Expansion of key polynomials in terms of non-symmetric Hall-Littlewood polynomials and charge-like statistics

Question

Edit: The problem I pose here is impossible to solve with the basis $H$, in the answer I made to this post I explain why. The only way I can think it to amend the situation would be to try with another family of "Non symmetric Hall Littlewood polynomials", this one doesn't generalize the theorem.

Original post:

Lascoux and Schützenberger proved on 1978 that the following identity holds: \begin{equation} \label{Eq1} \sum_{T} t^{ch(T)} HS_{weight(T)} (x;t) = s_{\lambda} (x) \end{equation} where the sum is over the tableaux $T$ of shape $\lambda$ and dominant weight, $HS$ indicates the symmetric Hall-Littlewood Polynomials and $s_{\lambda}(x)$ is a Schur function.

I am interested in a similar relationship which involves a family of Non-symmetric Hall-Littlewood Poylnomials which are defined inductively over the length of a permutation as a member of a Coxeter Group.

From now, I will work with $m$ variables $x_{1}, \ldots, x_{m}$ and a parameter $t$, we can think everything as living inside the polynomial ring $R = \mathbb{Z} [x_{1}, \ldots , x_{m} , t]$.

Let's stablish some notation: For a composition $\alpha = ( \alpha_{1} , \ldots , \alpha_{m} )$ and $\sigma \in \mathfrak{S}_{m}$ a member of the symmetric group, we will denote $\sigma (\alpha) = ( \sigma (\alpha_{1}) , \ldots , \sigma (\alpha_{m}) ) $.

We will use the Hecke Algebra generators, which for the polynomial ring $\mathbb{Z} [x_{1}, \ldots , x_{m} , t]$ are defined by

$T_{i} (f) = tf + \dfrac{t x_{i} - x_{i+1} }{ x_{i} - x_{i+1} } ( K_{i,i+1} f - f ) $

where $\quad K_{i,i+1} (f (x_{1} , \ldots ,x_{m}, t) = f (x_{1} , \ldots , x_{i+1} , x_{i}, \ldots ,x_{m}, t)$.

With that in mind, the definition of Non-symmetric Hall-Littlewood polynomials is the following:

1. For $\lambda$ a partition, $H_{\mu} (x;t) = x^{\mu}$.

2. Let $\alpha = \sigma (\mu)$ and $\beta = \tau_{i} \sigma (\mu)$ with $len (\tau_{i} \sigma) = len ( \sigma ) +1 $, then $H_{\beta} (x;t) = T_{i} H_{\alpha} (x;t)$.

We have the following relationship between symmetricic and non symmetric Hall-Littlewood polynomials:

$$ HS_{\mu} (x;t) = \sum_{\alpha} H_{\alpha} (x;t) $$ where $\alpha$ runs over all compositions which can be obtained rearranging $\mu$, that's it over the set $\mathfrak{S}_{m} (\mu)$.

With this context on mind, the first equation of the post can be rewritten as

$$ \sum_{T} t^{ch(T)} H_{weight(T)} (x;t) = s_{\lambda} (x) $$ where the sum is over the tableaux $T$ of shape $\lambda$

I am trying to find an invariant $ch^{*} (T)$ such that $$ \sum_{T} t^{ch^{*}(T)} H_{weight(T)} (x;t) = K_{\beta} (x) $$ where if $\beta = \sigma (\mu)$, $K_{\beta}(x) := \sum_{\tau \leq_{Coxeter} \sigma} H_{\tau (\mu)} (x;0)$ are key polynomials, for example $K_{(2,0,1)}= H_{(2,1,0)} (x;0) +H_{(2,0,1)} (x;0) = x_{1}^{2}x_{2} + x_{1}^{2}x_{3} $. And the the sum is over the tableaux $T$ with $R_{+} (T) \leq R_{\beta}$, where $R_{\beta}$ is the gith key tableau corresponding to $\beta$. For example for $\beta = (0,2,3)$ the right key would be $$ \begin{array}{lll} 2 &2 &3 \\ 3 &3 & \\ \end{array} $$ Due the the previously laid equations, this function $ch^{*}$ it would need to satisfy $$ ch^{*} ( T ) = ch ( \Gamma (T) ) $$ with $\Gamma: SSYT \rightarrow SSYT$ a bijection which satisfy $\Gamma ( SSYT( \lambda, \alpha ) ) = SSYT( \lambda, \alpha )$ for any shape $\lambda$ and weight $\alpha$.

Here is what I am looking making this post:

(1.) After lots of computations I think that $\Gamma$ must be an involution, but I wasn't able to find one. If this is already studied, I would like to know where it was published as I didn't find it

(2.) Currently don't have any combinatorial description of such function $ch^{*}$, I was thinking of programming it in sagemath, it would involve solving that polynomial equation.

(3.) Should I reformulate (2.) in a more programming specific manner and ask on another forum? If so, I would like to be pointed where.

Edit: Thanks to the user ArB for helping me to improve the post throught a clarification request.

Edit2: 21 of November: Modified the problem because it was way too ambitious. $$ T= \begin{matrix} 2 &3 \\ 3 & \\ \end{matrix} $$ is the only tableaux with $R_{+} (T) = \begin{matrix} 2 &3 \\ 3 & \\ \end{matrix}$ and it's sum it's not a partial key polynomial. So I reformulated it to complete key polynomials.

Answer 1

I have thought about this problem (well, a very similar one) a lot, and have some unpublished notes. There are several very annoying observations that seems to make defining charge very difficult. I think I have examples that strongly indicates that one cannot hope for just a 'reading word' type approach. One needs to take the full shape of the skyline fillings into account (assuming you want to do statistics on skyline fillings as in S. Mason's work).

Answer 2

I am posting this as an answer for better readibility and not to scare anyone trying to understand my original question. I hope this is the correct choice.

My attempt at a solution. I don't know if this is fine as a edit or if I should put it in an answer.

I have thought the following algorithm.

0. Let $T$ be the tableau for which we desire to compute the possible values for $ch^{*}(T)$. This algorithm will give a set as it doesn't distinguish between $T_{1},T_{2} \in SSYT ( \lambda, \alpha )$ with $R_{+} (T_{1}) = R_{+} (T_{2})$.

1. The first step on our algorithm is to compute $R_{+} (T)$. We call $\beta = weight ( R_{+}(T) ) $.

2. Observe that in the monomial expansion of $H_{\beta} (x;t)$, all the monomials on $x$ that appear are smaller than $\beta$ in the inverse lexicographic ordering. If we note $\overset{\leftarrow}{\beta}$ to the partition obtained by ordering $\beta$, I think that they are exactly the multiindex $\beta$ such that $\overset{\leftarrow}{\beta} \leq \gamma \leq \beta$ in the dominance order: $$ \delta \leq_{D} \epsilon \Longleftrightarrow \sum_{j=k}^{m} \delta_{j} \leq \sum_{j=k}^{m} \epsilon_{j} \: \forall \: 1 \leq k \leq m $$

3. Knowing this, we can look for the element of the largest degree on the inverse lexicographic order that it's on the expansion of $H_{\beta} (x;t)$, let's call $\gamma$ that multi-index. I will have a coefficient $p_{\gamma}(t)$, and I think it will be negative but I cannot prove it, altough I will be able to test it as soon as I manage to implement this. I think that $p_{\gamma}(t)$ will be the sum $\sum_{T} t^{ch^{*}(T)}$ with $T \in SSYT(\lambda, \gamma)$ and $R_{+} (T) = \beta$.

4. We repeat step (3.) with $H_{\beta}(x;t) + (-p_{\gamma}(t)) H_{\gamma} (x;t)$, as we reduced the degree of the polynomial this process it's bound to end. Then $-p_{\alpha} (t)$ will be $\sum_{T} t^{ch^{*}(T)}$ with $T \in SSYT(\lambda, \gamma)$ and $R_{+} (T) = \alpha$ which was what were looking for.

I should be able to find a way to ask sagemath to give me the highest degree monomial of a polynomial, and then to convert it's degree to a list in order to feed it to the script I have for computing $H_{\alpha}(x;t)$.

Note 1: This algorithm cannot distinguish between $T_{1},T_{2} \in SSYT ( \lambda, \alpha )$ with the same right key tableaux, as the algebraic problem which I posed doesn't distinguish them and it's not a problem of this particular algorithm. Of course the eventual solution should be combinatorial, but I want to implement this algorithm in order to give me the scaffoldings to find that kind of construction.

Note 2: We can do it on the inverse dominance order instead of inverse lexicographic order, but I think the lexicographic is easier to implement and the same for this purpose.

Edit: After wrestling an entire morning with Sagemath and it's different types of symbolic rings I was able to program such charge. Now I will be able to make my own table of charges.

Edit 2:

I have reached the following conclusions.

If one wants a charge that's defined for the entire coplactic class of a tableaux $T$, and is such that $$ k_{\alpha} (x) = \sum_{R_{+} (T) \leq R(\alpha)} t^{ch^{*} (T)} H_{T} (x;t) $$ It's impossible. I will illustrate this with two examples.

We can write $s_{211}(x)$ $HS_{211}(x,t)+(t^{3}+t^{4}+t^{5})HS_{1111}(x;t)$.

For the nonsymmetric version, for $$ T = \begin{matrix} 1 &2 \\ 3 & \\ 4 & \\ \end{matrix} $$ $$ k_{0211}(x) = H_{2110}(x,t)+H_{2101}(x,t)+H_{2011}(x,t)+H_{0211}(x,t)+H_{1210}(x,t)+H_{1201}(x,t)+t^{1} \cdot H_{1111}(x,t) $$ Where all the terms must have charge $0$ as the tableaux they come from are in the coplactic class of $T$. Also, the only $H$ which is not constant with $t$ is $H_{0211}(x,t) = x_{1}x_{2}x_{3}x_{4}+ x_{1}^{2}x_{2}x_{3} - \color{blue}{t x_{1}x_{2}x_{3}x_{4}}$. Moreover, it's the only one with a $x_{1}x_{2}x_{3}x_{4}$ coefficient aside from $H_{1111}(x,t)$. So the charge of $T$ must be $1$.

Now, let's repeat this process with $$ T' = \begin{matrix} 1 &3 \\ 2 & \\ 4 & \\ \end{matrix} $$ We obtain: $$ k_{1021}(x) = H_{2110}(x,t)+H_{2101}(x,t)+H_{2011}(x,t)+H_{1210}(x,t)+H_{1201}(x,t)+H_{1120}(x,t)+H_{1021}(x,t)+t^{1} \cdot H_{1111}(x,t) $$ Similarly, $H_{1021}(x,t) = x_{1}x_{2}x_{3}x_{4} + x_{1}x_{3}^{2}x_{4} - \color{blue}{t x_{1}x_{2}x_{3}x_{4}} $ is the only term in which $t$ makes an appeareance, and $x_{1}x_{2}x_{3}x_{4}$ does too.

So this charge must be $1$.

But this comes from $T'$.

If we call $$ T'' = \begin{matrix} 1 &4 \\ 2 & \\ 3 & \\ \end{matrix} $$

By the symmetric case we need that $$ t^{ch^{*}(T)}+t^{ ch^{*}(T') }+t^{ ch^{*}(T'') } = t^{ ch(T) }+ t^{ ch(T') }+t^{ ch(T'') } = t^{1}+t^{2}+t^{3} $$ Which is impossible as $t^{ch^{*}(T'')}$ would need to be $t^{3}+t^{2}-t$.

This example illustrates that the problem is too ambitious even if we try with the complete key polynomials.

Not all hope is lost for me, I need to work with a specific subset of key polynomials for which this seems to be possible.