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](https://mathoverflow.net/questions/196546/hard-lefschetz-theorem-for-the-flag-manifolds)?



Schubert calculus expressed in terms of the cotangent space of the Grassmannians
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 $2 |\lambda|$.


