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?


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 $2 |\lambda|$.

  • $\begingroup$ 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

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.


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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