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The basis of Schubert classes for the cohomology ring $H^*(\text{Gr}(m,N))$ of the Grassmannian of $m$-dimensional subspaces of $\mathbb{C}^N$ is indexed by $L(m,N-m)$, the poset of all partitions fitting inside a $m \times (N-m)$ box. This is the quotient of the powerset $2^{m(N-m)}$ by the action of the wreath product $S(m) \wr S(N-m)$. How does this come from the fact that $\text{Gr}(m,N) \cong U(N) / \big (U(m) \times U(N-m) \big )$? Can this be extended to other homogeneous spaces?

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Not the wreath product -- the direct product. (I suppose this is technically a particularly trivial wreath product.) – Allen Knutson Jul 6 '10 at 3:01
Allen, I believe you misunderstood his construction. I think one copy moves the rows, and then there's a symmetric group on each row. This lets you permute any set of boxes into a partition. – Ben Webster Jul 6 '10 at 8:18
Ouch! I did indeed! – Allen Knutson Jul 8 '10 at 5:33

The Schubert classes on G/P are the classes of the Schubert varieties, which are the closures of the Schubert cells, each of which contains a unique T-fixed point. The T-fixed points on G/P are the images of T-fixed points on G/B (since T acts on the fiber, which is a projective variety, hence itself has a T-fixed point by Borel's theorem).

Up on G/B, the T-fixed points are exactly of the form N_G(T)B/B, so indexed by the Weyl group W_G = N_G(T)/T. Down on G/P, they group together by the Weyl group W_P = N_P(T)/T, so they're indexed by W_G/W_P. Which is exactly what you observed in the G/P = Grassmannian case.

(Actually you asked about compact groups, so K/L where K is compact and L is compact of the same rank, which includes some cases like S^4 = SO(5)/SO(4) that is not of the form G/P for G complex and P a parabolic. Then there's still a basis of "Schubert classes", indexed by W_K/W_L similarly.)

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But that's not exactly what he observed. The Weyl group of U(N) is the symmetric group S(N), so the Schubert decomposition W_G/W_P indexes the Grassmannian by the free quotient S(N) / (S(m) x S(N-m)). Whereas the quotient he discusses is {0,1}^(m(N-m)) / (S(m) wreath S(N-m)), which is not free. Both are in natural correspondence with the Young diagrams inside an m by N-m box. To see this for the free quotient, observe that it indexes all possible orderings of m moves across and N-m moves up. – Michael Thaddeus Jul 6 '10 at 19:25

"Schur-Weyl duality" says that the permutation group S(k) is dual to the unitary group U(n). Following Wikipedia, Both of these groups act on the space $\mathbb{C}^n \otimes \mathbb{C}^n \otimes \dots \otimes \mathbb{C}^n$, with k factors.

  • S(k) acts by permuting the factors $\sigma(v_1 \otimes v_2 \otimes \dots \otimes v_n) = v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes \dots \otimes v_{\sigma(n)}$
  • U(n) acts like an n x n matrix on the $\mathbb{C}^n$, $g(v_1 \otimes v_2 \otimes \dots \otimes v_n) = g(v_1) \otimes g(v_2) \otimes \dots \otimes g(v_n) $

Schur-Weyl duality is the decomposition of this double-representation $$ \mathbb{C}^n \otimes \mathbb{C}^n \otimes \dots \otimes \mathbb{C}^n = \sum_D \pi_k^D \otimes \rho_n^D$$ into tensor products of irreducible representations of S(k) and U(n). D runs over young diagrams with k boxes and up to n rows. Each of the young diagrams appears once.

Your questions clearly involved the duality between the permutation and unitary groups, though I couldn't explain to you how. You might also see Fulton's book, Young Tableaux.

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I do not think that Schur-Weyl duality is relevant to this question. – Allen Knutson Jul 6 '10 at 3:04
Could be... certainly the Fulton is relevant. – john mangual Jul 7 '10 at 16:08

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