\documentclass[12pt]{article}
\linespread{1.4}
\usepackage{amsfonts,amsbsy}
\newcommand{\Epsilon}{{\cal E}}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\YM}{Yang-Mills}
\newcommand{\mh}{{\bf \hat{\mu}}}
\newcommand{\ihat}{\hat{\imath}}
\newcommand{\jhat}{\hat{j}}
\newcommand{\nh}{{\bf \hat{\nu}}}
\newcommand{\rh}{{\bf \hat{\rho}}}
\newcommand{\Amu}{{\bf A}_{\mu}(x)}
\newcommand{\Anu}{{\bf A}_{\nu}(x)}
\newcommand{\Amua}{A_{\mu}^a(x)}
\newcommand{\Amal}{\mathbf{A}_{\mu}^{(\alpha)}(x)}
\newcommand{\Amub}{\mathbf{A}_{\mu}^{(\beta)}(x)}
\newcommand{\LAa}{\boldsymbol{\lambda}_a}
\newcommand{\Lada}{\boldsymbol{\lambda}_a^{(adjoint)}}
\newcommand{\LAb}{\boldsymbol{\lambda}_b}
\newcommand{\LAc}{\boldsymbol{\lambda}_c}
\newcommand{\Fmunu}{{\bf F}_{\mu \nu}(x)}
\newcommand{\Ftmunu}{\tilde{\bf F}_{\mu \nu}(x)}
\newcommand{\Dmu}{\partial_{\mu}}
\newcommand{\Dnu}{\partial_{\nu}}
\newcommand{\BPL}{\boldsymbol{[}}
\newcommand{\BPR}{\boldsymbol{]}}
\newcommand{\SUN}{SU(N)}
\newcommand{\UN}{U(N)}
\newcommand{\SUT}{SU(2)}
\newcommand{\SUINF}{SU(\infty)}
\newcommand{\ZN}{\mathbf{Z}_N}
\newcommand{\Z}{\mathbf{Z}}
\newcommand{\TF}{T_4}
\newcommand{\RF}{{\bf R}^4}
\newcommand{\EPS}{\boldsymbol{\epsilon}}
\newcommand{\BET}{\boldsymbol{\beta}}
\newcommand{\Om}{\boldsymbol{[}\Omega(x) \boldsymbol{]}}
\newcommand{\Omegamu}{\Omega_{\mu}(x)} 
\newcommand{\Omeganu}{\Omega_{\nu}(x)}
\newcommand{\Omegatnu}{\Omega^+_{\nu}(x)}
\newcommand{\xOnurh}{\Omega_{\nu}(x + \rh) \Omega_{\rho}(x)}
\newcommand{\xOrhonh}{\Omega_{\rho}(x + \nh) \Omega_{\nu}(x)}
\newcommand{\Omu}{\BPL \Omega_{\mu}(x) \BPR}
\newcommand{\Onu}{\BPL\Omega_{\nu}(x) \BPR}
\newcommand{\Onuz}{\BPL \Omega_{\nu}(0) \BPR}
\newcommand{\Onurh}{\Omega_{\nu}(\rh) \Omega_{\rho}(0)}
\newcommand{\Orhonh}{\Omega_{\rho}(\nh) \Omega_{\nu}(0)}
\newcommand{\nmunu}{n_{\mu \nu}}
\newcommand{\dmunu}{\varphi_{\mu \nu}}
\newcommand{\mv}{\vec{m}}
\newcommand{\kv}{\vec{k}}
\newcommand{\ev}{\vec{e}}
\newcommand{\Gmu}{\Gamma_{\mu}}
\newcommand{\Gnu}{\Gamma_{\nu}}
\newcommand{\tbc}{twisted boundary conditions}
\newcommand{\tw}{twist}
\newcommand{\eps}{\epsilon_{\mu \nu \rho \sigma}}
\newcommand{\Anm}{\bar{A}_{\nu}^{(\mu)}}
\newcommand{\Arm}{\bar{A}_{\rho}^{(\mu)}}
\newcommand{\Asm}{\bar{A}_{\sigma}^{(\mu)}}
\newcommand{\Anr}{\bar{A}_{\nu}^{(\rho)}}
\newcommand{\kpn}{\kappa(\nmunu)}
\newcommand{\Dsl}{\not\!\!D}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newenvironment{statement}{\begin{itemize} \em}{\end{itemize}} 
%\usepackage{amstex}
%\usepackage{epsf}

% Anhado dos mas:

\newcommand{\bea}{\begin{eqnarray}}
\newcommand{\eea}{\end{eqnarray}}
\input{epsf.tex}










\begin{document}

\vskip -4cm


\begin{flushright}
FTUAM-98-20

IFT-UAM/CSIC-98-25
\end{flushright}

\vskip 0.2cm

{\Large
\centerline{{\bf On Nahm's transformation  }}
\centerline{{\bf   with twisted boundary conditions}}
\vskip 0.3cm

\centerline{\qquad  
  A. Gonz\'alez-Arroyo  }}
\vskip 0.3cm

\centerline{Departamento de F\'{\i}sica Te\'orica C-XI}
\centerline{ and Instituto de F\'{\i}sica Te\'orica C-XVI,}
\centerline{Universidad Aut\'onoma de Madrid,}
\centerline{Cantoblanco, Madrid 28049, SPAIN.}
\vskip 10pt
%\centerline{{$^\ddag$}Instituto de F\'{\i}sica Te\'orica C-XVI,}
%\centerline{Universidad Aut\'onoma de Madrid,}
%\centerline{Cantoblanco, Madrid 28049, SPAIN.}

\vskip 0.8cm

\begin{center}
{\bf ABSTRACT}
\end{center}
Following two different tracks, we arrive at a definition of
Nahm's transformation valid for self-dual fields on the
4-dimensional torus $T^4$ with non-zero twist tensor.
The  transform is again a self-dual gauge field defined on a
new torus $\widehat{T}^4$ and with non-zero twist tensor. It preserves 
the property of being an involution.
\vskip 1.5 cm
\begin{flushleft}
PACS: 11.15.-q, 11.15.Ha

Keywords: Nahm transformation, Instanton solutions, Twisted boundary
conditions.
\end{flushleft}

\newpage

\input{section1.tex}




\input{section2.tex}


\input{section3.tex}


\input{section4.tex}

\input{section5.tex}

\input{acknow.tex}


\newpage



\begin{thebibliography}{10}
\bibitem{polyakov}
A.M. Polyakov,
\newblock {\em Phys. Lett. }  59B (1975) 82.


\bibitem{BPST}
A.~Belavin, A.~M.~Polyakov, A.~S.~Schwartz and Yu.~S.~Tyupkin.
\newblock {\em Phys. Lett. }  59B (1975) 85.


\bibitem{ADHM}
M.~F.~Atiyah, N.~J.~Hitchin, V.~G.~Drinfeld and Y.~I.~Manin.
\newblock {\em Phys. Lett. } 65A (1978) 185.


\bibitem{thooft}
G.~'t~Hooft.
\newblock {\em Nucl. Phys.}  B153 (1979)  141.



\bibitem{stsf}
G.~'t~Hooft.
\newblock {\em Commun. Math. Phys. } 81 (1981) 267.





\bibitem{vb1}
P.~van~Baal.
\newblock {\em   Commun. Math. Phys.} 94 (1984) 397.


\bibitem{nahm}
 W.~Nahm.
 \newblock {\em Phys. Lett. } 90B (1980) 413.
 
E.~Corrigan and P.~Goddard.
\newblock {\em Ann. Phys. } 154 (1984) 253.

\bibitem{kraanvanbaal}
Th.~C.~Kraan and P.~van~Baal,
\newblock {\em    Phys. Lett. } 428B (1998) 268;

\newblock {\em Nucl. Phys.} B533 (1998) 627;\ 
\newblock {\em    Phys. Lett. } 435B (1998) 389.

\bibitem{lee}
K. Lee and C. Lu,
\newblock {\em    Phys. Rev. } D58 (1998) 025011.



\bibitem{cohen-gomez}
E.~Cohen and C.~Gomez.
\newblock {\em Nucl. Phys. } B223 (1983) 183.

\bibitem{ntl}
 A.~Gonz\'alez-Arroyo and C.~Pena.
  \newblock {\em Jour. of High Energy Phys. }  09 (1998) 013.
  
  

\bibitem{AS}
M.~F.~Atiyah and I.~M.~Singer.
\newblock {Ann. Math. } 87 (1968) 485 and 546;
\newblock {Ann. Math. } 93 (1971) 119 and 139.

\bibitem{PB2}
P.~J.~Braam and P.~van~Baal.
\newblock {\em   Commun. Math. Phys.} 122 (1989) 267.



\bibitem{torusrev}
A.~Gonz\'alez-Arroyo.
\newblock {Proceedings of the 1997 Pe\~n\'{\i}scola Advanced School in
Non-Perturbative Quantum Physics, World Scientific (1998);}
hep-th/9807108.

\bibitem{twisteat}
P.~van~Baal and B.~van~Geemen
\newblock {\em J. Math. Phys.}  27 (1986) 455-457;

D.~R.~Lebedev and M.~I.~Polikarpov.
\newblock {\em Nucl. Phys.} B269 (1986) 285.

\bibitem{bukow}
P.~van~Baal,
\newblock {\em Nucl. Phys. B (Proc. Suppl.)}  49 (1996) 238.



\end{thebibliography}
\end {document}


\section{The Nahm transform for non-zero twist}


Let us start with $\mathbf{R^4}$ with Euclidean metric. Now, we introduce a
rank four lattice of vectors $\Lambda$. The generators are the vectors
$e^{(\mu)}$ whose components define an invertible matrix $M_{\mu \nu}=
e^{(\mu)}_{\nu}$. The quotient $T^4=\mathbf{R^4}/\Lambda$ defines a 4-dimensional
torus. Now let us consider an $SU(N)$ self-dual gauge field $A_{\mu}(x)$
living in $T^4$. This is equivalent to a gauge field on $\mathbf{R^4}$ 
and satisfying the following periodicity property:
\be
\label{tbc}
A_{\nu}(x+e^{(\mu)})= [\Omega_{\mu}(x)]A_{\nu}(x) \equiv  
 \Omega_{\mu}(x)\, A_{\nu}(x)\, \Omega^{\dagger}_{\mu}(x) +
 \imath\ \Omega_{\mu}(x)\,  \partial_{\nu} \Omega^{\dagger}_{\mu}(x)\ \ ,
\ee
where the matrices $\Omega_{\mu}(x)$ belong to $SU(N)$. They must satisfy the
following consistency condition:
\be
\label{twist}
\Omega_{\mu}(x+e^{(\nu)})\, \Omega_{\nu}(x)= z_{\mu \nu}\,
\Omega_{\nu}(x+e^{(\mu)})\, \Omega_{\mu}(x)\ \ .
\ee
The constants  $z_{\mu \nu}$ are elements of  $Z(N)$ (the $SU(N)$ center),
and, hence,  can 
be written as follows:
\be
z_{\mu \nu} = \exp(2 \pi \imath \frac{n_{\mu \nu}}{N})\ ,
\ee
where $n_{\mu \nu}$ is an antisymmetric tensor of integers $\bmod\,  N$ ({\em the 
twist tensor}). 
With an appropriate choice of generators the twist tensor takes the form
\be
\label{canonical}
\mathbf{n}= \left(  \begin{array}{cc}
0 &
{\bf \Xi} \\ - {\bf \Xi} & 0  \end{array}   \right)\ ,
\ee
where $\mathbf{\Xi}= \mbox{diag}(q_1,q_2)$ and $q_1$,  $q_2$ are positive
integers ($1 \leq q_i \leq N$).

We may now introduce the following family of $U(N)$ self-dual gauge fields
on the torus:
\be
A^z_{\mu}(x)= A_{\mu}(x) + 2 \pi  z_{\mu}\, \mathbf{I}\ .
\ee
The arbitrary real numbers $z_{\mu}$ can be  arranged into a four-vector $z$, 
and $\mathbf{I}$ is the $N \times N$
identity matrix. The field strength,  and the topological charge $Q$ are
independent of $z$. 

 Let us now consider an orthonormal basis of the space of  solutions of 
 the Weyl equation in the fundamental
 representation with $A^z_{\mu}(x)$ as background field:
 \be
 \label{weyl}
 \overline{D}^z\Psi^{\alpha}(x,z)=0\ \ .
 \ee
 The positive chirality Weyl operator  is given by $\overline{D}^z \equiv D^z_{\mu} \bar{\sigma}_{\mu}$
  ($D^z_{\mu} = \partial_{\mu} - \imath A^z_{\mu}(x)$),
where   $\bar{\sigma}_{\mu}=(\mathbf{I_{2\times 2}}, \imath\, \vec{\sigma})$
and $\sigma_i$ are the Pauli matrices.

Now let $e(n)$ denote an element of the lattice $\Lambda$:
\be
e(n)=n_{\mu}e^{(\mu)}\ , 
\ee
where $n$ is a vector of integers. To this element we may associate an $SU(N)$
matrix $\Omega_n(x)$. These
  matrices   can be obtained by
composing in a given order the matrices $\Omega_{\mu}(x)$ $n_{\mu}$ times.
Changing the order of multiplications could change the  $\Omega_n(x)$ by 
multiplication by an 
element of the center. Once a choice is made,  we can write:
\be
A_{\nu}(x+e(n))= [\Omega_n(x)]A_{\nu}(x) \ \ .
\ee

Let us first assume that the twist is trivial $n_{\mu \nu} = 0$.
Then, by virtue of the Atiyah-Singer index theorem~\cite{AS} and the
assumption  that there are  no negative chirality zero-modes, one can find 
$Q$ orthonormal solutions of the Weyl equation~(\ref{weyl})
$\Psi^{\alpha}(x,z)$ ($\alpha=1,\ldots, Q$) satisfying:
\be
\label{bc}
\Psi^{\alpha}(x+e(n),z)= \Omega_n(x)\, \Psi^{\alpha}(x,z)
\ee
The previous equation is consistent because the  $\Omega_n(x)$ commute
among themselves.  From these solutions, we can construct the Nahm transform
of $A_{\nu}(x)$ as follows:
\be
\label{nahmeq} 
(\widehat{A}_{\mu}(z))_{\alpha \beta}=\imath \int_{T^4} d^4x\; \Psi^{\dagger \alpha
}(x,z)\,
\frac{\partial}{\partial z_{\mu}}  \Psi^{\beta}(x,z) 
\ee
It can be shown, that $\widehat{A}_{\mu}(z)$ is an $SU(Q)$ 
self-dual gauge field
with topological charge $N$~(See Ref.~\cite{PB2} for a proof of these and 
other properties of the Nahm transform).
Now, let us investigate the periodicity properties of $\widehat{A}_{\mu}(z)$.
For that we notice  that $\exp(- 2 \pi \imath \tilde{z}_{\mu} x_{\mu})\,
\Psi^{\alpha}(x,z+\tilde{z})$ satisfies the same equation than
$\Psi^{\alpha}(x,z)$. The boundary conditions are, however, different 
in general. In order to satisfy Eq.~\ref{bc}, one must have that $\tilde{z}_{\mu}
e^{(\nu)}_{\mu} \in \mathbf{Z}$. This is the condition that defines the dual
lattice $\widetilde{\Lambda}$. The  basis dual to  $\{e^{(\nu)}\}$ is given by
$\{\tilde{e}^{(\nu)}\}$ satisfying:
\be
\tilde{e}^{(\mu)}_{\rho}\, e^{(\nu)}_{\rho}\, = \delta_{\mu \nu}\ \ .
\ee
Now, since $\Psi^{\alpha}(x,z)$ is a basis of the space of solutions 
with boundary conditions~(\ref{bc}), we must have:
\be
\Psi^{\alpha}(x,z+\tilde{e}^{(\mu)})=  \Psi^{\beta}(x,z)
(\widehat{\Omega}^{\dagger}_{\mu}(z))_{\beta \alpha}\, \exp(2 \pi \imath\,
\tilde{e}^{(\mu)}_{\rho} x_{\rho}) \ .
\ee
Then one sees that $\widehat{A}_{\nu}(z)$ satisfy boundary conditions 
similar to
those of equation~(\ref{tbc}) with $\Omega_{\mu}(x)$ replaced by
$\widehat{\Omega}_{\mu}(z)$. In summary, the Nahm transformed field lives in the
dual torus $\widetilde{T}^4=\mathbf{R}^4/\widetilde{\Lambda}$.

Let us go back to the general twist case.  The previous procedure fails
because Eq.~(\ref{bc}) becomes inconsistent for non-trivial twist. The strategy
that we will put forward is the following. Consider a sublattice $\Lambda'$
of  $\Lambda$. The sublattice is chosen in such a way that for all $e(n)\in
\Lambda'$ the corresponding $\Omega_n(x)$ commute (Actually the $\Omega_n(x)$
must define a representation of $\Lambda'$). We will impose
Eq.~(\ref{bc}) only in $\Lambda'$. This is equivalent to
considering the fields as living in a wider torus  $T'^4=\mathbf{R}^4/\Lambda'$.
In this case the Nahm transform can be constructed in the same way as
described before, and lives in $\widetilde{T}'^4=\mathbf{R}^4/\widetilde{\Lambda'}$.
The rank of the Nahm transformed field is now $\widehat{N}=Q\,
|\Lambda/\Lambda'|$, where $|\Lambda/\Lambda'|$ is the order of the quotient
group of the two lattices. Henceforth, the basis of the space of
solutions is $\Psi^{\tilde{\alpha}}(x,z)$ with $\tilde{\alpha} =1, \ldots
,\widehat{N}$.


We want now  to explore the consequences  of the original stronger 
periodicity condition for the
fields. For any $n \in \mathbf{Z}^4$ we can define:
\be
\label{def1}
\Psi^{(n) \tilde{\alpha}}(x,z) =  \Omega^{\dagger}_n(x)\,
\Psi^{\tilde{\alpha}}(x+e(n),z)\ \ .
\ee
If $e(n)\in\Lambda'$, the boundary  conditions imply that the
$\Psi^{(n)\tilde{\alpha}}$
coincide with  $\Psi^{\tilde{\alpha}}$. Notice, however, that for any $n$
the previous functions are solutions of the Weyl equation~(\ref{weyl}).
The problem is that, if the twist tensor is non-zero,  the previous functions
satisfy boundary conditions which differ by multiplication by an element of
the center from those of $\Psi^{\tilde{\alpha}}(x,z)$. However, remember that
the same situation took place whenever we  displaced z by a vector
not belonging to the dual lattice and multiplied it by a corresponding phase.  
Henceforth, combining both operations, we arrive at our most 
important result, summarized in the
following formulas:
\begin{eqnarray}
\label{main}
&\Psi^{(n) \tilde{\alpha}}(x,z+\Delta) \exp(- 2 \pi \imath\, \Delta_{\mu}
x_{\mu}) =  \Psi^{\tilde{\beta}}(x,z)\,
(\widehat{\Omega}^{\dagger}_{n \Delta}(z))_{\tilde{\beta} \tilde{\alpha}}& \ \\
\label{cond}
&\frac{1}{N}n_{\mu \nu} n_{\nu} m_{\mu} - \Delta_{\mu} e(m)_{\mu}\, \in
\mathbf{Z}
\ \ \ \mbox{for}\  e(m) \in \Lambda' &           .
\end{eqnarray}
The second equation (\ref{cond}) is the necessary and sufficient condition 
for the left
hand side of Eq. (\ref{main}) to satisfy the same boundary conditions as
 $\Psi^{\tilde{\alpha}}(x,n)$. The first equation then expresses the fact
 that it can
be written as a linear combination of the basis of the space of solutions.
One can see from~(\ref{def1})  that the functions $\Psi^{(n)
\tilde{\alpha}}(x,z)$ are orthonormal irrespective of $n$, and hence the
$\widehat{\Omega}(z)$ are unitary.
Then,  one can conclude that for any $\Delta$ such that there exists an $n$
satisfying Eq.~(\ref{cond}) we have:
\be
\widehat{A}_{\nu}(z+\Delta)= [\widehat{\Omega}_{n \Delta}(z)]\, \widehat{A}_{\nu}(z)\ .
\ee
This formula is similar to Eq.~(\ref{tbc}), and is the main formula we were
looking for. For $\Delta$ belonging to $\widetilde{\Lambda'}$, it amounts to what
was known previously. But this formula makes the periodicity valid for a
larger lattice $\widehat{\Lambda}$. We must now study the consistency conditions
that follow from equation~(\ref{main}). If we consider two sets $(n,\Delta)$
and $(n',\Delta')$ satisfying Eq.~(\ref{cond}), we can deduce:
\begin{eqnarray}
\label{tbcc}
&\widehat{\Omega}_{n \Delta}(z+\Delta')\, \widehat{\Omega}_{n' \Delta'}(z)= \\
\nonumber
&\exp\{ \frac{2 \pi
\imath}{N} n_{\mu \nu} n_{\mu}  n'_{\nu} + 2 \pi \imath
(\Delta_{\rho}e(n')_{\rho} - \Delta'_{\rho}e(n)_{\rho})\}\ 
\widehat{\Omega}_{n' \Delta'}(z+\Delta)\, \widehat{\Omega}_{n \Delta}(z)\ .
\end{eqnarray}
This condition is the counterpart of Eq.~(\ref{twist}) for the Nahm transform.

Now, all we need is to straighten up  the conditions implied by Eqs.~
(\ref{main}-\ref{tbcc}). Let us stick to the form Eq.~(\ref{canonical}). Then,
if  $q_i \ne 0$ we can form the integers $p_i=N/\gcd(N,q_i)$. The case $q_i=0$
can be treated in the same fashion by setting $q_i=N$, $p_i=1$. The lattice
$\Lambda'$ is the one spanned by the vectors  $\{e^{(0)},e^{(1)},p_1 e^{(2)}, p_2 e^{(3)}\}$.
The elements of $\Lambda/\Lambda'$ can be labeled  by two integers
$n_2=0, \ldots, (p_1-1)$ and $n_3=0, \ldots, (p_2-1)$. Then an arbitrary
element\footnote{The corresponding element of $\Lambda/\Lambda'$ is given by
$n_2=s_1 k_0 \bmod p_1$ ; $n_3=s_2 k_1 \bmod p_2$, with $s_i$ defined after
Eq.~(\ref{nstar}).} of
the  lattice $\widehat{\Lambda}$
is given by $\Delta=\hat{e}^{ (\mu)} k_{\mu}$~($k_{\mu} \in
\mathbf{Z}$). 
The generators are given by:
\begin{eqnarray}
\nonumber
\hat{e}^{ (0)} = \tilde{e}^{ (0)}/p_1 & \\
\label{lambstar}
\hat{e}^{ (1)} = \tilde{e}^{ (1)}/p_2 &\\
\nonumber
\hat{e}^{ (2)} = \tilde{e}^{ (2)}/p_1 &\\
\nonumber
\hat{e}^{ (3)} = \tilde{e}^{ (3)}/p_2 &\ \ . 
\end{eqnarray}
From Eq.~(\ref{tbcc}) one can compute the twist matrix $\hat{n}_{\mu  \nu}$ 
associated to the 
Nahm transform:
\be
\label{nstar}
\mathbf{\hat{n}}=\left(  \begin{array}{cc}
0 &
{\bf \widehat{\Xi}} \\ - {\bf \widehat{\Xi}} & 0  \end{array}   \right)\ \ \mbox{with}\ {\bf
\widehat{\Xi}}= \mbox{diag}((p_1-s_1) p_2 Q\,,\,(p_2-s_2) p_1 Q)\ \ ,   
\ee
where $s_i$ is an  integer ($0 \leq s_i \leq (p_i-1)$) satisfying 
$s_i q_i =\gcd(q_i,N) \bmod N$. 
We have taken the rank of the Nahm
transform $\widehat{N}$ to be $Q\, p_1 p_2$. Taking into account that the topological
charge $Q$ has the form $l-\frac{q_1 q_2}{N}$ with $l$ an integer (see for
example Ref.~\cite{torusrev}) one can easily show  that the 
$\hat{n}_{\mu  \nu}$ are 
integers. 

To conclude this section, we mention that if we apply  Nahm's transformation
to the Nahm transform, we obtain back the original gauge field with twist
in $T^4$. If we label with a $\wedge$  all quantities referring to the 
Nahm transform, one can derive the following relations:
\begin{eqnarray}
\nonumber  \widehat{N}= Q p_1 p_2 \ \ & \ \ \widehat{Q}= \frac{N}{p_1 p_2} \\
 \hat{p}_i= p_i \ \ & \ \ \hat{s}_i=p_i- \frac{q_i}{\gcd(N,q_i)}\ .
\end{eqnarray}
With these relations  it is not hard to show that
$\widehat{\widehat{\Lambda}}=\Lambda$,
and that after applying the Nahm transform twice we recover the original
set of parameters. It is also easy to show that the Nahm transform of a
gauge field configuration with non-orthogonal twist ( $\frac{1}{8}
\epsilon^{\mu \nu \rho \sigma}n_{\mu \nu} n_{\rho \sigma} \neq 0 \bmod N$)
has also non-orthogonal twist.


\section{Replicas without twist }
We will now apply the formalism of section 2 to understand what happens
when we replicate the torus in the absence of twist. We start with the
self-dual $SU(N)$ gauge field $A_{\mu}(x)$ defined on the torus
$T^4=\mathbf{R^4}/\Lambda$. The topological charge is $Q$. Now we can
construct the Nahm transform, which is an $SU(Q)$ self-dual gauge field
defined in $\widetilde{T}^4=\mathbf{R^4}/\widetilde{\Lambda}$. 
On the other hand,
we might look at  $A_{\mu}(x)$ as a gauge field defined on the larger
torus $T'^4=\mathbf{R^4}/\Lambda'$, where $\Lambda'$ is a sublattice of
$\Lambda$. Applying the Nahm transform in this case gives an
$SU(Q\,|\Lambda/\Lambda'|)$ gauge field living in the dual torus
 $\widetilde{T}'^4=\mathbf{R^4}/\widetilde{\Lambda'}$. The question that we want to 
 answer is what is the relation between both Nahm transforms. 

 Without loss of generality we can restrict ourselves to the case in which we
 replicate the torus in one direction. By repeated application of this
situation we can deal with the general case. Let 
$e^{(\mu)}$ be the generators of $\Lambda$. The lattice $\Lambda'$ has the
same generators except for $e'^{(0)}=L\, e^{(0)}$ where $L=|\Lambda/\Lambda'|$ is a positive integer.
For the dual tori we have $\tilde{e}'^{(0)}=\frac{1}{L}\, \tilde{e}^{(0)}$.
Now consider the solutions of the Weyl equation in $T^4$. There are $Q$
orthonormal linearly independent solutions $\Psi^{\alpha}(x,z)$ satisfying the boundary
conditions Eq.~(\ref{bc}). With them we can construct the Nahm transform as
usual. Now let us consider the following functions:
\be
\Psi^{\alpha k}(x,z)=\Psi^{\alpha}(x,z+\frac{k}{L}\tilde{e}^{(0)})\, \exp(-
2 \pi \imath\, x_{\mu}\tilde{e}^{(0)}_{\mu} \frac{k}{L})\ \ ,
\ee
where $k$ is an integer ranging from $0$ to $L-1$. They are all solutions of
the Weyl equation. Let us see how these
functions behave under translations along $e^{(0)}$. We have 
\be
\Psi^{\alpha k}(x+l\,e^{(0)},z)= \exp(-2 \pi \imath \frac{k l}{L})\
\Omega_{n_l}(x)\,
\Psi^{\alpha k}(x,z)\ ,
\ee
where $n_l=(l,0,0,0)$.
Notice that, although only for $k=0$  the boundary conditions on $T^4$
are satisfied, they all satisfy the right boundary conditions on $T'^4$, i.e.
for $l=L$. Since there are $LQ$ different functions, they constitute a basis of
the space of solutions provided they are linearly independent. We will show
that indeed they are all mutually orthogonal in $T'^4$. The proof goes as
follows:
\begin{eqnarray}
\nonumber
&\int_{T'^4} dx\, \Psi^{\dagger \alpha' k'}(x,z)\,  \Psi^{ \alpha k}(x,z)=\\
\label{seq}
&\sum_{l=0}^{L-1} \int_{T^4} dx\, \Psi^{\dagger \alpha' k'}(x+l\,e^{(0)},z)\,  \Psi^{ \alpha
k}(x+l\,e^{(0)},z)=\\
\nonumber
&\left(\sum_{l=0}^{L-1} \exp(2 \pi \imath \frac{l(k-k')}{L})\right)
\int_{T^4} dx\, \Psi^{\dagger
\alpha' k'}(x,z)\, \Psi^{ \alpha k}(x,z)=\\
\nonumber
&L\, \delta_{k k'}\, \delta_{\alpha \alpha'}
\end{eqnarray}
In the last identity we used the orthogonality of the $\Psi^{\alpha}(x,z)$
for arbitrary $z$. 

Dividing the functions $\Psi^{ \alpha k}(x,z)$ by $\sqrt{L}$, we have the
necessary orthonormal basis of the space of solutions that we need to
construct the Nahm transform in $T'^4$. Applying the definition of the Nahm 
transform, and after similar manipulations as in Eq.~(\ref{seq}) we arrive at:
\be
\left(\widehat{A}(z)\right)_{(\alpha k)(\alpha' k')}= \delta_{k k'}\,
\left(\widehat{A}(z+\frac{k}{L}\tilde{e}^{(0)})\right)_{\alpha
\alpha'}\ \ ,
\ee
where the left hand side is the Nahm transform defined in $T'^4$ and it is
expressed in terms of the Nahm transform in $T^4$. It is clear from our
formulas that the former is reducible and invariant (modulo gauge
transformations) under translation of z by $\frac{1}{L}\tilde{e}^{(0)}$. 

