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\section{S-matrix}

For particles with generalized statistics (anyons) following \cite{KKS} we
relate the `physical' S-matrix $\tilde{S}_{\alpha\beta}^{\delta\gamma}%
(\theta)$ to an auxiliary one $S_{\alpha\beta}^{\delta\gamma}(\theta)$ by
\[
S_{\alpha\beta}^{\delta\gamma}(\theta)=\exp\left(  2\pi i\left(
s_{\delta\gamma}-\tfrac{1}{2}\right)  \varepsilon(\theta)\right)  \,\tilde
{S}_{\alpha\beta}^{\delta\gamma}(\theta)
\]
where $s_{\delta\gamma}$ is the `spin' of the particles of kind $\delta$ and
$\gamma$ which means that $\exp\left(  2\pi is_{\delta\gamma}\right)  $ is the
statistics factor. The auxiliary S-matrix $S$ describes the scattering of
purely fermionic particles and it satisfies the analytic properties which
follow from LSZ-assumptions, in contrast to the physical S-matrix $S$ which is
in general non-analytic at $\theta=0$. For the $SU(N)$-Gross-Neveu model
\cite{KKS} $s_{\delta\gamma}=\frac{1}{2}\left(  1-\frac{1}{N}\right)  $. As we
will see below for $N\rightarrow\infty$ the auxiliary as well as the physical
S-matrix satisfy $\tilde{S}\rightarrow\mathbf{1}$ and $S\rightarrow\mathbf{1}%
$. For $\theta\rightarrow\infty$ the auxiliary S-matrix behaves as
$S\rightarrow\mathbf{\exp}(i\pi/N)$, whereas the physical one satisfies
$\tilde{S}\rightarrow\mathbf{1}$. The latter behavior serves as a criterion
for a physical S-matrix.

For the scattering of particles we have \cite{BKKW,BW,KS} the matrix elements
\[
S_{\alpha\beta}^{\delta\gamma}(\theta)=\delta_{\alpha\gamma}\delta
_{\beta\delta}\,b(\theta)+\delta_{\alpha\delta}\delta_{\beta\gamma}%
\,c(\theta)\,.
\]
For later convenience we introduce in addition an $U(N)$-invariant auxiliary
S-matrix $\dot{S}$ which is $-S$ for particle-particle scattering (the minus
sign is due to the fermi statistics of the auxiliary particles)
\begin{equation}
\dot{S}_{\alpha\beta}^{\delta\gamma}(\theta)=
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(6,6) \put(1,1){\line(1,1){4}} \put(1,1){\vector(1,1){1.2}} \put(3,3){\vector(1,1){1.2}} \put(5,1){\line(-1,1){4}} \put(5,1){\vector(-1,1){1.2}} \put(3,3){\vector(-1,1){1.2}} \put(.5,.1){$\scriptstyle\alpha$} \put(5,.1){$\scriptstyle\beta$} \put(5,5.3){$\scriptstyle\gamma$} \put(.5,5.3){$\scriptstyle\delta$} \put(1.8,1){$\scriptstyle p_1$} \put(3.5,1){$\scriptstyle p_2$} \put(3.5,4.7){$\scriptstyle p_1$} \put(1.8,4.7){$\scriptstyle p_2$} \end{picture}
\end{array}
=-\delta_{\alpha\gamma}\delta_{\beta\delta}\,b(\theta)-\delta_{\alpha\delta
}\delta_{\beta\gamma}\,c(\theta)\,. \label{1.1}%
\end{equation}
The forward scattering of particles and anti-particles is obtained by
crossing
\begin{equation}
\dot{S}_{\beta\bar{\gamma}}^{\bar{\alpha}\delta}(\theta)=
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(6,6) \put(1,1){\line(1,1){4}} \put(1,1){\vector(1,1){1.2}} \put(3,3){\vector(1,1){1.2}} \put(5,1){\line(-1,1){4}} \put(1,5){\vector(1,-1){1.2}} \put(3,3){\vector(1,-1){1.2}} \put(.5,.1){$\scriptstyle\beta$} \put(5,.1){$\scriptstyle\bar\gamma$} \put(5,5.3){$\scriptstyle\delta$} \put(.5,5.3){$\scriptstyle\bar\alpha$} \put(1.8,1){$\scriptstyle p_1$} \put(3.5,1){$\scriptstyle p_2$} \put(3.5,4.7){$\scriptstyle p_1$} \put(1.8,4.7){$\scriptstyle p_2$} \end{picture}
\end{array}
=
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(8,6) \put(2,1){\line(1,1){4}} \put(2,1){\vector(1,1){1.2}} \put(4,3){\vector(1,1){1.2}} \put(5,1){\line(-1,2){2}} \put(5,1){\vector(-1,2){.6}} \put(4,3){\vector(-1,2){.6}} \put(6,1){\oval(3,8)[rt]} \put(2,5){\oval(3,8)[lb]} \put(.5,5){\vector(0,-1){1.5}} \put(7.5,2){\vector(0,-1){0}} \put(4.5,.1){$\scriptstyle\beta$} \put(7.2,.1){$\scriptstyle\bar\gamma$} \put(3,5.3){$\scriptstyle\delta$} \put(0,5.3){$\scriptstyle\bar\alpha$} \put(5,1.9){$\scriptstyle p_1$} \put(6.4,3){$\scriptstyle p_2$} \put(1.2,2.3){$\scriptstyle -p_2$} \end{picture}
\end{array}
=\delta_{\bar{\gamma}}^{\bar{\alpha}}\delta_{\beta}^{\delta}\,b(\hat{\theta
})-\mathbf{C}^{\bar{\alpha}\delta}\mathbf{C}_{\beta\bar{\gamma}}%
\,c(\hat{\theta}) \label{1.2}%
\end{equation}
with $\hat{\theta}=i\pi-\theta$. The charge conjugation matrix $\mathbf{C}$ is
anti-symmetric and its matrix elements are
\[
\mathbf{C}^{\alpha\bar{\beta}}=\mathbf{C}_{\alpha\bar{\beta}}=-\mathbf{C}%
^{\bar{\beta}\alpha}=-\mathbf{C}_{\bar{\beta}\alpha}=\mathbf{\delta}%
_{\alpha\beta}\,.
\]
The particle anti-particle backward scattering vanishes\footnote{It has been
argued in \cite{BW,KS} (see also \cite{Be} for $N=2$) that for the S-matrix of
the chiral $SU(N)$-Gross-Neveu model the backward scattering vanishes.}. We
also write
\[
\dot{S}_{12}(\theta)=\dot{a}(\theta)\frac{\mathbf{1}_{12}\theta-\mathbf{P}%
_{12}2\pi i/N}{\theta-2\pi i/N}\,,\quad\dot{S}_{1\bar{2}}(\theta)=\hat
{b}(\theta)\left(  \mathbf{1}_{1\bar{2}}+\mathbf{K}_{1\bar{2}}\frac{2\pi
i/N}{\theta-i\pi}\right)  \,.
\]
where $\dot{a}(\theta)=-a(\theta)$ and $\hat{b}(\theta)=b(i\pi-\theta)$
(similarly we write $\dot{b}(\theta)=-b(\theta)\,,\;\dot{c}(\theta
)=-c(\theta)$ and $\hat{c}(\theta)=c(i\pi-\theta)$). The permutation matrix
$\mathbf{P}_{12}$ is given by the second term in eq.~(\ref{1.1}) and the
annihilation-creation matrix $\mathbf{K}_{1\bar{2}}=-\mathbf{C}^{\bar{2}%
1}\mathbf{C}_{1\bar{2}}$ is given by the second term in eq.~(\ref{1.2}).

The Yang-Baxter equations imply the relations of the amplitudes
\[
\frac{b(\theta)}{a(\theta)}=\frac{\theta}{\theta-2\pi i/N}~,~~\frac{c(\theta
)}{a(\theta)}=-\frac{2\pi i/N}{\theta-2\pi i/N}~,~~\frac{c(\hat{\theta}%
)}{b(\hat{\theta})}=\frac{2\pi i/N}{\theta-i\pi}~.
\]
The general solution of unitarity and crossing relations have been obtained in
\cite{BKKW} and the amplitudes are given by
\[
a(\theta)=\frac{\Gamma\left(  1-\frac{\theta}{2\pi i}\right)  \Gamma\left(
1-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  }{\Gamma\left(  1+\frac{\theta
}{2\pi i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{\theta}{2\pi i}\right)  }
\]
and
\[
b(\hat{\theta})=\frac{\Gamma\left(  \frac{1}{2}+\frac{\theta}{2\pi i}\right)
\Gamma\left(  \frac{1}{2}-\frac{1}{N}-\frac{\theta}{2\pi i}\right)  }%
{\Gamma\left(  \frac{1}{2}-\frac{\theta}{2\pi i}\right)  \Gamma\left(
\frac{1}{2}-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  }
\]
There exist usefull integral representations
\[
\,a(\theta)=\exp\int_{0}^{\infty}\frac{dt}{t}e^{-t}\frac{1-e^{t\frac{2}{N}}%
}{\sinh t}\sinh t\frac{\theta}{i\pi}=1+\frac{\pi}{N}\left(  \coth\frac{\theta
}{2}-\frac{2}{\theta}\right)  +O(N^{-2})
\]
\[
b(\hat{\theta})=\exp\int_{0}^{\infty}\frac{dt}{t}\frac{e^{t\frac{2}{N}}%
-1}{\sinh t}\sinh t\frac{\theta}{i\pi}=1-\frac{\pi}{N}\tanh\frac{\theta}%
{2}+O(N^{-2})\,.
\]
Notice that for the particle-particle scattering the symmetric and the
anti-symmetric tensor S-matrix eigenvalues are
\begin{align*}
\dot{S}_{+}(\theta)  &  =-\left(  b(\theta)+c(\theta)\right)  =-a(\theta)\\
\dot{S}_{-}(\theta)  &  =-\left(  b(\theta)-c(\theta)\right)  =-a(\theta
)\frac{\theta+2\pi i/N}{\theta-2\pi i/N}\,.
\end{align*}
The anti-symmetric one possesses a single order pole at $\theta=2\pi i/N$ in
the physical strip $0<\operatorname{Im}\theta<\pi$ giving rise to a
bound-state particle.

The anti-symmetric tensor and the scalar particle-antiparticle scattering
eigenvalues
\begin{align*}
\bar{S}_{-}(\theta)  &  =b(\hat{\theta})\\
\bar{S}_{0}(\theta)  &  =b(\hat{\theta})+Nc(\hat{\theta})=b(\hat{\theta}%
)\frac{\theta+\pi i}{\theta-\pi i}\,.
\end{align*}
are non-singular in the physical strip. The unitaryty of the S-matrix
$S^{\dagger}(\theta)S(\theta)=1$ may be writen in terms of the eigenvalues as
\[
S(-\theta)S(\theta)=1
\]
for $S=\dot{S}_{\pm},\bar{S}_{-},\bar{S}_{0}$. In the following we will also
use the notations
\[
\dot{a}(\theta)=-a(\theta)\,,\;\dot{b}(\theta)=-b(\theta)\,,\;\dot{c}%
(\theta)=-c(\theta)\,,\;\hat{b}(\theta)=b(\hat{\theta})\,,\;\hat{c}%
(\theta)=c(\hat{\theta})\,.
\]


\setcounter{equation}{0}

\section{Form factors}

\subsection{The two particle form factors}

The most basic properties of the form factors are usually referred to as
Watson's equations \cite{Wa}, which have been already known in the fifties.
They follow from general principles. It is instructive at this point to
discuss them first for pure bosonic chargeless particles and a hermitian
scalar local operator $\mathcal{O}(x)$. We consider for $\mathcal{O}%
=\mathcal{O}(0)$%
\[
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(s_{12}+i\epsilon)=\langle
\,0\,|\,\mathcal{O}\,|\,p_{1},p_{2}\,\rangle_{\alpha_{1}\alpha_{2}}^{in}.
\]
Time inversion transformation implies that
\[
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(s_{12}-i\epsilon)=\langle
\,0\,|\,\mathcal{O}\,|\,p_{1},p_{2}\,\rangle_{\alpha_{1}\alpha_{2}}^{out}.
\]
Using the completeness of the $out$-states we have
\begin{equation}
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(s_{12}+i\epsilon)=\langle
\,0\,|\,\mathcal{O}\,|\,p_{1},p_{2}\,\rangle_{\alpha_{1}\alpha_{2}}^{in}%
=\sum_{out}\langle\,0\,|\,\mathcal{O}\,|\,out\,\rangle\langle\,out\,|\,p_{1}%
,p_{2}\,\rangle_{\alpha_{1}\alpha_{2}}^{in}. \label{2.1}%
\end{equation}
This formula holds in general models including particle production. For
$4m^{2}\le s_{12}<$ 'lowest inelastic threshold' only the two particle
S-matrix contributes
\begin{equation}
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(s_{12}+i\epsilon)=F_{\alpha_{1}%
^{\prime}\alpha_{2}^{\prime}}^{\mathcal{O}}(s_{12}-i\epsilon)\,S_{\alpha
_{1}\alpha_{2}}^{\alpha_{2}^{\prime}\alpha_{1}^{\prime}}(s_{12}) \label{2.2}%
\end{equation}
and analogously starting with $_{\alpha_{1}}\langle\,p_{1}\,|\,\mathcal{O}%
\,|\,p_{2}\,\rangle_{\alpha_{2}}$ in (\ref{2.1}) we obtain
\[
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(t_{12}-i\epsilon)=F_{\alpha_{1}%
\alpha_{2}}^{\mathcal{O}}(t_{12}+i\epsilon)\,,
\]
where the fact has been used that the one-particle S-matrix is always trivial.
In integrable theories there are no inelastic transitions, therefore
eq.~(\ref{2.2}) holds for all $s\ge4m^{2}$. Therefore we get in terms of
rapidity difference variables

\subparagraph{Watson's equations}%

\begin{equation}%
\begin{array}
[c]{l}%
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(\theta)=F_{\alpha_{1}^{\prime}%
\alpha_{2}^{\prime}}^{\mathcal{O}}(-\theta)\,S_{\alpha_{1}\alpha_{2}}%
^{\alpha_{2}^{\prime}\alpha_{1}^{\prime}}(\theta)\\
F_{\alpha_{1}\alpha_{2}}^{\mathcal{O}}(i\pi-\theta)=F_{\alpha_{1}\alpha_{2}%
}^{\mathcal{O}}(i\pi+\theta)\,.
\end{array}
\label{2.3}%
\end{equation}
where $\theta$ defined by $p_{1}p_{2}=m^{2}\cosh\theta$.

\label{s4}We consider 4 cases:

\begin{enumerate}
\item Let $\mathcal{O}$ be a chargeless operator which transforms as an
anti-symmetric $U\left(  N\right)  $-tensor, as for instance $\epsilon^{\mu
\nu}\partial_{\mu}J_{\nu}^{ab}(x)$, the psedo potential of the current. The
highest weight components of the two particle form factor function
$f_{1\bar{N}}^{\mathcal{O}}(\theta_{1},\theta_{2})=f_{1\bar{N}}^{\mathcal{O}%
}(\theta_{12})$ satisfies according to equation (\ref{2.3})
\[
f_{1\bar{N}}^{\mathcal{O}}(\theta)=b(\hat{\theta})\,f_{\bar{N}1}^{\mathcal{O}%
}(-\theta)=f_{\bar{N}1}^{\mathcal{O}}(2\pi i-\theta)\,.
\]
We obtain as a solution
\[
f_{1\bar{N}}^{\mathcal{O}}(\theta)=N^{\mathcal{O}}\bar{F}\left(
\theta\right)  \;,\quad f_{\bar{N}1}^{\mathcal{O}}(\theta)=-N^{\mathcal{O}%
}\bar{F}\left(  \theta\right)  \,.
\]
Here $N^{\mathcal{O}}$ is a constant which depends on the normalization of the
particular operator $\mathcal{O}$. The function $\bar{F}\left(  \theta\right)
$ denotes the so-called minimal form factor corresponding to $b(\hat{\theta})$
which satisfies
\[
\bar{F}\left(  \theta\right)  =\hat{b}(\theta)\bar{F}\left(  -\theta\right)
=\bar{F}(2\pi i-\theta)\,.
\]
We assume here minimality which means that we require that there are neither
poles nor zeros in the physical sheet.

\item Let $\mathcal{O}$ be a chargeless operator which transforms as a scalar
under $U\left(  N\right)  $, as for instance the trace of the energy momentum
tensor. The form factor function for two particles $f_{\alpha\bar{\beta}%
}^{\mathcal{O}}(\theta_{1},\theta_{2})$ $=f_{\alpha\bar{\beta}}^{\mathcal{O}%
}(\theta_{12})$ is proportional to $\delta_{\alpha\beta}$ and according to
equation (\ref{2.3}), it satisfies
\[
f_{\alpha\bar{\alpha}}^{\mathcal{O}}(\theta)=b(\hat{\theta})\frac{\theta+i\pi
}{\theta-i\pi}\,f_{\bar{\alpha}\alpha}^{\mathcal{O}}(-\theta)=f_{\bar{\alpha
}\alpha}^{\mathcal{O}}(2\pi i-\theta)\,.
\]
By direct inspection or from Expl. 5.5 in [BKZ2] we obtain as a solution for
the auxilary function of the two particle form factor
\[
f_{\alpha\bar{\beta}}^{\mathcal{O}}(\theta)=N^{\mathcal{O}}\bar{F}\left(
\theta\right)  \frac{\cosh\frac{\theta}{2}}{1-\frac{\theta}{i\pi}}%
\delta_{\alpha\beta}\,.
\]
Here $N^{\mathcal{O}}$ is again a normalization constant and again we have
assumed minimality. The pole in $F_{\bar{1}2}^{\mathcal{O}}(\theta)$ at
$\theta=i\pi$ occurs at the same position as in the related S-matrix.

\item Let $\mathcal{O}$ be a charge two operator which transforms as a
symmetric $U\left(  N\right)  $-tensor. The highest weight components of the
two particle form factor function $f_{11}^{\mathcal{O}}(\theta_{1},\theta
_{2})=f_{11}^{\mathcal{O}}(\theta_{12})$ satisfies according to equation
(\ref{2.3})
\[
f_{11}^{\mathcal{O}}(\theta)=\dot{a}(\theta)\,f_{11}^{\mathcal{O}}%
(-\theta)=f_{11}^{\mathcal{O}}(2\pi i-\theta)\,.
\]
We obtain as a solution
\[
f_{11}^{\mathcal{O}}(\theta)=N^{\mathcal{O}}F\left(  \theta\right)  .
\]
The function $F\left(  \theta\right)  $ is the minimal form factor
corresponding to $-a(\theta)$ satifying
\[
F\left(  \theta\right)  =\dot{a}(\theta)F\left(  -\theta\right)  =F(2\pi
i-\theta)
\]
Here $N^{\mathcal{O}}$ is again a normalization constant and again we have
assumed minimality.

\item Let $\mathcal{O}$ be a charge two operator which transforms as an
anti-symmetric $U\left(  N\right)  $-tensor. $f_{12}^{\mathcal{O}}(\theta
_{1},\theta_{2})=f_{12}^{\mathcal{O}}(\theta_{12})$ satisfies according to
equation (\ref{2.3})
\[
f_{12}^{\mathcal{O}}(\theta)=-a(\theta)\frac{\theta+2\pi i/N}{\theta-2\pi
i/N}\,f_{21}^{\mathcal{O}}(-\theta)=f_{21}^{\mathcal{O}}(2\pi i-\theta)\,.
\]
We obtain as a solution...From direct evaluation or from Expl. 5.3 in [BKZ1]
we obtain as a solution ???
\begin{align*}
F_{\alpha\beta}^{\mathcal{O}}(\theta)  &  =\prod\limits_{k=1}^{\infty}%
\frac{\Gamma\left(  \left(  2k-2\right)  i\pi+\phi-\frac{2}{N}\right)
\Gamma\left(  2ki\pi-\phi-\frac{2}{N}\right)  }{\Gamma\left(  \left(
2k-1\right)  i\pi+\phi-\frac{2}{N}\right)  \Gamma\left(  (2k+1)i\pi-\phi
-\frac{2}{N}\right)  }\cosh\left(  \frac{\theta}{2}\right) \\
&  \;\times H_{N}^{\mathcal{O}}F_{0}\left(  \theta\right)  K_{0}\left(
\theta\right)  \left(  |\beta\alpha\rangle-|\alpha\beta\rangle\right)  .
\end{align*}
Once again $H_{N}^{\mathcal{O}}$ is a constant which depends on the
normalization of the particular operator $\mathcal{O}$ and possibly on N.
Notice that the pole in $F_{\alpha\beta}^{\mathcal{O}}(\theta)$ at
$\theta=\frac{2\pi i}{N}$ occurs at the same position as in the related S-matrix.
\end{enumerate}

We will now give the explicit expressions for the minimal form factors and
list some of their properties which we will utilize below. There are several
alternative representations for the form factors functions $F\left(
\theta\right)  $ and $\bar{F}\left(  \theta\right)  $, which all serve for
different puposes. We have integral representations or one may represent them
in form of infinite products of Gamma functions $\Gamma\left(  1-\frac{\theta
}{2\pi i}\right)  $ There are several alternative representations for the
minimal form factors, which all serve for different puposes. First of all we
have an integral representation
\begin{align*}
F_{0}\left(  \theta\right)    & =\exp\int\limits_{0}^{\infty}\frac{dt}%
{t\sinh^{2}\left(  t\right)  }\left(  e^{-t}-1\right)  \sin^{2}\left(
\frac{t\left(  i\pi-\theta\right)  }{2\pi}\right)  \\
F_{\bar{0}}\left(  \theta\right)    & =\exp\int\limits_{0}^{\infty}\frac
{dt}{t\sinh^{2}\left(  t\right)  }e^{\left(  t-1\right)  }\sin^{2}\left(
\frac{t\left(  i\pi-\theta\right)  }{2\pi}\right)
\end{align*}
We may directly compute these integrals or solve Watson's equations via the
method provided in Appendix C.

Furthermore one may represent the minimal form factor in form of infinite
products of Gamma functions
\begin{align}
F_{0}\left(  \theta\right)    & =\prod\limits_{k=1}^{\infty}\frac
{\Gamma\left(  k-\frac{1}{N}+\frac{\phi}{2}\right)  \Gamma\left(  k+1-\frac
{1}{N}-\frac{\phi}{2}\right)  \Gamma\left(  k+\frac{1}{2}\right)  ^{2}}%
{\Gamma\left(  k+\frac{\phi}{2}\right)  \Gamma\left(  k+1-\frac{\phi}%
{2}\right)  \Gamma\left(  k+\frac{1}{2}-\frac{1}{N}\right)  ^{2}}%
\label{Wats1}\\
F_{\bar{0}}\left(  \theta\right)    & =\prod\limits_{k=1}^{\infty}\frac
{\Gamma\left(  k-\frac{1}{2}+\frac{\phi}{2}\right)  \Gamma\left(  k+\frac
{1}{2}-\frac{\phi}{2}\right)  \Gamma\left(  k-\frac{1}{N}\right)  ^{2}}%
{\Gamma\left(  k-\frac{1}{2}-\frac{1}{N}+\frac{\phi}{2}\right)  \Gamma\left(
k+\frac{1}{2}-\frac{1}{N}-\frac{\phi}{2}\right)  \Gamma\left(  k\right)  ^{2}%
}\label{Wats2}%
\end{align}
and by using the Weierstrass product representation for the Gamma function one
obtains. By using the Weierstrass product representation for the Gamma
function one obtains
\begin{align*}
F\left(  \theta\right)   &  =\sin\frac{\theta}{2i}\prod\limits_{k=1}^{\infty
}\left(  \frac{\left(  k+\frac{\theta}{2\pi i}\right)  \left(  k+1-\frac
{\theta}{2\pi i}\right)  \left(  k+\frac{1}{2}-\frac{1}{N}\right)  ^{2}%
}{\left(  k-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  \left(  k+1-\frac{1}%
{N}-\frac{\theta}{2\pi i}\right)  \left(  k+\frac{1}{2}\right)  ^{2}}\right)
^{k}\\
\bar{F}\left(  \theta\right)   &  =\prod\limits_{k=1}^{\infty}\left(
\frac{\left(  k-\frac{1}{2}-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  \left(
k+\frac{1}{2}-\frac{1}{N}-\frac{\theta}{2\pi i}\right)  \left(  k\right)
^{2}}{\left(  k-\frac{1}{2}+\frac{\theta}{2\pi i}\right)  \left(  k+\frac
{1}{2}-\frac{\theta}{2\pi i}\right)  \left(  k-\frac{1}{N}\right)  ^{2}%
}\right)  ^{k}.
\end{align*}
Here we have choosen as convenient normalization
\begin{equation}
F\left(  i\pi\right)  =\bar{F}\left(  i\pi\right)  =1\label{2.10}%
\end{equation}


Apart from Watson's equations the form factors satisfy the following
functional relations, which we will require in the course of our
argumentation.
\begin{equation}
F\left(  \theta\right)  \bar{F}\left(  \theta+i\pi\right)  =\frac{\sin
\frac{\theta}{2i}\Gamma\left(  1-\frac{\theta}{2\pi i}\right)  }{\Gamma\left(
1-\frac{1}{N}-\frac{\theta}{2\pi i}\right)  }\prod\limits_{k=1}^{\infty}%
\frac{\Gamma^{2}\left(  k\right)  \Gamma^{2}\left(  k+\frac{1}{2}\right)
}{\Gamma^{2}\left(  k-\frac{1}{N}\right)  \Gamma^{2}\left(  k-\frac{1}%
{N}+\frac{1}{2}\right)  } \label{2.11}%
\end{equation}
Asymptotically the minimal form factors behave as
\[
\lim\limits_{\theta\rightarrow\infty}F\left(  \theta\right)  \approx
\exp\left(  \frac{i\pi-\theta}{2N}\right)  \qquad\text{and}\qquad
\lim\limits_{\theta\rightarrow\infty}F_{\bar{0}}\left(  \theta\right)
\approx\exp\left(  \frac{\theta-i\pi}{2N}\right)  \;\;.
\]
In the $1/N$ expansion they behave as
\begin{align*}
F_{0}\left(  \theta\right)   &  =1+\frac{1}{N}\left(  \Psi\left(  \frac
{\theta}{2\pi i}\right)  -\Psi\left(  \frac{1}{2}\right)  +\left(
\frac{\theta}{2}-i\pi\right)  \coth\left(  \frac{\theta}{2}\right)  \right)
+O\left(  \frac{1}{N^{2}}\right) \\
F_{\bar{0}}\left(  \theta\right)   &  =1+\frac{1}{N}\left(  \left(
\frac{\theta}{2}-\frac{i\pi}{2}\right)  \tanh\left(  \frac{\theta}{2}\right)
-1\right)  +O\left(  \frac{1}{N^{2}}\right)  \;\;.
\end{align*}


\subsection{Generalized Form factors}

For a state of $n$ particles of kind $\alpha_{i}$ with momenta $p_{i}$ and a
local operator $\mathcal{O}(x)$ we define the form factor functions
$\mathcal{O}_{\alpha_{1},\dots,\alpha_{n}}(\theta_{1},\dots,\theta_{n})$ or
using a short notation $\mathcal{O}_{\underline{\alpha}}(\underline{\theta})$
by
\begin{equation}
\langle\,0\,|\,\mathcal{O}(x)\,|\,p_{1},\dots,p_{n}\,\rangle_{\underline
{\alpha}}^{in}=e^{-ix(p_{1}+\cdots+p_{n})}\mathcal{O}_{\underline{\alpha}%
}(\underline{\theta})~,~~\mathrm{for}~\theta_{1}>\dots>\theta_{n}.
\label{2.12}%
\end{equation}
where $\underline{\alpha}=(\alpha_{1},\dots,\alpha_{n})$ and $\underline
{\theta}=(\theta_{1},\dots,\theta_{n})$. For all other arrangements of the
rapidities the functions $\mathcal{O}_{\underline{\alpha}}({\underline{\theta
}})$ are given by analytic continuation. Note that the physical value of the
form factor, i.e. the left hand side of (\ref{2.12}), is given for ordered
rapidities as indicated above and the statistics of the particles. The
$\mathcal{O}_{\underline{\alpha}}({\underline{\theta}})$ are considered as the
components of a co-vector valued function $\mathcal{O}_{1\dots n}%
({\underline{\theta}})\in V_{1\dots n}=\left(  V^{1\dots n}\right)  ^{\dagger
}$which may be depicted as
\[
\mathcal{O}_{1\dots n}({\underline{\theta}})=
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(6,4) \put(3,2){\oval(6,2)} \put(3,2){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){1}} \put(5,0){\line(0,1){1}} \put(0,0){$\theta_1$} \put(5.3,0){$\theta_n$} \put(2.5,.5){$\dots$} \end{picture}
\end{array}
.
\]


Now we formulate the main properties of form factors in terms of the functions
$\mathcal{O}_{1\dots n}$ which follow from general LSZ-assumptions and
``maximal analyticity'', which means that $\mathcal{O}_{1\dots n}%
({\underline{\theta}})$ is a meromorphic function with respect to all $\theta
$'s and in the `physical' strips $0<\operatorname{Im}\theta_{ij}<\pi$%
~$(\theta_{ij}=\theta_{i}-\theta_{j}\,i<j)$ there are only poles of physical
origin as for example bound state poles.

The generalized forrm factor functions satify the following properties.

\paragraph{\textbf{Properties:\label{pf}}}

The co-vector valued auxiliary function $\mathcal{O}_{1\dots n}({\underline
{\theta}})$ is meromorphic in all variables $\theta_{1},\dots,\theta_{n}$ and

\begin{itemize}
\item[(i)] satisfies the symmetry property under the permutation of both, the
variables $\theta_{i},\theta_{j}$ and the spaces $i,j$ at the same time
\[
\mathcal{O}_{\dots ij\dots}(\dots,\theta_{i},\theta_{j},\dots)=\mathcal{O}%
_{\dots ji\dots}(\dots,\theta_{j},\theta_{i},\dots)\,\dot{S}_{ij}(\theta
_{i}-\theta_{j})
\]
for all possible arrangements of the $\theta$'s,

\item[(ii)] satisfies the periodicity property under the cyclic permutation of
the rapidity variables and spaces
\[
\mathcal{O}_{2\dots n1}(\theta_{2},\dots,\theta_{n},\theta_{1})=\sigma
_{\mathcal{O}1}\mathcal{O}_{1\dots n}(\theta_{1}+2\pi i,\theta_{2}%
,\dots,\theta_{n})
\]
where $\sigma_{\mathcal{O}1}$ takes into account the statistics and
$\sigma_{\mathcal{O}1}=-1$ if $\mathcal{O}$ and particle 1 are both fermionic
and $\sigma_{\mathcal{O}1}=1$ otherwise.

\item[(iii)] and has poles determined by one-particle states in each
sub-channel. In particular the function $\mathcal{O}_{\underline{\alpha}%
}({\underline{\theta}})$ has a pole at $\theta_{12}=i\pi$ such that
\[
\operatorname*{Res}_{\theta_{12}=i\pi}\mathcal{O}_{1\dots n}(\theta_{1}%
,\dots,\theta_{n})=2i\,\mathbf{C}_{12}\,\mathcal{O}_{3\dots n}(\theta
_{3},\dots,\theta_{n})\left(  \mathbf{1}-S_{2n}\dots S_{23}\right)
\]
where $\mathbf{C}_{12}$ is the charge conjugation matrix with matrix elements
$\mathbf{C}_{\alpha\alpha^{\prime}}=\delta_{\bar{\alpha}\alpha^{\prime}}$.

\item[(iv)] If there are also bound states in the model the function
$\mathcal{O}_{\underline{\alpha}}({\underline{\theta}})$ has additional poles.
If for instance the particles 1 and 2 form a bound state (12), there is a pole
at $\theta_{12}=\theta_{12}^{(12)}$ such that
\[
\operatorname*{Res}_{\theta_{12}=\theta_{12}^{(12)}}\mathcal{O}_{12\dots
n}(\theta_{1},\theta_{2},\dots,\theta_{n})\,=\mathcal{O}_{(12)\dots n}%
(\theta_{(12)},\dots,\theta_{n})\,\sqrt{2}\Gamma_{12}^{(12)}\,
\]
where the bound state intertwiner $\Gamma_{12}^{(12)}$ and the values of
$\theta_{1},~\theta_{2},~\theta_{(12)}$ and $\theta_{12}^{(12)}$ will be
discussed below in section \ref{sb}.

\item[(v)] Naturally, since we are dealing with relativistic quantum field
theories we finally have
\[
\mathcal{O}_{1\dots n}(\theta_{1}+\mu,\dots,\theta_{n}+\mu)=e^{s\mu
}\,\mathcal{O}_{1\dots n}(\theta_{1},\dots,\theta_{n})
\]
if the local operator transforms under Lorentz transformations as
$\mathcal{O}\to e^{s\mu}\mathcal{O}$ where $s$ is the ``spin'' of
$\mathcal{O}$.
\end{itemize}

The property (i) - (iv) may be depicted as
\[
\begin{array}
[c]{rrcl}%
\text{(i)} &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(7,3) \put(3.5,2){\oval(7,2)} \put(3.5,2){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){1}} \put(3,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(6,0){\line(0,1){1}} \put(1.4,.5){$\dots$} \put(4.4,.5){$\dots$} \end{picture}
\end{array}
& = &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(7,4) \put(3.5,3){\oval(7,2)} \put(3.5,3){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){2}} \put(3,0){\line(1,2){1}} \put(4,0){\line(-1,2){1}} \put(6,0){\line(0,1){2}} \put(1.4,1){$\dots$} \put(4.4,1){$\dots$} \end{picture}
\end{array}
\\
\text{(ii)} &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,4) \put(2.5,2){\oval(5,2)} \put(2.5,2){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){1}} \put(2,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
& = &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(7,4) \put(0,0){\line(0,1){1}} \put(6,1){\oval(2,2)[b]} \put(3.5,1){\oval(7,6)[t]} \put(3.5,2){\oval(5,2)} \put(3.5,2){\makebox(0,0){${\cal O}$}} \put(2,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
\;\sigma_{\mathcal{O}1}\\
\text{(iii)} & \displaystyle\frac{1}{2i}\,\operatorname*{Res}_{\theta
_{12}=i\pi}~~~
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(6,4) \put(3,2){\oval(6,2)} \put(3,2){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){1}} \put(2,0){\line(0,1){1}} \put(3,0){\line(0,1){1}} \put(5,0){\line(0,1){1}} \put(3.4,.5){$\dots$} \end{picture}
\end{array}
& = &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,4) \put(.5,0){\oval(1,2)[t]} \put(3,2){\oval(4,2)} \put(3,2){\makebox(0,0){${\cal O}$}} \put(2,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
~~-~~
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(6,5) \put(0,0){\line(0,1){3}} \put(3,3){\oval(6,4)[t]} \put(3,3){\oval(6,4)[br]} \put(3,0){\oval(4,2)[tl]} \put(3,3){\oval(4,2)} \put(3,3){\makebox(0,0){${\cal O}$}} \put(2,0){\line(0,1){2}} \put(4,0){\line(0,1){2}} \put(2.4,1.5){$\dots$} \end{picture}
\end{array}
\\
\text{(iv)} & \displaystyle\frac{1}{\sqrt{2}}\,\operatorname*{Res}%
_{\theta_{12}=\theta_{12}^{(12)}}~
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,3) \put(2.5,2){\oval(5,2)} \put(2.5,2){\makebox(0,0){${\cal O}$}} \put(1,0){\line(0,1){1}} \put(2,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
& = &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,4) \put(2.5,3){\oval(5,2)} \put(2.5,3){\makebox(0,0){${\cal O}$}}
\put(1.5,0){\oval(1,2)[t]}
\put(1.5,1){\line(0,1){1}} \put(4,0){\line(0,1){2}} \put(2.4,1){$\dots$} \end{picture}
\end{array}
\end{array}
\]


As was shown in \cite{BFKZ} the properties (i) -- (iii) follow from general
LSZ-assumptions and ``maximal analyticity''. The bound state form factor given
by (iv) was discussed in \cite{BFKZ} for special cases.

It has been shown \cite{Sm,Q} that functions satisfying the properties (i) --
(v) and a \cite{Q,BK1} general crossing relation represent\emph{\ local
}operators i.e. they are form factors of $x$-dependent operators which commute
(anti-commute) for space like differences of the arguments.

We will now provide a constructive and systematic way of how to solve the
properties (i) -- (v) for the co-vector valued function $f$ once the
scattering matrix is given.

We use the two particle form factor functions $F(x)$ and $\bar{F}(x)$ defined
by eqs. (\ref{2.8}) and (\ref{2.9}) which satisfy the Watson's equations
\[
F(x)=F(-x)\dot{a}(x)=F(2-x)
\]
\[
\bar{F}(x)=\bar{F}(-x)\hat{b}(x)=\bar{F}(2-x)
\]
They belong to the highest weights $w=(2,0,\dots,0)$ and $w=(1,0,\dots,-1)$, respectively.

We define the corresponding `Jost-functions'
\[
\phi(x)=\frac{1}{F(x)\bar{F}(1-x)}\propto\Gamma\left(  \frac{x}{2}\right)
\Gamma\left(  1-\frac{1}{N}-\frac{x}{2}\right)
\]
\[
\bar{\phi}(x)=\frac{1}{\bar{F}(x)F(1-x)}\propto\Gamma\left(  \frac{1}{2}%
-\frac{x}{2}\right)  \Gamma\left(  \frac{1}{2}-\frac{1}{N}+\frac{x}{2}\right)
\]
They satisfy
\[
\bar{\phi}(x)=\phi(1-x)
\]
\[
\phi(-x)=\phi(x)\dot{a}(x)
\]
\[
\bar{\phi}(-x)=\bar{\phi}(x)\hat{b}(x)
\]
and therefore
\begin{align*}
\phi(x+2)b(x+2)  &  =\bar{\phi}(-1-x)\hat{b}(-1-x)=\bar{\phi}(1+x)=\phi
(-x)=\phi(x)\dot{a}(x)\\
\bar{\phi}(x+2)  &  =\phi(-1-x)=\phi(1+x)\dot{a}(1+x)=\bar{\phi}(-x)\dot
{a}(1+x)=\bar{\phi}(x)\hat{b}(x)\dot{a}(1+x)
\end{align*}
\begin{align*}
\phi(x+2)  &  =\bar{\phi}(-1-x)=\bar{\phi}(1+x)\hat{b}(1+x)=\phi(-x)/\hat
{b}(-1-x)=\phi(x)\dot{a}(x)/b(x+2)\\
&  =\phi(x-1)\frac{\dot{S}(x-1)}{\bar{S}(-x)}=-\phi(x-1)\frac{S(x-1)}%
{u_{1}(x+1)}%
\end{align*}
\begin{align*}
\bar{\phi}(x+1)  &  =\phi(-x)=-\phi(x)S(x)=-\bar{\phi}(1-x)S(x)=-\bar{\phi
}(x-1)\bar{S}(x-1)S(x)\\
&  =-\bar{\phi}(x-1)\frac{\bar{S}(x-1)}{S(-x)}%
\end{align*}
Also the following unitarity and crossing relation have been used:
\[
S(x)=1/S(-x)\,,~~~~~\bar{S}(x)=1/\bar{S}(-x)
\]
\[
\bar{S}(x)=t_{1}(x)=u_{1}(1-x)=\frac{1-x}{1-x-2/N}\,S(1-x)
\]


%*****************************************************


\setcounter{equation}{0}

\section{Nested ``off-shell'' Bethe ansatz for $SU(N)$ and $U(N)$}

\subsection{The $SU(N)$-case}

As usual in the context of algebraic Bethe ansatz we define the monodromy
matrix
\begin{align}
T_{1\dots n,0}({\underline{\theta}},\theta_{0})  &  =\dot{S}_{10}(\theta
_{1}-\theta_{0})\,\cdots\dot{S}_{n0}(\theta_{n}-\theta_{0})\label{3.1}\\
T_{\underline{\alpha},\gamma}^{\underline{\alpha}^{\prime},\gamma^{\prime}%
}({\underline{\theta}},\theta_{0})  &  =
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(6.5,4) \put(1,2){\line(1,0){5}} \put(2,1){\line(0,1){2}} \put(5,1){\line(0,1){2}} \put(1.9,.5){$\alpha_1$} \put(1.9,3.3){$\alpha'_1$} \put(4.9,.5){$\alpha_n$} \put(4.9,3.3){$\alpha'_n$} \put(.3,1.8){$\gamma'$} \put(6.3,1.8){$\gamma$} \put(1.2,1.1){$\theta_1$} \put(4.2,1.1){$\theta_n$} \put(5.6,1.5){$z$} \put(3.2,2.6){$\dots$} \end{picture}
\end{array}
\nonumber
\end{align}
as a matrix acting in the tensor product of the ``quantum space'', a finite
dimensional vector space (whoise vectors correspond to the quantum numbers of
of $n$ particles) $V^{1\dots n}=V_{1}\otimes\cdots\otimes V_{n}$ and the
``auxiliary space'' $V_{0}$. The $V_{i}\cong\mathbf{C}^{N}$ denote the spaces
of all fundamental particles.

The Yang-Baxter algebra relations yield
\begin{equation}
T_{1\dots n,a}({\underline{\theta}},\theta_{a})\,T_{1\dots n,b}({\underline
{\theta}},\theta_{b})\,\dot{S}_{ab}(\theta_{a}-\theta_{b})=\dot{S}_{ab}%
(\theta_{a}-\theta_{b})\,T_{1\dots n,b}({\underline{\theta}},\theta
_{b})\,T_{1\dots n,a}({\underline{\theta}},\theta_{a}) \label{3.2}%
\end{equation}
\[
\unitlength4mm\begin{picture}(19,5)
\put(.3,0){$1$}
\put(4.3,0){$n$}
\put(0,2.3){$b$}
\put(0,4.3){$a$}
\put(7.5,2.3){$b$}
\put(7.5,.5){$a$}
\put(2.5,3){$\dots$}
\put(1,0){\line(0,1){5}}
\put(5,0){\line(0,1){5}}
\put(0,2){\line(1,0){8}}
\put(0,2){\oval(14,4)[rt]}
\put(8,2){\oval(2,4)[lb]}
\put(9.3,2.3){$=$}
\put(13.3,0){$1$}
\put(17.3,0){$n$}
\put(11,2){$b$}
\put(11,4.2){$a$}
\put(18.5,3.4){$b$}
\put(18.5,1.4){$a$}
\put(15.5,2){$\dots$}
\put(14,0){\line(0,1){5}}
\put(18,0){\line(0,1){5}}
\put(11,3){\line(1,0){8}}
\put(19,3){\oval(14,4)[lb]}
\put(11,3){\oval(2,4)[rt]}
\end{picture}
\]
which in turn implies the basic algebraic properties of the sub-matrices
$A,B,C,D$ with respect to the auxiliary space defined by
\begin{equation}
T_{1\dots n,0}({\underline{\theta}},z)\equiv\left(
\begin{array}
[c]{cc}%
A_{1\dots n}({\underline{\theta}},z) & B_{1\dots n,\beta}({\underline{\theta}%
},z)\\
C_{1\dots n}^{\beta}({\underline{\theta}},z) & D_{1\dots n,\beta}%
^{\beta^{\prime}}({\underline{\theta}},z)
\end{array}
\right)  ~~~~2\le\beta,\beta^{\prime}\le N \label{3.3}%
\end{equation}
where here and below $\beta$ and $\beta^{\prime}$ takes values in $\left\{
2,\dots,N\right\}  $. This means that $C$ is $N-1$-dimensional co-vector and
$D$ is a $(n-1)\times(N-1)$ matrix of operators acting in $V^{1\dots n}$.

In the context of the nested 'off-shell' Bethe ansatz one writes a co-vector
valued function in the dual space $V_{1\dots n}$ (in terms of its components)
as
\[
v_{\underline{\alpha}}(\underline{\theta})=\int\underline{dz}\,h(\underline
{\theta},{\underline{z}})\,w_{\underline{\beta}}(\underline{z})\,\Psi
_{\underline{\alpha}}^{\underline{\beta}}(\underline{\theta},{\underline{z}})
\]
where $\underline{z}=(z_{1},\dots,z_{m})$. The scalar function $h(\underline
{\theta},{\underline{z}})$ and the integrations over the $z$'s (or summations)
are to be specified. The 'next level' co-vector valued function $w_{\underline
{\beta}}(\underline{z})$ takes values a tensor product of smaller spaces of
dimension $N-1$. This procedure is continued until the spaces are
one-dimensional. The $\Psi_{\underline{\alpha}}^{\underline{\beta}}%
(\underline{\theta},{\underline{z}})$ are the basic Bethe ansatz co-vectors
(counted by $\underline{\beta}=(\beta_{1},\dots,\beta_{m})$) defined by
\begin{equation}%
\begin{array}
[c]{rcl}%
\Psi_{1\dots n}^{\underline{\beta}}({\underline{\theta}},\underline{z}) & = &
\Omega_{1\dots n}C_{1\dots n}^{\beta_{m}}({\underline{\theta},}z_{m})\cdots
C_{1\dots n}^{\beta_{1}}({\underline{\theta},}z_{1})\\
\Psi_{\underline{\alpha}}^{\underline{\beta}}(\underline{\theta}%
,{\underline{z}}) & = &
\begin{array}
[c]{c}%
\unitlength5mm\begin{picture}(10,7) \put(9,5){\oval(14,2)[lb]} \put(9,5){\oval(18,6)[lb]} \put(4,1){\line(0,1){4}} \put(8,1){\line(0,1){4}} \put(3.8,.2){$\alpha_1$} \put(7.8,.2){$\alpha_n$} \put(-.2,5.4){$\beta_1$} \put(1.8,5.4){$\beta_m$} \put(3.8,5.4){$1$} \put(7.8,5.4){$1$} \put(9.2,1.8){$1$} \put(9.2,3.8){$1$} \put(3.1,1){$\theta_1$} \put(7.1,1){$\theta_n$} \put(.8,2.5){$z_1$} \put(2,3.5){$z_m$} \put(5.5,3){$\dots$} \put(3.3,2.7){$\vdots$} \put(8.3,2.7){$\vdots$} \end{picture}
\end{array}
~~~
\begin{array}
[c]{l}%
2\le\beta_{i}\le N\\
1\le\alpha_{i}\le N~.
\end{array}
\end{array}
\label{3.4}%
\end{equation}


The co-vector $\Omega_{1\dots n}$ is the ``pseudo-vacuum'' consisting only
highest weight states
\[
\Omega_{1\dots n}=e^{1}\otimes\cdots\otimes e^{1}
\]
where the unit co-vectors $e^{\alpha}~(1\le\alpha\le N)$ belong to the
fundamental particles which transform as the vector representation of $SU(N)$.
The pseudo-vacuum satisfies
\begin{equation}%
\begin{array}
[c]{rcl}%
\Omega_{1\dots n}\,B_{1\dots n}^{\beta}({\underline{\theta}},z) & = & 0\\
\Omega_{1\dots n}\,A_{1\dots n}({\underline{\theta}},z) & = &
\displaystyle\prod_{i=1}^{n}\dot{a}(\theta_{i}-z)\Omega_{1\dots n}\\
\Omega_{1\dots n}\,D_{1\dots n,\beta}^{\beta^{\prime}}({\underline{\theta}%
},z) & = & \delta_{\beta}^{\beta^{\prime}}\displaystyle\prod_{i=1}^{n}\dot
{b}(\theta_{i}-z)\Omega_{1\dots n}\,.
\end{array}
\label{3.5}%
\end{equation}
Here the eigenvalues of the matrices $A$ and $D$, i.e. $\dot{a}(\theta
)=-a(\theta)$ and $\dot{b}(\theta)=-b(\theta)$ are given by the amplitudes of
the scattering matrix

\subsection{The $U(N)$-case}

We use the short notation $I=1\dots n_{+}\bar{1}\dots\bar{n}_{-}$ where
$n_{+}$ and $n_{-}$ ist the number of particles and anti-particles,
respectively. Similar as above we define the monodromy matrix
\begin{align}
T_{I,0}({\underline{\theta}},{\underline{\bar{\theta}}},\theta_{0})  &
=\dot{S}_{10}(\theta_{10})\,\cdots\dot{S}_{n_{+}0}(\theta_{n_{+}0})\dot
{S}_{\bar{1}0}(\bar{\theta}_{10})\cdots\dot{S}_{\bar{n}_{-}0}(\bar{\theta
}_{n_{-}0})\label{3.6}\\
&  =
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(17,4.5) \put(0,2){\line(1,0){17}} \put(2,0){\line(0,1){4}} \put(2,3){\vector(0,1){0}} \put(7,0){\line(0,1){4}} \put(7,3){\vector(0,1){0}} \put(1,0){$1$} \put(5.2,0){$ n_+$} \put(10,0){\line(0,1){4}} \put(10,2.6){\vector(0,-1){0}} \put(15,0){\line(0,1){4}} \put(15,2.6){\vector(0,-1){0}} \put(9,0){$\bar1$} \put(13.2,0){$\bar n_-$} \put(16.2,.7){$ 0$} \put(3.6,3){$\dots$} \put(11.6,3){$\dots$}\put(16.2,2){\vector(-1,0){0}} \end{picture}
\end{array}
\nonumber
\end{align}
as a matrix acting in the tensor product of the ``quantum space'', a space of
$n_{+}$particles and $n_{-}$ anti-particles (with respect to their quantum
numbers) $V^{I}=V_{1}\otimes\cdots\otimes V_{n_{+}}\otimes V_{\bar{1}}%
\otimes\cdots\otimes V_{\bar{n}_{-}}$ and the ``auxiliary space'' $V_{0} $.
The $V_{i}$ and $V_{\bar{\imath}}\cong\mathbf{C}^{N}$ denote the spaces of all
fundamental particles and anti-particles, respectively. Again we have the
sub-matrices $A,B,C,D$ defined by
\begin{equation}
T_{I,0}({\underline{\theta}\underline{,\bar{\theta}}},,z)\equiv\left(
\begin{array}
[c]{cc}%
A_{I}({\underline{\theta}},\bar{\theta},z) & B_{I,\beta}({\underline{\theta}%
},\bar{\theta},z)\\
C_{I}^{\beta}({\underline{\theta}},\bar{\theta},z) & D_{I,\beta}%
^{\beta^{\prime}}({\underline{\theta}},\bar{\theta},z)
\end{array}
\,,~~~~2\le\beta,\beta^{\prime}\le N\,.\right)  \label{3.7}%
\end{equation}


The nested 'off-shell' Bethe ansatz now writes as
\[
v_{\underline{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta
},{\underline{\bar{\theta}}})=\int\underline{dz}\,h(\underline{\theta
},{\underline{\bar{\theta}}},{\underline{z}})\,w_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},{\underline{\bar{\theta}}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},{\underline{\bar{\theta}%
},\underline{z}})\,.
\]
The basic Bethe ansatz co-vectors are for the case of $U(N)$
\begin{equation}%
\begin{array}
[c]{rcl}%
\Psi_{I}^{\underline{\bar{\beta}}}(\underline{\theta},{\underline{\bar{\theta
}},\underline{z}}) & = & \Omega_{I}^{\underline{\bar{\beta}}}C_{I}^{\beta_{m}%
}({\underline{\theta},\underline{\bar{\theta}}},z_{m})\cdots C_{I}^{\beta_{1}%
}({\underline{\theta},\underline{\bar{\theta}}},z_{1})\\
\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},{\underline{\bar{\theta}%
},\underline{z}}) & = &
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(16,7) \put(15,5){\oval(26,2)[lb]} \put(15,5){\oval(30,6)[lb]} \put(0,5){\vector(0,1){0}} \put(2,5){\vector(0,1){0}} \put(4,1){\vector(0,1){4}} \put(8,1){\vector(0,1){4}} \put(10,5){\vector(0,-1){4}} \put(14,5){\vector(0,-1){4}} \put(3.8,.2){$\alpha_1$} \put(7.8,.2){$\alpha_{n_+}$} \put(9.8,.2){$\bar\alpha_1$} \put(13.8,.2){$\bar\alpha_{n_-}$} \put(-.2,5.4){$\beta_1$} \put(1.8,5.4){$\beta_m$} \put(9.8,5.4){$\bar\beta_1$} \put(13.8,5.4){$\bar\beta_{n_-}$} \put(3.8,5.4){$1$} \put(7.8,5.4){$1$} \put(15.2,1.8){$1$} \put(15.2,3.8){$1$} \put(3,2.5){$\theta_1$} \put(6.5,2.5){$\theta_{n_+}$} \put(9,2.5){$\bar\theta_1$} \put(12.5,2.5){$\bar\theta_{n_-}$} \put(.8,2.5){$z_1$} \put(1.7,3.6){$z_m$} \put(5.4,4.5){$\dots$} \put(11.4,4.5){$\dots$} \put(14.3,2.6){$\vdots$} \end{picture}
\end{array}
~~
\begin{array}
[c]{l}%
2\le\beta_{i}\le N\\
2\le\bar{\beta}_{i}\le N\\
1\le\alpha_{i}\le N~.
\end{array}
\end{array}
\label{3.8}%
\end{equation}
Here there exist a space of ``pseudo-vacua'' with basic vectors $\Omega
_{I}^{\underline{\bar{\beta}}}$ ($\underline{\bar{\beta}}=(\bar{\beta}%
_{1},\dots,\bar{\beta}_{n_{-}}),\;2\leq\bar{\beta}_{i}\leq N$)
\[
\Omega_{1\dots n_{+}\bar{1}\dots\bar{n}_{-}}^{\underline{\bar{\beta}}}%
=e^{1}\otimes\cdots\otimes e^{1}\otimes e^{\bar{\beta}_{1}}\otimes
\cdots\otimes e^{\bar{\beta}_{\bar{n}_{-}}}
\]
fulfilling
\begin{equation}%
\begin{array}
[c]{rcl}%
\Omega_{I}^{\underline{\bar{\beta}}}\,B_{I}^{\beta}({\underline{\theta}},z) &
= & 0\\
\Omega_{I}^{\underline{\bar{\beta}}}\,A_{I}({\underline{\theta}},z) & = &
\displaystyle\prod_{i=1}^{n_{+}}\dot{a}(\theta_{i}-z)\prod_{i=1}^{n_{-}}%
\hat{b}(\bar{\theta}_{i}-z)\Omega_{I}^{\underline{\bar{\beta}}}\\
\Omega_{I}^{\underline{\bar{\beta}}}\,D_{I,\beta}^{\beta^{\prime}}%
({\underline{\theta}},z) & = & \displaystyle\prod_{i=1}^{n_{+}}\dot{b}%
(\theta_{i}-z)\Omega_{I}^{\underline{\bar{\beta}}}\left(  \dot{S}_{\bar{1}%
0}(\bar{\theta}_{1}-z)\cdots\dot{S}_{\bar{n}_{-}0}(\bar{\theta}_{n_{-}%
}-z)\right)  _{\beta}^{\beta^{\prime}}\,.
\end{array}
\label{3.9}%
\end{equation}
The amplitudes of the scattering matrices are given by eqs. (\ref{1.1}) and
(\ref{1.2}).

The general structure of a nested $U(N)$ Bethe ansatz depicted in Fig.
\ref{f3.1}, where every line stands for bunch of lines of the same type.
\begin{figure}[tbh]%
\[
\unitlength8mm\begin{picture}(7,7)
\put(6,6){\oval(12,2)[lb]}
\put(6,4.5){\oval(8,3)[lb]}
\put(6,3.4){\oval(6,2.8)[lb]}
\put(0,6){\vector(0,1){0}}
\put(2,4.5){\vector(0,1){0}}
\put(3,3.4){\vector(0,1){0}}
\put(5.5,2){\vector(-1,0){0}}
\put(5.5,3){\vector(-1,0){0}}
\put(5.5,5){\vector(-1,0){0}}
\put(4,1){\vector(0,1){1.4}}\put(4,1.4){\vector(0,1){0}}
\put(5,6){\vector(0,-1){5}}\put(5,5.6){\vector(0,-1){0}}
\put(.8,5.5){\dots}\put(5.5,3.8){\vdots}
\put(-.2,6.2){$N$}
%\put(1.9,6.2){$3$}
\put(2.9,3.5){$2$}
\put(3.9,2.5){$1$}
\put(4.8,6.2){$\bar N$}
\put(6.1,4.8){$N-1$}
\put(6.1,2.8){$2$}
\put(6.1,1.8){$1$}
\put(3.9,.5){$\underline\alpha$}
\put(4.9,.5){$\underline{\bar\alpha}$}
\put(3.5,1.2){$\underline\theta$}
\put(4.5,1.2){$\underline{\bar\theta}$}
\put(2.2,2){$\underline z^{(1)}$}
\put(1.2,3){$\underline z^{(2)}$}
\put(0,4.3){$\underline z^{(N-1)}$}
\end{picture}
\]
\caption{\textit{The general structure of a nested $U(N)$ Bethe ansatz. Every
line stands for bunch of lines of the same type. For the corresponding
}$SU(N)$\textit{-case the anti-particle line is missing.}}%
\label{f3.1}%
\end{figure}

\setcounter{equation}{0}

\section{The general form factor formula}

\subsection{Solutions of (i), (ii) and (iii)}

The aim of this subsection is to construct co-vector valued functions which
satisfy the properties of form factors (i), (ii) and (iii). For simplicity we
consider first states with only particles which correspond to the
$SU(N)$-Bethe ansatz and then general case of particles and anti-particles
which correspond to the $U(N)$-Bethe ansatz .

\subsubsection{Pure particle states}

For pure particle states only the properties $(i)$ and $(ii)$ are meaningful,
because there is no particle anti-particle annihilation.. By means of an
ansatz in terms of an integral representation we transform the properties
$(i),$ and $(ii)$ of a co-vector valued function $f_{\underline{\alpha}%
}(\underline{\theta})$ to properties $(i^{\prime}),\;$and $(ii^{\prime})$ of a
scalar function $h(\underline{\theta},{\underline{z}})$. In addition there
will be a next level Bethe ansatz co-vector valued function $g_{\underline
{\beta}}(\underline{z})$ (where the indices $\beta_{i}$ take only the values
$2,\dots,N$) which has to satisfy the properties $(i),$ and $(ii)$. Iterating
this procedure we solve the problem for $f_{\underline{\alpha}}(\underline
{\theta})$ where the result appears as a `nested off-shell' Bethe ansatz.

\paragraph{Ansatz:}

\label{an}Let a co-vector valued function be given by
\begin{equation}
f_{\underline{\alpha}}(\underline{\theta})=\int_{\mathcal{C}_{\underline
{\theta}}}d\underline{z}\,h(\underline{\theta},{\underline{z}})\,g_{\underline
{\beta}}(\underline{z})\,\Psi_{\underline{\alpha}}^{\underline{\beta}%
}(\underline{\theta},{\underline{z}})\,,\;(1\leq\alpha_{i}\leq N\,,\;2\leq
\beta_{i}\leq N) \label{4.1}%
\end{equation}
where we use the short notation $\int_{\mathcal{C}_{\underline{\theta}}%
}d\underline{z}=\int_{\mathcal{C}_{\underline{\theta}}}dz_{1}\cdots
\int_{\mathcal{C}_{\underline{\theta}}}dz_{m}$. The scalar function
$h(\underline{\theta},{\underline{z}})$ is meromorphic with poles where
$\operatorname{Re}(\theta_{i}-z_{j})=0\,,\;(i=1,\ldots,n\,,\;j=1,\ldots,m)$,
symmetric with respect to the $z$'s and
\[
\Psi_{\underline{\alpha}}^{\underline{\beta}}(\underline{\theta}%
,{\underline{z}})=\left[  \Omega C^{\beta_{m}}({\underline{\theta}}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta}},z_{1})\right]  _{\underline
{\alpha}}
\]
is an 'off-shell' Bethe ansatz state (c.f. eq. (\ref{3.4})). The co-vector
valued function
\[
g_{\underline{\beta}}(\underline{z})=\,g_{\beta_{1}\dots\beta_{m}}(z_{1}%
,\dots,z_{m})\,,\;(2\leq\beta_{i}\leq N)
\]
$\,$is a meromorphic function with poles where $\operatorname{Re}(z_{i}%
-z_{j})=const\,,\;(i,j=1,\ldots,m)$ (see below). The integration contour
$\mathcal{C}_{\underline{\theta}}$ depends on the $\underline{\theta}$ and
goes from $-\infty$ to $\infty$ avoiding all poles (detailes will be specified later).

\paragraph{Assumptions:\label{as}}

\begin{itemize}
\item[$(i^{\prime})$] The scalar function $h(\underline{\theta},{\underline
{z}})$ satisfies Watson's equations
\[
h(\dots,\theta_{i},\theta_{j},\dots,{\underline{z}})=h(\dots,\theta_{j}%
,\theta_{i},\dots,{\underline{z}})\dot{a}(\theta_{ij})\,.
\]


\item[$(ii^{\prime})$] The following `shift relations' of $h(\underline
{\theta},{\underline{z}})$ hold
\[
\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot
{a}(\theta_{1}-z_{j})}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,h(\underline
{\theta},{\underline{z}})=\sigma\,h(\underline{\theta}^{\prime},{\underline
{z}})
\]
where $\underline{\theta}^{\prime}=(\theta_{1}^{\prime}=\theta_{1}+2\pi
i,\dots,\theta_{n})$ (and $\sigma$-factor is a sign to be discussed later)
and
\[
\prod_{i=1}^{n}\dot{b}(\theta_{i}-z_{1})\prod_{j=1}^{m}\frac{1}{\dot{b}%
(z_{j}-z_{1})}\,h(\underline{\theta},{\underline{z}})=\prod_{i=1}^{n}\dot
{a}(\theta_{i}-z_{1}^{\prime})\prod_{j=1}^{m}\frac{\dot{a}(z_{1}^{\prime
}-z_{j})}{\dot{b}(z_{1}^{\prime}-z_{j})}\,h(\underline{\theta},{\underline{z}%
}^{\prime})
\]
where ${\underline{z}}^{\prime}=(z_{1}^{\prime}=z_{1}+2\pi i,\dots,z_{m})$.
Because of the symmetry with respect to the $z$'s and $(i^{\prime})$ similar
shift relations hold for $z_{1}\rightarrow z_{j}$ and $\theta_{1}%
\rightarrow\theta_{i}$.
\end{itemize}

Further we need more technical assumptions on the contour $\mathcal{C}%
_{\underline{\theta}}$ and on the analytic behavior of $h(\underline{\theta
},{\underline{z}})$ and $g_{\underline{\beta}}(\underline{z})$ which can be
found in appendix \ref{a1}.

\begin{theorem}
\label{t1}The co-vector valued function $f_{\underline{\alpha}}(\underline
{\theta})$ given by the ansatz (\ref{4.1}) satisfies $(i)$ and $(ii)$ on page
\pageref{pf} if $g_{\underline{\beta}}(\underline{z})\,$ satisfies $(i) $and
$(ii)$ and the scalar function $h(\underline{\theta},{\underline{z}})$
satisfies $(i^{\prime})$ and $(ii^{\prime})$. In addition some technical
assumptions on the integration contour $\mathcal{C}_{\underline{\theta}}$ and
analytic behavior must hold.
\end{theorem}

The technical assumptions are listed in the proof wich is deligated to
appendix \ref{a1}. It is easy to find the solutions for the scalar function
$h(\underline{\theta},{\underline{z}})$ wich satisfy the assumptions
$(i^{\prime})\,$ and $(ii^{\prime})$.

\begin{theorem}
\label{t2}The scalar function $h(\underline{\theta},{\underline{z}})$ satisfy
properties $(i^{\prime})$ and $(ii^{\prime})$ if it is of the form
\begin{equation}
h(\underline{\theta},{\underline{z}})=\prod_{1\le i<j\le n}F(\theta_{ij}%
)\prod_{i=1}^{n}\prod_{j=1}^{m}\phi(\theta_{i}-z_{j})\prod_{1\le i<j\le m}%
\tau(z_{i}-z_{j})\,\,p(\underline{\theta},{\underline{z}}) \label{4.2}%
\end{equation}
where functions $\tau(z)$ and $\phi(z)$ are defined as
\begin{equation}
\tau(z)=\frac{1}{F(z)\,\phi(z)\,\phi(-z)}=F(-z)\bar{F}^{2}(i\pi-z)\;,~~\phi
(z)=\frac{1}{F(z)\,\bar{F}(i\pi-z)} \label{4.3}%
\end{equation}
and the functions $F(\theta)$ and $\bar{F}(\theta)$ satisfy the Watson's
equations
\begin{align}
F(\theta)  &  =\dot{a}(\theta)F(-\theta)\,=F(2\pi i-\theta)\label{4.4}\\
\bar{F}(\theta)  &  =-\varepsilon\hat{b}(\theta)\bar{F}(-\theta)\,=\bar
{F}(2\pi i-\theta)\, \label{4.5}%
\end{align}
and if the function $p(\underline{\theta},{\underline{z}})$ is an entire
function of the $z$'s with
\begin{equation}
p(\underline{\theta},{\underline{z}})=\varepsilon^{-m}\sigma\,p(\underline
{\theta}^{\prime},{\underline{z}})=\varepsilon^{n}p(\underline{\theta
},{\underline{z}}^{\prime}) \label{4.6}%
\end{equation}
(where $\varepsilon$ is a sign factor to be discussed later).
\end{theorem}

%

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof
%EndExpansion
Watson's equations for $F(\theta)$ imply
\[
h(\dots,\theta_{i},\theta_{j},\dots,{\underline{z}})=h(\dots,\theta_{j}%
,\theta_{i},\dots,{\underline{z}})\dot{a}(\theta_{ij})
\]
which is $(i^{\prime})$. Watson's equations for $F(\theta)$ and $\bar
{F}(\theta)$ imply a shift relation for $\phi(z)$%
\begin{align}
\frac{\phi(z+2\pi i)}{\phi(z)}  &  =\frac{F(z)\bar{F}(i\pi-z)}{F(z+2\pi
i)\bar{F}(-i\pi-z)}=\frac{F(-z)\dot{a}(z)\bar{F}(i\pi-z)}{-F(-z)\bar{F}%
(i\pi+z)\varepsilon\hat{b}(-i\pi-z)}\label{4.7}\\
&  =\frac{\dot{a}(z)}{-\varepsilon b(z+2\pi i)}=\frac{\dot{a}(z)}%
{\varepsilon\dot{b}(z+2\pi i)}\nonumber
\end{align}
therefore together with eqs. (\ref{4.4}) and (\ref{4.6})
\begin{align*}
\frac{h(\underline{\theta}^{\prime},{\underline{z}})}{h(\underline{\theta
},{\underline{z}})}  &  =\prod_{i=2}^{n}\frac{F(\theta_{1i}+2\pi i)}%
{F(\theta_{1i})}\prod_{j=1}^{m}\frac{\phi(\theta_{1}^{\prime}-z_{j})}%
{\phi(\theta_{1}-z_{j})}\frac{p(\underline{\theta}^{\prime},{\underline{z}}%
)}{p(\underline{\theta},{\underline{z}})}\\
&  =\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot
{a}(\theta_{1}-z_{j})}{\varepsilon\dot{b}(\theta_{1}^{\prime}-z_{j})}%
\frac{\varepsilon^{m}}{\sigma}%
\end{align*}
holds which yields one equation of $(ii^{\prime})$. Similar
\begin{align*}
\frac{\tau(2\pi i-z)}{\tau(-z)}  &  =\frac{F(-z)\phi(z)\phi(-z)}{F(2\pi
i-z)\phi(2\pi i-z)\phi(z-2\pi i)}=\frac{\dot{a}(-z)F(z)}{F(z)}\frac{\dot
{b}(-z+2\pi i)}{\dot{a}(-z)}\frac{\dot{a}(z-2\pi i)}{\dot{b}(z)}\\
&  =\frac{\dot{b}(2\pi i-z)}{\dot{a}(2\pi i-z)}\frac{1}{\dot{b}(z)}%
\end{align*}
which yields together with the shift relation for $\phi(z)$ and (\ref{4.6})
\begin{align*}
\frac{h(\underline{\theta},{\underline{z}}^{\prime})}{h(\underline{\theta
},{\underline{z}})}  &  =\prod_{i=1}^{n}\frac{\phi(\theta_{i}-z_{1}^{\prime}%
)}{\phi(\theta_{i}-z_{1})}\prod_{j=1}^{m}\frac{\tau(z_{1}^{\prime}-z_{j}%
)}{\tau(z_{1}-z_{j})}\frac{p(\underline{\theta},{\underline{z}}^{\prime}%
)}{p(\underline{\theta},{\underline{z}})}\\
&  =\prod_{i=1}^{n}\frac{\varepsilon\dot{b}(\theta_{i}-z_{1})}{\dot{a}%
(\theta_{i}-z_{1}^{\prime})}\prod_{j=1}^{m}\frac{\dot{b}(z_{1}^{\prime}%
-z_{j})}{\dot{a}(z_{1}^{\prime}-z_{j})}\frac{1}{\dot{b}(z_{j}-z_{1})}\frac
{1}{\varepsilon^{n}}%
\end{align*}
which yields the other equation of $(ii^{\prime})$.%

%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof
%EndExpansion


The definitions (\ref{4.3}) yield the explicit formula for the
`Jost-function'
\begin{equation}
\phi(z)=\frac{1}{F(z)\,\bar{F}(i\pi-z)}=const\;\Gamma\left(  \frac{z}{2\pi
i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{z}{2\pi i}\right)  \label{4.8}%
\end{equation}
with $\phi(-z)/\phi(z)=\dot{a}(z)$. The $\tau$-funtion turns out to be
\begin{equation}
\tau(z)=\frac{1}{F(z)\,\phi(z)\,\phi(-z)}=const^{\prime}\frac{1}{F(z)}%
\frac{z\sinh\frac{z}{2}}{\Gamma\left(  1-\frac{1}{N}-\frac{z}{2\pi i}\right)
\Gamma\left(  1-\frac{1}{N}+\frac{z}{2\pi i}\right)  } \label{4.9}%
\end{equation}


+++++++++++++++++++++++++++++++++++++++++++++++++++++

\subsubsection{Particle anti-particle states}

We constuct co-vector valued functions $f_{I}(\underline{\theta}%
,\underline{\bar{\theta}})\,,\;(I=1\dots n,\bar{1}\dots\bar{n})$ with
components $f_{\underline{\alpha}\,\underline{\bar{\alpha}}}(\underline
{\theta},\underline{\bar{\theta}})\,,\;(\underline{\alpha}=(\alpha_{1}%
,\cdots,\alpha_{n})\,,\;\underline{\bar{\alpha}}=(\bar{\alpha}_{1},\cdots
,\bar{\alpha}_{\bar{n}})\,,\;\underline{\theta}=(\theta_{1},\cdots,\theta
_{n})\,,\;\underline{\bar{\theta}}=(\bar{\theta}_{1},\cdots,\bar{\theta}%
_{\bar{n}}))$ such that they satisfy the properties $(i),\;(ii)$ and $(iii)$
on page \pageref{pf}. Again by means of an ansatz in terms of an integral
representation we transform the properties $(i),\;(ii)$ and $(iii)$ of
$f_{\underline{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta
},\underline{\bar{\theta}})$ to properties $(i^{\prime\prime}),\;(ii^{\prime
\prime})$ and $(iii^{\prime\prime})$ of a scalar function $h(\underline
{\theta},\underline{\bar{\theta}},{\underline{z}})$. In addition there will be
a next level Bethe ansatz co-vector valued function $g_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}})$ (where
the indices $\beta_{i}$ and $\bar{\beta}_{i}$ take only the values $2,\dots
,N$) which has to satisfy properties $(i),\;(ii)$ and $(iii)^{(1)}$. Iterating
this procedure we solve the problem for $f_{\underline{\alpha}\,\underline
{\bar{\alpha}}}(\underline{\theta},\underline{\bar{\theta}})$ where the result
appears as a `nested off-shell' Bethe ansatz.

\paragraph{Ansatz:}

\label{anu}Let a co-vector valued function be given by
\begin{equation}
f_{\underline{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta}%
,\underline{\bar{\theta}})=\int_{\mathcal{C}_{{\underline{\theta}%
\,\underline{\bar{\theta}}}^{\prime}}}d\underline{z}\,h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})\,g_{\underline{\beta}%
\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})\,,\;(1\leq\alpha_{i},\bar{\alpha}_{i}\leq N\,,\;2\leq
\beta_{i},\bar{\beta}_{i}\leq N)\,. \label{4.10}%
\end{equation}
The scalar function $h(\underline{\theta},\underline{\bar{\theta}}%
,{\underline{z}})$ is symmetric with respect to the $z$'s and meromorphic with
poles at $\operatorname{Re}(\theta_{i}-z_{j})=0\,\,,\;(i=1,\ldots
,n\,,\;j=1,\ldots,m)$.
\[
\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})=\left[  \Omega C^{\beta_{m}}({\underline{\theta}%
},\underline{\bar{\theta}},z_{m})\cdots C^{\beta_{1}}({\underline{\theta}%
},\underline{\bar{\theta}},z_{1})\right]  _{\underline{\alpha}\,\underline
{\bar{\alpha}}}^{\,\underline{\bar{\beta}}}
\]
is an $U(N)$-'off-shell' Bethe ansatz state (c.f. eq. (\ref{3.8})). The
co-vector valued function
\[
g_{\underline{\beta}\,\underline{\bar{\beta}}}(\underline{z},\underline
{\bar{\theta}})=\,g_{\beta_{1}\dots\beta_{m}\bar{\beta}_{1}\ldots\bar{\beta
}_{\bar{n}}}(z_{1},\dots,z_{m},\bar{\theta}_{1},\dots,\bar{\theta}_{\bar{n}%
})\,,\;(2\leq\beta_{i},\bar{\beta}_{i}\leq N)
\]
$\,$is a meromorphic function with poles at $\operatorname{Re}(z_{i}%
-z_{j})=const\,,\;(i,j=1,\ldots,m)$ and $\operatorname{Re}(\bar{\theta}%
_{i}-z_{j})=const\,,\;(i=1,\ldots,\bar{n},\;j=1,\ldots,m)$ (see below). The
integration contour $\mathcal{C}_{\underline{\theta}}$ depends on the
$\underline{\theta}\,,\underline{\bar{\theta}}$ and goes from $-\infty$ to
$\infty$ avoiding all poles (detailes will be specified later).

\paragraph{Assumptions:\label{asu}}

\begin{itemize}
\item[$(i^{\prime\prime})$] The scalar function $h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})$ satisfies Watson's equations
$(i^{\prime})$ and in addition
\begin{align*}
h(\underline{\theta},\dots,\bar{\theta}_{i},\bar{\theta}_{j},\dots
,{\underline{z}})  &  =h(\underline{\theta},\dots,\bar{\theta}_{j},\bar
{\theta}_{i},\dots,{\underline{z}})\dot{a}(\theta_{ij})\,\\
h(\theta\dots,\theta_{i},\bar{\theta}_{j},\dots,{\underline{z}})  &
=h(\dots,\bar{\theta}_{j},\theta_{i},\dots,{\underline{z}})\hat{b}(\theta
_{ij}).
\end{align*}


\item[$(ii^{\prime\prime})$] The `shift relations' $(ii^{\prime})$ of
$h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})$ and in
addition the following ones hold
\[
\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{i=1}^{\bar{n}}\hat{b}%
(\bar{\theta}_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot{a}(\theta_{1}-z_{j}%
)}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,h(\underline{\theta},\underline
{\bar{\theta}},{\underline{z}})=\sigma\,h(\underline{\theta}^{\prime
},\underline{\bar{\theta}},{\underline{z}})
\]%
\[
\prod_{i=1}^{n}\dot{b}(\theta_{i}-\bar{\theta}_{\bar{n}}^{\prime\prime}%
)\prod_{j=1}^{m}\frac{1}{\dot{b}(z_{j}-\bar{\theta}_{\bar{n}}^{\prime
\prime\prime})}\,h(\underline{\theta},\underline{\bar{\theta}}^{\prime
},{\underline{z}})=\sigma\,h(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]
where $\underline{\theta}^{\prime}=(\theta_{1}^{\prime}=\theta_{1}+2\pi
i,\dots,\theta_{n})\,,$ $\underline{\bar{\theta}}^{\prime}=(\bar{\theta}%
_{1},\cdots,\bar{\theta}_{\bar{n}}+2\pi i)\,,\;\bar{\theta}_{\bar{n}}%
^{\prime\prime}=\bar{\theta}_{\bar{n}}+i\pi$ (and $\sigma$ is a sign factor as
above) and
\[
\prod_{i=1}^{n}\dot{b}(\theta_{i}-z_{1})\prod_{j=1}^{m}\frac{1}{\dot{b}%
(z_{j}-z_{1})}\,h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})=\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1}^{\prime})\prod_{i=1}^{\bar{n}}%
\hat{b}(\bar{\theta}_{i}-z_{1}^{\prime})\prod_{j=1}^{m}\frac{\dot{a}%
(z_{1}^{\prime}-z_{j})}{\dot{b}(z_{1}^{\prime}-z_{j})}\,h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}}^{\prime})
\]
where ${\underline{z}}^{\prime}=(z_{1}^{\prime}=z_{1}+2\pi i,\dots,z_{m})$.
Because of the symmetry with respect to the $z$'s and $(i^{\prime\prime})$
similar shift relations hold for $z_{1}\rightarrow z_{j}$ and $\theta
_{1}\rightarrow\theta_{i}$.

\item[$(iii^{\prime\prime})$] For $\theta_{1}-\bar{\theta}_{\bar{n}%
}\rightarrow i\pi$ there are `pinchings' of the integration contours at
\[
\begin{array}
[c]{lll}%
(1) &  & z_{j}=z^{(1)}=\theta_{1}\rightarrow\bar{\theta}_{\bar{n}}+i\pi\\
(2) &  & z_{j}=z^{(2)}=\theta_{1}-2\pi i/N\rightarrow\bar{\theta}_{\bar{n}%
}+i\pi-2\pi i/N\\
(3) &  & z_{j}=z^{(3)}=\theta_{1}-2\pi i\rightarrow\bar{\theta}_{\bar{n}}-i\pi
\end{array}
\]
(for $j=1,\ldots,m$) such that $f_{\underline{\alpha}\,\underline{\bar{\alpha
}}}(\underline{\theta},\underline{\bar{\theta}})$ has a pole at $\theta
_{1}-\bar{\theta}_{\bar{n}}=i\pi$ with
\[
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{\underline
{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta},\underline{\bar{\theta
}})=-2\pi i\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}%
\sum_{k=1}^{3}(-1)^{k}m\operatorname*{Res}_{z_{1}=z^{(k)}}h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})\,g_{\underline{\beta}%
\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]

\end{itemize}

Further we need more technical assumptions on the contour $\mathcal{C}%
_{\underline{\theta}\,\underline{\bar{\theta}}}$ and on the analytic behavior
of $h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})$ and
$g_{\underline{\beta}\,\underline{\bar{\beta}}}(\underline{z},\underline
{\bar{\theta}})$ which can be found in appendix \ref{a2}.

\begin{theorem}
\label{t3}The co-vector valued function $f_{\underline{\alpha}\,\underline
{\bar{\alpha}}}(\underline{\theta},\underline{\bar{\theta}})$ given by the
ansatz (\ref{4.1}) satisfies $(i),\;(ii)$ and $(iii)$ on page \pageref{pf} if
the scalar function satisfies $(i^{\prime\prime}),\;(ii^{\prime\prime})$ and
$(iii^{\prime\prime})$ and if $g_{\underline{\beta}\,\underline{\bar{\beta}}%
}(\underline{z},\underline{\bar{\theta}})\,$ satisfies $(i),\;(ii)$
$(iii)^{(1)}$ where the last property means
\[
(iii)^{(1)}\quad\quad\left\{
\begin{array}
[c]{l}%
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi}g_{\underline
{\beta}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}%
})=-2i\,\mathbf{C}_{\beta_{1}\bar{\beta}_{\bar{n}}}\,g_{\underline
{\tilde{\beta}}\,\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\beta}}}} }%
%BeginExpansion
{\tilde{\bar{\beta}}}
%EndExpansion
}}(\underline{\tilde{z}},\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\theta}}}} }%
%BeginExpansion
{\tilde{\bar{\theta}}}
%EndExpansion
})\\
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi}g_{I}%
(\underline{z},\underline{\bar{\theta}})=-2i\,\mathbf{C}_{1\bar{n}}%
\,g_{\tilde{I}}(\underline{\tilde{z}},\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\theta}}}} }%
%BeginExpansion
{\tilde{\bar{\theta}}}
%EndExpansion
})\\
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i?}g_{I}%
(\underline{z},\underline{\bar{\theta}})=2i\,\mathbf{C}_{1\bar{n}}%
\,g_{\tilde{I}}(\underline{\tilde{z}},\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\theta}}}} }%
%BeginExpansion
{\tilde{\bar{\theta}}}
%EndExpansion
})S_{12}\dots S_{1\bar{n}-1}%
\end{array}
\right.  \;.
\]
\[
(iii)^{(1)}\quad\left\{  \;
\begin{array}
[c]{l}%
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi}g_{1\dots
m\bar{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}})=-2i\,\mathbf{C}%
_{1\bar{n}}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2},\dots,\bar{\theta}%
_{\bar{n}-1})\\
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i?}g_{1\dots m\bar
{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}})=2i\,\mathbf{C}_{1\bar{n}%
}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2},\dots,\bar{\theta}_{\bar{n}%
-1})S_{12}\dots S_{1\bar{n}-1}\;.
\end{array}
\right.
\]
Here $\tilde{I}=(2\dots n,\bar{1}\dots\bar{n}-1)\,,\;\,\underline{\tilde
{\beta}}=(\beta_{2},\dots,\beta_{n})\,,\;$ $\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\beta}}}} }%
%BeginExpansion
{\tilde{\bar{\beta}}}
%EndExpansion
}=(\bar{\beta}_{1},\dots,\bar{\beta}_{\bar{n}-1})\,,\;\underline{\tilde{z}%
}=(z_{2},\dots,z_{n})\,,\;\underline{%
%TCIMACRO{\TeXButton{}{{\tilde{\bar{\theta}}}} }%
%BeginExpansion
{\tilde{\bar{\theta}}}
%EndExpansion
}=(\bar{\theta}_{1},\dots,\bar{\theta}_{\bar{n}-1})$. In addition some
technical assumptions must hold.
\end{theorem}

This theorem is proven in appendix \ref{a2}. It is easy to find the scalar
function $h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})$
which is a solution of the assumptions $(i^{\prime\prime})\,,\;(ii^{\prime
\prime})$ and $(iii^{\prime\prime})$.

\begin{theorem}
The scalar function $h(\underline{\theta},\underline{\bar{\theta}}%
,{\underline{z}})$ satisfy properties $(i^{\prime\prime})\,,\;(ii^{\prime
\prime})$ and $(iii^{\prime\prime})$ if it is of the form
\begin{align*}
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=\prod_{1\le i<j\le n}F(\theta_{ij})\prod_{i=1}^{n}\prod_{j=1}^{\bar{n}}%
\bar{F}(\theta_{i}-\bar{\theta}_{j})\prod_{i=1}^{n}\prod_{j=1}^{m}\phi
(\theta_{i}-z_{j})\\
&  \times\prod_{i=1}^{\bar{n}}\prod_{j=1}^{m}\frac{1}{\bar{F}(\bar{\theta}%
_{i}-z_{j})}\prod_{1\le i<j\le m}\tau(z_{i}-z_{j})\,\,p(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})
\end{align*}
where functions $\tau(z)$\thinspace,$\;\phi(z)$\thinspace,$\;F(\theta)$ and
$\bar{F}(\theta)$ are defined as above and if the function $p(\underline
{\theta},{\underline{z}})$ is an entire function of the $z$'s with
\[
p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})=(-\varepsilon
)^{\bar{n}}\varepsilon^{-m}\sigma p(\underline{\theta}^{\prime},\underline
{\bar{\theta}},{\underline{z}})=(-\varepsilon)^{n-m}\sigma^{-1}p(\underline
{\theta},\underline{\bar{\theta}}^{\prime},{\underline{z})}=\varepsilon
^{n}(-\varepsilon)^{\bar{n}}p(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}}^{\prime})
\]
(where $\varepsilon$ is a sign factor to be discussed later).
\end{theorem}

Note that the factors $1/\bar{F}(\bar{\theta}_{i}-z_{j})$ cancel in the
product $hg$ since $g$ satisfies Watson's equations (i).%

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof
%EndExpansion
Again Watson's equations for $F(\theta)$ and $\bar{F}(\theta)$ imply
\begin{align*}
h(\dots,\theta_{i},\theta_{j},\dots,{\underline{z}})  &  =h(\dots,\theta
_{j},\theta_{i},\dots,{\underline{z}})\dot{a}(\theta_{ij})\\
h(\underline{\theta},\dots,\bar{\theta}_{i},\bar{\theta}_{j},\dots
,{\underline{z}})  &  =h(\underline{\theta},\dots,\bar{\theta}_{j},\bar
{\theta}_{i},\dots,{\underline{z}})\dot{a}(\theta_{ij})\\
h(\theta\dots,\theta_{i},\bar{\theta}_{j},\dots,{\underline{z}})  &
=h(\dots,\bar{\theta}_{j},\theta_{i},\dots,{\underline{z}})\hat{b}(\theta
_{ij}).
\end{align*}
which is $(i^{\prime\prime})$. Watson's equations for $F(\theta)$ and $\bar
{F}(\theta)$
\begin{align*}
F(\theta)  &  =\dot{a}(\theta)F(-\theta)\,=F(2\pi i-\theta)\\
\bar{F}(\theta)  &  =-\varepsilon\hat{b}(\theta)\bar{F}(-\theta)\,=\bar
{F}(2\pi i-\theta)\,
\end{align*}
imply a shift relation for $\phi(z)$ eq. (\ref{4.7})
\[
\frac{\phi(z+2\pi i)}{\phi(z)}=\frac{\dot{a}(z)}{\varepsilon\dot{b}(z+2\pi i)}%
\]
therefore together with eqs. (\ref{4.4}) and (\ref{4.6})
\[
\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{i=1}^{\bar{n}}\hat{b}%
(\bar{\theta}_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot{a}(\theta_{1}-z_{j}%
)}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,h(\underline{\theta},\underline
{\bar{\theta}},{\underline{z}})=\sigma\,h(\underline{\theta}^{\prime
},\underline{\bar{\theta}},{\underline{z}})
\]%
\begin{align*}
\frac{h(\underline{\theta}^{\prime},\underline{\bar{\theta}},{\underline{z}}%
)}{\,h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})}  &
=\prod_{i=2}^{n}\frac{F(\theta_{1i}^{\prime})}{F(\theta_{1i})}\prod
_{j=1}^{\bar{n}}\frac{\bar{F}(\theta_{1}^{\prime}-\bar{\theta}_{j})}{\bar
{F}(\theta_{1}-\bar{\theta}_{j})}\prod_{j=1}^{m}\frac{\phi(\theta_{1}^{\prime
}-z_{j})}{\phi(\theta_{1}-z_{j})}\frac{p(\underline{\theta}^{\prime
},\underline{\bar{\theta}},{\underline{z}})}{\,p(\underline{\theta}%
,\underline{\bar{\theta}},{\underline{z}})}\\
&  =\prod_{i=2}^{n}\dot{a}(\theta_{i1})\prod_{j=1}^{\bar{n}}(-\epsilon)\hat
{b}(\bar{\theta}_{j}-\theta_{1})\prod_{j=1}^{m}\frac{\dot{a}(\theta_{1}%
-z_{j})}{\varepsilon\dot{b}(\theta_{1}^{\prime}-z_{j})}\frac{\varepsilon^{m}%
}{(-\varepsilon)^{\bar{n}}\sigma}%
\end{align*}
which is the first equation of $(ii^{\prime\prime})$ if $\,p(\underline
{\theta},\underline{\bar{\theta}},{\underline{z}})=(-\varepsilon)^{\bar{n}%
}\sigma\varepsilon^{-m}p(\underline{\theta}^{\prime},\underline{\bar{\theta}%
},{\underline{z}})$. Furthermore
\begin{align*}
\frac{h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)}{h(\underline{\theta},\underline{\bar{\theta}}^{\prime},{\underline{z}})}
&  =\prod_{i=1}^{n}\frac{\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})}{\bar
{F}(\theta_{i}-\bar{\theta}_{\bar{n}}^{\prime})}\prod_{j=1}^{m}\frac{\bar
{F}(\bar{\theta}_{\bar{n}}^{\prime}-z_{j})}{\bar{F}(\bar{\theta}_{\bar{n}%
}-z_{j})}\frac{p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)}{p(\underline{\theta},\underline{\bar{\theta}}^{\prime},{\underline{z}})}\\
&  =\prod_{i=1}^{n}(-\epsilon)\hat{b}(\bar{\theta}_{\bar{n}}^{\prime}%
-\theta_{i})\prod_{j=1}^{m}\frac{1}{(-\epsilon)\hat{b}(\bar{\theta}_{\bar{n}%
}-z_{j})}\frac{(-\epsilon)^{n-m}}{\sigma}%
\end{align*}
which is the second equation of $(ii^{\prime\prime})$
\[
\prod_{i=1}^{n}\dot{b}(\theta_{i}-\bar{\theta}_{\bar{n}}^{\prime\prime}%
)\prod_{j=1}^{m}\frac{1}{\dot{b}(z_{j}-\bar{\theta}_{\bar{n}}^{\prime
\prime\prime})}\,h(\underline{\theta},\underline{\bar{\theta}}^{\prime
},{\underline{z}})=\sigma\,h(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]
if $p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})=(-\varepsilon)^{n-m}\sigma^{-1}p(\underline{\theta},\underline{\bar{\theta
}}^{\prime},{\underline{z}})$.%

\begin{align*}
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=\prod_{1\le i<j\le n}F(\theta_{ij})\prod_{i=1}^{n}\prod_{j=1}^{\bar{n}}%
\bar{F}(\theta_{i}-\bar{\theta}_{j})\prod_{i=1}^{n}\prod_{j=1}^{m}\phi
(\theta_{i}-z_{j})\\
&  \times\prod_{i=1}^{\bar{n}}\prod_{j=1}^{m}\frac{1}{\bar{F}(\bar{\theta}%
_{i}-z_{j})}\prod_{1\le i<j\le m}\tau(z_{i}-z_{j})\,\,p(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})
\end{align*}
\begin{align*}
F(\theta)  &  =\dot{a}(\theta)F(-\theta)\,=F(2\pi i-\theta)\\
\bar{F}(\theta)  &  =-\varepsilon\hat{b}(\theta)\bar{F}(-\theta)\,=\bar
{F}(2\pi i-\theta)\,
\end{align*}
\begin{align*}
\frac{h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}^{\prime}%
)}{h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})}  &
=\prod_{i=1}^{n}\frac{\phi(\theta_{i}-z_{1}^{\prime})}{\phi(\theta_{i}-z_{1}%
)}\prod_{i=1}^{\bar{n}}\frac{\bar{F}(\bar{\theta}_{i}-z_{1})}{\bar{F}%
(\bar{\theta}_{i}-z_{1}^{\prime})}\prod_{j=1}^{m}\frac{\tau(z_{1}^{\prime
}-z_{j})}{\tau(z_{1}-z_{j})}\frac{p(\underline{\theta},\underline{\bar{\theta
}},{\underline{z}}^{\prime})}{p(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})}\\
&  =\prod_{i=1}^{n}\frac{\varepsilon\dot{b}(\theta_{i}-z_{1})}{\dot{a}%
(\theta_{i}-z_{1}^{\prime})}\prod_{i=1}^{\bar{n}}\frac{-\varepsilon}{\hat
{b}(\bar{\theta}_{i}-z_{1}^{\prime})}\prod_{j=1}^{m}\frac{\dot{b}%
(z_{1}^{\prime}-z_{j})}{\dot{a}(z_{1}^{\prime}-z_{j})}\frac{1}{\dot{b}%
(z_{j}-z_{1})}\frac{1}{\varepsilon^{n}(-\varepsilon)^{\bar{n}}}%
\end{align*}
which yields the third equation of $(ii^{\prime})$%
\[
\prod_{i=1}^{n}\dot{b}(\theta_{i}-z_{1})\prod_{j=1}^{m}\frac{1}{\dot{b}%
(z_{j}-z_{1})}\,h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})=\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1}^{\prime})\prod_{i=1}^{\bar{n}}%
\hat{b}(\bar{\theta}_{i}-z_{1}^{\prime})\prod_{j=1}^{m}\frac{\dot{a}%
(z_{1}^{\prime}-z_{j})}{\dot{b}(z_{1}^{\prime}-z_{j})}\,h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}}^{\prime})
\]
if $p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)=\varepsilon^{n}(-\varepsilon)^{\bar{n}}p(\underline{\theta},\underline
{\bar{\theta}},{\underline{z}}^{\prime})$.%

%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof
%EndExpansion


The definitions (\ref{4.3}) yield the explicit formula for the
`Jost-function'
\[
\phi(z)=\frac{1}{F(z)\,\bar{F}(i\pi-z)}=const\;\Gamma\left(  \frac{z}{2\pi
i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{z}{2\pi i}\right)
\]
with $\phi(-z)/\phi(z)=\dot{a}(z)$. The $\tau$-funtion turns out to be
\[
\tau(z)=\frac{1}{F(z)\,\phi(z)\,\phi(-z)}=const^{\prime}\frac{1}{F(z)}%
\frac{z\sinh\frac{z}{2}}{\Gamma\left(  1-\frac{1}{N}-\frac{z}{2\pi i}\right)
\Gamma\left(  1-\frac{1}{N}+\frac{z}{2\pi i}\right)  }
\]


+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

We shall now construct form factor functions given by eq.~(\ref{2.12}), that
means we construct co-vector valued functions which satisfy the properties (i)--(v).

\begin{definition}
\label{d1} We define the co-vector valued function
\begin{equation}
f_{\underline{\alpha}}(\underline{\theta})=\int_{\mathcal{C}_{\underline
{\theta}}}dz_{1}\cdots\int_{\mathcal{C}_{\underline{\theta}}}dz_{m}%
\,h(\underline{\theta},{\underline{z}})\,p(\underline{\theta},{\underline{z}%
})\,g_{\underline{\beta}}(\underline{z})\,\Psi_{\underline{\alpha}%
}^{\underline{\beta}}(\underline{\theta},{\underline{z}}) \label{4.12}%
\end{equation}
with the scalar function
\[
h(\underline{\theta},{\underline{z}})=\prod_{1\le i<j\le n}F(\theta_{ij}%
)\prod_{i=1}^{n}\prod_{j=1}^{m}\phi(\theta_{i}-z_{j})\prod_{1\le i<j\le m}%
\tau(z_{i}-z_{j})\,,
\]
and the Bethe ansatz state
\begin{equation}
\Psi_{\underline{\alpha}}^{\underline{\beta}}(\underline{\theta}%
,{\underline{z}})=\left[  \Omega C^{\beta_{m}}({\underline{\theta}}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta}},z_{1})\right]  _{\underline
{\alpha}}. \label{4.13}%
\end{equation}
The integration contour $\mathcal{C}_{\underline{\theta}}$ consists of several
pieces (see figure~\ref{f5.1}):

\begin{itemize}
\item[a)] A line from $-\infty$ to $\infty$ avoiding all poles such that
$\operatorname{Im}\theta_{i}-2\pi<\operatorname{Im}z_{j}<\operatorname{Im}%
\theta_{i}-2\pi/N$.

\item[b)] Clock wise oriented circles around all poles (of the $\phi
(\theta_{i}-z_{j})$) at $z_{j}=\theta_{i}$. \begin{figure}[tbh]%
\[
\unitlength4.5mm\begin{picture}(27,13)
\thicklines
\put(1,0){
\put(.19,3.2){\circle{.3}~$\theta_n-i\pi$}
\put(0,4.2){$\bullet~~\theta_n-2\pi i(\frac12-\frac1N)$}
\put(.2,4.4){\oval(1,1)}\put(.5,4.88){\vector(1,0){0}}
\put(0,9){$\bullet~\theta_n+i\pi$}
}
\put(8,6){\dots}
\put(12,0){
\put(0,0){$\bullet~\theta_2-2\pi i$}
\put(.19,5){\circle{.3}~$\theta_2-2\pi i\frac1N$}
\put(0,6){$\bullet~~\theta_2$}
\put(.2,6.2){\oval(1,1)}\put(.5,6.68){\vector(1,0){0}}
\put(0,11){$\bullet~\theta_2+2\pi i(1-\frac1N)$}
}
\put(20,1){
\put(0,0){$\bullet~\theta_1-2\pi i$}
\put(.19,5){\circle{.3}~$\theta_1-2\pi i\frac1N$}
\put(0,6){$\bullet~~\theta_1$}
\put(.2,6.2){\oval(1,1)}\put(.5,6.68){\vector(1,0){0}}
\put(0,11){$\bullet~\theta_1+2\pi i(1-\frac1N)$}
}
\put(9,2.5){\vector(1,0){0}}
\put(0,3){\oval(34,1)[br]}
\put(27,3){\oval(20,1)[tl]}
\end{picture}
\]
\caption{\textit{The integration contour $\mathcal{C}_{\underline{\theta}}$ .
The bullets belong to poles of the integrand resulting from $a(\theta
_{i}-u_{j})\,\phi(\theta_{i}-u_{j})$ and the small open circles belong to
poles originating from $b(\theta_{i}-u_{j})$ and $c(\theta_{i}-u_{j})$. }}%
\end{figure}
\end{itemize}
\end{definition}

The function $h(\underline{\theta},{\underline{z}})$ and the state
$\Psi_{1\dots n}(\underline{\theta},{\underline{z}})$ depending solely on the
S-matrix. They are specified in the following definitions.

\begin{definition}
\label{d2}The
\end{definition}

\subparagraph{Remarks:}

\begin{itemize}
\item Using Watson's equations for crossing and unitarity one derives the
following identities for the scalar functions $\phi(z)$ and $\tau(z)$
\begin{equation}
\phi(z)=\bar{\phi}(i\pi-z)=\frac{1}{\hat{b}(i\pi-z)}\,\bar{\phi}(z-i\pi
)=\frac{\dot{a}(z-2\pi i)}{b(z)}\,\phi(z-2\pi i)~, \label{4.14}%
\end{equation}
\[
\tau(z)=\tau(-z)=\frac{b(z)}{a(z)}\,\frac{a(2\pi i-z)}{b(2\pi i-z)}%
\,\tau(z-2\pi i)
\]
where $b(z)$ is the soliton-anti-soliton scattering amplitude related to
$a(z)$ by crossing $b(z)=a(i\pi-z)$.

\item The functions $\phi(z)$ and $\tau(z)$ are of the form
\[
\phi(z)=const.~\frac{1}{\sinh z}\exp\int_{0}^{\infty}\frac{dt}{t}\frac
{\sinh\frac{1}{2}(1-\nu)t\,\left(  \cosh t(\frac{1}{2}-z/(i\pi))-1\right)
}{\sinh\frac{1}{2}\,\nu t\sinh t}
\]
\[
\tau(z)=const.~\sinh z\sinh z/\nu
\]

\end{itemize}

In contrast to the functions $F(z),\phi(z)$ and $\tau(z)$ the `p-function' in
the integral representation (\ref{1.2}) depends on the spin, the charge and
the statistics of the local operator $\mathcal{O}(x)$. It is characterized by
the following

\paragraph{Conditions:\label{p}}

The p-function $p_{n}^{\mathcal{O}}(\underline{\theta},\underline{z})$ satisfies

\begin{itemize}
\item[(a)] $p_{n}^{\mathcal{O}}(\underline{\theta},\underline{z})$ is
symmetric with respect to the $\theta$'s and the $z$'s.

\item[(b)] $p_{n}^{\mathcal{O}}(\underline{\theta},\underline{z})$ is a
polynomial in $e^{\pm z_{j}}~(j=1,\dots,m)$

\item[(c)] $p_{n}^{\mathcal{O}}(\dots,\theta_{i}+2\pi i,\dots,\underline
{z})=\sigma_{\mathcal{O}i}p_{n}^{\mathcal{O}}(\underline{\theta},\underline
{z})$

where the statistics factor $\sigma_{\mathcal{O}i}$ is $-1$ if the operator
$\mathcal{O}(x)$ and the particle $i$ are both fermionic and $+1$ otherwise.

\item[(d)] $\left\{
\begin{array}
[c]{l}%
p_{n}^{\mathcal{O}}(\theta_{1}=\theta_{n}+i\pi,\tilde{\underline{\theta}%
},\theta_{n};\tilde{\underline{z}},z_{m}=\theta_{n}{)}=\dfrac{\left(
f_{ss}^{(0)}(0)\right)  ^{2}}{4\pi m}\,p_{n-2}^{\mathcal{O}}(\tilde
{\underline{\theta}},\tilde{\underline{z}})+\tilde{p}^{(1)}({\underline
{\theta}})\\
p_{n}^{\mathcal{O}}(\theta_{1}=\theta_{n}+i\pi,\tilde{\underline{\theta}%
},\theta_{n};\tilde{\underline{z}},z_{m}=\theta_{1}{)}=\sigma_{\mathcal{O}%
1}\dfrac{\left(  f_{ss}^{(0)}(0)\right)  ^{2}}{4\pi m}\,p_{n-2}^{\mathcal{O}%
}(\tilde{\underline{\theta}},\tilde{\underline{z}})+\tilde{p}^{(2)}%
({\underline{\theta}})
\end{array}
\right.  $

where $\tilde{p}^{(1,2)}({\underline{\theta}})$ are independent of the $z$'s
and nonvanishing only for charge less operators $\mathcal{O}(x)$.

\item[(e)] $p_{n}^{\mathcal{O}}(\underline{\theta}+\mu,\underline{z}%
+\mu)=e^{s\mu}p_{n}^{\mathcal{O}}(\underline{\theta},\underline{z})$ where $s$
is the `spin' of the operator $\mathcal{O}(x)$.

\item[(f)] The number of the particles $n$ and the number of integrations $m $
are related by $q=n-2m$ where $q$ is the charge of the operator $\mathcal{O}%
(x)$.
\end{itemize}

\begin{theorem}
The co-vector valued function $\mathcal{O}_{1\dots n}({\underline{\theta}})$
defined by (\ref{1.2}) fulfills the properties \textrm{(i), (ii)} and
\textrm{(iii)} on page \pageref{pf} if the functions $F(\theta),\phi(z)$ and
$\tau(z)$ are given by definition \ref{d2} and the p-function $p_{n}%
^{\mathcal{O}}(\underline{\theta},\underline{z})$ satisfies the conditions (a)
-- (f) on page \pageref{p}.
\end{theorem}

%

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof
%EndExpansion
The properties (i) and (ii) follow as in \cite{BFKZ}. The proof of (iii) is
also the same as in \cite{BFKZ}, if the functions $\tilde{p}^{(1)}%
({\underline{\theta}})$ and $\tilde{p}^{(2)}({\underline{\theta}})$ in (d)
vanish. For the case of chargeless operators they are in general nonvanishing.
Then the proof of (iii) in the form of
\[
\operatorname*{Res}_{\theta_{1n}=i\pi}\mathcal{O}_{1\dots n}(\theta_{1}%
,\dots,\theta_{n})=-2i\,\mathbf{C}_{1n}\,\mathcal{O}_{2\dots n-1}(\theta
_{2},\dots,\theta_{n-1})\Big( \mathbf{1}_{2\dots n-1}-S_{2n}\cdots
S_{n-1n}\Big)
\]
has to be modified as follows: Considering the $z_{m}$-integration as in
\cite{BFKZ} the terms involving $c_{n}p_{n-2}^{\mathcal{O}}(\tilde
{\underline{\theta}},\tilde{\underline{z}})$ yield the desired result and
those proportional to $\tilde{p}^{(1,2)}({\underline{\theta}})$ yield terms
with a factor $\mathbf{1}_{2\dots n-1}+S_{2n}\cdots S_{n-1n}$, however, they
vanish due to the following lemma.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof
%EndExpansion


\begin{itemize}
\item The three particle form factor
\end{itemize}

\label{s5} We calculate the $U(N)$ three particle form factor of the
fundamental fermi field. For convenience we multiply the field with the Dirac
operator
\[
{\mathcal{O}_{D\delta}(x)}=i(i\gamma\partial+m)\psi_{D\delta}(x).
\]
Defining
\[
\langle\,0\,|\,{\mathcal{O}_{\!D\delta}(0)}\,|\,p_{1},p_{2},\overline{p_{3}%
}\,\rangle_{\alpha\beta\bar{\gamma}}^{in}={F^{\mathcal{O}_{\!D\delta}(0)}%
}_{\alpha\beta\bar{\gamma}}(\theta_{12},\theta_{13},\theta_{23}),
\]
we may write this form factor as a linear combination of solutions of Watson's
equations as
%(.....up to kinematical factors)
(with $\theta_{i}=i\pi\phi_{i}$)
\[
{F^{\mathcal{O}_{\!D\delta}(0)}}_{\alpha\beta\bar{\gamma}}(\theta_{12}%
,\theta_{13},\theta_{23})=F_{0}(\theta_{12})\,\bar{F}_{0}(\theta_{13}%
)\,\bar{F}_{0}(\theta_{23})\sum_{\theta=\theta_{1},\theta_{2},\theta_{3}%
}K^{\theta}(\theta_{12},\theta_{13},\theta_{23})\,f_{\alpha\beta\bar{\gamma}%
}^{\theta}(\theta_{1},\theta_{2},\theta_{3}).
\]
The functions $K^{\theta}(\theta_{12},\theta_{13},\theta_{23})$ are trival
solutions of Watsons equations for $S=1$ and the $f_{\alpha\beta\bar{\gamma}%
}^{\theta}(\theta_{1},t_{2},t_{3})$ are solutions of Watsons given by (see
\cite{BKZ1} Example 5.6)
\[
f_{\alpha\beta\bar{\gamma}}^{\theta}(\theta_{1},\theta_{2},\theta_{3}%
)=\sum_{n=-\infty}^{\infty}\,\psi(\phi_{1}-\phi-2n)\,\psi(\phi_{2}%
-\phi-2n)\,\bar{\psi}_{1}(\phi_{3}-\phi-2n)g_{\alpha\beta\bar{\gamma}}^{\phi}
\]
with
\begin{align*}
g_{\alpha\beta\bar{\gamma}}^{\theta}(\theta_{1},\theta_{2},\theta_{3})  &
=\delta_{\alpha1}\delta_{\beta1}\delta_{\gamma1}\,\check{u}_{1}(\phi_{1}%
-\phi-2n)\,\check{u}_{1}(\phi_{2}-\phi-2n)\,\check{t}_{2}(\phi_{3}-\phi-2n)\\
&  ~~+~\!\!\!\!\!\!\delta_{\alpha1}\delta_{\beta\gamma}\,\check{u}_{1}%
(\phi_{1}-\phi-2n)\,\check{u}_{2}(\phi_{2}-\phi-2n)\;+\;\delta_{\beta1}%
\delta_{\alpha\gamma}\,\check{u}_{2}(\phi_{1}-\phi-2n)
\end{align*}
where
\[
\psi(x)=\frac{\Gamma\left(  1-\frac{1}{N}+\frac{x}{2}\right)  }{\Gamma\left(
1+\frac{x}{2}\right)  }~,~~~\bar{\psi}_{1}(x)=\frac{\Gamma\left(  \frac{1}%
{2}+\frac{x}{2}\right)  }{\Gamma\left(  \frac{3}{2}-\frac{1}{N}+\frac{x}%
{2}\right)  }.
\]
Therefore
\begin{align*}
f_{\alpha\beta\bar{\gamma}}^{u}(x,y,z)  &  =\sum_{u}\left\{  \frac
{\Gamma\left(  -\frac{1}{N}+\frac{x-u}{2\pi i}\right)  }{\Gamma\left(
\frac{x-u}{2\pi i}\right)  }\frac{\Gamma\left(  -\frac{1}{N}+\frac{y-u}{2\pi
i}\right)  }{\Gamma\left(  \frac{y-u}{2\pi i}\right)  }\frac{\Gamma\left(
-\frac{1}{2}+\frac{z-u}{2\pi i}\right)  }{\Gamma\left(  \frac{3}{2}-\frac
{1}{N}+\frac{z-u}{2\pi i}\right)  }\,(N-1)\,\delta_{\alpha1}\delta_{\beta
1}\delta_{\gamma1}~\right. \\
&  -~\frac{\Gamma\left(  -\frac{1}{N}+\frac{x-u}{2\pi i}\right)  }%
{\Gamma\left(  \frac{x-u}{2\pi i}\right)  }\frac{\Gamma\left(  -\frac{1}%
{N}+\frac{y-u}{2\pi i}\right)  }{\Gamma\left(  1+\frac{y-u}{2\pi i}\right)
}\frac{\Gamma\left(  \frac{1}{2}+\frac{z-u}{2\pi i}\right)  }{\Gamma\left(
\frac{3}{2}-\frac{1}{N}+\frac{z-u}{2\pi i}\right)  }\,\delta_{\alpha1}%
\delta_{\beta\gamma}\\
&  -~\left.  \frac{\Gamma\left(  -\frac{1}{N}+\frac{x-u}{2\pi i}\right)
}{\Gamma\left(  1+\frac{x-u}{2\pi i}\right)  }\frac{\Gamma\left(  1-\frac
{1}{N}+\frac{y-u}{2\pi i}\right)  }{\Gamma\left(  1+\frac{y-u}{2\pi i}\right)
}\frac{\Gamma\left(  \frac{1}{2}+\frac{z-u}{2\pi i}\right)  }{\Gamma\left(
\frac{3}{2}-\frac{1}{N}+\frac{z-u}{2\pi i}\right)  }\,\delta_{\beta1}%
\delta_{\alpha\gamma}\right\}
\end{align*}
where $u=\tilde{u}+{2\pi i}l$, $l\in\mathbf{Z}$. In the $\frac{1}{N}%
$-expansion this term behaves as
\begin{align*}
f_{\alpha\beta\bar{\gamma}}^{u}\left(  x,y,z\right)   &  =\frac{x+y-2z-2\pi
i}{\left(  y-z-i\pi\right)  \left(  x-z-i\pi\right)  }\tanh\left(  \frac
{z-u}{2}\right)  2\pi^{2}\delta_{\alpha1}\delta_{1\beta}\delta_{\gamma1}\\
&  -\left(  \frac{\coth\left(  \frac{x-u}{2}\right)  }{x-z-i\pi}+\frac
{\coth\left(  \frac{y-u}{2}\right)  }{y-z-i\pi}\right)  2\pi^{2}\delta
_{\alpha1}\delta_{1\beta}\delta_{\gamma1}\\
&  +\frac{2\pi^{2}}{y-z-i\pi}\left(  \coth\left(  \frac{y-u}{2}\right)
-\tanh\left(  \frac{z-u}{2}\right)  \right)  \delta_{\alpha1}\delta
_{\beta\gamma}\\
&  +\frac{2\pi^{2}}{x-z-i\pi}\left(  \coth\left(  \frac{x-u}{2}\right)
-\tanh\left(  \frac{z-u}{2}\right)  \right)  \delta_{\beta1}\delta
_{\alpha\gamma}+\mathcal{O}\left(  \frac{1}{N}\right)
\end{align*}


The coefficents in eq.~(\ref{2.11}) are determined by the condition that the
form factor has to yield the S-matrix if the fourth leg is put on mass shell,
or equivalently that it fulfills the pole formula (iii) of (\ref{2.3}). Also
the pole
%The pole factor $$K(\theta)=\frac1{\sinh\frac12(\theta-2i\pi/N)\,\sinh\frac12(\theta+2i\pi/N)}$$
corresponding to the bound state of two particles of mass $m_{12}=m\,\sin
(2\pi/N)/\sin(\pi/N)$ has to be present. Substituting now this Ansatz into
(iii) we obtain for the coefficient:
\[
K=
\]
As a consistency check we may also verify that by means of (iv) this
expression reduces precisely to $f_{(12)\bar{3}}\left(  \theta\right)  $.
Alternatively we may also represent this function as a contour integral, which
automatically fixes the coefficients K
\begin{align*}
f_{12\bar{3}}\left(  \theta_{1},\theta_{2},\theta_{3}\right)   &  =\frac
{1}{2\pi i}\sin\frac{\pi}{2}\left(  \phi_{12}\right)  F_{0}\left(  \theta
_{12}\right)  F_{\bar{0}}\left(  \theta_{13}\right)  F_{\bar{0}}\left(
\theta_{23}\right) \\
&  \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\times\int\limits_{\mathcal{C}%
}du\varphi(\theta_{1}-u)\varphi(\theta_{2}-u)\bar{\varphi}(\theta_{3}%
-u)\sin\frac{\pi}{2}\left(  u-\phi_{3}-\frac{1}{N}\right)  g_{12\bar{3}%
}\left(  \theta_{1},\theta_{2},\theta_{3},u\right)
\end{align*}
Here we have defined the functions
\[
\varphi(x)=\Gamma\left(  -\frac{x}{2}\right)  \Gamma\left(  1-\frac{1}%
{N}+\frac{x}{2}\right)  \qquad\hbox{and}\qquad\bar{\varphi}(x)=\Gamma\left(
\frac{1}{2}+\frac{x}{2}\right)  \Gamma\left(  \frac{1}{N}-\frac{1}{2}-\frac
{x}{2}\right)  \,.
\]
The countour is here to be understood in the way illustrated in figure ?

To be done

cross-check with 1/N-expansion

cross-check with asymptotic freedom

numerical calculation

\renewcommand{\theequation}{\mbox{\Alph{section}.\arabic{equation}}}\appendix

\section*{Appendix}

\setcounter{section}{0} \setcounter{equation}{0}

\section{Proof of Theorem \ref{t1}}

\label{a1}%

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof
%EndExpansion
Property (i) follows directly from the Yang-Baxter equations, the action of
the S-matrix on the pseudo-ground state $\Omega$ and (ii')
\begin{align*}
\,\Omega_{\dots ji\dots}\,C_{\dots ji\dots}({\dots\theta_{j},\theta_{i}\dots
})\,\dot{S}_{ij}(\theta_{ij})  &  =\Omega_{\dots ji\dots}\,\dot{S}_{ij}%
(\theta_{ij})\,C_{\dots ij\dots}({\dots\theta_{i},\theta_{j}\dots})\\
&  =\dot{a}(\theta_{ij})\,\Omega_{\dots ij\dots}\,C_{\dots ij\dots}%
({\dots\theta_{i},\theta_{j}\dots})\,.
\end{align*}


Using (i) the property (ii) may be rewritten as a difference equation
\[
f_{2\dots n1}(\theta_{2},\dots,\theta_{n},\theta_{1})=f_{12\dots
n}({\underline{\theta}})\dot{S}_{21}\cdots\dot{S}_{n1}=\sigma f_{1\dots
n}({\underline{\theta}}^{\prime})
\]
or more general
\begin{equation}
f_{1\dots n}({\underline{\theta}})\,Q_{1\dots n}({\underline{\theta}%
,}i)=\sigma f_{1\dots n}({\underline{\theta}}^{\prime}) \label{A.1}%
\end{equation}
where ${\underline{\theta}}^{\prime}=(\theta_{1},\dots,\theta_{i}^{\prime
}=\theta_{i}+2\pi i,\dots,\theta_{n})$ and $\sigma$ is a sign which depends on
the statistics and the particle $i$. The operator $Q({\underline{\theta}},i)$
is defined by
\begin{gather*}
Q_{1\dots n}({\underline{\theta}},i)=\dot{S}_{i+1i}(\theta_{i+1}-\theta
_{i})\cdots\dot{S}_{ni}(\theta_{n}-\theta_{i})\dot{S}_{1i}(\theta_{1}%
-\theta_{i}^{\prime})\cdots\dot{S}_{i-1i}(\theta_{i-1}-\theta_{i}^{\prime})\\
=\mathrm{tr}_{0}T_{Q,1\dots n,0}({\underline{\theta},i})=\mathrm{tr}%
_{0}\mathrm{\,}\dot{S}_{10}(\theta_{1}-\theta_{i}^{\prime})\,\cdots
\mathbf{P}_{i0}\cdots\dot{S}_{n0}(\theta_{n}-\theta_{i})\\
Q_{\underline{\alpha}}^{\underline{\alpha}^{\prime}}({\underline{\theta}%
},i)=\sum_{\alpha}
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(10,4) \put(1,1){\oval(8,2)[rt]} \put(9,3){\oval(8,2)[lb]} \put(2,1){\line(0,1){2}} \put(4,1){\line(0,1){2}} \put(6,1){\line(0,1){2}} \put(8,1){\line(0,1){2}} \put(1.9,.5){$\alpha_1$} \put(1.9,3.3){$\alpha'_1$} \put(4.9,.5){$\alpha_i$} \put(4.9,3.3){$\alpha'_i$} \put(7.9,.5){$\alpha_n$} \put(7.9,3.3){$\alpha'_n$} \put(.3,1.8){$\alpha$} \put(9.3,1.8){$\alpha$} \put(1.2,1.1){$\theta_1$} \put(5.1,1.1){$\theta'_i$} \put(4.4,2.4){$\theta_i$} \put(8.2,1.1){$\theta_n$} \put(2.6,2.5){\dots} \put(6.6,2.5){\dots} \end{picture}
\end{array}
.
\end{gather*}
Because of the symmetry $(i)$ it is sufficient to discuss the difference
equation for $i=1.$ For this case the matrix $Q({\underline{\theta}%
},1)=Q({\underline{\theta}})$ may be written as a trace of the ordinary
monodromy matrix (\ref{3.1}) over the auxiliary space for the specific value
of the spectral parameter $\theta_{0}=\theta_{1}$
\begin{equation}
Q_{1\dots n}({\underline{\theta}})=-\mathrm{tr}_{0}T_{1\dots n,0}%
({\underline{\theta}},\theta_{1})\,. \label{A.2}%
\end{equation}
Note that
\[
\dot{S}_{10}(0)=-\mathbf{P}_{10}=-~~
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(2,2) \put(0,0){\oval(2,2)[tr]} \put(2,2){\oval(2,2)[bl]} \end{picture}
\end{array}
\]
is minus the permutation operator. For $i=1$ the difference equation may be
depicted as
\[
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,4) \put(2.5,2){\oval(5,2)} \put(2.5,2){\makebox(0,0){$f$}} \put(1,0){\line(0,1){1}} \put(2,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
~~=\sigma~~
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(7,5) \put(4.5,2){\oval(5,2)[b]} \put(3.5,2){\oval(7,6)[t]} \put(1,2){\oval(2,2)[lb]} \put(1,0){\oval(2,2)[rt]} \put(3.5,3){\oval(5,2)} \put(3.5,3){\makebox(0,0){$f$}} \put(3,0){\line(0,1){2}} \put(5,0){\line(0,1){2}} \put(3.4,.5){$\dots$} \end{picture}
\end{array}
\]
where we use the rule that the rapidity of a line changes by $2\pi i$ if the
line bends by $360^{0}$ in the positive sense.

In the following we will suppress the indices $1\dots n$. The Yang-Baxter
relations (\ref{3.2}) imply the typical commutation rules for the matrices
$A,C,D$ defined in eq.~(\ref{3.3})
\begin{align}
C^{\beta}({\underline{\theta}},z)A({\underline{\theta}},\theta)  &
=\frac{\dot{a}(\theta-z)}{\dot{b}(\theta-z)}A({\underline{\theta}}%
,\theta)C^{\beta}({\underline{\theta}},z)-\frac{\dot{c}(\theta-z)}{\dot
{b}(\theta-z)}A({\underline{\theta}},z)C^{\beta}({\underline{\theta}}%
,\theta)\label{A.3}\\
C^{\beta}({\underline{\theta}},z)D_{\gamma}^{\gamma^{\prime}}({\underline
{\theta}},\theta)  &  =\frac{1}{\dot{b}(z-\theta)}\dot{S}_{\beta^{\prime
}\gamma^{\prime\prime}}^{\gamma^{\prime}\beta}(z-\theta)D_{\gamma}%
^{\gamma^{\prime\prime}}({\underline{\theta}},\theta)C^{\beta^{\prime}%
}({\underline{\theta}},z)-\frac{\dot{c}(z-\theta)}{\dot{b}(z-\theta)}%
D_{\gamma}^{\beta}({\underline{\theta}},z)C^{\gamma^{\prime}}({\underline
{\theta}},\theta)\nonumber
\end{align}
where $\beta,\beta^{\prime},\gamma,\gamma^{\prime},\gamma^{\prime\prime}%
\in\left\{  2,\ldots,N\right\}  $. In addition there are commutation rules
where also the matrices $A_{Q},C_{Q},D_{Q}$ defined by
\[
T_{Q}({\underline{\theta}})=\left(
\begin{array}
[c]{cc}%
A_{Q}({\underline{\theta}}) & B_{Q}^{\beta}({\underline{\theta}})\\
C_{Q}^{\beta}({\underline{\theta}}) & {D_{Q}}_{\beta}^{\beta^{\prime}%
}({\underline{\theta}})
\end{array}
\right)
\]
are involved \cite{BKZ}
\begin{align}
C^{\beta}({\underline{\theta}},z)A_{Q}({\underline{\theta}})  &  =\frac
{\dot{a}(\theta_{1}-z)}{\dot{b}(\theta_{1}^{\prime}-z)}A_{Q}({\underline
{\theta}})C({\underline{\theta}}^{\prime},z)-\frac{\dot{c}(\theta_{1}^{\prime
}-z)}{\dot{b}(\theta_{1}^{\prime}-z)}A({\underline{\theta}},z)C_{Q}%
({\underline{\theta}})\label{A.4}\\
C^{\beta}({\underline{\theta}},z){D_{Q}}_{\gamma}^{\gamma^{\prime}%
}({\underline{\theta}})  &  =\frac{1}{\dot{b}(z-\theta_{1})}\dot{S}%
_{\beta^{\prime}\gamma^{\prime\prime}}^{\gamma^{\prime}\beta}(z-\theta
_{1}^{\prime}){D_{Q}}_{\gamma}^{\gamma^{\prime\prime}}({\underline{\theta}%
})C^{\beta^{\prime}}({\underline{\theta}}^{\prime},z)-\frac{\dot{c}%
(z-\theta_{1})}{\dot{b}(z-\theta_{1})}D_{\gamma}^{\beta}({\underline{\theta}%
},z)C_{Q}^{\gamma^{\prime}}({\underline{\theta}})\,.\nonumber
\end{align}
Note that we assign to the auxiliary space of $T_{Q}({\underline{\theta}})$
corresponding to the horizontal line the spectral parameter $\theta_{1}$ on
the right hand side and $\theta_{1}^{\prime}=$ $\theta_{1}+2\pi i$ on the left
hand side. To analyze the right hand side of eq.~(\ref{A.1}) we proceed as
follows: We apply the trace of $T_{Q}({\underline{\theta}})$ to the co-vector
$f({\underline{\theta}})$ as given by eq.~(\ref{4.1}) and the Bethe ansatz
(\ref{3.4}).
\[
\Omega\,C^{\beta_{m}}({\underline{\theta}},z_{m})\cdots C^{\beta_{1}%
}({\underline{\theta}},z_{1}){T_{Q}}_{\gamma}^{\gamma^{\prime}}\,({\underline
{\theta}})=
\begin{array}
[c]{c}%
\unitlength5mm\begin{picture}(10,5.5) \put(10,5){\oval(17,2)[lb]} \put(10,5){\oval(20,6)[lb]} \put(4,0){\line(0,1){5}} \put(9,0){\line(0,1){5}} \put(10,5){\oval(14,8)[lb]} \put(2,0){\oval(2,2)[rt]} \put(-.5,5.3){$\beta_1$} \put(1,5.3){$\beta_m$} \put(2.8,5.3){1} \put(3.8,5.3){1} \put(8.8,5.3){1} \put(1.4,.8){$\gamma'$} \put(10.2,.8){$\gamma$} \put(10.2,1.8){1} \put(10.2,3.8){1} \put(4.3,0){$\theta_{2}$} \put(9.2,0){$\theta_{n}$} \put(7.5,2.3){$z_1$} \put(7.5,4.3){$z_m$} \put(5.5,3){$\dots$} \put(9.5,2.7){$\vdots$} \put(7.5,1.2){$\theta_1$} \put(2,0){$\theta_1'$} \end{picture}
\end{array}
\]
In the contribution from $A_{Q}({\underline{\theta}})$ which means
$\gamma^{\prime}=\gamma=1$ one may use Yang-Baxter relations to observe that
only the amplitudes $\dot{a}(\theta_{1}-z_{j})$ appear in the S-matrices
$\dot{S}(\theta_{1}-z_{j})$ which are constituents of the C-operators.
Therefore we may shift all $z_{j}$-integration contours $\mathcal{C}%
_{{\underline{\theta}}}$ to $\mathcal{C}_{{\underline{\theta}}^{\prime}}$
without changing the values of the integrals, because the by assumption in (a)
on page \pageref{as} there are no singularyties inside $\mathcal{C}%
_{\underline{\theta}}\cup-\mathcal{C}_{\underline{\theta}^{\prime}}$. Using a
short notation we have
\begin{align*}
f({\underline{\theta}})\,A_{Q}({\underline{\theta}})  &  =\int_{\mathcal{C}%
_{{\underline{\theta}}}}d\underline{z}\,h({\underline{\theta}},{\underline{z}%
})g_{\underline{\beta}}({\underline{z}})\,\Psi^{\underline{\beta}}%
({\underline{\theta}},{\underline{z}})\,A_{Q}({\underline{\theta}})\\
&  =\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline
{z}\,h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}%
}({\underline{z}})\,\Psi^{\underline{\beta}}({\underline{\theta}}%
,{\underline{z}})A_{Q}({\underline{\theta}})
\end{align*}
We now proceed as usual in the algebraic Bethe ansatz and push the
$A^{Q}({\underline{\theta}})$ and $D^{Q}({\underline{\theta}})$ through all
the C-operators using the commutation rules (\ref{A.4}) and obtain
\begin{align}
C^{\beta_{m}}({\underline{\theta}},z_{m})\cdots C^{\beta_{1}}({\underline
{\theta}},z_{1})\,A_{Q}({\underline{\theta}})  &  =\prod_{j=1}^{m}\frac
{\dot{a}(\theta_{1}-z_{j})}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,A_{Q}%
({\underline{\theta}})\,C^{\beta_{m}}({\underline{\theta}}^{\prime}%
,z_{1})\cdots C^{\beta_{1}}({\underline{\theta}}^{\prime},z_{1})\nonumber\\
&  +\sum uw_{A}\,, \label{A.5}%
\end{align}
\begin{align}
C^{\beta_{m}}({\underline{\theta}},z_{m})\cdots C^{\beta_{1}}({\underline
{\theta}},z_{1})\,{D_{Q}}_{\beta}^{\beta^{\prime}}({\underline{\theta}})  &
=\prod_{j=1}^{m}\frac{1}{\dot{b}(z_{j}-\theta_{n}^{\prime})}{T^{(1)}%
}_{\underline{\beta}^{\prime}\,\beta^{\prime\prime}}^{\underline{\beta}%
\,\beta^{\prime}}(\underline{z},\theta_{1}^{\prime}){D_{Q}}_{\beta}%
^{\beta^{\prime\prime}}({\underline{\theta}})\nonumber\\
&  \times C^{\beta_{m}^{\prime}}({\underline{\theta}},z_{m})\cdots
C^{\beta_{1}^{\prime}}({\underline{\theta}},z_{1})+\sum uw_{D}\,. \label{A.6}%
\end{align}
The ``wanted terms'' written explicitly originate from the first term in the
commutations rules (\ref{A.4}); all other contributions yield the so-called
``unwanted terms''. The next level monodromy matrix is
\[
{T^{(1)}}_{\underline{\beta}\,\beta}^{\underline{\beta^{\prime}}%
\,\beta^{\prime}}(\underline{z},\theta)=\left(  \dot{S}_{10}(z_{1}%
-\theta)\cdots\dot{S}_{m0}(z_{m}-\theta)\right)  _{\underline{\beta}\,\beta
}^{\underline{\beta^{\prime}}\,\beta^{\prime}}
\]
where the $\beta$'s and also the internal summation indices take the values
$2,\dots,N$. If we insert these equations into the representation (\ref{4.1})
of $f({\underline{\theta}})$ we find that the desired contribution from
$A_{Q}$ already gives the result we are looking for. The wanted contribution
from $D_{Q}$ applied to $\Omega$ gives zero. The unwanted contributions from
$A_{Q}$ and $D_{Q}$ cancel after integration over the $z_{j}$.

All these three facts can be seen as follows. We have
\begin{equation}
\Omega\,A_{Q}({\underline{\theta}})=\prod_{i=1}^{n}\dot{a}(\theta_{i}%
-\theta_{1})\,\Omega~,~~~\Omega\,{D_{Q}}_{\beta}^{\beta^{\prime}}%
({\underline{\theta}})=\delta_{\beta}^{\beta^{\prime}}\prod_{i=1}^{n}\dot
{b}(\theta_{i}-\theta_{1})\Omega=0 \label{A.6a}%
\end{equation}
which follow from eq.~(\ref{3.5}) and $b(0)=0$. By $\dot{a}(o)=-1$ and
asumption $(ii^{\prime})$
\[
\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot
{a}(\theta_{1}-z_{j})}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,h(\underline
{\theta},{\underline{z}})=\sigma h(\underline{\theta^{\prime}},{\underline{z}%
})
\]
the wanted term from $A_{Q}$ yields $\sigma f(\underline{\theta}^{\prime})$:
\begin{align*}
\left[  f({\underline{\theta}})\,A_{Q}({\underline{\theta}})\right]  ^{w}  &
=\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline{z}\,\prod
_{i=1}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{j=1}^{m}\frac{\dot{a}%
(\theta_{1}-z_{j})}{\dot{b}(\theta_{1}^{\prime}-z_{j})}\,h({\underline{\theta
}},{\underline{z}})g_{\underline{\beta}}({\underline{z}})\,\Psi^{\underline
{\beta}}({\underline{\theta}}^{\prime},{\underline{z}})\\
&  =\sigma\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline
{z}\,h({\underline{\theta}}^{\prime},{\underline{z}})g_{\underline{\beta}%
}({\underline{z}})\,\Psi^{\underline{\beta}}({\underline{\theta}}^{\prime
},{\underline{z}})\\
&  =\sigma f({\underline{\theta}}^{\prime})
\end{align*}
Similarly the wanted contribution from $D^{Q}$ vanish, since $\dot
{b}(0)=0\Rightarrow\Omega\,D^{Q}({\underline{\theta}})=0$ . Therefore it
remains to prove that the unwanted terms cancel.

Next we discuss the `unwanted terms'. The commutation relations (\ref{A.3})
and (\ref{A.4}) imply that the unwanted terms are proportional to a product of
$C$-operators, where one $C({\underline{\theta}},z_{j})$ is replaced by
$C_{Q}({\underline{\theta}})$. Because of the symmetry $(i)$ of $g_{\underline
{\beta}}({\underline{z}})$ it is sufficient to consider only the unwanted
terms for $j=1$ which are denoted by $uw_{A}^{1}$ and $uw_{D}^{1}$. They
originate from the second term in the commutation rules (\ref{A.4}) when
$A_{Q}({\underline{\theta}})$ is commuted with $C({\underline{\theta}},z_{1}%
)$. Then the resulting $A({\underline{\theta}},z_{1})$ pushed through the
other $C$'s and taking only the first terms in (\ref{A.3}) into account we
arrive at
\[
uw_{A}^{1}({\underline{z}})=-\frac{\dot{c}(\theta_{1}^{\prime}-z_{1})}{\dot
{b}(\theta_{1}^{\prime}-z_{1})}\,\prod_{1<j}\frac{\dot{a}(z_{1}-z_{j})}%
{\dot{b}(z_{1}-z_{j})}\,\Omega\,A({\underline{\theta}},z_{1})\,C^{\beta_{m}%
}({\underline{\theta}},z_{1})\dots C_{Q}^{\beta_{1}}({\underline{\theta}})
\]


Using (\ref{3.5}) and Yang-Baxter relations one obtains (taking into account
always the `$SU(N)$-ice rule')
\begin{gather*}
\left[  \Omega\,A({\underline{\theta}},z_{1})\,C^{\beta_{m}}({\underline
{\theta}},z_{1})\dots C_{Q}^{\beta_{1}}({\underline{\theta}})\right]
_{\underline{\alpha}}=\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\,\prod
_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod_{j=2}^{m}\dot{a}(\theta
_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline
{\tilde{\alpha}}}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}%
},\underline{\tilde{z}})\\%
\begin{array}
[c]{l}%
\unitlength4mm\begin{picture}(12,8) \put(11,7){\oval(19,4)[lb]} \put(11,7){\oval(21,7)[lb]} \put(3,7){\oval(6,10)[lb]} \put(3,6){\line(1,0){8}} \put(5,1){\line(0,1){6}} \put(10,1){\line(0,1){6}} \put(11,7){\oval(14,10)[lb]} \put(3,1){\oval(2,2)[rt]} \put(-.7,7.4){$\beta_1$} \put(.2,7.4){$\beta_2$} \put(1.2,7.4){$\beta_m$} \put(3.8,7.3){1} \put(4.8,7.3){1} \put(9.8,7.3){1} \put(3.8,0){$\alpha_1$} \put(4.8,0){$\alpha_2$} \put(9.8,0){$\alpha_n$} \put(2.4,5.8){1} \put(11.2,1.8){1} \put(11.2,3.3){1} \put(11.2,4.8){1} \put(11.2,5.8){1} \put(5.3,1){$\theta_{2}$} \put(10.2,1){$\theta_{n}$} \put(8.5,3.8){$z_2$} \put(8.5,6.3){$z_1$} \put(8.5,5.3){$z_m$} \put(6.8,4.3){$\dots$} \put(10.5,3.9){$\vdots$} \put(8.5,2.2){$\theta_1$} \end{picture}
\end{array}
=\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\,
\begin{array}
[c]{l}%
\unitlength4mm\begin{picture}(12,8) \put(11,6){\oval(19,4)[lb]} \put(11,6){\oval(21,7)[lb]} \put(2,6){\oval(4,10)[lb]} \put(9,6){\oval(12,2)[lb]} \put(9,2){\oval(2,6)[rt]} \put(11,2){\oval(2,2)[lb]} \put(4,0){\line(0,1){6}} \put(9,0){\line(0,1){6}} \put(2,0){\oval(2,2)[rt]} \put(-.7,6.4){$\beta_1$} \put(.2,6.4){$\beta_2$} \put(1.2,6.4){$\beta_m$} \put(2.8,6.3){1} \put(3.8,6.3){1} \put(8.8,6.3){1} \put(2.8,-1){$\alpha_1$} \put(3.8,-1){$\alpha_2$} \put(8.8,-1){$\alpha_n$} \put(11.2,.8){1} \put(11.2,2.3){1} \put(11.2,3.8){1} \put(4.3,0){$\theta_{2}$} \put(9.2,0){$\theta_{n}$} \put(7.5,2.8){$z_2$} \put(7.5,4.3){$z_m$} \put(5.8,3.3){$\dots$} \put(9.5,2.9){$\vdots$} \put(7.5,5.3){$\theta_1$} \end{picture}
\end{array}
\end{gather*}
where $\tilde{\Psi}_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}%
}(\underline{\tilde{\theta}},\underline{\tilde{z}})$ does not depend on
$\alpha_{1},\beta_{1},\theta_{1},z_{1}$. Therefore finally
\begin{align*}
\left[  f({\underline{\theta}})\,A_{Q}({\underline{\theta}})\right]
_{\underline{\alpha}}^{uw}  &  =-\int_{\mathcal{C}_{{\underline{\theta}%
}^{\prime}}}d\underline{z}\,h({\underline{\theta}},{\underline{z}%
})\,g_{\underline{\beta}}({\underline{z}})\frac{\dot{c}(\theta_{1}^{\prime
}-z_{1})}{\dot{b}(\theta_{1}^{\prime}-z_{1})}\,\prod_{j=2}^{m}\frac{\dot
{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\\
&  \times\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\,\prod_{i=2}^{n}\dot
{a}(\theta_{i}-\theta_{1})\prod_{j=2}^{m}\dot{a}(\theta_{1}-z_{j}%
)\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline{\tilde{\alpha}}%
}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}},\underline{\tilde{z}%
})\,.
\end{align*}


Similarly we obtain the first unwanted contribution from $\sum_{\beta}{D_{Q}%
}_{\beta}^{\beta}$ as
\[
uw_{D}^{1}=-\frac{\dot{c}(z_{1}-\theta_{1})}{\dot{b}(z_{1}-\theta_{1})}%
\prod_{j=2}^{m}\frac{1}{\dot{b}(z_{j}-z_{1})}\,\left(  -{T^{(1)}}%
_{\underline{\beta}^{\prime},\beta^{\prime\prime}}^{\underline{\beta},\beta
}(\underline{z},z_{1})\right)  D_{\beta}^{\beta^{\prime\prime}}({\underline
{\theta},z}_{1})C^{\beta_{m}^{\prime}}({\underline{\theta}},z_{m})\cdots
C_{Q}^{\beta_{1}^{\prime}}({\underline{\theta}})
\]
which may be depicted as
\[
\left(  -{T^{(1)}}_{\underline{\beta}^{\prime},\beta^{\prime\prime}%
}^{\underline{\beta},\beta}(\underline{z},z_{1})\right)  \Omega D_{\beta
}^{\beta^{\prime\prime}}({\underline{\theta},z}_{1})C^{\beta_{m}^{\prime}%
}({\underline{\theta}},z_{m})\cdots C_{Q}^{\beta_{1}^{\prime}}({\underline
{\theta}})=-\quad%
\begin{array}
[c]{l}%
\unitlength4mm\begin{picture}(12,8) \put(11,7){\oval(17,4)[lb]} \put(11,7){\oval(20,7)[lb]} \put(3,5){\oval(6,6)[lb]} \put(1,7){\oval(2,2)[lb]} \put(-1,5){\oval(2,2)[rt]} \put(1,6){\line(1,0){10}} \put(5,1){\line(0,1){6}} \put(10,1){\line(0,1){6}} \put(11,7){\oval(14,10)[lb]} \put(3,1){\oval(2,2)[rt]} \put(-.5,7.4){$\beta_1$} \put(.8,7.4){$\beta_2$} \put(2.2,7.4){$\beta_m$} \put(3.8,7.3){1} \put(4.8,7.3){1} \put(9.8,7.3){1} \put(3.8,0){$\alpha_1$} \put(4.8,0){$\alpha_2$} \put(9.8,0){$\alpha_n$} \put(11.2,1.8){1} \put(11.2,3.3){1} \put(11.2,4.8){1} \put(11.2,5.8){$\beta$} \put(-2,5.8){$\beta$} \put(5.3,1){$\theta_{2}$} \put(10.2,1){$\theta_{n}$} \put(8.5,3.8){$z_2$} \put(8.5,6.3){$z_1$} \put(8.5,5.3){$z_m$} \put(7,4.3){$\dots$} \put(10.5,3.9){$\vdots$} \put(8.5,2.2){$\theta_1$} \end{picture}
\end{array}
.
\]
Again (\ref{3.5}) and Yang-Baxter relations yield
\begin{multline*}
\left[  \Omega\,D_{\beta}^{\beta^{\prime\prime}}({\underline{\theta}}%
,z_{1})\,C^{\beta_{m}^{\prime}}({\underline{\theta}},z_{m})\dots C_{Q}%
^{\beta_{1}^{\prime}}({\underline{\theta}})\right]  _{\underline{\alpha}}\\
=\delta_{\beta}^{\beta^{\prime\prime}}\prod_{i=1}^{n}\left(  \dot{b}%
(\theta_{i}-z_{1})\,\dot{a}(\theta_{i}-\theta_{1})\right)  \prod_{j=2}^{m}%
\dot{a}(\theta_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}^{\prime}}\tilde
{\Psi}_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}^{\prime}%
}(\underline{\tilde{\theta}},\underline{\tilde{z}})
\end{multline*}
and assumption $(ii)$ for $g_{\underline{\beta}}(\underline{z})$ means
\[
g_{\underline{\beta}}(\underline{z}){Q^{(1)}}_{\underline{\beta}^{\prime}%
}^{\underline{\beta}}(\underline{z})=g_{\underline{\beta}}(\underline
{z}^{\prime})
\]
where analogously to eq. (\ref{A.2}) ${Q^{(1)}}_{\underline{\beta}^{\prime}%
}^{\underline{\beta}}(\underline{z})=-{T^{(1)}}_{\underline{\beta}^{\prime
}\beta}^{\underline{\beta}\beta}(\underline{z},z_{1})$ is the next level
$Q$-matrix and $\underline{z}^{\prime}=(z_{1}^{\prime}=z_{1}+2\pi
i,\dots,z_{m})$. Therefore using the relations of $h(\underline{\theta
},{\underline{z}})$ and $h(\underline{\theta},{\underline{z}}^{\prime})$ given
by assumtion $(ii^{\prime})$ we obtain
\begin{align*}
\left[  f({\underline{\theta}})\,{D_{Q}}_{\beta}^{\beta}({\underline{\theta}%
})\right]  _{\underline{\alpha}}^{uw}  &  =-\int_{\mathcal{C}_{{\underline
{\theta}}}}d\underline{z}\,h({\underline{\theta}},{\underline{z}}^{\prime
})\,g_{\underline{\beta}}({\underline{z}}^{\prime})\frac{\dot{c}(z_{1}%
-\theta_{1})}{\dot{b}(z_{1}-\theta_{1})}\prod_{j=2}^{m}\frac{\dot{a}%
(z_{1}^{\prime}-z_{j})}{\dot{b}(z_{1}^{\prime}-z_{j})}\\
&  \times\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1}^{\prime})\prod_{i=2}%
^{n}\,\dot{a}(\theta_{i}-\theta_{1})\prod_{j=2}^{m}\dot{a}(\theta_{1}%
-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline{\tilde
{\alpha}}}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}}%
,\underline{\tilde{z}})
\end{align*}
and rewrite the $z_{1}$-integral by replacing $z_{1}\rightarrow z_{1}-2\pi i$
such that $z_{1}^{\prime}\rightarrow z_{1}$ and at the same time the contour
$\mathcal{C}_{{\underline{\theta}}}\rightarrow\mathcal{C}_{{\underline{\theta
}}}+2\pi i$.

Finally we obtain for the sum of the unwanted term from $A_{Q}$ and $D_{Q}$
\begin{multline*}
\left[  f({\underline{\theta}})\,\left(  A_{Q}({\underline{\theta}}%
)+D_{Q}({\underline{\theta}})\right)  \right]  _{\underline{\alpha}}%
^{uw}=\prod_{i=2}^{n}\dot{a}(\theta_{i}-\theta_{1})\left(  \int_{\mathcal{C}%
_{{\underline{\theta}}^{\prime}}}d\underline{z}-\int_{\mathcal{C}%
_{{\underline{\theta}}}+2\pi i}dz_{1}\int_{\mathcal{C}_{{\underline{\theta}}}%
}d\underline{\tilde{z}}\,\right)  \frac{\dot{c}(z_{1}-\theta_{1}^{\prime}%
)}{\dot{b}(z_{1}-\theta_{1}^{\prime})}\\
\times h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}%
}({\underline{z}})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}%
(z_{1}-z_{j})}\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\prod_{j=2}^{m}\dot
{a}(\theta_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}%
_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}}(\underline
{\tilde{\theta}},\underline{\tilde{z}})\,.
\end{multline*}
Using $\frac{\dot{c}(z)}{\dot{b}(z)}=-\frac{2\pi i/N}{z}$ and the technical
assumptions (a) -- (d) we may replace for the $A_{Q}$ -term
\[
\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline{z}\cdots=\left(
-\operatorname*{Res}_{z_{1}=\theta_{1}^{\prime}}+\int_{\mathcal{C}%
_{{\underline{\theta}}}}dz_{1}\right)  \int_{\mathcal{C}_{{\underline{\theta}%
}}}d\underline{\tilde{z}}\cdots
\]
and for the $D_{Q}$-term
\[
\int_{\mathcal{C}_{{\underline{\theta}}}+2\pi i}dz_{1}\int_{\mathcal{C}%
_{{\underline{\theta}}}}d\underline{\tilde{z}}\,\cdots=\left(
-\operatorname*{Res}_{z_{1}=\theta_{1}^{\prime}}+\int_{\mathcal{C}%
_{{\underline{\theta}}}}dz_{1}\right)  \int_{\mathcal{C}_{{\underline{\theta}%
}}}d\underline{\tilde{z}}\cdots
\]
which proves the cancelation of the unwanted terms.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof
%EndExpansion


\paragraph{Technical assumtions for theorem \ref{t1}}

Further we need more technical assumptions on the contour $\mathcal{C}%
_{\underline{\theta}}$ and on the analytic behavior of $h(\underline{\theta
},{\underline{z}})$ and $g_{\underline{\beta}}(\underline{z})$. The following
integrals along closed contours have to vanish

\begin{itemize}
\item[(a)] \label{A.7}
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}}}}-\int_{\mathcal{C}%
_{{\underline{\theta}}^{\prime}}}\right)  d\underline{z}\,h({\underline
{\theta}},{\underline{z}})g_{\underline{\beta}}({\underline{z}})\prod
_{j=1}^{m}\dot{a}(\theta_{1}-z_{j})=0
\]
where $\mathcal{C}_{\underline{\theta}^{\prime}}$ is the contour corresponding
to $\underline{\theta}^{\prime}$ for $\underline{\theta}^{\prime}=(\theta
_{1}^{\prime}=\theta_{1}+2\pi i,\dots,\theta_{n})$;

\item[(b)] and more general
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline{z}%
-\int_{\mathcal{C}_{{\underline{\theta}}}}d\underline{z}\right)
h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}}({\underline{z}%
})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\prod
_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\prod_{j=2}^{m}\dot{a}(\theta_{1}%
-z_{j})=0\,;
\]


\item[(c)] further more
\[
\left[  \left(  \int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}%
-\int_{\mathcal{C}_{{\underline{\theta}}}}\right)  d\underline{\tilde{z}%
}h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}}({\underline
{z}})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}%
\prod_{j=2}^{m}\dot{a}(\theta_{1}-z_{j})\right]  _{z_{1}=\theta_{1}^{\prime}%
}=0
\]
where $\underline{\tilde{z}}=z_{2},\dots,z_{m}$;

\item[(d)] and
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}}}+2\pi i}-\int_{\mathcal{C}%
_{{\underline{\theta}}}}\right)  dz_{1}\,h({\underline{\theta}},{\underline
{z}})\,g_{\underline{\beta}}({\underline{z}})\prod_{j=2}^{m}\frac{\dot
{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\prod_{i=1}^{n}\dot{a}(\theta_{i}%
-z_{1})=0
\]
where $\mathcal{C}_{{\underline{\theta}}}+2\pi i$ is the contour
$\mathcal{C}_{{\underline{\theta}}}$ shifted by $2\pi i$.
\end{itemize}

\setcounter{equation}{0}

\section{Proof of Theorem \ref{t3}}

\label{a2}%

%TCIMACRO{\TeXButton{Proof}{\proof}}%
%BeginExpansion
\proof
%EndExpansion
Property (i) follows as above in appendix \ref{a1}; in addition one uses
$\Omega_{\dots ji\dots}\,\dot{S}_{i\bar{j}}(\theta_{ij})\,$ directly from the
Yang-Baxter equations, the action of the S-matrix on the pseudo-ground state
$\Omega$ and $(i^{\prime})$%
\begin{align*}
\,\Omega_{\dots ji\dots}\,C_{\dots ji\dots}({\dots\theta_{j},\theta_{i}\dots
})\,\dot{S}_{ij}(\theta_{ij})  &  =\Omega_{\dots ji\dots}\,\dot{S}_{ij}%
(\theta_{ij})\,C_{\dots ij\dots}({\dots\theta_{i},\theta_{j}\dots})\\
&  =\dot{a}(\theta_{ij})\,\Omega_{\dots ij\dots}\,C_{\dots ij\dots}%
({\dots\theta_{i},\theta_{j}\dots})\,.
\end{align*}


Using (i) the property (ii) for a particle state may be again rewritten as the
difference equation
\begin{equation}
f_{I}({\underline{\theta}}^{\prime},\underline{\bar{\theta}})=\sigma
f_{I}({\underline{\theta}},\underline{\bar{\theta}})\,Q_{I}({\underline
{\theta},\underline{\bar{\theta}},}i) \label{B.1}%
\end{equation}
if where ${\underline{\theta}}^{\prime}=(\theta_{1},\dots,\theta_{i}^{\prime
}=\theta_{i}+2\pi i,\dots,\theta_{n})$ and $\sigma$ is a sign which depends on
the statistics and the particle $i$. The operator $Q({\underline{\theta}},i)$
is defined by
\begin{gather*}
Q_{I}({\underline{\theta}},\underline{\bar{\theta}},i)=\mathrm{tr}%
_{0}T_{Q,I,0}({\underline{\theta},\underline{\bar{\theta}},i})\\
Q_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\alpha}^{\prime
}\,\underline{\bar{\alpha}}^{\prime}}({\underline{\theta}},i)=\sum_{\alpha}
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(10,4) \put(1,1){\oval(6,2)[rt]} \put(11,3){\oval(14,2)[lb]} \put(1.2,2){\vector(-1,0){0}} \put(10.6,2){\vector(-1,0){0}} \put(2,1){\vector(0,1){2}} \put(8,3){\vector(0,-1){2}} \put(10,3){\vector(0,-1){2}} \put(6,1){\vector(0,1){2}} \put(1.9,.5){$\alpha_1$} \put(1.9,3.3){$\alpha'_1$} \put(3.9,.5){$\alpha_i$} \put(3.9,3.3){$\alpha'_i$} \put(5.9,.5){$\alpha_n$} \put(5.9,3.3){$\alpha'_n$} \put(6.2,1.1){$\theta_n$} \put(.3,1.8){$\alpha$} \put(11.3,1.8){$\alpha$} \put(1.2,1.1){$\theta_1$} \put(4.1,1.1){$\theta'_i$} \put(3.4,2.4){$\theta_i$} \put(7.9,.5){$\bar\alpha_1$} \put(7.9,3.3){$\bar\alpha'_1$} \put(8.2,1.1){$\bar\theta_1$} \put(9.9,.5){$\bar\alpha_{\bar n}$} \put(9.9,3.3){$\bar\alpha'_{\bar n}$} \put(10.2,1.1){$\bar\theta_{\bar n}$} \put(2.3,2.5){\dots} \put(8.6,2.5){\dots} \put(4.6,2.5){\dots} \end{picture}
\end{array}
.
\end{gather*}
with
\[
T_{Q,I,0}({\underline{\theta},\underline{\bar{\theta}},}i)=\dot{S}_{10}%
(\theta_{1}-\theta_{i}^{\prime})\,\cdots\mathbf{P}_{i0}\cdots\dot{S}%
_{n0}(\theta_{n}-\theta_{i})\dot{S}_{\bar{1}0}(\bar{\theta}_{1}-\theta
_{i})\cdots\dot{S}_{\bar{n}0}(\bar{\theta}_{\bar{n}}-\theta_{i})\,.
\]
The proof of the difference equation (\ref{B.1}) is the analogous to that for
the pure particle case. In eq. (\ref{A.6a}) $\prod_{i=2}^{n}\dot{a}(\theta
_{i}-\theta_{1})$ has to be replaced by $\prod_{i=2}^{n}\dot{a}(\theta
_{i}-\theta_{1})\prod_{i=1}^{\bar{n}}\hat{b}(\bar{\theta}_{i}-\theta_{1})$
which corresponds to assumption $(ii^{\prime})$.

The property (ii) for an anti-particle state may be rewritten as the
difference equation
\begin{equation}
f_{I}({\underline{\theta}},\underline{\bar{\theta}})=\sigma f_{I}%
({\underline{\theta}},\underline{\bar{\theta}}^{\prime})\,\tilde{Q}%
_{I}({\underline{\theta},\underline{\bar{\theta}},}i) \label{B.2}%
\end{equation}
if where ${\underline{\bar{\theta}}}^{\prime}=(\bar{\theta}_{1},\dots
,\bar{\theta}_{i}^{\prime}=\bar{\theta}_{i}+2\pi i,\dots,\bar{\theta}_{n})$.
The operator $\tilde{Q}_{I}({\underline{\theta},\underline{\bar{\theta}},}%
\bar{\imath})$ is the inverse shift operator and is defined by
\begin{gather*}
\tilde{Q}_{I}({\underline{\theta},\underline{\bar{\theta}},}i)=\mathrm{tr}%
_{0}\tilde{T}_{Q,I,0}({\underline{\theta},\underline{\bar{\theta}},}i)\\
\tilde{Q}_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\alpha
}^{\prime}\,\underline{\bar{\alpha}}^{\prime}}({\underline{\theta}}%
,i)=\sum_{\alpha}
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(10,4) \put(1,3){\oval(14,2)[rb]} \put(11,1){\oval(6,2)[lt]} \put(1.2,2){\vector(-1,0){0}} \put(10.6,2){\vector(-1,0){0}} \put(2,1){\vector(0,1){2}} \put(4,1){\vector(0,1){2}} \put(6,3){\vector(0,-1){2}} \put(10,3){\vector(0,-1){2}} \put(1.9,.5){$\alpha_1$} \put(1.9,3.3){$\alpha'_1$} \put(2.2,1.1){$\theta_1$} \put(3.9,.5){$\alpha_n$} \put(3.9,3.3){$\alpha'_n$} \put(4.2,1.1){$\theta_n$} \put(.3,1.8){$\alpha$} \put(11.3,1.8){$\alpha$} \put(5.9,.5){$\bar\alpha_1$} \put(5.9,3.3){$\bar\alpha'_1$} \put(6.2,1.1){$\bar\theta_1$} \put(7.9,.5){$\bar\alpha_i$} \put(7.9,3.3){$\bar\alpha'_i$} \put(7.3,2.6){$\bar\theta'_i$} \put(8.4,1.1){$\bar\theta_i$} \put(1,2.3){$\bar\theta''_i$} \put(10.4,2.3){$\bar\theta'''_i$} \put(9.9,.5){$\bar\alpha_{\bar n}$} \put(9.9,3.3){$\bar\alpha'_{\bar n}$} \put(10.2,1.1){$\bar\theta_{\bar n}$} \put(2.3,2.5){\dots} \put(8.6,2.5){\dots} \put(4.6,2.5){\dots} \end{picture}
\end{array}
.
\end{gather*}
with
\[
\tilde{T}_{Q,I,0}({\underline{\theta},\underline{\bar{\theta}},}i)=\dot
{S}_{10}(\theta_{1}-\bar{\theta}_{i}^{\prime\prime})\,\cdots\dot{S}%
_{n0}(\theta_{n}-\bar{\theta}_{i}^{\prime\prime})\dot{S}_{\bar{1}0}%
(\bar{\theta}_{1}-\bar{\theta}_{i}^{\prime\prime})\cdots\mathbf{K}%
_{\bar{\imath}0}\cdots\dot{S}_{\bar{n}0}(\bar{\theta}_{\bar{n}}-\bar{\theta
}_{i}^{\prime\prime\prime})\,
\]
where $\bar{\theta}_{i}^{\prime\prime}=\bar{\theta}_{i}+i\pi=\bar{\theta}%
_{i}^{\prime}-i\pi$ and $\bar{\theta}_{i}^{\prime\prime\prime}=\bar{\theta
}_{i}-i\pi$. Note that
\[
\dot{S}_{\bar{\imath}0}(i\pi)=\mathbf{K}_{\bar{\imath}0}=~~
\begin{array}
[c]{c}%
\unitlength3mm\begin{picture}(2,2) \put(0,1){\vector(-1,0){0}} \put(1,0){\vector(0,-1){0}}\put(0,2){\oval(2,2)[br]} \put(2,0){\oval(2,2)[tl]} \end{picture}
\end{array}
\]
is the annihilation-creation operator. Because of the symmetry $(i)$ it is
sufficient to discuss the difference equation for $i=\bar{n}.$ For this case
the matrix $\tilde{Q}({\underline{\theta}},{\underline{\bar{\theta}},}\bar
{n})=\tilde{Q}({\underline{\theta}},{\underline{\bar{\theta}}})$ may be
written as a trace of the ordinary monodromy matrix (\ref{3.6}) over the
auxiliary space for the specific value of the spectral parameter $\theta
_{0}=\bar{\theta}_{\bar{n}}^{\prime\prime\prime}=\bar{\theta}_{\bar{n}%
}^{\prime}-i\pi$
\begin{equation}
\tilde{Q}_{I}({\underline{\theta}},{\underline{\bar{\theta}}})=\mathrm{tr}%
_{0}T_{I,0}({\underline{\theta},\underline{\bar{\theta}}}^{\prime}{,}%
\bar{\theta}_{\bar{n}}^{\prime\prime\prime}=\bar{\theta}_{\bar{n}}^{\prime
}-i\pi). \label{B.3}%
\end{equation}
For $i=\bar{n}$ the difference equation may be depicted as
\[
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(5,4) \put(2.5,2){\oval(5,2)} \put(2.5,2){\makebox(0,0){$f$}} \put(1,0){\line(0,1){1}} \put(3,0){\line(0,1){1}} \put(4,0){\line(0,1){1}} \put(1.4,.5){$\dots$} \put(1,.6){\vector(0,1){0}} \put(3,.2){\vector(0,-1){0}} \put(4,.0){\vector(0,-1){0}} \end{picture}
\end{array}
~~=\sigma~~
\begin{array}
[c]{c}%
\unitlength4mm\begin{picture}(7,5) \put(2.5,2){\oval(5,2)[b]} \put(3.5,2){\oval(7,6)[t]} \put(6,2){\oval(2,2)[rb]} \put(6,0){\oval(2,2)[lt]} \put(3.5,3){\oval(5,2)} \put(3.5,3){\makebox(0,0){$f$}} \put(2,0){\line(0,1){2}} \put(4,0){\line(0,1){2}} \put(2,.6){\vector(0,1){0}} \put(4,.2){\vector(0,-1){0}} \put(5,.0){\vector(0,-1){0}} \put(1,1){\vector(-1,0){0}} \put(2.4,.5){$\dots$} \end{picture}
\end{array}
\]
where we use the rule that the rapidity of a line changes by $2\pi i$ if the
line bends by $360^{0}$ in the positive sense.

In the following we will suppress the index $I$. The Yang-Baxter relations
again imply the commutation rules (\ref{A.3}) for the matrices $A,C,D$ defined
in eq.~(\ref{3.7}). These commutation rules are the same as those for the pure
particle case. Also the commutation rules for the matrices $A_{Q},C_{Q},D_{Q}$
are the same. Here exist the additional rules for the matrices $\tilde{A}%
_{Q},\tilde{C}_{Q},\tilde{D}_{Q}$ defined as usual.
\begin{align}
C^{\beta}({\underline{\theta}},{\underline{\bar{\theta}}}^{\prime},z)\tilde
{A}_{Q}({\underline{\theta}},{\underline{\bar{\theta}}})  &  =\frac{\dot
{a}(\bar{\theta}_{\bar{n}}^{\prime\prime\prime}-z)}{\dot{b}(\bar{\theta}%
_{\bar{n}}^{\prime\prime}-z)}\tilde{A}_{Q}({\underline{\theta}},{\underline
{\bar{\theta}}})C({\underline{\theta}},{\underline{\bar{\theta}}}%
,z)-\frac{\dot{c}(\bar{\theta}_{\bar{n}}^{\prime\prime}-z)}{\dot{b}%
(\bar{\theta}_{\bar{n}}^{\prime\prime}-z)}A({\underline{\theta}}%
,{\underline{\bar{\theta}}}^{\prime},z)\tilde{C}_{Q}({\underline{\theta}%
},{\underline{\bar{\theta}}})\label{B.4}\\
C^{\beta}({\underline{\theta}},{\underline{\bar{\theta}}}^{\prime},z)\tilde
{D}_{Q\,\gamma}^{~~\gamma^{\prime}}({\underline{\theta}},{\underline
{\bar{\theta}}})  &  =\frac{1}{\dot{b}(z-\bar{\theta}_{\bar{n}}^{\prime
\prime\prime})}\dot{S}_{\beta^{\prime}\gamma^{\prime\prime}}^{\gamma^{\prime
}\beta}(z-\bar{\theta}_{\bar{n}}^{\prime\prime})\tilde{D}_{Q\,\gamma
}^{~~\gamma^{\prime\prime}}({\underline{\theta}},{\underline{\bar{\theta}}%
})C^{\beta^{\prime}}({\underline{\theta}},{\underline{\bar{\theta}}%
},z)\nonumber\\
&  -\frac{\dot{c}(z-\bar{\theta}_{\bar{n}}^{\prime\prime\prime})}{\dot
{b}(z-\bar{\theta}_{\bar{n}}^{\prime\prime\prime})}D_{\gamma}^{\beta
}({\underline{\theta}},{\underline{\bar{\theta}}}^{\prime},z)\tilde{C}%
_{Q}^{\gamma^{\prime}}({\underline{\theta}},{\underline{\bar{\theta}}%
})\,.\nonumber
\end{align}
Note that we assign to the auxiliary space of $\tilde{T}_{Q}({\underline
{\theta}},{\underline{\bar{\theta}}})$ corresponding to the horizontal line
the spectral parameter $\bar{\theta}_{\bar{n}}^{\prime\prime\prime}%
=\bar{\theta}_{\bar{n}}-i\pi$ on the right hand side and $\bar{\theta}%
_{\bar{n}}^{\prime\prime}=\bar{\theta}_{\bar{n}}^{\prime}-i\pi=\bar{\theta
}_{\bar{n}}+i\pi$ on the left hand side. To analyze the right hand side of
eq.~(\ref{B.2}) we proceed as follows: We apply the trace of $\tilde{T}%
_{Q}({\underline{\theta},\underline{\bar{\theta}}})$ to the co-vector
$f({\underline{\theta},\underline{\bar{\theta}}}^{\prime})$ as given by
eq.~(\ref{4.10}) and the Bethe ansatz (\ref{3.8}).
\[
\Omega\,C^{\beta_{m}}({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},z_{m})\cdots C^{\beta_{1}}({\underline{\theta},\underline{\bar{\theta}}%
}^{\prime},z_{1})\,\tilde{T}_{Q\,\gamma}^{~~\gamma^{\prime}}({\underline
{\theta},\underline{\bar{\theta}}})=
\begin{array}
[c]{c}%
\unitlength5mm\begin{picture}(12,5.5) \put(11,5){\oval(19,2)[lb]} \put(11,5){\oval(22,6)[lb]} \put(3,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,5){\vector(0,-1){5}} \put(0,5){\vector(0,1){0}} \put(1.5,5){\vector(0,1){0}} \put(2,1){\vector(-1,0){0}} \put(11,1){\vector(-1,0){0}} \put(2,5){\oval(16,8)[rb]} \put(11,0){\oval(2,2)[lt]} \put(-.5,5.3){$\beta_1$} \put(1,5.3){$\beta_m$} \put(2.8,5.3){1} \put(5.8,5.3){1} \put(6.8,5.3){$\bar\beta_1$} \put(9.8,5.3){$\bar\beta_{\bar n}$} \put(1.4,.8){$\gamma'$} \put(11.4,.8){$\gamma$} \put(11.2,1.8){1} \put(11.2,3.8){1} \put(4.3,0.1){$\bar\theta''_{\bar n}$} \put(10.2,2.5){$\bar\theta'_{\bar n}$} \put(9.1,0){$\bar\theta_{\bar n}$} \put(7.5,2.3){$z_1$} \put(7.5,4.3){$z_m$} \put(4,3){$\dots$} \put(8,3){$\dots$} \put(9.5,2.7){$\vdots$} \put(2,0){$\theta_1$} \end{picture}
\end{array}
.
\]
We now proceed as usual in the algebraic Bethe ansatz and push the $\tilde
{A}_{Q}({\underline{\theta},\underline{\bar{\theta}}})$ and $\tilde{D}%
_{Q}({\underline{\theta},\underline{\bar{\theta}}})$ through all the
C-operators using the commutation rules (\ref{B.4}) and obtain
\begin{align}
C^{\beta_{m}}({\underline{\theta},\underline{\bar{\theta}}}^{\prime}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta},\underline{\bar{\theta}}%
}^{\prime},z_{1})\,\tilde{A}_{Q}({\underline{\theta},\underline{\bar{\theta}}%
})  &  =\prod_{j=1}^{m}\,\frac{\dot{a}(\bar{\theta}_{\bar{n}}^{\prime
\prime\prime}-z_{j})}{\dot{b}(\bar{\theta}_{\bar{n}}^{\prime\prime}-z_{j}%
)}\tilde{A}_{Q}({\underline{\theta},\underline{\bar{\theta}}})C^{\beta_{m}%
}({\underline{\theta},\underline{\bar{\theta}}},z_{m})\cdots C^{\beta_{1}%
}({\underline{\theta},\underline{\bar{\theta}}},z_{1})\nonumber\\
&  +\sum uw_{A}\,, \label{B.5}%
\end{align}
\begin{align}
C^{\beta_{m}}({\underline{\theta},\underline{\bar{\theta}}}^{\prime}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta},\underline{\bar{\theta}}%
}^{\prime},z_{1})\,\tilde{D}_{Q\,\beta}^{~~\beta^{\prime}}({\underline{\theta
},\underline{\bar{\theta}}})  &  =\prod_{j=1}^{m}\frac{1}{\dot{b}(z_{j}%
-\theta_{\bar{n}}^{\prime\prime\prime})}{T^{(1)}}_{\underline{\beta}^{\prime
}\underline{\bar{\beta}}^{\prime}\beta^{\prime\prime}}^{\underline{\beta
}\,\underline{\bar{\beta}}\beta^{\prime}}(\underline{z}{,\underline
{\bar{\theta}}},\bar{\theta}_{\bar{n}}^{\prime\prime})\tilde{D}_{Q\,\beta
}^{~~\beta^{\prime\prime}}({\underline{\theta},\underline{\bar{\theta}}%
})\nonumber\\
&  \times C^{\beta_{m}}({\underline{\theta},\underline{\bar{\theta}}}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta},\underline{\bar{\theta}}%
},z_{1}+\sum uw_{D}\,. \label{B.6}%
\end{align}
The ``wanted terms'' written explicitly originate from the first term in the
commutations rules (\ref{B.4}); all other contributions yield the so-called
``unwanted terms''. The next level monodromy matrix is
\[
{T^{(1)}}_{\underline{\beta}^{\prime}\underline{\bar{\beta}}^{\prime}%
\beta^{\prime}}^{\underline{\beta}\,\underline{\bar{\beta}}\beta}%
(\underline{z}{,\underline{\bar{\theta}}},\theta)=\left(  \dot{S}_{10}%
(z_{1}-\theta)\cdots\dot{S}_{m0}(z_{m}-\theta)\dot{S}_{\bar{1}0}(\bar{\theta
}_{1}-\theta)\cdots\dot{S}_{\bar{n}0}(\bar{\theta}_{\bar{n}}-\theta)\right)
_{\underline{\beta}^{\prime}\underline{\bar{\beta}}^{\prime}\beta^{\prime}%
}^{\underline{\beta}\,\underline{\bar{\beta}}\beta}
\]
where the $\beta$'s and also the internal summation indices take the values
$2,\dots,N$. If we insert these equations into the representation (\ref{4.10})
of $f({\underline{\theta},\underline{\bar{\theta}}}^{\prime})$ we find that
the desired contribution from $\tilde{A}_{Q}$ applied to $\Omega$ vanishes.
The wanted contribution from $\tilde{D}_{Q}$ gives the result we are looking
forgives zero. The unwanted contributions from $\tilde{A}_{Q}$ and $\tilde
{D}_{Q}$ cancel after integration over the $z_{j}$.

All these three facts can be seen as follows. We insert eqs.~(\ref{B.5}) and
(\ref{B.6}) into the ansatz (\ref{4.10})
\[
f_{\underline{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta}%
,\underline{\bar{\theta}}^{\prime})=\int_{\mathcal{C}_{{\underline{\theta
}\,\underline{\bar{\theta}}}^{\prime}}}d\underline{z}\,h(\underline{\theta
},\underline{\bar{\theta}}^{\prime},{\underline{z}})\,g_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}^{\prime
})\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
}^{\prime},{\underline{z}})
\]
$\,$and use the eigenvalue equations (\ref{3.9}) for $z=\bar{\theta}_{\bar{n}%
}^{\prime\prime}=\bar{\theta}_{\bar{n}}^{\prime}-i\pi$
\begin{align*}
\Omega\,\tilde{A}_{Q}({\underline{\theta},\underline{\bar{\theta}}})  &
=\prod_{i=1}^{n}\dot{a}(\theta_{i}-\bar{\theta}_{\bar{n}}^{\prime\prime}%
)\prod_{i=1}^{\bar{n}-1}\hat{b}(\bar{\theta}_{i}-\bar{\theta}_{\bar{n}%
}^{\prime\prime})\,\hat{b}(i\pi)\,\Omega=0\\
\Omega\,\tilde{D}_{Q\,\beta}^{~~\beta^{\prime}}({\underline{\theta}%
,\underline{\bar{\theta}}})  &  =\prod_{i=1}^{n}\dot{b}(\theta_{i}-\bar
{\theta}_{\bar{n}}^{\prime\prime})\Omega\left(  \dot{S}_{\bar{1}0}(\bar
{\theta}_{1}-\bar{\theta}_{\bar{n}}^{\prime\prime})\cdots\dot{S}_{\bar{n}%
0}(i\pi)\right)  _{\beta}^{\beta^{\prime}}%
\end{align*}
which implies $f({\underline{\theta},\underline{\bar{\theta}}}^{\prime
})\,\tilde{A}_{Q}({\underline{\theta},\underline{\bar{\theta}}})=\sum
uw_{A}\,$ because of $\hat{b}(i\pi)=0$. For $\tilde{D}_{Q}$ we have
graphically
\begin{align*}
f({\underline{\theta},\underline{\bar{\theta}}}^{\prime})\,\tilde{D}%
_{Q\,\beta}^{~~\beta}({\underline{\theta},\underline{\bar{\theta}}}%
)=\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}^{\prime}}%
}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},{\underline{z}})\,g_{\underline{\beta}\,\underline{\bar{\beta}}}%
(\underline{z},\underline{\bar{\theta}}^{\prime})\,\Psi_{\underline{\alpha
}\,\underline{\bar{\alpha}}}^{\underline{\beta}\,\underline{\bar{\beta}}%
}(\underline{\theta},\underline{\bar{\theta}}^{\prime},{\underline{z}%
})\,\tilde{D}_{Q\,\beta}^{~~\beta}({\underline{\theta},\underline{\bar{\theta
}}})  & \\
=\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}^{\prime}}%
}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},{\underline{z}})\;\;
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(14.5,8)(-2.5,0) \put(4.5,7){\oval(13,2)} \put(5.5,7){\makebox(0,0){$g$}} \put(-2,6){\oval(24,10)[rb]} \put(-2,1){\vector(-1,0){0}} \put(11,0){\oval(2,2)[lt]} \put(10,0){\vector(0,-1){0}} \put(11,6){\oval(20,4)[lb]} \put(11,6){\oval(24,8)[lb]} \put(-1,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(3,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,6){\vector(0,-1){6}} \put(3.7,3){$\dots$} \put(7.7,3){$\dots$} \put(2.2,2.6){$\vdots$} \put(-3.,1){$\beta$} \put(11.3,1){$\beta$} \put(10.3,2.5){$\bar\theta'_{\bar n}$} \put(4,0){$\bar\theta''_{\bar n}$} \end{picture}
\end{array}
& \\
=\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}^{\prime}}%
}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},{\underline{z}})\frac{1}{\prod\limits_{j=1}^{m}\dot{b}(z_{j}-\theta_{\bar
{n}}^{\prime\prime\prime})}\;\;
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(14.5,8)(-2.5,0) \put(4.5,7){\oval(13,2)} \put(5.5,7){\makebox(0,0){$g$}} \put(-2,6){\oval(24,4)[rb]} \put(-2,4){\vector(-1,0){0}} \put(11,0){\oval(2,8)[lt]} \put(10,0){\vector(0,-1){0}} \put(11,6){\oval(20,6)[lb]} \put(11,6){\oval(24,10)[lb]} \put(-1,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(3,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,6){\vector(0,-1){6}} \put(3.7,2){$\dots$} \put(7.7,2){$\dots$} \put(2.2,1.6){$\vdots$} \put(-3.,4){$\beta$} \put(11.3,4){$\beta$} \put(10.3,1.5){$\bar\theta_{\bar n}$} \put(4,4.5){$\bar\theta''_{\bar n}$} \end{picture}
\end{array}
&  +\sum uw_{D}\\
=\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}^{\prime}}%
}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},{\underline{z})}\frac{\prod\limits_{i=1}^{n}\dot{b}(\theta_{i}-\bar{\theta
}_{\bar{n}}^{\prime\prime})}{\prod\limits_{j=1}^{m}\dot{b}(z_{j}-\theta
_{\bar{n}}^{\prime\prime\prime})}\;\;
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(14.5,8)(-2.5,0) \put(4.5,7){\oval(13,2)} \put(5.5,7){\makebox(0,0){$g$}} \put(-2,6){\oval(24,2.4)[rb]} \put(-2,4.8){\vector(-1,0){0}} \put(11,0){\oval(2,9.6)[lt]} \put(10,0){\vector(0,-1){0}} \put(11,6){\oval(20,6)[lb]} \put(11,6){\oval(24,10)[lb]} \put(-1,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(3,0){\vector(0,1){3.7}} \put(6,0){\vector(0,1){3.7}} \put(7,6){\vector(0,-1){6}} \put(3.7,2){$\dots$} \put(7.7,2){$\dots$} \put(2.2,1.6){$\vdots$} \put(-3.,4.5){$\beta$} \put(11.3,4.5){$\beta$} \put(10.3,1.5){$\bar\theta_{\bar n}$} \put(4,3.5){$\bar\theta''_{\bar n}$} \end{picture}
\end{array}
&  +\sum uw_{D}\\
=\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}^{\prime}}%
}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}}^{\prime
},{\underline{z})}\frac{\prod\limits_{i=1}^{n}\dot{b}(\theta_{i}-\bar{\theta
}_{\bar{n}}^{\prime\prime})}{\prod\limits_{j=1}^{m}\dot{b}(z_{j}-\theta
_{\bar{n}}^{\prime\prime\prime})}g_{\underline{\beta}\,\underline{\bar{\beta}%
}}(\underline{z},\underline{\bar{\theta}})\,\Psi_{\underline{\alpha
}\,\underline{\bar{\alpha}}}^{\underline{\beta}\,\underline{\bar{\beta}}%
}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &  +\sum
uw_{D}\\
=\sigma\int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}%
}^{\prime}}}d\underline{z}\,\,h({\underline{\theta},\underline{\bar{\theta}}%
},{\underline{z})}g_{\underline{\beta}\,\underline{\bar{\beta}}}(\underline
{z},\underline{\bar{\theta}})\,\Psi_{\underline{\alpha}\,\underline
{\bar{\alpha}}}^{\underline{\beta}\,\underline{\bar{\beta}}}(\underline
{\theta},\underline{\bar{\theta}},{\underline{z}})  &  +\sum uw_{D}\\
=\sigma f({\underline{\theta},\underline{\bar{\theta}}})  &  +\sum uw_{D}%
\end{align*}
where $(ii)$ for $g_{\underline{\beta}\,\underline{\bar{\beta}}}%
({\underline{z},\underline{\bar{\theta}})}$ in the form of
\[
g({\underline{\theta}},\underline{\bar{\theta}}^{\prime})\,\tilde{Q}%
^{(1)}({\underline{z},\underline{\bar{\theta}}})=g({\underline{z}}%
,\underline{\bar{\theta}})
\]
and asumption $(ii^{\prime\prime})$ (with $\bar{\theta}_{\bar{n}}%
^{\prime\prime\prime}=\bar{\theta}_{\bar{n}}-i\pi\,,\;\bar{\theta}_{\bar{n}%
}^{\prime\prime}=\bar{\theta}_{\bar{n}}+i\pi\,,\;\theta_{\bar{n}}^{\prime
}=\bar{\theta}_{\bar{n}}+2\pi i$)
\[
\prod_{i=1}^{\bar{n}}\dot{b}(\theta_{i}-\bar{\theta}_{\bar{n}}^{\prime\prime
})\prod_{j=1}^{m}\frac{1}{\dot{b}(z_{j}-\bar{\theta}_{\bar{n}}^{\prime
\prime\prime})}\,h(\underline{\theta},\underline{\bar{\theta}}^{\prime
},{\underline{z}})=\sigma\,h(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]
\[
\Leftarrow\quad\left\{
\begin{array}
[c]{l}%
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})=\prod
\limits_{1\le i<j\le n}F(\theta_{ij})\prod\limits_{i=1}^{n}\prod
\limits_{j=1}^{\bar{n}}\bar{F}(\theta_{i}-\bar{\theta}_{j})\prod
\limits_{i=1}^{n}\prod\limits_{j=1}^{m}\phi(\theta_{i}-z_{j})\\
~~~~~~~~~~~~~~\times\prod\limits_{i=1}^{\bar{n}}\prod\limits_{j=1}^{m}%
\dfrac{1}{\bar{F}(\bar{\theta}_{i}-z_{j})}\prod\limits_{1\le i<j\le m}%
\tau(z_{i}-z_{j})\,\,p(\underline{\theta},\underline{\bar{\theta}}%
,{\underline{z}})\\
\bar{F}(\theta-2\pi i)\varepsilon\dot{b}(\theta-i\pi)=\bar{F}(\theta-2\pi
i)\hat{b}(2\pi i-\theta)=\bar{F}(2\pi i-\theta)=\bar{F}(\theta)\\
\bar{F}(\theta)\varepsilon\dot{b}(i\pi-\theta)=\bar{F}(\theta)\hat{b}%
(\theta)=\bar{F}(-\theta)\\
p(\underline{\theta},\underline{\bar{\theta}}^{\prime},{\underline{z}%
})=\varepsilon^{\bar{n}-m}p(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\end{array}
\right.
\]
has been used.??? $\int_{\mathcal{C}_{{\underline{\theta}\,\underline
{\bar{\theta}}}^{\prime}}}d\underline{z}\rightarrow\int_{\mathcal{C}%
_{{\underline{\theta}\,\underline{\bar{\theta}}}}}d\underline{z}$ ???

Therefore it remains to prove that the unwanted terms cancel.

Next we discuss the `unwanted terms'. The commutation relations (\ref{A.3})
and (\ref{A.4}) imply that the unwanted terms are proportional to a product of
$C$-operators, where one $C({\underline{\theta}},z_{j})$ is replaced by
$C_{Q}({\underline{\theta}})$. Because of the symmetry $(i)$ of $g_{\underline
{\beta}}({\underline{z}})$ it is sufficient to consider only the unwanted
terms for $j=1$ which are denoted by $uw_{A}^{1}(${$\underline{z}$}$)$ and
$uw_{D}^{1}(${$\underline{z}$}$)$. They originate from the second term in the
commutation rules (\ref{A.4}) when $A_{Q}({\underline{\theta}}) $ is commuted
with $C({\underline{\theta}},z_{1})$. Then the resulting $A({\underline
{\theta}},z_{1})$ pushed through the other $C$'s and taking only the first
terms in (\ref{A.3}) into account we arrive at
\[
uw_{A}^{1}({\underline{z}})=-\frac{\dot{c}(\theta_{1}^{\prime}-z_{1})}{\dot
{b}(\theta_{1}^{\prime}-z_{1})}\,\prod_{1<j}\frac{\dot{a}(z_{1}-z_{j})}%
{\dot{b}(z_{1}-z_{j})}\,\Omega\,A({\underline{\theta}},z_{1})\,C^{\beta_{m}%
}({\underline{\theta}},z_{1})\dots C_{Q}^{\beta_{1}}({\underline{\theta}})
\]


Using (\ref{3.5}) and Yang-Baxter relations one obtains (taking into account
always the `$SU(N)$-ice rule')
\[
\left[  \Omega\,A({\underline{\theta}},z_{1})\,C^{\beta_{m}}({\underline
{\theta}},z_{1})\dots C_{Q}^{\beta_{1}}({\underline{\theta}})\right]
_{\underline{\alpha}}=\prod_{i=1}^{n}\left(  \dot{a}(\theta_{i}-z_{1}%
)\,\dot{a}(\theta_{i}-\theta_{1})\right)  \prod_{j=2}^{m}\dot{a}(\theta
_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline
{\tilde{\alpha}}}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}%
},\underline{\tilde{z}})
\]
where $\tilde{\Psi}_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}%
}(\underline{\tilde{\theta}},\underline{\tilde{z}})$ does not depend on
$\alpha_{1},\beta_{1},\theta_{1},z_{1}$. Therefore finally
\begin{align*}
\left[  f({\underline{\theta}})\,A_{Q}({\underline{\theta}})\right]
_{\underline{\alpha}}^{uw}  &  =-\int_{\mathcal{C}_{{\underline{\theta}%
}^{\prime}}}d\underline{z}\,h({\underline{\theta}},{\underline{z}%
})\,g_{\underline{\beta}}({\underline{z}})\frac{\dot{c}(\theta_{1}^{\prime
}-z_{1})}{\dot{b}(\theta_{1}^{\prime}-z_{1})}\,\prod_{j=2}^{m}\frac{\dot
{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\\
&  \times\prod_{i=1}^{n}\left(  \dot{a}(\theta_{i}-z_{1})\,\dot{a}(\theta
_{i}-\theta_{1})\right)  \prod_{j=2}^{m}\dot{a}(\theta_{1}-z_{j}%
)\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline{\tilde{\alpha}}%
}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}},\underline{\tilde{z}%
})
\end{align*}


Similarly we obtain the unwanted contribution from $D_{Q}({\underline{\theta}%
})$%
\begin{align*}
C^{\beta_{m}}({\underline{\theta}},z_{m})\cdots C^{\beta_{1}}({\underline
{\theta}},z_{1})\,{D_{Q}}_{\beta}^{\beta}({\underline{\theta}})  &
=-\frac{\dot{c}(z_{1}-\theta_{1})}{\dot{b}(z_{1}-\theta_{1})}\prod_{j=2}%
^{m}\frac{1}{\dot{b}(z_{j}-z_{1})}\,{T^{(1)}}_{\underline{\beta}^{\prime
},\beta^{\prime\prime}}^{\underline{\beta},\beta}(\underline{z},z_{1}%
)D_{\beta}^{\beta^{\prime\prime}}({\underline{\theta},z}_{1})\\
&  \times C^{\beta_{m}^{\prime}}({\underline{\theta}},z_{m})\cdots
C_{Q}^{\beta_{1}^{\prime}}({\underline{\theta}})+uw_{D}\,.
\end{align*}
Again (\ref{3.5}) and Yang-Baxter relations yield
\begin{multline*}
\left[  \Omega\,D_{\beta}^{\beta^{\prime\prime}}({\underline{\theta}}%
,z_{1})\,C^{\beta_{m}^{\prime}}({\underline{\theta}},z_{m})\dots C_{Q}%
^{\beta_{1}^{\prime}}({\underline{\theta}})\right]  _{\underline{\alpha}}\\
=\delta_{\beta}^{\beta^{\prime\prime}}\prod_{i=1}^{n}\left(  \dot{b}%
(\theta_{i}-z_{1})\,\dot{a}(\theta_{i}-\theta_{1})\right)  \prod_{j=2}^{m}%
\dot{a}(\theta_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}^{\prime}}\tilde
{\Psi}_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}^{\prime}%
}(\underline{\tilde{\theta}},\underline{\tilde{z}})
\end{multline*}
and assumption $(ii)$ for $g_{\underline{\beta}}(\underline{z})$ means
\[
g_{\underline{\beta}}(\underline{z}){Q^{(1)}}_{\underline{\beta}^{\prime}%
}^{\underline{\beta}}(\underline{z})=g_{\underline{\beta}}(\underline
{z}^{\prime})
\]
where ${Q^{(1)}}_{\underline{\beta}^{\prime}}^{\underline{\beta}}%
(\underline{z})={T^{(1)}}_{\underline{\beta}^{\prime}\beta}^{\underline{\beta
}\beta}(\underline{z},z_{1})$ is the next level $Q$-matrix and $\underline
{z}^{\prime}=(z_{1}^{\prime}=z_{1}+2\pi i,\dots,z_{m})$. Therefore using the
relations of $h(\underline{\theta},{\underline{z}})$ and $h(\underline{\theta
},{\underline{z}}^{\prime})$ given by assumtion $(ii^{\prime})$ we obtain
\begin{align*}
\left[  f({\underline{\theta}})\,{D_{Q}}_{\beta}^{\beta}({\underline{\theta}%
})\right]  _{\underline{\alpha}}^{uw}  &  =-\int_{\mathcal{C}_{{\underline
{\theta}}}}d\underline{z}\,h({\underline{\theta}},{\underline{z}}^{\prime
})\,g_{\underline{\beta}}({\underline{z}}^{\prime})\frac{\dot{c}(z_{1}%
-\theta_{1})}{\dot{b}(z_{1}-\theta_{1})}\prod_{j=2}^{m}\frac{\dot{a}%
(z_{1}^{\prime}-z_{j})}{\dot{b}(z_{1}^{\prime}-z_{j})}\\
&  \times\prod_{i=1}^{n}\left(  \dot{a}(\theta_{i}-z_{1}^{\prime})\,\dot
{a}(\theta_{i}-\theta_{1})\right)  \prod_{j=2}^{m}\dot{a}(\theta_{1}%
-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}_{\underline{\tilde
{\alpha}}}^{\underline{\tilde{\beta}}}(\underline{\tilde{\theta}}%
,\underline{\tilde{z}})
\end{align*}
and rewrite the $z_{1}$-integral by replacing $z_{1}\rightarrow z_{1}-2\pi i$
such that $z_{1}^{\prime}\rightarrow z_{1}$ and at the same time the contour
$\mathcal{C}_{{\underline{\theta}}}\rightarrow\mathcal{C}_{{\underline{\theta
}}}+2\pi i$.

Finally we obtain for the sum of the unwanted term from $A_{Q}$ and $D_{Q}$
\begin{multline*}
\left[  f({\underline{\theta}})\,\left(  A_{Q}({\underline{\theta}}%
)+D_{Q}({\underline{\theta}})\right)  \right]  _{\underline{\alpha}}%
^{uw}=\prod_{i=1}^{n}\dot{a}(\theta_{i}-\theta_{1})\left(  \int_{\mathcal{C}%
_{{\underline{\theta}}^{\prime}}}d\underline{z}-\int_{\mathcal{C}%
_{{\underline{\theta}}}+2\pi i}dz_{1}\int_{\mathcal{C}_{{\underline{\theta}}}%
}d\underline{\tilde{z}}\,\right)  \frac{\dot{c}(z_{1}-\theta_{1}^{\prime}%
)}{\dot{b}(z_{1}-\theta_{1}^{\prime})}\\
\times h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}%
}({\underline{z}})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}%
(z_{1}-z_{j})}\prod_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\prod_{j=2}^{m}\dot
{a}(\theta_{1}-z_{j})\,\delta_{\alpha_{1}}^{\beta_{1}}\tilde{\Psi}%
_{\underline{\tilde{\alpha}}}^{\underline{\tilde{\beta}}}(\underline
{\tilde{\theta}},\underline{\tilde{z}})\,.
\end{multline*}
Using $\frac{\dot{c}(z)}{\dot{b}(z)}=-\frac{2\pi i/N}{z}$ and the technical
assumptions (a) -- (d) we may replace for the $A_{Q}$ -term
\[
\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline{z}\cdots=\left(
-\operatorname*{Res}_{z_{1}=\theta_{1}^{\prime}}+\int_{\mathcal{C}%
_{{\underline{\theta}}}}dz_{1}\right)  \int_{\mathcal{C}_{{\underline{\theta}%
}}}d\underline{\tilde{z}}\cdots
\]
and for the $D_{Q}$-term
\[
\int_{\mathcal{C}_{{\underline{\theta}}}+2\pi i}dz_{1}\int_{\mathcal{C}%
_{{\underline{\theta}}}}d\underline{\tilde{z}}\,\cdots=\left(
-\operatorname*{Res}_{z_{1}=\theta_{1}^{\prime}}+\int_{\mathcal{C}%
_{{\underline{\theta}}}}dz_{1}\right)  \int_{\mathcal{C}_{{\underline{\theta}%
}}}d\underline{\tilde{z}}\cdots
\]
which proves the cancelation of the unwanted terms.

+++++++++++++++++++++++++++++++++++++++++

\subparagraph{The proof of (iii)}

in the form of
\[
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{I}(\theta
_{1},\dots,\bar{\theta}_{\bar{n}})=-2i\,\mathbf{C}_{1\bar{n}}\,f_{\tilde{I}%
}(\theta_{2},\dots,\bar{\theta}_{\bar{n}-1})\Big( \mathbf{1}_{2\dots
n-1}-S_{2\bar{n}}\cdots S_{\bar{n}-1\bar{n}}\Big)
\]


which is equivalent to (iii) due to (i). For $\theta_{1}-\bar{\theta}_{\bar
{n}}\rightarrow i\pi$ there are `pinchings' of the integration contours at
\[
\begin{array}
[c]{lll}%
(1) &  & z_{j}=z^{(1)}=\theta_{1}\rightarrow\bar{\theta}_{\bar{n}}+i\pi\\
(2) &  & z_{j}=z^{(2)}=\theta_{1}-2\pi i/N\rightarrow\bar{\theta}_{\bar{n}%
}+i\pi-2\pi i/N\\
(3) &  & z_{j}=z^{(3)}=\theta_{1}-2\pi i\rightarrow\bar{\theta}_{\bar{n}}-i\pi
\end{array}
\]
(for $j=1,\ldots,m$) such that $f_{\underline{\alpha}\,\underline{\bar{\alpha
}}}(\underline{\theta},\underline{\bar{\theta}})$ has a pole at $\theta
_{1}-\bar{\theta}_{\bar{n}}=i\pi$ with
\[
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{\underline
{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta},\underline{\bar{\theta
}})=-2\pi i\int_{\mathcal{C}_{{\underline{\theta}}}}d\underline{\tilde{z}%
}\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}\sum_{k=1}%
^{3}(-1)^{k}m\operatorname*{Res}_{z_{1}=z^{(k)}}h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})\,g_{\underline{\beta}%
\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]


\subparagraph{(1) $z_{1}=z^{(1)}$}

First we assume that $\bar{\alpha}_{\bar{n}}>1$ and calculate the contribution
from the `pinching' of the integration contours at $z_{1}=z^{(1)}=\theta
_{1}\rightarrow\bar{\theta}_{\bar{n}}+i\pi$, then in $\Psi_{\underline{\alpha
}\,\underline{\bar{\alpha}}}^{\underline{\beta}\,\underline{\bar{\beta}}%
}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})$ only $\dot
{S}(\theta_{1}-z_{1})=-\mathbf{P}\dot{c}(\theta_{1}-z_{1})$ contributes.
Therefore using the first reation of $(iii)^{(1)}$ for $g_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}})$
\[
(iii)^{(1)}\quad\left\{  \;
\begin{array}
[c]{l}%
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi}g_{1\dots
m\bar{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}})=-2i\,\mathbf{C}%
_{1\bar{n}}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2},\dots,\bar{\theta}%
_{\bar{n}-1})\\
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i?}g_{1\dots m\bar
{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}})=2i\,\mathbf{C}_{1\bar{n}%
}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2},\dots,\bar{\theta}_{\bar{n}%
-1})S_{2\bar{n}}\cdots S_{\bar{n}-1\bar{n}}\;.
\end{array}
\right.
\]
i.e.
\[
\operatorname*{Res}_{z_{1}=\bar{\theta}_{\bar{n}}+i\pi}g_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}%
})=-2i\mathbf{C}_{\beta_{1}\bar{\beta}_{\bar{n}}}g_{\underline{\tilde{\beta}%
}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
)
\]
and the graphical representations
\begin{align*}%
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(12,8)(-1.5,0) \put(4.5,7){\oval(13,2)} \put(5.5,7){\makebox(0,0){$g$}} \put(11,6){\oval(20,4)[lb]} \put(11,6){\oval(22,8)[lb]} \put(11,6){\oval(24,10)[lb]} \put(-1,5.7){\vector(0,1){0}} \put(0,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(3,0){\vector(0,1){5}} \put(4,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,6){\vector(0,-1){6}} \put(9,6){\vector(0,-1){6}} \put(10,6){\vector(0,-1){6}} \put(4.3,3){$\dots$} \put(7.3,3){$\dots$} \put(2.2,2.6){$\vdots$} \end{picture}
\end{array}
&  \approx\frac{-2i}{z_{1}-\bar{\theta}_{\bar{n}}-i\pi}\;
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(11,9) \put(5,7){\oval(10,2)} \put(5,7){\makebox(0,0){$g$}} \put(5,6){\oval(11,6)[t]} \put(11,6){\oval(19,4)[lb]} \put(11,6){\oval(21,8)[lb]} \put(11,6){\oval(23,10)[lb]} \put(-.5,5.7){\vector(0,1){0}} \put(.5,5.7){\vector(0,1){0}} \put(1.5,5.7){\vector(0,1){0}} \put(3,0){\vector(0,1){5}} \put(4,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,6){\vector(0,-1){6}} \put(9,6){\vector(0,-1){6}} \put(10.5,6){\vector(0,-1){6}} \put(4.3,3){$\dots$} \put(7.3,3){$\dots$} \put(2.2,2.6){$\vdots$} \end{picture}
\end{array}
\\
&  =\frac{-2i\,\dot{c}(0)}{z_{1}-\bar{\theta}_{\bar{n}}-i\pi}
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(11,9) \put(5,7){\oval(9,2)} \put(5,7){\makebox(0,0){$g$}} \put(5,6){\oval(10,6)[t]} \put(11,6){\oval(18,4)[lb]} \put(11,6){\oval(20,8)[lb]} \put(2,6){\oval(4,10)[lb]} \put(2,0){\oval(2,2)[rt]} \put(0,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(2,5.7){\vector(0,1){0}} \put(9,5.6){\oval(12,1.6)[lb]} \put(9,3){\oval(1,3.6)[rt]} \put(11,3){\oval(3,4)[lb]} \put(3,5.6){\vector(0,1){0}} \put(4,0){\vector(0,1){5.6}} \put(6,0){\vector(0,1){5.6}} \put(7,6){\vector(0,-1){6}} \put(9,6){\vector(0,-1){6}} \put(10,6){\vector(0,-1){6}} \put(4.3,3){$\dots$} \put(7.3,3){$\dots$} \put(2.2,2.7){$\vdots$} \end{picture}
\end{array}
\\
&  =\frac{-2i\,R(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)}{z_{1}-\bar{\theta}_{\bar{n}}-i\pi}
\begin{array}
[c]{l}%
\unitlength3.5mm\begin{picture}(11,9) \put(5,7){\oval(9,2)} \put(5,7){\makebox(0,0){$g$}} \put(5.5,6){\oval(11,6)[t]} \put(10,6){\oval(16,4)[lb]} \put(10,6){\oval(18,8)[lb]} \put(2,6){\oval(4,10)[lb]} \put(2,0){\oval(2,2)[rt]} \put(0,5.7){\vector(0,1){0}} \put(1,5.7){\vector(0,1){0}} \put(2,5.7){\vector(0,1){0}} \put(4,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,6){\vector(0,-1){6}} \put(9,6){\vector(0,-1){6}} \put(11,6){\vector(0,-1){6}} \put(4.3,3){$\dots$} \put(7.3,3){$\dots$} \put(2.2,2.7){$\vdots$} \end{picture}
\end{array}
\end{align*}
where $R(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)=-\prod\limits_{i=1}^{n}\dot{a}(\theta_{i}-\theta_{1})\prod\limits_{i=1}%
^{\bar{n}}\hat{b}(\bar{\theta}_{i}-\theta_{1})\prod\limits_{j=2}^{m}\left(
\dot{a}(\theta_{1}-z_{j})\hat{b}(\bar{\theta}_{\bar{n}}-z_{j})\right)  $we
obtain
\begin{align*}
g_{\underline{\beta}\,\underline{\bar{\beta}}}(\underline{z},\underline
{\bar{\theta}})\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}%
^{\underline{\beta}\,\underline{\bar{\beta}}}(\underline{\theta}%
,\underline{\bar{\theta}},{\underline{z}})  &  =g_{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}})\left[
\Omega C^{\beta_{m}}({\underline{\theta}},\underline{\bar{\theta}}%
,z_{m})\cdots C^{\beta_{1}}({\underline{\theta}},\underline{\bar{\theta}%
},z_{1})\right]  _{\underline{\alpha}\,\underline{\bar{\alpha}}}%
^{\underline{\beta}\,\,\underline{\bar{\beta}}}\\
&  \approx\frac{R(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})}{z_{1}-\bar{\theta}_{\bar{n}}-i\pi}\mathbf{C}_{\alpha_{1}\bar{\alpha}%
_{\bar{n}}}g_{\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
)\left[  \tilde{\Omega}\tilde{C}^{\beta_{m}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{m})\cdots\tilde{C}^{\beta_{2}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{2})\right]  _{\underline{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}^{\,\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}
%EndExpansion
}%
\end{align*}
where $\tilde{C}$ is the $C$-matrix acting in $V_{2\dots n\bar{1}\dots\bar
{n}-1}$%
\[
\left[  \tilde{C}^{\beta}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,z)\right]  _{\underline{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}^{\underline{\tilde{\beta}}\,\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}
%EndExpansion
}=\left[  \dot{S}_{20}(\theta_{2}-z)\cdots\dot{S}_{\bar{n}-10}(\bar{\theta
}_{\bar{n}-1}-z)\right]  _{\underline{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
\,1}^{\underline{\tilde{\beta}}\,\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}
%EndExpansion
\;\beta}\;.
\]
depending only on the particles $2,\ldots,n,1,\ldots,\bar{n}-1$. Also $\dot
{c}(0)=\dot{a}(0)=-1$ has been used. Therefore the contribution from
$z_{1}=z_{1}^{(1)}=\bar{\theta}_{\bar{n}}+i\pi$ is
\begin{align*}
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{\underline
{\alpha}\,\underline{\bar{\alpha}}}^{(1)}({\underline{\theta}},\underline
{\bar{\theta}})  &  =-2i\mathbf{C}_{\alpha_{1}\bar{\alpha}_{\bar{n}}}%
\int_{\mathcal{C}_{{\underline{\theta}}}}d\underline{\tilde{z}}h(\underline
{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,\underline{\tilde{z}}{)}g_{\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
)\left[  \tilde{\Omega}\tilde{C}^{\beta_{m}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{m})\cdots\tilde{C}^{\beta_{2}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{2})\right]  _{\underline{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}^{\underline{\tilde{\beta}}\,\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}
%EndExpansion
}\\
&  =-2i\mathbf{C}_{\alpha_{1}\bar{\alpha}_{\bar{n}}}f_{\underline
{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
)
\end{align*}
since for $z_{1}=z_{1}^{(1)}=\bar{\theta}_{\bar{n}}+i\pi$ by assumption
\[
-2\pi i\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi
}h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)\,\,R(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})=h(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}}{)}
\]


Note that with
\begin{align*}
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=\prod_{1\le i<j\le n}F(\theta_{ij})\prod_{i=1}^{n}\prod_{j=1}^{\bar{n}}%
\bar{F}(\theta_{i}-\bar{\theta}_{j})\prod_{i=1}^{n}\prod_{j=1}^{m}\phi
(\theta_{i}-z_{j})\\
&  \times\prod_{i=1}^{\bar{n}}\prod_{j=1}^{m}\frac{1}{\bar{F}(\bar{\theta}%
_{i}-z_{j})}\prod_{1\le i<j\le m}\tau(z_{i}-z_{j})\,\,p(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})
\end{align*}
and therefore
\begin{align*}
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=h(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,\underline{\tilde{z}})\phi(\theta_{1}-z_{1})\prod_{i=2}^{n}\left(
F(\theta_{1i})\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\phi(\theta_{i}%
-z_{1})\right) \\
&  \times\prod_{i=1}^{\bar{n}}\frac{\bar{F}(\theta_{1}-\bar{\theta}_{i})}%
{\bar{F}(\bar{\theta}_{i}-z_{1})}\prod_{j=2}^{m}\left(  \frac{\phi(\theta
_{1}-z_{j})\tau(z_{1}-z_{j})}{\bar{F}(\bar{\theta}_{\bar{n}}-z_{j})}\,\right)
\end{align*}
we get
\begin{multline*}
-2\pi i\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}%
\frac{h(\underline{\theta},\underline{\bar{\theta}},z_{1}^{(1)},{\underline
{\tilde{z}}})}{h(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,\underline{\tilde{z}}{)}}\,\,R(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})=-2\pi i\dot{c}(0)\operatorname*{Res}_{\theta_{1}%
-\bar{\theta}_{\bar{n}}=i\pi}\phi(\theta_{1}-z_{1}^{(1)})\frac{p(\underline
{\theta},\underline{\bar{\theta}},z_{1}^{(1)},{\underline{\tilde{z}}}%
)}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,\underline{\tilde{z}}{)}}\\
\times\prod_{i=2}^{n}\left(  (\dot{a}\phi)(\theta_{i}-\theta_{1})F(\theta
_{1i})\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\right)  \prod_{i=1}^{\bar{n}%
}\frac{\hat{b}(\bar{\theta}_{i}-\theta_{1})\bar{F}(\theta_{1}-\bar{\theta}%
_{i})}{\bar{F}(\bar{\theta}_{i}-\theta_{1})}\\
\prod_{j=2}^{m}\left(  (\dot{a}\phi)(\theta_{1}-z_{j})\frac{\hat{b}%
(\bar{\theta}_{\bar{n}}-z_{j})\tau(\theta_{1}-z_{j})\,}{\bar{F}(\bar{\theta
}_{\bar{n}}-z_{j})}\right) \\
=-2\pi i\dot{c}(0)\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi
}\phi(\theta_{1}-z_{1}^{(1)})(-\varepsilon)^{-m+1-\bar{n}}\frac{p(\underline
{\theta},\underline{\bar{\theta}},z_{1}^{(1)},{\underline{\tilde{z}}}%
)}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,\underline{\tilde{z}}{)}}=1
\end{multline*}
where eqs. (\ref{4.4}) (\ref{4.5}) (\ref{4.8}) (\ref{4.9}) $\tau(z)=\frac
{1}{F(z)\,\phi(z)\,\phi(-z)}$, $\phi(z)=\frac{1}{F(z)\,\bar{F}(i\pi-z)}$have
been used
\begin{align*}
(\dot{a}\phi)(\theta_{i}-\theta_{1})F(\theta_{1i})\bar{F}(\theta_{i}%
-\bar{\theta}_{\bar{n}})  &  =1\\
\frac{\hat{b}(\bar{\theta}_{i}-\theta_{1})\bar{F}(\theta_{1}-\bar{\theta}%
_{i})}{\bar{F}(\bar{\theta}_{i}-\theta_{1})}  &  =\frac{1}{-\varepsilon}\\
\frac{(\dot{a}\phi)(\theta_{1}-z_{j})\hat{b}(\bar{\theta}_{\bar{n}}-z_{j}%
)}{\bar{F}(\bar{\theta}_{\bar{n}}-z_{j})}\tau(\theta_{1}-z_{j})  &
=\frac{\phi(z_{j}-\theta_{1})}{-\varepsilon\bar{F}(z_{j}-\bar{\theta}_{\bar
{n}})}\frac{1}{F(\theta_{1}-z_{j})\,\phi(\theta_{1}-z_{j})\,\phi(z_{j}%
-\theta_{1})}\\
&  =\frac{1}{-\varepsilon}%
\end{align*}
++++++++++++++++++++++++++++++++++++

\subparagraph{(2) $z_{1}=z^{(2)}$}

Next calculate the contribution from the `pinching' of the integration
contours at $z_{1}=z^{(2)}=\theta_{1}-2\pi i/N\rightarrow\bar{\theta}_{\bar
{n}}+i\pi-2\pi i/N$.
\begin{align*}
\operatorname*{Res}\limits_{z_{m}=z^{(2)}}g_{\underline{\beta}\,\underline
{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}})\Psi_{\underline{\alpha
}\,\underline{\bar{\alpha}}}^{\underline{\beta}\,\underline{\bar{\beta}}%
}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=\operatorname*{Res}\limits_{z_{m}=z^{(2)}}g_{\underline{\beta}\,\underline
{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}})\left[  \Omega
C^{\beta_{m}}({\underline{\theta}},\underline{\bar{\theta}},z_{m})\cdots
C^{\beta_{1}}({\underline{\theta}},\underline{\bar{\theta}},z_{1})\right]
_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta}%
\,\underline{\bar{\beta}}}\\
&  =R_{2}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})\mathbf{C}_{\alpha_{1}\bar{\alpha}_{\bar{n}}}g_{\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
)\\
&  \times\left[  \tilde{\Omega}\tilde{C}^{\beta_{m-1}}({\underline
{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{m-1})\cdots\tilde{C}^{\beta_{1}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}
%EndExpansion
,z_{1})S_{12}\cdots S_{1\bar{n}-1}\mathbf{C}_{1\bar{n}}\right]  _{\underline
{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}
%EndExpansion
}^{\,\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}} }%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}
%EndExpansion
}%
\end{align*}


(as $\dot{S}_{1\bar{n}}(\theta_{1}-\bar{\theta}_{\bar{n}})\rightarrow
-\mathbf{C}^{\bar{n}1}\mathbf{C}_{1\bar{n}}$ for $\theta_{1}\rightarrow
\bar{\theta}_{\bar{n}}+i\pi)$ where
\[
R_{2}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)=-X_{\bar{\alpha}\beta}\mathbf{C}^{\bar{\alpha}\alpha}\dot{S}_{\alpha
1}^{\beta1}(\theta_{1}-z^{(2)})\prod_{i=2}^{n}\left(  \dot{a}(\theta
_{i}-\theta_{1})\dot{b}(\theta_{i}-z^{(2)})\right)  \prod_{i=1}^{\bar{n}}%
\hat{b}(\bar{\theta}_{i}-\theta_{1})
\]
\[
(iii)^{(1)}\quad\left\{
\begin{array}
[c]{l}%
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi}g_{1\dots
m\bar{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}})=-2i\,\mathbf{C}%
_{1\bar{n}}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2},\dots,\bar{\theta}%
_{\bar{n}-1})\\
\operatorname*{Res}\limits_{z_{1}-\bar{\theta}_{\bar{n}}=i\pi-2\pi
i/N}g_{1\dots m\bar{1}\dots\bar{n}}(z_{1},\dots,\bar{\theta}_{\bar{n}%
})=2i\,\mathbf{X}_{1\bar{n}}\,g_{2\dots m\bar{1}\dots\bar{n}-1}(z_{2}%
,\dots,\bar{\theta}_{\bar{n}-1})S_{2\bar{n}}\cdots S_{\bar{n}-1\bar{n}}\;.
\end{array}
\right.
\]
\begin{align*}
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{\underline
{\alpha}\,\underline{\bar{\alpha}}}^{(2)}({\underline{\theta}},\underline
{\bar{\theta}})  &  =-2\pi i\int_{\mathcal{C}_{{\underline{\theta}}}%
}d\underline{\tilde{z}}\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}%
}=i\pi}(-1)m\operatorname*{Res}_{z_{1}=z^{(2)}}h(\underline{\theta}%
,\underline{\bar{\theta}},{\underline{z}})\,g_{\underline{\beta}%
\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})\\
&  =2i\mathbf{C}_{\alpha_{1}\bar{\alpha}_{\bar{n}}^{\prime}}\int
_{\mathcal{C}_{{\underline{\theta}}}}d\underline{\tilde{z}}h(\underline
{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}}{)}g_{\underline{\tilde{\beta}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}%
%EndExpansion
}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
)\\
&  \times\left[  \tilde{\Omega}\tilde{C}^{\beta_{m-1}}({\underline
{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,z_{m-1})\cdots\tilde{C}^{\beta_{1}}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,z_{1})\left(  S_{2\bar{n}}\cdots S_{\bar{n}-1\bar{n}}\right)  _{\bar{\alpha
}_{\bar{n}}}^{\bar{\alpha}_{\bar{n}}^{\prime}}\right]  _{\underline
{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}%
%EndExpansion
}^{\underline{\tilde{\beta}}\,\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\beta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\beta}}}%
%EndExpansion
}\\
&  =2i\mathbf{C}_{\alpha_{1}\bar{\alpha}_{\bar{n}}^{\prime}}f_{2\dots n\bar
{1}\dots\bar{n}-1}({\underline{\tilde{\theta}}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
)\left(  \left(  S_{2\bar{n}}\cdots S_{\bar{n}-1\bar{n}}\right)
_{\underline{\tilde{\alpha}}\,%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\alpha}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\alpha}}}%
%EndExpansion
}\right)  _{\bar{\alpha}_{\bar{n}}}^{\bar{\alpha}_{\bar{n}}^{\prime}}%
\end{align*}
since for $z_{m}=z^{(2)}=\theta_{1}-2\pi i/N\rightarrow\bar{\theta}_{\bar{n}%
}+i\pi-2\pi i/N$ by assumption
\[
\left.  -2\pi i(-1)m\operatorname*{Res}_{\theta_{1}=\bar{\theta}_{\bar{n}}+\pi
i}X_{\bar{\alpha}\beta}\mathbf{C}^{\bar{\alpha}\alpha}\dot{S}_{\alpha1}%
^{\beta1}(\theta_{1}-z^{(2)})h(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})\,\,R_{2}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})\right|  _{z_{m}=z^{(2)}}=2ih(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}}{)}
\]
Note that with
\begin{align*}
h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})  &
=\prod_{1\le i<j\le n}F(\theta_{ij})\prod_{i=1}^{n}\prod_{j=1}^{\bar{n}}%
\bar{F}(\theta_{i}-\bar{\theta}_{j})\prod_{i=1}^{n}\prod_{j=1}^{m}\phi
(\theta_{i}-z_{j})\\
&  \times\prod_{j=1}^{m}\prod_{i=1}^{\bar{n}}\frac{1}{\bar{F}(\bar{\theta}%
_{i}-z_{j})}\prod_{1\le i<j\le m}\tau(z_{i}-z_{j})\,\,p(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})
\end{align*}
\begin{align*}
\frac{h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)}{h(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}  &  =\phi(\theta_{1}-z_{m})\prod_{i=2}^{n}\left(
F(\theta_{1i})\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\phi(\theta_{i}%
-z_{m})\right) \\
&  \times\prod_{i=1}^{\bar{n}}\frac{\bar{F}(\theta_{1}-\bar{\theta}_{i})}%
{\bar{F}(\bar{\theta}_{i}-z_{m})}\prod_{j=1}^{m-1}\left(  \frac{\phi
(\theta_{1}-z_{j})\tau(z_{j}-z_{m})}{\bar{F}(\bar{\theta}_{\bar{n}}-z_{j}%
)}\right)  \frac{p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}%
})}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}%
\end{align*}
\[
R_{2}(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)=-X_{\bar{\alpha}\beta}\mathbf{C}^{\bar{\alpha}\alpha}\dot{S}_{\alpha
1}^{\beta1}(\theta_{1}-z^{(2)})\prod_{i=2}^{n}\left(  \dot{a}(\theta
_{i}-\theta_{1})\dot{b}(\theta_{i}-z^{(2)})\right)  \prod_{i=1}^{\bar{n}}%
\hat{b}(\bar{\theta}_{i}-\theta_{1})
\]
we have
\begin{align*}
X_{\bar{\alpha}\beta}(\underline{\tilde{z}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
)  &  =2i\mathbf{C}_{\bar{\alpha}\beta}\left\{  2\pi
i(-1)m(N-1)\operatorname*{Res}_{\theta=2\pi i/N}\dot{c}(\theta)\phi
(\theta)(-1)^{n-1}\right. \\
&  \times\left.  \prod_{i=1}^{\bar{n}}f(\bar{\theta}_{i}-\bar{\theta}_{\bar
{n}})\prod_{j=1}^{m-1}\frac{1}{f(z_{j}-\bar{\theta}_{\bar{n}}+i\pi)}%
\frac{p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)_{z_{m}=z^{(2)}}}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}\right\}  ^{-1}%
\end{align*}
where $f(\theta)=\frac{1}{-\varepsilon}\bar{F}(\theta-i\pi)/\bar{F}%
(\theta-i\pi+2\pi i/N)$. This follows from
\begin{multline*}
\operatorname*{Res}_{\theta_{1}=\bar{\theta}_{\bar{n}}+\pi i}\frac
{h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})}%
{h(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}R_{2}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})=-\operatorname*{Res}_{\theta_{1}=\bar{\theta}_{\bar{n}}+\pi
i}X_{\bar{\alpha}\beta}\mathbf{C}^{\bar{\alpha}\alpha}\dot{S}_{\alpha1}%
^{\beta1}(\theta_{1}-z_{m})\phi(\theta_{1}-z_{m})\\
\times\prod_{i=2}^{n}\left(  \dot{a}(\theta_{i}-\theta_{1})\dot{b}(\theta
_{i}-z_{m})F(\theta_{1i})\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\phi
(\theta_{i}-z_{m})\right) \\
\times\prod_{i=1}^{\bar{n}}\left(  \hat{b}(\bar{\theta}_{i}-\theta_{1}%
)\frac{\bar{F}(\theta_{1}-\bar{\theta}_{i})}{\bar{F}(\bar{\theta}_{i}-z_{m}%
)}\right)  \prod_{j=1}^{m-1}\left(  \frac{\phi(\theta_{1}-z_{j})\tau
(z_{j}-z_{m})}{\bar{F}(\bar{\theta}_{\bar{n}}-z_{j})}\right)  _{z_{m}=z^{(2)}%
}\frac{p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}\\
=-\operatorname*{Res}_{\theta_{1}=\bar{\theta}_{\bar{n}}+\pi i}X_{\bar{\alpha
}\beta}\mathbf{C}^{\bar{\alpha}\alpha}\dot{S}_{\alpha1}^{\beta1}(\theta
_{1}-z_{m})\phi(\theta_{1}-z_{m})\\
\times(-1)^{n-1}\prod_{i=1}^{\bar{n}}f(\bar{\theta}_{i}-\bar{\theta}_{\bar{n}%
})\prod_{j=1}^{m-1}\frac{1}{f(z_{j}-\bar{\theta}_{\bar{n}}+i\pi)}%
\frac{p(\underline{\theta},\underline{\bar{\theta}},{\underline{z}}%
)_{z_{m}=z^{(2)}}}{p(\underline{\tilde{\theta}},%
%TCIMACRO{\TeXButton{}{\underline{\tilde{\bar{\theta}}}}}%
%BeginExpansion
\underline{\tilde{\bar{\theta}}}%
%EndExpansion
,\underline{\tilde{z}})}=\frac{2i}{-2\pi i(-1)m}%
\end{multline*}
We have used $\dot{a}(-\theta)F(\theta)=F(-\theta)$ , $\dot{b}(\theta
)\phi(\theta)=\dot{a}(\theta-2\pi i)\phi(\theta-2\pi i)=\phi(2\pi
i-\theta)=-\phi(\theta-2\pi iN)$ which follows from
\begin{align*}
\phi(\theta-2\pi iN)  &  =const.\Gamma\left(  \frac{\theta-2\pi iN}{2\pi
i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{\theta-2\pi iN}{2\pi i}\right) \\
&  =const.\Gamma\left(  \frac{\theta}{2\pi i}-\frac{1}{N}\right)
\Gamma\left(  1-\frac{\theta}{2\pi i}\right) \\
&  =const.\frac{-\frac{\theta}{2\pi i}}{\frac{\theta}{2\pi i}-\frac{1}{N}%
}\Gamma\left(  1+\frac{\theta}{2\pi i}-\frac{1}{N}\right)  \Gamma\left(
-\frac{\theta}{2\pi i}\right) \\
&  =-\frac{b}{a}(\theta)\phi(-\theta)=-\dot{b}(\theta)\phi(\theta
)=-\varepsilon\dot{a}(\theta-2\pi i)\phi(\theta-2\pi i)=-\varepsilon\phi(2\pi
i-\theta)
\end{align*}
also $z^{(2)}=\theta_{1}-2\pi i/N$ and get
\[
\dot{a}(\theta_{i}-\theta_{1})\dot{b}(\theta_{i}-z^{(2)})F(\theta_{1i})\bar
{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\phi(\theta_{i}-z^{(2)})=-F(-\theta
_{1i})\bar{F}(\theta_{i}-\bar{\theta}_{\bar{n}})\phi(\theta_{i}-z^{(2)}+2\pi
i)=-1
\]
and
\begin{align*}
\hat{b}(\bar{\theta}_{i}-\theta_{1})\frac{\bar{F}(\theta_{1}-\bar{\theta}%
_{i})}{\bar{F}(\bar{\theta}_{i}-\theta_{1}+2\pi i/N)}  &  =\frac
{1}{-\varepsilon}\frac{\bar{F}(\bar{\theta}_{i}-\theta_{1})}{\bar{F}%
(\bar{\theta}_{i}-\theta_{1}+2\pi i/N)}\\
&  =\frac{1}{-\varepsilon}\frac{\bar{F}(\bar{\theta}_{i}-\bar{\theta}_{\bar
{n}}-i\pi)}{\bar{F}(\bar{\theta}_{i}-\bar{\theta}_{\bar{n}}-i\pi+2\pi
i/N)}=f(\bar{\theta}_{i}-\bar{\theta}_{\bar{n}})
\end{align*}
and $\tau(z)=\frac{1}{F(z)\,\phi(z)\,\phi(-z)}=\frac{\bar{F}(i\pi-z)}%
{\,\phi(-z)}$%
\begin{align*}
\frac{\phi(\theta_{1}-z_{j})\tau(z_{j}-z^{(2)})}{\bar{F}(\bar{\theta}_{\bar
{n}}-z_{j})}  &  =\frac{\phi(\theta_{1}-z_{j})}{\phi(-z_{j}+\theta_{1}-2\pi
i/N)}\frac{\bar{F}(i\pi-z_{j}+\theta_{1}-2\pi i/N)}{\bar{F}(\bar{\theta}%
_{\bar{n}}-z_{j})\,}\\
&  =-\frac{\phi(\theta_{1}-z_{j})\bar{F}(i\pi-z_{j}+\theta_{1}-2\pi i/N)}%
{\dot{b}(\theta_{1}-z_{j})\phi(\theta_{1}-z_{j})\bar{F}(\bar{\theta}_{\bar{n}%
}-z_{j})\,}=\frac{\bar{F}(i\pi-z_{j}+\theta_{1}-2\pi i/N)}{\hat{b}%
(-\bar{\theta}_{\bar{n}}+z_{j})\bar{F}(\bar{\theta}_{\bar{n}}-z_{j})\,}\\
&  =-\varepsilon\frac{\bar{F}(\bar{\theta}_{\bar{n}}-z_{j}-2\pi i/N)}{\bar
{F}(z_{j}-\bar{\theta}_{\bar{n}})\,}=1/f(z_{j}-\bar{\theta}_{\bar{n}}+i\pi)
\end{align*}
.

++++++++++++++++++++++++++++++++++++

, t
\[
\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}f_{\underline
{\alpha}\,\underline{\bar{\alpha}}}(\underline{\theta},\underline{\bar{\theta
}})=-2\pi i\operatorname*{Res}_{\theta_{1}-\bar{\theta}_{\bar{n}}=i\pi}%
\sum_{k=1}^{3}(-1)^{k}m\operatorname*{Res}_{z_{1}=z^{(k)}}h(\underline{\theta
},\underline{\bar{\theta}},{\underline{z}})\,g_{\underline{\beta}%
\,\underline{\bar{\beta}}}(\underline{z},\underline{\bar{\theta}}%
)\,\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]
\[
\Psi_{\underline{\alpha}\,\underline{\bar{\alpha}}}^{\underline{\beta
}\,\underline{\bar{\beta}}}(\underline{\theta},\underline{\bar{\theta}%
},{\underline{z}})
\]


We consider the $n$-particle form factor function given by eq.~(\ref{4.10})
\[
f_{1\dots n}^{\mathcal{O}}({\underline{\theta}})=\prod_{i=1}^{m}\left(
\int_{\mathcal{C}_{\underline{\theta}}}du_{i}\right)  \,g_{n}({\underline
{\theta}},{\underline{u}})\,\Omega_{1\dots n}C_{1\dots n}({\underline{\theta}%
},u_{1})\cdots C_{1\dots n}({\underline{\theta}},u_{m})
\]
with the scalar function
\begin{align*}
g_{n}({\underline{\theta}},{\underline{u}})  &  =\frac{N_{n}^{\mathcal{O}}%
}{N_{n-2}^{\mathcal{O}}}\,g_{n-2}(\tilde{\underline{\theta}},\tilde
{\underline{u}})\\
&  \times F(\theta_{1}-\theta_{n})\,\prod_{i=2}^{n-1}\Big(F(\theta_{1}%
-\theta_{i})\,F(\theta_{i}-\theta_{n})\Big) \prod_{i=1}^{n}\phi(\theta
_{i}-u_{m})\\
&  \times\prod_{j=1}^{m-1}\Big(\phi(\theta_{1}-u_{j})\,\phi(\theta_{n}%
-u_{j})\,\tau(u_{j}-u_{m})\Big)\ e^{\pm\tilde{s}(2u_{m}-\theta_{1}-\theta
_{n})}%
\end{align*}
where $\tilde{\underline{\theta}}=\theta_{2},\dots,\theta_{n-1}$ and
$\tilde{\underline{u}}=u_{1},\dots,u_{m-1}$. We calculate the residue of this
function at $\theta_{1}=\theta_{n}+i\pi$. It consists of three terms
\[
\operatorname*{Res}_{\theta_{1}=\theta_{n}+i\pi}f_{1\dots n}^{\mathcal{O}%
}({\underline{\theta}})=R_{1}+R_{2}+R_{3}
\]
This is because each of the $m$ integration contours will be ``pinched'' at
three points:

\begin{itemize}
\item[(1)] $u_{j}=\theta_{n}=\theta_{1}-i\pi$,

\item[(2)] $u_{j}=\theta_{n}+i\pi=\theta_{1}$

\item[(3)] $u_{j}=\theta_{n}-i\pi=\theta_{1}-2i\pi$.
\end{itemize}

Due to symmetry it is sufficient to determine the contribution from the
$u_{m}$-integration and multiply the result by $m$.

The contribution of (1) is given by $u_{m}$-integration along the small circle
around $u_{m}=\theta_{n}$ (see figure~\ref{f5.1}). The S-matrix $S(\theta
_{n}-u_{m})$ yields the permutation operator $S(0)=P$ and $S(\theta_{1}%
-u_{m})$ the annihilation-creation operator $S(i\pi)=K$
\[
S_{\alpha\beta}^{\delta\gamma}(0)=\delta_{\alpha\delta}\,\delta_{\beta\gamma
}=
\begin{array}
[c]{c}%
\unitlength1.5mm
\begin{picture}(8,8) \put(2,2){\oval(4,4)[rt]} \put(6,6){\oval(4,4)[lb]} \put(3.5,0){$\alpha$} \put(7,3.5){$\beta$} \put(3.5,7){$\gamma$} \put(0,3.5){$\delta$} \end{picture}
\end{array}
~~,~~~~ S_{\alpha\beta}^{\delta\gamma}(i\pi)= \mathbf{C}_{\alpha\beta
}\,\mathbf{C}^{\delta\gamma}= \delta_{\alpha\bar\beta}\,\delta_{\delta
\bar\gamma}=
\begin{array}
[c]{c}%
\unitlength1.5mm
\begin{picture}(8,8) \put(2,6){\oval(4,4)[rb]} \put(6,2){\oval(4,4)[lt]} \put(3.5,0){$\alpha$} \put(7,3.5){$\beta$} \put(3.5,7){$\gamma$} \put(0,3.5){$\delta$} \end{picture}~.
\end{array}
\]
Therefore we have for $u_{m}=\theta_{n}=\theta_{1}-i\pi$
\begin{align*}
\Omega_{1\dots n}\,C_{1\dots n}({\underline\theta},u_{1})\cdots C_{1\dots
n}({\underline\theta},\theta_{n}) =
\begin{array}
[c]{c}%
\unitlength6mm
\begin{picture}(8,5.5) \put(0,2){\line(1,0){8}} \put(0,2){\vector(1,0){.5}} \put(1,2){\vector(1,0){.5}} \put(7,2){\vector(-1,0){.8}} \put(8,2){\vector(-1,0){.8}} \put(0,4){\line(1,0){8}} \put(0,4){\vector(1,0){.5}} \put(1,4){\vector(1,0){.5}} \put(7,4){\vector(-1,0){.8}} \put(8,4){\vector(-1,0){.5}} \put(2,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(4,0){\oval(6,2)[t]} \put(.4,5){\oval(1.2,8)[rb]} \put(1,4){\vector(0,1){1}} \put(0,1){\vector(1,0){.5}} \put(7.6,5){\oval(1.2,8)[lb]} \put(7,4){\vector(0,1){1}} \put(8,1){\vector(-1,0){.5}} \put(.3,0){$\theta_1$} \put(2.2,0){$\theta_2$} \put(4.7,0){$\theta_{n-1}$} \put(7.2,0){$\theta_n$} \put(7.4,1.2){$u_m$} \put(7.4,4.3){$u_1$} \put(2.6,1.2){$u_m=\theta_n$} \put(7.4,1.2){$v$} \put(3.5,3){$\dots$} \put(.3,2.7){$\vdots$} \end{picture}
\end{array}
~~~~~~~~~~~~~~~\\
=\prod_{j=1}^{m-1}\Big(\dot b(\theta_{1}-u_{j})\dot a(\theta_{n}-u_{j})\Big)\,
\mathbf{C}_{1n}\,\Omega_{2\dots n-1}C_{2\dots n-1}(\tilde{\underline\theta
},u_{1})\cdots\\
\cdots C_{2\dots n-1}(\tilde{\underline\theta},u_{m-1})\,\dot S_{2n}\cdots\dot
S_{n-1n}%
\end{align*}
where $\mathbf{C}_{1n}$ is the charge conjugation matrix with $\mathbf{C}%
_{\alpha\beta}= \delta_{\bar\alpha\beta}$. We have used the fact that because
of charge conservation the amplitude $b(\cdot)$ only contributes to the
S-matrices $S(\theta_{1}-u_{j})$ and $a(\cdot)$ to the S-matrices
$S(\theta_{n}-u_{j})$.

We combine this with the scalar function $g_{n}$ and after having performed
the remaining $u_{j}$-integrations we obtain
\begin{align*}
R_{1\dots n}^{(1)}  &  =\mathbf{C}_{1n}\,f_{2\dots n-1}^{\mathcal{O}}%
(\tilde{\underline{\theta}})\,\dot{S}_{2n}\cdots\dot{S}_{n-1n}\\
&  \times m\,\frac{N_{n}^{\mathcal{O}}}{N_{n-2}^{\mathcal{O}}}%
\,\operatorname*{Res}_{\theta_{1}=\theta_{n}+i\pi}(-2\pi i)\operatorname*{Res}%
_{u_{m}=\theta_{n}}\dot{a}(\theta_{n}-u_{m})\phi(\theta_{n}-u_{m})\,\dot
{b}(\theta_{1}-u_{m})\phi(\theta_{1}-u_{m})\\
&  \times F(i\pi)\,\prod_{i=2}^{n-1}\Big(F(\theta_{1}-\theta_{i})F(\theta
_{i}-\theta_{n})\phi(\theta_{i}-u_{m})\Big)\\
&  \times\prod_{j=1}^{m-1}\Big(\dot{b}(\theta_{1}-u_{j})\phi(\theta_{1}%
-u_{j})\dot{a}(\theta_{n}-u_{j})\phi(\theta_{n}-u_{j})\tau(u_{j}%
-u_{m})\Big)\,e^{\pm\tilde{s}(2u_{m}-\theta_{1}-\theta_{n})}\\
&  =2i\,\sigma_{\mathcal{O}n}\,\mathbf{C}_{1n}\,f_{2\dots n-1}^{\mathcal{O}%
}(\tilde{\underline{\theta}})\,S_{2n}\cdots S_{n-1n}%
\end{align*}
if we relate the normalization constants by the recursion relation
\begin{equation}
N_{n}^{\mathcal{O}}=N_{n-2}^{\mathcal{O}}e^{\pm i\pi\tilde{s}}\frac{\left(
f_{ss}^{min}(0)\right)  ^{2}}{4\pi m}~. \label{B.7}%
\end{equation}
We have also used that $\sigma_{\mathcal{O}n}=(-1)^{n}$ and
\[
F(u)\,F(u+i\pi)\,\phi(u)=1~~~\mathrm{and}~~~~b(u+i\pi)\,\phi(u+i\pi
)\,a(u)\,\phi(u)\,\tau(-u)=1
\]
which follows from the definitions (\ref{4.8}) and (\ref{4.9}). Finally we
have used the normalization $F(i\pi)=1$ and $\operatorname*{Res}_{u=0}\dot
{a}(u)\phi(u)=\operatorname*{Res}_{u=i\pi}\dot{b}(u)\phi(u)=-2i/f_{ss}%
^{min}(0)$ because of eq.~(\ref{4.11}). Note also that the signs from the
$\dot{a}$s and $\dot{b}$s cancel and $\sigma_{\mathcal{O}n}=(-1)^{n-2}$, since
all particles are fermions.

The remaining contribution to (iii) is due to $R_{2}$ and $R_{3}$
\[
R_{1\dots n}^{(2)}+R_{1\dots n}^{(3)} =2i\,\mathbf{C}_{1n}\,f^{\mathcal{O}%
}_{2\dots n-1}(\tilde{\underline\theta}).
\]
If both particles at 1 and n are solitons both vanish. If one particle at 1 or
n is an soliton and the other an anti-soliton one term gives the desired
expression and the other vanish. If both particles at 1 and n are
anti-solitons both terms cancel. These fact can be proven as follows.

The contribution of (2) is given by the $u_{m}$-integration along the small
circle around $u_{M}=\theta_{1}$ (see again figure~\ref{f5.1}). Now
$S(\theta_{1}-u_{m})$ yields the permutation operator $S(0)=P$ and the
co-vector part of this contribution for $u_{m}=\theta_{1}=\theta_{n}+i\pi$ is
\begin{align*}
\lefteqn{\Omega_{1\dots n}\,C_{1\dots n}(\underline{\theta},u_{1})\cdots
C_{1\dots n}(\underline{\theta},u_{m}=\theta_{1})\,P_{n}(s)}\\
&  =
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(8,5.5) \put(8,2){\line(-1,0){8}} \put(8,2){\vector(-1,0){.5}} \put(8,4){\line(-1,0){8}} \put(8,4){\vector(-1,0){0.5}} \put(0,2){\vector(1,0){0.6}} \put(0,4){\vector(1,0){0.6}} \put(7,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(2,0){\vector(0,1){5}} \put(1,4){\vector(0,1){1}} \put(.4,0.00){\oval(1.2,2)[rt]} \put(0,1){\vector(1,0){.5}} \put(8,5){\oval(14,8)[lb]} \put(8,1){\vector(-1,0){.5}} \put(.3,0){$\theta_1$} \put(1.3,0){$\theta_2$} \put(4.7,0){$\theta_{n-1}$} \put(7.4,4.3){$u_1$} \put(7.6,1.2){$u_m$} \put(3.6,1.2){$u_m$} \put(3.5,3){$\dots$} \put(.3,2.7){$\vdots$} \put(7.3,-.1){$\theta_n$} \end{picture}
\end{array}
~~~=~~~
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(8,5.5) \put(0,2){\line(1,0){8}} \put(0,2){\vector(1,0){.5}} \put(0,3.5){\line(1,0){8}} \put(0,3.5){\vector(1,0){.5}} \put(8,2){\vector(-1,0){.6}} \put(8,3.5){\vector(-1,0){.6}} \put(6,0){\vector(0,1){5}} \put(5,0){\vector(0,1){5}} \put(2,0){\vector(0,1){5}} \put(1,4.6){\vector(0,1){.4}} \put(.4,0){\oval(1.2,2)[rt]} \put(0,1){\vector(1,0){.5}} \put(7.6,3){\oval(1.2,4)[lb]} \put(8,1){\vector(-1,0){.5}} \put(4,3){\oval(6,2.2)[rt]} \put(4,4.6){\oval(6,1)[lb]} \put(.3,0){$\theta_1$} \put(7.1,2.5){$\theta_1$} \put(1.3,0){$\theta_2$} \put(3.7,0){$\theta_{n-1}$} \put(7.4,3.8){$u_1$} \put(7.6,1.2){$u_m$} \put(3,4.4){$u_m$} \put(3,2.7){$\dots$} \put(.3,2.5){$\vdots$} \put(6.3,-.1){$\theta_n$} \end{picture}
\end{array}
\\
&  =\prod_{i=1}^{n}\dot{a}(\theta_{i}-u_{m})\,\prod_{j=1}^{m-1}\Big(\dot
{a}(\theta_{1}-u_{j})\dot{a}(\theta_{n}-u_{j})\Big)\\
&  ~~~~~\times\mathbf{C}_{1n}\,\Omega_{2\dots n-1}C_{2\dots n-1}%
(\tilde{\underline{\theta}},u_{1})\cdots C_{2\dots n-1}(\tilde{\underline
{\theta}},u_{m-1})\,P_{1}(\bar{s})\,P_{n}(s)
\end{align*}
where the Yang-Baxter relation () has been used iteratively. $P_{1}(\bar{s})$
and $P_{n}(s)$ project onto the components where the particle at 1 is a
soliton and at $n$ is an anti-soliton, respectively. The remaining components
if both particles at 1 and n are anti-solitons are calculated below.

We combine this with the scalar function $g_{n}$ and after having performed
the remaining $u_{j}$-integrations we obtain
\begin{align*}
&  R_{1\dots n}^{(2)}\,P_{n}(s)\ =\ \mathbf{C}_{1n}\,f_{2\dots n-1}%
^{\mathcal{O}}(\tilde{\underline{\theta}})\,P_{1}(\bar{s})\,P_{n}(s)\\
&  ~~\times m\,\frac{N_{n}^{\mathcal{O}}}{N_{n-2}^{\mathcal{O}}}%
\,\operatorname*{Res}_{\theta_{1}=\theta_{n}+i\pi}(-2\pi i)\operatorname*{Res}%
_{u_{m}=\theta_{1}}\dot{a}(\theta_{1}-u_{m})\phi(\theta_{1}-u_{m})\,\dot
{a}(\theta_{n}-u_{m})\phi(\theta_{n}-u_{m})\\
&  ~~\times F(i\pi)\,\prod_{i=2}^{n-1}\Big(F(\theta_{1}-\theta_{i}%
)F(\theta_{i}-\theta_{n})\dot{a}(\theta_{i}-u_{m})\phi(\theta_{i}%
-u_{m})\Big)\\
&  ~~\times\prod_{j=1}^{m-1}\Big(\dot{a}(\theta_{1}-u_{j})\phi(\theta
_{1}-u_{j})\dot{a}(\theta_{n}-u_{j})\phi(\theta_{n}-u_{j})\tau(u_{j}%
-u_{m})\Big)\,e^{\pm\tilde{s}(2u_{m}-\theta_{1}-\theta_{n})}\\
&  ~=-2i\,\sigma_{\mathcal{O}n}\,\mathbf{C}_{1n}\,f_{2\dots n-1}^{\mathcal{O}%
}(\tilde{\underline{\theta}})\,P_{1}(\bar{s})\,P_{n}(s)~.
\end{align*}
if we apply the condition $\exp(2\pi\tilde{s})=(-1)^{n}=\sigma_{\mathcal{O}n}
$ and if we relate the normalization constants as above. We have used the
identities
\begin{equation}
F(-u)\,F(u+i\pi)\,\dot{a}(u)\,\phi(u)=1~~\mathrm{and}~~a(u)\,\phi
(u)\,a(u-i\pi)\,\phi(u-i\pi)\,\tau(-u)=1. \label{B.8}%
\end{equation}


The contribution of (3) is given by $u_{m}$-integration along the small circle
around $u_{m}=\theta_{n}-i\pi$ (see again figure~\ref{f5.1}). Now
$S(\theta_{n}-u_{m})$ yields the annihilation-creation operator $S(i\pi)=K$
and the co-vector part of this contribution for $u_{m}=\theta_{1}-2\pi
i=\theta_{n}-i\pi$ is
\begin{align*}
\lefteqn{\Omega_{1\dots n}\,C_{1\dots n}(\underline{\theta},u_{1})\cdots
C_{1\dots n}(\underline{\theta},u_{m}=\theta_{n}-i\pi)\,P_{1}(s)}\\
&  =
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(8,5.5) \put(0,2){\line(1,0){8}} \put(0,2){\vector(1,0){0.5}} \put(0,4){\line(1,0){8}} \put(0,4){\vector(1,0){0.5}} \put(8,2){\vector(-1,0){0.6}} \put(8,4){\vector(-1,0){0.6}} \put(1,0){\vector(0,1){5}} \put(2,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,4){\vector(0,1){1}} \put(7.6,0){\oval(1.2,2)[lt]} \put(8,1){\vector(-1,0){0.5}} \put(0,5){\oval(14,8)[rb]} \put(0,1){\vector(1,0){.5}} \put(.3,0){$\theta_1$} \put(1.3,0){$\theta_2$} \put(4.7,0){$\theta_{n-1}$} \put(7.4,4.3){$u_1$} \put(7.6,1.2){$u_m$} \put(3.6,1.2){$u_m$} \put(3.5,3){$\dots$} \put(.3,2.7){$\vdots$} \put(7.3,-.1){$\theta_n$} \end{picture}
\end{array}
~~~=~~~
\begin{array}
[c]{c}%
\unitlength6mm\begin{picture}(8,5.5) \put(0,2){\line(1,0){8}} \put(0,2){\vector(1,0){.5}} \put(0,3.5){\line(1,0){8}} \put(0,3.5){\vector(1,0){.5}} \put(8,2){\vector(-1,0){.6}} \put(8,3.5){\vector(-1,0){.6}} \put(2,0){\vector(0,1){5}} \put(3,0){\vector(0,1){5}} \put(6,0){\vector(0,1){5}} \put(7,4.6){\vector(0,1){.4}} \put(7.6,0){\oval(1.2,2)[lt]} \put(8,1){\vector(-1,0){.5}} \put(.4,3){\oval(1.2,4)[rb]} \put(0,1){\vector(1,0){.5}} \put(4,3){\oval(6,2.2)[lt]} \put(4,4.6){\oval(6,1)[rb]} \put(1.3,0){$\theta_1$} \put(2.3,0){$\theta_2$} \put(4.7,0){$\theta_{n-1}$} \put(7.4,3.8){$u_1$} \put(7.6,1.2){$u_m$} \put(4,4.4){$u_m$} \put(4,2.7){$\dots$} \put(.3,2.5){$\vdots$} \put(7.3,-.1){$\theta_n$} \put(1.1,2.5){$\theta_n$} \end{picture}
\end{array}
\\
&  =\prod_{i=1}^{n}\dot{b}(\theta_{i}-u_{m})\,\prod_{j=1}^{m-1}\Big(\dot
{b}(\theta_{1}-u_{j})\dot{b}(\theta_{n}-u_{j})\Big)\\
&  ~~~~\times\mathbf{C}_{1n}\,\Omega_{2\dots n-1}C_{2\dots n-1}(\tilde
{\underline{\theta}},u_{1})\cdots C_{2\dots n-1}(\tilde{\underline{\theta}%
},u_{m-1})\,P_{1}(s)\,P_{n}(\bar{s})
\end{align*}
if the particle at 1 is a soliton and that at n an anti-soliton. This
contribution obviously vanishes if n is a soliton. Again the remaining
components if both particles at 1 and n are anti-solitons are calculated below.

Again we combine this with the scalar function $g_{n}$ and after having
performed the remaining $u_{j}$-integrations we obtain
\begin{align*}
&  R_{1\dots n}^{(3)}\,P_{1}(s)\ =\ \mathbf{C}_{1n}\,f_{2\dots n-1}%
^{\mathcal{O}}(\tilde{\underline{\theta}})\,P_{1}(s)\,P_{n}(\bar{s})\\
&  ~~\times m\,\frac{N_{n}^{\mathcal{O}}}{N_{n-2}^{\mathcal{O}}}%
\,\operatorname*{Res}_{\theta_{1}=\theta_{n}+i\pi}(-2\pi i)\quad_{u_{m}%
=\theta_{n}-i\pi}\dot{b}(\theta_{n}-u_{m})\phi(\theta_{n}-u_{m})\,\dot
{b}(\theta_{1}-u_{m})\phi(\theta_{1}-u_{m})\\
&  ~~\times F(i\pi)\,\prod_{i=2}^{n-1}\Big(F(\theta_{1}-\theta_{i}%
)F(\theta_{i}-\theta_{n})\dot{b}(\theta_{i}-u_{m})\phi(\theta_{i}%
-u_{m})\Big)\\
&  ~~\times\prod_{j=1}^{m-1}\Big(\dot{b}(\theta_{1}-u_{j})\phi(\theta
_{1}-u_{j})\dot{b}(\theta_{n}-u_{j})\phi(\theta_{n}-u_{j})\tau(u_{j}%
-u_{m})\Big)\,e^{\pm\tilde{s}(2u_{m}-\theta_{1}-\theta_{n})}\\
&  ~=-2i\,\sigma_{\mathcal{O}n}\,\mathbf{C}_{1n}\,f_{2\dots n-1}^{\mathcal{O}%
}(\tilde{\underline{\theta}})\,P_{1}(s)\,P_{n}(\bar{s})
\end{align*}
provided that we fix the normalization constants as above. We have used the
identities
\begin{equation}
\dot{b}(u)\,F(-u)\,F(u+i\pi)\,\phi(u)=1~~\mathrm{and}~~a(u)\,\phi
(u)\,a(u-i\pi)\,\phi(u-i\pi)\,\tau(-u)=1. \label{B.9}%
\end{equation}
%

%TCIMACRO{\TeXButton{End Proof}{\endproof}}%
%BeginExpansion
\endproof
%EndExpansion


\paragraph{Technical assumtions for theorem \ref{t3}}

Further we need more technical assumptions on the contour $\mathcal{C}%
_{\underline{\theta}\,\underline{\bar{\theta}}}$ and on the analytic behavior
of $h(\underline{\theta},\underline{\bar{\theta}},{\underline{z}})$ and
$g_{\underline{\beta}\,\underline{\bar{\beta}}}(\underline{z},\underline
{\bar{\theta}})$. The following integrals along closed contours have to
vanish\label{atu}

\begin{itemize}
\item[(a)]
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}}}%
}-\int_{\mathcal{C}_{{\underline{\theta}}^{\prime}{\underline{\bar{\theta}}}}%
}\right)  d\underline{z}\,h({\underline{\theta}},\underline{\bar{\theta}%
},{\underline{z}})g_{\underline{\beta}\,\underline{\beta}^{\prime}%
}({\underline{z}},\underline{\bar{\theta}})\prod_{j=1}^{m}\dot{a}(\theta
_{1}-z_{j})=0
\]
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}\,\underline{\bar{\theta}}%
}^{\prime}}}-\int_{\mathcal{C}_{{\underline{\theta}\underline{\bar{\theta}}}}%
}\right)  d\underline{z}\,h({\underline{\theta}},\underline{\bar{\theta}%
},{\underline{z}})g_{\underline{\beta}\,\underline{\beta}^{\prime}%
}({\underline{z}},\underline{\bar{\theta}})\prod_{j=1}^{m}\hat{b}(\bar{\theta
}_{\bar{n}}-z_{j})=0
\]
where $\mathcal{C}_{{\underline{\theta}}^{\prime}{\underline{\bar{\theta}}}}$
is the contour corresponding to $\underline{\theta}^{\prime}$ for
$\underline{\theta}^{\prime}=(\theta_{1}^{\prime}=\theta_{1}+2\pi
i,\dots,\theta_{n})$ and $\mathcal{C}_{{\underline{\theta}\,\underline
{\bar{\theta}}}^{\prime}}$ that for $\underline{\bar{\theta}}^{\prime}%
=(\bar{\theta}_{1},\cdots,\bar{\theta}_{\bar{n}}+2\pi i)$;

\item[(b)] and more general
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}d\underline{z}%
-\int_{\mathcal{C}_{{\underline{\theta}}}}d\underline{z}\right)
h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}}({\underline{z}%
})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\prod
_{i=1}^{n}\dot{a}(\theta_{i}-z_{1})\prod_{j=2}^{m}\dot{a}(\theta_{1}%
-z_{j})=0\,;
\]


\item[(c)] further more
\[
\left[  \left(  \int_{\mathcal{C}_{{\underline{\theta}}^{\prime}}}%
-\int_{\mathcal{C}_{{\underline{\theta}}}}\right)  d\underline{\tilde{z}%
}h({\underline{\theta}},{\underline{z}})\,g_{\underline{\beta}}({\underline
{z}})\prod_{j=2}^{m}\frac{\dot{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}%
\prod_{j=2}^{m}\dot{a}(\theta_{1}-z_{j})\right]  _{z_{1}=\theta_{1}^{\prime}%
}=0
\]
where $\underline{\tilde{z}}=z_{2},\dots,z_{m}$;

\item[(d)] and
\[
\left(  \int_{\mathcal{C}_{{\underline{\theta}}}+2\pi i}-\int_{\mathcal{C}%
_{{\underline{\theta}}}}\right)  dz_{1}\,h({\underline{\theta}},{\underline
{z}})\,g_{\underline{\beta}}({\underline{z}})\prod_{j=2}^{m}\frac{\dot
{a}(z_{1}-z_{j})}{\dot{b}(z_{1}-z_{j})}\prod_{i=1}^{n}\dot{a}(\theta_{i}%
-z_{1})=0
\]
where $\mathcal{C}_{{\underline{\theta}}}+2\pi i$ is the contour
$\mathcal{C}_{{\underline{\theta}}}$ shifted by $2\pi i$.
\end{itemize}

\newpage

\section{Formulae}

%

\[
\dot{S}_{\alpha\beta}^{\delta\gamma}(\theta)=-\delta_{\alpha\gamma}%
\delta_{\beta\delta}\,b(\theta)-\delta_{\alpha\delta}\delta_{\beta\gamma
}\,c(\theta)\,,\quad\dot{S}_{\beta\bar{\gamma}}^{\bar{\alpha}\delta}%
(\theta)=\delta_{\bar{\gamma}}^{\bar{\alpha}}\delta_{\beta}^{\delta}%
\,b(\hat{\theta})-\mathbf{C}^{\bar{\alpha}\delta}\mathbf{C}_{\beta\bar{\gamma
}}\,c(\hat{\theta})
\]
$\left(  \hat{\theta}=i\pi-\theta\;,\quad\mathbf{C}^{\alpha\bar{\beta}%
}=\mathbf{C}_{\alpha\bar{\beta}}=-\mathbf{C}^{\bar{\beta}\alpha}%
=-\mathbf{C}_{\bar{\beta}\alpha}=\delta_{\alpha\beta}\right)  \,.$%
\[
\dot{S}_{12}(\theta)=\dot{a}(\theta)\frac{\mathbf{1}_{12}\theta-\mathbf{P}%
_{12}2\pi i/N}{\theta-2\pi i/N}\,,\quad\dot{S}_{1\bar{2}}(\theta)=\hat
{b}(\theta)\left(  \mathbf{1}_{1\bar{2}}+\mathbf{K}_{1\bar{2}}\frac{2\pi
i/N}{\theta-i\pi}\right)  \,.
\]
\[
\frac{b(\theta)}{a(\theta)}=\frac{\theta}{\theta-2\pi i/N}~,~~\frac{c(\theta
)}{a(\theta)}=-\frac{2\pi i/N}{\theta-2\pi i/N}~,~~\frac{c(\hat{\theta}%
)}{b(\hat{\theta})}=\frac{2\pi i/N}{\theta-i\pi}~.
\]
\[
a(\theta)=\frac{\Gamma\left(  1-\frac{\theta}{2\pi i}\right)  \Gamma\left(
1-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  }{\Gamma\left(  1+\frac{\theta
}{2\pi i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{\theta}{2\pi i}\right)
}\;,\quad b(\hat{\theta})=\frac{\Gamma\left(  \frac{1}{2}+\frac{\theta}{2\pi
i}\right)  \Gamma\left(  \frac{1}{2}-\frac{1}{N}-\frac{\theta}{2\pi i}\right)
}{\Gamma\left(  \frac{1}{2}-\frac{\theta}{2\pi i}\right)  \Gamma\left(
\frac{1}{2}-\frac{1}{N}+\frac{\theta}{2\pi i}\right)  }
\]
\[
\,a(\theta)=\exp\int_{0}^{\infty}\frac{dt}{t}e^{-t}\frac{1-e^{t\frac{2}{N}}%
}{\sinh t}\sinh t\frac{\theta}{i\pi}=1+\frac{\pi}{N}\left(  \coth\frac{\theta
}{2}-\frac{2}{\theta}\right)  +O(N^{-2})
\]
\[
b(\hat{\theta})=\exp\int_{0}^{\infty}\frac{dt}{t}\frac{e^{t\frac{2}{N}}%
-1}{\sinh t}\sinh t\frac{\theta}{i\pi}=1-\frac{\pi}{N}\tanh\frac{\theta}%
{2}+O(N^{-2})\,.
\]
\begin{align*}
\dot{S}_{+}(\theta)  &  =-\left(  b(\theta)+c(\theta)\right)  =-a(\theta)\\
\dot{S}_{-}(\theta)  &  =-\left(  b(\theta)-c(\theta)\right)  =-a(\theta
)\frac{\theta+2\pi i/N}{\theta-2\pi i/N}\\
\bar{S}_{-}(\theta)  &  =b(\hat{\theta})\\
\bar{S}_{0}(\theta)  &  =b(\hat{\theta})+Nc(\hat{\theta})=b(\hat{\theta}%
)\frac{\theta+\pi i}{\theta-\pi i}\,.
\end{align*}
\[
\fbox{$F(\theta)=\dot{a}(\theta)F(-\theta)\,=F(2\pi i-\theta)\;,\quad\bar
{F}(\theta)=-\varepsilon\hat{b}(\theta)\bar{F}(-\theta)\,=\bar{F}(2\pi
i-\theta)$}
\]
\[
F(\theta)=\dot{a}(\theta)F(-\theta)\,=F(2\pi i-\theta)\;,\quad\bar{F}%
(\theta)=-\varepsilon\hat{b}(\theta)\bar{F}(-\theta)\,=\bar{F}(2\pi
i-\theta)\,
\]
\[
\tau(z)=\frac{1}{F(z)\,\phi(z)\,\phi(-z)}=F(-z)\bar{F}^{2}(i\pi-z)\;,~~\phi
(z)=\frac{1}{F(z)\,\bar{F}(i\pi-z)}
\]
\[
\tau(-z)=\frac{\dot{a}(z)}{F(z)\,\phi(z)\,\phi(-z)}=\dot{a}(z)\tau
(z)\;,\quad\phi(-z)=\frac{\dot{a}(z)}{F(z)\,\bar{F}(i\pi+z)}=\dot{a}%
(z)\phi(z)
\]
\begin{align*}
\frac{\phi(z+2\pi i)}{\phi(z)}  &  =\frac{F(z)\bar{F}(i\pi-z)}{F(z+2\pi
i)\bar{F}(-i\pi-z)}=\frac{F(-z)\dot{a}(z)\bar{F}(i\pi-z)}{-F(-z)\bar{F}%
(i\pi+z)\varepsilon\hat{b}(-i\pi-z)}\\
&  =\frac{\dot{a}(z)}{-\varepsilon b(z+2\pi i)}=\frac{\dot{a}(z)}%
{\varepsilon\dot{b}(z+2\pi i)}%
\end{align*}
\begin{align*}
\frac{\tau(2\pi i-z)}{\tau(-z)}  &  =\frac{F(-z)\phi(z)\phi(-z)}{F(2\pi
i-z)\phi(2\pi i-z)\phi(z-2\pi i)}=\frac{\dot{a}(-z)F(z)}{F(z)}\frac{\dot
{b}(-z+2\pi i)}{\dot{a}(-z)}\frac{\dot{a}(z-2\pi i)}{\dot{b}(z)}\\
&  =\frac{\dot{b}(2\pi i-z)}{\dot{a}(2\pi i-z)}\frac{1}{\dot{b}(z)}%
\end{align*}
\begin{align*}
\phi(\theta-2\pi iN)  &  =const.\Gamma\left(  \frac{\theta-2\pi iN}{2\pi
i}\right)  \Gamma\left(  1-\frac{1}{N}-\frac{\theta-2\pi iN}{2\pi i}\right) \\
&  =const.\Gamma\left(  \frac{\theta}{2\pi i}-\frac{1}{N}\right)
\Gamma\left(  1-\frac{\theta}{2\pi i}\right) \\
&  =const.\frac{-\frac{\theta}{2\pi i}}{\frac{\theta}{2\pi i}-\frac{1}{N}%
}\Gamma\left(  1+\frac{\theta}{2\pi i}-\frac{1}{N}\right)  \Gamma\left(
-\frac{\theta}{2\pi i}\right) \\
&  =-\frac{b}{a}(\theta)\phi(-\theta)=-\dot{b}(\theta)\phi(\theta
)=-\varepsilon\dot{a}(\theta-2\pi i)\phi(\theta-2\pi i)=-\varepsilon\phi(2\pi
i-\theta)
\end{align*}


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