Openness of $\varphi(U_Q \cap U_{Q'})$ in the definition of Grassmannian Manifolds (Lee: Introduction to Smooth Manifolds)

Question

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 $L(P,Q)$?)

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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.

Answer 1

First, we should clarify how the topology on $G_k(V)$ is defined. One way to do this is to consider the subset $V_k \subset V^k$ consisting of all linearly independent $k-$tuples $(v_1, \dots ,v_k)$ of $V^k$. Then we define a projection $\pi: V_k \mapsto G_k(V)$ by $(v_1,\dots ,v_k) \mapsto span(v_1, \dots,v_k) $ and give $G_k(V)$ the quotient topology. With this, we can see that the sets $U_Q$ are open for any $(n-k)-$dimensional subspace $Q$ of $V$. This is exactly the case when $\pi^{-1}(U_Q)$ is open in $V^k$ and this set can be written as

$\pi^{-1}(U_Q) = \{ (v_1,\dots,v_k) \in V_k | (u_1,\dots,u_{n-k},v_1,\dots,v_k) \text{ is linearly independent}\} $

where $(u_1,\dots u_{n-k})$ is any linearly independent subset of $Q$. This is open by the continuity of the determinant function. So the set $U_Q$ is open and hence $U_Q \cap U_{Q'}$ is open as well. Then it follows by continuity of $\psi$ that $\varphi(U_Q \cap U_{Q'}) = \psi^{-1}(U_Q \cap U_{Q'})$ is also open.

For a more extensive treatment of Grassmanian manifolds you could also check Spivak's A Comprehensive Introduction to Differential Geometry, Volume Five, chapter 13, section 2.

Answer 2

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.)