I'm trying to understand the HodgeIndex Theorem at the moment. What does it say explicitly for the case of the complex Grassmannian Gr($n,k$), and can this be established without recourse to the theorem?
1 Answer
The answer is a happy surprise for me: The Hodge index theorem for a Grassmannian matches a special case of John Stembridge's $q=1$ phenomenon, that I also studied in an old paper.
First, some generalities about what is going on, and about when you do or don't "need" the Hodge Index Theorem. I am following the Hodge Index theorem described in Claire Voisin's book, that says that the signature of a complex $n$manifold is a certain alternating sum of Hodge numbers. A Grassmannian is a type of partial flag variety, and flag varieties have Schubert cell decompositions. That makes it easy to compute both the Hodge numbers, and the signature of the manifold separately so that the Hodge Index Theorem becomes an identity between two calculations.
More precisely, the Schubert cells are a basis for the cohomology: they make a CW complex with evendimensional cells, which forces the CW differential to vanish. For complex geometry reasons, the Schubert cells are all in $H^{k,k}$ for some $k$, i.e., all on the diagonal of the Hodge diamond. The Hodge Index theorem then says that the signature is the alternating sum of these numbers. At the same time, the intersection matrix of such a manifold is just a symmetric permutation matrix connecting dual Schubert cells. So the signature is also the number of selfdual Schubert cells. The Hodge Index theorem thus equates the number of selfdual cells with signed sum of all of the cells.
In the case of a Grassmannian $\mathrm{Gr}(a,a+b)$, the Schubert cells are labelled by partitions in an $a \times b$ rectangle. The Hodge index formula is the $q$enumeration of these partitions with $q = 1$. On the other hand, the direct calculation is the number of selfcomplementary partitions, where you replace a partition with its complement in the rectangle and turn the rectangle upsidedown. It is easy to calculate the selfcomplementary partitions in the case. The lattice path of the partition is symmetric about the middle of the rectangle, so these are just partitions in an $\lfloor a/2 \rfloor \times \lfloor b/2 \rfloor$ rectangle. Or there are none if $a$ and $b$ are both odd: in this case the signature is trivial a priori because the middle dimension has odd degree. This combinatorial equality is exactly the $q=1$ phenomenon.
What I wonder now is whether every case of the $q=1$ phenomenon, for the $q$dimension of an irreducible representation of a complex simple or compact simple Lie group, can be explained by a Hodge Index theorem with coefficients, combined with some geometric model of the representation. This would be sophisticated (to the point of overkill, if you just want the combinatorics), because the involution in general is something called "evacuation", which John related to canonical bases.

$\begingroup$ hello, professor kuperberg. I considered your abovementioned materials several years ago when I was a graduate student. Maybe the following papers maybe of interest to you. 1 ams.org/mathscinet/search/… 2 ams.org/mathscinet/search/… 3 ams.org/mathscinet/search/… $\endgroup$– PingMay 3, 2012 at 14:02