# Schubert Calculus

I recently gave a talk on Schubert calculus in the Student Algebraic Geometry Seminar at UC Berkeley. As a combinatorialist, this is a branch of enumerative geometry that particularly strikes my fancy. I also made a handout for the talk called “Combinatorics for Algebraic Geometers,” and I thought I’d post it here in blog format.

# Motivation

## Enumerative geometry

In the late 1800’s, Hermann Schubert investigated problems in what is now called enumerative geometry, or more specifically, Schubert calculus. As some examples, where all projective spaces are assumed to be over the complex numbers:

2. Answer: One, as long as the points are distinct.

3. How many planes in $\PP^3$ contain a given line $l$ and a given point $P$?
4. Answer: One, as long as $P\not\in l$.

5. How many lines in $\PP^3$ intersect four given lines $l_1,l_2,l_3,l_4$?
6. Answer: Two, as long as the lines are in sufficiently “general” position.

7. How many $(r-1)$-dimensional subspaces of $\PP^{m-1}$ intersect each of $r\cdot (m-r)$ general subspaces of dimension $m-r-1$ nontrivially?
8. Answer: $$\frac{(r(m-r))!\cdot (r-1)!\cdot (r-2)!\cdot \cdots\cdot 1!}{(m-1)!\cdot(m-2)!\cdot\cdots\cdot 1!}$$

The first two questions are not hard, but how would we figure out the other two? And what do we mean by “sufficiently general position”?

Schubert’s 19th century solution to problem 3 above would have invoked what he called the “Principle of Conservation of Number,” as follows. Suppose the four lines were arranged so that $l_1$ and $l_2$ intersect at a point $P$, $l_3$ and $l_4$ intersect at $Q$, and none of the other pairs of lines intersect. Then the planes formed by each pair of crossing lines intersect at another line $\alpha$, which necessarily intersects all four lines. The line $\beta$ through $P$ and $Q$ also intersects all four lines, and it is not hard to see that these are the only two in this case.

Schubert would have said that since there are two solutions in this configuration and it is a finite number of solutions, it is true for every configuration of lines for which the number is finite by continuity. Unfortunately, due to degenerate cases involving counting with multiplicity, this led to many errors in computations in harder questions of enumerative geometry. Hilbert’s 15th problem asked to put Schubert’s enumerative methods on a rigorous foundation. This led to the modern-day theory known as Schubert calculus.

## Describing moduli spaces

Schubert calculus can also be used to describe intersection properties in simpler ways. As we will see, it will allow us to easily prove statements such as:

The variety of all lines in $\PP^4$ that are both contained in a general $3$-dimensional hyperplane $S$ and intersect a general line $l$ nontrivially is isomorphic to the variety of all lines in $S$ passing through a specific point in that hyperplane.

(Here, the specific point in the hyperplane is the intersection of $S$ and $L$.)

## 6 thoughts on “Schubert Calculus”

1. Hey Maria, thanks for posting this! I’ve been looking to learn a little bit about Schubert calculus, and this looks like a fantastic introduction – I’ll try to get around to reading it in detail this weekend.

• Anyway – do you suggest any references for a systematic treatment of the subject? Preferably something that isn’t too technical (everything you use in your post would be fine, I think; I think I can fake an understanding of algebraic geometry up to a point).

• Hey Carl, glad you found this useful! I have a few references scattered throughout the post, but I’d say the main one I learned from is Fulton’s book, “Young tableaux”. I’d say that one is pretty accessible for those who don’t have a strong background in algebraic geometry/topology, because he neatly pushes all the technical details about cohomology into the appendix. But then, the appendix is also detailed enough if you do want to understand those details.

I’ve also heard good things about the new book by Eisenbud and Harris, “3264 and all that.” It can be found online here: http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf
and it certainly goes into much more detail on the geometric side of things, because it’s building up a more general theory than just intersections of Schubert varieties. I plan to read (parts of) it at some point soon.

2. One combinatorial way to look into the Schubert decomposition is:
1.We associate a matrix to each k-subset as you said.
2.Then in each matrix consider the set of subsets of columns (numbered from left to right) which are bases.
3.Put a total order on such sets by considering the induce lexicographic order.
Each schubert cell then correspond to all subspaces with an specified subset of the columns being the maximal basis. So in the example of page 2 we want all matrices with the subset (2,4,7) being the biggest basis lexicographically. This is a complicated way of saying the same but now:
– If we consider ALL possible orders of the columns (schubert is very dependent on ordering from left to right) and intersect them all, in the common refinement we would be grouping subspaces by specifying the complete set of bases. This is the Matroid Strata decomposition. Bad thing is, Mnevs Universality theorem says that such cells can be “as bad as possible”, and while Im never sure of the precise meaning, it certainly means that working with them is much worse than working with the schubert cells which are just affine spaces.
-If we consider all cyclic orders and intersect them, the cells in the refinement are called the Positroid decomposition, and there are lots of combinatorics associated to them. In particular they have something called a Le diagram associated
-Lauren Williams and Kelli Talaska consider something similar, the Go diagrams, to give a refinement of the positroid decomposition. The open cells are not bad, but their closures and how the intersect was (or maybe still is) a bit mysterious.

3. Hi Maria, thanks for the post and the link of the video lecture about Schubert calculus in your other post, they have been so helpful!
I have one question which might be stupid. When we want to find which Schubert cell a point of the Grassmannian belongs to, we cut off the upside-down staircase from the matrix, all entries in the staircase should be 0, right? Why must the first column of the matrix be a zero column?