text:
\section{Introduction}

The D1D5 system has been a very rich area for exploring black hole
physics: one can account for the black hole entropy by counting states
of the dual CFT~\cite{sen1, sen2, stromvafa}; one can compute the
spectrum and rate of absorption and emission from the gravitational
description and match it to a CFT description~\cite{radiation-1,
  radiation-2, radiation-3, radiation-4, radiation-5, callan96,
  emparan,cv,cm1,cm2,cm3,acm1,ac1}; and one can create and study
explicit microstate geometries (see~\cite{fuzzballs2-1, fuzzballs2-2,
  fuzzballs2-3, fuzzballs2-4, BalaDeBoer} for reviews of older work,
and~\cite{bena10, bena-bolt, bena-branes, deBoer10} for some recent
developments).

There is believed to be an orbifold point in the D1D5 moduli space
where a free two-dimensional CFT with orbifolded target space becomes
a good description~\cite{swD1D5, fmD1D5, deBoerD1D5, dijkgraafD1D5,
  frolov-1, frolov-2, Jevicki}.  Because calculations are easier at
the orbifold point, all of the above mentioned CFT calculations are
done there. At the orbifold point, however, the gravity approximation
is not valid. While the entropy-counting calculations are protected on
account of supersymmetry, the agreement of so many absorption and
emission calculations at the orbifold point with supergravity
calculations at a different point in moduli space strongly suggests
some kind of ``non-renormalization'' theorem~\cite{dmw02}.

To understand more complicated black hole physics, it becomes
necessary to deform the CFT off of the orbifold point by blowing up
the fixed points of the orbifold. We do this by introducing four
marginal deformation operators.\footnote{The four deformation
  operators form a 2-index tensor of $SU(2)_2$, so we sometimes call
  this object ``the deformation operator.''} While the supergravity
point in moduli space is far from the orbifold point, we take a
perturbative approach and start by looking at the action of a single
application of the deformation operator. We hope that this approach
will yield insight into black hole physics. To this end, \cite{acm2}
computes the effect of the deformation operator on the vacuum,
and~\cite{acm3} shows an algorithm for computing the effect of the
deformation operator on more general states of the CFT.

In this paper, we introduce a new and more direct method for computing
the effect of the deformation operator on general states. This method
may also be of theoretical interest, because it makes the physics of
twist operators closer to that of Bogolyubov transformations. We start
by looking for a Bogolyubov-type relation between operators ``before''
the deformation and ``after'' the deformation. Unfortunately, the
method suffers from multidimensional series whose value depends
sensitively on the way one evaluates them. We resolve these
ambiguities by introducing a physically motivated prescription for
evaluating the infinite series; however, the prescription makes our
relation a weaker statement, since the prescription depends on what
other excitations one has. Thus, we term the connection between
operators before the deformation and after the deformation
``intertwining relations'' instead of Bogolyubov coefficients.

In Section~\ref{sec:CFT}, we briefly introduce the deformation
operator for the D1D5 CFT and review results of~\cite{acm2}. In
particular, the deformation operator contains a 2-twist operator,
whose effect on the vacuum is to produce a ``squeezed state.'' The
squeezed state, written as an exponential of pairs of bosonic and
fermionic modes acting on the twisted vacuum, is reminiscent of the
state one generically gets from a Bogolyubov transformation of the
vacuum. The twist operator is the only difficult part of working with
the deformation operator, and for this paper we restrict our attention
to it.

In Section~\ref{sec:intertwining}, we make the above analogy more
precise by introducing the analogue of the Bogolyubov coefficients,
that is we show how the 2-twist operator maps modes of the untwisted
sector into twisted sector modes. We first give a simple, formal
derivation of these intertwining coefficients. Then we give an example
of the order-of-summation issue that results from using the
coefficients. We can understand the problem by making a more rigorous
derivation of the intertwining relations, which allows us to give a
prescription that gives the correct answer. In~\ref{sec:prescription}
we concisely state the prescription.

In Section~\ref{sec:example}, we demonstrate the prescription and the
validity of the more rigorous derivation by computing a particular
state in two distinct ways and showing agreement.

Finally, we recapitulate the main points and discuss future directions
in the conclusion. Our notation and conventions for the CFT are
outlined in Appendix~\ref{ap:CFT-notation}. Appendix~\ref{ap:series}
contains a list of series that we use. Some of them we could not prove
directly, but we are confident are correct after numerical
comparisons.

\section{The Deformation Operator}\label{sec:CFT}

Recall that the D1D5 system is IIB string theory compactified
on\footnote{One could consider also K3 instead of $T^4$; however, for
  simplicity we consider only $T^4$.}
\begin{equation}
M_{9,1}\rightarrow M_{4,1}\times S^1\times T^4,
\end{equation}
with a bound state of $N_1$ D1 branes wrapping $S^1$ and $N_5$ D5
branes wrapping $T^4\times S^1$. For $S^1$ large compared to the
$T^4$, the low energy description of the resulting bound state is a
1+1-dimensional CFT. The CFT has $SU(2)_L\times SU(2)_R$ R-symmetry
that corresponds to the isometry of the near-horizon $S^3$.
Additionally, we find it convenient to label fields with the $SO(4)_I
\simeq SU(2)_1\times SU(2)_2$ symmetry corresponding to the torus
directions~\cite{Jevicki}. This symmetry is broken by the torus, but is still useful
to organize the fields.

It has been conjectured that we can move to a point in the moduli
space of the D1D5 system called the ``orbifold point,'' where the CFT
is a 1+1-dimensional sigma model with target space
$(T^4)^{N_1N_5}/S_{N_1N_5}$~\cite{swD1D5, fmD1D5,deBoerD1D5,
  dijkgraafD1D5,frolov-1,frolov-2, Jevicki}. Each of the $N_1N_5$
copies consist of 4 real bosonic fields $X$ and four real fermionic
fields $\psi$, see Appendix~\ref{ap:CFT-notation} and~\cite{acm2,
  acm3} for more details.  We parameterize the base space with one
complex coordinate $z$, using radial quantization.

The orbifold theory has different twist sectors, which correspond to
circling $S^1$ and coming back only up an element of the orbifold
group, $S_{N_1N_5}$. The twist operators $\sigma_n$ map between
different twist sectors. We do not need most of the technical details
of the twist operators in this paper, but we do use the fact that one
can map to a covering space which looks like the untwisted sector of
the theory via a meromorphic mapping $z=z(t)$~\cite{lm1}.

For this paper, we only need the chiral primary 2-twist operator
$\sigma_2^{++}(z)$, where the plusses indicate that it is the top
member of both a $SU(2)_L$ doublet and a $SU(2)_R$ doublet. If the
only twist operators are a $\sigma_2^{++}(z_0)$ and another 2-twist at
$z=\infty$, then
\begin{equation}
z = z_0 + t^2
\end{equation}
maps to a single-valued covering space parameterized by the complex
coordinate $t$. This map removes the twisting, but the $SU(2)_L\times
SU(2)_R$ charge is preserved by inserting spin fields,
{$S^+(t=0)\bar{S}^+(\bar{t}=0)$}, which also ensure the correct
fermion boundary conditions~\cite{lm2}.

The D1D5 CFT has 20 exactly marginal deformation
operators~\cite{dmw99, dmw02,gomis,gava} that correspond to different
directions in moduli space.  Four of those operators are the blow-up
modes of the orbifold. These are the deformation operators that move
us to the supergravity point in the moduli space and hence the ones
which we are most interested in.

\subsection{The operator}

The deformation operator is a singlet under $SU(2)_L\times SU(2)_R$.
To obtain such a singlet we apply modes of $G^\mp_{\dot A}$ to
$\sigma_2^\pm$. In~\cite{acm2} it was shown that we can write the
deformation operator(s) as
\begin{equation}\label{eq:def-op}
\widehat O_{\dot A\dot B}(w_0)=
  \Big[\int_{w_0} \frac{\drm w}{2\pi i} G^-_{\dot A} (w)\Big]
  \Big[\int_{\bar w_0} \frac{\drm \bar{w}}{2\pi i} \bar G^-_{\dot B} (\bar w)\Big]
   \sigma_2^{++}(w_0)
\end{equation}
The left and right movers separate out completely for all the
computations that we perform. Thus from now on we work with the left
movers only; in particular, we write the twist operator only with its
left spin: $\sigma_2^+$.

The operator $\sigma^+_2$ is normalized to have a unit OPE with its
conjugate
\begin{equation}
\sigma_{2,+}(z')\sigma_2^{+}(z)\sim {1\over (z'-z)}.
\end{equation}
This implies that acting on the Ramond vacuum~\cite{acm2}
\begin{equation}
\sigma_2^+(z)|0_R^-\rangle^{(1)} |0_R^-\rangle^{(2)}=|0_R^-\rangle+\bigO(z).
\label{qwthree}
\end{equation}
Here $|0_R^-\rangle$ is the spin down Ramond vacuum of the CFT on the
doubly wound circle produced after the twist. The normalization
\eqref{qwthree} has given us the coefficient unity for the first term
on the RHS, and the $\bigO(z)$ represent excited states of the
CFT on the doubly wound circle.

In this paper, we focus on the effect of the twist operator since the
effect of the susy-current, $G$, is easy to compute: one can break the
$G$-contours in Equation~\eqref{eq:def-op} into contours acting
before the twist operator and contours acting after the twist
operator~\cite{acm2, acm3}.

\subsection{The action of the twist operator on the vacuum}

Since we are mainly interested in better understanding the twist
operator, it is useful to define
\begin{equation}
\ket{\chi} = \sigma_2^+(z_0)\ket{0_R^-}^{(1)}\ket{0_R^{-}}^{(2)},
\end{equation}
and recall that in~\cite{acm2} it was found that the twist operator
acting on the Ramond vacuum may be written as the squeezed state
\begin{equation}
\ket{\chi} = \exp\left[
  -\frac{1}{2}\sum_{m, n}\gamma^B_{mn}\epsilon^{AB}\epsilon^{\dot{A}\dot{B}}
                                             \alpha_{A\dot{A},-m}\alpha_{B\dot{B},-n}
+\sum_{m,n}\gamma^F_{mn}\epsilon_{AB}\psi^{+A}_{-m}\psi^{-B}_{-n}\right]\ket{0^-_R}
\label{pfive}
\end{equation}
where
\begin{subequations}
\begin{align}\label{gammaB}
\gamma^B_{mn} &= \begin{cases}
\frac{4 z_0^{\frac{m+n}{2}}}{mn(m+n)\pi}\frac{\Gamma(\frac{m}{2}+1)\Gamma(\frac{n}{2}+1)}{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})} & m,n\,\text{odd, positive}\\
0 & \text{otherwise}\end{cases}
\\
\gamma^F_{mn} &= \begin{cases}
-\frac{2z_0^\frac{m+n}{2}}{n(m+n)\pi}\frac{\Gamma(\frac{m}{2}+1)\Gamma(\frac{n}{2}+1)}{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}
 & m, n\,\text{odd, positive}\\
0 & \text{otherwise}.\end{cases}
\end{align}
\end{subequations}

\section{Intertwining Relations}\label{sec:intertwining}

We first present a formal, intuitive derivation of the intertwining
relations, which relate modes of the untwisted sector to modes of the
twisted sector. In Section~\ref{sec:subtleties}, we show how the
derived relations lead to multi-dimensional infinite series whose
value depends on how one takes the limit of the partial sums going to
infinity. We then give a more rigorous derivation of the relations,
which suggests a prescription on how to evaluate the ambiguous series.

\subsection{Basic Derivation}

We are interested in finding a relationship between modes in the
untwisted sector, ``before the twist,'' and modes in the twisted
sector, ``after the twist.'' More precisely we would like to know in
general how to write
\begin{calc}
\sigma^+_2(z_0) \big(\text{excitations before the twist}\big)\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
  &= \big(\text{excitations after the twist}\big)\sigma^+_2(z_0)\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}\\
  &= \big(\text{excitations after the twist}\big)\ket{\chi}.
\end{calc}
In~\cite{acm3}, an algorithm was found for doing just that; however,
the question this paper addresses is whether there is a more general
relation between individual modes before the twist operator and after
the twist operator.

Since our argument does not depend on $SU(2)_L$ or
$SU(2)_1\times SU(2)_2$, we suppress the indices on the bosons and
fermions. We begin by noting that before the twist the correct field
expansions are
\begin{equation}\label{eq:bosons-before}
i\partial X^{(1)}(z) = \sum_n \frac{\alpha^{(1)}_n}{z^{n+1}}\qquad
i\partial X^{(2)}(z) = \sum_n \frac{\alpha^{(2)}_n}{z^{n+1}}\qquad
|z|<|z_0|,
\end{equation}
and
\begin{equation}\label{eq:fermions-before}
\psi^{(1)}(z) = \sum_{n\in\mathbb{Z} + \frac{1}{2}} \frac{\psi^{(1)}_n}{z^{n+\frac{1}{2}}}\qquad
\psi^{(2)}(z) = \sum_{n\in\mathbb{Z} + \frac{1}{2}} \frac{\psi^{(2)}_n}{z^{n+\frac{1}{2}}}\qquad
|z|<|z_0|.
\end{equation}
After the twist, the correct expansions are given by
\begin{equation}\label{eq:bosons-after}
i\partial X^{(1)}(z) = \frac{1}{2}\sum_n \frac{\alpha_n}{z^{\frac{n}{2}+1}}\qquad
i\partial X^{(2)}(z) = \frac{1}{2}\sum_n \frac{(-1)^n\alpha_n}{z^{\frac{n}{2}+1}}\qquad
|z|>|z_0|,
\end{equation}
and
\begin{equation}\label{eq:fermions-after}
\psi^{(1)}(z) = \frac{1}{2}\sum_n\frac{\psi_n}{z^{\frac{n}{2}+\frac{1}{2}}}\qquad
\psi^{(2)}(z) = \frac{1}{2}\sum_n\frac{(-1)^n\psi_n}{z^{\frac{n}{2}+\frac{1}{2}}}\qquad
|z|>|z_0|.
\end{equation}
Note that after the twist, there is no unique way of distinguishing
copies $(1)$ and $(2)$, but the above expansions correspond to
\emph{a} way of defining $(1)$ and $(2)$. Following~\cite{acm2,acm3},
we define the modes in the twisted sector with an extra factor of 2 so
that we can work with integers.

Now, recall that the fields $\partial X(z)$ and $\psi(z)$ should be
holomorphic functions except at isolated points where there are other
operator insertions. In particular, there is nothing special that
occurs on the circle $|z|=|z_0|$ (there \emph{is} something special at
the isolated point $z_0$). The curve is the boundary between our
twisted and untwisted mode expansions about the origin $z=0$, but if
one were to do mode expansions about a different point in the complex
plane then the expansions would change across a different curve (that
still passes through $z_0$).  Therefore, \emph{at least away from the
  twist operator at $z_0$}, we expect that the fields $\partial X(z)$ and
$\psi(z)$ should be continuous across $|z|=|z_0|$.

Thus, on the circle $|z|=|z_0|$ ({\em  excluding some neighborhood around
$z=z_0$}) we may identify
\begin{equation}\label{eq:boson-exp-equality}
\sum_n \frac{\alpha^{(1)}_n}{z^{n+1}} = \frac{1}{2}\sum_n \frac{\alpha_n}{z^{\frac{n}{2}+1}}
\qquad
\sum_n \frac{\alpha^{(2)}_n}{z^{n+1}} = \frac{1}{2}\sum_n \frac{(-1)^n\alpha_n}{z^{\frac{n}{2}+1}},
\qquad
|z|=|z_0|
\end{equation}
and
\begin{equation}\label{eq:fermion-exp-equality}
\sum_{n\in\mathbb{Z} + \frac{1}{2}} \frac{\psi^{(1)}_n}{z^{n+\frac{1}{2}}}
 = \frac{1}{2}\sum_n\frac{\psi_n}{z^{\frac{n}{2}+\frac{1}{2}}}
\qquad
\sum_{n\in\mathbb{Z} + \frac{1}{2}} \frac{\psi^{(2)}_n}{z^{n+\frac{1}{2}}}
  = \frac{1}{2}\sum_n\frac{(-1)^n\psi_n}{z^{\frac{n}{2}+\frac{1}{2}}},
\qquad
|z|=|z_0|.
\end{equation}
Multiplying \eqref{eq:boson-exp-equality} by $z^m$ and integrating
along the circle $|z|=|z_0|$, we
get
\begin{calc}\label{eq:boson-coef}
\alpha_m^{(1)} &= \frac{1}{2}\sum_n\alpha_n \int\frac{dz}{2\pi i} z^{m-\frac{n}{2}-1}\\
  &= \frac{1}{2}\sum_n\alpha_n \int_0^{2\pi}\frac{d\theta}{2\pi} (z_0e^{i\theta})^{m-\frac{n}{2}}\\
  &= \frac{1}{2}\alpha_{2m} + \frac{i}{2\pi}
     \sum_{n\,\text{odd}}\frac{z_0^{m-\frac{n}{2}}}{m-\frac{n}{2}}\alpha_n,
\end{calc}
and similarly
\begin{equation}\label{eq:boson-coef-2}
\alpha_m^{(2)} = \frac{1}{2}\alpha_{2m}
             - \frac{i}{2\pi}\sum_{n\,\text{odd}}\frac{z_0^{m-\frac{n}{2}}}{m-\frac{n}{2}}\alpha_n,
\end{equation}
where the sum over the odds is both positive and negative.  These are
the desired relations between modes before the twist and modes after
the twist. The relations are analogous to the more general Bogolyubov
transformations discussed in condensed matter in~\cite{deGennes}. Note
that the contour in~\ref{eq:boson-coef} is open, since we must exclude
some infinitesimal neighborhood around $z_0$. It is straightforward to
find the analogous relation for fermions in the R sector:
\begin{equation}\label{eq:fermion-coef}
\psi^{(1,2)}_n =
\frac{1}{2}\psi_{2n}
  \pm \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\psi_k.
\end{equation}
Given the delicate nature of the above argument, in particular with
regard to what is happening around $z_0$, one may not be surprised
that there are some hidden subtleties with these relations. Note also
that the derivation would seem to work for any holomorphic field
$\mathcal{O}(z)$, which \emph{cannot be correct} since some modes have
a nontrivial commutator with the twist operator, e.g. $J_0^-$. We should be
careful in what we mean when we write ``$=$'' in the above
expressions.  For example in Equation~\ref{eq:boson-coef}, we
implicitly mean the operator relation
\begin{equation}
\sigma_2^+(z_0)\alpha_m^{(1)} = \left[\frac{1}{2}\alpha_{2m} + \frac{i}{2\pi}
     \sum_{n\,\text{odd}}\frac{z_0^{m-\frac{n}{2}}}{m-\frac{n}{2}}\alpha_n\right]\sigma_2^+(z_0)
\end{equation}
with the above radial ordering. The usage should be clear from the
context.

Before showing what goes wrong, let us first explore what goes right.
First of all this is the kind of relation we were hoping for: it
relates positive and negative modes before the twist to positive and
negative modes after the twist directly. The method given
in~\cite{acm3} can only relate states to states; it cannot relate an
individual mode before the twist to modes after the twist without
knowing what other excitations one has before the twist.

An important requirement for Bogolyubov coefficients is that they
respect the commutation relations. We can check that the above
relations are consistent with the commutation relations by computing,
for instance,
\begin{calc}\label{eq:comm-check}
\com{\alpha_m^{(1)}}{\alpha_n^{(1)}} &=
\com{\frac{1}{2}\alpha_{2m}
   + \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{m-\frac{k}{2}}}{m-\frac{k}{2}}\alpha_k}
    {\frac{1}{2}\alpha_{2n}
   + \frac{i}{2\pi}\sum_{l\,\text{odd}}\frac{z_0^{n-\frac{l}{2}}}{n-\frac{l}{2}}\alpha_l}\\
 &= \frac{m}{2}\delta_{m+n,0}
   - \frac{z_0^{m+n}}{4\pi^2}\sum_{k\,\text{odd}}\frac{k}{(m-\frac{k}{2})(n+\frac{k}{2})}.
\end{calc}
The sum is divergent; however, if we make the relatively modest
assumption that we should cutoff the sum symmetrically for positive
and negative $k$, then we find\footnote{This mild UV ambiguity might
  be seen as a hint of the other UV issues with the intertwining
  relations; however, this issue does not arise for the fermions, and
  it is of a different character. The UV issues that we discuss at
  length arise with multidimensional series; whereas the above is
  arguably the only reasonable regularization of the series
  in~\ref{eq:comm-check}.  For example, if one cuts off the positive
  modes at $L$ and the negative modes at $-2L$ then one \emph{does}
  get a different answer, but it is not a consistent truncation of the
  Hilbert space to have a creation operator without its corresponding
  annihilation operator.}
\begin{equation}
\lim_{L\to\infty}\sum_{k\,\text{odd}}^{|k|<L}\frac{k}{(m-\frac{k}{2})(n+\frac{k}{2})}
 = - 2m\pi^2\delta_{m+n,0},
\end{equation}
which gives the correct answer. Similar calculations go through for
the other (anti-)commutations.

Second, it gets the right answer for moving a single mode through the
twist operator. For instance,
\begin{calc}
\sigma^+_2(z_0)\alpha^{(1)}_{A\dot{A},n}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
 &= \left(\frac{1}{2}\alpha_{A\dot{A},2n} + \frac{i}{2\pi}
    \sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\alpha_{A\dot{A},k}\right)
    \sigma^+_2(z_0)\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}\\
 &= \left(\frac{1}{2}\alpha_{A\dot{A}, 2n} + \frac{i}{2\pi}
    \sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\alpha_{A\dot{A},k}\right)
    \ket{\chi}\\
 &= \left[\frac{1}{2}\alpha_{A\dot{A}, 2n} + \frac{i}{2\pi}
          \left(\sum_{k\,\text{odd}^+} \frac{z_0^{n+\frac{k}{2}}}{n+\frac{k}{2}}\alpha_{A\dot{A},-k}
      + \sum_{k,l\,\text{odd}^+} \frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}
                              \gamma^B_{kl}k\alpha_{A\dot{A}, -l}\right)\right]\ket{\chi}.\\
\end{calc}
Making use of the identity in~\eqref{eq:id-1} one sees that
\begin{equation}\label{eq:gamma-id}
\sum_{k\,\text{odd}^+} \frac{z_0^{n-\frac{k}{2}}k\gamma^B_{kl}}{n-\frac{k}{2}}
 = \frac{z_0^{n+\frac{l}{2}}}{n+\frac{l}{2}}
      \left(\frac{\Gamma(\frac{l}{2})\Gamma(-n+\frac{1}{2})}{\Gamma(\frac{l+1}{2})\Gamma(-n)}
              - 1\right),
\end{equation}
and thus
\begin{equation}\label{eq:single-boson}
\sigma^+_2(z_0)\alpha^{(1)}_{A\dot{A},n}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
 = \left[\frac{1}{2}\alpha_{A\dot{A},2n} + \frac{i}{2\pi}\sum_{l\,\text{odd}^+}
         \frac{z_0^{n+\frac{l}{2}}}{n+\frac{l}{2}}
    \frac{\Gamma(\frac{l}{2})\Gamma(-n+\frac{1}{2})}{\Gamma(\frac{l+1}{2})\Gamma(-n)}
    \alpha_{A\dot{A},-l}
  \right]\ket{\chi}.
\end{equation}
For $n$ positive the above vanishes as it should, since the positive
even mode in the first term annihilates $\ket{\chi}$ and $\Gamma(-n)$
kills the second term. For $n$ negative this reproduces the result
found in~\cite{acm3} for a single mode in the
initial state.

Similarly, if one performs the analogous calculation for
fermions with~\eqref{eq:fermion-coef}, then one can use
\begin{equation}
\psi^{\alpha A}_{+k}\ket{\chi} = 2\sum_{p\,\text{odd}^+}
                              \left(\gamma^F_{pk}\delta^\alpha_+\psi^{+A}_{-p}
                                -\gamma^F_{kp}\delta^\alpha_-\psi^{-A}_{-p}\right)
                              \ket{\chi}\qquad k\,\text{odd, positive},
\end{equation}
to find
\begin{subequations}\label{eq:single-fermion}
\begin{align}
\sigma_2^+(z_0)\psi^{(1)+A}_{n}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
 &= \left[\frac{1}{2}\psi^{+A}_{2n} + \frac{i}{2\pi}
        \sum_{p\,\text{odd}^+}\frac{z_0^{n+\frac{p}{2}}}{n+\frac{p}{2}}
           \frac{\Gamma(\frac{p}{2}+1)\Gamma(-n+\frac{1}{2})}{\Gamma(\frac{p+1}{2})\Gamma(-n+1)}
           \psi^{+A}_{-p}\right]\ket{\chi}\\
\sigma_2^+(z_0)\psi^{(1)-A}_{n}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
 &= \left[\frac{1}{2}\psi^{-A}_{2n} + \frac{i}{2\pi}
        \sum_{p\,\text{odd}^+}\frac{z_0^{n+\frac{p}{2}}}{n+\frac{p}{2}}
           \frac{\Gamma(\frac{p}{2})\Gamma(-n+\frac{1}{2})}{\Gamma(\frac{p+1}{2})\Gamma(-n)}
           \psi^{-A}_{-p}\right]\ket{\chi}.
\end{align}
\end{subequations}
This agrees with~\cite{acm3}.

\subsection{Problems}\label{sec:problems}

There are two problems with the above derivation. One is that this
formal derivation only makes use of the holomorphicity of the fields,
which means that one could make the same argument for any other
holomorphic field. For instance, consider $J^a(z)$, one would get
\begin{equation}
J^{a(1,2)}_n \overset{?}{=} \frac{1}{2}J^a_{2n}
             \pm \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}} J^a_k,
\end{equation}
which leads to
\begin{calc}\label{eq:wrong-J-com}
\com{J^-_0}{\sigma_2^+(z_0)} &= J^-_0\sigma_2^+(z_0)
              - \sigma_2^+(z_0)\big(J_0^{(1)} + J_0^{(2)}\big)\\
 &\overset{?}{=} 0.
\end{calc}
One should find $\sigma_2^-(z_0)$, not zero. One finds similar
contradictions if one tries to use the same argument for $T(z)$, too.

The second problem, alluded to above, concerns using the intertwining
relations with more than one mode. The simplest instance may be to
consider
\begin{equation}
\sigma_2^+(z_0)\alpha^{(1)}_{++,m}\alpha^{(1)}_{--,-n}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}
 = -m\delta_{m,n}\ket{\chi}\qquad m,n>0.
\end{equation}
If we use Equation~\ref{eq:boson-coef}, then we get
\begin{calc}\label{eq:two-boson-problem}
- m \delta_{m,n}\ket{\chi}
 &\overset{?}{=}
\left(\frac{1}{2}\alpha_{++, 2m}
        + \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{m-\frac{k}{2}}}{m-\frac{k}{2}}
            \alpha_{++, k}\right)
\left(\frac{1}{2}\alpha_{--, -2n}
        + \frac{i}{2\pi}\sum_{l\,\text{odd}}\frac{z_0^{-n-\frac{l}{2}}}{-n-\frac{l}{2}}
            \alpha_{--,l}\right)\ket{\chi}\\
 &= -\frac{m}{2}\delta_{m,n}\ket{\chi}
    -\frac{1}{4\pi^2}\left(\sum_{k\,\text{odd}}\frac{z_0^{m-\frac{k}{2}}}{m-\frac{k}{2}}
                          \alpha_{++, k}\right)
                   \left(\sum_{l\,\text{odd}}\frac{z_0^{-n-\frac{l}{2}}}{-n-\frac{l}{2}}
                         \alpha_{--,l}\right)\ket{\chi}\\
 &= -\frac{m}{2}\delta_{m,n}\ket{\chi}
    -\frac{1}{4\pi^2}\left(\sum_{k\,\text{odd}}\frac{z_0^{m-\frac{k}{2}}}{m-\frac{k}{2}}
                          \alpha_{++, k}\right)
                   \left[\sum_{j\,\text{odd}^+}\alpha_{--,-j}\left(
                       \frac{z_0^{-n+\frac{j}{2}}}{-n+\frac{j}{2}} + \sum_{l\,\text{odd}^+}
                       \frac{z_0^{-n-\frac{l}{2}}l\gamma_{lj}}{-n-\frac{l}{2}}
                         \right)\right]\ket{\chi}\\
\end{calc}
Note that the even--odd cross-terms vanish since they commute and
either the $\alpha_{++,2m}$ or the $\alpha_{++,k}$-sum kills
$\ket{\chi}$ (from~\eqref{eq:single-boson}). Similarly, we need only
look at the commutator of the $\alpha_{++,k}$-sum and the
square-bracketed expression, which gives
\begin{calc}\label{eq:ambig-sum}
&-\left(\sum_{k\,\text{odd}^+}\frac{kz_0^{m-\frac{k}{2}}}{m-\frac{k}{2}}\right)
 \left(\frac{z_0^{-n+\frac{k}{2}}}{-n+\frac{k}{2}} + \sum_{l\,\text{odd}^+}
                       \frac{z_0^{-n-\frac{l}{2}}l\gamma_{lk}}{-n-\frac{l}{2}}
                         \right)\\
&= z_0^{m-n}\sum_{k\,\text{odd}^+}\left(\frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
  + \sum_{l\,\text{odd}^+}\frac{z_0^{-\frac{k}{2}-\frac{l}{2}}kl\gamma_{kl}}
                                          {(m-\frac{k}{2})(n+\frac{l}{2})}\right).
\end{calc}
We would like the above expression to evaluate to $2\pi^2 m
\delta_{m,n}$ in order to get the correct answer; however, the above
summations depend sensitively on the order in which one adds the
infinite number of terms. For instance, if we attempt to perform the
$k$-sum \emph{first}, then the first term is divergent and the second
term, using~\eqref{eq:gamma-id}, gives
\begin{equation}
-\sum_{l\,\text{odd}^+}\frac{l}{(m+\frac{l}{2})(n+\frac{l}{2})},
\end{equation}
which is also divergent.

On the other hand, if we perform the $l$ sum first,
using~\eqref{eq:gamma-id} we find
\begin{calc}
& z_0^{m-n}\sum_{k\,\text{odd}^+}\left(\frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
  + \sum_{l\,\text{odd}^+}\frac{z_0^{-\frac{k}{2}-\frac{l}{2}}kl\gamma_{kl}}
                                          {(m-\frac{k}{2})(n+\frac{l}{2})}\right)\\
&= z_0^{m-n}\sum_{k\,\text{odd}^+}\left(\frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
  + \frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
     \left[\frac{\Gamma(\frac{k}{2})\Gamma(n+\frac{1}{2})}{\Gamma(\frac{k+1}{2})\Gamma(n)}
            - 1\right]\right)\\
&= z_0^{m-n}\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n)}
  \sum_{k\,\text{odd}^+}\left(\frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
  \frac{\Gamma(\frac{k}{2})}{\Gamma(\frac{k+1}{2})}\right),
\end{calc}
which using another identity,
\begin{equation}
\sum_{k\,\text{odd}^+}\frac{k}{(m-\frac{k}{2})(n-\frac{k}{2})}
      \frac{\Gamma(\frac{k}{2})}{\Gamma(\frac{k+1}{2})}
 = 2\pi^2 m \frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\delta_{m,n}\qquad m,n>0,
\end{equation}
gives the correct answer.

How should we think of the ambiguity in Equation~\ref{eq:ambig-sum}?
Any infinite series is implicitly evaluated by determining the limit
of a sequence of \emph{partial sums}. In our case, higher values of
$k$ and $l$ correspond to higher modes, so it is natural to think of
imposing UV cutoffs on the sums, $k<L_1$ and $l<L_2$. We then wish to
take the limit as $L_1, L_2\to\infty$, but there are many different
ways to do that. If we define $b=L_1/L_2$ to parameterize the
different ways of evaluating~\ref{eq:ambig-sum}, then evaluating the
$k$-sum first corresponds to $b=\infty$, while evaluating the $l$-sum
first corresponds to $b=0$. These are just two of an infinite number
of ways to evaluate the double-sum.

Given that these ambiguous multi-dimensional series are rampant in
this formalism and that frequently the correct method of evaluating
them may be much less obvious,\footnote{For instance, there are cases
  involving triple-sums, where more than one order of evaluating the
  sums give distinct, finite results.} we need a well-motivated
principle that determines the correct way to handle the UV physics.

\subsection{A More Rigorous Derivation}\label{sec:subtleties}

We now present a more rigorous derivation of the intertwining
relations in Equations~\ref{eq:boson-coef}, \eqref{eq:boson-coef-2},
and \eqref{eq:fermion-coef}. By continuously deforming the contour
integral for an initial state mode outward only where the integrand is
holomorphic, we can treat the point $z_0$ more carefully. This
resolves the two problems outlined above.

\begin{figure}[ht]
\subfigure[]
  {\includegraphics[height=3cm, clip=true, trim=0 48 0 48]{mode1-z}\label{fig:mode1-z}}
\subfigure[]
  {\includegraphics[height=3cm, clip=true, trim=0 48 0 48]{open-contour-z}}
  \caption{In the $z$-plane, showing how the contour $C_1$ (solid,
    red) in (a) may be deformed out and around the branch cut into
    contours $C_2$ (solid, blue) and $C_3$ (dashed, orange) in (b).
    The gray circular region is the ``before the twist'' region,
    $|z|<|z_0|$.  The branch cut is indicated by the dashed black line
    extending out from the circle.\label{fig:z-plane-contours}}
\end{figure}

\subsubsection{Bosons}

Working with the bosons first, let us note that
\begin{equation}
\alpha_n^{(1)} = \oint_{C_1}\frac{\drm z}{2\pi i}i\partial X^{(1)}(z) z^n,
\end{equation}
where $C_1$ is a circular contour with radius less than $|z_0|$ shown
in Figure~\ref{fig:mode1-z}, and $i\partial X^{(1)}(z)$ is a holomorphic
function except at $z=z_0$ (and excluding any other operator
insertions). Thus, we may deform the contour into an open circle $C_2$
of radius larger than $|z_0|$ and a contour, $C_3$, sneaking around
the branch cut starting at $z=z_0$, as shown in
Figures~\ref{fig:z-plane-contours} and~\ref{fig:close-up}.  We take
$C_3$ to be a circle of radius $\veps$, which we eventually take to
zero. The orientations of the contours are shown in the figures.

\begin{figure}[tb]
\begin{center}
\includegraphics[height=3cm]{close-up-z}
\caption{A close-up depiction of contours $C_2$ and $C_3$ meeting.
  Note that we have added an artifical gap around the branch cut for
  illustrative purposes only---in fact, both $C_2$ and $C_3$ are full
  circles.\label{fig:close-up}}
\end{center}
\end{figure}

We write the above contour integral as
\begin{equation}
\alpha_n^{(1)} = \int_{C_2}\frac{\drm z}{2\pi i}i\partial X^{(1)}(z) z^n
          + \int_{C_3}\frac{\drm z}{2\pi i}i\partial X^{(1)}(z) z^n.
\end{equation}
The $C_2$ term, is what we have been calculating and is given by (with
$\veps$ corrections)
\begin{calc}
\int_{C_2}\frac{\drm z}{2\pi i} i\partial X(z) z^n &=
\frac{i}{2}\sum_k\alpha_k \int_{C_2}\frac{\drm z}{2\pi i} z^{n - \frac{k}{2} -1}\\
  &= \frac{1}{4\pi}\sum_k \alpha_k(z_0+\veps)^{n-\frac{k}{2}}
           \int_0^{2\pi}\drm \theta\,e^{i(n-\frac{k}{2})\theta}\\
  &= \frac{1}{2}\alpha_{2n}
    + \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{(z_0+\veps)^{n-\frac{k}{2}}}
                                             {n-\frac{k}{2}}\alpha_k.
\end{calc}

At this point, it becomes necessary to introduce the covering space
where the fields are well-defined. We map to the covering space
coordinate $t$ via
\begin{equation}
z = z_0 + t^2\qquad a^2 = z_0.
\end{equation}
The points $ia$ and $-ia$ are the two images of the origin $z=0$, one
corresponding to each copy of the fields. Mapping the $z$-plane
contours in Figure~\ref{fig:z-plane-contours} to the $t$-plane results
in Figure~\ref{fig:t-plane-contours}.

\begin{figure}
\subfigure[]
 {\includegraphics[height=6cm]{mode1-only}}
\qquad
\subfigure[]
 {\includegraphics[height=6cm]{large-top-mode}}
\qquad
\subfigure[]
 {\includegraphics[height=6cm]{open-contour}}
 \caption{We show the $t$-plane where there is no branch cut and
   the fields are single-valued.  We start in~(a) with the $\tilde{C}_1$
   contour (solid, red), which we finally deform out into $\tilde{C}_2$ (solid,
   blue) and $\tilde{C}_3$ (dashed, orange) in~(c).
   \label{fig:t-plane-contours}}
\end{figure}

We compute the $C_3$ term by going to the $t$-plane
\begin{calc}
\int_{C_3}\frac{\drm z}{2\pi i}\partial X(z)z^n =
  \int_{\tilde{C}_3}\frac{\drm t}{2\pi i} \partial X(t)(z_0 + t^2)^n
  \xrightarrow[\tilde{\veps}\to 0]{} 0,
\end{calc}
where the radius of the semicircular contour in the $t$-plane,
$\tilde{C}_3$, is given by $\tilde{\veps}$. The contour gives zero
contribution since there is no bosonic insertion at $t=0$ and the
integrand is therefore analytic.  As the length of the contour goes to
zero, therefore, so does the integral. Thus, we see that the story for
the bosons is exactly as stated, and in the limit as $\veps\to 0$ we
reproduce our previous result,
\begin{equation}
  \alpha_n^{(1,2)} = \frac{1}{2}\alpha_{2n}
  \pm \frac{i}{2\pi}\sum_{k\,\text{odd}}
  \frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\alpha_k.
\end{equation}

\subsubsection{Fermions}

For the fermions we have
\begin{equation}
\psi^{(1)}_n = \oint_{C_1}\frac{\drm z}{2\pi i}\psi(z) z^{n-\frac{1}{2}},
\end{equation}
which becomes
\begin{equation}
\psi^{(1)}_n = \int_{C_2}\frac{\drm z}{2\pi i}\psi(z) z^{n-\frac{1}{2}}
              +\int_{C_3}\frac{\drm z}{2\pi i}\psi(z) z^{n-\frac{1}{2}}
\end{equation}
The $C_2$ term is what we have computed for the fermions previously,
and is given by
\begin{calc}
\int_{C_2}\frac{\drm z}{2\pi i}\psi(z) z^{n-\frac{1}{2}} &=
   \frac{1}{2}\sum_k\psi_k\int_{C_2}\frac{\drm z}{2\pi i} z^{n-\frac{k}{2}-1}\\
  &= \frac{1}{2}\psi_{2n}
  \pm \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\psi_k.
\end{calc}
The $C_3$ contribution also goes to zero for the fermions. When one
goes to the $t$-plane, one finds
\begin{equation}
  \int_{\tilde{C}_3}\frac{\drm t}{2\pi i}\psi(t)\sqrt{2t}(z_0 + t^2)^{n-\frac{1}{2}},
\end{equation}
which acts on the spin field $S(t=0)$. The most singular term in the
OPE between $\psi(t)$ and $S(0)$ is proportional to $1/\sqrt{t}$, and
so one again finds that the semi-circular contour vanishes.

At this point, we see the resolution of the first problem we found
with our formal derivation. If the field has an OPE with the
spin-field in the covering space which is singular enough, then the
$C_3$-contribution is nonvanishing in the limit as $\tilde{\veps}\to
0$. This extra contribution gives exactly the correct answer for
$J_0^-$, for instance, as we demonstrate in
Section~\ref{sec:contour-J}.

\subsubsection{Multiple contours}

Having resolved the first problem with the formal derivation, we
should now discuss the UV issues that arise with multiple
contours.\footnote{Recall that the definition of the twist operator
  $\sigma_2$ involves a hole in the $z$-plane, whose radius is
  carefully taken to zero~\cite{lm1}. One might suspect that this
  limit has important consequences for these UV issues and that the
  size of the hole plays the role of a UV cutoff. In fact, the hole is
  taken to zero size \emph{before} any of the issues discussed in this
  paper, and does not play any role here. The actual issue is the
  interaction between neighboring contours, as discussed.}

Let us write $\veps_i$ for the radius of the $i$th mode's $C_3$
semi-circle. For instance, consider an initial state
\begin{equation}
\alpha^{(1)}_{n_1}\alpha^{(1)}_{n_2}\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}.
\end{equation}
Then the $C_3$ part coming from $\alpha^{(1)}_{n_1}$ has radius
$\veps_1$ in the $z$-plane and therefore the semi-circle in the
$t$-plane has radius $\sqrt{\veps_1}$. Similarly, for
$\alpha^{(1)}_{n_2}$. If we require that the $C_2$ parts of
$\alpha_{n_1}$ and $\alpha_{n_2}$ preserve the same ordering, then the
semi-circles in the $t$-plane satisfy
\begin{equation}
\veps_2 < \veps_1.
\end{equation}

We now argue that, in fact, we should have $\veps_2 \ll \veps_1$ in
order to use the intertwining relations as we want to. Consider the
two semi-circular $C_3$ contours at leading order in $\veps_1$ and
$\veps_2$:
\begin{calc}
\int_{\tilde{C}_3(\veps_1)}\frac{\drm t}{2\pi i} \partial X(t)(z_0 + t^2)^{n_1}
\int_{\tilde{C}_3(\veps_2)}\frac{\drm t'}{2\pi i} \partial X(t') (z_0 + {t'}^2)^{n_2}
 &\sim \int_{\tilde{C}_3(\veps_1)}\frac{\drm t}{2\pi i} (z_0 + t^2)^{n_1}
\int_{\tilde{C}_3(\veps_2)}\frac{\drm t'}{2\pi i} \frac{ (z_0 + {t'}^2)^{n_2}}{(t-t')^2}\\
 &\sim \int_{\tilde{C}_3(\veps_1)}\frac{\drm t}{2\pi i} z_0^{n_1}
\int_{\tilde{C}_3(\veps_2)}\frac{\drm t'}{2\pi i} \frac{ z_0^{n_2}}{(t-t')^2}\\
 &\sim z_0^{n_1+n_2}\log \frac{1 - \sqrt{\frac{\veps_2}{\veps_1}}}
                              {1+ \sqrt{\frac{\veps_2}{\veps_1}}}\\
 &\sim z_0^{n_1 + n_2} \sqrt{\frac{\veps_2}{\veps_1}} + \cdots.
\end{calc}
This vanishes only if $\veps_2\ll \veps_1$. Note that this argument is
unaffected if the modes are on different copies (the integrals work
out in essentially the same way).

Taking $\veps_2\ll \veps_1$ suggests a particular way to take the
limit for double-sums: if $L_i$ is the cutoff on the
$\alpha_{n_i}^{(1)}$-sum, then we should take
\begin{equation}
L_2 \gg L_1,
\end{equation}
that is, evaluate the $L_2$-sum first and then the $L_1$-sum. This is
exactly the way that we got the correct answer, when the issue was
demonstrated in~\ref{eq:two-boson-problem}.

\subsubsection{The prescription}\label{sec:prescription}

We now develop the precise prescription that resolves the UV
ambiguities. Before we state the prescription, we should mention that
there are two kinds of sums over modes. There are the after-the-twist
intertwining sums, which have ordering ambiguities among themselves,
and there can also be before-the-twist sums on modes before the twist
operator.  For example, consider a composite operator like
\begin{equation}\label{eq:Ln}
J_n^{a(1)} = -\frac{1}{4}{(\sigma^{aT})^\alpha}_\beta\sum_{j =-\infty}^\infty
                \psi^{(1)\dagger}_{\alpha A, n-j}\psi^{(1)\beta A}_j.
\end{equation}
In fact, these sums also have UV-limit issues when combined with the
intertwining relations as is demonstrated in
Section~\ref{sec:composite}. If we look at the intertwining relation
in~\ref{eq:boson-coef}, for example, with the implicit cutoff on the
sum,
\begin{equation}\label{eq:boson-coef-cutoff}
\alpha_m^{(2)} = \frac{1}{2}\alpha_{2m}
                  - \frac{i}{2\pi}\sum_{n\,\text{odd}}^{|n|<L}
                     \frac{z_0^{m-\frac{n}{2}}}{m-\frac{n}{2}}\alpha_n,
\end{equation}
we see that we are approximating a mode $m$ as a linear combination of
modes with UV cutoff $L$. In order for this approximation to become an
exact expansion we must take $L\to\infty$ \emph{with $m$ fixed.} That
is we need to have much higher frequency modes in our sum than the
mode that we are expanding. Therefore, we need to cutoff the
before-the-twist sum in~\eqref{eq:Ln} and ensure that its cutoff is
much less than the after-the-twist cutoff
in~\eqref{eq:boson-coef-cutoff}.

Before proceeding, let us consider more generally what sort of
multi-dimensional series we can get in this formalism. If we have a
bunch of modes in the initial state that we pull across using the
intertwining relations, then there are several different types of
terms that can arise.  There is a term in which all of the positive
modes act on $\ket{\chi}$ separately, and we are left with a product
of one-dimensional sums that result in either~\eqref{eq:single-boson}
or~\eqref{eq:single-fermion}. Then, there are terms where the positive
modes from one sum contract with negative modes of another. This gives
a double sum, like the one in~\ref{eq:ambig-sum}. If there is a sum
on the before-the-twist modes, then we can get a triple sum if those
two modes contract. This is the most complicated sum possible.

Finally, we are ready to state the prescription that ensures that the
multi-dimensional series converge to the correct answer. The
prescription is
\begin{enumerate}
\item The after-the-twist intertwining sums should be performed from
  innermost contour (right-most mode) to outermost contour (left-most
  mode). This ensures that the $C_3$-terms can be dropped.
\item Any sums on before-the-twist modes should be performed
  \emph{last}. There is no UV-ambiguity among the
  before-the-twist sums.
\end{enumerate}

Note that with this prescription we have a weakened version of our
goal. Because of the UV sensitive series, one cannot directly map
operators to operators since one requires knowledge of what other
modes are around in order to correctly evaluate the series.  While the
modes that we start with before the twist operator may commute, we
need to think about them pulling across the twist operator in a
particular order. It is in this sense, that we have intertwining
relations and not Bogolyubov coefficients.

\section{An example: intertwining relations for $J^a_n$}\label{sec:example}

There are two equivalent ways of calculating the effect of the twist
operator on composite operators such as $J^a_n$. One way is to use the
contour deformation method described in this paper, being careful not
to throw away the small contour $C_3$. The other way is to write $J^a$
as the product of fermion modes and use the intertwining relations for
the fermions, being careful to use the prescriptions described in
Section~\ref{sec:prescription}.

\subsection{The contour method}\label{sec:contour-J}

We first describe the contour method mentioned above. If we go through
the argument described in Section~\ref{sec:subtleties} then
\begin{equation}
J^{a(1)}_n = \frac{1}{2}J^a_{2n} + \frac{i}{2\pi}\sum_{k\,\text{odd}}
        \frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^a_k
        + \lim_{\veps\to 0}\int_{C_3}\frac{\drm z}{2\pi i} J^{a(1)}(z)z^n,
\end{equation}
but this time we find a nonvanishing contribution coming from the
$C_3$ contour. As we pull the $J^{a(1)}_n$ contour out, it acts on the
twist operator (or equivalently, the spin field in the covering space)
and can switch a $\sigma_2^+$ to $\sigma_2^-$. Therefore, let us
consider $\sigma_2^{\alpha}$.

We can evaluate the $C_3$ contour by going to the $t$-plane, where
$\sigma_2^\alpha(z_0)$ leaves only a spin field $S^\alpha(t=0)$. The
image of $C_3$ in the $t$-plane, $\tilde{C}_3$, is a semi-circle
around the origin as shown in Figure~\ref{fig:t-plane-contours}. Thus,
the $\tilde{C}_3$ contour implicitly acts on the spin field:
\begin{calc}
\int_{C_3}\frac{\drm z}{2\pi i} J^{(1)a}(z)z^n
 &= \left[\int_{\tilde{C}_3}\frac{\drm t}{2\pi i} J^a(t)(z_0 + t^2)^n\right]S^\alpha(0)\\
 &= \frac{1}{2}{(\sigma^{aT})^\alpha}_\beta S^\beta(0)
    \int_{\tilde{C}_3}\frac{\drm t}{2\pi i}\frac{(z_0 + t^2)^n}{t}\\
 &= \frac{1}{2}{(\sigma^{aT})^\alpha}_\beta S^\beta(0)
     \frac{1}{2\pi i}\left(z_0^n \log t \bigg|^{\veps}_{-\veps} + \bigO(\veps)\right)\\
 &=  -\frac{z_0^n}{4}{(\sigma^{aT})^\alpha}_\beta S^\beta(0),
\end{calc}
where we have given the result after taking $\veps \to 0$. When we go
back to the $z$-plane we should write the above result as
\begin{equation}
\sigma_2^\alpha(z_0)J^{a(1)}_n  = \left[\frac{1}{2}J^a_{2n}
          +\frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^a_k\right]
            \sigma_2^\alpha(z_0)
             - \frac{z_0^n}{4}{(\sigma^{aT})^\alpha}_\beta\sigma_2^\beta(z_0).
\end{equation}
If we had considered $J_n^{a(2)}$, then we obtain a similar result. We
can summarize the two relations and write them in a suggestive form as
\begin{equation}\label{eq:J-intertwining}
\sigma_2^\alpha(z_0)J^{a(1,2)}_n  = \left[\frac{1}{2}J^a_{2n}
          \pm\frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^a_k\right]
            \sigma_2^\alpha(z_0)
             - \frac{z_0^n}{2}\com{J^a_0}{\sigma_2^\alpha(z_0)}.
\end{equation}
We see that if we use these intertwining relations, then we get the
correct answer in~\ref{eq:wrong-J-com}.

While the above intertwining relation is correct, it may not be the
most useful form. Because we have switched the charge on the twist
operator, we now must think about a new state $\ket{\chi}$ created by
the negatively-charged operator acting on the vacuum.

\subsubsection{An example}

For concreteness, let us consider
\begin{equation}
\sigma_2^+(z_0) J_n^{-(1)}\ket{0^-}^{(1)}\ket{0^-}^{(2)},\qquad n<0,
\end{equation}
then we have
\begin{equation}
\left[\frac{1}{2}J^-_{2n} + \frac{i}{2\pi}
         \sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^-_k
- \frac{z_0^n}{2}J_0^-\right]\ket{\chi}.
\end{equation}
Thus, our first task is to compute the three terms,
\begin{equation}
J_{2n}^-\ket{\chi}\qquad
J_k^-\ket{\chi}\,k\,\text{odd}\qquad
J_0^-\ket{\chi}.
\end{equation}

There are no real complications in working the terms out. For
instance, for the first term, one starts with\footnote{Note that the
  $J$ after the twist has an extra factor of $1/2$ from before the
  twist, which arises from the fermion--fermion anticommutator having
  an extra factor of $2$ after the twist.}
\begin{equation}
J^-_{2n} = -\frac{1}{4}\sum_j\psi^\dagger_{+A, 2n - j}\psi^{-A}_{j},
\end{equation}
then breaks the sum into terms with both modes negative and terms with
odd positive modes that act on $\ket{\chi}$. One can write the result
in the form
\begin{equation}
J^-_{2n}\ket{\chi} =
-\frac{1}{4}\sum_{2n+1\leq j \leq -1} \psi^\dagger_{+A, 2n-j}\psi^{-A}_j\ket{\chi}
 + \sum_{j,p\,\text{odd}^+}\gamma^F_{jp}\psi^\dagger_{+A, -p}\psi^{-A}_{2n-j}\ket{\chi}.
\end{equation}
Similarly,  $J^-_0$ becomes
\begin{equation}
J_0^-\ket{\chi} = \sum_{j,p\,\text{odd}^+}\gamma^F_{jp}\psi^\dagger_{+A, -j}\psi^{-A}_{-p}\ket{\chi}.
\end{equation}

The $J_k^-$ term is not much work, but there are two distinct cases,
corresponding to whether or not there are terms where both $\psi$s are
raising operators: $k\leq -3$ and $k\geq -1$. One finds
\begin{equation}
J^-_k\ket{\chi} = -\frac{1}{4}\sum_{k+1\leq j \leq -1} \psi^\dagger_{+A, k-j}\psi^{-A}_j\ket{\chi}
         + \sum_{j,p\,\text{odd}^+} \gamma^F_{jp}\psi^\dagger_{+A, k-j}\psi^{-A}_{-p}\ket{\chi}
         \qquad k\leq -3, \text{odd}
\end{equation}
and
\begin{equation}
J^-_k\ket{\chi} = \sum_{j,p\,\text{odd}^+}\gamma^F_{k+j+1, p}\psi^\dagger_{+A, -j-1}\psi^{-A}_{-p}
                 \ket{\chi}
\qquad k\geq -1,\text{odd}.
\end{equation}
The slightly more difficult task is the sum over $k$, which can be written as
\begin{calc}
\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^-_k\ket{\chi}
 &= \sum_{k\,\text{odd}}^{k\leq -3}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^-_k\ket{\chi}
   +\sum_{k\,\text{odd}}^{k\geq -1}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^-_k\ket{\chi}\\
 &=
-\frac{1}{2}\sum_{j,p\,\text{odd}^+}\frac{z_0^{n+\frac{j+p+1}{2}}}{n+\frac{j+p+1}{2}}
       \psi^\dagger_{+A, -j-1}\psi^{-A}_{-p}\\
&\qquad+\sum_{j\,\text{odd}}^{j\geq 3}\sum_{p\,\text{odd}^+}
     \left(
       \sum_{k\,\text{odd}}^{-j\leq k\leq -3}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}
            \gamma^F_{k+j+1,p}
     \right)\psi^\dagger_{+A, -j-1}\psi^{-A}_{-p}\\
&\qquad+\sum_{j,p\,\text{odd}^+}
       \left(
         \sum_{k\,\text{odd}}^{k\geq -1}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}\gamma^F_{k+j+1,p}
       \right)\psi^\dagger_{+A, -j-1}\psi^{-A}_{-p},
\end{calc}
after a few manipulations. Note that all of the terms on the right
implicitly act on $\ket{\chi}$. Finally, using~\eqref{eq:gammaF-sum-1}
one arrives at
\begin{calc}
\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}J^-_k\ket{\chi}
 =-\sum_{j,p\,\text{odd}^+}
    \frac{z_0^{n+\frac{j+p+1}{2}}}{2(n+\frac{j+p+1}{2})}
     \frac{\Gamma(\frac{p}{2})\Gamma(-n-\frac{j}{2})}
                {\Gamma(\frac{p+1}{2})\Gamma(-n-\frac{j+1}{2})}
              \psi^\dagger_{+A,-j-1}\psi^{-A}_{-p}\ket{\chi}
\end{calc}

Putting it all together, we have
\begin{equation}\begin{split}\label{eq:J-on-vac}
\sigma_2^+(z_0) J_n^{-(1)}\ket{0^-}^{(1)}\ket{0^-}^{(2)}
 &= -\frac{1}{8}\sum_{j=n+1}^{-1}\psi^\dagger_{+A, 2n-2j}\psi^{-A}_{2j}\ket{\chi}\\
&\quad-\frac{i}{4\pi}
\sum_{j=1}^\infty\sum_{p\,\text{odd}^+}
    \frac{z_0^{n+j+\frac{p}{2}}}{n+j+\frac{p}{2}}
     \frac{\Gamma(\frac{p}{2})\Gamma(-n-j+\frac{1}{2})}
                {\Gamma(\frac{p+1}{2})\Gamma(-n-j)}
              \psi^\dagger_{+A,-2j}\psi^{-A}_{-p}\ket{\chi}\\
&\quad+\frac{1}{2}\sum_{j,p\,\text{odd}^+}
\left[-\frac{1}{4}\delta_{j+p+2n,0} + \frac{\gamma^F_{2n+p,j}+\gamma^F_{2n+j,p}}{2}
      - z_0^n \frac{\gamma^F_{jp} + \gamma^F_{pj}}{2}\right]\psi^\dagger_{+A, -j}\psi^{-A}_{-p}
\ket{\chi},
\end{split}\end{equation}
where for ease of comparison we have broken the result into
even--even, even--odd, and odd--odd terms. In the above expression, we
define $\gamma^F$ with negative indices to be zero. Furthermore, we
have explicitly symmetrized over $j$ and $p$ in the last term since
\begin{equation}
\psi^\dagger_{+A, -j}\psi^{-A}_{-p} = \psi^\dagger_{+A, -p}\psi^{-A}_{-j}.
\end{equation}
Below, we compare this result with what one finds when one breaks the
$J^{-(1)}_{n}$ into fermions and uses the fermion intertwining
relations.

\subsection{The composite method}\label{sec:composite}

We now show how to reproduce the above result by using the fermion
intertwining relations and our prescription.

We start by writing
\begin{equation}
J^{a(1)}_n = -\frac{1}{4}{(\sigma^{aT})^\alpha}_\beta\sum_{j =-\infty}^\infty
                \psi^{(1)\dagger}_{\alpha A, n-j}\psi^{(1)\beta A}_j,
\end{equation}
and note that the sum on $j$ is a sum over ``before-the-twist'' modes
and therefore should be evaluated last according to the prescription.
We then can use our intertwining relations to write this directly as
\begin{equation}\label{eq:J-sum}
J^{a(1)}_n = -\frac{1}{4}{(\sigma^{aT})^\alpha}_\beta\sum_{j =-\infty}^\infty
  \left[\frac{1}{2}\psi^\dagger_{\alpha A, 2n-2j}
           + \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-j-\frac{k}{2}}}{n-j-\frac{k}{2}}
                     \psi^\dagger_{\alpha A, k}\right]
  \left[\frac{1}{2}\psi^{\beta A}_{2j} + \frac{i}{2\pi}\sum_{l\,\text{odd}}
                             \frac{z_0^{j-\frac{l}{2}}}{j-\frac{l}{2}}\psi_l^{\beta A}\right].
\end{equation}
At this point, we can say nothing further until we know what
$J^{a(1)}_n$ acts on. Of course, when faced with an expression like
the above it is rather tempting to evaluate the $j$-sum using
\begin{equation}
\sum_{j=-\infty}^\infty \frac{1}{(n-j-\frac{k}{2})(j-\frac{l}{2})} =
   -\pi^2\delta_{\frac{l}{2}, n-\frac{k}{2}} \qquad k,l\,\text{odd},
\end{equation}
which immediately leads to the false relation
\begin{equation}
J^{a(1)}_n = \frac{1}{2}J^a_{2n} +
  \frac{i}{2\pi}\sum_{k\,\text{odd}}\frac{z_0^{n-\frac{k}{2}}}{n-\frac{k}{2}}
  J^a_k
\qquad \text{(false relation)}.
\end{equation}
This demonstrates the need for the restriction on before-the-twist
sums.

In order to proceed and compare to~\eqref{eq:J-intertwining}, let us
again consider the state
\begin{equation}\label{eq:J-state}
\sigma_2^+(z_0)J^{-(1)}_n\ket{0_R^-}^{(1)}\ket{0_R^-}^{(2)}.
\end{equation}
Now, one could make the $j$-sum in~\eqref{eq:J-sum} finite by
considering the above; however, our prescription ensures that one gets
the correct answer even if one leaves it as an infinite series.

If we act on the vacuum as in~\eqref{eq:J-state}, then we
get~\eqref{eq:J-sum} acting on $\ket{\chi}$. The $\psi^{\beta A}$ is
the rightmost mode and so the $l$-sum should be evaluated first,
followed by the $k$-sum, and then finally the $j$-sum. We can
use~\eqref{eq:single-fermion} to quickly read off the result of the
$l$-sum. There are now two distinct terms to consider for the $k$-sum.
There is the possibility of the $\psi^\dagger_{\alpha A}$ contracting with
the result of the $l$-sum, and the $\psi^\dagger$ can pass through and act
on $\ket{\chi}$ (and we can again use~\eqref{eq:single-fermion}). The
contraction gives zero. For other composite operators, however, the
contraction term can be nonzero.

Following the above procedure, we get
\begin{equation}\begin{split}
-\frac{1}{2}\sum_{j=-\infty}^\infty
  &\left[\frac{1}{2}\psi^\dagger_{+A, 2(n-j)} + \frac{i}{2\pi}\sum_{p\,\text{odd}^+}
    \frac{z_0^{n-j+\frac{p}{2}}}{n-j+\frac{p}{2}}
    \frac{\Gamma(\frac{p}{2})\Gamma(-n+j+\frac{1}{2})}{\Gamma(\frac{p+1}{2})\Gamma(-n+j)}
    \psi^\dagger_{+A, -p}\right]\\
  &\left[\frac{1}{2}\psi^{-A}_{2j} + \frac{i}{2\pi}\sum_{q\,\text{odd}^+}
    \frac{z_0^{j+\frac{q}{2}}}{j+\frac{q}{2}}
    \frac{\Gamma(\frac{q}{2})\Gamma(-j+\frac{1}{2})}{\Gamma(\frac{q+1}{2})\Gamma(-j)}
    \psi^{-A}_{-q}\right].
\end{split}\end{equation}
There are four terms from the above multiplication.

The even--even term is
\begin{equation}
(\text{even--even})=
-\frac{1}{8}\sum_{j=n+1}^{-1}\psi^\dagger_{+A, 2n-2j}\psi^{-A}_{2j}.
\end{equation}
One can show that the two even--odd cross-terms are identical, and sum
to
\begin{equation}
(\text{even--odd}) =
-\frac{i}{4\pi}\sum_{j=-\infty}^{-1}
\sum_{p\,\text{odd}^+}\frac{z_0^{n-j+\frac{p}{2}}}{n-j+\frac{p}{2}}
    \frac{\Gamma(\frac{p}{2})\Gamma(-n+j+\frac{1}{2})}{\Gamma(\frac{p+1}{2})\Gamma(-n+j)}
    \psi^\dagger_{+A, -p}\psi^{-A}_{2j}
\end{equation}
One may write the odd--odd term as
\begin{equation}
(\text{odd--odd}) = \frac{1}{8\pi^2}\sum_{p,q\,\text{odd}^+}\psi^\dagger_{+A, -p}\psi^{-A}_{-q}
        z_0^{n+\frac{p+q}{2}}
        \frac{\Gamma(\frac{p}{2})\Gamma(\frac{q}{2})}{\Gamma(\frac{p+1}{2})\Gamma(\frac{q+1}{2})}
        S(n,p,q),
\end{equation}
where
\begin{equation}
S(n,p,q) = \sum_{j=n+1}^{-1}\frac{1}{(n-j+\frac{p}{2})(j+\frac{q}{2})}
                       \frac{\Gamma(-n+j+\frac{1}{2})\Gamma(-j+\frac{1}{2})}
               {\Gamma(-n+j)\Gamma(-j)}.
\end{equation}

Comparing the above to the three terms in Equation~\eqref{eq:J-on-vac},
one finds agreement provided
\begin{equation}
\frac{z_0^{n+\frac{j+p}{2}}}{8\pi^2}\frac{\Gamma(\frac{j}{2})\Gamma(\frac{p}{2})}
                                          {\Gamma(\frac{j+1}{2})\Gamma(\frac{p+1}{2})}
                                          S(n,j,p)
 = \frac{1}{2}\left[-\frac{1}{4}\delta_{j+p+2n,0} + \frac{\gamma^F_{2n+p,j}+\gamma^F_{2n+j,p}}{2}
      - z_0^n \frac{\gamma^F_{jp} + \gamma^F_{pj}}{2}\right].
\end{equation}
This equation follows from the identity
\begin{multline}\label{eq:hard-id}
\sum_{k=0}^\mu\frac{1}{(\mu-k-\frac{\alpha}{2})(k-\frac{\beta}{2})}
   \frac{\Gamma(\mu -k + \frac{3}{2})\Gamma(k+\frac{3}{2})}
        {\Gamma(\mu - k + 1)\Gamma(k+1)}\\
=\pi - \frac{\pi}{\frac{\alpha + \beta}{2} -\mu}
\left(\frac{\Gamma(\frac{\alpha+3}{2})\Gamma(\frac{\alpha}{2}-\mu)}
           {\Gamma(\frac{\alpha}{2}+1)\Gamma(\frac{\alpha-1}{2}-\mu)}
+\frac{\Gamma(\frac{\beta+3}{2})\Gamma(\frac{\beta}{2}-\mu)}
           {\Gamma(\frac{\beta}{2}+1)\Gamma(\frac{\beta-1}{2}-\mu)}
\right),
\end{multline}
where the left-hand side is $S(n,p,q)$ with $\mu = -n-2$, $\alpha= p
-2$, and $\beta = q-2$. If we call the above sum $F(\mu, \alpha,
\beta)$, then one can prove the identity by showing that both sides of
Equation~\eqref{eq:hard-id} obey the four-term recursion relation
\begin{multline}
  f_0(\mu,\alpha, \beta) F(\mu, \alpha, \beta)
+ f_1(\mu,\alpha, \beta) F(\mu+1, \alpha, \beta)\\
+ f_2(\mu,\alpha, \beta) F(\mu+2, \alpha, \beta)
+ f_3(\mu,\alpha, \beta) F(\mu+3, \alpha, \beta)
 = 0,
\end{multline}
with
\begin{subequations}
\begin{align}
f_0(\mu, \alpha,\beta) &= (-3+\alpha -2 \mu ) (\alpha +\beta -2 \mu ) (3-\beta +2 \mu ) (\alpha +\beta -6 (3+\mu ))\\
f_1(\mu, \alpha,\beta) &= 3 (\alpha +\beta -2 (1+\mu )) \big(-2 \beta ^2 (2+\mu )+\alpha ^2 (\beta -2 (2+\mu ))+\beta  (79+8 \mu  (9+2 \mu ))\notag\\
&+\alpha  \left(79+\beta ^2+8 \mu  (9+2 \mu )-2 \beta  (12+5 \mu )\right)-2 (128+\mu  (175+4 \mu  (20+3 \mu )))\big)\\
f_2(\mu, \alpha,\beta) &= 3 (\alpha +\beta -2 (2+\mu )) \big(\beta ^2 (5+2 \mu )+\alpha ^2 (5-\beta +2 \mu )+4 (2+\mu ) \left(43+32 \mu +6 \mu ^2\right)\notag\\
&\quad-2 \beta  (47+\mu  (39+8 \mu ))-\alpha  \left(94+\beta ^2-2 \beta  (12+5 \mu )+2 \mu  (39+8 \mu )\right)\big)\\
f_3(\mu, \alpha,\beta) &= (\alpha +\beta -6 (2+\mu )) (\alpha -2 (3+\mu )) (\beta -2 (3+\mu )) (\alpha +\beta -2 (3+\mu )).
\end{align}
\end{subequations}
Thus the identity holds by induction once one confirms that it holds
for $\mu=0,1,2$. It may be helpful to use an algebraic manipulation
program such as \textit{Mathematica} to show the above.

\section{Conclusion}\label{sec:conclusion}

We presented a method for computing the effect of the blow-up mode
deformation operator on untwisted states of the D1D5 CFT. An
alternative method was developed in~\cite{acm3}; however, the method
developed here comes in the form of intertwining relations for the
2-twist operator, $\sigma_2^+(z_0)$. This method, therefore, is more
direct and of theoretical interest since it makes the physics more
transparent. We would be remiss, if we did not note that both the
method in~\cite{acm3} and that presented here should generalize to
higher order twist operators.

Using the technology and understanding developed in~\cite{acm2, acm3}
and here, we hope to address some important outstanding questions
concerning the D1D5 CFT and black hole physics. In particular, we hope
to elucidate the nature of the proposed
``non-renormalization'' theorem~\cite{dmw02}; how states of the CFT fragment,
thereby ``scrambling'' information~\cite{susskind,preskill}; black
hole formation; and the in-falling observer. All of these are
important and deep issues that will probably require lots of work to
resolve; however, this paper should serve as a step toward these
goals. All of the discussion here and in~\cite{acm2, acm3} has been
off-shell. A more modest next step would be to analyze the physics
that arise on-shell. Another task is to systematically analyze the
combinatorial factors that arise at different orders in perturbation
theory, perhaps using the technology developed in~\cite{rastelli1,
  rastelli2}.

\section*{Acknowledgments}

We are grateful for insightful conversations with and sundry
contributions from S.~Mathur.  We thank K.~Bobkov, B.~Dundee,
M.~Randeria, M.~Shigemori, and Y.~K.~Srivastava for several helpful
conversations.  The work of SGA is supported in part by DOE grant
DE-FG02-91ER-40690.  The work of BDC is supported by the Foundation
for Fundamental Research on Matter.

redpajama_set_name: RedPajamaArXiv


text:
New funding available to support research projects of early-career EAS members
One of the aims of the Society is to ensure that research in the atherosclerosis field continues and thrives for years to come. To do this, we need to encourage talented young scientists in their careers in the atherosclerosis field, and for many years EAS have done this through various travel grant & fellowship schemes – and through scientific programmes dedicated to young investigators.
In 2022 EAS were able to go one step further, with funds available to support research projects of early-career EAS members, on a competitive basis. By a competitive application process we have selected the three grant recipients for 2022.
Congratulations to EAS grant recipients 2022!
Mariëtte Boon
Leiden University Medical Center, Leiden, the Netherlands
Project: Improving cardiovascular risk factors in South Asians: Keep it cool
"I am very happy and honoured to receive the EAS grant! I cannot wait to expand my research by studying the effects of cold treatment in the metabolically vulnerable South Asian population"
Frida Emanuelsson
Dept. of Clinical Biochemistry, Rigshospitalet, Copenhagen, Denmark
Project: Causal effects of incretin signaling on lipid metabolism and cardiovascular disease.
"I'm so happy and grateful to the EAS for this grant and for the opportunities that come with it – I can't wait to get on with the project."
Mia Ståhle
Turku PET Centre, University of Turku, Finland
and Amsterdam UMC, University of Amsterdam, The Netherlands
Project: Evaluating new translational targets to detect and treat aortic valve stenosis
"I am very excited for this opportunity to explore and develop new diagnostic imaging tools that will ultimately benefit patients with aortic valve stenosis"
Call for grants 2023
We will open the Call for grant application process again in 2023 and 2024 – details about the application deadlines for 2023 will be announced in February.
You are eligible to apply for a grant if you meet all of these criteria:
You are an Individual Member of EAS
You are a researcher holding a PhD or MD title
It is no more than 10 years since you completed your PhD/MD (adjusted for parental leave)
Your workplace confirms that you can carry out the project, that you are in a status of employment or affiliation, and that you have access to the resources you will need to carry out the project.
What type of projects will be considered?
EAS will consider applications for funding for research projects focused on basic research, preclinical & translational studies.
Our terms and conditions of funding
The maximum total budget we will award for each research project is 20,000 €/year for up to 3 years.
The maximum duration of the proposed project is 36 months.
If you receive funding, you will be expected to provide a progress report at 12-month intervals from the start date of the project. EAS continuing to fund the project into Years 2 and 3 is subject to the progress report for the previous year.
The funds can cover salaries, personal grants, consumables, travel expenses.
The funds must be paid to an institutional bank account.
EAS will not pay additional overheads.
EAS makes no claim on Intellectual property resulting from the project.
As part of your application, you will need to provide
A copy of your PhD/MD certificate
A letter of recommendation from your supervisor/tutor/head of dept. This should confirm that you can carry out the project, that you are in a status of employment and that you have access to the resources you need to carry out the project.
Your CV (2 pages max, including publications. Highlight 3 of your publications that are most important to the proposed project)
A letter of motivation (1 page max)
A description of the project (5 pages max). Use these headings: background, hypothesis, study design, rational & preliminary work, methodology & experimental plan – including schedule, budget and equipment needed
How we will evaluate the applications
Applications will be reviewed by a Working Group appointed by the EAS Executive Committee.
The main criteria for evaluation of the applications is Scientific quality.
Applications will be evaluated on: Scientific rationale, Preliminary data, Design and methods, Project feasibility & PI competence.
Where several applications have equally high scientific quality, other considerations (age, gender, geography) will be taken into account.
What we expect from grant recipients
You should present the project results at EAS congress
At least one publication to originate from the project
You must acknowledge EAS in all presentations and publications of the supported work and results from it.
You must provide to EAS progress reports on the project every 12 months, with a final report at the conclusion of the project.
How we handle Conflict of Interest
In case a Working Group member has COI with an applicant, she/he steps out of the review of that application, and the other members of the WG grade the application.
One of the aims of the Society is to ensure that research in the atherosclerosis field continues and thrives for years to come. To do this, we need to encourage talented young scientists in their careers in the atherosclerosis field, and for many years EAS have done this through various travel grant & fellowship schemes - and through scientific programmes dedicated to young investigators.
"I am very happy and honoured to receive the EAS grant! I cannot wait to expand my research by studying the effects of cold treatment in the metabolically vulnerable South Asian population"
"I'm so happy and grateful to the EAS for this grant and for the opportunities that come with it – I can't wait to get on with the project."
"I am very excited for this opportunity to explore and develop new diagnostic imaging tools that will ultimately benefit patients with aortic valve stenosis"
We will open the Call for grant application process again in 2023 and 2024 - details about the application deadlines for 2023 will be announced in February.
A description of the project (5 pages max). Use these headings: background, hypothesis, study design, rational & preliminary work, methodology & experimental plan - including schedule, budget and equipment needed

redpajama_set_name: RedPajamaCommonCrawl


text:
Five in a row for Liverpool in Champions League
Liverpool's Mohamed Salah (left) celebrates with Jordan Henderson after scoring his team's second goal during the Champions League Group B match against FC Porto at Anfield stadium in Liverpool yesterday. Liverpool won 2-0.
LIVERPOOL, England (AP):
Thiago Alcantara's sublime long-range strike and Mohamed Salah's latest individual goal maintained Liverpool's perfect record in its Champions League group with a 2-0 win over Porto on Wednesday.
Liverpool was assured of advancing to the last 16 even before their fifth straight victory in Group B, while Porto stayed in second place ahead of a match against Atletico Madrid in two weeks that will help to determine which team qualifies with the English club.
Thiago's 52nd-minute opener was a thing of beauty, the Spain midfielder meeting a headed clearance with a first-time shot off his laces from 30 metres out that stayed low and flew into the corner.
Salah then scored a 10th goal in his last 10 Champions League games - and his 17th in all competitions this season - by dropping his shoulder as he drifted into the area from the right and shooting low inside the near post in the 70th.
Liverpool will look to complete a sweep of victories in the group stage, a feat never achieved by an English team, in their closing game against AC Milan.
On the same night, Porto host Atletico, who, like Milan, have four points. A win would guarantee Porto's progression to the knockout stage.
Other Sports Stories
Arsenal player in spot fixing investigation
Reifer hails Windies U-19 character
Double Crown back to dominate mile
Simmons plans new batting approach
New strikers deliver for Mount Pleasant
STETHS, Edwin Allen in Ben Francis final


The table below shows Australia's scores across the various indicators, from lowest to highest. We score quite negatively (-1.75) on making information on development funding publicly accessible. Why is this? We haven't been able to obtain the underlying data that CGD used for the calculations for the scores (they come from a separate report), but the methodology explains at least some of the score. The indicator is based on reporting to the OECD Creditor Reporting System (CRS), OECD Forward Spending Survey, and International Aid Transparency Initiative (IATI). Although Australia's score is helped by it producing IATI data (it's efforts were assessed as 'fair' in this regard), Australia doesn't report to the OECD's Forward Spending Survey at all, which means that it automatically loses one third of its score for this indicator. Its reporting to CRS was assessed as needing improvement, likely because of a mixture of lack of timeliness in providing data and lack of accuracy in reported data.
Australia's commitment to development: glass half-full or half-empty?


text:
Home/Featured/Virginia Forces Christian Ministries to Adopt 'Government Ideology' or Pay $100K
Virginia Forces Christian Ministries to Adopt 'Government Ideology' or Pay $100K
A Lawsuit filed in order to prevent Virginia from implementing two pro-LGBT laws that force "people of faith to adopt a particular government ideology under threat of punishment."
Written by Tyler O'Neil | Tuesday, October 6, 2020
The so-called Virginia Values Act (S.B. 868), which Gov. Ralph Northam (D-Va.) signed on Holy Saturday (the day before Easter Sunday) in the middle of a pandemic, compels churches, religious schools, and Christian ministries to hire employees who do not share their stated beliefs on marriage, sexuality, and gender identity. A companion law (H.B. 1429) requires ministries and others like them to pay for transgender surgery in employee health care plans, a procedure that violates these ministries' convictions.
Three Christian schools and a Christian network of pregnancy centers are suing Attorney General Mark Herring (D-Va.) in order to prevent Virginia from implementing two pro-LGBT laws that force "people of faith to adopt a particular government ideology under threat of punishment." The two laws purport to prevent "discrimination" against LGBT people but, in reality, they force Christian ministries to choose between violating their sincerely held religious beliefs or paying hefty fines, as much as $100,000 per offense.
"The faith of many Americans inspires them to act for the good of their neighbors and also requires them to abide by its teachings," Denise Harle, senior counsel at Alliance Defending Freedom (ADF), the law firm representing the Christian ministries, said in a statement. "Our clients offer spiritual guidance, education, pregnancy support, and athletic opportunities to their communities because of the religious beliefs that motivate them. But Virginia's new law forces these ministries to abandon and adjust their convictions or pay crippling fines—in direct violation of the Virginia Constitution and other state laws."
"Such government hostility toward people of faith has no place in a free society," Harle argued.
The lawsuit
The ministries filed a pre-enforcement challenge, a lawsuit designed to convince a court to prevent the government from implementing an unconstitutional law.
In the lawsuit, ADF is representing Calvary Road Christian School in Alexandria (preschool through 6th grade, 250 students), Grace Christian School in Staunton (preschool through 12th grade, 320 students), Community Christian Academy in Charlottesville (K-9, 50 students), and Care Net, a 501(c)3 nonprofit that supports a network of 1,100 pregnancy centers, churches, and other ministry organizations and has approximately 22,000 volunteers.
"For the Ministries, personnel is policy; and so they intentionally employ staff and recruit volunteers who further their respective Christian missions. Virginia's new laws, however, make this free religious exercise and association impossible — and label these liberties 'discrimination,'" the lawsuit argues.
S.B. 868 and H.B. 1429 require the ministries to hire employees who disagree with their beliefs on marriage, sexuality, and gender; mandate that the ministries hire employees whose beliefs and lifestyles are "antagonistic to the ministries' convictions"; prohibit the ministries from firing employees who oppose their missions; require the ministries to provide services in a manner that violates their beliefs; ban the ministries from even communicating their biblical beliefs; make the ministries use their facilities in a way that contradicts their beliefs; and force the ministries to pay for "gender reassignment" procedures in their health plans, even though the ministries object to these procedures.
These laws put the ministries "in an impossible position: they must either abandon the religious convictions they were founded upon, or be ready to face investigations, an onerous administrative process, fines up to $100,000 for each violation, unlimited compensatory and punitive damages and attorney-fee awards, and court orders forcing them to engage in actions that would violate their consciences."
The lawsuit brings five claims against AG Herring. It claims the LGBT laws violate the ministries' right of free exercise of religion under the Virginia Religious Freedom Restoration Act; violate their right to free exercise of religion under the Virginia Constitution; violate their right to free speech under the Virginia Constitution; violate the Virginia Constitution's Establishment Clause; and violate the state constitution's Due Process Clause.
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The Trans Teen Industrial Complex
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A Gasp of Pain, A Sigh of Relief


text:
Two Brothers Save a Cat
By Daniel Winters • August 20th, 2013
I'm a sucker for writing craft – the stuff that makes a good plot into something you can truly care about. I don't simply mean, "I like it. It's got a nice plot." I mean I truly care about skillful writing.
That may be why I like Brothers: A Tale of Two Sons so much. The game focuses on two brothers. Before we meet them they have already lost their mother, drowned despite the younger brother's best efforts. Now their father, too, is ill. The doctor points them in a direction for a cure, and we follow them on their quest. From here on we're stuck with these guys, so we'd better like them.
In movie scripts – the good movie scripts – there's usually a particular moment written early in the running time. Blake Snyder calls it the "save the cat" moment, but the idea is the same regardless of whether or not a cat is stuck up a tree.
"It's the scene where we meet the hero and the hero does something – like saving a cat – that defines who he is, and makes us, the audience, like him," writes Snyder.
With the condensed running time and lack of player control in film, you only need one of these scenes. An actor can constantly act in a way that tells us about his character – the way he walks, the way he talks, the way he does things – but a videogame character can't. Aside from cutscenes, the character is forced to act according to the player's whim. And it's hard to characterize someone who's being played by someone else.
Brothers knows how to save cats, though. This is unsurprising when you find that it was directed by a film director. Game designers have cribbed off movies since their inception, but they rarely do it when building their characters.
In Snyder's books he gives numerous examples of "save the cat" moments, one being from Aladdin. How do you write a film where you're supposed to root for a thief? He is a thief, no matter how charming his smile is. I don't want to be friends with a thief, but I could think about hanging out with a guy who steals bread and gives it to starving children instead of eating it himself – especially if he then saves the kids from getting whipped by a bad guy. Now he's more than a thief; now we can hope good things happen to him and that he gets what he desires.
Instead of one moment like this, in a sea of exposition, action scenes and careful explanations of motivations, Brothers gives us many.
You control the brothers independently – one analogue stick and one button for each. Maneuvering them next to something and then pressing the button has them interact with it. Instead of one short scene in the film, there's a short vignette for each thing you discover.
You find a woman is sweeping. The older brother can take her broom and sweep up for her. The younger brother, though, tries to impress her by balancing the broom on the palm of his hand. Earlier, they come across an old woman in her rocking chair. The older brother asks her for directions before thanking her. The younger brother turns her chair into a seesaw and gets everyone laughing.
And while these moments are just small cutscenes, they are not presented in the same way. Instead, they require you to hold down the button to carry on watching the interactions. It's gameplay disguised as a cutscene. A player interaction disguised as a character action. A character action that's designed to endear them to us.
Shockingly simple and shockingly effective.
It's impossible not to like the two brothers. And while the game crams their personalities down your throat, it never becomes boring or invasive – you're only reminded by how much you like these guys and how much you care for them.
This from a game without a recognized language. This from a game that purports to be a puzzle/platformer.
There's a reason why I like Two Brothers so much and why there's a buzz around the game. Blake Snyder has that covered, too: "Liking the person we go on a journey with is the single most important element in drawing us into the story."
In this case, it's two people. And they're damn good at saving cats.
Blake Snyder, Brothers: A Tale of Two Sons, Save the Cat
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The All New, All Different Last Week's Comics 8/21/2013


It might be getting chilly in November but the restaurant scene in London is hotting up (sorry, not sorry).
Ahead of Christmas, the capital is filling up with exciting new openings and pop-ups, including some hotly anticipated projects.
There are sweet spots from Michelin-starred chefs, cosy neighbourhood spots and plant-based concepts with an indulgent flare.
So forget about that diet, put your festive shopping on the back burner and book into one of these to warm up against the cold front.
Sushisamba is opening its second London site in Covent Garden, serving up its signature blend of Japanese, Brazilian and Peruvian cuisine.
As well as the main restaurant and bar, the new site will also have a 14-seat sushi counter.
Expect to see some exclusive menu items too, including the Samba Covent Garden Roll – a blend of tuna, salmon, yellowtail, white fish, avocado, cucumber, red onion, sweet potato corn, aji amarillo and leche de tigre.
Biff's Jack Shack, who serve self-billed 'filthy AF vegan junk food', have a new spot in Boxpark Shoreditch.
All their usual plant-based burgers, wingz and filthy fries are on the menu.
And while officially opening on November 1, they will be giving away 20 free burgers between October 29 and November 2 to celebrate the launch.
Clapham is getting a vegan pop-up in the shape of The Meet.
The plant-based menu will feature everything from sharing plates and burgers to late night bites.
And at the weekend, there's even bottomless vegan brunch and Sunday roast.
Portbobello Road is getting a new neighbourhood restaurant dishing up contemporary Italian food.
Artisan will offer the likes of grilled octopus with artichoke salad and their take on vitello tonnato.
And of course, there's fresh pasta and pizzas cooked in the wood-fired oven.
The long-awaited project from Albert Adria is finally landing on Regent Street.
Dessert-only concept Cakes & Bubbles will offer a taste of his Barcelona Michelin-starred restaurant, Tickets, alongside a whole host of other sweet treats.
And to celebrate their opening, they will be giving away free cheesecake and champagne to the first 40 guests on opening day, plus an extra 100 free cheesecake for those who didn't quite make it as the first through the doors.
Richard Falk, formerly head chef at The Dairy, will be heading up new all-day eatery Lino.
There's a low-waste approach to the menu, which includes using left-over croissants from the morning for ice creams later in the day, with everything made from scratch on site.
The ethos translates to the drinks too, with cocktails being made using ferments made in-house.
Coming to Hammersmith is new French bistro Le Petit Citron.
The focus will be on Provençal cooking, with dishes such as lemon and garlic poussin with French beans, persillade and pine nuts on the menu for lunch and dinner.
And at the weekend, it's all about brunch, featuring an abundance of (of course) croissants.
After successfully launching Market Hall Fulham earlier this year, the team behind the project are back with a second site – this time in Victoria.
The idea is the same – get a group of great vendors round an open indoor seating space.
There's Koya Ko by Koya, BaoziInn, Kerbisher & Malt, Super Tacos from Breddos and more vying for your attention this time.
The team behind Sager + Wilde have been putting their energy behind opening a new spot in Old Street.
Fare Bar + Canteen will be an all-day eatery combining a cafe, cocktail and wine bar, and restaurant.
Food-wise, it will be sharing plates with a blend of British and Mediterranean influences.
Hackney Arts Centre has had a bit of a make-over – becoming EartH – and with it comes a new eating space, EartH Kitchen.
The restaurant bit is taking over what was once the cinema foyer, and will have Chris Gillard, formerly of St John, at the helm.
You'll find dishes like griddled mackerel, with beetroot, leaves and horseradish, while for dessert, it's all about modern takes on traditional tarts and puddings.
The team behind Salon have gone to Peckham for their second concept, Levan.
It's an all-day affair, with freshly baked pastries kicking off the morning and simple bistro-inspired lunches.
For dinner, look out for dishes such as Comté fries with saffron aioli and smoked pollack with spinach, asparagus and vin jaune.
Bao-hungry Londoners have a new spot to try in Haymarket, where Brighton's Baby Bao is currently running a pop-up.
While inspired by Taiwanese bao buns, it also takes inspiration from further afield.
You can find flavours such as Korean BBQ prawns and cod tempura for the buns, and there are also lettuce wraps for those who are gluten-free.
Gamma Gamma has recently opened in Soho, serving up small plate fusion dishes.
Although billed as pan-Asian, you will find flavours from South America, such as ceviche and chimichurri, on the menu as well.
A bit left field is perhaps its focus on exotic meats, such as zebra, kangaroo and ostrich.
For a taste of the Philippines in Hoxton, there's now Kinilaw & Buko.
Kinilaw is all about a Filipino dish that's similar to ceviche, using palm vinegar and kalamansi limes.
On the sweet side is Buko, the Filipino take on ice cream, which are served in coconut shells or cones.


Does this mean we do not need teachers?
This is an exploration of how connection to the Internet has affected children's experience of the classroom.
In the UK the government has set targets for connection - these mean that effectively nearly all schools in all education authorities in England, for example, now have a connection to the Internet, while most secondary schools have, or soon will have, a broadband connection. In the East Riding of Yorkshire all schools have a connection of some kind; all secondary schools and a few other schools have broadband connections. The DfES wants us to have 20% of schools with a Broadband connection by August 2002 and 40%: by August 2004 - these are targets that the LEA can realistically expect to achieve.
Domestic connections are less evenly spread and students are sensitive to this - when I asked my daughter about her use at school, she said she had declined a chance to work online at school, so that friends who do not have access at home could do so instead.
And approaches to teaching with computer technologies are still evolving - a few years ago, some teachers may have aspired to a classroom with workstations for every child, each one networked and able to connect to the Internet. Now teachers may prefer one or two computers, a projector and a suitable input device (anything from a touch-sensitive board with interactive software to a keyboard or mouse), so that teaching and learning can be a more collective experience. Neither of these is simply better in all situations - both are more or less appropriate to particular ways of learning.
a child in a small East Riding primary school with two classes either of which will usually be in the room that has an interactive white board and data projector. Use of the technology is perhaps limited only by the teacher's plans and inclinations.
a GCSE student at a secondary school who has used Internet technology in school to support three subjects (geography, French and technology). Closer questioning reveals that the English teacher uses resources from the Web, but prints them out for use in a classroom without computer equipment. This student, meanwhile, makes extensive use of the technology at home, for study and communication.
a GCE student at a secondary school who has made use in the upper 6th year of Internet technology in one subject only at school, but who says he always uses it for research in any work he undertakes. He is more or less wholly autonomous and sees Internet technologies as giving the quickest route to much of what he wants to know.
The Internet has enabled many different technologies to evolve - some of which (newsgroups and bulletin boards) may already have passed out of common use. Others - Web browsing, e-mail, file transfer and chat are in very good health. If they become less popular it will perhaps be only because an alternative technology has improved on what they do - as Web-based forums have replaced bulletin boards.
Distinguishing these different technologies may be important for product developers and programmers, but if learners use them all, as and when they think they need to, this is perhaps no bad thing - in the same way that a learner with a print text might write notes in pencil, photocopy a page and highlight text without reflecting on the difference in these activities, because all are part of a joined-up programme of learning. We don't (normally) tell children, except the very young, what to use from the things in a pencil case.
Even without thinking of modern computer technologies, one could distinguish the classroom as a physical space from the classroom as a place occupied by a group of learners, with some kinds of social relationship and interaction, regulated and directed (some would say "led") by a teacher.
Internet technologies challenge our understanding because the classroom may become successively less coherent to our thinking - students may be in different physical spaces, but part of an organized group or they may have less in common, other than some shared experiences.
Within a traditional classroom a teacher may use computer technologies to reinforce methods of control, as much as to encourage independent thinking. But where students are able to use the technology without continual technical support, then it can enable some learners to do things for themselves, or give them limited but significant help with things like surface features of presentation, organizing writing and producing legible text with standard spelling forms.
Hi, I was wondering if u could help me!!!?
For yr 10 essay i have to compare to of the short stories from the anthology!!!
The thing is i dont know which 2 to compare! Can U help! And what should i include in the comparing??
The first student seems not to be confident about what to do next - but with a simple clear answer, may become more personally committed to the work. This could of course happen without the intervention of the technology - but it gives the student another chance of finding help.
I thought I'd let you know how pleased I am with my English resit result.
You gave me a lot of help before my exam and my marks have increased from 23 to 61. This changes my AS from an E to a C.
Thank you very much. I did everything you suggested and it seems to have worked. I feel much more confident about my A2 now.
But even with the first group, one can say that the students are independent in making the request - that is, in wanting to learn, for whatever reason. And they have found their way to someone they think, or hope, will be able to help.
Internet technologies give students an opportunity to display their work to wide - potentially worldwide - audiences. In this way, they can also help other students who may learn from their work, as well as from things that their teachers produce.
The first example is a word-processed document. This may look unremarkable - but becomes more impressive if one sees that it is written in English, and is the result of a collaboration, conducted entirely by use of Internet technologies, between two pairs of children - one pair in the Netherlands and the other in Italy. They have produced a document in English and their teacher has sent it to me to be displayed on a Web site that teaches about Anne Frank and the Holocaust. In studying this subject, it seems entirely appropriate to work in a way that does not recognize physical frontiers.
The second example is a graphic presentation, in Microsoft™ PowerPoint format. Students in an East Riding school produced this and other documents and presentations, following a visit to Anne Frank Huis in Amsterdam, and a summer school for able children, with Anne Frank as the focus.
The third example is a study of a musical interlude in a play by Arthur Miller. The writer is a student (of moderate ability, according to her teacher) who used a study-guide of mine, and produced this response as a text document. In collaboration with her teacher, I made it into a Web page. The guide is quite impressive in its own right - much more so when one knows that the author is not a native English speaker, but attends an international school in Switzerland.
The fourth example is an investigation for A-level English language - the student achieved a very high mark, but his work is of a kind that others could emulate and which therefore works as a model - often not the case with the best coursework. After the exam process was complete the student who wrote it gave me permission to publish it on a Web site, with appropriate contextual information.
The last example is a Web site, which I am currently building for a secondary school in the East Riding. The pupils work with their history teachers to find people to interview - they produce text files and locate documents and images to scan or photograph. I coordinate all this, organize it into a coherent site and publish on the Web. The editor of Spartacus, John Simkin, is publicizing it through a history newsletter, which has some 25,000 subscribers. Already I have had a message from a teacher in Romania, who says she will use it with her students.
I find your oral history web site great! It is very interesting and I plan to use it in my history classes as resource for my students.
Imagine this - a group of teachers in different countries, from Finland to Portugal, of the European Union create a project to write learning resources to teach languages of the community to non-native speakers. They have occasional meetings but mostly work from their schools and homes, to publish, jointly, Web documents in real time - the site is hosted in Austria and many of the group's members have never been there. This is a reality and you can see the sites by clicking on these links.
The association primarily gives some quality assurance - the sites are the work of qualified teachers and have substantial content. But the traffic on these sites also reflects the choices of the end users - students and teachers who know what they like. Although the resources are free to users, it makes sense for schools or LEAs to use modest finances to commission more content from the authors, as this will continue to be free in the future, as opposed to subscriptions, which need renewal.
The Cecil Slack archive was once a collection of letters. These letters twice appeared in books - heavily edited. In the East Riding a literacy consultant and adviser worked together to produce teaching materials, using a few of the letters as sources. In creating a Web site, I have had the pleasure of bringing the full collection of letters and diaries to a huge audience worldwide. I was able to work closely with my colleagues to publish their teaching resources and to develop some new ones. The site reaches beyond our LEA, but also has an effect in it - the head of history at one of our secondary schools was inspired to organize a visit to the battlefields of the Somme.
For the Anne Frank exhibition, I was able to plan before the event, and involve more colleagues. As well as the adviser for history, I was able to work closely with the project manager for healthy schools on creating teaching materials for citizenship and with a project manager and a primary head teacher, to develop resources for teaching Religious Education, based partly on our LEA's agreed syllabus. I also obtained permission from its author, a teacher, to reproduce an excellent resource from the Schoolhistory.co.uk Web site. And finally, the Web site records the activities of students at a secondary school, who focused on the exhibition for a summer school for gifted and talented pupils.
The idea of a machine or robot to take the teacher's place and work indefatigably without pay is as old as the earliest automata. You can see it wonderfully in Isaac Asimov's 1951 story (set in 2157) The Fun They Had.
The part Margie hated most was the slot where she had to put homework and test papers. She always had to write them out in a punch code they made her learn when she was six years old, and the mechanical teacher calculated the mark in no time...Margie went into the schoolroom. It was right next to her bedroom, and the mechanical teacher was on and waiting for her. It was always on at the same time every day except Saturday and Sunday, because her mother said little girls learned better if they learned at regular hours.
This idea remains a fantasy - the reality is that the technology does not replace the teacher but can lead to new ways of working. If we continue to want to assess students (which seems likely in the UK for years to come), then we can do so - but use Internet technologies to submit it (e-mail attachment, file transfer and so on). Students will still make excuses for things they have not done - but the excuses change: teachers will be familiar with claims of printer and hard-disk failure which are implausible, given the real reliability of these components.
What does not change ultimately is that a student still places trust in a qualified teacher - though there may be little or no direct real world contact. I have worked informally with students in this way - in one case for a year of a GCE course - and it is quite manageable. For all she knows I have two heads; for all I know so does she?
Teachers who publish learning resources in this way will necessarily only be able to help waifs and strays who ask for help. But we can be proactive if we are employed to work - whether as subject teachers or support tutors - with a designated group. And students can easily miss out on close attention in traditional classrooms - in fact, using computer technology could more easily show us if we are spending time unevenly or if we are favouring some students over others.
The potential of computer technology to help children who wish to learn outside the conventional classroom - some or all of the time - is massive. I find this exciting, but it has barely begun to happen yet in schools in the UK.
But it is not only pupils - there are many teachers with disabilities, and could be in the future, who could teach remotely, using computer technology. Of course, I do not suggest that those who want to work in schools and challenge prejudices should do anything else. But what of those who have done this and do not want to continue, or who want to work remotely from the start? The teacher's initial invisibility might disarm prejudice - and if the student were to learn later that his or her excellent teacher was partially sighted or had cerebral palsy, then he or she might form a deeper judgement of the teacher's abilities - and the PC term "differently able" would suddenly make sense.
There are many reasons why some young people do not want to be in school - there may even be some schools or classes in schools where this wish is healthy. Some children find it difficult to attend, or to do so every day. Some are excluded because their behaviour in school is not always appropriate. Others may feel they are bullied or may have a poor self-image. Where, for good reasons, schools, parents and students have accepted that full time attendance at school is not right, then Internet technologies can be very helpful in sustaining the learner's contact with, and inclusion in, the wider society they may one day be able to join or re-join.
Why should teachers not have a class of remote students, perhaps making occasional visits to a school or learning centre, but otherwise working from home? Of course there are some practical difficulties for some learners in this but there are also solutions. For example, there are many good teachers who no longer feel able to work in schools - or who are themselves frustrated by the same things that drive the children away. And there may be schools or departments where teachers are not familiar with the UK curriculum or exam courses - but these may be good teachers, who need only some help in adapting their knowledge and teaching style to a new situation.