Doing mathematics in a pandemic – Part I: AlCoVE

I’ll be writing up a series of posts on what I’ve learned so far about adapting my work to a pandemic-compatible lifestyle. This is the first, and focuses on math conferences. Stay safe out there!

It was March 15, 2020, and suddenly everything stopped.

This story likely sounds familiar, because the same thing probably happened to you. Classes went online. Conferences were cancelled. No more chatting with colleagues at department tea. Home life suddenly became radically different and also much more central.

The world had grinded to a halt, and yet… there was one thing that began. And that was an overwhelming sense of community and solidarity, because everyone else in the world had stopped too. And it seemed to me to be an excellent opportunity to try to create something new together.

Mathematics and community

It is said that the most important aspect of conferences is not the talks, but the coffee breaks between them.

It sounds at first like a joke about how dependent mathematicians are on caffeine. But there is a real truth to it in a different sense. The coffee breaks are where connections are made, where new ideas are spawned, where the speaker meets the one person who just might have the right tools to crack that open problem that they posed on their last slide. They’re where pairs of mathematicians who find themselves in a deep conversation comparing each of their latest tableaux insertion algorithms awkwardly check their watches and schedules and both sheepishly admit that they weren’t really looking forward to the conference banquet anyway. They then grin and scurry off to an unoccupied whiteboard to make a new joint discovery.

When everything stopped, that stopped too. But did it have to, entirely?

This was a question I posed on the Facebook group for mathematicians who specialize in symmetric functions and related algebraic combinatorics (yes, there is a Facebook group for that!). I asked if anyone would want to help me organize an online conference that tried to re-create as many of those in-person networking aspects as possible. Something that could even potentially continue into the future, as flying to so many conferences all the time, while good for mathematical progress, is not really environmentally sustainable.

I got three enthusiastic responses within an hour. Laura Colmenarejo, Oliver Pechenik, and Liam Solus were on board, and we had an organizing committee!

AlCoVE: an Algebraic Combinatorics Virtual Expedition

In order to capture the essential aspects of the conference, namely that it is about algebraic combinatorics and that it aims to capture as many of the in-person advantages of conferences as possible, we named it the Algebraic Combinatorics Virtual Expedition, or AlCoVE. It didn’t hurt that alcove walks are a highly useful and modern combinatorial construction that arise in the study of Coxeter groups, symmetric functions, and geometry (see these slides by Elizabeth Millićević for excellent illustrations of alcove walks). We had a name, and we had a pun. It was a good start.

Then came the design phase. Laura, Liam, Oliver, and I met on Zoom weekly to start planning, and started by trying to answer some of the basics:

  1. What days should the conference be held?
    We initially thought of holding a weekend conference, but then we considered that with home life being more central during the pandemic, perhaps we should have it during “work hours” so as not to overlap with participants’ family/life plans. So we decided on Monday and Tuesday, June 15-16, on a week in which participants at universities with either a semester or quarter schedule would be unlikely to be teaching.

    I think it was the right choice in the end; in our post-conference feedback form, only 8 of the 71 respondents said they would have preferred a weekend conference. Another 24 were neutral, and the remaining 39 said they preferred the weekdays over weekend.

  2. How do we account for differing time zones, given that participants are going to be in many different locations around the world?
    Our solution to this was perhaps a bit biased towards the West, as our organizers were all in either America or Europe. But we planned the conference to be from 11 AM to 5 PM Eastern time, so that on the west coast of the USA it would be from 8 AM to 2 PM, and in Europe it would be an evening conference, for instance from 4 PM to 10 PM in London.

    That being said, we had participants from India, South Africa, Australia, China, New Zealand, and more. The time zone barrier just didn’t matter as much as we thought it would. And according to the feedback form, most participants were happy with the time and scheduling of the conference.

  3. How many speakers should we have and how long should each talk be?
    Zoom fatigue is real, and it’s just harder to focus when staring at a screen than sitting in a lecture hall. In light of this fact, we decided to have talks be on the short side, a total of 30 minutes each including questions. This gave us space in the schedule for 12 talks (6 per day) with plenty of breaks and exciting social events in between.

    We then came up with a list of potential speakers to invite. We were lucky to have a team four organizers with a diverse set of interests and geographical networks within algebraic combinatorics, and we tried to come up with a good balance of mathematical and geographical diversity among the speakers. While we didn’t initially consider gender diversity while creating our list, we were pleased to see that 6 out of 12 of the mathematicians we naturally thought of first were female. It was perhaps a reflection of the friendliness and diversity that already exists in the algebraic combinatorics community.

    To our delight, everyone that we invited to speak accepted our invitation. There are perhaps some advantages of organizing a conference at a time when literally everything else is cancelled.

  4. Should we have a poster session?
    This took us a long time to decide on and subsequently plan; indeed, a virtual talk is one thing, but how do you run a virtual poster session? Then again, poster sessions are a great way to give younger participants, especially graduate students, the opportunity to share their work and ideas.

    We did end up organizing a poster session, and limited the number of posters to 12 so that it would be more manageable in a virtual setting. We had a ton of excellent submissions that were very hard to choose between. The way we implemented it was by assigning one breakout room for each poster in Zoom, and then give every single participant “co-host” power in the meeting so that they can freely move between breakout rooms as if they are walking from one poster to another. (Non-co-hosts do not have this power in Zoom.)

    It went well overall. See “Conference Day 2” below for details on how the poster session went, and ideas on how to make a poster session potentially run even more smoothly at future conferences.

  5. What should “coffee breaks” or “lunch breaks” consist of, in order to optimize social and mathematical connection in a virtual environment?
    I’m glad you asked! This was by far the most fun part of planning the conference, and there were many bouts of doubled-over, tears-streaming-down-face laughter among the organizing committee during our Zoom meetings as we brainstormed potential fun ideas for conference activities. Here was what we came up with, and the surprises involved in planning each:
    • Polls. Fun, meaningless pseudo-mathematical polls, with multiple-choice questions like “What is the worst Coxeter group?” and “Do you consider yourself a combinatorist, a combinatorialist, or a combinatoricist?” were our first idea for a social event during the breaks.

      We were inspired by Zoom’s “poll” feature, but we quickly realized that using Zoom’s built-in poll system was not ideal. We wanted to split participants into breakout rooms to take the poll, so that smaller discussions of the questions could take place. But Zoom’s polls do not show up when participants are in breakout rooms. So that eliminated Zoom’s feature as an option pretty quickly.

      Instead, we used Google Forms to put together the polls. Here is one example:
      AlCoVE Poll 1
      We simply shared the link in the Zoom chat, then split participants up into breakout rooms randomly and gave them time to participate. We then called everyone back at the end to discuss the poll results, and it served as a predictably hilarious and relaxing break between talks.

    • Escape Rooms. We created one short “virtual escape room”, again in Google Forms, which has a regular expression matching feature to check answers, so that you could prevent participants from going to the next “room” (page) until they have solved the riddle in the previous “room”.

      (Tip: To enable this feature on a given question when creating a Google form, simply click on the three dots in the lower right of a question frame and click “Response Validation”. There are then options to make the answer have to match a regular expression of your choice, and return an error message if it is incorrect.)

      Here was our conference escape room. Clearly none of us were professional puzzle writers, but when Team 2 escaped their breakout room and came back into the main Zoom room before any other team, they punched the air and cheered in victory, and we knew the social event had achieved its purpose.

    • Scavenger hunt. We created a scavenger hunt, again in Google forms, that asked participants to find things in their home, such as math textbooks or conference T-shirts, to try to match or differ from their teammates in their breakout room to score the most possible points. We got some excellent pictures submitted to the scavenger hunt challenges, and it made for great “conference photos”.

    • Virtual Excursions. These were intended as true breaks from participants’ home office desks, in which participants were split into breakout rooms and encouraged to walk around their house with their phone or laptop on Zoom to show their breakout room their local surroundings and just generally stretch their legs. The aim was to re-create the aspect of conferences in which participants walk from the conference building to the banquet hall and end up walking with a little group and chatting on their way.

      It didn’t quite end up truly re-creating what we were hoping for, but it was an easy excursion to organize and was one step up from just putting participants in breakout rooms with no direction as to what would happen in that break, which can lead to awkward silences and a lot of turned-off videos.

    • Make-Your-Own-Lunch breaks. There was a half hour “lunch break” in the middle of each conference day, in which participants were again split into breakout rooms and encouraged to make and eat lunch together over Zoom. It did lead to more interaction – who doesn’t like to talk about food? – and was the only official meal we scheduled for participants to have together.

    • Happy Hour. At the end of the first day of the conference, we made everyone co-hosts (see Conference Day 1 or 2 below for some details on this process) and set up 11 breakout rooms. You can name breakout rooms manually in Zoom, and we called one room the “Lobby” and put everyone in the lobby to start out. The other rooms were called “Table 1” through “Table 10”. Since participants had co-host powers, they were able to go “sit down” at any table they chose.

      This worked very well to mimic an actual happy hour in which there are a number of tables in a large conference room and participants mingle by moving from table to table to see old friends and meet new acquaintances. The only things we couldn’t provide virtually were drinks and appetizers!

Conference Day 1: Success or disaster?

With all the talks, poster sessions, and social events planned out, it was finally time for the conference! Laura, Liam, Oliver, and I had several last-minute meetings to test everything and everything seemed to be in order.

Naturally, a major issue arose within the first half hour of the conference. As participants were signing in, we quickly realized it was capping the number of participants at 100, even though I had already bought the Zoom ability for my account to host 500 participants. Meanwhile, over 400 people had registered. The first talk was 5 minutes away, and I had no idea why Zoom was capping us at 100. What were we going to do?

We quickly sent emails directing everyone to switch to a different Zoom meeting number on Oliver’s university account, which had a 300 person capacity, and crossed our fingers and hoped that the number of participants did not exceed 300 at any given time that morning.

Luckily we capped out at about 290 during the first talk. Disaster averted!

In the meantime, I poked around Zoom and found the switch I needed to flip. Apparently even if your personal Zoom account is listed as being able to host large meetings, you are considered a “user” on your own Zoom account and you have to enable that “user” (yourself) to be able to use that power that the entire account bought. It makes no sense, but there it is.

I flipped the switch on and we switched back to the original planned Zoom link after the lunch break. It was still glitchy; on both Oliver’s and my accounts, we had delays in the Zoom chat when people tried to post links and other information. It seemed that 200+ participants was simply getting a bit too large for Zoom to handle in one meeting, and their “large meeting” option was not entirely without issues yet.

The last thing that was awkward on the first day was the preparation for the happy hour. There is no way to assign users as co-hosts of the meeting before they log on, which means we had to manually make users co-hosts in advance of the happy hour. But there is also no button that makes everyone co-hosts at once on Zoom, so the only option is to manually make every participant a co-host one by one.

To make matters even more awkward, every time a participant is made a co-host, a little notification shows up on everyone’s screen. So the aim is to make them slowly enough that you don’t overwhelm the talk slides with notifications, but fast enough that everyone is a co-host by the time the happy hour starts so that they can all sit down at the “tables” (breakout rooms) of their choice. It was a tricky business but we got it done.

Aside from the technical issues, the first day went well. There were fantastic talks and funny polls and virtual excursions and a happy hour to cap it off at the end of the day.

Conference Day 2: Success!

While we didn’t have beginner’s luck, we did learn from Day 1, because Day 2 went much more smoothly. The conference didn’t cap our participants at 100. There were fewer participants overall and therefore fewer glitches in the chat window. The talks were incredible again, and the social activities went smoothly.

The main new challenge was the poster session. This was far harder to prepare for than the happy hour, because not only did I have to make everyone co-hosts during the talk preceding the poster session, but I had to create breakout rooms according to the posters. I created one “Lobby” room and then one room per poster, and tried to put the speaker and name of the poster as the name of the breakout room.

What I didn’t realize was that Zoom has a character limit on the breakout room names. What that meant was that I couldn’t just copy and paste the names of the presenters and titles of the posters from our website into Zoom. I had to first abbreviate and edit the titles so that they were under Zoom’s character limit, in a way that the content of the poster would still be clear to a participant browsing the titles from within Zoom. And naturally if I was editing a title in Zoom but tabbed over to glance at the title again before hitting “save”, it would delete my work and I’d have to start over. It was an unbelievable pain and I’d definitely prepare the abbreviations in advance next time.

Luckily I just barely finished the naming and assigning and co-hosting by the time the poster session was about to begin. And it began! I mostly stayed in the main room and directed lost souls who lost internet connection for a bit, but my co-organizers said that the poster session went very smoothly overall.

Video recordings and wrap-up

We recorded all the talks, and after the conference we used iMovie to do some basic processing (such as a title slide for each), and uploaded them to the new AlCoVE YouTube channel. We hope to add to this channel in future years!

Indeed, what was magical about AlCoVE is how it brought together so many mathematicians from all around the world so easily, and still re-created some of the social and networking advantages of in-person conferences. Moving more conferences online can not only drastically reduce the carbon footprint of academia, but even help with inclusivity and diversity in the community, as even those who ordinarily would not be able to travel were able to participate.

All in all, I believe AlCoVE was a very positive thing to come out of the worldwide shutdowns. I’m grateful to everyone who helped organize or speak or participate, and I hope (and will try to ensure) that it continues to run in future years, pandemic or no pandemic.

Addendum: FPSAC 2020

A few weeks after AlCoVE, I participated in FPSAC 2020 Online, the online pandemic version of an existing annual international conference called Formal Power Series in Algebraic Combinatorics (FPSAC). It was designed quite differently and also worked very well, and I learned about alternatives to Zoom breakout rooms like and Unhangout that could potentially be better for a happy hour or poster session than Zoom was. I’m excited to see where all of these recent virtual technologies lead the mathematical community in the long run.

A linear algebra-free proof of the Matrix-Tree Theorem

As a new assistant professor at Colorado State University, I had the privilege this fall of teaching Math 501, the introductory graduate level course in combinatorics. We encountered many ‘mathematical gemstones’ in the course, and one of my favorites is the Matrix-Tree theorem, which gives a determinantal formula for the number of spanning trees in a graph. In particular, there is a version for directed graphs that can be stated as follows.

Consider a directed graph $D=(V,E)$, consisting of a finite vertex set $V=\{v_1,\ldots,v_n\}$ and a set of directed edges $E\subseteq V\times V$. An oriented spanning tree of $D$ is a subset $T\subset E$ of the edges, along with a chosen root vertex $v_k$, such that there is a unique path in $T$ from any vertex $v_j\in V$ to the root $v_k$. Such a tree is said to be oriented towards $v_k$, since all the edges are `pointing towards’ the root. The term spanning indicates that $T$ is incident to every vertex in $V$. For example, in the digraph $D$ at left below, an oriented spanning tree rooted at $v_9$ is shown using red edges in the graph at right.

Define $\tau(D,v_k)$ to be the number of oriented spanning trees of $D$ rooted at $v_k$. One can check that, in the above graph, we have $\tau(D,v_9)=16$. Now, let $m_{i,j}$ be the number of directed edges from $v_i$ to $v_j$ in $D$, so that $m_{i,j}$ is equal to $1$ if $(v_i,v_j)$ is an edge and $0$ otherwise. Define the Laplacian of the digraph $D$ to be the matrix $$L(D)=\left(\begin{array}{ccccc} \mathrm{out}(v_1) & -m_{1,2} & -m_{1,3} & \cdots & -m_{1,n} \\ -m_{2,1} & \mathrm{out}(v_2) & -m_{2,3} & \cdots & -m_{2,n} \\ -m_{3,1} & -m_{3,2} & \mathrm{out}(v_3) & \cdots & -m_{3,n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -m_{n,1} & m_{n,2} & -m_{n,3} & \cdots & \mathrm{out}(v_n) \end{array}\right)$$ where $\mathrm{out}(v_i)$ is the outdegree of $v_i$, the number of non-loop edges having starting vertex $v_i$ (that is, the number of edges from $v_i$ to a vertex other than $v_i$). Then the (directed) Matrix-Tree theorem states that $$\tau(D,v_k)=\det(L_0(D,k))$$ where $L_0(D,k)$ is the deleted Laplacian obtained by deleting the $k$th row and column from $L(D)$. For instance, in the above graph, we have $$\det L_0(D,9)=\det \left(\begin{array}{cccccccc} 2 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & -1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 2 & 0 & 0 & -1 & 0 & 0 \\ 0 & -1 & 0 & 2 & -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 2 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & -1 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)=16$$ There are several known proofs of the Matrix-Tree theorem. One of the more `standard’ proofs is by induction on the number of edges in the digraph, combined with a bit of linear algebra and row reduction. But it got me thinking:

Is there be a way to prove that the determinant formula holds directly, without relying on induction or linear algebra?

In particular, the determinant of a matrix $A=(a_{ij})$ can be defined explicitly as $$\det(A)=\sum_{\pi\in S_n} \mathrm{sgn}(\pi)\prod_{i} a_{i\pi(i)}$$ where $\pi:\{1,2,\ldots,n\}\to \{1,2,\ldots,n\}$ ranges over all permutations (bijections) in the symmetric group $S_n$. For instance, \begin{align*} \det\left(\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)&=a_{11}a_{22}a_{33}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32} \\ &\phantom{=}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}. \end{align*} It is natural to ask whether applying this formula to the deleted Laplacian gives any combinatorial insight into why the Matrix-Tree theorem should hold. And indeed, there is a direct proof using this combinatorial definition of the determinant!

A combinatorial proof

For simplicity we set $k=n$, so that we are deleting the $n$th row and column to create the deleted Laplacian $\det(L_0(D,n))$. It is sufficient to consider this case since we can always relabel the vertices to have the deleted vertex be the $n$th.

We now give a combinatorial interpretation of each of the terms of the determinant $\det(L_0(D,n)$ as a sum over permutations of $\{1,2,\ldots,n-1\}$. The term corresponding to the identity permutation is the product of the diagonal entries of $L_0(D,n)$, which is $$\prod_{i\neq n} \mathrm{out}(v_i).$$ This counts the number of ways of choosing a non-loop edge starting at each vertex $v_i\neq v_n$; we call such a choice an out-edge subgraph $G$ of $D$. Note that all oriented spanning trees with root $v_n$ are out-edge subgraphs, but in general an out-edge subgraph may have cycles among the vertices other than $v_n$. In fact, it is not hard to see that every out-edge subgraph consists a number of nontrivial directed cycles among non-$v_n$ vertices, along with a unique directed path from every other vertex into either one of the cycles or into $v_n$. Two examples of out-edge subgraphs which are not trees are shown below.

Now, for a general term corresponding to a permutation $\pi$ of $\{1,2,\ldots,n-1\}$, consider the decomposition of $\pi$ into disjoint cycles. Suppose there are $p$ fixed points and $r$ nontrivial cycles; let $a_1,\ldots,a_p$ be the fixed points of $\pi$ and $(a_{1}^{(j)}\cdots a_{c_j}^{(j)})$ are the other cycles of lengths $c_1,\ldots,c_r$. Then the sign of $\pi$ is $$\mathrm{sgn}(\pi)=(-1)^{(c_1-1)+\cdots+(c_r-1)}=(-1)^{(n-1-p)-r}.$$ The entries multiplied together in the term corresponding to $\pi$ are the outdegrees of $v_{a_1},\ldots, v_{a_p}$ along with the values $-m_{a_{t}^{(i)},a_{t+1}^{(i)}}$. Their product is $(-1)^{n-1-p}$ times the number of ways to choose an edge from $v_{a_t^{(i)}}$ to $v_{a_{t+1}^{(i)}}$ for each $i$ and $t$. Putting this all together, the entire term of the determinant corresponding to $\pi$ is $(-1)^{r}$ times the number of subgraphs formed by choosing a cyclic path on the vertices corresponding to each nontrivial cycle in $\pi$, as well as an out edge for each fixed point. Such a choice is an out-edge subgraph that is compatible with $\pi$ in the sense that any cycle of $\pi$ corresponds to a cycle on the subgraph.

For some examples of compatibility, the permutations $(123)$, $(123)(57)$, $(57)$, and the identity are compatible with the out-edge subgraph drawn above at left. The permutations $(365)$ and the identity are compatible with the subgraph above at right.

It follows that we can rewrite the determinant as: $$\det L_0(D,n)=\sum_{(G,\pi)} (-1)^{r(\pi)}$$ where $r(\pi)$ is the number of nontrivial cycles of the $\pi$, and where the sum ranges over all pairs $(G,\pi)$ where $G$ is an out-edge subgraph and $\pi$ is a permutation compatible with $G$. (Note that the same out-edge subgraph $G$ may occur several times, paired with different permutations $\pi$.)

We finally construct a sign-reversing involution on the compatible pairs $(G,\pi)$ that cancel all the negative entries in the sum above. In particular, if $G$ has no cycles then send $(G,\pi)$ to itself, and otherwise consider the cycle $C$ in $G$ containing the vertex with the smallest label among all cycles in $G$. Define $\pi’$ by removing $C$ from $\pi$ if $\pi$ contains the cycle $C$, and otherwise adding $C$ to $\pi$ (in other words, toggle whether the elements of $C$ form a cycle or are all fixed points in the permutation). Then $\pi’$ is still compatible with $G$, so we can map $(G,\pi)$ to $(G,\pi’)$ in this case. This forms a sign-reversing involution in which the only non-canceling terms come from the pairs $$(T,\mathrm{id})$$ where $T$ is an out-edge subgraph with no cycles and $\mathrm{id}$ is the identity permutation.

Since a non-cyclic out-edge subgraph on $v_1,\ldots,v_{n-1}$ must be rooted at $v_n$ (for otherwise it would have a cycle), we can conclude that $\det L_0(D,n)$ is the number of spanning trees of $D$ rooted at $v_n$.

On Raising Your Hand

A few weeks ago I attended the AWM (Association of Women in Mathematics) Research Symposium in Houston, TX. I gave a talk in my special session, speaking on queer supercrystals for the first time, to a room full of female mathematicians.

I was a bit disappointed when, at the end of my talk, no one raised their hand to ask any questions.  It’s usually the classic sign of an uninteresting or inappropriately aimed talk, so I figured that maybe I had to revisit my slides and make them more accessible for the next time I spoke on the subject.

Afterwards, however, several of the women in my session came up to me privately to ask specific questions about my research.  When I told my husband about this after the conference, he pointed out that perhaps they just were the kind of people to prefer asking questions one-on-one rather than raising their hands during or after the lecture.

“Did anyone in your session ask questions after the other talks?” he asked me, testing his theory.

I thought about it, and was surprised when I realized the answer.  “Woah, I think you’re right,” I said.  “I asked at least one question after nearly every talk.  But I think I was the only one.  Once in a while one other woman would ask something too.  But the rest kept their hands down and went up to the speaker during the break to ask their questions.”

Upon further reflection, I realized that this was even true during the plenary talks.  During an absolutely fantastic lecture by Chelsea Walton, I was intrigued by something she said.  She mentioned that the automorphism group of the noncommutative ring $$\mathbb{C}\langle x,y\rangle/(xy-qyx)$$ is $\mathbb{C}^{\times} \times \mathbb{C}^{\times}$ for all $q\neq \pm 1$, but the answer is different at $q=1$ and $q=-1$.  I knew that many of the standard $q$-analogs arise naturally in computations in this particular ring, such as the $q$-numbers $$[n]_q=1+q+q^2+\cdots +q^{n-1}.$$ So, I wondered if the exceptions at $q=1$ and $q=-1$ were happening because $q$ was a root of unity, making some of the $q$-numbers be zero.  So maybe she was considering $q$ as a real parameter?  I raised my hand to ask.

“Is $q$ real or complex in this setting?”

“It’s complex,” Chelsea answered.  “Any nonzero complex parameter $q$.”

“Really?” I asked. “And there are no exceptions at other roots of unity?”

“Nope!” she replied with a smile, getting excited now.  “Just at $1$ and $-1$.  The roots of unity get in your way when looking at the representation theory.  But for the automorphism group, there are only two exceptional values for $q$.”  Fascinating!

No one else asked any mathematical questions during or after that talk.

Now, I have the utmost faith in womankind.  And I would normally have chalked the lack of questions and outspokenness up to it being a less mathematically cohesive conference than most, because the participants were selected from only a small percentage of mathematicians (those that happened to be female).  But it reminded me of another time, several years ago, that I had been surprised to discover the same phenomenon among a group of women in mathematics.

One summer I was visiting the Duluth REU, a fantastic research program for undergraduates run by Joe Gallian in the beautiful and remote city of Duluth, Minnesota.  As a former student at the program myself, I visited for a couple of weeks to hang out and talk math with the students.  I attended all the weekly student talks, and as usual, participated heavily, raising my hand to ask questions and give suggestions.

The day before I left, Joe took me aside.  “I wanted to thank you for visiting,” he said.  “Before you came, the women never raised their hand during the other students’ talks.  But after they saw you doing it, suddenly all of them are participating and raising their hands!”

I was floored.  I didn’t know that being a woman had anything to do with asking questions.

I have always felt a little out of place at AWM meetings.  They are inevitably host to many conversations about the struggles faced by women in competitive male-dominant settings, which I have never really related to on a personal level.  I love the hyper-competitive setting of academia.  I live for competition; I thrive in it.  And it never occurs to me to hold back from raising my hand, especially when I’m genuinely curious about why $q$ can be a complex root of unity without breaking the computation.

But, clearly, many women are in the habit of holding back, staying in the shadows, asking their questions in a one-on-one setting and not drawing attention to themselves.  And I wonder how much this phenomenon plays a role in the gender imbalance and bias in mathematics.

At the reception before the dinner at the AWM conference, I spotted Chelsea.  She was, unsurprisingly, quite popular, constantly engaged in conversation with several people at once.  I eventually made my way into a conversation in a group setting with her in it, and I introduced myself.

“Hi, I just wanted to say I really enjoyed your talk!  I was the one asking you whether $q$ was real.”

Her expression suddenly shifted from ‘oh-no-not-another-random-person-I-have-to-meet’ to a warm, smiling face of recognition.  “Oh!  I liked your question!” she exclaimed.  The conversation immediately turned to math, and she was nice enough to walk me through enough computations to convince me that $q=\pm 1$ were special cases in computing the automorphism group of the noncommutative ring.  (See Page 2 of this post for the full computation!)

The entire experience got me thinking.  It was because I raised my hand that Chelsea recognized me, that she was happy to talk to me and mathematics was communicated.  It was because I raised my hand that I got the question out in the open so that other participants could think about it as well.  It was because I raised my hand that women were doing mathematics together.  And perhaps it is because I raise my hand that I have no problem interacting in a male-dominant environment.  After all, they raise their hands all the time.

It is tempting to want to ask the men in mathematics to take a step back and let the women have the limelight once in a while.  But I don’t think that’s the answer in this case.  Men should keep raising their hands.  It’s part of how mathematics gets done.  It helps to communicate ideas more efficiently, to the whole room at once rather than only in private one-on-one settings.  It draws visibility to the interesting aspects of a talk that other participants may not have thought of.

What we really need is for women to come out of the shadows.  So, to my fellow women in mathematics: I’m calling on all of us to ask all our questions, to engage with the seminar room, to not hold back in those immensely valuable times when we are confused.  And raise our hands!

Addicted to Crystal Math

Some of my recent research has centered on crystals, so here’s an introductory post, from a combinatorial point of view, on what crystals are and where they come up.

First, a clarification: the crystals I have in mind are not the chemical kind that you find in nature, nor are they any of these other uses of the word crystal on the Wikipedia disambiguation page. There is yet another use of the word, sometimes referred to as a crystal base, introduced by Kashiwara in the mid-90’s. Kashiwara developed crystal base theory as a way of understanding the $q\to 0$ limit of the representation theory of quantum groups $U_q(\mathfrak{g})$ where $\mathfrak{g}$ is a Lie algebra and $U_q(\mathfrak{g})$ is a $q$-analog of the universal enveloping algebra $U(\mathfrak{g})$. In the quantized statistical mechanics models in which quantum groups come up, the parameter $q$ is a measure of temperature, so the $q\to 0$ limit is exploring what happens in the situation of absolute zero temperature. Hence the word “crystal”, referring to what happens to many forms of matter when frozen.

Although the original motivation for crystals came from quantum groups, the combinatorial structure is also very useful in symmetric function theory. One example of a crystal is a certain weighted, edge-labeled directed graph structure on the summands that comprise a Schur polynomial. Recall that the Schur function $s_{\lambda/\mu}(x_1,\ldots,x_n)$ is given by the sum $$s_{\lambda/\mu}=\sum_{T\in \mathrm{SSYT}(\lambda/\mu)} x^T$$ where $\mathrm{SSYT}(T)$ is the set of semistandard Young tableaux of skew shape $\lambda/\mu$, and $x^T=x_1^{m_1}x_2^{m_2}\cdots$ where $m_i$ is the number of $i$’s in $T$. For instance, we have $$s_{(2,1)}(x_1,x_2,x_3)=x_1^2x_2+x_1x_2^2+x_1^2x_3+2x_1x_2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2,$$ since the eight semistandard Young tableaux of straight shape $(2,1)$ are:

We can organize these tableaux by weight in a 2-dimensional grid, since each weight is a tuple $(m_1,m_2,m_3)$ where $m_1+m_2+m_3=3$. We angle the grid so that either changing the weight by $(-1,1,0)$ or by $(0,-1,1)$ is “lowering” the weight, going downwards at a 45 degree angle to the right or left. We will then connect some of the tableaux along the edges of the grid with edges labeled by weight-lowering operators $F_1$ and $F_2$ (with inverses $E_1$ and $E_2$), shown in red and blue below:

In general $F_i$ changes the weight by $(0,0,\ldots,-1,1,0,\ldots,0)$ with the $-1$ in the $i$-th position, and we think of $F_i$ as an operator on tableaux that changes an $i$ into an $i+1$. To define $F_i(T)$, we consider the reading word $w$ of the tableau $T$, formed by concatenating the rows from bottom to top, and consider just the $i$ and $i+1$ entries of $w$. Replace each $i$ with a closed bracket ‘)’, and each $i+1$ with an open bracket ‘(‘, and cancel the brackets in matching pairs until only a word of the form $$)))\cdots ))(((\cdots (($$ (or alternatively, of the form $ii\cdots ii(i+1)(i+1)\cdots(i+1)$) remains. Then $F_i$ changes the last $i$ in this subsequence to $i+1$. For example, if $T$ looks like:

then to apply $F_1$ to $T$, we consider the reading word of $1$s and $2$s and cancel matching pairs as follows:

2 & 2 & 1 & 1 & 1 & 1 & 2 & 1 & 2 & 2 & 1 \\
( & ( & ) & ) & ) & ) & ( & ) & ( & ( & ) \\
( & & & ) & ) & ) & & & ( & & \\
& & & & ) & ) & & & ( & &

Then $F_1$ changes the rightmost $)$ that was not cancelled to $($, changing that $1$ to $2$. The word becomes:

2 & 2 & 1 & 1 & 1 & 2 & 2 & 1 & 2 & 2 & 1.

Note that $F_i$ may not be defined, if there is no closed bracket that is not cancelled. If no $F_i$ operator may be applied to a tableau $T$, we say that $T$ is lowest weight. The bottom tableau on the crystal diagram above is lowest weight for the operators $F_1,F_2$. Similarly, we can define raising operators $E_i$ as the natural partial inverses of the $F_i$ operators (simply by reversing the arrows in the crystal graph), and we define highest weight elements as those for which all $E_i$ operators are undefined.

Now here’s the interesting thing: highest weight tableaux or words are precisely the Littlewood-Richardson tableaux that we have encountered in several previous posts. Recall that a tableau is Littlewood-Richardson if its reading word $w=w_1\cdots w_n$ is ballot, meaning that every tail of the word (of the form $w_k\cdots w_n$ for some $k$) has at least as many $1$’s as $2$’s, at least as many $2$’s as $3$’s, and so on. In other words, when reading the word from right to left, the smaller letters always occur more frequently than larger letters. For instance, the word $$12122111$$ is ballot, and it is not hard to see that ballotness in a word with just $1$’s and $2$’s is equivalent to the condition that every $2$ is bracketed with a $1$, so that $E_1$ is undefined.

The key fact about tableau crystals is as follows:

Every connected component of a crystal graph has a unique highest weight element.

For instance, here is the crystal graph for the skew shape $(3,1)/(1)$:

Note that each connected component has a Littlewood-Richardson tableau at the top, and no other elements are killed by all raising operators. Notice also that the first graph above has the same structure as that of the shape $(2,1)$ crystal drawn above. This is no coincidence: another key fact about the operators $F_i$ and $E_i$ is that they commute with Jeu de Taquin rectification, which is easily proven using Knuth equivalence. The tableaux in the shape $(2,1)$ crystal are simply the Jeu de Taquin rectifications of those in the corresponding component of the $(3,1)/(1)$ graph. Similarly, the tableaux in the second component rectify to form the tableau crystal for shape $(3)$. It follows that, if we sum the monomial weights in the graph, we obtain the equation $$s_{(3,1)/(1)}=s_{(2,1)}+s_{(3)},$$ which is an instance of the famous Littlewood-Richardson rule.

In general, let $c^{\lambda}_{\mu\nu}$ be the number of Littlewood-Richardson tableaux of shape $\lambda/\mu$ and content $\nu$. Then recall that the Littlewood-Richardson rule for multiplying Schur functions states:

$$s_\mu s_\nu = \sum_{\lambda} c^{\lambda}_{\mu\nu} s_\lambda,$$

or equivalently, after a bit of symmetric function theory manipulation,

$$s_{\lambda/\mu}=\sum_{\nu} c^{\lambda}_{\mu\nu} s_\nu.$$

The latter equation now follows immediately from the facts about crystals listed above, as we saw for the shape $(3,1)/(1)$. We can also prove the product version directly using crystals, by introducing a tensor product. There is a natural tensor product on crystals coming from their correspondence with Lie algebra representations, which we will define in general on the next page. In combinatorial terms, we can define the tensor product of crystals of tableaux by defining the concatenation of two tableaux $T$ and $S$ to be the tableau $T\cdot S$ formed by drawing $S$ above and to the right of $T$, with the lower left corner of $S$ matching the upper right corner of $T$. For instance, tableaux of skew shape $(3,1)/(1)$ can be formed by concatenating a tableau of shape $(1)$ with a tableau of shape $(2)$. (Note that this concatenates the reading words of $T$ and $S$ as well, and induces the plactic monoid structure on Knuth equivalence classes of words.)

Then the tensor product $\mathcal{B}_\mu \otimes \mathcal{B}_\nu$ of two tableaux crystals on shapes $\mu$ and $\nu$ is the crystal on all words of the form $T\cdot S$ for $T\in \mathcal{B}_\mu$ and $S\in \mathcal{B}_\lambda$, with operators $F_i$ connecting the nodes appropriately. It turns out that every connected component in the resulting crystal are again full crystals of tableaux, with Schur functions as their weight generating functions. We thus obtain a crystal-theoretic interpretation of the multiplicative Littlewood-Richardson rule. Namely, we have
$$\mathcal{B}_\mu \otimes \mathcal{B}_\nu=\bigcup c^\lambda_{\mu\nu} \mathcal{B}_{\lambda}.$$

For instance, $\mathcal{B}_{(1)}\otimes \mathcal{B}_{(2)}$ is the crystal whose graph is precisely the diagram of $\mathcal{B}_{(3,1)/(1)}$ shown above, so $$\mathcal{B}_{(1)}\otimes \mathcal{B}_{(2)}=\mathcal{B}_{(2,1)}\cup \mathcal{B}_{(3)},$$ matching the Schur expansion $s_{(1)}s_{(2)}=s_{(2,1)}+s_{(3)}.$

With these examples in mind, on the next page we will define the tensor category of crystals over any semisimple Lie algebra.

Hockey sticks on planet ABBABA

Question: On the planet ABBABA, the inhabitants have a binary language where the only two letters in their alphabet are A and B. The language is incredibly efficient and complex in that every finite sequence of A’s and B’s is a valid word. How many of the words in this language have exactly five A’s and at most five B’s?

For instance, ABAAABA, AAAAA, and BBBBBAAAAA are all valid such words, since they all have five A’s and no more than five B’s.

For the sake of readers who might want to grapple with this problem on their own, I’ll leave the answer and discussion to the next page. We’ll start with a straightforward counting method, and then go on to find a much faster method that will also lead us to an elegant combinatorial proof of a well-known binomial coefficient identity. Once you think you have the answer, turn to page 2!

The CW complex structure of the Grassmannian

In a previous post, we briefly described the complex Grassmannian $\mathrm{Gr}(n,k)$ as a CW complex whose cells are the Schubert cells with respect to a chosen flag. We’ll now take a closer look at the details of this construction, along the lines of the exposition in this master’s thesis of Tuomas Tajakka (chapter 3) or Hatcher’s book Vector Bundles and $K$-theory (page 31), but with the aid of concrete examples.

Background on CW complexes

Let’s first review the notion of a CW complex (or cell complex), as described in Hatcher’s Algebraic Topology.

An $n$-cell is any topological space homeomorphic to the open ball $B_n=\{v\in\mathbb{R}^n:|v|<1\}$ in $\mathbb{R}^n$. Similarly an $n$-disk is a copy of the closure $\overline{B_n}=\{v\in \mathbb{R}^n:|v|\le 1\}$.

To construct a cell complex, one starts with a set of points called the $0$-skeleton $X^0$, and then attaches $1$-disks $D$ via continuous boundary maps from the boundary $\partial D$ (which simply consists of two points) to $X^0$. The result is a $1$-skeleton $X^1$, which essentially looks like a graph:

This an then be extended to a $2$-skeleton by attaching $2$-disks $D$ via maps from the boundary $\partial D$ (which is a circle) to $X^1$. Note that, as in the picture below, the attaching map may collapse the entire boundary to a point, making the circle into a balloon-like shape.

In general the $n$-skeleton $X^n$ is formed from $X^{n-1}$ by attaching a set of $n$-disks by maps from their boundaries to $X^{n-1}$.

More precisely, to form $X^n$ from $X^{n-1}$, we start with a collection of $n$-disks $D^n_\alpha$ and continuous attaching maps $\varphi_\alpha:\partial D_\alpha^n\to X^{n-1}$. Then $$X^n=\frac{X^{n-1}\sqcup \bigsqcup_\alpha D_\alpha^n}{\sim}$$ where $\sim$ is the identification $x\sim \varphi_\alpha(x)$ for $x\in \partial D^n_\alpha$. The cell complex is $X=\bigcup_n X^n$, which may be simply $X=X^n$ if the process stops at stage $n$.

By the construction, the points $X^0$ along with the open $i$-cells associated to the $i$-disks in $X^i$ for each $i$ are disjoint and cover the cell complex $X$. The topology on $X$ is given by the rule that $A\subset X$ is open if and only if $A\cap X^n$ is open in $X^n$ for all $n$.

The real projective plane

As an example, consider the real projective plane $\mathbb{P}^2_{\mathbb{R}}$. It has a cell complex structure in which $X^0=\{(0:0:1)\}$ is a single point, and $X^1=X^0\sqcup \{(0:1:\ast)\}$ is topologically a circle formed by attaching a $1$-cell to the point at both ends.

Then, $X^2$ is formed by attaching a $2$-cell $\mathbb{R}^2$ to the circle such that the boundary wraps around the $1$-cell twice. Intuitively, this is because the points of the form $(1:xt:yt)$ as $t\to \infty$ and as $t\to -\infty$ both approach the same point in $X^1$, so the boundary map must be a $2$-to-$1$ mapping.

To make this analysis more rigorous, consider $\mathbb{P}^2_{\mathbb{R}}$ as the quotient $(\mathbb{R}^3\setminus\{(0,0,0)\})/\sim$ where $\sim$ is the equivalence relation given by scalar multiplication. In particular there is a natural continuous quotient map $$q:\mathbb{R}^3\setminus \{0\}\to \mathbb{P}^2_{\mathbb{R}}.$$ We can now take $n$-disks in $\mathbb{R}^3\setminus\{0\}$ and map them into $\mathbb{P}^2_{\mathbb{R}}$ via $q$, which will give the CW complex structure described above.

In particular, the point $(0,0,1)$ maps to $(0:0:1)$, giving the $0$-skeleton. Then consider the set of points $$\{(0,a,b):a^2+b^2=1, a\ge 0\}\subseteq \mathbb{R}^3\setminus \{0\},$$ which is a half-circle and thus homeomorphic to a $1$-disk. The quotient map $q$ gives an attaching map of the boundary of this $1$-disk to $(0:0:1)$, by mapping both endpoints $(0,0,1)$ and $(0,0,-1)$ to $(0:0:1)$. It also maps the interior of this disk bijectively to the $1$-cell $(0:1:\ast)$ in $\mathbb{P}^2_{\mathbb{R}}$.

Finally, consider the set $$D=\{(a,b,c):a^2+b^2+c^2=1,a\ge 0\},$$ which is a half-sphere and thus homeomorphic to a $2$-disk. Then under $q$, the interior of $D$ maps to the $2$-cell $(1:\ast,\ast)$, and $q$ gives a $2$-to-$1$ attaching map of the boundary $\partial D=\{(0,b,c): b^2+c^2=1\}$ onto the $1$-skeleton $\{(0:x:y)\}$. Indeed, given a desired ratio $(0:x:y)$, there are two points $(0,b,c)\in \partial D$ mapping to it, since a line with a given slope intersects the unit circle in exactly two points.

The complex projective plane

A nearly identical construction to the one above can help us realize the complex projective plane $\mathbb{P}_{\mathbb{C}}^2$ as a CW complex as well, though the structure is somewhat different.

In particular, we can again start with the point $(0:0:1)$ as our $0$-skeleton, and attach the $1$-cell $(0:1:\ast)$ as follows. Let $$q:\mathbb{C}^3\setminus\{0\}\to \mathbb{P}_{\mathbb{C}}^2$$ be the natural quotient map, and define $$D=\{(0,b,c):|(0,b,c)|=1, b\in \mathbb{R}_{\ge 0}\}\subseteq \mathbb{C}^3\setminus \{0\}.$$ Then we can write $c=x+yi$ and the conditions defining $D$ are equivalent to $b,x,y\in \mathbb{R}$, $b\ge 0$, $b^2+x^2+y^2=1$ (since $|c|^2=c\overline{c}=x^2+y^2$). Thus $D$ is a real half-sphere and hence a $2$-disk. The quotient map takes the interior of $D$, on which $b>0$, bijectively and continuously to $(0:1:\ast)$, while collapsing its entire circular boundary to the $0$-skeleton as the attaching map.

Finally, we can similarly take the image of $$E=\{(a,b,c): |(a,b,c)|=1, a\in \mathbb{R}_{\ge 0}\},$$ which is homeomorphic to a 4-disk, to obtain the remaining points in $\mathbb{P}_{\mathbb{C}}^2$. (Indeed, setting $b=x+yi$ and $c=z+wi$ for real numbers $x,y,z,w$, the space $E$ is cut out by the equations $a^2+x^2+y^2+z^2+w^2=1$, $a\ge 0$ in $5$-dimensional real space, so it is a $4$-hemisphere.) The attaching map on the boundary is “circle-to-one”: every point in the $1$-skeleton has a circle as its preimage in $E$. (Can you prove this?)

Note that these constructions can be continued for higher dimensional projective spaces over $\mathbb{C}$ or $\mathbb{R}$.

The Grassmannian

We now can analyze the complex Grassmannian $\mathrm{Gr}(n,k)$ – the space of $k$-dimensional subspaces of $\mathbb{C}^n$ – in a similar manner to that of a complex projective space. In fact, $\mathbb{P}_{\mathbb{C}}^n=\mathrm{Gr}(n+1,1)$, so we have already done this for some Grassmannians.

For $\mathrm{Gr}(n,k)$ with $k\ge 2$, we use a similar quotient map, but this time from the Stiefel manifold $V_k=V_k(\mathbb{C}^n)$, the space of orthonormal $k$-frames. That is, $V_k$ is the space of all tuples of linearly independent vectors $(v_1,\ldots,v_k)\in (\mathbb{C}^n)^k$ for which $$\langle v_i, v_j\rangle=\delta_{ij}$$ for all $i$ and $j$. (Here $\langle-,-\rangle$ can be taken as the standard Hermitian inner form $\langle (z_1,\ldots,z_n), (w_1,\ldots,w_n) \rangle=\sum z_i \overline{w_i}$.)

Now, there is a natural quotient map $q:V_k\to \mathrm{Gr}(n,k)$ given by row-reducing the matrix formed by the $k$-frame, as we did in this post. Note that $V_1=\mathbb{C}^n\setminus\{0\}$ is the space we used above for projective space, and that $q$ matches our map above in this case.

We will not go into full details of the CW complex construction here, since Tajakka’s paper does it so nicely already. Instead, let’s outline the construction for one of the nontrivial attaching maps in $\mathrm{Gr}(4,2)$.

Recall that in $\mathrm{Gr}(4,2)$, the Schubert varieties are indexed by partitions whose Young diagram fits inside a $2\times 2$ rectangle. The Schubert variety $\Omega_{(2,2)}$ (with respect to the standard flag) consists of the single point of the Grassmannian represented by the row-reduced matrix

$$\left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right]$$

Use this point as the $0$-skeleton of the Grassmannian. For the $2$-skeleton, we attach the $2$-cell given by $$\Omega_{(2,1)}^{\circ}=\left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & \ast & 0 \end{array}\right]$$ to the $0$-skeleton, which is the unique boundary point in its closure. So the $1$-skeleton is actually homeomorphic to the complex projective line.

We now construct the attaching maps that show that if we include the two $4$-dimensional (over $\mathbb{R}$) Schubert cells given by $\Omega_{(1,1)}^{\circ}$ and $\Omega_{(2)}^{\circ}$, we get a $4$-dimensional CW complex. Consider the cell $$\Omega_{(1,1)}=\left[\begin{array}{cccc} 0 & 0 & 1 & \ast \\ 0 & 1 & 0 & \ast \end{array}\right].$$ Define $E$ to be the following subspace of $V_2(\mathbb{C}^4)$:
$$D=\{(v_1,v_2)\in V_2: v_1=(0,0,a,b), v_2=(0,c,d,e)\text{ for some }b,d,e\in \mathbb{C}, a,c\in \mathbb{R}_{\ge 0}\}.$$ Then $D$ maps under $q$ to $\Omega_{(1,1)}=\overline{\Omega_{(1,1)}^\circ}$, its interior mapping to $\Omega_{(1,1)}^\circ$.

We now wish to show that $D$ is topologically a $4$-disk, making $q$ an attaching map. This is the tricky part in general, but we’ll describe the proof explicitly in this special case.

Let $$D_1=\{(0,0,a,b):a\ge 0, b\in \mathbb{C}, a^2+b\overline{b}=1\}$$ and
$$D_2=\{(0,x,0,z):x\ge 0, z\in \mathbb{C}, x^2+z\overline{z}=1\}.$$ Then both are hemispheres in $\mathbb{R}^3$ and hence homeomorphic to $2$-disks, so their Cartesian product $D_1\times D_2$ is homeomorphic to a $4$-disk. So it suffices to show that $D_1\times D_2$ is homeomorphic to $D$.

In particular, given a pair $(v,u)\in D_1\times D_2$, we can apply a certain rotation $T_v$ to $u$ (depending continuously on $v$) that transforms $(v,u)$ into an independent, orthonormal pair of vectors $(v,T_v(u))\in D$. We can define $T_v$ to be the unique rotation that takes $(0,0,1,0)$ to $v$ and fixes all vectors orthogonal to both $(0,0,1,0)$ and $v$. Then since $(0,0,1,0)$ is already orthogonal to $u$ by the definition of $D_2$, this will give a pair $(v,T_v(u))$ of orthonormal vectors, which are independent since $u$ and $v$ are. Tajakka proves (in much more generality) that this map gives the required homeomorphism.

Back to $\mathbb{R}$eality

The real Grassmannian also has a CW complex structure, given by an almost identical construction to the one above (see Hatcher, page 31). Let’s analyze the map described above, replacing all complex coordinates with real ones.

Let $v=(0,0,a,b)$ with $a^2+b^2=1$ and $a\ge 0$ as above. The map $T_v$, in coordinates, sends $(0,x,0,w)$ to $(0,x,-wb,wa)$. Keeping in mind that $a^2+b^2=1$ and $x^2+w^2=1$, it is easy to see that the vectors $(0,0,a,b)$ and $(0,x,-wb,wa)$ form an orthonormal $2$-frame. The disk $D$ in this setting is the set of all orthonormal $2$-frames of this form with $a,x\ge 0$.

Taking the image of $D$ under the quotient map $q:V_2(\mathbb{R}^4)\to \mathrm{Gr}_{\mathbb{R}}(4,2)$, it’s not hard to show that the restriction to the boundary is a $2$-to-$1$ map, as in the construction of the real projective plane. Indeed, if $q((0,0,a,b),(0,x,0,w))$ is the point of the Grassmannian represented by $$\left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & c & 0 \end{array}\right],$$ this forces $a=0$, $b=\pm 1$, $x=1$, $w=-c/b$. So there are two preimages of this point, determined by the two possible values for $b$.

I haven’t worked out (or found in a reference) what the degrees of the attaching maps are for the higher dimensional Schubert cells. If you work out a higher dimensional example, or know of a reference for this, please post it in the comments below!

Thanks to Jake Levinson for enlightening mathematical discussions that motivated me to write this post.

Higher specht polynomials

There are polynomials. There are Specht polynomials. And then there are higher Specht polynomials.

A colleague recently pointed out to me the results of this paper by Ariki, Terasoma, and Yamada, or more concisely summarized here. The authors give a basis of the ring of coinvariants $$R_n=\mathbb{C}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$$ which, unlike the other bases we’ve discussed on this blog so far, respects the decomposition into irreducible $S_n$-modules. These basis elements include the ordinary Specht polynomials, but also some polynomials of higher degrees, hence the name “higher” Specht polynomials.

Background: What is a Specht polynomial?

The construction of the irreducible representations of the symmetric group $S_n$ (called Specht modules) is often described in terms of “polytabloids”, but can be equivalently described in terms of a basis of Specht polynomials.

Before we describe this basis, recall that the dimension of the irreducible representation $V_\lambda$ of $S_n$ indexed by the partition $\lambda$ is equal to the number $f^\lambda$ of Standard Young Tableaux (SYT) of shape $\lambda$, i.e., the number of fillings of the Young diagram of $\lambda$ with the numbers $1,\ldots,n$ in such a way that the entries are increasing down rows and across columns. An example is shown below for $n=7$ and $\lambda=(3,3,1)$:

Given a standard Young tableau $T$, the Specht polynomial $F_T$ is defined as the following product taken over all of its columns $C$
$$F_T(x_1,\ldots,x_n)=\prod_{C} \prod_{i\lt j\in C}(x_j-x_i)$$

For instance, in the above tableau, $$F_T(x_1,\ldots,x_7)=(x_3-x_1)(x_4-x_1)(x_4-x_3)(x_6-x_2)(x_7-x_5).$$

In other words, the Specht polynomial for $T$ is the product of the Vandermonde determinants given by each of its columns’ entries. It is known that the set of polymomials $F_T$, where $T$ ranges over all SYT’s of shape $\lambda$, form a basis of their span, and their span is isomorphic to $V_\lambda$ as an $S_n$-module (here the $S_n$-action is the usual action on the variables $x_1,\ldots,x_n$).

Specht polynomials in the coinvariant ring

As described in detail in this post, the coinvariant ring is the $S_n$-module $$R_n=\mathbb{C}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$$ where $e_i$ is the $i$th elementary symmetric function in the variables $x_1,\ldots,x_n$.

The ring $R_n$ inherits the action of $S_n$ on the polynomial ring, and we saw in the previous post on the coinvariant ring that it is isomorphic to the regular representation as an $S_n$-module. It is also graded by polynomial degree, with Hilbert series equal to $(n)_q!$. There are many known bases of $n!$ elements that respect this grading, but the ones we have described in previous posts do not respect the $S_n$-action. In particular, can we find an explicit basis which partitions into bases of the irreducible components in $R_n$?

To start, the Specht polynomials $F_T$ defined above are nonzero in $R_n$ and span one copy of each Specht module. But we know that in the regular representation, there are $f^\lambda$ copies of the irreducible representation $V_\lambda$ for each $\lambda$, and this construction only gives one copy of each. In fact, since the copies of $V_\lambda$ must be alternating in the column variables, they are the lowest-degree copies of each. Hence, we need higher Specht polynomials.

To define them, we first need to understand the RSK correspondence and the charge statistic.

The tools: RSK and charge

Somehow, in all of my years of blogging, I have never actually described the RSK (Robinson-Schensted-Knuth) correspondence. By the nature of the regular representation of $S_n$, we must have $$\sum_{\lambda \vdash n} (f^{\lambda})^2=n!$$ since each irreducible representation $V_\lambda$ of dimension $f^\lambda$ occurs exactly $f^{\lambda}$ times, and the regular representation has dimension $n!$. Recall also that $f^\lambda$ is the number of standard Young tableaux of shape $\lambda$.

The RSK correspondence gives a combinatorially described bijection between pairs of standard Young tableaux of the same partition shape $\lambda$ of $n$ (for some $\lambda$) and permutations of $n$. It’s a beautiful bijection, but since the readers of this post will most likely already be familiar with it, I will simply refer the reader to Simon Rubenstein-Salzedo’s clearly-written introduction to RSK.

The upshot is that there is one higher Specht polynomial for each pair of standard Young tableaux $(S,T)$ of the same shape, denoted $F_T^S$. The degree of $F_T^S$ is given by the charge statistic on $S$, defined as follows. Let $w(S)$ be the reading word of $S$, formed by concatenating the rows of $S$ from bottom to top, and assign subscripts $s(i)$ to the letters $i$ of the reading word recursively as follows. Assign the letter $1$ the subscript $s(1)=0$, and for each $i\ge 1$, if $i+1$ is to the right of $i$ then assign it $s(i+1)=s(i)$, and otherwise $s(i+1)=s(i)+1$. The charge of $S$ is the sum of the subscripts $s(i)$.

For example, let $S$ and $T$ be the tableaux given below:

Then the charge of $S$ is computed by first forming the reading word $$w(S)=3267145$$ and then placing subscripts starting from the letter $1$ according to the rules above:
$$3_2 2_1 6_3 7_3 1_0 4_2 5_2$$
Then the charge of $S$ is $2+1+3+3+0+2+2=13$.

The lowest-degree (minimal charge) case occurs when $S=S_0$ is the canonical choice having elements $1,2,3,\ldots,n$ written in order left to right in each row from top to bottom, as in:

In this case we will recover the standard Specht polynomials, and the others are the higher Specht polynomials.

Construction of the Higher Specht polynomials

We now have the tools to describe the construction. Given a pair $(S,T)$ of standard Young tableaux of the same shape $\lambda$, let $x_T^{\mathrm{charge}(S)}$ be the monomial $x_1^{i_1}\cdots x_n^{i_n}$ where $i_k$ is the charge subscript on the cell in $S$ corresponding to the cell containing $k$ in $T$. For example, if we fill the boxes of $S$ in the example above with their corresponding subscripts in the charge computation, it looks like:

and we have $x_T^{\mathrm{charge}(S)}=x_2^2x_3x_4^2x_5^2x_6^3x_7^3$.

Then the higher Specht polynomial $F_T^S$ is defined as $$F_T^S(x_1,\ldots,x_n)=\sum_{\tau\in \mathrm{Col}(T)}\sum_{\sigma\in \mathrm{Row}(T)} \mathrm{sgn}(\tau)\tau\sigma x_T^{\mathrm{charge}(S)}$$ where $\mathrm{Row}(T)$ and $\mathrm{Col}(T)$ are the subgroups of $S_n$ generated by the permutations of the rows of $T$ and the columns of $T$, respectively.

Notice that this polynomial has degree given by the charge of $S$, which is minimal when $S=S_0$ is the canonical tableau of shape $\lambda$ as described above. And when $S=S_0$, $x_T^{\mathrm{charge}(S)}$ is the product $\prod_{i} x_i^{h_i}$ where $h_i$ is the row of the entry $i$ in $T$. It’s not hard to see that antisymmetrizing this monomial over the column permutations recovers the ordinary Specht polynomial $F_T$. Furthermore, symmetrizing across rows keeps the monomial fixed, and so the higher Specht polynomial $F_T^{S_0}$ is simply a constant multiple of the ordinary Specht polynomial $F_T$.

In general the Specht module $V_\lambda$ occurs once for each tableau $S$ of shape $\lambda$, giving the right multiplicities for each irreducible representation.

Thanks to some Sage code by Nicolas Thiery, we can use this formula to efficiently compute the higher Specht polynomials. Here are the polynomials for $n=3$:

6 \\
2(y-x) \\
2(z-x) \\
z(y-x) \\
y(z-x) \\

Note that the first three and the last are (scalar multiples of) ordinary Specht polynomials, and $z(y-x)$ and $y(z-x)$ are two higher Specht polynomials giving another copy of $V_{(2,1)}$.


In their paper, Ariki, Terasoma, and Yamada actually construct a more general version of the higher Specht polynomials that respects the restricted $S_\mu$-module structure of $R_n$ where $S_{\mu_1}\times \cdots \times S_{\mu_k}\subseteq S_n$ is any Young subgroup. It’s also natural to ask whether there are generalizations of the higher Specht modules to related rings such as the Garsia-Procesi modules or even (we can dream!) the Garsia-Haiman modules.

Schubert Calculus mini-course

I’ve written a lot about Schubert calculus here over the last few years, in posts such as Schubert Calculus, What do Schubert Curves, Young tableaux, and K-theory have in common? (Part II) and (Part III), and Shifted partitions and the Orthogonal Grassmannian.

I soon found out that writing a lot about your favorite topics on a blog can sometimes get you invited to give talks on said topics. This past June, I gave one of the three graduate mini-courses at the Equivariant Combinatorics workshop at the Center for Mathematics Research (CRM) in Montreal.

Naturally, a lot of what I covered came from the blog posts I already had written, but I also prepared more material on flag varieties and Schubert polynomials, more details on the cohomology of the Grassmannian, and more problems and examples suitable for the workshop. So I organized all of this material into a single lecture notes document available here:

Variations on a Theme of Schubert Calculus

The coolest part was that the CRM had good quality audio/video setup for all three workshops, and so you can also view my lecture videos that accompany these notes at the following five links, and take the entire course yourself:

Lecture 1: Introduction, Projective varieties, Schubert cells in the Grassmannian

Lecture 2: Duality theorem, CW complexes, Homology/cohomology

Lecture 3: Littlewood-Richardson rule, Flag variety, Schubert polynomials

Lecture 4: Cohomology of flag variety, generalized flag varieties

Lecture 5: The orthogonal Grassmannian, recap

You can also find the videos for the other two mini-courses from the summer – Steven Griffeth’s lectures on Cherednik algebras and Jeffrey Remmel’s lectures on symmetric function theory – at this link.

Hope someone out there finds these lectures useful or interesting. I certainly had a blast teaching the course!

Pythagorean triples on a sphere?

It’s 8/15/17, which means it’s time to celebrate! The three numbers making up the date today form a Pythagorean triple, a triple of positive integers $(a,b,c)$ with $a^2+b^2=c^2$. Indeed, $8^2+15^2=64+225=289=17^2$.

Alternatively, by the Pythagorean theorem, a Pythagorean triple is any triple of positive integers which make up the sides of a right triangle:

It’s exciting when all three sides are integers, since many common right triangles’ lengths involve square roots: $(1,1,\sqrt{2})$, $(1,2,\sqrt{5})$, and $(1,\sqrt{3},2)$, to name a few. And these sides aren’t even rational, which the poor Pythagorean scholar Hippasus discovered by proving that $\sqrt{2}$ is irrational and was subsequently drowned to death by his colleagues, according to some historical accounts. So the ancient Pythagoreans in fact only believed in right triangles having rational side lengths.

Of course, given one Pythagorean triple, like $(3,4,5)$, we can construct infinitely many by scaling the sides: $(6,8,10)$, $(9,12,15)$, etc. (In fact, 12/9/15 was the previous Pythagorean triple day, and 12/16/20 will be the next.) So to classify all Pythagorean triples, it suffices to find the reduced triples, those with no common factors greater than $1$.

So the last reduced Pythagorean triple day was back in 2013 on 12/5/13, and the next one won’t be until 7/24/25!

Constructing Pythagorean triples

It’s well known that there are infinitely many reduced Pythagorean triples as well. One beautiful, famous proof of this derives a parameterization of all triples via geometric means:

Consider the unit circle and the point $P=(-1,0)$ on the circle. Let $Q=(0,r/s)$ be a rational point on the $y$-axis inside the circle. Then line $PQ$ intersects the circle at a second point $R$, and it turns out $R$ has rational coordinates as well. Some simple algebra with similar triangles (try it, if you haven’t seen it before!) gives us $$R=\left(\frac{r^2-s^2}{r^2+s^2},\frac{2rs}{r^2+s^2}\right).$$ But since $R$ lies on the unit circle, if $(x,y)$ are the coordinates of $R$ then $x^2+y^2=1$, and substituting and clearing denominators we have $$(r^2-s^2)^2+(2rs)^2=(r^2+s^2)^2$$ (which can also be checked with direct algebraic computation). It follows that $(r^2-s^2,2rs,r^2+s^2)$ is a Pythagorean triple for any integers $r$ and $s$, giving us infinitely many Pythagorean triples. And in fact, for $r$ and $s$ relatively prime of different parity, these triples are reduced.

Spherical considerations

Given that this is a global day to celebrate (assuming you use the standard world calendar), and the Earth is a sphere, I have to wonder whether Pythagorean triples actually exist if drawn on a perfect-sphere approximation of the Earth. First, we’d have to define what we even mean by that – are we measuring in meters? In feet? And what is a right triangle on the surface of a sphere anyway?

The most common definition of a triangle on a sphere is one formed by great circles. A great circle is any circle of maximal diameter around a sphere, in other words, whose plane contains the center of the sphere. So, the equator, the prime meridian, and all longitude lines are great circles on the Earth, but latitude lines are not. A line segment on a sphere is a segment of a great circle, and a (spherical) triangle is a shape formed by three points connected by three (spherical) line segments. The angle between two great circles that intersect is the angle between their planes. (Thanks to Wikipedia for the excellent image below.)

Since our world now has size rather than being a flat plane, let’s set the radius of the Earth to be $1$ unit for simplicity. So we’re working with triangles $ABC$ with a right angle at $C$, and asking when they have rational lengths:

Are there any Pythagorean triples on the unit sphere? Are there infinitely many?

These questions suddenly aren’t very easy. If we scale our sphere down by $\pi/2$ we can get at least one, by taking the equator, the prime meridian, and the $90^\circ$ longitudinal line. This forms a right triangle with all three lengths equal (and all angles right!) and so we can simply scale the Earth to make it any rational lengths we want. But this still doesn’t answer anything about the unit sphere.

There is some hope, however. In this paper by Hartshorne and Van Luijk, the authors show that there are infinitely many Pythagorean triples in the hyperbolic plane, using the Poincare Disk model and some nice hyperbolic trig formulas combined with some Eulerian number theory tricks. So Pythagorean triples are not the sole property of flat Euclidean space.

In addition, there is a “spherical Pythagorean theorem”, which in our notation, since the radius of our sphere is $1$, says that $$\cos(c)=\cos(a)\cos(b)$$ where $a,b,c$ are the side lengths of the triangle and $c$ is opposite the right angle. And yet, this offers little insight into whether this equation has rational solutions. Trig functions are generally much harder to deal with than polynomials, especially when it comes to solving equations over the rationals.

So, if you have any insights on this problem (or references to papers that have worked on it – I couldn’t find any in an initial search, but I am not very familiar with it), please let me know in the comments! And until then, happy reduced-Pythagorean-triple-in-flat-Euclidean-space day!

A faster divisibility rule for $7$ (and $13$)

I occasionally teach evening online classes for Art of Problem Solving (, a fantastic resource for high school students interested in learning mathematics), and one of the lessons I taught recently was on fast mental arithmetic tricks for testing divisibility by various small integers.

How do you tell if a number is divisible by $2$? Easy: look at the last digit. If that digit is even, the whole number is divisible by $2$, otherwise it’s not.

How do you tell if a number is divisible by $3$? Easy: Add up the digits and see if the sum is divisible by $3$. For instance, $1642$ is not divisible by $3$ because $1+6+4+2=13$ is not, but $1644$ is because $1+6+4+4=15$ is divisible by $3$.

The list goes on; there are nice, well-known divisibility rules for $4$, $5$, $6$, $8$, $9$, $10$, $11$, and $12$. But $7$ turns out to be rather annoying. Many of the existing texts on this subject use a recursive rule that goes something like this: Cross off the last digit, double that digit, and subtract it from the number that remains. Then keep doing that until you get a small number and see if the result is divisible by $7$.


Not only is this method somewhat cumbersome, the small number you get in the end does not tell you the remainder mod $7$ (see this post for an introduction to modular arithmetic), only whether or not that remainder is $0$. But it turns out that one can derive a different method along the same lines as the divisibility rule for $3$. (See this Wikipedia article for many, many divisibility rules!)

First, let’s recall the proof of the rule for $3$. Given a number in base ten, like $1642$, we can write it out as a sum of powers of $10$:

$$1642=1\cdot 10^3+6\cdot 10^2+4\cdot 10+2$$
Notice that $10$ has a remainder of $1$ when divided by $3$, so when looking at remainders mod $3$, we can replace all these $10$’s in the expression above with $1$’s:
$$1642\equiv 1+6+4+2 \pmod{3}$$ and we have recovered the sum of the digits expression.

Similarly, for mod $7$, we can write things out in terms of powers of $10$, and reduce these powers modulo $7$. The number $10$ itself is $3$ mod $7$, then $100\equiv 10^2\equiv 3^2$ is $2$ mod $7$. The next remainders, for $10^3,10^4, 10^5$ are $6,4,5$, and then the remainder when $10^6$ is divided by $7$ is $1$ again, wrapping around to the same value as $10^0$. Writing this out in a table, we see a pattern start to emerge:

n & 1 & 10 & 10^2 & 10^3 & 10^4 & 10^5 & 10^6 & 10^7 & 10^8 \\
n \bmod 7 & 1 & 3 & 2 & 6 & 4 & 5 & 1 & 3 & 2

Indeed, the remainders $1,3,2,6,4,5$ repeat every six steps, since at each step we are simply multiplying the previous remainder by $10$ and taking the result mod $7$. But we can make this pattern even simpler by replacing $6$ with $-1$, $4$ with $-3$, and $5$ with $-2$, which still have the same remainders mod $7$ (they differ by $7$ from the originals):

n & 1 & 10 & 10^2 & 10^3 & 10^4 & 10^5 & 10^6 & 10^7 & 10^8 \\
n \bmod 7 & 1 & 3 & 2 & -1 & -3 & -2 & 1 & 3 & 2

So, the pattern simply continues $1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,\ldots$. We can therefore replace the powers of $10$ in a base ten expansion by these smaller digits to compute the number mod $7$. For instance, $$1642=10^3+6\cdot 10^2+4\cdot 10+2\equiv -1+6\cdot 2+4\cdot 3+2=25\equiv 4\pmod 7,$$ so $1642$ has a remainder of $4$ when divided by $7$.

As a larger example, starting with the number $133251140379$, we can make a simple table with the pattern $1,3,2,-1,-3,-2,\ldots$ underneath the digits from right to left (we reverse the direction since the powers of $10$ increase to the left):

1 & 3 & 3 & 2 & 5 & 1 & 1 & 4 & 0 & 3 & 7 & 9 \\
-2 & -3& -1& 2 & 3 & 1 & -2& -3& -1& 2 & 3 & 1

Then we can simply multiply the pairs of numbers in column together (and take the results mod 7) and add the results mod $7$. We can write each product below the pairs:

1 & 3 & 3 & 2 & 5 & 1 & 1 & 4 & 0 & 3 & 7 & 9 \\
-2 & -3& -1& 2 & 3 & 1 & -2& -3& -1& 2 & 3 & 1 \\\hline
-2 & -2& -3& 4 & 1 & 1 & -2& -5& 0 & 6 & 0 & 2

(Notice that in the computation of the products we sometimes automatically take the result mod $7$: for instance, $5\cdot 3=15\equiv 1\pmod 7$, so under the fifth column we simply write $1$.) Finally, the sum of the numbers in the bottom row is $-2-2-3+4+1+1-2-5+6+2=0$, so $133251140379$ is indeed divisible by $7$.

I tested this method by hand on a few large numbers and compared my calculation time to the iterative subtraction method, and it saves a good chunk of time. The particular number above took me 35 seconds to check with the $1,3,2$-pattern method, and 69 seconds to check with the double-and-subtract rule. Try it out, and let me know in the comments whether you find this method easier!

Addendum: A similar rule for $13$

While this method may become more cumbersome for larger moduli (which will have a longer pattern to memorize), it turns out that powers of $10$ mod $13$ also have an easily-memorizable repeating pattern:

n & 1 & 10 & 10^2 & 10^3 & 10^4 & 10^5 & 10^6 & 10^7 & 10^8 \\
n \bmod 13 & 1 & -3 & -4 & -1 & 3 & 4 & 1 & -3 & -4

The pattern is $1,-3,-4,-1,3,4$ repeating, which can be memorized by taking the length $3$ string $1,3,4$ and then writing negative signs on every other block of three entries starting at the tens digit. Let’s try it out on our example number above.

1 & 3 & 3 & 2 & 5 & 1 & 1 & 4 & 0 & 3 & 7 & 9 \\
4 & 3 &-1 &-4 &-3 & 1 & 4 & 3 & -1&-4 &-3 & 1 \\\hline
4 &-4 &-3 & 5 &-2 & 1 & 4 & -1& 0 & 1 & 5 & -4
We have $4-4-3+5-2+1+4-1+1+5-4=6$, so $133251140379$ has a remainder of $6$ when divided by $13$.