#the alternating sign matrix conjecture

[thousandmaths]

Day 27: Talk with Vic IV

Today I met with Vic, as my recently-chosen advisor, to discuss possible topics of study. Today’s citation is “Conjectures on the Quotient Ring by Diagonal Invariants” by Mark Haiman.

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Vic

We began with some broad-scale discussions about my areas of interest. I was surprised how much I was able to specify, even though a lot of it came up later in the meeting than was strictly useful.

Coxeter-Catalan combinatorics

Alternating Sign Matrix-adjacent things

Order theory, especially its topological features

After some discussion about the Equivariant Combinatorics seminar, Vic tipped his hand and talked about stuff that came up in one of the REU projects that the students decided not to explore (and “they were right to not go that way”, he admits). He figured that it might serve as a nice problem to start getting engaged with the literature especially around symmetric functions and the Coxeter generalizations.

The story starts with $(q,t)$-Catalan polynomials but quickly degenerates into something a lot more comprehensible, so I’ll lop off a little bit of the motivation. Basically, by setting $t=1/q$ in the $(q,t)$-Catalan polynomial, you end up with a classical $q$-analogue of the Catalan polynomials first considered by MacMahon:

$$ C_n(q) = \sum_{\pi\in\text{Dyck}(n)} q^{\text{maj}(\pi)}, $$

where $\text{Dyck}(n)$ is the set of Dyck paths of length $n$, and $\text{maj}(\pi)$ is the major index (in the extended sense— instead of permutation we use binary words: $0$s representing the up-steps and $1$s representing the down-steps).

It’s worth noting that this is also a $q$-analogue in the analytic sense: that is, the ordinary Catalan numbers are $\frac{1}{n+1}\binom{2n}{n} = \frac{(2n)!}{n!(n+1)!}$, and the reason why MacMahon was considering them is because he wanted to prove that

$$ C_n(q) = \frac{1}{[n+1]_q} \begin{bmatrix} 2n \\\ n \end{bmatrix}_q = \frac{(2n)!_q}{n!_q(n+1)!_q}, $$

where the $!_q$’s are intended to denote the usual $q$-analogues of factorials. This formula is a lot easier for calculation purposes.

Anyway a kind of annoying feature about $C_n(q)$ is that it’s not unimodal: for instance, $C_3(q) = 1+q^2+q^3+q^4+q^6$. This is not unimodal because the coefficients in front of the $q^k$ do not increase, and then decrease. Instead they jump around a little bit: first decreasing from $1$ to $0q$, then increasing to $1q^2$, and then staying the same for a while, then decreasing from $1q^4$ to $0q^5$ and then increasing back to $1q^6$.

However, what seems like a coincidence is that if you break this polynomial into its even and odd powers, and then set $p=q^2$:

$$C_3(q) = (1+q^2+q^4+q^6) + q^3 = (1+p+p^2+p^3) + q(0+p+0p^2),$$

then as polynomials in $p$, both parenthetical expressions actually are unimodal (because “increase” and “decrease” are meant in the weak sense, where staying the same can count as either increasing or decreasing).

It turns out that this “coincidence” is not one at all: it happens for all $C_n(q)$. There is a very good (and very difficult*) reason for this phenomenon. One way to describe what Vic’s suggested project is, is to provide a combinatorial interpretation for it, instead of the difficult algebraic option that currently exists.** There is some hope in that this might be possible, since when we specialize to $q=1$, and consider the resulting two sequences in $n$, both of these are recognized by the Online Encyclopedia of Integer Sequences.

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[ * The experts in the crowd may be interested to know that this has to do with the interpretation of the $(q,t)$-Catalan polynomial as the graded Frobenius character of the diagonal coinvariant algebra. The latter admits an $\mathfrak{sl}_2$-action and hence has the requisite raising and lowering operators, and there exist standard tools to leveraging such operators into a unimodality statement. (This last remark, it seems to me, is a big reason why you might want crystals in your life, beyond simply enjoying frustration.) ]

[ ** Note that this is not a hoplessly difficult problem: since we are specialized to the $t=q^{-1}$ case, our aims would not, in particular, imply a combinatorial proof of $q,t$-symmetry. ]