On Grassmann manifold

Question

In most references about Grassmann manifold, we usually introduce the following map: suppose that $S$ is a fixed subspace of codimension $k$, and let $\Omega_S \subset Gr(k,\mathbb{K})$ be all subspaces $W$ such that $W \cap S = \{0\}$. $\Omega_S$ is an open set.

Fix some $U \in \Omega_S$. One can show that $W \in \Omega_S$ if and only if there exists some linear map $\beta_W \in Hom(U,S)$ such that $W = \{x + \beta_W x : x \in U \}$. From there, one can easily check that there exists a bijection $\Omega_S \simeq Hom(U,S)$.

I'm interested in show that this bijection is actually an homeomorphism of topological spaces!

So far, I haven't been able to make a lot about this claim. What bugs me is that the topologies at stake are not necessarily intuitive to work with: $Gr(r, \mathbb{K})$ is endowed with the quotient topology, and $Hom(U,S)$ is really just an euclidean space, so might just take some random norm/distance.

Any help greatly appreciated !

EDIT : Look here Lee's page 22 for more informations.

Answer 1

This is becoming old, but anyway, I was able to find an answer myself and never posted it here, so here we go. Define $$ t : \Omega_S \to Hom(U, S). $$ as above. Let $\pi : V_0^k \to Gr(k, \mathbb{K})$ where $V_0^k := \{ (v_1, \dots, v_k) : v_i\text{'s linearly independent} \}$, with $$ \pi(v_1, \dots, v_k) = Vect(v_1, \dots, v_k), $$defines the quotient topology over $Gr(k, \mathbb{K})$, in which $\Omega_S$ is open. By definition, it suffices to show that $$ t' : \pi^{-1}(\Omega_S) \to Hom(U,S) $$ in continuous, where $t'(v_1, \dots, v_k) = -pr_{S/Vect(v_1, \dots, v_k)}$ ($pr_{A/B}$ is the projection on $A$ w.r.t. to $B$). Let $w_1, \dots, w_{n-k}$ be a base for $S$. Define the (clearly) continuous application $$A : \pi^{-1}(\Omega_S) \ni (v_1, \dots, v_k) \mapsto [v_1, \dots, v_k, w_1, \dots, w_{n-k} ] \in \mathcal{M}_{n}(\mathbb{K})$$ (i.e. the matrix with columns $v_i's$ and $w_i's$). Then $$ t'(v_1, \dots, v_k) = \left( x \mapsto \sum_{i=1}^{n-k} \left\langle x, (A(v_1, \dots, v_k)A^t(v_1, \dots, v_k))^{-1}w_i \right\rangle w_i\right) $$which is easily seen to be continuous.