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INLO-PUB-02/03\\
FAU-TP3-03-02\\
%hep-th/0212xxx
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\par
\vskip .5 truecm
\large \centerline{\bf Doubly Periodic Instantons and their Constituents}
%\vskip .3 truecm
%\large \centerline{\bf and their Constituents}
\par
\vskip 0.5 truecm
\normalsize
\begin{center}
{\bf C.~Ford} ${}^{a}$ and
{\bf  J.~M.~Pawlowski} ${}^{b}$
\\
\vskip 0.5 truecm
${}^a$\it{
Instituut-Lorentz for Theoretical Physics\\
Niels Bohrweg 2, 2300 RA Leiden, The Netherlands\\ 
{\small\sf ford@lorentz.leidenuniv.nl}\\
}
                                            
\vskip 0.5 truecm
${}^b$\it{Institut f\"ur Theoretische Physik III,
Universit\"at Erlangen\\
Staudtstra\ss e 7,
D-91058 Erlangen, Germany \\
{\small\sf jmp@theorie3.physik.uni-erlangen.de}
}
%\vskip 0.5 truecm



\vskip 0.5 true cm

\end{center}

\vskip .7 truecm\normalsize
\begin{abstract} 
  Using the Nahm transform we investigate doubly periodic charge one
  $SU(2)$ instantons with radial symmetry. Two special points where
  the Nahm zero modes have softer singularities are identified as the
  locations of instanton core constituents. For a square torus this
  constituent picture is closely reflected in the action density.  In
  rectangular tori with large aspect ratios the cores merge to form
  monopole-like objects. For particular values of the parameters the
  torus can be cut in half yielding two copies of a twisted charge
  ${1\over2}$ instanton.  These findings are illustrated with plots of
  the action density within a two-dimensional slice containing the
  constituents.



\end{abstract}
\baselineskip=18.9pt
\section{Introduction}

Monopoles are at the heart of various kinds of electric-magnetic
duality in extended super Yang-Mills theory, which have led to an
impressive body of quantitative results.  The role of monopoles and
other topological constructs in non-supersymmetric Yang-Mills theory
is less clear.  Using such objects, particularly monopoles and
vortices, models of confinement and chiral symmetry breaking have been
advanced.  A weakness of these models is that classical Yang-Mills
does not possess monopole solutions, rather they have instanton
solutions.  However, in recent years it has become clear that \it
periodic instantons \rm have a substructure that can be identified
with monopoles and other extended objects.  In particular, charge one
$SU(N)$ calorons (finite temperature instantons) can be viewed as
bound states of $N$ monopole constituents
\cite{Lee:1998bb,Kraan:1998pm}.  Provided the `size' of the instanton
is larger than the period (or inverse temperature) tubes of action
density form. These can be identified with the worldlines of the
constituent monopoles.  It would be interesting to extend these
results to higher charge sectors and higher tori, i.e.  more than one
period so that we are considering instantons on $\T^n\times\R^{4-n}$,
$n>1$.  Indeed, lattice based numerical studies indicate that doubly
periodic instantons ($n=2$) have remarkable properties including
fractionally charged solutions, vortex-like objects and an
exponentially decaying action density \cite{Gonzalez-Arroyo:1998ez}.
None of these attributes are shared by the calorons. Nevertheless, it
would be desirable to have a unified description of calorons and
multiply-periodic instantons and their constituents.

The calorons are best understood within a formalism due to Nahm
\cite{Nahm:1979yw,Nahm:1983sv}.  This is an extension of the ADHM
construction.  Nahm's approach was generalised to other four
manifolds, notably the four-torus, $\T^4$ \cite{Braam:1988qk}.  Here
it becomes a duality, mapping $U(N)$ charge $k$ instantons on the four
torus to $U(k)$ charge $N$ instantons on the dual torus, $\tilde\T^4$
(defined by inverting the four periods).  The caloron construction and
even ADHM can be understood as a limiting case of the four torus
duality.  One therefore expects that doubly periodic instantons can be
treated in a similar fashion.  In \cite{Ford:2000zt} doubly periodic
charge one instantons with radial symmetry in the non-compact $\R^2$
directions were considered. Under the Nahm transformation they are
mapped to abelian potentials on the dual torus $\tilde\T^2$.  Starting
with these rather simple Nahm potentials the original instantons can
be `recovered' via the inverse Nahm transform.  Technically, this
involves solving certain Weyl-Dirac equations; for each
$x\in\T^2\times\R^2$ one has a different Weyl equation on
$\tilde\T^2$. The Weyl zero modes were determined explicitly for a two
dimensional subspace (corresponding to the origin of $\R^2$).

In a recent letter \cite{Ford:2002pa} we used these zero modes to
compute the action density, analytically and numerically, within the
subspace.  We found that there are points in $\T^2\times\R^2$ where
the solution of the Weyl equations have softer singularities.  These
are obvious candidates for constituent locations.  We interpreted
these constituents as overlapping \it instanton cores.\rm\, For square
tori our constituent picture correlates closely to the action density.
In this paper more details of the calculations are provided and action
density plots for rectangular tori are presented.  Starting with a
square torus we stretch one of the sides to see how this affects the
cores.  At first they become stretched with the torus.  On further
stretching the cores begin to overlap with periodic copies of
themselves in the short direction to form monopole-like objects much
as in the caloron solutions.  Some of the charge one solutions we
considered can be cut in half yielding two copies of a charge
$\frac{1}{2}$ instanton.  This procedure is only consistent if the
gauge potential has specific asymptotic boundary conditions which are
analogous to a four torus twist.  Here we argue that the gauge
potential has suitable asymptotics and show that the field strength
decays exponentially.

The paper is organised as follows.  In section 2 the basics of the
Nahm transform are briefly reviewed, and its application to doubly
periodic instantons is discussed.  Sections 3 to 5 deal with the Nahm
potentials, zero modes and constituents corresponding to charge one
instantons with gauge group $SU(2)$. Section 6 describes some
technical results regarding Green's function that crop up in field
strength calculations.  Our numerical action density computations are
detailed in section 7.  Section 8 considers charge $\frac{1}{2}$
solutions and the asymptotics of the gauge field and field strength
are discussed in chapter 9.



\section {Nahm Transform for Doubly Periodic Instantons}

Before we specialise to $\T^2\times\R^2$ let us briefly recall how the
Nahm transform is formulated on the four torus \cite{Braam:1988qk}
(for a string-theoretic approach see \cite{Hori:1999me}).  Start with
a self dual anti-hermitian $SU(N)$ potential, $A_\mu(x)$, on a
euclidean four-torus with topological charge $k$.  A gauge field on
$\T^4$ is understood to be an $\R^4$ potential which is periodic
(modulo gauge transformations) with respect to $x_\mu\rightarrow
x_\mu+L_\mu \quad (\mu=0,1,2,3)$, the $L_\mu$ being the four periods
of the torus. Self-duality is the requirement that
$F_{\mu\nu}=\frac{1}{2} \epsilon_{\mu\nu\alpha\beta}F_{\alpha\beta}$,
and we have taken the convention $\epsilon_{0123}=-1$.  The next step
is to turn the $SU(N)$ instanton into a $U(N)$ instanton by adding a
constant $U(1)$ potential, $A_\mu(x)\rightarrow A_\mu(x)-iz_\mu$,
where the $z_\mu$ are real numbers.  We can regard the $z_\mu$ as
coordinates of the \it dual torus \rm, $\tilde\T^4$, since the shifts
$z_\mu\rightarrow z_\mu+2\pi/L_\mu$ can be effected via \it periodic
\rm $U(1)$ gauge transformations.

Now consider the $U(N)$ Weyl operator
\begin{equation}
D_z(A)=\sigma_\mu D^\mu_z(A),~~~~
D_z^\mu(A)=\partial^\mu+A^\mu(x)-iz^\mu 
\end{equation}
with $\sigma_\mu=(1,i\tau_1,i\tau_2,i\tau_3)$ where the $\tau_i$ are Pauli
matrices.  Provided certain mathematical technicalities are met
$D^\dagger_z(A)=-\sigma_\mu^\dagger D_z^\mu(A)$ has $k$
square-integrable zero modes $\psi^i(x;z)$ with $i=1,2,...,k$. The
Nahm potential is defined as
\begin{equation}\label{nahm}
\hat A_\mu^{ij}(z)=\int_{T^4}d^4x\,
{\psi^i}^\dagger(x;z)\frac{\partial}{\partial z^\mu}
\psi^j(x;z).\label{basicnahm}
\end{equation}
Here the zero modes are taken to be orthonormal.  Remarkably, $\hat
A(z)$ is a $U(k)$ instanton on the dual torus with topological charge
$N$.  The associated field strength can be written as follows
\begin{equation}\label{associateF}
\hat F^{ij}_{\mu\nu}(z)
=\int_{T^4}d^4x\int_{T^4}d^4x'\,
\psi^i{}^\dagger(x;z)\sigma_\mu (D_z^\dagger D_z)^{-1}
(x,x';z)\sigma_\nu^\dagger
\psi^j(x;z)-[\mu\leftrightarrow\nu].
\end{equation}
The self-duality of $F_{\mu\nu}$ implies $D_z^\dagger D_z=\sigma_0 
(D^\mu_z)^2$.  Accordingly, $(D^\dagger_z D_z)^{-1}(x,x';z)$ commutes with
all $\sigma_\mu$, from which the self-duality of $\hat F$ follows.
The `original' gauge potential, $A_\mu(x)$, can be recovered by Nahm
transforming $\hat A_\mu(z)$
\begin{equation}\label{t4inverse}
A_\mu^{pq}(x)=\int_{\tilde T^4} d^4 z
~\psi^p{}^\dagger(x;z)
\frac{\partial}{\partial x^\mu}
\psi^q(z;x),\end{equation}
where the
$\psi^p(z;x)$, $p=1,2,...,N$ are an orthonormal set of zero
modes of
$D_x^\dagger(\hat A)=-\sigma_\mu^\dagger D_x^\mu(\hat A)$
with
$D_x^\mu={\partial}/{\partial z_\mu}+\hat A^\mu(z)-ix^\mu$.
In other words the four torus Nahm transform is involutive.

Formally, one can obtain the $\T^2\times\R^2$ Nahm transform by taking
two of the periods, say $L_0$ and $L_3$, to be infinite.  Given an
$SU(N)$ instanton periodic with respect to $x_1\rightarrow x_1+L_1$,
$x_2\rightarrow x_2+L_2$ its Nahm transform is
\begin{eqnarray}
\hat A_\mu^{ij}(z)&=&\int_{T^2\times R^2} d^4x\, 
\psi^i{}^\dagger(x;z)\frac{\partial}{\partial z^\mu}\psi^j(x;z),
~~~~~~~\mu=1,2, \label{t2nahm}\nonumber
\\
\hat A_\mu^{ij}(z)&=&\int_{T^2\times R^2} d^4x\,
\psi^i{}^\dagger(x;z)ix_\mu \psi^j(x;z),
~~~~~~~\mu=0,3, 
\end{eqnarray}
where the $\psi^i(x;z)$ with $i=1,...,k$ are orthonormal zero modes of
$D_z^\dagger(A)$.  Note that we may gauge $z_0$ and $z_3$ to zero.  By
assumption the zero modes $\psi^i(x;z)$ are square integrable.  This
does not guarantee that the integrals in (\ref{t2nahm}) exist.
However, the integrals only diverge at special $z_\mu$ values where
the zero modes do not decay exponentially.  The number of these points
turns out to be $N$, i.e. it is determined by the gauge group of the
instanton.  To summarise, the Nahm transform maps doubly periodic
$SU(N)$ instantons with topological charge $k$ to self dual $U(k)$
potentials on the dual torus, $\tilde \T^2$, with $N$ singularities.

The dimensionally reduced self-duality equations (or Hitchin
equations) take a particularly simple form in complex coordinates
\begin{eqnarray}\label{complex}
y=z_1+iz_2, & & \bar y=z_1-i z_2,
\end{eqnarray} 
with derivatives $\partial_y=\h(\partial_{z_1}-i\partial_{z_2})$,
$\partial_{\bar y}=\h(\partial_{z_1}+i\partial_{z_2})$.  Combine $\hat
A_1(z)$ and $\hat A_2(z)$ into a complex gauge potential, $\hat
A_y=\frac{1}{2}(\hat A_1(z)-i\hat A_2(z))$ and $\hat A_{\bar
  y}=\frac{1}{2}(\hat A_1(z)+i\hat A_2(z))$, and form a `Higgs' field
out of the remaining components, $\Phi(z)=\h(\hat A_0(z)-i\hat
A_3(z))$.  The self-duality equations read
\begin{equation}
[D_y,\Phi]=0,~~~~~~~
\hat F_{y\bar y}=[D_y,D_{\bar y}]=[\Phi,\Phi^\dagger].
\end{equation}
The reader should be aware that we will occasionally use both the
cartesian $z$ and complex $y$ coordinates in the same equation. The
Nahm transform (\ref{t2nahm}) is a straightforward limit of the $\T^4$
version.  The inverse transform is a different matter.  That is, given
a (singular) solution of the Hitchin equations on the dual torus,
$\tilde \T^2$, how does one recover the corresponding doubly periodic
instanton?  To our knowledge there is not a mathematically rigorous
treatment of this issue in the literature.  Our approach will be to
study specific simple solutions of the Hitchin equations and
investigate whether the number of normalisable zero modes of
$D_x^\dagger(\hat A)$ matches the number of singularities of $\hat
A(z)$.  When this is the case we can then ask whether the natural
analogue of (\ref{t4inverse}), namely
\begin{equation}\label{t2inverse}
A^{pq}_\mu(x)=\int_{\tilde T^2} d^2z\,
\psi^p{}^\dagger(z;x)\frac{\partial}{\partial x^\mu}
\psi^q(z;x),
\end{equation}
provides doubly periodic instantons.




\section{Charge One}

In the one-instanton case the Nahm potential is abelian and the
Hitchin equations reduce to
\begin{equation}
\hat F_{y\bar y}=0,~~~~\partial_y\Phi=0,\label{abhitchin}
\end{equation}
so that $\Phi$ is anti-holomorphic in $y$.  As we shall see, $A_y$ can
be taken as holomorphic.  Since $A_y$ is periodic it is an elliptic
function.  It is well known that they have at least two singularities.
We expect the number of singularities to correspond to $N$.  Let us
stick to the simplest case $N=2$. Thus $\hat A_y$ will have two simple
poles \footnote{The Weyl equations for Nahm potentials with higher
  poles do not admit normalisable zero modes.}.  Consider the ansatz
\begin{equation}\label{ansatz}
\hat A_y=\partial_y\phi,~~~~~~~~
\hat A_{\bar y}=-\partial_{\bar y}\phi,
\end{equation}
which gives $\hat F_{y\bar y}=-2\partial_y\partial_{\bar y}\phi$.  It
follows from \eq{abhitchin} that $\phi$ must be harmonic except at two
singularities.  A suitable $\phi$ satisfies
\begin{equation}
\left(
\partial_{z_1}^2+\partial_{z_2}^2\right)\phi(z)
=-2\pi\kappa\left[
\delta^2(z-\omega)-\delta^2(z+\omega)\right],
\end{equation}
where $\kappa$ is a constant and $\pm\omega$ are the positions of the
two singularities (we have used translational invariance to shift the
`centre of gravity' of the singularities to the origin).  The delta
functions should be read as periodic (with respect to $z_1\rightarrow
z_1 +2\pi/L_1$ and $z_2\rightarrow z_2+2\pi/L_2$).  Physically, the
Nahm potential describes two Aharonov-Bohm fluxes of strength $\kappa$
and $-\kappa$ threading the dual torus.  They must have equal and
opposite strength to ensure a periodic $\hat A$.  We may assume that
$\kappa$ lies between $0$ and $1$ since it is possible via a
(singular) gauge transformation to shift $\kappa$ by an integer amount
(under such a transformation the \sl total \rm flux through
$\tilde\T^2$ remains zero). One can write $\phi$ explicitly in terms
of Jacobi theta functions (see for example\ \cite{mumford})
\begin{equation}\label{phi}
\phi(z)=\frac{\kappa}{2}\left(
\log
\frac{\left|\theta\left(
(y+\omega_1+i\omega_2)\frac{L_1}{2\pi}
+\frac{1}{2}+\frac{iL_1}{2L_2},\frac{iL_1}{L_2}\right)\right|^2}
{\left|\theta\left(
(y-\omega_1-i\omega_2)\frac{L_1}{2\pi}
+\frac{1}{2}+\frac{iL_1}{2L_2},\frac{iL_1}{L_2}\right)\right|^2}
+\frac{iL_1 L_2 \omega_2}{\pi} (y-\bar y)-
2\omega_2 L_1
\right),
\end{equation}
where $y$ and $\bar y$ are the complex coordinates introduced in
\eq{complex}.  The theta function is defined as
\begin{equation}
\theta(w,\tau)=
\sum_{n=-\infty}^\infty
e^{i\pi n^2\tau+2\pi in w},
~~~~~~~\hbox{Im}\,\tau>0,\end{equation}
and has the periodicity properties
$
\theta(w+1,\tau)=\theta(w,\tau)$ and
$\theta(w+\tau,\tau)=e^{-i\pi \tau-2\pi i w}
\theta(w,\tau)$.
In each cell $\theta(w,\tau)$ has a single zero
located at the centre of the torus ($w=\h+\h\tau$).
We have chosen the constant term in (\ref{phi}) so that
the integral of $\phi$ over the dual torus is zero.
This renders $\phi(z)$ an odd function, $\phi(-z)=-\phi(z)$.
It is easy to see that inserting (\ref{phi}) into
(\ref{ansatz}) yields a holomorphic $A_y$ with simple poles
at the fluxes.


What about the Higgs field?  Since we are aiming for an $SU(2)$
instanton it must, like $\hat A(z)$, have two singularities. The
singularities must be at the same two positions as in $A_{\bar y}$ or
else we have a total of four singularities which means we are
considering a (rather special) $SU(4)$ instanton.  The simplest way to
arrange for $A_{\bar y}$ and the Higgs to have singularities at the
same points is to choose them to be proportional
\begin{equation}
\Phi=\alpha \partial_{\bar y} \phi,
\end{equation}
where $\alpha$ is a complex constant.  Not all possibilities are
exhausted, since while the poles must coincide the zeroes need not.
This remaining ambiguity corresponds to the freedom to add to $\Phi$ a
complex constant.  However when we insert our Nahm potential into the
Weyl equation such a shift is equivalent to a translation of $x_0$ and
$x_3$.  This Nahm potential was derived in \cite{Ford:2000zt} via the
ADHM construction.  Here the parameters $\kappa$ and $\alpha$ are
related to the `size', $\lambda$, of an instanton centred at $x_\mu=0$,
\begin{equation}\label{adhm}
\sqrt{\kappa^2+|\alpha|^2}=\frac{\pi \lambda^2}{L_1 L_2}.
\end{equation}
Although we do not directly use the ADHM construction in the present
paper this relation proves useful in interpreting our results.  The
Nahm potential we have described has five parameters; $\omega _1$,
$\omega_2$ fixing the flux separations, $\kappa$ the flux strength and
the complex parameter $\alpha$ specifying the Higgs field. Then, with
the four translations in $\T^2\times \R^2$, the corresponding doubly
periodic instanton has a total of nine parameters.

To reconstruct the charge one $SU(2)$ instantons corresponding to our
simple Nahm potential we require the two zero modes of the Weyl
operator
\begin{equation}\label{weylop}
-\frac{i}{2} D_x^\dagger(\hat A)=
\left(
\begin{array}{cc}
\h \bar x_\perp +i\alpha\partial_{\bar y}\phi&
\partial_y+\partial_y\phi-\frac{i}{2}\bar x_{||}
\\
\partial_{\bar y}-\partial_{\bar y}\phi-\frac{i}{2}x_{||}
&\h x_\perp-i \bar\alpha\partial_y\phi
\end{array}\right). 
\end{equation}
In addition to the complex coordinates $y$, $\bar y$ on $\tilde\T^2$
we have introduced two sets of complex coordinates for
$\T^2\times\R^2$; in the `parallel' directions $x_{||}=x_1+ix_2$,
$\bar x_{||}=x_1-ix_2$, and in the `transverse' non-compact directions
$x_\perp=x_0+ix_3$, $\bar x_\perp=x_0-ix_3$.  In this paper we shall
concentrate on the special case $\alpha=0$, i.e. zero Higgs.  This
means that the corresponding instanton will be radially symmetric: the
action density depends on $x_1$, $x_2$ and $r=|x_\perp|$ only. These
radially symmetric solutions are a seven parameter subset of the
doubly periodic $SU(2)$ one-instantons.





\section{Nahm Zero Modes}

From now on we stick to the zero Higgs case (i.e.\ $\alpha=0$).  Even
with this specialisation the Weyl equations are still rather
forbidding.  However, there is one special case which is immediately
tractable: when $x_\perp=0$ the Weyl equation decouples and the two
zero modes have a simple form \cite{Ford:2000zt}
\begin{equation}\label{modes}
\psi^1(z;x)=\left(\begin{array}{cr}
0 \\ e^{-\phi(z)}G_+(z-\omega)
\end{array}\right),~~~~
\psi^2(z;x)=\left(\begin{array}{cr}
e^{\phi(z)}G_-(z+\omega)\\
0
\end{array}\right),
\end{equation}
where $G_\pm(z)$ are periodic Green's functions satisfying
\begin{equation}
\left(-i\partial_{y}-\h\bar x_{||}\right)G_+(z)
=\h\delta^2(z),~~~~
\left(-i\partial_{\bar y}-\h x_{||}\right)
G_-(z)=\h\delta^2(z).
\end{equation}
Here we have used both the $z$ and $y$ coordinates in the same
equation; this possibly confusing usage will continue until the end of
section~\ref{sec:soft}.  $G_-(z)$ has a theta function representation
\begin{equation}\label{szeboe}
G_-(z)=\frac{iL_1}{4\pi^2}
e^{\h ix_{||}(\bar y-y)}
\frac{\theta'(\h+\frac{iL_1}{2L_2},\frac{iL_1}{L_2})\,\theta
(\frac{L_1}{2\pi} 
y +\h+\frac{iL_1}{2L_2}-\frac{i}{L_2}x_{||},\frac{iL_1}{L_2})}{
\theta(\h+\frac{iL_1}{2L_2}-\frac{i}{L_2}x_{||},\frac{iL_1}{L_2})\,
\theta(\frac{L_1}{2\pi} y+\h+\frac{iL_1}{2L_2},\frac{i L_1}{L_2})},
\end{equation}
with $\theta'(w,\tau)=\partial_w \theta(w,\tau)$.  The corresponding
result for $G_+(z)$ can be obtained via $G_+(z)=G_-^*(-z)$.  The zero
modes have square-integrable singularities at both fluxes: near
$z\sim\omega$ we have
$|\psi^1|^2\propto|y-\omega_1-i\omega_2|^{2(\kappa-1)}$ and
$|\psi^2|^2\propto |y-\omega_1-i\omega_2|^{-2\kappa}$.  Near the other
flux the singularity profiles of the two modes are exchanged in that
$|\psi^1|^2\propto|y+\omega_1+i\omega_2|^{-2\kappa}$ and
$|\psi^2|^2\propto |y+\omega_1+i\omega_2|^{2(\kappa-1)}$.  By virtue
of the second theta function in the numerator of (\ref{szeboe})
$\psi^2(z;x)$ has a single zero in $\tilde \T^2$ at
$y=-\omega_1-i\omega_2-2\pi i x_{||}/(L_1L_2) $ (see also
\cite{Reinhardt:2002cm}).  Similarly, $\psi^1(z;x)$ has a zero at $y=
\omega_1+i\omega_2-2\pi i x_{||}/(L_1L_2)$.




\section{Soft zero modes}\label{sec:soft}

When $x_{||}=0$ the Green's functions $G_\pm$ do not exist.  This does
not mean there are no zero modes, indeed one can see that
\begin{equation}
\psi^1(z;x=0)=
\left(\begin{array}{cr}
0\\e^{-\phi(z)}\end{array}\right),~~~~~~~
\psi^2(z;x=0)=
\left(\begin{array}{cr}
e^{\phi(z)}\\
0
\end{array}\right),
\end{equation}
are solutions of the $x_{||}=0$ Weyl equation.  $\psi^1$ has the
expected square-integrable singularity at $z=-\omega$, but for
$z=\omega$, $\psi^1(z;x=0)$ is zero.  On the other hand
$\psi^2(z,x=0)$ diverges at $z=\omega$ but not at $z=-\omega$. By
choosing $x_{||}=0$ we have effectively moved the zeroes of the zero
modes to flux locations.  There is another way of doing this; choosing
$x_{||}=-iL_1 L_2(\omega_1+i\omega_2)/\pi$ brings the zero of $\psi^1$
to the other flux (and similarly for $\psi^2$).  Thus there are two
$x_{||}$ values where each Nahm zero mode diverges at only one flux.
We wish to investigate whether these `soft' points can be interpreted
as the locations of some kind of constituent.


The singularity profiles of $\psi^1$ and $\psi^2$ are exchanged under
the replacement $\kappa\rightarrow 1-\kappa$ suggesting that the
constituents are exchanged under this mapping. That is, if there are
indeed lumps at the two points, then $\kappa\rightarrow 1-\kappa$
swaps the two lumps.  The following result formalises this idea
\begin{equation} \label{duality}
F_{\mu\nu}(x_{||},x_\perp,\kappa)=
V^{-1}(x)
F_{\mu\nu}\left(-x_{||}+\frac{L_1 L_2}{i\pi}(\omega_1+i\omega_2),-x_\perp,
1-\kappa\right)V(x),
\end{equation}
where $V(x )$ is some $U(2)$ gauge transformation.  The proof of
\Eq{duality} goes as follows: make the change of variables
$z\rightarrow -z$ in (\ref{t2inverse}). The zero mode $\psi^p(-z;x)$
satisfies the same Weyl equation as $\psi^p(z;x)$ except that the
signs of $\kappa$ and the $x_\mu$ are flipped.  However, $-\kappa$
does not lie between $0$ and $1$.  Under periodic gauge
transformations $\hat A(z,-\kappa)$ and $\hat A(z,1-\kappa)$ are
equivalent up to a constant potential
\begin{equation}
\hat A_\mu(z,-\kappa)= U^{-1}(z)\left(
\partial_{\mu}+\hat A_\mu(z,1-\kappa)+B_\mu\right)U(z), 
\end{equation}
where $\pi B_1=-iL_1 L_2\omega_2$ and $\pi B_2=iL_1 L_2 \omega_1$.
Explicitly
\begin{eqnarray}
U(z)&=&
e^{\h\omega_1 L_1 L_2(y-\bar y)/\pi}\\ 
&\times&\left[
\frac{
\theta\left(\frac{L_1}{2\pi}(\bar y-\omega_1+i\omega_2)+\h-\frac{iL_1}{2L_2}
,\frac{iL_1}{L_2}\right)
\theta\left(\frac{L_1}{2\pi}( y+\omega_1+i\omega_2)+\h+\frac{iL_1}{2L_2}
,\frac{iL_1}{L_2}\right)}
{\theta\left(\frac{L_1}{2\pi}( y-\omega_1+i\omega_2)+\h+\frac{iL_1}{2L_2}
,\frac{iL_1}{L_2}\right)
\theta\left(\frac{L_1}{2\pi}( \bar y+\omega_1-i\omega_2)+\h-\frac{iL_1}{2L_2}
,\frac{iL_1}{L_2}\right)}\right]^{\h}.\nonumber
\end{eqnarray}
In the Weyl equation $B_\mu$ can be absorbed into $x_1$ and $x_2$.
Thus $U^{-1}(z)\psi^p(z,-x_{||}-i L_1 L_2(\omega_1+i\omega_2)/\pi,
-x_\perp)$ satisfies the same Weyl equation as
$\psi^p(-z;x_{||},x_\perp)$.

 

\section{Green's Function Approach}\label{Greens}

For the $\T^4$ transform, the original field strength can be written
\begin{equation}\label{t4strength}
F_{\mu\nu}^{pq}(x)=\int_{\tilde T^4}d^4z
\int_{\tilde T^4} d^4z'\, 
\psi^p{}^\dagger(z;x)\sigma_\mu~(D_x^\dagger D_x)^{-1}(z,z')~\sigma_\nu^\dagger
\psi^q(z';x)-[\mu\leftrightarrow\nu],\end{equation}
and as with the corresponding expression for $\hat F_{\mu\nu}^{ij}
(z)$, equation \eq{associateF}, the self-duality of $F_{\mu\nu}^{pq}(x)$
is manifest because the Green's function $(D_x^\dagger D_x)^{-1}(z,z')$
commutes with all $\sigma_\mu$.
Starting directly from the inverse Nahm transform (\ref{t4inverse}) we have
\begin{eqnarray}\label{direct}
F_{\mu\nu}^{pq}(x)&=&
\int_{\tilde T^4}d^4z\,
\partial_\mu\psi^p{}^\dagger(z;x)
\partial_\nu\psi^q(z;x)\\ \nonumber
&&+\int_{\tilde T^4}d^4z\,
\psi^p{}^\dagger(z;x)\partial_\mu\psi^r(z;x)
\int_{\tilde T^4} d^4z'\,
\psi^r{}^\dagger(z';x)\partial_\nu\psi^q(z';x)
\psi-[\mu\leftrightarrow\nu], 
\end{eqnarray}
where $\partial_\mu=\partial/\partial x^\mu$. To get from
(\ref{direct}) to (\ref{t4strength}) we require the derivative of the
Nahm zero modes with respect to the coordinates of $\T^4$
\begin{equation}\label{modederivative}
\frac{\partial}{\partial x^\mu}\psi^q=- i\,
D_x(\hat A)\left(
D_x^\dagger(\hat A)
D_x(\hat A)\right)^{-1}\sigma_\mu^\dagger\psi^q+
\psi^r~R_\mu^{rq}(x),\end{equation}
for some $N\times N$ matrix valued
vector field on the four torus, $R_\mu^{qr}(x)$.
This clearly satisfies the equation obtained
by differentiating the Weyl equation, $D_x^\dagger(\hat A)\,\psi^q(z;x)=0$,
with respect to $x^\mu$.
Multiplying \eq{modederivative} from the left with $\psi^p{}^\dagger$ and 
integrating over the dual torus yields
\begin{equation}\label{t4R}
R^{pq}(x)=A^{pq}(x),
\end{equation}
which together with equations (\ref{direct}) and 
(\ref{modederivative}) give (\ref{t4strength}).

Our analysis of doubly periodic instantons has been based on the Weyl
equation and the inverse Nahm transform (\ref{t2inverse}).  These are
the exact counterparts of four torus equations.  Therefore one might
expect all the results quoted above to carry over directly to
$\T^2\times\R^2$.  There are, however, a number of important
differences between the two cases.  With regard to the
$\T^2\times\R^2$ construction, the Green's function $(D_x^\dagger(\hat
A)D_x(\hat A))^{-1}(z,z')$ does not commute with the $\sigma_\mu$
matrices.  Away from the $N$ singularities of $\hat A(z)$,
$D_x^\dagger(\hat A)D_x(\hat A)$ commutes with the $\sigma_\mu$.  At
the singularities it has source terms which are not proportional to
$\sigma_0$.  For the $U(1)$ Nahm potential considered in section 3 (as
usual we set $\alpha=0$)
\begin{equation}D_x^\dagger D_x=
\sigma_0 (D_x^\mu)^2+2\pi i \kappa\sigma_3\left[
\delta^2(z-\omega)-\delta^2(z+\omega)\right],
\end{equation}
indicating that $(D_x^\dagger D_x)^{-1}$ does not commute with
$\sigma_1$ and $\sigma_2$.

We will show that the naive $\T^2\times\R^2$ limit of
(\ref{t4strength}) does not apply, so that $(D_x^\dagger D_x)^{-1}$
not being proportional to $\sigma_0$ does not signal a breakdown of
the inverse Nahm transform.  Starting with the inverse transform
(\ref{t2inverse}), the field strength can be written in exactly the
same way as in (\ref{direct}) except that the integral is over $\tilde
\T^2$ rather than $\tilde\T^4$.  The derivative formula
(\ref{modederivative}) also applies without modification to the
$\tilde\T^2$ zero modes.  What does not carry through is equation
(\ref{t4R}).  Much as in the four torus analysis $R_\mu$ can be
determined by multiplying the $\T^2\times \R^2$ analogue of
\eq{modederivative} by $\psi^p{}^\dagger$ and integrating over $\tilde
\T^2$.
\begin{equation}\label{t2R}
A_\mu^{pq}(x)
=-i \int_{\tilde T^2}  
\psi^p{}^\dagger 
D_x(\hat A)\left(D_x^\dagger(\hat A)D_x(\hat A)\right)^{-1}
\sigma^\dagger_\mu\psi^q +R_\mu^{pq}.
\end{equation}
Naively one would expect the integral on the right hand side to vanish
as $\psi^p{}^\dagger$ is a (left) zero mode of $D_x(\hat A)$. However,
this property only leads to vanishing $\int_{\tilde T^2}
\psi^p{}^\dagger D_x(\hat A) \phi$ for sufficiently smooth functions
$\phi$.  In \eq{t2R} $\psi^p{}^\dagger$ and the Green's function
$(D_x^\dagger D_x)^{-1}$ have coincident singularities.

Retaining the integral yields a modified field strength formula
\begin{equation}\label{fF}
F_{\mu\nu}^{pq}(x)
=\int_{\tilde T^2}d^2z\int_{\tilde T^2}d^2z'\,
\psi^p{}^\dagger(z;x)\sigma_\mu\,
f(z,z';x)\,\sigma_\nu^\dagger\psi^q(z';x)-[\mu\leftrightarrow\nu],
\end{equation}
with
\begin{eqnarray}\label{fdef}
f(z,z';x)
&=&\left(D_x^\dagger D_x\right)^{-1}(z,z')
\\ \nonumber
&&-\int_{\tilde T^2}
d^2s\left(D_x^\dagger D_x\right)^{-1}(z,s)
\,D_x^\dagger(\hat A)\psi^p(s;x)\\ \nonumber
&&~~~~~~~~\times\int_{\tilde T^2}
d^2s'\,
\psi^p{}^\dagger(s';x)D_x(\hat A)\,
\left(D_x^\dagger D_x\right)^{-1}(s',z').
\end{eqnarray}
This object does commute with the $\sigma_\mu$ consistent with
self-duality.  The field strength formula \eq{fF} is similar to those
one can derive for $\R^4$ instantons and calorons via the ADHM
formalism.

To make these considerations more concrete we see how they apply to
our simple $U(1)$ Nahm potential. In this case
\begin{equation}
\frac{1}{4}D_x^\dagger D_x=
\left(\begin{array}{cr}
\frac{1}{4}|x_\perp|^2+
(-\partial_y-\partial_y\phi+\frac{i}{2}\bar x_{||})
(\partial_{\bar y}-\partial_{\bar y}\phi-\frac{i}{2}x_{||})~~~~~~~~~~~0~~~~~
~~~~~~~~\\
~~~~~~~~~~~~~0~~~~~~~~~~~
\frac{1}{4}|x_\perp|^2+(-\partial_{\bar y}+\partial_{\bar y}
\phi+\frac{i}{2}x_{||})
(\partial_y+\partial_y\phi-\frac{i}{2}\bar x_{||})
\end{array}\right).
\end{equation}
When $x\neq0$ this is invertible. The inverse has the form
\begin{equation}\label{ddagdinverse}
2\left(D_x^\dagger D_x\right)^{-1}(z,z')=
(\sigma_0+i\sigma_3)e^{-\phi(z)}K_+(z,z';x)e^{-\phi(z')}+
(\sigma_0-i\sigma_3)e^{\phi(z)}K_-(z,z';x)e^{\phi(z')},
\end{equation}
where the $K_\pm(z,z';x)$ are finite for all $z$ and $z'$
(unless $x_{||}=x_\perp=0$).
Because of the exponentials
 $(D_x^\dagger D_x)^{-1}(z,z')$ is singular if $z$ or $z'$
equals $\pm\omega$.
In the $x_\perp=0$ slice the $K_\pm$ have simple integral representations
\begin{equation}\label{kxperpzero}
K_\pm(z,z';x_\perp=0)=
\int_{\tilde T^2}d^2 s\,G_\pm(z-s)e^{\pm2\phi(s)}G_\mp(s-z'), 
\end{equation}
where $G_\pm$ are the Green's functions defined in section 4. The
$x_\perp=0$ zero modes of section 4 can be rewritten as follows
\begin{eqnarray}\label{zeromodes}
\psi^1(z;x)& =& -D_x(\hat A)\left(\begin{array}{cr}
e^{\phi(z)}K_-(z,\omega;x)\\
0\end{array}\right)\\[1mm] \nonumber
\psi^2(z;x)&= & -D_x(\hat A)\left(\begin{array}{cr}
0\\
e^{-\phi(z)}K_+(z,-\omega;x)\end{array}\right).
\end{eqnarray}
In fact, these are also valid for $x_\perp\neq0$.  The modes are
orthogonal and can be normalised by dividing them by $\sqrt {\rho}$,
where
\begin{equation}
\rho(x)=K_+(-\omega,-\omega;x)=K_-(\omega,\omega;x).\end{equation}
Inserting the zero modes into (\ref{fdef}) enables us to prove that 
$f$ commutes with $\sigma_1, \sigma_2$. This is detailed in 
appendix~\ref{app:proof}. One can also show that $f(z,z';x)$, 
unlike $(D_x^\dagger D_x)^{-1}(z,z')$,
is well behaved at $x=0$. As $z$ or $z'$ approaches $\pm\omega$,
$f$ tends to zero.

We have seen that it is possible to construct the Nahm zero modes out
of objects, $K_\pm$, entering the inverse of $D_x^\dagger D_x$. The
gauge potential can also be computed from these zero modes. The final
result is very simple
\begin{eqnarray}\label{fullpot}
A_{x_{||}}&=&-\frac{\tau_3}{2}
\partial_{x_{||}}\log\rho-2\pi i(\tau_1-i\tau_2)\kappa\rho
\partial_{\bar x_\perp}\frac{\nu^*}{\rho},\\ \nonumber
A_{x_\perp}&=&-\frac{\tau_3}{2}
\partial_{x_{\perp}}\log\rho+2\pi i(\tau_1-i\tau_2)\kappa\rho
\partial_{\bar x_{||}}\frac{\nu^*}{\rho},
\end{eqnarray}
where
\begin{equation}
\label{defofnu}
\nu(x)=K_-(\omega,-\omega;x),~~~~~~~~~~~~
\nu^*(x)=K_-(-\omega,\omega;x).
\end{equation}
Note that $\rho$ is dimensionless, real and periodic (with respect to
$x_1\rightarrow x_1+L_1$, $x_2\rightarrow x_2+L_2$), while $\nu$ is
dimensionless, complex and periodic up to constant phases.
\Eq{fullpot} can be checked component-wise (see
appendix~\ref{app:outline}).

We have given closed forms for the $K_\pm$ in the special case
$x_\perp=0$. An explicit construct valid beyond this two-dimensional
slice would immediately provide the exact Nahm zero modes and gauge
potential. Note that the Green's functions $K_\pm$ are radially
symmetric.  Therefore the functions $\rho$ and $\nu$ are as well.
They can be expressed as power series
\begin{equation}\label{series}
\rho=\sum_{n=0}^\infty\rho_n|x_\perp|^{2n},~~~~~~~~~
\nu=\sum_{n=0}^\infty\nu_n|x_\perp|^{2n}.\end{equation}




\section{ Field Strength Components}

In section 5 we saw that the Nahm transform singled out two points in
the $x_\perp=0$ slice. This suggests that they may be the locations of
some kind of constituent. As they are isolated points in $\T^2\times
\R^2$ rather than lines it appears that we are \it not \rm dealing
with monopole constituents.  Indeed if we take (\ref{adhm}) at face
value the constituents appear to be BPST instantons, one with size
$\lambda=\sqrt{\kappa L_1 L_2/\pi}$ at $x_{||}=x_\perp=0$, the other
with size $\lambda=\sqrt{(1-\kappa) L_1 L_2/\pi}$ at the second soft
point.  This description shares some features with early attempts to
describe $SU(2)$ instantons in terms of $2k$ `instanton quarks'
\cite{Belavin:fb}.

To see if this is realised we must compute the field strengths.  This
can be done explicitly in the $x_\perp=0$ slice.  Inserting the power
series (\ref{series}) into the ansatz (\ref{fullpot}) it is
straightforward to compute $F_{x_{||}\bar x_{||}}$ and $F_{x_{||}\bar
  x_\perp}$
\begin{equation}
\left.F_{x_{||}\bar x_{||}}\right|_{x_\perp=0}=
\tau_3\partial_{x_{||}}\partial_{\bar x_{||}}
\log\rho_0,~~~~~~~~~
\left.F_{x_{||}\bar x_\perp}\right|_{x_\perp=0}
=2\pi i(\tau_1+i\tau_2)\,
\kappa\rho_0\,\partial^2_{x_{||}}\frac{\nu_0}{\rho_0}. 
\end{equation}
The other components are fixed by self-duality, i.e. $F_{x_{||}\bar
  x_{||}}+F_{x_\perp\bar x_\perp}=0$ and $F_{x_{||}x_\perp}=0$.  Both
$\rho_0$ and $\nu_0$ diverge at $x_{||}=0$, but the field strengths
should not.  A careful analysis of $\rho_0$ and $\nu_0$ in the
neighbourhood of $x_{||}=0$ shows that the field strengths are well
defined at this point.  Alternatively, one can just note that $\rho$
and $\nu$ are well behaved at the second soft point and invoke
(\ref{duality}).  Using these results the action density can be
computed
\begin{equation}
\label{actiondensity}
-\h \hbox{Tr} \left.F_{\mu\nu}F^{\mu\nu}\right|_{x_\perp=0}=
16\left(\partial_{x_{||}}\partial_{\bar x_{||}}\log\rho_0
\right)^2+
128\pi^2\kappa^2\rho^2_0\left|
\partial_{x_{||}}^2\frac{\nu_0}{\rho_0}\right|^2.
\end{equation}
From (\ref{kxperpzero}) one can derive integral representations of
$\rho_0$ and $\nu_0$ where the integrands are expressible in terms of
standard functions.  These representations allow us to plot the action
density within the $x_\perp=0$ slice for different values of the
parameters $\kappa$, $\omega_1$ and $\omega_2$. The two periods $L_1$
and $L_2$ can also be varied. However, the physically important
quantity is the ratio of the lengths
\begin{equation}
a=\frac{L_1}{L_2},
\end{equation}
since scaling identically both lengths equates to a trivial scaling of
the action density:
\begin{eqnarray}\label{ident} 
\beta^4\,
\Tr\, F^2(\beta x;\,\beta L_1, \beta  L_2 ;\,\beta^{-1} {\omega_1}, 
\beta^{-1}{\omega_2};\, \kappa)= 
\Tr\, F^2(x;\,L_1,L_2;\,\omega_1,\omega_2;\, \kappa). 
\end{eqnarray}
In \cite{Ford:2002pa} plots for different $\kappa$ in the equal length
case, $a=1$, were presented.  There we found that the action density
has either one or two peaks.  A single peak is observed when the two
soft points are quite close or when one size is somewhat larger than
the other. In the latter case the smaller sized instanton dominates.
Even when two peaks are resolved there is quite a strong overlap. In
fact such an overlap is `necessary' since otherwise the instanton
would have charge two rather than one.  The constituent locations then
correspond to the `cores' of the overlapping instantons.

Here we also consider rectangular tori with $a\neq 1$. Without loss of
generality we set $L_2=1$ and $L_1=a$.  To begin with consider
$\kappa=\frac{1}{2}$ where the two constituents are identical. Now
choose $(\omega_1,\omega_2)=(\frac{1}{2}\pi a^{-1},\frac{1}{2}\pi)$.
This means that there is a core at the centre of the torus and another
of equal size at the corners.  Starting with $a=1$ increase the value
of $a$ thereby stretching the torus in the $x_1$ direction.

In fig.~\ref{history} the action density is plotted for some values of
$a$ ranging from $1$ to $3$.  The first plot ($a=1$) is just the equal
length case clearly showing two instanton-like peaks exactly at the
expected locations.  The second plot ($a=\frac{3}{2}$) shows that as
the torus is stretched the two cores become stretched as well; they no
longer have the approximate spherical symmetry of the $a=1$ case.  On
stretching the torus further the now deformed cores seek to overlap
with the periodic copies of themselves.  This has already started to
happen in the third plot where $a=2$; the soft point locations are no
longer maxima of the action density.  Finally, at $a=3$ the
self-overlap has generated two monopole like objects; the solution has
a very weak dependence on $x_2$ and there are peaks at two values of
$x_1$.  Note that the two monopole `worldlines' do not cross the soft
points.  Increasing $a$ more weakens further the dependence on $x_2$.
In the limit $a\rightarrow\infty$ a periodic monopole (i.e. a self
dual solution on $S^1\times\R^2$) will form.  However, this is a
singular limit.

\begin{figure}[h]
\begin{picture}(0,400)(-250,0)
\put(-250,230){\epsfig{file=history1.eps,width=.4\hsize}}
\put(-250,375){$a=1$} 
\put(-202,232){$\bf  x_2$}  
\put(-65,266){ $\bf x_1$}
\put(0,230){\epsfig{file=history1.5.eps,width=.4\hsize}}
\put(10,375){$a=\s032$} 
\put(48,228){$\bf  x_2$}  
\put(175,269){ $\bf x_1$}
\put(-250,20){\epsfig{file=history2.eps,width=.4\hsize}}
\put(-240,180){$a=2$} 
\put(-207,14){$\bf  x_2$}  
\put(-75,76){ $\bf x_1$}
\put(0,20){\epsfig{file=history3.eps,width=.4\hsize}}
\put(10,180){$a=3$} 
\put(31,13){$\bf  x_2$}  
\put(175,76){ $\bf x_1$}
\end{picture}
\caption{action density $-\s012 \Tr\, F^2$ for $x_\perp=0$,  
$\kappa=\s012$  and $\omega_1=\s012\pi, \omega_2=\s012 \pi a$ for 
$a=1,\s032,2,3$.}
\label{history}
\end{figure}

What is happening is that the doubly periodic instantons are becoming
more caloron-like (caloron-like objects have been observed recently in
lattice-based approaches \cite{Gattringer:2002tg,Ilgenfritz:2002qs} ).
It is helpful to recall some of the basic features of the simplest
caloron solutions. An SU(2) caloron with charge one comprises two
monopole constituents separated by a distance $\pi\lambda^2/L_2$
(taking $x_2$ to be the compact coordinate).  Here $\omega_2$ (there
is no $\omega_1$) determines the mass ratio of the two monopoles.  For
the $\T^2\times\R^2$ problem $\pi\lambda^2=\kappa L_1 L_2$ suggesting
that the separation of the monopoles should be simply $\kappa\,L_1$
for $\kappa\leq \s012$ and $(1-\kappa)\,L_1$ for $\kappa\geq \s012$.
In fig.~\ref{history} we took $\kappa=\frac{1}{2}$ and the worldlines
are indeed separated by a half-period.  In fig.~\ref{3over8} we have
taken $\omega$ as above but increased $\kappa$ to $\frac{5}{8}$. The
first plot is the equal length case showing a peak at the centre of
the torus. At the origin of the torus, the location of the larger
constituent, there is a broader but much shallower peak. The second
plot is for $a=3$ which clearly shows two monopoles.  Moreover, the
distance between them is consistent with $\frac{3}{8}L_1$. Though the
dependence on $x_2$ is not quite as weak as for the corresponding plot
of fig.~\ref{history}; away from $\kappa=\frac{1}{2}$ more stretching
is required to access the monopole regime. \\[-15ex]

\begin{figure}[h]
\begin{picture}(0,250)(-250,20)
\put(-250,30){\epsfig{file=actionIII.eps,width=.4\hsize}}
\put(-245,170){$a=1$} 
\put(-190,28){$\bf  x_2$}  
\put(-63,60){ $\bf x_1$}
\put(0,30){\epsfig{file=3over8.eps,width=.4\hsize}}
\put(5,170){$a=3$} 
\put(39,30){$\bf  x_2$}  
\put(170,80){ $\bf x_1$}
\end{picture}
\caption{action density $-\s012 \Tr\, F^2$ for $x_\perp=0$,  
$\kappa=\s058$  and $\omega_1=\s012\pi, \omega_2=\s0{1}{2}\pi a$ with $a=1,3$.}
\label{3over8}
\end{figure}

We have argued that for $a$ at or near to $1$ the constituent core
picture best describes the action density while for larger (or
smaller) $a$ almost static monopole-like objects form.  In that
respect, the instanton core and monopole constituent pictures are
complementary.  However, we have an example displaying features of
both descriptions: \\[1ex]

\begin{figure}[h]
\begin{picture}(150,150)(-250,-270)
\put(-170,-280){\epsfig{file=monopole.eps,width=.50
\hsize}}
\put(-230,-180){$\bf -\s012 \Tr\,
 F^2$}
\put(-100,-265){$\bf  x_2$}  
\put(70,-230){ $\bf  x_1$}
\end{picture}
\caption{Plot of action density for $x_\perp=0$,  
$\kappa=\s012$, $a=1$ and $\omega_1=\s012 \pi, \omega_2=0$.
}
\label{monopole}
\end{figure} 

Above we plot the action density for $\kappa=\s012$, $\omega_1=\s012
\pi$ and $\omega_2=0$.  This configuration shows essentially no
dependence on $x_2$ even though the lengths are equal.  What seems to
have occurred is that the two cores at $x_{||}=0$ and
$x_{||}=\frac{1}{2}i$ have merged into a \it single \rm monopole.
Unlike in the previous examples the monopole worldline passes through
the soft points.  That there is only one worldline is not inconsistent
with the constituent monopole picture since by virtue of $\omega_2=0$
the `second' monopole should be massless. The monopoles observed in
figs.~\ref{history} and \ref{3over8} have equal masses in accord with
the symmetric value of $\omega_2$.


\section{ Charge One Half}

When $\kappa=\frac{1}{2}$ the two instanton cores are identical.  This
has an interesting consequence.  If we choose the constituent
locations so that they are separated by half periods the charge one
instanton can be `cut' to yield two copies of a charge $\frac{1}{2}$
instanton (see also \cite{Gonzalez-Arroyo:1998ia}).  This happens when
$(\omega_1,\omega_2)$ is $(\frac{1}{2}\pi/L_1,0)$,
$(0,\frac{1}{2}\pi/L_2)$ or $(\frac{1}{2}\pi/L_1,\frac{1}{2}\pi/L_2)$,
see fig.~\ref{constituents}. After cutting we have a twist
$Z_{12}=-\id$ in the half torus.  But to produce a half integer
topological charge another (non-orthogonal) twist is required; this is
most simply achieved with $Z_{03}=-\id$.  Such a twist would have the
novel feature of being associated with the non-compact $x_0$ and $x_3$
directions. Far away from $x_\perp=0$ the potential it must be pure
gauge
\begin{equation} 
A_\mu(x)\sim
V^{-1}(x_1,x_2,\theta)\partial_\mu V(x_1,x_2,\theta),~~~~~r\rightarrow\infty,
\end{equation}
where $x_\perp=re^{i\theta}$.  The non-compact twist translates into a
double valued gauge function,
$V(x_1,x_2,\theta+2\pi)=-V(x_1,x_2,\theta)$.
\begin{figure}[h]
\centering
\begin{picture}(100,250)(100,0)
\epsfxsize=10cm\epsffile{constituents.eps}
\end{picture}
\caption{When $\kappa=\h$, the
choices $\omega_1=\h\pi/L_1$,
$\omega_2=0$ and
$\omega_1=\h\pi/L_1$,
$\omega_2=\h\pi/L_2$
yield constituent separations
allowing the torus to be cut (dashed line) to
yield charge $\h$ instantons.
}
\label{constituents}
\end{figure}

The examples considered in fig.~\ref{history} are all `doubled' half
instantons.  Here the cut tori are diamond shaped (see fig. 4) apart
from the first $(a=1)$ plot, where it is square.  This case has
another interesting feature.  Compare the constituent locations with
the half instanton obtained via $L_1=\sqrt{2}$, $L_2=1/\sqrt{2}$,
$\omega_1=0$, $\omega_2=\pi/\sqrt{2}$.  On cutting both yield charge
$\frac{1}{2}$ instantons on square tori with length $1/\sqrt{2}$.  By
analogy with the four torus case, for a given set of twists and
periods, charge $\frac{1}{2}$ instantons are expected to be unique up
to translations.  Yet these two cases have different values of $a$,
one has $a=1$ where the core picture should work while the other
($a=2$) is expected to have some monopole characteristics.  As can
clearly be seen from the first plot in fig.~\ref{history} the action
density has maxima at the soft points.  We have computed the $a=2$
case and found maxima {\it not} coincident with the soft points.  The
soft points correspond to saddle points of the action density.  But
allowing for translations the two half instantons are identical.  What
seems to be happening is that when $a=2$ the monopoles are still not
developed, and in this intermediate state there are lumps of action
density which while not at the soft points are identical to those at
the soft points of the $a=1$ case.  More precisely the two action
densities satisfy
\begin{eqnarray}\label{relation}
\Tr\, F^2_{a=2}(x_1,x_2 )=
 \Tr\, F^2_{a=1}(\s0{1}{\sqrt{2}}(x_1+x_2)-\s014,\s0{1}{\sqrt{2}}
(x_2-x_1)-\s014). \end{eqnarray}
The $a=1$ plot is obtained by a 45 degree rotation  
of the $a=2$ plot followed by a translation. This reflects the fact that 
the two Nahm potentials are also related by a $45$ degree rotation.


\section{ Asymptotics}

In this section we consider the large $|x_\perp|$ limit of the zero
mode equations, and deduce the large $|x_\perp|$ behaviour of the
gauge potential and field strength.  This allows us to check that the
charge $\frac{1}{2}$ instantons have the expected properties.  In
particular, we see that the gauge potential has asymptotic behaviour
commensurate with a twist, $Z_{03}=-\id$.  Furthermore, the
exponential decay of the action density is derived.  In fact, all the
$\kappa=\frac{1}{2}$ solutions have both the twist and exponential
decay even if they are not `doubled' $\frac{1}{2}$ instantons.

When $|x_\perp|$ is much larger than the two periods the zero modes of
$D^\dagger_x(\hat A)$ become strongly localised about the two flux
singularities.  In this regime we can approximate the zero modes with
$\R^2$ solutions.  A single flux in $\R^2$ corresponds to
$\phi=-\h\kappa \log y\bar y$.  The Weyl operator is then
\begin{equation}
-\frac{i}{2}D_x^\dagger
=\left(\begin{array}{cc}
\frac{1}{2}\bar x_\perp&
\partial_y-\frac{\kappa}{2y}-\frac{i}{2}\bar x_{||}\\
\partial_{\bar y}+\frac{\kappa}{2\bar y}-\frac{i}{2}x_{||}&
\frac{1}{2}x_\perp
\end{array}\right).
\end{equation}
When $x_\perp\neq 0$, $D_x^\dagger$ has a  normalisable
zero mode. In the special cases $\kappa=\pm \h$ the normalised modes 
 take rather simple forms:
\begin{eqnarray}
\!\!\!\psi_{\frac{1}{2}} (z)=
\frac{e^{i x\cdot z-|x_\perp|\sqrt{y\bar y}}}{\sqrt{2\pi|x_\perp|}}
\left(
\begin{array}{cr}
(y\bar y)^{-\frac{1}{4}}x_\perp\\
\frac{(y\bar y)^{\frac{1}{4}}}{\bar y}|x_\perp|
\end{array}\right)
, &\quad\!\!\!\!  & \psi_{-\frac{1}{2}}(z)=\frac{
e^{i x\cdot z-|x_\perp|\sqrt{y\bar y}}}{\sqrt{2\pi|x_\perp|}}
\left(
\begin{array}{cr}
\frac{(y\bar y)^{\frac{1}{4}}}{y}x_\perp\\
(y\bar y)^{-\frac{1}{4}}|x_\perp|
\end{array}\right),  
\end{eqnarray}
where $x\cdot z= x_1 z_1+ x_2 z_2$. For sufficiently large $|x_\perp|$
we can take
\begin{equation}\label{approx}
\psi^1(z)=\psi_{\frac{1}{2}}(z-\omega),~~~~~~~~~~~~~~~
\psi^2(z)=\psi_{-\frac{1}{2}}(z+\omega),\end{equation}
and extend the integration region from $\tilde\T^2$ to $\R^2$.
For this approximation to work 
it is important that the flux separation $2|\omega|$
is less then both dual periods, $2\pi/L_1$ and $2\pi/L_2$.
This is because we assume that the `interference'
between the fluxes dominates
the effect of the periodicity for large $|x_\perp|$. 
Note that
\begin{equation}
\int_{R^2}d^2z\,
\psi_{\frac{1}{2}}^\dagger(z-\omega)
\psi_{-\frac{1}{2}}(z+\omega)=0.\end{equation}
Inserting the zero modes (\ref{approx})
into the inverse Nahm transform (\ref{t2inverse}) produces a $U(2)$
potential; we discard the pure gauge $U(1)$ part.
The remaining $SU(2)$ piece  turns out to be in a different gauge to that
implicit in section~\ref{Greens}.
An explicit computation (see appendix C)
shows that the $SU(2)$ potential has asymptotics corresponding
to a twist and the action density  has the exponential decay
\begin{equation}\label{actiondecay}
-\frac{1}{2}\Tr F_{\mu\nu}F_{\mu\nu}\sim
\frac{4 \pi |\omega|}{|x_\perp|^3}e^{-4|\omega||x_\perp|}.
\end{equation}
We can apply this to the charge $\frac{1}{2}$ case.  Consider such an
instanton in a square box of length $L$.  This is a one instanton in
the doubled box with $L_1=L$, $L_2=2L$. Here $\omega_1=\pi/(2L)$,
$\omega_2=0$, and so the $\frac{1}{2}$ instanton has the fall-off
\begin{equation}
-\frac{1}{2}\Tr F_{\mu\nu}F_{\mu\nu}\sim
\frac{2\pi^2}{L|x_\perp|^3}e^{-2\pi||x_\perp|/L}.
\end{equation}
For a rectangular box simply replace $L$ with the longest length.


One can repeat the analysis for $\kappa\neq \frac{1}{2}$.  The
corresponding $\R^2$ zero modes are more complicated; they can be
written in terms of modified Bessel functions.  Here the asymptotics
of the gauge potential does not correspond to a four torus twist.
More precisely, $\rho$ has the power decay
\begin{equation}
\rho=\frac{C}{|x_\perp|^{2(1-\kappa)}}+\hbox{
exponentially decaying remainder},
\end{equation}
where $C$ is a (dimensionful) constant.  This translates into a twist
only if $\kappa=\frac{1}{2}$.



\section{Outlook}

We have developed an instanton core picture of the simplest doubly
periodic instantons.  Numerical calculations of the action density
correlate well with this picture for square tori.  However, in
rectangular tori the solutions become caloron-like when the ratio of
the two periods is increased; the cores merge with periodic copies of
themselves in the short direction to form monopole-like tubes of
action density.  The basic properties of these monopoles, such as
their mass ratios and spatial separation, follow the pattern of the
charge one $SU(2)$ calorons.  It would be interesting to develop this
monopole constituent description in a more direct manner.  One
approach would be to exploit the fact that the radially symmetric
solutions considered here fall into a class of axially-symmetric
multi-calorons discussed recently \cite{Bruckmann:2002vy} if one
allows for infinite topological charge.

There are a number of obvious ways to extend the results in this
paper. An explicit treatment of the $x_\perp\neq 0$ zero mode equation
is still lacking.  In the non-radially symmetric case ($\alpha\neq0$)
the zero mode equations have not yielded solutions for any points in
$\T^2\times\R^2$. Even in the absence of explicit zero modes it might
be possible to extract some information about possible instanton core
or monopole constituents.  A more straightforward extension would be
to generalise the results regarding radially symmetric one instantons
to $SU(N)$.  That these solutions decay exponentially indicates they
can be compactified further to $\T^3\times\R$ or even $\T^4$.  The
latter is only possible for the $\kappa=\frac{1}{2}$ case where the
asymptotics of the gauge field correspond to
a twist.\\[-2ex]

\noindent{\bf Acknowledgements} 

We thank F. Bruckmann, G. Dunne and P. van Baal for helpful
discussions.  C.F. was supported through a European
Community Marie Curie Fellowship (contract HPMF-CT-2000-00841).\\[.2ex]

\appendix 

\section{Properties of $f$} \label{app:proof}
Inserting the zero modes \eq{zeromodes} into the definition of $f$
(\ref{fdef}) yields
\begin{equation}
f(z,z';x)=\s012 (\sigma_0+i\sigma_3)e^{-\phi(z)}g_+(z,z';x)e^{-\phi(z')}+
\s012 (\sigma_0-i\sigma_3)e^{\phi(z)}g_-(z,z';x)e^{\phi(z')},
\end{equation}
where
\begin{equation}\label{gdef}
g_\pm(z,z';x)=K_\pm(z,z';x)-\frac{1}{\rho(x)}
K_\pm(z,\mp\omega;x)K_\pm(\mp\omega,z';x).
\end{equation}
It is not obvious that $f$ commutes with $\sigma_1$ and $\sigma_2$.
This only holds if the coefficients of the projectors $\sigma_0\pm
i\sigma_3$ match, that is if
\begin{equation}\label{nonquat}
e^{-\phi(z)}g_+(z,z';x)e^{-\phi(z')}
=e^{\phi(z)}g_-(z,z';x)e^{\phi(z')}.
\end{equation}
A proof of (\ref{nonquat}) for $x_\perp=0$ was given in
\cite{Ford:2000zt}.  It is sufficient to show that the left and right
hand sides of (\ref{nonquat}) have the same asymptotics at the fluxes
since when $z,z'\neq \pm \omega$ they satisfy the same differential
equation.  Consider $g_\pm(z,z';x)$ in the neighbourhood of
$z=\omega$.  It follows immediately from (\ref{gdef}) that
$g_-(\omega,z';x)=0$ whereas $g_+(\omega,z';x)$ is non-zero.  This
does not contradict (\ref{nonquat}) since $e^{\phi(z)}$ diverges
($\propto|y-\omega_1-i\omega_2|^{-\kappa}$) at $z=\omega$.  A short
calculation gives
\begin{equation}
g_-(z,z';x)\sim\frac{
|y-\omega_1-i\omega_2|^\kappa
}{4\pi\kappa\rho(x)}
K_-(\omega,z';x).\end{equation}
This is compatible with (\ref{nonquat})
if
\begin{equation}\label{key}
g_+(\omega,z';x)=
\frac{e^{2\phi(z')}}{4\pi\kappa\rho(x)}K_-(\omega,z';x).
\end{equation}
To check this, note that for $z'\neq\pm\omega$ both sides satisfy the
same differential equation and then examine them in the neighbourhoods
of $z'=\pm\omega$.  From this discussion of the asymptotics of $g_\pm$
it should be clear that $f(z,z';x)$, unlike $(D_x^\dagger
D_x)^{-1}(z,z')$ is always finite. Indeed $f(z,z';x)$ tends to zero as
$z$ or $z'$ approaches a flux point.




\section{Computation of \eq{fullpot}}\label{app:outline} 
Here we outline the computation of $A_{x_{||}}$; the other components
can be dealt with in much the same way. Inserting the zero modes into
the inverse Nahm transform yields
\begin{equation}
A_{x_{||}}^{11}=\frac{1}{\sqrt{\rho}}
\int_{\tilde T^2} d^2z\,
K_-(\omega,z;x)e^{\phi(z)}
\left(
D^\dagger_x(\hat A)\partial_{x_{||}} D_x(\hat A)
\right)^{11}\frac{e^{\phi(z)}}{\sqrt{\rho}}K_-(z,\omega;x).
\end{equation}
Using $(D^\dagger\partial_{x_{||}}D)^{11}=\partial_{x_{||}}(D^\dagger
D)^{11}$ and the representation (33) of $(D^\dagger D)^{-1}$
\begin{equation}
A_{x_{||}}^{11}={1\over{\sqrt{\rho}}}\int d^2z
K_-(\omega,z;x)\partial_{x_{||}}\delta^2(z-\omega){1\over{\sqrt{\rho}}}=
-{1\over2}\partial_{x_{||}}\log\rho,
\end{equation}
as required. For $A_{x_{||}}^{21}$ we proceed similarly
\begin{eqnarray}
A_{x_{||}}^{21}&=&\frac{1}{\sqrt{\rho}}
\int_{\tilde T^2} d^2z\,
K_+(-\omega,z;x)e^{-\phi(z)}
\left(
D^\dagger_x(\hat A)\partial_{x_{||}} D_x(\hat A)
\right)^{21}\frac{e^{\phi(z)}}{\sqrt{\rho}}K_-(z,\omega;x)\\ \nonumber
&=& i\frac{x_\perp}{\rho}\int_{\tilde T^2} d^2z\,
K_+(-\omega,z;x)K_-(z,\omega;x).
\end{eqnarray}
The key step is to use (\ref{key})
\begin{equation}
A_{x_{||}}^{21}=4\pi i\kappa x_\perp
\int_{\tilde T^2} d^2z
K_+(-\omega,z;x)e^{-2\phi(z)}g_+(z,\omega;x).
\end{equation}
Now use (\ref{gdef}) to write $g_+$ in terms of $K_+$ and the formula
\begin{equation}
\int _{\tilde T^2}
d^2s K_+(z,s;x)e^{-2\phi(s)}K_+(s,z';x)
=-\frac{\partial}{\partial|x_\perp|^2}K_+(z,z';x),
\end{equation}
to get the result $A_{x_{||}}^{21}=-4\pi i\kappa \partial_{\bar x_\perp}
(\nu^*/\rho)$.



\section{ Exponential Decay of the Action Density}

We consider the $A_{x_\perp}$ derived from the asymptotic zero modes
given in section 9.  A straightforward calculation gives
\begin{equation}
A_{x_\perp}^{11}-A_{x_\perp}^{22}=0.\end{equation}
The off-diagonal components are more involved:
\begin{equation}
A_{x_\perp}^{21}=
\int_{R^2}d^2z\,
\psi_{-\frac{1}{2}}^\dagger(z+\omega)\frac{\partial}{\partial x_\perp}
\psi_{\frac{1}{2}}(z-\omega)=\frac{\bar x_\perp}{2\pi |x_\perp|}
e^{2ix\cdot\omega} (I+J).
\end{equation}
Here
\begin{equation}
\label{Iint}
I=
\int_{R^2} d^2z\frac{(y_2\bar y_2)^{\frac{1}{4}}}
{\bar y_2}(y_1\bar y_1)^{-\frac{1}{4}}
e^{-|x_\perp|(\sqrt{y_1\bar y_1}+\sqrt{y_2\bar y_2})}
\end{equation}
and 
\begin{eqnarray}\label{Jint}
\lefteqn{J = {1\over2}
\int d^2z e^{-|x_\perp|(\sqrt{y_1\bar y_1}+\sqrt{y_2\bar y_2})}}\hspace{2cm}
\\\nonumber
& & \times\left[
(y_2\bar y_2)^{-{1\over4}}{(y_1\bar y_1)^{{1\over4}}\over{
\bar y_1}}-|x_\perp|{(y_2\bar y_2)^{{1\over4}}\over{
\bar y_2}}(y_1\bar y_1)^{{1\over4}}-
|x_\perp|(y_2\bar y_2)^{-{1\over4}}
{(y_1\bar y_1)^{{3\over4}}\over{
\bar y_1}}\right],
\end{eqnarray}
where $y_1=y-\omega_1-i\omega_2$ and $y_2=y+\omega_1+i\omega_2$.  As
$|x_\perp|$ is increased the integral $I$ is dominated by a small
neighbourhood about $y_1=0$, so we may approximate $y_2$ with
$2(\omega_1+i\omega_2)$
\begin{equation}
I\sim\frac{(2|\omega|)^{\frac{1}{2}}
e^{-2|\omega||x_\perp|}
}{2(\omega_1-i\omega_2)}
\int_{R^2}d^2z(y_1\bar y_1)^{-\frac{1}{4}}e^{-|x_\perp|\sqrt{y_1\bar y_1}}
=\frac{(2|\omega|)^{\frac{1}{2}}
e^{-2|\omega||x_\perp|}
}{2(\omega_1-i\omega_2)}\cdot
\frac{\pi^{\frac{3}{2}}}{|x_\perp|^{\frac{3}{2}}}.
\end{equation}
$J$ exhibits a similar exponential decay. 
$A_{\bar x_\perp}^{21}$ is simply $x_\perp e^{2ix\cdot\omega}J/
(2\pi |x_\perp|)$, and so
\begin{equation}
F_{x_\perp \bar x_\perp}^{21}=- 
\frac{1}{2\pi |x_\perp|}
e^{2ix\cdot\omega}\partial_{\bar x_\perp}(\bar x_\perp I)
\sim e^{2ix\cdot \omega}
e^{-2|\omega||x_\perp|}
{(\omega_1+i\omega_2)\pi^{{1\over2}}\over
{2^{{3\over2}}|\omega|^{{1\over2}}|x_\perp|^{{3\over2}}}}
\end{equation}
We also have
\begin{equation}
F_{x_\perp \bar x_\perp}^{11}=O(e^{-4|\omega||x_\perp|}).
\end{equation}
Now compare with the corresponding field strengths generated by the
ansatz (\ref{fullpot}) in section~\ref{Greens}.
\begin{equation}
F_{x_\perp\bar x_\perp}^{21}=2\pi i\rho\partial_{\bar x_{||}}\partial_{
\bar x_\perp}
\frac{\nu^*}{\rho},~~~~~~~~~~~~
F_{x_\perp\bar x_\perp}^{11}=\partial_{x_\perp}
\partial_{ \bar x_\perp}\log\rho
+(2\pi\rho)^2\partial_{x_{||}}\frac{\nu}{\rho}\partial_{\bar x_{||}}
\frac{\nu^*}{\rho}.\end{equation}
Consider
\begin{equation}\label{fit}
\rho=\frac{C}{|x_\perp|^t},~~~~~~~~~~
\nu^*={1\over4}(2\pi)^{-{1\over2}}
e^{2i\omega\cdot x} e^{-2|\omega||x_\perp|}|\omega|^{-\frac{3}{2}}
|x_\perp|^{-\frac{3}{2}},\end{equation}
where $t$ is some real number and $C$ is a  constant. 
Neglecting sub-leading terms the expressions for $F_{x_\perp\bar
x_\perp}$ agree up to a phase,
$\sqrt{x_\perp/\bar x_\perp}=e^{i\theta}$,
which can be compensated for by the double-valued gauge 
transformation $V=e^{{1\over2}i\theta\tau_3}$. Under this gauge transformation
 the diagonal part of $A_{x_\perp}$ becomes non-trivial
\begin{equation}
A_{x_\perp}=\frac{\tau_3}{4x_\perp}+
O(e^{-4|\omega||x_\perp|}).
\end{equation}
which agrees with \eq{fullpot} if  $t=1$ in (\ref{fit}).
This is exactly what is required to generate the twist $Z_{03}=-\id$.
Using the large $|x_\perp|$ form of $\rho$ and $\nu$
it is straightforward to obtain the decay formula
(\ref{actiondecay}).
\eject 

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%\cite{Gonzalez-Arroyo:1998ia}
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%\cite{Bruckmann:2002vy}
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