Hikita's proof of the Stanley-Stembridge Conjecture

In this arXiv paper, Hikita recently posted a proof of the famous Stanley-Stembridge conjecture! I went through the proof with my Advanced Combinatorics class, and wrote up lecture notes here:

Lecture Notes on Stanley-Stembridge Part I

Lecture Notes on Stanley-Stembridge Part II

which I will summarize in this post.

The Springer Correspondence Part IV: Affine Springer Fibers

I have written about the flag variety, the Springer resolution, and the relation between the type A Springer correspondence and Hall-Littlewood polynomials in a previous sequence of posts.

Time to extend this construction to the (type A) affine flag variety, corresponding to affine Lie type $\widetilde{A}_n$! We’ll also see how, due to a result of Hikita in 2012, this construction gives rise to a geometric interpretation of the famous symmetric functions $\nabla e_n$ from the Shuffle theorem.

How many trivalent trees does it take to hold up a moduli space?

After a bit of a hiatus due to difficulties with Wordpress, Mathematical Gemstones is back! Feel free to browse the new layout.

Today’s post introduces moduli spaces of curves, and one of the many ways combinatorial gemstones arise in this vast area of algebraic geometry.

A brief motivating question for studying families of curves: Given 4 general points in the plane, how can you write down all conics that pass through them? (Try it for the four points $(\pm 1, \pm 1)$!)

Moduli of Curves IV: Getting classy

Young tableaux are to planes as labeled trees are to curves. Continuing from the previous post in this series, we now show how to think of $\overline{M}_{0,n}$ as a projective variety.

We now show how to look from the perspective of a single point, and combine this with the forgetting maps, to construct a projective embedding of $\overline{M}_{0,n}$ and thereby realize it as a geometric space itself, rather than just as a set of curves. The key new construction we need is the Kapranov map.

The Kapranov map

Given a curve $C\in \overline{M}_{0,n}$, consider the $\mathbb{P}^1$ component that the marked point $n$ is on. The other special points on this component are either nodes, each leading to a distinct branch of the dual tree, or marked points (which can be considered branches with just one leaf). Choose coordinates on this $\mathbb{P}^1$ in a way so that point $n$ is at coordinate $\infty=(1:0)$, and the branch containing marked point $1$ is attached at coordinate $0=(0:1)$. Then define the Kapranov map \[ \psi_n: \overline{M}_{0,n}\to \mathbb{P}^{n-3} \] that sends a curve $C$ to the tuple $(x_2:x_3:\cdots : x_{n-1})$ where $(x_i:1)$ is the coordinate at which the branch containing $i$ is attached to $n$’s component. (Note that this means that all $x_i$’s for $i$ in a given branch are equal.)

Sorting sets of binary strings into threes

Time for a just-for-fun combinatorial problem! Thanks to my brother Kenneth Monks for suggesting it.

Consider the numbers of the form $2^{2^n}-1$ for positive integers $n$. The first few, for $n=1,2,3,4,\ldots$, are:

\[3, 15, 255, 65535, \ldots\]

One thing that all of these numbers have in common is that they are divisible by $3$. This is not hard to prove by induction; the first entry is divisible by $3$, and if $2^{2^n}-1$ is divisible by $3$, then $2^{2^{n+1}}-1=(2^{2^n})^2-1=(2^{2^n}-1)(2^{2^n}+1)$ is also divisible by $3$.

But is there a combinatorial proof?