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Let $T^*_{\mathbb{C}}(Gr_{n,r})$ denote the cotangent space of the Grassmannian of $r$-planes in $\mathbb{C}^n$. Moreover, let $\Lambda^\bullet$ denote the exterior algebra of $T^*_{\mathbb{C}}(Gr_{n,r})$. Condsidering $Gr_{n,r}$ as the homogeneous space $U_n/(U_r \times U_{n-r})$, we have a unique representation of $U_n/(U_r \times U_{n-r})$ on $\Lambda^{\bullet}$ for which the associated homogeneous vector bundle is the direct sum $\bigoplus_{k \in \mathbb{N}} \Omega^k$.

(i) Just as for any homogeneous space, every de Rham cohomology class of $Gr_{n,r}$ has a $G$-invariant representative. Moreover, every $G$-invariant element must be harmonic, and so, gives by Hodge decomposition a cohomology class. Is it correct to conclude from this that the cohomology group $H^\bullet$ is isomorphic as a vector space to the space of $U(r) \times U(n-r)$-invariant elements in $\Lambda^\bullet$?

(ii) With respect to a standard weight basis of $T^*(Gr_{n,r})$, what do the $U(r) \times U(n-r)$-invariant elements look like, and how does this presentation of Schubert calculus relate to the partition presentation given in this question?

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The tangent space to the Grassmanian corresponds to the following representation of $U(r)\times U(n-r)$, call it $\rho$: it is the $r\times (n-r)$ matrices, with $U(r)$ acting on the left and $U(n-r)$ acting on the right, so if we denote by $A$ the standard representation of $U(r)$ and by $B$ the standard representation of $U(n-r)$ we obtain $$ \rho = A\otimes B^*. $$ Now remember that the tangent space has Hodge decomposition into holomorphic and anti-holomorphic part, so the actual representation we are dealing with is $$ A\otimes B^* + A^*\otimes B. $$ So if we want to find invariant differential $k$-forms we need $U(r)\times U(n-r)$-invariants of $$ \Lambda^k(A\otimes B^* + A^*\otimes B) = \sum_{i=0}^k \Lambda^i(A\otimes B^*) \Lambda^{k-i}(A^*\otimes B). $$ Using the formula $$ \Lambda^n(A\otimes B) = \sum_{\lambda \vdash n} s_\lambda(A)\otimes s_{\lambda'}(B), $$ where $\lambda'$ stands for the conjugate partition, we obtain $$ \sum_{\lambda,\mu:|\lambda|+|\mu|=n} s_\lambda(A) s_{\lambda'}(B^*) s_\mu(A^*) s_{\mu'}(B). $$ Taking invariants we see that only terms with $\lambda'=\mu$ survive, and each one of them produces a one-dimensional space of invariants. Moreover, $s_\lambda(A)$ is non-trivial only if $\lambda$ has $\leq r$ rows, similarly $\mu$ has $\leq n-r$ rows. So we conclude that there is a one-dimensional space of invariants for each partition $\lambda$ in a $r\times (n-r)$ rectangle, and it is a one-dimensional space of differential forms of degree $2 |\lambda|$.

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  • $\begingroup$ Thanks for the great answer! Just to be sure I understand what's happening, in the last displayed equation, is it clear that each of the summands are irreducible modules? If they're not irreducible then I don't understand why Taking invariants we see that only terms with $\lambda = \mu$ survive, $\endgroup$
    – Han Jin Ma
    Aug 14 '17 at 9:54
  • $\begingroup$ Sorry, I should have made it more explicit. It is known that irreducible modules of $U(n)$ are precisely the Schur functors applied to the standard representation. That's why these are irreducible modules. $\endgroup$ Aug 15 '17 at 18:23
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For your first question: yes. See Stoll, Invariant forms on Grassman manifolds, p. 15. I think your second question is answered in the same book.

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Regarding your second question, I think the answer is in the famous Kostant "Lie Algebra Cohomology and the Generalized Borel-Weil Theorem" or rather its second part.

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