This question is extensively rewritten by David Speyer; the original version is below.

The Grassmannian $G(k,n)$ is the quotient $SU(n)/S[U(k) \times U(n-k)]$. Let's write $\pi$ for the map $SU(n) \to G(k,n)$.

Suppose that I have an $k(n-k)-1$ form $\omega$ on $SU(n)$ which is invariant under the $S[U(k) \times U(n-k)]$ action. Then the restriction of $\omega$ to each $S[U(k) \times U(n-k)]$ orbit will be closed. So $d \omega$ will restrict to $0$ on each orbit, and will be $S[U(k) \times U(n-k)]$ invariant. This implies that there is a differential form $\eta$ on $G(k,n)$ so that $d \omega = \pi^{\ast} \eta$. (Actually, I'm not sure whether this implies it in general. But it happens in the cases that the OP gives in comments below, and in any case we can assume that there is an $\eta$ with $d \omega = \pi^{\ast} \eta$.) We can abuse notation and refer to $\eta$ as $d \omega$.

Is there a variant of Stokes theorem which will allow us to compute $\int_{G(k,n)} \eta$ by integrating $\omega$ over some $k(n-k)-1$ cycle? What is this $k(n-k)-1$ cycle?

What is the compact manifold that we regard as the boundary of complex Grassmanians when applying Stokes theorem to the integral of a 2-form over the Grassmanian?

In case of $\mathbb{CP}^k$, for instance, this becomes $S^{2k-1}$ so that $$\int_{\mathbb{CP}^k} F = \int_{S^{2k-1}} A$$ where $F:=dA$.

Thanks in advance AB

butthis is a local expression only: yr vector potential is not globally defined as a $1$-form on $\mathbb{CP}^1$. $\endgroup$ – Fran Burstall Jul 18 '13 at 22:46