*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 questionfor 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 $(\pm 1, \pm 1)$!)

*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?**

*What if we didn’t have to emit quite as much carbon to do mathematics effectively?*

As the latest heat wave sweeps the globe, it’s nice and cool down here in the basement of the mathematics building at Colorado State University. Thanks to a combination of my NSF grant and contributions from the CSU College of Natural Sciences, we have set up an easy-to-use, high quality virtual conferencing room here!

While there is certainly a benefit to in-person networking at conferences, some other aspects of academic work can arguably be *improved* by building high quality virtual spaces. For one, there is a significant carbon footprint caused by academic travel, and the COVID-19 pandemic has given us the opportunity to think about ways to slash our carbon footprint. For another, there are disabilities, economic disparities between regions, and caregiving situations that make it difficult for some academics to travel as much as others. Improving virtual options could improve accessibility and enable such academics to more easily share their ideas.

And finally, playing with virtual technology is *fun*!

*This is a long overdue followup post to Garsia-Procesi Modules: Part 1 that I finally got around to editing and posting. Enjoy!*

In this post, I talked about the combinatorial structure of the Garsia-Procesi modules $R_\mu$, the cohomology rings of the type A Springer fibers. Time to dive even further into the combinatorics!

Recall that, for a partition $\mu$ of $n$, the graded $S_n$-module $R_\mu$ can be constructed (due to Tanisaki) as the quotient ring \[\mathbb{C}[x_1,\ldots,x_n]/I_\mu\] where $I_\mu$ is generated by certain partial elementary symmetric functions called *Tanisaki generators*.

Define $d_k(\mu)=\mu’_{n-k+1}+\mu’_{n-k+2}+\cdots$ to be the sum of the last $k$ columns of $\mu$, where we pad the conjugate partition $\mu’$ with $0$’s in order to think of it as as a partition of $n$ having $n$ parts. Then the partial elementary symmetric function $e_r(x_{i_1},\ldots,x_{i_k})$ is a Tanisaki generator if and only if $k-d_k(\mu)\lt r\le k$. Tanisaki and Garsia and Procesi both use this notation, but I find $d_k$ hard to remember and compute with, especially since it involves adding zeroes to $\mu$ and adding parts of its transpose in reverse order, and *then* keeping track of an inequality involving it to compute the generators.

In my graduate Advanced Combinatorics class last semester, I covered the combinatorics of crystal base theory. One of the concepts that came up in this context was *ballot sequences*, which are motivated by the following elementary problem about voting:

Suppose two candidates, A and B, are running for local office. There are 100 voters in the town, 50 of whom plan to vote for candidate A and 50 of whom plan to vote for candidate B. The 100 voters line up in a random order at the voting booth and cast their ballots one at a time, and the votes are counted real-time as they come in with the tally displayed for all to see. What is the probability that B is never ahead of A in the tally?