I have been reading Differential geometry of smooth manifolds from two different books: one by Jeffrey M Lee and the other by John M Lee. In the first chapter of both books, the authors construct what is called a Grassmannian manifold. Since the proof given in the book is almost complete, there are no "doubts" in the proof. However, here, I will be sharing a few concerns while constructing the manifold structure on the Grassmannian.
For reference, we will be using the following result which tells us that a set can be given a smooth structure under certain conditions:
Theorem: Let $M$ be a set and $\left\lbrace U_{\alpha} \right\rbrace_{\alpha \in \Delta}$ be a collection of subsets of $M$ together with injective maps $\phi_{\alpha}: U_{\alpha} \to \mathbb{R}^n$. Assume the following:
- $\left\lbrace \left( U_{\alpha}, \phi_{\alpha} \right) \right\rbrace_{\alpha \in \Delta}$ is a smooth atlas for $M$.
- There is a countable subcollection $\left\lbrace U_{\alpha_n} \right\rbrace_{n \in \mathbb{N}}$ of $\left\lbrace U_{\alpha} \right\rbrace_{\alpha \in \Delta}$ which covers $M$.
- For distinct points $p, q \in M$, either there is some $\alpha \in \Delta$ such that $p, q \in U_{\alpha}$ or there are $\alpha, \beta \in \Delta$ with $\alpha \neq \beta$ such that $p \in U_{\alpha}$, $q \in U_{\beta}$ and $U_{\alpha} \cap U_{\beta} = \emptyset$. Then, $M$ is a smooth manifold with the topology induced by the atlas.
Here, again, when we say a "smooth atlas", we mean independently (without assuming a topology on $M$). That definition is given as follows.
Definition: Let $M$ be a set, $\left\lbrace U_{\alpha} \right\rbrace_{\alpha \in \Delta}$ be a collection of subsets of $M$, and $\phi_{\alpha}: U_{\alpha} \rightarrow \mathbb{R}^n$ be injective maps onto open sets in $\mathbb{R}^n$. Then, the collection $\left\lbrace \left( U_{\alpha}, \phi_{\alpha} \right) \right\rbrace_{\alpha \in \Delta}$ is a smooth chart on $M$ if
- $\bigcup\limits_{\alpha \in \Delta} U_{\alpha} = M$.
- For each $\alpha, \beta \in \Delta$, the set $\phi_{\alpha} \left( U_{\alpha} \cap U_{\beta} \right)$ is open in $\mathbb{R}^n$.
- For each $\alpha, \beta \in \Delta$ with $U_{\alpha} \cap U_{\beta} \neq \emptyset$, the map $\phi_{\beta} \circ \phi_{\alpha}^{-1} : \phi_{\alpha} \left( U_{\alpha} \cap U_{\beta} \right) \rightarrow \phi_{\beta} \left( U_{\alpha} \cap U_{\beta} \right)$ is a (smooth) diffeomorphism.
Using this definition and the theorem, the authors try to construct a smooth structure on the Grassmannian. They start with a finite-dimensional real vector space $V$, say of dimension $n$, and consider the set of all $k$-dimensional subspaces of $V$. We call it $G_{k} \left( V \right)$. This set will precisely become the Grassmannian manifold. To do so, we would need to construct a suitable atlas on $G_k \left( V \right)$ such that the conditions of the Theorem are satisfied.
To achieve this end, we see that for any $k$-dimensional subspace $P$ of $V$, there is an $\left( n - k \right)$-dimensional complementary subspace $Q$ such that $V = P \oplus Q$. Then, for any linear map $T: P \rightarrow Q$, if we consider its graph $\Gamma \left( T \right) = \left\lbrace \left( x, Tx \right) | x \in P \right\rbrace$, then it can be considered as a subspace of $V$, by identifying $\left( x, Tx \right)$ with $x + Tx$. Moreover, it is also easy to see that $\Gamma \left( T \right) \cap Q = \left\lbrace 0 \right\rbrace$. In fact, every $k$-dimensional subspace of $V$ which intersects $Q$ trivially, can be seen as a graph of a linear map from $P$ to $Q$. To see this, let $W \subseteq V$ be a $k$-dimensional subspace such that $W \cap Q = \left\lbrace 0 \right\rbrace$. Then, for each $w \in W$, there is a unique $p \in P$ and $q \in Q$ such that $w = p + q$. We define $T: P \rightarrow Q$ as $Tp = q$.
That is, if we consider $U_Q$ to be the set of all $k$-dimensional subspaces which intersect $Q$ trivially, then we have a bijection $\psi_Q: L \left( P, Q \right) \rightarrow U_Q$. Here, $L \left( P, Q \right)$ is the space of all linear transformations from $P$ to $Q$, and can be identified with $R^{k \left( n - k \right)}$.
So, the collection $\left\lbrace \left( U_Q, \phi_Q \right) \right\rbrace$, where $\phi_Q = \psi_Q^{-1}$ can be the required smooth atlas. Again, to see this, we have to prove point (2) and (3) of the definition.
If $\left( P', Q' \right)$ is another pair of complementary subspaces of $V$, where $P$ is $k$-dimensional, we first want to prove that $\phi_Q \left( U_Q \cap U_{Q'} \right) = \left\lbrace T: P \rightarrow Q | \Gamma \left( T \right) \cap Q = \left\lbrace 0 \right\rbrace \text{ and } \Gamma \left( T \right) \cap Q' = \left\lbrace 0 \right\rbrace \right\rbrace$ is empty. However, because $\phi_Q$ is a bijection, we know that $\phi_Q \left( U_{Q} \cap U_{Q'} \right) = \left\lbrace T: P \rightarrow Q | \Gamma \left( T \right) \cap Q' = \emptyset \right\rbrace$. What I don't understand is why should this be an open set in $L \left( P, Q \right)$? To understand what is happening, I took a look at the three-dimensional case, when $\Gamma \left( T \right)$ could be $2$-dimensional. Since there is a one-to-one correspondence between $T$ and $\Gamma \left( T \right)$, we might as well work with a $2$-dimensional subspace in $\mathbb{R}^3$ which intersects $Q$ and $Q'$ trivially.
To get more of an idea, I have considered that $Q$ is the $X$-axis, $Q'$ is the $Y$-axis. Suppose that $T \in \phi_Q \left( U_Q \cap U_{Q'} \right)$. Then, it corresponds to a plane which does not contain both $X$- and $Y$-axes. Now, this plane can be sort of "rotated" by a small amount so that the resulting plane(s) do not contain the $X$- and the $Y$-axes. See Figure for more intuitive understanding of my thoughts.
What I have been stuck on is that how should we use these thoughts to construct open balls around each $T \in \phi_Q \left( U_Q \cap U_{Q'} \right)$?
Choose a basis $p_1,\dots,p_k$ for $P$ and some basis $q'_{k+1}, \dots,q'_{n}$ for $Q'$. Let $T \colon P \rightarrow Q$ be a linear map such that $\Gamma(T) \cap Q = \{ 0 \}$. Since $P$ and $Q'$ have complimentary dimensions, the condition $\Gamma(T) \cap Q = \{ 0 \}$ is equivalent to the condition that the $n \times n$ matrix whose columns are $$ p_1 + T(p_1), \dots, p_k + T(p_k), q'_{k+1}, \dots, q'_n $$ is invertible. Let's call the map $L(P,Q) \rightarrow \mathbb{R}$ given by $$T \mapsto \det \left( p_1 + T(p_1), \dots, p_k + T(p_k), q'_{k+1}, \dots, q'_n \right) $$ $G$. This is a continuous map and so if you have some $T_0 \in L(P,Q)$ with $\Gamma(T_0) \cap Q' = \{ 0 \}$ then $G(T_0) \neq 0$ and so by continuity you can find an open neighborhood of $T_0$ such that $G(T) \neq 0$ in that neighborhood, or, equivalently, $\Gamma(T) \cap Q' = \{ 0 \}$ for all $T$ in that neighborhood.