Stabilizer, Cosets, homeomorphism and Compact groups : proving things in The Structure of Compact Groups by Hofmann and Morris

Question

I'm currently struggling trying to prove a few things in the book The Structure of Compact Groups by Hofmann and Morris. The first one would be Proposition 1.10.i (or E1.4) :

If the topological group G acts on the Hausdorff topological space $X$ then each stability subgroup $G_x$ is closed. The bijective function $f_x : G/G_x \to Gx$ arising from the canonical decomposition of the function $g \mapsto gx: G \to Gx$ is continuous. If G is compact, it is a homeomorphism. In particular, if G is compact and acts transitively on a Hausdorff space $X$, then for any $x \in X$, the spaces $G/G_x$ and $X$ are naturally homeomorphic and $X$ may be considered as a homogeneous space of $G$ (modulo the stability group $G_x$).

What I'd like to prove is the part "If G is compact, it is a homeomorphism".

Here is what I did already:

Let $g\in G$ be arbitrary, and suppose that $h\in gG_x$. Then $h=g k$ for some $k\in G_x$, and $$h\cdot x=(gk)\cdot x=g\cdot(k\cdot x)=g\cdot x\;.$$ Let $G/G_x=\{gG_x:g\in G\}$, the set of left cosets of $G_x$, and let

$$\varphi:G/G_x \to Gx:gG_x\mapsto g\cdot x\;.$$

The function $\varphi$ is well-defined: if $hG_x=gG_x$, then $h\in gG_x$, and we just showed that in that case $h\cdot x=g\cdot x$.

It’s clear that $\varphi$ is a surjection: if $y\in Gx$, then $y=g\cdot x=\varphi(gG_x)$ for some $g\in G$. Now to see that $\varphi$ is injective: we have $ g\cdot x = h\cdot x$ if and only if $ h^{-1}\cdot (g\cdot x) = x$, if and only if $h^{-1}g \in G_x$, if and only if $gG_x=hG_x$.

Now, $G \times X \rightarrow X$ by $(g,y) \mapsto g\cdot y$ is continuous by definition of the group action, thus the map $f_y: G \rightarrow X$ where $f_y(g)=g\cdot y$ is continuous. And the quotient topology on $G/G_x$ is the unique topology on the set of cosets $G/G_x$ such that any continuous map $G\rightarrow Z$ constant on $G_x$ cosets factors through the quotient map $G \rightarrow G/G_x$. That is, we have the following commutative diagram:

Thus, the induced map $\varphi:G/G_x \to Gx:gG_x\mapsto g\cdot x$ is continuous.

But now, it remains to show that $\varphi$ is open to show that it is an homeomorphism and I don't really see how I could do it, I'm more used to work with just sets, without group's structure in Topology...

And my second problem comes with the Exercise E1.5, just after that Proposition :

Show that the sphere $\mathbb{S}^{n−1}$ may be identified with a homogeneous space of $SO(n)$ modulo a subgroup isomorphic to $SO(n−1)$.

Well, here is how I would like to go : show that the $(n-1)$-spheres may be identified with a space of left cosets $SO(n)/SO(n)_x$ for $x$ an element of $\mathbb{R}^n$, given the natural $SO(n)$ action and then show that $SO(n)_x \cong SO(n−1)$ as top. groups and from there I've read more than once that $SO(n)/SO(n-1) \cong \mathbb{S}^{n-1}$, so I should be able to find a proof... but how could I do that?

Note that I'm more used to Topology than to Algebra, so "transitive action" properties, and so on, shouldn't be assumed. For example when the book says "G acts transitively on..." I'm already lost and have to Google and all. I've tried extensively to Google those things I'm asking here, but wasn't able to find anything helpful.

Answer 1

In order to show that $\varphi$ is open, it suffices to show that $G/G_x$ is compact (since $X$ is Hausdorff). The cosets $G/N$ of a compact group $G$ are compact if $N$ is a closed subgroup, but this is the case here.

For the second part, you apply the first part. In order to do this, you have to show that the action of $SO(n)$ on $S^{n-1}$ is transitive, meaning that for every two points on the sphere, you can find $A\in SO(n)$ such that $Ax=y$. In order to do this, it suffices to prove the result for $x$ and $e_1$ (convince yourself). For this, you can take $x$ and complete it into an orthonormal basis $\lbrace x,v_2,\cdots,v_n\rbrace$ of $\mathbb{R}^n$ (by Gram-Schmidt). The corresponding matrix $A=(x v_2 \cdots v_3)$ will send $e_1$ to $x$ and belongs to $O(n)$. Change if necessary the sign of one vector in order to have a matrix in $SO(n)$.

The fact that $SO(n)/SO(n)_x\cong SO(n)/SO(n-1)$ is similar. You take the stabilized point $x$, you complete it into an orthonormal basis of $\mathbb{R}^n$. In this, basis a matrix $A\in SO(n)$ will have $(1,0,\cdots,0)$ in the first colum and $(1,0,\cdots,0)$ in the first raw as well (because $AA^T=I_n$). The remaining block will correspond to a matrix of $SO(n-1)$. You apply now the first part of the exercice.

Answer 2

Any continuous map from a compact space to a Hausdorff space is a closed map, because a closed subset of a compact space is compact, the continuous image of a compact set is compact, and a compact subset of a Hausdorff space is closed. This answers your first question, since a bijective continuous map is a homeomorphism if it is a closed map.

To answer your second question, we can follow the strategy that you outlined. First, let's assume $n>1$ (as the result is false for the trivial case $n=1$). Let $e_1,\dots,e_n$ be the standard basis of $\mathbb R^n$. I claim that the orbit of $e_1$ under the action of $SO(n)$ is all of $\mathbb S^{n-1}=\{x\in\mathbb R^n : \|x\|=1\}$. Since orthogonal matrices preserve the lengths of vectors, it is clear that the orbit is contained in $\mathbb S^{n-1}$. Given an arbitrary vector $u\in\mathbb S^{n-1}$, we can extend $u$ to an orthonormal basis $f_1,f_2,\dots,f_n$ of $\mathbb R^n$, where $u=f_1$. Let $Q$ be the matrix whose columns are $f_1,f_2,\dots,f_n$. Then $Q$ is an orthogonal matrix, and if $\det(Q)=-1$, then we can modify $Q$ by negating, say, its second column, to obtain $Q\in SO(n)$. Then $Qe_1=f_1=u$, which shows that $u$ is in the orbit of $e_1$, as desired. Thus the orbit of $e_1$ under the action of $SO(n)$ is all of $\mathbb S^{n-1}$.

The only other thing to show is that, under this action, the stabilizer of $e_1$ is a subgroup isomorphic to $SO(n-1)$. To do this, notice first that if a matrix $Q\in SO(n)$ stabilizes $e_1$, this means that the first column of $Q$ is $e_1$. The orthogonality of columns now implies that the first row of $Q$ is all zeros except for the first entry. Thus $Q$ must have the form $Q=\pmatrix{1 & 0 \\ 0 & Q_0}$ for some $(n-1)\times(n-1)$ matrix $Q_0$. And in order for $Q$ to be orthogonal, $Q_0$ must be orthogonal, and in order for $Q$ to have determinant 1, $Q_0$ must have determinant one. Thus we must have $Q_0\in SO(n-1)$. Conversely, any matrix $Q_0\in SO(n-1)$ gives rise to a matrix $Q\in SO(n)$ stabilizing $e_1$ by the formula $Q=\pmatrix{1 & 0 \\ 0 & Q_0}$. Thus the stabilizer of $e_1$ consists precisely of the matrices $Q=\pmatrix{1 & 0 \\ 0 & Q_0}$, where $Q_0\in SO(n-1)$, and these clearly form a topological group isomorphic to $SO(n-1)$.