# Tau Functions, Birkhoff Factorizations & Difference Equations

Darlayne Addabbo  and  Maarten Bergvelt<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

$Q$-systems and $T$-systems are integrable difference equations that recently
have attracted much attention, and have wide applications in representation
theory and statistical mechanics. We show that certain $\tau$-functions, given
as matrix elements of the action of the loop group of $\mathbf{GL}_{2}$ on two
component Fermionic Fock space, give solutions of $Q$-systems.

An obvious generalization using the loop group of $\mathbf{GL}_{3}$ acting on
three component Fermionic Fock space leads to a new system of 4 difference
equations, with hopefully also interesting applications.

###### Contents

  1. 1 Introduction
  2. 2 The $2\times 2$ Case
    1. 2.1 $2\times 2$ $\tau$-functions and $Q$-system
    2. 2.2 Birkhoff Factorization
    3. 2.3 Matrix Baker Functions and Connection Matrices
  3. 3 $3\times 3$ Case
    1. 3.1 $\tau$-functions
    2. 3.2 Examples of $3\times 3$ tau functions
    3. 3.3 Birkhoff Factorization for $3\times 3$
    4. 3.4 Matrix Baker Functions and Connection Matrices, $3\times 3$ case
    5. 3.5 Explicit Formulae for Connection Matrices
    6. 3.6 Difference Equations from Factorizations
  4. A Multi Component Fermions and Semi Infinite Wedge Space
    1. A.1 Introduction
    2. A.2 Semi Infinite Wedge Space
    3. A.3 Fermionic Translation Operators and Translation Group.
    4. A.4 The Lie Algebra $\operatorname{\widetilde{gl}_{n}}$ and Fermions.
    5. A.5 Root Lattice
  5. B Expressions for the $\tau$-functions
    1. B.1 The case of $n=2$, Theorem 2.1.
    2. B.2 Proof of $n=3$, Theorem 3.1.
  6. C Birkhoff Factorization and Matrix Elements of Semi Infinite Wedge Space
    1. C.1 Birkhoff Factorization and $n$-component Fermions.
    2. C.2 The $2\times 2$-case; Proof of Theorem 2.2.
    3. C.3 Birkhoff Factorization in the $3\times 3$ case, Proof of Theorem 3.3
  7. D Factorization and Reduction to One Component Fermions
  8. E One Component Fermion Correlation Functions
  9. F Proof of Determinantal Identities

##  1\. Introduction

Many integrable differential equations can be transformed in simpler,
bilinear, form, by introducing new dependent variables called
$\tau$-functions. In practice, these $\tau$-functions are given as matrix
elements of infinite dimensional groups or Lie algebras, etc.

For instance, the famous Korteweg-de Vries equation

(1.1) $\partial_{t}u+u_{xxx}+6u\partial_{x}u=0$

is transformed by the substitution

$u=2\partial_{x}^{2}\ln(\tau)$

into Hirota bilinear form

$(D_{x}D_{t}+D_{x}^{4})\tau\cdot\tau=0,$

where $D_{u}$ is the Hirota operator so that
$D_{u}\sigma\cdot\tau=(\partial_{u}\sigma)\tau-\sigma\partial_{u}\tau$. See
[6] for details and many more examples. In the case of the KdV equation the
$\tau$-function is a matrix element for the action of the loop group of
$\mathbf{GL}_{2}$ on one component Fermionic Fock space, see for instance [4],
[13], [8].

To produce the integrable equations from $\tau$-functions one introduces an
intermediate object, the Baker function. It satisfies linear equations, and
the compatibility of these equations gives the integrable hierarchy.

For instance, for the KdV case the $\tau$-function is a scalar function
$\tau(t_{1},t_{3},t_{5},\dots)$ of odd times ($t_{1}=x,t_{3}=t$), and the
Baker function is also a scalar function $\Psi$ of the times $t_{2k+1}$ and of
an extra variable $z$, the spectral parameter. It is defined by

$\Psi(z;t_{1},t_{3},\dots)=\Gamma(z,t)\tau(t)/\tau,$

where $\Gamma(z,t)=e^{\sum
z^{k}t_{k}}e^{-\sum\frac{1}{kz^{k}}\partial_{t_{k}}}$, essentially the vertex
operator for the free fermion vertex algebra, see e.g., [9]. The Baker
function satisfies linear equations

(1.2) $\partial_{t_{k}}\Psi(z;t)=B_{k}(\partial_{x})\Psi(z;t),$

where $B_{k}(\partial_{x})$ is a degree $k$ differential operator in
$\partial_{x}$. The compatibility of (1.2) for $k=1$ and $k=3$ turns out to
give precisely the Korteweg-de Vries equation (1.1).

In this paper we are interested in integrable _difference_ (as opposed to
differential) equations. Still, we follow very much the setup sketched above
for the KdV hierarchy.

In the first part we introduce a collection of $\tau$-functions as matrix
elements of the action of loop group elements for $\mathbf{GL}_{2}$, depending
on discrete variables111These variables can be thought of as coordinates on
the lower triangular subgroup of the loop group of $\mathbf{GL}_{2}$ $c_{k}$,
which play a similar role as the higher KdV times $t_{2k+1}$, $k>1$. These
$\tau$-functions are of the form $\tau_{k}^{(\alpha)}(c_{k})$, where
$k,\alpha$ are discrete variables. In fact, these $\tau$-functions turn out to
be (see Theorem 2.1) Hankel determinants, well known since the 19th century in
the theory of orthogonal polynomials, see e.g.,[7].

We then define Baker functions. They turn out to be in this case $2\times 2$
matrices depending on a spectral parameter $z$ and on the discrete variables
(and on the $c_{k}$):

$\Psi_{k}^{(\alpha)}(z)=\frac{1}{\tau_{k}^{(\alpha)}}\begin{bmatrix}z^{k}&0\\\
0&z^{-k}\end{bmatrix}\begin{bmatrix}S^{+}(z)&0\\\
0&S^{-}(z)\end{bmatrix}\begin{bmatrix}\tau_{k}^{(\alpha)}&\tau_{k-1}^{(\alpha)}/z\\\
\tau_{k+1}^{(\alpha)}/z&\tau_{k}^{(\alpha)}\end{bmatrix},$

where $S^{\pm}(z)=(1-S/z)^{\pm 1}$ are the shift fields, constructed from the
elementary shift $S:\mathbb{C}[c_{k}]\to\mathbb{C}[c_{k}]$, defined as the
multiplicative map such that $S(1)=0,S(c_{k})=c_{k+1}$. The shift fields
$S^{\pm}(z)$ play a similar role here as the vertex operator $\Gamma(z)$ does
in the theory of the KdV hierarchy.

Next we introduce linear equations for the $2\times 2$ Baker functions:

$\Psi^{(\alpha+1)}_{k}=\Psi^{(\alpha)}_{k}V^{(\alpha)}_{k},\quad\Psi^{(\alpha+1)}_{k-1}=\Psi^{(\alpha)}_{k}W^{(\alpha)}_{k}.$

Then we show that compatibility of these equations leads to the discrete zero
curvature equations

$V_{k}^{(\alpha)}(W_{k+1}^{(\alpha)})^{-1}=(W_{k+1}^{(\alpha-1)})^{-1}V_{k+1}^{(\alpha-1)}.$

Since we can give explicit expressions for the connection matrices
$V_{k}^{(\alpha)},W_{k}^{(\alpha)}$ in terms of the $\tau$-functions we obtain
the following basic equation:

$\boxed{(\tau_{k}^{(\alpha)})^{2}=\tau_{k}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}-\tau_{k+1}^{(\alpha-1)}\tau_{k-1}^{(\alpha+1)},\quad
k=0,1,\dots.}$

This is upon a change of variables the $Q$-system, see e.g.,[5].

In the second part of the paper we generalize our derivation of the $Q$-system
by using the loop group of $\mathbf{GL}_{3}$, obtaining $\tau$-functions
$\tau_{k,l}^{(\alpha,\beta)}(c_{k},d_{k},e_{k})$, where
$k,l,\alpha,\beta\in\mathbb{Z}$ and the $c_{k},d_{k},e_{k}$ are coordinates on
the lower triangular subgroup of the loop group of $\mathbf{GL}_{3}$. We can
explicitly calculate these $\tau$-functions, see Theorem 3.1, but the answer
is much more complicated than the simple Hankel determinants in the $2\times
2$ case. Next we introduce Baker functions, now $3\times 3$ matrices depending
on a spectral parameter, and the linear equations for the Baker functions.
Again we can explicitly calculate the connection matrices in terms of
$\tau$-functions, see Lemma 3.6 and Lemma 3.7. Compatibility now gives us a
system of _four_ equations (Theorem 3.8)

(1.3) $\boxed{\begin{aligned}
\tau_{k,l-1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}&+\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k,l-1}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}=0,\\\
\tau_{k+1,l+1}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}&-\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta+1)}+\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta)}=0,\\\
(\tau_{k,l}^{(\alpha,\beta)})^{2}&=\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}+\tau_{k+1,l+1}^{(\alpha-1,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)},\\\
(\tau_{k,l}^{(\alpha,\beta)})^{2}&=\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta-1)}-\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l-1}^{(\alpha,\beta+1)}-\tau_{k-1,l}^{(\alpha,\beta-1)}\tau_{k+1,l}^{(\alpha,\beta+1)}.\end{aligned}}$

The first two of these new equations are generalizations of $T$-system
equations, more precisely the first equation is for fixed $\beta$ a
$T$-system, after a change of variables, and for the second equation one fixes
$\alpha$. See e.g., [10] for information on $T$-systems.

The last two equations seem to be related to the fusion rules of quantum
transfer matrices for Young diagrams consisting of two rectangular blocks, see
[12], [11].

##  2\. The $2\times 2$ Case

###  2.1. $2\times 2$ $\tau$-functions and $Q$-system

We have an action of the central extension $\operatorname{\widehat{GL}_{2}}$
of the loop group
$\operatorname{\widetilde{GL}_{2}}=GL_{2}(\mathbb{C}((z^{-1})))$ on two
component semi-infinite wedge space $F^{(2)}$, see Appendix A.4. Let
$\pi\colon\operatorname{\widehat{GL}_{2}}\to\operatorname{\widetilde{GL}_{2}}$
be the projection on the loop group and consider the action on the vacuum
vector of $F^{(2)}$, see (A.4), of $g_{C}\in\operatorname{\widehat{GL}_{2}}$,
where

(2.1) $\pi(g_{C})=\begin{bmatrix}1&0\\\ C(z)&1\end{bmatrix}.$

Here

$C(z)=\sum_{i\in\mathbb{Z}}c_{i}z^{-i-1},$

where $c_{i}\in\mathbb{C}$; sometimes it is also useful to think of the
$c_{i}$ as formal variables.

Recall (see (A.12)) the projection of the fermionic translation operators to
the loop group $\operatorname{\widetilde{GL}_{2}}$: in particular we get the
matrices

$\pi(Q_{0})=\begin{bmatrix}z^{-1}&0\\\
0&1\end{bmatrix},\quad\pi(Q_{1})=\begin{bmatrix}1&0\\\ 0&z^{-1}\end{bmatrix}.$

We also have the translation element

$T=Q_{1}Q_{0}^{-1}\overset{\pi}{\mapsto}\begin{pmatrix}z&0\\\
0&z^{-1}\end{pmatrix}.$

We define shifts on the series $C(z)$ by

(2.2)
$C^{(\alpha)}(z)=z^{\alpha}C(z)=\sum_{i\in\mathbb{Z}}c_{i+\alpha}z^{-i-1}.$

We define similarly the shifted group element
$g^{(\alpha)}=Q_{0}^{\alpha}g_{C}Q_{0}^{-\alpha}$ so that

(2.3) $\pi(g_{C}^{(\alpha)})=\begin{bmatrix}1&0\\\
C^{(\alpha)}(z)&1\end{bmatrix}=\begin{bmatrix}1&0\\\
z^{\alpha}C(z)&1\end{bmatrix}.$

We then have

(2.4) $Q_{0}^{-1}g_{C}^{(\alpha+1)}=g_{C}^{(\alpha)}Q_{0}^{-1},$

and the same relation with $\pi$ applied,

(2.5)
$\pi(Q_{0})^{-1}\pi(g_{C}^{(\alpha+1)})=\pi(g_{C}^{(\alpha)})\pi(Q_{0})^{-1}.$

The fundamental objects in the theory of the Toda lattice and Q-systems are
the $\tau$-functions defined by

(2.6) $\boxed{\tau_{k}^{(\alpha)}=<T^{k}v_{0},g_{C}^{(\alpha)}v_{0}>.}$

The $\tau$-functions in the $2\times 2$ case are determinants of Hankel
matrices.

######  Theorem 2.1.

$\displaystyle\tau_{k}^{(\alpha)}$
$\displaystyle=\frac{1}{k!}\operatorname{Res}_{\bf{w}}\left(\prod_{i=1}^{k}C^{(\alpha)}(w_{i})\prod_{k\geq
j>i\geq 0}(w_{i}-w_{j})^{2}\right)=$
$\displaystyle=\det\begin{bmatrix}c_{\alpha}&c_{\alpha+1}&\cdots&c_{\alpha+k-1}\\\
c_{\alpha+1}&c_{\alpha+2}&\cdots&c_{\alpha+k}\\\
\vdots&\vdots&\vdots&\vdots\\\
c_{\alpha+k-1}&c_{\alpha+k}&\cdots&c_{\alpha+2k-2}\end{bmatrix}.$

For the proof see Appendix B.1.

The simple form of the $\tau$-functions allows us to apply the Desnanot-Jacobi
identity (cf., [3]) to obtain ([5]) the following difference equations,
satisfied by our tau functions:

(2.7)
$\tau_{k}^{(\alpha)}\tau_{k-2}^{(\alpha+2)}=\tau_{k-1}^{(\alpha+2)}\tau_{k-1}^{(\alpha)}-(\tau_{k-1}^{(\alpha+1)})^{2}.$

The disadvantage of obtaining the difference equations in this way is that it
is not at all apparent how to generalize to the $3\times 3$ situation, in
which the formulas for the tau functions are much more complicated. We thus
present another way of obtaining our $2\times 2$ difference equations.

###  2.2. Birkhoff Factorization

Define an element of the central extension of the loop group of
$\mathbf{GL}_{2}$:

(2.8) $g^{[k](\alpha)}=T^{-k}g^{(\alpha)}_{C},$

and assume that it has a Birkhoff Factorization([14], see also Appendix C.1):

$g^{[k](\alpha)}=g^{[k](\alpha)}_{-}g^{[k](\alpha)}_{0,+},$

where $\pi(g^{[k](\alpha)}_{-})=1+\mathcal{O}(z^{-1})$ and
$\pi(g^{[k](\alpha)}_{0+})=A_{k}^{(\alpha)}+\mathcal{O}(z)$, for
$A_{k}^{(\alpha)}$ an invertible $z$ independent matrix. This assumption is
justified precisely when $\tau^{(\alpha)}_{k}(g)=\langle
v_{0},g^{[k](\alpha)}v_{0}\rangle$ is not zero.

Now we want display the negative component of $\pi(g^{[k](\alpha)})$. To
calculate this we make some extra structure explicit.

Let $\mathcal{N}$ be the subgroup of elements of
$\operatorname{\widetilde{SL}_{2}}$ of the form (2.1). We can think of the
coefficients $c_{k}$ as coordinates on $\mathcal{N}$, so

$B=\mathbb{C}[c_{k}]_{k\in\mathbb{Z}}$

is the coordinate ring of $\mathcal{N}$.

Define first shifts acting on $B$: these are multiplicative maps given on
generators by

(2.9) $S^{\alpha}\colon B\to B,\quad S^{\alpha}(1)=0,\quad
S^{\alpha}(c_{k})=c_{k+\alpha},\quad\alpha\in\mathbb{Z}.$

We will often write $S^{\pm}$ for $S^{\pm 1}$.

Define also _shift fields_. These are multiplicative maps

(2.10) $S^{\pm}(z)\colon B\to B[[z^{-1}]],$

given by

$S^{\pm}(z)=(1-\frac{S^{+}}{z})^{\pm 1},$

######  Theorem 2.2.

For $k\geq 0$ and all $\alpha\in\mathbb{Z}$

$\pi(g^{[k](\alpha)}_{-})=\begin{bmatrix}S^{+}(z)\tau^{(\alpha)}_{k}&S^{+}(z)\tau^{(\alpha)}_{k-1}/z\\\
S^{-}(z)\tau^{(\alpha)}_{k+1}/z&S^{-}(z)\tau^{(\alpha)}_{k}\end{bmatrix}/\tau^{(\alpha)}_{k}.$

We sketch the proof in Appendix C.2.

Now we have expressed the negative component of the Birkhoff factorization in
terms of matrix elements of the centrally extended loop group we have no
longer need for the central extension and we will simplify notation: we will
write $g^{[k](\alpha)}_{-}$ for $\pi(g^{[k](\alpha)}_{-})$, $T$ for $\pi(T)$
and similarly $Q_{a}$ for $\pi(Q_{a})$, $a=0,1$. In particular in the rest of
this section we write

$T=\begin{bmatrix}z&0\\\ 0&z^{-1}\end{bmatrix}=Q_{1}Q_{0}^{-1},\quad
Q_{0}=\begin{bmatrix}z^{-1}&0\\\ 0&1\end{bmatrix},\quad
Q_{1}=\begin{bmatrix}1&0\\\ 0&z^{-1}\end{bmatrix}.$

###  2.3. Matrix Baker Functions and Connection Matrices

Next we define the Baker functions. These are elements of the loop group
defined by

(2.11) $\Psi^{[k](\alpha)}=T^{k}Q_{0}^{-\alpha}g_{-}^{[k](\alpha)}.$

Since the Baker functions are all invertible, they are related by _connection
matrices_ belonging to the loop group: define
$\Gamma_{[k],(\alpha)}^{[l],(\beta)}\in\operatorname{\widetilde{GL}_{2}}$ by

$\Psi^{[l](\beta)}=\Psi^{[k](\alpha)}\Gamma_{[k],(\alpha)}^{[l],(\beta)}.$

We are in particular interested in connection matrices that implement nearest
neighbor steps on the lattice of Baker functions. So define

$U_{k}^{(\alpha)}=\Gamma_{[k](\alpha)}^{[k+1](\alpha)},\quad
V_{k}^{(\alpha)}=\Gamma_{[k](\alpha)}^{[k](\alpha+1)},\quad
W_{k}^{(\alpha)}=\Gamma_{[k](\alpha)}^{[k-1](\alpha+1)},$

so that we have

(2.12) $\displaystyle\Psi^{[k+1](\alpha)}$
$\displaystyle=\Psi^{[k](\alpha)}U_{k}^{(\alpha)},$
$\displaystyle\Psi^{([k](\alpha+1)}$
$\displaystyle=\Psi^{[k](\alpha)}V_{k}^{(\alpha)},$
$\displaystyle\Psi^{[k-1](\alpha+1)}$
$\displaystyle=\Psi^{[k](\alpha)}W_{k}^{(\alpha)}.$

Pictorially:

${\Psi^{[k](\alpha-1)}}$${\Psi^{[k+1](\alpha-1)}}$${\Psi^{[k+2](\alpha-1)}}$${\Psi^{[k](\alpha)}}$${\Psi^{[k+1](\alpha)}}$${\Psi^{[k-1](\alpha+1)}}$${\Psi^{[k](\alpha+1)}}$${\Psi^{[k+1](\alpha+1)}}$$\scriptstyle{U_{k}^{(\alpha-1)}}$$\scriptstyle{V_{k}^{(\alpha-1)}}$$\scriptstyle{U_{k+1}^{(\alpha-1)}}$$\scriptstyle{V_{k+1}^{(\alpha-1)}}$$\scriptstyle{W_{k+1}^{(\alpha-1)}}$$\scriptstyle{W_{k+2}^{(\alpha-1)}}$$\scriptstyle{U_{k}^{(\alpha)}}$$\scriptstyle{V_{k}^{(\alpha)}}$$\scriptstyle{W_{k}^{(\alpha)}}$$\scriptstyle{W_{k+1}^{(\alpha)}}$$\scriptstyle{V_{k+1}^{(\alpha)}}$$\scriptstyle{U_{k-1}^{(\alpha+1)}}$$\scriptstyle{U_{k}^{(\alpha+1)}}$

Walking around the triangles in this diagram we see that we get factorizations
of all basic connection matrices. In particular

(2.13)
$U_{k}^{(\alpha)}=V_{k}^{(\alpha)}(W_{k+1}^{(\alpha)})^{-1}=(W_{k+1}^{(\alpha-1)})^{-1}V_{k+1}^{(\alpha-1)}.$

Such factorizations are well known in the theory of integrable systems, see
for instance Adler[1], Sklyanin[15].

We study the connection matrices more explicitly.

######  Lemma 2.3.

(2.14) $\displaystyle U_{k}^{(\alpha)}$
$\displaystyle=(g^{[k](\alpha)}_{-})^{-1}Tg_{-}^{[k+1](\alpha)}=g_{0+}^{[k](\alpha)}(g_{0+}^{[k+1](\alpha)})^{-1},$
(2.15) $\displaystyle V_{k}^{(\alpha)}$
$\displaystyle=(g_{-}^{[k](\alpha)})^{-1}Q_{0}^{-1}g_{-}^{[k](\alpha+1)}=g_{0+}^{[k](\alpha)}Q_{0}^{-1}(g_{0+}^{[k](\alpha+1)})^{-1},$
(2.16) $\displaystyle W_{k}^{(\alpha)}$
$\displaystyle=(g_{-}^{[k](\alpha)})^{-1}Q_{1}^{-1}g_{-}^{[k-1](\alpha+1)}=g_{0+}^{[k](\alpha)}Q_{0}^{-1}(g_{0+}^{[k-1](\alpha+1)})^{-1}.$

###### Proof.

The first expression for the connection matrices (in terms of negative
components $g^{[k](\alpha)}_{-}$) follows from the definition (2.12) of the
connection matrices and the definition (2.11) of the Baker functions. It also
uses $Q_{0}T=Q_{1}$.

For the second expression for the connection matrices in term of positive
components $g^{[k](\alpha)}_{0+}$ we use (see also (2.4), or rather (2.5))

$\displaystyle Tg^{[k+1](\alpha)}_{-}g^{[k+1](\alpha)}_{0+}$
$\displaystyle=g^{[k](\alpha)}_{-}g^{[k](\alpha)}_{0+},$ $\displaystyle
Q_{0}^{-1}g^{[k](\alpha+1)}_{-}g^{[k](\alpha+1)}_{0+}$
$\displaystyle=g^{[k](\alpha)}_{-}g^{[k](\alpha)}_{0+}Q_{0}^{-1},$
$\displaystyle Q_{1}^{-1}g^{[k-1](\alpha+1)}_{-}g^{[k-1](\alpha+1)}_{0+}$
$\displaystyle=g^{[k](\alpha)}_{-}g^{[k](\alpha)}_{0+}Q_{0}^{-1}.$

Rearranging factors proves then the second form for the connection matrices. ∎

Note that the second equality in the above Lemma tells us that the elementary
connection matrices are positive in $z$, i.e., contain only $z^{k}$ for $k\geq
0$. This allows us to easily calculate the connection matrices in terms of
$\tau$-functions.

First note that Theorem 2.2 allows us to expand $g^{[k](\alpha)}$ and its
inverse up to order $z^{-1}$ as

(2.17) $\displaystyle g_{-}^{[k](\alpha)}$ $\displaystyle=1_{2\times
2}+\frac{1}{z}\begin{bmatrix}S^{+}[1]\tau_{k}^{(\alpha)}/\tau_{k}^{(\alpha)}&\frac{1}{h_{k-1}^{(\alpha)}}\\\
&\\\
h_{k}^{(\alpha)}&S^{-}[1]\tau^{(\alpha)}_{k}/\tau^{(\alpha)}_{k}\end{bmatrix}+\mathcal{O}(z^{-2}),$
$\displaystyle(g_{-}^{[k](\alpha)})^{-1}$ $\displaystyle=1_{2\times
2}+\frac{1}{z}\begin{bmatrix}-S^{+}[1]\tau_{k}^{(\alpha)}/\tau_{k}^{(\alpha)}&\frac{-1}{h_{k-1}^{(\alpha)}}\\\
&\\\
-h_{k}^{(\alpha)}&-S^{-}[1]\tau^{(\alpha)}_{k}/\tau^{(\alpha)}_{k}\end{bmatrix}+\mathcal{O}(z^{-2}).$

Here

$h_{k}^{(\alpha)}=\frac{\tau_{k+1}^{(\alpha)}}{\tau_{k}^{(\alpha)}},$

and we expand the shift fields in partial shifts:

$S^{\pm}(z)f=\sum_{n=0}^{\infty}S^{\pm}[n]fz^{-n},$

######  Lemma 2.4.

We have

$U^{(\alpha)}_{k}=\begin{bmatrix}z-b^{(\alpha)}_{k}&\frac{1}{h^{(\alpha)}_{k}}\\\
-h^{(\alpha)}_{k}&0\end{bmatrix}.$

Similarly,

$\displaystyle V_{k}^{(\alpha)}$
$\displaystyle=\begin{bmatrix}z-\alpha^{(\alpha)}_{k-1}&\frac{1}{h_{k-1}^{(\alpha+1)}}\\\
-h_{k}^{(\alpha)}&1\end{bmatrix},\quad
W^{(\alpha)}_{k}=\begin{bmatrix}1&-\frac{1}{h^{(\alpha)}_{k-1}}\\\
h_{k-1}^{(\alpha+1)}&z-\beta^{(\alpha)}_{k-1}\end{bmatrix},\quad k=0,1,\dots$

Here

$\alpha_{k}^{(\alpha)}=\frac{h_{k+1}^{(\alpha)}}{h_{k}^{(\alpha+1)}},\quad\beta_{k}^{(\alpha)}=\frac{h_{k}^{(\alpha+1)}}{h_{k}^{(\alpha)}},\quad
b_{k}^{(\alpha)}=\alpha_{k-1}^{(\alpha)}+\beta_{k-1}^{(\alpha)},$

and we put $\alpha_{-1}^{(\alpha)}=\frac{1}{h_{-1}^{(\alpha)}}=0$.

Note that each $U^{(\alpha)}_{k}$ has determinant 1, and that
$\det(V^{(\alpha)}_{k})=z=\det(W^{(\alpha)}_{k})$.

###### Proof.

As an example we calculate
$V_{k}^{(\alpha)}=\left(g^{[k](\alpha)}_{-}\right)^{-1}Q_{0}^{-1}g^{[k](\alpha+1)}$,
using (2.17):

$\displaystyle V_{k}^{(\alpha)}$
$\displaystyle=\begin{bmatrix}1+x/z+\mathcal{O}(z^{-2})&\mathcal{O}(z^{-1})\\\
-\frac{h_{k}^{(\alpha)}}{z}+\mathcal{O}(z^{-2})&1+\mathcal{O}(z^{-1})\end{bmatrix}\begin{bmatrix}z&0\\\
0&1\end{bmatrix}\begin{bmatrix}1+y/z+\mathcal{O}(z^{-2})&\frac{1}{zh_{k-1}^{(\alpha+1)}}\\\
\mathcal{O}(z^{-1})&1+\mathcal{O}(z^{-1})\end{bmatrix}=$
$\displaystyle=\begin{bmatrix}z+x+y&\frac{1}{h_{k-1}^{(\alpha+1)}}\\\
-h_{k}^{(\alpha)}&1\end{bmatrix},$

dropping all terms containing $z^{-1}$ or lower. Here $x,y$ are some
expressions in the $\tau$-functions we will determine by noting that
$\det(V_{k}^{(\alpha)})=z$. We see that
$x+y=\frac{-h_{k}^{(\alpha)}}{h_{k-1}^{(\alpha+1)}}=-\alpha_{k-1}^{(\alpha)}$.
This proves the Lemma for $V_{k}^{(\alpha)}$, $k=1,2,\dots$. The proof for
$W_{k}^{(\alpha)}$ and $V_{0}^{(\alpha)}$ is similar. To get the formula for
$U_{k}^{(\alpha)}$ use
$U_{k}^{(\alpha)}=V_{k}^{(\alpha)}\left(W_{k+1}^{(\alpha)}\right)^{-1}$, again
discarding terms involving $z^{-1}$ and lower. ∎

Now we return to the two factorization (2.13) of the connection matrix. Using
the expressions for $V_{k}^{(\alpha)},W_{k}^{(\alpha)}$ from Lemma 2.4 we find
two expressions for the coefficient $b_{k}^{(\alpha)}$ in $U_{k}^{(\alpha)}$:

$b_{k}^{(\alpha)}=\alpha_{k-1}^{(\alpha)}+\beta_{k-1}^{(\alpha)}=\alpha_{k}^{(\alpha-1)}+\beta_{k}^{(\alpha-1)},$

giving equations for the $h_{k}^{(\alpha)}$ variables:

(2.18)
$\frac{h^{(\alpha)}_{k}}{h_{k-1}^{(\alpha+1)}}+\frac{h_{k-1}^{(\alpha+1)}}{h^{(\alpha)}_{k-1}}=\frac{h_{k+1}^{(\alpha-1)}}{h_{k}^{(\alpha)}}+\frac{h_{k}^{(\alpha)}}{h_{k}^{(\alpha-1)}}.$

######  Theorem 2.5.

The equations (2.18) are equivalent to the $Q$-system

(2.19)
$(\tau_{k}^{(\alpha)})^{2}=\tau_{k}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}-\tau_{k+1}^{(\alpha-1)}\tau_{k-1}^{(\alpha+1)},\quad
k=0,1,\dots.$

###### Proof.

Write (2.18) out in terms of $\tau$-functions:

$\frac{\tau_{k+1}^{(\alpha)}\tau_{k-1}^{(\alpha+1)}}{\tau_{k}^{(\alpha)}\tau_{k}^{(\alpha+1)}}+\frac{\tau_{k}^{(\alpha+1)}\tau_{k-1}^{(\alpha)}}{\tau_{k-1}^{(\alpha+1)}\tau_{k}^{(\alpha)}}=\frac{\tau_{k+2}^{(\alpha-1)}\tau_{k}^{(\alpha)}}{\tau_{k+1}^{(\alpha-1)}\tau_{k+1}^{(\alpha)}}+\frac{\tau_{k+1}^{(\alpha)}\tau_{k}^{(\alpha-1)}}{\tau_{k}^{(\alpha)}\tau_{k+1}^{(\alpha-1)}}.$

Writing all terms under the same denominator and rearranging terms, we see
that this is equivalent to:

(2.20)
$(\tau_{k}^{(\alpha)})^{2}(\tau_{k+2}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}-\tau_{k+1}^{(\alpha-1)}\tau_{k+1}^{(\alpha+1)})=(\tau_{k+1}^{(\alpha)})^{2}(\tau_{k+1}^{(\alpha-1)}\tau_{k-1}^{(\alpha+1)}-\tau_{k}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}).$

Notice that if

$(\tau_{k}^{\alpha})^{2}=\tau_{k}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}-\tau_{k+1}^{(\alpha-1)}\tau_{k-1}^{(\alpha+1)},$

then (2.20) implies

$(\tau_{k+1}^{\alpha})^{2}=\tau_{k+1}^{(\alpha-1)}\tau_{k+1}^{(\alpha+1)}-\tau_{k+2}^{(\alpha-1)}\tau_{k}^{(\alpha+1)}.$

We thus need only prove that the equality holds for $k=0$. But this is just

$(\tau_{0}^{\alpha})^{2}=\tau_{0}^{(\alpha-1)}\tau_{0}^{(\alpha+1)}-\tau_{1}^{(\alpha-1)}\tau_{-1}^{(\alpha+1)},$

which is true since $\tau_{-1}^{(\alpha)}=0$ and $\tau_{0}^{(\alpha)}=1$ for
all $\alpha$. So the theorem follows. ∎

So we have rederived the equations (2.7) using the Birkhoff factorization.

##  3\. $3\times 3$ Case

###  3.1. $\tau$-functions

We proceed very much as in the $2\times 2$ case.

We have an action of the central extension $\operatorname{\widehat{GL}_{3}}$
of the loop group
$\operatorname{\widetilde{GL}_{3}}=GL_{3}(\mathbb{C}((z^{-1})))$ on three
component semi-infinite wedge space $F^{(3)}$, see Appendix A.4. Let
$\pi\colon\operatorname{\widehat{GL}_{3}}\to\operatorname{\widetilde{GL}_{3}}$
be the projection on the loop group and consider the action on the vacuum
vector of $F^{(3)}$, see (A.4), of $g_{C}\in\operatorname{\widehat{GL}_{2}}$,
where

(3.1) $\pi(g_{C})=\begin{bmatrix}1&0&0\\\ C(z)&1&0\\\
D(z)&E(z)&1\end{bmatrix}.$

Here

$X(z)=\sum_{i\in\mathbb{Z}}x_{i}z^{-i-1},\quad X=C,D,E,\quad x=c,d,e.$

where $x_{i}\in\mathbb{C}$; sometimes it is also useful to think of the
$x_{i}$ as formal variables.

Recall (see (A.12)) the projection of the fermionic translation operators to
the loop group $\operatorname{\widetilde{GL}_{3}}$: in particular we get the
(commuting) matrices

$\pi(Q_{0})=\begin{bmatrix}z^{-1}&0&0\\\ 0&1&0\\\
0&0&1\end{bmatrix},\quad\pi(Q_{1})=\begin{bmatrix}1&0&0\\\ 0&z^{-1}&0\\\
0&0&1\end{bmatrix},\quad\pi(Q_{2})=\begin{bmatrix}1&0&0\\\ 0&1&0\\\
0&0&z^{-1}\end{bmatrix}.$

We also have the translation elements

$T_{1}=Q_{1}Q_{0}^{-1}\overset{\pi}{\mapsto}\begin{pmatrix}z&0&0\\\
0&z^{-1}&0\\\ 0&0&1\end{pmatrix},\quad
T_{2}=Q_{2}Q_{1}^{-1}\overset{\pi}{\mapsto}\begin{pmatrix}1&0&0\\\ 0&z&0\\\
0&0&z^{-1}\end{pmatrix}$

We define shifts on the series $X(z)$ by

(3.2)
$X^{(\alpha)}(z)=z^{\alpha}X(z)=\sum_{i\in\mathbb{Z}}x_{i+\alpha}z^{-i-1},\quad
X=C,D,E,\quad x=c,d,e.$

We define similarly the shifted group element
$g^{(\alpha,\beta)}=Q_{0}^{\alpha}Q_{1}^{\beta}g_{C}Q_{1}^{-\beta}Q_{0}^{-\alpha}$
so that

(3.3) $\pi(g_{C}^{(\alpha,\beta)})=\begin{bmatrix}1&0&0\\\
C^{(\alpha-\beta)}(z)&1&0\\\ D^{(\alpha)}(z)&E^{(\beta)}(z)&1\end{bmatrix}.$

We then have (using $Q_{0}Q_{1}=-Q_{1}Q_{0}$ twice)

(3.4) $Q_{0}^{-1}g^{(\alpha+1,\beta)}=g^{(\alpha,\beta)}Q_{0}^{-1},\quad
Q_{1}^{-1}g^{(\alpha,\beta+1)}=g^{(\alpha,\beta)}Q_{1}^{-1},$

and the same relations with $\pi$ applied,

(3.5)
$\pi(Q_{0})^{-1}\pi(g^{(\alpha+1,\beta)})=\pi(g^{(\alpha,\beta)})\pi(Q_{0})^{-1},\quad\pi(Q_{1})^{-1}\pi(g^{(\alpha,\beta+1)})=\pi(g^{(\alpha,\beta)})\pi(Q_{1})^{-1},$

The fundamental objects in the $3\times 3$ theory are the $\tau$-functions
defined by

(3.6)
$\boxed{\tau_{k,l}^{(\alpha\beta)}=<T_{1}^{k}T_{2}^{l}v_{0},g^{(\alpha,\beta)}v_{0}>.}$

Note that if we introduce another translation group element
$T_{3}=Q_{2}Q_{0}^{-1}$ then we can write non uniquely

$T_{1}^{k}T_{2}^{l}=(\pm 1)T_{1}^{n_{c}}T_{2}^{n_{e}}T_{3}^{n_{d}},$

where $k=n_{c}+n_{d},l=n_{d}+n_{e}$, and we take $n_{c},n_{d},n_{e}\geq 0$.

######  Theorem 3.1.

(3.7) $\tau^{\alpha,\beta}_{k,l}=\sum_{n_{c}+n_{d}=k,n_{d}+n_{e}=l\atop
n_{c},n_{d},n_{e}\geq 0}c^{(\alpha,\beta)}_{n_{c},n_{d},n_{e}}$

where

$c^{(\alpha,\beta)}_{n_{c},n_{d},n_{e}}=\frac{(-1)^{\frac{{n_{d}(n_{d}+1)}}{2}}}{n_{c}!n_{d}!n_{e}!}\operatorname{Res}_{\mathbf{x},\mathbf{y},\mathbf{z}}\left(\prod_{i=1}^{n_{c}}C^{(\alpha-\beta)}(x_{i})\prod_{i=1}^{n_{d}}D^{(\alpha)}(y_{i})\prod_{i=1}^{n_{e}}E^{(\beta)}(z_{i})p_{n_{c},n_{d},n_{e}}\right),$

and

$p_{n_{c},n_{c},n_{e}}=\\\ =\frac{\prod\limits_{1\leq i<j\leq
n_{c}}(x_{i}-x_{j})^{2}\prod\limits_{1\leq i<j\leq
n_{d}}(y_{i}-y_{j})^{2}\prod\limits_{1\leq i<j\leq
n_{e}}(z_{i}-z_{j})^{2}\prod\limits_{i=1}^{n_{c}}\prod\limits_{j=1}^{n_{d}}(x_{i}-y_{j})\prod\limits_{i=1}^{n_{d}}\prod\limits_{j=1}^{n_{e}}(y_{i}-z_{j})}{\prod\limits_{i=1}^{n_{c}}\prod\limits_{j=1}^{n_{e}}(x_{i}-z_{j})}$

We discuss the proof in Appendix B.2.

###  3.2. Examples of $3\times 3$ tau functions

  1. (1)

$\tau_{k,\ell}^{(\alpha,\beta)}=0$ if $k<0$ or $\ell<0$

  2. (2)

$\tau_{0,0}^{(\alpha,\beta)}=1$

  3. (3)

$\tau_{k,0}^{(\alpha,\beta)}=\det\begin{bmatrix}c_{\alpha-\beta}&c_{\alpha-\beta+1}&\cdots&c_{\alpha-\beta+k-1}\\\
c_{\alpha-\beta+1}&c_{\alpha-\beta+2}&\cdots&c_{\alpha-\beta+k}\\\
\vdots&\vdots&\cdots&\vdots\\\
c_{\alpha-\beta+k-1}&c_{\alpha-\beta+k}&\cdots&c_{\alpha-\beta+2k-2}\\\
\end{bmatrix}$

  4. (4)

$\tau_{0,k}^{(\alpha,\beta)}=\det\begin{bmatrix}e_{\beta}&e_{\beta+1}&\cdots&e_{\beta+k-1}\\\
e_{\beta+1}&e_{\beta+2}&\cdots&e_{\beta+k}\\\ \vdots&\vdots&\cdots&\vdots\\\
e_{\beta+k-1}&e_{\beta+k}&\cdots&e_{\beta+2k-2}\\\ \end{bmatrix}$

  5. (5)

$\tau_{1,1}^{(\alpha,\beta)}=-d_{\alpha}+\displaystyle\sum_{i=0}^{\infty}e_{\beta+i}c_{\alpha-\beta-i-1}$

  6. (6)

$\tau_{1,2}^{(\alpha,\beta)}=e_{\beta}\displaystyle\sum_{i=1}^{\infty}e_{\beta+i+1}c_{\alpha-\beta-i-1}-e_{\beta+1}\displaystyle\sum_{i=0}^{\infty}e_{\beta+i+1}c_{\alpha-\beta-i-2}+e_{\beta+1}d_{\alpha}-e_{\beta}d_{\alpha+1}$

  7. (7)

$\tau_{2,1}^{(\alpha,\beta)}=c_{\alpha-\beta+1}\displaystyle\sum_{i=0}^{\infty}e_{\beta+i}c_{\alpha-\beta-i-1}-c_{\alpha-\beta}\displaystyle\sum_{i=0}^{\infty}e_{\beta+i}c_{\alpha-\beta-i}+c_{\alpha-\beta}d_{\alpha+1}-c_{\alpha-\beta+1}d_{\alpha}$

######  Remark 3.2.

Note that the summands $c^{(\alpha,\beta)}_{n_{c},n_{d},n_{e}}$ of the
$\tau$-functions are of degree $n_{x}$ in the coefficients $x_{k}$ of the
series $X(z)=\sum x_{k}z^{-k-1}$, for $x=c,d,e$, $X=C,D,E$.

###  3.3. Birkhoff Factorization for $3\times 3$

Define centrally extended loop group elements

(3.8) $g^{[k,l](\alpha,\beta)}=T^{-l}_{2}T_{1}^{-k}g^{(\alpha,\beta)},$

and assume that they have a Birkhoff Factorization([14], see also Appendix
C.1):

$g^{[k,l](\alpha,\beta)}=g^{[k,l](\alpha,\beta)}_{-}g^{[k,l](\alpha,\beta)}_{0,+},$

where $\pi(g^{[k,l](\alpha,\beta)}_{-})=1+\mathcal{O}(z^{-1})$ and
$\pi(g^{[k,l](\alpha,\beta)}_{0+})=A_{k,l}^{(\alpha,\beta)}+\mathcal{O}(z)$,
for $A_{k,l}^{(\alpha,\beta)}$ an invertible $z$ independent matrix. This
assumption is justified precisely when $\tau^{(\alpha,\beta)}_{k,l}(g)=\langle
v_{0},g^{[k,l](\alpha,\beta)}v_{0}\rangle$ is not zero.

Now we want display the negative component of $\pi(g^{[k,l](\alpha,\beta)})$.
To calculate this we make some extra structure explicit.

Let $\mathcal{N}$ be the subgroup of elements of
$\operatorname{\widetilde{SL}_{3}}$ of the form (3.3). We can think of the
coefficients $x_{k}$, $x=c,d,e$ as coordinates on $\mathcal{N}$, so

$B=\mathbb{C}[[c_{k},d_{k},e_{k}]]_{k\in\mathbb{Z}}$

is the coordinate ring of $\mathcal{N}$.

Define first shifts acting on $B$: these are multiplicative maps given on
generators by ($x,y=c,d,e$)

(3.9) $S_{x}^{\alpha}\colon B\to B,\quad S_{x}^{\alpha}(1)=0,\quad
S_{x}^{\alpha}(x_{k})=x_{k+\alpha},\quad
S_{x}^{\alpha}\left(y_{k}\right)=y_{k},y\neq x,\quad\alpha\in\mathbb{Z}.$

We will often write $S_{x}^{\pm}$ for $S_{x}^{\pm 1}$.

Define also _shift fields_. These are multiplicative maps

(3.10) $S_{x}^{\pm}(z)\colon B\to B[[z^{-1}]],$

given by

$S_{x}^{\pm}(z)=(1-\frac{S_{x}^{+}}{z})^{\pm 1},$

######  Theorem 3.3.

For $k\geq 0$ and all $\alpha\in\mathbb{Z}$

$\pi(g^{[k,l](\alpha,\beta)}_{-})=\left(\Sigma\mathcal{T}_{k,l}^{(\alpha,\beta)}\right)/\tau_{kl}^{(\alpha,\beta)},$

where

$\displaystyle\Sigma=$
$\displaystyle\begin{bmatrix}S_{c}^{+}(z)S_{d}^{+}(z)&0&0\\\ &&\\\
0&S^{-}_{c}(z)S^{+}_{e}(z)&0\\\ &&\\\
0&0&S^{-}_{d}(z)S_{e}^{-}(z)\end{bmatrix},$
$\displaystyle\mathcal{T}_{k,l}^{(\alpha,\beta)}=$
$\displaystyle\begin{bmatrix}\tau_{k,l}^{(\alpha,\beta)}&\frac{\tau_{k-1,l}^{(\alpha,\beta)}}{z}&(-1)^{k}\frac{\tau_{k-1,l-1}^{(\alpha,\beta)}}{z}\\\
&&\\\
\frac{\tau_{k+1,l}^{(\alpha,\beta)}}{z}&\tau_{k,l}^{(\alpha,\beta)}&(-1)^{k}\frac{\tau_{k,l-1}^{(\alpha,\beta)}}{z}\\\
&&\\\
(-1)^{k+1}\frac{\tau_{k+1,l+1}^{(\alpha,\beta)}}{z}&(-1)^{k}\frac{\tau_{k,l+1}^{(\alpha,\beta)}}{z}&\tau_{k,l}^{(\alpha,\beta)}\\\
\end{bmatrix}$

We sketch the proof in Appendix C.3

Now we have expressed the negative component of the Birkhoff factorization in
terms of matrix elements of the centrally extended loop group we have no
longer need for the central extension and we will simplify notation: we will
write $g^{[k,l](\alpha,\beta)}_{-}$ for $\pi(g^{[k,l](\alpha,\beta)}_{-})$,
$T_{i}$ for $\pi(T_{i})$, $i=1,2$ and similarly $Q_{a}$ for $\pi(Q_{a})$,
$a=0,1,3$. In particular in the rest of this section we write

$T_{1}=\begin{bmatrix}z&0&0\\\ 0&z^{-1}&0\\\
0&0&1\end{bmatrix}=Q_{1}Q_{0}^{-1},\quad T_{2}=\begin{bmatrix}1&0&0\\\
0&z&0\\\ 0&0&z^{-1}\end{bmatrix}=Q_{2}Q_{1}^{-1},$

and

$Q_{0}=\begin{bmatrix}z^{-1}&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix},\quad
Q_{1}=\begin{bmatrix}1&0&0\\\ 0&z^{-1}&0\\\ 0&0&1\end{bmatrix},\quad
Q_{2}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&z^{-1}\end{bmatrix}.$

###  3.4. Matrix Baker Functions and Connection Matrices, $3\times 3$ case

Next we define the Baker functions: these are now elements of the loop group
of $\mathbf{GL}_{3}$ defined by

$\displaystyle\Psi^{[k,l](\alpha,\beta)}$
$\displaystyle=T_{1}^{k}T_{2}^{l}Q_{0}^{-\alpha}Q_{1}^{-\beta}g_{-}^{[k,l](\alpha,\beta)}.$

Since the Baker functions are all invertible, they are related by (right)
multiplication by connection matrices belonging to
$\operatorname{\widetilde{GL}_{3}}$. In particular, define

(3.11)
$\Psi^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}=\Psi^{[k,l](\alpha,\beta)}\Gamma^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}_{[k,l](\alpha,\beta)},$

so that

$\Gamma^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}Q_{0}^{x_{0}}Q_{1}^{x_{1}}Q_{2}^{x_{2}}g_{-}^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})},$

where

$\displaystyle x_{0}$ $\displaystyle=k-k^{\prime}+\alpha-\alpha^{\prime},$
$\displaystyle x_{1}$
$\displaystyle=k^{\prime}-k+l-l^{\prime}+\beta-\beta^{\prime},$ $\displaystyle
x_{2}$ $\displaystyle=l^{\prime}-l.$

The simplest connection matrices are those where $(x_{0},x_{1},x_{1})$ has two
zero components and the other absolute value 1. Define therefore the
elementary connection matrices

(3.12) $\displaystyle V_{[k,l]}^{(\alpha_{+},\beta)}$
$\displaystyle=\Gamma^{[k,l](\alpha+1,\beta)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}Q_{0}^{-1}g_{-}^{[k,l](\alpha+1,\beta)},$
$\displaystyle V_{[k,l]}^{(\alpha,\beta_{+})}$
$\displaystyle=\Gamma^{[k,l](\alpha,\beta+1)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}Q_{1}^{-1}g_{-}^{[k,l](\alpha,\beta+1)},$
$\displaystyle W_{[k_{-},l]}^{(\alpha_{+},\beta)}$
$\displaystyle=\Gamma^{[k-1,l](\alpha+1,\beta)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}Q_{1}^{-1}g_{-}^{[k-1,l](\alpha+1,\beta)},$
$\displaystyle W_{[k,l_{-}]}^{(\alpha,\beta_{+})}$
$\displaystyle=\Gamma^{[k,l-1](\alpha,\beta+1)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}Q_{2}^{-1}g_{-}^{[k,l-1](\alpha,\beta+1)}.$

Also define translation matrices

(3.13) $\displaystyle U^{(\alpha,\beta)}_{[k_{+},l]}$
$\displaystyle=\Gamma^{[k+1,l](\alpha,\beta)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}T_{1}(g_{-}^{[k+1,l](\alpha,\beta)}),$
$\displaystyle U^{(\alpha,\beta)}_{[k,l_{+}]}$
$\displaystyle=\Gamma^{[k,l+1](\alpha,\beta)}_{[k,l](\alpha,\beta)}=(g_{-}^{[k,l](\alpha,\beta)})^{-1}T_{2}(g_{-}^{[k,l+1](\alpha,\beta)})$

Pictorially, fixing $k,\alpha$

${\Psi^{[k,l](\alpha,\beta-1)}}$${\Psi^{[k,l+1](\alpha,\beta-1)}}$${\Psi^{[k,l+2](\alpha,\beta-1)}}$${\Psi^{[k,l](\alpha,\beta)}}$${\Psi^{[k,l+1](\alpha,\beta)}}$${\Psi^{[k,l-1](\alpha,\beta+1)}}$${\Psi^{[k,l](\alpha,\beta+1)}}$${\Psi^{[k,l+1](\alpha,\beta+1)}}$$\scriptstyle{U_{k,l_{+}}^{(\alpha,\beta-1)}}$$\scriptstyle{V_{[k,l]}^{(\alpha,(\beta-1)_{+})}}$$\scriptstyle{U_{k,l+1}^{(\alpha,\beta-1)}}$$\scriptstyle{V_{[k,l+1]}^{(\alpha,(\beta-1)_{+})}}$$\scriptstyle{W_{[k,(l+1)_{-}]}^{(\alpha,(\beta-1)_{+})}}$$\scriptstyle{W_{[k,(l+2)_{-}]}^{(\alpha,(\beta-1)_{+})}}$$\scriptstyle{U_{k,l_{+}}^{(\alpha,\beta)}}$$\scriptstyle{V_{[k,l]}^{(\alpha,\beta_{+})}}$$\scriptstyle{W_{[k,l_{-}]}^{(\alpha,\beta_{+})}}$$\scriptstyle{W_{[k,(l+1)_{-}]}^{(\alpha,\beta_{+})}}$$\scriptstyle{V_{[k,(l+1)_{-}](\alpha,\beta_{+})}}$$\scriptstyle{U_{k,l_{+}}^{(\alpha,\beta+1)}}$$\scriptstyle{U_{k,l_{+}}^{(\alpha,\beta+1)}}$

Walking around the triangles in this and the similar diagram where $l,\beta$
are fixed we see that we get factorizations of the basic translation matrices
$U_{[k_{+},l]}^{(\alpha,\beta)},U_{[k,l_{+}]}^{(\alpha,\beta)}$: for instance

(3.14) $\displaystyle U_{[k,l_{+}]}^{(\alpha,\beta)}$
$\displaystyle=V_{[k,l]}^{(\alpha,\beta_{+})}(W_{[k,(l+1)_{-}]}^{(\alpha,\beta_{+})})^{-1}=(W_{[k,(l+1)_{-}]}^{(\alpha,(\beta-1)_{+})})^{-1}V_{[k,l+1]}^{(\alpha,(\beta-1)_{+})},$
$\displaystyle U_{[k_{+},l]}^{(\alpha,\beta)}$
$\displaystyle=V_{[k,l]}^{(\alpha_{+},\beta)}(W_{[(k+1)_{-},l]}^{(\alpha_{+},\beta)})^{-1}=(W_{[(k+1)_{-},l]}^{((\alpha-1)_{+},\beta)})^{-1}V_{[k+1,l]}^{((\alpha-1)_{+},\beta)}.$

Such factorizations are well known in the theory of integrable systems, see
for instance Adler[1], Sklyanin[15].

We will argue that all identities for the connection matrices are the result
of those in (3.14).

######  Lemma 3.4.

Any connection matrix
$\Gamma^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}_{[k,l](\alpha,\beta)}$
is a product of the four types of elementary matrices (3.12).

###### Proof.

We need to show that you can move from $\Psi^{[k,l](\alpha,\beta)}$ to
$\Psi^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}$ just using
the elementary connections matrices. First of all the translation matrices
(3.13) are products of elementary connection matrices: see (3.14). Now we can
move from $\Psi^{[k,l](\alpha,\beta)}$ to
$\Psi^{[k^{\prime},l^{\prime}](\alpha,\beta)}$ using just
$U_{[(k_{i})_{+},l_{i}]}$ and/or $U_{[k_{i},(l_{i})_{+}]}$, keeping
$(\alpha,\beta)$ fixed. Then we use the $V^{(\alpha_{+},\beta)}$ and/or
$V^{(\alpha,\beta_{+})}$ to adjust the $(\alpha,\beta)$ to
$(\alpha^{\prime},\beta^{\prime})$ to reach
$\Psi^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}$. ∎

The expression in the Lemma of
$\Gamma=\Gamma^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}_{[k,l](\alpha,\beta)}$
as a product of elementary connection matrices is of course not unique. Each
path from $\Psi^{[k,l](\alpha,\beta)}$ to
$\Psi^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}$ in the
lattice of Baker functions with diagonal or antidiagonal steps gives a product
expression for $\Gamma$: the diagonal steps give a factor $V$ and the anti-
diagonal steps give $W$ factors. Now it should be clear that any two paths
from $\Psi^{[k,l](\alpha,\beta)}$ to
$\Psi^{[k^{\prime},l^{\prime}](\alpha^{\prime},\beta^{\prime})}$ can be
deformed into each other by moves

(3.15)
${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$

or

(3.16)
${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$${\bullet}$

The moves (3.16) correspond to identities (3.14) and moves (3.15) correspond
to similar equations of the form $VW=WV$ (without inverses on $W$). These last
equations will be equivalent to those in (3.14). So the upshot is that all
equations obtained by writing an arbitrary connection matrix $\Gamma$ as a
product of elementary connection matrices follow from (3.14).

We will therefore concentrate on (3.14). In particular we will see that these
equations will give equations for our $\tau$ functions.

We first check that our elementary connection matrices and translation
matrices do not contain any negative powers of $z$, which is not obvious from
the definitions (3.12) and (3.13).

######  Lemma 3.5.

For the elementary connection matrices we have the positive expressions

(3.17) $\displaystyle V_{[k,l]}^{(\alpha_{+},\beta)}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{0+}Q_{0}^{-1}(g^{[k,l](\alpha+1,\beta)}_{0+})^{-1},$
$\displaystyle V_{[k,l]}^{(\alpha,\beta_{+})}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{0+}Q_{1}^{-1}(g^{[k,l](\alpha,\beta+1)}_{0+})^{-1},$
$\displaystyle W_{[k_{-},l]}^{(\alpha_{+},\beta)}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{0+}Q_{0}^{-1}(g^{[k-1,l](\alpha+1,\beta)}_{0+})^{-1},$
$\displaystyle W_{[k,l_{-}]}^{(\alpha,\beta_{+})}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{0+}Q_{1}^{-1}(g^{[k,l-1](\alpha,\beta+1)}_{0+})^{-1}$

Similarly, for the translation matrices

(3.18) $\displaystyle U^{(\alpha,\beta)}_{k_{+},l}$
$\displaystyle=g_{0+}^{[k+1,l](\alpha,\beta)}(g_{0+}^{[k,l](\alpha,\beta)})^{-1}$
$\displaystyle U^{(\alpha,\beta)}_{k,l_{+}}$
$\displaystyle=g_{0+}^{[k,l+1](\alpha,\beta)}(g_{0+}^{[k,l](\alpha,\beta)})^{-1}$

###### Proof.

From (3.5) it follows that

$\displaystyle
Q_{0}^{-1}g^{[k,l](\alpha+1,\beta)}_{-}g^{[k,l](\alpha+1,\beta)}_{0+}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{-}g^{[k,l](\alpha,\beta)}_{0+}Q_{0}^{-1},$
$\displaystyle
Q_{1}^{-1}g^{[k,l](\alpha,\beta+1)}_{-}g^{[k,l](\alpha,\beta+1)}_{0+}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{-}g^{[k,l](\alpha,\beta)}_{0+}Q_{1}^{-1},$

from which the result for the $V$ matrices follows by rearranging factors.

Similarly, using $Q_{1}^{-1}T_{1}=Q_{0}^{-1},Q_{2}^{-1}T_{2}=Q_{1}^{-1}$ and
again (3.5) we find

$\displaystyle
Q_{1}^{-1}g^{[k-1,l](\alpha+1,\beta)}_{-}g^{[k-1,l](\alpha+1,\beta)}_{0+}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{-}g^{[k,l](\alpha,\beta)}_{0+}Q_{0}^{-1},$
$\displaystyle
Q_{2}^{-1}g^{[k,l-1](\alpha,\beta+1)}_{-}g^{[k,l-1](\alpha,\beta+1)}_{0+}$
$\displaystyle=g^{[k,l](\alpha,\beta)}_{-}g^{[k,l](\alpha,\beta)}_{0+}Q_{1}^{-1},$

giving the result for the $W$ matrices.

Finally the positive expression for $U$ matrices follows from

$\displaystyle g^{[k,l](\alpha+1,\beta)}_{-}g^{[k,](\alpha)}_{0+}$
$\displaystyle=g^{[k](\alpha)}_{-}g^{[k](\alpha)}_{0+}Q_{0}^{-1},$
$\displaystyle Q_{1}^{-1}g^{[k](\alpha+1)}_{-}g^{[k](\alpha+1)}_{0+}$
$\displaystyle=g^{[k](\alpha)}_{-}g^{[k+1](\alpha)}_{0+}Q_{1}^{-1},$

and rearranging factors. ∎

The above Lemma tells us that the elementary connection matrices and
translation matrices are positive in $z$, i.e., contain only $z^{k}$ for
$k\geq 0$. This allows us to calculate them easily in terms of
$\tau$-functions.

First note that Theorem 3.3 allows us to expand $g_{-}^{[k,l](\alpha,\beta)}$
and its inverse up to order $z^{-1}$ as (we suppress the shift
$(\alpha,\beta)$)

(3.19)
$g_{-}^{[k,\ell]}=\begin{bmatrix}1+\mathcal{O}(z^{-1})&\frac{1}{zh_{\underline{k-1},\ell}}+\mathcal{O}(z^{-2})&\frac{(-1)^{k}}{zh_{\underline{k-1},\underline{\ell-1}}}+\mathcal{O}(z^{-2})\\\
&&&\\\
\frac{h_{\underline{k},\ell}}{z}+\mathcal{O}(z^{-2})&1+\mathcal{O}(z^{-1})&\frac{(-1)^{k}}{zh_{k,\underline{\ell-1}}}+\mathcal{O}(z^{-2})\\\
&&&\\\
\frac{(-1)^{k+1}h_{\underline{k},\underline{\ell}}}{z}+\mathcal{O}(z^{-2})&\frac{(-1)^{k}h_{k,\underline{\ell}}}{z}+\mathcal{O}(z^{-2})&1+\mathcal{O}(z^{-1})\end{bmatrix},$

where $\mathcal{O}(z^{i})$ are terms with power of $z$ equal to $i$ or lower,
and we define quotients of $\tau$ functions as

$h_{\underline{k},l}=\frac{\tau_{k+1,l}}{\tau_{k,l}},\quad
h_{{k},\underline{l}}=\frac{\tau_{k,l+1}}{\tau_{k,l}},\quad
h_{\underline{k},\underline{l}}=\frac{\tau_{k+1,l+1}}{\tau_{k,l}}.$

This formula then gives the following formula for $(g_{-}^{[k,\ell]})^{-1}$

(3.20)
$(g_{-}^{[k,\ell]})^{-1}=\begin{bmatrix}1+\mathcal{O}(z^{-1})&-\frac{1}{zh_{\underline{k-1},\ell}}+\mathcal{O}(z^{-2})&\frac{(-1)^{k+1}}{zh_{\underline{k-1},\underline{\ell-1}}}+\mathcal{O}(z^{-2})\\\
&&&\\\
\frac{-h_{\underline{k},\ell}}{z}+\mathcal{O}(z^{-2})&1+\mathcal{O}(z^{-1})&\frac{(-1)^{k+1}}{zh_{k,\underline{\ell-1}}}+\mathcal{O}(z^{-2})\\\
&&&\\\
\frac{(-1)^{k}h_{\underline{k},\underline{\ell}}}{z}+\mathcal{O}(z^{-2})&\frac{(-1)^{k+1}h_{k,\underline{\ell}}}{z}+\mathcal{O}(z^{-2})&1+\mathcal{O}(z^{-1})\end{bmatrix}.$

###  3.5. Explicit Formulae for Connection Matrices

######  Lemma 3.6.

(3.21) $\displaystyle V^{(\alpha_{+},\beta)}_{[k,l]}$
$\displaystyle=\begin{bmatrix}z+A&\frac{1}{h_{\underline{k-1},\underline{\ell}}^{(\alpha+1,\beta)}}&\frac{(-1)^{k}}{h_{\underline{k-1},\underline{\ell-1}}^{(\alpha+1,\beta)}}\\\
&&&\\\ -h_{\underline{k},l}^{(\alpha,\beta)}&1&0\\\ &&&\\\
(-1)^{k}h_{\underline{k},\underline{l}}^{(\alpha,\beta)}&0&1\end{bmatrix},\quad
A=-\frac{h_{\underline{k},\ell}^{(\alpha,\beta)}}{h_{\underline{k-1},\ell}^{(\alpha+1,\beta)}}+\frac{h_{\underline{k},\underline{\ell}}^{(\alpha,\beta)}}{h_{\underline{k-1},\underline{\ell-1}}^{(\alpha+1,\beta)}},$
$\displaystyle V^{(\alpha,\beta_{+})}_{[k,l]}$
$\displaystyle=\begin{bmatrix}1&\frac{-1}{h^{(\alpha,\beta)}_{\underline{k-1},l}}&0\\\
h^{(\alpha,\beta+1)}_{\underline{k},l}&z+B&\frac{(-1)^{k}}{h^{(\alpha,\beta+1)}_{k,\underline{l-1}}}\\\
0&(-1)^{k+1}h_{k,\underline{l}}^{(\alpha,\beta)}&1\end{bmatrix},\quad
B=\frac{-h_{{k},\underline{l}}^{(\alpha,\beta)}}{h_{k,\underline{l-1}}^{(\alpha,\beta+1)}}+\frac{-h_{\underline{k},l}^{(\alpha,\beta+1)}}{h_{\underline{k-1},l}^{(\alpha,\beta)}},$
$\displaystyle W^{(\alpha_{+},\beta)}_{[k_{-},l]}$
$\displaystyle=\begin{bmatrix}1&\frac{-1}{h_{\underline{k-1},l}^{(\alpha,\beta)}}&0\\\
&&&\\\
h_{\underline{k-1},l}^{(\alpha+1,\beta)}&z+C&\frac{(-1)^{k-1}}{h_{k-1,\underline{l-1}}^{(\alpha+1,\beta)}}\\\
&&&\\\ 0&(-1)^{k+1}h_{k,\underline{l}}^{(\alpha,\beta)}&1\end{bmatrix},\quad
C=\frac{h_{k,\underline{l}}^{(\alpha,\beta)}}{h_{k-1,\underline{l-1}}^{(\alpha+1,\beta)}}+\frac{-h_{\underline{k-1},l}^{(\alpha+1,\beta)}}{h_{\underline{k-1},l}^{(\alpha,\beta)}},$
$\displaystyle W^{(\alpha,\beta_{+})}_{[k,l_{-}]}$
$\displaystyle=\begin{bmatrix}1&0&\frac{(-1)^{k+1}}{h_{\underline{k-1},\underline{l-1}}^{(\alpha,\beta)}}\\\
&&&\\\ 0&1&\frac{(-1)^{k+1}}{h_{k,\underline{l-1}}^{(\alpha,\beta)}}\\\ &&&\\\
(-1)^{k+1}h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}&(-1)^{k}h_{k,\underline{l-1}}^{(\alpha,\beta+1)}&z+D\end{bmatrix},\quad
D=\frac{h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}}{h_{\underline{k-1},\underline{l-1}}^{(\alpha,\beta)}}-\frac{h_{k,\underline{l-1}}^{(\alpha,\beta+1)}}{h_{k,\underline{l-1}}^{(\alpha,\beta)}}.$

###### Proof.

For the off-diagonal entries use the expressions (3.19) and (3.20) for $g_{-}$
and $(g_{-})^{-1}$ in the definitions (3.17) of the elementary connection
matrices, and then use positivity Lemma 3.5 to discard all terms with $z^{-1}$
or lower. For the diagonal terms use that the determinant of all elementary
connection matrices is $z$. ∎

######  Lemma 3.7.

The inverses of the $W$ matrices are

(3.22) $\displaystyle(W^{(\alpha_{+},\beta)}_{[k_{-},l]})^{-1}$
$\displaystyle=\frac{1}{z}\begin{bmatrix}z-\frac{h_{\underline{k-1},l}^{(\alpha+1,\beta)}}{h_{\underline{k-1},l}^{(\alpha,\beta)}}&\frac{1}{h_{\underline{k-1},l}^{(\alpha,\beta)}}&\frac{(-1)^{k}}{h_{\underline{k-1},l}^{(\alpha,\beta)}h_{k-1,\underline{l-1}}^{(\alpha+1,\beta)}}\\\
&&&\\\
-h_{\underline{k-1},l}^{(\alpha+1,\beta)}&1&\frac{(-1)^{k}}{h_{k-1,\underline{l-1}}^{(\alpha+1,\beta)}}\\\
&&&\\\
(-1)^{k+1}h_{\underline{k-1},l}^{(\alpha+1,\beta)}h_{k,\underline{l}}^{(\alpha,\beta)}&(-1)^{k}h_{k,\underline{l}}^{(\alpha,\beta)}&z+\frac{h_{k,\underline{l}}^{(\alpha,\beta)}}{h_{k-1,\underline{l-1}}^{(\alpha+1,\beta)}}\end{bmatrix},$
(3.23) $\displaystyle(W^{(\alpha,\beta_{+})}_{[k,l_{-}]})^{-1}$
$\displaystyle=\frac{1}{z}\begin{bmatrix}z+\frac{h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}}{h_{\underline{k},\underline{l-1}}^{(\alpha,\beta)}}&-\frac{h_{k,\underline{l-1}}^{(\alpha,\beta+1)}}{h_{\underline{k-1},\underline{l-1}}^{(\alpha,\beta)}}&\frac{(-1)^{k}}{h_{\underline{k-1},\underline{l-1}}^{(\alpha,\beta)}}\\\
&&&\\\
\frac{h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}}{h_{k,\underline{l-1}}^{(\alpha,\beta)}}&z+\frac{h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}}{h_{\underline{k-1},\underline{l-1}}^{(\alpha,\beta)}}&\frac{(-1)^{k}}{h_{k,\underline{l-1}}^{(\alpha,\beta)}}\\\
&&&\\\
(-1)^{k}h_{\underline{k},\underline{l-1}}^{(\alpha,\beta+1)}&(-1)^{k+1}h_{k,\underline{l-1}}^{(\alpha,\beta+1)}&1\end{bmatrix}.$

For the translation matrices we similarly obtain the following expressions

(3.24) $\displaystyle U_{[k_{+},l]}$
$\displaystyle=\begin{bmatrix}z+A&\frac{1}{h_{\underline{k},l}}&\frac{(-1)^{k+1}}{h_{\underline{k},\underline{l-1}}}\\\
&&&\\\ -h_{\underline{k},l}&0&0\\\ &&&\\\
(-1)^{k}h_{\underline{k},\underline{l}}&0&1\end{bmatrix},$ (3.25)
$\displaystyle U_{[k,l_{+}]}$
$\displaystyle=\begin{bmatrix}1&\frac{-1}{h_{\underline{k-1},l}}&0\\\ &&&\\\
h_{\underline{k},l+1}&z+B&\frac{(-1)^{k+1}}{h_{k,\underline{l}}}\\\ &&&\\\
0&(-1)^{k+1}h_{k,\underline{l}}&0\end{bmatrix}.$

Here we suppress the shifts $(\alpha,\beta)$, and we postpone the calculation
of the diagonal terms $A,B$.

###  3.6. Difference Equations from Factorizations

######  Theorem 3.8.

The $\tau$-functions satisfy the following four systems of equations

([2])
$\displaystyle\tau_{k,l-1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}$
$\displaystyle+\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k,l-1}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}=0,$
([1])
$\displaystyle\tau_{k+1,l+1}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}$
$\displaystyle-\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta+1)}+\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta)}=0,$
([0]) $\displaystyle(\tau_{k,l}^{(\alpha,\beta)})^{2}$
$\displaystyle=\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}+\tau_{k+1,l+1}^{(\alpha-1,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)},$
([1,1]) $\displaystyle(\tau_{k,l}^{(\alpha,\beta)})^{2}$
$\displaystyle=\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta-1)}-\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l-1}^{(\alpha,\beta+1)}-\tau_{k-1,l}^{(\alpha,\beta-1)}\tau_{k+1,l}^{(\alpha,\beta+1)}.$

###### Proof.

We substitute in the two factorization (2.13) the results of Lemma 3.6 and
Lemma 3.7. We find rather complicated rational expressions in the
$\tau$-functions. For $U_{k_{+},l}^{(\alpha,\beta)}$ we get non trivial
equation in 3 components:

([0,0])
$\displaystyle-\frac{\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l-1}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k+1,l}^{(\alpha,\beta)}}$
$\displaystyle+\frac{\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}}{\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}-\frac{\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)}}{\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}+$
$\displaystyle+\frac{\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}}{\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}$
$\displaystyle+\frac{\tau_{k+2,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k+1,l}^{(\alpha,\beta)}}-\frac{\tau_{k+1,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}}=0$
([0,2])
$\displaystyle\frac{\tau_{k,l-1}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha,\beta)}}$
$\displaystyle+\frac{\tau_{k-1,l-1}^{(\alpha+1,\beta)}}{\tau_{k,l}^{(\alpha+1,\beta)}}-\frac{\tau_{k,l-1}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}}=0$
([2,0])
$\displaystyle\frac{\tau_{k+2,l+1}^{(\alpha-1,\beta)}}{\tau_{k+1,l}^{(\alpha-1,\beta)}}$
$\displaystyle+\frac{\tau_{k+1,l+1}^{(\alpha,\beta)}}{\tau_{k,l}^{(\alpha,\beta)}}-\frac{\tau_{k+1,l+1}^{(\alpha-1,\beta)}\tau_{k+1,l}^{(\alpha,\beta)}}{\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)}}=0$

For $U_{k,l_{+}}^{(\alpha,\beta)}$ we get

([1,0])
$\displaystyle\frac{\tau_{k+1,l+1}^{a,\beta+1}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta+1)}}$
$\displaystyle-\frac{\tau_{k+1,l+1}^{(\alpha,\beta)}}{\tau_{k,l+1}^{(\alpha,\beta)}}+\frac{\tau_{k+1,l}^{(\alpha,\beta+1)}}{\tau_{k,l}^{(\alpha,\beta+1)}}=0$
([0,1])
$\displaystyle\frac{\tau_{k-1,l}^{a,\beta-1}\tau_{k,l+1}^{(\alpha,\beta)}}{\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta)}}$
$\displaystyle+\frac{\tau_{k-1,l+1}^{(\alpha,\beta-1)}}{\tau_{k,l+1}^{(\alpha,\beta-1)}}-\frac{\tau_{k-1,l}^{(\alpha,\beta)}}{\tau_{k,l}^{(\alpha,\beta)}}=0$
([1,1])
$\displaystyle\frac{\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k-1,l+1}^{(\alpha,\beta-1)}}{\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l+1}^{(\alpha,\beta)}}-\frac{\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k-1,l}^{(\alpha,\beta)}}{\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}}$
$\displaystyle-\frac{\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k,l-1}^{(\alpha,\beta+1)}}{\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}}+$
$\displaystyle+\frac{\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta-1)}}{\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta)}}$
$\displaystyle+\frac{\tau_{k,l+2}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l+1}^{(\alpha,\beta)}}-\frac{\tau_{k,l+1}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}}{\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta+1)}}=0$

The two equations [0,2] and [2,0] can be brought under common denominator, and
then the numerator is forced to vanish. (We think here of the coefficients
$c_{k},d_{k},e_{k}$ as formal variables, so the denominators do not vanish.)
The resulting equations are the same, up to a shift in the variables. This
gives the equation [2] in the theorem. In the same way [1] follows from
[0,1],[1,0].

Next we bring [0,0] under common denominator. The vanishing of the numerator
gives

(3.26)
$\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l-1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}+\\\
+\left(\tau_{k+2,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}-\tau_{k+1,l}^{(\alpha+1,\beta)}\tau_{k+1,l}^{(\alpha-1,\beta)}\right)(\tau_{k,l}^{(\alpha,\beta)})^{2}-\\\
\left(\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)}-\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}\right)(\tau_{k+1,l}^{(\alpha,\beta)})^{2}=0$

In the first term we substitute

$\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k+1,l}^{(\alpha-1,\beta)}=\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k+1,l+1}^{(\alpha-1,\beta)}-\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)},$

which follows from [2] by change of variables $k\mapsto k+1,l\mapsto
l+1,\alpha+1\mapsto\alpha$. The first term then becomes

$\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}\left(\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k+1,l+1}^{(\alpha-1,\beta)}-\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)}\right).$

In the second term of (3.26) we use [2] in the form

$\tau_{k,l-1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}=\tau_{k,l-1}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}-\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}$

so that the second term becomes

$-\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha,\beta)}\left(\tau_{k,l-1}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha,\beta)}-\tau_{k+1,l}^{(\alpha,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}\right).$

After cancellation of two terms and collecting like terms we find for (3.26)

(3.27)
$(\tau_{k,l}^{(\alpha,\beta)})^{2}\left(\tau_{k+2,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}-\tau_{k+1,l}^{(\alpha+1,\beta)}\tau_{k+1,l}^{(\alpha-1,\beta)}-\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l-1}^{(\alpha+1,\beta)}\right)=\\\
(\tau_{k+1,l}^{(\alpha,\beta)})^{2}\left(\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)}-\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}-\tau_{k+1,l+1}^{(\alpha-1,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}\right).$

Now observe that the square factor on the right in (3.27) is obtained from the
square factor on the left by a shift $k\mapsto k+1$, and similarly, mutatis
mutandis, for the terms in the big parentheses. Therefore, if we know that

(3.28)
$(\tau_{k,l}^{(\alpha,\beta)})^{2}=\tau_{k,l}^{(\alpha+1,\beta)}\tau_{k,l}^{(\alpha-1,\beta)}+\tau_{k+1,l+1}^{(\alpha-1,\beta)}\tau_{k-1,l-1}^{(\alpha+1,\beta)}-\tau_{k+1,l}^{(\alpha-1,\beta)}\tau_{k-1,l}^{(\alpha+1,\beta)},$

then (3.27) tells us that also

$(\tau_{k+1,l}^{(\alpha,\beta)})^{2}=\tau_{k+1,l}^{(\alpha+1,\beta)}\tau_{k+1,l}^{(\alpha-1,\beta)}+\tau_{k+2,l+1}^{(\alpha-1,\beta)}\tau_{k,l-1}^{(\alpha+1,\beta)}-\tau_{k+2,l}^{(\alpha-1,\beta)}\tau_{k,l}^{(\alpha+1,\beta)}.$

So it suffice to check (3.28) for $k=0$, which reads

$(\tau_{0,l}^{(\alpha,\beta)})^{2}=\tau_{0,l}^{(\alpha+1,\beta)}\tau_{0,l}^{(\alpha-1,\beta)}.$

But this last equation is true by Example 4 of section 3.2: the
$\tau$-function $\tau_{0,l}^{(\alpha,\beta)}$ is independent of $\alpha$, see
also (3.3). This proves part [0] of the theorem.

Finally we use the [1,1] component of the two expressions of
$U^{(\alpha,\beta)}_{k,l_{+}}$. Bringing it under common denominator, the
vanishing of the numerator gives

(3.29)
$\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k-1,l}^{(\alpha,\beta)}\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l+1}^{(\alpha,\beta)}-\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k-1,l+1}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}+\\\
\left(\tau_{k,l+1}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta-1)}-\tau_{k,l+2}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta+1)}\right)(\tau_{k,l}^{(\alpha,\beta)})^{2}+\\\
\left(\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l-1}^{(\alpha,\beta+1)}-\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta-1)}\right)(\tau_{k,l+1}^{(\alpha,\beta)})^{2}=0$

In the first term substitute

$\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k-1,l}^{(\alpha,\beta)}=\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k-1,l}^{(\alpha,\beta-1)}+\tau_{k,l}^{(\alpha,\beta)}\tau_{k-1,l+1}^{(\alpha,\beta-1)},$

which is obtained from [1] by shifts $k\mapsto k-1,\beta\mapsto\beta-1$. The
first term becomes

$\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta)}\left(\tau_{k,l+1}^{(\alpha,\beta)}\tau_{k-1,l}^{(\alpha,\beta-1)}+\tau_{k,l}^{(\alpha,\beta)}\tau_{k-1,l+1}^{(\alpha,\beta-1)}\right).$

In the second term we also use [1], in the form

$\tau_{k+1,l+1}^{(\alpha,\beta)}\tau_{k,l}^{(\alpha,\beta+1)}=\tau_{k+1,l+1}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}+\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta)},$

transforming the second term to

$-\tau_{k-1,l+1}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta)}\left(\tau_{k+1,l+1}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta)}+\tau_{k+1,l}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta)}\right).$

After cancellation of two terms and collecting like terms (3.29) becomes

(3.30)
$\left(\tau_{k,l+1}^{(\alpha,\beta+1)}\tau_{k,l+1}^{(\alpha,\beta-1)}-\tau_{k,l+2}^{(\alpha,\beta-1)}\tau_{k,l}^{(\alpha,\beta+1)}-\tau_{k-1,l+1}^{(\alpha,\beta-1)}\tau_{k+1,l+1}^{(\alpha,\beta+1)}\right)(\tau_{k,l}^{(\alpha,\beta)})^{2}=\\\
\left(\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta-1)}-\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l-1}^{(\alpha,\beta+1)}-\tau_{k-1,l}^{(\alpha,\beta-1)}\tau_{k+1,l}^{(\alpha,\beta+1)}\right)(\tau_{k,l+1}^{(\alpha,\beta)})^{2}.$

As before, (3.30) implies that if

(3.31)
$(\tau_{k,l}^{(\alpha,\beta)})^{2}=\tau_{k,l}^{(\alpha,\beta+1)}\tau_{k,l}^{(\alpha,\beta-1)}-\tau_{k,l+1}^{(\alpha,\beta-1)}\tau_{k,l-1}^{(\alpha,\beta+1)}-\tau_{k-1,l}^{(\alpha,\beta-1)}\tau_{k+1,l}^{(\alpha,\beta+1)},$

the same equation holds after a shift $l\mapsto l+1$. So we reduce to the case
of $l=0$ of (3.31):

$(\tau_{k,0}^{(\alpha,\beta)})^{2}=\tau_{k,0}^{(\alpha,\beta+1)}\tau_{k,0}^{(\alpha,\beta-1)}--\tau_{k-1,0}^{(\alpha,\beta-1)}\tau_{k+1,0}^{(\alpha,\beta+1)},$

using Item 1 of the examples in subsection 3.2. Now by Item 3 of these
examples $\tau_{k,0}^{(\alpha,\beta)}=\tau_{k}^{(\alpha-\beta)}$. So the case
$l=0$ is just the $Q$-system of Theorem 2.5. This proves the final part [1,1]
of the theorem.

∎

##  Appendix A Multi Component Fermions and Semi Infinite Wedge Space

###  A.1. Introduction

In the main text we will work with $n\times n$ matrices (depending on a
spectral parameter $z$) for $n=2$ or $3$. In this appendix we will not specify
$n$, as the theory of $n$-component fermions and associated semi-infinite
wedge space does not significantly depend on $n$. A convenient reference for
background and more details is ten Kroode and van der Leur [16].

###  A.2. Semi Infinite Wedge Space

Let $\\{e_{0}=\begin{bmatrix}1\\\ 0\\\ \vdots\\\
0\end{bmatrix},e_{1}=\begin{bmatrix}0\\\ 1\\\ \vdots\\\
0\end{bmatrix},\dots,e_{n-1}=\begin{bmatrix}0\\\ \vdots\\\ 0\\\
1\end{bmatrix}\\}$ denote the standard basis of $\mathbb{C}^{n}$. Denote the
corresponding elementary matrices by $E_{ab}$ (such that
$E_{ab}e_{c}=\delta_{bc}e_{a}$); they are also indexed by integers
$0,1,\dots,n-1$. We will also need the loop space of $\mathbb{C}^{n}$, denoted
by

(A.1) $H^{(n)}=\mathbb{C}^{n}\otimes\mathbb{C}[z,z^{-1}],$

with basis $e_{a}^{k}=e_{a}z^{k}$, for $a=0,\dots,n-1$ and $k\in\mathbb{Z}$.
Let $F^{(n)}$ be the $n$-component fermionic Fock space, the semi infinite
wedge space based on $H^{(n)}$. It is spanned by semi-infinite wedges

$\omega=w_{0}\wedge w_{1}\wedge w_{3}\wedge\dots,\quad w_{i}\in H^{(n)},$

where the $w_{i}$ satisfy some restrictions that we will presently discuss.
Semi-infinite wedges obey the usual rules of exterior algebra, like
multilinearity in each factor and antisymmetry under exchange of two factors.

To formulate the restrictions on the $w_{i}$ that can appear in the wedge
$\omega$ above we introduce the Clifford algebra $Cl^{(n)}$ acting on
$F^{(n)}$: it is generated by exterior and interior products with
$e_{a}^{k}=e_{a}z^{k}$, denoted by $e(e_{a}^{k})$ and $i(e_{a}^{k})$, defined
as wedging and unwedging:

$e(e_{a}^{k})\alpha=e_{a}^{k}\wedge\alpha,\quad
i(e_{a}^{k})\alpha=\beta,\text{ if }\alpha=e_{a}^{k}\wedge\beta.$

It is useful to collect the generators of the Clifford algebra in generating
series. Therefore, define _fermion fields_

$\psi_{a}^{\pm}(w)=\sum_{k\in\mathbb{Z}}{}_{a}\psi_{(k)}^{\pm}w^{-k-1},\quad
a=0,1,\dots,n-1$

where

(A.2)
${\prescript{}{a}{\psi}_{(k)}^{+}}=e(e_{a}^{k})=e_{a}z^{k}\wedge,\quad{\prescript{}{a}{\psi}_{(k)}^{-}}=i(e_{a}z^{-k-1}).$

The fermionic fields satisfy commutation relations

$[\psi_{a}^{\pm}(z),\psi^{\pm}_{b}(w)]_{+}=0,\quad[\psi_{a}^{+}(w_{1}),\psi_{b}^{-}(w_{2})]_{+}=\delta_{ab}\delta(w_{1},w_{2}),$

where the formal delta distribution is defined by

(A.3) $\delta(z,w)=\sum_{k\in\mathbb{Z}}z^{k}w^{-k-1}.$

Let $v_{0}$ be the vacuum vector

(A.4) $v_{0}=\begin{bmatrix}1\\\ 0\\\ 0\\\
\vdots\end{bmatrix}\wedge\begin{bmatrix}0\\\ 1\\\ 0\\\
\vdots\end{bmatrix}\wedge\dots\wedge\begin{bmatrix}0\\\ 0\\\ \vdots\\\
1\end{bmatrix}\wedge\begin{bmatrix}z\\\ 0\\\ 0\\\
\vdots\end{bmatrix}\wedge\begin{bmatrix}0\\\ z\\\ 0\\\
\vdots\end{bmatrix}\wedge\dots\wedge\begin{bmatrix}0\\\ 0\\\ \vdots\\\
z\end{bmatrix}\wedge\begin{bmatrix}z^{2}\\\ 0\\\ 0\\\
\vdots\end{bmatrix}\wedge\begin{bmatrix}0\\\ z^{2}\\\ 0\\\
\vdots\end{bmatrix}\wedge\dots\,.$

Then we define $F^{(n)}$ to be span of the wedges obtained by acting on the
vacuum $v_{0}$ by monomials in the wedging and unwedging operators. To get a
basis for $F^{(n)}$ we specify an ordering on the wedging/unwedging operators
acting on $F^{(n)}$.

######  Definition A.1.

An _elementary wedge_ in $F^{(n)}$ is an element $\omega=Mv_{0}$, where

$M=M_{n-1}\dots M_{1}M_{0},\quad M_{a}=M_{a}^{+}M_{a}^{-},\quad
a=n-1,n-2,\dots,2,1$

where

$M_{a}^{\pm}={\prescript{}{a}{\psi}_{(k_{1})}^{\pm}}{\prescript{}{a}{\psi}_{(k_{2})}^{\pm}}\dots\prescript{}{a}{\psi}_{(k_{s})}^{\pm},\quad
k_{1}<k_{2}<\dots<k_{s}\leq 1,$

is a monomial in ${\prescript{}{a}{\psi}_{(k)}^{\pm}}$ for $k\leq-1$, ordered
in in increasing order form left to right.

The statement that the elementary wedges form a basis for $F^{(n)}$ follows
from the Poincaré-Birkhoff-Witt theorem for the Lie superalgebra underlying
the Clifford algebra.

We define a bilinear form, denoted $\langle\,,\rangle$, on $F^{(n)}$ by
declaring the elementary wedges to be orthonormal. We have then

(A.5) $\langle{\prescript{}{a}{\psi}_{(k)}^{+}}v,w\rangle=\langle
v,{\prescript{}{a}{\psi}_{(-k-1)}^{-}}w\rangle.$

and

(A.6) $\langle\psi_{a}^{+}(z)v,w\rangle=\langle
v,\psi_{a}^{-}(z^{-1})z^{-1}w\rangle.$

The $n$-component fermionic Fock space $F^{(n)}$ has a grading by the Abelian
group $\mathbb{Z}^{n}$, i.e., we have a decomposition
$F^{(n)}=\oplus_{\alpha\in\mathbb{Z}^{n}}F^{(n)}_{\alpha}$. The vacuum has
degree $\alpha=0$. To describe the grading introduce a basis in
$\mathbb{Z}^{n}$ by

$\delta_{a}=\begin{pmatrix}0&0&\dots&1&\dots&0\end{pmatrix},\quad\text{ where
the $1$ is in position $a=0,1,\dots,n-1$}$

The grading on $F^{(n)}$ induces a grading on linear maps on $F^{(n)}$: if
$L\colon F^{(n)}\to F^{(n)}$ has the property that there exist a
$\delta\in\mathbb{Z}^{n}$ so that for all $\omega\in\mathbb{Z}^{n}$ $L$
restricts to a map $F^{(n)}_{\omega}\to F_{\omega+\delta}$, then we say that
$L$ has degree $\delta$. Then the grading is uniquely determined by declaring
wedging operators $e_{a}z^{k}\wedge$ to have degree $\delta_{a}$, and the
unwedging operators $i(e_{a}z^{k})$ to have degree $-\delta_{a}$. The fields
$\psi^{\pm}_{a}(z)$ have degree $\pm\delta_{a}$.

The _total degree_ of an element $\omega$ of degree $\alpha$ is just the sum
of the entries in the degree row vector $\alpha$.

###  A.3. Fermionic Translation Operators and Translation Group.

Besides the action of fermion operators ${\prescript{}{a}{\psi}_{(k)}^{\pm}}$
we also have on $F^{(n)}$ the action of fermionic translation operators
$Q_{a}\colon F^{(n)}\to F^{(n)}$, $a=0,1,\dots,n-1$, given by

(A.7)
$Q_{a}v_{0}=\psi_{a}^{+}(z)v_{0}\mid_{z=0}={\prescript{}{a}{\psi}_{(-1)}^{+}}v_{0},$

and

(A.8) $\displaystyle\psi_{a}^{\pm}(z)Q_{a}$ $\displaystyle=z^{\pm
1}Q_{a}\psi_{a}^{\pm 1}(z),$ (A.9) $\displaystyle\psi_{a}^{\pm}(z)Q_{b}$
$\displaystyle=-Q_{b}\psi_{a}^{\pm 1}(z),\quad a\neq b,$ (A.10) $\displaystyle
Q_{a}Q_{b}$ $\displaystyle=-Q_{b}Q_{a},\quad a\neq b.$

The $Q_{a}$ are invertible. The $Q_{a}^{\pm 1}$ have degree $\pm\delta_{a}$.

The $Q_{a}$ are unitary for the standard bilinear form of $F^{(n)}$:

(A.11) $\langle Q_{a}v,w\rangle=\langle v,Q_{a}^{-1}w\rangle,\quad
a=0,1,\dots,n-1.$

The fermionic translation operators belong to the central extension of the
loop group $\operatorname{\widetilde{GL}_{n}}$ (acting on $F^{(n)}$). They are
lifts of commuting elements of the non extended loop group, see for instance
[2], Proposition 5.3.4. We indicate these elements of the loop group also by
$Q_{a}$, and we have

(A.12) $Q_{a}=\sum_{b\neq a}E_{bb}+z^{-1}E_{aa}.$

The group generated by $Q_{a},a=0,1,\dots,n-1$, contains a subgroup of
elements of total degree zero, generated by the translation operators
$T_{s}=Q_{s}Q_{s-1}^{-1}$, $s=1,2,\dots,n$, of degree
$\delta_{s}-\delta_{s-1}$. Another set of generators for this subgroup is also
useful: define

$T_{ab}=Q_{a}Q_{b}^{-1},$

of degree $\delta_{a}-\delta_{b}$.

######  Lemma A.2.

  1. (1)

$T_{ab}Q_{c}=Q_{c}T_{ab}$ if $c\neq a,b$.

  2. (2)

For $a\neq b$

$(Q_{a}Q_{b}^{-1})^{m}=(-1)^{\frac{m(m-1)}{2}}Q_{a}^{m}Q_{b}^{-m}.$

  3. (3)

For all $k,l\in\mathbb{Z}$ we have

$T_{2}^{k}T_{1}^{l}=(-1)^{\frac{k(k-1)}{2}+\frac{l(l-1)}{2}}Q_{2}^{k}Q_{1}^{l-k}Q_{0}^{-l}.$

  4. (4)

$T_{10}^{\alpha}T_{20}^{\beta}T_{21}^{\gamma}=(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\beta(\beta-1)}{2}+\frac{\gamma(\gamma-1)}{2}+\alpha\gamma}Q_{2}^{\alpha+\beta}Q_{1}^{\alpha-\gamma}Q_{0}^{-\alpha-\beta}.$

  5. (5)

$T_{10}^{\alpha}T_{20}^{\beta}T_{21}^{\gamma}=(-1)^{\frac{\beta(\beta-1)}{2}+\alpha\beta+\alpha\gamma+\beta\gamma}T_{2}^{\beta+\gamma}T_{1}^{\alpha+\beta}$

###### Proof.

Part (1) is clear. Part (2) is a simple induction. For part (3) we get

$\displaystyle T_{2}^{k}T_{1}^{l}$
$\displaystyle=(Q_{2}Q_{1}^{-1})^{k}(Q_{1}Q_{0}^{-1})^{l}=(-1)^{\frac{k(k-1)}{2}}(-1)^{\frac{l(l-1)}{2}}Q_{2}^{k}Q_{1}^{-k}Q_{1}^{l}Q_{0}^{-l}.$

Similarly for part (4) we have

$\displaystyle T_{10}^{\alpha}T_{20}^{\beta}T_{21}^{\gamma}$
$\displaystyle=(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\beta(\beta-1)}{2}+\frac{\gamma(\gamma-1)}{2}}Q_{1}^{\alpha}Q_{0}^{-\alpha}Q_{2}^{\beta}Q_{0}^{-\beta}Q^{\gamma}_{2}Q_{1}^{-\gamma}=$
$\displaystyle=(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\beta(\beta-1)}{2}+\frac{\gamma(\gamma-1)}{2}}Q_{2}^{\beta}Q_{1}^{\alpha}Q_{2}^{\gamma}Q_{1}^{-\gamma}Q_{0}^{-\alpha}Q_{0}^{-\beta}=$
$\displaystyle=(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\beta(\beta-1)}{2}+\frac{\gamma(\gamma-1)}{2}+\alpha\gamma}Q_{2}^{\beta+\gamma}Q_{1}^{\alpha-\gamma}Q_{0}^{-\alpha-\beta}$

Finally, for part (5) we substitute

$Q_{2}^{\beta+\gamma}Q_{1}^{\alpha-\gamma}Q_{0}^{-\alpha-\beta}=(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\gamma(\gamma-1)}{2}+\alpha\beta+\beta\gamma}T_{2}^{\beta+\gamma}T_{1}^{\alpha+\beta},$

(from part (3)), in the right hand side of part (4). The result then follows
from

$(-1)^{\frac{\alpha(\alpha-1)}{2}+\frac{\beta(\beta-1)}{2}+\frac{\gamma(\gamma-1)}{2}+\alpha\gamma+\frac{\alpha(\alpha-1)}{2}+\frac{\gamma(\gamma-1)}{2}+\alpha\beta+\beta\gamma}=(-1)^{\frac{\beta(\beta-1)}{2}+\alpha\beta+\alpha\gamma+\beta\gamma}.$

∎

Define the ordered product of $k\geq 0$ fermions by

(A.13)
$\vec{\prod}_{l=1}^{k}{}_{a}\psi_{(-l)}^{\pm}={}_{a}\psi^{\pm}_{(-k)}\dots{}_{a}\psi_{(-1)}^{\pm}.$

The empty product is as usual the identity.

######  Lemma A.3.

  1. (1)

For all $k\in\mathbb{Z}$, $k\neq 0$

$Q_{a}^{k}v_{0}=\begin{cases}v_{0}&k=0\\\
\vec{\prod}_{l=1}^{k}{\prescript{}{a}{\psi}_{(-l)}^{+}}v_{0}&k>0,\\\
\vec{\prod}_{l=1}^{-k}{\prescript{}{a}{\psi}_{(-l)}^{-}}v_{0}&k<0.\end{cases}$

  2. (2)

For all $\alpha,\beta\geq 0$

$\displaystyle Q_{1}^{\beta}$ $\displaystyle
Q_{0}^{\alpha}v_{0}=(-1)^{\beta\gamma}\vec{\prod}_{l=1}^{\alpha}{\prescript{}{0}{\psi}_{(-l)}^{+}}\vec{\prod}_{m=1}^{\beta}{\prescript{}{1}{\psi}_{(-m)}^{+}}v_{0}=$
$\displaystyle=\vec{\prod}_{m=1}^{\beta}{\prescript{}{1}{\psi}_{(-m)}^{+}}\vec{\prod}_{l=1}^{\gamma}{\prescript{}{0}{\psi}_{(-l)}^{+}}v_{0}=$
$\displaystyle=e_{1}z^{-\beta}\wedge e_{1}z^{1-\beta}\wedge\dots\wedge
e_{1}z^{-1}\wedge e_{0}z^{-\alpha}\wedge e_{0}z^{1-\alpha}\wedge\dots\wedge
e_{0}z^{-1}\wedge v_{0}.$

∎

The translation operators $T_{s}$ or $T_{ab}$ are also unitary, just as the
fermionic translation operators: from (A.11) it follows that

(A.14) $\langle Tv,w\rangle=\langle v,T^{-1}w\rangle.$

###  A.4. The Lie Algebra $\operatorname{\widetilde{gl}_{n}}$ and Fermions.

The loop algebra $\operatorname{\widetilde{gl}_{n}}$ is defined as the Lie
subalgebra of $gl(H^{(n)})$ generated by $E_{ab}z^{k}$, $a,b=0,1,\dots,n-1,$,
$k\in\mathbb{Z}$, where

$\left(E_{ab}z^{k}\right)\cdot\left(e_{c}z^{m}\right)=\delta_{bc}e_{a}z^{k+m}.$

The loop algebra $\operatorname{\widetilde{gl}_{n}}$ does not quite act on
$F^{(n)}$. One would like to define on $F^{(n)}$

$E_{ab}z^{k}\mapsto\sum_{l\in\mathbb{Z}}(e_{a}z^{k+l}\wedge)(i(e_{b}z^{l}))=\sum_{l\in\mathbb{Z}}{\prescript{}{a}{\psi}_{(k+l)}^{+}}{\prescript{}{b}{\psi}_{(-l-1)}^{-}}.$

However, considering the action of $E_{aa}z^{0}$ on the vacuum $v_{0}$ we
would get:

$E_{aa}v_{0}=\sum_{l\geq 0}v_{0},$

(since $(e_{a}z^{l}\wedge)(i(e_{a}z^{l}))v_{0}=v_{0}$ for $l\geq 0$), so that
these diagonal elements would have a divergent action. Therefore we introduce
a normal ordering on fermion fields ([9]) by

$\colon\psi_{a}^{+}(z)\psi_{b}^{-}(w)\colon=\psi_{a,\text{cr}}^{+}(z)\psi_{b}^{-}(w)-\psi_{b}^{-}(w)\psi_{a,\text{ann}}^{+}(z),$

where the creation and annihilation parts of a fermion field are given by

$\psi_{\text{cr}}(z)=\sum_{k\geq
0}\psi_{(-k-1)}z^{k},\quad\psi_{\text{ann}}(z)=\sum_{k\geq
0}\psi_{(k)}z^{-k-1}.$

In terms of the components the normal ordering means

$\colon{\prescript{}{a}{\psi}_{(k)}^{+}}{\prescript{}{b}{\psi}_{(l)}^{-}}\colon=\begin{cases}{\prescript{}{a}{\psi}_{(k)}^{+}}{\prescript{}{b}{\psi}_{(l)}^{-}}&k\geq
0,\\\
-{\prescript{}{b}{\psi}_{(l)}^{-}}{\prescript{}{a}{\psi}_{(k)}^{+}}&k<0.\end{cases}$

Then we can define an action on $F^{(n)}$ by

$E_{ab}z^{k}\mapsto\colon\sum_{l\in\mathbb{Z}}(e_{a}z^{k+l}\wedge)(i(e_{b}z^{l}))\colon=\sum_{l\in\mathbb{Z}}\colon{}_{a}\psi^{+}_{(k+l)}{}_{b}\psi^{-}_{(-l-1)}\colon.$

This gives a central extension

(A.15)
$0\to\mathbb{C}\to\hat{\operatorname{\widetilde{gl}_{n}}}\to\operatorname{\widetilde{gl}_{n}}\to
0$

Introduce a generating series of loop algebra elements by:

(A.16) $E_{ab}(z_{1})=\sum_{l\in\mathbb{Z}}E_{ab}z^{l}z_{1}^{-l-1}.$

Acting on $F^{(n)}$ it can be represented by a normal ordered product of
fermion fields:

(A.17) $E_{ab}(z_{1})=\colon\psi_{a}^{+}(z_{1})\psi_{b}^{-}(z_{1})\colon.$

Indeed,

$\displaystyle E_{ab}(z_{1})$
$\displaystyle=\sum_{k,l\in\mathbb{Z}}\colon{\prescript{}{a}{\psi}_{(k+l)}^{+}}{\prescript{}{b}{\psi}_{(-l-1)}^{-}}\colon
z_{1}^{-k-1}=$
$\displaystyle=\sum_{k,l\in\mathbb{Z}}\colon({\prescript{}{a}{\psi}_{(k+l)}^{+}}z_{1}^{-k-l-1})({\prescript{}{b}{\psi}_{(-l-1)}^{-}}z_{1}^{l})\colon=\colon\psi_{a}^{+}(z_{1})\psi_{b}^{-}(z_{1})\colon.$

As field on $F^{(n)}$ $E_{ab}(z_{1})$ has degree $\delta_{a}-\delta_{b}$.

Equation (A.17) is the reason we chose the encoding wedging and unwedging
operators as coefficients of fermion fields according to (A.2).

We will need the commutator of the generating series of Lie algebra elements
with fermionic translation operators.

######  Lemma A.4.

For all $\alpha,\beta\in\mathbb{Z}$ we have

$Q_{0}^{\beta}Q_{1}^{-\alpha}E_{1,0}(z)Q_{1}^{\alpha}Q_{0}^{-\beta}=(-1)^{\alpha+\beta}z^{\alpha+\beta}E_{1,0}(z).$

###  A.5. Root Lattice

Recall the group $\mathbb{Z}^{n}$ that gives a grading for fermionic Fock
space $F^{(n)}$. It contains as a subgroup the root lattice $A_{n-1}$,
generated by

$\alpha_{i}=\delta_{i+1}-\delta_{i},\quad i=1,2,\dots,n-1$

So

$A_{n-1}=\mathbb{Z}^{n-1}\alpha\subset\mathbb{Z}^{n}.$

We will call elements in $A_{n-1}$ of the form
$\alpha=\sum_{i=1}^{n-1}n_{i}\alpha$ positive roots if $n_{i}\geq 0$.

The translation group is also graded by $A_{n-1}$: the generator
$T_{s}=Q_{s}Q_{s-1}^{-1}$ has degree $\alpha_{s}$. Similarly the Lie algebra
generating fields $E_{ab}(z)$ have

$\deg(E_{aa-1}(z))=\alpha,\quad\deg(E_{01}(z))=-\alpha.$

##  Appendix B Expressions for the $\tau$-functions

In this appendix we prove Theorem 2.1 and Theorem 3.1. This gives expressions
for the $\tau$-functions in terms of coordinates on the lower triangular
subgroup $\mathcal{N}$ of $\operatorname{\widetilde{GL}_{n}}$, for $n=2,3$.

###  B.1. The case of $n=2$, Theorem 2.1.

The $\tau$-functions for $\operatorname{\widehat{GL}_{2}}$ are given as matrix
elements on $F^{(2)}$:

(B.1) $\tau_{k}^{(\alpha)}=<T^{k}v_{0},g_{C}^{(\alpha)}v_{0}>,$

where the element $g_{C}^{(\alpha)}$ of the lower triangular sub group
$\mathcal{N}$ of $\operatorname{\widehat{GL}_{2}}$ has projection given by
(2.3). We write it as

$g^{(\alpha)}_{C}=\exp(\Gamma^{(\alpha)}_{C})=\sum_{l=0}^{\infty}\frac{1}{l!}(\Gamma^{(\alpha)}_{C})^{l},$

where

(B.2)
$\Gamma^{(\alpha)}_{C}=\operatorname{Res}_{z_{1}}\left(C^{(\alpha)}(z_{1})E_{10}(z_{1})\right),$

and $C^{(\alpha)}(z)$ is given by (2.2), and the generating series of loop
algebra elements $E_{10}(z_{1})$ by (A.16).

Now $T^{k}v_{0}$ has degree $k(\delta_{1}-\delta_{0})$ in $F^{(2)}$ and
$\Gamma_{C}^{{\alpha}}$ has degree $\delta_{1}-\delta_{0}$, since
$E_{10}(z_{1})$ has. In $F^{(2)}$ homogeneous elements of different degree are
orthogonal for $\langle\,,\rangle$. Hence only the $l=k$ term contributes to
(B.1) and

$\tau_{k}^{(\alpha)}=\frac{1}{k!}\langle
T^{k}v_{0},(\Gamma_{C}^{(\alpha)})^{k}v_{0}\rangle.$

Recall that $C^{(\alpha)}(z)=\sum_{n\in\mathbb{Z}}c_{n+\alpha}z^{-n-1}$.
Associated to the series $C^{(\alpha)}(z)$ we have a $\mathbb{C}$-linear map

$c^{(\alpha)}\colon\mathbb{C}[[z,z^{-1}]]\to
B=\mathbb{C}[[c_{i}]]_{i\in\mathbb{Z}},\quad
f(z)\mapsto\operatorname{Res}_{z}(C^{(\alpha)}(z)f(z)).$

We will need multiple copies of the map $c^{(\alpha)}$ acting on series in
variables $z_{1},z_{2},\dots$. We put
$c^{(\alpha)}_{i}(f(z_{i}))=c^{(\alpha)}(f(z))$, for $i=1,2,\dots$, and impose
linearity in the variables $z_{j}$, $j\neq i$, i.e., the condition that (for
instance)
$c^{(\alpha)}_{i}(f(z_{i})z_{j}^{m})=z^{m}_{j}c^{(\alpha)}_{i}(f(z_{i}))$.

Then we can write

$(\Gamma^{(\alpha)}_{C})^{l}=\prod_{i=1}^{l}c^{(\alpha)}_{i}\prod_{j=1}^{l}E_{10}(z_{j}),$

and so

$\displaystyle\tau_{k}^{(\alpha)}$
$\displaystyle=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\langle
T^{k}v_{0},\prod_{j=1}^{k}E_{10}(z_{j})v_{0}\rangle\right)=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\langle
T^{k}v_{0},\prod_{j=1}^{k}\psi_{1}^{+}(z_{j})\psi_{0}^{-}(z_{j})v_{0}\rangle\right)=$
(B.3) $\displaystyle=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\langle
Q_{1}^{k}Q_{0}^{-k}v_{0},\prod_{s=1}^{k}\psi_{1}^{+}(z_{s})\prod_{t=1}^{k}\psi_{0}^{-}(z_{t})v_{0}\rangle\right).$

Here we use that in the expression (A.17) of $E_{10}(z_{1})$ in fermions we
don’t need normal ordering, and also Lemma A.2, part 2.

Next we use the factorization Lemma D.1 and the Lemma E.2 to calculate the
factors involving $\psi_{0}(z)$ and $\psi_{1}(z)$ separately; we find

$\tau_{k}^{(\alpha)}=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\det(V^{(k)}_{\\{z_{i}\\}})^{2}\right).$

Here $V^{(k)}_{\\{z\\}}$ is the Vandermonde matrix (D.1).

This proves the first part of Theorem 2.1, since
$\det(V^{(k)}_{\\{z_{i}\\}})=\prod_{k\geq j>i\geq 1}(z_{i}-z_{j})$.

For the second part we need a formula for the square of a Vandermonde
determinant. Let the permutation group $S_{k}$ act on
$\mathbb{C}[z_{1},z_{2},\dots,z_{k}]$ by permuting the subscripts.

######  Lemma B.1.

(B.4) $\det(V^{(k)}_{\\{z_{i}\\}})^{2}=\sum_{\sigma\in
S_{k}}\det(z_{\sigma(i)}^{i+j-2})_{i,j=1}^{k}.$

###### Proof.

The right hand side of (B.4) can be written as

$\sum_{\sigma\in
S_{k}}\sigma\det\begin{pmatrix}1&z_{1}&z_{1}^{2}&\dots&z_{1}^{k-1}\\\
z_{2}&z_{2}^{2}&z_{2}^{3}&\dots&z_{2}^{k}\\\
z_{3}^{2}&z_{3}^{3}&z_{3}^{4}&\dots&z_{3}^{k+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
z_{k}^{k-1}&z_{k}^{k}&z_{k}^{k+1}&\dots&z_{k}^{2k-2}\end{pmatrix}.$

From this we see that the $\sigma=e$ term on the RHS of the identity we want
to prove is

$A=\det(z_{i}^{i+j-2})=\prod_{s=1}^{k}z_{s}^{s-1}\det(z_{i}^{j-1})=\prod_{s=1}^{k}z_{s}^{s-1}\det(V^{(k)}_{\\{z_{i}\\}}).$

Now for every permutation $\sigma$ in $S_{k}$ we have

$\sigma(A)=(-1)^{\lvert\sigma\rvert}\prod_{s=1}^{k}z_{\sigma(s)}^{s-1}\det(V^{(k)}_{\\{z_{i}\\}}).$

Sum over all permutations

$\sum_{\sigma}\sigma(A)=\sum_{\sigma\in
S_{k}}(-1)^{\lvert\sigma\rvert}\prod_{s=1}^{k}z_{\sigma(s)}^{s-1}\det(V^{(k)}_{\\{z_{i}\\}})=\det(V^{(k)}_{\\{z_{i}\\}})^{2}.$

∎

Introduce a monomial column vector $v(z)=\begin{pmatrix}1\\\ z\\\ z^{2}\\\
\vdots\\\ z^{k-1}\end{pmatrix}$. Define the Hankel matrix (with coefficients
in $B$)

$\displaystyle T^{(\alpha)}_{k}$
$\displaystyle=c^{(\alpha)}(v(z)v(z)^{T})=c^{(\alpha)}\begin{pmatrix}1&z&z^{2}&\dots&z^{k-1}\\\
z&z^{2}&z^{3}&\dots&z^{k}\\\ z^{2}&z^{3}&z^{4}&\dots&z^{k+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
z^{k-1}&z^{k}z&z^{k+1}&\dots&z^{2k-1}\end{pmatrix}=$
$\displaystyle=\begin{pmatrix}c_{\alpha}&c_{(\alpha+1)}&c_{\alpha+2}&\dots&c_{\alpha+k-1}\\\
c_{(\alpha+1)}&c_{\alpha+2}&c_{\alpha+3}&\dots&c_{\alpha+k}\\\
c_{\alpha+2}&c_{\alpha+3}&c_{\alpha+4}&\dots&c_{\alpha+k+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
c_{\alpha+k-1}&c_{\alpha+k}&c_{\alpha+k+1}&\dots&c_{\alpha+2k-2}\end{pmatrix},$

where we apply $c^{(\alpha)}$ component wise. Then the value of
$\det(T^{(\alpha)}_{k})\in B$ can be calculated in terms of Vandermonde
determinants.

######  Lemma B.2.

If $V^{(k)}_{\\{z\\}}$ is the Vandermonde matrix (D.1) then

$\det(T^{(\alpha)}_{k})=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\det(V^{(k)}_{\\{z_{i}\\}})^{2}\right).$

###### Proof.

By the trivial observation that

$c(f(z))\cdot c(g(z))=c_{1}(f(z_{1}))\cdot c_{2}(g(z_{2}))$

we have

$\displaystyle\det(T^{(\alpha)}_{k})$
$\displaystyle=\prod_{i=1}^{k}c^{(\alpha)}_{i}\det\begin{pmatrix}1&z_{1}&z_{1}^{2}&\dots&z_{1}^{k-1}\\\
z_{2}&z_{2}^{2}&z_{2}^{3}&\dots&z_{2}^{k}\\\
z_{3}^{2}&z_{3}^{3}&z_{3}^{4}&\dots&z_{3}^{k+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
z_{k}^{k-1}&z_{k}^{k}&z_{k}^{k+1}&\dots&z_{k}^{2k-2}\end{pmatrix}=$
$\displaystyle=\prod_{i=1}^{k}c^{(\alpha)}_{i}\det(z_{i}^{i+j-2}).$

Now for any polynomial $f(z_{1},z_{2},\dots,z_{k})$ and any permutation
$\sigma\in S_{k}$ we have
$\prod_{i=1}^{k}c^{(\alpha)}_{i}(\sigma
f(z_{1},z_{2},\dots,z_{k}))=\prod_{i=1}^{k}c^{(\alpha)}_{i}(f(z_{1},z_{2},\dots,z_{k}))$.
Hence

$\det(T^{(\alpha)}_{k})=\frac{1}{k!}\prod_{i=1}^{k}c^{(\alpha)}_{i}\left(\sum_{\sigma}\sigma\det(z_{i}^{i+j-2})\right),$

and the lemma follows from the previous Lemma B.1. ∎

This finishes the proof of the second part of Theorem 2.1.

###  B.2. Proof of $n=3$, Theorem 3.1.

###### Proof.

In this proof we suppress for typographical simplicity the shift superscripts
(α,β).

We write

$g=\exp(\Gamma_{c})\exp(\Gamma_{d})\exp(\Gamma_{e}),$

where

$\Gamma_{c}=\operatorname{Res}_{z}(C(z_{1})E_{10}(z_{1})),\quad\Gamma_{d}=\operatorname{Res}_{z_{1}}(D(z_{1})E_{20}(z_{1})),\quad\Gamma_{e}=\operatorname{Res}_{z_{1}}(E(z_{1})E_{21}(z_{1})).$

This implies that $\tau_{k,l}$ is the sum of

(B.5)
$c_{n,r,s}=\operatorname{Res}_{\mathbf{x},\mathbf{y},\mathbf{z}}\left(\prod_{i=1}^{s}C(x_{i})\prod_{i=1}^{r}D(y_{i})\prod_{i=1}^{n}E(z_{i})p_{n,r,s}\right),$

where

(B.6)
$p_{n,r,s}=\langle{T_{1}^{k}T_{2}^{l}v_{0},\prod_{i=1}^{s}\psi_{1}^{+}(x_{i})\psi_{0}^{-}(x_{i})\prod_{i=1}^{r}\psi_{2}^{+}(y_{i})\psi_{0}^{-}(y_{i})\prod_{i=1}^{n}\psi_{2}^{+}(z_{i})\psi_{1}^{-}(z_{i})v_{0}}\rangle,$

where

(B.7) $n+r=\ell,\quad s+r=k.$

We can factorize this using Appendix D as

$p_{n,r,s}=(-1)^{\left(\frac{k(k-1)+l(l-1)}{2}+kl+\frac{s(s-1)+r(r-1)+n(n-1)}{2}+ns\right)}\times\\\
\times\langle{Q_{2}^{l}v_{0},\prod_{i=1}^{r}\psi_{2}^{+}(y_{i})\prod_{i=1}^{n}\psi_{2}^{+}(z_{i})v_{0}}\rangle\times\\\
\times\langle{Q_{1}^{k-l}v_{0},\prod_{i=1}^{s}\psi_{1}^{+}(x_{i})\prod_{i=1}^{n}\psi_{1}^{-}(z_{i})v_{0}}\rangle\langle{Q_{0}^{-k}v_{0},\prod_{i=1}^{s}\psi_{0}^{-}(x_{i})\prod_{i=1}^{r}\psi_{0}^{-}(y_{i})}\rangle.$

Since, using (B.7),

(B.8)
$(-1)^{\frac{k(k-1)+l(l-1)}{2}+kl+\frac{s(s-1)+r(r-1)+n(n-1)}{2}+ns}=(-1)^{\frac{r(r+1)}{2}},$

the theorem follows from the calculation of correlation functions in Appendix
E. ∎

##  Appendix C Birkhoff Factorization and Matrix Elements of Semi Infinite
Wedge Space

In this Appendix we sketch proofs of Theorem 2.2 and 3.3. First we will
discuss a more general statement about the Birkhoff factorization in
$\operatorname{\widetilde{GL}_{n}}$.

###  C.1. Birkhoff Factorization and $n$-component Fermions.

Let ${\operatorname{\widetilde{GL}_{n}}}_{-}$ be the subgroup of
$\operatorname{\widetilde{GL}_{n}}$ of elements $g_{-}=1+\mathcal{O}(z^{-1})$,
and let ${\operatorname{\widetilde{GL}_{n}}}_{0+}$ be the subgroup of elements
$g_{0+}=A+\mathcal{O}(z)$, where $A$ is invertible (and independent of $z$).
Then most elements in $\operatorname{\widetilde{GL}_{n}}$ have a factorization

(C.1) $g=g_{-}g_{0+},\quad g\in\operatorname{\widetilde{GL}_{n}}.$

We will express $g_{-}$ in terms fermion matrix elements in semi-infinite
wedge space $F^{(n)}$.

Recall that, just as the loop algebra $\operatorname{\widetilde{gl}_{n}}$, the
loop group $\operatorname{\widetilde{GL}_{n}}$ does not actually act on
$F^{(n)}$, we have a central extension (cf. [14])

$1\to\mathbb{C}^{\times}\to\operatorname{\widehat{GL}_{n}}\overset{\pi}{\to}\operatorname{\widetilde{GL}_{n}}\to
1,$

and we have an action of $\operatorname{\widehat{GL}_{n}}$ on $F^{(n)}$. There
are subgroups
${\operatorname{\widehat{GL}_{n}}}_{-},{\operatorname{\widehat{GL}_{n}}}_{0+}$
of $\operatorname{\widehat{GL}_{n}}$ projecting under $\pi$ to the
corresponding subgroups of $\operatorname{\widetilde{GL}_{n}}$; the image of
$\mathbb{C}^{\times}$ belongs to ${\operatorname{\widehat{GL}_{n}}}_{0+}$.
Most elements $\hat{g}\in\operatorname{\widetilde{GL}_{n}}$ will also have a
Birkhoff factorization,

(C.2) $\hat{g}=\hat{g}_{-}\hat{g}_{0+}.$

In fact the central extension is trivial over
${\operatorname{\widehat{GL}_{n}}}_{-}$ and we can and will identify
${\operatorname{\widehat{GL}_{n}}}_{-}$ with
${\operatorname{\widetilde{GL}_{n}}}_{-}$, and we will write $g_{-}$ for
either an element of ${\operatorname{\widehat{GL}_{n}}}_{-}$ or
${\operatorname{\widetilde{GL}_{n}}}_{-}$.

To calculate the negative component of $g$ in the factorization (C.1) we
choose a lift $\hat{g}$ of $g$, i.e., $\pi(\hat{g})=g$, and study the action
of $\hat{g}$ on $F^{(n)}$. If $v_{0}$ is the vacuum (A.4) of $F^{(n)}$, the
$\tau$-function is defined as the matrix element

(C.3) $\tau(\hat{g})=\langle v_{0},\hat{g}v_{0}\rangle.$

We have

$\hat{g}_{0+}v_{0}=\tau(\hat{g})v_{0},$

and hence (assuming $\hat{g}$ has a Birkhoff factorization, or
$\tau(\hat{g})\neq 0$)

(C.4)
$g_{-}v_{0}=g_{-}\hat{g}_{0+}v_{0}/\tau(\hat{g})=\hat{g}v_{0}/\tau(\hat{g}).$

Now write $g_{-}$ in terms of matrix elements:

$g_{-}=\sum_{a,b=0}^{n-1}g_{ab}(z)E_{ab},$

where

(C.5) $g_{ab}(z)=\sum_{k\in\mathbb{Z}}g_{ab}^{(k)}z^{-k-1},\quad
g_{ab}^{(-1)}=\delta_{ab},\quad g_{ab}^{(l)}=0,\text{ if }l<-1.$

We calculate, if $v_{0}=e_{0}\wedge e_{1}\wedge e_{2}\wedge\dots$,

$g_{-}v_{0}=(g_{-}e_{0})\wedge(g_{-}e_{1})\wedge(g_{-}e_{2})\wedge\dots=v_{0}+\sum_{k\geq
0}g_{ab}^{(k)}E_{ab}z^{-k-1}v_{0}+\dots,$

where the omitted terms are quadratic and higher in the $E_{ab}$. Now on
$F^{(n)}$ (see section A.4)

(C.6)
$E_{ab}z^{-k-1}=\colon\sum_{l}(e_{a}z^{l-k-1}\wedge)(i(e_{b}z^{l}))\colon=\colon\sum_{l\in\mathbb{Z}}{\prescript{}{a}{\psi}_{(l-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-l-1)}^{-}}\colon$

For $k\geq 0$ the normal ordering in (C.6) can be omitted and

$g_{ab}^{(k)}E_{ab}z^{-k-1}v_{0}=g_{ab}^{(k)}\left({\prescript{}{a}{\psi}_{(-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-1)}^{-}}v_{0}+{\prescript{}{a}{\psi}_{(1-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-2)}^{-}}v_{0}+\dots\right)$

is a (finite) sum of elementary wedges
${\prescript{}{a}{\psi}_{(l-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-l-1)}^{-}}v_{0}$,
each with coefficient $g_{ab}^{(k)}$. To calculate $g_{ab}^{(k)}$ we just pick
one of these elementary wedges, say the $l=0$ term, and use orthogonality of
elementary wedges to find ($k\geq 0$)

(C.7) $\displaystyle g_{ab}^{(k)}$
$\displaystyle=\langle{\prescript{}{a}{\psi}_{(-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-1)}^{-}}v_{0},g_{-}v_{0}\rangle=$
$\displaystyle=\langle{\prescript{}{a}{\psi}_{(-k-1)}^{+}}{\prescript{}{b}{\psi}_{(-1)}^{-}}v_{0},\hat{g}v_{0}\rangle/\tau(\hat{g})=$
$\displaystyle=\langle
Q_{b}^{-1}v_{0},{\prescript{}{a}{\psi}_{(k)}^{-}}\hat{g}v_{0}\rangle/\tau(\hat{g}),$

using (A.5) and (A.7).

Now observe that

$\colon{\prescript{}{a}{\psi}_{(0)}^{+}}{\prescript{}{b}{\psi}_{(-1)}^{-}}\colon
v_{0}=\delta_{ab}v_{0},$

and

$\colon{\prescript{}{a}{\psi}_{(l)}^{+}}{\prescript{}{b}{\psi}_{(-1)}^{-}}\colon
v_{0}=0,\quad l>0$

This allows us to calculate $g_{ab}^{(k)}$ for $k<0$ in the same way as for
$k\geq 0$, see (C.7).

These remarks prove the following theorem.

######  Theorem C.1.

Let $g\in\operatorname{\widetilde{GL}_{n}}$ admit a Birkhoff factorization
$g=g_{-}g_{0+}$. Then

$g_{-}=\sum_{a,b=0}^{n-1}g_{ab}(z)E_{ab},$

where

(C.8) $g_{ab}(z)=\langle
Q_{b}^{-1}v_{0},\psi_{a}^{-}(z)\hat{g}v_{0}\rangle/\tau(\hat{g}),$

and $\hat{g}\in\operatorname{\widehat{GL}_{n}}$ is any lift of $g$, so that
$\pi(\hat{g})=g$. The $\tau$ function is given by (C.3).

###  C.2. The $2\times 2$-case; Proof of Theorem 2.2.

Now we specialize in Theorem C.1 $n$ and $g$: in this subsection $n=2$ and

$g=\pi(g^{[k](\alpha)})=\pi(T^{-k})\pi(g^{(\alpha)}_{C}),$

where $\pi(g^{(\alpha)}_{C})$ is given by (2.3). In this case our calculation
in Theorem C.1 of the Birkhoff factorization gives us

$g^{[k](\alpha)}_{-}=\sum_{a,b=0}^{1}g^{[k](\alpha)}_{ab}(z)E_{ab},\quad
g^{[k](\alpha)}_{ab}(z)=\langle
Q_{b}^{-1}v_{0},\psi_{a}^{-}(z)T^{-k}g_{C}^{(\alpha)}v_{0}\rangle/\tau_{k}^{(\alpha)}E_{ab},$

where the $\tau$-function is defined in (2.6). We’ll proceed to rewrite
$g_{ab}^{[k](\alpha)}(z)$.

First of all, we will expand $g_{C}^{(\alpha)}$ in fermion operators. Note
that

$g^{(\alpha)}_{C}=\exp(\Gamma^{(\alpha)}_{C}),$

where

(C.9)
$\Gamma^{(\alpha)}_{C}=\operatorname{Res}_{z_{1}}\left(C^{(\alpha)}(z_{1})E_{10}(z_{1})\right),$

and $C^{(\alpha)}(z)$ is given by (2.2), and the generating series
$E_{10}(z_{1})$ by (A.16). This means that $g^{(\alpha)}_{C}=\sum_{l\geq
0}(\Gamma^{(\alpha)}_{C})^{l}/l!$, both acting on $H^{(2)}$ and on $F^{(2)}$.
Hence

(C.10) $g^{[k](\alpha)}_{ab}(z)=\sum_{l\geq 0}\frac{1}{l!}\langle
Q_{b}^{-1}v_{0},\psi_{a}^{-}(z)T^{-k}(\Gamma^{(\alpha)}_{C})^{l}v_{0}\rangle/\tau_{k}^{(\alpha)}$

Next we use the standard grading on $F^{(2)}$. Note that
$\Gamma_{C}^{(\alpha)}$ has degree $\delta_{1}-\delta_{0}$ (since
$E_{10}(z_{1})$ has). So $Q_{b}^{-1}$ has degree $-\delta_{b}$ and
$\psi_{a}^{-}(z)T^{-k}(\Gamma^{(\alpha)}_{C})^{l}$ has degree
$k(\delta_{0}-\delta_{1})+l(\delta_{1}-\delta_{0})-\delta_{a}$. Hence, by
orthogonality of terms of different degree in $F^{(2)}$, we find that the only
non zero contribution to the sum (C.10) arises when $l=k+a-b$ and

$g^{[k](\alpha)}_{ab}(z)=\frac{1}{l!}\langle
Q_{b}^{-1}v_{0},\psi_{a}^{-}(z)T^{-k}(\Gamma^{(\alpha)}_{C})^{l}v_{0}\rangle/\tau_{k}^{(\alpha)}.$

From now on we will use often in formulas for $g^{[k](\alpha)}_{ab}(z)$ the
abbreviation $l=k+a-b$.

Next we need to commute $T^{-k}$ through $\psi_{a}^{-}(z)$.

######  Lemma C.2.

If $T=Q_{1}Q_{0}^{-1}$ then

$\psi_{a}^{-}(z)T^{-k}=(-1)^{k}z^{k(2a-1)}T^{-k}\psi_{a}^{-}(z).$

###### Proof.

By (A.8), (A.9) we have

$\displaystyle\psi_{a}^{-}(z)T^{-1}=\psi_{a}^{-}(z)Q_{0}Q_{1}^{-1}$
$\displaystyle=\begin{cases}z^{-1}Q_{0}\psi_{a}^{-}(z)Q_{1}^{-1}&\text{ if
}a=0,\\\ -Q_{0}\psi_{a}^{-}(z)Q_{1}^{-1}&\text{ if }a=1.\\\ \end{cases}$
$\displaystyle=\begin{cases}-z^{-1}T^{-1}\psi_{a}^{-}(z)&\text{ if }a=0,\\\
-zT^{-1}\psi_{a}^{-}(z)&\text{ if }a=1.\\\ \end{cases}$
$\displaystyle=-z^{2a-1}T^{-1}\psi_{a}^{-}(z).$

∎

This implies that

$\displaystyle g_{ab}^{[k](\alpha)}(z)$
$\displaystyle=\frac{(-1)^{k}z^{k(2a-1)}}{l!}\langle
Q_{b}^{-1}v_{0},T^{-k}\psi_{a}^{-}(z)(\Gamma_{C}^{(\alpha)})^{l}v_{0}\rangle/\tau_{k}^{(\alpha)}=$
(C.11) $\displaystyle=\frac{(-1)^{k}z^{k(2a-1)}}{l!}\langle
T^{k}Q_{b}^{-1}v_{0},\psi_{a}^{-}(z)(\Gamma_{C}^{(\alpha)})^{l}v_{0}\rangle/\tau_{k}^{(\alpha)},$

by unitarity, (A.14).

Next we want to apply the factorization Lemma D.1. We need to write
$T^{k}Q_{b}^{-1}$ in standard form $Q_{1}^{\alpha}Q_{0}^{\beta}$.

######  Lemma C.3.

$T^{k}Q_{b}^{-1}=(-1)^{\frac{k(k-1)}{2}}(-1)^{bk}Q_{1}^{k-b}Q_{0}^{-k-1+b}$

###### Proof.

By Lemma A.2, Part 2

$\displaystyle T^{k}Q_{b}^{-1}$
$\displaystyle=(-1)^{\frac{k(k-1)}{2}}Q_{1}^{k}Q_{0}^{-k}Q_{b}^{-1}=$
$\displaystyle=(-1)^{\frac{k(k-1)}{2}}\begin{cases}Q_{1}^{k}Q_{0}^{-k-1}&\text{
if }b=0,\\\ (-1)^{k}Q_{1}^{k-1}Q_{0}^{-k}&\text{ if }b=1.\end{cases}$

∎

Now we are going to express $(\Gamma_{C}^{(\alpha)})^{l}$ in terms of fermion
fields, see (C.9) and (A.17). Recall that
$C^{(\alpha)}=\sum_{n\in\mathbb{Z}}c_{n+\alpha}z^{-n-1}$. Associated to the
series $C^{(\alpha)}$ we have a map

$c^{(\alpha)}\colon\mathbb{C}[[z,z^{-1}]]\to
B=\mathbb{C}[[c_{i}]]_{i\in\mathbb{Z}},\quad
f(z)\mapsto\operatorname{Res}_{z}(C^{(\alpha)}(z)f(z)).$

We will need multiple copies of the map $c^{(\alpha)}$ acting on series in
variables $z_{1},z_{2},\dots$. We put
$c^{(\alpha)}_{i}(f(z_{i}))=c^{(\alpha)}(f(z))$, for $i=1,2,\dots$, and impose
linearity in the variables $z_{j}$, $j\neq i$, i.e., the condition that (for
instance)
$c^{(\alpha)}_{i}(f(z_{i})z_{j}^{m})=z^{m}_{j}c^{(\alpha)}_{i}(f(z_{i}))$.

Then we can write

$\displaystyle(\Gamma^{(\alpha)}_{C})^{l}$
$\displaystyle=\prod_{i=1}^{l}c^{(\alpha)}_{i}\prod_{j=1}^{l}E_{10}(z_{j})=$
$\displaystyle=\prod_{i=1}^{l}c^{(\alpha)}_{i}\prod_{j=1}^{l}\psi_{1}^{+}(z_{j})\psi_{0}^{-}(z_{j})=$
(C.12)
$\displaystyle=(-1)^{\frac{l(l-1)}{2}}\prod_{i=1}^{l}c^{(\alpha)}_{i}\prod_{s=1}^{l}\psi_{1}^{+}(z_{s})\prod_{t=1}^{l}\psi_{0}^{-}(z_{t}).$

Here we use that in the expression of $E_{10}(z_{1})$ in fermions we don’t
need normal ordering, and also Lemma A.2, part 2.

Now combine in (C.11) Lemma C.3 and (C.12) to get

$g_{ab}^{[k](\alpha)}(z)=\epsilon_{ab}^{[k](\alpha)}(z)\prod_{i=1}^{l}c^{(\alpha)}_{i}\langle
Q_{1}^{k-b}Q_{0}^{-k-1+b}v_{0},\psi_{a}^{-}(z)\prod_{s=1}^{l}\psi_{1}^{+}(z_{s})\prod_{t=1}^{l}\psi_{0}^{-}(z_{t})v_{0}\rangle/\tau_{k}^{(\alpha)},$

where

$\epsilon_{ab}^{[k](\alpha)}(z)=\frac{(-1)^{k}z^{k(2a-1)}}{l!}(-1)^{\frac{l(l-1)}{2}}(-1)^{\frac{k(k-1)}{2}}(-1)^{kb}.$

Here still $l=k+a-b$. Now use the factorization Lemma D.1; we get

$\displaystyle g_{ab}^{[k](\alpha)}(z)$
$\displaystyle=\frac{\epsilon_{ab}^{[k](\alpha)}(z)}{\tau_{k}^{(\alpha)}}\prod_{i=1}^{l}c_{i}\langle
Q_{1}^{k-b}v_{0},\left.\begin{cases}\prod_{j=1}^{l}\psi_{1}^{+}(z_{j})v_{0}\rangle&a=0,\\\
\psi_{1}^{-}(z)\prod_{j=1}^{l}\psi_{1}^{+}(z_{j})v_{0}\rangle,&a=1\end{cases}\right\\}\times$
$\displaystyle\times\langle
Q_{0}^{-k-1+b}v_{0},\left.\begin{cases}\psi_{0}^{-}(z)\prod_{j=1}^{l}\psi_{0}^{-}(z_{j})v_{0}\rangle&a=0,\\\
\prod_{j=1}^{l}\psi_{0}^{-}(z_{j})v_{0}\rangle,&a=1\end{cases}\right\\}$

Now using Lemma E.4 to calculate the factors involving $\psi_{0}(z)$ and
$\psi_{1}(z)$ separately we find

$g^{[k](\alpha)}_{ab}(z)=\frac{\epsilon_{ab}^{[k](\alpha)}(z)}{\tau_{k}^{(\alpha)}}\prod_{i=1}^{l}c_{i}\left(\det(V^{(l)}_{\\{z_{i}\\}})^{2}\prod_{j=1}(z-\underline{z}_{j})^{1-2a}\right).$

Comparing with the expression for the $\tau$-function in Theorem 2.1 and the
definition of the shift fields (2.10) gives Theorem 2.2.

###  C.3. Birkhoff Factorization in the $3\times 3$ case, Proof of Theorem 3.3

The proof of Theorem 3.3 is similar to that of 2.2 sketch in the previous
subsection. We leave the details to the reader.

##  Appendix D Factorization and Reduction to One Component Fermions

Often we want to calculate a matrix element in $F^{(n)}$ of fermion fields of
the form

$\langle Q_{n-1}^{\alpha_{n-1}}\dots
Q_{1}^{\alpha_{1}}Q_{0}^{\alpha_{0}}v_{0},P(\psi_{a}^{\pm}(z_{a}))v_{0}\rangle,$

where $P$ is some polynomial in the fermion fields $\psi_{a}^{\pm}(z_{a})$,
$a=0,1\dots,n-1$. By linearity we can reduce to the case where $P=M$ is a
monomial, and then we can in the monomial rearrange the factors as in
Definition A.1: $M=M_{n-1}\dots M_{1}M_{0}$, $M_{a}=M_{a}^{+}M_{a}^{-}$, where
$M_{a}^{\pm}$ is a monomial in a single type of fermions, ordered according to
the subscript of the arguments of the fields:

(D.1)
$M_{a}^{\pm}=\overleftarrow{\prod}_{i=1}^{t}\psi_{a}^{\pm}(z_{i})=\psi_{a}^{\pm}(z_{t})\psi_{a}^{\pm}(z_{t-1})\dots\psi^{\pm}_{a}(z_{2})\psi_{a}^{\pm}(z_{1}).$

This defines the ordered product of fermion fields.

We calculate such matrix elements using the following factorization Lemma.

######  Lemma D.1.

Let $M_{a}=M_{a}(\psi_{a}^{\pm}(z_{a}))$, $a=0,\dots,n-1$, be monomials in
fermion fields of type $a$. Then

$\langle
Q_{n-1}^{\alpha_{n-1}}Q_{1}^{\alpha_{1}}Q_{0}^{\alpha_{0}}v_{0},M_{n-1}\dots
M_{1}M_{0}v_{{}_{0}}\rangle=\langle
Q_{n-1}^{\alpha_{n-1}},M_{n-1}v_{0}\rangle\dots\langle
Q_{1}^{\alpha_{1}},M_{1}v_{0}\rangle\langle
Q_{0}^{\alpha_{0}},M_{0}v_{0}\rangle.$

So the correlation functions we want to calculate reduce to products of
correlation functions containing only one type of fermionic translation
operator $Q_{a}$ and one type of fermion fields $\psi_{a}^{\pm}(z)$. Such
correlation functions (on $F^{(n)}$) are the same as the corresponding
correlation function calculated on $F$, one component semi-infinite wedge
space, according to the following Lemma.

######  Lemma D.2.

$\langle Q_{a}^{\alpha}v_{0},F(\psi_{a}^{\pm})v_{0}\rangle_{F^{(n)}}=\langle
Q^{\alpha}v_{0},F(\psi^{\pm})v_{0}\rangle_{F}.$

In Appendix E we review some formulas for one component fermions.

##  Appendix E One Component Fermion Correlation Functions

In this Appendix we collect some results of one component fermions. In other
words we are dealing with the fermionic Fock space $F=F^{(1)}$, based on
$H=H^{(1)}$. The whole discussion of Appendix A transfers to the present one
component context. For typographical convenience we will write $\psi^{\pm}(z)$
for $\psi_{0}^{\pm}(z)$ and similarly we write $Q^{\pm 1}$ for $Q_{0}^{\pm
1}$.

The correlation functions are matrix elements in $F=F^{(1)}$ of the form
$\langle Q^{k}v_{0},M(z_{i},w_{j})v_{0}\rangle$, where $M$ is some monomial in
$\psi(z_{i})$ and $\psi(w_{j})$. For reasons of orthogonality of distinct
charges we need to insert a power of the fermionic translation operator $Q$.

The simplest case of a correlation function is

$\langle Q^{\pm
1}v_{0},\psi^{\pm}(z)v_{0}\rangle=\langle\psi^{\pm}_{(-1)}v_{0},\sum_{k\in\mathbb{Z}}\psi^{\pm}_{(-k-1)}v_{0}\rangle
z^{k}=1,$

since only the $k=0$ terms contributes.

For more general correlation functions we use the following Lemma recursively.

######  Lemma E.1.

Let $M(z_{i},w_{j})$ be a monomial in $\psi^{+}(z_{i})$, $i=1,2,\dots,s$ and
$\psi^{-}(w_{j})$, $j=1,2,\dots,t$, and put $k=s-t$. Then

$\displaystyle\langle Q^{k\pm
1}v_{0},\psi^{\pm}(z)M(z_{i},w_{j})v_{0}\rangle=\frac{\prod\limits_{i=1}^{s}(z-z_{i})}{\prod\limits_{j=1}^{t}(z-\underline{w_{j}})}\langle
Q^{k\pm 1}v_{0},M(z_{i},w_{j})v_{0}\rangle,$

where we expand in positive powers of the underlined variables.

For example, if all fermion fields are of the same type we get simple
determinantal expressions for the correlation functions. Recall the ordered
product of fermions, (D.1).

######  Lemma E.2.

For all $k\geq 1$ we have

$\langle Q^{\pm
k}v_{0},\overleftarrow{\prod}_{i=1}^{k}\psi^{\pm}(z_{i})v_{0}\rangle=\det(V^{(k)}_{\\{z_{i}\\}}).$

Here the Vandermonde matrix in $\\{z_{i}\\}=\\{z_{1},z_{2},\dots,z_{k}\\}$ is
given by

$V^{(k)}_{\\{z_{i}\\}}=\begin{pmatrix}1&1&\dots&1\\\
z_{1}&z_{2}&\dots&z_{k}\\\ z_{1}^{2}&z_{2}^{2}&\dots&z_{k}^{2}\\\
\vdots&\vdots&\ddots&\vdots\\\ z_{1}^{k-1}&z_{2}^{k-1}&\dots&z_{k}^{k-1}\\\
\end{pmatrix}=\begin{pmatrix}a_{ij}\end{pmatrix}_{i,j=1}^{k},$

where $a_{ij}=z^{i-1}_{j}$. Then we have

$\displaystyle\det(V^{k}_{\\{z_{i}\\}})$ $\displaystyle=\sum_{\sigma\in
S_{k}}(-1)^{\lvert\sigma\rvert}\prod_{i=1}^{k}a_{i\sigma(i)}=$
$\displaystyle=\sum_{\sigma\in_{k}}(-1)^{\lvert\sigma\rvert}\prod_{i=1}^{k}z_{\sigma(i)}^{i-1}=$
$\displaystyle=\prod_{k\geq\alpha>\beta\geq 1}(z_{\alpha}-z_{\beta}).$

We need

######  Lemma E.3.

$<Q^{m-n}v_{0},\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}>=\\\
=\frac{\prod_{1\leq i<j\leq m}(w_{i}-w_{j})\prod_{1\leq i<j\leq
n}(y_{i}-y_{j})}{\prod_{i=1}^{m}\prod_{j=1}^{n}(w_{i}-y_{j})}$

###### Proof.

We first consider the case that $m<n$. So

$\langle
Q^{m-n}v_{0},\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}\rangle=\\\
=\langle\psi_{(m-n)}^{-}\psi_{(m-n+1)}^{-}\cdots\psi_{(-1)}^{-}v_{0},\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}\rangle,$

which is equal to the sum of the coefficients corresponding to all ways of
pulling out $\psi_{(m-n)}^{-}\psi_{(m-n+1)}^{-}\cdots\psi_{(-1)}^{-}v_{0},$
from the product of fermionic fields acting on the vacuum,

$\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}.$

Given a $\sigma\in\mathfrak{S}_{n}$, we reorder the fermionic fields and
record the sign obtained from doing so:

$\operatorname{sgn}(\sigma)\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{\sigma}(1))\cdots\psi^{-}(y_{\sigma}(n))v_{0}$

We then pull out terms in such a way that no additional sign changes occur
from permuting the operators: We pull out $\psi_{(-1)}^{-}$ from
$\psi^{-}(y_{\sigma(n)})$, $\psi_{(-2)}^{-}$ from $\psi^{-}(y_{\sigma(n-1)})$,
$\cdots$, and $\psi_{(m-n)}^{-}$ from $\psi^{-}(y_{\sigma(m-n)})$. The product
of the coefficients corresponding to these choices is
$y_{\sigma(n)}^{0}y_{\sigma(n-1)}^{1}\cdots y_{\sigma(m+1)}^{n-m-1}$. The
remaining contributions for this choice of $\sigma$ come from pulling out
coefficients of products of wedging and contracting operators from

$\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{\sigma(1)})\cdots\psi^{-}(y_{\sigma(m)})$

whose actions cancel with each other. We again pull out these terms such a way
that no additional sign changes occur: We count only contributions coming from
terms in
$\psi^{+}(w_{m})\psi^{-}(y_{\sigma(1)})$ that cancel with each other, terms in
$\psi^{+}(w_{m-1})\psi^{-}(y_{\sigma(2)})$ that cancel with each other,
$\cdots$ and terms in $\psi^{+}(w_{1})\psi^{-}(y_{\sigma(m)})$ that cancel
with each other.

We claim that we can count each pair,
$\psi^{+}(w_{m-i})\psi^{-}(y_{\sigma(i+1)})$, $0\leq i\leq m-1$, as
contributing
$\displaystyle\frac{1}{w_{m-i}-y_{\sigma(i+1)}}=\displaystyle\sum_{\ell=0}^{\infty}\displaystyle\frac{y_{\sigma(i+1)}^{\ell}}{w_{m-i}^{\ell+1}}$.
We know we are not omitting any nontrivial terms in doing this, since any
$\displaystyle\frac{y_{\sigma(i+1)}^{\ell}}{w_{m-i}^{\ell+1}}$ with $\ell<0$
corresponds to $\psi^{+}_{(\ell)}\psi^{-}_{(-\ell-1)}$ and
$\psi^{-}_{(-\ell-1)}$ kills the vacuum or any vector obtained by acting by
contracting operators on the vacuum. We must therefore only prove that we are
not including any extra nontrivial terms. Towards this end, consider some
monomial,

$\operatorname{sgn}(\sigma)y_{\sigma(n)}^{0}y_{\sigma(n-1)}\cdots
y_{\sigma(m+1)}^{n-m-1}\frac{y_{\sigma(1)}^{\ell_{1}}}{w_{m}^{\ell_{1}+1}}\frac{y_{\sigma(2)}^{\ell_{2}}}{w_{m-1}^{\ell_{2}+1}}\cdots\frac{y_{\sigma(m)}^{\ell_{m}}}{w_{1}^{\ell_{m}+1}},$

corresponding to a product of wedges acting on the vacuum vector which give 0.
Since all of the wedges, $\psi^{-}_{(\ell)}$, is such that $\ell<0$, the only
way this is possible is if two of the wedges are the same. But this means that
two of the $y_{i}$s in the above expression are being raised to the same
power. Define a new element, $\gamma\in\mathfrak{S}_{n}$ by composing $\sigma$
with the transposition that interchanges these two $y_{i}$s. The sign of this
new element is $-\operatorname{sgn}(\sigma)$. So there a monomial in the
expansion of

$\operatorname{sgn}(\gamma)\displaystyle\frac{y_{\gamma(n)}^{0}y_{\gamma(n-1)}\cdots
y_{\gamma(m+1)}^{n-m-1}}{\displaystyle\prod_{i=0}^{m-1}(w_{m-i}-y_{\gamma(i+1)})}$

which cancels with the above monomial.

Summing over all $\sigma\in\mathfrak{S}_{n}$, we have that

$\displaystyle\langle Q^{m-n}v_{0},\psi^{+}(w_{1})\cdots$
$\displaystyle\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}\rangle=$
$\displaystyle=\sum_{\sigma\in\mathfrak{S}_{n}}\operatorname{sgn}(\sigma)\frac{y_{\sigma(n)}^{0}y_{\sigma(n-1)}\cdots
y_{\sigma(m+1)}^{n-m-1}}{\prod_{i=0}^{m-1}(w_{m-i}-y_{\sigma(i+1)})}.$

Using Leibniz’s formula to expand this as a determinant and then computing the
determinant, we find that this is exactly equal to

$\frac{\prod_{1\leq i<j\leq m}(w_{i}-w_{j})\prod_{1\leq i<j\leq
n}(y_{i}-y_{j})}{\prod_{i=1}^{m}\prod_{j=1}^{n}(w_{i}-y_{j})}$

The proof in the case that $m\geq n$ is similar. Here, we need to argue that

$\displaystyle\langle Q^{m-n}v_{0},$
$\displaystyle\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}\rangle=$
$\displaystyle=\sum_{\sigma\in\mathfrak{S}_{m}}$
$\displaystyle\operatorname{sgn}(\sigma)\frac{w_{\sigma(1)}^{m-n-1}w_{\sigma(2)}^{m-n-2}\cdots
w_{\sigma(m-n)}^{0}}{\prod_{i=0}^{n-1}(w_{\sigma({m-i})}-y_{i+1})}.$

∎

We also need the following case.

######  Lemma E.4.

$\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle=$
$\displaystyle\frac{\displaystyle\prod_{i=1}^{m}(z-y_{i})\displaystyle\prod_{1\leq
i<j\leq m}(w_{i}-w_{j})\displaystyle\prod_{1\leq i<j\leq
n}(y_{i}-y_{j})}{\displaystyle\prod_{i=1}^{n}(z-w_{i})\displaystyle\prod_{i=1}^{m}\displaystyle\prod_{j=1}^{n}(w_{i}-y_{j})}$

###### Proof.

Consider the case that $m>n$. The proof here is analogous to LEMMA E.3.

$\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle$

is the sum of the products of coefficients corresponding to all of the ways of
pulling out $Q^{m-n-1}v_{0}=\psi^{+}_{(m-n-1)}\cdots\psi^{+}_{(-1)}v_{0}$ from

$\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}.$

Since $Q^{m-n-1}v_{0}$ is a product of wedging operators acting on the vacuum,
the actions of the contracting operators in the negative fermion fields,
$\psi^{-}(z)$ and the $\psi^{-}(y_{i})$s, must cancel with the actions of the
wedging operators in the positive fermion fields, the $\psi^{+}(w_{i})$s. The
remaining positive fermion fields contribute to the

$\psi^{+}_{(m-n-1)}\cdots\psi^{+}_{(-1)}v_{0}.$

Given an element $\sigma\in\mathfrak{S}_{m}$, we reorder the positive fermion
fields and record the sign obtained by doing so:

$\operatorname{sgn}(\sigma)\psi^{-}(z)\psi^{+}(w_{\sigma(1)})\cdots\psi^{+}(w_{\sigma(m)})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}.$

We then pull out wedging and contracting operators in a fixed way such that no
additional sign changes occur. Since $m>n$, taking the sum over all
$\sigma\in\mathfrak{S}_{m}$, we obtain all terms that contribute to

$\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0}\rangle.$

In more detail, given $\sigma\in\mathfrak{S}_{m}$, we have

$\operatorname{sgn}(\sigma)\psi^{-}(z)\psi^{+}(w_{\sigma(1)})\cdots\psi^{+}(w_{\sigma(m)})\psi^{-}(y_{1})\cdots\psi^{-}(y_{n})v_{0},$

and we collect all of the coefficients in $\psi^{-}(z)\psi^{+}(w_{\sigma(1)})$
corresponding to wedging and contracting operators whose actions cancel with
each other. We pull out $\psi^{+}_{(n-m+1)}$ from
$\psi^{+}(w_{\sigma(2)})$, $\psi^{+}_{(n-m)}$ from $\psi^{+}(w_{\sigma(3)})$,
$\cdots,$ and $\psi^{+}_{(-1)}$ from $\psi^{+}(w_{\sigma(m-n)})$. We then
collect the coefficients corresponding to all pairs of wedging and contracting
operators in $\psi^{+}(w_{\sigma(m)})\psi^{-}(y_{1})$ whose actions cancel
with each other, all pairs of wedging and contracting operators in
$\psi^{+}(w_{\sigma(m-1)})\psi^{-}(y_{2})$ whose actions cancel with each
other, $\cdots$, and all pairs of wedging and contracting operators in
$\psi^{+}(w_{\sigma(m-n+1)})\psi^{-}(y_{n})$ whose actions cancel with each
other.

We claim that each pair, $\psi^{+}(a)\psi^{-}(b)$ corresponding to wedging and
contracting operators whose actions cancel with each other can be thought of
as contributing $\displaystyle\frac{1}{a-b}$. To see why this is true, we
comment that each monomial in the expansion of

$\langle
Q^{m-m-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{-}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle$

corresponds to a choice of fermionic operators. Given
$\sigma\in\mathfrak{S}_{m}$, a monomial corresponds to a choice of $k_{i}$s
such that

$\operatorname{sgn}\sigma\times\psi^{-}_{(-k_{0}-1)}\psi^{+}_{(k_{0})}\psi^{+}_{(n-m+1)}\cdots\psi^{+}_{(-1)}\psi^{+}_{(-k_{1}-1)}\psi^{+}_{(-k_{2}-1)}\cdots\psi^{+}_{(-k_{n}-1)}\psi^{-}_{(k_{n})}\psi^{-}_{(k_{n-1})}\cdots\psi^{-}_{(k_{1})}v_{0}=v_{0}.$

We first note that $k_{1},\cdots,k_{n}$ must all be less than $0$, since if
$k_{i}>0$, the corresponding contracting operator, $\psi^{-}_{(k_{i})}$ kills
the vacuum or any vector obtained from the vaacuum by acting by contracting
operators. The only other possible way of choosing $k_{i}$s, $1\leq i\leq n$,
so that the above product of fermion operators acting on the vacuum is $0$, is
if two $k_{i}$s are equal or some $k_{i}$ is $n-m\leq k_{i}\leq-1$. Using the
same argument that we used in LEMMA E.3, we see that the monomials
corresponding to such choices cancel with each other. So we are not including
any extra terms in allowing each of the $k_{i}$s to range over all negative
integers. So each pair, $\psi^{+}(w_{\sigma(m-i)})\psi^{-}(y_{i+1}),$ $0\leq
i\leq n-1,$ from which we pull out coefficients corresponding to wedging and
contracting operators whose actions cancel with each other can be thought of
as contributing $\displaystyle\frac{1}{w_{\sigma(m-i)}-y_{i+1}}$. Similarly,
$k_{0}<0$ and even though choosing $k_{0}=i$ for some $n-m+1\leq i\leq-1$ will
result in killing the vacuum, we can include such choices since the
corresponding monomials all cancel with each other. Summing over all
$\sigma\in\mathfrak{S}_{m}$, we obtain

$\displaystyle\sum_{\sigma\in\mathfrak{S}_{m}}\operatorname{sgn}(\sigma)\displaystyle\frac{w_{\sigma(2)}^{m-n-2}\cdots
w_{\sigma(m-n)}^{0}}{(z-w_{\sigma(1)})(w_{\sigma(m)}-y_{1})(w_{\sigma(m-1)}-y_{2})\cdots(w_{\sigma(m-n+1)}-y_{n})},$

which is precisely the Leibniz form of

$\det\begin{bmatrix}\frac{1}{z-w_{1}}&\frac{1}{z-w_{2}}&\cdots&\frac{1}{z-w_{m}}\\\
&&&\\\ w_{1}^{m-n-2}&w_{2}^{m-n-2}&\cdots&w_{m}^{m-n-2}\\\
\vdots&\vdots&\cdots&\vdots\\\ 1&1&\cdots&1\\\
\frac{1}{w_{1}-y_{n}}&\frac{1}{w_{2}-y_{n}}&\cdots&\frac{1}{w_{m}-y_{n}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{2}-y_{1}}&\cdots&\frac{1}{w_{m}-y_{1}}\end{bmatrix},$

and the result holds by LEMMA G.3.

Next, consider the case that $m\leq n$. $Q^{m-n-1}v_{0}$ here is a product of
contracting operators acting on the vacuum.

Since $m\leq n$, there are more negative fermion fields than positive fermion
fields. We would like to proceed exactly as we did in the $m>n$ case. Ideally,
we would want to fix an order of the $n+1$ negative fermion fields given by
some permutation $\sigma\in\mathfrak{S}_{n+1}$, pull out (in a fixed way)
wedging and contracting operators whose product acting on the vacuum is
$Q^{m-n-1}v_{0}$, and sum over all permutations of the negative fermion fields
to obtain all ways of pulling out $Q^{m-n-1}v_{0}$ from
$\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}$.
This cannot be done as easily as in the $m>n$ case though, since the positive
and negative fermion fields do not commute and the positive fermion fields are
sandwiched between $\psi^{-}(z)$ and
$\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i}).$ We would thus obtain more than
a sign change if we chose a permutation that did not fix $\psi^{-}(z)$.

By writing

$\displaystyle\langle Q^{m-n-1}v_{0},$
$\displaystyle\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle=$
$\displaystyle=\langle v_{0},$
$\displaystyle\psi^{+}_{(0)}\psi^{+}_{(1)}\cdots\psi^{+}_{(n-m)}\psi^{-}(z)\psi^{+}(w_{1})\cdots\psi^{+}(w_{m})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle,$

we see that, in order to obtain a product of the vacuum vector from

$\psi^{+}_{(0)}\psi^{+}_{(1)}\cdots\psi^{+}_{(n-m)}\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0},$

the actions of any contracting operators in $\psi^{-}(z)$ must either cancel
with the actions of wedging operators in the $\psi^{+}(w_{i})$s to the right
of $\psi^{-}(z)$, or must cancel with the wedging operators, $\psi^{+}_{(i)}$,
$0\leq i\leq n-m$ to the left of $\psi^{-}(z)$. There are thus $n+1$ choices,
$\psi^{+}_{(0)}$, $\psi^{+}_{(1)}$, $\cdots$, $\psi^{+}_{(n-m)}$,
$\psi^{+}(y_{1})$, $\cdots$, $\psi^{+}(y_{n})$, for terms to which
$\psi^{-}(z)$ can be paired. Given one of these $n+1$ choices, we then can
proceed as in the $m>n$ case, fixing an order of negative fermion fields given
by some $\sigma\in\mathfrak{S}_{n}$, pulling out fermion operators from these
ordered terms in a fixed way, and reocording the corresponding monomials.
Summing over all $\sigma\in\mathfrak{S}_{n}$ and all $n+1$ choices, we obtain
our result.

Since the fermion operators in $\psi^{-}(z)$ can be paired only with the
$\psi^{+}_{(i)}$s to the left of $\psi^{-}(z)$ or with the $\psi^{+}_{(i)}$s
in the $\psi^{+}(w_{i})$s, our choices of fermion operators have only two
possible forms:

  1. (1)

$(-1)^{n-m-k_{0}}\operatorname{sgn}\sigma\times$ $\langle
v_{0},\psi^{+}_{(0)}\psi^{+}_{(1)}\cdots\psi^{+}_{(k_{0}-1)}\psi^{+}_{(k_{0}+1)}\cdots\psi^{+}_{(n-m)}\psi^{+}_{(k_{0})}$
$\psi^{-}_{(-k_{0}-1)}\psi^{+}_{(k_{1})}\psi^{+}_{(k_{2})}\cdots\psi^{+}_{(k_{m})}\psi^{-}_{(-k_{m}-1)}\cdots\psi^{-}_{(-k_{1}-1)}\psi^{-}_{(m-n-1)}\cdots\psi^{-}_{(-k_{0}-2)}\psi^{-}_{(-k_{0})}\cdots\psi^{-}_{(-2)}\psi^{-}_{(-1)}v_{0}\rangle$

(The additional sign appearing in front of $\operatorname{sgn}\sigma$ comes
from shifting the $\psi^{+}_{(k_{0})}$, $0\leq k_{0}\leq n-m$ past the $n-m-
k_{0}$ $\psi^{+}_{(i)}$s to its right, so that it appears immediately next to
$\psi^{-}(z)$.)

  2. (2)

$(-1)^{i-1}\operatorname{sgn}\sigma\times$ $\langle
v_{0},\psi^{+}_{(0)}\psi^{+}_{(1)}\cdots\psi^{+}_{(n-m)}\psi^{-}_{(-k_{i}-1)}\psi^{+}_{(k_{i})}\psi^{+}_{(k_{1})}\cdots\psi^{+}_{(k_{i-1})}\psi^{+}_{(k_{i+1})}\cdots\psi^{+}_{(k_{m})}$
$\psi^{-}_{(-k_{m}-1)}\psi^{-}_{(-k_{m-1}-1)}\cdots\psi^{-}_{(-k_{i+1}-1)}\psi^{-}_{(-k_{i-1}-1)}\cdots\psi^{-}_{(-k_{1}-1)}\psi^{-}_{(m-n-1)}\cdots\psi^{-}_{(-2)}\psi^{-}_{(-1)}v_{0}\rangle$

(Here, the additonal sign appearing in front of $\operatorname{sgn}\sigma$
comes from shifting the $\psi^{+}(w_{i})$ past the $i-1$ $\psi^{(w_{j})}$s to
its left so that $\psi^{+}(w_{i})$ appears next to $\psi^{-}(z)$.)

We now argue which choices of $k_{i}$s contribute to $\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle$.

For choices of the form $(1)$, each $k_{j}$ $1\leq j\leq m$ must be
nonnegative, since if $k<0$, $\psi^{-}_{(-k-1)}$ kills the vacuum or any
vector obtained from the vacuum by acting by contracting operators. Given
$k_{0}$, $0\leq k_{0}\leq n-m$,

$\psi^{-}_{(-k_{m}-1)}\cdots\psi^{-}_{(-k_{1}-1)}\psi^{-}_{(m-n-1)}\cdots\psi^{-}_{(-k_{0}-2)}\psi^{-}_{(-k_{0})}\cdots\psi^{-}_{(-1)}v_{0}=0$

if any of the $k_{j}$s, $1\leq j\leq m$, $j\neq k_{0}$ are equal to each other
or to some $\ell$, $0\leq\ell\leq n-m$. Using the same argument used above and
in LEMMA E.3 though, any such monomials cancel with each other. So we can sum
over all $k_{j}\geq 0$, $j\neq i$. Then summing over all $k_{0}$, $0\leq
k_{0}\leq n-m$ and all $\sigma\in\mathfrak{S}_{n}$, we obtain

$\sum_{k_{0}=0}^{n-m}(-1)^{n-m-
k_{0}}z^{k_{0}}\displaystyle\sum_{\sigma\in\mathfrak{S}_{n}}\operatorname{sgn}{\sigma}\displaystyle\frac{y^{n-m}_{\sigma(m+1)}\cdots
y^{k_{0}+1}_{\sigma(n-k_{0})}y_{\sigma(n+1-k_{0})}^{k_{0}-1}\cdots
y_{\sigma(n)}^{0}}{(w_{m}-y_{\sigma(1)})(w_{m-1}-y_{\sigma(2)})\cdots(w_{1}-y_{\sigma(m)})}.$

We next sum over all possible values of $k_{i}$s for choices of the form
$(2)$, making sure not to include any choices equivalent to those already
counted in $(1)$.

When $k<0$, $\psi^{-}_{(-k-1)}$ kills the vacuum or any vector obtained from
the vacuum by acting by contracting operators. Given a choice of $i$, we have

$\psi^{+}_{(k_{1})}\cdots\psi^{+}_{(k_{i-1})}\psi^{+}_{(k_{i+1})}\cdots\psi^{+}_{(k_{m})}$
$\psi^{-}_{(-k_{m}-1)}\psi^{-}_{(-k_{m-1}-1)}\cdots\psi^{-}_{(-k_{i+1}-1)}\psi^{-}_{(-k_{i-1}-1)}\cdots\psi^{-}_{(-k_{1}-1)}\psi^{-}_{(m-n-1)}\cdots\psi^{-}_{(-2)}\psi^{-}_{(-1)}v_{0}=$
$=\psi^{-}_{(m-n-1)}\cdots\psi^{-}_{(-2)}\psi^{-}_{(-1)}v_{0}.$

So we can choose $k_{i}<0$ or $0\leq k_{i}\leq n-m$. We claim though, that
$0\leq k_{i}\leq n-m$ correspond to choices of fermion operators already made
in $(1)$. Note that the $k_{i}<0$ cannot correspond to choices from $(1)$
since $k_{i}$ is the power of $z$ in a given monomial and every monomial
coming from $(1)$ has power of $z$ greater than or equal to $0$ and less than
or equal to $n-m$. As in our previous arguments, given a choice of $k_{i}$, we
can allow our $k_{j}$s, $j\neq i$ to range over all nonnegative integers,
since any terms corresponding to choices of $k_{j}$s that kill the vacuum will
cancel with each other. Summing over all choices of $k_{i}$s we obtain

$\sum_{i=1}^{m}(-1)^{i-1}\displaystyle\sum_{\sigma\in\mathfrak{S}_{n}}\operatorname{sgn}{\sigma}\displaystyle\frac{y^{n-m}_{\sigma(m)}y^{n-m-1}_{\sigma(m+1)}\cdots
y_{\sigma(n)}^{0}}{(z-w_{i})(w_{1}-y_{\sigma(m-1)})\cdots(w_{i-1}-y_{\sigma(m-i+1)})(w_{i+1}-y_{\sigma(m-i)})\cdots(w_{m}-y_{\sigma(1)})}.$

(To see why choices of $k_{i}$, $0\leq k_{i}\leq n-m$ correspond to choices
already made in $(1)$, note that, up to a sign, the choices corresponding to
$(1)$ give monomials

$\frac{z^{k_{0}}y_{\sigma(1)}^{k_{m}}\cdots
y_{\sigma(m)}^{k_{1}}y_{\sigma(m+1)}^{n-m}\cdots
y_{\sigma(m-k_{0})}^{k_{0}+1}y_{\sigma(m-k_{0}+1)}^{k_{0}-1}\cdots
y_{\sigma(n-1)}y_{\sigma(n)}^{0}}{w_{1}^{k_{1}+1}w_{2}^{k_{2}+1}\cdots
w_{m}^{k_{m}+1}}$

where $0\leq k_{0}\leq n-m$ and the remaining $k_{i}$s are nonnegative
integers.

The choices corresponding to $(2)$ give

$\frac{z^{k_{i}}y_{\sigma(1)}^{k_{m}}\cdots
y_{\sigma(m-i)}^{k_{i+1}}y_{\sigma(m-i+1)}^{k_{i-1}}\cdots
y_{\sigma(m-1)}^{k_{1}}y_{\sigma(m)}^{n-m}\cdots
y_{\sigma(n-1)}y_{\sigma(n)}^{0}}{w_{1}^{k_{1}+1}w_{2}^{k_{2}+1}\cdots
w_{m}^{k_{m}+1}}$

where $k_{i}<0$ or $0\leq k_{i}\leq n-m$ and the remaining $k_{j}$s range over
all nonnegative integers. We comment that we do not need to know the signs for
these monomials, since we can read from them the corresponding choice of
fermion operators and this choice determines the sign. Thus, if any monomials
corresponding to $(1)$ or $(2)$ are the same up to a sign, their corresponding
signs must also match. Given a monomial of the form $(2)$, whose $k_{i}$ is
$0\leq k_{i}\leq n-m$, we can set $k_{i}=k_{0}$ for some $k_{0}$ corresponding
to a choice in $(1)$. If the choice of $k_{1},\cdots k_{m}$ in $(1)$ also
agree with those in $(2)$, the powers of each $w_{i}$s in these monomials
match and the powers of the $y_{i}$s match up to a permutation of their
indices. We can then modify $(1)$ by choosing a $\sigma\in\mathfrak{S}_{n}$
such that the monomials match exactly. So choices of $k_{i}$, $0\leq k_{i}\leq
n-m$ in $(2)$ correspond to terms already counted in $(1)$.)

Summing the contributions from $(1)$ and $(2)$, we have

$\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle=$

$\sum_{k_{0}=0}^{n-m}(-1)^{n-m-
k_{0}}z^{k_{0}}\displaystyle\sum_{\sigma\in\mathfrak{S}_{n}}\operatorname{sgn}{\sigma}\displaystyle\frac{y^{n-m}_{\sigma(m+1)}\cdots
y^{k_{0}+1}_{\sigma(n-k_{0})}y_{\sigma(n+1-k_{0})}^{k_{0}-1}\cdots
y_{\sigma(n)}^{0}}{(w_{m}-y_{\sigma(1)})(w_{m-1}-y_{\sigma(2)})\cdots(w_{1}-y_{\sigma(m)})}+$

$\sum_{i=1}^{m}(-1)^{i-1}\displaystyle\sum_{\sigma\in\mathfrak{S}_{n}}\operatorname{sgn}{\sigma}\displaystyle\frac{y^{n-m}_{\sigma(m)}y^{n-m-1}_{\sigma(m+1)}\cdots
y_{\sigma(n)}^{0}}{(z-w_{i})(w_{1}-y_{\sigma(m-1)})\cdots(w_{i-1}-y_{\sigma(m-i+1)})(w_{i+1}-y_{\sigma(m-i)})\cdots(w_{m}-y_{\sigma(1)})}.$

This then gives us that

$\langle
Q^{m-n-1}v_{0},\psi^{-}(z)\displaystyle\prod_{i=1}^{m}\psi^{+}(w_{i})\displaystyle\prod_{i=1}^{n}\psi^{-}(y_{i})v_{0}\rangle=$

$(-1)^{m}\det\begin{bmatrix}\displaystyle\frac{-1}{z-w_{m}}&\displaystyle\frac{1}{w_{m}-y_{1}}&\displaystyle\frac{1}{w_{m}-y_{2}}&\cdots&\displaystyle\frac{1}{w_{m}-y_{n}}\\\
\displaystyle\frac{-1}{z-w_{m-1}}&\displaystyle\frac{1}{w_{m-1}-y_{1}}&\displaystyle\frac{1}{w_{m-1}-y_{2}}&\cdots&\displaystyle\frac{1}{w_{m-1}-y_{n}}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\displaystyle\frac{-1}{z-w_{2}}&\displaystyle\frac{1}{w_{2}-y_{1}}&\displaystyle\frac{1}{w_{2}-y_{2}}&\cdots&\displaystyle\frac{1}{w_{2}-y_{n}}\\\
\displaystyle\frac{-1}{z-w_{1}}&\displaystyle\frac{1}{w_{1}-y_{1}}&\displaystyle\frac{1}{w_{1}-y_{2}}&\cdots&\displaystyle\frac{1}{w_{1}-y_{n}}\\\
&&&&\\\ z^{n-m}&y_{1}^{n-m}&y_{2}^{n-m}&\cdots&y_{n}^{n-m}\\\ &&&&\\\
z^{n-m-1}&y_{1}^{n-m-1}&y_{2}^{n-m-1}&\cdots&y_{n}^{n-m-1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\ z&y_{1}&y_{2}&\cdots&y_{n}\\\
1&1&1&\cdots&1\end{bmatrix}.$

(To see this, expand the determinant along the first column and express the
corresponding minors using the Leibniz formula for the determinant.) The
result then holds by LEMMA G.4. ∎

##  Appendix F Proof of Determinantal Identities

The proofs here are very similar to the proof of the Cauchy determinant
identity. In each of the proofs, we perform elementary row and column
operations, pull terms out, and show that the determinant reduces to some
product times a Cauchy determinant.

Here, we will use the convention that
$\frac{1}{a-b}=\displaystyle\sum_{i=0}^{\infty}\frac{b^{n}}{a^{n+1}}.$ For
example, when we write $\frac{-1}{a-b}$, we mean
$-\displaystyle\sum_{i=0}^{\infty}\frac{b^{n}}{a^{n+1}}$ and not
$\frac{1}{b-a}=\displaystyle\sum_{i=0}^{\infty}\frac{a^{n}}{b^{n+1}}.$

######  Lemma F.1.

When $m\geq n,$

$\det\begin{bmatrix}\frac{1}{w_{n}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{n}-y_{m}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{1}-y_{2}}&\cdots&\frac{1}{w_{1}-y_{m}}\\\
&&&\\\ y_{1}^{m-n-1}&y_{2}^{m-n-1}&\cdots&y_{m}^{m-n-1}\\\
\vdots&\vdots&\vdots&\vdots\\\ 1&1&\cdots&1\\\
\end{bmatrix}=\frac{\displaystyle\prod_{1\leq i<j\leq
m}(y_{i}-y_{j})\prod_{1\leq i<j\leq
n}(w_{i}-w_{j})}{\displaystyle\prod_{i=1}^{n}\prod_{j=1}^{m}(w_{i}-y_{j})}.$

###### Proof.

We first comment that if $m=n$, we obtain

$\det\begin{bmatrix}\frac{1}{w_{n}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{n}-y_{n}}\\\
\frac{1}{w_{n-1}-y_{1}}&\frac{1}{w_{n-1}-y_{2}}&\cdots&\frac{1}{w_{n-1}-y_{n}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{1}-y_{2}}&\cdots&\frac{1}{w_{1}-y_{n}}\end{bmatrix}$

and the result holds by the formula for a Cauchy determinant. So in the
following, we may assume that $m>n$.

We subtract the last column from each of the previous columns, expand the
determinant of the resulting matrix along the last row, and then pull out
$\frac{\displaystyle\prod_{i=1}^{m-1}(y_{i}-y_{m})}{\displaystyle\prod_{i=1}^{n}(w_{i}-y_{m})}.$
This gives us

$\frac{\displaystyle\prod_{i=1}^{m-1}(y_{i}-y_{m})}{\displaystyle\prod_{i=1}^{n}(w_{i}-y_{m})}\det\begin{bmatrix}\frac{1}{w_{n}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{n}-y_{m-1}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{1}-y_{2}}&\cdots&\frac{1}{w_{1}-y_{m-1}}\\\
&&&&\\\ \displaystyle\sum_{j=0}^{m-n-2}\displaystyle
y_{1}^{m-n-2-j}y_{m}^{j}&\displaystyle\sum_{j=0}^{m-n-2}y_{2}^{m-n-2-j}y_{m}^{j}&\cdots&\displaystyle\sum_{j=0}^{m-n-2}y_{3}^{m-n-2-j}y_{m}^{j}y_{m}^{m-n-1}\\\
&&&&\\\ \vdots&\vdots&\vdots&\vdots\\\
y_{1}^{2}+y_{m}^{2}+y_{1}y_{m}&y_{2}^{2}+y_{m}^{2}+y_{2}y_{m}&\cdots&y_{m-1}^{2}+y_{m}^{2}+y_{m-1}y_{m}\\\
y_{1}+y_{m}&y_{2}+y_{m}&\cdots&y_{m-1}+y_{m}\\\ 1&1&\cdots&1\\\
\end{bmatrix}.$

We then subtract $y_{m}$ times the $(n+2)$nd row from the $(n+1)$st row,
subtract $y_{m}$ times the $(n+3)$rd row from the $(n+2)$nd row, $\cdots$, and
subtract $y_{m}$ times the last row from the second to last row. Since
$\displaystyle\sum_{j=0}^{\ell}y_{i}^{\ell-j}y_{m}^{j}-\displaystyle\sum_{j=0}^{\ell-1}y_{i}^{\ell-1-j}y_{m}^{j+1}=y_{i}^{\ell}$,
this gives us

$\frac{\displaystyle\prod_{i=1}^{m-1}(y_{i}-y_{m})}{\displaystyle\prod_{i=1}^{n}(w_{i}-y_{m})}\det\begin{bmatrix}\frac{1}{w_{n}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{n}-y_{m-1}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{1}-y_{2}}&\cdots&\frac{1}{w_{1}-y_{m-1}}\\\
&&&&\\\ y_{1}^{m-n-2}&y_{2}^{m-n-2}&\cdots&y_{m-1}^{m-n-2}\\\
\vdots&\vdots&\vdots&\vdots\\\ y_{1}^{2}&y_{2}^{2}&\cdots&y_{m-1}^{2}\\\
y_{1}&y_{2}&\cdots&y_{m-1}\\\ 1&1&\cdots&1\\\ \end{bmatrix}.$

We then repeat this process $m-n-1$ more times, obtaining

$\frac{\displaystyle\prod_{j=n+1}^{m}\displaystyle\prod_{i<j}(y_{i}-y_{j})}{\displaystyle\prod_{i=1}^{n}\displaystyle\prod_{j=n+1}^{m}(w_{i}-y_{j})}\det\begin{bmatrix}\frac{1}{w_{n}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{n}-y_{n}}\\\
&&&\\\
\frac{1}{w_{n-1}-y_{1}}&\frac{1}{w_{n-1}-y_{2}}&\cdots&\frac{1}{w_{n-1}-y_{n}}\\\
&&&\\\ \vdots&\vdots&\cdots&\vdots\\\ &&&\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{n}-y_{2}}&\cdots&\frac{1}{w_{1}-y_{n}}\end{bmatrix}.$

Since the determinant here is a Cauchy determinant equal to

$\frac{\displaystyle\prod_{1\leq i<j\leq
n}(y_{i}-y_{j})\displaystyle\prod_{1\leq i<j\leq
n}(w_{i}-w_{j})}{\displaystyle\prod_{i=1}^{n}\displaystyle\prod_{j=1}^{n}(w_{i}-y_{j})},$

we have that the determinant with which we started is equal to

$\frac{\displaystyle\prod_{j=n+1}^{m}\displaystyle\prod_{i<j}(y_{i}-y_{j})\displaystyle\prod_{1\leq
i<j\leq n}(y_{i}-y_{j})\displaystyle\prod_{1\leq i<j\leq
n}(w_{i}-w_{j})}{\displaystyle\prod_{i=1}^{n}\displaystyle\prod_{j=n+1}^{m}(w_{i}-y_{j})\displaystyle\prod_{i=1}^{n}\displaystyle\prod_{j=1}^{n}(w_{i}-y_{j})}=$
$=\frac{\displaystyle\prod_{1\leq i<j\leq
n}(w_{i}-w_{j})\displaystyle\prod_{1\leq i<j\leq
m}(y_{i}-y_{j})}{\displaystyle\prod_{i=1}^{n}\displaystyle\prod_{j=1}^{m}(w_{i}-y_{j})}.$

∎

######  Lemma F.2.

When $m<n,$

$\det\begin{bmatrix}\displaystyle
w_{1}^{n-m-1}&w_{2}^{n-m-1}&\cdots&w_{n-1}^{n-m-1}&w_{n}^{n-m-1}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\ w_{1}&w_{2}&\cdots&w_{n-1}&w_{n}\\\
1&1&\cdots&1&1\\\ &&&\\\
\frac{1}{w_{1}-y_{m}}&\frac{1}{w_{2}-y_{m}}&\cdots&\frac{1}{w_{n-1}-y_{m}}&\frac{1}{w_{n}-y_{m}}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{2}-y_{1}}&\cdots&\frac{1}{w_{n-1}-y_{1}}&\frac{1}{w_{n}-y_{1}}\\\
\end{bmatrix}=\frac{\displaystyle\prod_{1\leq i<j\leq
m}(y_{i}-y_{j})\prod_{1\leq i<j\leq
n}(w_{i}-w_{j})}{\displaystyle\prod_{i=1}^{n}\prod_{j=1}^{m}(w_{i}-y_{j})}.$

###### Proof.

We prove this identity in the same way that we proved our previous identity,
by applying elementary row and column operations to reduce the determinant to
a product times a Cauchy determinant. We begin by subtracting the last column
from each of the previous columns. This gives us

$\det\begin{bmatrix}w_{1}^{n-m-1}-w_{n}^{n-m-1}&w_{2}^{n-m-1}-w_{n}^{n-m-1}&\cdots&w_{n-1}^{n-m-1}-w_{n}^{n-m-1}&w_{n}^{n-m-1}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
w_{1}-w_{n}&w_{2}-w_{n}&\cdots&w_{n-1}-w_{n}&w_{n}\\\ 0&0&\cdots&0&1\\\
\frac{-(w_{1}-w_{n})}{(w_{1}-y_{m})(w_{n}-y_{m})}&\frac{-(w_{2}-w_{n})}{(w_{2}-y_{m})(w_{n}-y_{m})}&\cdots&\frac{-(w_{n-1}-w_{n})}{(w_{n-1}-y_{m})(w_{n}-y_{m})}&\frac{1}{w_{n}-y_{m}}\\\
\vdots&\vdots&\cdots&\vdots&\\\
\frac{-(w_{1}-w_{n})}{(w_{1}-y_{1})(w_{n}-y_{1})}&\frac{-(w_{2}-w_{n})}{(w_{2}-y_{1})(w_{n}-y_{1})}&\cdots&\frac{-(w_{n-1}-w_{n})}{(w_{n-1}-y_{1})(w_{n}-y_{1})}&\frac{1}{w_{n}-y_{1}}\end{bmatrix}.$

We then expand along the $(n-m)$th row to obtain

$\det\begin{bmatrix}w_{1}^{n-m-1}-w_{n}^{n-m-1}&w_{2}^{n-m-1}-w_{n}^{n-m-1}&\cdots&w_{n-1}^{n-m-1}-w_{n}^{n-m-1}\\\
\vdots&\vdots&\cdots&\vdots\\\ w_{1}-w_{n}&w_{2}-w_{n}&\cdots&w_{n-1}-w_{n}\\\
&&&&\\\
\frac{w_{1}-w_{n}}{(w_{1}-y_{m})(w_{n}-y_{m})}&\frac{w_{2}-w_{n}}{(w_{2}-y_{m})(w_{n}-y_{m})}&\cdots&\frac{w_{n-1}-w_{n}}{(w_{n-1}-y_{m})(w_{n}-y_{m})}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{w_{1}-w_{n}}{(w_{1}-y_{1})(w_{n}-y_{1})}&\frac{w_{2}-w_{n}}{(w_{2}-y_{1})(w_{n}-y_{1})}&\cdots&\frac{w_{n-1}-w_{n}}{(w_{n-1}-y_{1})(w_{n}-y_{1})}\end{bmatrix}.$

(We comment that expanding along the $(n-m)$th row gives us a sign of
$(-1)^{m}$. This sign cancels with the factors of $-1$ pulled out from each of
our last $m$ rows.) We then pull out
$\frac{\displaystyle\prod_{i=1}^{n-1}(w_{i}-w_{n})}{\displaystyle\prod_{i=1}^{m}(w_{n}-y_{i})}$
to obtain

$\frac{\displaystyle\prod_{i=1}^{n-1}(w_{i}-w_{n})}{\displaystyle\prod_{i=1}^{m}(w_{n}-y_{i})}\det\begin{bmatrix}\displaystyle\sum_{j=0}^{n-m-2}w_{1}^{n-m-2-j}w_{n}^{j}&\displaystyle\sum_{j=0}^{n-m-2}w_{2}^{n-m-2-j}w_{n}^{j}&\cdots&\displaystyle\sum_{j=0}^{n-m-2}w_{n-1}^{n-m-2-j}w_{n}^{j}\\\
\vdots&\vdots&\cdots&\vdots\\\ w_{1}+w_{n}&w_{2}+w_{n}&\cdots&w_{n-1}+w_{n}\\\
&&&\\\ 1&1&\cdots&1\\\ &&&\\\
\frac{1}{w_{1}-y_{m}}&\frac{1}{w_{2}-y_{m}}&\cdots&\frac{1}{w_{n-1}-y_{m}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{2}-y_{1}}&\cdots&\frac{1}{w_{n-1}-y_{1}}\end{bmatrix}.$

Similarly to the proof of our previous identity, we subtract $w_{n}$ times row
$2$ from row $1$, subtract $w_{n}$ times row $3$ from row $2$, $\cdots$, and
subtract $w_{n}$ times row $n-m-1$ from row $n-m-2$. This gives

$\frac{\displaystyle\prod_{i=1}^{n-1}(w_{i}-w_{n})}{\displaystyle\prod_{i=1}^{m}(w_{n}-y_{i})}\det\begin{bmatrix}w_{1}^{n-m-2}&w_{2}^{n-m-2}&\cdots&w_{n-1}^{n-m-2}\\\
\vdots&\vdots&\cdots&\vdots\\\ w_{1}&w_{2}&\cdots&w_{n-1}\\\ &&&\\\
1&1&\cdots&1\\\ &&&\\\
\frac{1}{w_{1}-y_{m}}&\frac{1}{w_{2}-y_{m}}&\cdots&\frac{1}{w_{n-1}-y_{m}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{2}-y_{1}}&\cdots&\frac{1}{w_{n-1}-y_{1}}\end{bmatrix}.$

We then repeat the above process $n-m-1$ more times, obtaining

$\frac{\displaystyle\prod_{j=m+1}^{n}\displaystyle\prod_{i<j}(w_{i}-w_{j})}{\displaystyle\prod_{i=m+1}^{n}\displaystyle\prod_{j=1}^{m}(w_{i}-y_{j})}\det\begin{bmatrix}\frac{1}{w_{1}-y_{m}}&\frac{1}{w_{2}-y_{m}}&\cdots&\frac{1}{w_{m}-y_{m}}\\\
\frac{1}{w_{1}-y_{m-1}}&\frac{1}{w_{2}-y_{m-1}}&\cdots&\frac{1}{w_{m}-y_{m-1}}\\\
\vdots&\vdots&\cdots&\vdots\\\
\frac{1}{w_{1}-y_{1}}&\frac{1}{w_{2}-y_{1}}&\cdots&\frac{1}{w_{m}-y_{1}}\end{bmatrix}.$

The Cauchy determinant here is equal to

$\frac{\displaystyle\prod_{1\leq i<j\leq