Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:01:31Z http://mathoverflow.net/feeds/question/14219 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? Steven Sam 2010-02-05T02:53:31Z 2012-12-13T03:52:10Z <p>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? </p> <p>Maybe a more specific question: is it possible to prove that these coefficients are the same without calculating them beforehand?</p> <p>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.</p> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/14223#14223 Answer by David Speyer for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? David Speyer 2010-02-05T03:09:48Z 2010-02-05T04:53:38Z <p>The easier part is to see that <code>$\lim_{\infty \leftarrow n} H^*(G(d,n))$</code> is very close to <code>$\mathrm{Rep}(GL_d)$</code> as rings. (As Allen points out below, we don't want all the representations of $GL_d$, only the polynomial ones. I'll continue glossing over that point, but it is another sign that any answer has to be somewhat complex.)</p> <p>The representation ring of $GL_d$ is the same as the $GL_d$-equivariant $K$-theory of a point. Let <code>$\mathrm{Mat}_{d \times n}^{\circ}$</code> be the full rank $d \times n$-matrices. I won't define this rigorously, but the ind-scheme <code>$\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ}$</code> is "contractible", so the <code>$GL_d$</code>-equivariant $K$-theory of a point is <code>$K^0(\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$</code>. One can justify turning this into <code>$\lim_{n \leftarrow \infty} K^0( \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$</code>. Of course, <code>$\mathrm{Mat}_{d \times n}^{\circ} / GL_d = G(d,n)$.</code> </p> <p>The Chern character map <strike>is an isomorphism between</strike> sends $K^0(G(d,n))$ to the completion of $H^*(G(d,n))$, at least once we tensor with $\mathbb{Q}$. (That completion is related to the issue of whether we work with polynomial representations or all representations.)</p> <p>The hard thing is to explain why there should be an isomorphism which takes irreps to Schubert cycles, and works over $\mathbb{Z}$. This is especially difficult because Chern character is <strong>not</strong> that isomorphism. I don't know any really short answer to this question, but <a href="http://arxiv.org/abs/math/0306414" rel="nofollow">Harry Tamvakis</a> has an expository paper which does a pretty good job. </p> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/14226#14226 Answer by Ben Webster for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? Ben Webster 2010-02-05T03:36:04Z 2010-02-05T03:36:04Z <p>This is slightly off-topic (I don't have a lot to add to David's answer), but let me just mention that the phenomenon of type A Lie theoretic objects coinciding with other Lie theoretic objects also of type A but for a different rank in a way that just doesn't seem to work for other groups is shockingly common. I'm not sure there's a high concept explanation for it, other than that GL_n is a really important object.</p> <p>This business with Grassmannians and classifying spaces may be the best known example, but it just happens over and over. Schur-Weyl duality and Howe duality are great examples, though I think my favorite may be that type A slices between nilpotent orbits are the same as type A quiver varieties are the same as type A slices between $\mathrm{Gr}_\lambda$'s in the affine Grassmannian. These varieties seemingly have little or nothing to do with each other in other types, but somehow they magically match up in type A.</p> <p>Another example I like a lot is that all representation categories of quantum groups are braided <em>with the type A braid group</em>.</p> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/14231#14231 Answer by Allen Knutson for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? Allen Knutson 2010-02-05T04:32:55Z 2010-02-05T21:06:48Z <p>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).</p> <p>The wrong one is $Rep(GL_d)$, whose basis is indexed by decreasing sequences in ${\mathbb Z}^d$.</p> <p>That has a subring $Rep(M_d)$, representations of the Lie monoid of <em>all</em> $d\times d$ matrices, whose basis is indexed by decreasing sequences in ${\mathbb N}^d$, or partitions with at most $d$ rows.</p> <p>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!</p> <ol> <li><p>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.</p></li> <li><p>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 $Alt^* (n_+) \otimes Alt^* (n_-)$.</p></li> </ol> <p>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 $H^* (G/P)$).</p> <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.</p> <p>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.</p> <p>I believe the degenerate product on $H^*(G/P)$ is exactly the one described by [Belkale-Kumar].</p> <p>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).</p> <p>I think this is going to be the closest to what you want, for other groups' Grassmannians.</p> <ol> <li>(No, 3. Silly site software!) Belkale has <a href="http://front.math.ucdavis.edu/0208.5107v2" rel="nofollow">the best (least decategorified) proof</a> 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. </li> </ol> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/38173#38173 Answer by Joel Kamnitzer for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? Joel Kamnitzer 2010-09-09T13:49:41Z 2010-09-09T13:49:41Z <p>This is certainly one of the most vexing questions of all time! There really is no simple answer, though I agree that Allen has presented some reasonable ones.</p> <p>In their paper <a href="http://front.math.ucdavis.edu/0711.4079" rel="nofollow"> Schubert calculus and representations of general linear group </a> Mukhin-Tarasov-Varchenko give another answer. They construct any algebra $\mathcal{B}$ called the Bethe algebra, which acts on a tensor product multiplicity space $Hom(V_\lambda, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_m})$, depending on parameters $b_1, \dots, b_n$. They prove that the image $A$ of the Bethe algebra acting on this vector space has the same dimension of this vector space (for generic $b_i$ you get all diagonal matrices with respect to some basis). Then they prove this algebra $A$ is isomorphic to the functions on an scheme-theorectic intersection of $n+1$ Schubert varieties corresponding to the $\lambda_i$, with respect to flags given by the $b_1, \dots, b_n$.</p> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/79733#79733 Answer by David Speyer for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? David Speyer 2011-11-01T17:46:09Z 2012-12-13T03:52:10Z <p>Another paper which should be added to the list of references here is Derksen-Weyman, <a href="http://arxiv.org/abs/math.AG/0507393" rel="nofollow">On the number of subrepresentations of a general quiver representation</a>. They generalize the Intersection Multiplicity = Tensor Product Multiplicity equation to a general statement about quivers, and prove it in a very elegant manner.</p> <p>I have long had a vague feeling that Mukhin-Tarasov-Varchenko, Belkale and Derksen-Weyman are describing the same construction in different languages. I haven't been able to confirm this but, if so, I recommend Derksen-Weyman's version as particularly readable.</p> http://mathoverflow.net/questions/14219/why-do-littlewood-richardson-coefficients-describe-the-cohomology-of-the-grassman/116228#116228 Answer by Richard Stanley for Why do Littlewood-Richardson coefficients describe the cohomology of the Grassmannian? Richard Stanley 2012-12-13T03:13:29Z 2012-12-13T03:13:29Z <p>There is some additional information on page 399 of <em>Enumerative Combinatorics</em>, vol. 2. The first conceptual explanations are due to Horrocks in 1957 and Carrell in 1978. See also pages 278-279 of Fulton's <em>Intersection Theory</em>.</p>