I am reading Lee's Introduction to Smooth Manifolds and I have some problems with definition of Grassmannian manifold given in Example 1.24, p.22. I'll write the details below.
My question is:
- Why is the set $\varphi(U_Q \cap U_{Q'}) \subset L(P, Q)$ (i.e., the set of all $A \in L(P, Q)$ whose graphs intersect both $Q$ and $Q'$ trivially) an open set? (Which topology is used on $L(P,Q)$. Should I take the standard topology from $\mathbb R^{k(n-k)}$ and an identification of this space with $L(P,Q)$?)
Here is the relevant part from the book:
Example 1.24 (Grassmann Manifolds). Let $V$ be an $n$-dimensional real vector space. For any integer $0 \le k \le n$, we let $G_k(V)$ denote the set of all $k$-dimensional linear subspaces of $V$ . We will show that $G_k(V)$ can be naturally given the structure of a smooth manifold of dimension $k(n-k)$. The construction is somewhat more involved than the ones we have done so far, but the basic idea is just to use linear algebra to construct charts for $G_k(V)$ and then use the smooth manifold construction lemma (Lemma 1.23).
Let $P$ and $Q$ be any complementary subspaces of $V$ of dimensions $k$ and $(n-k)$, respectively, so that $V$ decomposes as a direct sum: $V = P \oplus Q$. The graph of any linear map $A \colon P \to Q$ is a $k$-dimensional subspace $\Gamma(A) \subset V$, defined by $\Gamma(A)=\{x+Ax: x\in P\}.$ Any such subspace has the property that its intersection with $Q$ is the zero subspace. Conversely, any subspace with this property is easily seen to be the graph of a unique linear map $A\colon P \to Q$.
Let $L(P, Q)$ denote the vector space of linear maps from $P$ to $Q$, and let $U_Q$ denote the subset of $G_k(V)$ consisting of $k$-dimensional subspaces whose intersection with $Q$ is trivial. Define a map $\psi \colon L(P, Q) \to U_Q$ by $\psi(A)=\Gamma(A).$ The discussion above shows that $\psi$ is a bijection. Let $\varphi = \psi^{-1} \colon U_Q \to L(P, Q)$. By choosing bases for $P$ and $Q$, we can identify $L(P, Q)$ with $M((n-k)\times k, \mathbb R)$ and hence with $\mathbb R^{k(n-k)}$, and thus we can think of $(U_Q,\varphi)$ as a coordinate chart. Since the image of each chart is all of $L(P, Q)$, condition (i) of Lemma 1.23 is clearly satisfied.
Now let $(P', Q')$ be any other such pair of subspaces, and let $\psi'$, $\varphi'$ be the corresponding maps. The set $\varphi(U_Q \cap U_{Q'} ) \subset L(P, Q)$ consists of all $A \in L(P, Q)$ whose graphs intersect both $Q$ and $Q'$ trivially, which is easily seen to be an open set, so (ii) holds.