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.]
Solution
A reflection of \(S^n\subset\mathbb R^{n+1}\) across the hyperplane perpendicular to a unit vector \(u\) is \[ R_u(x)=x-2\langle x,u\rangle u. \] Note that \(R_u=R_{-u}\).
1. Given two reflections \(R_u\) and \(R_v\), choose a continuous path of unit vectors\[ u_t\in S^n, \qquad u_0=u,\quad u_1=v, \]when \(n\ge1\).
The sphere \(S^n\) is path connected for \(n\ge1\). If necessary one may replace \(v\) by \(-v\), which defines the same reflection.
2. The formula\[ H_t(x)=R_{u_t}(x)=x-2\langle x,u_t\rangle u_t \]defines a homotopy through reflections from \(R_u\) to \(R_v\).
The displayed expression depends continuously on \((x,t)\). For each fixed \(t\), it is exactly reflection across the hyperplane \(u_t^\perp\), hence restricts to a self-map of \(S^n\). At \(t=0\) and \(t=1\) it equals the two prescribed reflections.
For \(n=0\) there is only one hyperplane in \(\mathbb R\), so the assertion is trivial. Hence any two reflections of \(S^n\) are homotopic, indeed homotopic through reflections.