Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.

Question

Real compact Lie group $SO_n$ acts smoothly and transitively on $\mathbb{S}^{n-1} \subseteq \mathbb{R}^n$ with obvious action. Isotropy subgroup of each point in $\mathbb{S}^{n-1}$ is isomoprhic to $SO_{n-1}$. Let $\{e_1,\ldots,e_n\}$ be standard orthonormal base for $\mathbb{R}^n$. I want to see that map $$p \colon SO_n \to \mathbb{S}^{n-1}$$ $$A \mapsto A e_n$$ is fibre bundle.

Map $p$ is obviously continuous surjection. I suppose one could take open cover $\{U^+,U^-\}$ of $\mathbb{S}^{n-1}$, where $U^\pm := \mathbb{S}^{n-1} \setminus \{\pm e_n\}$. Now, I should construct two homeomorphisms $\alpha_1$, $\alpha_2$, such that the following diagrams commute:

_{(source: presheaf.com)}

_{(source: presheaf.com)}

I cannot make this happen. How to define $\alpha_1$ and $\alpha_2$? Or should I use some other cover? Any help would be appreciated.

The reason I want this is that I would like to use homotopy lifting property (or exact sequence of homotopy groups) to calculate fundamental groups of $SO_n$ by induction.

Answer 1

If $G$ is a compact Lie group and $H$ is a closed subgroup, then $G\rightarrow G/H$ is a principal $H$-bundle. In particular, it is a fibre bundle with fibre $H$.

So, $SO(n)\rightarrow SO(n)/SO(n-1)$ is a fibre bundle with fibre $SO(n-1)$. We then compose this quotient map with the diffeomorphism $SO(n)/SO(n-1)\rightarrow S^{n-1}$ to get a fibre bundle $SO(n)\rightarrow S^{n-1}$. It then just remains to compute this map. It is precisely $A\mapsto[A]\mapsto Ae_n$, as desired.

Answer 2

$\newcommand{\R}{\mathbf{R}}$Your cover of $S^{n-1}$ is suitable.

Let $x = (x_{1}, \dots, x_{n-1})$ denote an element of $\R^{n-1}$, let $\Pi:U^{+} \to \R^{n-1}$ be stereographic projection from $e_{n}$, i.e., $$ \Pi(x, x_{n}) = \frac{x}{1 - x_{n}},\quad \|x\|^{2} + x_{n}^{2} = 1,\ x_{n} \neq 1, $$ and let $\Sigma:\R^{n-1} \to U^{+}$ be the inverse map $$ \Sigma(x) = \left(\frac{2x}{\|x\|^{2} + 1}, \frac{\|x\|^{2} - 1}{\|x\|^{2} + 1}\right). $$ Since $\Sigma$ and $\Pi$ are diffeomorphisms, they define a trivialization $\alpha: p^{-1}(U^{+}) \to U^{+} \times SO(n-1)$.

More explicitly, the columns $v_{1}, \dots, v_{n-1}$ of the Jacobian $D\Sigma(x)$ are a basis for $T_{\Sigma(x)} U^{+}$. Conveniently, they're also mutually orthogonal and of equal length since stereographic projection is conformal. Consequently, if $A \in SO(n - 1)$, and if $e = \{e_{1}, \dots, e_{n-1}\}$ denotes the Cartesian frame on $\R^{n-1}$, then $$ \Sigma_{*}(x, Ae) = \bigl(\Sigma(x), D\Sigma(x)(Ae)\bigr) = \bigl(\Sigma(x), A\, D\Sigma(x)(e)\bigr) = \bigl(\Sigma(x), Av)\bigr). $$ In words, under the diffeomorphism $\Sigma$, the standard left action of $SO(n-1)$ on the frame bundle of $\R^{n-1}$ induces the left action of $SO(n-1)$ on the frame bundle of $U^{+}$.

If $\tilde{A}$ is an element of $SO(n)$ whose last column is not $e_{n}$, and if the columns of $\tilde{A}$ are denoted $f_{1}, \dots, f_{n-1}, f_{n}$, there exists a unique $x$ in $\R^{n-1}$ such that $\Sigma(x) = f_{n}$, and there exists a unique $A$ in $SO(n-1)$ such that the orthogonal frames $f = (f_{1}, \dots, f_{n-1})$ and $Av = (Av_{1}, \dots, Av_{n-1})$ in $T_{f_{n}}(U^{+})$ are proportional. This correspondance defines the trivialization $\alpha_{1}$.

To trivialize this bundle over $U^{-}$, proceed similarly, but instead of stereographic projection $\Pi^{-}$ from the south pole (which reverses orientation), use $\Pi^{-}$ followed by the map $x_{1} \mapsto -x_{1}$.