ag.algebraic geometry - Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian?

There are several rings-with-bases to get straight here. I'll explain that, then describe three serious connections (not just Ehresmann's proof as recounted in the OP).

The wrong one is $Rep(GL_d)$, whose basis is indexed by decreasing sequences in ${mathbb Z}^d$.

That has a subring $Rep(M_d)$, representations of the Lie monoid of all $dtimes d$ matrices, whose basis is indexed by decreasing sequences in ${mathbb N}^d$, or partitions with at most $d$ rows.

That is a quotient of $Rep({bf Vec})$, the Grothendieck ring of algebraic endofunctors of ${bf Vec}$, whose basis (coming from Schur functors) is indexed by all partitions. Obviously any such functor will restrict to a rep of $M_d$ (not just $GL_d$); what's amazing is that the irreps either restrict to $0$ (if they have too many rows) or again to irreps!

1. Harry Tamvakis' proof is to define a natural ring homomorphism $Rep({bf Vec}) to H^*(Gr(d,infty))$, applying a functor to the tautological vector bundle, then doing a Chern-Weil trick to obtain a cohomology class. (It's not just the Euler class of the resulting huge vector bundle.)
   The Chern-Weil theorem is essentially the statement that Harry's map takes alternating powers to special Schubert classes. So then it must do the right thing, but to know that he essentially repeats the Ehresmann proof.




2. Kostant studied $H^* (G/P)$ in general, in "Lie algebra cohomology and
   something something Schubert cells" (sorry!), by passing to the compact
   picture $H^* (K/L)$, then to de Rham cohomology, then taking $K$-invariant
   forms, which means $L$-invariant forms on the tangent space $Lie(K)/Lie(L)$.
   Then he complexifies that space to $Lie(G)/Lie(L_C)$, and identifies that
   with $n_+ oplus n_-$, where $n_+$ is the nilpotent radical of $Lie(P)$.
   Therefore forms on that space is $Alt^* (n_+) otimes Alt^* (n_-)$.

Now, there are two things left to do to relate this space to $H^* (G/P)$. One is to take cohomology of this complex (which is hard, but he describes the differential), and the other
is to take $L$-invariants as I said. Luckily those commute. Kostant degenerates the differential so as to make sense on each factor separately (at the cost of not quite getting $H^* (G/P)$).

Theorem: (1) Once you take cohomology, $Alt^* (n_+)$ is a multiplicity-free $L$-representation. So when you tensor it with its dual and take $L$-invariants, you get a canonical basis by Schur's lemma. (2) This basis is the degeneration of the Schubert basis.

Theorem: (1) If $P$ is (co?)minuscule, the differential is zero, so you can skip the take-cohomology step. That is, $Alt^* (n_+)$ is already a multiplicity-free $L$-rep. The Schur's lemma basis has structure constants coming from representation theory. (2) In the Grassmannian case, the degeneration doesn't actually affect the answer, so the product of Schubert classes does indeed come from representation theory.

I believe the degenerate product on $H^*(G/P)$ is exactly the one described by [Belkale-Kumar].

It's fun to see what's going on in the Grassmannian case -- $L = U(d) times U(n-d)$, $n_+ = M_{d,n-d}$, and $Alt^* (n_+)$ contains each partition (or rather, the $U(d)$-irrep corresponging) fitting inside that rectangle tensor its transpose (or rather, the $U(n-d)$-irrep).

I think this is going to be the closest to what you want, for other groups' Grassmannians.

1. (No, 3. Silly site software!)
   Belkale has the best (least decategorified) proof I've seen. He takes three Schubert cycles meeting transversely, and for each point of intersection, constructs an actual invariant vector inside the corresponding triple product of representations. The set of such vectors is then a basis.