Show that the orthogonal group acts transitively on the sphere $S^n.$

Question

Show that the orthogonal group $$ O(n + 1) = \{ A \in GL(n+1 , \mathbb{R}) \mid A^{-1} = A^{T}\}$$acts transitively on the sphere $S^n,$ with stabilizer subgroup $O(n).$ Then use this to determine, with justification, the relationship between $i$ and $n$ so that $$\pi_i(O(n)) \cong \pi_i(O(n + 1)) \cong \pi_i(O(n + 2)) \cong \cdots . $$

I got the following hint:

You need the Orbit Stabilizer Theorem from algebra, and the fact that if $G$ is a Lie group, then $G_{x} \rightarrow G \rightarrow Gx $ is a fibre sequence.

Still, I am unable to fill the details, could anyone help me in doing so, please?

Answer 1

Assume that we already know that the action is transitive. Then what is the stabilizer of the point $v=(0,...,0,1) \in S^n?$ I claim that if we let $O(n)$ be identified with the subgroup of $O(n+1)$ of all matrices of the form

$$\begin{pmatrix} A & \vdots \\ & 0\\ \dots 0 & 1 \end{pmatrix}$$

where $A \in O(n)$, then $O(n)$ is the stabilizer at $v \in S^n$.

(This new block matrix is clearly orthogonal since the dot product of two different columns are still zero and dot product of a column with itself is $1$)

$\textbf{Proof:}$

Let $B \in O(n+1)$ and $Bv = v$ so that $B$ is in the stabilizer at $v$. Then the rightmost column of $B$ has to be

$$\begin{matrix} \vdots \\ 0\\ 1 \end{matrix}$$

(we just make a direct computation of $Bv$)

and for $B$ to be orthogonal two different columns dot producted with eachother has to be zero. In particular for the rightmost culumn $ (b_{i,1},...,b_{i,n+1}) \ \cdot \ (0,...,0,1) = 0$ if $i \neq n+1$ which implies that $b_{i,n+1} = 0$ for all $i \neq n+1$. What this says is that $B$ is of the form

$$\begin{pmatrix} B' & \vdots \\ & 0\\ \dots 0 & 1 \end{pmatrix}$$

for some $n \times n$ matrix $B'$. Verifying that $B'$ is orthogonal is easy using the dot product of columns definition (dot product of two different columns has to be $0$ and the dot product of a column with itself is $1$).

It is simple to check that any matrix of the form

$$\begin{pmatrix} B' & \vdots \\ & 0\\ \dots 0 & 1 \end{pmatrix}$$

where $B' \in O(n)$ fixes $v$.

This proves that $O(n+1)_v = O(n)$.

$$\tag*{$\blacksquare$}$$

Now on to proving the result about homotopy groups.

The long exact sequence in homotopy groups associated to the fiber sequence

$$O(n)=O(n+1)_v \rightarrow O(n+1) \rightarrow O(n+1)v = S^n$$

(we used transitivity for last equality)

then becomes

$$...\rightarrow \pi_{i+1}(S^n) \rightarrow \pi_{i}(O(n)) \rightarrow \pi_i(O(n+1)) \rightarrow \pi_{i}(S^n) \rightarrow ...$$

Then if $i+1 < n,$ $\pi_{i+1}(S^n) = \pi_{i}(S^n) = 0$ so the map $$\pi_{i}(O(n)) \rightarrow \pi_i(O(n+1))$$ is an isomorphism if $i \leq n-2$. Then for induction it is simple to prove that $$\pi_{i}(O(n+k)) \rightarrow \pi_i(O(n+k+1))$$ is an isomorphism for positive $k$ if $i \leq n-2$ since $i \leq n-2$ implies that $i \leq n+k-2$.

$$\tag*{$\blacksquare$}$$

Hope you find this helpful!

Answer 2

I don't know if I count as a reputable source, but here is an outline of the proof that the action is transitive. Then Noel's answer finishes off the rest of the problem.

Lemma 1 The set $O(n)$ is a group under matrix multiplication. Further, $A\in O(n)$ iff the columns of $A$ form an orthonormal basis of $\mathbb{R}^n$.

Proof hint: Recall that $( A^{-1})^t = (A^t)^{-1}$ for any invertible matrix $A$. Also, note that the entries of $A^t A$ are dot products of columns of $A$, and $A^t A = A^{-1} A = I$. $\square$

Let $\vec{p}\in S^n$ denote the north pole: $\vec{p} = (1,0,...,0)^t$.

Lemma 2 Suppose for every $\vec{x}\in S^n$, there is a matrix $A \in O(n)$ with $A\vec{p} = \vec{x}$. Then the action is transitive.

Proof hint: If $A\in O(n)$ moves $\vec{p}$ to $\vec{x}$, then $A^{-1}\in O(n)$ and $A^{-1}$ moves $\vec{x}$ to $\vec{p}$. $\square$

Lemma 3: For any $n\times n$ matrix $A$, $A\vec{p}$ is the first column of $A$.

Lemma 4: Given any $\vec{x}\in S^n$, the set $\{\vec{x}\}$ can be extended to a orthonormal basis $\{ \vec{x}, \vec{v}_1,... \vec{v}_{n-1}\}$ of $\mathbb{R}^n$.

Proof hint: Induction on dimension $n$.

Lemma 5 Given any $\vec{x}\in S^n$, there is a matrix $A$ for which $A\vec{p} =\vec{x}$.

Proof hint: Use Lemma 4 and 1.