General Information

---

Project Description

Schubert Calculus was created in order to solve problems in projective geometry. For example, if we were to take four arbitrary lines in $\mathbb{C}^3_{\text{ ,}}$ we might want to figure out how many lines in $\mathbb{C}$ intersect all four lines. It turns out that the set of lines intersecting a particular line can be written in terms of a Schubert Variety. Since there is a correspondence between Schubert Varieties and Young Tableaux, we can change this into a simple problem about box diagrams. From this, we can find that there are two lines that intersect the four lines in $\mathbb{C}^3$.

---

Weekly Log

Week 1:

During my first meeting with Professor Buch, we spent a few hours covering background material. This week was spent filling in gaps in prerequisite knowledge, from subjects such as Topology, Algebra and Algebraic Geometry.

Week 2:

We had another meeting regarding more geometric insight on Young Tableaux. Heman and I both wrote code to deal with some algebra regarding Grothendieck stable polynomials. We also tinkered a little bit with possible properties of these polynomials, so as to get closer to solving the main conjecture.

Week 3:

After realizing my Java code was really slow, I spent some time figuring out an efficient algorithm for multiplying partitions by each other. I ended up discovering that most of the time spent during execution went towards creating new objects, and as it turns out, copying elements is significantly faster than creating them from scratch. With this in mind, I got a very large speedup.
Once I was satisfied with my code, I started playing around with basic ideas of multiplying partitions. There are a couple of nice properties of fillings that we were able to figure out, which were good for two reasons. Primarily, it allowed for a more intuitive understanding of the algebra, and secondly, it directly corresponded to speedups in code.

Week 4:

While listening to a talk during our IBM visit, inspiration struck and magically produced our first general formula. Once we were able to solve L-shapes, the ball kept on rolling. We then solved two-row shapes in general, which was extremely nice.
The idea is that, hopefully, by examining three row shapes, we can end up figuring out an induction on the number of rows in a partition.

Week 5:

We were so wrong. There isn't such an induction. Whoever I angered, please forgive me.
One thing that seems to work is reducing an arbitrary 3 row shape to one in which the first and second row are of the same length. Maybe something will work out there. More likely, it won't, because partitions are actively trying to ruin my life.

Week 6:

By means of caffei- I mean, induction, I was able to find a wide variety of formulas for three for partitions. The problem is that partitions are still out for me, which means that nothing generalizes nicely for arbitrary three row shapes. My suspicion is that without involving four row shapes, you cannot solve three row shapes. But since I am stubborn, I will ignore this suspicion for a while.

Week 7:

Since three row shapes aren't working out, I've been considering shapes of form $(n, n - 1, ..., 1)$ and shapes with $n$ boxes in the first row and $n$ rows in total that fit within the first kind of shape. In general, we call the first kind of shape an $``n$-triangle".
Spoiler alert: This doesn't go anywhere😅

Week 8:

I'm rapidly approaching the point of "nothing is working anymore I don't know what to do." Since it's time to write reports and our paper, I'm focusing more on picking out the important results we've found until now. While there were a number to choose from, many were redundant due to a later generalization, or more or less useless as they never contributed to anything else.

Week 9:

I must've pleased some divine entity somewhere, as a series of fortunate events occurred. Following some suspicions, a formula to solve for shapes in form $(a, b, 1)$ was found. And then, following some more suspicions and copious amounts of caffe- Uhhh, induction, we found a similar formula for $(a, b, 2)$.
At this point, two things were bothering me. I couldn't figure out a proof for $(a, b, 2)$, as it was significantly different from $(a, b, 1)$. Moreover, I couldn't even figure out a formula from which to start to prove the solution of $(a, b, 3)$. But then, at the prime moment of 2:43 AM, I discovered a formula for $(a, b, c)$, which is to say that I solved three row shapes. The relevant part of the formula is $$\left(\sum_{i = 1}^{b - 1} G_{(c,c)} G_{(b - 1, i)}\right) - \left( \sum_{i = 1}^c G_{(a - 1)} G_{(c,c,i)} \right) = G_{(a, a, c)} + \sum_{\nu} c_{\nu} G_{\nu},$$ where the sum ranges over partitions $\nu$ that are rectangular, have two rows, or have three rows with second row shorter than $a$.
Despite this, I was unable to prove the formula, leaving our results incomplete. Regardless, it was quite satisfying to have found a way to solve three row partitions in general.

---

Additional Information

• My Mentor
  • Professor Anders Buch