Information about the behavior of zeros of classical families of multiple or Hermite–Padé orthogonal polynomials as functions of the intrinsic parameters of the family is scarce. We establish the interlacing properties of the zeros of Angelesco-Jacobi polynomials when one of the three main parameters is increased by $1$, extending the work of [5]. We also show their monotonicity with respect to (large values) of the parameter $\gamma$, representing in the electrostatic model of the zeros the size of the positive charge fixed at the origin, as well as monotonicity with respect to the endpoint of the interval of orthogonality. These results are extended to zeros of multiple Jacobi-Laguerre and Laguerre-Hermite polynomials using asymptotic relations between these families.

§ INTRODUCTION

The Jacobi polynomials $P_n^{(\alpha,\beta)}$ can be defined using the  Rodrigues formula,

\begin{equation}%\label{RodrJac}
\label{Rodrigues}
P_n^{(\alpha,\beta)} (z)=\frac{1}{2^n n!} (z-1)^{-\alpha}
(z+1)^{-\beta} \left( \frac{d}{dz} \right)^n \left[
(z-1)^{n+\alpha} (z+1)^{n+\beta}\right].
\end{equation}

For $\alpha, \beta>-1$, they are orthogonal on $[a,b]=[-1,1]$ with respect to the weight

w_{\alpha,\beta}(x)=(1-x)^\alpha (1+x)^\beta, %, $\alpha, \beta>-1

that is,

\int_{-1}^1 P_n^{(\alpha,\beta)}(x) x^k w_{\alpha,\beta}(x)\, dx=0, \quad k=0, 1, \dots, n-1.

In consequence, all zeros

\begin{equation}
\label{zerosJacobi}
-1< x_{n,1}(\alpha,\beta)< \dots < x_{n,n}(\alpha,\beta)<  1
\end{equation}

of $P_n^{(\alpha,\beta)}(x)$ are real, simple, and belong to $(-1,1)$. Their electrostatic interpretation, found by Stieltjes in 1885, is an elegant result that offers intuition about their behavior as functions of the parameters $\alpha, \beta$. Namely, zeros (<ref>) provide a unique minimizer of the logarithmic energy

\begin{equation*}
E_{total}(X)=E_{mutual}(X)+\sum_{x\in X}\varphi(x), \quad X=\{x_1,\dots,x_n\} \subset [-1,1]^n,
\end{equation*}

among all $n$ point configurations on $[-1,1]$, where

\begin{equation*}
E_{mutual}(X):=\sum_{1\leq k<j\leq n}\log\frac{1}{\lvert x_k - x_j\rvert},
\end{equation*}

and the external field $\varphi(x)$ acting on $X$ is given by two fixed charges of mass $(\beta+1)/2$ and $(\alpha+1)/2$ at $-1$ and $1$, respectively:

\begin{equation*}
\varphi(x)=\frac{\alpha+1}{2}\log\frac{1}{\lvert x-1\rvert} + \frac{\beta+1}{2}\log\frac{1}{\lvert x+1\rvert}.
\end{equation*}

Since we conclude that the size of $\alpha$ (resp., $\beta$) is proportional to the strength of repulsion of the end point $1$ (resp., $-1$), it is a basis for conjecture that each zero $x_{n,j}(\alpha,\beta)$ should behave monotonically with respect to $\alpha$ and $\beta$. This intuition was rigorously substantiated by Szegő in <cit.>:

Given $n\in\mathbb{N}$, the zeros (<ref>) of $P_n^{(\alpha,\beta)}(x)$ decrease with respect to $\alpha$ and increase with respect to $\beta$.

The monotonicity with respect to the parameters can also be recast in terms of interlacing of zeros of polynomials from the same family. In what follows, we denote by $\P_n(\R)$ the family of real–rooted algebraic polynomials of
degree $n$.
Let $(a,b)\subset \R$, $p$ and $q$ two polynomials, and let $ a<\lambda_1(p) < \cdots < \lambda_k(p)<b$ and $a<\lambda_1(q)< \cdots< \lambda_{j}(q)<b$ be the zeros of $p$ and $q$, respectively, all simple, that that belong to $(a,b)$. We say that $q$ interlaces $p$ on $(a,b)$, and denote it by $p \prec q$ on $ (a,b)$, if  $j=k$  and

\begin{equation}\label{SamedegreeInter}
\lambda_1(p) < \lambda_1(q) < \lambda_2(p) < \lambda_2(q) <\cdots < \lambda_k(p) <\lambda_k(q),
\end{equation}

or if $j=k-1$  and

\begin{equation}\label{LessdegreeInter}
\lambda_1(p) < \lambda_1(q) < \lambda_2(p) < \lambda_2(q) < \cdots <  \lambda_{n-1}(p) < \lambda_{k-1}(q) < \lambda_k(p).
\end{equation}

Moreover, if $(a,b)=\R$, we simplify the notation by simply writing $ p \prec q$.

It is clear that the monontonicity of zeros of Jacobi polynomials (Theorem <ref>) is equivalent to interlacing of families like $P_n^{(\alpha,\beta)}$ and $P_n^{(\alpha+t,\beta)}$ for some range of $t>0$ (and similarly, for the parameter $\beta$). A result is this direction was found in [6]:

Let  $\alpha,\beta>-1$ and $t\in (0,2)$. Then

\begin{equation*}
P_n^{(\alpha+t,\beta)}(x)\prec P_n^{(\alpha,\beta)}(x), \qquad  P_n^{(\alpha,\beta)}(x)\prec P_n^{(\alpha,\beta+t)}(x).
\end{equation*}

The Laguerre polynomials $L_n^{(\gamma)}$ can be defined by their explicit representation in terms of the terminating generalized hypergeometric series,

L_n^{(\gamma)}(x)=\frac{(\gamma+1)_n}{n !}{ }_1 F_1(-n ; \gamma+1 ; x),

and for $\gamma >-1$ satisfy the orthogonality relations

\int_0^{\infty} x^j L_n^{(\gamma)}(x)  x^\gamma e^{-x} \, d x \\
=0, \quad j=0, 1, \dots, n-1.

The Hermite polynomials $H_n$ is the third classical family of polynomials; they satisfy the orthogonality relations

\int_{\mathbb{R}} x^j  H_n(x) e^{-x^2} d x=0,  \quad j=0, 1, \dots, n-1,

see <cit.>. There are well-known limit relations that connect all three classical families; see, e.g. <cit.>.

Multiple (or Hermite–Padé) polynomials are a natural generalization of the classical notion of orthogonal polynomials. Typically, they come in two flavors, but in this paper we focus on the so-called Type II polynomials. To define them, we need a multiindex $\Vec{n}=\left( n_1,\dots,n_r\right)\in \mathbb N^r$ and a $r$-tuple of positive Borel measures $\left\{ \mu_j\right\}_{j=1}^r$, with an additional assumption that all the moments

\int |x|^k \, d\mu_j(x), \quad j=1,\dots, r, \quad k\ge 0,

exist. A monic polynomial $P_{\Vec{n}}(x)$ of degree $\left|\Vec{n} \right |:= n_1+\cdots+n_r$, if it exists, is called a Type II multiple orthogonal polynomial if it satisfies the following orthogonality conditions:

\begin{equation*}
\int x^k P_{\Vec{n}}(x)d\mu_j(x)=0 \quad \text{for} \quad k=0,1,\dots,n_j-1 \quad \text{and} \quad j=1,\dots,r .
\end{equation*}

In general, existence of such a polynomial cannot be guaranteed  for all multi-indices $\Vec{n}$ (in which case we say that the system of measures $\left\{ \mu_j\right\}_{j=1}^r$ is perfect). One of the simplest examples of a perfect system is the so-called Angelesco case, when the convex hull of the support of each $\mu_j$ is a closed interval $I_j\subset \mathbb R$ such that the interiors of $\left\{ I_j\right\}_{j=1}^r$ are pairwise disjoint. If measures $\mu_j$ are additionally classical weights of orthogonality, the corresponding systems receive the name of multiple Jacobi-Jacobi (or Angelesco-Jacobi), Jacobi-Laguerre and Laguerre-Hermite orthogonal polynomials; see [8] and the definitions below for details. As their classical counterparts, they may also depend on a number of parameters.

One of the features of the Angelesco system is that all zeros in $P_{\Vec{n}}(x)$ are real and simple, and each interval $I_j$ contains exactly $n_j$ of these zeros. However, their electrostatic model, in the spirit of Stieltjes' work, was given only recently, in [10]. Unlike the standard case, the model involves charges of opposite sign, some of them located at the zeros of the so-called electrostatic partner $S$ of $P_{\Vec{n}} $, a polynomial that can be constructed using an explicit formula. In the Angelesco case, the zeros of $S_j$ partially interlace with those of $P_{\Vec{n}} $ (and the location of the zeros corresponds to a critical point of the gradient of the vector energy, rather than a global minimum). The complexity of this vector equilibrium makes the analysis of the behavior of the zeros of $P_{\Vec{n}} $ as functions of the parameters of the weight highly nontrivial.

In this paper, we will present results about interlacing and monotonicity of the zeros a the multiple Angelesco-Jacobi, Jacobi-Laguerre and Laguerre-Hermite orthogonal polynomials. All main results are collected in Section <ref>. Section <ref> contains some structure relations for Angelesco-Jacobi polynomials, as well as the proof of the interlacing properties of their zeros. The behavior of the arithmetic and geometric means of these zeros is proven in Section <ref>, and these results are used in the proof of the monotonicity properties in Sections <ref> and <ref>. Finally, the extension of some of these results to the zeros of the Jacobi-Laguerre and Laguerre-Hermite polynomials is established in Section <ref>.

§ MAIN RESULTS

Angelesco-Jacobi, known also as Jacobi-Jacobi polynomials (see [1]), are Hermite–Padé polynomials $P_{\vec n}$, $\vec n=(n_1, n_2)$, satisfying orthogonality relations

\begin{equation*}
\begin{split}
&\int_{a}^{0} x^{k} P_{\vec n}(x)(1-x)^{\alpha}(x-a)^{\beta}|x|^{\gamma}  \, d x=0, \quad k=0,1,2, \dots, n_1-1, \\
&\int_{0}^{1} x^{k} P_{\vec n}(x)(1-x)^{\alpha}(x-a)^{\beta}x^{\gamma} \, d x=0, \quad k=0,1,2, \dots, n_2-1,
\end{split}
\end{equation*}

with $a<0$ and $\alpha,\beta,\gamma>-1$. To make the dependence on the parameters explicit, we denote these polynomials by $P_{\vec n}^{(\alpha,\beta,\gamma)}(x;a)$.[  Notice that the role of the parameters $\alpha, \beta, \gamma$ can differ in the literature.] Thus, this is a type II multiple orthogonal polynomial for the Angelesco system  given by the same weight function $\omega (x)=(1-x)^\alpha(x-a)^\beta\abs{x}^\gamma$ on the intervals $[a,0]$ and $[0,1]$.

We will consider the diagonal case

\begin{equation}
\label{diagcase}
\vec n = (n, n);
\end{equation}

to simplify notation, in what follows and under assumption (<ref>), we will write $P_{  n}^{(\alpha,\beta,\gamma)}(x;a)$ instead of $P_{\vec n}^{(\alpha,\beta,\gamma)}(x;a)$, keeping in mind that $\deg P_{  n}^{(\alpha,\beta,\gamma)}(x;a)=2n$.

These polynomials can be defined via a Rodrigues formula

\begin{equation}
\label{RodriguezForm}
\begin{split}
(1-x)^\alpha(x-a)^\beta x^\gamma c_n(\alpha,\beta,\gamma) & P^{(\alpha,\beta,\gamma)}_{n}(x;a)\\
& =\frac{(-1)^n}{n!}\left( \frac{d}{dx} \right)^n\left[(1-x)^{\alpha+n}(x-a)^{\beta+n}x^{\gamma+n}\right]
\end{split}
\end{equation}

\begin{equation}
c_n(\alpha,\beta, \gamma):=\binom{3n+\alpha+\beta+\gamma}{n};
\end{equation}

here and in what follows, we use the generalized definition of the binomial coefficients,

\binom{a}{k} := \frac{\Gamma(a+1)}{\Gamma(a-k+1) k! }, \quad k\in \mathbb N \cup \{0\}, \quad  a> k.

Among consequences of (<ref>) are the raising operator identity

\begin{equation}\label{raisingJA}
\begin{split}
(2n+\alpha+&\beta+\gamma+1) (1-x)^\alpha  (x-a)^\beta x^\gamma   P^{(\alpha,\beta,\gamma)}_{n}(x;a)\\
& =- \frac{d}{dx} \left[(1-x)^{\alpha}(x-a)^{\beta}x^{\gamma} P^{(\alpha+1,\beta+1,\gamma+1)}_{n-1}(x;a)\right],
\end{split}
\end{equation}

and an explicit expression (see [8]),

\begin{equation}
\label{explicitJA}
c_n(\alpha,\beta,\gamma) P_{n}^{(\alpha,\beta,\gamma)}(x;a)=\sum_{k=0}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta,\gamma) (x-1)^{n-k}x^{k+j}(x-a)^{n-j},
\end{equation}

where for $0\le j \le n-k$ and $  0\le k \le n$,

\begin{equation}
\label{explicitCoefficientsJA}
d^{(n)}_{j,k}( \alpha,\beta,\gamma):=  \binom{n+\alpha}{k}\binom{n+\beta}{j}\binom{n+\gamma}{n-k-j}.
\end{equation}

The zero interlacing of consecutive $ P_n^{(\alpha,\beta,\gamma)}(x;a)$, in the spirit of Theorem <ref>, is not known (for the interlacing of zeros of the nearest neighbor polynomials, see [7]). However, de Santos [5] proved the following result (see <cit.>), that we formulate using Definition <ref>:

Let $\alpha, \beta, \gamma >-1$, $a<0$, and $n\in\N$, $n\ge 2$. Then

\begin{align*}
P_n^{(\alpha,\beta,\gamma)}(x;a)  &\prec  P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \text{ on } (a,0),   \\
P_n^{(\alpha,\beta,\gamma)}(x;a)  &\prec  P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \text{ on } (0,1).
\end{align*}

It is convenient to remind the reader that $P_n$ (resp., $P_{n-1}$) has exactly $n$ (resp., $n-1$) zeros on $(a,0)$ and on $(0,1)$.

A simple observation is that the assertion of this theorem is equivalent to

x (x-a) P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \prec P_n^{(\alpha,\beta,\gamma)}(x;a) .

This result shows that the zeros of two consecutive diagonal Angelesco-Jacobi polynomials interlace at each subinterval if the decrease in degree is accompanied by an increase in the value of each parameter $\alpha, \beta, \gamma$ in $1$, see Figure <ref>.

Graph of the polynomials $P_5^{(-1/2,1,0)}(x;-2)$ (dashed line) and $P_4^{(1/2,2,1)}(x;-2)$ (thick line) on $[-2,1]$.

One of the main results of this paper is the following refinement of Theorem <ref>:

Let $\alpha, \beta, \gamma >-1$,  $a<0$, and $n\in \N$, $n\ge 2$. Then

$  x(x-a)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)\prec P_n^{(\alpha+1,\beta,\gamma)}(x;a) \prec  P_n^{(\alpha,\beta,\gamma)}(x;a)$;

$ P_n^{(\alpha,\beta,\gamma)}(x;a)\prec P_n^{(\alpha,\beta+1,\gamma)}(x;a) \prec x(x-1)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)$;

\begin{align*}
& P_n^{(\alpha,\beta,\gamma+1)}(x;a)  \prec P_n^{(\alpha,\beta,\gamma)}(x;a) \prec P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \quad \text{on } (a,0), \\
& P_n^{(\alpha,\beta,\gamma)}(x;a)  \prec P_n^{(\alpha,\beta,\gamma+1)}(x;a)\prec P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)  \quad \text{on } (0,1).
\end{align*}

Although the electrostatic model for $ P_{n}^{(\alpha,\beta,\gamma)}(x; a)$, mentioned above, involves both positive and negative moving charges and is quite complicated to analyze, it still provides a correct intuition on the behavior of zeros as functions of the parameters $\alpha$ and $\beta$ (representing the size of the positive charges fixed at the endpoints $x=a$ and $x=1$, respectively), as the result found in [5] shows:

The zeros of $ P_{n}^{(\alpha,\beta,\gamma)}(x; a)$ decrease with respect to $\alpha$ and increase with respect to $\beta$. Furthermore, if $\alpha=\beta$ and $a=-1$, then the negative zeros decrease with respect to $\gamma$, and the positive zeros increase with respect to $\gamma$.

This theorem claims the monotonicity of zeros as functions of the parameter $\gamma$ (the “size” of the positive charge at $x=0$) in the simplest symmetric setting. When the symmetry is broken, the electrostatic model allows us to expect monotonicity, at least for large values of $\gamma$, that is, when the charge at the origin outpowers the interaction of individual positive and negative unit charges. This is indeed the case, as the following result shows:

Let $\alpha,\beta>-1$,  $a<0$, and $n\in\mathbb{N}$. Then, for all sufficiently large values of $\gamma$, the negative zeros of $P^{(\alpha,\beta,\gamma)}_{n}(x;a)$ decrease with respect to $\gamma$, and its positive zeros increase with respect to $\gamma$.

We conjecture that the assertion is valid for all values of the parameter $\gamma>-1$.

Finally, although intuitively obvious, we establish monotonicity of all zeros of the Angelesco-Jacobi polynomials as functions of the left endpoint $a$:

Let $n\in\mathbb{N}$,  $a<0$ and  $\alpha, \beta,\gamma>-1$. The zeros of the polynomial $P_{n}^{(\alpha,\beta+1,\gamma)}(x;a)$ are increasing with respect to $a$.

This result, although predictable for the zeros in $(a,0)$, is less obvious for the whole interval in $(a,1)$.

Given $a<0$ and $\beta,\gamma>-1$, the Jacobi-Laguerre polynomials $L_{\vec{n}}^{(\beta,\gamma)}(x;a)$, $\vec n=(n_1, n_2)$,  are Hermite–Padé polynomials that satisfy the orthogonality relations

\begin{equation*}
\begin{split}
&\int_{a}^{0} x^{k} L^{(\beta,\gamma)}_{\vec{n}}(x)(x-a)^{\beta}|x|^{\gamma}e^{-x}  \, d x=0, \quad k=0,1,2, \dots, n_1-1, \\
&\int_{0}^{\infty} x^{k} L^{(\beta,\gamma)}_{\vec{n}}(x)(x-a)^{\beta}x^{\gamma}e^{-x} \, d x=0, \quad k=0,1,2, \dots, n_2-1.
\end{split}
\end{equation*}

The Laguerre-Hermite polynomials  $H_{\vec{n}}^{(\gamma)}(x;a)$, for $\gamma>-1$, satisfy

\begin{equation*}
\begin{split}
&\int_{-\infty}^{0} x^{k} H^{(\gamma)}_{\vec{n}}(x)|x|^{\gamma}e^{-x^2}  \, d x=0, \quad k=0,1,2, \dots, n_1-1, \\
&\int_{0}^{\infty} x^{k} H^{(\gamma)}_{\vec{n}}(x)x^{\gamma}e^{-x^2} \, d x=0, \quad k=0,1,2, \dots, n_2-1.
\end{split}
\end{equation*}

As in the classical case, these families of multiple orthogonal polynomials are connected by the following asymptotic relations:

\begin{align}
\label{HermiteLimit}
H_{\vec{n}}^{(\gamma)}(x;a) & =\lim_{\alpha\to\infty} \alpha^{(n_1+n_2)/2}P_{\vec{n}}^{(\alpha,\alpha,\beta)}(x/\sqrt{\alpha};-1),
\\
\label{LaguerreLimit}
L_{\vec{n}}^{(\beta,\gamma)}(x;a) & =\lim_{\alpha\to\infty}\alpha^{n_1+n_2}P_{\vec{n}}^{(\alpha,\beta,\gamma)}(x/\alpha;a/\alpha).
\end{align}

As for Angelesco-Jacobi polynomials, we will focus on the diagonal case (<ref>), adopting the notation $L_n^{(\beta,\gamma)}(x;a)$ and $ H_n^{(\gamma)}(x)$ instead of $ L_{\vec{n}}^{(\beta,\gamma)}(x;a)$ and $ H_{\vec{n}}^{(\gamma)}(x)$, respectively.

Using the asymptotic relation above, we extend the interlacing result of Theorem <ref> to the Jacobi-Laguerre and Laguerre-Hermite polynomials:

Let $a<0$ and $\beta,\gamma>-1$, the following interlacing holds:

\begin{equation*}
L_n^{(\beta,\gamma)}(x;a)\prec  L_n^{(\beta+1,\gamma)}(x;a) \prec  xL_{n-1}^{(\beta+1,\gamma+1)}(x;a)    .
\end{equation*}

\begin{equation*}
\begin{split}
L_n^{(\beta,\gamma+1)}(x;a) \prec L_n^{(\beta,\gamma)}(x;a) \prec L_{n-1}^{(\beta+1,\gamma+1)}(x;a) \quad & \text{on }   (a,0), \\
L_n^{(\beta,\gamma)}(x;a) \prec L_n^{(\beta,\gamma+1)}(x;a) \prec L_{n-1}^{(\beta+1,\gamma+1)}(x;a)\quad & \text{on } (0,\infty).
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
H_n^{(\gamma+1)}(x) \prec H_n^{(\gamma)}(x)\prec H_{n-1}^{(\gamma+1)}(x) \quad & \text{on } (-\infty,0), \\
H_n^{(\gamma)}(x) \prec H_n^{(\gamma+1)}(x)\prec  H_{n-1}^{(\gamma+1)}(x) \quad & \text{on } (0,\infty).
\end{split}
\end{equation*}

Finally, monotonicity of the zeros of the Jacobi-Laguerre polynomials as functions of $\gamma$ for large values of this parameters also takes place:

Let $\beta>-1$,  $a<0$, and $n\in\mathbb{N}$. Then, for all sufficiently large values of $\gamma$, the negative zeros of $L^{(\beta,\gamma)}_{n}(x;a)$ decrease with respect to $\gamma$, and its positive zeros increase with respect to $\gamma$.

§ PROOF OF THEOREM <REF>

Theorem <ref> is a consequence of the following structure relations for Angelesco-Jacobi polynomials. To the best of our knowledge, they are new and can have an independent interest. To formulate this result, we define for $n\in \N$ and $\alpha, \beta, \gamma>-1$ the coefficients

\begin{equation} \label{defParameters}
\begin{split}
\mathfrak A_n & =  \mathfrak A_n(\alpha, \beta, \gamma):=\frac{2n+\alpha+\beta+\gamma+1}{3n+\alpha+\beta+\gamma+1}>0, \\
\mathfrak  B_n & =\mathfrak B_n(\alpha, \beta, \gamma) := 1- \mathfrak  A_n =\frac{n}{3n+\alpha+\beta+\gamma+1}>0
\end{split}
\end{equation}

(where we omit the dependence on the rest of the parameters when these values are clear).

Let $n\in\mathbb{N}$, $\alpha, \beta,\gamma>-1$, and $\mathfrak A_n   =  \mathfrak A_n(\alpha, \beta, \gamma)$ and $\mathfrak B_n  =  \mathfrak B_n(\alpha, \beta, \gamma)$ be as defined above.  Then

P_n^{(\alpha+1,\beta,\gamma)}(x;a)= \mathfrak A_n P_n^{(\alpha,\beta,\gamma)}(x;a)+\mathfrak B_n x(x-a)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a);

P_n^{(\alpha,\beta+1,\gamma)}(x;a)= \mathfrak A_n P_n^{(\alpha,\beta,\gamma)}(x;a)+ \mathfrak B_n (x-1)xP_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a);

P_n^{(\alpha,\beta,\gamma+1)}(x;a)=\mathfrak A_n P_n^{(\alpha,\beta,\gamma)}(x;a)+\mathfrak  B_n (x-a)(x-1)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a).

From the explicit expression (<ref>) and the identity

\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}

it follows that for $k\geq 1$,

\begin{equation*}
d^{(n)}_{j,k}( \alpha+1,\beta,\gamma)=d^{(n)}_{j,k}( \alpha,\beta,\gamma)+d^{(n-1)}_{j,k-1}( \alpha+1,\beta+1,\gamma+1).
\end{equation*}

Using it in (<ref>) we get that

\begin{align*}
c_{n}(\alpha+1,\beta,\gamma)P_n^{(\alpha+1,\beta,\gamma)}(x;a)   = & \sum_{k=1}^{n}\sum_{j=0}^{n-k}d^{(n-1)}_{j,k-1}( \alpha+1,\beta+1,\gamma+1)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}\\ & +
\sum_{j=0}^{n}d^{(n)}_{j,0}( \alpha+1,\beta,\gamma)(x-1)^{n}x^{j}(x-a)^{n-j} \\
& +\sum_{k=1}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}.
\end{align*}

Notice that $d^{(n)}_{j,0}( \alpha+1,\beta,\gamma)=d^{(n)}_{j,0}( \alpha,\beta,\gamma)$, so that

\begin{multline*}
c_{n}(\alpha+1,\beta,\gamma)P_n^{(\alpha+1,\beta,\gamma)}(x;a)=\sum_{k=0}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j} \\
+\sum_{k=1}^{n}\sum_{j=0}^{n-k}d^{(n-1)}_{j,k-1}( \alpha+1,\beta+1,\gamma+1)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}.
\end{multline*}

Shifting the summation variable $k$ we get

\begin{multline*}
c_{n}(\alpha+1,\beta,\gamma)P_n^{(\alpha+1,\beta,\gamma)}(x;a)=\sum_{k=0}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}\\
+x(x-a)\sum_{k=0}^{n-1}\sum_{j=0}^{n-1-k}d^{(n-1)}_{j,k}( \alpha+1,\beta+1,\gamma+1)(x-1)^{n-1-k}x^{k+j}(x-a)^{n-1-j}.
\end{multline*}

After dividing the identity by $ c_{n}(\alpha+1,\beta,\gamma)$ and using (<ref>) we get (i).

The rest of the assertions are proved using similar arguments.

By Remark <ref>, we have that

x (x-a) P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \prec P_n^{(\alpha,\beta,\gamma)}(x;a).

These are polynomials with zeros in $[a,1)$ and degree exactly $2n$. It is a well-known and easy to prove fact that their convex combination preserves interlacing (see, e.g., [2, 3] or <cit.>), which implies that with $
\mathfrak A_n  >0$, $\mathfrak  B_n   >0$ defined in (<ref>), we have

\begin{align*}
x(x-a) & P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)\prec \\
& \mathfrak A_n P_n^{(\alpha,\beta,\gamma)}(x;a)+\mathfrak B_n x(x-a)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)\prec  P_n^{(\alpha,\beta,\gamma)}(x;a).
\end{align*}

The identity (i) of Theorem <ref> implies part (i) of the theorem, that is,

$$  x(x-a)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)\prec P_n^{(\alpha+1,\beta,\gamma)}(x;a) \prec  P_n^{(\alpha,\beta,\gamma)}(x;a).

The assertion  (ii) of the theorem is proved in a similar way.

Let us prove (iii), and specifically, the interlacing on $(a,0)$. By Theorem <ref>,

\begin{equation} \label{eq:ordering}
P_n^{(\alpha,\beta,\gamma)}(x;a)   \prec  P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \text{ on } (a,0).
\end{equation}

In what follows, for $\alpha,\beta,\gamma>-1$, we denote by

\begin{equation}
\label{def:Zeros}
x_{n,1}(\alpha,\beta,\gamma;a) < x_{n,2}(\alpha,\beta,\gamma;a) < \dots < x_{n,2n}(\alpha,\beta,\gamma;a)
\end{equation}

the zeros of $P^{(\alpha,\beta,\gamma)}_{n }(x;a)$. Recall that $a<x_{n,j}(\alpha,\beta,\gamma;a)<0$ for $j=1,\dots,n$, and $0<x_{n,j}(\alpha,\beta,\gamma;a)<1$ for $j=n+1,\dots,2n$. With this notation, (<ref>) means that

\begin{align*}
a   & <   x_{n,1}(\alpha,\beta,\gamma;a)   < x_{n-1,1}(\alpha+1,\beta+1,\gamma+1;a)< x_{n,2} (\alpha,\beta,\gamma;a) < \\ &   \dots < x_{n,n-1}(\alpha,\beta,\gamma;a)   < x_{n-1,n-1}(\alpha+1,\beta+1,\gamma+1;a) \\
& < x_{n-1,n}(\alpha,\beta,\gamma;a)<0.
\end{align*}

Since $P_n^{(\alpha,\beta,\gamma)}(x;a)$ and $P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)$ are both of even degree with all their zeros on $(a,1)$, we get that

\begin{equation}\label{Signata}
\sign \left(P_n^{(\alpha,\beta,\gamma)}(a;a)\right)=\sign \left(P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(a;a)\right)=1 ,
\end{equation}

and that for $j=1,\dots,n-1$,

\begin{equation}\label{notchangeofsign}
\begin{split}
\sign &\left(P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x_{n,j}(\alpha,\beta,\gamma;a);a)\right)   \\
& = -\sign \left(P_n^{(\alpha,\beta,\gamma)}(x_{n-1,j}(\alpha+1,\beta+1,\gamma+1;a);a)\right)   \\
& = -\sign   \left(P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x_{n,j+1}(\alpha,\beta,\gamma;a);a)\right)=(-1)^{j+1}.
\end{split}
\end{equation}

Recall that by (iii) of Theorem <ref>,

P_n^{(\alpha,\beta,\gamma+1)}(x;a)=\mathfrak A_n P_n^{(\alpha,\beta,\gamma )}(x;a)+\mathfrak  B_n (x-a)(x-1)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a)

with $\mathfrak A_n   >0$, $\mathfrak  B_n   >0$ defined in (<ref>). Since $(x-a)(x-1)<0$ on $(a,1)$, we have by (<ref>) that for $j=1, \dots, n-1$,

\begin{align*}
\sign & (  P_{n}^{(\alpha,\beta,\gamma+1)}(x_{n-1,j }(\alpha+1,     \beta+1, \gamma+1;a);a) ) \\
& = \sign\left(  P_{n}^{(\alpha,\beta,\gamma )}(x_{n-1,j }(\alpha+1,\beta+1,\gamma+1;a);a) \right),
% \\
% & = -\sign \left(P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x_{n,j}(\alpha,\beta,\gamma;a);a)\right).
\end{align*}

\begin{align*}
\sign & \left( P_{n}^{(\alpha,\beta,\gamma+1)} (x_{n,j+1} (\alpha,\beta,\gamma;a);a) \right)   = -\sign\left( P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x_{n,j+1}(\alpha,\beta,\gamma;a);a)\right) \\
& = -\sign \left(P_n^{(\alpha,\beta,\gamma)}(x_{n-1,j}(\alpha+1,\beta+1,\gamma+1;a);a)\right) .
\end{align*}

In other words, $P_{n}^{(\alpha,\beta,\gamma+1)}(x;a)$ presents a sign change (and hence a root) in every subinterval

\left( x_{n-1,j}(\alpha,\beta,\gamma;a), x_{n,j+1}(\alpha+1,\beta+1,\gamma+1;a) \right) , \quad j=1, \dots, n-1.

Moreover, by (<ref>),

\begin{align*}
\sign\left( P_{n}^{(\alpha,\beta,\gamma+1)}(a;a) \right)&=   \sign \left(P_n^{(\alpha,\beta,\gamma)}(a;a)\right) =1,
\end{align*}

while by (<ref>),

\begin{align*}
\sign & \left( P_{n}^{(\alpha,\beta,\gamma+1)}(x_{n,1}(\alpha,\beta,\gamma;a) ;a) \right)\\
&=   -\sign\left( P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x_{n,1}(\alpha,\beta,\gamma;a) ;a) \right) =-1,
\end{align*}

which means that another sign change (and the remaining root) of $P_{n}^{(\alpha,\beta,\gamma+1)}(x ;a)$ on $(a,0)$ is located in $(a, x_{n,1}(\alpha,\beta,\gamma;a))$.

In summary,

\begin{align*}
a & < x_{n,1}(\alpha,\beta,\gamma+1;a)      <   x_{n,1}(\alpha,\beta,\gamma;a)    <   x_{n-1,1}(\alpha+1,\beta+1,\gamma+1;a) \\
& < x_{n,2}(\alpha,\beta,\gamma+1;a)       < x_{n,2} (\alpha,\beta,\gamma;a) < x_{n-1,2}(\alpha+1,\beta+1,\gamma+1;a) < \\
&   \dots < x_{n-1,n-1 }(\alpha+1,\beta+1,\gamma+1;a)   < x_{n ,n }(\alpha,\beta,\gamma+1;a) \\
& < x_{n ,n}(\alpha,\beta,\gamma;a)<0,
\end{align*}

which exactly means that

P_n^{(\alpha,\beta,\gamma+1)}(x;a)  \prec P_n^{(\alpha,\beta,\gamma)}(x;a) \prec P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a) \quad \text{on } (a,0),

as stated.

The proof of the interlacing on $(0,1)$ follows similar arguments, using that

\begin{equation*}
\sign \left(P_n^{(\alpha,\beta,\gamma)}(0;a)\right)= - \sign \left( P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(0;a)\right)=1.
\end{equation*}

§ MEAN BEHAVIOR OF THE ZEROS

Here we gather some technical lemmas that will be useful in the next section.

For $\alpha,\beta,\gamma>-1$, the absolute value of the geometric mean of the zeros of $P_n^{(\alpha,\beta,\gamma)}(x;a)$ is an increasing function of  $\gamma$.

By (<ref>),

\begin{equation}\label{Geometric}
P_n^{(\alpha,\beta,\gamma)}(0;a) = \frac{d^{(n)}_{0,0}( \alpha,\beta,\gamma) }{c_n(\alpha,\beta,\gamma)} \, a^n = \binom{3n+\alpha+\beta+\gamma}{n}^{-1}\binom{n+\gamma}{n}\, a^n,
\end{equation}

so that

\begin{equation}\label{Geometric2}
\left| P_n^{(\alpha,\beta,\gamma)}(0;a)\right|=\frac{\Gamma(\alpha+\beta+\gamma+2n+1)\Gamma(n+\gamma+1)}{\Gamma(\alpha+\beta+\gamma+3n+1)\Gamma(\gamma+1)}\left| \, a\right|^n.
\end{equation}

Taking the logarithmic derivative of this identity with respect to $\gamma$ and using the following known identity <cit.>,

\frac{\Gamma'(z)}{\Gamma(z)}=\psi(z)=-%
\gamma_e+\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+z}\right),

(where $\gamma_e$ is Euler's constant) we obtain

\begin{align*}
\frac{\frac{\partial }{\partial \gamma}\left| P_n^{(\alpha,\beta,\gamma)}(0)\right|}{\left| P_n^{(\alpha,\beta,\gamma)}(0)\right|} & =-\sum_{k=0}^{\infty}\frac{1}{k+\alpha+\beta+\gamma+2n+1}   +\sum_{k=0}^{\infty}\frac{1}{k+\alpha+\beta+\gamma+3n+1} \\ &
\end{align*}

After simplifying, we arrive at

\begin{align*}
\frac{\frac{\partial }{\partial \gamma}\left| P_n^{(\alpha,\beta,\gamma)}(0)\right|}{\left| P_n^{(\alpha,\beta,\gamma)}(0)\right|} & =\sum_{k=1}^\infty\frac{1}{k+\beta}-\frac{1}{k+\alpha+\beta+\gamma+2n} \\
& =\sum_{k=1}^\infty\frac{k+\alpha+\gamma+2n}{(k+\beta)(k+\alpha+\beta+\gamma+2n)}.
\end{align*}

Since by assumption, $\alpha,\beta,\gamma>-1$,  each  term in the sum on the right hand side is positive, so that

\begin{equation*}
\frac{\partial}{\partial \gamma}\left| P_n^{(\alpha,\beta,\gamma)}(0)\right|>0.
\end{equation*}

Under the assumptions above, and with notation (<ref>),

\begin{equation} \label{limits}
\begin{split}
\lim_{\gamma\to+\infty}x_{n,j}(\alpha,\beta,\gamma;a)=a,& \quad j=1,\dots,n\\
\lim_{\gamma\to +\infty}x_{n,j}(\alpha,\beta,\gamma;a)=1,&\quad j= n+1,\dots,2n.
\end{split}
\end{equation}

Since  $| x_{n,j}(\alpha,\beta,\gamma;a)|< |a|$ for $j=1,\dots,n$, and $| x_{n,j}(\alpha,\beta,\gamma;a)|< 1$ for $j=n+1,\dots,2n$, we conclude that

\begin{equation} \label{bound1}
\prod_{j=1}^{2n}  \left|x_{n,j}(\alpha,\beta,\gamma;a)\right|\leq \left| a^n\right|.
\end{equation}

On the other hand, taking the limit in (<ref>) as $\gamma\to +\infty$ and using Stirling's formula <cit.>, we get that

\begin{equation*}
\lim_{\gamma\to +\infty}\left|P_n^{(\alpha,\beta,\gamma)}(0;a)\right|=\lim_{\gamma\to+ \infty}\prod_{j=1}^{2n} \left|x_{n,j}(\alpha,\beta,\gamma;a)\right|=\left|a^n\right|.
\end{equation*}

Combining it with the upper bound (<ref>) we obtain (<ref>).

Our next result concerns the arithmetic mean of the zeros of $P_n^{(\alpha,\beta,\gamma)}(x;a)$,

A_n^{(\alpha,\beta,\gamma)}(a):=\frac{1}{2n}\, \sum_{j=1}^{2n}x_{n,j}(\alpha,\beta,\gamma;a).

For $\alpha,\beta,\gamma>-1$,

\begin{equation}
\label{eq:arithmeticmean}
A_n^{(\alpha,\beta,\gamma)}(a) = \frac{1}{2} \frac{(2n+\beta+\gamma) + a\left(2n+\gamma+\alpha \right) }{3n+\alpha+\beta+\gamma} .
\end{equation}

Using (<ref>) and the relation between the arithmetic mean of the zeros and the coefficient of $x^{2n-1}$ of $P_n^{(\alpha,\beta,\gamma)}(x)$ we get that

\begin{equation} \label{arithmeticMean1}
c_n(\alpha,\beta,\gamma)A_n^{(\alpha,\beta,\gamma)}(a)=\frac{1}{2n}\sum_{k=0}^n\sum_{j=0}^{n-k} d^{(n)}_{j,k}( \alpha,\beta,\gamma)\left((n-k)+(n-j)a\right).
\end{equation}

By the identity

\sum_{j=0}^m\binom{a}{m}\binom{b}{m-j}=\binom{a+b}{m}

we have that

\begin{align*}
\sum_{k=0}^n\sum_{j=0}^{n-k} d^{(n)}_{j,k}( \alpha,\beta,\gamma)(n-k) & =n\frac{2n+\beta+\gamma}{3n+\alpha+\beta+\gamma}\binom{3n+\alpha+\beta+\gamma}{n},
\\
\sum_{k=0}^n\sum_{j=0}^{n-k} d^{(n)}_{j,k}( \alpha,\beta,\gamma)(n-j) &=n\frac{2n+\gamma+\alpha}{3n+\alpha+\beta+\gamma}\binom{3n+\alpha+\beta+\gamma}{n}.
\end{align*}

Combining these two formulas in (<ref>) finally yields (<ref>).

For $\alpha,\beta >-1$ and $n\in \N$,

\begin{equation}\label{eq;derivativearit}
\lim_{\gamma\to +\infty} \frac{\partial}{\partial\gamma} A_n^{(\alpha,\beta,\gamma)}(a)=
\lim_{\gamma\to +\infty}\sum_{j=1}^{2n}\frac{\partial}{\partial\gamma}x_{n,j}(\alpha,\beta,\gamma;a)=0.
\end{equation}

This is an immediate consequence of differentiating (<ref>)  with respect to $\gamma$:

\frac{\partial}{\partial\gamma}A^{(\alpha,\beta,\gamma)}(a)  =    \frac{1}{2n}\, \sum_{j=1}^{2n}\frac{\partial}{\partial\gamma}x_{n,j}(\alpha,\beta,\gamma;a)= \frac{n+\alpha+ (n+\beta)a}{2(3n+\alpha+\beta+\gamma)^2} .

§ PROOF OF THEOREM <REF>

One of the main tools in the proof of Theorems <ref> and <ref>
in [5] is the function (we preserve the notation of [5])

\begin{multline}\label{auxiliaryfunction}
f_{n-1}(x,a,\alpha+1,\beta+1,\gamma+1) :=\\  \sum_{j=1}^{2(n-1)}\frac{1}{x-x_{n-1,j}(a,\alpha+1,\beta+1,\gamma+1)}
\end{multline}

where $x_{n,j}(\alpha,\beta,\gamma;a)$  are the zeros of $P^{(\alpha,\beta,\gamma)}_{n }(x)$, see (<ref>), which has the property that

\begin{equation}
\label{DesanosF}
f_{n-1}(x,a,\alpha+1,\beta+1,\gamma+1)=0 \quad \text{for } x= x_{n,j}(\alpha, \beta,\gamma; a), \quad j=1, \dots, 2n.
\end{equation}

With a shorter notation $x_{n,j}(\gamma)$ instead of $x_{n,j}(\alpha,\beta,\gamma;a)$, $y_{n-1,j}(\gamma)$ instead of $x_{n-1,j}(\alpha+1,\beta+1,\gamma+1;a)$, and $f_{n-1}(x; \gamma+1)$ instead of $ f_{n-1}(x,a,\alpha+1,\beta+1,\gamma+1)$, we use implicit differentiation in (<ref>) to obtain

\begin{equation}
\label{eq:derivative1}
\frac{\partial}{\partial\gamma}\, x_{n,j}(\gamma) =-\frac{\frac{\partial}{\partial \gamma}f_{n-1}(x_{n,j}(\gamma) ;\gamma+1)}{\frac{\partial}{\partial x}f_{n-1}(x_{n,j} (\gamma);\gamma+1)}, \qquad j=1, \dots, 2n.
\end{equation}

In [5] it was shown that $f_{n-1}(x;  \gamma+1)$ is a strictly decreasing function of $x$ in each open interval limited by consecutive zeros of $x(x-a)(x-1)P^{(\alpha+1,\beta+1,\gamma+1)}_{n-1}(x;a)$, so that by (<ref>) and $(ii)$ of Lemma <ref>,

\sign\left(\frac{\partial}{\partial\gamma}x_{n,j}(\gamma) \right)=\sign\left( \frac{\partial}{\partial \gamma}f_{n-1}(x_{n,j}(\gamma);\gamma+1)\right), \quad j=1, \dots, 2n.

Thus, we can prove the theorem by analyzing the sign of the derivative $\frac{\partial}{\partial \gamma}f_{n-1}(x;\gamma+1)$, evaluated at the zeros $x_{n,j}(\gamma)$ of $P^{(\alpha,\beta,\gamma)}_{n }(x)$: we need to show that

\begin{equation*}
\begin{split}
%\frac{\partial}{\partial\gamma}x_{n,j} <0\iff
\frac{\partial}{\partial\gamma}f_{n-1}( x_{n,j}(\gamma) ;\gamma+1)&<0\quad \text{for } j=1,\dots,n, \\
%\frac{\partial}{\partial\gamma}x_{n,j}(\gamma)&>0 \iff
\frac{\partial}{\partial\gamma}f_{n-1}( x_{n,j}(\gamma);\gamma+1)&>0\quad \text{for } j=n+1,\dots,2n.
\end{split}
\end{equation*}

Observe that

\begin{equation}
\label{derivativeOfF}
\frac{\partial}{\partial\gamma}f_{n}(x;\gamma+1)=\begin{cases}
\dfrac{1}{x}, & n=1, \\
\sum_{j=1}^{2(n-1)}\dfrac{ \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))}{(x-y_{n-1,j}(\gamma))^2}+\dfrac{1}{x}, & n\ge 2.
\end{cases}
\end{equation}

We proceed by induction on $n$. The case of $n=1$  is trivial: by (<ref>),

\begin{equation*}
\frac{\partial}{\partial\gamma}f_{n-1}(x_{1,j}(\gamma);\gamma+1)=\frac{1}{x_{1,j}(\gamma)}\begin{cases}
<0 & \text{if } j=1,\\
>0 & \text{if } j=2,
\end{cases}
\end{equation*}

as needed.

Assume that for an $n\in \N$, $n\ge 2$, we have that there exists $\Gamma_1>-1$ such that for $\gamma>\Gamma_1$,

\begin{equation} \label{negativedr}
\begin{split}
\frac{\partial}{\partial\gamma}\, y_{n-1,j}(\gamma)<0& \quad \text{for } j=1,\dots,n-1, \\
\frac{\partial}{\partial\gamma} \, y_{n-1,j}(\gamma)>0& \quad \text{for } j=n,\dots,2(n-1)  .
\end{split}
\end{equation}

By Corollary <ref>, there exists $\Gamma_2>-1$ such that for $\gamma> \Gamma_2$,

\begin{equation}
\label{bounds}
\begin{split}
a<x_{n,i}(\gamma)<\frac{1}{2}a  & \qquad \text{for } i=1,\dots,n, \\
\frac{1}{2}<x_{n,i}(\gamma)<1  & \qquad \text{for } j=n+1,\dots,2n;
\end{split}
\end{equation}

notice that by Theorem <ref>, we also have that for $\gamma> \Gamma_2$,

\begin{equation}
\label{bounds2}
\begin{split}
a<y_{n-1,i}(\gamma)<\frac{1}{2}a  & \qquad \text{for } i=1,\dots,n-1, \\
\frac{1}{2}<y_{n-1,i}(\gamma)<1  & \qquad \text{for } j=n+1,\dots,2n-2.
\end{split}
\end{equation}

Finally,  by Corollary <ref> and since $a<0$, there exists $\Gamma_3>-1$ such that for $\gamma> \Gamma_3$,

\begin{equation} \label{boundAverageDr}
-\frac{(1-a)^2}{4}<\sum_{j=1}^{2(n-1)} \frac{\partial}{\partial\gamma}(y_{n-1,j}(\gamma))<-\frac{(1-a)^2}{4a}.
\end{equation}

Set $\Gamma= \max \left\{\Gamma_1, \Gamma_2, \Gamma_3 \right\}$ and assume $\gamma>\Gamma$. By (<ref>),

\begin{align}
\label{ineq1}
\frac{4}{(1-a)^2} \sum_{j=n}^{2(n-1)} \frac{\partial}{\partial\gamma}(y_{n-1,j}(\gamma)) &<-\frac{1}{a}- \frac{4}{(1-a)^2}\sum_{j=1}^{ n-1} \frac{\partial}{\partial\gamma}(y_{n-1,j}(\gamma)) ,
\\
\label{ineq2}
\frac{4}{(1-a)^2}\sum_{j=1}^{n-1}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)) &
> -1- \frac{4}{(1-a)^2}\sum_{j=n}^{2(n-1)}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)) .
\end{align}

Using (<ref>)–(<ref>) we conclude that for $i=1, 2, \dots, n$,

\begin{align}
\label{Bound:negative1}
\sum_{j=1}^{n-1}\frac{ \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))}{(x_{n,i}(\gamma)-y_{n-1,j}(\gamma))^2} & <\frac{4}{a^2}\sum_{j=1}^{n-1} \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)),
\end{align}

as well as

\begin{equation}
\label{Bound:negative3}
\begin{split}
\sum_{j=n}^{2(n-1)}\frac{ \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))}{(x_{n,i}(\gamma)-y_{n-1,j}(\gamma))^2} & <\frac{4}{(1-a)^2}\sum_{j=n}^{2(n-1)}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))\\
&< -\frac{1}{a}- \frac{4}{(1-a)^2}\sum_{j=1}^{n-1}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)),
\end{split}
\end{equation}

where we have used (<ref>). Combining (<ref>) and (<ref>) in (<ref>) and taking into account that $a<0$ and $1/x_{n,i}(\gamma)< 1/a$, we get
that for $i=1, 2, \dots, n$,

\frac{\partial}{\partial\gamma}f_{n}(x_{n,i}(\gamma);\gamma)<4\left( \frac{1}{a^2}-\frac{1}{(1-a)^2}\right)\sum_{j=1}^{n-1}\frac{\partial}{\partial\gamma}(y_{n-1,j}(\gamma))-\frac{1}{a}+\frac{1}{x_{n,i}(\gamma)}<0.

In the same fashion, for $i=n+1, \dots, 2n$, from (<ref>)–(<ref>) we conclude that

\begin{equation}\label{Bound: positve2}
\sum_{j=n}^{2(n-1)}\frac{ \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))}{(x_{n,i}(\gamma)-y_{n-1,j}(\gamma))^2}>4\sum_{j=n}^{2(n-1)} \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)),
\end{equation}

as well as

\begin{multline}\label{Bound:negative2}
\sum_{j=1}^{n-1}\frac{ \frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))}{(x_{n,i}(\gamma)-y_{n-1,j}(\gamma))^2}>\frac{4}{(1-a)^2}\sum_{j=1}^{n-1}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma))\\
> -1- \frac{4}{(1-a)^2}\sum_{j=n}^{2(n-1)}\frac{\partial}{\partial \gamma}(y_{n-1,j}(\gamma)),
\end{multline}

where we have used (<ref>). Combining  (<ref>) and (<ref>) in (<ref>) and taking into account that $a<0$ and $1/x_{n,i}(\gamma)> 1$, we get finally that for $i=n+1, \dots, 2n$,

\frac{\partial}{\partial\gamma}f_{n}(x_{n,i}(\gamma);\gamma)> 4 \left( 1-\frac{1}{(1-a)^2}\right)\sum_{j=n}^{2(n-1)}\frac{\partial}{\partial\gamma}(y_{n-1,j}(\gamma))-1+\frac{1}{x_{n,i}(\gamma)}>0.

§ PROOF OF THEOREM <REF>

We need the following result to understand the behavior of the zeros of $P_n^{(\alpha,\beta,\gamma)}(x;a)$ as functions of the end point $a$:

Let $n\in\mathbb{N}$, and $\alpha,\beta, \gamma>-1$. Then

\begin{align*}
& (a-x)\frac{\partial}{\partial a}   P_n^{(\alpha,\beta+1,\gamma)}(x;a)  \\
& = \frac{(n+\beta+1)(1+\alpha+\beta+\gamma+2n)}{1+\alpha+\beta+\gamma+3n} P_n^{(\alpha,\beta ,\gamma)}(x;a)-(\beta+1) P_{n}^{(\alpha,\beta+1,\gamma)}(x;a) \\
& = n P_n^{(\alpha,\beta+1,\gamma)}(x;a)-   \frac{n(n+\beta+1)}{1+\alpha+\beta+\gamma+3n} x(x-1)P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x;a).
\end{align*}

We prove the first identity by using the Rodrigues formula (<ref>): differentiating it with respect to $a$ yields

\begin{align}
\nonumber
(1-x)^{\alpha} & (x-a)^{\beta-1} x^{\gamma}c_n(\alpha,\beta,\gamma) \left( (a-x)\frac{\partial}{\partial a}P_n^{(\alpha,\beta,\gamma)}(x;a)-\beta P_n^{(\alpha,\beta,\gamma)}(x;a)\right)\\
\label{RodriguezIden1}
& =(n+\beta)\frac{(-1)^n}{n!} \left( \frac{d}{dx} \right) ^n\left[ (1-x)^{\alpha+n}(x-a)^{\beta+n-1}x^{\gamma+n}\right] \\
&  =(n+\beta)(1-x)^\alpha(x-a)^{\beta-1} x^\gamma c_n(\alpha,\beta-1,\gamma) P^{(\alpha,\beta-1,\gamma)}_{n}(x;a).
\label{RodriguezIden2}
\end{align}

In other words,

\begin{multline*}
(a-x)\frac{\partial}{\partial a}P_n^{(\alpha,\beta,\gamma)}(x;a)-\beta P_n^{(\alpha,\beta,\gamma)}(x;a)\\=(n+\beta)\frac{c_n(\alpha,\beta-1,\gamma) }{c_n(\alpha,\beta,\gamma) }P^{(\alpha,\beta-1,\gamma)}_{n}(x;a),
\end{multline*}

which is the first equality claimed in the theorem.

The second identity can be derived from the explicit expression (<ref>).
Differentiating it with respect to $a$, we get

\begin{align*}
c_n(\alpha,\beta+1,\gamma)& \frac{\partial}{\partial a} P_n^{(\alpha,\beta+1,\gamma)}(x;a) \\
& =-\sum_{k=0}^n\sum_{j=0}^{n-k}(n-j)d^{(n)}_{j,k}( \alpha,\beta+1,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-1-j},
\end{align*}

so that

\begin{equation}\label{separetedsums}
\begin{split}
c_n(\alpha,\beta+1,\gamma)&(a-x)\frac{\partial}{\partial a}  P_n^{(\alpha,\beta+1,\gamma)}(x;a) \\
& = n\sum_{k=0}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta+1,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}\\
&  -\sum_{k=0}^{n-1}\sum_{j=1}^{n-k}jd^{(n)}_{j,k}( \alpha,\beta+1,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}.
\end{split}
\end{equation}

By (<ref>),

\begin{equation}\label{firstterm}
\begin{split}
nc_n(\alpha,\beta+1,\gamma) & P_n^{(\alpha,\beta+1,\gamma)}(x;a) \\
& =n\sum_{k=0}^n\sum_{j=0}^{n-k}d^{(n)}_{j,k}( \alpha,\beta+1,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}.
\end{split}
\end{equation}

On the other hand, by (<ref>), for $j>0$ we have

\begin{equation*}
\end{equation*}

which implies that

\begin{equation*}
\begin{split}
\sum_{k=0}^{n-1} & \sum_{j=1}^{n-k}jd^{(n)}_{j,k}( \alpha,\beta,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j} = (n+\beta+1) \times \\ &\sum_{k=0}^{n-1}\sum_{j=0}^{n-1-k}d^{(n-1)}_{j,k}( \alpha+1,\beta+1,\gamma+1)(x-1)^{n-1-k}x^{k+j+1}(x-1)^{n-j},
\end{split}
\end{equation*}

or equivalently, by (<ref>),

\begin{equation}\label{secondterm}
\begin{split}
\sum_{k=0}^{n-1} & \sum_{j=1}^{n-k}jd^{(n)}_{j,k}( \alpha,\beta,\gamma)(x-1)^{n-k}x^{k+j}(x-a)^{n-j}=\\
\end{split}
\end{equation}

Using the equations (<ref>) and (<ref>) in (<ref>) we get

\begin{equation*}
\begin{split}
& c_n(\alpha,\beta+1,\gamma)(a-x)\frac{\partial}{\partial a}  P_n^{(\alpha,\beta+1,\gamma)}(x;a)= n c_n(\alpha,\beta+1,\gamma) P_n^{(\alpha,\beta+1,\gamma)}(x;a)\\
\end{split}
\end{equation*}

which is equivalent to the second identity of the theorem.

From the implicit differentiation of

with respect to the parameter $a$ we get

\begin{equation*}
\frac{\partial}{\partial a}x_{n,j}(\alpha,\beta+1,\gamma;a)=-\frac{\frac{\partial}{\partial a}P_{n}^{(\alpha,\beta+1,\gamma)}(x_{n,j}(\alpha,\beta+1,\gamma;a);a)}{\frac{\partial}{\partial x}P_{n}^{(\alpha,\beta+1,\gamma)}(x_{n,j}(\alpha,\beta+1,\gamma;a);a)}
\end{equation*}

Using Lemma <ref> to replace the numerator on the right-hand side, we obtain

\begin{multline} \label{finalidentityXroots}
\frac{\partial}{\partial a}x_{n,j}(\alpha,\beta+1,\gamma;a)= \frac{(n+\beta+1)(1+\alpha+\beta+\gamma+2n)}{1+\alpha+\beta+\gamma+3n} \times \\ \frac{ P_n^{(\alpha,\beta ,\gamma)}(x_{n,j}(\alpha,\beta+1,\gamma;a);a)}{\left(x_{n,j}(\alpha,\beta+1,\gamma;a)-a\right)\frac{\partial}{\partial x}P_{n}^{(\alpha,\beta+1,\gamma)}(x_{n,j}(\alpha,\beta+1,\gamma;a);a)}.
\end{multline}

Recall that
$x_{n,j}(\alpha,\beta+1,\gamma;a) >a$ for all $j=1,\dots,2n$. Furthermore, $P_{n}^{(\alpha,\beta+1,\gamma)}(x ;a) $ is a polynomial of degree $2n$, with $P_{n}^{(\alpha,\beta,\gamma)}(a;a)>0$ and simple zeros on $(a,1)$, which implies that

\sign\left(\frac{\partial}{\partial x}P_{n}^{(\alpha,\beta+1,\gamma)}(x_{n,j}(a,\alpha,\beta+1,\gamma;a)\right) =(-1)^j, \quad j=1, \dots, 2n.

Analogously, $P_{n}^{(\alpha,\beta+1,\gamma)}(x;a)$ is a polynomial of degree $2n$, with $P_{n}^{(\alpha,\beta+1,\gamma)}(a)>0$. The assertion (ii) of Theorem <ref> implies that

\sign\left(P_{n}^{(\alpha,\beta,\gamma)}(x_{n,j}(\alpha,\beta+1,\gamma;a);a)\right)=(-1)^j, \quad j=1, \dots, 2n.

Gathering these facts in (<ref>) yields

\frac{\partial}{\partial a}x_{n,j}(\alpha,\beta+1,\gamma;a)>0, \quad j=1, \dots, 2n.

§ PROOF OF THEOREMS <REF> AND <REF>

The results obtained for the Angelesco-Jacobi polynomials can be extended to other families of multiple orthogonal polynomials of type II, namely, the Jacobi-Laguerre and Laguerre-Hermite polynomials. The extension is based on the following raising operator identities, which can be obtained from (<ref>) using the limits (<ref>)-(<ref>):

\begin{equation}\label{raisingJL}
(x-a)^\beta x^\gamma  e^{-x} L^{(\beta,\gamma)}_{n}(x;a)
=-\frac{d}{dx} \left[(x-a)^{\beta}x^{\gamma}e^{-x} L^{(\beta+1,\gamma+1)}_{n-1}(x;a)\right],
\end{equation}

\begin{equation}\label{raisingHL}
x^\gamma  e^{-x^2} H^{(\gamma)}_{n}(x)
=-\frac{d}{dx} \left[x^{\gamma}e^{-x^2} H^{(\gamma+1)}_{n-1}(x)\right].
\end{equation}

Since the roots of $L^{(\beta+1,\gamma+1)}_{n-1}(x;a)$ and $H^{(\gamma+1)}_{n-1}(x)$ are simple, using (<ref>)-(<ref>), we conclude that $L^{(\beta+1,\gamma+1)}_{n-1}(x;a)$ and $L^{(\beta,\gamma)}_{n}(x;a)$ do not have common zeros. The same assertion is valid for the polynomials $H^{(\gamma+1)}_{n-1}(x)$ and $H^{(\gamma)}_{n}(x)$.

The following generalization of the structure relations, obtained in Theorem <ref>, take place:
Let $n\in\mathbb{N}$ and $\beta,\gamma>-1$. Then

L_n^{(\beta+1,\gamma)}(x;a)=  L_n^{(\beta,\gamma)}(x;a)- xL_{n-1}^{(\beta+1,\gamma+1)}(x;a).

L_n^{(\beta,\gamma+1)}(x;a)=  L_n^{(\beta,\gamma)}(x;a)- (x-a)L_{n-1}^{(\beta+1,\gamma+1)}(x;a).

By $(ii)$ of Theorem <ref>,

\begin{equation}\label{LaguerreH1}
\begin{split}
\alpha^{2n}P_n^{(\alpha,\beta+1,\gamma)}&(x/\alpha;a/\alpha)= \mathfrak A_n\alpha^{2n} P_n^{(\alpha,\beta,\gamma)}(x/\alpha;a/\alpha)+\\ &\mathfrak B_n\alpha^{2n-1} (x/\alpha-1)xP_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x/\alpha;a/\alpha),
\end{split}
\end{equation}

where $\mathfrak A_n = \mathfrak A_n(\alpha,\beta,\gamma)$ and $\mathfrak B_n=\mathfrak B_n(\alpha,\beta,\gamma)= 1- \mathfrak A_n $ were defined in (<ref>). From their explicit expression,

\lim_{\alpha\to\infty}\mathfrak A_n(\alpha,\beta,\gamma)=\lim_{\alpha\to\infty} \alpha \mathfrak B_n(\alpha,\beta,\gamma)=1.

Thus, taking the limit in (<ref>) as $\alpha \to \infty$ we get assertion (1) of the theorem. Similar arguments, now with assertion (iii) of Theorem <ref>, yield (2).

Let $n\in\mathbb{N}$ and $\gamma>-1$. Then

\begin{equation}
H_n^{(\gamma+1)}(x)=  H_n^{(\gamma)}(x;a)- \frac{1}{2}H_{n-1}^{(\gamma+1)}(x;a).
\end{equation}

By $(iii)$ of Theorem <ref>,

\begin{multline}\label{Hermite}
\alpha^n P_n^{(\alpha,\alpha,\gamma+1)}(x/\sqrt{\alpha};-1)=  \mathfrak A_n \alpha^n P_n^{(\alpha,\alpha,\gamma)}(x/\sqrt{\alpha};-1) + \\ \mathfrak B_n (x+1)(x-1)\alpha^n P_{n-1}^{(\alpha+1,\alpha+1,\gamma+1)}(x/\sqrt{\alpha};-1),
\end{multline}

with $\mathfrak A_n = \mathfrak A_n(\alpha,\beta,\alpha)$ and $\mathfrak B_n(\alpha,\beta,\alpha)= 1- \mathfrak A_n$.
By (<ref>).

\lim_{\alpha\to\infty}\mathfrak A_n(\alpha,\beta,\alpha)=1 \quad \text{and }\quad\lim_{\alpha\to\infty} \alpha \mathfrak B_n(\alpha,\beta,\alpha)=1/2,

and it remains to take the limit in (<ref>) as $\alpha\to \infty$.

Part (<ref>) of Theorem <ref> can be restated as

\begin{equation}
\begin{split}
\alpha^{2n} P_n^{(\alpha,\beta,\gamma)}(x/\alpha;a/\alpha) &     \prec\alpha^{2n} P_n^{(\alpha,\beta+1,\gamma)}(x/\alpha;a/\alpha) \\ &\prec x(x/\alpha-1)\alpha^{2n-2}P_{n-1}^{(\alpha+1,\beta+1,\gamma+1)}(x/\alpha;a/\alpha).
\end{split}
\end{equation}

With the notation (<ref>), this means that

\begin{equation}\label{Auxiliarinterlacing}
\begin{split}
a   & <   x_{n,1}(\alpha,\beta,\gamma;a/\alpha)   < x_{n,1}(\alpha,\beta+1,\gamma;a/\alpha)
\\
&  < x_{n-1,1} (\alpha+1,\beta+1,\gamma+1;a/\alpha)  <\dots < x_{n-1,n-1}(\alpha+1,\beta+1,\gamma+1;a/\alpha)    \\
& < x_{n,n}(\alpha,\beta,\gamma;a/\alpha) < x_{n,n}(\alpha,\beta+1,\gamma;a/\alpha)<0 <   x_{n,n+1}(\alpha,\beta,\gamma;a/\alpha) \\
&  < x_{n,n+1}(\alpha,\beta+1,\gamma;a/\alpha)< x_{n-1,n} (\alpha+1,\beta+1,\gamma+1;a/\alpha)  \\  &  < \dots < x_{n-1,2n-2}(\alpha+1,\beta+1,\gamma+1;a/\alpha)   < x_{n,2n}(\alpha,\beta,\gamma;a/\alpha) \\
&< x_{n,2n}(\alpha,\beta+1,\gamma;a/\alpha)<1.
\end{split}
\end{equation}

Let us denote by

\begin{equation}
\label{def:ZerosL}
l_{n,1}(\beta,\gamma;a) < l_{n,2}(\beta,\gamma;a) < \dots < l_{n,2n}(\beta,\gamma;a)
\end{equation}

the zeros of $L^{(\beta,\gamma)}_n(x;a)$. Recall that $a<l_{n,j}(\beta,\gamma;a)<0$ for $j=1,\dots,n$ and $0<l_{n,j}(\beta,\gamma;a)$ for $j=n+1,\dots,2n$. Also, by (<ref>),

\begin{equation}\label{JatoJL}
\lim_{\alpha\to\infty} \alpha x_{n,j}(\alpha,\beta,\gamma;a/\alpha)=l_{n,j}(\beta,\gamma;a) \quad j=1,\dots,2n.
\end{equation}

Multiplying (<ref>) by $\alpha$ and taking the limit as $\alpha\to \infty$ we get

\begin{equation}
\label{interlacingL}
\begin{split}
a   & <  l_{n,1}(\beta,\gamma;a)   \leq l_{n,1}(\beta+1,\gamma;a)\leq l_{n-1,1} (\beta+1,\gamma+1;a) \leq \\  &   \dots \leq l_{n-1,n-1}(\beta+1,\gamma+1;a)   \leq l_{n,n}(\beta,\gamma;a)  \leq l_{n,n}(\beta+1,\gamma;a)<0 \\
&\leq   l_{n,n+1}(\beta,\gamma;a)   \leq l_{n,n+1}(\beta+1,\gamma;a)\leq l_{n-1,n} (\beta+1,\gamma+1;a) \leq \\  &   \dots \leq l_{n-1,2n-2}(\beta+1,\gamma+1;a)   \leq l_{n,2n}(\beta,\gamma;a)  \leq l_{n,2n}(\beta+1,\gamma;a).
\end{split}
\end{equation}

From Remark <ref> we know that $ L_n^{(\beta,\gamma)}(x;a)$ and $ L_n^{(\beta+1,\gamma)}(x;a)$ do not have common zeros. Furthermore, by (1) of Theorem <ref>, neither $ L_n^{(\beta+1,\gamma)}(x;a)$ has common zeros with $ L_n^{(\beta,\gamma)}(x;a)$ or $ L_n^{(\beta+1,\gamma)}(x;a)$. Thus, all the inequalities above are strict, which implies

\begin{equation*}
L_n^{(\beta,\gamma)}(x;a)\prec  L_n^{(\beta+1,\gamma)}(x;a) \prec  xL_{n-1}^{(\beta+1,\gamma+1)}(x;a)    .
\end{equation*}

The proof of assertion (2) is similar, as is for (3), although in this case strict inequalities follow from Remark <ref> and Theorem <ref>.

Consider the polynomial $\alpha^{2n}P_n^{(\alpha,\beta,\gamma)}(x/\alpha;a/\alpha)$. By Theorem <ref>, for $\delta>0$ and for sufficiently large values of $\gamma$ its zeros  exhibit the following monotonic behavior:

\begin{equation*}
\begin{split}
\alpha x_{n,j}(\alpha,\beta,\gamma+\delta;a/\alpha)<\alpha x_{n,j}(\alpha,\beta,\gamma;a/\alpha) &\quad \text{for } j=1,\dots n; \\
\alpha x_{n,j}(\alpha,\beta,\gamma;a/\alpha)<\alpha x_{n,j}(\alpha,\beta,\gamma+\delta;a/\alpha)& \quad \text{for } j=n+1,\dots 2n  .
\end{split}
\end{equation*}

Taking the limit as $\alpha\to \infty$, (<ref>), yields

\begin{equation*}
\begin{split}
l_{n,j}(\beta,\gamma+\delta;a)\leq l_{n,j}(\beta,\gamma;a) &\quad \text{for } j=1,\dots n, \\
l_{n,j}(\beta,\gamma;a)\leq l_{n,j}(\beta,\gamma+\delta;a)& \quad \text{for } j=n+1,\dots 2n .
\end{split}
\end{equation*}

This is the monotonic behavior of the zeros of $L^{(\beta,\gamma)}_{n}(x;a)$ with respect to $\gamma$, as claimed.

§ ACKNOWLEDGMENTS

The first author was partially supported by Simons Foundation Collaboration Grants for Mathematicians (grant 710499).
He also acknowledges the support of Junta de Andalucía (research group FQM-229 and Instituto Interuniversitario Carlos I de Física Teórica y Computacional) and of the project PID2021-124472NB-I00, funded by MCIN/AEI/10.13039/501100011033 and by “ERDF A way of making Europe”.

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M. A. McClain, eds.

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# Multiplicity one theorem over characteristic 2

Dor Mezer

###### Abstract.

It is shown for all local fields ${\mathbb{F}}$ which are of characteristic
different from $2$ that any distribution on $GL_{n+1}({\mathbb{F}})$ which is
invariant under conjugation by $GL_{n}({\mathbb{F}})$ is also invariant under
transposition. In this paper we give an adaptation of the proof of this
theorem to fields of characteristic 2.

###### Key words and phrases:

Distribution, Multiplicity one, Gelfand pair, invariant distribution

###### 2010 Mathematics Subject Classification:

20G05, 20G25, 22E50, 46F10

##  1\. Introduction

Let ${\mathbb{F}}$ be a local field of characteristic $2$. In this paper we
prove the following theorem:

######  Theorem 1.1.

Any distribution on $GL_{n+1}({\mathbb{F}})$ invariant under conjugation by
$GL_{n}({\mathbb{F}})$ is also invariant under transposition.

For non-archimedean fields of characteristic zero it is proven in [4], for
archimedean fields in [8] and [3],and for fields of odd characteristic in [6].
In this paper we will give an adaptation of the proof in [6] to characteristic
2.

It is shown in [4, section 1] that Theorem 1.1 has the following corollary,
already known by different methods (see [1]).

######  Theorem 1.2.

Let $\pi$ be an irreducible smooth representation of $GL_{n+1}$, and let
$\rho$ be an irreducible smooth representation of $GL_{n}$. Then

$\dim\operatorname{Hom}_{GL_{n}}(\pi,\rho)\leq 1$

Let $V$ be an $n$-dimensional vector space over ${\mathbb{F}}$. Let
${\widetilde{G}}:=GL(V)\rtimes\\{\pm 1\\}$ be the semidirect product with the
respect to the action of $\\{\pm 1\\}$ on $GL(V)$ by
$A\mapsto\left(A^{t}\right)^{-1}$. The group ${\widetilde{G}}$ acts on
$\mathfrak{gl}(V)$ by $(g,1).(A,v,\phi)=(gAg^{-1},gv,(g^{*})^{-1}\phi)$, and
$(g,-1).(A,v,\phi)=(\left(gAg^{-1}\right)^{t},g\phi^{t},(g^{*})^{-1}v^{t})$.
Let $\chi$ be the character of ${\widetilde{G}}$ defined by
$(g,\delta)\mapsto\delta$. It is shown in [2] (the same proof works verbatim)
that Theorem 1.1 reduces to the following theorem:

######  Theorem 1.3.

Any $({\widetilde{G}},\chi)$-equivariant distribution on
$\mathfrak{gl}(V)\times V\times V^{*}$ is 0.

We will prove this theorem by induction on the dimension of $V$, and so we
will assume this theorem for all smaller $n$. Throughout the paper, let $\xi$
be a $({\widetilde{G}},\chi)$-equivariant distribution on
$\mathfrak{gl}(V)\times V\times V^{*}$.

There are two points in the proof in [6] in which the assumption
$\mathrm{char}({\mathbb{F}})\neq 2$ was made use of. The first and more
significant one is the proof of Proposition 4.6. The main goal of this paper
is to prove this theorem over fields of characteristic 2 (Corollary 3.4). In
section 2 we use the techinique of Harish-chandra descent to a further extent
than was used in [6], to get a stronger restriction on the support of $\xi$,
which will be used in the proof of Corollary 3.4 in section 3. The second
usage of the assumption $\mathrm{char}({\mathbb{F}})\neq 2$ in [6] was the
usage of a theorem by Rallis and Schiffman ([6, Theorem 2.9]), which is relied
on the theory of the Weil representation, a theory which is a bit different
when working over a field of characteristic 2. In Appendix A, we prove a
version of the theorem over a field of characteristic 2, which is sufficient
for the proof as given in [6].

###  1.1. Acknowledgements

I would like to deeply thank my advisor, Dmitry Gourevitch, for helping me
with my work on this paper, for exposing me to this fascinating area of
mathematics, and for the guidance and support along the way. I would also like
to thank him for the exceptional willingness to help.
I would also like to thank Guy Henniart for fruitful discussions he had with
my advisor laying the foundations for this project, and for his continued
interest in my research along the way.
I am also grateful to Shamgar Gurevich, for the help he gave me when writing
this paper.
D.M. was partially supported by ERC StG grant 637912.

##  2\. Reduction to the purely inseparable locus

######  Notation 2.1.

Use $\Delta:\mathfrak{gl}(V)\times A\times A^{*}\to{\mathbb{F}}[x]$ to denote
the map which sends $(A,v,\phi)$ to the characteristic polynomial of $A$.

######  Theorem 2.2.

For any point $(A,v,\phi)$ in the support of $\xi$, the characteristic
polynomial of $A$ is a power of an irreducible polynomial.

###### Proof.

Assume that a polynomial $f$ of degree $n$ has two coprime components
$f=f_{1}f_{2}$. By the localization principle (see [6, Theorem 2.4]), it is
enough for us to show that for any such polynomial $f$, the fiber of $\Delta$
above $f$ has no non-zero $({\widetilde{G}},\chi)$-equivariant distributions.
Let $\zeta$ be such a distribution on $\Delta^{-1}(f)$. Let $d_{1}=\deg
f_{1},d_{2}=\deg f_{2}$. Denote by $\Lambda$ the space of all pairs of
subspaces $V_{1},V_{2}$ of $V$ such that
$\dim(V_{1})=d_{1},\dim(V_{2})=d_{2},V=V_{1}\oplus V_{2}$. $\Lambda$ has a
natural action of $G$ on it, which extends to an action of ${\widetilde{G}}$
by the involution $(V_{1},V_{2})\mapsto(V_{2}^{\perp},V_{1}^{\perp})$with
respect to the quadratic form $v\mapsto v^{t}v$. These two actions are (both)
transitive. There is a ${\widetilde{G}}$-equivariant map
$\Delta^{-1}(f)\to\Lambda$, given by taking the (unique) pair of $A$-invariant
subspaces of $V$ on which $A$ acts with characteristic polynomials $f_{1}$ and
$f_{2}$. The fiber of this map above $(V_{1},V_{2})$ is a closed subspace of
$(\mathfrak{gl}(V_{1})\times V_{1}\times
V_{1}^{*})\times(\mathfrak{gl}(V_{2})\times V_{2}\times V_{2}^{*})$, and the
stabilizer of this point is equal to
${\widetilde{G}}(V_{1})\times{\widetilde{G}}(V_{2})$. By the localization
principle (see [6, Theorem 2.4]) and induction hypothesis, there are no non-
zero $({\widetilde{G}}(V_{1})\times{\widetilde{G}}(V_{2}),\chi)$-equivariant
distributions on $(\mathfrak{gl}(V_{1})\times V_{1}\times
V_{1}^{*})\times(\mathfrak{gl}(V_{2})\times V_{2}\times V_{2}^{*})$, and so it
follows from Frobenius descent (see [6, Theorem 2.7]) that $\zeta=0$ too. ∎

######  Theorem 2.3.

For any point $(A,v,\phi)$ in the support of $\xi$, the irreducible factor in
the characteristic polynomial of $A$ is purely inseparable.

###### Proof.

By the localization principle (see [6, Theorem 2.4]) it is enough to consider
a $({\widetilde{G}},\chi)$-equivariant distribution $\zeta$ on $F\times
V\times V^{*}$, where $F$ is the fiber of $\Delta$ over some $f^{m}$, $f$
being irreducible and not purely separable, and show that $\zeta=0$. We have
$f(x)=g(x^{2^{k}})$ for $g(x)$ irreducible and separable. By assumption, $\deg
g>1$. For any $A\in F$, the characteristic polynomial of $B:=A^{2^{k}}$ is
equal to $g(x)^{2^{k}m}$, as it is the polynomial whose roots are $2^{k}$
powers of the roots of $g(x^{2^{k}})^{m}$. All of its irreducible factors are
separable, and so $B$ has a well defined Jordan decomposition $B=B_{s}+B_{n}$.
Moreover, the map $h:A\mapsto B_{s}$ is continuous on $F$. Note that since
$B_{s}$ is expressable as a polynomial in $B$ (and thus as a polynomial in
$A$), it commutes with $A$. Use Frobenius descent (see [6, Theorem 2.7]) with
respect to $h$ \- the stabilizer of a point is isomorphic to
$\widetilde{GL}_{2^{k}m}(\mathbb{E})$, where
$\mathbb{E}:={\mathbb{F}}[x]/g(x)$. The fiber of $h$ above a point is
isomorphic to a closed subspace of $\mathfrak{gl}_{2^{k}m}(\mathbb{E})$, and
so by induction hypothesis applied to the Frobenius descent of $\zeta$, we get
that $\zeta=0$. ∎

##  3\. Vanishing of linear invariants

The following proposition is proved in [2, Lemma 7.2] over a field of
characteristic $0$, and the proof there applies verbatim over arbitrary
characteristic.

######  Proposition 3.1.

For any point $(A,v,\phi)$ in the support of $\xi$, we have $<v,\phi>=0$.

######  Definition 3.2.

Let $\mu\in{\mathbb{F}}$. Define $\rho_{\mu}$ as the following
$\mathrm{GL}(V)$-equivariant automorphism on $\mathfrak{gl}(V)\times V\times
V^{*}$:

$(A,v,\phi)\mapsto(A+\mu v\otimes\phi,v,\phi).$

######  Theorem 3.3.

For any point $(A,v,\phi)$ in the support of $\xi$, we have $<Av,\phi>=0$.

###### Proof.

By applying Theorem 2.3 to $\rho_{\mu}(\xi)$ for some $\mu\in{\mathbb{F}}$, we
get that the characteristic polynomial of $A+\mu v\otimes\phi$ must be a power
of an irreducible purely inseparable polynomial too. Denote the characteristic
polynomial of $A+\mu v\otimes\phi$ by $\sum_{i=0}^{n}c_{i}(A+\mu
v\otimes\phi)x^{n-i}$.

######  Case 1.

$n$ is odd

The characteristic polynomial of $A+\mu v\otimes\phi$ is of the form
$(x+\lambda_{\mu})^{n}$. Since $\lambda_{\mu}=c_{1}(A+\mu
v\otimes\phi)=c_{1}(A)-\mu<v,\phi>=c_{1}(A)=\lambda_{0}$, we get that
$\lambda_{\mu}$ (and so also the characteristic polynomial of $A+\mu
v\otimes\phi$) is independent of $\mu$. Thus we get that

$c_{2}(A)=c_{2}(A+\mu
v\otimes\phi)=c_{2}(A)-\mu(<Av,\phi>+c_{1}(A)<v,\phi>)=c_{2}(A)-\mu<Av,\phi>,$

and so $<Av,\phi>=0$.

######  Case 2.

$n$ is divisible by $4$

In this case $n=\binom{n}{2}=0$ in $\mathbb{F}$. The characteristic polynomial
of $A+\mu v\otimes\phi$ is always of the form
$(x^{2^{k}}+\lambda)^{\frac{n}{2^{k}}}$ for some $\lambda$ dependent on $\mu$.
By maybe changing $\lambda$, we can assume that $2^{k}$ is the maximal power
of $2$ that divides $n$. In particular our polynomial is a polynomial in
$x^{4}$, and so we have $c_{2}(A+\mu v\otimes\phi)=0$ for all $\mu$. However,

$c_{2}(A+\mu
v\otimes\phi)=c_{2}(A)-\mu(<Av,\phi>+c_{1}(A)<v,\phi>)=c_{2}(A)-\mu<Av,\phi>.$

Thus we must have $<Av,\phi>=0$.

######  Case 3.

$n=2\mod 4$ and $n>2$

The irreducible factor of the characteristic polynomial is either linear or
quadratic. Thus the characteristic polynomial must be either $(x+\lambda)^{n}$
or $(x^{2}+\lambda)^{n/2}$. Allowing $\lambda$ to be a square, we assume it is
of the second form. So $c_{2}$ is equal $(n/2)\lambda=\lambda$. Let
$\lambda_{\mu}$ be such that the characteristic polynomial of $A+\mu
v\otimes\phi$ is $(x^{2}+\lambda_{\mu})^{n/2}$. We have

$\lambda_{\mu}=c_{2}(A+\mu v\otimes\phi)=c_{2}(A)-\mu<Av,\phi>$
$(c_{2}(A)-\mu(<Av,\phi>))^{n/2}=\lambda_{\mu}^{n/2}=c_{n}(A+\mu
v\otimes\phi)=c_{n}(A)-\mu(\dots)$

The right hand side is linear in $\mu$ while the left hand side is polynomial
of degree $n/2$ unless $<Av,\phi>=0$. Thus assuming $n>2$ we are done.

######  Case 4.

$n=2$

We can use the localization principle (see [6, Theorem 2.4]) with respect to
the map $(A,v,\phi)\mapsto<Av,\phi>$, to be left with proving that if $\xi$ is
supported on $\\{(A,v,\phi)|<Av,\phi>=m\\}$ (for some $m\neq 0$) then $\xi=0$.
We already know that for any point $(A,v,\phi)$ in the support of $\xi$, we
must have $\mathrm{tr}A=0$, and that $<v,\phi>=0$. Recalling that by the
assumption $<Av,\phi>=m$ we necessarily have $v\neq 0$, we can write
explicitly $A=\begin{pmatrix}a&b\\\ c&a\end{pmatrix},v=\begin{pmatrix}x\\\
y\end{pmatrix},\phi=\begin{pmatrix}ty&tx\end{pmatrix}$. This yields
$<Av,\phi>=t(cx^{2}+by^{2})$. Let $\sigma:=\left(\begin{pmatrix}0&1\\\
1&0\end{pmatrix},-1\right)\in{\widetilde{G}}$. It acts by

$\left(\begin{pmatrix}a&b\\\ c&a\end{pmatrix},\begin{pmatrix}x\\\
y\end{pmatrix},\begin{pmatrix}ty&tx\end{pmatrix}\right)\mapsto\left(\begin{pmatrix}a&b\\\
c&a\end{pmatrix},\begin{pmatrix}tx\\\
ty\end{pmatrix},\begin{pmatrix}y&x\end{pmatrix}\right).$

Use Frobenius descent (see [6, Theorem 2.7]) with respect to the
${\widetilde{G}}$-equivariant map

$\left\\{\left(\begin{pmatrix}a&b\\\ c&a\end{pmatrix},\begin{pmatrix}x\\\
y\end{pmatrix},\begin{pmatrix}ty&tx\end{pmatrix}\right)|\begin{pmatrix}ty&tx\end{pmatrix}\neq
0\right\\}\to\left\\{\left(\begin{pmatrix}x\\\
y\end{pmatrix},\begin{pmatrix}ty&tx\end{pmatrix}\right)|\begin{pmatrix}ty&tx\end{pmatrix}\neq
0\right\\}$

given in the above coordinates. It is easy to see that the action on the
target is transitive. The stabilizer of the point $\left(\begin{pmatrix}1\\\
0\end{pmatrix},\begin{pmatrix}0&1\end{pmatrix}\right)$ inside $G$ is the
unimodular subgroup $N:=\left\\{\begin{pmatrix}1&*\\\
0&1\end{pmatrix}\right\\}$. The stabilizer inside ${\widetilde{G}}$ is
$\tilde{N}:=N\rtimes\\{1,\sigma\\}$, and it is indeed also unimodular. The
fiber above $\left(\begin{pmatrix}1\\\
0\end{pmatrix},\begin{pmatrix}0&1\end{pmatrix}\right)$ is
$\left\\{\begin{pmatrix}a&b\\\ m&a\end{pmatrix}\right\\}$, on which $\sigma$
acts trivially. Thus, since our distribution is $\chi$-equivariant (and so
$\sigma$-anti-invariant), it must be 0.

∎

The following Corollary of the previous theorem appears in [6, Proposition
4.6] and is proved there in a way which fails over a field of characteristic
$2$.

######  Corollary 3.4.

For any point $(A,v,\phi)$ in the support of $\xi$ and any $k\geq 0$, we have
$<A^{k}v,\phi>=0$.

###### Proof.

Let $\Delta:\mathfrak{gl}(V)\times V\times V^{*}\to{\mathbb{F}}[x]$ be the
characteristic polynomial map. Recall the automorphism
$\rho_{g}:\Delta^{-1}(f)\to\Delta^{-1}(f)$ defined for every $g$ coprime to
$f$ by $\rho_{g}((A,v,\phi)):=(A,g(A)v,g(A^{*})\phi)$. By using the
localization principle (see [6, Theorem 2.4]), we can reduce the claim to a
distribution on a single fiber of $\Delta$ over a polynomial $f$. Then for any
$g$ coprime to $f$, we can apply $\rho_{g}$, then extend the distribution back
to $\mathfrak{gl}(V)\times V\times V^{*}$ and apply Theorem 3.1 to get that
$<g(A)v,g(A^{*})\phi>=0$ and Theorem 3.3 to get that
$<Ag(A)v,g(A^{*})\phi>=0$. Since this is true for a Zariski dense set of
polynomials $g$, it is true for all polynomials. Thus we get that for any $g$,
we have $<g(A)^{2}v,\phi>=<g(A),g(A^{*})\phi>=0$ and
$<Ag(A)^{2}v,\phi>=<Ag(A),g(A^{*})\phi>=0$. In particular for any $k\geq 0$ we
can take $g(x)=x^{k}$ to get that $<A^{2k}v,\phi>=0$ and $<A^{2k+1}v,\phi>=0$.
∎

Once this Corollary is proven, the rest of the proof of Theorem 1.3 given in
[6] applies almost verbatim also over a field of characteristic $2$. The only
point in which there is a difference is the usage of [6, Theorem 2.9] of which
the proof relies on the theory of the Weil representation. This theory is a
bit different over characteristic 2. We give in Appendix A the necessary
adaptations, and prove Theorem A.1 which plays the same role as [6, Theorem
2.9].

##  Appendix A The Weil representation over characteristic 2

In this section we prove the following theorem, due to [7] in the case where
$\mathrm{char}{\mathbb{F}}\neq 2$:

######  Theorem A.1.

Let ${\mathbb{F}}$ be a local field with charateristic $2$. Let $V$ be a
finite dimensional linear space over ${\mathbb{F}}$, and let $V^{*}$ be its
dual space. Define $Z:=\\{(v,\phi)\in V\times V^{*}|<v,\phi>=0\\}$. Then for
every distribution $\xi$ on $V\times V^{*}$ such that both $\xi$ and its
Fourier transform ${\mathcal{F}}(\xi)$ are supported on $Z$, we have that
$\xi$ must be ”abs-homogeneous” of degree $\dim V$. That is, for any
$t\in{\mathbb{F}}$ and $\Phi\in{\mathcal{S}}(V\times V^{*})$, we have
$|(t\xi)(\Phi)|=|t|^{\dim V}\cdot|\xi(\Phi)|$.

Let us recall the facts which we will need about the Weil representation in
characteristic $2$ and the pseudo-symplectyc group (see [9] and [5]).

######  Definition A.2.

Let ${\mathbb{F}},V$ as above. We define $X:=V\oplus V^{*}$. Define a bilinear
form $B$ on $X\oplus X^{*}\cong V\oplus V^{*}\oplus V\oplus V^{*}$ by

$B((u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1}),(u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))=<u_{1},v^{\prime}_{2}>+<v_{2},u^{\prime}_{1}>.$

Let $Q$ be the quadratic form on $X\oplus X^{*}$ defined by $B$, and denote by
$\mathfrak{Q}(X\oplus X^{*})$ the linear space of all quadratic forms on
$X\oplus X^{*}$.
Define the pseudo-symplectic group

$\displaystyle PSp:=\\{$ $\displaystyle(\sigma,f)\in
O(Q)\times\mathfrak{Q}(X\oplus X^{*})|\forall w,w^{\prime}\in X\oplus X^{*},$
$\displaystyle B(w,w^{\prime})+B(\sigma w,\sigma
w^{\prime})=f(w+w^{\prime})+f(w)+f(w^{\prime})\\}$

with group law
$(\sigma_{1},f_{1})(\sigma_{2},f_{2})=(\sigma_{1}\sigma_{2},f_{1}\sigma_{2}+f_{2})$.

######  Proposition A.3.

We have an embedding $j:SL_{2}({\mathbb{F}})\to PSp$ defined by

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\mapsto\left(\begin{pmatrix}aI&bI\\\
cI&dI\end{pmatrix},f\right)$

Where $\sigma$ above is defined in coordinates $V\oplus V^{*}\oplus V\oplus
V^{*}$, $I$ being the identity matrix of $V\oplus V^{*}$, and $f$ is defined
by
$f(u,u^{\prime},v,v^{\prime})=ac<u,u^{\prime}>+bd<v,v^{\prime}>+bc(<u,v^{\prime}>+<v,u^{\prime}>)$.

###### Proof.

Indeed:

$\displaystyle
B(\sigma(u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1}),\sigma(u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))+B((u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1}),(u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))=$
$\displaystyle=B((au_{1}+bv_{1},au^{\prime}_{1}+bv^{\prime}_{1},cu_{1}+dv_{1},cu^{\prime}_{1}+dv^{\prime}_{1}),(au_{2}+bv_{2},au^{\prime}_{2}+bv^{\prime}_{2},cu_{2}+dv_{2},cu^{\prime}_{2}+dv^{\prime}_{2}))+$
$\displaystyle+B((u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1}),(u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))=$
$\displaystyle=<au_{1}+bv_{1},cu^{\prime}_{2}+dv^{\prime}_{2}>+<cu_{2}+dv_{2},au^{\prime}_{1}+bv^{\prime}_{1}>+<u_{1},v^{\prime}_{2}>+<v_{2},u^{\prime}_{1}>=$
$\displaystyle=ac(<u_{1},u^{\prime}_{2}>+<u_{2},u^{\prime}_{1}>)+bd(<v_{1},v^{\prime}_{2}>+<v_{2},v^{\prime}_{1}>)+bc(<v_{1},u^{\prime}_{2}>+<u_{2},v^{\prime}_{1}>)+$
$\displaystyle+(ad+1)(<u_{1},v^{\prime}_{2}>+<v_{2},u^{\prime}_{1}>)=$
$\displaystyle=ac(<u_{1},u^{\prime}_{2}>+<u_{2},u^{\prime}_{1}>)+bd(<v_{1},v^{\prime}_{2}>+<v_{2},v^{\prime}_{1}>)+$
$\displaystyle+bc(<v_{1},u^{\prime}_{2}>+<u_{2},v^{\prime}_{1}>+<u_{1},v^{\prime}_{2}>+<v_{2},u^{\prime}_{1}>)=$
$\displaystyle=f((u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1})+(u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))+f((u_{1},u^{\prime}_{1},v_{1},v^{\prime}_{1}))+f((u_{2},u^{\prime}_{2},v_{2},v^{\prime}_{2}))$

To check that this is a morphism of groups, one needs to check that given
$g_{1},g_{2}\in SL_{2}({\mathbb{F}})$ which map into
$(\sigma_{1},f_{1}),(\sigma_{2},f_{2})$, the quadratic form associated to
$\sigma_{1}\sigma_{2}$ is $f_{1}\sigma_{2}+f_{2}$. Indeed

$\displaystyle(f_{1}\sigma_{2}+f_{2})(u,u^{\prime},v,v^{\prime})=$
$\displaystyle=a_{1}c_{1}<a_{2}u+b_{2}v,a_{2}u^{\prime}+b_{2}v^{\prime}>+b_{1}d_{1}<c_{2}u+d_{2}v,c_{2}u^{\prime}+d_{2}v^{\prime}>+$
$\displaystyle+b_{1}c_{1}(<a_{2}u+b_{2}v,c_{2}u^{\prime}+d_{2}v^{\prime}>+<c_{2}u+d_{2}v,a_{2}u^{\prime}+b_{2}v^{\prime}>)+$
$\displaystyle+a_{2}c_{2}<u,u^{\prime}>+b_{2}d_{2}<v,v^{\prime}>+b_{2}c_{2}(<u,v^{\prime}>+<v,u^{\prime}>)=$
$\displaystyle=(a_{1}c_{1}a_{2}^{2}+b_{1}d_{1}c_{2}^{2}+a_{2}c_{2})<u,u^{\prime}>+(a_{1}c_{1}b_{2}^{2}+b_{1}d_{1}d_{2}^{2}+b_{2}d_{2})<v,v^{\prime}>+$
$\displaystyle+(a_{1}c_{1}a_{2}b_{2}+b_{1}d_{1}c_{2}d_{2}+b_{1}c_{1}a_{2}d_{2}+b_{1}c_{1}b_{2}c_{2}+b_{2}c_{2})(<u,v^{\prime}>+<v,u^{\prime}>)=$
$\displaystyle=(a_{1}c_{1}a_{2}^{2}+b_{1}d_{1}c_{2}^{2}+a_{2}c_{2}a_{1}d_{1}+a_{2}c_{2}b_{1}c_{1})<u,u^{\prime}>+$
$\displaystyle+(a_{1}c_{1}b_{2}^{2}+b_{1}d_{1}d_{2}^{2}+a_{1}d_{1}b_{2}d_{2}+b_{1}c_{1}b_{2}d_{2})<v,v^{\prime}>+$
$\displaystyle+(a_{1}c_{1}a_{2}b_{2}+b_{1}d_{1}c_{2}d_{2}+b_{1}c_{1}a_{2}d_{2}+a_{1}d_{1}b_{2}c_{2})(<u,v^{\prime}>+<v,u^{\prime}>)=$
$\displaystyle=(a_{1}a_{2}+b_{1}c_{2})(c_{1}a_{2}+d_{1}c_{2})<u,u^{\prime}>+(a_{1}b_{2}+b_{1}d_{2})(c_{1}b_{2}+d_{1}d_{2})<v,v^{\prime}>+$
$\displaystyle+(a_{1}b_{2}+b_{1}d_{2})(c_{1}a_{2}+d_{1}c_{2})(<u,v^{\prime}>+<v,u^{\prime}>)$

which is indeed the quadratic form associated to $\sigma_{1}\sigma_{2}$. ∎

######  Theorem A.4 (see [9] and [5]).

Fix an additive character $\psi$ of ${\mathbb{F}}$. There is a projective
representation $\rho$ of $PSp$ on ${\mathcal{S}}(X)$, such that for any
$a,b\in PSp$, we have $\rho(ab)^{-1}\rho(a)\rho(b)=\pm 1$. We also have the
explicit formulas, with $x=(v,v^{\prime})\in X$:

$\left(\left(\rho j\begin{pmatrix}t&0\\\
0&t^{-1}\end{pmatrix}\right)\Phi\right)(x)=|t|^{\dim V}\Phi(tx),$
$\left(\left(\rho j\begin{pmatrix}1&u\\\
0&1\end{pmatrix}\right)\Phi\right)(x)=\psi(u<v,v^{\prime}>)\Phi(x),$
$\left(\left(\rho j\begin{pmatrix}0&1\\\
1&0\end{pmatrix}\right)\Phi\right)(x)={\mathcal{F}}(\Phi)(x).$

In the last equation ${\mathcal{F}}$ denotes the Fourier transform on $X$ with
respect to the symmetric bilinear non-degenerate form
$((u,u^{\prime}),(v,v^{\prime})):=<u,v^{\prime}>+<v,u^{\prime}>$.

###### Proof of Theorem A.1.

The projective action of $PSp$ on ${\mathcal{S}}(V\oplus V^{*})$ gives a
projective representation of $PSp$ on ${\mathcal{S}}^{*}(V\oplus V^{*})$.
Looking in the formulas for this action, we see that if
$\xi\in{\mathcal{S}}^{*}(V\oplus V^{*})$ is as in the formulation of the
theorem, we have

$\rho j\begin{pmatrix}1&u\\\ 0&1\end{pmatrix}(\xi)=\pm\xi$

and also

$\rho j\begin{pmatrix}1&0\\\ u&1\end{pmatrix}(\xi)=\rho
j\left(\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\begin{pmatrix}1&u\\\
0&1\end{pmatrix}\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\right)(\xi)=\pm\xi.$

Since elements of the form $\begin{pmatrix}1&u\\\ 0&1\end{pmatrix}$ and
$\begin{pmatrix}1&0\\\ u&1\end{pmatrix}$ generate $SL_{2}({\mathbb{F}})$, we
get that $\xi$ is $SL_{2}({\mathbb{F}})$ $\pm$-invariant. I particular $\rho
j\begin{pmatrix}t&0\\\ 0&t^{-1}\end{pmatrix}(\xi)=\pm\xi$. So we get that

$|t|^{-\dim V}\cdot|(t\xi)(\Phi)|=|\xi(\Phi)|$

as desired. ∎

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