Why graph of a linear map, gives a bijection in the proof of Grassmannian as a manifold?

Question

In the proof of Grassmannian $(G_k(\mathbb{R}^n))$ as a manifold, we take an open neighbourhood $U_L = \{L^\prime\;|\; L^\prime\cap L^\perp=\{0\}\}\subset G_k(\mathbb{R}^n)$ of a $k$-dimensional subspace $L\in G_k(\mathbb{R}^n)$ and define a map $\phi_L:U_L\rightarrow \text{Hom}(\mathbb{R}^k,\mathbb{R}^{n-k})\cong \mathbb{R}^{k(n-k)}$ and we also have $\mathbb{R}^n = L \oplus L^\perp$, orthogonal projection $p:\mathbb{R}^n\rightarrow L$.

Now, I am having trouble understanding the proof of $\phi_L$ being bijection. The proof says, thinking $L^\prime\in U_L$ as a graph of a linear map $f:L\rightarrow L^\perp$, makes $\phi_L$ a bijection from $U_L$ to $\text{Hom}(\mathbb{R}^k,\mathbb{R}^{n-k})$.

Since $L^\prime$ is a graph of linear map $f$, we have $L^\prime = \{v+f(v)|v\in L\}$. From here I cannot find connection that will say $\phi_L$ is a bijection. Please clarify the doubt. Thanks in advance.

Answer 1

Given $f \colon L \rightarrow L^{\perp}$, consider the subspace

$$ \operatorname{graph}(f) := \{ v + f(v) \, | \, v \in L \} \subseteq \mathbb{R}^n. $$

This is a $k$-dimensional subspace of $\mathbb{R}^n$. If $w \in \operatorname{graph}(f) \cap L^{\perp}$ then $w = v + f(v)$ for some $v \in L$ but then $v = f(v) - w \in L^{\perp} \cap L$ so $v = 0$ which implies that $w = 0$. This shows that $\operatorname{graph}(f) \in U_L$ so we have a map $\psi \colon \operatorname{Hom}(L,L^{\perp}) \rightarrow U_L$ given by $\psi(f) = \operatorname{graph}(f)$.

Given $L' \in U_L$, note that $p|_{L'} \colon L' \rightarrow L$ is one-to-one (and onto). The reason is that if $v \in L'$ and $p(v) = 0$ then $v \in L^{\perp}$ so $v \in L' \cap L^{\perp} = \{ 0 \}$. Define a linear map $f_{L'} \colon L \rightarrow L^{\perp}$ by the formula

$$ f_{L'}(v) = \left( p|_{L'} \right)^{-1}(v) - v. $$

Note that

$$ p(f_{L'}(v)) = p \left( \left( p|_{L'} \right)^{-1}(v) \right) - p(v) = v - p(v) = 0 $$

because $v \in L$ so $f_{L'}(v)$ is indeed in $L^{\perp}$. This gives us a map $\varphi \colon U_L \rightarrow \operatorname{Hom}(L,L^{\perp})$ given by $\varphi(L') = f_{L'}$.

Finally, note that $p(v + f(v)) = p(v) = v$ for all $v \in L$ and so

$$ \varphi(\psi(f))(v) = \varphi(\operatorname{graph}(f))(v) = \left( p|_{\operatorname{graph}(f)} \right)^{-1}(v) - v = v + f(v) - v = f(v) $$

for all $v \in V$ so $\varphi \circ \psi = \operatorname{id}|_{\operatorname{Hom}(L,L^{\perp})}$.

In addition

$$ \psi(\varphi(L')) = \operatorname{graph}(f_{L'}) = \{ v + f_{L'}(v) \, | \, v \in L \} = \\\{ v + \left(p|_{L'}\right)^{-1}(v) - v \, | \, v \in L \} = \{ \left(p|_{L'}\right)^{-1}(v) \, | \, v \in L \} = L'$$ so $\psi \circ \varphi = \operatorname{id}|_{U_L}$.

Answer 2

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Zero}{\mathbf{0}}$Fix a $k$-plane $L \subset \Reals^{n}$ and let $L^{\perp}$ be its orthogonal complement. Up to a linear change of coordinates, we may as well take \begin{align*} L &= \Reals^{k} \oplus \Zero^{n-k} = \{(x^{1}, \dots, x^{k}, 0, \dots, 0)\}, \\ L^{\perp} &= \Zero^{k} \oplus \Reals^{n-k} = \{(0, \dots, 0, x^{k+1}, \dots, x^{n})\}. \end{align*} It should now be easy to see that each $k$-plane $L'$ transverse to $L^{\perp}$ (i.e., satisfying $L' \cap L^{\perp} = \{\Zero^{n}\}$, i.e., so that $L'$ is "non-vertical" with respect to $L$) is the graph of a unique linear transformation $f:L \to L^{\perp}$, and conversely that every linear transformation $f:L \to L^{\perp}$ arises in this way (by associating the graph of $f$ with $f$).

In case it helps, notice that if $v \in L$, then $v + f(v) \in L \oplus L^{\perp}$ is the decomposition of a vector into a component in $L$ and a component orthogonal to $L$.

(Incidentally, note carefully that the construction can be carried out functorially in a vector space $V$ without a metric by replacing $L^{\perp}$ with the quotient space $V/L$. This allows one to construct holomorphic bundles of complex Grassmannians starting from a holomorphic vector bundle.)