The orbit space $\mathbb{R}^n/O(n)$ is homeomorphic to $[0,\infty)$

Question

Here, $O(n)$ is the group of orthogonal $n\times n$ matrices with real entries.

To show this, I tried defining a map $f:\mathbb{R}^n\to[0,\infty)$ by $f(x)=|x|$. Note that $f$ makes exactly the same identifications as the quotient map $\mathbb{R}^n\to\mathbb{R}^n/O(n)$ (this needs a bit of an argument to show but I know how to do it and I am comfortable with it). Thus if $f$ is a quotient map, we have $\mathbb{R}^n/O(n)\cong[0,\infty)$. But this is where I am stuck: how do we show that $f$ is a quotient map? The closed map lemma cannot be used, since $\mathbb{R}^n$ is not compact.

Answer 1

The map $f$ induces a continuous bijection $F:\Bbb R^n/O(n)\to[0,\infty)$. We need to show this has an inverse. Try $g:[0,\infty)\to\Bbb R^n/O(n)$ induced by $x\mapsto (x,0,\ldots,0)$. This is the composite of a continuous map from $[0,\infty)$ to $\Bbb R^n$ followed by the quotient map from $\Bbb R^n$ to $\Bbb R^n/O(n)$, so is continuous. It is fairly clear that this is a continuous inverse to $F$.

Answer 2

You can do this directly from the definition of a quotient map. Since $f$ is evidently continuous and surjective, you just need to do one more thing: for any subset $U \subset [0,\infty)$, assuming that $f^{-1}(U)$ is open, prove that $U$ is open.

Underlying this proof is the fact that the orbits of the group action are exactly the spheres centered on the origin $\mathcal{O}$, plus an orbit consisting of the origin $\mathcal{O}$ itself.

So choose $x \in f^{-1}(U)$ and let $B(x,r) \subset f^{-1}(U)$ be an open ball around $x$ that is contained in $f^{-1}(U)$.

Let's show that $(|x|-r,|x|+r) \cap [0,\infty) \subset U$. That is, for each $t \in (|x|-r,|x|+r) \cap [0,\infty)$, let's find $y \in B(x,r)$ such that $f(y)=t$.

1. If $x = \mathcal{O}$, choose $y$ to be any point such that $|y|=t$;
2. If $x \ne \mathcal{O}$ let $y$ to be the unique point on the ray $\mathcal{O}x$ such that $|y|=t$, that is, $$y = t \frac{x}{|x|} $$

In each of these cases one can easily show that $|y-x|=| t - |x| \,| < r$ and so $y \in B(x,r)$.

Answer 3

You can show directly that $f$ is closed. Consider a sequence of $x_n\in F$, where $F$ is arbitrary closed in ${\bf R}^n$. If $x_n$ is unbounded, so is $f(x_n)$, so $f(x_n)$ is not convergent. Otherwise, all $x_n\in K$ for some compact $K\subseteq F$.

More generally, if $f\colon X\to Y$ is a continuous function between metric spaces such that images of unbounded sets are unbounded, and balls in $X$ are compact, then $f$ is closed.

Answer 4

Your map $\mathbb R^n \to [0,\infty)$ is the composition $G \circ F$ of the two maps:

$F: (x_1,x_2,\ldots,x_n) \mapsto x_1+x_2+\ldots + x_n$

$G : x \mapsto |x|$

$F$ is open by the open mapping theorem..

It's not hard to show $G$ is open.

The image of the interval $(a,b)$ is one of $(a,b)$, $(-b,-a)$ or $[0, \max{|a|,|b|})$.

Therefore the composition is open and that implies it's a quotient map.