# The symmetric functions catalog

An overview of symmetric functions and related topics

2021-06-26

## Minor problems

### Interlacing roots and key polynomials

Let $\key_{\lambda,\sigma}$ denote a key polynomial, and define

$P_{\sigma}(\lambda_1,\dotsc,\lambda_n;k) \coloneqq \key_{k\lambda,\sigma}(1^n).$

This is a polynomial in $\setQ[\lambda_1,\dotsc,\lambda_n,k],$ and for fixed $\lambda,$ this is an Ehrhart polynomial of a certain union of faces in a GT-polytope. See [AA19a] for more background.

We may compute the corresponding $h^*$-polynomial, and get some interesting properties.

Let $\lambda$ and $\sigma$ be fixed and define $H_{\lambda,\sigma}(t) \in \setN[t]$ via

$\sum_{k\geq 0} P_{\lambda,\sigma}(k) t^k = \frac{H_{\lambda,\sigma}(t)}{(1-t)^{d+1}}$

where $d$ is the degree (in $k$) of $P_{\lambda,\sigma}(k).$ Then $H_{\lambda,\sigma}(t)$ is a real-rooted polynomial. Moreover, if $\sigma_1,\sigma_2,\dotsc,\sigma_\ell$ is a saturated chain in the Bruhat order, then

$P_{\lambda,\sigma_1}(k), P_{\lambda,\sigma_2}(k),\dotsc,P_{\lambda,\sigma_\ell}(k)$

is a sequence of interlacing polynomials.

I have checked this for some small cases.

### A $q$-generalization of an inequality

In [AA19b], we used the following inequality, where $a,b\geq 0$ and $k \geq j \ge 0.$

$\binom{ka+kb}{ka}^j \geq \binom{ja+jb}{ja}^k.$

This is not very hard to prove. I realized that there might be a $q$-analogue of this inequality.

Suppose $a,b\geq 0$ and $k \geq j \ge 0.$ Then

$q^{kab \binom{j}{2}} \qbinom{ka+kb}{ka}_q^j - q^{jab \binom{k}{2}} \qbinom{ja+jb}{ja}_q^k$

is a polynomial in $\setN[q].$

This was poset on MathOverflow, and there I sketched a proof that shows that this is true whenever $j$ divides $k.$

### A Schur-positive expansion?

Let $\BST(\lambda,\mu)$ be the set of border-strip tableaux of shape $\lambda$ and strip-sizes $\mu.$ Define

$T_\lambda(\xvec) \coloneqq \sum_{\mu} |\BST(\lambda,\mu)| \powerSum_\mu(\xvec).$

Show that $T_\lambda(\xvec)$ is Schur-positive. Note the close resemblance with the usual power-sum expansion of Schur polynomials.

See this MO-post also

### On A189912

The sequence A189912 is defined as

$a_n \coloneqq \sum_{k=0}^n \frac{n!}{(n-k)! (\lfloor k/2 \rfloor!)^2 (\lfloor k/2 \rfloor +1)}.$

Let us split this sum into even and odd $k.$ We get

\begin{align*} &\sum_{k=0}^n \frac{n!}{(n-2k)! (\lfloor 2k/2 \rfloor!)^2 (\lfloor 2k/2 \rfloor +1)} + \\ &\sum_{k=0}^n \frac{n!}{(n-(2k+1))! (\lfloor (2k+1)/2 \rfloor!)^2 (\lfloor (2k+1)/2 \rfloor +1)}. \end{align*}

$\sum_{k=0}^n \left( \frac{n!}{(n-2k)! (k!)^2 (k+1)} + \frac{n!}{(n-2k-1)! (k!)^2 (k+1)} \right).$

Rewriting gives

$\sum_{k=0}^n \frac{n!}{ (k!) (k+1)!}\left( \frac{1}{(n-2k)!} + \frac{1}{(n-2k-1)!} \right) = \sum_{k=0}^n \frac{n!}{(k!) (k+1)!}\left( \frac{1}{(n-2k)!} + \frac{n-2k}{(n-2k)!} \right)$

so we end up with

$\sum_{k=0}^n (n+1-2k) \frac{n!}{ (k!) (k+1)! (n-2k)!}.$

The expression $\frac{n!}{(k!) (k+1)! (n-2k)!}$ is exactly A055151, so this verifies the conjecture by Werner Schulte, Oct 23 2016.

### Schubert charge

The Schubert polynomials generalize the Schur polynomials, so for Grassmann permutations, there should be a bijection to SSYT from pipe dreams. What is the notion of (co)charge on pipe dreams? Does it generalize to arbitrary permutations?

### Roots of staircase Schur polynomials (solved)

I wrote down the following observation here, and Vasu Tewari pointed out the straightforward proof presented below.

Proposition.

Let $\delta_n$ be the staircase partition $(n,n-1,\dotsc,1,0).$ Consider the Schur polynomial indexed by the stretched straircase,

$P_{n,k}(t) \coloneqq \schurS_{k \delta_n}(t,1^{n}).$

Then

$P_{n,k}(t) = (k+1)^{\binom{n}{2}} ([k+1]_t)^{n}.$
Proof.

By the Vandermonde determinant formula,

$\schurS_{k \delta_n}(x_1,\dotsc,x_n) = \prod_{1 \leq i \lt j \leq n+1} \frac{x_i^{k+1}-x_j^{k+1}}{x_i-x_j}.$

This then implies the claim.