TITLE: Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian?
QUESTION [41 upvotes]: I'm looking for a "conceptual" explanation to the question in the title. The standard proofs that I've seen go as follows: use the Schubert cell decomposition to get a basis for cohomology and show that the special Schubert classes satisfy Pieri's formula. Then use the fact that basic homogeneous symmetric functions $h_n$ are algebraically independent generators of the ring of symmetric functions, to get a surjective homomorphism from the ring of symmetric functions to the cohomology ring. Pieri's rule can be shown with some calculations, but is there any reason a priori to believe that tensor product multiplicities for the general linear group should have anything at all to do with the Grassmannian? 
Maybe a more specific question: is it possible to prove that these coefficients are the same without calculating them beforehand?
One motivation for asking is that the cohomology ring of the type B and type C Grassmannians ${\bf OGr}(n,2n+1)$ and ${\bf IGr}(n,2n)$ are described by (modified) Schur P- and Q-functions which seem to have nothing(?) to do with the representation theory of the orthogonal and symplectic groups ${\bf SO}(2n+1)$ and ${\bf Sp}(2n)$. So as far as I can tell the answer isn't just because general linear groups and Grassmannians are "type A" objects.

REPLY [4 votes]: There is some additional information on page 399 of Enumerative Combinatorics, vol. 2. The first conceptual explanations are due to Horrocks in 1957 and Carrell in 1978. See also pages 278-279 of Fulton's Intersection Theory.