Supergroup $OSP(2,2n)$ and super Jacobi polynomials

TL;DR

This paper explores super Jacobi polynomials related to the Lie supergroup OSP(2,2n), revealing how their coefficients behave under parameter transformations and connecting them to supercharacters of various modules.

Contribution

It introduces a new family of super Jacobi polynomials obtained via blow-up procedures, linking them to supercharacters of Kac modules, irreducible modules, and projective covers.

Findings

Polynomials depend rationally on parameters k,p,q

Blow-up at (-1,0) yields a family parameterized by t

Supercharacters are obtained as special cases for specific t values

Abstract

Coefficients of super Jacobi polynomials of type $B(1,n)$ are rational functions in three parameters $k,p,q$. At the point $(-1,0,0)$ these coefficient may have poles. Let us set $q=0$ and consider pair $(k,p)$ as a point of $\Bbb A^2$. If we apply blow up procedure at the point $(-1,0)$ then we get a new family of polynomials depending on parameter $t\in \Bbb P$. If $t=\infty$ then we get supercharacters of Kac modules for Lie supergroup $OSP(2,2n)$ and supercharacters of irreducible modules can be obtained for nonnegative integer $t$ depending on highest weight. Besides we obtained supercharcters of projective covers as specialisation of some nonsingular modification of super Jacobi polynomials.

Equations

$$\begin{gathered} \mathcal{L}=\sum _{i=1}^m{\partial }_{x_i}^2+k\sum _{j=1}^n{\partial }_{y_j}^2-k\sum _{i<j}^m\left(\frac{x_i+x_j}{x_i-x_j}({\partial }_{x_i}-{\partial }_{x_j})+\frac{x_i{x}_j+1}{x_i{x}_j-1}({\partial }_{x_i}+{\partial }_{x_j})\right) \\ -\sum _{i<j}^n\left(\frac{y_i+y_j}{y_i-y_j}({\partial }_{y_i}-{\partial }_{y_j})+\frac{y_i{y}_j+1}{y_i{y}_j-1}({\partial }_{y_i}+{\partial }_{y_j})\right) \\ -\sum _{i=1}^m\left(p\frac{x_i+1}{x_i-1}+2q\frac{x_i^2+1}{x_i^2-1}\right){\partial }_{x_i}-k\sum _{j=1}^n\left(r\frac{y_j+1}{y_j-1}+2s\frac{y_j^2+1}{y_j^2-1}\right){\partial }_{y_j} \\ -\sum _{i,j}\left(\frac{x_i+y_j}{x_i-y_j}({\partial }_{x_i}-k{\partial }_{y_j})+\frac{x_i{y}_j+1}{x_i{y}_j-1}({\partial }_{x_i}+k{\partial }_{y_j})\right), \end{gathered}$$

$$n(\lambda )={\lambda }_2+2{\lambda }_3+\dots ,c_{\lambda }=2n({\lambda }^{\prime })+2kn(\lambda )+|\lambda |(2h+2k+1),$$

$$a_i={\lambda }_i+ki,i=1,2,\dots ,c_{\lambda }^0(\square ,x)=j-1+k(i-1)+x,$$

$$c_{\lambda }^{-}(\square ,x)={\lambda }_i-j-k({\lambda }_j^{\prime }-i)+x,c_{\lambda }^{+}(\square ,x)={\lambda }_i+j+k({\lambda }_j^{\prime }+i)+x,$$

$$C_{\lambda }^0(x)=\prod _{\square \in \lambda }{c}_{\lambda }^0(\square ,x),C_{\lambda }^{-}(x)=\prod _{\square \in \lambda }{c}_{\lambda }^{-}(\square ,x),C_{\lambda }^{+}(x)=\prod _{\square \in \lambda }{c}_{\lambda }^{+}(\square ,x),$$

$$J_{\lambda }(1)=4^{\lambda }\frac{C_{\lambda }^0(h+\frac{1}{2}p+q)}{C_{\lambda }^{-}(-k)}\frac{C_{\lambda }^0(k+h-\frac{1}{2}p+\frac{1}{2})}{C_{\lambda }^{+}(2h-1)}.$$

$$V_{\mu }(\lambda )=\prod _{j\ne i}^{l(\lambda )+1}\frac{(a_i-a_j-k)}{(a_i-a_j)}\frac{(a_i+a_j+2h-k)}{(a_i+a_j+2h)}$$

$$\times \frac{(a_i-k+h+\frac{1}{2}p+q)}{(a_i-k(l(\lambda )+2))}\frac{(a_i+k(l(\lambda )+1)+2h)}{(a_i+h)}\frac{(a_i+h-\frac{1}{2}p+\frac{1}{2})}{(a_i+h+\frac{1}{2})}.$$

$$V_{\mu }(\lambda )=\prod _{j\ne i}^{l(\lambda )}\frac{(a_i-a_j+k)}{(a_i-a_j)}\frac{(a_i+a_j+k+2h)}{(a_i+a_j+2h)}$$

$$\times \frac{(a_i+k+h-\frac{1}{2}p-q)}{(a_i+k(l(\lambda )+1)+2h)}\frac{(a_i-kl(\lambda ))}{(a_i+h)}\frac{(a_i+h+\frac{1}{2}p-\frac{1}{2})}{(a_i+h-\frac{1}{2})},$$

$$a_{\lambda ,\mu }=V_{\mu }(\lambda )\frac{J_{\lambda }(1)}{J_{\mu }(1)},\mu \in S(\lambda )\setminus \{\lambda \},a_{\lambda ,\lambda }=-k^{-1}(2h+p+2q)-\sum _{\mu \in S(\lambda )\setminus \{\lambda \}}{V}_{\mu }(\lambda ).$$

$$a_{\lambda ,\mu }^{(2)}=\frac{1}{4}\prod _{r\ne i}^{l(\lambda )+1}\frac{(a_i+a_r+2h-k)}{(a_i+a_r+2h)}$$

$$\times \frac{(a_i+k(l(\lambda )+1)+2h)}{(a_i+h)}\frac{(a_i+h-\frac{1}{2}p+\frac{1}{2})}{(a_i+h+\frac{1}{2})}\frac{2j-1+2ki+2h}{(j-1+ki+h-\frac{1}{2}p+\frac{1}{2})}$$

$$\prod _{s=1}^{j-1}\frac{a_i+k{\lambda }_s^{\prime }+s+2h}{a_i+k{\lambda }_s^{\prime }+s+2h-1}\prod _{r=1}^{i-1}\frac{a_r+j+k{\lambda }_j^{\prime }+2h-1+k}{a_r+j+k{\lambda }_j^{\prime }+2h-1}$$

$$a_{\lambda ,\mu }^{(2)}=4\prod _{r\ne i}^{l(\lambda )}\frac{(a_i+a_r+2h+k)}{(a_i+a_r+2h)}$$

$$\times \frac{(a_i+k+h-\frac{1}{2}p-q)}{(a_i+k(l(\lambda )+1)+2h)}\frac{1}{(a_i+h)}\frac{(a_i+h+\frac{1}{2}p-\frac{1}{2})}{(a_i+h+\frac{1}{2})}$$

$$\times \frac{(j-1+k(i-1)+h+\frac{1}{2}p+q)(j-1+ki+h-\frac{1}{2}p+\frac{1}{2})}{2j-1+2ki+2h}$$

$$\prod _{s=1}^{j-1}\frac{a_i+k{\lambda }_s^{\prime }+s+2h-2}{a_i+k{\lambda }_s^{\prime }+s+2h-1}\prod _{r=1}^{i-1}\frac{a_r+j+k{\lambda }_j^{\prime }+2h-k-1}{a_r+j+k{\lambda }_j^{\prime }+2h-1}$$

$$J_{\varnothing }=1,\mathcal{L}{J}_{\lambda }=c_{\lambda }{J}_{\lambda },p_1{J}_{\lambda }=\sum _{\mu \in S(\lambda )}{a}_{\lambda ,\mu }{J}_{\mu },$$

$${\mathcal{L}}_{\mu }^{\lambda }=\prod _{\nu \in S(\lambda )\setminus \{\mu \}}\frac{\mathcal{L}-c_{\nu }}{c_{\mu }-c_{\nu }},$$

$${\mathcal{L}}_{\mu }^{\lambda }(p_1{J}_{\lambda })=a_{\lambda ,\mu }{J}_{\mu },$$

$$V=\underset{i\in \mathbb{Z}}{⨁}{V}_i,V_i=<J_{\lambda }\mid c_{\lambda }(-1,0,0)=i>$$

$$F_i(f)=P_i(p_{1}f),f\in V$$

$$p_1{V}_j\subset W_i\oplus W_{i_1}\oplus W_{i_2}\oplus \cdots \oplus W_{i_r}$$

$$W_i=<J_{\lambda }\mid \lambda \in S>,W_{i_1}=<J_{\lambda }\mid \mu \in T>$$

$$f_1(t)=\prod _{\mu \in T}(t-c_{\mu }(k,p))$$

$${\mathcal{C}}_1=(-1{)}^{N+1}\frac{1}{{\sigma }_N}\left({\mathcal{D}}_1^N-{\sigma }_1{\mathcal{D}}_1^{N-1}+\cdots +(-1{)}^{N-1}{\sigma }_{N-1}{\mathcal{D}}_1\right)$$

$$\mathcal{C}={\mathcal{C}}_1{\mathcal{C}}_2\dots {\mathcal{C}}_r.$$

$$p_{1}f=g+g_1+\cdots +g_r$$

$$\mathcal{C}(p_{1}f)=g=P{r}_F(p_{1}f).$$

$$c_{\lambda }=\sum _{(i,j)\in \lambda }[2(j-1+k(i-1))+1-2n-p],{\tilde{c}}_{\lambda }=\sum _{(ij)\in \lambda }(2j-2i+1-2n)$$

$$\mu =\lambda \cup \square ,\nu =\lambda \setminus \tilde{\square },j-i+\tilde{j}-\tilde{i}=2n-1$$

$$(n({\mu }^{\prime })-n({\nu }^{\prime }))-(n(\mu )-n(\nu ))+(|\mu |-|\nu |)(\frac{1}{2}-n)=0$$

$$\mu =\lambda \cup \square ,\nu =\lambda \cup \tilde{\square }\text{or}\mu =\lambda \setminus \square ,\nu =\lambda \setminus \tilde{\square }$$

Full text

1. 1 Introduction
2. 2 Super Jacobi Polynomials
3. 3 Translation functors
4. 4 Nonsingular basis
5. 5 Specialisation
6. 6 Supercharacters
7. 7 Acknowledgements

1. Introduction

Let $J_{\lambda}(x,k,p,q)$ be the family of Jacobi polynomials for root system $BC_{N}$ (see [2]). It easily follows from the orthogonality relations that at the point $(-1,0,0)$ Jacobi polynomials are well defined and coincide with the corresponding characters of irreducible finite dimensional modules over symplectic Lie group $SP(2N)$.

In this paper we investigate the same problem for super Jacobi polynomials. The main difficulty in this case is that super Jacobi polynomials are not well defined at the point $(-1,0,0)$. One partial result on this problem was obtained in the paper [4]. Namely let $J_{\lambda}(x,y,k,p,q)$ be the family of super Jacobi polynomials in $m+n$ indeterminates (see [3]) labeled by the set of partitions $\lambda$ such that $\lambda_{m+1}\leq n$. It has been proved in the paper [4] that $\lim_{(p,q)\rightarrow(0,0)}\lim_{k\rightarrow-1}J_{\lambda}(x,y,k,p,q)$ coincides up to sign with Euler supercharacters (with a special choice of the parabolic subalgebras) of the Lie supergroup $OSP(2m,2n)$. The goal of this paper is to investigate further in particular case $m=1$ possible relations super Jacobi polynomials with representation theory of the Lie supergroup $OSP(2,2n)$. The main result of the paper can be formulated in the following way. Let us make a substitution $p=t(k+1)$ and take a limit as $k\to-1$ of super Jacobi polynomials. Then we get a new family of polynomials which depends rationally on $t$ and which we will denote by $SJ_{\lambda}(t)$. Let $H(1,n)$ be the set of partitions such that $\lambda_{2}\leq n$. We will call a diagram $\lambda$ singular if $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$ for some $1\leq j\leq n$ and otherwise we will call it regular. The main result of the paper can be formulated in the following way:

$1)$ if diagram $\lambda$ is regular then $SJ_{\lambda}(t)$ does not depend on $t$ and coincides (up to sign) with the supercharacter of irreducible module $L(\lambda)$ (see Theorem 6.6 and Remark 6.8);

$2)$ if diagram $\lambda$ is singular (and $j$ is the same as above) then $SJ_{\lambda}(\lambda^{\prime}_{j})$ is well defined and coincides (up to sign) with the supercharacter of irreducible module $L(\lambda)$ (see Corollary 6.9);

In this paper we use two main properties of super Jacobi polynomials. The first one is that they are eigenfunctions of the deformed Calogero - Moser - Sutherland operator and the second property is that they satisfy the Pieri identity. So instead of calculating the limit of the super Jacobi polynomials we calculate the limit of the CMS operator (which is trivial) and the limit of the coefficients of the Pieri formulae. Our main tool is translation functors which were defined in the paper [5].

2. Super Jacobi Polynomials

In this section we define super Jacobi polynomials using the fact that they satisfy the Pieri formula and that they are eigenfunctions of the deformed Calogero - Moser - Sutherland operator. We will always suppose in this paper that $q=0$. The deformed CMS operator of type $BC_{m,n}$ has the following form (see [3], page 1712)

[TABLE]

where $\partial_{x_{i}}=x_{i}\frac{\partial}{\partial{x_{i}}},\,\partial_{y_{j}}=y_{i}\frac{\partial}{\partial{y_{j}}}$ and the parameters $k,p,q,r,s$ satisfy the relations $p=kr,\quad 2q+1=k(2s+1).$ In the formulae below we always suppose that $h=-km-n-\frac{1}{2}p-q$ where $m,n$ are non negative integer numbers. In order to define coefficients of the Pieri formulae let us introduce the following notations: let $H(m,n)$ be the set of partitions $\lambda$ such that $\lambda_{m+1}\leq n$ and

[TABLE]

where $\lambda^{\prime}$ denote the conjugate partition to $\lambda$ and $\Box$ denote the box $(ij)$. Let us set

[TABLE]

Let us also denote by $S^{+}(\lambda)$ the set of partitions $\mu$ which can be obtained from $\lambda$ by adding one box and by $S^{-}(\lambda)$ we will denote the set of partitions $\mu$ which can be obtained from $\lambda$ by deleting one box. Let us also set $S(\lambda)=S^{+}(\lambda)\cup S^{-}(\lambda)\cup\{\lambda\}.$ If $\mu\in S(\lambda)^{+}$ and $\mu_{i}=\lambda_{i}+1$ then we set

[TABLE]

[TABLE]

If $\mu\in S^{-}(\lambda)$ and $\mu_{i}=\lambda_{i}-1$ then we set

[TABLE]

[TABLE]

and

[TABLE]

We also need some more simple expression for coefficient $a_{\lambda,\mu}$. We have $a_{\lambda,\mu}=a^{(1)}_{\lambda,\mu}a_{\lambda,\mu}^{(2)}$ where $a^{(1)}_{\lambda,\mu}$ does not depend on $p$ and $(i,j)$ is the added box and

[TABLE]

Let $\mu\in S^{-}(\lambda)$ then we have (where $(ij)$ is the deleted box)

[TABLE]

Let $P_{n,m}=\mathbb{C}[x^{\pm 1}_{1},\dots,x^{\pm 1}_{m},y^{\pm 1}_{1},\dots,y^{\pm 1}_{n}]$ be the algebra of Laurent polynomials in $m+n$ indeterminates. Now we are ready to define super Jacobi polynomials.

Theorem 2.1**.**

Let $ak+bh+c\neq 0$ for any $a,b,c\in\mathbb{Z}$. Then there exists a unique family of polynomials $J_{\lambda}=J_{\lambda}(x,y,k,p,q)\,\in P_{n,m},\lambda\in H(m,n)$ such that:

[TABLE]

where $p_{1}=x_{1}+x_{1}^{-1}+\dots+x_{m}+x_{m}^{-1}+k^{-1}(y_{1}+y_{1}^{-1}+\dots+y_{n}+y_{n}^{-1}).$

Proof.

Let $ak+bh+c\neq 0$ for any $a,b,c\in\mathbb{Z}$. Then it is not difficult to verify that from the conditions $c_{\nu}=c_{\mu}$ and $\mu,\nu\in S(\lambda)$ it follows that $\mu=\nu$. Therefore the operator

[TABLE]

is well defined. So if a family of polynomials $\{J_{\lambda}\}$ satisfy the conditions of the Theorem then from the last formula in (3) it follows that

[TABLE]

and uniqueness can be proved by induction on the number of boxes in $\mu$. Existence follows from [3] section 7. ∎

3. Translation functors

Introduce some linear transformations which we call translation functors. Let $V$ be the linear span of the super Jacobi polynomials. Then we have the decomposition

[TABLE]

in other words $V_{i}$ is the eigenspace of the of the operator $\mathcal{L}$ corresponding to the eigenvalue $i$. Let us denote by $P_{i}$ the projector onto subspace $V_{i}$ with respect to the above decomposition and define the linear transformation

[TABLE]

Proposition 3.1**.**

Let $f\in V_{j}$ and suppose that $f$ has no poles at $(-1,0,0)$. Then the same is true for $F_{i}(f)$ for any $i\in\mathbb{Z}$.

Proof.

We have

[TABLE]

where $W_{i},W_{i_{1}},W_{i_{2}},\dots,W_{i_{r}}$ are finite dimensional subspaces in $V,V_{i_{1}},\dots,V_{i_{r}}$. Let

[TABLE]

and set

[TABLE]

Then operator $\mathcal{D}_{1}=f_{1}(\mathcal{L})$ acts as zero in $W_{i_{1}}$ and as a diagonal operator in $W_{i}$ with diagonal elements $d_{\lambda}=f_{1}(c_{\lambda}),\lambda\in S$. Now having in mind Cayley-Hamilton theorem we can define

[TABLE]

where $\sigma_{1},\dots,\sigma_{N}$ stand for the elementary symmetric polynomials in $d_{\lambda},\lambda\in S$. From our assumptions we see that $\sigma_{N}\neq 0$ when $k=-1,p=0$. We see that $\mathcal{C}_{1}(W_{i_{1}})=0$ and by the Cayley-Hamilton theorem $\mathcal{C}_{1}$ acts as the identity in $W_{i}$.

In the same way we can construct operators $\mathcal{C}_{2},\dots\mathcal{C}_{r}$ and define

[TABLE]

Let

[TABLE]

be the decomposition according to (6). Applying to both sides of this equality the operator $\mathcal{C}$ we get

[TABLE]

But $\mathcal{C}$ is a differential operator with coefficients that have no poles at $k=-1,p=0,q=0$, so both sides must be regular at this point. ∎

From now on we suppose that $m=1$. The following formulae give eigenvalue of super Jacobi polynomial $J_{\lambda}$.

[TABLE]

where $\tilde{c}_{\lambda}$ is the value of $c_{\lambda}$ for $k=-1,p=q=0$. For convenience we will use below a notation $F_{i}=F_{\lambda}$ if $\tilde{c}_{\lambda}=i$.

Lemma 3.2**.**

Let $\mu,\nu\in S(\lambda)$ and $\mu\neq\nu$. Then $\tilde{c}_{\mu}=\tilde{c}_{\nu}$ if and only if the following conditions are fulfilled

[TABLE]

where $\Box=(i,j),\quad\tilde{\Box}=(\tilde{i},\tilde{j})$.

Proof.

The conditions $\tilde{c}_{\mu}=\tilde{c}_{\nu}$ is equivalent to the following equation

[TABLE]

Therefore two cases are possible: $|\mu|-|\nu|=0$ or $|\mu|-|\nu|=2$. Let us consider the first case. Then we have

[TABLE]

besides

[TABLE]

or in the equivalent form $j-\tilde{j}-(i-\tilde{i})=0$ or $j-i=\tilde{j}-\tilde{i}$. This means that boxes $\Box,\tilde{\Box}$ can be added to $\lambda$ and both are located on the same diagonal. Therefore $\Box=\tilde{\Box}$. In the same way we can consider the case $\mu=\lambda\setminus\square,\,\,\nu=\lambda\setminus\tilde{\square}.$

Let us consider the second case $|\mu|-|\nu|=2$. So $\mu=\lambda\cup\square$ and $\nu=\lambda\setminus\tilde{\Box}$ and we also have

[TABLE]

and

[TABLE]

So we have $j-i+\tilde{j}-\tilde{i}+2-2n=1.$ ∎

We need some combinatorics related to translation functors.

Definition 3.3**.**

Let $\lambda,\mu\in H(1,n)$. Let us set

[TABLE]

Definition 3.4**.**

A diagram $\lambda\in H(1,n)$ is called singular if there exists $1\leq j\leq n$ such that $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$. Otherwise the diagram $\lambda$ is called regular.

Definition 3.5**.**

Let $\lambda$ be a singular diagram and $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$. Define the number

[TABLE]

and denote by $\lambda^{\sharp}$ the diagram which can be obtained from $\lambda$ by deleting $r(\lambda)$ boxes from the first row and $r(\lambda)$ boxes from the row of index $\lambda^{\prime}_{j}$.

Definition 3.6**.**

Let $\lambda_{1}\geq n$. Let us define by induction the set $\pi_{\lambda}$: If $\lambda_{1}\leq n$ then $\pi_{\lambda}=\{\lambda\}$; if $\lambda_{1}>n$ then $\pi_{\lambda}=F_{\lambda}(\pi_{\mu})$, where $\mu$ can is obtained from $\lambda$ by deleting the box from the first row.

Theorem 3.7**.**

The following statements hold true

$1)$* Let $\lambda_{1},\mu_{1}\leq n$ and $\mu$ can be obtained from $\lambda$ by deleting one box then*

[TABLE]

$2)$* Let $\lambda$ be a regular diagram and $\lambda_{1}>n$ and $\mu$ can be obtained from $\lambda$ by deleting one box from the first row then*

[TABLE]

$3)$* Let $\lambda$ is singular $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$ and $\mu$ can be obtained from $\lambda$ by deleting one box from the first row then*

[TABLE]

where $\nu$ can be obtained from $\mu$ by deleting one box from column of index $j$.

$4)$* Let $\mu$ be a singular diagram and $\lambda$ be the diagram which can be obtained from $\mu$ by adding one box to the first row. Then we have*

[TABLE]

$5)$* Let $\lambda_{1}>n$ then*

[TABLE]

Proof.

Let us prove statement $1)$. Clearly $\lambda\in S(\mu)$. If $\nu\in S(\mu)\setminus\{\lambda\}$ and $\tilde{c}_{\lambda}=\tilde{c}_{\nu}$ then by Lemma 3.2 $j-i+\tilde{j}-\tilde{i}=2n-1$. By our assumptions $j,\,\tilde{j}\leq n$ therefore $j-i+\tilde{j}-\tilde{i}-2n+1<0$ and we got a contradiction. The case $\tilde{c}_{\lambda}=\tilde{c}_{\mu}$ implies that $2(\lambda_{1}-1)+1-2n=0$. But this is impossible and we proved the first statement.

Now let us prove statement $2)$. Again as before $\lambda\in S(\mu)^{+}$ and if $\tilde{c}_{\lambda}=\tilde{c}_{\nu}$ then by Lemma 3.2 $\lambda_{1}-1+j-i=2n-1$, where $\nu=\mu\setminus(ij)$. So $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$. If $j>n$ then we have $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j\leq 1+n-j\leq 0.$ this is a contradiction with the condition $\lambda_{1}>n$. Therefore $j\leq n$ but this contradicts regularity and we proved the second statement.

Let us prove statement $3)$. As before if there exists $\nu\in S(\lambda)$ such that $\tilde{c}_{\lambda}=\tilde{c}_{\nu}$ then $\nu=\lambda\setminus(ij)$. But if $\lambda^{\prime}_{j}=\lambda^{\prime}_{j+1}$ then we cannot delete the box from column of index $j$. So we have $F_{\lambda}(\mu)=\{\lambda\}$. And if $\lambda^{\prime}_{j}>\lambda^{\prime}_{j+1}$ we can delete the box from column of index $j$. Therefore in this case we have $F_{\lambda}(\mu)=\{\lambda,\,\nu\}$ and we proved the third statement.

Now let us prove statement $4)$. Let $r=r(\mu)$ then $\mu^{\sharp}$ can be obtained from $\mu$ by deleting the sels

[TABLE]

and it is easy to check that $\tilde{c}_{\mu}=\tilde{c}_{\mu^{\sharp}}$. Therefor we see that $\tilde{c}_{\mu^{\sharp}}=\tilde{c}_{\mu}\neq\tilde{c}_{\lambda}$. So $\mu^{\sharp}\notin F_{\lambda}(\mu^{\sharp})$. Now let $\nu\in S^{+}(\mu^{\sharp})$ then

[TABLE]

Therefore $\mu_{1}=\tilde{j}-\tilde{i}$. Since $\mu_{1}>n$ then $\tilde{j}>n$ and therefore $\tilde{i}=1$ and $\tilde{j}=\mu_{1}^{\sharp}+1$ So we come to equality $\mu_{1}=\mu_{1}^{\sharp}$ which is impossible since $\mu$ is singular. Now consider case when $\nu\in S^{-}(\mu^{\sharp})$. In the same way as before we come to equality

[TABLE]

Therefore $\tilde{j}<j$ and $(\mu^{\sharp})^{\prime}_{\tilde{j}}=\mu^{\prime}_{\tilde{j}}$. So $\tilde{j}=j-1$ and $\mu^{\prime}_{j-1}=\mu^{\prime}_{j}$. But the last equality contradicts to the regularity of $\lambda$. So we have $F_{\lambda}(\mu^{\sharp})=\emptyset$. If $\lambda$ is not regular then in the same way we get $F_{\lambda}(\mu^{\sharp})=\{\mu^{\sharp}\setminus(\mu^{\prime}_{j},j-1)$. And it is easy to check that $\mu^{\sharp}\setminus(\mu^{\prime}_{j},j-1)=\lambda^{\sharp}$ and we proved the forth statement. Now let us prove the statement $5)$ induction on $|\lambda|$. Let $\lambda$ is regular an $\mu$ can be obtained from $\lambda$ by deleting box from the first row. If $\mu$ is regular the by induction $\pi_{\mu}=\{\mu\}$. Therefore by statement $2)$ we have $F_{\lambda}(\pi_{\mu})=\{\lambda\}=\pi_{\lambda}$. If $\mu$ is singular the by induction $\pi_{\mu}=\{\mu.\mu^{\sharp}\}$. Then by statement $2)$ we again have $F_{\lambda}(\mu)=\lambda$ and by statement $4)$ we have $F_{\lambda}(\mu^{\sharp})=\emptyset$. Therefore $F_{\lambda}(\pi_{\mu})=\pi_{\lambda}=\{\lambda\}$. At last consider the case when $\lambda$ is singular. If $\mu$ is regular than by statement $3)$ we have $\pi_{\lambda}=\{\lambda,\nu\}$ and as it is easy to see $\nu=\lambda^{\sharp}$. If $\mu$ is singular then by statement $4)$ we have $\pi_{\lambda}=\{\lambda,\lambda^{\sharp}\}$. Theorem is proved. ∎

Definition 3.8**.**

Let us denote by $X_{\lambda}$ for singular diagram $\lambda$ the set

[TABLE]

Lemma 3.9**.**

Let $\lambda$ be a singular diagram such that $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$. Then set $X_{\lambda}$ consist of $\lambda^{\prime}_{j}+1$ elements.

Proof.

Let us induct on $\lambda^{\prime}_{j}$. If $\lambda^{\prime}_{j}=1$ then $\lambda^{\prime}_{j+1}=\dots=\lambda^{\prime}_{n}=1$ and $r(\lambda)=n-j+1$ and $\lambda_{1}=j-1$. Therefore $\lambda^{\sharp}$ is not singular and $X_{\lambda}=\{\lambda,\lambda^{\sharp}\}$. So we check the first step of induction. Suppose that $\lambda^{\prime}_{j}>1$. Then by definition $(\lambda^{\sharp})^{\prime}_{j}=\lambda^{\prime}_{j}-1$ and Lemma follows from inductive assumption. ∎

4. Nonsingular basis

Definition 4.1**.**

Let $\lambda\in H(1,n)$ then we define by induction the family of polynomials $I_{\lambda}$

[TABLE]

where $\mu$ is the diagram which can be obtained from $\lambda$ by deleting the last box from the first row.

Theorem 4.2**.**

Polynomials $I_{\lambda}$ have no poles at the point $k=-1,p=q=0$.

Proof.

Let us prove the Theorem in the case when $\lambda_{1}\leq n$. Then by definition we have $I_{\lambda}=J_{\lambda}$ and we need to prove that these polynomials are well defined. We will use induction on $|\lambda|$. If $|\lambda|=0$ then $J_{\lambda}=1$ and the Theorem is obviously true. Let $|\lambda|>0$ and $\mu$ be the diagram obtaining from $\lambda$ by deleting the last box from the last row. Then from Pieri formula (3) and by statement $1)$ of Theorem 3.7 we have

[TABLE]

and we only need to prove that $a_{\mu,\lambda}$ has no poles or zeroes at the point $(-1,0,0)$. Let us check that. We have explicit expression (1) for $a_{\mu,\lambda}$ (we need to permute $\mu$ and $\lambda$ in that formula). Case $l(\lambda)=1$ can be easily checked. So suppose that $l(\lambda)=l>1$. We need to verify that all factors in the nominator and denominator of $a^{(2)}_{\mu,\lambda}$ are non zero. Consider for example the product

[TABLE]

We have $(a_{l}+a_{1}+2h-k)(-1,0,0)=\mu_{l}+\mu_{1}-2n+2-l<0$. Therefore for $r>1$ we have

[TABLE]

and for any $r$

[TABLE]

All other factors can be checked in the same manner. So we checked that $a_{\mu,\lambda}$ is well defined at the point $(-1,0,0)$ and is not zero.

And by induction we proved the case when $\lambda\leq n$. The second statement follows from the previous one and the fact that translation functors map regular polynomial to regular polynomial. Theorem is proved. ∎

Now we are going to calculate explicitly the polynomials $I_{\lambda}$ in case $\lambda_{1}>n$.

Lemma 4.3**.**

Let $\lambda_{1}>n$. Then the following formulae hold true

[TABLE]

where

[TABLE]

where $r=r(\lambda)$ and $\lambda^{(0)}$ is the diagram which can be obtained from $\lambda$ by deleting $r$ boxes from the first row and $\lambda^{(s)}$ can be obtained from $\lambda^{(s-1)}$ by deleting one box from the row of index $\lambda^{\prime}_{j}$ for $s=1,2,\dots,r$.

Proof.

First it is not difficult to verify that $a_{\lambda,\mu}=1$ if $\mu$ can be obtained from $\lambda$ by addind box to the first row. Then formula (8) follows from Theorem 4.2 statement $5)$ with some coefficient $b_{\lambda}$. If $\mu$ is the diagram which can be obtained from $\lambda$ by deleting one box from the first row then by Theorem 3.7 statement $4)$ we have $F_{\lambda}(\mu^{\sharp})=\{\lambda^{\sharp}\}$. Therefore $b_{\lambda}=b_{\mu}a_{\mu^{\sharp},\lambda^{\sharp}}$ and we can apply inductive assumption. ∎

Corollary 4.4**.**

Let $\lambda$ be a singular diagram and let us define by induction $\lambda^{s\sharp}=(\lambda^{(s-1)\sharp})^{\sharp}$. Then the following equality holds true

[TABLE]

where $l=\lambda^{\prime}_{j}$.

Proof.

We have

[TABLE]

and Corollary follows. ∎

5. Specialisation

In order to define and calculate explicitly polynomials $SJ_{\lambda}(t)$ we need some additional preliminary results about rational functions. Let $\varphi(k,p)$ be a rational function of the form

[TABLE]

where $\alpha_{i},\beta_{j}$ are linear functions in $k$. We are going to calculate the following rational function

[TABLE]

Let us represent $I=I_{0}\cup I_{1}$ where for $i\in I_{0}$ we have $\alpha_{i}=d_{i}(k+1)$ and for $i\in I_{1}$ $\alpha_{i}$ is not divisible on $k+1$. In the same way let us represent $J=J_{0}\cup J_{1}$ where for $j\in J_{0}$ we have $\beta_{j}=e_{j}(k+1)$ and for $j\in J_{1}$ $\beta_{j}$ is not divisible on $k+1$. So we can represent $\varphi(k,p)=\varphi_{0}(k,p)\varphi_{1}(k,p)$ where

[TABLE]

By definition the function $\varphi_{1}(k,p)$ is well defined at the point $(-1,0)$.

Lemma 5.1**.**

‘The following equalities hold true:

$1)$* if $\mid I_{0}\mid<\mid J_{0}\mid$ then $\varphi(t)=\infty$*

$2)$* if $\mid I_{0}\mid>\mid J_{0}\mid$ then $\varphi(t)=0$*

$3)$* if $\mid I_{0}\mid=\mid J_{0}\mid$ then*

[TABLE]

Proof.

A proof easily follows from the definitions. ∎

Remark 5.2**.**

Suppose that function $\varphi(-1,p)=f(p)$ is well defined and we know it $f(p)=p^{c}g(p)$ where $g(p)$ does not have zeros or poles at $p=0$. Therefore we see that $c=\mid I_{0}\mid-\mid J_{0}\mid$, and if $c=0$, then $\varphi_{1}(-1,0)=g(-1)$ and

[TABLE]

Remark 5.3**.**

It is easy to check that

[TABLE]

For a function $F(k)$ let us define $\tilde{F}=\lim_{k\rightarrow-1}F(k)$. The following Theorem was proved in [6].

Theorem 5.4**.**

The following statements hold true

$1)$* If $\mu\in S^{+}(\lambda)$ then $\tilde{a}_{\lambda,\mu}=1$,*

$2)$* if $\mu\in S^{-}(\lambda)$ and $\mu_{i}=\lambda_{i}-1$ then*

[TABLE]

[TABLE]

$3)$**

[TABLE]

Lemma 5.5**.**

Let $\lambda,\mu$ be such diagrams that $\mu=\lambda\setminus(i,j),\,\,1\leq j\leq n$.

$1)$* If $\lambda_{1}>n,$ and $\mu_{1}-n=\mu^{\prime}_{r}+n-r,\,\lambda^{\prime}_{r}>1$ for some $1\leq r\leq n$ then*

[TABLE]

$2)$* If $\lambda_{1}\leq n$ and $i=1$ then*

[TABLE]

Proof.

Let us prove statement $1)$. By Lemma 5.1 we need to calculate $(a_{\lambda,\mu})(-1,0,0)$ and numbers $d,e$. But $(a_{\lambda,\mu})(-1,0,0)=1$ by Theorem 2 from [6]. Let us consider all factors in $a_{\lambda,\mu}$ which depend on $p$. Let $(i,j)$ be the box such that $\mu=\lambda\setminus(i,j)$. Then we have $i=\lambda^{\prime}_{j},\,j=\lambda_{i}$. Since $i>1$ then it is not difficult to verify that $a_{\lambda,\mu}$ has one pole at $a_{i}+a_{1}+2h$ and one zero at $a_{1}+j+k\lambda^{\prime}_{j}+2h-k-1$. Therefore we have

[TABLE]

If $r>j$ then at the point $(-1,0,0)$ we have

[TABLE]

Therefore at the same point we have

[TABLE]

Further at the point $(-1,0,0)$ we have

[TABLE]

Therefore at the same point we have

[TABLE]

Also at the point $(-1,0,0)$ we have

[TABLE]

Therefore at the same point we have

[TABLE]

And it is also easy to check that all other factors strictly less then zero. Therefore we proved the first statement. The second statement can be proved in the same manner. ∎

Lemma 5.6**.**

Let $\lambda$ be a singular diagram such that $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$ then

[TABLE]

Proof.

It easily follows from the previous Lemma. ∎

Definition 5.7**.**

For any $t\in\mathbb{C}$ and $\lambda\in H(1,n)$ let us define

[TABLE]

Corollary 5.8**.**

Let $\lambda\in H(1,n)$. Then

$1)$* $SI_{\lambda}(t)$ does not depend on $t$*

$2)$* If $\lambda$ is a regular diagram then $SJ_{\lambda}(t)$ does not depend on $t$.*

$2)$* If $\lambda$ is a singular diagram then for $t\notin\mathbb{Z}_{\geq 0}$ the polynomial $SJ_{\lambda}(t)$ is well defined and we have the following equality*

[TABLE]

where $l=\lambda^{\prime}_{j}$.

Proof.

Let us prove the first statement. Polynomial $I_{\lambda}$ is well defined at the point $(-1,0,0)$. Therefore in the notations of the Lemma 5.1 for any its coefficient $\varphi(k,p,q)$ we have $J_{0}=\emptyset$. Therefore we always have $|I_{0}|=|J_{0}|$ or $|I_{0}|>|J_{0}|$. So by Lemma 5.1 we have $\varphi(t)=const$ in the first case and $\varphi(t)=0$ in the second case and we proved the first statement.

By Theorem 4.2 and from the Definition 4.1 it follows that for regular $\lambda$ we have $J_{\lambda}=I_{\lambda}$ and we get the second statement.

The third statement follows from the Corollary 4.4 and Lemma 5.6.

∎

Let us denote by $SJ_{\lambda}(\infty)$ the limit $SJ_{\lambda}(t)$ when $t\to\infty$ and from the Corollary 5.8 we see that

[TABLE]

And for singular diagram we have

[TABLE]

From the previous formulae it is also easy to deduce that

[TABLE]

6. Supercharacters

Let $\pm\varepsilon,\pm\delta_{1},\dots,\pm\delta_{n}$ be the weights of identical representation of the Lie superalgebra $\mathfrak{osp}(2,2n)$. The root system of the Lie superalgebra $\mathfrak{osp}(2,2n)$ is

[TABLE]

with the bilinear form

[TABLE]

The Weyl group $W_{0}$ is semi-direct product of $S_{n}$ and $\mathbb{Z}_{2}^{n}$. It acts on the wights by permuting and changing the signs of $\delta_{j},\,j=1,\dots,n$. Let us chose the following system of simple roots

[TABLE]

We will consider only integer weights

[TABLE]

and the subset of the highest weights

[TABLE]

We will denote for any $\chi\in P^{+}$ by $L(\chi)$ the corresponding finite dimensional irreducible module and by $K(\chi)$ the corresponding Kac module (see ([7])). We also set

[TABLE]

[TABLE]

A highest weight $\chi$ is called typical if $(\chi+\rho,\alpha)\neq 0$ for any $\alpha\in R^{+}_{1}$. A highest weight $\chi$ is called atypical if $(\chi+\rho,\alpha)=0$ for some $\alpha\in R^{+}_{1}$. In the case of Lie superalgebra $\mathfrak{osp}(2,2n)$ there is at most one such $\alpha\in R^{+}_{1}$. We need the following formula for supercharacter of irreducible module $L(\chi)$ by Van Der Jeugt [7]. Let us denote by $\{f\}$ the alternation operation over $W_{0}$.

For any $\chi\in P$ we set

[TABLE]

and if $(\chi+\rho,\alpha)=0$ then we set

[TABLE]

Lemma 6.1**.**

(see [7]) The following equalities hold true:

$1)$* If $\chi\in P^{+}$ is typical then*

[TABLE]

$2)$* If $\chi\in P^{+}$ is atypical such that $(\chi+\rho,\alpha)=0,\,\alpha\in R^{+}_{1}$ then*

[TABLE]

Lemma 6.2**.**

The following statements hold true:

$1)$* If $(\chi+\rho,\varepsilon+\delta_{j})=(\chi+\rho,\varepsilon+\delta_{j+1})=0$ then*

[TABLE]

$2)$* If $(\chi+\rho,\varepsilon+\delta_{j})=(\chi+\rho,\varepsilon-\delta_{j})=0$ then*

[TABLE]

$3)$* if $(\chi+\rho,\varepsilon-\delta_{j})=(\chi+\rho,\varepsilon-\delta_{j+1})=0$ then*

[TABLE]

Proof.

Let us prove the first statement. Denote by $A$ the following expression

[TABLE]

Then we have

[TABLE]

[TABLE]

And from the conditions of the Lemma it is easy to see that expression in brackets is symmetric with respect to transposition $(j,j+1)$. Therefore the result of alternation is zero.

Let us prove the second statement. Denote by $B$ the following expression

[TABLE]

Then we have

[TABLE]

[TABLE]

And from the conditions of the Lemma it is easy to see that expression in brackets is symmetric with respect to transformation $\delta_{j}\rightarrow-\delta_{j}$. Therefore the result of alternation is zero.

Let us prove the third statement. Denote by $A$ the following expression

[TABLE]

Then we have

[TABLE]

[TABLE]

And from the conditions of the Lemma it is easy to see that expression in brackets is symmetric with respect to transposition $(j,j+1)$. Therefore the result of alternation is zero. Lemma is proved. ∎

For any diagram $\lambda\in H(1,n)$ let us define the highest weight $\chi_{\lambda}$ by the formula

[TABLE]

where $\mu$ is the diagram obtaining from $\lambda$ by deleting the first row.

Proposition 6.3**.**

Let $\lambda$ be a singular diagram then

[TABLE]

where for any for diagram $\lambda\in H(1,n)$ $s(\lambda)=\lambda_{2}+\lambda_{3}+\dots$.

Proof.

Let us consider two cases. First one is when $\lambda^{\prime}_{j}>1$ and the second case is when $\lambda^{\prime}_{j}=1$. Consider now the first case. Let us temporary denote $\chi_{\lambda}=\chi$. It is easy to check that

[TABLE]

Therefore

[TABLE]

If $\chi-\varepsilon-\delta_{j}\in P^{+}$ then $\chi-\varepsilon-\delta_{j}=\chi_{\lambda^{\sharp}}$ and we proved the proposition. If $\chi-\varepsilon-\delta_{j}\notin P^{+}$ then

[TABLE]

and by previous Lemma we have $\{L_{\chi-\varepsilon-\delta_{j}}\}=-\{L_{\chi-2\varepsilon-\delta_{j}-\delta_{j+1}}\}.$ And we repeat this procedure until we arrive to $\chi_{\lambda^{\sharp}}$. Besides it is easy to see that $r=s(\lambda)-s(\lambda^{\sharp})$.And we prove the Proposition in the first case.

Now let us consider the second case. It is easy to check that

[TABLE]

As before using the first statement of Lemma 6.1 we get

[TABLE]

Then using the second statement of the same Lemma we get

[TABLE]

and by the third statement of the same Lemma we get

[TABLE]

Proposition is proved. ∎

Corollary 6.4**.**

If $\lambda$ is singular and $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j$ then the following equality hold true

[TABLE]

[TABLE]

where $l=\lambda^{\prime}_{j}$.

Corollary easily follows from the Proposition 6.3.

In order to connect super Jacobi polynomials with representation theory we need to consider the outer automorphism $\theta$ (see [8]) of the Lie superalgebra $\mathfrak{osp}(2,2n)$ which acts on weights the by $\theta(\varepsilon)=-\varepsilon$ and $\theta(\delta_{j})=\delta_{j},\,j=1,\dots,n$.

Definition 6.5**.**

Let $\lambda\in H(1,n)$ then we set

[TABLE]

and

[TABLE]

We will also consider supercharacters as polynomials in indeterminates

[TABLE]

Theorem 6.6**.**

Polynomials $sch\,E(\lambda)$ satisfy the following Pieri identity

[TABLE]

where

[TABLE]

Proof.

We need a formula for supercharacter of irreducible module over Lie superalgebra $\mathfrak{osp}(2,2n)$ in case $\lambda_{1}\leq n$. Let use denote $u=x+x^{-1}$ and $v_{j}=y_{j}+y^{-1}_{j},\,j=1,\dots,n$. Then by [1] Proposition 3.1. the following formulae holds true

[TABLE]

where $L_{0}=\prod_{i<j}(v_{i}-v_{j})$ and the brackets $\left\{\right\}$ mean the alternation over the Weyl group $S_{n}\ltimes Z_{2}^{n}$.

Let us denote by $B_{\lambda}$ the expression in the brackets in the formula (16). Then we have

[TABLE]

It is easy to see that

[TABLE]

[TABLE]

and for $j=\lambda_{1}$ and $\mu^{\prime}_{j}>0$ we have

[TABLE]

If $\mu^{\prime}_{j}=0$ Then we have

[TABLE]

and

[TABLE]

Therefore

[TABLE]

Now let us consider the case $\lambda_{1}\geq n$. In this case a proof much easy since the formula for Kac module is more simple. The condition $d_{\lambda,\lambda}=0$ follows from the fact that characters of $E(\Box)$ has no constant term. The condition $d_{\lambda,\mu}=0$ follows from the fact that $\theta(sch\,K(\chi_{\lambda}))=sch\,K(\chi_{\mu})$. In the third case condition $d_{\lambda,\mu}=1$ follows from the fact that $\theta(K(\chi_{\mu})=K(\chi_{\mu})$. ∎

Corollary 6.7**.**

$SJ_{\lambda}(\infty)=(-1)^{s(\lambda)}sch\,E(\lambda)$**

Proof.

Let us take limit as $p\to 0$ in the formulae of the Theorem 5.4 then we get

[TABLE]

Therefore $(-1)^{s(\lambda)}sch\,E(\lambda)$ satisfy the same Pieri formulae as $J_{\lambda}(\infty)$ and Corollary follows. ∎

Remark 6.8**.**

Conditions $\lambda_{1}>n$ and $\lambda_{1}-n\neq\lambda^{\prime}_{j}+n-j$ for any $1\leq j\leq n$ are equivalent to the typicality of $\chi_{\lambda}$. Therefore under such conditions $K(\chi_{\lambda})$ is irreducible module over Lie superalgebra $\mathfrak{osp}(2,2n)$ and module $L(\lambda)=K(\chi_{\lambda})\oplus\theta(K(\chi_{\lambda}))$ is irreducible module over Lie supergroup $OSP(2,2n)$.

We also should note that from the formula (12) and Corollary 6.7 it follows that

[TABLE]

Therefore from the BGG duality in this case (see [9]) it follows that up to sign the polynomials $SI_{\lambda}$ coincide with supercharacters of the projective coves of irreducible finite dimensional modules over supergroup $OSP(2,2n)$.

Corollary 6.9**.**

Let $\lambda\in H(1,n)$ such that $\lambda_{1}-n=\lambda^{\prime}_{j}+n-j,\,1\leq j\leq n$. Then

[TABLE]

Proof.

The equality follows from Corollary 6.4 and Corollary 6.7. ∎

7. Acknowledgements

This work by A.N. Sergeev was supported by Russian Ministry of Education and Science (grant 1.492.2016/1.4) (sections 2,3,4) and by the Russian Academic Excellence Project ’5-100’ (sections 5,6).