Gamma positivity of the Descent based Eulerian polynomial in positive elements of Classical Weyl Groups

TL;DR

This paper investigates the gamma positivity of positive Eulerian polynomials derived from classical Weyl groups, revealing specific modular conditions under which these polynomials are gamma positive or can be expressed as sums of gamma positive polynomials.

Contribution

It establishes gamma positivity criteria for positive Eulerian polynomials in classical Weyl groups, including type-A and type-D, based on the congruence class of n modulo 4.

Findings

$A_n^+(t)$ is gamma positive iff n ≡ 0,1 (mod 4)

For n ≡ 2 (mod 4), $A_n^+(t)$ is a sum of two gamma positive polynomials

For n ≡ 3 (mod 4), $A_n^+(t)$ is a sum of three gamma positive polynomials

Abstract

The classical Eulerian polynomials $A_n(t)$ are known to be gamma positive. Define the positive Eulerian polynomial $A_n^+(t)$ as the polynomial obtained when we sum descents over the alternating group. We show that $A_n^+(t)$ is gamma positive iff $n \equiv 0,1$ (mod 4). When $n \equiv 2$ (mod 4) we show that $A_n^+(t)$ can be written as a sum of two gamma positive polynomials while if $n \equiv 3$ (mod 4), we show that $A_n^+(t)$ can be written as a sum of three gamma positive polynomials. Similar results are shown when we consider the positive type-D and type-D Eulerian polynomials.

Equations

$$A_n(t)$$

$$A_n^{+}(t)$$

$$A_n^{-}(t)$$

$$A_n(p,q,t)=\sum _{\pi \in {\mathfrak{S}}_n}{p}^{2-13(\pi )}{q}^{31-2(\pi )}{t}^{\mathsf{des}}(\pi )=\sum _{i=0}^{\lfloor n/2\rfloor }{a}_{n,i}(p,q)t^i(1+t{)}^{n-2i}$$

$$A_{n,\mathsf{exc}}(t)=\sum _{\pi \in {\mathfrak{S}}_n}{t}^{\mathsf{exc}(\pi )}\text{mbox}and{A}_{n,\mathsf{exc}}^{+}(t)=\sum _{\pi \in {\mathcal{A}}_n}{t}^{\mathsf{exc}(\pi )}$$

$$f(s,t)$$

$$Df(s,t)$$

$$f(t)=\sum _i^{\lfloor n/2\rfloor }{\gamma }_i{t}^i(1+t{)}^{n-2i-1}(1+t)$$

$$A_n(s,t)=\sum _{i=0}^{\lfloor (n-1)/2\rfloor }{\gamma }_{n,i}(st{)}^i(s+t{)}^{n-1-2i}$$

$$A_{n+1}(s,t)=(s+t)A_n(s,t)+stD{A}_n(s,t).$$

$${\gamma }_{n+1,i}=(i+1){\gamma }_{n,i}+2(n+1-2i){\gamma }_{n,i-1}.$$

$$\sum _{\pi \in S}{t}^{\mathsf{des}(\pi )}{s}^{\mathsf{asc}(\pi )}=\sum _{\pi \in T}{t}^{\mathsf{des}(\pi )}{s}^{\mathsf{asc}(\pi )}.$$

$$f(a_1,a_2,\dots ,a_{2r+1},2m+1,a_{2r+2},\dots ,a_{2m})=a_{2r+2},\dots ,a_{2m},2m+1,a_1,a_2,\dots ,a_{2r+1}.$$

$$\mathsf{inv}(\pi )$$

$$\mathsf{inv}(f(\pi ))$$

$$\mathsf{inv}(\pi )-\mathsf{inv}(f(\pi ))$$

$$\equiv$$

$$g(a_1,\dots ,a_{2m-2r-1},2m+1,a_{2m-2r},\dots ,a_{2m})=a_{2m-2r},\dots ,a_{2m},2m+1,a_1,\dots ,a_{2m-2r-1}.$$

$$P_r^{2m+1}(s,t)$$

$$Q_r^{2m+1}(s,t)$$

$$\sum _{r=0}^{m-1}{P}_r^{2m+1}(s,t)=\sum _{r=0}^{m-1}{Q}_r^{2m+1}(s,t)$$

$$A_{2m+1}^{+}(s,t)$$

$$A_{2m+1}^{-}(s,t)$$

$$\sum _{\pi \in K}{t}^{\mathsf{des}(\pi )}{s}^{\mathsf{asc}(\pi )}$$

$$\sum _{\pi \in U}{t}^{\mathsf{des}(\pi )}{s}^{\mathsf{asc}(\pi )}$$

$$G_r^{2m}(s,t)$$

$$I_r^{2m}(s,t)$$

$$\sum _{r=0}^{m-1}{G}_r^{2m}(s,t)=\sum _{r=0}^{m-1}{H}_r^{2m}(s,t)\text{mbox}and\sum _{r=0}^{m-1}{I}_r^{2m}(s,t)=\sum _{r=0}^{m-1}{J}_r^{2m}(s,t).$$

$$A_{2m}^{+}(s,t)$$

$$A_{2m}^{-}(s,t)$$

$$a_{n,i}^{+}=(n-k)a_{n-1,k-1}^{+}+(k+1)a_{n-1,k}^{+}\text{mbox}and{a}_{n,i}^{-}=(n-k)a_{n-1,k-1}^{-}+(k+1)a_{n-1,k}^{-}.$$

$$2a_{n,i}^{+}$$

$$2a_{n,i}^{-}$$

$$A_{4k+4}^{+}(s,t)$$

$$A_{4m+2}^{+}(s,t)$$

$$A_{4m+2}^{+}(t)$$

Full text

Gamma positivity of the Descent based Eulerian

polynomial in positive elements of Classical Weyl Groups

Hiranya Kishore Dey

Department of Mathematics

Indian Institute of Technology, Bombay

Mumbai 400 076, India.

email: hkdey@math.iitb.ac.in

  

Sivaramakrishnan Sivasubramanian

Department of Mathematics

Indian Institute of Technology, Bombay

Mumbai 400 076, India.

email: krishnan@math.iitb.ac.in

1 Introduction

Let $f(t)\in\mathbb{Q}[t]$ be a degree $n$ univariate polynomial with $f(t)=\sum_{i=0}^{n}a_{i}t^{i}$ where $a_{i}\in\mathbb{Q}$ with $a_{n}\not=0$. Let $r$ be the least non-negative integer such that $a_{r}\not=0$. Define $\mathsf{len}(f)=n-r$. $f(t)$ is said to be palindromic if $a_{r+i}=a_{n-i}$ for $0\leq i\leq\left\lfloor(n-r)/2\right\rfloor$. Define the center of symmetry of $f(t)$ to be $\mathrm{\operatorname{c-o-s}}(f(t))=(n+r)/2$. Note that for a palindromic polynomial $f(t)$, its center of symmetry $\mathrm{\operatorname{c-o-s}}(f(t))$ could be half integral.

Let $\mathrm{PalindPoly}_{(n+r)/2,r}(t)$ denote the vector space of palindromic univariate polynomials $f(t)$ with least non zero exponent of $t$ being at least $r$ and with $\mathrm{\operatorname{c-o-s}}(f(t))=(n+r)/2$,. Clearly, $\Gamma=\{t^{r+i}(1+t)^{n-r-2i}:0\leq i\leq\lfloor(n-r)/2\rfloor\}$ is a basis of $\mathrm{PalindPoly}_{(n+r)/2,r}(t)$. Hence, if $f(t)\in\mathrm{PalindPoly}_{(n+r)/2,r}(t)$, then we can write $f(t)=\sum_{i=0}^{\left\lfloor(n-r)/2\right\rfloor}\gamma_{n,i}t^{r+i}(1+t)^{n-r-2i}$. $f(t)$ is said to be gamma positive if $\gamma_{n,i}\geq 0$ for all $i$ (that is, if $f(t)$ has positive coefficients when expressed in the basis $\Gamma$).

For a positive integer $n$, let $[n]=\{1,2,\ldots,n\}$. Denote by $\mathfrak{S}_{n}$ the symmetric group and let $\mathcal{A}_{n}\subseteq\mathfrak{S}_{n}$ be the alternating group on $[n]$. For $\pi\in\mathfrak{S}_{n}$, let $\mathsf{Des\_Set}(\pi)=\{i\in[n-1]:\pi_{i}>\pi_{i+1}\}$ be its set of descents and let $\mathsf{des}(\pi)=|\mathsf{Des\_Set}(\pi)|$ be its number of descents. Let $\mathsf{asc}(\pi)=n-1-\mathsf{des}(\pi)$ denote its number of ascents. Further, let

[TABLE]

It is well known that the Eulerian polynomials $A_{n}(t)$ are palindromic (see Graham, Knuth and Patashnik [8]). Gamma positivity of $A_{n}(t)$ was first proved by Foata and Schützenberger (see [6] and see Theorem 8 as well). Foata and Strehl [7] later used a group action based proof which has been termed as “valley hopping” by Shapiro, Woan and Getu [14]. The same proof shows that the Narayana polynomials are gamma positive (see Petersen’s book [13, Chapter 4]).

It is a well known result of MacMahon [11] that both descents and excedances are equidistributed over $\mathfrak{S}_{n}$. Several refinements of this result are known both when enumeration is done with respect to descents $\mathsf{des}()$ and with respect to excedance $\mathsf{exc}()$. For two statistics $\operatorname{2-13},\operatorname{31-2}:\mathfrak{S}_{n}\mapsto\mathbb{Z}_{\geq 0}$, Brändén [3] and Shin-and-Zeng [15, 16] have shown a $p,q$-refinement by showing that the polynomial

[TABLE]

where the $a_{n,i}(p,q)$’s are polynomials with positive coefficients.

The survey [2] by Athanasiadis and the Handbook chapter [4] by Brändén contain a wealth of information related to gamma positivity of polynomials arising in combinatorics. The polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ have been studied by Tanimoto [19] who gave recurrences for $a_{n,k}^{+},a_{n,k}^{-}$, the coefficients of $t^{k}$ in these polynomials. To the best of our knowledge, there are no results on gamma positivity of the polynomials $A_{n}^{+}(t)$. In this work, we show the following results.

Theorem 1

For $n\geq 1$, both $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ are gamma positive iff $n\equiv 0,1$ (mod 4). Further, both the polynomials have the same center of symmetry.

Theorem 2

When $n\equiv 2$ (mod 4), the polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ can be written as a sum of two gamma positive polynomials.

Theorem 3

Let $n\equiv 3$ (mod 4). Then, $A_{4m+3}^{+}(t)$ and $A_{4m+3}^{-}(t)$ can be written as a sum of three gamma positive polynomials.

Theorem 1 thus gives a different refinement of Foata and Schützenberger’s Gamma positivity result (see Theorem 8) when $n\equiv 0,1$ (mod 4). Theorems 2 and 3 refine the univariate version of Theorem 8.

We generalize our results to the case when descents are summed up over the elements with positive sign in classical Weyl groups. Let $\mathfrak{B}_{n}$ denote the group of signed permutations on $T_{n}=\{-n,-(n-1),\ldots,-1,1,2,\ldots,n\}$, that is $\sigma\in\mathfrak{B}_{n}$ consists of all permutations of $T_{n}$ that satisfy $\sigma(-i)=-\sigma(i)$ for all $i\in[n]$. Similarly, let $\mathfrak{D}_{n}\subseteq\mathfrak{B}_{n}$ denote the subset consisting of those elements of $\mathfrak{B}_{n}$ which have an even number of negative entries. As both $\mathfrak{B}_{n}$ and $\mathfrak{D}_{n}$ are Coxeter groups, they have a natural notion of descent associated to them. Similar to the classical Eulerian polynomials, we thus have the Eulerian polynomials of type-B and type-D. Gamma positivity of the type-B Eulerian polynomial was shown by Chow [5] and Petersen [12]. Gamma positivity of the type-D Eulerian polynomial was shown by Stembridge [18] and by Chow [5]. There is a natural notion of length in these groups and we get results when we enumerate descents in these groups but restrict the sum over elements with even length. Our results for Type-B Weyl groups are Theorems 28 and 29. Similarly for Type-D Weyl groups, our main result is Theorem 35.

For $\pi\in\mathfrak{S}_{n}$, let $\mathsf{exc}(\pi)=|\{i\in[n-1]:\pi_{i}>i\}|$ denote its number of excedances. Define

[TABLE]

As mentioned earlier, it is well known that both descents and excedances are equidistributed over $\mathfrak{S}_{n}$. It is also easy to see that they are not equidistributed over $\mathcal{A}_{n}$. That is, for all $n$, we have $A_{n}(t)=A_{n,\mathsf{exc}}(t)$ while $A_{n}^{+}(t)\not=A_{n,\mathsf{exc}}^{+}(t)$. In this paper we deal with gamma positivity of the descent based polynomials $A_{n}^{+}(t)$. Similar results can be proved about the polynomials $A_{n,\mathsf{exc}}^{+}(t)$. We plan to do that in a subsequent paper.

2 Preliminaries on Gamma positive polynomials

Most of the univariate polynomials in this work have homogeneous bivariate counterparts with the following slightly more general definition of gamma positivity. Let $f(s,t)=\sum_{i=0}^{n}a_{i}s^{n-i}t^{i}$ be a homogeneous bivariate polynomial with $a_{n}\not=0$ and let $r$ be the smallest non-negative integer such that $a_{r}\not=0$. As before, define $f(s,t)$ to be palindromic if $a_{r+i}=a_{n-i}$ for all $i$ and define $\mathrm{\operatorname{c-o-s}}(f(s,t))=(n+r)/2$.

Let $f(s,t)$ have $\mathrm{\operatorname{c-o-s}}(f(s,t))=n/2$. $f(s,t)$ is said to be bivariate gamma positive if $f(s,t)=\sum_{i=0}^{n/2}\gamma_{n,i}(st)^{r+i}(s+t)^{n-r-2i}$ with $\gamma_{n,i}\geq 0$ for all $i$. Clearly, if $f(s,t)$ is a bivariate gamma positive polynomial, then $f(t)=f(s,t)|_{s=1}$ is clearly a univariate gamma positive palindromic polynomial with the same center of symmetry. We need the following lemmas. Some of them are elementary and so we omit proofs.

Lemma 4

Let $f_{1}(s,t)$ and $f_{2}(s,t)$ be two bivariate gamma positive polynomials with respective centers of symmetry $\mathrm{\operatorname{c-o-s}}(f_{1}(s,t))=r_{1}+(n-r_{1})/2$ and $\mathrm{\operatorname{c-o-s}}(f_{2}(s,t))=r_{2}+(n-r_{2})/2$. Then $f_{1}(s,t)f_{2}(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(f_{1}(s,t)f_{2}(s,t))=(r_{1}+r_{2})+(2n-r_{1}-r_{2})/2$.

Let $D$ be the operator $\displaystyle\left(\frac{d}{ds}+\frac{d}{dt}\right)$ in $\mathbb{Q}[s,t]$.

Lemma 5

Let $f(s,t)$ be a bivariate gamma positive polynomial with $\mathrm{\operatorname{c-o-s}}(f(s,t))=n/2$. Then, $g(s,t)=Df(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(g(s,t))=(n-1)/2$.

Proof:   Let,

[TABLE]

Hence $\mathrm{\operatorname{c-o-s}}(Df(s,t))=(n-1)/2$ completing the proof.

The following are easy corollaries.

Corollary 6

Let $f(s,t)$ be a bivariate gamma positive polynomial with $\mathrm{\operatorname{c-o-s}}(f(s,t))=n/2$.

Then, for a natural number $\ell$, $\displaystyle g(s,t)=D^{\ell}f(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(g(s,t))=(n-\ell)/2$. 2. 2.

Then, $h(s,t)=(st)^{i}f(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(h(s,t))=i+n/2$ 3. 3.

Then, $h(s,t)=(s+t)f(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(h(s,t))=(n+1)/2$

Lemma 7

Let $f(t)$ be gamma positive with $\mathrm{\operatorname{c-o-s}}(f(t))=n/2$ and with $\mathsf{len}(f(t))$ being odd. Then, $f(t)$ is the sum of two gamma positive sequences $p_{1}(t)$ and $p_{2}(t)$ with centers of symmetry $(n-1)/2$ and $(n+1)/2$ respectively. Further, both $p_{1}(t)$ and $p_{2}(t)$ have even length.

Proof:   Let the degree of $f(t)$ be $d$ and let $f(t)=\sum_{i=n-d}^{\lfloor n/2\rfloor}\gamma_{i}t^{i}(1+t)^{n-2i}$. We are given $\mathrm{\operatorname{c-o-s}}(f(t))=k=n/2$. Thus,

[TABLE]

It is easy to see that $p_{1}(t)$ and $p_{2}(t)$ are gamma positive with respective centers of symmetry $(n-1)/2$ and $(n+1)/2$. Note that oddness of length is required in the proof and that $p_{1}(t)$ and $p_{2}(t)$ have even length.

2.1 Gamma positivity of the Eulerian numbers

We sketch a proof that the bivariate Eulerian polynomials $A_{n}(s,t)$ are gamma positive. We do this as it sets up the stage for proofs of other results in this paper. The result is due to Foata and Schützenberger (see [6]) and the proof given below is a modification of the proof given in [20].

Theorem 8** (Foata and Schützenberger)**

With $A_{n}(s,t)$ as defined in (1), we have

[TABLE]

where the $\gamma_{n,i}$ are positive integers for all $n,i$.

Proof:  (Sketch) By induction on $n$. The base case when $n=2$ is easy to see. Foata and Schützenberger showed the following recurrence

[TABLE]

This can be seen by adding the symbol $(n+1)$ in the $n+1$ places of each permutation $\pi\in\mathfrak{S}_{n}$. The first term $(s+t)A_{n}(s,t)$ accounts for those permutations in $\mathfrak{S}_{n+1}$ in which the letter $(n+1)$ appears in the first or the last position. The term $\displaystyle stDA_{n}(s,t)$ is the contribution of all $\pi\in\mathfrak{S}_{n+1}$ in which the letter $(n+1)$ appears in positions $r$ for $2\leq r\leq n$.

Let $\displaystyle T=(s+t)+stD$ be an operator in $\mathbb{Q}[s,t]$. By induction, $A_{n}(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(A_{n}(s,t))=(n-1)/2$. By Corollary 6, both $(s+t)A_{n}(s,t)$ and $stDA_{n}(s,t)$ are gamma positive with centers of symmetry $n/2$. Thus their sum is also gamma positive with the same center of symmetry $n/2$. From the above argument, the following recurrence is easy to obtain

[TABLE]

Note that the above recurrence together with the initial conditions $\gamma_{1,0}=1$ or $\gamma_{2,0}=1,\gamma_{2,1}=0$ settles non-negativity of $\gamma_{n,i}$ for all $n,i$, completing the proof.

Corollary 9

From (7), we get that $\gamma_{n,i}$ is even for all $n\geq 1$ and for all $i\geq 1$ while $\gamma_{n,0}=1$ for all $n$. Thus, for all $n$, the only odd gamma coefficient is $\gamma_{n,0}$.

3 Recurrences for $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$

Recall $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ were defined in (2) and (3). We first consider palindromicity of $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$.

Lemma 10

For $n\geq 1$, the polynomials $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ are palindromic iff $n\equiv 0,1$ mod 4.

Proof:   Let $\pi=x_{1},x_{2},\ldots,x_{n}\in\mathfrak{S}_{n}$. Define $f:\mathfrak{S}_{n}\mapsto\mathfrak{S}_{n}$ by $f(\pi)=n+1-x_{1},n+1-x_{2},\ldots,n+1-x_{n}$. Clearly, $f$ is a bijection and $\mathsf{inv}_{A}(\pi)+\mathsf{inv}_{A}(f(\pi))=\binom{n}{2}$. Hence, if $n\equiv 0,1$ mod 4 and $\pi\in\mathcal{A}_{n}$, then $f(\pi)\in\mathcal{A}_{n}$.

If $n\equiv 2,3$ mod 4, then the map $f$ flips the parity of $\mathsf{inv}(\pi)$ and hence $a_{n,k}^{+}=a_{n,n-1-k}^{-}$. Further, $a_{n,0}^{+}=1$ and $a_{n,n-1}^{+}=a_{n,0}^{-}=0$ and hence $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ are not palindromic.

For $\pi\in\mathcal{A}_{n}$ let $\pi^{\prime}=\pi$ restricted to $[n-1]$. It is easy to see that both $\pi^{\prime}\in\mathcal{A}_{n-1}$ and $\pi^{\prime}\in\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}$ are possible. A similar statement is true when we restrict $\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}$ to $[n-1]$.

In subsequent results, we will need to keep track of the position of the letter $n$ in $\pi$. Let $\pi=x_{1},x_{2},\ldots,x_{n}\in\mathfrak{S}_{n}$. We term the left-most position before $x_{1}$ as the initial or ‘zero’-th gap and the right-most position after $x_{n}$ as the final gap. For $1\leq i\leq n$, we denote the gap between $x_{i}$ and $x_{i+1}$ as the $`i$’-th gap of a permutation. Define $\mathsf{pos\_n}(\pi)$ to be the index $i$ such that $\pi(i)=n$ and recall $\pi^{\prime}$ is $\pi$ restricted to $[n-1]$. We start with the following definitions.

$\mathcal{A}_{n}^{i}$= $\{\pi\in{\mathcal{A}_{n}}:\mathsf{pos\_n}(\pi)=i+1\}$. That is, $\mathcal{A}_{n}^{i}$ consists of those $\pi\in{\mathcal{A}_{n}}$ which arise from all possible $\pi^{\prime}\in\mathfrak{S}_{n-1}$ by inserting the letter $``n"$ in the $i$-th gap for $0\leq i\leq n-1$. 2. 2.

$[\mathcal{A}_{n-1}\rightarrow\mathcal{A}_{n}^{i}]=\{\pi\in\mathcal{A}_{n}:\mathsf{pos\_n}(\pi)=i+1,\pi^{\prime}\in\mathcal{A}_{n-1}\}$. That is, $[\mathcal{A}_{n-1}\rightarrow\mathcal{A}_{n}^{i}]$ consists of those $\pi\in\mathcal{A}_{n}$ which arise from $\pi^{\prime}\in\mathcal{A}_{n-1}$ by inserting the letter $``n"$ in the $i$-th gap for $0\leq i\leq n-1$. 3. 3.

$[(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})\rightarrow\mathcal{A}_{n}^{i}]=\{\pi\in{\mathcal{A}_{n}}:\mathsf{pos\_n}(\pi)=i+1,\pi^{\prime}\in(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})\}$. We think of $[(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})\rightarrow\mathcal{A}_{n}^{i}]$ as those $\pi\in\mathcal{A}_{n}$ which arise from $\pi^{\prime}\in(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})$ by inserting the letter $``n"$ in the $i$-th gap for $0\leq i\leq n-1$. 4. 4.

$(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}$= $\{\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}:\mathsf{pos\_n}(\pi)=i+1,\pi^{\prime}\in\mathfrak{S}_{n-1}\}$. We think of $(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}$ as those $\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}$ which arise from all $\pi^{\prime}\in\mathfrak{S}_{n-1}$ by inserting the letter $`n^{\prime}$ in the $i$-th position for $0\leq i\leq n-1$. 5. 5.

$[\mathcal{A}_{n-1}\rightarrow(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}]=\{\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}:\mathsf{pos\_n}(\pi)=i+1,\pi^{\prime}\in\mathcal{A}_{n-1}\}$ . We think of $[\mathcal{A}_{n-1}\rightarrow(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}]$ as those $\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}$ which arise from all $\pi^{\prime}\in\mathcal{A}_{n-1}$ by inserting the letter $``n"$ in the $i$-th position for $0\leq i\leq n-1$. 6. 6.

$[(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})\rightarrow(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}]=\{\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}:\mathsf{pos\_n}(\pi)=i+1,\pi^{\prime}\in\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}\}.$ We can think of $[(\mathfrak{S}_{n-1}-\mathcal{A}_{n-1})\rightarrow(\mathfrak{S}_{n}-\mathcal{A}_{n})^{i}]$ as those $\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}$ which arise from all $\pi^{\prime}\in\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}$ by inserting the letter $`n^{\prime}$ in the $i$-th position for $0\leq i\leq n-1$.

The way permutations in $\mathcal{A}_{n}$ arise from elements in $\mathcal{A}_{n-1}$ and $\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}$ depends on the parity of $n$ and is given by the following two lemmas. Since the proofs are simple, we omit them.

Lemma 11

For a positive integer $m$, let $n=2m+1$. Then, $\pi\in\mathcal{A}_{n}$ arises by placing $``n"$ in either the first or the last position of elements of $\mathcal{A}_{2m}$, or placing $``n"$ in the even gaps of elements of $\mathcal{A}_{2m}$. This way, after insertion, $``n"$ will appear in an odd position in $\mathcal{A}_{n}$. Or we place $``n"$ in the odd gaps of elements of $\mathfrak{S}_{2m}-\mathcal{A}_{2m}$. This way, after insertion, $``n"$ will appear in an even position in $\mathcal{A}_{n}$.

Lemma 12

For a positive integer $m$, let $n=2m$. Then, $\pi\in\mathcal{A}_{n}$ arises either by placing $``n"$ in the last position of elements of $\mathcal{A}_{n-1}$, or by placing $``n"$ in the first position of elements of $\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}$ or by placing $``n"$ in the odd gaps of elements of $\mathcal{A}_{n-1}$. This way, after insertion, $``n"$ will appear in an even position of $\mathcal{A}_{n}$. Or, we could place $``n"$ in the even gaps of elements of $\mathfrak{S}_{n-1}-\mathcal{A}_{n-1}$. This way, after insertion, $``n"$ will appear in an odd position in $\mathcal{A}_{n}$.

Our recurrence relation for the polynomials $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ depend on the parity of $n$ and so we bifurcate the remaining part into two cases.

3.1 When $n=2m+1$

We begin with the following.

Lemma 13

For $0\leq r\leq m-1$, let $S=[(\mathfrak{S}_{2m}-\mathcal{A}_{2m})\rightarrow\mathcal{A}_{2m+1}^{2r+1}]$ and $T=[\mathcal{A}_{2m}\rightarrow(\mathfrak{S}_{2m+1}-\mathcal{A}_{2m+1})^{2m-2r-1}]$. Then, the following is true.

[TABLE]

Proof:   It is clear that if $\pi\in\mathcal{A}_{2m+1}^{2r+1}$ then $\pi^{\prime}\in\mathfrak{S}_{2m}-\mathcal{A}_{2m}$ and that if $\pi\in(\mathfrak{S}_{2m+1}-\mathcal{A}_{2m+1})^{2r+1}$ then $\pi^{\prime}\in\mathcal{A}_{2m}$. We give a bijection $f:S\mapsto T$ such that $t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}=t^{\mathsf{des}(f(\pi))}s^{\mathsf{asc}(f(\pi))}$ for all $\pi\in S$. Define

[TABLE]

Let $\pi\in\mathfrak{S}_{2m+1}$ and let $K,L$ be restrictions of $\pi$. Define $\mathsf{invpair}_{\pi}(K,L)=|\{(i,j):\pi_{i}\in K,\pi_{j}\in L,i<j,\pi_{i}>\pi_{j}\}|$. That is, $\mathsf{invpair}_{\pi}(K,L)$ is the number of inversions in $\pi$ between elements of $K$ and $L$. Clearly,

[TABLE]

[TABLE]

The last line follows as

$\mathsf{invpair}_{\pi}([a_{2r+2},\ldots,a_{2m}],[a_{1},\ldots,a_{2r+1}])-\mathsf{invpair}_{\pi}([a_{1},\ldots,a_{2r+1}],[a_{2r+2},\ldots,a_{2m}])$

$\equiv(2m-2r-1)*(2r+1)$ (mod 2) $\equiv$ 1 (mod 2). Thus, $\pi\in\mathcal{A}_{2m+1}$ iff $f(\pi)\in\mathfrak{S}_{2m+1}-\mathcal{A}_{2m+1}$. Define $g:T\mapsto S$ by

[TABLE]

Clearly, $f$ is a bijection with $f^{-1}=g$. Further, for $\pi\in S$, it is clear that $t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}=t^{\mathsf{des}(f(\pi))}s^{\mathsf{asc}(f(\pi))}$, completing the proof.

For $0\leq r\leq m$, define

[TABLE]

Summing Lemma 13 when $r$ runs from [math] to $m-1$, we get the following.

Corollary 14

For any odd natural number $2m+1$, the following equality holds.

[TABLE]

With these preliminary results, we can now prove the following.

Theorem 15

Similar to the recurrence (6) for $A_{n}(s,t)$, we get the following two recurrences for $A_{2m+1}^{+}(s,t)$ and $A_{2m+1}^{-}(s,t)$ respectively.

[TABLE]

Proof:   We prove (15) first. Note from the proof of Theorem 8, that the right hand side equals $\sum_{\pi\in\mathcal{A}_{2m+1}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}$ where the summation is over $\pi$ with $2m+1$ coming in all the positions of $\mathcal{A}_{2m}$.

Recall $A_{2m+1}^{+}(s,t)=\sum_{\pi\in\mathcal{A}_{2m+1}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}$. We consider the contribution to the right hand side from $\pi\in\mathcal{A}_{2m+1}$ with the letter $2m+1$ occurring in position $\ell$ for all possible choices of $\ell$.

Contribution from $\pi$ where $2m+1$ appears in the first or the last position are accounted for by the terms $sA_{2m}^{+}(s,t)$ and $tA_{2m}^{+}(s,t)$ respectively. The remaining $\pi\in\mathcal{A}_{2m+1}$ arise when $2m+1$ appears in even positions of $\mathcal{A}_{2m}$ or when $\pi\in\mathcal{A}_{2m+1}$ arise when $2m+1$ appears in odd positions of $\mathfrak{S}_{2m}-\mathcal{A}_{2m}$. The contribution of these two possibilities are accounted for by the two terms $\sum_{r=1}^{m-1}\left(\sum_{\pi\in\mathcal{A}_{2m}\rightarrow\mathcal{A}_{2m+1}^{2r}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}\right)$ and $\sum_{r=1}^{m-1}\left(\sum_{\pi\in(\mathfrak{S}_{2m}-\mathcal{A}_{2m})\rightarrow\mathcal{A}_{2m+1}^{2r+1}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}\right)$ respectively.

By (13), the second term above $\sum_{r=1}^{m-1}\left(\sum_{\pi\in(\mathfrak{S}_{2m}-\mathcal{A}_{2m})\rightarrow\mathcal{A}_{2m+1}^{2r+1}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}\right)=P_{r}^{2m+1}(s,t)$. By Corollary 14, this equals $Q_{r}^{2m+1}(s,t)$. This equals the sum over all $\pi$ with $2m+1$ coming in all the positions of $\mathcal{A}_{2m}$, completing the proof. An identical proof shows (16).

3.2 When $n=2m$

The moves we make are similar to the case when $n=2m+1$, though there are minor differences. Hence, we omit some proofs. We start with the following counterpart of Lemma 13.

Lemma 16

For $0\leq r\leq m-1$, let $K=[\mathcal{A}_{2m-1}\rightarrow\mathcal{A}_{2m}^{2r+1}]$ and $L=[\mathcal{A}_{2m-1}\rightarrow(\mathfrak{S}_{2m}-\mathcal{A}_{2m})^{2m-2r-2}]$. Let $U=[(\mathfrak{S}_{2m-1}-\mathcal{A}_{2m-1})\rightarrow(\mathfrak{S}_{2m-1}-\mathcal{A}_{2m-1})^{2r+1}]$ and $V=[(\mathfrak{S}_{2m-1}-\mathcal{A}_{2m-1})\rightarrow\mathcal{A}_{2m-1}^{2m-2r-2}]$. Then, we assert the following.

[TABLE]

Proof:   A bijection similar to the one used to prove Lemma 13 works. We omit the details.

For $n=2m$ and $0\leq r\leq m-2$ define

[TABLE]

Summing over Lemma 16 when $0\leq r\leq m-2$ gives the following corollary.

Corollary 17

For any even natural number $2m$, the following two equalities hold.

[TABLE]

Theorem 18

Similar to the recurrence (6) for $A_{n}(s,t)$, we get the following recurrences for $A_{2m}^{+}(s,t)$ and $A_{2m}^{-}(s,t)$.

[TABLE]

Proof:   We prove (21) first. Recall $A_{2m}^{+}(s,t)=\sum_{\pi\in\mathcal{A}_{2m}}t^{\mathsf{des}(\pi)}s^{\mathsf{asc}(\pi)}$. Consider the contribution to the right hand side from $\pi\in\mathcal{A}_{2m}$ with the letter $2m$ occurring in position $\ell$ for all possible choices of $\ell$.

Contribution from $\pi$ where $2m$ appears in the first or the last position are accounted for by the terms $tA_{2m-1}^{-}(s,t)$ and $sA_{2m-1}^{+}(s,t)$ respectively. The remaining $\pi\in\mathcal{A}_{2m}$ arise when $2m$ appears in even positions of $\mathcal{A}_{2m-1}$ or when $2m$ appears in odd positions of $\mathfrak{S}_{2m-1}-\mathcal{A}_{2m-1}$. By Corollary 17, such permutations contribute $\frac{1}{2}\sum_{r=0}^{m-2}\left(G_{r}^{2m}(s,t)+H_{r}^{2m}(s,t)+I_{r}^{2m}(s,t)+J_{r}^{2m}(s,t)\right)$, completing the proof.

Tanimoto gave the following recurrence for the numbers $a_{n,k}^{+}$ and $a_{n,k}^{-}$. These easily follow from Theorems 15 and 18.

Corollary 19** (Tanimoto)**

Let $n=2m+1$ be an odd positive integer. Then,

[TABLE]

Let $n=2m$ be an even positive integer. Then,

[TABLE]

4 Gamma positivity of $A_{n}^{+}(s,t)$ when $n\equiv 0,1$ (mod 4)

We are now ready to prove one of our main results.

Proof:   (Of Theorem 1) By Lemma 10, the polynomials $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ are palindromic iff $n\equiv 0,1$ (mod 4). Our proof is by induction on $n$ with the base case being $n=4$. In this case, it is easy to see that $A_{4}^{+}(s,t)=s^{3}+5s^{2}t+5st^{2}+t^{3}$ and that $A_{4}^{-}(s,t)=6s^{2}t+6st^{2}$. Both these are clearly gamma positive.

When $n\equiv 1$ (mod 4), the recurrence in Theorem 15 is identical to Foata and Schützenberger’s recurrence in Theorem 8. Thus, the same proof of Theorem 8 shows that both $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$ are gamma positive with the same center of symmetry. To go from $n=4k+1$ to $n=4k+4$, we use both Theorem 18 and Theorem 15. Recall $D$ is the operator $\displaystyle\left(\frac{d}{ds}+\frac{d}{dt}\right)$. The following is easy to see.

[TABLE]

where the following table lists $L_{i}(s,t)$ and its center of symmetry.

\begin{array}[]{l|r}f(s,t)&\mathrm{\operatorname{c-o-s}}(f(s,t))\\ \hline\cr L_{1}(s,t)=(s+t)^{3}+2ts(s+t)&3/2\\ \hline\cr L_{2}(s,t)=6ts(s+t)&3/2\\ \hline\cr L_{3}(s,t)=2(st)^{2}+4st(s+t)^{2}&2\\ \hline\cr L_{4}(s,t)=3st(s+t)^{2}+6(st)^{2}&2\\ \hline\cr L_{5}(s,t)=3(st)^{2}(s+t)&5/2\\ \hline\cr L_{6}(s,t)=1/2(st)^{3}&3\end{array}

Using Corollary 6, we get that each of the six terms in (4) has center of symmetry $2m+3/2$. Since the terms all have the same center of symmetry, the polynomial $A_{4k+4}^{+}(s,t)$ is gamma positive. An indentical proof works for $A_{4k+4}^{-}(s,t)$ and one can check that both polynomials have the same center of symmetry. The proof is complete.

Several interpretations for the numbers $\gamma_{n,i}$ as cardinalities of appropriate sets are known (see Athanasiadis [2, Theorem 2.1]). We end this section with the following.

Question 20

Let $n\equiv 0,1$ (mod 4). By Theorem 1, both $A_{n}^{+}(s,t)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}\gamma_{n,i}^{+}(st)^{i}(s+t)^{n-1-2i}$ and $A_{n}^{-}(s,t)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}\gamma_{n,i}^{-}(st)^{i}(s+t)^{n-1-2i}$. Can we get an interpretation for the gamma coeffients $\gamma_{n,i}^{+}$ or $\gamma_{n,i}^{-}$?

5 When $n\equiv 2$ (mod 4)

When $n\equiv 2$ (mod 4), the polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ are not palindromic and so cannot be expressed in the gamma basis. Nonetheless, we show that both the above polynomials can be written as a sum of two gamma positive polynomials with different centers of symmetry. We note that this result only holds for the univariate polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ and not for the bivariate counterparts $A_{n}^{+}(s,t)$ and $A_{n}^{-}(s,t)$.

5.1 Sum of gamma positive polynomials

Proof:   (Of Theorem 2) We prove the result for $A_{4m+2}^{+}(t)$. An identical proof works for $A_{4m+2}^{-}(t)$. As $n=4m+2$, by Theorem 18 we have

[TABLE]

Let $p(s,t)=\frac{1}{2}stDA_{4m+1}(s,t)$. By Corollary 6, $p(s,t)$ is gamma positive with $\mathrm{\operatorname{c-o-s}}(p(s,t))=2m-1/2+1$. Further, by Corollary 9, all coefficients of $A_{4m+1}(s,t)$ except the coefficient of $(st)^{0}(s+t)^{4m}$ are even. As the exponent $4m$ will appear in $D(s+t)^{4m}$ that coefficient will also be even. Hence $p(s,t)$ and $p(t)=p(s,t)|_{s=1}$ are gamma positive with integral coefficients in the gamma basis and we have $\mathrm{\operatorname{c-o-s}}(p(t))=2m+1/2$.

Further, it is simple to see that $p(t)$ has odd length and thus by Lemma 7, it can be written as $p(t)=p_{1}(t)+p_{2}(t)$ with $\mathrm{\operatorname{c-o-s}}(p_{1}(t))=2m$ and $\mathrm{\operatorname{c-o-s}}(p_{2}(t))=2m+1$. Thus, (24) becomes

[TABLE]

where $w_{1}(t)=A_{4m+1}^{+}(t)+p_{1}(t)$ has $\mathrm{\operatorname{c-o-s}}(w_{1}(t))=2m$ and $w_{2}(t)=tA_{4m+1}^{-}(t)+p_{2}(t)$ has $\mathrm{\operatorname{c-o-s}}(w_{2}(t))=2m+1$. The proof is complete.

The bivariate Eulerian polynomials restricted to the alternating group are not gamma positive in the stronger sense. We postpone the example to after the next subsection to avoid repeating the data (see Remark 22).

5.2 A curious identity

When $n\equiv 2$ (mod 4), Theorem 2 states that both the polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ can be written as a sum of two gamma positive polynomials while Theorem 8 states that $A_{n}(t)$ is gamma positive. Below, we see how these four gamma-positive polynomials arising from $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ add up to give a single gamma positive polynomial. We illustrate this with an example when $n=6$. It is easy to see that

[TABLE]

[TABLE]

In the gamma basis, we have

\begin{array}[]{l|l|l|l}A_{6}(t)&(1+t)^{5}&{\color[rgb]{1,0,0}52}t(1+t)^{3}&{\color[rgb]{0,0,1}136}t^{2}(1+t)\\ \hline\cr g_{1}(t)&t(1+t)^{4}&{\color[rgb]{1,0,0}25}t^{2}(1+t)^{2}&{\color[rgb]{0,0,1}64}t^{3}\\ g_{2}(t)&&{\color[rgb]{1,0,0}27}t(1+t)^{2}&{\color[rgb]{0,0,1}72}t^{2}\\ \hline\cr f_{1}(t)&(1+t)^{4}&25t(1+t)^{2}&64t^{2}\\ f_{2}(t)&&27t^{2}(1+t)^{2}&72t^{3}\end{array}

If $A_{4m+2}^{+}(t)=f_{1}(t)+f_{2}(t)$ then, each gamma coefficient of $A_{4m+2}(t)$ is the sum of a gamma coefficient of $f_{1}(t)$ and a gamma coefficient of $f_{2}(t)$ (and independently a sum of two gamma coefficients, one each of $g_{1}(t)$ and $g_{2}(t)$). We prove this below.

Let $\gamma_{n,i}^{\mathfrak{S}}$ denote the $i$-th gamma coefficient when $A_{n}(t)$ is expressed in the $\Gamma$ basis. Similarly, let $A_{4m+2}^{+}(t)=f_{1}(t)+f_{2}(t)$ with $f_{1}(t)=\sum_{i=0}^{2m}\gamma_{4m+2,i}^{\mathcal{A},1}t^{i}(1+t)^{4m-2i}$ and with $f_{2}(t)=\sum_{i=0}^{2m+1}\gamma_{4m+2,i}^{\mathcal{A},2}t^{i}(1+t)^{4m-2i+2}$. When $n\equiv 0,1$ (mod 4), let $A_{n}^{+}(t)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}\gamma_{n,i}^{\mathcal{A}}t^{i}(1+t)^{n-1-2i}$ and let $A_{n}^{-}(t)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}\gamma_{n,i}^{\mathfrak{S}-\mathcal{A}}t^{i}(1+t)^{n-1-2i}$.

Lemma 21

Let $n=4m+2$ where $m$ is a positive integer. With the above notation,

[TABLE]

Proof:   Equation (26) in the proof of Theorem 2 gives $A_{4m+2}^{+}(t)=f_{1}(t)+f_{2}(t)$. The proof of Theorem 2 also shows that

[TABLE]

Thus,

[TABLE]

Hence, $\gamma_{4m+2,i-1}^{\mathcal{A},1}+\gamma_{4m+2,i}^{\mathcal{A},2}=[4m-2(i-2)]\gamma_{4m+1,i-1}^{\mathfrak{S}}+i\gamma_{4m+1,i}^{\mathfrak{S}}=\gamma_{4m+2,i-1}^{\mathfrak{S}}$, completing the proof.

Remark 22

When summed over $\mathcal{A}_{n}$, the bivariate Eulerian polynomials are not gamma positive in the stronger sense. When $n=6$, our proof gives

[TABLE]

Thus, though $A_{6}^{+}(t)$ is a sum of two gamma positive polynomials, $A_{6}^{+}(s,t)$ is not.

6 When $n\equiv 3$ (mod 4)

Our main result of this section is the following counterpart of Theorem 2.

Proof:   (Of Theorem 3) Recall from (15) that $A_{4m+3}^{+}(s,t)=(s+t)A_{4m+2}^{+}(s,t)+stDA_{4m+2}^{+}(s,t)$. By (18),

[TABLE]

Let

[TABLE]

By Corollary 6, the three polynomials $p_{1}(t)$, $p_{2}(t)$ and $p_{3}(t)$ have respective centers of symmetry $2m+\frac{1}{2}$, $2m+\frac{3}{2}$ and $2m+1$.

[TABLE]

Define

[TABLE]

By Corollary 6, $\mathrm{\operatorname{c-o-s}}(p_{4m+3}^{1}(t))=2m+\frac{1}{2}$, $\mathrm{\operatorname{c-o-s}}(p_{4m+3}^{2}(t))=2m+1$ and $\mathrm{\operatorname{c-o-s}}(p_{4m+3}^{3}(t))=2m+\frac{3}{2}$. Clearly adding (30) and (31) is equivalent to adding (32), (33) and (34). Thus, $A_{4m+3}^{+}(t)$ can be written as a sum of three gamma positive polynomials. A very similar proof can be given to show that $A_{4m+3}^{-}(t)$ can be written as a sum of three gamma positive polynomials. The proof is complete.

7 Type-B Coxeter Groups

Recall that $\mathfrak{B}_{n}$ is the set of permutations $\pi$ of $\{-n,-(n-1),\ldots,-1,1,2,\ldots n\}$ that satisfy $\pi(-i)=-\pi(i)$. $\mathfrak{B}_{n}$ is referred to as the hyperoctahedral group or the group of signed permutations on $[n]$. For $\pi\in\mathfrak{B}_{n}$, for $1\leq i\leq n$, we alternatively denote $\pi(i)$ as $\pi_{i}$. For $\pi\in\mathfrak{B}_{n}$, define $\mathsf{Negs}(\pi)=\{\pi_{i}:i>0,\pi_{i}<0\}$ be the set of elements which occur with a negative sign. Define $\mathsf{inv}_{B}(\pi)=|\{1\leq i<j\leq n:\pi_{i}>\pi_{j}\}|+|\{1\leq i<j\leq n:-\pi_{i}>\pi_{j}\}+|\mathsf{Negs}(\pi)|$. We refer to $\mathsf{inv}_{B}(\pi)$ alternatively as the length of $\pi\in\mathfrak{B}_{n}$. Let $\mathfrak{B}^{+}_{n}\subseteq\mathfrak{B}_{n}$ denote the subset of even length elements of $\mathfrak{B}_{n}$. Let $\mathfrak{B}_{n}^{-}=\mathfrak{B}_{n}-\mathfrak{B}_{n}^{+}$. For all $\pi\in\mathfrak{B}_{n}$, let $\pi_{0}=0$ and let $\mathsf{des}_{B}(\pi)=|\{i\in\{0,1,2\ldots,n-1\}:\pi_{i}>\pi_{i+1}\}|$ and $\mathsf{asc}_{B}(\pi)=|\{i\in\{0,1,2\ldots,n-1\}:\pi_{i}<\pi_{i+1}\}|$. Define

[TABLE]

The polynomial $B_{n}(t)$ is called the type-B Eulerian polynomial. Hyatt in [9] has given recurrences for the type-B and type-D Eulerian polynomials. We could not find a type-B counterpart of Equation (6) to the best of our knowledge. We start with the following

Theorem 23

With $B_{n}(s,t)$ as defined in (35), we have

[TABLE]

where the $\gamma_{n,i}^{B}$ are non-negative integers for all $n,i$.

Proof:   Recall $\displaystyle D=\left(\frac{d}{ds}+\frac{d}{dt}\right)$. We have the following version of (6).

[TABLE]

We observe that the term $(s+t)B_{n}(s,t)$ is the contribution of all those permutations $\pi\in\mathfrak{B}_{n}$ where the letter ‘$n+1$’ or $\overline{n+1}$, is at the last position. It is easy to see that $\displaystyle 2stDB_{n}(s,t)$ is the contribution of all $\pi\in\mathfrak{B}_{n}$ in which the letter ‘$n+1$’ or $\overline{n+1}$ appears in position $r$ for $1\leq r\leq n$.

It is straightforward to check that $B_{1}(s,t)=s+t$. Hence, by induction and Corollary 6, we get that $B_{n+1}(s,t)$ is gamma-positive. Further, it is simple to see that the following recurrence is satisfied by the coefficients $\gamma_{n+1,i}^{B}$.

[TABLE]

Alternatively, the above recurrence together with the initial conditions $\gamma^{B}_{1,0}=1$ or $\gamma^{B}_{2,0}=1,\gamma^{B}_{2,1}=4$ settles non-negativity of $\gamma^{B}_{n,i}$ for all $n,i$.

Similar to Lemma 10, the following result shows palindromicity of the polynomials $B_{n}(s,t)$.

Lemma 24

For $n\geq 2$, the polynomials $B_{n}^{+}(s,t)$ and $B_{n}^{-}(s,t)$ are palindromic iff $n\equiv 0$ mod 2.

Proof:   Let $\pi=x_{1},x_{2},\ldots,x_{n}\in\mathfrak{B}_{n}$. Define $f:\mathfrak{B}_{n}\mapsto\mathfrak{B}_{n}$ by $f(\pi)=\overline{x_{1}},\overline{x_{2}},\ldots,\overline{x_{n}}$. Clearly, $f$ is a bijection. It is easy to see that (or see [17, Lemma 3]) that flipping the sign of a single $x_{i}$ changes the parity of the number of type-B inversions. Hence, if $n\equiv 0$ mod 2, then $\pi\in\mathfrak{B}_{n}^{+}$ iff $f(\pi)\in\mathfrak{B}_{n}^{+}$.

If $n\equiv 1$ mod 2, then the map $f$ flips the parity of $\mathsf{inv}_{B}(\pi)$ and hence $b_{n,k}^{+}=b_{n,n-k}^{-}$. Further, $b_{n,0}^{+}=1,b_{n,n}^{+}=0$ and $b_{n,n}^{-}=1,b_{n,0}^{-}=0$ and hence $B_{n}^{+}(s,t)$ and $B_{n}^{-}(s,t)$ are not palindromic.

Define

[TABLE]

Clearly, $B_{n}^{+}(s,t)=\frac{1}{2}\left(B_{n}(s,t)+\mathrm{SgnB}_{n}(s,t)\right)$. Motivated by this, we determine $\mathrm{SgnB}_{n}(s,t)$ in the next Subsection.

7.1 Enumeration of descent with signs in $\mathfrak{B}_{n}$

Let $a_{1},a_{2},\ldots,a_{n}$ be $n$ distinct positive integers with $a_{1}<a_{2}<\cdots<a_{n}$. Let $\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}$ be the Type-B group of $2^{n}n!$ permutations on the letters $a_{1},a_{2},\ldots,a_{n}$. Clearly, when $a_{i}=i$ for all $i$, we get $\mathfrak{B}_{n}=\mathfrak{B}_{\{1,2,\ldots,n\}}$. We write $\pi\in\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}$ in two line notation with $a_{1},a_{2},\ldots,a_{n}$ above and $\pi_{a_{i}}$ below $a_{i}$. Thus, $\pi_{a_{i}}$ is defined for all $i$. For any permutation $\pi\in\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}$ define $\mathsf{inv}_{B}(\pi)=|\{1\leq i<j\leq n:\pi_{a_{i}}>\pi_{a_{j}}\}|+|\{1\leq i<j\leq n:-\pi_{a_{i}}>\pi_{a_{j}}\}|\}+|\mathsf{Negs}(\pi)|$. Let $\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}^{+}$ denote the subset of even length elements of $\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}$ and $\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}^{-}=\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}-\mathfrak{B}_{\{a_{1},a_{2},\ldots,a_{n}\}}^{+}$. Let $a_{0}=0$ and $\pi(0)=0$ for all $\pi$. Define $\mathsf{des}_{B}(\pi)=|\{i\in\{0,1,2\ldots,n-1\}:\pi_{a_{i}}>\pi_{a_{i+1}}\}|$ and $\mathsf{asc}_{B}(\pi)=|\{i\in\{0,1,2\ldots,n-1\}:\pi_{a_{i}}<\pi_{a_{i+1}}\}|$. Also define $\mathsf{pos}_{a_{i}}(\pi)=k$ if $|\pi_{a_{k}}|=a_{i}$. Define

[TABLE]

We start with the following.

Lemma 25

For positive integers $a_{1}<a_{2}<\ldots<a_{n-1}<a_{n}$, let

$S=\mathfrak{B}^{+}_{{{\{a_{1},a_{2},\ldots,a_{n}\}}},\mathsf{pos}_{a_{n-1}}(\pi)=n,\mathsf{pos}_{a_{n}}(\pi)=n-1}$* and $T=\mathfrak{B}^{-}_{{{\{a_{1},a_{2},\ldots,a_{n}\}}},\mathsf{pos}_{a_{n-1}}(\pi)=n,\mathsf{pos}_{a_{n}}(\pi)=n-1}$.*

Then, the following holds:

[TABLE]

Proof:   Define $h:S\mapsto T$ by $h(\pi_{a_{1}},\pi_{a_{2}},\ldots,\pi_{a_{n-1}},\pi_{a_{n}})=\pi_{a_{1}},\pi_{a_{2}},\ldots,\pi_{a_{n-1}},\overline{\pi_{a_{n}}}$. It is clear that $h$ is a bijection that preserves descents, $\mathsf{pos}_{a_{n}}(\pi)$ and flips the inversion number, completing the proof.

Lemma 26

For all positive integers $a_{1}<a_{2}<\ldots<a_{n}$,

[TABLE]

Proof:   We induct on $n$. When $n=1$, the sum is clearly [math]. For any set of $n-1$ distinct integers, let the sum be [math]. Note that the set $\{\pi\in\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n}\}}},\mathsf{pos}_{a_{n}}(\pi)\neq n\}$ is the disjoint union of the following three sets $S^{1}=\{\pi\in\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n}\}}},\mathsf{pos}_{a_{n-1}}(\pi)=n,\mathsf{pos}_{a_{n}}(\pi)=n-1\}$ , $S^{2}=\{\pi\in\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n}\}}},\mathsf{pos}_{a_{n-1}}(\pi)=n,\mathsf{pos}_{a_{n}}(\pi))\neq n-1\}$ and $S^{3}=\{\pi\in\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-2},a_{n-1}\}}},\mathsf{pos}_{a_{n-1}}(\pi)\neq n,\mathsf{pos}_{a_{n}}(\pi)\neq n\}$. By Lemma 25, enumeration over $S^{1}$ gives the required sum to be zero. We need the following two claims when enumeration is over $S^{2}$ and $S^{3}$ respectively.

Claim 1: For all $n$ positive integers $a_{1}<a_{2}<\ldots<a_{n}$, the following holds:

[TABLE]

Proof of Claim 1: By induction on the integers $a_{1}<a_{2}<\ldots<a_{n-2}<a_{n}$ we get

[TABLE]

Hence, there exists $g_{n-1}:\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-2},a_{n}\}},\mathsf{pos}_{a_{n}}(\pi)\neq n-1}^{+}\rightarrow\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-2},a_{n}\}},\mathsf{pos}_{a_{n}}(\pi)\neq n-1}^{-}$ such that $g_{n-1}(\pi)$ and $\pi$ has same number of descents. Now we put ‘$a_{n-1}$’ or ‘$\overline{a_{n-1}}$’ in the last position of $\pi$ and $g_{n-1}(\pi)$ to get

[TABLE]

The proof of Claim 1 is complete.

Claim 2: For any $n$ positive integers $a_{1},a_{2},\ldots,a_{n}$,

[TABLE]

Proof of Claim 2: Consider $n-1$ positive integers, $a_{1}<a_{2}<\ldots<a_{n-2}<a_{n-1}$ and by induction, we assume that

[TABLE]

Hence, there exists $f_{n-1}:\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-1}\}},\mathsf{pos}_{a_{n-1}}(\pi)\neq n-1}^{+}\rightarrow\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-1}\}},\mathsf{pos}_{a_{n-1}}(\pi)\neq n-1}^{-}$ such that for all $\pi\in\mathfrak{B}_{{\{a_{1},a_{2},\ldots,a_{n-1}\}},\mathsf{pos}_{a_{n-1}}(\pi)\neq n-1}^{+}$, $\mathsf{des}_{B}(f_{n-1}(\pi))=\mathsf{des}_{B}(\pi)$ and $\mathsf{pos}_{a_{n-1}}(f_{n-1}(\pi))=\mathsf{pos}_{a_{n-1}}(\pi)$. So, we can put ‘$a_{n}$’or ‘$\overline{a_{n}}$’ in the same position (except the last position) of $\pi$ and $f_{n-1}(\pi)$ to get

[TABLE]

The proof of Claim 2 is complete.

We sum up the results for $S^{1},S^{2}$ and $S^{3}$. Summing Lemma 25, (45) and (46), we get

[TABLE]

The proof is complete.

In Lemma 26, we set $u=1$ and $a_{i}=i$ for $1\leq i\leq n$ and get the following corollary.

Corollary 27

For any positive integer $n$, the following is true:

[TABLE]

7.2 Main results for $\mathfrak{B}_{n}$

Theorem 28

Let $n\geq 2$ be a positive integer. The polynomials $B_{n}^{+}(s,t)$ and $B_{n}^{-}(s,t)$ satisfy the following recurrence.

[TABLE]

Proof:   We prove (49) first. Recall $B_{n}^{+}(s,t)=\sum_{\pi\in\mathfrak{B}_{n}^{+}}t^{\mathsf{des}_{B}(\pi)}s^{\mathsf{asc}_{B}(\pi)}$. Consider the contribution to the right hand side from $\pi\in\mathfrak{B}_{n}+$ with ‘$n$’ or ‘$\overline{n}$’ occurring in position $k$ for all possible choices of $k$. It is easy to see that

[TABLE]

Here, $sB_{n-1}^{+}(s,t)$ accounts for the contribution of all $\pi\in\mathfrak{B}_{n}^{+}$ in which the letter ‘$n$’ appears in the last position and $tB_{n-1}^{-}(s,t)$ is the contribution of all $\pi\in\mathfrak{B}_{n}^{+}$ in which the letter ‘$\overline{n}$’ appears in the last position. Further, from corollary 27 we get

[TABLE]

The last line follows from the fact that $2stDB_{n-1}(s,t)$ is the contribution of all the permutations in $\mathfrak{B}_{n}$ where ‘$n$’ or ‘$\overline{n}$’ is not in the final position, completing the proof of (49). The proof of (50) is identical and hence omitted.

The following is the main result of this Section.

Theorem 29

For all positive integers $n\geq 2$, both $B_{n}^{+}(s,t)$ and $B_{n}^{-}(s,t)$ are gamma-positive with the same center of symmetry $\frac{n}{2}$ iff $n\equiv 0$ (mod 2).

Proof:   By Lemma 24, we clearly need $n\equiv 0$ (mod 2). We use inductionon $n$. The base case is when $n=2$. We have $B_{2}^{-}(s,t)=4st$ and $B_{2}^{+}(s,t)=s^{2}+2st+t^{2}$, both of which are gamma positive with center of symmetry $1$. By induction, assume that for $n-2=2m-2$, both $B_{n-2}^{-}(s,t)$ and $B_{n-2}^{+}(s,t)$ are gamma positive with center of symmetry $m-1$. By applying the recurrence relations (49) and (50) twice, we get both

[TABLE]

Each of the four individual terms in (51) and (52) clearly has center of symmetry $m$. Thus, both $B_{2m}^{+}(s,t)$ and $B_{2m}^{-}(s,t)$ are gamma positive with center of symmetry $m$, completing the proof.

Theorem 30

For all positive odd integers $n=2m+1$, $n\geq 3$, $B_{n}^{-}(t)$ and $B_{n}^{+}(t)$ are the sum of two gamma positive polynomials.

Proof:   $B_{n}^{+}(s,t)$ and $B_{n}^{-}(s,t)$ satisfy the recurrences (49) and (50) which are identical to the recurrences (21) and (22) which we used to prove Theorem 2. Hence, the proof of this result follows in an identical manner to the proof of Theorem 2.

Similar to Corollary 19, we get the following recurrences for the numbers $b_{n,k}^{+}$ and $b_{n,k}^{-}$ using Theorem 28.

Corollary 31

For all positive integers $n$ and $0\leq k\leq n$, $b_{n,k}^{+}$ and $b_{n,k}^{-}$ satisfy the following recurrence relations.

$b_{n,k}^{+}=2kb_{n-1,k}^{-}+(2n-2k+1)b_{n-1,k-1}^{-}+b_{n-1,k}^{+},$** 2. 2.

$b_{n,k}^{-}=2kb_{n-1,k}^{+}+(2n-2k+1)b_{n-1,k-1}^{+}+b_{n-1,k}^{-}.$**

8 Type-D Coxeter groups

Recall that $\mathfrak{D}_{n}\subseteq\mathfrak{B}_{n}$ is the subset of type-B permutations that have an even number of negative signs. Let $w=w_{1},w_{2},\ldots,w_{n}\in\mathfrak{D}_{n}$. The following combinatorial definition of type-D inversions is well known, see for example Petersen’s book [13, Page 302] $\mathsf{inv}_{D}(w)=\mathsf{inv}_{A}(w)+|\{1\leq i<j\leq n:-w_{i}>w_{j}\}|$. Here $\mathsf{inv}_{A}(w)$ is computed with respect to the usual order on $\mathbb{Z}$. Let $\mathfrak{D}_{n}^{+}=\{w\in\mathfrak{D}_{n}:\mathsf{inv}_{D}(w)\>\mbox{ is even}\}$ and let $\mathfrak{D}_{n}^{-}=\mathfrak{D}_{n}-\mathfrak{D}_{n}^{+}$.

We next give the combinatorial definition of type-D descents. Let $w=w_{1},w_{2},\ldots,w_{n}\in\mathfrak{D}_{n}$. Then $\mathsf{des}_{D}(w)=\mathsf{des}(w)+\chi(w_{1}+w_{2}<0)$ where $\chi(Z)$ is a 0/1 indicator function taking value 1 if $Z$ is true and taking value 0 otherwise. Here, $\mathsf{des}(w)$ is as defined in the type-A case. See Petersen’s book [13, Page 302] for this definition. Let $\mathsf{asc}_{D}(w)=n-\mathsf{des}_{D}(w)$ and define

[TABLE]

The polynomials $D_{n}(t)$ are called the type-D Eulerian polynomials. Stembridge [18] showed that the polynomials $D_{n}(t)$ are gamma positive for all $n\geq 1$ (see Petersen [13, Section 13.4]). For $D_{n}(s,t)$ as defined (53), we are unable to get recurrences similar to (6) or (39). Our proof of the gamma positivity of $D_{n}^{+}(t)$ follows the proof that type-D Eulerian polynomials are gamma positive from Petersen’s book [13, Section 13.4].

Since $\mathfrak{D}_{n}\subseteq\mathfrak{B}_{n}$, each $w\in\mathfrak{B}_{n}$ has both the type-B and the type-D inversion numbers (though the type-D inversion number does not have Coxeter theoretic meaning). Define $\mathfrak{B}_{n,\mathfrak{D}}^{+}=\{w\in\mathfrak{B}_{n}:\mathsf{inv}_{D}(w)\>\mbox{ is even}\}$ and $\mathfrak{B}_{n,\mathfrak{D}}^{-}=\{w\in\mathfrak{B}_{n}:\mathsf{inv}_{D}(w)\>\mbox{ is odd}\}$.

Lemma 32

With the above notation, we have

[TABLE]

Proof:   We prove (56) first. For this, consider the bijection from $\mathfrak{D}_{n}^{+}$ to $\mathfrak{B}_{n,\mathfrak{D}}^{+}-\mathfrak{D}_{n}^{+}$ by flipping the first element i.e consider the map $f:\mathfrak{D}_{n}^{+}\mapsto\mathfrak{B}_{n,\mathfrak{D}}^{+}-\mathfrak{D}_{n}^{+}$ given by $f(u_{1},u_{2},\ldots,u_{n})=\overline{u_{1}},u_{2},\ldots,u_{n}$. Clearly $f^{2}=id$ and $f$ preserves the number of descents and parity of $\mathsf{inv}_{D}u$. The proof of (57) is similar and hence omitted.

Let $u\in\mathfrak{S}_{n}$ and let $\beta(u)$ denote the set of $2^{n}$ elements of $\mathfrak{B}_{n}$ obtained by adding negative signs to the entries of $w$ in all possible ways.

Lemma 33

Let $u\in\mathfrak{S}_{n}$. Then, for all $w\in\beta(u)$, either $w\in\mathfrak{B}_{n,\mathfrak{D}}^{+}$ or $w\in\mathfrak{B}_{n,\mathfrak{D}}^{-}$. Hence, for any $u\in\mathfrak{S}_{n}$, $u\in\mathfrak{B}_{n,\mathfrak{D}}^{+}$ if and only if $u\in\mathcal{A}_{n}$.

Proof:   The point to prove is that all the $2^{n}$ elements of $\beta(u)$ have the same parity of the number of type-D inversions. For any element $w=w_{1},w_{2},\ldots,w_{n}\in\beta(u)$, let $w_{\bar{l}}=w_{1},w_{2},\ldots,-w_{l},\ldots,w_{n}$ denote the permutation obtained by flipping the sign of the $l$-th entry of $w$. Note that we only need to show that for any $l$ $w$ and $w_{\bar{l}}$ have the same parity of number of type-D inversions. Note that

$\mathsf{inv}_{D}(w)-\mathsf{inv}_{D}(w_{\bar{l}})=|\{1\leq i<j=l:w_{i}>w_{j}\}|+|\{1\leq i<j=l:-w_{i}>w_{j}\}|-\{|1\leq i<j=l:w_{i}>-w_{j}\}|-|\{1\leq i<j=l:-w_{i}>-w_{j}\}|$

We show that $|\{1\leq i<j=l:w_{i}>w_{j}\}|+|\{1\leq i<j=l:-w_{i}>w_{j}\}|$ and $\{|1\leq i<j=l:w_{i}>-w_{j}\}|+|\{1\leq i<j=l:-w_{i}>-w_{j}\}|$ have the same parity. For this, let $i<l$. We show that $\chi(w_{i}>w_{l})+\chi(-w_{i}>w_{l})$ and $\chi(w_{i}>-w_{l})+\chi(-w_{i}>-w_{l})$ have the same parity. This follows from the simple observation below which we divide in cases:

case i: If $w_{i}>w_{l}$ and $-w_{i}\leq w_{l}$, then $-w_{i}\leq-w_{l}$ and $w_{i}>-w_{l}$.

case ii: If $w_{i}>w_{l}$ and $-w_{i}>w_{l}$, then $-w_{i}\leq-w_{l}$ and $w_{i}\leq-w_{l}$.

case iii: If $w_{i}\leq w_{l}$ and $-w_{i}\leq w_{l}$, then $-w_{i}>-w_{l}$ and $w_{i}>-w_{l}$.

case iv: If $w_{i}\leq w_{l}$ and $-w_{i}>w_{l}$, then $-w_{i}>-w_{l}$ and $w_{i}\leq-w_{l}$.

Hence, $w$ and $w_{\bar{l}}$ have same parity of Type-D inversions and the proof is complete.

By Lemma 33, we get

[TABLE]

The proof of gamma positivity of $D_{n}(t)$ given in Petersen’s book [13, Page 305] shows that it is possible to assign values $c_{i}(u)$ for $1\leq i\leq n$ to $u\in\mathfrak{S}_{n}$ such that $\sum_{w\in\beta(u)}t^{\mathsf{des}_{D}w}=c_{1}(u)c_{2}(u)\ldots c_{n}(u).$ We work with the following bivariate version.

[TABLE]

and for all $j\geq 3$

[TABLE]

For a permutation $u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathfrak{S}_{n}$ with $u_{1}<u_{2}<u_{3}$, let $u^{\prime}=u_{3},u_{1},u_{2},\ldots,u_{n}$. It is easy to see that $u\in\mathfrak{B}_{n,\mathfrak{D}}^{+}$ if and only if $u^{\prime}\in\mathfrak{B}_{n,\mathfrak{D}}^{+}$. We partition $\mathcal{A}_{n}$ as union of the following five disjoint sets $\mathcal{A}_{n}^{1}=\{u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathcal{A}_{n}:u_{1}<u_{2}>u_{3}\}$ , $\mathcal{A}_{n}^{2}=\{u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathcal{A}_{n}:u_{1}>u_{2}>u_{3}\}$, $\mathcal{A}_{n}^{3}=\{u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathcal{A}_{n}:u_{1}<u_{2}<u_{3}\}$, $\mathcal{A}_{n}^{4}=\{u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathcal{A}_{n}:u_{1}>u_{2}<u_{3},u_{1}>u_{3}\}$, $\mathcal{A}_{n}^{5}=\{u=u_{1},u_{2},u_{3},\ldots,u_{n}\in\mathcal{A}_{n}:u_{1}>u_{2}<u_{3},u_{1}<u_{3}\}$.

For a permutation $u\in\mathfrak{S}_{n}$, define $\mathsf{lpk}(u)=|\{1\leq i<n:u_{i-1}<u_{i}>u_{i+1}\}|$ where $u_{0}=0$.

Lemma 34

For positive integers $n\geq 3$, $|\mathcal{A}_{n}^{3}|=|\mathcal{A}_{n}^{4}|$. Moreover, $\displaystyle\sum_{u\in\mathcal{A}_{n}^{3}\cup\mathcal{A}_{n}^{4}}\sum_{w\in\beta(u)}t^{\mathsf{des}_{D}(w)}s^{\mathsf{asc}_{D}(w)}\displaystyle$ is a gamma positive polynomial with all gamma coefficients being even.

Proof:   Define $f:\mathcal{A}_{n}^{3}\mapsto\mathcal{A}_{n}^{4}$ by $f(u_{1},u_{2},u_{3},u_{4},\ldots,u_{n})=u_{3},u_{1},u_{2},u_{4},\ldots,u_{n}$. Clearly, $\pi\in\mathcal{A}_{n}$ iff $f(\pi)\in\mathcal{A}_{n}$. Further, $f$ is a bijection as its inverse is clearly $g:\mathcal{A}_{n}^{4}\mapsto\mathcal{A}_{n}^{3}$ given by $g(u_{1},u_{2},u_{3},u_{4},\ldots,u_{n})=u_{2},u_{3},u_{1},u_{4},\ldots,u_{n}$. Hence, we have

[TABLE]

In the last line, we used bivariate version of [13, Eqn (13.10)] from Petersen’s book. The proof is complete.

Theorem 35

For all positive integers $n\geq 3$, the polynomials $D_{n}^{+}(s,t)$ and $D_{n}^{-}(s,t)$ are gamma non-negative.

Proof:   We first prove that the polynomials $D_{n}^{+}(s,t)$ are gamma positive. From Lemma 32, we get

[TABLE]

By (58), we get

[TABLE]

Clearly, if $u_{1}<u_{2}>u_{3}$, or $u_{1}>u_{2}>u_{3}$ or $u_{3}>u_{1}>u_{2}$, then $\sum_{w\in\beta(u)}t^{\mathsf{des}_{D}(w)}s^{\mathsf{asc}_{D}(w)}=(4st)^{\mathsf{lpk}(u)}(s+t)^{n-2\mathsf{lpk}(u)}$. So, the distribution of $\mathsf{des}_{D}(w)$ over these in each of the three cases i.e over $\mathcal{A}_{n}^{1}$,$\mathcal{A}_{n}^{2}$ and $\mathcal{A}_{n}^{5}$ is gamma positive and further, $u$ in any of these three sets has at least one left peak. Thus, the gamma-coefficients are even. Hence, the first term in (62) is gamma positive. The gamma positivity of the second term follows from Lemma 34. The proof for $\mathfrak{D}_{n}^{-}(s,t)$ follows identical steps and hence is omitted. The proof is complete.

9 Questions and Conjectures

In this section, we raise a few questions and make a few conjectures. The most important question is to find an interpretation for gamma coefficients.

Question 36

Can we find an interpretation for the gamma coefficients that occur in Theorems 2, 3, 29, 30 and 35?

For $\pi\in\mathfrak{S}_{n}$, denote $\mathsf{ides}(\pi)=\mathsf{des}(\pi^{-1})$ and define the two-sided Eulerian polynomial by $\mathrm{TSA}_{n}(s,t)=\sum_{\pi\in\mathfrak{S}_{n}}t^{\mathsf{des}(\pi)+1}s^{\mathsf{ides}(\pi)+1}=\sum_{i,j}a_{n,i,j}t^{i}s^{j}$. The polynomial $\mathrm{TSA}_{n}(s,t)$ is known to satisfy $a_{n,i,j}=a_{n,j,i}$ and $a_{n,i,j}=a_{n,n-i,n-j}$. Gessel (see [3, Conjecture 10.2]) conjectured that $\mathrm{TSA}_{n}(s,t)=\sum_{i,j}\gamma_{n,i,j}(st)^{i}(s+t)^{j}(1+st)^{n+1-2i-j}$ with $\gamma_{n,i,j}\geq 0$. This was recently proved by Zhicong Lin [10]. See the paper by Adin et al [1] as well.

When $n\equiv 0,1$ (mod 4), define $\mathrm{TSA}_{n}^{+}(s,t)=\sum_{\pi\in\mathcal{A}_{n}}t^{\mathsf{des}(\pi)+1}s^{\mathsf{ides}(\pi)+1}=\sum_{i,j}a_{n,i,j}^{+}t^{i}s^{j}$ and likewise define $\mathrm{TSA}_{n}^{-}(s,t)=\sum_{\pi\in\mathfrak{S}_{n}-\mathcal{A}_{n}}t^{\mathsf{des}(\pi)+1}s^{\mathsf{ides}(\pi)+1}=\sum_{i,j}a_{n,i,j}^{-}t^{i}s^{j}$.

Similarly, when we sum over $\mathfrak{B}_{n}^{+}$ and $\mathfrak{B}_{n}^{-}$, define $\mathrm{TSB}_{n}^{+}(s,t)=\sum_{\pi\in\mathfrak{B}_{n}^{+}}t^{\mathsf{des}_{B}(\pi)}s^{\mathsf{ides}_{B}(\pi)}=\sum_{i,j}a_{n,i,j}^{+}t^{i}s^{j}$ and $\mathrm{TSB}_{n}^{-}(s,t)=\sum_{\pi\in\mathfrak{B}_{n}^{-}}t^{\mathsf{des}_{B}(\pi)}s^{\mathsf{ides}_{B}(\pi)}=\sum_{i,j}a_{n,i,j}^{+}t^{i}s^{j}$.

Similarly, when we sum over $\mathfrak{D}_{n}^{+}$ and $\mathfrak{D}_{n}^{-}$, define $\mathrm{TSD}_{n}^{+}(s,t)=\sum_{\pi\in\mathfrak{D}_{n}^{+}}t^{\mathsf{des}_{D}(\pi)}s^{\mathsf{ides}_{D}(\pi)}=\sum_{i,j}a_{n,i,j}^{+}t^{i}s^{j}$ and $\mathrm{TSD}_{n}^{-}(s,t)=\sum_{\pi\in\mathfrak{D}_{n}^{-}}t^{\mathsf{des}_{D}(\pi)}s^{\mathsf{ides}_{D}(\pi)}=\sum_{i,j}a_{n,i,j}^{+}t^{i}s^{j}$.

Conjecture 37

When $n\equiv 0,1$ (mod 4), both polynomials $\mathrm{TSA}_{n}^{+}(s,t)$ and $\mathrm{TSA}_{n}^{-}(s,t)$ can be written as $\mathrm{TSA}_{n}^{+}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{+}(st)^{i}(s+t)^{j}(1+st)^{n+1-2i-j}$ and $\mathrm{TSA}_{n}^{-}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{-}(st)^{i}(s+t)^{j}(1+st)^{n+1-2i-j}$ with all $\gamma_{n,i,j}^{+},\gamma_{n,i,j}^{-}\geq 0$. 2. 2.

When $n\equiv 0$ (mod 2), both the polynomials $\mathrm{TSB}_{n}^{+}(s,t)$ and $\mathrm{TSB}_{n}^{-}(s,t)$ can be written as $\mathrm{TSB}_{n}^{+}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{B,+}(st)^{i}(s+t)^{j}(1+st)^{n-2i-j}$ and $\mathrm{TSB}_{n}^{-}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{B,-}(st)^{i}(s+t)^{j}(1+st)^{n-2i-j}$ with all $\gamma_{n,i,j}^{B,+},\gamma_{n,i,j}^{B,-}\geq 0$. 3. 3.

For all positive integers $n$, both $\mathrm{TSD}_{n}^{+}(s,t)$ and $\mathrm{TSD}_{n}^{-}(s,t)$ can be written as $\mathrm{TSD}_{n}^{+}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{D,+}(st)^{i}(s+t)^{j}(1+st)^{n-2i-j}$ and $\mathrm{TSD}_{n}^{-}(s,t)=\sum_{i,j}\gamma_{n,i,j}^{D,-}(st)^{i}(s+t)^{j}(1+st)^{n-2i-j}$ with all $\gamma_{n,i,j}^{D,+},\gamma_{n,i,j}^{D,-}\geq 0$.

It is well known that if a univariate polynomial $f(t)$ is palindromic, has positive coefficients and is real rooted, then it is gamma positive (see [13, Chapter 4]). Several of the polynomials in this paper seem to be real rooted.

Conjecture 38

When $n\geq 4$ with $n\equiv 0,1$ (mod 4), the polynomials $A_{n}^{+}(t)$ and $A_{n}^{-}(t)$ are real rooted. When $n\geq 2$, with $n\equiv 0$ (mod 2), the polynomials $B_{n}^{+}(t)$ and $B_{n}^{-}(t)$ are real rooted. When $n\geq 2$, the polynomials $D_{n}^{+}(t)$ and $D_{n}^{-}(t)$ are real rooted.

Acknowledgement

The first author acknowledges support from a CSIR-SPM Fellowship. The second author acknowledges support from project grant P07 IR052, given by IRCC, IIT Bombay.