59001(21) & - & 0.0 \\
& 2457334.46774(15) & - & 3.0 \\
& - & 2457338.41610(11) & 16.5 \\
& 2457338.56296(13) & - & 17.0 \\
& 2457339.44062(12) & - & 20.0 \\
& - & 2457340.46366(14) & 23.5 \\
\hline\hline
\end{tabular}
\end{scriptsize}
\end{center}
\end{table}
\section{Light curve solutions}
The light curves of our targets were solved by the code
\textsc{PHOEBE} (\citealt{Prsa+Zwitter+2005}). It is based on the
Wilson--Devinney (WD) code (\citealt{Wilson+Devinney+1971},
\citealt{Wilson+1979}) but also provides a graphical user
interface and modeling in Sloan filters of our observations. We
used the traditional convention the Min. I (phase 0.0) to be the
deeper light minimum and the star that is eclipsed at Min. I to be
a primary component.
Mean temperatures $T_{m}$ of the binaries were determined in
advance (see Table 6 further) on the basis of their infrared color
indices \emph{(J-K)} from the 2MASS catalog and the calibration
color-temperature of \cite{Tokunaga+2000}. The preliminary runs
revealed that all targets are overcontact systems. Hence, we
applied mode ''Overcontact binary not in thermal contact'' of the
code. Firstly we fixed $T_{1}$ = $T_{m}$ and varied the initial
epoch $T_{0}$ and period \emph{P} to search for fitting the phases
of light minima and maxima. After that we fixed their values and
varied simultaneously secondary temperature $T_{2}$, orbital
inclination $i$, mass ratio $q$ and potential $\Omega$ to search
for reproducing of the whole light curves. The data in \emph{i'}
and \emph{g'} bands were modelled simultaneously.
We adopted coefficients of gravity brightening $g_1=g_2$ = 0.32
and reflection effect $A_1=A_2$ = 0.5 appropriate for late-type
stars while the linear limb-darkening coefficients for each
component and each color were updated according to the tables of
\cite{vanHamme+1993}. Solar metallicity was assumed for the
targets because they consist of late stars from the solar
vicinity. In order to reproduce the light curve distortions we
used cool spots whose parameters (longitude $\lambda$, angular
size $\alpha$ and temperature factor $\kappa$) were adjusted.
After reaching the best solution we varied together all parameters
($T_{2}$, $i$, $q$, $\Omega$, $T_{0}$ and \emph{P}) around the
values from the last run and obtained the final model. In order to
determine stellar temperatures $T_{1}$ and $T_{2}$ around the mean
value T$_{m}$ we used the formulae
(\citealt{Kjurkchieva+etal+2016a}):
\begin{equation}
T_{1}=T_m + \frac{\Delta T}{c+1},
\end{equation}
\begin{equation}
T_{2}=T_1 -\Delta T,
\end{equation}
where $c=l_2/l_1$ (luminosity ratio) and $\Delta T=T_m-T_2^{PH}$
were taken from the last \textsc{PHOEBE} fitting.
Although \textsc{PHOEBE} (as WD) works with potentials, it gives a
possibility to calculate all values (polar, point, side, and back)
of relative radius $r_i=R_i/a$ of each component ($R_i$ is linear
radius and \emph{a} is orbital separation). In the absence of
radial velocity curves we put as default \emph{a} = 1 because from
photometry only we cannot determine binary separation. Moreover,
\textsc{PHOEBE} yields as output parameters bolometric magnitudes
$M_{bol}^i$ of the two components in conditional units (when
radial velocity data are not available). But their difference
$M_{bol}^2-M_{bol}^1$ determines the true luminosity ratio
$c=L2/L1=l2/l1$. Fillout factor $f = [\Omega -
\Omega(L_1)]/[\Omega(L_2) - \Omega(L_1)]$ can be also calculated
from the output parameters of PHOEBE solution.
In order to take into account the effect of expected correlation
between the mass ratio and orbital inclination we carried out
\emph{q}-search analysis as described in
\cite{Kjurkchieva+etal+2016a}. For this aim we fixed the component
temperatures and radii as well as the spot parameters and
calculated the normalized $\chi^{2}$ for a two-dimensional grid
along $i$ and $q$. Figure 1 illustrates the result from this
\emph{q}-search procedure for the target V796 Cep.
Table~5 contains final values of the fitted stellar parameters and
their \textsc{PHOEBE} uncertainties: initial epoch $T_{0}$; period
\emph{P}; mass ratio \emph{q}; inclination \emph{i}; potential
$\Omega$; secondary temperature $T_{2}^{PH}$. Table 6 exhibits the
calculated parameters: stellar temperatures $T_{1, 2}$; stellar
radii $r_{1, 2}$ (back values); fillout factor \emph{f}; ratio of
relative stellar luminosities $l_2/l_1$. Their errors are
determined from the uncertainties of output parameters used for
their calculation. Table~7 gives information for the spot
parameters. The synthetic light curves corresponding to our
solutions are shown in Fig.~2 as continuous lines.
\begin{table}
\centering
\caption[]{Values of the fitted parameters}\vspace{0.1in}
\label{Tab5}
\begin{scriptsize}
\begin{tabular}{cllcccc}
\hline
Star & $T_0$ & \emph{P} & \emph{q} &\emph{i} &$\Omega$ & $T_2^{PH}$ \\
\hline
V796 Cep & 2457321.43582(9) & 0.392966 & 0.948(2) & 70.7(1) & 3.612(7) & 6400(19) \\
V797 Cep & 2457321.31715(74) & 0.270416 & 0.886(2) & 64.7(1) & 3.525(2) & 4625(42) \\
CSS J015341.9+381641& 2457333.61496(31) & 0.347518 & 0.892(2) & 70.0(2) & 3.490(1) & 5607(28) \\
NSVS 3853195 & 2457333.59001(21) & 0.292524(4) & 0.899(2) & 69.8(1) & 3.539(3) & 5592(30) \\
\hline
\end{tabular}
\end{scriptsize}
\end{table}
\begin{table}
\centering
\caption{Calculated parameters }\vspace{0.1in}
\label{Tab5}
\begin{scriptsize}
\begin{tabular}{cccccccc}
\hline
Target & $T_m$& $T_1$ & $T_2$ & $r_1$ & $r_2$ & \emph{f} & $l_2/l_1$ \\
\hline
V796 Cep & 6407 & 6410(19) & 6403(19) & 0.421(1) & 0.412(1) & 0.101 & 0.951 \\
V797 Cep & 4770 & 4833(44) & 4688(42) & 0.424(1) & 0.403(1) & 0.075 & 0.771 \\
CSS J015341.9+381641& 5715 & 5765(29) & 5657(28) & 0.434(1) & 0.414(1) & 0.166 & 0.867 \\
NSVS 3853195 & 5688 & 5733(31) & 5637(30) & 0.425(1) & 0.406(1) & 0.089 & 0.865 \\
\hline
\end{tabular}
\end{scriptsize}
\end{table}
\begin{table}
\begin{minipage}[t]{\columnwidth}
\caption{Parameters of the cool spots of the targets} \label{tab6}
\centering
\begin{scriptsize}
\begin{tabular}{ccrrc}
\hline\hline
Star & $\beta$ & $\lambda$ & $\alpha$ & \emph{k } \\
\hline
V796 Cep & 90(5) & 35(1) & 5.0(1) & 0.90(1) \\
CSS J015341.9+381641& 90(5) & 90(1) & 20.0(1) & 0.80(1) \\
NSVS 3853195 & 80(5) & 120(1) & 25.0(2) & 0.95(1) \\
\hline\hline
\end{tabular}
\end{scriptsize}
\end{minipage}
\end{table}
The mean (\emph{g', i'}) residuals for the final fittings are:
(0.005, 0.007) for V796 Cep; (0.021, 0.022) for V797 Cep; (0.009,
0.012) for CSS J015341.9+381641; (0.009, 0.013) for NSVS 3853195.
The mean (\emph{g', i'}) residuals of the standard stars (Table 3)
for the first and second pairs of targets are correspondingly
(0.010, 0.011) and (0.017, 0.015). Hence, our fittings are
excellent for the three targets and very good for the faint star
V797 Cep (Fig. 2). The small imperfectness of our modeling may due
to inadequate treatment of the overcontact binaries
(\citealt{Prsa+etal+2016}) and to long exposures
(\citealt{Kipping+2010}).
\begin{figure}
\begin{center}
\includegraphics[width=6cm,scale=1.00]{0253fig4.eps}
\caption[]{Distribution fillout factor-- mass ratio of W UMa
stars: red triangles are for shallow contact high mass ratio
targets; black circles are for targets with decreasing periods
from the sample of Yang (2013)} \label{Fig4}
\end{center}
\end{figure}
\section{Conclusions}
The main results from the light curve solutions of our data are as
follows.
(1) We determined the initial epochs $T_0$ of the four targets
(Table 5).
(2) We improved the period of NSVS 3853195 (Table 5) on the base
of all photometric data: CRTS, NSVS, SWASP and IRIDA. The previous
period values of the other three targets fitted well our data.
(3) Our observations revealed that CSS~J015404.1+382805 and NSVS
3853195 are the same star (while VSX identified two stars).
(4) The components of each target are almost the same in mass,
temperature, radius and luminosity (Tables 5-6).
(5) The stellar components of all targets are of G and K spectral
types and they undergo partial eclipses.
(6) All targets have overcontact configurations with small fillout
factor (Fig. 3, Table 6). This means that they probably
are newly formed contact binaries (\citealt{Qian+etal+2014}).
(7) Three binaries revealed O'Connell effect that was reproduced
by cool spots (Table~7) on their primary components. They are
appearances of the magnetic activity of these targets.
(8) All our targets have mass ratio $q \geq$ 0.88 (Table~5), i.e.
they belong to the H subtype W UMa systems (with $q \geq$ 0.72).
\cite{Csizmadia+Klagyivik+2004} revealed that the different
subtypes of W UMa's are located into different regions on the mass
ratio -- luminosity ratio diagram (their fig. 1) but above the
line $l_2/l_1 = q^{4.6}$ representing the mass-luminosity relation
for MS detached stars. Our targets support this conclusion and the
relation between their mass ratio and luminosity ratio is $l_2/l_1 = q^{1.5}$, i.e. close
to that of \cite{Lucy+1968a}.
\begin{table}
\begin{minipage}[t]{\columnwidth}
\caption{Parameters of the cool spots of the targets} \label{tab6}
\centering
\begin{scriptsize}
\begin{tabular}{ccrcc}
\hline\hline
Star & $q$ & $l_2/l_1$ & $f$ & References \\
\hline
AD Cnc & 0.77 & 1.00 & 0.08 & \cite{Qian+etal+2007} \\
BI Vul & 0.97 & 1.22 & 0.04 & \cite{Qian+etal+2013} \\
CSTAR 038663 & 0.89 & 1.13 & 0.10 & \cite{Qian+etal+2014} \\
1SWASP J174310.98+432709.6 & 1.00 & 0.65 & 0.23 & \cite{Kjurkchieva+etal+2015a}\\
NSVS 11234970 & 0.99 & 0.55 & 0.21 & \cite{Kjurkchieva+etal+2015a}\\
NSVS 11504202 & 0.98 & 0.71 & 0.00 & \cite{Kjurkchieva+etal+2015a}\\
NSVS 11534299 & 0.87 & 0.77 & 0.00 & \cite{Kjurkchieva+etal+2015a}\\
NSVS 1776195 & 0.83 & 0.96 & 0.00 & \cite{Kjurkchieva+etal+2015b}\\
NSVS 113026 & 0.79 & 1.00 & 0.07 & \cite{Kjurkchieva+etal+2015b}\\
NSVS 2244206 & 0.73 & 0.53 & 0.26 & \cite{Kjurkchieva+etal+2016b}\\
NSVS 908513 & 0.71 & 0.60 & 0.15 & \cite{Kjurkchieva+etal+2016b}\\
VSX J062624.4+570907 & 0.77 & 0.63 & 0.16 & \cite{Kjurkchieva+etal+2016b}\\
CSS J171508.5+350658 & 0.89 & 0.64 & 0.00 & \cite{Kjurkchieva+etal+2016a}\\
USNO-B1.0-1395-0370184 & 0.97 & 0.90 & 0.01 & \cite{Kjurkchieva+etal+2016c}\\
USNO-B1.0-1395-0370731 & 0.85 & 0.83 & 0.25 & \cite{Kjurkchieva+etal+2016c}\\
NSVS 2459652 & 0.786& 0.73 & 0.17 & \cite{Kjurkchieva+etal+2016d}\\
NSVS 7377875 & 0.898& 0.84 & 0.11 & \cite{Kjurkchieva+etal+2016d}\\
V796 Cep & 0.95 & 0.95 & 0.10 & this paper \\
V797 Cep & 0.89 & 0.77 & 0.07 & this paper \\
CSS J015341.9+381641 & 0.89 & 0.87 & 0.17 & this paper \\
NSVS 3853195 & 0.90 & 0.86 & 0.09 & this paper \\
\hline\hline
\end{tabular}
\end{scriptsize}
\end{minipage}
\end{table}
(9) The investigation of shallow-contact binary stars with
high mass ratios is important for the modern astrophysics because
they are considered as newly formed contact configurations, at the
beginning of contact evolution (\citealt{Qian+etal+2014}). The most
detailed studies of this type refer to the binaries AD Cnc (\citealt{Qian+etal+2007}), BI Vul (\citealt{Qian+etal+2013}) and CSTAR 038663 (\citealt{Qian+etal+2014}). They revealed that these cool, short-period (0.25--0.28
d), shallow-contact binaries exhibit strong magnetic activity
(including optical 0.2 mag flares of CSTAR 038663) and multiple
period changes. Recently we observed and modelled (in the same
way) shallow-contact W UMa's of H subtype (Table 8). On the
diagram fillout factor--mass ratio (Fig. 4) the targets from Table
8 fall into the bottom right (red triangles) due to their small
fillout factors (0.0-0.25) and high mass ratios (0.7--1.0). On the
same diagram the contact binaries with decreasing periods from the
sample of \cite{Yang+etal+2013} constitute cluster (black circles) to
the upper-left from our sample because they have intermediate
fillout factors (0.05--0.30) and moderate mass ratios (0.3--0.6).
One could guess that the deep-contact W UMa's would form a third
cluster more left and upwards from the first two clusters on the
diagram. So, the diagram fillout factor--mass ratio acquires
evolutional meaning: through the common envelope phase the
position of a given star will describe a trace starting from the
bottom right and ending to the upper left side of the diagram.
More investigations of shallow-contact binary stars with
high mass ratios will provide more statistics of their global
parameters and opportunity for further study of the rapid
evolution of binary stars reached the contact stage. The presented
study is only a step in that direction.
\normalem
\begin{acknowledgements}
This work was supported partly by grants H08/21 and M08/02 of the Fund for
Scientific Research of the Bulgarian Ministry of Education and Science.
This publication makes
use of data products from the Two Micron All Sky Survey, which is
a joint project of the University of Massachusetts and the
Infrared Processing and Analysis Center/California Institute of
Technology, funded by the National Aeronautics and Space
Administration and the National Science Foundation. This research
also has made use of the SIMBAD database, operated at CDS,
Strasbourg, France, NASA's Astrophysics Data System Abstract
Service, the USNOFS Image and Catalogue Archive operated by the
United States Naval Observatory, Flagstaff Station
(http://www.nofs.navy.mil/data/fchpix/) and the photometric
software VPHOT operated by the AAVSO, Cambridge, Massachusetts.
The research was supported partly by funds of project RD 02-81 of
the Shumen University.
The authors are grateful to the anonymous referee for the
valuable notes and recommendations.
\end{acknowledgements}
\bibliographystyle{raa}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,407 |
Directly adapted from the acclaimed film, Nativity! The Musical centres around the chaos of underperforming school St Bernadette's. Headteacher Mrs Bevan sees this year's nativity as her last opportunity to turn around the school's reputation, and puts failed actor Paul Maddens in charge. Still suffering heartache from being left by a girlfriend at Christmas five years earlier, Mr Maddens' fears for the festive season are realised when a quick lie escalates into a city-wide affair thanks to childish school assistant Mr Poppy. Yet, with rivalry against the local preparatory school stirring him on, Mr Maddens is determined not to let St Bernadette's down.
Simon Lipkin brings all the same ridiculousness and amusement to Mr Poppy as his manic movie counterpart. Mr Poppy's wild antics and outlandish aspirations for the nativity muster many laughs from both young and old audience members alike, as does the subsequent exasperation from Mr Maddens [Daniel Boys]. Andy Brady makes the perfect rival as Oakmoor teacher Gordon Shakespeare, conveying the arrogance and outrageous competitiveness of the character with ease. Likewise, Jamie Chapman masters the flamboyancy of theatre critic Patrick Burns.
However, the children playing the pupils of St Bernadette's and Oakmoor are the stars of the show. Their individual quirks and talents make the musical all the more endearing and help to portray the familiar mayhem of school nativities that originally gave Debbie Isitt the idea for the film.
The composition and songwriting by Isitt and Nicky Ager are exemplary. In addition to lovable hits "Sparkle and Shine", "She's the Brightest Star" and "Nazareth", there is short but necessary relief through more sombre, heartfelt songs – often sweetly assisted by Pepper the poodle- and a hilariously exuberant "Welcome to Hollywood" Act Two opener. The lyrics are ingenious and responded to extremely well by the cast.
The production also offers remarkable scenery. Set and costume designer David Woodhead does a stellar job of boosting the vibrancy of the performance and is particularly clever with his manipulation of the set for the climactic cathedral scene. The costumes are equally as bright and aid the feel-good atmosphere of the musical.
Along with its charming storyline, Nativity! The Musical is tremendously entertaining and creative. Forget pantomimes this year: there is simply no better show to get an audience into the festive spirit. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,031 |
With only one game on the American Hockey League schedule tonight, the AHL added news to the schedule as well as they announced the divisional alignment for the 2015-16 season. We already knew that the California teams would be a part of a new Pacific Division, but we weren't quite sure what the makeup of it would be. Today, we got to find out.
A six-division league for the past few seasons, the biggest change now has the league returning to a four-division set-up. Gone are the West, East, Northeast, and Midwest divisions as they get replaced by the Pacific and Central divisions. The North division moves to the Eastern Conference and joins the Atlantic division.
As Manchester, Norfolk, and Worcester move to the West, the league moved Rochester, Utica, and Toronto to the East. The other interesting move has San Antonio and Texas joining the California teams in the Pacific Division.
Atlantic Division: Bridgeport, Hartford, Hershey, Lehigh Valley, Portland, Providence, Springfield, Wilkes-Barre/Scranton.
North Division: Albany, Binghamton, Rochester, St. John's, Syracuse, Toronto, Utica.
Central Division: Charlotte, Chicago, Grand Rapids, Iowa, Lake Erie, Manitoba, Milwaukee, Rockford.
Pacific Division: Bakersfield, Ontario, San Antonio, San Diego, San Jose, Stockton, Texas.
With the rumors of the Pacific Division only playing 68 games next season, it'll be interesting to see if Texas and San Antonio are included or if it will apply to the entire league. Interesting offseason on the way.
A goaltender that the Oklahoma City Barons have faced multiple times over the last five seasons will be pulling up his goalposts and heading to Germany next season. Announced today on YouTube and their website, the Straubing Tigers of the German DEL have signed Climie to a contract.
The 32-year old Climie began his career in the Dallas Stars organization, spending a couple seasons between the Idaho Steelheads of the ECHL and the Iowa/Texas Stars (and even a playoff run with the Houston Aeros when Dallas was between AHL affiliates in 2008-09).
In 2010-11, Climie moved just a few hours south when he made the jump to join the then-Phoenix Coyotes organization, affiliated with the San Antonio Rampage. After one season there, Climie signed with Vancouver, joining the Chicago Wolves. When Vancouver left the Chicago affiliation, Climie stayed, signing AHL contracts to stay with the Wolves.
This season, Climie became the Chicago Wolves all-time leader in wins by a goaltender, passing Kari Lehtonen. Climie also holds the record for most wins in a single season for the Rampage.
Climie has played in 249 regular season games in the AHL, posting a career 2.57 goals against and a .914 saves percentage. Climie has also earned 15 shutouts throughout his seven years in the American Hockey League.
In Straubing, Climie replaces another former AHLer in Jason Bacashihua as he looks to take over the Tigers number one goaltender slot.
What will the AHL divisions look like next year?
This map will be changing fairly soon.
The biggest debate now that the AHL West movement is about to be announced, is what will the divisions look like within the AHL next season. There's a lot of possibilities of course, so I'll tackle the ones that are being debated with my reasoning on why they'll be in that spot. This is all going under the assumption that the league keeps the 15/15 split with the conferences, which could always change.
The California division (working name of course) will feature the five new teams making the trip to the West Coast. San Diego, San Jose II, Bakersfield, Ontario, and Stockton. Being as far west as they are, they'll be seeing a lot of each other.
The current Midwest division features OKC, Texas, San Antonio, Iowa, and Charlotte. With as many teams moving west, I'm putting Charlotte in the east (one of the highly debated topics, I'll go into it later). And we know by now that Oklahoma City is gone. Joining the Midwest, I've got Milwaukee and Rockford. As much as I tried to keep those two with Chicago, the only way that it probably could happen is if the league went with a 16/14 western favor split.
The North loses Rochester and Utica to the east and Adirondack to the move west, they gain Grand Rapids, Lake Erie, and Chicago to join Toronto and Hamilton. If you're bussing from the east to play in Rockford or Milwaukee, you have to hit Chicago first, prompting my putting Chicago there.
And now we look at the East. In the East division, I'm putting Charlotte. Biggest reason I put Charlotte there, is they're now an island in the southeast part of the country. This way, they go into a conference where teams are still close and they can get those road trips in easily. They replace the moving Norfolk team to join Wilkes-Barre/Scranton, Hershey, Lehigh Valley, and Binghamton.
The Northeast division, as much as I'd have loved to put Binghamton here to resurrect the "Empire" division, it didn't fit well logistically. Rochester and Utica move east to join Syracuse, Albany, and Bridgeport.
And finally in the Atlantic, with Worcester and Manchester leaving, Springfield and Hartford shift over to join St. John's, Portland, and Providence.
Any changes you'd make? Let us know in the comments.
The Finnish Liiga's Kärpät announced today that forward Nicolas Deschamps has been signed for the upcoming season. Deschamps spent last season with the Hershey Bears, scoring 40 points in 65 games. Deschamps made his NHL debut last season as well, playing in three games for the Washington Capitals, going scoreless. Deschamps was left unqualified this season, leaving him a free agent.
Hershey was the third AHL team that Deschamps played with during the past four seasons. His rookie campaign was with the Syracuse Crunch, where in 80 games he picked up 46 points. Just after New Year's in 2012, Deschamps was shipped from the Anaheim Ducks to the Toronto Maple Leafs. In 71 games combined, Deschamps scored 37 points that season.
Deschamps was on the move again in 2012-13 as the Maple Leafs traded him to the Washington Capitals organization. That season, Deschamps only managed to score 23 points in 66 games, but his seven points in 16 games with Hershey earned him a one-year extension with the Caps.
The Milwaukee Admirals have signed forward Triston Grant and goaltender Rob Madore to one-year AHL contracts. Triston Grant has spent the past two seasons with the Grand Rapids Griffins, winning the Calder Cup in the 2012-13 season. Rob Madore backstopped the Cincinnati Cyclones to the Kelly Cup Finals, earning playoff MVP honors.
The 30-year old Grant returns to the Admirals after four seasons, having spent time with the Rochester Americans, Oklahoma City Barons, and the Grand Rapids Griffins. This will mark Grant's tenth season in the American Hockey League, where he has 127 points in 561 career AHL games. With 1565 career AHL penalty minutes, Grant has hit at least 100 PIM in each season he has played.
Prior to first stint with the Admirals, Grant spent three seasons with the Philadelphia Phantoms. He has 11 NHL games under his belt with both the Flyers and Predators. He earned one assist and 19 penalty minutes.
Madore joins the Admirals for his third full professional season. Last year, Madore earned call-ups to Charlotte and Binghamton, playing one game for the Checkers. He then signed an AHL deal for the remainder of last season with the San Antonio Rampage, playing in seven games for them. Madore earned a 17-12-1 record for Cincinnati during the regular season, while going 14-10 during the playoffs.
Madore spent the 2012-13 season splitting time between the Checkers, Florida Everblades, and the Evansville IceMen.
The Rockford IceHogs have signed defenseman Kirill Gotovets to an AHL contract for the 2014-15 season. Gotovets has spent the last four seasons with Cornell University, scoring 19 points in 112 games throughout those four seasons. The 23-year old was drafted in 2009 by the Tampa Bay Lightning as the 183rd overall pick in the 7th round.
Gotovets made his professional debut last season with the Milwaukee Admirals, appearing in one game while on an Amateur Try-Out. The Minsk, Belarus-native played for the national team during the World Championships, going scoreless in seven games. In two previous stints with the team during the World Championships, he went scoreless in six games.
Prior to joining Cornell, Gotovets played with Shattuck St. Mary's high school, scoring 59 points in 98 games. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,001 |
\section{Introduction}
The first goal of this paper is to push, as far as possible, the idea that inverting so-called ``Bott elements" on certain cohomology theories for schemes results in \'etale descent. The cohomology theories that we are interested in are those which are representable in Morel--Voevodsky's stable motivic homotopy category, and thus only satisfy Nisnevich descent. Results of this form are extremely useful when one tries to approximate the values of these cohomology theories by studying their \'etale sheafifications which are usually more tractable due to the presence of finer covers. For example, given appropriate finiteness hypotheses, there is a spectral sequence whose $E_2$-page involves \'etale cohomology, and converges to the \'etale sheafified version of these theories.
The story for these motivic cohomology theories parallels the one for algebraic $K$-theory. Having proved the Nisnevich descent theorem for algebraic $K$-theory \cite{tt}, Thomason, in his ICM address \cite{thomason-icm}, explained how this ``(unleashed) a pack of new fundamental results for $K$-theory". One of them was the comparison result with \'etale $K$-theory \cite[\S 4]{thomason-icm}, referring to his seminal paper \cite{aktec}. More precisely, Thomason proved that Bott-inverted algebraic $K$-theory satisfy \'etale (hyper)descent. Whence, combined with the rigidity results of \cite{gabber-rigidity, suslin-closed}, there is a spectral sequence whose $E_2$-terms are \'etale cohomology groups, abutting to Bott-inverted algebraic $K$-theory. A recent treatment of this result using completely different methods was carried out in \cite{clausen-mathew}.
The second goal of this paper, and a \emph{necessary ingredient} for the first, is to prove a variant of an important conjecture within the subject of stable motivic homotopy theory. In \cite{voe-open}, Voevodsky listed a collection of open problems in the subject, many of which are focused on understanding the slice filtration of a motivic spectrum.
On the motivic spectrum representing algebraic $K$-theory, the slice filtration gives the most definitive construction of the elusive motivic spectral sequence whose $E_2$-page is motivic cohomology and abutting to algebraic $K$-theory. Such a spectral sequence was the subject of prior work of Bloch--Lichtenbaum, Levine and Friedlander--Suslin \cite{bloch-moving, levine-techniques, friedlander-suslin}. In contrast to previous constructions, this spectral sequence is constructed natively within the framework of the motivic stable homotopy category, analogous to the topological theory of Postnikov towers. It therefore extends to more general cohomology theories beyond algebraic $K$-theory and does not \emph{a priori} depend on difficult constructions involving moving algebraic cycles. We refer the reader to the survey article \cite[Section 2]{levine-appreciation} for an outline of Voevodsky's slice approach to the motivic spectral sequence.
As for any spectral sequence, an important issue is the problem of convergence. Voevodsky stated, as Conjecture 13 in \cite{voe-open}, that he expects convergence of his slice spectral sequence on the sphere spectrum, over any perfect field. Moreover, he expects that the slice spectral sequence induces a separated filtration on its homotopy sheaves; see also \cite[Conjecture 13]{voe-open}. This conjecture turns out to be false in general \cite[Remark on page 909]{levine2013convergence} and has since been modified by Levine to cover only fields of finite virtual cohomological dimension \cite[Conjecture 5]{levine2013convergence}. Roughly speaking, Levine conjectures that for such a field $k$, it is the \emph{$I(k)$-completed} slice tower which admits good convergence properties. Here, $I(k)$ is the so-called \emph{fundamental ideal}, defined to be the kernel of the rank map of the Grothendieck-Witt group of symmetric bilinear forms over $k$. Its action on any motivic spectrum is given by Morel's fundamental identification of $\pi_{0,0}$ of the motivic sphere spectrum with the Grothendieck-Witt group \cite{morel-pi0}. We will refer to \cite[Conjecture 5]{levine2013convergence} as \emph{Levine-Voevodsky's slice convergence conjecture}.
\subsection{Slice convergence} In this paper, we offer a resolution of a version of Levine-Voevodsky's slice convergence conjecture. To state our results, recall that we have a map in the stable motivic homotopy category over a field $k$:
\[
\rho
\colon
\mathbbm{1}
\rightarrow
{\mathbb{G}_m},
\]
classifying the unit $-1\in k^{\times}$.
\begin{theorem}[Theorem~\ref{thm:convergence}]
\label{thm:conv-intro}
Let $k$ be a field of exponential characteristic $e$ and $\ell > 0$ coprime to $e$ such that $\mathrm{vcd}_\ell(k) < \infty$.
Suppose that $E/\ell \in \mathcal S\mathcal H(k)$ is bounded below\footnote{Throughout this paper we use homological notation for $t$-structures, as in \cite[\S1.2.1]{HA}.} in the homotopy $t$-structure and suppose that there exists $R \gg 0$ for which
\[
E \xrightarrow{\rho^R} E \wedge \Gmp{R}
\] is zero; in other words it \emph{has bounded $\rho$-torsion}. Then, the filtration on $\pi_{i,j}(E)$ induced by Voevodsky's slice tower $f_{\bullet} E \rightarrow E$ is separated and exhaustive,
i.e., the filtration is convergent.
\end{theorem}
We comment on how this is related to Levine-Voevodsky's slice convergence conjecture. There are two phenomenon which led Levine to formulate \cite[Conjecture 5]{levine2013convergence} in place of \cite[Conjecture 13]{voe-open}:
\begin{enumerate}
\item Levine showed in \cite{levine-gw} that the induced filtration on the zero-th homotopy sheaf of the sphere spectrum is the $I$-adic filtration. Therefore there is no chance for the induced filtration on homotopy sheaves to be separated if this is not the case for the $I$-adic filtration;
\item as observed by Kriz \cite[Remark 1.1]{levine2013convergence} this shows that Voevodsky's original conjecture is false for the field $\mathbb{R}$; the point here is that $\mathbb{R}$ has \emph{infinite cohomological dimension} at the prime $2$ and thus is not $I$-adically complete.
\end{enumerate}
Motivated by these two considerations, it is natural to consider the element $\rho$ which interpolates between the layers of the $I$-adic filtration:
\[
I^1(k) \xrightarrow{\rho} I^2(k) \xrightarrow{\rho} \cdots I^j(k) \xrightarrow{\rho} I^{j+1}(k) \xrightarrow{\rho} \cdots,
\]
where we were led to guess that $\rho$-completion on the spectrum level is closely related to $I$-adic completion on the homotopy sheaves level. As a ``best hope" one could guess that $\rho$-completion would lead to slice convergence. However, as already observed in \cite{levine2013convergence}, the slice convergence property is not stable under various categorical operations. Therefore, in place of a completed statement, we offer Theorem~\ref{thm:conv-intro} where convergence does hold if $E$ is $\rho$-complete in a strong sense: that it has bounded $\rho$-torsion. We note that being $\rho$-complete means that it is an inverse limit of $\rho$-torsion objects.
Theorem~\ref{thm:convergence} turns out to be good enough for many applications in motivic homotopy theory, beyond this paper. Most notably, it renders the computations of \cite{ormsby-rondigs} valid without having to slice complete. It also gives a new, streamlined proof of Thomason's homotopy limit problem by the first author and Hopkins \cite[Appendix A]{bachmann-hopkins} based on the slice spectral sequence. We may also view Theorem~\ref{thm:conv-intro} as a \emph{derived} version of Levine-Voevodsky's slice conjecture and we refer the interested reader to Remark~\ref{rmk:levine-conjecture} for more details.
\subsection{\'Etale descent} Back to the main subject of this paper, we now formulate our main \'etale descent results. Any motivic spectrum $E \in \mathcal S\mathcal H(S)$ comes equipped with a canonical map $E \rightarrow L_{\mathrm{\acute et}}E$, the \emph{\'etale localization}, witnessing $L_{\mathrm{\acute et}}E$ as the initial motivic spectrum receiving a map from $E$ and satisfying \'etale hyperdescent; this map is just the unit map of the usual adjunction between $\mathcal S\mathcal H(S)$ and its \'etale local version $\mathcal S\mathcal H_{\mathrm{\acute et}}(S)$. Based on a version of the map $\rho$ over $S$, we prove:
\begin{theorem} \label{thm:main-intro} (see Theorem~\ref{thm:main})
Suppose $S$ is a scheme locally of finite dimension, and let $1 \le m, n \le \infty$.
Assume the following conditions hold.
\begin{enumerate}
\item $1/\ell \in S$,
\item For every $s \in S$ we have $\mathrm{vcd}_\ell(s) < \infty$, and
\item There exists a \emph{good $\tau$-self map} (in the sense discussed in \S\ref{sec:bott-sph}):
\[
\tau: \mathbbm{1}/(\ell^n,\rho^m) \to \mathbbm{1}/(\ell^n,\rho^m)(r).
\]
\end{enumerate}
Then for every $E \in \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$ the map
\[
E/(\ell^n,\rho^m) \to E/(\ell^n,\rho^m)[\tau^{-1}]_{\ell,\rho}^\wedge
\] is an \'etale localization.
\end{theorem}
We note that if $m, n < \infty$, then the $(\ell,\rho)$-completion is unnecessary; see Remark~\ref{rmk:p-completion-unnec}. A summary of the good $\tau$-self maps that we managed to construct can be found in \S\ref{subsec:summary} and we will discuss this point further below. We point out some specializations of Theorem~\ref{thm:main-intro} at different primes powers.
\begin{enumerate}
\item Suppose that $\ell$ is an odd prime. Then recall that (as discussed in e.g., \S\ref{sec:standard}) for any motivic spectrum $E$, we have a splitting:
\[
E/\ell^n \simeq E^+/\ell^n \vee E^-/\ell^n.
\]
In this situation, \'etale localization only involve the ``$+$'' part of $E/\ell^n$: the map \[ E^+/\ell^n \to E^+/\ell^n[\tau^{-1}] \] is an étale localization.
\item If $\ell =2$ then the situation is more complicated. In the presence of a square root of $-1$ (e.g., over an algebraically closed field), $\rho$-completion is harmless and we get that
\[ E/2^n \to E/2^n[\tau^{-1}] \] is an étale localization. Otherwise we have to contend ourselves with $\rho$-completion.
\end{enumerate}
\begin{remark}\label{rem:hyp} Thomason's theorem \cite{aktec} in fact proves that Bott inverted algebraic $K$-theory satisfies not only satisfies \'etale descent but also \'etale \emph{hyperdescent} (this has been reproved by \cite{clausen-mathew}). In particular there is no need to hypersheafify in order to obtain a conditionally convergent spectral sequence abutting to \'etale sheafified $K$-theory from \'etale cohomology. This is useful as \'etale sheafification is much less drastic process than hypersheafificaiton. Variations on this theme in motivic homotopy theory are being explored in \cite{bbx} where it is proved that under noetherian, finite dimensional and finite virtual cohomological dimension assumptions the $\infty$-category obtained by formally imposing \'etale hyperdescent, $\mathcal S\mathcal H_{\mathrm{\acute et}}(S)$, coincides with one obtained by formally imposing just \'etale descent. The input, just as in \cite{clausen-mathew}, is a Quillen-Lichtenbaum style result for the motivic sphere spectrum, analogous to Rost--Voevodsky's results for $K$-theory.
Granting this result, the reader may feel free to interpret the ``\'etale localization" appearing in Theorem~\ref{thm:main-intro} as inverting all desuspensions of \v{C}ech nerves of \'etale covers (as opposed to \'etale hypercovers).
\end{remark}
\subsection{Strategy} Dealing with the prime $2$ is one of the key technical challenges of this paper that we were able to overcome in many cases.
We will now explain this point and the overall strategy of the proof of Theorem~\ref{thm:main-intro}. We proceed by examining the slice spectral sequence for the Bott inverted sphere spectrum over a field (Corollary~\ref{cor:sphere-inversion}), following the general strategy in \cite{elso} for the case of algebraic cobordism. An examination of the form of the slices \cite[Theorem 2.16]{1-line} reveals that they are just motivic cohomology and hence, modulo certain cases which are dealt with using the Beilinson--Lichtenbaum conjectures, satisfy \'etale descent after inverting these Bott elements due to Levine's results \cite{levine2000inverting}. However, as already elaborated in \cite[Section 4]{elso}, examining the slice tower of the Bott inverted sphere spectrum requires a delicate analysis --- the crux point is that the process of Bott inversion is a \emph{colimit}, while the slices (or, more accurately, the co-slices) try to approximate the sphere spectrum as a \emph{limit}. At the prime $2$, and for fields with non-finite $2$-cohomological dimension (such as $\bb R$), the situation is worse: even the slice filtration itself is \emph{not} convergent \cite[Remark]{levine2013convergence}. From this point of view, the situation seems hopeless.
On the other hand, previous attempts to cope with this infinitary phenomenon to still obtain a Thomason-style descent theorem have found success under the assumption of finite \emph{virtual cohomological dimension}. To our knowledge, the first paper of this sort was written by the third author in \cite{pa-thesis} and later generalized in joint work with Rosenschon \cite{ro}. The case of hermitian $K$-theory was settled in \cite{bkso}.
It is at this spot that we employ a different analysis from \cite{elso}, which also led to substantial improvements even for the case of algebraic cobordism, other oriented theories and also recovers the descent results of the papers in the previous paragraph. In \emph{loc. cit.}, particularly in \cite[Section 6.5]{elso} and \cite[Section 4]{elso}, the last two authors (with Levine and Spitzweck) examined the resulting slice spectral sequences on Bott-inverted algebraic cobordism and on \'etale algebraic cobordism. This relied on subtle convergence results on inverting elements in a spectral sequence and an analysis of the constituent spaces and bonding maps of an \'etale-localized motivic spectra in a range. In this paper, we work \emph{directly} with the slice filtration and prove a convergence result in the form of Lemma~\ref{lemm:key} which cleanly isolates the role of the convergence of the slice filtration. At the prime $2$ and with the weaker assumption that the ambient field $k$ has finite virtual cohomological dimension, it turns out that convergence holds after completion with respect to the map
\[
\rho
\colon
\mathbbm{1}
\rightarrow
{\mathbb{G}_m}
\]
induced by the unit $-1\in k^{\times}$. This last statement heavily relies on our Theorem~\ref{thm:conv-intro} on slice convergence.
\subsection{Bott elements and multiplicativity of the Moore spectrum} There are two further, related, issues which we address in the paper: (1) construction of a suitable ``spherical" Bott element and (2) the lack of a multiplicative structure on the motivic Moore spectrum. What is at stake in point (1) is the fact that the sphere spectrum is \emph{not} oriented. In previous iterations of Thomason-style descent results like in \cite{aktec}, \cite{jardine}, \cite{elso}, one produces Bott elements out of $n$-th roots of unity present in the ambient scheme/ring/field which is naturally an $n$-torsion element in an appropriate group; see \cite[Section 6.4]{elso} or \cite[Appendix A]{aktec}. For the sphere spectrum, due to so-called ``$\eta$-logarithmic relation" in Milnor-Witt $K$-theory
\[
[ab] = [a] + [b] + \eta[a][b],
\]
a root of unity is $n_{\epsilon}$-torsion (see Proposition~\ref{prop:powers-n-epsilon}), rather than $n$-torsion.
At odd primes, the right thing to do is to look at the part of the sphere spectrum where the Milnor-Witt number $n_{\epsilon} = n$, i.e., the so-called ``$+$" part.
At the prime $2$ the story gets more interesting, coupled with the usual complication that the mod-$2$ Moore spectrum \emph{does not} admit a multiplication. Producing a mod-$2$ Bott element is the first time where $\rho$-completion enters the picture: see \S\ref{subsec:quad}. To address the multiplicativity issues around the mod-$2$ Moore spectra, various authors have considered Oka's action of the mod-$4$ Moore spectrum on the mod-$2$ Moore spectrum. This is actually \emph{insufficient} for our reduction steps (see the argument in Theorem~\ref{thm:main}), primarily because we do not know that this module structure satisfies the usual associativity axiom. In the appendix, we use an idea originating in the work of Davis--Lawson and Hopkins, that this action can be made \emph{asymptotically associative} if we are willing to consider the pro-system $\{ \mathbbm{1}/2^n \}$; see \S\ref{sec:moore-mult}.
In any case, \S\ref{subsec:summary} summarizes the Bott elements/self-maps that we were able to construct in motivic homotopy theory. This adds to the ever-growing list of canonical and vastly interesting elements in stable motivic homotopy theory (such as $\rho$ and $\eta$).
\subsection{Some applications} We now summarize some applications of our main results. These results unveil certain surprising properties of the $\infty$-category (rather, the premotivic functor) $\mathcal S\mathcal H_{\mathrm{\acute et}}$ which were not a priori visible without our main result.
\begin{theorem} \label{thm:apps} Assumptions as in Theorem \ref{thm:main-intro}, where $m,n<\infty$:
\begin{enumerate}
\item If $S$ is defined over a field containing a primitive $\ell$-th root of unity and satisfying $k^\times/\ell = \{1\}$, then there exists a Bott element $\tau$ in $\mathbbm{1}_{\ell}^\wedge$ and \[ \mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge \simeq \mathcal S\mathcal H(S)_\ell^\wedge[\tau^{-1}]. \]
\item In general, $(\ell,\rho)$-complete \'etale localization is equivalent to Bousfield localization at the homology theory $\mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]$.
\item \'Etale localization is smashing on $\mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$.
\item If $f: T \rightarrow S$ is finite type, the base change functor $f^*$ on $\mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$ preserves \'etale local objects so that the \'etale local sphere (in the $(\rho,\ell)$-complete category) is stable under base change.
\end{enumerate}
\end{theorem}
These corollaries are discussed right after Theorem~\ref{thm:main}. The first statement should be considered the ``model statement" --- the limitation comes from the fact that the construction of higher Bott elements requires more multiplicative structure on the $\rho$-complete sphere than what we already know. In light of this, the second and third statement are weaker but also pleasant consequences of our results. In particular, they tell us that on the $(\rho,\ell)$-complete categories the inclusion of \'etale-local objects preserves colimits. Lastly, base change results in motivic homotopy theory are rewarding but often hard to come by --- our results proves this by way of knowing that the $\tau$-self maps are manifestly stable under base change.
\subsection{Overview of \'etale motivic cohomology theories} This paper completes, in many cases, our structural understanding of how \'etale motivic theories behave; we now sketch this. Suppose that $\mathcal{T}$ is a premotivic functor in the sense of \cite{triangulated-mixed-motives} or a functor satisfying Ayoub's axioms as in \cite[1.4.1]{ayoubthesis}. The most prominent examples are $\mathcal{T} = \mathcal S\mathcal H, \mathcal D\mathcal M\footnote{For this section, we take this in the sense of \cite[Chapter 11]{triangulated-mixed-motives}}$ or $\mathcal M od_{E}$ where $E$ is a (highly structured) motivic ring spectrum such as $\mathrm{MGL}$ which is defined over $\bb Z$. We can also consider the \'etale local version of $\mathcal{T}$, which we denote by $\mathcal{T}_{\mathrm{\acute et}}$ and comes equipped with a premotivic adjunction
\[
L_{\mathrm{\acute et}}: \mathcal{T} \rightleftarrows \mathcal{T}_{\mathrm{\acute et}}: i_{\mathrm{\acute et}}.
\]
Consider the following categorified version of the arithemetic fracture square, which is a cartesian square of stable $\infty$-categories:
\[
\begin{tikzcd}
\mathcal{T}_{\mathrm{\acute et}}' \ar{r} \ar{d} & \prod_\ell (\mathcal{T}_{\mathrm{\acute et}})^{\wedge}_{\ell} \ar{d}\\
\mathcal{T}_{\mathrm{\acute et},\mathbb{Q}} \ar{r} & (\prod_\ell (\mathcal{T}_{\mathrm{\acute et}})^{\wedge}_{\ell})_{\mathbb{Q}}.
\end{tikzcd}
\]
We have a natural induced functor $\mathcal{T}_{\mathrm{\acute et}} \rightarrow \mathcal{T}'$ which is fully faithful but not necessarily essentially surjective. Nonetheless, this does mean that the values of invariants evaluated on schemes can be computed in $\mathcal{T}'_{\mathrm{\acute et}}$, and thus an understanding of $\mathcal{T}'_{\mathrm{\acute et}}$ gives us a lot of new information on $\mathcal{T}_{\mathrm{\acute et}}$ as we now explain.
The rational part $\mathcal{T}_{\mathrm{\acute et},\bb Q}$ coincides with $\mathcal{T}_{\mathbb{Q}}$ in many cases. In fact, a result of Cisinski-D\'eglise (summarized in, say, \cite[Theorem 12.2]{elso}) furnishes equivalences
\[
\mathcal S\mathcal H_{\mathrm{\acute et}}(S)_{\bb Q} \simeq \mathcal S\mathcal H(S)^+_{\bb Q} \simeq \mathcal D\mathcal M(S)_{\bb Q},
\]
for any Noetherian, geometrically unibranch scheme $S$. Furthermore, as discussed in Remark~\ref{rem:recoll}, the \emph{prime-at-$\ell$} of $\mathcal{T}_{\mathrm{\acute et}}$ part is always zero. Hence, what remain are the $\ell$-complete parts of $\mathcal{T}_{\mathrm{\acute et}}$ where $\ell$ is coprime to the residual characteristics. In this situation, we seek two types of theorems concerning the induced adjunction
\[
L_{\mathrm{\acute et}}: \mathcal{T}^\wedge_\ell \rightleftarrows (\mathcal{T}_{\mathrm{\acute et}})^\wedge_\ell: i_{\mathrm{\acute et}}.
\]
\begin{enumerate}
\item[Suslin-style rigidity] If $\ell$ is coprime to the residual characteristics, then $(\mathcal{T}_{\mathrm{\acute et}})^\wedge_\ell$ is described as a certain category of sheaves over the \emph{small} \'etale site.
\item[Thomason-style descent] If $\ell$ is coprime to the residual characteristics, then the endofunctor $i_{\mathrm{\acute et}}L_{\mathrm{\acute et}}$ is computed explicitly by an inversion of a ``Bott-element" $\tau$.
\end{enumerate}
In conjunction, these two results give us access to the values of the cohomology theories represented in $\mathcal{T}_{\mathrm{\acute et}}$. One way to make this concrete, at least in cases where we have ``Postnikov completeness" of the \'etale site of a scheme $X$ (see \cite[Section 2]{clausen-mathew} for a modern discussion when this happens) we obtain a conditionally convergent spectral sequence:
\[
H^p_{\mathrm{et}}(X, \ul{\pi}^{\mathrm{\acute et}}_{q,w}(E)) \Rightarrow [X(w)[q-p] , E[\tau^{-1}]] = (E[\tau^{-1}])^{p-q,-w}(X)
\]
The input of this spectral sequence is obtained as consequence of Suslin-rigidity; we note that it consists of usual \'etale cohomology groups with coefficients in a (torsion/$\ell$-complete) sheaf of abelian groups on the small \'etale site of $X$. The target is obtained as a consequence of Thomason-style descent. This philosophy was already known to Thomason at the beginning in \cite{aktec}. This should be contrasted with the slice spectral sequence, available in the Nisnevich/Zariski setting, where the input consists of motivic cohomology groups which are largely unknown.
We briefly recall Suslin-style rigidity. In \cite{bachmann-SHet,bachmann-SHet2}, the first author has established (again, in many cases) Suslin-style rigidity for $\mathcal{T} =\mathcal S\mathcal H$. Earlier, analogous results were established by R{\"o}ndigs-{\O}stv{\ae}r \cite{rigidity-in-motivic-homotopy-theory} and for $\mathcal{T} = \mathcal D\mathcal M$ by Ayoub \cite{ayoub2014realisation} and Cisinski-D\'eglise \cite{etalemotives}, building on the case of fields where we have Suslin's eponymous result (see \cite[Theorem 7.20]{mvw} for an exposition).
In this light, what one needs to understand is the precise gap between $\mathcal{T}$ and $\mathcal{T}_{\mathrm{\acute et}}$. For $\mathcal{T} = \mathcal S\mathcal H$ (and up to certain $\rho$-completions), this is exactly the $\tau$-complete category, which remains mysterious.
\subsection{Terminology and notation}
We freely use the language of $\infty$-categories, as set out in \cite{HTT,HA}.
\subsubsection{Motivic homotopy theory}
\begin{itemize}
\item We denote by $\mathcal S\mathcal H_\mathrm{\acute et}(S)$ the localization of $\mathcal S\mathcal H(S)$ at the étale hypercovers and by $\mathcal S\mathcal H(S_\mathrm{\acute et})$ the stabilization of the \emph{small} hypercomplete étale $\infty$-topos of $S$.
We denote by $L_\mathrm{\acute et}$ the (various) étale hyperlocalization functors.
We call a map $E \to F$ an étale localization if $L_\mathrm{\acute et} E \simeq L_\mathrm{\acute et} F \simeq F$.
\item We denote by $E_\ell^\wedge = \lim_n E/\ell^n$ (or sometimes $E/\ell^\infty$) the $\ell$-completion of a spectrum (see e.g. \cite[Example 2.3]{bachmann-SHet}; we write $\mathcal S\mathcal H(S)_\ell^\wedge$ for the category of $\ell$-complete motivic spectra.
\item We denote by $\mathcal S\mathcal H(k)_{\ge 0}, \mathcal S\mathcal H(k)_{\le 0}$ the homotopy $t$-structure on $\mathcal S\mathcal H(k)$ defined in \cite[Section 5.2]{morel-trieste}.
\item We put $\mathbbm{1}(1) = {\mathbb{G}_m}[-1]$ and $\mathbbm{1}(n) = \mathbbm{1}(1)^{\wedge n}$.
For $E \in \mathcal S\mathcal H(S)$ we put $E(n) = E \wedge \mathbbm{1}(n)$.
\item As is standard we write $T = \bb A^1/\bb A^1 - 0$ for the Tate object.
\item We denote by $H\bb Z$ Spitzweck's motivic cohomology spectrum \cite{spitzweck2012commutative} and write $\mathcal D\mathcal M(S)$ for the $\infty$-category of modules over $H\bb Z$ \cite{RondigsModules}.
\end{itemize}
\subsubsection{Field theory} We will adopt the following terminology concerning field theory.
Let $k$ be a field, $k^{\mathrm{sep}}$ a separable closure of $k$ and $G_k:=\mathrm{Gal}(k^{\mathrm{sep}}/k)$ the absolute Galois group.
Moreover, we write:
\begin{itemize}
\item $\mathrm{cd}_\ell(k)$ for the \emph{$\ell$-cohomological dimension} of $k$ in the sense of \cite[Section 3.1]{serre-gal}.
\item $\mathrm{vcd}_2(k):= \mathrm{cd}_2(k[\sqrt{-1}])$ for the \emph{virtual $2$-cohomological dimension}.
\item More generally, for an integer $s$ we put \[ \mathrm{vcd}_s(k) := \max\{\mathrm{vcd}_\ell(k) \mid p|s\}, \text{ where } \mathrm{vcd}_\ell(k) := \mathrm{cd}_\ell(k(\sqrt{-1})). \]
Note that $\mathrm{vcd}_\ell(k) = \mathrm{cd}_\ell(k)$ unless possibly if $\ell=2$ \cite[Proposition II.10']{serre-gal}.
\item The \emph{cohomological dimension} $\mathrm{cd}(k)$ of $k$ is defined as $\mathrm{cd}(k) = \sup_\ell\{ \mathrm{cd}_\ell(k)\}$.
\end{itemize}
\subsection{Acknowledgements}
We would like to thank Joseph Ayoub, Jeremy Hahn, Mike Hopkins, Tyler Lawson, Marc Levine, Denis Nardin, and Markus Spitzweck for helpful discussions. We also thank an anonymous referee for comments on a previous draft.
The authors thank the Isaac Newton Institute for Mathematical Sciences for support and hospitality during the programme $K$-theory, algebraic cycles and motivic homotopy theory.
This work was supported by EPSRC Grant Number EP/R014604/1 and the RCN Frontier Research Group Project no. 250399 ``Motivic Hopf Equations" and no. 312472 ``Equations in Motivic Homotopy Theory."
Part of this work was carried out while Elmanto was a postdoc at the Center for Symmetry and Deformation at the University of Copenhagen,
which also supported Bachmann's visit in 2019.
We are grateful for the Center's support.
{\O}stv{\ae}r was partially supported by the Humboldt Foundation,
Professor Ingerid Dal and sister Ulrikke Greve Dals prize for excellent research in the humanities,
and a Guest Professorship under the auspices of The Radbound Excellence Initiative.
\section{Preliminaries}
\subsection{Endomorphisms of the motivic sphere} \label{sec:standard}
We denote by \[ \lra{-1}: \mathbbm{1} \to \mathbbm{1} \in \mathcal S\mathcal H(S) \] the map corresponding to the switch map $\bb P^1 \wedge \bb P^1 \to \bb P^1 \wedge \bb P^1$.
Clearly we have $\lra{-1}^2 = 1$.
It follows that if $E \in \mathcal S\mathcal H(S)$ is such that $E \xrightarrow{2} E$ is an equivalence, then there is a canonical decomposition \[ E \simeq E^+ \vee E^- \] which is characterised by the fact that $\lra{-1}$ acts as the identity $\mathrm{id}$ on $E^+$ and as $-\mathrm{id}$ on $E^-$ \cite[\S16.2.1]{triangulated-mixed-motives}.
We denote by \[ \mathcal S\mathcal H(S)[1/2]^+, \mathcal S\mathcal H(S)[1/2]^- \subset \mathcal S\mathcal H(S)[1/2] \] the full subcategories of those objects with $E \simeq E^{\pm}$. For $n \in \bb N$ we define the \emph{$n$-th Milnor-Witt number} as \[ n_\epsilon = \sum_{i = 1}^n \lra{(-1)^{i-1}} \in [\mathbbm{1}, \mathbbm{1}]_{\mathcal S\mathcal H(S)}, \] where $\lra{1} := 1$.
We will make use of the following result.
\begin{proposition} \label{prop:powers-n-epsilon} Suppose that $S$ is a base scheme.
\begin{enumerate}
\item If $-1$ is a square on $S$, then $\lra{-1} = 1$.
\item Denote by $p_n: {\mathbb{G}_m} \to {\mathbb{G}_m}$ the map corresponding to $x \mapsto x^n$ (i.e., the $n$-th power map), and by $\bar{p}_n: \mathbbm{1} \to \mathbbm{1}$ the desuspended map.
Then $\bar{p}_n = n_\epsilon$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1) We claim the following more general statement: for $a \in \mathcal{O}^{\times}(S)$, denote by $\langle a \rangle: \mathbbm{1} \rightarrow \mathbbm{1}$ the endomorphism in $\mathcal S\mathcal H(S)$ induced by the endomorphism of $S$-schemes $\bb P^1 \rightarrow \bb P^1, [x:y] \mapsto [ax:y]$. Then we have
\[
\langle a^2 \rangle = 1.
\]
To prove the claim, we first note that the map $(x:y) \mapsto (a^2x:y)$ is equal to $(x:y) \mapsto (ax:a^{-1}y)$. We have a map $\mathrm{GL}_2(S) \to \mathrm{Map}(\bb P^1, \bb P^1)$. It is thus enough to connect the matrix $A = \begin{bmatrix}a^{-1} & 0 \\ 0 & a\end{bmatrix}$ to the identity matrix in $\mathrm{GL}_2(S)$ via $\bb A^1$-paths. By ``Whitehead's Lemma'', $A$ is a product of elementary matrices: \[ \begin{bmatrix}a^{-1} & 0 \\ 0 & a\end{bmatrix} = \begin{bmatrix}1 & 1/a \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 1-a & 1\end{bmatrix}\begin{bmatrix}1 & -1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1-a^{-1} & 1\end{bmatrix}. \]
Since the space of elementary matrices is $\bb A^1$-path connected, the result follows.
(2) For this proof, we will need some rudiments of the theory of framed correspondences in the sense of \cite{EHKSY, garkusha2014framed}. Under the equivalence $\Sigma{\mathbb{G}_m} \simeq \bb P^1$, the map $p_n$ suspends to the map $q_n: \bb P^1 \to \bb P^1, (x:y) \mapsto (x^n:y^n)$; hence it suffices to prove the claim for this map. Further note that the projection $c: \bb P^1 \to \bb P^1/\bb P^1 \setminus 0 \simeq T$ is an equivalence, $\bb P^1 \setminus 0 \simeq \bb A^1$ being contractible. It follows that it suffices to determine the map
\[
c \circ q_n: \bb P^1 \rightarrow T
\] which is precisely the map induced via Voevodsky's Lemma \cite[Corollary A.1.7]{EHKSY} from the equationally framed transfer with support $0\in \bb A^1$ and framing $t^n$ (in the sense of \cite[Definition 2.1.1]{EHKSY}). It follows from the main result of \cite{EHKSY2} (in particular, the agreement results \cite[Theorems 3.2.11, 3.3.6]{EHKSY2}) that this is stably the same as the map $\mathbbm{1} \to \mathbbm{1}$ induced by the corresponding framed correspondence via the reconstruction theorem of \cite[Theorem 3.5.11]{EHKSY} and its generalization over any base \cite[Theorem 16]{framed-loc}. The result now follows from \cite[Proposition B.1.4]{EHKSY}, provided that their $n_\epsilon$ is the same as ours. Their $\lra{a}$ comes from the canonical action of $\scr O^\times(S) \to \Omega K(S)$ on framed correspondences, which coincides with the canonical action via fundamental classes by \cite[Theorem 3.2.11]{EHKSY2}, which coincides with our definition essentially by construction.
\end{proof}
\begin{remark} Given $u \in \scr O^\times(S)$ we let $[u]$ denote the resulting map $\mathbbm{1} \to {\mathbb{G}_m}$.
One may show that \[ \lra{-1} = 1 + \eta[-1], \] where $\eta: {\mathbb{G}_m} \to \mathbbm{1}$ is the geometric Hopf map \cite[\S6.2]{morel-trieste}.
\end{remark}
\subsection{$\rho$-completion}
In this section, we discuss $\rho$-completed motivic spectra in view of certain convergence results that we will use later; we use \cite[Section 2.2]{mnn} as a reference for the formalism of complete objects and completions; see also \cite{rigidity-in-motivic-homotopy-theory} for a previous reference in the motivic context. We put \[ \rho := [-1]: \mathbbm{1} \to {\mathbb{G}_m} \in \mathcal S\mathcal H(S). \]
By abuse of notation, we also denote the map $\Gmp{-1} \to \mathbbm{1}$ (obtained by smashing $\rho$ with $\mathrm{id}_{\Gmp{-1}}$) by $\rho$.
We write $\mathbbm{1}/\rho$ for the cofiber of $\rho: \Gmp{-1} \to \mathbbm{1}$.
Recall that a morphism $\alpha: E \to F \in \mathcal S\mathcal H(S)$ is a \emph{$\rho$-equivalence} if $\alpha \wedge \mathbbm{1}/\rho: E/\rho \to F/\rho$ is an equivalence. We denote by \[ \mathcal S\mathcal H(S)_\rho^\wedge \] the localization of $\mathcal S\mathcal H(S)$ at the $\rho$-equivalences. Recall that $E \in \mathcal S\mathcal H(S)_\rho^\wedge$, i.e., it is \emph{$\rho$-complete}, if and only if for all $F$ such that $F \otimes \mathbbm{1}/\rho \simeq 0$, the space $\mathrm{Map}(F, E)$ is contractible.
From general principles the localization $\mathcal S\mathcal H(S) \to \mathcal S\mathcal H(S)_\rho^\wedge$ has a fully faithful right adjoint, and the composite \emph{localization functor} is given by \cite[Formula (2.22)]{mnn} \[ E \mapsto E_\rho^\wedge := \lim_n E/\rho^n. \]
\begin{remark}
Functors between premotivic categories that preserve cofibers and smashing with ${\mathbb{G}_m}$ preserve $\rho$-equivalences. Since ${\mathbb{G}_m}$ is invertible the latter condition is in particular satisfied by symmetric monoidal functors and their adjoints. It follows that the motivic base change functors $f^*, f_\#$, and $f_*$ preserve $\rho$-equivalences.
\end{remark}
\begin{warning}
On the other hand, the functors related to the slice filtration, like $f_n$ and $s_n$, do not interact well with $\rho$-equivalences.
See for example Remark \ref{rmk:fM-rho}.
\end{warning}
In order to streamline the exposition, in what follows we will extensively use the category $\mathcal S\mathcal H(S)_\rho^\wedge$.
In many cases, however, this has very little effect:
\begin{remark} \label{rmk:rho-completion}
\begin{enumerate}
\item Suppose $RS = \emptyset$\footnote{For a scheme $X$, we denote by $RX$ the set of pairs $(x, \alpha)$ with $x \in X$ and $\alpha$ an ordering of $k(x)$.},
i.e., $-1$ is a sum of squares locally in $\scr O_S$ \cite[Theorem 4.3.7]{bochnak2013real}.
Then $\rho \in \pi_0(\mathbbm{1})_*$ is nilpotent, as follows, for example, from the main result of \cite{bachmann-real-etale}.
It follows that \[ \mathcal S\mathcal H(S)_\rho^\wedge = \mathcal S\mathcal H(S). \]
\item Similarly, over a general base, $\rho$ is nilpotent on $\mathbbm{1}[1/2]^+$ and a unit on $\mathbbm{1}[1/2]^-$, see \cite[Lemma 39]{bachmann-real-etale}.
It follows that \[ \mathcal S\mathcal H(S)[1/2]_\rho^\wedge = \mathcal S\mathcal H(S)[1/2]^+. \]
\end{enumerate}
\end{remark}
\begin{example} \label{ex:HZ-rho-complete}
Let $S$ be essentially smooth over a Dedekind domain.
Then $H\bb Z \in \mathcal S\mathcal H(S)$ is $\rho$-complete.
Indeed it suffices to show that for $X \in {\cat{S}\mathrm{m}}_S$, $i, j \in \bb Z$, we have \[ [\Sigma^{i,j} X_+, H\bb Z \wedge \Gmp{-d}] = 0 \] for $d$ sufficiently large.
This follows from the vanishing of motivic cohomology in negative weights over such bases \cite[Corollary 7.19]{spitzweck2012commutative}.
\end{example}
\subsection{Virtual étale cohomological dimension} \label{subsec:virtual-coh-dim} We will need the following result about motivic cohomology in large degrees.
\begin{lemma} \label{lemm:vcd-iso}
For $m > \mathrm{vcd}_\ell(k)$ and $a \in \bb Z$ the map \[ H^m(k, \bb Z/\ell(a)) \xrightarrow{[-1]} H^{m+1}(k, \bb Z/\ell(a+1)) \] is an isomorphism.
\end{lemma}
\begin{proof}
If $a < m$ then both groups vanish by \cite[Lemma 3.2(2)]{MR1744945} (\cite[Lemma 5.2]{mvw}), so there is nothing to prove.
Thus let $a \ge m$,
so that \[ H^m(k, \bb Z/\ell(a)) \simeq H^m_\mathrm{\acute et}(k, \mu_\ell^{\otimes a}) \] by the Beilinson-Lichtenbaum conjecture \cite[Theorem 6.17]{voevodsky-BK} (\cite[Theorem 10.2]{mvw}).
If $\ell$ is odd then $\mathrm{cd}_\ell(k) = \mathrm{vcd}_\ell(k)$ and again both groups are zero.
If $\ell=2$ then $\mu_\ell^{\otimes a} = \bb Z/2$ and the claim follows from the Gysin sequence (see e.g., Lemma \ref{lemm:vcd-vanishing}).
\end{proof}
\section{Bott-inverted spheres}\label{sec:bott-sph}
\subsection{\texorpdfstring{$\tau$}{tau}-self maps}
\begin{definition} \label{def:tau-self-map}
Let $\ell$ be a prime with $\ell \in \mathcal{O}^\times(S)$ and $1 \le m, n \le \infty$.
If $m, n < \infty$ denote by $\mathbbm{1}/(\ell^n, \rho^m)$ the obvious cofiber; if $n = \infty$ and $m<\infty$ we set $\mathbbm{1}/(\ell^\infty, \rho^m) = (\mathbbm{1}/\rho^m)_\ell^\wedge$,
and similarly if $m = \infty$.
Suppose given a map \[ \tilde\tau: \mathbbm{1}/(\ell^n, \rho^m) \to \mathbbm{1}/(\ell^n, \rho^m)(r). \]
\begin{enumerate}
\item We write \[ \mathbbm{1}/(\ell^n,\rho^m)[\tilde\tau^{-1}] = \colim \left[ \mathbbm{1}/(\ell^n,\rho^m) \xrightarrow{\tilde\tau} \mathbbm{1}/(\ell^n,\rho^m)(r) \xrightarrow{\tilde\tau(r)}\mathbbm{1}/(\ell^n,\rho^m)(2r) \to \dots \right]. \]
More generally, for $E \in \mathcal S\mathcal H(S)$ we set \[ E/(\ell^n, \rho^m)[\tilde\tau^{-1}] = E \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tilde\tau^{-1}]. \]
\item We call $\tilde\tau$ a $\tau$-self map if for every map $x \to S$, where $x$ is the spectrum of a field, the map \[ H\bb Z_x/(\ell^n, \rho^m) \to H\bb Z_x/(\ell^n, \rho^m)[\tilde\tau^{-1}]_{\ell,\rho}^\wedge \in \mathcal S\mathcal H(x)_{\ell,\rho}^\wedge \] is an étale localization.
\item We call $\tilde\tau$ a good $\tau$-self map if it is a $\tau$-self map and, for every $E, F \in \mathcal S\mathcal H(S)$, every morphism $\alpha: E \to F/(\ell^n,\rho^m)[\tilde\tau^{-1}]_{\ell,\rho}^\wedge$ factors through $E \to E/(\ell^{n'},\rho^{m'})[\tilde\tau'^{-1}]$, where $\tilde\tau'$ is another $\tau$-self map.
\end{enumerate}
\end{definition}
\begin{remark} \label{rmk:p-completion-unnec}
Unless $m=\infty$ or $n=\infty$, $E/(\ell^n, \rho^m)[\tilde\tau^{-1}]$ is already $(\rho, \ell)$-complete.
\NB{Call $E$ $a$-torsion if $a: E \to E$ is the zero map.
Then sums of $a$-torsion spectra are $a$-torsion, and an extension of $a$-torsion spectra is $a^2$-torsion. Hence directed colimits of $a$-torsion spectra are $a^2$-torsion.}
\end{remark}
\begin{lemma} \label{lemm:bott-element-etale-invertible}
Let $\tau: \mathbbm{1}/(\ell^n, \rho^m) \to \mathbbm{1}/(\ell^n, \rho^m)(r)$ be a $\tau$-self map.
Then $L_\mathrm{\acute et}(\tau)_{\ell,\rho}^\wedge$ is an equivalence.
\end{lemma}
\begin{proof}
Suppose that $n=\infty$.
The map $L_\mathrm{\acute et}(\tau)_{\ell,\rho}^\wedge$ is an equivalence if and only if $L_\mathrm{\acute et}(\tau/\ell)_{\ell,\rho}^\wedge$ is (by $\ell$-completeness).
Since $\tau/\ell$ is also a $\tau$-self map, replacing $\tau$ by $\tau/\ell$ we may assume that $n<\infty$.
Similarly we may assume that $m<\infty$.
By \cite[Theorem 3.1]{bachmann-SHet2} we have $\mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge \simeq \mathcal S\mathcal H(S_\mathrm{\acute et}^\wedge)_\ell^\wedge$.
This implies that pulling back along maps of the form $Spec(k) \to S$, where $k$ is separably closed, is a conservative family for $\mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge$.
We may thus replace $S$ by the spectrum of a separably closed field $k$.
\NB{The only reason for this is to know that isos are detected in homology. Easy to see ab initio, too.}
Since $L_\mathrm{\acute et} \mathbbm{1}/(\ell^n,\rho^m) \in \mathcal S\mathcal H_{\ge -1}$, in order to show that $L_\mathrm{\acute et}(\tau)_{\ell,\rho}^\wedge$ is an equivalence, it suffices to show this in homology.
In other words, we need to show that the map \[
H\bb Z/(\ell^n,\rho^m) \xrightarrow{\tau} H\bb Z/(\ell^n,\rho^m)(r)
\] is an étale equivalence.
This is true since by part (2) of Definition~\ref{def:tau-self-map}, we demand that $H\bb Z/(\ell^n,\rho^m) \to H\bb Z/(\ell^n,\rho^m)[\tau^{-1}]_{\ell,\rho}^\wedge$ is an étale localization.
\end{proof}
\subsection{Cohomological Bott elements}
We shall in the next subsection construct $\tau$-self maps as multiplication by suitable elements.
In preparation, we study the analogous question for $H\bb Z$.
\begin{definition} \label{def:cohomological-bott-element}
Let $S = Spec(k)$, $k$ a field.
Let $1 \le m, n \le \infty$.
By a \emph{cohomological Bott element} we mean an element $\tau \in \pi_{0,-r} H\bb Z/(\ell^n, \rho^m)$ for some $r>0$ such that for every field $l/k$ and every choice of multiplication on $H\bb Z/(\ell^n, \rho^m)$ under the standard multiplication on $H\bb Z_\rho^\wedge/\ell^n$ (in the sense of \S\ref{sec:moore-mult}), the map \[ H\bb Z/(\ell^n,\rho^m)|_l \to H\bb Z/(\ell^n,\rho^m)|_l[\tau^{-1}]_{\ell,\rho}^\wedge \] is an étale localization.
Here,
$H\bb Z/(\ell^n,\rho^m)[\tau^{-1}]$ denotes the mapping telescope of the self map given by multiplication by $\tau$.
\end{definition}
\begin{remark}
In light of Example \ref{ex:HZ-rho-complete}, $H\bb Z/\ell^m$ is already $\rho$-complete, so we shall suppress the additional completion from now on.
\end{remark}
\begin{remark} \label{rmk:product-bott-elts}
If $\tau_1, \tau_2$ are cohomological Bott elements, then so is $\tau_1 \tau_2$.
Indeed by cofinality, $E[(\tau_1\tau_2)^{-1}]$ (where $E=H\bb Z/(\ell^n, \rho^m)$) can be computed as a colimit over $\bb N \times \bb N$, with horizontal maps given by multiplication by $\tau_1$ and vertical maps by $\tau_2$.
Computing the horizontal colimit first, it suffices to show that multiplication by $\tau_2$ induces an equivalence on $E[\tau_1^{-1}]$; this follows from the assumption that $\tau_1$ and $\tau_2$ are both cohomological Bott elements.
\end{remark}
\begin{lemma} \label{lemm:bott:powers-reduction}
Let $\tau_1 \in \pi_{0,-r} H\bb Z/\ell$ and $\tau_2 \in \pi_{0,-r'} H\bb Z/\ell^{n}$ be elements such that $\tau_2$ reduces to a power of $\tau_1$ modulo $\ell$ (here $n=\infty$ is allowed).
\begin{enumerate}
\item $\tau_1$ is a cohomological Bott element if and only if $\tau_2$ is a cohomological Bott element.
\item If $\tau_2$ is a cohomological Bott element, then its image $\tau_3$ in $\pi_{0,-r'}H\bb Z/(\ell^n,\rho^m)$ is a cohomological Bott element.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) Suppose first that $n < \infty$.
We have a cofiber sequence $H\bb Z/\ell^n \xrightarrow{\ell} H\bb Z/\ell^n \to H\bb Z/\ell \oplus \Sigma H\bb Z/\ell^{n-1}$ in $\mathcal S\mathcal H(k)$.
Thus if $H\bb Z/\ell^n[\tau_2^{-1}]$ is étale local then so is $H\bb Z/\ell[\tau_2^{-1}] \simeq H\bb Z/\ell[\tau_1^{-1}]$.
Moreover, the converse holds if also $H\bb Z/\ell^{n-1}[\tau_2^{-1}]$ is étale local; this will hold by induction on $n$.
Finally suppose $n=\infty$.
Then $H\bb Z/\ell[\tau_1^{-1}] \simeq H\bb Z_\ell^\wedge[\tau_2^{-1}]_\ell^\wedge/\ell$, so $\tau_1$ is a cohomological Bott element if $\tau_2$ is.
Conversely, if $\tau_1$ is a cohomological Bott element then $H\bb Z_\ell^\wedge[\tau_2^{-1}]_\ell^\wedge \simeq \lim_n H\bb Z/\ell^n[\tau^{-1}]$ is a limit of étale local spectra, by what we already established, so it is étale local.
In other words, $\tau_2$ is a cohomological Bott element.
(2) Write $\tau_3$ for the image.
Then for any choice of compatible multiplication, the map $H\bb Z/(\ell^n,\rho^m) \to H\bb Z/(\ell^n,\rho^m)[\tau_3^{-1}]$ is the cofiber of multiplication by $\rho^m$ on the map $H\bb Z/\ell^n \to H\bb Z/\ell^n[\tau_2^{-1}]$.
The latter is an étale localization by (1), and hence so is the former.
This concludes the proof.
\end{proof}
\begin{lemma} \label{lemm:bott-descent}
Let $l/k$ be a finite separable extension of degree coprime to $\ell$, and $\tau \in \pi_{0,-r} H\bb Z/\ell^n$.
Then $\tau$ is a cohomological Bott element if and only if $\tau|_l$ is a cohomological Bott element.
\end{lemma}
\begin{proof}
It suffices to show that $H\bb Z/\ell^n[\tau^{-1}]$ is a summand of $(l/k)_* H\bb Z/\ell^n|_l[\tau|_l^{-1}]$.
This is clear by the existence of transfers in $\mathcal D\mathcal M(k)$ (see e.g., Corollary \ref{cor:galois-descent}).
\NB{this is total overkill...}).
\end{proof}
\begin{lemma} \label{lemm:bott-zeta}
Let $\zeta \in k$ be a primitive $\ell^n$-th root of unity and $\tau \in \pi_{0,-1} H\bb Z/\ell^n$.
Suppose that $\beta(\tau) = [\zeta]$, where \[ \beta: H^{0,1}(k, \bb Z/\ell^n) \to H^{1,1}(k, \bb Z) \] is the integral Bockstein and $[\zeta] \in H^{1,1}(k, \bb Z) \simeq k^\times$ is the element corresponding to $\zeta$. Then $\tau$ is a cohomological Bott element.
\end{lemma}
\begin{proof}
Under the equivalence $\bb Z(1) \simeq {\mathbb{G}_m}$ \cite[Lemma 3.2(1)]{MR1744945} (\cite[Theorem 4.1]{mvw}),
the Bockstein corresponds to the inclusion $\ul{\pi}_0(\bb Z/\ell^n(1)) \simeq \mu_{\ell^n} \hookrightarrow {\mathbb{G}_m}$.
It follows that $L_\mathrm{\acute et}(\bb Z/\ell^n(1)) \simeq \mu_{\ell^n}$ and that $\tau$ defines an equivalence $\bb Z/\ell^n \simeq L_\mathrm{\acute et}(\bb Z/\ell^n(1))$; in particular $\tau$ is an étale local equivalence.
It remains to show that for $X \in {\cat{S}\mathrm{m}}_k$ the induced map \[ H^{**}(X, \bb Z/\ell^n)[\tau^{-1}] = \colim_r H^{*,*+r}(X, \bb Z/\ell^n) \to H^{**}_\mathrm{\acute et}(X, \bb Z/\ell^n) \] is an isomorphism.
This is immediate from the solution of the Beilinson-Lichtenbaum conjecture \cite[Theorem 6.17]{voevodsky-BK}.
\end{proof}
\subsection{Spherical Bott elements}
\begin{definition} \label{def:spherical-bott-element}
Let $1 \le m, n \le \infty$, $S$ a scheme.
By a \emph{spherical Bott element} we mean an element $\tau \in \pi_{0,-r}(\mathbbm{1}/(\ell^n, \rho^m))$ such that for every map $x \to S$ where $x$ is the spectrum of a field, the induced element $H\bb Z \wedge \tau|_x \in \pi_{0,-r}(H\bb Z_x/(\ell^n,\rho^m))$ is a cohomological Bott element (see Definition \ref{def:cohomological-bott-element}).
\end{definition}
\begin{remark} \label{rmk:spherical-bott-base-change}
Spherical Bott elements are stable under base change, essentially by definition.
\end{remark}
\begin{lemma} \label{lemm:tau-self-map-construct}
Choose a multiplication\footnote{The existence of this structure will be extensively discussed in Appendix~\ref{sec:oka}.} on $\mathbbm{1}/\ell^n \in \mathcal S\mathcal H$ and one on $\mathbbm{1}_\ell^\wedge/\rho^m \in \mathcal S\mathcal H(S)$.
Let $n' \le n$ and choose a $\mathbbm{1}/\ell^n$-module structure $m$ on $\mathbbm{1}/\ell^{n'}$.
Let $\tau \in \pi_{0,-r} \mathbbm{1}/(\ell^n, \rho^m)$ be a spherical Bott element.
Then the composite \[ \tilde\tau: \mathbbm{1}/(\ell^{n'},\rho^m) \xrightarrow{\tau \wedge \mathrm{id}} \Sigma^{0,r} \mathbbm{1}/(\ell^{n},\rho^m) \wedge \mathbbm{1}/(\ell^{n'},\rho^m) \xrightarrow{m} \Sigma^{0,r} \mathbbm{1}/(\ell^{n'},\rho^m) \] is a $\tau$-self map.
If the multiplication is chosen to be homotopy commutative, and the module structure homotopy associative, then the $\tau$-self map is good.
\end{lemma}
\begin{proof}
We need to check that forming the mapping telescope of $\tilde\tau \wedge H\bb Z|_k$ is an étale localization, for every field $k$.
This is true basically by definition, as soon as we know that the multiplication on $H\bb Z/\ell^n$ induced by the one on $\mathbbm{1}/\ell^n$ is the standard one.
This is indeed the case, by Corollary \ref{cor:multn-HZ-correct}.
Goodness of the $\tau$-self map follows from Corollary \ref{cor:factor-through-free}.
\end{proof}
\section{Construction of spherical Bott elements} \label{sec:construction}
\subsection{Construction via roots of unity} \label{sec:bott-primitive}
\begin{lemma} \label{lem:bott-exists}
Let $\ell^n \ne 2$ and suppose $S$ contains a primitive $\ell^n$-th root of unity $\zeta$.
Then there exists a spherical Bott element mod $(\ell^n, \rho^\infty)$.
\end{lemma}
\begin{proof}
Note that either $\ell$ is odd or $-1$ is a square on $S$.
In both cases, using the notation $n_\epsilon$ from \S\ref{sec:standard}, we find that
\[ \ell^n_\epsilon = \ell^n: \mathbbm{1}_{\ell,\rho}^\wedge \to \mathbbm{1}_{\ell,\rho}^\wedge.\]
By construction we have a cofibration sequence \[ \mathbbm{1}_{\ell,\rho}^\wedge/\ell^n(1) \xrightarrow{\beta} {\mathbb{G}_m}_{\ell,\rho}^\wedge \xrightarrow{\ell^n} {\mathbb{G}_m}_{\ell,\rho}^\wedge, \] whence by Lemma \ref{lemm:bott-zeta} a spherical Bott element exists as soon as $\ell^n[\zeta]_{\rho,\ell}^\wedge = 0$.
This is immediate from Proposition \ref{prop:powers-n-epsilon}.
\end{proof}
\subsection{Lifting to higher $\ell$-powers} \label{sec:bott-power}
Given an object $E$ in a triangulated symmetric monoidal category and a map $m: E \otimes E \to E$, by a \emph{derivation} $\delta: E \to E[1]$ we mean a map such that $\delta \circ m = m \circ (\mathrm{id} \otimes \delta + \delta \otimes \mathrm{id})$.
\begin{lemma} \label{lemm:power-derivation}
Let $\scr C$ be a symmetric monoidal triangulated category. Suppose $E \in \scr C$ is a non-commutative, non-associative and non-unital algebra, and let $\delta: E \to E[1]$ be a derivation.
Then for $X \in \scr C$ and $t: X \to E$ we have \[ \delta(t^n) = \sum_{i = 1}^n t \dots t \delta(t) t \dots t: X^{\otimes n} \to E[1]. \]
Here the sum is over all possible ways of replacing one instance of $t$ in the string $t^n$ by $\delta(t)$, and both sides are associated from the left (i.e., $t^n = t(t(\dots(tt)\dots))$ and so on).
\end{lemma}
\begin{proof}
We write the map $\delta(t^{n+1})$ as \[ X^{\otimes n+1} \simeq X \otimes X^{\otimes n} \xrightarrow{t \otimes t^{\otimes n}} E \otimes E \xrightarrow{\mathrm{id} \otimes t^n} E \otimes E \xrightarrow{m} E \xrightarrow{\delta} E[1]. \]
The definition of a derivation implies that this is the sum of the two maps $t\delta(t^n)$ and $\delta(t)t^n$.
The result thus follows by induction, starting at $n=2$ where we use the definition of a derivation.
\end{proof}
Recall the notion of a regular multiplication from \S\ref{sec:def-set}.
\begin{proposition} \label{prop:bott-power}
Let $q$ be a prime power, fix a regular multiplication on $\Mp{q} \in \mathcal S\mathcal H$ and assume there exists a spherical Bott element mod $(q,\rho^\infty)$.
Then for every $n \ge 1$ there exists a spherical Bott element mod $(q^n,\rho^\infty)$.
\end{proposition}
\begin{proof}
Throughout the proof, all spectra are implicitly $(\rho,q)$-completed.
As a first step, choose regular multiplications on $\Mp{q^n}$ for all $n \ge 1$ such that the sequence of reduction maps \[ \dots \Mp{q^3} \to \Mp{q^2} \to \Mp{q} \] consists of morphisms of homotopy unital ring spectra. This is possible by \cite[Lemma 5]{oka1984multiplications}.
We shall proceed by induction on $n$.
Thus let $\tau$ be a spherical Bott element modulo $q^n$.
We claim that $\tau^q$ lifts to an element modulo $q^{n+1}$; this will be a spherical Bott element by Lemma \ref{lemm:bott:powers-reduction}(1).
To see this, we begin with the cofiber sequence in \cite[(3.3)]{rigidity-in-motivic-homotopy-theory}
\[ \Mp{q} \to \Mp{q^{n+1}} \to \Mp{q^n} \xrightarrow{\bar\delta} \Mp{q}[1]. \]
We need to show that $\bar\delta(\tau^q) = 0$.
Let $r: \Mp{q^n} \to \Mp{q}$ be the reduction map; then $\bar\delta = r\delta$, where $\delta: \Mp{q^n} \to \Mp{q^n}[1]$ is the coboundary.
By our choice $r$ is multiplicative, and so by Lemma \ref{lemm:power-derivation} we have
\[ \bar\delta(\tau^q) = r\left[ \delta(\tau)\tau^{q-1} + \tau\delta(\tau)\tau^{q-2} + \dots + \tau^{q-1}\delta(\tau) \right], \] where the sum consists of $q$ terms.
Note that $q = 0$ on $\Mp{q}$, since $\Mp{q}$ has a multiplication.
Suppose first that the multiplication on $\Mp{q^n}$ is associative and commutative.
Then each of the $q$ terms in our sum is the same\footnote{One might be concerned here about an absence of signs. We are given various maps $\mathbbm{1}^{\wedge q} \to \mathbbm{1}/q[1]$, differing by permutations \emph{of the source only}. But the switch map on $\mathbbm{1} \wedge \mathbbm{1}$ is the identity, whence there are no signs.}, so the sum is zero, and we are done.
To finish the proof, we observe that we do not actually need the multiplication to be homotopy associative or commutative.
Since $\delta = u\partial$ (where $u: \mathbbm{1} \to \Mp{q^n}$ is the unit map) we get $\partial \delta = 0$.
This implies that any element commutes with $\delta(\tau)$, and any two elements associate with $\delta(\tau)$ (see Section \ref{sec:oka}).
Thus \[ \delta(\tau)\tau^{m+1} \stackrel{(1)}{=} \delta(\tau)(\tau\tau^m) \stackrel{(2)}{=} (\delta(\tau) \tau)\tau^m \stackrel{(3)}{=} (\tau \delta(\tau))\tau^m \stackrel{(2)}{=} \tau(\delta(\tau)\tau^m), \] where $(1)$ is by definition (everything being associated from the left), $(2)$ is because everything associates with $\delta(\tau)$ and $(3)$ is because everything commutes with $\delta(\tau)$.
This implies by induction that $\delta(\tau)\tau^m$ is independent of the order of multiplication, for any $m$, and so all the terms are the same, as before.
\end{proof}
\subsection{Construction by descent} \label{sec:Bott-descent}
Let $f: S' \to S$ be a morphism of schemes and $\sigma: S' \to S'$ be an automorphism over $S$.
Then $f \sigma = f = f \sigma^{-1}$ and so $\sigma^* f^* \stackrel{\alpha}{\simeq} f^*$, $f_* \sigma^{-1}_* \stackrel{\beta}{\simeq} f_*$.
This provides us with a sequence of equivalences \[ f_*f^* \stackrel{f_*\alpha}{\simeq} f_* \sigma^* f^* \simeq f_* \sigma^{-1}_* f^* \stackrel{\beta f^*}{\simeq} f_*f^*, \]
where we have used that an adjoint of an equivalence is canonically equivalent to the inverse.
Hence for every object $E \in \mathcal S\mathcal H(S)$ we get an automorphism \[ \sigma_E: f_*f^* E \to f_*f^* E. \]
This construction is natural in $E$.
The following result is proved in Appendix \ref{sec:galois}; see Corollary \ref{cor:galois-descent} and the preceding paragraphs.
\begin{proposition} \label{prop:galois-descent}
Let $f:S' \to S$ be a finite Galois cover with group $G$.
\begin{enumerate}
\item The above incoherent construction refines to a coherent action, i.e., a functor
\[BG \to \mathcal S\mathcal H(S); * \mapsto f_*f^* E.
\]
\item The unit of adjunction $E \to f_*f^*E$ refines to a $G$-equivariant map (for the trivial action of $G$ on $E$).
\item Suppose that $E \in \mathcal S\mathcal H(S)[1/2, 1/|G|]^+$.
Then $E \to f_*f^*E$ exhibits $E$ as the homotopy fixed points of the $G$-action on $f_*f^* E$, and this limit diagram is preserved by any additive functor.
\end{enumerate}
\end{proposition}
\begin{remark} \label{rmk:descent-summand}
In the situation of Proposition \ref{prop:galois-descent}(3), the map $E \to f_*f^*E$ is a split injection, i.e., $E$ is a summand of $f_*f^* E$.
See Corollary \ref{cor:compute-fixed} for details.
\end{remark}
\begin{corollary} \label{cor:bott-descent}
Let $q=\ell^n$ be an odd prime power, $q \ne 3$.
Assume $1/\ell \in S$ and let $S'$ be obtained from $S$ by adjoining a primitive $\ell$-th root of unity.
If $S'$ affords a spherical Bott element mod $(q,\rho^\infty)$ then so does $S$.
\end{corollary}
\begin{proof}
We again complete everything implicitly at $(\rho,\ell)$.
If $S=S'$ there is nothing to prove, so assume $S' \ne S$.
Thus $f: S' \rightarrow S$ is a Galois cover with Galois group $G \subset \bb Z/(\ell-1)$ \cite[Corollary 10.4]{neukirch2013algebraic}.
Consequently \[ \Mp{q} \in \mathcal S\mathcal H(S)[1/2, 1/|G|]^+ \] and so by Proposition \ref{prop:galois-descent} we get $[\mathbbm{1}, \Mp{q}(m)]_S = [\mathbbm{1}, \Mp{q}(m)]_{S'}^G$.
Let $\tau \in [\mathbbm{1}, \Mp{q}(m)]_{S'}$ be a spherical Bott element and put \[ \tau' = \prod_{g \in G} (g \tau) \in [\mathbbm{1}, \Mp{q}(|G|m)]_{S'}. \]
Then for $h \in G$ we have \[ h\tau' = \prod_g (hg \tau) = \tau', \] since the multiplication in $\Mp{q}$ is commutative and associative (here we use that $q \ne 3$).
In other words $\tau'$ is fixed by $G$ and so defines an element $\tau_S \in [\mathbbm{1}, \Mp{q}(|G|m)]_S$.
By construction, $f^*\tau_S = \prod_g (g^* \tau)$ is a product of spherical Bott elements and hence a spherical Bott element (see Remark \ref{rmk:product-bott-elts}).
By Lemma \ref{lemm:bott-descent} it follows that $\tau_S$ is also a spherical Bott element.
\end{proof}
\subsection{Construction over special fields} \label{subsec:quad}
\subsubsection{Quadratically closed fields}
\begin{proposition} \label{prop:tau-p-closed}
Let $\ell$ be a prime, $S = Spec(k)$.
Assume that $k$ affords a primitive $\ell$-th root of unity,
\NB{so $char(k) \ne \ell$...},
and $k^\times/\ell \simeq \{1\}$ (i.e., every element of $k$ admits an $\ell$-th root).
Then there exists a spherical Bott element \[ \tau \in \pi_{0,-1}(\mathbbm{1}_{\ell,\rho}^\wedge). \]
\end{proposition}
\begin{proof}
Lemma \ref{lem:bott-exists} and its proof show that there exists $\tau_1 \in \pi_{0,-1}(\mathbbm{1}_\rho^\wedge/q)$, where $q=4$ if $\ell=2$ and $q=\ell$ else.
We shall show by induction that for each $n$: \begin{equation} \label{eq:ind-step} \text{there exists a lifting } \tau_{n+1} \in \pi_{0,-1}(\mathbbm{1}_\rho^\wedge/q^{n+1}) \text{ of } \tau_n. \tag{$*$} \end{equation}
By the Milnor exact sequence \cite[Proposition VI.2.15]{goerss2009simplicial} there is a surjection \[ \pi_{0,-1}(\mathbbm{1}_{\ell,\rho}^\wedge) \to \lim_n \pi_{0,-1}(\mathbbm{1}_\rho^\wedge/q^n); \]
hence there is a (non-canonical) lift $\tau \in \pi_{0,-1}(\mathbbm{1}_{\ell,\rho}^\wedge)$.
This will be a spherical Bott element by Lemma \ref{lemm:bott:powers-reduction}(1).
It hence remains to prove \eqref{eq:ind-step}.
The cofiber sequence \[ \mathbbm{1}(1)_\rho^\wedge/q \to \mathbbm{1}(1)_\rho^\wedge/q^{n+1} \to \mathbbm{1}(1)_\rho^\wedge/q^n \to \mathbbm{1}[1](1)_\rho^\wedge/q \simeq {\mathbb{G}_m}_\rho^\wedge/q \] implies that it is enough to prove the vanishing $\pi_0({\mathbb{G}_m}_\rho^\wedge/q) = 0$.
Suppose first that $\ell=2$.
Then $\rho$ is nilpotent by Remark~\ref{rmk:rho-completion}(1), so ${\mathbb{G}_m}_\rho^\wedge \simeq {\mathbb{G}_m}$ and $\pi_0({\mathbb{G}_m}_\rho^\wedge/q) \simeq K_1^{MW}(k)/4$ \cite[Corollary 6.43]{A1-alg-top}.
Recall that we have the fiber product decomposition \cite[Theorem 5.4]{gsz} (\cite{morel-ideal})
\[ K_1^{MW}(k) \simeq I(k) \times_{I(k)/I^2(k)} K_1^M(k). \]
Since $k$ is quadratically closed, $I(k) = 0$ \cite[Lemma 31.1]{MR2427530} and hence $K_1^{MW}(k) \simeq K_1^M(k)$.
Thus \[ K_1^{MW}(k)/4 \simeq K_1^M(k)/4 \simeq k^\times/4 \simeq \{1\}, \] and we are done (here the last isomorphism again follows from the fact that $k$ is quadratically closed).
Now suppose that $\ell$ is odd.
Then ${\mathbb{G}_m}_\rho^\wedge/q \simeq {\mathbb{G}_m}[1/2]^+/\ell$ and so \[ \pi_0({\mathbb{G}_m}_\rho^\wedge/q) \simeq K_1^{MW}(k)[1/2]^+/\ell \simeq K_1^M(k)/\ell \simeq k^{\times}/\ell \simeq \{1 \}. \] Here we have used the fact that for any field $k$, the group $I/I^2(k)[1/2] = 0$ since it is a module over $W/I(k) \simeq \bb Z/2$ and thus the fiber product decomposition for $K_1^{MW}$ is just a product.
This concludes the proof.
\end{proof}
\subsubsection{The real numbers} \label{subsec:reals}
If $S = Spec(\bb R)$, then there exists $\tau_n \in \pi_{0,-r}(\mathbbm{1}_2^\wedge/\rho^n)$ lifting a power of $\tau \in \pi_{0,-1}H\bb Z/2$ \cite[Theorem 7.10 and its proof]{behrens2019c_2}.
Moreover $\mathbbm{1}_2^\wedge/\rho^n$ is an $\scr E_\infty$-ring \cite[Lemma 7.8]{behrens2019c_2}.
Consequently this defines a spherical Bott element modulo $(2^\infty, \rho^n)$ by Lemma \ref{lemm:bott-zeta}(2).
\subsection{Summary of \texorpdfstring{$\tau$}{tau}-self maps} \label{subsec:summary}
Using Lemma \ref{lemm:tau-self-map-construct} and the multiplications and module structures on Moore spectra \cite{oka1984multiplications}, as reviewed (and slightly extended in Corollary~\ref{cor:asymp}) in Appendix \ref{sec:moore-mult}, we find that there is a good $\tau$-self map modulo $(\ell^n, \rho^m)$ as soon as there is a spherical Bott element modulo $(\ell^{n'}, \rho^m)$ for some $n' \ge n$.
In particular we have said elements in the following cases:
\begin{enumerate}
\item $m = \infty$, $\ell$ odd, $n < \infty$ (use Corollary \ref{cor:bott-descent}, Proposition \ref{prop:bott-power}, and Lemma \ref{lem:bott-exists}).
\item $m = \infty$, $\ell=2$, $n<\infty$, $\sqrt{-1} \in S$ (using the same results).
\item $m = \infty$, $n=\infty$, $\ell$ arbitrary, $S$ defined over a field $k$ containing a primitive $\ell$-th root of unity and satisfying $k^\times/\ell \simeq \{1\}$ (use Proposition \ref{prop:tau-p-closed} and Remark \ref{rmk:spherical-bott-base-change}).
\item $m < \infty$, $\ell=2$, $n \le \infty$, $S$ defined over $\bb R$ (use \S\ref{subsec:reals} and Remark \ref{rmk:spherical-bott-base-change}).
\end{enumerate}
\section{Slice convergence} \label{sect:s-conv}
In this section we provide an extension of Levine's results on the convergence of the slice spectral sequence \cite{levine2013convergence} or, more precisely, convergence of the \emph{slice tower} for motivic spectra satisfying certain $\ell$-torsion and $\rho$-torsion hypotheses; see Theorem~\ref{thm:convergence} for a precise statement. For this we use a very deep result: Voevodsky's resolution of the Milnor and Bloch-Kato conjectures \cite{ovv,voevodsky-BK}. We first set out our conventions on towers.
\begin{definition} \label{def:sep-conv}
Let $\scr C$ be a category and $E \in \scr C$.
\begin{enumerate}
\item By a \emph{tower over $E$} we mean an object $E_\bullet \in \mathrm{Fun}(\bb Z^\mathrm{op}, \scr C_{/E})$.
We typically display towers as \[ \dots E_2 \to E_1 \to E_0 \to E_{-1} \to \dots \to E \quad\text{or}\quad E_\bullet \to E. \]
\item Suppose $\scr C$ is an abelian $1$-category.
Given a tower $E_\bullet \to E$ in $\scr C$ we define the descending filtration \[ F_i E = \mathrm{im}(E_i \to E) \subset E. \]
We call the tower $E_\bullet \to E$ \emph{separated} if \[ 0 = \cap_i F_i E \] and \emph{convergent} if in addition it is \emph{exhaustive}, i.e., \[ E = \cup_i F_i E. \]
\end{enumerate}
\end{definition}
Clearly if $F: \scr C \to \scr D$ is any functor and $E_\bullet \to E$ is a tower in $\scr C$, then $FE_\bullet \to FE$ is a tower in $\scr D$.
We shall utilize this definition of convergence to detect when maps are null.
\begin{lemma} \label{lemm:convergence-application}
Let $E_\bullet \to E$ be a tower in the category $\mathcal S\mathcal H$ of spectra.
Denote by $E_i/E_{i+1}$ the cofiber of the canonical morphism $E_{i+1} \to E_i$.
Let $k \in \bb Z$ and assume that (a) the tower $\pi_k(E_\bullet \to E)$ is convergent, and (b) $\pi_k(E_i/E_{i+1}) = 0$ for every $i$.
Then $\pi_k(E) = 0$.
\end{lemma}
\begin{proof}
It suffices to prove the result for $k =0$. Let $f: \mathbbm{1} \to E$ be any map.
We need to show that $f = 0$.
We shall show that $f \in F_n \pi_0 E$ for all $n$; then we are done by separatedness.
By definition of exhaustiveness, we have $f \in F_N \pi_0 E$ for some $N$; hence it suffices to show that $f \in F_n \pi_0 E$ implies $f \in F_{n+1} \pi_0 E$.
Hence suppose $f \in F_n \pi_0 E$, and pick $f_n: \mathbbm{1} \to E_n$ such that the composite $\mathbbm{1} \to E_n \to E$ is homotopic to $f$; this is possible by definition of $F_n \pi_0 E$.
Since $\pi_0(E_n/E_{n+1}) = 0$ by assumption, the composite $\mathbbm{1} \to E_n \to E_n/E_{n+1}$ is homotopic to zero, and hence $f_n$ lifts to a map $f_{n+1}: \mathbbm{1} \to E_{n+1}$.
It follows that $f \in F_{n+1} \pi_0 E$.
\end{proof}
In order to apply this result, we need a good supply of convergent towers.
We shall produce them from Voevodsky's slice tower, using a strengthening of Levine's convergence theorem for the slice filtration \cite[Theorem 7.3]{levine2013convergence} that we will establish next.
Recall that there is a functorial tower \[ \mathcal S\mathcal H(S) \rightarrow \mathcal S\mathcal H(S)^{\bb Z \cup \{ \infty\} }; E \mapsto (f_\bullet E \to E) \] called the \emph{slice tower} \cite{voe-open}; see also \cite[Section 3]{1-line} for a more extensive discussion and references therein.
Recall the definition of virtual cohomological dimension from \S\ref{subsec:virtual-coh-dim}.
In order to state our result, we shall make use of the following assumptions on $E \in \mathcal S\mathcal H(k)$ and $t \in \bb Z$:
\begin{enumerate}[(a)]
\item $E \in \mathcal S\mathcal H(k)_{\ge c}$ for some $c \in \bb Z$.
\item For $i, j \in \bb Z$, $K/k$ any finitely generated, separable field extension, and $a \in \ul{\pi}_i(E)_j(K)$ we have $t^r a = 0$ for $r$ sufficiently large.
\item There exists an integer $R \gg 0$ such that the endomorphism $\rho^{R}\colon E \rightarrow E \wedge \Gmp{R}$ is homotopic to zero.
\end{enumerate}
\begin{theorem} \label{thm:convergence}
Let $k$ be a field\footnote{Not necessarily perfect, contrary to Levine's assumption in \cite{levine2013convergence}.} of exponential characteristic $e$ and $t > 0$ coprime to $e$ such that $\mathrm{vcd}_t(k) < \infty$.
There exists a function
\begin{equation} \label{eq:fn}
\bb Z^6 \to \bb N, (c, d, R, i, j, M) \mapsto N(c, d, R, i, j, M)
\end{equation}
such that for every $x \in X \in {\cat{S}\mathrm{m}}_k$ with $\dim X \le d$, $(i,j,M) \in \bb Z^3$ and $E \in \mathcal S\mathcal H(k)$ satisfying (a) and (b), the following hold:
\begin{enumerate}
\item For $n > N(c,d,R,i,j,0)$ we have\footnote{Here for a sheaf $F$ on ${\cat{S}\mathrm{m}}_k$ we denote by $F_x$ its stalk at $x \in X$.} \[ \ul{\pi}_{i,j}(f_n(E)/\rho^R)_x = 0. \]
In particular, the tower
\begin{equation} \label{eq:pi-tower1}
\ul\pi_{i,j}(f_\bullet(E)/\rho^R)_x \rightarrow \ul\pi_{i,j}(E/\rho^R)_x
\end{equation} is separated.
\item In addition, if $E$ also satisfies (c), then the morphism \[ \ul{\pi}_{i,j}(f_{M+N(c,d,R,i,j,M)} E)_x \to \ul{\pi}_{i,j}(f_{M} E)_x \] is zero.
In particular, the tower
\begin{equation} \label{eq:pi-tower2}
\ul\pi_{i,j}(f_\bullet E)_x \rightarrow \ul\pi_{i,j}(E)_x
\end{equation} is separated.
\end{enumerate}
\end{theorem}
A few remarks are in order.
\begin{remark} \label{rmk:levine-conjecture}
This result is closely related to \cite[Conjecture 5]{levine2013convergence}.
Indeed the main idea in our argument is that for fields with $\mathrm{vcd}_2(k) < \infty$, $\rho$ is the only obstruction to nilpotence of the ideal $I=I(k) \subset GW(k)$.
In particular, under assumption (c), $I$ acts nilpotently on each $\ul{\pi}_{i,j}(E)_x$, and thus in particular this module is $I$-adically complete.
Levine's conjecture thus predicts our separatedness result.
Conversely, over fields of finite $\mathrm{vcd}_2$, the $I$-adic and $\rho$-adic filtrations are commensurate on $K_*^{MW}$; it thus seems justified to think (over such fields) of derived $\rho$-completion as a form of $I$-adic completion.
We thus view our results as establishing a derived version of Levine's conjecture.
\end{remark}
\begin{remark} \label{rm:fn} The function~\eqref{eq:fn} indicates the dependence of the number $N(c, d, R, i, j, M)$ on the connectivity of $E$ (given by $c$), the bidegrees we are interested in (given by $(i, j)$), the effective cover of $E$ we are taking (given by $M$) and the dimension of the scheme (or, rather, the point; this is given by $d$). Note that this function does not depend on the number $r$ that appears in condition (b).
\end{remark}
\begin{remark} \label{rmk:exhaustive}
The slice tower is always exhaustive (see, for example, \cite[Lemma 3.1]{rso-solves}); hence the theorem implies that the towers \eqref{eq:pi-tower1}, \eqref{eq:pi-tower2} are convergent.
\end{remark}
\begin{remark} \label{rmk:invertible}
If $d$ is coprime to $t$, then (b) implies multiplication by $d$ is an isomorphism on $\ul{\pi}_{i,j}(E)$ for all $i,j$, and hence so is $d\colon E \rightarrow E$.
In particular, under the hypotheses of Theorem~\ref{thm:convergence}, we have that $E \in \mathcal S\mathcal H(k)[1/e]$.
\end{remark}
\begin{remark} \label{rmk:convergence-simp}
It follows from Remark \ref{rmk:rho-completion} that condition (c) is vacuous if $k$ is unorderable (e.g., $\mathrm{cd}_2(k) < \infty$), or if $t$ is odd and $E \in \mathcal S\mathcal H(k)[1/2]^+$.
In these cases the statement of Theorem \ref{thm:convergence} is not quite optimal; in fact the proof shows that \[ \ul{\pi}_{i,j}(f_{M} E)_x = 0 \] for $M \gg 0$ (depending on $i,j,c,d$).
\end{remark}
\begin{remark} \label{rmk:nilpotence}
If $a \in \pi_{p,q}\mathbbm{1}$ and $F \in \mathcal S\mathcal H(k)$, then standard arguments show that $a^2: F/a \to \Sigma^{2p,2q} F/a$ is the zero map (see e.g., \cite[Lemma 5.2]{rondigs2019remarks}).
It follows that for $E \in \mathcal S\mathcal H(k)_{\ge c}$ Theorem \ref{thm:convergence} applies to $E/(\rho^a,t^b)$ (and also $E/t^b$ or $(E/t^b)^+$, if Remark \ref{rmk:convergence-simp} applies).
\end{remark}
Our result implies convergence of the slice spectral sequence in novel cases.
We record the following, even though we do not use it in the rest of the article.
\begin{corollary}
Let $k$ be a field of exponential characteristic $e$ and $t > 0$ coprime to $e$ such that $\mathrm{vcd}_t(k) < \infty$.
Suppose $E \in \mathcal S\mathcal H(k)_{\ge c}$ for some $c \in \bb Z$.
\begin{enumerate}
\item The map $E_{t,\rho}^\wedge \xrightarrow{\mathrm{sc}} \mathrm{sc}(E)_{t,\rho}^\wedge$ induces an isomorphism on $\pi_{**}$;\NB{It is also the case that $E_{t,\rho}^\wedge \simeq \lim_{m,n}\mathrm{sc}(E/(t^n,\rho^m))$, but this seems like a less useful statement.}
here $\mathrm{sc}$ denotes the slice completion functor \cite[\S 3]{1-line}.
\item There is a conditionally convergent spectral sequence \[ \pi_{p,n}(s_q(E)_{t,\rho}^\wedge) \Rightarrow \pi_{p,n}(E_{t,\rho}^\wedge), \] using the indexing conventions of \cite{1-line}.
\end{enumerate}
\end{corollary}
\begin{proof}
(1) It is enough to show that $E/(\rho^n,t^m) \to \mathrm{sc}(E)/(\rho^n,t^m)$ is a $\pi_{**}$-isomorphism.
This is immediate from Theorem \ref{thm:convergence}(1) and Remark \ref{rmk:exhaustive} (and Remark \ref{rmk:nilpotence}, which tells us that Theorem \ref{thm:convergence} applies).
(2) We consider the completed slice tower \[ \dots \to f_n(E)_{t, \rho}^\wedge \to f_{n-1}(E)_{t,\rho}^\wedge \to \dots \to E_{t,\rho}^\wedge. \]
The cones of the maps in this tower are given by the $(t,\rho)$-completed slices $s_n(E)_{t, \rho}^\wedge$.
To prove conditional convergence for the corresponding spectral sequence displayed in (2), we need to show that \[ \pi_{**}(\lim_n(f_n(E)_{t,\rho}^\wedge)) \simeq 0 \quad\text{and}\quad \pi_{**}(\colim_n (f_n(E)_{t,\rho}^\wedge)) \simeq \pi_{**}(E_{t,\rho}^\wedge). \]
Since limits commute we have the fiber sequence \[ \lim_n(f_n(E)_{t,\rho}^\wedge) \simeq (\lim_n f_n(E))_{t,\rho}^\wedge \to E_{t,\rho}^\wedge \xrightarrow{\mathrm{sc}} \mathrm{sc}(E)_{t,\rho}^\wedge. \]
Hence the claim about limits reduces to (1).
For the claim about colimits, observe that for $X \in {\cat{S}\mathrm{m}}_k$ and $n<w$ we have
\begin{gather*}
\mathrm{map}(X(w), f_n(E)_{t,\rho}^\wedge) \\
\simeq \lim_m \mathrm{cof}(\mathrm{map}(X(w), \Sigma^{-m,-m} f_n(E)/t^m) \xrightarrow{\rho^m} \mathrm{map}(X(w), f_n(E)/t^m)) \\
\simeq \lim_m \mathrm{cof}(\mathrm{map}(\Sigma^{m,m} X(w), f_n(E/t^m)) \xrightarrow{\rho^m} \mathrm{map}(X(w), f_n(E/t^m))) \\
\simeq \lim_m \mathrm{cof}(\mathrm{map}(\Sigma^{m,m} X(w), E/t^m) \xrightarrow{\rho^m} \mathrm{map}(X(w), E/t^m)) \\
\simeq \mathrm{map}(X(w), E_{t,\rho}^\wedge),
\end{gather*}
where $X(w) := \Sigma^{0,w}\Sigma^\infty_+ X$.
Since the $X(w)$ are compact generators of $\mathcal S\mathcal H(k)$, we deduce that in fact \[ \colim_n (f_n(E)_{t,\rho}^\wedge) \simeq E_{t,\rho}^\wedge. \]
\end{proof}
\begin{remark}
Under the assumptions of the corollary, it is in fact the case that $E_{t,\rho}^\wedge \to \mathrm{sc}(E)_{t,\rho}^\wedge$ is an equivalence.
This requires a slightly more elaborate argument and will be treated elsewhere.
\end{remark}
In the rest of this section we prove Theorem \ref{thm:convergence}, adapting the argument of Levine \cite[Theorem 7.3]{levine2013convergence}.
Without loss of generality, we make the following standing assumptions:
\begin{enumerate}
\item $c = 0$,
\item $k$ is \emph{perfect} (using \cite[Corollary 2.1.7]{elmanto2018perfection} and Remark \ref{rmk:invertible}),
\item $k$ is infinite (using standard transfer arguments; see \cite[Appendix A]{levine2013convergence} for details),
\item $x$ is a generic point, so $X_x$ is the spectrum of a field of transcendence degree $\le d$ over $k$ (using unramifiedness of homotopy sheaves \cite[Lemma 6.4.4]{morel-conn}).
\end{enumerate}
With these assumptions at play, we quickly review Levine's approach to studying the slice filtration via the \emph{simplicial filtration} as in \cite[Section 4]{levine2013convergence}. For $E \in \mathcal S\mathcal H(k)$ and for any $M \ge 0, X \in {\cat{S}\mathrm{m}}_k$, consider the mapping spectrum \[ f_ME(X)= \mathrm{map}(\Sigma^{\infty}X_+, f_ME) \in \mathcal S\mathcal H; \] the functor $X \mapsto f_ME(X)$ is an $\bb A^1$-invariant Nisnevich sheaf of spectra. As elaborated in \cite[Section 3]{levine2013convergence}, we have an augmented simplicial spectrum of the form
\[
E^{(M)}(X, \bullet) \rightarrow f_ME(X),
\]
which is a colimit diagram.
Under the Dold-Kan correspondence \cite[Theorem 1.2.4.1]{HA} we get the associated filtered spectrum
\begin{equation} \label{eq:simpfilt}
\mathrm{sk}_0 E^{(M)}(X, \bullet) \rightarrow \mathrm{sk}_1 E^{(M)}(X, \bullet) \rightarrow \cdots \to \mathrm{sk}_k E^{(M)}(X, \bullet) \rightarrow \cdots \to f_ME(X).
\end{equation}
Noting that \[ \ul{\pi}_i(E)_{-j}(K) = [\Sigma^i \Gmp{j} \wedge K_+, E] \] (where $K/k$ is a field extension) we obtain for $j \ge 0$ a spectral sequence of the form
\begin{equation} \label{eq:simpss}
E^1_{p,q}(K, E, M, j) \Rightarrow \ul{\pi}_{p+q}(f_ME)_{-j}(K).
\end{equation}
As usual, defining
\[
F^{\mathrm{simp}}_{k}\ul{\pi}_{i}(f_ME)_{-j}(K):= \mathrm{Im}(\pi_{i}\mathrm{sk}_k(E^{(M)}(\Gmp{j} \wedge K_+, \bullet)) \rightarrow \ul{\pi}_{i}(f_ME)_{-j}(K)),
\]
we have the exhaustive increasing filtration
\begin{equation} \label{eq:pi-simpss}
F^{\mathrm{simp}}_{0}\ul{\pi}_{i}(f_ME)_{-j}(K) \rightarrow \cdots F^{\mathrm{simp}}_{k}\ul{\pi}_{i}(f_ME)_{-j}(K) \rightarrow F^{\mathrm{simp}}_{k+1}\ul{\pi}_{i}(f_ME)_{-j}(K) \cdots \rightarrow \ul{\pi}_{i}(f_ME)_{-j}(K),
\end{equation}
and the associated graded identifies with the $E^{\infty}$ page \cite[sentence before Lemma 4.4]{levine2013convergence}:
\[
gr_p^{\mathrm{simp}}\ul{\pi}_{i}(f_ME)_{-j}(K) \cong E^{\infty}_{p,i-p}(K,E,M,j).
\]
The next lemma concerns the spectral sequence~\eqref{eq:simpss} and the corresponding filtration on sections of homotopy sheaves. To state the result we need some notation. Suppose that $F, G, H$ are strictly $\bb A^1$-invariant sheaves on ${\cat{S}\mathrm{m}}_k$, $H$ has transfers in the sense of \cite[Chapter 4]{morel-book}, and there is a pairing $F \otimes G \rightarrow H$. Then for a finitely generated separable extension $K'$ of $k$, recall that we have the subgroup
\[
[ F G]^{tr}(K) := \langle tr_{K'/K} F(K')G(K') \rangle_{K'/K\,\text{finite}} \subset H(K).
\]
We refer to \cite[Section 4]{bachmann-hurewicz} or \cite[Section 7]{levine-gw} for details on this construction. The case that we are interested in will be $F = \ul{K}^{MW}_n \cong \ul{\pi}_0(\mathbbm{1})_n$ acting on $G=\ul{\pi}_{i}(E)_j$.
\begin{lemma}[Levine] \label{lemm:levine-simp} Suppose that $E \in \mathcal S\mathcal H(k)_{\geq 0}$, i.e., (a) holds. Then for all $M \geq 0, j \ge 0$ and all $i \in \bb Z$ we have:
\begin{enumerate}
\item The spectral sequence~\eqref{eq:simpss} is convergent.
\item There is an inclusion of abelian groups
\[
E^1_{p,q}(K, E, M, j) \subset \bigoplus_{w \in (\Delta^p_K, \partial \Delta_K^p)^{(M)}} \ul{\pi}_{q+M}(E)(K(w))_{-M-j}
\]
\item \[ F^{\mathrm{simp}}_{*}\ul{\pi}_{i}(f_ME)_{-j}(K) = 0 \text{ if } * < M, \] and \[ F^{\mathrm{simp}}_{*}\ul{\pi}_{i}(f_ME)_{-j}(K) = \ul{\pi}_{i}(f_ME)_{-j}(K) \text{ if } * > M+i. \]
\item Under the canonical map $\ul{\pi}_i(f_ME)_{-j} \rightarrow \ul{\pi}_i(E)_{-j}$, the image of $F^{\mathrm{simp}}_{M}\ul{\pi}_{i}(f_ME)_{-j}(K)$ in $\ul{\pi}_i(E)_{-j}(K)$ is the subgroup
\[
[K_M^{MW} \ul{\pi}_i(E)_{-M-j}]^{tr}(K).
\]
\end{enumerate}
\end{lemma}
\begin{proof} The first two points are covered in \cite[Lemma 4.4]{levine2013convergence}, the third is immediate from (2) (see also \cite[Lemma 5.1]{levine2013convergence}) and the last point is \cite[Theorem 5.3]{levine2013convergence}.
\end{proof}
\begin{lemma} \label{lemm:convergence-permanence}
Let $E \in \mathcal S\mathcal H(k)$ and $M \in \bb Z$. In the notation of Theorem~\ref{thm:convergence}:
\begin{enumerate}
\item[(a)] If $E$ satisfies (a) then so does $f_M E$.
\item[(b)] If $E$ satisfies (b) then so does $f_M E$.
\end{enumerate}
\end{lemma}
\begin{proof}
(a) We may assume (replacing $E$ by $E \wedge \Gmp{-M}$) that $M=0$.
The claim is then immediate from \cite[Proposition 4(3)]{bachmann-very-effective}.
(b) We may assume (replacing $E$ by $E \wedge \Sigma^{-i}\Gmp{j}$) that $i=j=0$.
If $M < 0$ there is nothing to prove, since $\ul{\pi}_0(f_ME)_0 \simeq \ul{\pi}_0(E)_0$.
Thus we may assume that $M \ge 0$. Consider the filtration~\eqref{eq:pi-simpss} of $\ul{\pi}_0(f_M E)_0(K)$.
Since this filtration is finite and exhaustive, it suffices to prove the claim on the level of associated graded groups, which are the $E^{\infty}$-terms of the spectral sequence~\eqref{eq:simpss}. The claim follows since the $E^{\infty}$-terms are subquotients of the $E^1$-terms, which by Lemma~\ref{lemm:levine-simp} are in turn subgroups of sums of groups of the form $\ul{\pi}_{a}(E)(K(w))_{b}$, which satisfy (b) by assumption.
\end{proof}
\begin{remark} \label{rmk:fM-rho}
It is not clear that if $E$ satisfies condition (c), then so does $f_M E$.
Instead the composite \[ f_ME \xrightarrow{\rho^R} f_M(E) \wedge \Gmp{R} \to f_{M-R}(E) \wedge \Gmp{R} \] is zero.
We are not going to use this observation (explicitly).
\end{remark}
\begin{lemma} \label{lemm:convergence-1}
Let $E$ satisfy conditions (b) and (c) of Theorem \ref{thm:convergence}.
There exists a function $M_0(d, R)$ (independent of $E$) such that for $n \ge M_0(d, R)$, $\operatorname{trdeg}(K/k) \leq d$, and $i, j \in \bb Z$ we have \[ [\ul{K}_n^{MW} \ul{\pi}_i(E)_j]^{tr}(K) = 0. \]
\end{lemma}
\begin{proof}
We define the function by
\begin{equation} \label{eq:modr}
M_0 := M_0(d, R) := \mathrm{vcd}_t(k) + d + R + 1.
\end{equation}
Let $K/k$ be of transcendence degree $\le d$ and $K'/K$ finite.
Assumption (b) implies that every element of $\ul{\pi}_i(E)_j(K')$ is a sum of elements which are $\ell^m$-torsion, for various $\ell|t$ and $m \ge 0$.
Put $F_* = K_*^{MW}(K')/(\rho^R, \ell^m)$.
Together with assumption (c), the above observation implies that it suffices to establish the vanishing $F_{M_0} = 0$.
If $\ell$ is odd then $F_*$ is a quotient of $K_*^{MW}(K')[1/2]$ and so splits into $+$ and $-$ parts; moreover $\rho$ is an isomorphism on the $-$ part (see Remark \ref{rmk:rho-completion}(2)) and thus $F_*^-$ is zero. Hence, it is enough to show that $K_{M_0}^M(K')/\ell^m = 0$.
This is true by choice of $M_0$, the fact that $\mathrm{cd}_\ell(K') \leq d + \mathrm{cd}_\ell(k) = d + \mathrm{vcd}_\ell(k)$ \cite[Tag 0F0T]{stacks}, and the comparison between Milnor $K$-theory and \'etale cohomology \cite{voevodsky-BK}.
Now let $\ell=2$.
Note that if $M$ is an $A$-module and $a \in A$ satisfies $a^N M = 0$ for some $N$, then $M=0$ if and only if $M/a = 0$.
Hence we may assume that $m=1$. Note that $h$ is nilpotent on $F_*$: \[ 0 = (\eta \rho)^R = (\lra{-1} - 1)^R = (h-2)^R = h^R. \] It thus suffices to show that $(F_{M_0})/h = 0$.
But $K_*^{MW}/h = I^*$ \cite[Th\'eor\'eme 2.4.]{morel-ideal}, so it is enough to show the equality \[ \rho^R I(K')^{\mathrm{vcd}_2(K')} = I(K')^{\mathrm{vcd}_2(K')+R}. \] Under the isomorphism of \emph{loc. cit.}, the element $\rho$ is sent to the element $2 \in I(k) \subset W(k)$. Hence we need to check that $I(K')^{\mathrm{vcd}_2(K')+R} = 2^R I(K')^{\mathrm{vcd}_2(K')}$.
For a proof see p.\ 619 in \cite{elman-lum-2cohom}.
\end{proof}
The following result is the key step in the proof of Theorem \ref{thm:convergence}.
It establishes a vanishing region in the homotopy sheaves of sufficiently effective (and connective) spectra.
\begin{lemma} \label{lemm:convergence-2}
There exists a function $M(d, R, r, s)$ such that for \[ E \in \mathcal S\mathcal H(k)^\mathrm{eff}(M(d,R,r,s)) \] satisfying (a)\footnote{Recall our standing convention that $c=0$, so this means $E \in \mathcal S\mathcal H(k)_{\ge 0}$.}, (b) and (c) of Theorem \ref{thm:convergence}, and $i \le r, 0 \le j \le s$ we have \[ \ul{\pi}_i(E)_{-j}(K) = 0, \]
whenever $\operatorname{trdeg}(K/k) \leq d$.
\end{lemma}
\begin{proof}
We shall define $M(d,R, r, s)$ by induction on $r$.
If $i < 0$ then $\ul{\pi}_i(E)_{-j}(K)=0$ for any $j, K$ by assumption, and hence $M(d,R, r, s)=0$ works for all $r < 0$, $s \in \bb Z$.
Now suppose that $M(d,R, r',s)$ has been defined for all $r' < r$.
We first want to define $M(d,R,r,0)$; so we need to investigate $\ul{\pi}_r(E)_0(K)$. Consider the filtration~\eqref{eq:pi-simpss} on $\pi_r(f_n E)_0(K)$ with associated graded $gr_p^{\mathrm{simp}}\ul{\pi}_{r}(f_nE)_{0}(K)$.
It follows from Lemma \ref{lemm:levine-simp}(2) that these groups vanish when:
\begin{enumerate}
\item $p < n$ by codimension reasons, or
\item $p > n+r$ by connectivity of $E$.
\end{enumerate}
Conversely, whenever there is a non-zero contribution we must have $(*)$ $n \le p \le n+r$, and the contribution comes from some $\ul{\pi}_{r'}(E)_{-n}(K(w))$, where \[ \operatorname{trdeg}(K(w)/K) = p-n \le r \text{ and } r' = n+r-p \le r. \]
Here the last two inequalities just come from $(*)$.
We define \[ M(d,R,r,0) = \max\{M(d+r,R, r-1, M_0(d,R)), M_0(d,R)\}, \]
where $M_0(d,R)$ is the function from~\eqref{eq:modr}.
Set $n=M_0(d,R)$.
By assumption, $E$ is at least $n$-effective, so $f_nE \simeq E$ and the map \[ \alpha_n: \pi_r(f_nE)_0(K) \to \pi_r(E)_0(K) \] is an isomorphism.
We claim that for $p>n$, we have $gr_p^{\mathrm{simp}}\ul{\pi}_{r}(f_nE)_{0}(X)= 0$.
Indeed by the above discussion any contribution to this group arises from $\ul{\pi}_{r'}(E)_{-n}(L)$, where $r'<r$ and $\operatorname{trdeg}(L/k) \le d+r$.
This vanishes by induction and the construction of $M$.
Now, part (3) of Lemma~\ref{lemm:levine-simp} implies that the image of $\alpha_n$ is given by $F_n^{\mathrm{simp}}\ul{\pi}_{i}(f_nE)_{0}(K)$, which is identified in part (4) as the group \[ [K_n^{MW} \ul{\pi}_r(E)_{-n}]^{tr}(K), \] which vanishes by Lemma \ref{lemm:convergence-1}.
Hence our definition of $M(d,R,r,0)$ has the desired property.
Finally, since $\ul{\pi}_i(E)_{-j} = \ul{\pi}_i(E \wedge \Gmp{-j})_0$, we can define $M(d,R,r,s) = M(d,R,r,0) + s$. This concludes the construction of the desired function.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:convergence}.]
Let us call a tower of sheaves $\ul{F}_\bullet$ \emph{locally null} if there exists a function $N(d)$ such that for $\operatorname{trdeg}(K/k) \le d$ and $i \ge N(d)$ we have $\ul{F}_{i}(K) = 0$.
(1) Under our standing assumptions (in particular (4)), the claim is equivalent to showing that $\ul{\pi}_i((f_\bullet E)/\rho^R)_j$ is locally null (with $N$ independent of $E$).
Replacing $E$ by $E \wedge \Gmp{j}$ (which still satisfies (a), (b)) replaces $\ul{\pi}_i(f_M(E)/\rho^R)_0$ by $\ul{\pi}_i(f_{M-j}(E)/\rho^R)_j$.
It thus suffices to treat the case $j=0$.
Note that $f_M(E)/\rho^R$ is $(M-R)$-effective, satisfies conditions (a) and (b) by Lemma \ref{lemm:convergence-permanence}, and satisfies condition (c) by Remark \ref{rmk:nilpotence} (with $2R$ in place of $R$).
Lemma \ref{lemm:convergence-2} thus shows that $\ul{\pi}_i(f_M(E)/\rho^R)_0(K) = 0$ as soon as $M \ge M(d,2R, i, 0) + R$.
The claim follows.
(2)
Consider the following commutative diagram
\begin{equation*}
\begin{CD}
f_M(E) @>>> f_M(E)/\rho^R @>a>> f_{M-R}(E) \\
@VVV @VVV @VVV \\
E @>>> E/\rho^R @>b>> E.
\end{CD}
\end{equation*}
Here the vertical maps are the canonical ones, and the horizontal maps in the left hand square are the projections.
The map $b$ is a splitting $E/\rho^R \simeq E \vee \Sigma E \wedge \Gmp{-R} \to E$ (using condition (c)), and the map $a$ is the unique one making the right hand square commute
(using that $f_M(E)/\rho^R$ is $(M-R)$-effective).
The bottom horizontal composite is $\mathrm{id}_E$ by construction, and hence the top horizontal composite is the canonical map $f_M(E) \to f_{M-R}(E)$.
Since $\ul{\pi}_i(f_\bullet(E)/\rho^R)_j$ is locally null by (1), we deduce that for $M$ sufficiently large depending only on $i, j, d$, the map \[ \ul{\pi}_i(f_{M+R} E)_j(K) \to \ul{\pi}_i(f_{M} E)_j(K) \] factors through \[ \ul{\pi}_i(f_{M+R}(E)/\rho^R)_j(K) = 0. \]
This implies the claim.
Separatedeness of both towers follows immediately.
This concludes the proof.
\end{proof}
\section{Spheres over fields}
In this section we treat a special case of our main result to which we will reduce the general case.
Throughout $S=Spec(k)$, where $k$ is a field of exponential characteristic $e$, $\ell \ne e$ is a prime and
\[ \tau: \mathbbm{1}/(\ell^{\nu}, \rho^{\mu}) \to \mathbbm{1}/(\ell^{\nu}, \rho^{\mu})(r) \]
is a $\tau$-self map (see Definition \ref{def:tau-self-map}).
Here $0 \le \mu,\nu \le \infty$.
The following is the key technical result.
\begin{lemma} \label{lemm:key}
Suppose that $\mathrm{vcd}_\ell(k) < \infty$.
Then \[ \lim_n \left[ f_n(\mathbbm{1})/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho) \right] = 0. \]
\end{lemma}
\begin{proof}
Write $L = \lim_n L_n$ for the limit in question.
For $X \in {\cat{S}\mathrm{m}}_k$, $w \in \bb Z$, spectra of the form $X(w) := \Sigma^{0, w} \Sigma^\infty X_+$ generate $\mathcal S\mathcal H(k)$ (see, for example, \cite[Theorem 9.2]{cell} or \cite[Proposition 6.4]{hoyois-sixops} when $G$ is the trivial group),
and hence it suffices to show that \[ [X(w)[i], L] = 0 \] for all $w, i$.
By the Milnor exact sequence \cite[Proposition VI.2.15]{goerss2009simplicial} \[ 0 \to \mathrm{lim}^1_n[X(w)[i+1], L_n] \rightarrow [X(w)[i], L] \rightarrow \mathrm{lim}_n [X(w)[i], L_n] \to 0, \] it is enough to show that for $(i, w)$ fixed and $n$ sufficiently large we have $[X(w)[i], L_n] = 0$.
Consider the descent spectral sequence
\[
H^p_{\mathrm{Nis}}(X, \ul{\pi}_{q,w}(L_n)) \Rightarrow [X(w)[q-p] , L_n],
\]
which is strongly convergent due to the finite cohomological dimension of the Nisnevich site (see e.g., the proof of \cite[Proposition 4.3]{elso}).
This implies that it suffices to prove the following claim:
\begin{itemize}
\item For any $N \in \bb Z$ and $n$ sufficiently large (depending on $N$), all $w \in \bb Z$ and $k \le N$ we have $\ul{\pi}_{k,w}(L_n)|_{X_\mathrm{Nis}} = 0$.
\end{itemize}
\tombubble{Or use that Zariski and Nisnevich cohomology coincide.}
By unramifiedness of homotopy sheaves \cite[Lemma 6.4.4]{morel-conn}, it is enough to show that $\ul{\pi}_{k,w}(L_n)_\xi = 0$ for generic points $\xi$ of étale extensions of $X$.
Since homotopy sheaves commute with colimits, and $w \in \bb Z$ is arbitrary, for this it suffices to show that \[ \ul{\pi}_{k,w}(f_n(\mathbbm{1})/(\ell^\nu,\rho^\mu,\ell,\rho))_\xi = 0. \]
Up to adding a constant to $n$ (which only depends on $\mu$), it is thus enough to show the vanishing \[ \ul{\pi}_{k,w}(f_n(\mathbbm{1})/(\ell,\rho))_\xi = 0. \]
Using Theorem \ref{thm:convergence}(1), we find that the tower of abelian groups
\[
\ul{\pi}_{k,w}(f_{\bullet}(\mathbbm{1})/(\ell,\rho))_{\xi} \rightarrow \ul{\pi}_{k,w}(f_n(\mathbbm{1})/(\ell,\rho))_\xi
\]
is convergent (here $\bullet \ge n$, and the tower is trivially exhaustive).
Hence by Lemma \ref{lemm:convergence-application}, it suffices to prove the following claim:
\begin{itemize}
\item For any $N \in \bb Z$ there exists $n$ (depending on $N$ and $\operatorname{trdeg}(\xi/k)$) such that for all $m \ge n$, $k \le N$ and $w \in \bb Z$ we have
\[ \ul{\pi}_{k,w}(s_m(\mathbbm{1})/(\ell,\rho))_\xi = 0. \]
\end{itemize}
By \cite[Theorem 2.12]{1-line}, each slice $s_m(\mathbbm{1})[1/e]$ is a finite sum of suspensions of motivic cohomology spectra \[ \Sigma^{m+s,m} H\bb Z/l, \] for certain $s \ge 0, l \ge 0$.
As before, up to possibly adding a constant to $n$, it is thus enough to show that \[\ul{\pi}_{k,w}(\Sigma^{m+s,m} H\bb Z/(\ell,\rho))_\xi = 0\] for $m$ sufficiently large.
Now \[ \ul{\pi}_{k,w}(\Sigma^{m+s,m} H\bb Z/\ell)_\xi = H^{m+s-k}(\xi, \bb Z/\ell(m-w)) \] and so we need to show that multiplication by $\rho$ induces an isomorphism on these groups,
for $m$ sufficiently large (depending on $N$ and $\operatorname{trdeg}(\xi/k)$), all $w \in \bb Z$ and all $k \le N$, $s \ge 0$.
In particular $s-k \ge -N$.
Note also that we have $\mathrm{vcd}_\ell(\xi) \le \mathrm{vcd}_\ell(k) + \operatorname{trdeg}(\xi/k)$ by \cite[Tag 0F0T]{stacks}.
The required vanishing thus follows from Lemma \ref{lemm:vcd-iso}: we may put $m \ge n := \mathrm{vcd}_\ell(k) + \operatorname{trdeg}(\xi/k) + N + 1$. This concludes the proof.
\end{proof}
\begin{corollary} \label{cor:sphere-inversion}
If $\mathrm{vcd}_\ell(k) < \infty$, then $\mathbbm{1}/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho)$ is étale local.
\end{corollary}
\begin{proof}
We have a cofiber sequence of towers
\begin{equation*}
\begin{CD}
\dots @>>> f_2(\mathbbm{1}) @>>> f_1(\mathbbm{1}) @>>> f_0(\mathbbm{1})=\mathbbm{1} \\
@. @VVV @VVV @VVV \\
\dots @>>> \mathbbm{1} @>{\mathrm{id}}>> \mathbbm{1} @>{\mathrm{id}}>> \mathbbm{1} \\
@. @VVV @VVV @VVV \\
\dots @>>> f^2(\mathbbm{1}) @>>> f^1(\mathbbm{1}) @>>> f^0(\mathbbm{1}),
\end{CD}
\end{equation*}
where $f^i = \mathrm{id}/f_i$.
Smashing with $\mathbbm{1}/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho)$ and applying the (exact) inverse limit functor, we obtain a cofiber sequence \[ \lim_n \left[ f_n(\mathbbm{1})/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho) \right] \to \mathbbm{1}/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho) \xrightarrow{\alpha} \lim_n \left[ f^n(\mathbbm{1})/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho) \right]. \]
By Lemma \ref{lemm:key}, the map $\alpha$ is an equivalence.
Since étale local spectra are closed under limits and extensions and each $f^n(\mathbbm{1})$ is a finite extension of the slices $s_m(\mathbbm{1})$ (recall that $\mathbbm{1}$ is effective), it suffices to show that $s_i(\mathbbm{1})/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho)$ is étale local for every $i$.
By the form of the slices of $\mathbbm{1}$ recalled in the proof of Lemma \ref{lemm:key}, it suffices to show that \[ H\bb Z/(\ell^\nu,\rho^\mu)[\tau^{-1}]/(\ell,\rho) \] is étale local.
This holds by the definition of a $\tau$-self map.
\end{proof}
\section{Main result}
In this section we establish our main étale localization results.
Before doing so, we need some preliminaries.
\begin{proposition} \label{prop:compact-gen}
Let $1/\ell \in k$, $\mathrm{vcd}_\ell(k) < \infty$ and $S$ a finite type $k$-scheme.
\NB{Not minimal assumptions...}
If $X \in {\cat{S}\mathrm{m}}_S$ is quasi-separated and $w \in \bb Z$, then $\Sigma^\infty_+ X/(\ell, \rho) \wedge \Gmp{w} \in \mathcal S\mathcal H_\mathrm{\acute et}(S)_{\ell,\rho}^\wedge$ is compact.
\end{proposition}
\begin{proof}
Since the pullback $\mathcal S\mathcal H_\mathrm{\acute et}(S)_{\ell,\rho}^\wedge \to \mathcal S\mathcal H_\mathrm{\acute et}(X)_{\ell,\rho}^\wedge$ preserves colimits, we may assume that $S=X$.
First suppose that $\mathrm{cd}_\ell(k) < \infty$ (e.g., $\ell$ odd).
It is enough to show that $\Sigma^\infty_+ S/\ell$ is compact in $\mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge$.
This is proved in \cite[Corollary 5.7 and Example 5.9]{bachmann-SHet}.
Now let $\ell=2$.
\tombubble{this argument is a bit awkward, I wonder if there is a better formulation...}
Write $\mathcal S\mathcal H(S_\mathrm{\acute et})$ for the category of hypersheaves of spectra on the small étale site of $S$ (see e.g., \cite[\S2.2]{bachmann-SHet} for definitions).
We have the functors \[ E \mapsto \tau_{\ge n}E, \tau_{\le n}E, \pi_n(E): \mathcal S\mathcal H(S_\mathrm{\acute et}) \to \mathcal S\mathcal H(S_\mathrm{\acute et}) \] coming from the standard $t$-structure; they all preserve filtered colimits (for $\tau_{\le n}$ this follows from \cite[Proposition 1.3.2.7(2)]{SAG} and the other cases follow from this).
We thus have restricted functors \[ E \mapsto \tau_{\ge n}(E/\ell), \tau_{\le n}(E/\ell), \pi_n(E/\ell): \mathcal S\mathcal H(S_\mathrm{\acute et})_\ell^\wedge \to \mathcal S\mathcal H(S_\mathrm{\acute et})_\ell^\wedge \] with the same property.
Moreover $E/\ell \simeq \lim_n \tau_{\le n}(E/\ell)$ \cite[Lemma 2.16]{bachmann-SHet}.
Via \cite[Theorem 6.6]{bachmann-SHet} we transplant these functors to $\mathcal S\mathcal H_\mathrm{\acute et}(k)_\ell^\wedge$.
In order to prove that $S/(\ell,\rho)$ is compact, it is enough to show that the functor $E \mapsto [S/\rho, E/\ell]$ preserves filtered colimits.
Using the Postnikov completeness result just recalled, there is a conditionally convergent spectral sequence
\[ [S/\rho, \pi_i(E/\ell)[j]] \Rightarrow [S/\rho, E/\ell[i+j]]. \]
By standard arguments it is enough to show that the spectral sequence converges strongly for every $E$, and that $[S/\rho, \pi_i(E/\ell)[j]]$ is compatible with filtered colimits in $E$.
Consider $F = \pi_0(E/\ell) \in \mathcal S\mathcal H(S_\mathrm{\acute et})^\heartsuit$.
This is a sheaf of $\bb Z/\ell^2$-modules on the small étale site of $S$.
There is a canonical filtration $0 \to K \to F \to F/\ell \to 0$ with $K, F/\ell$ sheaves of $\bb Z/\ell$-modules and an associated long exact sequence
\[ \dots \to [S/\rho, K] \to [S/\rho, F] \to [S/\rho, F/\ell] \to \dots. \]
It will thus be enough to show that the functor \[ F \mapsto [S/\rho, F[i]] \] is compatible with filtered colimits of sheaves of $\bb Z/\ell$-vector spaces on the small étale site of $S$ and vanishes for $i<0$ or $i > \mathrm{cd}_2(S[\sqrt{-1}])$ (here we note that $\mathrm{cd}_2(S[\sqrt{-1}]) < \infty$, e.g., by \cite[Example 5.9]{bachmann-SHet}).
We claim that there is a long exact sequence \[ \dots \to H^i(S, F) \xrightarrow{(-1)} H^{i+1}(S, F) \to [S/\rho, F[i]] \to H^{i+1}(S, F) \to \dots. \]
Since étale cohomology commutes with filtered colimits (see e.g., \cite[A.2.3.2(1)]{SAG}), the result will follow from Lemma \ref{lemm:vcd-vanishing} below.
To prove the claim, consider the sequence of functors and adjoints \[ \mathcal S\mathcal H(S_\mathrm{\acute et})_\ell^\wedge \leftrightarrows \mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge \leftrightarrows \mathcal D\mathcal M_\mathrm{\acute et}(S, \bb Z/\ell). \]
There is an object $F \in \mathcal D\mathcal M_\mathrm{\acute et}(S, \bb Z/\ell)$ with image $F$ in $\mathcal S\mathcal H(S_\mathrm{\acute et})_\ell^\wedge$; we are thus reduced to proving the analogous result in $\mathcal D\mathcal M_\mathrm{\acute et}(S, \bb Z/\ell)$.
To conclude we note that $\rho\colon \Gmp{-1} \rightarrow \mathbbm{1}$ induces multiplication by $(-1)$ in $\mathcal D\mathcal M_\mathrm{\acute et}(k, \bb Z/2) \simeq D(k_\mathrm{\acute et}, \bb Z/2)$.\tombubble{reference?}.
\end{proof}
Our next result follows from \cite[Theorem 99.13]{MR2427530} in the case of fields.
\begin{lemma}
\label{lemm:vcd-vanishing}
Let $X$ be a scheme with $1/2 \in X$ and set $X' = X[\sqrt{-1}]$.
Suppose $F$ is an étale sheaf of $\bb Z/2$-vector spaces on $X$.
Then for $i > \mathrm{cd}_2(X')$ the map \[ (-1): H^i_\mathrm{\acute et}(X, F) \to H^{i+1}_\mathrm{\acute et}(X, F) \] is an isomorphism.
\end{lemma}
\begin{proof}
If $X=X'$ or $\mathrm{cd}_2(X') = \infty$, there is nothing to prove.
We may thus assume that $Aut(X'/X)$ is the group $C_2$ of order $2$, and that $\mathrm{cd}_2(X') < \infty$.
Consider the strongly convergent Hochschild--Serre spectral sequence \cite[Theorem III.2.20]{MilneEtaleCohomology} \[ H^p(C_2, H^q(X', F)) \Rightarrow H^{p+q}(X, F). \]
It is a module over the same spectral sequence with $F=\bb Z/2$.
The class $(-1) \in H^1(X, \bb Z/2)$ is detected by a class $\alpha \in H^1(C_2, H^0(X, \bb Z/2))$ coming from the isomorphism $H^*(C_2, \bb Z/2) \simeq H^*(\bb R\bb P^\infty, \bb Z/2) \simeq \bb Z/2[\alpha]$.
The class $\alpha$ has the property that if $G$ is any $\bb Z/2[C_2]$-module and $i > 0$, the multiplication map $\alpha: H^i(C_2, G) \to H^{i+1}(C_2, G)$ is an isomorphism.\footnote{Indeed $A := \bb Z/2[C_2] \simeq \bb Z/2[\epsilon]/\epsilon^2$, so any finitely generated $A$-module is a sum of copies of $A$ and $\bb Z/2$. For such modules the claim holds. A general $A$-module is a filtered colimit of finitely generated ones, and cohomology commutes with filtered colimits, so the claim follows in general.}
We deduce that $\alpha$ induces an isomorphism on all columns of the spectral sequence, except possibly the one containing $H^0(C_2, H^*(X', F))$.
The result follows.
\end{proof}
We deduce a weak form of Theorem \ref{thm:apps}(3) from the introduction, which we will use in the proof of our main result.
(The strong form will be deduced later.)
\begin{corollary} \label{cor:etale-locn-smashing}
Let $k$ be a field of exponential characteristic $e \ne \ell$ and $\mathrm{vcd}_\ell(k) < \infty$.
Then $L_\mathrm{\acute et}: \mathcal S\mathcal H(k)_{\ell,\rho}^\wedge \to \mathcal S\mathcal H(k)_{\ell,\rho}^\wedge$ is a smashing localization, or in other words for $E \in \mathcal S\mathcal H(k)_{\ell,\rho}^\wedge$ we have $L_\mathrm{\acute et}(E) \simeq E \wedge L_\mathrm{\acute et}(\mathbbm{1})$.
\end{corollary}
\begin{proof}
The functor $L_\mathrm{\acute et}: \mathcal S\mathcal H(k)_{\ell,\rho}^\wedge \to \mathcal S\mathcal H_\mathrm{\acute et}(k)_{\ell,\rho}^\wedge$ will identify the target as (highly structured) modules over the étale-local sphere
if we can show that both the source and target are compact-rigidly generated,
see e.g., \cite[Lemma 22]{bachmann-hurewicz} .
This is indeed the case; see \cite[Example 2.3]{bachmann-SHet} (to obtain a family of generators), \cite[Corollary B.2]{levine2013algebraic} (to see that the generators are rigid)
and Proposition \ref{prop:compact-gen} (to see that the generators are compact).
The result follows.
\end{proof}
With these preparations out of the way, we come to our main theorem.
We refer to \S\ref{subsec:summary} for a summary of the good $\tau$-self maps we managed to construct.
\begin{theorem} \label{thm:main}
Let $S$ be a scheme locally of finite dimension, $1 \le m, n \le \infty$.
Suppose $\tau: \mathbbm{1}/(\ell^n,\rho^m) \to \mathbbm{1}/(\ell^n,\rho^m)(r)$ is a $\tau$-self map.
Assume that $1/\ell \in S$ and for every $s \in S$ we have $\mathrm{vcd}_\ell(s) < \infty$.
Assume further that either $S$ is the spectrum of a field or $\tau$ is good.
Then for every $E \in \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$ the map \[ E/(\ell^n,\rho^m) \to E/(\ell^n,\rho^m)[\tau^{-1}]_{\ell,\rho}^\wedge \] is an étale localization.
\end{theorem}
\begin{proof}
By Lemma \ref{lemm:bott-element-etale-invertible}, the map in question is an étale equivalence.
We thus need to prove that $E' := E/(\ell^n,\rho^m)[\tau^{-1}]_{\ell,\rho}^\wedge$ is étale local.
Let $\scr X \to X \in {\cat{S}\mathrm{m}}_S$ be an étale hypercover and $K = \Sigma^\infty_+ cof(\scr X \to X)$.
We need to show that $[K, E'[i](j)] = 0$ for all $i, j$.
Since $E$ was arbitrary, replacing it by $E(i)[j]$ we may as well assume that $i=j=0$.
By the definition of goodness, see Definition \ref{def:tau-self-map}(3), any map $K \to E'$ factors through $K \to K'' := K/(\ell^{n'},\rho^{m'})[\tau^{-1}]_{\ell,\rho}^\wedge$.
It is thus enough to show that $K'' = 0$.
It follows from \cite[Proposition B.3]{norms} that the collection of functors $\{s^*: \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge \to \mathcal S\mathcal H(s)_{\ell,\rho}^\wedge \mid s \in S\}$ is conservative;
also each of these functors preserves colimits.
It is thus enough to show that $s^*(K'') = 0$.
By construction $s^*(K)$ is étale-locally equivalent to zero; hence it suffices to show that $s^*(K'') \simeq s^*(K)/(\ell^{n'},\rho^{m'})[\tau^{-1}]_{\ell,\rho}^\wedge$ is étale local.
We have thus reduced to the case of fields.
Assume now that $S = Spec(k)$.
Then $\mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}]/(\ell,\rho)$ is étale local by Corollary \ref{cor:sphere-inversion},
and hence so is $\mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}]_{p, \rho}^\wedge$, being a limit of extensions of the former term.
Note that \[ E' \simeq E_{\ell,\rho}^\wedge \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}] \in \mathcal S\mathcal H(k)_{\ell,\rho}^\wedge, \] with the smash product being formed in $\mathcal S\mathcal H(k)_{\ell,\rho}^\wedge$ (i.e., completed).
Since étale localization is smashing on $\mathcal S\mathcal H(k)_{\ell,\rho}^\wedge$ by Corollary \ref{cor:etale-locn-smashing}, we deduce that $E'$ is étale local in $\mathcal S\mathcal H(k)_{\ell,\rho}^\wedge$,
and hence also in $\mathcal S\mathcal H(k)$.
\end{proof}
\begin{remark} \label{rem:recoll}
Suppose that $\ell$ is not invertible on $S$.
Let $j: U = S[1/\ell] \hookrightarrow S$ and $Z = S \setminus U$ the closed complement.
Since $\mathcal S\mathcal H_\mathrm{\acute et}(Z)_\ell^\wedge = 0$ \cite[Theorem A.1]{bachmann-SHet2}
This implies that for $E \in \mathcal S\mathcal H(S)$ we have
\NB{$\mathcal S\mathcal H_\mathrm{\acute et}(\mathord-)$ satisfies localization, so $L_\mathrm{\acute et} E/\ell^n$ can be glued from $i_!^\mathrm{\acute et} i^*_\mathrm{\acute et} L_\mathrm{\acute et} E/\ell^n \simeq i_!^\mathrm{\acute et} L_\mathrm{\acute et} i^* E/\ell^n = 0$
and $j_*^\mathrm{\acute et} j^*_\mathrm{\acute et} L_\mathrm{\acute et} E/\ell^n \simeq j_* L_\mathrm{\acute et} j^*E/\ell^n$.}
\[ L_\mathrm{\acute et} E/\ell^n \simeq j_* L_\mathrm{\acute et} j^*E/\ell^n. \]
Consequently the assumption that $1/\ell \in S$ in Theorem \ref{thm:main} is essentially harmless.
\end{remark}
We conclude with three immediate applications.
\begin{corollary} \label{cor:main}
Assumptions as in Theorem \ref{thm:main}.
\begin{enumerate}
\item Let $\ell$ be odd, $n < \infty$ and $m=\infty$.
Then \[ E/\ell^n \to (E/\ell^n)^+[\tau^{-1}] \] is an étale localization.
\item Let $\ell=2$, $n < \infty$, $m=\infty$ and assume that $-1$ a square in $S$.
Then \[ E/2^n \to E/2^n[\tau^{-1}] \] is an étale localization.
\item Suppose that $S$ is defined over a field containing a primitive $\ell$-th root of unity and satisfying $k^\times/\ell = \{1\}$.
Then we have \[ \mathcal S\mathcal H_\mathrm{\acute et}(S)_\ell^\wedge \simeq \mathcal S\mathcal H(S)_\ell^\wedge[\tau^{-1}]. \]
\end{enumerate}
\end{corollary}
\begin{proof}
In each case the $\rho$-completion is unnecessary (or simplified to $(\mathord-)^+$ in case (1)), by Remark \ref{rmk:rho-completion}.
In (1), (2) also the $\ell$-completion is unnecessary, by Remark \ref{rmk:p-completion-unnec}.
\end{proof}
Denote by $L_\mathrm{\acute et}^\wedge: \mathcal S\mathcal H(S) \to \mathcal S\mathcal H_\mathrm{\acute et}(S)_{\ell,\rho}^\wedge$ the left adjoint to the inclusion $\iota: \mathcal S\mathcal H_\mathrm{\acute et}(S)_{\ell,\rho}^\wedge \subset \mathcal S\mathcal H(S)$.
\begin{corollary} \label{cor:integral}
Assumptions as in Theorem \ref{thm:main}.
Suppose that $m,n<\infty$.
Then the functor $\iota \circ L_\mathrm{\acute et}^\wedge: \mathcal S\mathcal H(S) \to \mathcal S\mathcal H(S)$ is equivalent to Bousfield localization at the homology theory $\mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]$.
\end{corollary}
\begin{proof}
Let $E \in \mathcal S\mathcal H(S)$ and $\alpha: E \to \iota L_\mathrm{\acute et}^\wedge E$ be the $(\ell,\rho)$-complete étale localization map.
We first show that $\alpha$ is a $\mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]$-equivalence, or in other words that $\alpha \wedge\mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]$ is an equivalence.
By Theorem \ref{thm:main}, both sides are étale local, so it is enough to show that
\[ L_\mathrm{\acute et}(\alpha \wedge \mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]) \simeq L_\mathrm{\acute et}(\alpha) \wedge L_\mathrm{\acute et}(\mathbbm{1}/(\ell^n, \rho^m)[\tau^{-1}]) \] is an equivalence.
This is clear.
It remains to show that $\mathcal S\mathcal H_\mathrm{\acute et}(S)_{\ell,\rho}^\wedge$ consists of $\mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}]$-local objects.
In other words, if $E \in \mathcal S\mathcal H(S)$ with $E \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}] = 0$, then we need to show that $\iota L_\mathrm{\acute et}^\wedge(E) = 0$.
Again by Theorem \ref{thm:main} we have \[ \iota L_\mathrm{\acute et}^\wedge(E)/(\ell^n,\rho^m) \simeq \iota L_\mathrm{\acute et}^\wedge(E/(\ell^n,\rho^m)) \simeq E \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tau^{-1}] = 0. \]
Since $\iota L_\mathrm{\acute et}^\wedge(E)$ is $(\ell, \rho)$-complete, we deduce that $\iota L_\mathrm{\acute et}^\wedge(E) = 0$, as desired.
\end{proof}
\begin{corollary} \label{cor:smashing}
Assumptions as in Theorem \ref{thm:main}.
Then étale localization is smashing on $\mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$.
\end{corollary}
\begin{proof}
Denote by $L_\mathrm{\acute et}: \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge \to \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$ the étale localization functor.
It suffices to show that, for $E \in \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$, the spectrum $E' := E \wedge L_\mathrm{\acute et}(\mathbbm{1}) \in \mathcal S\mathcal H(S)_{\ell,\rho}^\wedge$ is étale local.
By definition $E'$ is $(\ell,\rho)$-complete, so $E' \simeq \lim_{n,m} E'/(\ell^n,\rho^m)$, and it suffices to show that $E'/(\ell^n,\rho^m)$ is étale local (for $n, m$ sufficiently large).
Theorem \ref{thm:main} implies that \[ E'/(\ell^n,\rho^m) \simeq E \wedge L_\mathrm{\acute et}(\mathbbm{1}/(\ell^n,\rho^m)) \simeq E \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tau_{n,m}^{-1}], \]
where $\tau_{n,m}$ denotes a $\tau$-self map mod $(\ell^n,\rho^m)$ (it follows from our assumptions that this exists, at least for $n>1$).
Again by Theorem \ref{thm:main}, $E \wedge \mathbbm{1}/(\ell^n,\rho^m)[\tau_{n,m}^{-1}]$ is étale local.
This concludes the proof.
\end{proof}
We also obtain the following new base change result, generalizing \cite[Theorem 7.5]{elso}.
\begin{theorem} \label{thm:basechange} Assumptions as in Theorem \ref{thm:main}. Suppose that $f:T \rightarrow S$ is a morphism of finite type. Then the functor
\[
f^*:\mathcal S\mathcal H(S)_{\ell,\rho}^\wedge \rightarrow \mathcal S\mathcal H(T)_{\ell,\rho}^\wedge,
\]
preserves \'etale local spectra. In particular the canonical map
\[
f^*(\mathbbm{1}_{\mathrm{\acute et}})_S \rightarrow (\mathbbm{1}_{\mathrm{\acute et}})_T
\]
is an equivalence in $\mathcal S\mathcal H(T)_{\ell,\rho}^\wedge$.
\end{theorem}
\begin{proof} Follows from Theorem~\ref{thm:main} and the fact that the Bott elements are stable under base change by definition.
\end{proof}
\begin{remark} \label{rem:fracture}
If $S$ is finite dimensional, then étale localization is smashing on $\mathcal S\mathcal H(S)_\bb Q$, since it just corresponds to ``passing to the plus part''\footnote{The minus part vanishes étale locally (see e.g. the proof of \cite[Theorem 7.2]{bachmann-SHet}), and the plus part has étale hyperdescent by \cite[Corollary 4.39]{clausen-mathew}: condition (1) holds by \cite[\S10.2]{norms} and for (2) we can take $A=H\bb Q$ (this is where we need to be in the plus part).}, or equivalently $\rho$-completion.
In particular Corollary \ref{cor:smashing} holds rationally (instead of completed at $\ell$) as well.
Using Remark~\ref{rem:recoll}, we see that it also holds completed at $\ell$ which fail to be invertible on $S$ (provided that $\mathrm{vcd}_p(s) < \infty$ for all $s \in S$).
One may deduce that étale localization is smashing on $\mathcal S\mathcal H(S)_{\rho}^\wedge$, provided that $\mathrm{vcd}(s) < \infty$ for all $s \in S$ (in addition to $S$ being finite dimensional).
Similarly \[ f^*:\mathcal S\mathcal H(S)_{\rho}^\wedge \rightarrow \mathcal S\mathcal H(T)_{\rho}^\wedge \] preserves \'etale local spectra (under the given assumption).
\end{remark}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,588 |
\section{Introduction}
Let $\mathcal{A}_{g}[2]$ denote the moduli space of principally polarized
abelian varieties of dimension $g$ together with a full level $2$ structure.
Similarly, let $\mathcal{M}_g[2]$ denote the moduli space of smooth
curves of genus $g$ together with a full level $2$ structure.
Note that we are considering these as \emph{coarse} moduli spaces.
The two
types of moduli spaces are connected through the Torelli
morphism $t_g: \mathcal{M}_g[2] \to \mathcal{A}_{g}[2]$ sending a
smooth curve to its Jacobian. There is an action of $\Sp{2g}$, the symplectic group
of degree $2g$ over the field of two elements, on $\mathcal{A}_{g}[2]$ and
$\mathcal{M}_{g}[2]$ via its action on the level $2$ structure (for some more details
see for instance \cite{bergvallq2}).
For $g \leq 2$, $\mathcal{M}_{g}[2]$ is isomorphic to the locus $\mathcal{H}_{g}[2]$ consisting of
hyperelliptic curves. The space $\mathcal{H}_{g}[2]$ can in turn be described as a disjoint union
of copies of $\mathcal{M}_{0,2g+2}$ (see \cite{dolgachevortland}, \cite{runge} and \cite{tsuyumine}), the moduli space of smooth genus $0$ curves together with
$2g+2$ marked points. The cohomology of $\mathcal{M}_{0,2g+2}$,
together with the action of the symmetric group $S_{2g+2}$,
can (because of purity) be computed using counts of points over finite fields (see for instance \cite{dimcalehrer} and \cite{kisinlehrer}).
In Section~\ref{sec-g1} respectively
Section~\ref{sec-g2} below, we compute in this way the $\Sp{2}$-action and Hodge structure of the cohomology
of $\mathcal{M}_{1}[2] \cong \mathcal{A}_{1}[2]$ respectively the $\Sp{4}$-action and Hodge structure
of the cohomology of $\mathcal{M}_{2}[2]$. By adding the complement of
$t_2(\mathcal{M}_{2}[2])$ inside $\mathcal{A}_{2}[2]$ consisting of products of elliptic curves we
also compute the $\Sp{4}$-equivariant weighted Euler characteristic of $\mathcal{A}_{2}[2]$. For a definition of
this type of Euler characteristic see Section~\ref{sec-euler}.
The main result of this article is Table~\ref{a3table} which contains the $\Sp{6}$-equivariant weighted Euler characteristic of
$\mathcal{A}_{3}[2]$. This is based upon the work of the second author in \cite{bergvallq2} in which the
$\Sp{6}$-equivariant cohomology of $\mathcal{M}_{3}[2]$ is computed, see Section~\ref{sec-decomp} below.
There are two other loci consisting of either products of an indecomposable abelian surface and an elliptic curve,
or products of three elliptic curves. The cohomology of these loci are computed in Section~\ref{sec-A21} and
Section~\ref{sec-A111} respectively.
We note in Section~\ref{sec-main} that the weighted Euler characteristic of $\mathcal{A}_{3}[2]$ contains much fewer
classes than the weighted Euler characteristic of its different loci. This cancellation property was noted also in \cite{jbgvdg}
for the integer valued Euler characteristic of local systems upon the corresponding strata inside $\mathcal{A}_{3}$,
the moduli space of principally polarized abelian threefolds with no level structure. The Hodge structure of the
cohomology of $\mathcal{A}_{3}$ was previously known, see \cite{Hain}.
\section{Euler characteristics} \label{sec-euler}
For a quasi-projective variety $X$ defined over $\mathbb{C}$,
let $W_{k}H^i(X)$ denote the weight $k$-part of $H^i(X)$, the $i$th Betti cohomology group
with complex coefficients.
For an action of a finite group $G$ on $X$, let the
$G$-equivariant weighted Euler characteristic of $X$ be the virtual representation of $G$ defined as
\begin{equation*}
e_X(v) = \sum_{i,k \geq 0} (-1)^i \, W_kH^i(X) \, v^k.
\end{equation*}
This Euler characteristic is additive in the sense that if $X=Y \sqcup Z$, where $Y$ and $Z$ are preserved by $G$,
then $e_{X}(v) =v^{2 \mathrm{codim}_X(Y)} e_Y(v) + v^{2 \mathrm{codim}_X(Z)} e_Z(v)$. Note that if $X$ fullfils
purity, in the sense of Dimca and Lehrer in~\cite{dimcalehrer}, then one can from this Euler characteristic
determine the individual cohomology groups as representations of $G$.
Say now that $X$ is a variety defined over $\mathcal{O}[\frac{1}{N}] $, where $\mathcal{O}$ is a ring of integers of
an algebraic number field, together with an action of a finite group $G$. Say furthermore that there is a polynomial
$P(t)$, with complex coefficients and of degree $2\dim X$, such that $P(q)=|X(\mathbb{F}_q)|$ for almost all prime powers $q$.
We can, using the Lefschetz fixed point theorem, from this information determine the weighted Euler characteristic of $X(\mathbb{C})$.
The set $X(\mathbb{F}_q)$ consists of the fixed points of Frobenius. By counting the fixed points of Frobenius
composed with elements of $G$ we can in the same way determine the $G$-equivariant weighted Euler characteristic
of $X(\mathbb{C})$. This will be called a twisted point count. For a reference, see \cite[Appendix~A]{dimcalehrer2}.
\section{Decomposable and indecomposable abelian threefolds} \label{sec-decomp}
We say that an abelian threefold is indecomposable if it is not isomorphic to
a product of abelian varieties of lower dimension. We denote the corresponding
locus in $\mathcal{A}_3[2]$ by $\mathcal{A}_3^{\mathrm{in}}[2]$.
The Torelli morphism $t_3$ gives
an isomorphism $\mathcal{M}_3[2] \cong \mathcal{A}_3^{\mathrm{in}}[2]$
(on the level of coarse moduli spaces). The moduli space $\mathcal{M}_3[2]$ can
be decomposed as a disjoint union
\begin{equation*}
\mathcal{M}_3[2] = \mathcal{Q}[2] \sqcup \mathcal{H}_3[2]
\end{equation*}
where $\mathcal{Q}[2]$ denotes the locus consisting of curves whose canonical model
is a plane quartic curve and where $\mathcal{H}_3[2]$
denotes the hyperelliptic locus.
The cohomology groups of $\mathcal{Q}[2]$ and $\mathcal{H}_3[2]$ were
determined as representations of $\Sp{6}$ by the second author in \cite{bergvallq2}.
For completeness, we repeat the results in Table~\ref{Qtable} and Table~\ref{Hyptable}.
There are two types of decomposable abelian threefolds.
The threefold can either be isomorphic to a product of an indecomposable abelian surface and an elliptic curve
or to a product of three elliptic curves.
We denote the corresponding loci in $\mathcal{A}_3[2]$ by $\mathcal{A}_{2,1}[2]$
and $\mathcal{A}_{1,1,1}[2]$ respectively.
\section{The main result} \label{sec-main}
We have the decomposition
\begin{equation*}
\mathcal{A}_3[2] =
t_3(\mathcal{Q}[2]) \sqcup
t_3(\mathcal{H}_3[2]) \sqcup
\mathcal{A}_{2,1}[2] \sqcup
\mathcal{A}_{1,1,1}[2]
\end{equation*}
and below we compute the cohomology groups of each of the spaces on the right hand side as representations of $\Sp{6}$.
Moreover, we will see that each cohomology group $H^i$ of a space on the right hand side is pure of weight $2i$
and Tate type $(i,i)$.
By the additivity of the weighted Euler characteristic,
\begin{equation*}
e_{\mathcal{A}_3[2]}(v) =
e_{t_3(\mathcal{Q}[2])}(v) +
v^2 \, e_{t_3(\mathcal{H}_3[2])}(v) +
v^4 \, e_{\mathcal{A}_{2,1}[2]}(v) +
v^6 \, e_{\mathcal{A}_{1,1,1}}(v).
\end{equation*}
Putting the results together for the different strata we get the $\Sp{6}$-equivariant weighted Euler characteristic of
$\mathcal{A}_3[2]$, see Table~\ref{a3table}. Each column in this table corresponds to an irreducible representation of $\Sp{6}$.
The irreducible representations are denoted $\phi_{dn}$ where $d$ is the dimension of
the representation and $n$ is letter used to distinguish different representations of the same
dimension, see \cite{conwayetal}.
\begin{table}[ht]
\begin{equation*}
\resizebox{0.9\textwidth}{!}{$
\begin{array}{c|cccccccccc}
\, & \phi_{1a} & \phi_{7a} & \phi_{15a} & \phi_{21a} & \phi_{21b} & \phi_{27a} & \phi_{35a} & \phi_{35b} & \phi_{56a} & \phi_{70a} \\
\hline
e_{\mathcal{A}_3[2]}(v) & 1+v^2+v^4+v^6+v^{12} & 0 & v^{12} & 0 & 0 & -v^6-v^8 & 0 & -v^6-v^8+v^{12} & 0 & 0 \\
\hline
\, & \phi_{84a} & \phi_{105a} & \phi_{105b} & \phi_{105c} & \phi_{120a} & \phi_{168a} & \phi_{189a} & \phi_{189b} & \phi_{189c} & \phi_{210a} \\
\hline
e_{\mathcal{A}_3[2]}(v) & v^{12} & 0 & v^4 & 0 & v^{10} & v^{10} & 0 & 0 & 0 & -v^6 \\
\hline
\, & \phi_{210b} & \phi_{216a} & \phi_{280a} & \phi_{280b} & \phi_{315a} & \phi_{336a} & \phi_{378a} & \phi_{405a} & \phi_{420a} & \phi_{512a} \\
\hline
e_{\mathcal{A}_3[2]}(v) & -v^6 & 0 & 0 & v^{10} & 0 & 0 & 0 & 0 & v^8 & -v^{12}
\end{array}$}
\end{equation*}
\caption{The $\Sp{6}$-equivariant weighted Euler characteristic of $\mathcal{A}_3[2]$.}
\label{a3table}
\end{table}
Note that the results for $\phi_{1a}$ agree with the computation of the cohomology groups of $\mathcal{A}_3$ together with their Hodge structure in \cite{Hain}. Note also that only 13 of the 30 irreducible representations of $\Sp{6}$ occur in $e_{\mathcal{A}_3[2]}(v)$
and that for each irreducible representation the coefficients of $v^i$ are all either zero or $\pm 1$.
This is in sharp contrast to the cohomology of the individual pieces - all irreducible representations
except $\phi_{7a}$ occur in some cohomology group of some piece and they occur with multiplicities
up to $14$.
\section{An indecomposable abelian surface and an elliptic curve} \label{sec-A21}
As in the genus $3$ case, $t_2$ gives an isomorphism $\mathcal{M}_2[2] \cong \mathcal{A}_2^{\mathrm{in}}[2]$,
where $ \mathcal{A}_2^{\mathrm{in}}[2]$ denotes the indecomposable locus inside $ \mathcal{A}_2[2]$.
There is a close relationship between $\mathcal{A}_{2,1}[2]$ and
the product space $\mathcal{M}_2[2] \times \mathcal{A}_1[2]$.
Let $C$ be a genus $2$ curve with level $2$ structure represented
by the symplectic basis $(e_1,e_2,f_1,f_2)$ of $\mathrm{Jac}(C)[2]$
and let $E$ be an elliptic curve with level
$2$ structure $(e_3,f_3)$. Then $t_2(C) \times E$ is an abelian threefold and
$t_2(C)[2] \times E[2]$ is a six dimensional vector space over $\mathbb{F}_2$ with a symplectic pairing given by
\begin{equation*}
e_i \cdot e_j = f_i \cdot f_j = 0
\end{equation*}
and
\begin{equation*}
e_i \cdot f_j = \delta_{i,j}
\end{equation*}
for all $i$ and $j$, where we identify $e_i$ with $t_2(e_i)$ and $f_i$ with $t_2(f_i)$.
Clearly, not all level $2$ structures on $t_2(C) \times E$ arise in this way
but those that do are permuted by the group $\Sp{4} \times \Sp{2}$. Let $\mathscr{C}$ be the
quotient set $\Sp{6}/(\Sp{4} \times \Sp{2})$. We may then describe the locus $\mathcal{A}_{2,1}[2]$ as
\begin{equation*}
\mathcal{A}_{2,1}[2] \cong \coprod_{c \in \mathscr{C}} (\mathcal{M}_2[2] \times \mathcal{A}_1[2])_c,
\end{equation*}
where $(\mathcal{M}_2[2] \times \mathcal{A}_1[2])_c$ is an isomorphic copy of $\mathcal{M}_2[2] \times \mathcal{A}_1[2]$ indexed by $c$ and the
components are permuted as
\begin{equation*}
g(\mathcal{M}_2[2] \times \mathcal{A}_1[2])_c = (\mathcal{M}_2[2] \times \mathcal{A}_1[2])_{gc}
\end{equation*}
for $g \in \Sp{6}$. In terms of cohomology groups this means that
\begin{equation*}
H^i(\mathcal{A}_{2,1}[2]) = \mathrm{Ind}_{\Sp{4} \times \Sp{2}}^{\Sp{6}} H^i(\mathcal{M}_2[2] \times \mathcal{A}_1[2]).
\end{equation*}
By the K\"unneth theorem we have that
\begin{equation*}
H^i(\mathcal{M}_2[2] \times \mathcal{A}_1[2]) \cong \bigoplus_{p+q=i} H^p(\mathcal{M}_2[2]) \otimes H^q(\mathcal{A}_1[2]).
\end{equation*}
Thus, in order to understand the action of $\Sp{4} \times \Sp{2}$ on $H^i(\mathcal{M}_2[2] \times \mathcal{A}_1[2])$ it is
enough to understand the action of $\Sp{4}$ on $H^i(\mathcal{M}_2[2])$ and the action of $\Sp{2}$ on $H^i(\mathcal{A}_1[2])$ for all $i$.
\subsection{The moduli space of elliptic curves with level two structure} \label{sec-g1}
In order to understand the action of $\Sp{2}$ on $H^i(\mathcal{A}_1[2])$ we note that
$\Sp{2}$ is isomorphic to the symmetric group $S_3$ and that $\mathcal{A}_1[2]$ is isomorphic
to $\mathcal{M}_{0,4}$, the moduli space of four ordered points on $\mathbb{P}^1$. Under these
identifications, the action of $\Sp{2}$ is given by permuting the first three points.
Since
$\mathcal{M}_{0,4}$ is pure in the sense of Dimca and Lehrer \cite{dimcalehrer} we can
deduce the action of $\Sp{2}$ on the cohomology groups by
a twisted point count, see Section~\ref{sec-euler}.
Simple computations give, where $F$ is the Frobenius,
\begin{equation*}
\resizebox{0.55\hsize}{!}{$
\arraycolsep=8pt\def1.5{1.5}
\begin{array}{lll}
|\mathcal{M}_{0,4}^{F \circ \mathrm{id}}| & = \frac{(q+1) q (q-1) (q-2)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q-2 \\
|\mathcal{M}_{0,4}^{F \circ (12)}| & = \frac{(q+1) (q^2-q) q}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q \\
|\mathcal{M}_{0,4}^{F \circ (123)}| & = \frac{(q+1) (q^3-q)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q+1.
\end{array}$}
\end{equation*}
Thus, the traces of $(\mathrm{id},(12),(123))$ on
$H^0(\mathcal{A}_1[2])$ and $H^1(\mathcal{A}_1[2])$ are
$(1,1,1)$ and $(2,0,-1)$, respectively. In other words, $H^0(\mathcal{A}_1[2])$ is the trivial representation
of $\Sp{2}$ while $H^1(\mathcal{A}_1[2])$ is the standard representation.
\subsection{The moduli space of genus two curves with level two structure} \label{sec-g2}
In order to understand the action of $\Sp{4}$ on $H^i(\mathcal{M}_2[2])$ we note that
$\Sp{4}$ is isomorphic to the symmetric group $S_6$ and that $\mathcal{M}_2[2]$ is isomorphic
to $\mathcal{M}_{0,6}$, the moduli space of six ordered points on $\mathbb{P}^1$.
Under these identifications,
the action of $\Sp{4}$ on $\mathcal{M}_2[2]$ is given by permuting the points.
Also, $\mathcal{M}_{0,6}$ is pure so we can again deduce the action of $\Sp{4}$ on the cohomology
via twisted point counts. Simple computations give, where $F$ is the Frobenius,
\begin{equation*}
\resizebox{0.83\hsize}{!}{$
\arraycolsep=8pt\def1.5{1.5}
\begin{array}{lll}
|\mathcal{M}_{0,6}^{F \circ \mathrm{id}}| & = \frac{(q+1) q (q-1) (q-2) (q-3) (q-4)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3 - 9q^2 + 26q - 24\\
|\mathcal{M}_{0,6}^{F \circ (12)}| & = \frac{(q+1) q (q-1) (q-2) (q^2-q)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3 - 3q^2 + 2q \\
|\mathcal{M}_{0,6}^{F \circ (12)(34)}| & = \frac{(q+1) q (q^2-q) (q^2-q-2)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3 - q^2 - 2q\\
|\mathcal{M}_{0,6}^{F \circ(12)(34)(56)}| & = \frac{(q^2-q) (q^2-q-2) (q^2-q-4)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3 - 3q^2 - 2q + 8 \\
|\mathcal{M}_{0,6}^{F \circ (123)}| & = \frac{(q+1) q (q-1) (q^3-q)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3-q \\
|\mathcal{M}_{0,6}^{F \circ (123)(45)}| & = \frac{(q+1) (q^2-q) (q^3-q)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3-q \\
|\mathcal{M}_{0,6}^{F \circ (123)(456)}| & = \frac{(q^3-q) (q^3-q-3)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3-q-3 \\
|\mathcal{M}_{0,6}^{F \circ (1234)}| & = \frac{(q+1) q (q^4-q^2)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3+q^2 \\
|\mathcal{M}_{0,6}^{F \circ (1234)(56)}| & = \frac{(q^2-q) q (q^4-q^2)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3 - q^2 \\
|\mathcal{M}_{0,6}^{F \circ (12345)}| & = \frac{(q+1) (q^5-q)}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3+q^2+q+1 \\
|\mathcal{M}_{0,6}^{F \circ (123456)}| & = \frac{q^6-q^3-q^2+q}{|\mathrm{PGL}_2(\mathbb{F}_q)|} & = q^3+q-1.
\end{array}$}
\end{equation*}
We may read off $\mathrm{Tr}(\sigma,H^i(\mathcal{M}_{0,6}))$ as the coefficient of $(-1)^i q^{3-i}$ in
$|\mathcal{M}_{0,6}^{F \circ \sigma'}|$, where $\sigma'$ is the element in the above list which is conjugate to $\sigma$.
We now know $H^i(\mathcal{M}_2[2])$ as a representation of $\Sp{4}$ for all $i$ and we know
$H^i(\mathcal{A}_1[2])$ as a representation of $\Sp{2}$ for all $i$ so we therefore know
$H^{i}(\mathcal{M}_2[2] \times \mathcal{A}_1[2])$ as a representation of $\Sp{4} \times \Sp{2}$.
Inducing this representation to $\Sp{6}$ gives us $H^i(\mathcal{A}_{2,1}[2])$ as a representation of
$\Sp{6}$. We give the result in
Table~\ref{a21table}.
As an aside we note that the complement of $t_2(\mathcal{M}_{2}[2])$
inside $\mathcal{A}_{2}[2]$ consists of products of elliptic curves. The cohomology of this
locus can be computed using the same techniques as in Section~\ref{sec-A111}, but in this
case we will
omit the details. Let us denote
the irreducible representations of $\Sp{4} \cong S_6$ by $s_{\lambda}$, which are indexed
in the standard way by $\lambda$, a partition of~$6$. Adding the contributions from the two loci we find that,
$$ e_{\mathcal{A}_2[2]}(v) =(1+v^2)s_{6} -v^4(s_{5,1}+s_{4,2})+v^6s_{3,2,1}.
$$
\section{Products of three elliptic curves} \label{sec-A111}
There is a close relationship between the product space $(\mathcal{A}_1[2])^3$ and the locus
$\mathcal{A}_{1,1,1}[2]$. Let $E_1$, $E_2$ and $E_3$ be three elliptic curves
with level $2$ structures $(e_{1},f_{1})$, $(e_{2},f_2)$ and $(e_3,f_3)$, respectively.
Then $E_1 \times E_2 \times E_3$ is an abelian threefold and $E_1[2] \times E_2[2] \times E_3[2]$
is a six dimensional vector space over $\mathbb{F}_2$ with a symplectic pairing given by
\begin{equation*}
e_i \cdot e_j = f_i \cdot f_j = 0
\end{equation*}
for all $i$ and $j$ and
\begin{equation*}
e_i \cdot f_j = \delta_{i,j}.
\end{equation*}
Clearly, not all level $2$ structures on $E_1 \times E_2 \times E_3$ arise in this way
but those that do are permuted by the group $(\Sp{2})^3$ while the three curves themselves
are permuted by the symmetric group $S_3$. Let $\mathscr{C}$ be the
quotient set $\Sp{6}/(S_3 \ltimes (\Sp{2})^3)$. We may describe the locus $\mathcal{A}_{1,1,1}[2]$ as
\begin{equation*}
\mathcal{A}_{1,1,1}[2] \cong \coprod_{c \in \mathscr{C}} (\mathcal{A}_1[2])^3_c,
\end{equation*}
where $(\mathcal{A}_1[2])^3_c$ is an isomorphic copy of $(\mathcal{A}_1[2])^3$ indexed by $c$ and the
components are permuted as
\begin{equation*}
g(\mathcal{A}_1[2])^3_c = (\mathcal{A}_1[2])^3_{gc}
\end{equation*}
for $g \in \Sp{6}$. In terms of cohomology groups this means that
\begin{equation*}
H^i(\mathcal{A}_{1,1,1}[2]) = \mathrm{Ind}_{S_3 \ltimes (\Sp{2})^3}^{\Sp{6}} H^i((\mathcal{A}_1[2])^3).
\end{equation*}
By the K\"unneth theorem we have that
\begin{equation*}
H^i((\mathcal{A}_1[2])^3) \cong \bigoplus_{p+q+r=i} H^p(\mathcal{A}_1[2]) \otimes H^q(\mathcal{A}_1[2]) \otimes H^r(\mathcal{A}_1[2]).
\end{equation*}
Thus, in order to understand the action of $S_3 \ltimes (\Sp{2})^3$ on $H^i((\mathcal{A}_1[2])^3$ it is
enough to understand the action of $\Sp{2}$ on $H^i(\mathcal{A}_1[2])$ for all $i$ and the action of $S_3$ on the
factors.
Since the action of $\Sp{2}$ was described in Section~\ref{sec-g1} we only consider the action of $S_3$ on the factors.
Let $\alpha \otimes \beta \otimes \gamma \in H^p(\mathcal{A}_1[2]) \otimes H^q(\mathcal{A}_1[2]) \otimes H^r(\mathcal{A}_1[2]) \subseteq H^{p+q+r}((\mathcal{A}_1[2])^3)$.
We have
\begin{align*}
(12).\alpha \otimes \beta \otimes \gamma = (-1)^{pq}\beta \otimes \alpha \otimes \gamma, \\
(13).\alpha \otimes \beta \otimes \gamma = (-1)^{pr}\gamma \otimes \beta \otimes \alpha, \\
(23).\alpha \otimes \beta \otimes \gamma = (-1)^{qr}\alpha \otimes \gamma \otimes \beta,
\end{align*}
where the signs are a consequence of the K\"unneth isomorphism.
Since $S_3$ is generated by transpositions and $H^{p+q+r}((\mathcal{A}_1[2])^3)$ is generated by
elements of the form $\alpha \otimes \beta \otimes \gamma$ for all possible choices of $p$, $q$ and
$r$, this determines the action of $S_3$ on $H^{p+q+r}((\mathcal{A}_1[2])^3)$.
We now have all the information we need in order to understand the action of $S_3 \ltimes (\Sp{2})^3$ on $H^i((\mathcal{A}_1[2])^3)$.
\pagebreak[2]
\begin{example}
\label{a1example}
Let $u$ be a basis vector for the trivial representation of $\Sp{2}$ and
let $v_1$ and $v_2$ be basis vectors for the standard representation of $\Sp{2}$.
Let $\sigma \in \Sp{2}$ be an element of order $3$ acting as
\begin{align*}
\sigma.u & = u, \\
\sigma.v_1 & = v_2, \\
\sigma.v_2 & = -v_1-v_2,
\end{align*}
and let $g \in S_3 \ltimes (\Sp{2})^3$ on $H^i((\mathcal{A}_1[2])^3$ be the element $g=((23),(\sigma,\sigma,\mathrm{id}))$.
We of course have $g.u\otimes u \otimes u = u \otimes u \otimes u$, so $\mathrm{Tr}(g,H^0((\mathcal{A}_1[2])^3)=1$.
In cohomological degree $1$ we have
\begin{equation*}
g.v_2 \otimes u \otimes u = -(v_1 + v_2) \otimes u \otimes u
\end{equation*}
while $g.\alpha$ has no component in the direction of $\alpha$ for all other choices of $\alpha \in H^1((\mathcal{A}_1[2])^3)$.
Thus, $\mathrm{Tr}(g,H^1((\mathcal{A}_1[2])^3)=-1$. In degree $2$ we have
\begin{equation*}
g.u \otimes v_2 \otimes v_2 = u \otimes v_2 \otimes (v_1+v_2)
\end{equation*}
while $g.\alpha$ has no component in the direction of $\alpha$ for all other choices of $\alpha \in H^2((\mathcal{A}_1[2])^3)$.
We conclude that $\mathrm{Tr}(g,H^2((\mathcal{A}_1[2])^3)=1$. Finally, in degree $3$ we have
\begin{equation*}
g.v_2 \otimes v_2 \otimes v_2 = -(v_1+v_2) \otimes v_2 \otimes (v_1+v_2)
\end{equation*}
while $g.\alpha$ has no component in the direction of $\alpha$ for all other choices of $\alpha \in H^3((\mathcal{A}_1[2])^3)$.
Hence, $\mathrm{Tr}(g,H^3((\mathcal{A}_1[2])^3)=-1$. We thus have
\begin{equation*}
\sum_{i=0}^3 \mathrm{Tr}(g,H^i((\mathcal{A}_1[2])^3) \, t^i = 1-t+t^2-t^3.
\end{equation*}
\end{example}
Similar computations for the other conjugacy classes of $S_3 \ltimes (\Sp{2})^3$ give
the results in Table~\ref{a1table}, where
$$P_g((\mathcal{A}_1[2])^3,t):=\sum_{i=0}^3 \mathrm{Tr}(g,H^i((\mathcal{A}_1[2])^3)) \, t^i,$$
is called the equivariant Poincar\'e polynomial.
See Chapter 4 of \cite{jameskerber} for a beautiful description
of how to compute representatives of $S_3 \ltimes (\Sp{2})^3$. In Table~\ref{a1table}, $\sigma$ is
the element of $\Sp{2}$ described in Example~\ref{a1example} while $\tau$ is the element of order $2$
acting as
\begin{equation*}
\tau.v_1=-v_1, \quad \tau.v_2=v_1+v_2
\end{equation*}
where $v_1$ and $v_2$ are the same basis vectors of the standard representation considered in Example~\ref{a1example}.
\begin{table}[htbp]
\begin{equation*}
\resizebox{0.95\textwidth}{!}{$
\begin{array}{l|l|l|l|l|l}
g & P_g((\mathcal{A}_1[2])^3,t) & g & P_g((\mathcal{A}_1[2])^3,t) & g & P_g((\mathcal{A}_1[2])^3,t) \\
\hline
(\textrm{id},(\textrm{id},\textrm{id},\textrm{id})) & 1 + 6t+12t^2+8t^3 & ((23),(\textrm{id},\textrm{id},\textrm{id})) & 1+2t-2t^2-4t^3 & ((123),(\textrm{id},\textrm{id},\textrm{id})) & 1+2t^3 \\
(\textrm{id},(\textrm{id},\textrm{id},\tau)) & 1+4t+4t^2 & ((23),(\textrm{id},\tau,\textrm{id})) & 1+2t &
((123),(\tau,\textrm{id},\textrm{id})) & 1 \\
(\textrm{id},(\textrm{id},\textrm{id},\sigma)) & 1+3t-4t^3 & ((23),(\textrm{id},\sigma,\textrm{id})) & 1+2t+t^2+2t^3 & ((123),(\sigma,\textrm{id},\textrm{id})) & 1-t^3 \\
(\textrm{id},(\textrm{id},\tau,\tau)) & 1+2t & ((23),(\tau,\textrm{id},\textrm{id})) & 1-2t^2 & \, & \, \\
(\textrm{id},(\textrm{id},\tau,\sigma)) & 1+t-2t^2 & ((23),(\tau,\tau,\textrm{id})) & 1 & \, & \,\\
(\textrm{id},(\textrm{id},\sigma,\sigma)) & 1 -3t^2+2t^3 & ((23),(\tau,\sigma,\textrm{id})) & 1+t^2 & \, & \, \\
(\textrm{id},(\tau,\tau,\tau)) & 1 & ((23),(\sigma,\textrm{id},\textrm{id})) & 1-t-2t^2+2t^3 & \, & \,\\
(\textrm{id},(\tau,\tau,\sigma)) & 1-t & ((23),(\sigma,\tau,\textrm{id})) & 1-t & \, & \,\\
(\textrm{id},(\tau,\sigma,\sigma)) & 1-2t+t^2 & ((23),(\sigma,\sigma,\textrm{id})) & 1-t+t^2-t^3 & \, & \,\\
(\textrm{id},(\sigma,\sigma,\sigma)) & 1-3t+3t^2-t^3 & \, & \,& \, & \,
\end{array}$}
\end{equation*}
\caption{Equivariant Poincar\'e polynomials of $(\mathcal{A}_1[2])^3$ for a representative $g$ of every conjugacy class of $S_3 \ltimes (\Sp{2})^3$.}
\label{a1table}
\end{table}
By inducing the corresponding representations from $S_3 \ltimes (\Sp{2})^3$ to $\Sp{6}$ we obtain
the cohomology of $\mathcal{A}_{1,1,1}[2]$ as a representation of $\Sp{6}$. We give the result in
Table~\ref{a111table}.
\section{Cohomology groups of strata}
In this section we give the cohomology groups of $\mathcal{Q}[2]$, $\mathcal{H}_3[2]$,
$\mathcal{A}_{2,1}[2]$ and $\mathcal{A}_{1,1,1}[2]$ as representations of $\Sp{6}$.
The results are presented in Table~\ref{Qtable}-\ref{a111table}.
Each column in these tables corresponds to an irreducible representation of $\Sp{6}$.
The irreducible representations are denoted $\phi_{dn}$ where $d$ is the dimension of
the representation and $n$ is letter used to distinguish different representations of the same
dimension, see~\cite{conwayetal}.
\begin{table}[htbp]
\begin{equation*}
\resizebox{0.62\textwidth}{!}{$
\begin{array}{r|rrrrrrrrrr}
\, & \phi_{1a} & \phi_{7a} & \phi_{15a} & \phi_{21a} & \phi_{21b} & \phi_{27a} & \phi_{35a} & \phi_{35b} & \phi_{56a} & \phi_{70a} \\
\hline
H^0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
H^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^3 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
H^5 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\
H^6 & 1 & 0 & 2 & 0 & 1 & 1 & 1 & 3 & 0 & 0 \\
\hline
\, & \phi_{84a} & \phi_{105a} & \phi_{105b} & \phi_{105c} & \phi_{120a} & \phi_{168a} & \phi_{189a} & \phi_{189b} & \phi_{189c} & \phi_{210a} \\
\hline
H^0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
H^3 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 2 \\
H^4 & 0 & 0 & 2 & 0 & 2 & 1 & 2 & 1 & 0 & 3 \\
H^5 & 1 & 2 & 2 & 1 & 2 & 4 & 3 & 3 & 3 & 4 \\
H^6 & 5 & 1 & 1 & 4 & 0 & 3 & 2 & 2 & 5 & 3 \\
\hline
\, & \phi_{210b} & \phi_{216a} & \phi_{280a} & \phi_{280b} & \phi_{315a} & \phi_{336a} & \phi_{378a} & \phi_{405a} & \phi_{420a} & \phi_{512a} \\
\hline
H^0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
H^3 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 \\
H^4 & 4 & 0 & 3 & 1 & 3 & 2 & 3 & 6 & 5 & 4 \\
H^5 & 4 & 4 & 4 & 6 & 5 & 6 & 6 & 6 & 8 & 9 \\
H^6 & 1 & 6 & 3 & 6 & 1 & 6 & 4 & 2 & 6 & 6
\end{array}$}
\end{equation*}
\caption{The cohomology groups of $\mathcal{Q}[2]$ as a representation of $\Sp{6}$.}
\label{Qtable}
\end{table}
\begin{table}[htbp]
\begin{equation*}
\resizebox{0.62\textwidth}{!}{$
\begin{array}{r|rrrrrrrrrr}
\, & \phi_{1a} & \phi_{7a} & \phi_{15a} & \phi_{21a} & \phi_{21b} & \phi_{27a} & \phi_{35a} & \phi_{35b} & \phi_{56a} & \phi_{70a} \\
\hline
H^0 & 1&0&0&0&0&0&0&1&0&0\\
H^1 & 0&0&0&0&0&1&0&1&0&0\\
H^2 & 0&0&0&1&0&0&0&0&0&0\\
H^3 & 0&0&0&1&0&0&0&0&0&1\\
H^4 & 0&0&0&0&0&1&1&1&0&1\\
H^5 & 0&0&1&0&1&1&1&2&0&0 \\
\hline
\, & \phi_{84a} & \phi_{105a} & \phi_{105b} & \phi_{105c} & \phi_{120a} & \phi_{168a} & \phi_{189a} & \phi_{189b} & \phi_{189c} & \phi_{210a} \\
\hline
H^0 & 0&0&0&0&0&0&0&0&0&0\\
H^1 & 0&0&0&0&0&1&0&0&0&1\\
H^2 & 0&0&1&0&2&1&1&0&0&3\\
H^3 & 0&0&3&1&3&2&4&1&0&5\\
H^4 & 2&2&3&2&3&6&5&4&4&6\\
H^5 & 4&2&1&4&1&4&3&3&6&4 \\
\hline
\, & \phi_{210b} & \phi_{216a} & \phi_{280a} & \phi_{280b} & \phi_{315a} & \phi_{336a} & \phi_{378a} & \phi_{405a} & \phi_{420a} & \phi_{512a} \\
\hline
H^0 & 0&0&0&0&0&0&0&0&0&0\\
H^1 & 0&0&0&1&0&0&0&0&0&0\\
H^2 & 1&0&0&2&0&0&1&3&2&2\\
H^3 & 5&1&3&3&4&3&5&10&7&7\\
H^4 & 6&5&7&8&7&9&9&10&12&14\\
H^5 & 2&7&4&8&3&8&6&4&8&9
\end{array}$}
\end{equation*}
\caption{The cohomology groups of $\mathcal{H}_{3}[2]$ as a representation of $\Sp{6}$.}
\label{Hyptable}
\end{table}
\begin{table}[ht]
\begin{equation*}
\resizebox{0.62\textwidth}{!}{$
\begin{array}{c|cccccccccc}
\, & \phi_{1a} & \phi_{7a} & \phi_{15a} & \phi_{21a} & \phi_{21b} & \phi_{27a} & \phi_{35a} & \phi_{35b} & \phi_{56a} & \phi_{70a} \\
\hline
H^0 & 1&0&0&0&0&1&0&1&0&0 \\
H^1 & 0&0&0&0&0&2&0&2&0&0 \\
H^2 & 0&0&0&1&0&0&0&1&0&0 \\
H^3 & 0&0&0&1&0&0&0&0&0&2 \\
H^4 & 0&0&0&0&0&0&0&0&1&1 \\
\hline
\, & \phi_{84a} & \phi_{105a} & \phi_{105b} & \phi_{105c} & \phi_{120a} & \phi_{168a} & \phi_{189a} & \phi_{189b} & \phi_{189c} & \phi_{210a} \\
\hline
H^0 & 0&0&1&0&0&1&0&0&0&0 \\
H^1 & 1&0&1&0&2&2&0&0&0&2 \\
H^2 & 1&0&2&2&3&3&3&0&0&5 \\
H^3 & 1&0&3&2&2&3&5&2&1&6 \\
H^4 & 0&1&1&0&2&1&2&2&1&2 \\
\hline
\, & \phi_{210b} & \phi_{216a} & \phi_{280a} & \phi_{280b} & \phi_{315a} & \phi_{336a} & \phi_{378a} & \phi_{405a} & \phi_{420a} & \phi_{512a} \\
\hline
H^0 & 0&0&0&0&0&0&0&0&0&0 \\
H^1 & 1&0&0&3&0&0&0&1&1&1 \\
H^2 & 2&2&0&6&1&2&3&6&5&5 \\
H^3 & 4&2&4&4&4&5&6&10&8&10 \\
H^4 & 4&1&3&2&5&3&5&6&5&5
\end{array}$}
\end{equation*}
\caption{The cohomology groups of $\mathcal{A}_{2,1}[2]$ as representations of $\Sp{6}$.}
\label{a21table}
\end{table}
\begin{table}[htbp]
\begin{equation*}
\resizebox{0.62\textwidth}{!}{$
\begin{array}{c|cccccccccc}
\, & \phi_{1a} & \phi_{7a} & \phi_{15a} & \phi_{21a} & \phi_{21b} & \phi_{27a} & \phi_{35a} & \phi_{35b} & \phi_{56a} & \phi_{70a} \\
\hline
H^0 & 1&0&0&0&0&1&0&1&0&0 \\
H^1 & 0&0&0&0&0&1&0&2&0&0 \\
H^2 & 0&0&0&1&0&0&0&0&0&1 \\
H^3 & 0&0&0&0&0&0&0&0&1&1 \\
\hline
\, & \phi_{84a} & \phi_{105a} & \phi_{105b} & \phi_{105c} & \phi_{120a} & \phi_{168a} & \phi_{189a} & \phi_{189b} & \phi_{189c} & \phi_{210a} \\
\hline
H^0 & 1&0&1&0&0&1&0&0&0&0 \\
H^1 & 1&0&1&1&2&2&1&0&0&3 \\
H^2 & 0&0&2&1&2&2&3&1&0&4 \\
H^3 & 0&0&1&0&1&0&1&1&0&1 \\
\hline
\, & \phi_{210b} & \phi_{216a} & \phi_{280a} & \phi_{280b} & \phi_{315a} & \phi_{336a} & \phi_{378a} & \phi_{405a} & \phi_{420a} & \phi_{512a} \\
\hline
H^0 & 0&0&0&1&0&0&0&0&1&0 \\
H^1 & 1&1&0&4&0&1&1&2&2&2 \\
H^2 & 2&1&1&3&2&2&3&6&4&5 \\
H^3 & 3&0&2&0&3&1&3&4&3&3
\end{array}$}
\end{equation*}
\caption{The cohomology groups of $\mathcal{A}_{1,1,1}[2]$ as representations of $\Sp{6}$.}
\label{a111table}
\end{table}
\clearpage
\bibliographystyle{plain}
\renewcommand{\bibname}{References}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,021 |
Víztisztító berendezésnek nevezzük azt a készüléket, amely a különböző szennyezőanyagok vízből való eltávolítására, koncentrációjuk csökkentésére szolgál alapvetően fizikai, kémiai vagy biológiai módszerrel.
A víztisztításnak több területe is van, például öntözővíz, uszoda vagy ivóvíz tisztítás. A háztartásban a gépek védelme (mechanikus szennyezőanyagok, vízlágyítás) és az ivóvíz minőségének javítása miatt szokás kiegészítő vízszűrést alkalmazni.
Története
A 19-20. századig az úgynevezett lassú és gyors mechanikai homokszűrők voltak elterjedtek. Az 1800-as években százasával épültek Európában a homokszűrők. A szűrési folyamat felgyorsítása érdekében alumínium-szulfátot kezdtek alkalmazni a szűrő előtt, amivel hatvanszorosára növelték a szűrési sebességet, kisebb méret mellett.
Az első modern szűrőüzem az Egyesült Államokban épült 1902-ben, New Jersey-ben. Meg kell említeni John R. Baylis (1885-1963) amerikai kémikust, mérnököt, aki az ivóvízszűrési eljárások minőségének javításában elért kutatásokról ismert. A nevéhez kapcsolódik például az ivóvíz illat és íz minőségét befolyásoló tényezők megértése, és nagy átbocsátású szűrők kifejlesztése is az 1930-as, 1940-es években.
Szűrési módszerek
Szűrővel, adszorpcióval, ioncseréléssel, biológiai anyagcserével és egyéb módszerekkel oldható meg.
Néhány elterjedt típus:
Homokszűrő
Cserélhető betétes vízszűrő (aktív szén, szövet, kerámia stb.)
Visszaöblíthető vízszűrő
Média-szűrők pl. vas, mangán, nehézfémek eltávolítására
Ülepítéses
Desztilláció
Otthoni ivóvíz utótisztítás
Ebben a kategóriában helyt kapnak a specifikus vízszűrők, amelyek egy jól behatárolható szennyező anyag megszűrésére képesek. Ilyen például a vas és mangánmentesítő, vagy az arzénmentesítő szűrőbetét.
A háztartási ivóvíz tisztításban alkalmazott főbb módszerek:
Szűrőbetétes víztisztító
Ozmózis víztisztító
Specifikus (speciális) vízszűrő
A szűrőbetétes háztartási ivóvíztisztító berendezések minimum egy, de akár korlátlan számú vízszűrő egységből állhatnak. Ezeknek a víztisztító berendezéseknek az alap szűrési mechanizmusa az aktívszenes víztisztítás. Ezt a víztisztítási fajtát egészítik ki többlépcsős vízszűrés esetén PP szűrőbetétekkel. Az általános PP szűrőbetét 5 mikronos, de 1 mikronos szűrés is előfordul a háromfázisos változatoknál. Előfordul, hogy kapilláris membránnal vagy ultraszűrővel egészítik ki a víztisztítót, ezzel növelve annak hatásfokát, hiszen a szűrési méret 0,02 mikronra csökkenthető segítségével.
Az új generációs integrált szűrőbetétek képviselik ebben a kategóriában a legújabb innovációt. Egy szűrőbetéten belül több szűrési funkciót (többlépcsős szűrés) valósítanak meg. Ezek a szűrőbetétek felépítésüknek köszönhetően költséghatékonyak, környezetkímélők, helytakarékosak. A szűrőbetétek 4. és 5. szűrési lépcsői specifikus szűrőanyagok (nehézfémek, nitrát, arzén, fluorid stb. eltávolítására alkalmasak).
Kockázatok:
nem megfelelő szűrőbetét választás (a vezetékes vízben lévő valós szennyezők ismerete nélkül választott szűrőanyag többet árthat, mint használ);
az aktívszén nem megfelelő bakteriológiai védelme miatt bekövetkező bakteriológiai és kémiai minőségromlás (denitrifikáció).
Előfordulhatnak olyan szennyezők is a vízben, melyek eltávolítása csak teljes sótalanítással oldható meg. Erre az esetre a desztilláció mellett az ozmózison alapuló víztisztító berendezés jelent megoldást. Hátránya többek között az, hogy a kezelésre szánt víz nagy részét hulladékvízként a szennyvízcsatornába vezeti, másrészt a vízben lévő szennyezőkön kívül a hasznos ásványi anyagokat is eltávolítja. Az RO membrán típusától függően szükség lehet gyűjtőtartályra is a szűrési folyamat lassúsága miatt, így a háztartási víztisztításban a tartályban lévő pangó víz kiemelt kockázatot jelent.
Az ozmózis víztisztító előtt található előszűrők az ozmózis membrán védelmére szolgálnak, durva vízszűrést végeznek, ezzel kímélik az ozmózis membránt a nagyobb szennyeződésektől.
Az RO víztisztítás főbb kockázatai:
baktériumok elszaporodhatnak a szűrő utáni blokkban, ami könnyen túllépheti az egészségügyi határértéket. Erre megoldást jelenthet a rendszeres, előírás szerinti karbantartás, valamint UV szűrő elhelyezése (csírátlanítás).
Visszasózó berendezések esetében a pontatlan kalibrálásból adódó túl alacsony/magas sómennyiség.
Olcsóbb, nem bevizsgált berendezések esetében a készülékházból, szűrőkből kioldódó vegyi anyagok magas koncentrációja.
Kockázatok minimalizálása:
Készüléktípustól és a vízzel érintkező anyagoktól függő – az üzembe helyezéshez tartozó átöblítési és fertőtlenítési eljárás, a 2-3 napnál hosszabb üzemszünet utáni teendők, a szűrők cseréjére, regenerálására, illetve a berendezés fertőtlenítésére vonatkozó előírások szerepelnek.
A kezdeti fertőtlenítés és öblítés, a hosszabb üzemszünetek utáni alapos átöblítés, a tartállyal ellátott berendezés esetén a tartály leürítése, tisztítása és fertőtlenítése, valamint a szűrő-patronok rendszeres cseréje a baktériumok elszaporodásának lehetőségét csökkenthetik. A berendezések rendszeres (3-6 havonta történő) fertőtlenítést, karbantartást, szűrő- vagy egyéb alkatrész (pl. UV-lámpa) - cserét igényelnek, amely jelentős költséget jelenthet. A biztonságos üzemeltetés csak a használati útmutatóban leírtak pontos betartása mellett lehetséges. A berendezések beüzemelésekor, illetve 2-3 napos üzemszünetet követően egyes esetekben jelentős mennyiségű vízzel át kell öblíteni. A "kancsós" berendezéseket célszerű hűtőben tárolni.
A maradék fertőtlenítőszer eltávolítása miatt a kezelt víz "romlandó"-nak minősül, ezért célszerű hűtőben tartani, és 24 órán belül elfogyasztani.
Mivel Magyarországon szinte mindenhol megfelel a vezetékes ivóvíz minősége az előírásoknak, így az utólagos kezelés a nem megfelelő íz és illat javítása érdekében lehet indokolt.
Külső hivatkozások
HVG Nagy víztisztító teszt
ANTSZ GYIK
OKI Ivóvíz Kiskáté (GYIK)
OKI 2014 Ivóvíz minőség
NNK tájékoztató: tények és tévhitek a víztisztítókról
A Magyar Honvédség víztisztító századának lehetséges szerepvállalása az afrikai kontinensen
Háztartási ivóvíztisztító kalauz
Források
NNK tájékoztató a víztisztítókról
BWT Vízkezelés
Blog a vízről
Vízellátás | {
"redpajama_set_name": "RedPajamaWikipedia"
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{"url":"https:\/\/grindskills.com\/comparing-two-distributions-in-fourier-space\/","text":"# Comparing two distributions in Fourier space\n\nThere exist a number of tools that provide a distance between two continuous probability distributions. Most (semi)distances, like the Kullback-Leibler divergence, use probability density functions. However, the literature is quite sparse when it comes to comparing two distributions in Fourier space, i.e. via their characteristic function. Is there an elegant way to do so?\n\nOne notable distance that can be considered in Fourier space is the maximum mean discrepancy (MMD). One first selects a positive semi-definite kernel $$k:X\u00d7X\u2192Rk : \\mathcal X \\times \\mathcal X \\to \\mathbb R$$ corresponding to a reproducing kernel Hilbert space (RKHS) $$Hk\\mathcal H_k$$. The MMD is then\n\nMMDk(P,Q)=sup\\begin{align} \\operatorname{MMD}_k(\\mathbb P, \\mathbb Q) &= \\sup_{f \\in \\mathcal H_k : \\lVert f \\rVert_{\\mathcal H_k} \\le 1} \\mathbb E_{X \\sim \\mathbb P}[ f(X)] - \\mathbb E_{Y \\sim \\mathbb Q}[ f(Y)] \\\\&= \\left\\lVert \\mathbb E_{X \\sim \\mathbb P}[k(X, \\cdot)] - \\mathbb E_{Y \\sim \\mathbb Q}[k(Y, \\cdot)] \\right\\rVert \\\\&= \\sqrt{ \\mathbb{E}_{\\substack{X, X' \\sim \\mathbb P\\\\Y, Y' \\sim \\mathbb Q}}\\left[ k(X, X') + k(Y, Y') - 2 k(X, Y) \\right] } .\\end{align}\nYou might be familiar with the energy distance; that is a special case of the MMD for a particular choice of kernel.\nThis is a proper metric for many choices of kernel, called characteristic kernels; it is always a semimetric.\n\nWhat does this have to do with the Fourier transform? Well, if $$\\mathcal X = \\mathbb R^d\\mathcal X = \\mathbb R^d$$ and $$k(x, y) = \\psi(x \u2013 y)k(x, y) = \\psi(x - y)$$, so that $$\\psi : \\mathbb R^d \\to \\mathbb R\\psi : \\mathbb R^d \\to \\mathbb R$$ is a positive-definite function, then it turns out the MMD can also be written as\n$$\\operatorname{MMD}_k(\\mathbb P, \\mathbb Q) = \\sqrt{\\int \\left\\lvert \\varphi_{\\mathbb P}(\\omega) \u2013 \\varphi_{\\mathbb Q}(\\omega) \\right\\rvert^2 \\, \\mathrm{d}\\hat\\psi(\\omega)} \\operatorname{MMD}_k(\\mathbb P, \\mathbb Q) = \\sqrt{\\int \\left\\lvert \\varphi_{\\mathbb P}(\\omega) - \\varphi_{\\mathbb Q}(\\omega) \\right\\rvert^2 \\, \\mathrm{d}\\hat\\psi(\\omega)}$$\nwhere $$\\varphi\\varphi$$ denotes the characteristic function, and $$\\hat\\psi\\hat\\psi$$ is the Fourier transform of $$\\psi\\psi$$ in the measure sense. (It will always be a finite nonnegative measure; you can see from this definition that a translation-invariant kernel is characteristic iff its Fourier transform is everywhere positive.)\nFor a proof, see Corollary 4 of\n\nSriperumbudur et al., Hilbert space embeddings and metrics on probability measures, JMLR 2010.\n\nThe MMD \u2013 which you can easily estimate via the third form above \u2013\u00a0thus compares distributions by the $$L_2L_2$$ distance between their full characteristic functions, with frequencies weighted according to the choice of kernel. For example, the common Gaussian kernel $$k(x, y) = \\exp\\left( -\\frac{1}{2\\sigma^2} \\lVert x \u2013 y \\rVert^2 \\right)k(x, y) = \\exp\\left( -\\frac{1}{2\\sigma^2} \\lVert x - y \\rVert^2 \\right)$$ will weight the frequencies with a Gaussian with mean 0 and variance $$1\/\\sigma^21\/\\sigma^2$$.\n\nSometimes it\u2019s better, and sometimes computationally faster \/ more informative, to instead compare the characteristic functions at particular locations rather than everywhere. It turns out it\u2019s better to make a slight tweak, evaluating differences in smoothed characteristic functions at random locations:\n\nChwialkowski et al., Fast Two-Sample Testing with Analytic Representations of Probability Measures, NeurIPS 2015.\n\nA follow-up work finds the most informative frequencies to test, rather than random:\n\nJitkrittum et al., Interpretable Distribution Features with Maximum Testing Power, NeurIPS 2016.\n\nThese are all closely connected to classical tests based on empircal characteristic functions as mentioned by kjetil in the comments.","date":"2022-09-26 22:27:51","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 13, \"wp-katex-eq\": 0, \"align\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9957951903343201, \"perplexity\": 771.4790977284105}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030334942.88\/warc\/CC-MAIN-20220926211042-20220927001042-00302.warc.gz\"}"} | null | null |
"Remind yourself of who you are, as it's essential to being able to own a space" – Alexis Bustos, Interaction Designer, UX Researcher, Advocate for Inclusivity in Design & Tech
by krystlfabella | Apr 30, 2020 | Podcast | 0 comments
https://media.blubrry.com/filipinaontherise/p/content.blubrry.com/filipinaontherise/Alexis-Bustos-4_30_20-11_00-AM.m4a
Alexis Bustos is a San Francisco-based Interaction Designer, who is making waves while promoting Pinay Excellence in the world of User Experience Design & Research! She is consistently advocating for inclusive design and accessibility in tech. She holds a BFA in Interaction Design from California College of the Arts and is currently an interaction designer at Star, a a global team of product creators. Alexis is often a host and guest of Adobe Live, empowering communities of designers and talking about the future of design itself.
On this episode, Alexis' journey into the design world and how a trip to the Philippines and seeing a society of women and their voices influenced her path and her accountability of being loud and proud and a challenging voice in design and tech spaces
And how powerful it can be to go back to your roots, see where you come from to, use the path to enlighten your future
She shares how she challenges thinking and brings a socially conscious lens to dialogue in the tech/design world.
We also cover the power of representation, of being a face in a platform like Adobe live, is a catalyst for more diversity
Alexis also shares her experiences as a bi-racial woman, and how those mixed identities, while often a challenge of belonging growing up, fashioned her skills in code-switching (and why mixed kids are the best code-switchers!)
At the end, she shares a message to "little Alexis" and just about anyone who ever feels like "the only one" in a space, field or a path, to take ownership and that your voice matters
Co-hosting this episode talk Krystel Reyes, a member on the Filipina on the Rise team, who went to school with Alexis and is currently also an interaction designer.
Find her on: https://www.alexisbustos.design/
(Royalty Free Music from Bensound) | {
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Q: Gulp doesn't generate index.html I'm using the Gulp with the following settings:
var gulp = require('gulp');
var sass = require('gulp-sass');
var inject = require('gulp-inject');
var wiredep = require('wiredep').stream;
var del = require('del');
gulp.task('clean', function(cb){
del(['dist'], cb);
});
gulp.task('styles', function(){
var injectAppFiles = gulp.src('src/styles/*.scss', {read: false});
var injectGlobalFiles = gulp.src('src/global/*.scss', {read: false});
function transformFilepath(filepath) {
return '@import "' + filepath + '";';
}
var injectAppOptions = {
transform: transformFilepath,
starttag: '// inject:app',
endtag: '// endinject',
addRootSlash: false
};
var injectGlobalOptions = {
transform: transformFilepath,
starttag: '// inject:global',
endtag: '// endinject',
addRootSlash: false
};
return gulp.src('src/main.scss')
.pipe(wiredep())
.pipe(inject(injectGlobalFiles, injectGlobalOptions))
.pipe(inject(injectAppFiles, injectAppOptions))
.pipe(sass())
.pipe(gulp.dest('dist/styles'));
});
gulp.task('default', ['clean', 'styles'], function(){
var injectFiles = gulp.src(['dist/styles/main.css']);
var injectOptions = {
addRootSlash: false,
ignorePath: ['src', 'dist']
};
return gulp.src('src/index.html')
.pipe(inject(injectFiles, injectOptions))
.pipe(gulp.dest('dist'));
});
I expect to get dist/styles/main.css and dist/index.htm, but the second one doesn't generate. What can I do? Packets that I use: gulp, gulp-sass, gulp-inject, bootstrap-sass (bower), wiredep, del
A: Finally, I solved my problem. I just removed a variable cb from the clean callback
gulp.task('clean', function(cb){
del(['dist'], cb);
});
->
gulp.task('clean', function(){
del(['dist']);
});
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,622 |
\section{mm-wave $\gamma$-ray bright blazars}
{\em CGRO}-EGRET established the predominance of flat spectrum radio quasars
(FSRQ) as extragalactic sources of high-energy $\gamma$-rays~\cite{3egcat}.
These are believed to be powered by supermassive black holes ejecting large
amounts of material along relativistic jets perpendicular to accretion disks and
producing high energy particles in these jets. Observations at the highest photon
energies shown a predominance of Bl Lac objects over FSRQ at $E\sim 1~\rm TeV$.
Although these observations may be biased by the horizon limitation due to pair absorption
with extragalactic background light (EBL), the {\em Fermi} $\gamma$-ray Space Telescope,
more sensitive and more responsive to GeV photons than EGRET, reported a larger
fraction of Bl Lac objects among blazars in its bright source list release
(0FGL)~\cite{fermi-0fgl,fermi-0fgl-agn}. The wider spectral coverage of {\em Fermi} is
allowing detailed studies of the intrinsic properties of various types of active galactic
nuclei (AGN), cosmic evolution and their use as EBL tracers~\cite{dermer2007}.
The extrapolation of FSRQ to high radio frequencies makes the coincidence of
foreground {\em WMAP} sources with $\gamma$-ray blazars expectable. Furthermore,
it is clear than most of the 390 {\em WMAP} sources are AGNs~\cite{wmap-cat}. These
emitters are dominated in the mm-wave band by the same synchrotron component
observed at lower frequencies and can be expected to emit photons of very high energy,
related to the hardest part of the synchrotron emission. {\em WMAP} sources have fluxes
above 0.1~Jy at frequencies $\ga 40~\rm GHz$, making them suitable positional references
for large millimeter telescopes. Known common {\em WMAP}/EGRET sources, found up to
$z\ga 2.3$, are relatively nearby analogs of more distant blazars detectable in principle by
{\em Fermi} up to $z\ga 7$ and by the Large Millimeter Telescope (LMT~\cite{gtm})
at $z\ga 30$, if such an object could have existed at such an early phase of the Universe.
The combination of {\em Fermi} and LMT data promises the availability of a sample of
thousands of objects up to the highest redshifts ideally suited to study radio loud AGN
evolution and the connection of the EBL with the star formation history of the Universe.
\begin{table}[b]
\begin{center}
\caption{{\em WMAP} and EGRET source matches by type.\\
($\dagger$ = excludes P, G and S types~\cite{3egcat}).\label{g2wmask}}
\begin{tabular}{|c|c|c|c|c|}
\hline\hline \textbf{Source} & \textbf{3EG} & \textbf{{\em WMAP}} &
\multicolumn{2}{c|}{\textbf{Matches}} \\
\textbf{Type} & \textbf{catalog} & \textbf{mask} & \textbf{Observed} &
\textbf{Expected} \\
\hline A & 67 & 54 & 40 & 6.6 \\
\hline a & 28 & 23 & 17 & 3.3 \\
\hline U & 170 & 61 & 11 & 10.0 \\
\hline Total & 265$^{\dagger}$ & 138 & 68 & 20 \\
\hline\hline
\end{tabular} \end{center} \end{table}
\begin{table*}[t]
\begin{center}
\caption{{em WMAP} sources in the 3EG catalog and 0FGL bright source list -- Part I.\\
$D$ denotes the integer part of the distance in standard deviations.\label{muestra1}}
\begin{tabular}{|l|c|c|c|c|l|c|c|c|c|}
\hline \textbf{WMAP} & \textbf{3EG / 0FGL} & \textbf{$D$} & \textbf{Counterpart} &&
\textbf{WMAP} & \textbf{3EG / 0FGL} & \textbf{$D$} & \textbf{Counterpart} \\
\hline
\hline J0050--0649 & 0FGL J0051.1--0647 & [0] & PKS 0048--071 & ~ &
J0334--4007 & 0FGL J0334.1--4006 & [0] & PKS 0332--403 \\
\hline J0051--0927 & 0FGL J0050.5--0928 & [0] & PKS 0048--097 &&
J0339--0143 &3EG J0340--0201& [0] & CTA 026 \\
\hline J0108+0135 & 3EG J0118+0248 & [2] & 4C 01.02 &&
J0407--3825 &0FGL J0407.6--3829& [0] & PKS 0405--385 \\
\hline J0132--1653 & 3EG J0130--1758 &[1] & QSO B0130--171 &&
J0416--2051 &3EG J0412--1853 & [1] & (QSO B0413--21)?? \\
\hline J0137+4753 & 0FGL J0137.1+4751 & [0] & QSO B0133+47 &&
J0423--0120 & 0FGL J0423.1--0112 & [0] & QSO~B0420--015 \\
&&&&& & 3EG J0422--0102 & [0] & \\
\hline J0204+1513 & 3EG J0204+1458 & [0] & 4C +15.05 &&
J0428--3757 & 0FGL J0428.7--3755 & [0] & PKS 0426--380 \\
\hline J0205--1704 & 0FGL J0204.8--1704 & [0] & PKS 0202--17 &&
J0442--0017 & 3EG J0442--0033& [0] & QSO B0440--004 \\
\hline J0210--5100 & 0FGL J0210.8--5100 & [0] & QSO B0208--5115 &&
J0455--4617 & 3EG J0458--4635& [0] & 0454--463 \\
& 3EG J0210--5055 & [0] & && &&& \\
\hline J0218+0138 & 0FGL J0217.8+0146 & [1] & PKS 0215+015 &&
J0456--2322 & 0FGL J0457.1--2325 & [0] & QSO B0454--234 \\
&&&&&& 3EG J0456--2338 & [0] & \\
\hline J0220+3558 & 0FGL J0220.9+3607 & [0] & B2 0218+35 &&
J0501--0159 & 3EG J0500--0159 & [0] & 4C --02.19 \\
\hline J0223+4303 & 0FGL J0222.6+4302 & [1] & 3C 66A &&
J0506--6108 & 3EG J0512--6150 & [1] & 0506--612? \\
& 3EG J0222+4253 & [0] & && &&& \\
\hline J0237+2848 & 0FGL J0238.4+2855 & [0] & 4C +28.07 &&
J0523--3627 & 3EG J0530--3626 & [2] & QSO J0522--3627 \\
& 3EG J0239+2815 & [1] & && &&& \\
\hline J0238+1637 & 0FGL J0238.6+1636 & [0] & AO 0235+164 &&
J0538--4405 & 0FGL J0538.8--4403 & [0] & PKS 0537--441 \\
& 3EG J0237+1635 & [1] & QSO B0235+16 &&
& 3EG J0540--4402 & [0] & \\
\hline J0319+4131 & 0FGL J0320.0+4131 & [0] & NGC 1275 &&
J0539--2844 & 3EG J0531--2940 & [2] & (QSO~J0539--2839)?? \\
\hline\hline
\end{tabular}\end{center}\end{table*}
\section{Matching {\em WMAP} with $\gamma$-ray sources}
As a by product of the study of the cosmic microwave background, the {\em Wilkinson
Microwave Anisotropy Probe (WMAP)} produced a catalog of 390 bright sources detected
at frequencies between 23 and 94 GHz, outside a mask defined by the mm-wave emission
from the Galactic plane~\cite{wmap-cat}. These objects constitute a fairly homogeneous
sample suitable for comparison with all-sky $\gamma$-ray catalogs.
We compared the {\em WMAP} foreground source catalog~\cite{wmap-cat} with:
(1) the Third EGRET (3EG) catalog of high-energy $\gamma$-ray sources~\cite{3egcat};
(2) the $E>10~\rm GeV$ EGRET photons compiled by~\cite{egret-vhe}; (3) the list of bright
sources detected with signal-to-noise ($\sqrt{TS}$) larger than 10 by the {\em Fermi}
$\gamma$-ray Space Telescope in its first three months of operations~\cite{fermi-0fgl}.
The positional uncertainty of {\em WMAP} equals $4^\prime$, similar to that of {\em Fermi},
but much better than that of EGRET. Preferring to accept false positives than to reject
real coincidences, we used a relaxed matching criterion of 2.5 times the combined positional
accuracy, $(\sigma_{wmap}^{2}+\sigma_{\gamma}^{2})^{1/2}$, where $\sigma_{\gamma}$
refers to 3EG ($\sim 1^{\circ}$), EGRET-VHE ($0.5^\circ$) or 0FGL ($2-10^{\prime}$).
For each comparison we estimated the expected number of random coincidences quantifying
the solid angle covered by the $\gamma$-ray sources (or events) outside the {\em WMAP}
Galactic mask. For each {\em WMAP}/$\gamma$ pair we blindly listed potential radio, optical
and X-ray counterparts from SIMBAD / Vizier.
The results are listed in tables~\ref{muestra1} and~\ref{muestra4}.
\subsection{Matching {\em WMAP} with EGRET}
The comparison between the {\em WMAP} and 3EG catalogs produced 69 matches out of
the 390 {\em WMAP} and 138 EGRET sources outside the {\em WMAP} Galactic mask.
The number expected randomly is 20; the probability of having 69 matches among the
390 {\em WMAP} sources is $P\la 10^{-17}$, indicating that most -but not all- the matches
are real.
When accounting for the source type we count 40 matches out of the 54 high confidence
blazar association, labelled A in the 3EG catalog, compared to 6.6 expected randomly.
The low confidence "a" associations have 17 matches out of 23 tries and 3.3 expected
chance coincidences. On the other hand we have 11 matches among the 61 unidentified
sources out of the {\em WMAP} mask, expecting 10.0 by chance. On statistical grounds,
we can confirm the physical association of foreground {\em WMAP} sources with 3EG
blazars, accounted by both "A" and "a" classes, but not with unidentified EGRET sources
(table~\ref{g2wmask}).
We also compared the {\em WMAP} positions with the list of very high energy (VHE)
photons, $E>10\,\rm GeV$~\cite{egret-vhe}: 510 out of the 1506 VHE photons
are outside the {\em WMAP} Galactic mask. The combined positional accuracy of
VHE photons and {\em WMAP} sources is $30.2^\prime$. We obtained coincidences
between 33 VHE photons and 29 {\em WMAP} sources as follows:\\
-- 20 {\em WMAP} sources coincide with a single isolated VHE photon;\\
-- 4 {\em WMAP} sources coincide with a single VHE photon and a 3EG,
with no 0FGL counterpart;\\
-- 1 {\em WMAP} source (WMAP J1408--0749) coincides with a 3EG source
(3EG J1409--0745) and 4 VHE photons (VHE 494, 498, 1058 and 1061) -- but
with no 0FGL counterpart:\\
-- 3 {\em WMAP} sources coincide with VHE photons, 3EG and
0FGL sources; of these WMAP~J0210-5100 has two VHE photons;\\
-- 1 {\em WMAP} source (WMAP~J0137+4753) coincides with a VHE photon
and a 0FGLsource (0FGL~0137.1+4751), with no 3EG counterpart.\\
We note that given the number and error box sizes of the VHE photons, we expect
30 random matches under our 2.5$\sigma$ criterion. Most or all of the single
{\em WMAP}-VHE coincidences are likely to be spurious. We note the case of
WMAP~J0137+4753, which belonged to the {\em WMAP}\& VHE only category prior
to the publication of the 0FGL and turned out to be a real $\gamma$-ray source.
\subsection{Matching the {\em WMAP} catalog with the {\em Fermi} bright source list}
The {\em Fermi} bright source list (0FGL), made public in February 2009, consists
of 205 bright $\gamma$-ray sources detected with significance $\sqrt{TS}> 10$
in the first three months of observations~\cite{fermi-0fgl}. Of these, 121 are of the
AGN class, mostly blazars~\cite{fermi-0fgl-agn}.
The 0FGL list has 122 objects outside the {\em WMAP} Galactic mask, which we
compared with the respective {\em WMAP} and {\em Fermi} positions. The improved
positional accuracy of {\em Fermi}, in the $4^{\prime}-10^{\prime}$ range, results in
only 0.82 spurious coincidences expected. We
found 54 matches between {\em WMAP} and the 0FGL, 25 of which are common with
EGRET sources and 29 are independent.
\subsection{Sample overview}
The sample is presented in tables~\ref{muestra1} and~\ref{muestra4}. The $D$ column
expresses the distance between the {\em WMAP} and the $\gamma$-ray event in terms
of the integer part of the combined positional accuracy; associations with [0] have
intersecting error countours and are more likely to be real than those with [2], separated
by $\geq 2\sigma$. Most of the sources have a suitable radio, optical and/or X-ray counterpart,
often in the intersection of the {\em WMAP}/$\gamma$-ray error boxes. In a few cases the
candidate counterpart is not unique and the one displayed is a subjective election.
Counterparts in parenthesis are tentative.
We note the following:
\begin{itemize}
\item WMAP J0051--0927 :: 0FGL J0050.5--0928 has a preferred association with
the Bl Lac object PKS 0048--071
= PHL 856; the radiogalaxy FIRST J005051.9--092529 is an alternative.
\item WMAP J0237+2848 :: 0FGL J0238.4+2855 :: 3EG J0239+2815.
This is listed as a low confidence {\em Fermi} association with 4C +28.07
in~\cite{fermi-0fgl}. The {\em WMAP} and {\em Fermi} boxes match, both containing
4C +28.07. All are somewhat outside the EGRET box.
\item WMAP~J0319+4131 :: 0FGL J0320.0+4131 is identified with the radiogalaxy
NGC 1275. We note that the radiogalaxy Cen~A is excluded of this study by
the {\em WMAP} Galactic mask.
\item WMAP J0423--0120 :: 0FGL J0423.1--0112 :: 3EG J0422--0102.
This is a low confidence {\em Fermi} association with PKS~0420--014, which is
at the center of the {\em WMAP} circle.
\item WMAP J0909+0119 :: 0FGL J0909.7+0145. The 0FGL error radius is
$17^{\prime}$, the {\em WMAP} and 0FGL positions differing by $25^\prime$.
PKS 0907+022 is a low confidence {\em Fermi} association not compatible with
{\em WMAP}. 4C+01.24 (PKS = QSO B0906+015) is inside the {\em WMAP} box and
marginally compatible with the {\em Fermi} source.
\item WMAP~J1517--2421 has two possible EGRET counterparts but only one
(3EG J1517--2538) is compatible with the 0FGL source.
\item WMAP~J1642+3948 is $19^\prime$ away from 0FGL~J1641.4+3939, which has
an positional error of $9.5^\prime$, making the association tentative. There are suitable
candidates for both sources, with 3C345 being positionally the best {\em common}
counterpart, as discussed in \S\ref{3c345-id}.
While there is no 3EG source association, we note the revised EGRET counterpart
EGR~J1642+3940~\cite{egrcat}.
\end{itemize}
\section{Near infrared monitoring}
On August 2007, we started a dedicated monitoring program
with up to 60 nights per semester awarded on the 2.1m telescope of the Observatorio
Astrof\'{\i}sico Guillermo Haro (OAGH), in Cananea, Sonora, Mexico
(lat=$+31.05$, long=$-110.38$). The program consists of optical photometry
(BRVI), low resolution spectroscopy (4000 - 7500 \AA) and near infrared JHKs imaging
of the sample above, with the addition of high priority {\em GLAST/Fermi} sources and
targets notified via the multi-wavelength {\em Fermi} group. The current study is to lead
to programs of follow-up and identification of $\gamma$-ray sources using near infrared,
optical and mm-wave facilities.
Particularly successful has been the JHKs photometric survey using the CAnanea
Near Infrared CAmera (CANICA). CANICA is equiped with a Rockwell $1024\times
1024$ pixel Hawaii infrared detector working at 75.4~K with standard near infrared filters.
The scale plate is 0.32$^{\prime\prime}$/pixel. Observations are usually carried out in
series of dithered frames in each filter. Datasets are coadded after correcting for bias and
flat-fielding using IRAF based macros. Figure~\ref{fotometria} shows the photometric
magnitudes measured with CANICA, with detection thresholds around magnitudes 19, 18
and 17 for J, H and Ks respectively, i.e. about two magnitudes fainter than 2MASS.
\begin{figure}[t]
\centering\includegraphics[width=\hsize]{acarrami_fig1.eps}
\caption{CANICA photometric measurements of mm-wave/$\gamma$-ray sources.
We show only detections.\label{fotometria}}
\end{figure}
\section{Correlated infrared/$\gamma$-ray activity}
We have found correlated infrared and $\gamma$-ray correlations with little
or no evidence for time delays in our sample. We show here joint CANICA
H band (1.6$\mu\rm m \Rightarrow 0.76 eV$) and {\em Fermi} ($1-300 \,\rm GeV$) light
curves, normalizing the maximum $\gamma$-ray flux to the maximum H-band flux and
setting both zeros at the same level. We are currently studying
the infrared color behavior of our sample during flares and constraining possible
delays in the timing of near infrared flares relative to the $\gamma$-ray ones.
Our current results on 3C 454.3 are presented in~\cite{g2mm-3c454}.
\subsection{PKS 0235+164}
\begin{figure}[t]
\centering\includegraphics[width=\hsize]{acarrami_fig2.eps}
\caption{Joint CANICA - {\em Fermi} light curve of PKS 0235+164. CANICA is given
by the (black) points while {\em Fermi} fluxes form the (red) histogram with error bars.
The discontinuous horizontal line marks the 2MASS flux. The dotted vertical lines
indicate the change of year. \label{h2f-pks0235+164}}
\end{figure}
PKS 0235+164 is positionally coincident with WMAP~J0238+1637, mm-wave
source coincident with 3EG~J0237+1635 = 0FGL~J0238.6+1636. The {\em WMAP}
source is labelled as probable variable. Figure~\ref{h2f-pks0235+164} shows the
joint H band and 1-300~GeV fluxes scaled.
The near infrared flux at the end of 2008 and beginning of 2009 matched the
2MASS values. On JD2454715 we found PKS~0235+164 almost three magnitudes
brighter, prior to peaking 25 days later - in coincidence with the {\em Fermi} flare.
\subsection{PKS 0454--234}
The {\em WMAP} and {\em Fermi} error circles in this region of the sky intersect close to
the center of the 3EG~J0456--2338 error box. PKS 0454--234 is within the intersection,
in a neat positional match. We started monitoring on JD 2454748, when the near
infrared flux was 0.85 magnitudes above the 2MASS reference value. The source
flared by a factor of nearly 3 in the next 50 days to drop to a half 18 days later.
The near infrared peak seems delayed by about a week relative to the 1-300 GeV
maximum. The AGN has been in relative quiescence since (fig.~\ref{h2f-pks0454-234}).
\begin{figure}[t]
\centering\includegraphics[width=\hsize]{acarrami_fig3.eps}
\caption{Joint CANICA - {\em Fermi} light curve of PKS 0454--234. The discontinuous
horizontal line marks the 2MASS flux. The dotted vertical lines indicate the change
of year.\label{h2f-pks0454-234}}
\end{figure}
\subsection{PKS 1510--089}
PKS 1510--89 is the high confidence counterpart of 3EG J1512--0849, WMAP
J1512--0904 and 0FGL~J1512.7--0905, with an excellent positional match
between all data. We started monitoring PKS 1510--89 in early 2008, a few months
prior to the {\em Fermi} launch, when the flux fluctuated around the 2MASS
reference value. We caught a simultaneous infrared - $\gamma$-ray flare
on JD 2454924 and the subsequent decay a month later.
\begin{figure}[t]
\centering\includegraphics[width=\hsize]{acarrami_fig4.eps}
\caption{CANICA - {\em Fermi} light curve of PKS 1510--089. The discontinuous
horizontal line marks the 2MASS flux. The dotted vertical lines indicate the change
of year.\label{h2f-pks1510-089}}
\end{figure}
\section{Source identifications}
\subsection{The identification of QSO B0133+476}
QSO B0133+476 is a rather active quasar first catalogued originally as DA55
in the Dominion DA 1420~MHz survey~\cite{da1968}. Also known as Mis V1436,
archival optical data exists since at least 1953, when it was around magnitude R=18.7.
This object has been monitored by other groups since 2007, when it was found to be
4.5 magnitudes brighter. Its optical variability is well documented in~\cite{misao}.
This object has no EGRET counterpart, the only evidence for $\gamma$-ray emission
prior to the {\em Fermi} launch being its marginal closeness to a $85(\pm 38)\rm GeV$ photon.
The unprovable association of this single photon with QSO B0133+476 is of interest
as its energy is close to the pair absorption EBL limit for the redshift of the object,
$z=0.859$. If true, this association would indicate the potential of this source for testing
the EBL horizon using {\em Fermi}, specially under phases of high emission activity.
Our report of near infrared flaring from data taken between JD 2454788 and
2454795~\cite{qso0133-nir} was followed by the {\em Fermi} detection report, at a flux
level above the EGRET limiting sensitivity~\cite{qso0133-gam}. The infrared light curve
is shown in figure~\ref{go-qso0133+476}. The object was caught undergoing a rapid flare
which has declined relatively slowly during 2009, but always at levels above 2MASS.
\begin{figure}[t]
\centering\includegraphics[width=0.49\hsize]{acarrami_fig5a.eps}
\includegraphics[width=0.49\hsize]{acarrami_fig5b.eps}
\caption{Positional circumstance of QSO B0133+476. {\em Left:} the VHE photon error
circle is shown at $\alpha\sim 23^{\circ}$, $\delta\sim 48.6^{\circ}$, with the almost concentric
{\em WMAP} and {\em Fermi} circles in the middle. {\em Right:} a zoom panel showing
the $4^\prime$ {\em WMAP} circle contained in the $7^\prime$ {\em Fermi} position.
QSOB0133+476 is indicated by the open hexagon and an asterix.\label{j0137}}
\end{figure}
\begin{figure}[t]
\centering\includegraphics[width=\hsize]{acarrami_fig6.eps}
\caption{CANICA light curve of QSO B0133+47. The flaring at the end of 2008 is
clearly visible, at a factor of 10 above the 2MASS flux levels indicated by the dashed
horizontal line.\label{go-qso0133+476}}
\end{figure}
\subsection{QSO J0808--0751}
The EGRET source 3EG J0812--0646 was not in the high priority {\em GLAST/Fermi}
list, with the association with QSO~J0808--0751 been only tentative ($D=2$).
WMAP~J0808--0750 is somewhat displaced from the 3EG location, but contains
this $z=1.84$ QSO. The rapid flare observed simultaneously in the near infrared and
by {\em Fermi} of this source, not in the 0FGL, confirms the identity of the QSO as the
$\gamma$-ray source. We missed a second flare occurring while the source was on the
daylight. The flux increase in the near infrared reached a factor of ten within 50 days.
\begin{figure}[t]
\centering\includegraphics[width=0.98\hsize]{acarrami_fig7.eps}
\caption{CANICA - {\em Fermi} QSO J0808--0751 light curve. The dots indicate the H
band fluxes in mJy, while the histogram and upper limit arrows indicate the
1-300~GeV fluxes and limits in scaled units. The discontinuous horizontal line marks
the 2MASS flux. The dotted vertical lines indicate the change of year.
\label{h2f-qsoj0808-0751}}
\end{figure}
\subsection{The identification of 0FGL~J1641.4+3939 with 3C 345 \label{3c345-id}}
The angular distance between WMAP~J1642+3948 and 0FGL~J1641.4+3939 is
$19.2^\prime = 1.9\sigma$ times the combined positional uncertainty, the
association between both objects being tentative only. This is a rather populated
region of the sky, where several potential counterparts can be found for each object:\\
- candidate counertparts of 0FGL~J1641.4+3939 include 3C 345, QSO B1641.5+3956,
QSO B1641.6+3949, QSO B1640+398, QSO B1641.8+3956, QSO B1641+3958,
QSO B1640.5+3944, QSO B1640+396, three more QSOs, a Seyfert 1 and a radiogalaxy
from SDSS. \\
- potential counterparts for WMAP~J1642+3948 are GB6 J1642+3948 = 3C 345,
FIRST J164304.3+394836, QSO B1641+3949, ...\\
3C~345 has a lower $\chi^{2}$ relative to the combined {\em WMAP} and {\em Fermi}
positions (fig.~\ref{h2f-3c345}), being the best candidate under the assumption of a common
association. Even though more data are desirable, the simultaneous flux changes measured
provide evidence for the physical association between 0FGL~1641.4+3939 and 3C345.
\section{Summary}
We have selected a sample of mm-wave/$\gamma$-ray bright blazars from the
{\em WMAP}, 3EG catalogs and 0FGL list. We have monitored these in the near
infrared finding correlated variability, which has also allowed the identification or
confirmation of some of these objects.
\begin{figure}[t]
\vspace*{1mm}
\centering\includegraphics[width=0.49\hsize]{acarrami_fig8a.eps}
\includegraphics[width=0.49\hsize]{acarrami_fig8b.eps}
\caption{{\em Left:} position of 0FGL~J1641.4+3939 (larger circle) and WMAP~J1642+3948
(smaller circle) with a few potential counterparts (AGN class - open hexagons).
{\em Right:} CANICA H band light curve of 3C345 compared with the (1-300 GeV) fluxes
from 0FGL~J1641.4+3939 by {\em Fermi}. The discontinuous horizontal lines mark the 2MASS
flux and error. The dotted vertical line indicates the change of year.\label{h2f-3c345}}
\end{figure}
\bigskip
\begin{acknowledgments}
We acknowledge the use of 2MASS, WMAP, Fermi, SIMBAD, Vizier and POSS
databases. We appreciate the support of the technical staff at the Observatorio
Astrof\'{\i}sico Guillermo Haro.
\end{acknowledgments}
\bigskip
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,773 |
module MapCreator {
export interface ILayerMap {
Layer: number[][];
Name: string;
}
export interface ISettings {
Rows: number;
Columns: number;
TileSize: eg.Size2d;
SpriteSheetUrl: string;
Layers: ILayerMap[];
}
export class PersistenceManager {
constructor(utilities: JQuery, private _getLayers: () => ILayer[], private _spriteSheetUrl: string, private _buildResourceMap: (grid: eg.Graphics.Grid) => number[][]) {
var saveButton = $("#save"),
saveName = $("#saveName"),
saveDialog = $("#saveDialog"),
layerMaps: ILayerMap[] = [],
that = this;
saveButton.click(() => {
this.SaveMap(saveName.val());
saveDialog.html("Last saved at: " + new Date().toTimeString());
});
}
public BuildResourceMaps(): ILayerMap[] {
var layerMaps: ILayerMap[] = [],
layers = this._getLayers();
for (var i = 0; i < layers.length; i++) {
layerMaps.push({
Layer: this._buildResourceMap(layers[i].Layer),
Name: layers[i].Name
});
}
return layerMaps;
}
public SaveMap(saveName: string, namespace: string = "mapBuilder"): void {
var layerMaps: ILayerMap[] = this.BuildResourceMaps(),
layers = this._getLayers();
var settings: ISettings = {
Rows: layers[0].Layer.Rows,
Columns: layers[0].Layer.Columns,
TileSize: layers[0].Layer.TileSize,
SpriteSheetUrl: this._spriteSheetUrl,
Layers: layerMaps
},
maps = localStorage.getItem(namespace);
if (!maps) {
maps = {}
}
else {
maps = JSON.parse(maps);
}
maps[saveName] = settings;
localStorage.setItem(namespace, JSON.stringify(maps));
}
public static GetSavedMaps(namespace: string = "mapBuilder"): { [name: string]: ISettings; } {
var strMaps = localStorage.getItem(namespace),
maps: { [name: string]: ISettings; } = JSON.parse(strMaps);
return maps;
}
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 4,294 |
Q: How to Change the Database Source from within an application I'm developing an app on xammp that allows you add multiple schools, and with each new school added, a new database is added. All that works fine.
The issue is in switching from database name to another database name in my db config. I have a form action "set_source" that triggers the setDBNAME action but get an uncaught error.
I tried the following code
class dbConfig
{
public $hostname = 'localhost'; //hostname for db connection
public $username = 'root'; // username for db connection
public $password = 'password'; //password for db conection
public $dbname = 'school1'; //this must be dynamically changeable
public $mysqli;
function setDBNAME($id)
{
$sql = "SELECT `school_name`, `subdomain`, `selected`, `url` FROM board.school WHERE school_id = $id AND selected = 'True'";
$data = $this->mysqli->execute_query($sql); //line triggering error
if($data) {
return new dbConfig(localhost, root, password, $data['url']);
}
else {
return null;
}
}
}
But Got the following error
Fatal error: Uncaught Error: Call to a member function execute_query()
on null in path\to\file\dbConfig.php:15 Stack trace: #0 path\to
\file\myaccountController.php(50): dbConfig->setDBNAME('1') #1
path\to\file\Controller11.php(29): myaccountController->set_source()
#2 path\to\file\myaccount.php(20): Controller11->run() #3 {main}
thrown in path\to\file\dbConfig.php on line 15
Any pointers and or solutions will be muchos gracias
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,470 |
Q: on click event on JQuery chosen plugin I´m using JQuery chosen, and I read the API and I see that only this events are publish
change: Chosen triggers the standard DOM
chosen:ready Triggered after Chosen has been fully instantiated.
chosen:maxselected Triggered if max_selected_options is set and that total is broken.
chosen:showing_dropdown Triggered when Chosen's dropdown is opened.
chosen:hiding_dropdown Triggered when Chosen's dropdown is closed.
chosen:no_results Triggered when a search returns no matching results.
But in my case what I want is an event when I click in one of the elements of my dropdown, even if the value has not change.
Any suggestions?
A: Yes, you can! Check the jsFiddle: https://jsfiddle.net/qjjcx2bj/3/
By getting the index of the selected list item, you can retrieve the value of the selected item.
var ind = $(this).attr('data-option-array-index');
$("#select1").children('option:eq('+ind+')').val();
Hope this helps.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,502 |
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