Home » Mathematics » Algebraic combinatorics » Schubert Calculus III: Cohomology and Enumerative Geometry

Schubert Calculus III: Cohomology and Enumerative Geometry


Last post set up the Schubert decomposition of X = G(k,n). 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 GL_n representation theory and symmetric polynomials.

All of these computations are in A(X) \cong H^*(X), the Chow/cohomology ring.

(1) First of all, \lambda \cdot \mu = 0 if and only if \lambda and \mu overlap when placed opposite each other in the k \times (n-k) 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, \lambda \cdot \lambda^c = 1 \cdot [\text{pt}]. Here \lambda^c means the complementary partition in the k \times (n-k) box.

(2) Let [\text{box}] denote the single-box partition. For any \lambda,

\lambda \cdot [\text{box}] = \sum \lambda' ,

where the sum is over all possible ways of adding a single box to \lambda, while remaining a partition contained in the k \times (n-k) rectangle.

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

\lambda \cdot \mu = \sum \lambda' ,

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

(2,1) \cdot (1,1) = (4,1) + (3,2) + (3,1,1) + (2,2,1),

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 \Lambda \to A(X) defined by s_\lambda \mapsto \lambda (or to 0 if \lambda doesn’t fit in the rectangle) is a ring homomorphism. (Here \Lambda is the ring of symmetric polynomials.)

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

This follows from the Jacobi-Trudi rule, which lets us write \mu as a polynomial in terms of the single-row partitions. We expand \mu 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 \Lambda. So, we can “import” the Littlewood-Richardson rule for multiplying Schur polynomials:

(4) For any \lambda, \mu, we have \lambda \cdot \mu = \sum_\nu c_{\lambda \mu}^\nu \nu,, where c_{\lambda \mu}^\nu is the Littlewood-Richardson number, but the sum includes only those \nu 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 C \subset \mathbb{P}^3 be a smooth curve of degree d. The locus Z \subset G(2,4) of lines passing through C is codimension 1. So, given four such curves in general position, only finitely-many lines will touch all four. How many?

Since we know Z is codimension one, we know it is cohomologous to a linear combination of codimension-one Schubert classes — in particular, [Z] = k \cdot [\text{box}], the only partition of size 1, for some multiple k.

To find k, we use “the method of undetermined coefficients”, and multiply both sides of the equation with the complementary partition to [\text{box}], which is (2,1): the locus of “lines contained in a given 2-plane H and passing through a given point p \in H.”

So [Z] \cdot (2,1) = k \cdot [\text{pt}], where the left-hand side is now the intersection of Z with a generic (2,1) Schubert locus: lines contained in H, passing through p \in H, and passing through the curve C.

Well, already C \cap H is, for general H, exactly d reduced points, where d = \deg(C). And generically, none of these points will coincide with the required point p. So there are exactly d choices of line satisfying all our conditions. So k = d. (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 C_i of degrees d_i, the locus in G(2,4) of lines passing through all four curves is the intersection product [Z_1] \cdot [Z_2] \cdot [Z_3] \cdot [Z_4]. This will be the product of the degrees, times [\text{box}]^4. By rule (2) applied three times, [\text{box}]^4 is two times the class of a point, so:

Solution: There are 2d_1d_2d_3d_4 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 C \subset \mathbb{P}^3 be a smooth curve of genus g and degree d. The locus Z of chords through C (lines passing through C at two points) is codimension 2 in G(2,4). 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 (2) and (1,1), say

[Z] = a (2) + b (1,1).

First, let’s multiply both sides by (2), the locus “lines contained in a given plane”. By rule (1), the (1,1) term vanishes, so we get

[Z] \cdot (2) = a \cdot [\text{pt}].

So we need to count chords to C that are contained in a general plane H. Since the curve is degree d, the intersection H \cap C consists of d points. By genericity, no three are collinear, so there are exactly {d \choose 2} lines passing through two points at a time. So that’s a.

For b, we instead multiply by (1,1), the locus “lines containing a given point”. By rule (1), it’s now the (2) term that vanishes, so we get

[Z] \cdot (1,1) = b \cdot [\text{pt}].

So we need to count chords to C that contain a given point P. We’ll think of this as follows: we consider the morphism C \to \mathbb{P}^2 by projecting away from P. This doesn’t change the degree (since P is not a point on the curve itself).

Now, a chord through P yields two points of C that project to the same point in the plane, that is, they result in a node in the image curve in \mathbb{P}^2. Since our point P 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 {d+1 \choose 2} - g. So this is b.

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

[Z_1] \cdot [Z_2] = (a_1 (2) + b_1 (1,1))(a_2 (2) + b_2 (1,1)) \cdot [\text{pt}].

When we expand, the cross terms cancel, leaving (a_1a_2 + b_1b_2) times the class of a point. So:

Solution. There are {d_1 \choose 2}{d_2 \choose 2} + ({d_1+1 \choose 2} - g_1)({d_2 + 1 \choose 2} - g_2) 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.)


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: