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 $L(m,Nm)$, the poset of all partitions fitting inside a $m \times (Nm)$ box. This is the quotient of the powerset $2^{m(Nm)}$ by the action of the wreath product $S(m) \wr S(Nm)$. How does this come from the fact that $\text{Gr}(m,N) \cong U(N) / \big (U(m) \times U(Nm) \big )$? Can this be extended to other homogeneous spaces?

$\begingroup$ Not the wreath product  the direct product. (I suppose this is technically a particularly trivial wreath product.) $\endgroup$ – Allen Knutson Jul 6 '10 at 3:01

$\begingroup$ 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. $\endgroup$ – Ben Webster♦ Jul 6 '10 at 8:18

$\begingroup$ Ouch! I did indeed! $\endgroup$ – 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 $P$ parabolic. Then there's still a basis of "Schubert classes", indexed by $W_K/W_L$ similarly.)

$\begingroup$ 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(Nm)). Whereas the quotient he discusses is {0,1}^(m(Nm)) / (S(m) wreath S(Nm)), which is not free. Both are in natural correspondence with the Young diagrams inside an m by Nm box. To see this for the free quotient, observe that it indexes all possible orderings of m moves across and Nm moves up. $\endgroup$ – Michael Thaddeus Jul 6 '10 at 19:25
"SchurWeyl 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) $
SchurWeyl duality is the decomposition of this doublerepresentation $$ \mathbb{C}^n \otimes \mathbb{C}^n \otimes \dots \otimes \mathbb{C}^n = \sum_D \pi_k^D \otimes \rho_n^D$$ 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.

4$\begingroup$ I do not think that SchurWeyl duality is relevant to this question. $\endgroup$ – Allen Knutson Jul 6 '10 at 3:04

$\begingroup$ Could be... certainly the Fulton is relevant. $\endgroup$ – john mangual Jul 7 '10 at 16:08