Skip to main content
2 of 2
deleted 6 characters in body
Han Jin Ma
  • 443
  • 2
  • 10

Schubert calculus expressed in terms of the cotangent space of the Grassmannians

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?

Han Jin Ma
  • 443
  • 2
  • 10