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)?

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    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. – Richard Stanley Dec 13 '14 at 17:41
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    Isn't your assertion easy to prove from the combinatorial formula for Macdonald polynomials? en.wikipedia.org/wiki/… – Per Alexandersson Dec 14 '14 at 9:26
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    @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.) – imakhlin Dec 14 '14 at 18:24
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    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/…. – Richard Stanley Dec 21 '14 at 4:07
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    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. – Richard Stanley Dec 23 '14 at 3:07

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.

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