Quantum Cohomology

Now that we know what a Gromov-Witten invariant is (at least, for nice spaces…we’re avoiding stacks for this series, and assuming that everything behaves nicely and that these actually count things…), we can start talking about Quantum Cohomology, which organizes all these invariants together.  In the next (and potentially last) post in the series, we’ll use quantum cohomology to compute the Gromov-Witten invariants of , and talk about what it means.

Anyway, we need to, once and for all, choose a basis of .  We’ll take , a basis of , and a basis for the rest.  Now, as we’ve assumed that is homogeneous, these can be the Schubert classes (we’ll talk more about these in a future post), but we won’t be making use of that, specifically.

Because Gromov-Witten invariants containing can be reduced to those that don’t by the rules from last time, so we define , because these are the only ones we need to know all of them.  However, we’re going to abandon this notation for the moment, though we’ll use it when we actually compute an example next time.

We now define a matrix with entries , and using standard physics notation, we set the entries of the inverse.  These inverses exist becuase choosing a different basis for the cohomology is just a change of basis for the matrix, and if we choose the Schubert classes, we get a permutation matrix.

One thing this gives us is a way to write the cup product:

So we can write cup product in terms of Gromov-Witten invariants.  But the only ones that show up have ! This suggests that we can try to define a new multiplication somehow using the higher Gromov-Witten invariants.  We’ll do this in several steps:

Define, for , .  This will encode all the Gromov-Witten invariants only containing .

Lemma: Fix .  There are only finitely many effective classes such that .

Proof: By our homogeneity assumption, the effecitve classes are nonnegative linear combinations of , eftective and nonzero.  Now, it’s a fact that , adn so for any fixed , there are only finitely many with .

Now, for , we must have that , and so (recall the equation we had last time which was required for them to be nonzero).  QED

Now, if we describe , then this lemma makes a formal power series in .  We can write it as .

Next, we set , which would then be .

Finally, we can define the quantum product, which is the linear extension of

on .

The first thing we can see about this product is that it is commutative.  This is just because mixed partials commute.  The MUCH harder theorem, which is the real theorem lurking at the base of this subject, is that is ASSOCIATIVE.  And next time, we’ll see why this is so valuable.