3
$\begingroup$

Here's the new, more thought through version.

Consider a sequence of nonnegative integers $\lambda=(\lambda_1,\ldots,\lambda_n)$ with $\lambda_i\ge \lambda_{i+1}+2$ (the weight $\lambda-2\rho$ is dominant, in terms of $\mathfrak{gl}_n$ weights). Next, consider the polynomial $$P_\lambda(x_1,\ldots,x_n;t)=\sum\limits_{\sigma\in S_n} \sigma\left(x_1^{\lambda_1}\ldots x_n^{\lambda_n}\prod\limits_{i<j}\dfrac{x_i-tx_j}{x_i-x_j}\right).$$ This can, probably, be rightfully referred to as the Hall-Littlewood polynomial (all parts are distinct, so no normalization is needed). I'm just trying to accentuate the fact that I'm concerned with this specific polynomial, not an element of $\Lambda_{\mathbb{C}[t]}$. (Or is then "Hall-Littlewood polynomial" not the appropriate term?)

Anyway, I strongly believe that with our assumption on $\lambda$ in place the polynomial $P_\lambda(x_1,\ldots,x_n;-t)$ has positive coefficients. As I mentioned in the first version, this is confirmed by my observations and a certain geometrical argument.

My assumption can be somewhat weakened, but even as is this looks to me like a very basic fact in the theory of these well-studied expressions. Now, my questions are:

1) Can someone confirm that this is true and provide a reference to some down-to-earth (combinatorial) proof?

2) Why is this absent from all (almost all?) surveys on the subject of Hall-Littlewood polynomials? Just because this is a statement about the polynomials themselves rather than symmetric functions? Is it really not mentioned in Macdonald's book?

3) My real question. Is there a proof expressing $P_\lambda(x_1,\ldots,x_n;-t)$ as a sum of visibly positive summands enumerated by some combinatorial set (hopefully, SSYTs or Gelfand-Tsetlin patterns)?

$\endgroup$
12
  • 3
    $\begingroup$ We have $P_{2,2}(x;-t)=m_{2,2}+(t+1)m_{2,1,1}+(-t^3+3t+2)m_{1,1,1,1}$. Doesn't this contradict your assertion? On the other hand, it is well-known that $P_\lambda(x;-1)$ is Schur-positive. $\endgroup$ Dec 13, 2014 at 17:41
  • 1
    $\begingroup$ Isn't your assertion easy to prove from the combinatorial formula for Macdonald polynomials? en.wikipedia.org/wiki/… $\endgroup$ Dec 14, 2014 at 9:26
  • 1
    $\begingroup$ @PerAlexandersson Right, thanks for bringing this up. I remember this passage being confusing. Obviously, most Hall-Littlewood polynomials do have negative coefficients. How can then Macdonald polynomials be positive and reduce to Hall-Littlewood poynomials at $q=0$? (I don't really know what I'm talking about, just trying to use my common sense.) $\endgroup$ Dec 14, 2014 at 18:24
  • 1
    $\begingroup$ One has to distinguish between the $P$-Macdonald basis (which specializes to Hall-Littlewood by setting $q=0$) and the $H$-Macdonald basis (for which Haglund, Haiman, and Loehr gave a combinatorial interpretation of the expansion into monomials). See sagemath.org/doc/reference/combinat/sage/combinat/sf/…. $\endgroup$ Dec 21, 2014 at 4:07
  • 1
    $\begingroup$ I found a simple proof that if $\lambda=(\lambda_1,\dots,\lambda_n)$, where $\lambda_i\geq \lambda_{i+1}+n-1$ for all $1\leq i\leq n-1$, then the coefficients of the Schur function expansion of $P_\lambda(x_1,\dots,x_n;t)$ are polynomials in $t$ with nonnegative coefficients (which can be described combinatorially). I don't see any way of extending the proof to answer Igor's question. $\endgroup$ Dec 23, 2014 at 3:07

2 Answers 2

2
$\begingroup$

There is a nice formula for Hall-Littlewood polynomials that follows from the combinatorial formula due to Haglund, Haiman and Loehr:

In this paper, eqution (69), we have that the Macdonald P-polynomial is given as $$ P_\lambda(x;q,t)= \left[\prod_{u \in dg'(\lambda)} (1-q^{l(u)+1}t^{a(u)})\right] \sum_{\gamma \sim \lambda} \frac{E_\gamma(x;1/q,1/t)}{\prod_{s\in \gamma}(1-q^{l(s)+1}t^{a(s)})} $$ where the sum is over all permutations (as compositions) of $\lambda$.

Putting $q=0$ then gives an expression for the Hall-Littlewood polynomials, See eq. (7.8) in this paper.

Thus, $$ P_\lambda(x;t) = \sum_{\gamma \sim \lambda} \sum_{F \in NAF(\gamma)} x^F t^{coinv(F)} (1-t)^{dn(F)} $$ where $dn(F)$ is the number of boxes $u$ in the filling $F$, such that the box to the left of $u$ is filled with a different entry from $u$. There are some details to be filled in, regarding a 0th column, that is referred to as the basement. Here, $NAF(\gamma)$ is a certain set of non-attacking fillings with weakly decreasing rows, and the 0th column has an $i$ in row $i$.

So, $$ P_\lambda(x;-t) = \sum_{\gamma \sim \lambda} \sum_{F \in NAF(\gamma)} x^F (-t)^{coinv(F)} (1+t)^{dn(F)} $$ It is tempting to hope that each inner sum is positive, but that is not the case. However, it should not be too tricky to find a sign-reversing involution to cancel negative signs. One can for example apply certain operators that preserve symmetric functions, but does something predictable on the $E_\gamma$, see this article, (apologizes for self-referencing) which also has some other similar-looking expressions for the Hall-Littlewood polynomials.

$\endgroup$
1
$\begingroup$

The specialization $P_\lambda(x;-1)$ is what is referred to as Schur's P functions. They can be described combinatorially using shifted tableaux, and are Schur-positive.

See also slides here by S. Cho.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.