# Canonical bilinear form and Euler characters

A.N. Sergeev  Department of Mathematics, Saratov State University,
Astrakhanskaya 83, Saratov 410012 Russian Federation<EMAIL_ADDRESS>
###### Abstract.

An explicit formula for the canonical bilinear form on the Grothendieck ring
of the Lie supergroup $GL(n,m)$ is given. As an application we get an
algorithm for the decomposition Euler characters in terms of charctrers of
irreducible modules in the category of partially polynomial modules.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 A bilinear form on the ring $P_{m,n}$
  4. 4 Kac modules and Euler characters
  5. 5 Partially polynomial representations
  6. 6 Projective covers and Euler characters
  7. 7 Some special classes of irreducible modules
  8. 8 Acknowledgments

##  1\. Introduction

Let $\mathcal{F}$ be the category of finite dimensional modules over complex
algebraic group $GL(n)$ and $K(\mathcal{F})$ be its Grothendieck ring. There
exists a natural pairing on the ring $K(\mathcal{F})$

$([U],[V])=\dim Hom_{GL(n)}(U,V)$

Let us identify the ring $K(\mathcal{F})$ with the ring of symmetric Laurent
polynomials $\mathbb{Z}[x_{1}^{\pm 1},\dots,x_{n}^{\pm 1}]^{S_{n}}$ then the
above pairing in terms of characters can be expressed in the following form

$([U],[V])=\left[(ch\,U)^{*}ch\,V\prod_{i\neq
j}\left(1-\frac{x_{i}}{x_{j}}\right)\right]_{0}$

where $[f]_{0}$ means the constant term of Laurent polynomial and
$f^{*}(x_{1},\dots,x_{n})=f(x_{1}^{-1},\dots,x_{n}^{-1})$ (see [10, 11]). This
formula is very interesting from many points of view. On one side it allows to
connect problems in representation theory to some problems with symmetric
functions and their generalisations including Jack and Macdonald polynomials.
On the other side the above formula can be extended to the root system of any
semisimple Lie algebra and the corresponding analogues of symmetric
polynomials. The main goal of this paper is to prove the same kind of formula
for complex algebraic Lie supergroup $GL(m,n)$ and to illustrate some of its
applications. The category of finite dimensional representations of $GL(m,n)$
is not semisimple. Therefore in this case we have the natural pairing only
between projective modules $P(\mathcal{F})$ and finite dimensional modules
$K(\mathcal{F})$ (see [1, 7])

$P(\mathcal{F})\times
K(\mathcal{F})\longrightarrow\mathbb{Z},\quad([U],[V])=\dim
Hom_{GL(m,n)}(U,V)$

where $[L]$ means the class of module $L$ in appropriate version of
Grothendieck ring. We also should mention that the category of partially
polynomials modules defined below is a convenient object from the point of
view of our bilinear form.

##  2\. Preliminaries

Instead of general linear complex algebraic supergroup $GL(m,n)$ it is more
convenient to deal with the complex Lie superalgebra $\mathfrak{gl}(m,n)$. So
we will consider only finite dimensional representations of
$\mathfrak{gl}(n,m)$ such that every one of them can be lifted to the
representation of $GL(m,n)$.

Let us remind that the Lie superalgebra $\mathfrak{gl}(m,n)$ is the Lie
superalgebra of the linear transformations of a $\mathbb{Z}_{2}$ graded vector
space $V=V_{0}\oplus V_{1}$ ($V$ is also called the standard representation of
$\mathfrak{g}$). We have

$\mathfrak{g}_{\bar{0}}=\mathfrak{gl}(m)\oplus\mathfrak{gl}(n),\quad\mathfrak{g}_{\bar{1}}=V_{0}\otimes
V_{1}^{*}\oplus V_{1}\otimes V_{0}^{*}$

We also have $\mathbb{Z}$ graded decomposition
$\mathfrak{g}=\mathfrak{g}_{-1}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{1}$
where

$\mathfrak{g}_{-1}=V_{1}\otimes V^{*}_{0},\,\,\mathfrak{g}_{1}=V_{0}\otimes
V^{*}_{1},\,$

Let us fix bases in $V_{0}=<e_{1},\dots,e_{m}>$ and
$V_{1}=<f_{1},\dots,f_{n}>$ respectively. Let $\mathfrak{b}$ be the subalgebra
of upper triangular matrix in $\mathfrak{gl}(m,n)$ and $\mathfrak{k}$ be the
subalgebra of diagonal matrix in $\mathfrak{gl}(m,n)$ in the above basis. By
$\varepsilon_{1},\dots,\varepsilon_{m},\delta_{1},\dots,\delta_{n}$ we will
denote the weights of standard representation with respect to $\mathfrak{k}$.
The corresponding system of positive roots $R^{+}=R^{+}_{0}\cup R^{+}_{1}$ of
$\mathfrak{gl}(m,n)$ can be described in the following way

$R^{+}_{0}=\\{\varepsilon_{i}-\varepsilon_{j}\,:\,1\leq i<j\leq
m\,:\,\delta_{k}-\delta_{l},\,1\leq k<l\leq n\\}$
$R_{1}^{+}=\\{\varepsilon_{i}-\delta_{k},\,1\leq i\leq m,\,1\leq k\leq n\\}$

Let also

$P=\\{\chi=\lambda_{1}\varepsilon_{1}+\dots+\lambda_{m}\varepsilon_{m}+\mu_{1}\delta_{1}+\dots+\mu_{n}\delta_{n},\mid
n_{i},m_{j}\in\mathbb{Z}\\}$

be the weight lattice and

$P^{+}=\\{\chi\in P\mid\lambda_{i}-\lambda_{j}\geq
0,\,i<j\,:\mu_{k}-\mu_{l}\geq 0,\,k<l\\}$

be the set of highest weights.

We will use the following parity on the weight lattice due to C. Gruson and V.
Serganova [5] and Brundan and Stroppel [3] by saying that $\varepsilon_{i}$
(resp. $\delta_{j}$) is even (resp. odd). It is easy to check that every
finite dimensional module $L$ can be represented in the form

$L=L^{+}\oplus L^{-}$

where $L^{+}$ is the submodule of $L$ in which weight space has the same
parity as the corresponding weight and $L^{-}$ is the submodule in which the
parities differ. We should note that this construction is a particular case of
Deligne construction category $Rep(G,z)$ from the paper [4] for $G=GL(m,n)$
and $z=diag(\underbrace{1,\dots,1}_{m},\underbrace{-1,\dots,-1}_{n}).$

Let us denote by $\mathcal{F}$ the category of finite dimensional modules over
$\mathfrak{gl}(n,m)$ such that every module in $\mathcal{F}$ is semisimple
over Cartan subalgebra $\mathfrak{k}$ and and all its weights are in $P$. By
$K(\mathcal{F})$ we will denote the quotient of the Grothendieck ring of
$\mathcal{F}$ by the relation $[L]-[\Pi(L)]=0$ where $\Pi(L)$ is the module
with the shifted parity $\Pi(L)_{0}=L_{1},\Pi(L)_{1}=L_{0}$ and
$x*v=(-1)^{p(x)}xv,x\in\mathfrak{gl}(m,n).$ For every $L\in\mathcal{F}$ we can
define

$ch\,L=\sum_{\chi}\dim L_{\chi}e^{\chi}$

where the sum is taken over all weights of $L$. It is easy to see that $ch\,L$
is well defined function on $K(\mathcal{F})$.

The ring $K(\mathcal{F})$ can be describe explicitly in the following way. Let

$P_{m,n}=\mathbb{Z}[x_{1}^{\pm 1},\dots,x_{m}^{\pm 1},\,y_{1}^{\pm
1},\dots,y_{n}^{\pm 1}]$

be the ring of Laurent polynomials in variables $x_{1},\dots,x_{m}$ and
$y_{1},\dots,y_{n}.$

If we set $x_{i}=e^{\varepsilon_{i}},\,y_{j}=e^{\delta_{j}}$ then we get a
character map

$ch:K(\mathcal{F})\longrightarrow P_{m,n}$

Let also

$\Lambda^{\pm}_{m,n}=\\{f\in P_{m,n}^{S_{m}\times S_{n}}\mid
x_{i}\frac{\partial f}{\partial x_{i}}+y_{j}\frac{\partial f}{\partial
y_{j}}\in(x_{i}+y_{j})\\}$

be the subring of $P_{m.n}$ of supersymmetric Laurent polynomials.

######  Theorem 2.1.

[15] The ring $K(\mathcal{F})$ is isomorphic to the ring $\Lambda^{\pm}_{m,n}$
under the character map.

######  Remark 2.2.

Actually in the paper [15] slightly different versions of Grothen-dieck ring
and the algebra $\Lambda^{\pm}_{m,n}$ were considered. But it is easy to check
that they are isomorphic to our ones. We prefer to use characters instead of
supercharacters in this paper in order to avoid some unnecessary signs.

It will be needed later an explicit description of the projective covers of
the irreducible finite dimensional modules due to Brundan [1]. We give the
description here in a slightly different way.

First let us for any $\chi\in P^{+}$ define a pair of sets

$A=\\{(\chi+\rho,\varepsilon_{1}),\dots,(\chi+\rho,\varepsilon_{m})\\},\quad
B=\\{(\chi+\rho,\delta_{1}),\dots,(\chi+\rho,\delta_{n})\\}$

where

$\rho=\frac{1}{2}\sum_{\alpha\in R_{0}^{+}}\alpha-\frac{1}{2}\sum_{\alpha\in
R_{1}^{+}}\alpha+\frac{1}{2}(n-m+1)(\sum_{i=1}^{m}\varepsilon_{i}-\sum_{j=1}^{n}\delta_{j})$
$=\sum_{i=1}^{m}(1-i)\varepsilon_{i}+\sum_{j=1}^{n}(m-j)\delta_{j}$

Our $\rho$ is slightly different from the standard one but it is more
convenient since the elements of $A$ and $B$ are integers. So instead of
highest weights we will use the set of pairs $(A,B)$ such that
$A,B\subset\mathbb{Z}$ and $|A|=m,\,|B|=n$. We will also use the language of
diagrams which is due to Brundun and Stroppel [3] but we will use it here in a
form due to I. Musson and V. Serganova [9].

######  Definition 2.3.

Let $(A,B)$ be a pair of subsets in $\mathbb{Z}$ such that $|A|=m,\,|B|=n$.
Then the corresponding diagram is the following function on $\mathbb{Z}$

$f(x)=\begin{cases}\times,\,\,x\in A\cap B\\\ \circ,\,\,x\in A^{\prime}\cup
B^{\prime}\\\ >,\,\,x\in A\setminus B\\\ <,\,\,x\in B\setminus A\end{cases}$

Let us also set

$\varphi(\times)=1,\,\varphi(\circ)=-1,\,\,\varphi(>)=\varphi(<)=0$
$[a,b]=\\{c\in\mathbb{Z}\mid a\leq c\leq b\\},\,[a,b)=\\{c\in\mathbb{Z}\mid
a\leq c<b\\},$ $\,(a,b)=\\{c\in\mathbb{Z}\mid a<c<b\\}$

and for integers $a<b$ let us define a transposition

$\pi_{a}^{b}:\mathbb{Z}\longrightarrow\mathbb{Z},\quad\pi_{a}^{b}(x)=\begin{cases}x,\,x\neq
a,b\\\ b,\,x=a\\\ a,\,x=b\end{cases}$

######  Definition 2.4.

We will call a transposition $\pi_{a}^{b}$ an admissible for $f$ if $a\in
f^{-1}(\times),\,b\in f^{-1}(\circ)$ and the following conditions are
fulfilled

$b>a,\,\,\sum_{i\in[a,b]}(\varphi\circ
f)(i)=0,\,\,\sum_{i\in[a,c]}(\varphi\circ f)(i)>0,\,\text{for
any}\,\,c\in[a,b).$

Since $b$ is uniquely defined by $f$ and $a$ we sometimes will omit $b$.

The following Lemma easily follows from the definition above.

######  Lemma 2.5.

The following statements hold true

$1)$ If $\pi_{a}^{b},\,\pi_{c}^{d}$ are two admissible transpositions for $f$
then one of the following conditions is fulfilled

$[a,b]\cap[c,d]=\emptyset,\,\,\,[a,b]\subset(c,d),\,\,\,[c,d]\subset(a,b)$ (1)

$2)$ Let $\pi_{a}^{b}$ be an admissible transposition for $f$ and $d\in[a,b]$
be such that $f(d)=\circ$. Then there exist an admissible transposition for
$f$ of the form $\pi_{c}^{d}$.

######  Corollary 2.6.

Admissible transpositions pairwise commute.

###### Proof.

It easily follows from Lemma 2.5. ∎

Now let us define for a diagram $f$ and any $C\subset f^{-1}(\times)$ the
permutation of $\mathbb{Z}$ by the formula

$\pi_{C}=\prod_{c\in C}\pi_{c}$ (2)

We should mention that the above product is well defined since admissible
transpositions commute with each other.

######  Definition 2.7.

Let $P(f)$ be the projective cover of irreducible module $L(f)$. We will
denote by $\mathcal{P}(f)$ the set of $g$ such that $K(g)$ is a subquotient of
$P(f)$.

Now we can formulate the main result of Brundan [1].

######  Theorem 2.8.

$P(f)$ has a multiplicity free Kac flag and

$\mathcal{P}(f)=\\{g\mid g=\pi_{C}(f),C\subset f^{-1}(\times)\\}$

In order to get an algorithm for the decomposition Kac modules into the sum of
irreducible modules we need the following combinatorial Lemma.

######  Lemma 2.9.

Let $f,g$ be such diagrams that

$g=\tau_{r}\circ\tau_{2}\circ\dots\circ\tau_{1}(f)$

where $\tau_{i}=\pi_{a_{i}}^{b_{i}}$ is a transposition and
$f_{i}=\tau_{i-1}\circ\dots\tau_{1}(f),\,i=1,\dots,r$. Suppose also that for
any pair of $i>j$ we have

$[a_{i},b_{i}]\cap[a_{j},b_{j}]=\emptyset,\,\text{or}\,\,\,[a_{i},b_{i}]\subset(a_{j},b_{j})$
(3)

Then for any $i=1,\dots r$ the transposition $\tau_{i}$ is admissible for $f$
if and only if $\tau_{i}$ is admissible for $f_{i},i=1,\dots,r$.

###### Proof.

Let us prove first that functions $f_{i}$ and $f$ coincide on the segment
$[a_{i},b_{i}]$ for $1\leq i\leq r$. The following equalities are easy to
check

$\varphi\circ f_{i}=\varphi\circ
f+2\sum_{j=1}^{i-1}(\delta_{b_{j}}-\delta_{a_{j}}),\,\,\,i=1,\dots,r$
$f_{i}^{-1}(\times)=\left(f^{-1}(\times)\setminus\\{a_{1},\dots,a_{i-1}\\}\right)\cup\\{b_{1},\dots,b_{i-1}\\}$

If $t\in[a_{i},b_{i}]$ then from the conditions of the Lemma it follows that
$\delta_{a_{j}}(t)=\delta_{b_{j}}(t)=0$ for any $1\leq j<i$. Therefore
$\varphi\circ f_{i}(t)=\varphi\circ f(t)$ on the segment $[a_{i},b_{i}]$. Now
Lemma follows from Definition 2.4. ∎

######  Corollary 2.10.

We will keep the notations from Lemma 2.9. Suppose that
$\tau_{1}=\pi_{a_{1}}^{b_{1}},\dots,\tau_{r}=\pi_{a_{r}}^{b_{r}}$ is a set of
transpositions such that $a_{1}<a_{2}<\dots<a_{r}$ and $a_{i}\neq
b_{j},\,1\leq i,j\leq r$. Suppose also that for any $i=1,\dots r$
transposition $\tau_{i}$ is admissible for $f_{i}$. Then all transpositions
$\tau_{1},\dots,\tau_{r}$ are admissible for $f$.

###### Proof.

Let us prove by induction that conditions of Lemma 2.9 are fulfilled. We will
use induction on $r$. If $r=1$ then the statement of the Lemma is trivial. Let
$r>1$. By inductive assumption transpositions $\tau_{1},\dots,\tau_{r-1}$ are
admissible for $f$. Therefore

$f_{r}^{-1}(\times)=\left(f^{-1}(\times)\setminus\\{a_{1},\dots,a_{r-1}\\}\right)\cup\\{b_{1},\dots,b_{r-1}\\}$

Since $\tau_{r}$ is admissible for $f_{r}$ we have $a_{r}\in
f_{r}^{-1}(\times)$. By assumptions of the Lemma $a_{r}\neq
b_{1},\dots,b_{r-1}$ therefore $a_{r}\in f^{-1}(\times)$. Let
$\pi_{a_{r}}^{c}$ be the corresponding admissible transposition for $f$. Then
for any $i\leq r$ one of the following conditions holds true

$[a_{i},b_{i}]\cap[a_{r},c]=\emptyset,\,\,[a_{r},c]\subset(a_{i},b_{i}),\,\,[a_{i},b_{i}]\subset(a_{r},c)$

The last condition is impossible since $a_{r}>a_{i}$. Therefore by Lemma 2.9
transposition $\pi_{a_{r}}^{c}$ is admissible for $f_{r}$. Therefore
$\pi_{a_{r}}^{b_{r}}=\pi_{a_{r}}^{c}$ is admissible for $f$. ∎

######  Corollary 2.11.

Suppose that
$\tau_{1}=\pi_{a_{1}}^{b_{1}},\,\dots,\tau_{r}=\pi_{a_{r}}^{b_{r}}$ is a set
of transpositions such that $b_{1}>b_{2}>\dots>b_{r}$ and $a_{i}\neq
b_{j},\,1\leq i,j\leq r$. Suppose also that $\tau_{i}$ is admissible for
$f_{i},\,i=1,\dots,n$. Then $\tau_{i},\,i=1,\dots,r$ is admissible for $f$.

###### Proof.

Let us prove by induction that conditions (3) are fulfilled. We will use
induction on $r$. If $r=1$ then the statement of the Lemma is trivial. Let
$r>1$. By inductive assumption the conditions (3) are fulfilled therefore by
Lemma 2.9 transpositions $\tau_{1},\dots,\tau_{r-1}$ are admissible for $f$.
Therefore

$f_{r}^{-1}(\times)=\left(f^{-1}(\times)\setminus\\{a_{1},\dots,a_{r-1}\\}\right)\cup\\{b_{1},\dots,b_{r-1}\\}$

Since $\tau_{r}$ is admissible for $f_{r}$ we have $a_{r}\in
f_{r}^{-1}(\times)$. By our assumptions $a_{r}\neq b_{1},\dots,b_{r-1}$
therefore $a_{r}\in f^{-1}(\times)$. Besides since $b_{r}\neq
a_{1},\dots,a_{r-1}$ we have $b_{r}\in f^{-1}(\circ)$. Let $\pi_{a_{r}}^{c}$
be the corresponding admissible transposition for $f$. Suppose that $b_{r}<c$,
then $b_{r}\in[a_{r},c]$. Therefore by Lemma 2.5 there exist an admissible for
$f$ transposition $\pi_{a}^{b_{r}}$. Then for any $i<r$ one of the following
conditions holds true

$[a_{i},b_{i}]\cap[a,b_{r}]=\emptyset,\,\,[a,b_{r}]\subset(a_{i},b_{i}),\,\,[a_{i},b_{i}]\subset(a,b_{r})$

The last condition is impossible since $b_{r}<b_{i}$. Therefore by Lemma 2.9
transposition $\pi_{a}^{b_{r}}$ is admissible for $f_{r}$. Therefore
$\pi_{a_{r}}^{b_{r}}=\pi_{a}^{b_{r}}$ is admissible for $f$. If $b_{r}\geq c$
then again condition $[a_{i},b_{i}]\subset(a_{r},c)$ is impossible and
$\pi_{a_{r}}^{b_{r}}=\pi_{a}^{c}$ is admissible for $f$. ∎

######  Corollary 2.12.

Irreducible module $L(f)$ is a subquotient of Kac module $K(g)$ if and only if
there exist a sequence of transpositions

$\sigma_{1}=\pi_{c_{1}}^{,d_{1}},\,\dots,\sigma_{r}=\pi_{c_{r}}^{d_{r}}$

where $c_{i}<d_{i},\,i=1,\dots,r$ such that

$1)$ $\sigma_{i}$ is admissible for $\sigma_{i}\circ\dots\circ\sigma_{1}(g)$,
$i=1,\dots,r$ and $\sigma_{r}\circ\dots\circ\sigma_{1}(g)=f$

$2)$ $c_{1}>c_{2}>\dots>c_{r}$

$3)$ $c_{i}\neq d_{j},\,1\leq i,j\leq r$

###### Proof.

Suppose that all conditions of the Corollary are fulfilled. Then

$g=\sigma_{1}\circ\sigma_{2}\circ\dots\circ\sigma_{r}(f)$

If we set $\tau_{i}=\sigma_{r+1-i},\,a_{i}=c_{r-i+1},\,b_{i}=d_{r-i+1}$ where
$i=1,\dots,r$ then it is easy to see that all conditions of Corollary 2.10 are
fulfilled. Therefore $K(g)$ is a subquotient of $P(f)$. Therefore by $BGG$
reciprocity [16] $L(f)$ is a sub quotient of $K(g)$.

Now let us suppose that $L(f)$ is a subquotient of Kac module $K(g)$. Then
again by $BGG$ reciprocity $K(g)$ is a sub quotient of $P(f)$. Therefore by
Theorem 2.8 $g=\pi_{A}(f),\,A\subset f^{-1}(\times).$ Let
$A=\\{a_{1},a_{2},\dots,a_{r}\\}$ where $a_{1}<a_{2}<\dots<a_{r}$. Since
admissible transpositions pairwise commute we have

$g=\tau_{r}\circ\tau_{2}\circ\dots\circ\tau_{1}(f)$

where $\tau_{i}=\pi_{a_{i}}^{b_{i}}$. Let us check that conditions (3) are
fulfilled. It is enough to verify that inclusion
$[a_{j},b_{j}]\subset(a_{i},b_{i})$ is impossible if $i>j$. Indeed if it is so
then $a_{j}>a_{i}$ and we get a contradiction. Therefore by Lemma 2.9
$\tau_{i}$ is admissible for $f_{i}=\tau_{i-1}\circ\circ\dots\circ\tau_{1}(f)$
and we can set

$\sigma_{i}=\tau_{r-i+1},\,c_{i}=a_{r-i+1},\,d_{i}=b_{r-i+1}\,\,\,i=1,\dots,r.$

∎

In the same way we can prove the following Corollary.

######  Corollary 2.13.

Irreducible module $L(f)$ is a subquotient of Kac module $K(g)$ if and only if
there exist a sequence of transpositions

$\sigma_{1}=\pi_{c_{1}}^{,d_{1}},\,\dots,\sigma_{r}=\pi_{c_{r}}^{d_{r}}$

where $c_{i}<d_{i},\,i=1,\dots,r$ such that

$1)$ $\sigma_{i}$ is admissible for $\sigma_{i}\circ\dots\circ\sigma_{1}(g)$,
$i=1,\dots,r$ and $\sigma_{r}\circ\dots\circ\sigma_{1}(g)=f$

$2)$ $d_{1}<d_{2}<\dots<d_{r}$

$3)$ $c_{i}\neq d_{j},\,1\leq i,j\leq r$

The above corollaries can be used to calculate the irreducible subquotients of
Kac modules.

######  Definition 2.14.

$\mathcal{K}(g)=\\{f\mid Hom_{\mathfrak{g}}(P(f),K(g))\neq 0\\}$

######  Example 2.15.

Let $m=n=2$ and $g^{-1}(\times)=\\{2,3\\}$. We are going to describe the set
$\mathcal{K}(g)$.

As the first step we are going to find a transpositions $\pi_{a}^{b}$ such
that $b\in g^{-1}(\times),$ and $\pi_{a}^{b}$ is admissible for
$\pi_{a}^{b}(g).$ And it is easy to see that there exists only one such
transposition $\pi_{1}^{2}$.

The next step is to find a transposition $\pi_{a}^{b}$ such that
$b\in\pi_{1}^{2}(g)^{-1}(\times),\,\pi_{a}^{b}$ is admissible for
$\pi_{a}^{b}\circ\pi_{1}^{2}(g)$ and $a<1$. It is easy to check that there
exists only one such transposition $\pi_{0}^{3}$. So we have

$\mathcal{K}(g)=\left\\{g,\,\pi_{1}^{2}(g),\,\pi_{0}^{3}\circ\pi_{1}^{2}(g)\right\\}$

######  Remark 2.16.

We should mention that our algorithm is essentially the same as in the paper
[9]. Legal move of weight zero $g\xrightarrow{[b,a]}f,\,\,a<b$ in the sense of
[9] is the same as $\sigma=\pi_{a}^{b}$ is an admissible transposition for
$f=\sigma(g)$. And a regular increasing pass from $g$ to $f$ is the same as
the sequence of transpositions

$\sigma_{1}=\pi_{a_{1}}^{b_{1}},\dots,\sigma_{r}=\pi_{a_{r}}^{b_{r}},\quad
a_{1}<b_{1},\dots,a_{r}<b_{r}$

such that

$1)$ $\sigma_{i}$ is admissible for $\sigma_{i}\circ\dots\circ\sigma_{1}(g)$,
$i=1,\dots,r$ and $\sigma_{r}\circ\dots\circ\sigma_{1}(g)=f$

$2)$ $b_{1}<b_{2}<\dots<b_{r}$

$3)$ $a_{i}\neq b_{j},1\leq i,j\leq r$.

##  3\. A bilinear form on the ring $P_{m,n}$

In this section we are going to define a bilinear form on the ring of Laurent
polynomials $P_{m,n}$ and connect this bilinear form with the canonical
bilinear form on the Grothendieck ring of Lie superalgebra
$\mathfrak{g}=\mathfrak{gl}(m,n)$. Let $p\rightarrow p^{*}$ be the followinng
automorphism of $P_{m,n}$

$x_{i}^{*}=x_{i}^{-1},\,i=1,\dots,m,\quad y_{j}^{*}=y_{j}^{-1},\,j=1,\dots,n$

######  Definition 3.1.

Let us set

$\Delta(x)=\prod_{i>j}\left(1-\frac{x_{i}}{x_{j}}\right),\,\Delta(y)=\prod_{i>j}\left(1-\frac{y_{i}}{y_{j}}\right),\,\,\Delta(x,y)=\prod_{i,j}\left(1+\frac{y_{j}}{x_{i}}\right)$

and for $p,q\in P_{m,n}$ let us define

$(p,q)=\frac{1}{m!}\frac{1}{n!}\left[p^{*}q\frac{\Delta(x)\Delta(x)^{*}\Delta(y)\Delta(y)^{*}}{\Delta(x,y)\Delta(x,y)^{*}}\right]_{0}$
(4)

where $[\,,\,]_{0}$ means the constant term and
$(\Delta(x,y)\Delta(x,y)^{*})^{-1}$ should be understood as

$(\Delta(x,y)\Delta(x,y)^{*})^{-1}=\frac{(y_{1}\dots y_{n})^{m}}{(x_{1}\dots
x_{m})^{n}}\left[\prod_{i.j}\left(1+\frac{y_{j}}{x_{i}}\right)\right]^{-2}.$

######  Theorem 3.2.

The following equality hold true

$\dim Hom_{\mathfrak{g}}(P,L)=(chP,chL)$

where $P$ is a finite dimensional projective module, $L$ is any finite
dimensional module.

###### Proof.

We are going to prove the Theorem in several steps. First we are going to
prove that characters of Kac modules are pairwise orthogonal with respect to
the pairing $(\,,\,)$. Let $K(f),K(g)$ be two Kac modules and
$\chi=(\lambda,\mu)$ and $\tilde{\chi}=(\nu,\tau)$ are the corresponding
highest weights, where $\lambda,\nu$ are highest weights of $\mathfrak{gl}(m)$
and $\mu,\tau$ are highest weights of $\mathfrak{gl}(n)$. Then we have

$chK(f)=\Delta(x,y)s_{\lambda}(x)s_{\mu}(y),\,\,chK(g)=\Delta(x,y)s_{\nu}(x)s_{\tau}(y),\,\,$

where $s_{\lambda},s_{\mu},s_{\nu},s_{\tau}$ are Schur functions. Therefore we
have

$(f,g)=\frac{1}{m!}\frac{1}{n!}\left[s^{*}_{\lambda}s^{*}_{\mu}\Delta(x)^{*}\Delta(y)^{*}s_{\nu}s_{\tau}\Delta(x)\Delta(y)\right]_{0}=\delta_{\lambda,\nu}\delta_{\mu,\tau}$

according to the orthogonality of Schur polynomials.

Now let $P(f)$ be the projective cover of the irreducible module $L(f)$ and
$K(g)$ be a Kac module. Then we are going to prove that

$\dim Hom_{\mathfrak{g}}(P,K)=(chP,chK)$ (5)

We can suppose that $P^{+}(f)=P(f),\,K^{+}(g)=K(g)$. In other words the parity
of every weight vector coincides with the parity of the weight. We have

$\dim Hom_{\mathfrak{g}}(P(f),K(g))=n_{g,f}$

where $n_{g,f}$ is the multiplicity of irreducible module $L(f)$ in the Jordan
- Helder series of the module $K(g)$. On the other hand from the
orthonormality of Kac modules it follows that $(P(f),K(g))=m_{f,g}$, where
$m_{f,g}$ is the multiplicity of Kac module $K(g)$ in the Kac flag of the
module $P(f)$. But by BGG reciprocity $m_{f,g}=n_{g,f}$ and we proved equality
(5).

To complete the proof, it just remains to show that the following equality

$\dim Hom_{\mathfrak{g}}(P(f),L)=(ch\,P(f),ch\,L)$

is true for any finite dimensional module $L$. For this, we give two different
arguments, the first based on a fact proved by Serganova in [13] and the
second using instead completion in the spirit of Brundan ([2] §, 4c).

Now let $L$ be a module which has a Kac flag. Then

$\dim Hom_{\mathfrak{g}}(P(f),L)=\sum_{g}\dim Hom_{\mathfrak{g}}(P(f),K(g))=$
$\sum_{g}(ch\,P(f),ch\,K(g))=(ch\,P(f),ch\,L)$

where $K(g)$ runs over all subquotients of $L$

Now let $L$ be any finite dimensional module and $P$ be a projective module.
By Serganova [13] there exist a resolvent of $L$

$\dots\rightarrow K_{i}\rightarrow K_{i-1}\rightarrow\dots\rightarrow
K_{1}\rightarrow L\rightarrow 0$ (6)

where every $K_{i}$ has a flag of Kac modules. Therefore we have an exact
sequence of vector spaces

$\dots\rightarrow Hom_{\mathfrak{g}}(P,K_{i})\rightarrow\dots\rightarrow
Hom_{\mathfrak{g}}(P,K_{1})\rightarrow Hom_{\mathfrak{g}}(P,L)\rightarrow 0$

For any finite dimensional module $V$ let us denote by $wt(V)$ the set of the
weights of the module $V$. Let $N$ be such that for any $i>N$ we have
$wt(P)\cap wt(K_{i})=\emptyset.$ Then for any $i>N$ we have
$Hom_{\mathfrak{g}}(P,K_{i})=0$ and

$dim(P,L)=dim(P,K_{1})-dim(P,K_{2})+\dots+(-1)^{i+1}dim(P,K_{i})$ (7)

On the other hand from equality (6) we have

$schL-schK_{1}+schK_{2}-\dots+(-1)^{i}schK_{i}+\dots=0$

The above sum makes sense since every weight entries the sum with finite
multiplicity.

Now let us calculate $(chP,chK_{i})$. We have by definition

$(chP,chK_{i})=\frac{1}{m!}\frac{1}{n!}\left[(chP)^{*}chK_{i}\frac{\Delta^{*}(x)\Delta(y)^{*}\Delta(x)\Delta(y)}{\Delta(x,y)^{*}\Delta(x,y)}\right]_{0}=$
$=\frac{1}{m!}\frac{1}{n!}\left[(chP)^{*}chK_{i}\Delta^{*}(x)\Delta(y)^{*}\Delta(x)\Delta(y)\sum\prod\left(\frac{y_{j}}{x_{i}}\right)^{n_{ij}}\right]_{0}$

Now let us take $M$ such that for any $i>M$ all monomials of the polynomial
$(chP)^{*}chK_{i}\Delta^{*}(x)\Delta(y)^{*}\Delta(x)\Delta(y)$ were negative
degree with respect to $x_{1},\dots,x_{m}$. Therefore all monomials in the
above expansion have negative degree with respect to $x_{1},\dots,x_{m}$.
Therefore $(chP,chK_{i})=0$. Therefore for $i>M$ we have

$(chP,chL)-(chP,chK_{1})+\dots+(-1)^{i}(chP,chK_{i})=0$ (8)

Therefore if we take $i>\max\\{N,M\\}$ then from the equalities (7), (8) we
have $\dim Hom_{\mathfrak{g}}(P,L)=(chP,chL)$ and Theorem 3.2 is proved.

Now let us use a completion. For
$\chi=\lambda_{1}\varepsilon_{1}+\dots+\lambda_{m}\varepsilon_{m}+\mu_{1}\delta_{1}+\dots+\mu_{n}\delta_{n}$
let us set $m(\chi)=\mu_{1}+\dots+\mu_{n}$. Let $K(\mathcal{F})_{d}$ be the
subgroup of the Grothendieck group $K(\mathcal{F})$ generated by
$\\{L(\chi)\\}$ for $\chi\in P^{+}$ with $m(\chi)\geq d$. We know that
$[K(\chi)]$ is the finite linear combination of $L(\chi)$ and
$[L(\tilde{\chi})]$ where $\tilde{\chi}<\chi$. Therefore we can find the
sequence $\\{A_{i}\\}_{i\geq 1}$ of finite subsets in $P^{+}$ such that
$A_{i}\subset A_{i+1}$ and

$[L(\chi)]-\sum_{\tilde{\chi}\in A_{i}}c_{\tilde{\chi}}[K(\tilde{\chi})]\in
K(\mathcal{F})_{d_{i}}$

where $d_{1}<d_{2}<d_{3}\dots.$ It is easy to see that for given projective
module $P$ there exists $N_{1}$ such that for any $d\geq N_{1}$ we have
$\dim_{\mathfrak{g}}(P,L)=0$ for any irreducible module $L\in K_{d}$. And it
follows from formula (4) that there exists $N_{2}$ such than for any $d\geq
N_{2}$ we have $(chP,chL)=0$ for any irreducible module $L\in K_{d}$.
Therefore for $d_{i}\geq\max\\{N_{1},N_{2}\\}$ we have

$\dim Hom(P,L(\chi))=\sum_{\tilde{\chi}\in A_{i}}c_{\tilde{\chi}}\dim
Hom(P,K(\tilde{\chi}))=$ $=\sum_{\tilde{\chi}\in
A_{i}}c_{\tilde{\chi}}(P,K(\tilde{\chi}))=(P,L(\chi))$

and we proved the Theorem in this way. ∎

######  Corollary 3.3.

$\mathcal{P}(f)=\\{g\mid(ch\,P(f),ch\,K(g))\neq 0\\}$

##  4\. Kac modules and Euler characters

Now we are going to calculate the number $(ch\,K(f),ch\,E(g))$ where $K(f)$ is
a Kac module and $E(g)$ is an Euler virtual module. General formula for Euler
characters was given by V. Serganova in [13]. For any parabolic subalgebra
$\mathfrak{p}\subset\mathfrak{g}$ and any finite dimensional module $M$ of
$\mathfrak{p}$ by a super version of Borel - Weil - Bott construction one can
define the virtual Euler module $E^{\mathfrak{p}}(M)$. According to the
general formula due to Serganova [13]

$chE^{\mathfrak{p}}(M)=\sum_{w\in
W}w\left(\frac{De^{\rho}chM}{\prod_{\alpha\in R_{\mathfrak{p}}\cap
R_{1}^{+}}(1-e^{\alpha})}\right)$

with

$D=\frac{\prod_{\alpha\in
R_{1}^{+}}(e^{\alpha/2}-e^{-\alpha/2})}{\prod_{\alpha\in
R_{0}^{+}}(e^{\alpha/2}-e^{-\alpha/2})}$

Here $\rho$ is the half-sum of the even positive roots minus the half-sum odd
positive roots, $R_{\mathfrak{p}}$ is the set of roots $\alpha$ such that
$\mathfrak{g}_{\pm\alpha}\subset\mathfrak{p}$. Consider now
$\mathfrak{g}=\mathfrak{gl}(m,n)$ and let $(r,s)$ be a pair of integers such
that $0\leq r\leq m,\,$ $0\leq s\leq n,\,\,r-s=m-n$. We will denote the set of
such pairs as $P(m,n)$. Next let us choose for $(r,s)\in P(m,n)$ the following
system of simple roots

$\\{\varepsilon_{i}-\varepsilon_{i+1},\delta_{j}-\delta_{j+1},\varepsilon_{r}-\delta_{1},\delta_{s}-\varepsilon_{r+1},\varepsilon_{m}-\delta_{s+1}\\},\,i\in[1,m]\setminus\\{r\\},\,j\in[1,n]\setminus\\{s\\}.$

So we have the corresponding set of positive even and odd roots. Consider now
the parabolic subalgebra $\mathfrak{p}$ with

$R_{\mathfrak{p}}=\\{\varepsilon_{i}-\varepsilon_{j},\,\delta_{p}-\delta_{q},\,\pm(\varepsilon_{i}-\delta_{p})\\},$

where $r+1\leq i,j\leq m,\,i\neq j$ and $s+1\leq p,q\leq n,\,p\neq q$.

If we set

$\chi_{r,s}=\sum_{i=1}^{r}\tau_{i}\varepsilon_{i}+\sum_{j=1}^{s}\nu_{j}\delta_{j}$

where

$\tau=(\tau_{1},\dots,\tau_{r}),\,\,\nu=(\nu_{1},\dots,\nu_{s})$

are non increasing sequences of integers then $\chi$ defines one dimensional
representation of $\mathfrak{p}$. For any function
$f(x_{1},\dots,x_{m},y_{1},\dots,y_{n})$ let us define the following
alternation operation

$\\{f(x,y)\\}=\sum_{w\in S_{m}\times S_{n}}\varepsilon(w)w(f(x,y)).$

Then it is easy to check that Euler character is given by the following
formula

$ch\,E(\chi_{r,s})\Delta(x)x^{\rho_{m}}\Delta(y)y^{\rho_{n}}$
$=\left\\{\prod_{(ij)\in
D_{+}}\left(1+\frac{y_{j}}{x_{i}}\right)\prod_{(ij)\in
D^{-}}\left(1+\frac{x_{i}}{y_{j}}\right)x^{\tau}y^{\nu}x^{\rho_{m}}y^{\rho_{n}}\right\\}$
(9)

where

$D_{+}=[1,r]\times[1,n],\quad D_{-}=[r+1,m]\times[1,s].$

######  Remark 4.1.

If we apply to the formula (9) the automorphism $\omega$ which acts
identically on $x_{1},\dots,x_{n}$ and acts multiplication by $-1$ on
$y_{1},\dots,y_{m}$ then we get the Euler supercharacter (see [14] Proposition
5.10). And it was proved in [14] that Euler supercharacters
$\omega(E(\chi_{r,s}))$ where $(r,s)\in P(m,n)$ form a basis in the ring of
superchsracters. Therefore $chE(\chi_{r,s})$ where $(r,s)\in P(m,n)$ form a
basis in the ring $K(\mathcal{F})$.

As before we can use diagram $g=(A,B)$ where

$A=\\{\tau_{1},\tau_{2}-1,\dots,\tau_{r}+1-r\\},\,$
$B=\\{s-r-\nu_{s},s-r-\nu_{s-1}-1,\dots,-r-\nu_{1}\\}$

As a particular case we have the formula for character of Kac module
$K(\tilde{\chi})$ where $\tilde{\chi}=(\lambda,\mu)$ and
$\lambda=(\lambda_{1},\dots,\lambda_{m}),\,\,\mu=(\mu_{1},\dots,\mu_{n}),$ are
non increasing sequences of integers. In this case we have

$chK(\tilde{\chi})=\Delta(x,y)s_{\lambda}(x)s_{\mu}(y)$

and the corresponding diagram $f=(\tilde{A},\tilde{B})$ where

$\tilde{A}=\\{\lambda_{1},\lambda_{2}-1,\dots,\lambda_{m}+1-m\\},\,$
$\,\tilde{B}=\\{n-m-\mu_{n},n-m-\mu_{n-1}-1,\dots,-m-\mu_{1}\\}$

######  Definition 4.2.

Let $X,Y$ be two sets of integers such that $X\cap Y=\emptyset$. Let
$x_{1}>x_{2}>\dots,x_{m}$ be the elements of $X$ in decreasing order and
$y_{1}>y_{2}>\dots>y_{n}$ be the elements of $Y$ in decreasing and
$z_{1}>x_{2}>\dots,z_{m+n}$ be the elements of $Z=X\cup Y$ in decreasing
order. The sign of a permutation $\sigma$ such that

$\sigma(x_{1},\dots,x_{m},y_{1},\dots,y_{n})=(z_{1},\dots,z_{n+m})$

will be denoted by $\varepsilon(X,Y)$.

Let us keep the notation of the above definition. Then the following Lemma can
be easily proved.

######  Lemma 4.3.

Let us set

$a_{i}=|X\cap(-\infty,x_{i})|,\,\,i=1,\dots,m\quad
b_{j}=[Y\cap(y_{j},+\infty)|,j=1,\dots,n.$

where $|A|$ means the cardinality of $A$. Then the following equalities hold
true

$\varepsilon(X,Y)=(-1)^{a_{1}+\dots+a_{m}}=(-1)^{b_{1}+\dots+b_{n}}$

######  Definition 4.4.

Let $h$ be a diagram and $C\subset h^{-1}(\circ)$. Then by $h*C$ we will
denote the following diagram

$(h*C)^{-1}(x)=\begin{cases}h^{-1}(x),\,x=<,>\\\ h^{-1}(x)\cup C,\,x=\times\\\
h^{-1}(x)\setminus C,\,x=\circ\end{cases}$

Now we can formulate the main result of this section.

######  Theorem 4.5.

The following statement holds true:

$(ch\,K(f),ch\,E(g))=\begin{cases}\varepsilon(f,g),\,f=g*C,C\subset
g^{-1}({\circ})\cap\mathbb{Z}_{\leq n-m}\\\ 0,\,\,\text{otherwise}\end{cases}$
(10)

where

$\varepsilon(f,g)=(-1)^{\frac{1}{2}r(r-1)+\frac{1}{2}m(m-1)+s(m-r)+S(C)}\varepsilon(A,C)\varepsilon(C,B)$

and $S(C)$ is equal to the sum of the elements of $C$.

###### Proof.

We have from the definition of Kac module that

$\frac{ch\,K(f)}{\Delta(x,y)}=s_{\lambda}(x)s_{\mu}(y)$

and from the definition of Euler character

$\frac{ch\,E(g)\Delta(x)x^{\rho_{m}}\Delta(y)y^{\rho_{n}}}{\Delta(x,y)}=\left\\{\frac{\prod_{(ij)\in
D_{+}}\left(1+\frac{y_{j}}{x_{i}}\right)\prod_{(ij)\in
D_{-}}\left(1+\frac{x_{i}}{y_{j}}\right)}{\Delta(x,y)}x^{\tau}y^{\nu}x^{\rho_{m}}y^{\rho_{n}}\right\\}$
$=\frac{(x_{r+1}\dots x_{m})^{s}}{(y_{1}\dots
y_{s})^{m-r}}\left\\{\prod_{(i,j)\in
D_{r,s}}\left(1+\frac{y_{j}}{x_{i}}\right)^{-1}x^{\tau}y^{\nu}x^{\rho_{m}}y^{\rho_{n}}\right\\}$

where $D_{r,s}=[r+1,m]\times[s+1,n]$. Therefore

$\frac{ch\,E(g)\Delta(x)x^{\rho_{m}}\Delta(y)y^{\rho_{n}}}{\Delta(x,y)}=$
$\sum_{a_{1}\geq a_{2}\geq\dots\geq a_{m-r}\geq
0}(-1)^{|a|}\left\\{\frac{(x_{r+1}\dots x_{m})^{s}}{(y_{1}\dots
y_{s})^{m-r}}s_{a}(y_{s+1},\dots,y_{n})s_{a}(x^{-1}_{r+1},\dots,x^{-1}_{m})\right\\}$

where $|a|=a_{1}+\dots+a_{m-r}$. Further we have

$\left\\{x^{s}_{r+1}\dots
x^{s}_{m}s_{a}(x^{-1}_{r+1},\dots,x_{m}^{-1})x_{1}^{\tau_{1}}\dots
x_{r}^{\tau_{r}}x^{\rho_{m}}\right\\}=$ $\\{x_{1}^{\tau_{1}}\dots
x_{r}^{\tau_{r}}x_{r+1}^{s-a_{m-r}}\dots
x_{m}^{s-a_{1}}x^{\rho_{m}}\\}=s_{\tau_{1},\dots,\tau_{r},s-a_{m-r},\dots,s-a_{1}}\Delta(x)x^{\rho_{m}}.$

In the same way it is easy to see that

$\left\\{y_{1}^{r-m}\dots
y_{s}^{r-m}s_{a}(y_{s+1},\dots,y_{n})y_{1}^{\nu_{1}}\dots
y_{s}^{\nu_{s}}y^{\rho_{n}}\right\\}=$
$=s_{\nu_{1}+s-n,\dots,\nu_{s}+s-n,\,a_{1},\dots,a_{n-s}}\Delta(y)y^{\rho_{n}}$

Therefore

$(ch\,K(f),ch\,E(g))=$
$\sum_{a}(-1)^{|a|}(s_{\lambda},s_{\tau,s-a_{m-r},\dots,s-a_{1}})(s_{\mu},s_{\nu_{1}+r-m,\dots,\nu_{s}+r-m,a})$

It is easy to check that for given $\lambda$ there exists a unique sequence
$a$ and a permutation $\sigma\in S_{m}$ such that

$(\lambda_{1},\dots,\lambda_{m})+\rho_{m}=\sigma((\tau_{1},\dots,\tau_{r},s-a_{m-r},\dots,s-a_{1})+\rho_{m})$

or in an equivalent form $\tilde{A}=\sigma(A,C)$ where

$C=\\{s-r-a_{m-r},\dots,s+1-m-a_{1}\\}.$

In the same way there exists a permutation $\tau\in S_{n}$ such that

$(\mu_{1},\dots,\mu_{n})+\rho_{n}=\tau((\nu_{1}+r-m,\dots,\nu_{s}+r-m,\,a_{1},\dots,a_{n-s})+\rho_{n})$

or in the equivalent form $\tilde{B}=w_{n}\circ\tau\circ w_{n}(CB)$, where
$w_{n}(i)=n-i+1,\,i=1,\dots,n$. Therefore

$(K(f),E(g))=(-1)^{|a|}sign(\sigma)sign(\tau).$

But

$S(C)=s(m-r)+\frac{1}{2}r(r-1)-\frac{1}{2}m(m-1)-|a|$

and the Theorem is proved. ∎

######  Corollary 4.6.

Let $f,g$ be such diagrams that $(K(f),E(g))\neq 0$ and $g=(A,B)$. Let us also
suppose that for transposition $\tau=\pi_{a}^{b}$ we have $\tau(g)=g,\,a\in
f^{-1}(\times)$ and $a,b\leq n-m$. Then

$(ch\,K(\tau(f)),ch\,E(g))=(-1)^{n_{ab}+m_{ab}+a-b}(ch\,K(f),ch\,E(g))$

where $n_{a,b}=|A\cap(a,b)|,\,m_{a,b}=|B\cap(a,b)|$.

###### Proof.

By Theorem 4.5 $f=g*C$ where $C\subset g^{-1}({\circ})\cap\mathbb{Z}_{\leq
n-m}$. Since $\tau(g)=g$ we have $\tau(f)=g*\tau(C)$ and by Theorem 4.5 we
have

$(K(f),E(g))=(-1)^{\frac{1}{2}r(r-1)+\frac{1}{2}m(m-1)+s(m-r)+S(C)}\varepsilon(A,C)\varepsilon(C,B)$
$(K(\tau(f)),E(g))=(-1)^{\frac{1}{2}r(r-1)+\frac{1}{2}m(m-1)+s(m-r)+S(\tau(C))}\varepsilon(A,\tau(C))\varepsilon(\tau(C),B).$

Further we have the following equalities in $\mathbb{Z}_{2}$:
$S(C)-S(\tau(C))=a-b$ and by Lemma 4.3

$\varepsilon(A,C)-\varepsilon(A,\tau(C))=|A\cap(a,b)|,\,\,\varepsilon(C,B)-\varepsilon(\tau(C),B)=|B\cap(a,b)|.$

Lemma is proved. ∎

######  Corollary 4.7.

Let $f,g,h$ be such diagrams that

$(ch\,P(f),ch\,K(g))\neq 0,\quad(ch\,K(g),ch\,E(h))\neq 0$

and $\tau=\pi_{a}^{b},\,a,b\leq n-m$ be an admissible transposition for $f$
such that $a,b\notin h^{-1}(\times)$. Then

$(ch\,K(\tau(g)),ch\,E(h))+(ch\,K(g),ch\,E(h))=0$

###### Proof.

By Corollary 4.6 it is enough to prove that $n_{ab}+m_{ab}+a-b$ is an odd
number. Let us denote by $(a,b)_{x}=g^{-1}(x)\cap(a,b)$. Then we have

$(a,b)=(a,b)_{>}\cup(a,b)_{<}\cup(a,b)_{\times}\cup(a,b)_{\circ}$

Therefore

$b-a-1=|(a,b)_{>}|+|(a,b)_{<}|+|(a,b)_{\times}|+|(a,b)_{\circ}|.$

where $|A|$ means the cardinality of the set $A$. But

$n_{a,b}=|(a,b)_{>}|+|(a,b)_{\times}|,\,m_{a,b}=|(a,b)_{<}|+|(a,b)_{\times}|,\,$

Therefore it is enough to prove that $|(a,b)_{\times}|+|(a,b)_{\circ}|$ is an
even number. Let $C=\\{c_{1},\dots,c_{r}\\}$. We have

$\varphi\circ g=\varphi\circ f+2\sum_{i=1}^{r}(\delta_{d_{i}}-\delta_{c_{i}})$

and by definition admissible transposition we have
$\sum_{i\in(a,b)}\varphi\circ f(i)=0$. Therefore

$|(a,b)_{\times}|-|(a,b)_{\circ}|=\sum_{i\in(a,b)}\varphi\circ
g(i)=2\sum_{i=1}^{r}(\delta_{d_{i}}-\delta_{c_{i}}).$

Corollary is proved. ∎

##  5\. Partially polynomial representations

######  Definition 5.1.

A weight $\chi\in P$

$\chi=\lambda_{1}\varepsilon_{1}+\dots+\lambda_{m}\varepsilon_{m}+\mu_{1}\delta_{1}+\dots+\mu_{n}\delta_{n}$

is called partially polynomial (in $y_{1},\dots,y_{m}$) if
$\mu_{1},\dots,\mu_{m}\in\mathbb{Z}_{\geq 0}$.

######  Corollary 5.2.

A diagram $f=(A,B)$ corresponds to the partially polynomial highest weight if
and only if all elements of $B$ are not grater than $n-m$.

######  Definition 5.3.

A representation $V$ of $\mathfrak{gl}(m,n)$ is called partially polynomial
(in $y_{1},\dots,y_{n}$) if all its weights are partially polynomials or its
character is a polynomial in $y_{1},\dots,y_{n}$.

We should note that there is no loss of generality in restricting our
attention to partially polynomial representations, since an arbitrary finite
dimensional irreducible representation of $\mathfrak{gl}(m,n)$ can be obtained
from some partially polynomal representation by tensoring with a one
dimensional representation.

######  Example 5.4.

Standard representation with the character
$x_{1}+\dots+x_{n}+y_{1}+\dots+y_{m}$ is partially polynomial. One dimensional
representation with character $\frac{y_{1}\dots y_{n}}{x_{1}\dots x_{m}}$ is
also partially polynomial representation.

The subcategory of the modules with partially polynomials weights will be
denote by $\mathcal{F}^{+}$.

######  Definition 5.5.

For any $M\in\mathcal{F}$ let us denote by $M^{-}$ the submodule generated by
all weight vectors with non partially polynomials weights. Let us also define
a functor $F^{+}:\mathcal{F}\rightarrow\mathcal{F}^{+}$ by the following
formula

$F^{+}(M)=M/M^{-}$

######  Lemma 5.6.

$1)$ Functor $F^{+}$ is right exact.

$2)$ Functor $F^{+}$ maps projective objects in $\mathcal{F}$ to projective
objects in $\mathcal{F}^{+}$.

###### Proof.

$1)$ By definition of $M^{-}$ we have the following equality for all
$N\in\mathcal{F}^{+}$

$Hom_{\mathcal{F}}(M,N)=Hom_{\mathcal{F}^{+}}(F(M),N)$

Consider a functor $G:\mathcal{F}^{+}\rightarrow\mathcal{F}$ such that
$G(N)=N$. Then the above equality means that $G$ is right adjoint to $F$.
Therefore $F$ is right exact.

$2)$ follows from $1)$. ∎

######  Lemma 5.7.

The following statements hold true

$1)$ Let $\chi\in P^{+}$ and $K(\chi)$ be the corresponding Kac module, then
$F^{+}(K(\chi))=K(\chi)$ if $\lambda$ is a partially polynomial weight and $0$
otherwise.

$2)$ Let $L(\chi)$ be the irreducible finite dimensional module corresponding
to the weight $\chi$. Then $F^{+}(L(\chi))=L(\chi)$ if $\chi$ is a partially
polynomial weight and $0$ otherwise.

###### Proof.

$1)$ Let $\chi$ be a partially polynomial highest weight. Therefore
$\chi-\alpha$ is also partially polynomial for any positive root $\alpha$.
Therefore all weights of the module $K(\chi)$ are partially polynomials, so
$K(\chi)^{-}=0$ and $F^{+}(K(\chi))=K(\chi)$. If $\chi$ is not partially
polynomial then $K(\chi)^{-}=K(\chi)$ since it is generated by the vector of
the weight $\chi$. Therefore $F^{+}(K(\chi))=0$.

$2)$ Let $\chi$ be a partially polynomial weight. Since $L(\chi)$ is a
quotient of $K(\chi)$ then by the first statement we have
$F^{+}(L(\chi))=L(\chi)$. If $\chi$ is not partially polynomial then by the
first statement of the Lemma and by Lemma 5.6 $F^{+}(L(\lambda))=0$. ∎

######  Corollary 5.8.

Let $M\in\mathcal{F}$ and suppose that it has composition series of Kac
modules and in the Grothendieck group of $\mathcal{F}$ we have

$[M]=\sum_{\lambda\in I}m_{\chi}[K(\chi)]$

Then in the Grothendieck group of $\mathcal{F}^{+}$ we have

$[F^{+}([M])=\sum_{\chi\in I^{pol}}m_{\chi}[K(\chi)]$

where $I^{pol}$ is a subset of partially polynomial weights of $I$.

##  6\. Projective covers and Euler characters

In this section we are going to give an algorithm to represent Euler
characters as the sum of characters of irreducible modules. For any finite
dimensional module $V$ we have the following formula in the ring
$\Lambda^{\pm}_{m,n}$

$ch\,V=\sum_{P}(ch\,P,\,ch\,V)ch\,L_{P}$

where sum is taken over all projective covers and $L_{P}$ is the irreducible
module corresponding to $P$.

So if $V=E(h)$ is an Euler character then we need to calculate
$(ch\,P(f),\,ch\,E(h))$ for fixed $h$ and all $f$. In order to do so we
calculate first $(ch\,P(f),\,ch\,E(h))$ for fixed $f$ and all $h$.

######  Definition 6.1.

Let $f$ be a diagram. Let us set

$f^{-1}_{0}(\times)=\\{a\in f^{-1}(\times)\mid\pi_{a}^{b}\,\,\text{is
admissible and}\,\,b\leq n-m\\},$ $f^{-1}_{1}(\times)=\\{a\in
f^{-1}(\times)\mid\pi_{a}^{b}\,\,\text{is admissible and}\,\,b>n-m\\},$

######  Definition 6.2.

Let us set

$\mathcal{E}(f)=\\{h\mid(ch\,P(f),ch\,E(h))\neq 0\\}$

######  Definition 6.3.

Let $f$ be a diagram and $B\subset f^{-1}(\times)$. We will denote by $f_{B}$
the following diagram

$f^{-1}_{B}(x)=\begin{cases}f^{-1}(x),\,x=<,>\\\ f^{-1}(\times)\setminus B\\\
f^{-1}(\circ)\cup B\\\ \end{cases}$

And for a diagram $f$ we will denote by $f_{>d}$ the diagram $f_{B}$ in the
case when $B=f^{-1}(\times)\cap\mathbb{Z}_{>d}$. In other words $f_{>d}$ is
the diagram which can be obtained from $f$ by deleting from $f^{-1}(\times)$
all the numbers which are strictly grater that $d$ and adding them to
$f^{-1}(\circ)$.

The following Theorem describes the pairing between projective covers and
Euler characters.

######  Theorem 6.4.

The following equalities hold true

$1)$

$\mathcal{E}(f)=\\{h\mid h*A\in\mathcal{P}(f),\,A\subset
f_{1}^{-1}(\times)\cap\mathbb{Z}_{\leq n-m}\\}$ (11)

$2)$ If $h\in\mathcal{E}(f)$ then

$(ch\,P(f),ch\,E(h))=(ch\,K(h*A),ch\,E(h))$

###### Proof.

Let us prove the first statement. We will denote by $\mathcal{Q}$ the write
hand side of the equality (11). Let $h\in\mathcal{E}(f)$. We are going to
prove that $h\in\mathcal{Q}$. Let us denote by $\mathcal{A}(h)$ the set of all
$g$ such that

$(ch\,P(f),\,ch\,K(g))\neq 0,\,\,\,(ch\,K(g),ch\,E(h))\neq 0$

Then we have

$(ch\,P(f),\,ch\,E(h))=\sum_{g\in\mathcal{A}(h)}(ch\,K(g),ch\,E(h))$

Since $h\in\mathcal{E}(f)$ the set $\mathcal{A}(h)$ is not empty and there
exists $g\in\mathcal{A}(h)$. By Theorem 4.5 we have

$g=h*A,A\subset\mathbb{Z}_{\leq n-m}$

So we only need to prove that $A\subset f_{1}^{-1}(\times)$. If $A=\emptyset$
then $A\subset f_{1}^{-1}(\times)$. Let $A\neq\emptyset$ and $a\in A$. There
are two cases $a\notin f^{-1}(\times)$ and $a\in f^{-1}(\times)$.

Let us consider the first case. Let $\tau=\pi_{c}^{a}$ be the corresponding
admissible transposition then $c\notin g^{-1}(\times)$ and therefore $a\notin
h^{-1}(\times)$. By Corollary 4.7 the set $\mathcal{A}(h)$ is invariant under
the action of $\tau$ and for any $\tilde{g}\in\mathcal{A}(h)$ we have

$(ch\,K(\tilde{g}),chE(h))+(ch\,K(\tau(\tilde{g})),chE(h))=0$

Therefore $(ch\,P(f),\,ch\,E(h))=0$ and the first case is impossible.

Consider the second case $a\in f^{-1}(\times)$. Let $\tau=\pi_{a}^{b}$ be the
corresponding admissible transposition then $b\notin g^{-1}(\times)$ and
therefore $b\notin h^{-1}(\times)$. We have two possibilities $b\leq n-m$ and
$b>n-m$. If $b\leq n-m$ then in the same way as above we can prove that
$(ch\,P(f),\,ch\,E(h))=0$. So the only possibility left $b>n-m$. Therefore
$A\subset f^{-1}_{1}(\times)$ and $\mathcal{E}(f)\subset\mathcal{Q}$.

Now let us prove the opposite inclusion. Let $h*A\in\mathcal{P}(f)$ and
$A\subset f^{-1}(\times)\cap\mathbb{Z}_{\leq n-m}$. Suppose that
$h*\tilde{A}\in\mathcal{P}(f)$. Clearly $(K(h*A),E(h))\neq 0$. Let
$g\in\mathcal{P}(f)$ such that $(ch\,K(g),ch\,E(h))\neq 0$. Then by Theorem
4.5 we have $g=h*\tilde{A},\,\tilde{A}\subset\mathbb{Z}_{n-m}$. Let $a\in A$
and $\pi_{a}^{b}$ be the corresponding admissible transposition. Then one of
the elements $a,b$ belongs to $\tilde{A}$. Suppose that $a\notin\tilde{A}$.
Therefore $b\in\tilde{A}$ but it is impossible since $b>n-m$. So
$a\in\tilde{A}$ . Therefore $A=\tilde{A}$. So we see that

$(ch\,P(f),ch\,E(h))=(ch\,K(h*A),ch\,E(h))\neq 0$

So we proved the inclusion $\mathcal{E}(f)\supset\mathcal{Q}$ and the second
statement. The Theorem is proved. ∎

Now we are going to investigate the case of partially polynomial
representations in more details.

######  Definition 6.5.

Let us denote by $\mathcal{E}^{+}(f)$ the set of partially polynomial diagrams
$h$ such that $(ch\,P(f),ch\,E(h))\neq 0$.

######  Corollary 6.6.

Let $f$ be a partially polynomial diagram then the following equality holds
true

$\mathcal{E}^{+}(f)=\\{h\mid h=\pi_{C}(f)_{>n-m},\,C\subset f^{-1}(\times)\\}$

###### Proof.

Let us denote the right hand side the above equality by $\mathcal{R}$ and by
$\mathcal{Q}^{+}$. we will denote the set of partially polynomial diagrams in
$\mathcal{Q}$, where $\mathcal{Q}$ is the same as in the proof of Theorem 6.4.
By definition we have

$\mathcal{E}^{+}(f)=\mathcal{Q}^{+}$

and we need to prove that $\mathcal{R}=\mathcal{Q}^{+}$. Let
$h\in\mathcal{Q}^{+}$ then by Theorem 6.4 we have
$g=h*A\in\mathcal{P}(f),\,A\subset f_{1}^{-1}(\times)\cap\mathbb{Z}_{\leq
n-m}$ and since $h,f$ are partially polynomial diagrams then $g$ is a
partially polynomial diagram too. Futher we have

$h=g_{A}=(\pi_{A}(g))_{>n-m}$

Besides since $g\in P(f)$ we have $g=\pi_{B}(f),\,B\subset
f_{0}^{-1}(\times)$. Therefore

$h=(\pi_{A}\pi_{B}(f))_{>n-m}=(\tau_{C}(f))_{>n-m},\,\,C=A\cup B$

So $h\in\mathcal{R}$. Now let us take $h\in\mathcal{R}$. Then by definition

$h=\pi_{C}(f)_{>n-m},C\subset f^{-1}(\times)$

Let us set $B=f_{0}^{-1}(\times)\cap C,A=f_{1}^{-1}(\times)\cap C$. Then
$h*A=\pi_{B}(f)\in\mathcal{P}(f).$ Therefore $h\in\mathcal{Q}^{+}$ and we
proved the Corollary. ∎

######  Corollary 6.7.

Let $h$ be a partially polynomial diagram and

$ch\,E(h)=\sum_{f}b_{f,h}ch\,L(f)$

be the decomposition of Euler character $E(h)$ in terms of characters of
irreducible modules. Then $b_{f,h}=0,\pm 1$ and it is nonzero if and only if
there exists the sequence of transpositions

$\sigma_{1}=\pi_{c_{1}}^{d_{1}},\dots,\sigma_{s}=\pi_{c_{s}}^{d_{s}},\,\sigma_{s+1}=\pi_{c_{s+1}}^{d_{s+1}},\dots,\sigma_{r}=\pi_{c_{r}}^{d_{r}}$

such that

$f=\sigma_{r}\circ\dots\circ\sigma_{s+1}\left(\left(\sigma_{s}\circ\dots\sigma_{1}(h)\right)*\\{d_{s+1},\dots,d_{r}\\}\right)$

and

$1)$ $\sigma_{i}$ is admissible for
$h_{i}=\sigma_{i}\circ\dots\circ\sigma_{1}(h)$, $i=1,\dots,s$

$2)$ $c_{1}>c_{2}>\dots>c_{s}$, $d_{1},\dots,d_{s}\leq n-m$ and $c_{i}\neq
d_{j},\,1\leq i,j\leq s$

$3)$ $c_{s+1}>\dots>c_{r}$, $d_{s+1},\dots,d_{r}>n-m$

$4)$ $\sigma_{i}$ is admissible for $h_{i}=\sigma_{i}(h_{i-1}*\\{d_{i}\\})$,
$i=s+1,\dots,r$

$5)$
$\\{c_{1},\dots,c_{s}\\}\cap\\{c_{s+1},\dots,c_{r}\\}=\emptyset,\,\,\,\\{c_{1},\dots,c_{r}\\}\cap\\{d_{1},\dots,d_{s}\\}=\emptyset$

###### Proof.

Suppose that all conditions of the Corollary are fulfilled. Then

$f=\sigma_{r}\circ\dots\circ\sigma_{s+1}\left(\left(\sigma_{s}\circ\dots\sigma_{1}(h)\right)*\\{d_{s+1},\dots,d_{r}\\}\right)$

Since $d_{1},\dots,d_{s}\leq n-m$ and $d_{s+1},\dots,d_{r}>n-m$ we can rewrite
the above equality in the form

$f=\sigma_{r}\circ\dots\circ\sigma_{s+1}\circ\sigma_{s}\circ\dots\sigma_{1}(h*\\{d_{s+1},\dots,d_{r}\\})$

Now we are going to prove that $\sigma_{1},\dots,\sigma_{r}$ are admissible
for $f$. Let us set

$\tau_{i}=\sigma_{r-i+1},a_{i}=c_{r-i+1},\,b_{i}=d_{r-i+1}\,\,\,1\leq i\leq r$

Then we have

$h*\\{b_{r-s},\dots,b_{1}\\}=\left(\tau_{r}\circ\dots\circ\tau_{r-s+1}\circ\tau_{r-s}\circ\dots\circ\tau_{1}(f)\right)$

Again from the conditions of the Lemma it follows that
$\tau_{i},\,i=1,\dots,r$ is admissible for
$f_{i}=\tau_{i-1}\circ\dots\circ\tau_{1}(f)$ We have
$a_{r-s+1}<a_{r-s+2}<\dots<a_{r}$ and by our assumptions $a_{r-s+i}\neq
b_{r-s+j},\,1\leq 1,j\leq s$. Therefore by Corollary 2.10
$\tau_{r-s+1},\dots,\tau_{r}$ are admissible for $f_{r-s+1}$. Further again by
our assumptions $a_{1}<\dots<a_{r-s}$ and since $b_{1},\dots,b_{r-s}>n-m$ we
have $b_{1}>\dots>b_{r-s}$ Let us take $r-s+1\leq j\leq r$. Then
$b_{j}<b_{r-s}$. Therefore by Corollary 2.11
$\tau_{1},\dots,\tau_{r-s+1},\tau_{j}$ are admissible for $f$.

Now let us suppose that $(P(f),E(h))\neq 0$. Therefore by the proof of
Corollary 6.6

$h=(\pi_{A_{1}}\pi_{A_{0}}(f))_{>n-m},\,A_{1}\subset
f_{1}^{-1}(\times),\,A_{0}\subset f_{0}^{-1}(\times)$

Let $A_{0}=\\{a_{1},a_{2},\dots,a_{r-s}\\}$ where $a_{1}<a_{2}<\dots<a_{r-s}$
and also
$A_{1}=\\{a_{r-s+1},\dots,a_{r}\\}$ where $a_{r-s+1}<\dots<a_{r}$. Since
admissible transpositions pairwise commute we have

$h*\\{b_{r-s},\dots,b_{1}\\}=\tau_{r}\circ\tau_{2}\circ\dots\circ\tau_{1}(f)$

where $\tau_{i}=\pi_{a_{i}}^{b_{i}}$. Let us check that conditions (3) are
fulfilled. It is enough to verify that inclusion
$[a_{j},b_{j}]\subset(a_{i},b_{i})$ is impossible if $i>j$. Indeed if $i\leq
r-s$ or $j>r-s$ then $a_{j}>a_{i}$ and we get a contradiction. If
$i>r-s,\,j\leq r-s$ then $b_{i}\leq n-m$ and $b_{j}>n-m$ and this is again a
contradiction. Therefore conditions $(\ref{cond})$ are fulfilled and by
Theorem 2.9 $\tau_{i}$ is admissible for
$f_{i}=\tau_{i-1}\circ\circ\dots\circ\tau_{1}(f)$ and we can set

$\sigma_{i}=\tau_{r-i+1},\,c_{i}=a_{r-i+1},\,d_{i}=b_{r-i+1}\,\,\,i=1,\dots,r.$

∎

######  Example 6.8.

Let $n=m=2$ and $h^{-1}(\times)=-1,\,h^{-1}(<)=h^{-1}(>)=\emptyset$. We are
going to describe the set $\mathcal{E}^{+}(h).$

In this case there are two possibilities for the first step.

$1)$ We are going to find a transpositions $\pi_{a}^{b}$ such that $b\in
h^{-1}(\times),$ and $\pi_{a}^{b}$ is admissible for $\pi_{a}^{b}(h).$ And it
is easy to see that there exists only one such transposition $\pi_{-2}^{-1}$.

$2)$ We also need to find a transposition $\pi_{a}^{b}$ such that
$\pi_{a}^{b}$ is admissible for $\pi_{a}^{b}(h*\\{b\\})$ and $b>0.$ It is also
easy to check that there exist two such transpositions
$\pi_{0}^{1},\pi_{-2}^{1}$.

In the second step there are also two possibilities.

$1)$ We are going to find a transpositions $\pi_{a}^{b}$ such that
$b\in(\pi_{-2}^{-1}h)^{-1}(\times)$ and $\pi_{a}^{b}$ is admissible for
$\pi_{a}^{b}\circ\pi_{-2}^{-1}(h).$ And it is easy to see that there is no
such a transposition which satisfies conditions $1),2),3)$ of Corollary 6.7.

$2)$ We are going to find a transpositions $\pi_{a}^{b}$ such that
$\pi_{a}^{b}$ is admissible for $\pi_{a}^{b}\circ\pi_{-2}^{-1}(h*\\{b\\})$ and
$b>0$. And it is easy to see that there is only one such transposition
$\pi_{0}^{1}.$

So we have

$\mathcal{E}^{+}(h)=\\{\pi_{0}^{1}((\pi_{-2}^{-1}(h))*\\{1\\}),\,\,\pi_{0}^{1}(h*{1}),\,\,\pi_{-2}^{1}(h*{1})\\}$

And it is easy to see that

$E(h)=-L(\pi_{0}^{1}((\pi_{-2}^{-1}(h))*\\{1\\}))-L(\pi_{0}^{1}(h*{1}),)-L(\pi_{-2}^{1}(h*{1}))$

##  7\. Some special classes of irreducible modules

In this section we will only consider diagrams of the form $f=(A,A)$ where
$A\subset\mathbb{Z}_{\leq 0}$ and instead of $E(f)$ we will write $E(A)$ for
Euler virtual module and $L(A),\,P(A)$ for irreducible module and for
projective indecomposable module correspondently. Below $|A|$ means the number
of elements in the set $A$. Our aim in this section is to give an explicit
formula for characters of irreducible modules for the most atypical block of
Lie superalgebra $\mathfrak{gl}(2,2)$.

######  Definition 7.1.

Let us set

$\mathcal{P}_{n}^{(m)}=\\{A\subset\mathbb{Z}_{\leq 0},|A|=n\mid\exists\
B\subset\mathbb{Z}_{\leq 0},\,|B|\leq m,\,(ch\,P(A),ch\,E(B))\neq 0\\}$

and denote by $\omega$ the following shift

$\omega:\mathbb{Z}\rightarrow\mathbb{Z},\,\,\,\omega(x)=x-1$

In the following Lemma we give an inductive description of the set
$\mathcal{P}_{n}^{(m)}$.

######  Lemma 7.2.

The following formulae hold true for $m<n$

$\mathcal{P}_{n}^{(m)}=\bigcup_{i=0}^{m}\mathcal{P}_{n,i}^{(m)},\,\quad\mathcal{P}_{n,i}^{(m)}=\left\\{\\{-i,C\\}\mid
C\in\omega^{i+1}(\mathcal{P}_{n-1}^{(m-i)})\right\\}$

###### Proof.

It is easy to see that

$\mathcal{P}_{n}^{(m)}=\bigcup_{i=0}^{m}\mathcal{A}_{i},\quad\mathcal{A}_{i}=\\{A\in\mathcal{P}_{n}^{(m)}\mid\max_{a\in
A}a=i\\}$

and we only need to show that $\mathcal{A}_{i}=\mathcal{P}_{n,i}^{(m)}$.

First let us note that $A\in\mathcal{P}_{n}^{(m)}$ if and only if there exist
at least $n-m$ elements from $A$ such that every corresponding admissible
transposition has one positive element. If in addition $A$ contains $0$ then
transposition $\pi_{0}^{1}$ is admissible for $A$. Therefore for the set
$A\setminus\\{0\\}$ there must be at least $n-m-1$ admissible transpositions
with one positive element. This proves the equality
$\mathcal{A}_{0}=\mathcal{P}_{n,0}^{(m)}$. The same arguments work for any
$0<i\leq m$. Lemma is proved. ∎

In the case $m=1$ we can give an explicit description of the set
$\mathcal{P}_{n}^{(m)}$.

######  Lemma 7.3.

We have $\mathcal{P}^{(1)}_{n}=S_{1}\cup S_{2}$ where

$S_{1}=\\{\\{0,-1,-2,\dots,2-n,a\\}\mid a\leq 1-n\\},\,$
$S_{2}=\\{\\{0,-1,-2,\dots,-n\\}\setminus\\{b\\}\mid b=0,-1,\dots,1-n,-n\\}$

###### Proof.

Use of Lemma 7.2 and induction on $n$. ∎

The following Lemma gives the value of our bilinear form on some pairs of
projective modules and Euler characters.

######  Lemma 7.4.

Let $A_{0}=\\{0,-1,-2,\dots,1-n,\,-n\\}$. Then we have

$\mathcal{E}^{+}(A_{0}\setminus\\{-n\\})\cap\mathcal{P}^{(1)}_{n}=\\{E(\emptyset),E(0),\dots,E(1-n)\\},\quad(1,1,-1\dots,(-1)^{n-1})$
$\mathcal{E}^{+}(A_{0}\setminus\\{b\\}))\cap\mathcal{P}^{(1)}_{n}=\\{E(b),\,E(b-1)\\},\quad((-1)^{n-1},(-1)^{n-1}),\,\,b=0,\dots,1-n$
$\mathcal{E}^{+}(0,-1,\dots,2-n,b)\cap\mathcal{P}^{(1)}_{n}=\\{E(b+1),\,E(b)\\},\,((-1)^{n-1},\,(-1)^{n-1}),\,b\leq-n$

we also indicate on the right the corresponding value $(ch\,P(A),ch\,E(B))$.

###### Proof.

It easily follows from Theorem 6.4. ∎

######  Corollary 7.5.

The following formulae hold true

$1)$

$ch\,L(A_{0}\setminus\\{-n\\})=ch\,E(\emptyset),\,\,$

$2)$ if $b\in[1-n,0]$ then

$ch\,L(A_{0}\setminus\\{b\\})=(-1)^{n-1}[ch\,E(b)-ch\,E(b+1)+\dots+$
$(-1)^{b}ch\,E(0)+(-1)^{b+1}(1-b)ch\,E(\emptyset)]$

$3)$ if $b\leq-n$ then

$ch\,L(0,-1,\dots,2-n,b)=(-1)^{n-1}[ch\,E(b+1)-\dots+$
$(-1)^{b+1}ch\,E(0)+(-1)^{b}n\,ch\,E(\emptyset)]$

###### Proof.

Let us prove the first statement. It is enough to check that

$(P(A_{0}\setminus\\{-n\\}),E(\emptyset))=1,$ (12)

and

$(P(B),E(\emptyset))=0,\,\text{for any}\,B,\,|B|=n,\,B\subset\mathbb{Z}_{\geq
0},\,B\neq A.$ (13)

Equality (12) follows from Lemma 7.4. Besides if
$B\notin\mathcal{P}^{(1)}_{n}$ then equality (13) follows from the definition
of the set $\mathcal{P}^{(1)}_{n}$ and if $B\in\mathcal{P}^{(1)}_{n}$ then
this equality follows from Lemma 7.4. The other two statements can be proved
in the same manner. ∎

######  Definition 7.6.

Let us define a linear operator on the characters of Kac modules by the
formula

$T(ch\,K(A))=ch\,K(\omega(A)),\quad\omega(a)=a-1$

It is easy to see that in the category $\mathcal{F}^{+}$ the operator $T$
corresponds to tensor multiplication on one dimensional module with the
character $\frac{y_{1}\dots y_{n}}{x_{1}\dots x_{m}}$. Now we are going to
describe the action of the linear operator $T$ on the Euler characters.

######  Lemma 7.7.

The following formulae hold true (we suppose that $A,B\subset\mathbb{Z}_{\leq
0})$

$T(ch\,E(B))=(-1)^{n+p}\left[ch\,E(\omega(B))-(-1)^{p}ch\,E(\\{0\\}\cup\omega(B))\right]$
(14)

where $p=|B|$ and we suppose that $E(B)=0$, if the number of elements in $B$
is strictly grater than $n$.

###### Proof.

It is enough to prove the following equality for any set $A$ such that $|A|=n$

$(ch\,K(A),ch\,T(E(B)))=$
$(ch\,K(A),\,(-1)^{n+p}\left[ch\,E(\omega(B))-(-1)^{p}ch\,E(\\{0\\}\cup\omega(B))\right])$

So let us calculate separately the left hand side and the right hand side. We
have

$(ch\,K(A),ch\,TE(B))=\left(ch\,K(A),\frac{y_{1}\dots y_{n}}{x_{1}\dots
x_{n}}ch\,E(B)\right)=$ $\left(\frac{x_{1}\dots x_{n}}{y_{1}\dots
y_{n}}ch\,K(A),ch\,E(B)\right)$ $=(ch\,K(\omega^{-1}(A)),ch\,E(B))$

But by Theorem 4.5 we have

$(ch\,K(\omega^{-1}(A)),ch\,E(B))=\begin{cases}0,\,\,\text{if}\,\,B\not\subset\omega^{-1}(A),\,\text{or}\,\,0\in
A\\\ (-1)^{\frac{1}{2}p(p-1)+\frac{1}{2}n(n-1)+S(\omega^{-1}(A)\setminus
B)}\,\text{otherwise}\end{cases}$

Let $B\not\subset\omega^{-1}(A)$ then $\omega(B)\not\subset A$ and
$\omega(B)\not\subset\\{0\\}\cup A$. Therefore

$\left(ch\,K(A),ch\,E(\omega(B)\right)=0,\,\,\left(ch\,K(A),ch\,E(\\{0\\}\cup\omega(B)\right)=0$

Suppose that $0\in A$ and $B\subset\omega^{-1}(A)$. Then we have
$\\{0\\}\cup\omega(B)\subset A$ and therefore

$\left(ch\,K(A),ch\,E(\omega(B)\right)=(-1)^{\frac{1}{2}p(p-1)+\frac{1}{2}n(n-1)+S(A\setminus\omega(B))}$
$\left(ch\,K(A),ch\,E(\\{0\\}\cup\omega(B)\right)=(-1)^{\frac{1}{2}p(p+1)+\frac{1}{2}n(n-1)+S(A\setminus\omega(B)\cup\\{0\\})}$

So formulae (14) holds true in this case.

Now consider the last possible case $0\notin A$ and $B\subset\omega^{-1}(A)$.
In this case we have $\\{0\\}\cup\omega(B)\not\subset A$. Therefore

$\left(ch\,K(A),ch\,E(\\{0\\}\cup\omega(B)\right)=0$

and

$\left(ch\,K(A),ch\,E(\omega(B)\right)=(-1)^{\frac{1}{2}p(p-1)+\frac{1}{2}n(n-1)+S(A\setminus\tau(B))}$

Lemma is proved. ∎

Now consider the case of the most atypical bloc for Lie superalgebra
$\mathfrak{gl}(2,2)$ in the category $\mathcal{F}^{+}$. In order to give a
reasonable description of the irreducible characters in this bloc we need some
special type of graphs.

######  Definition 7.8.

Let $n\in Z_{\geq 0},\,m\in\mathbb{Z}_{\geq 1}$. Let us denote by
$\Gamma_{n,m}$ the graph with $n+m$ vertices that are integers from the
segment $[1-n-m,0]$ such that :

$1)$ there exists exactly one edge containing any two vertices from $[1-n,0]$;

$2)$ there exists exactly one edge joining every vertex from $[1-n,0]$ with
every vertex from $[-n,1-n-m]$.

######  Definition 7.9.

For every graph $\Gamma=\Gamma_{n,m}$ let us define the following element of
the Grothendieck ring by the formula

$\chi(\Gamma)=ch\,E(\emptyset)-\sum_{v}\varepsilon(v)ch\,E(v)-\sum_{e}\varepsilon(e)ch\,E(e)$

where $\varepsilon(v)=(-1)^{i}$ if $v=\\{i\\}$ and
$\varepsilon(e)=\varepsilon(v)\varepsilon(u)$ if $e$ contains $v,u$.

######  Remark 7.10.

It is convenient to define $\chi(\Gamma_{n,m})$ in the case when $n=-1$. In
such a case we set $\chi(\Gamma_{n,m})=E(\emptyset)$.

######  Example 7.11.

$\chi(\Gamma_{-1,3})=ch\,E(\emptyset)$
$\chi(\Gamma_{2,1})=ch\,E(\emptyset)-ch\,E(0)+ch\,E(-1)-ch\,E(-2)+$
$ch\,E(0,-1)-ch\,E(0,-2)+ch\,E(-1,-2)$
$\chi(\Gamma_{0,2})=ch\,E(\emptyset)-ch\,E(0)+ch\,E(-1)$

######  Theorem 7.12.

The following equalities hold true

$ch\,L(a,a-1)=\chi(\Gamma_{|a|-1,1}),\,\,a\leq 0$ (15)
$ch\,L(a,b)=(-1)^{a-b-1}\left[\chi(\Gamma_{|a|-1,1})+\chi(\Gamma_{|a|,a-b})\right],\,\,a-b\geq
2,\,a\leq 0$ (16)

###### Proof.

We are going to use the functor $T$. In the case of $n=2$ the functor acts by
the following formulae

$T(ch\,E(\emptyset))=ch\,E(\emptyset)-ch\,E(0),\,\,$
$T(ch\,E(a))=-ch\,E(a-1)-ch\,E(0,a-1)$

and

$T(ch\,E(a,b))=ch\,E(a-1,b-1)$

It is not difficult to verify the following equality

$T(\chi(\Gamma_{n,m}))=\chi(\Gamma_{n+1,m})$

Further we see that $ch\,L(a,a-1)=T^{|a|}(ch\,E(\emptyset))$. We will prove
equality (15) induction on $|a|$ . If $a=0$ the equality is trivial
$ch\,L(0,-1)=ch\,E(\emptyset)$. Let $|a|>0$ then we have

$T^{|a|}(ch\,E(\emptyset))=T(T^{|a|-1}(ch\,E(\emptyset)))=T(\chi(\Gamma_{|a|-2,1}))=\chi(\Gamma_{|a|-1,1})$

Now let us prove equality (16) also induction on $|a|$. If $a=0$ and $b\leq-2$
then by corollary (7.5) we have

$ch\,L(0,b)=(-1)^{b+1}(2ch\,E(\emptyset)-ch\,E(0)+\dots+$
$(-1)^{b}ch\,E(b+1))=(-1)^{b+1}[ch\,E(\emptyset)+\chi(\Gamma_{0,|b|})]$

If we apply to both sides of the above formula functor $T^{r}$ then we get

$ch\,L(-r,b-r)=(-1)^{b+1}T^{r}[ch\,E(\emptyset)+\chi(\Gamma_{0,|b|})]=$
$(-1)^{b+1}[\chi(\Gamma_{r-1,1})+\chi(\Gamma_{r,|b|})]$

If we replace $r$ by $-a$ and $b$ by $b-a$ we get the statement. ∎

##  8\. Acknowledgments

This work was supported by the Ministry of Science and Higher Education of the
Russian Federation in the framework of the basic part of the scientific
research state task, project FSRR-2020-0006.

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