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.
Best Answer
We are asking the question, what is the set of all matrices $A$ fulfilling these conditions: $\Gamma(A) \cap Q = \{0\}$ and $\Gamma(A) \cap Q' = \{0\}$.
From the facts that $V = P \oplus Q$ and $A$ is a linear map $A\colon P \to Q$ follows, that any matrix $A_{(n-k)\times k}$ fulfills the first condition.
Let $p_1, \ldots, p_k, q_1, \ldots , q_{n-k}$ be the base of $V$, created of bases of $P$ and $Q$. Let $M$ be the next corresponding matrix to matrix $A$: $M=\left(\begin{matrix} 0_{k \times k} & 0_{k \times (n-k)} \\ A_{(n-k)\times k} & 0_{(n-k)\times (n-k)} \end{matrix} \right)$. In fact, $M\colon V \to V$ is a linear map that extends the map $A$ from the domain $P$ to $V$. This extension is done for using the square matrices in the next. The vector $x$ belongs to the set $\Gamma(A) \cap Q'$ if $M.x + I.x = B.c$ and $I.x=C.d$, where columns of matrix $B$ ($C$) consists of base vectors of $Q'$ ($P$) and $c$ and $d$ are vector of coefficients in the linear combination. It is system of linear equations ($2n$ equations). Entries of vectors $x, c, d$ are unknowns. The system can be formally written as: $$ \left(\begin{matrix} (M+I)_{n \times n} & -B_{n \times (n-k)} & O_{n \times k} \\ I_{n\times n} & 0_{n\times (n-k)} & -C_{n \times k} \end{matrix} \right).\left( \begin{matrix} x_{n \times 1} \\ c_{(n-k)\times 1} \\ d_{k \times 1} \end{matrix} \right) = 0$$ This system is required to have the only one solution. So the determinant has to be nonzero.
Since matrix $ \left(\begin{matrix} M+I & -B & O \\ I & 0 & -C \end{matrix} \right)$ consists of constants (entries of $I, B, C$, and zeroes) and variables (entries of matrix $A$, resp. $M$) and determinant function is continuous, the set of all searched matrices $A$ is open. (It is the preimage of $\mathbb{R}-\{0\}$ in this determinant function.)