I am trying to understand the solution of the following problem:
Show that any two reflections of $S^n$ across different $n$-dimensional hyperplanes are homotopic, in fact homotopic through reflections. [The linear algebra formula for a reflection in terms of inner products may be helpful.]
And here is the solution I found here on this site:
My questions are:
1 - How the answer take into account the following statement in the question "across different $n$-dimensional hyperplanes"?
2- Is there a more detailed proof of that $F$ is a homotopy? like including the maps explicitly?
Could anyone help me answer this questions please?
Each $v \in S^n$ determines the hyperplane $H(v)$ perpendicular to $v$. This is given by $$H(v) = \{ x \in \mathbb R^{n+1} \mid v \cdot x = 0 \} .$$ Here $\cdot$ denotes the standard scalar product. The reflection at $H(v)$ is the linear map
$$\rho_v : \mathbb R^{n+1} \to \mathbb R^{n+1}, \rho_v(x) = x - 2(v \cdot x) v .$$
This is an orthogonal map since $\rho_v(x) \cdot \rho_v(y) = x \cdot y$, thus it restricts to a map $$r_v : S^n \to S^n .$$
Hatcher claims that for any two $v, w \in S^n$ the maps $r_v$ and $r_w$ are homotopic through reflections. This means that there exists a homotopy $H : S^n \times I \to S^n$ such that $H_0 = r_v, H_1 = r_w$ and each $H_t$, $t \in I$, has the form $H_t = r_u$ for some $u \in S^n$. The linked solution argues as follows:
Choose a path $\gamma : I \to S^n$ such that $\gamma(0) = v$ and $\gamma(1) = w$ and define $$H(x,t) = x - 2(\gamma(t) \cdot x)\gamma(t) .$$ This is the desired homotopy.