Last post set up the Schubert decomposition of . We’re going to use it to do some intersection theory computations.

Since Schubert varieties are indexed by partitions, this is really about “multiplying partitions” — and the ring structure will coincide with the structure for multiplying partitions in the contexts of representation theory and symmetric polynomials.

All of these computations are in , the Chow/cohomology ring.

(1) First of all, if and only if and overlap when placed opposite each other in the box. (If they overlap, then some dimension count is too high; if they don’t overlap, then the intersection includes the “identity matrices” described in the last post, hence is nonempty.)

(1a) In particular, . Here means the complementary partition in the box.

(2) Let denote the single-box partition. For any ,

,

where the sum is over all possible ways of adding a single box to , while remaining a partition contained in the rectangle.

(3) Similarly to the previous rule, we have the Pieri rule: if is the single-row partition with boxes, then

,

where the sum is over all the ways of adding boxes “horizontally” to : they don’t have to be connected, but they must be horizontally spread out (no two in the same column). For example,

,

assuming all the partitions fit in the rectangle. The proof of this rule is by direct computation in linear algebra (details in Fulton’s *Young Tableaux*).

This is actually the same as the Pieri rule for multiplying Schur polynomials (indexed by partitions). So we have the following:

**Theorem**. The map defined by (or to if doesn’t fit in the rectangle) is a ring homomorphism. (Here is the ring of symmetric polynomials.)

*Proof*: Since the Schur polynomials are an additive basis for , the above map is well-defined additively. We just need to check that it’s multiplicative, that is, it sends to .

This follows from the Jacobi-Trudi rule, which lets us write as a polynomial in terms of the single-row partitions. We expand and use the Pieri rule (which is satisfied in both rings) repeatedly. QED.

Indeed, this theorem shows that **any** ring with an additive basis indexed by partitions, satisfying the Pieri rule, has a surjective map from . So, we can “import” the Littlewood-Richardson rule for multiplying Schur polynomials:

(4) For any , we have , where is the Littlewood-Richardson number, but the sum includes only those that fit in the rectangle.

Since the Littlewood-Richardson rule is more or less fully understood in combinatorics, this gives a complete description of the ring structure of the cohomology of Grassmannians!

**Interlude: Enumerative Geometry**

I want to give some examples to show why this abstract description is actually useful for general geometry problems. I will do two examples — in both cases, we’ll solve an enumerative geometry problem by reducing it to a computation in Schubert calculus — even though the statement of the problem will have nothing apparent to do with combinatorics!

**Problem 1**. Let be a smooth curve of degree . The locus of lines passing through is codimension 1. So, given four such curves in general position, only finitely-many lines will touch all four. How many?

Since we know is codimension one, we know it is cohomologous to a linear combination of codimension-one Schubert classes — in particular, , the only partition of size , for some multiple .

To find , we use “the method of undetermined coefficients”, and multiply both sides of the equation with the complementary partition to , which is : the locus of “lines contained in a given 2-plane and passing through a given point .”

So , where the left-hand side is now the intersection of with a generic Schubert locus: lines contained in , passing through , *and* passing through the curve .

Well, already is, for general , exactly reduced points, where . And generically, none of these points will coincide with the required point . So there are exactly choices of line satisfying all our conditions. So . (Implicitly, I have used the fact that the cohomology class of a point is torsion-free — but this is true of any effective cycle on any smooth projective variety.)

Now, we can answer our question: given four general curves of degrees , the locus in of lines passing through all four curves is the intersection product . This will be the product of the degrees, times . By rule (2) applied three times, is two times the class of a point, so:

**Solution**: There are lines that intersect all four of the curves.

That’s that! In similar spirit, but with more geometry coming in the picture:

**Problem 2**. Let be a smooth curve of genus and degree . The locus of chords through (lines passing through at two points) is codimension 2 in . So, given two such curves in general position, only finitely-many lines are chords to both. How many?

As before, the easy dimension count tells us that our locus is cohomologous to a linear combination of the partitions and , say

First, let’s multiply both sides by , the locus “lines contained in a given plane”. By rule (1), the term vanishes, so we get

So we need to count chords to that are contained in a general plane . Since the curve is degree , the intersection consists of points. By genericity, no three are collinear, so there are exactly lines passing through two points at a time. So that’s .

For , we instead multiply by , the locus “lines containing a given point”. By rule (1), it’s now the term that vanishes, so we get

So we need to count chords to that contain a given point . We’ll think of this as follows: we consider the morphism by projecting away from . This doesn’t change the degree (since is not a point on the curve itself).

Now, a chord through yields two points of that project to the same point in the plane, that is, they result in a *node* in the image curve in . Since our point is general, the only singularities on the image curve will be simple nodes (see Hartshorne for this fact). And we can count the singularities, using Riemann-Roch and the degree-genus formula for plane curves. The end result is that the number of nodes is So this is .

Now we can answer our question! For our two curves, we multiply the classes to find:

When we expand, the cross terms cancel, leaving times the class of a point. So:

**Solution**. There are shared chords between the two curves.

It’s a little amazing that this computation works — the auxiliary computations, about “chords contained in a plane” and “chords containing a point”, are a little mysterious to think about. Somehow it’s enough to consider these linear-algebra-flavoured deformations of the problem. Anyway, it certainly wouldn’t be easy to predict this answer in advance!

(Reference for this content: both of these problems are out of Eisenbud and Harris’s unpublished book on intersection theory.)