# Proof behind $S^n\cong SO(n+1)/SO(n)$

I have been trying to understand the fact that $S^n \cong SO(n+1)/SO(n)$. I believe I have the intuition correct at this point; consider the case when $n=2$ as we have $S^2 \cong SO(3)/SO(2)$.:

We are trying to find the correspondence between rotations in $\mathbb{R}^3$ and points on $S^2$. At first I incorrectly thought that these spaces were isomorphic, however one then realizes that there are more rotations than points on a sphere, in the following sense:

Consider the point $p$ at the "north pole" of the sphere $S^2$. We can correspond $p$ with any point on the sphere by rotating the sphere, that is, by applying elements on $SO(3)$ so that $p$ ends up at any place upon it. However, we can first apply any rotation around the $z$-axis (through $p$). So, by "modding out" these rotations, which are exactly the elements of $SO(2)$, we have our isomorphism.

I am looking for someone to help me formalize this into a proof. Thank you!

## 1 Answer

It is basically the orbit–stabilizer theorem.

$SO(3)$ acts by rotations on $\mathbb R^3$. This action restricts to a transitive action on $S^2$. Fix a vector in $S^2$, say $e_1 = (1,0,0)$. One has a continuous map $SO(3) \to S^2$ given by $A \mapsto Ae_1$. The subgroup of $SO(3)$ stabilizing $e_1$, the "kernel" of this map, to abuse language, is the block-diagonal subgroup $H = \{1\} \times SO(2)$. It follows that the quotient $SO(3)/H$ is in continuous bijection with $S^2$. Because both spaces are compact Hausdorff, it is a homeomorphism.

As you seem to have noticed, there is nothing special about $n=3$ in this result.

• Why is it that the map given by the orbit-stabilizer theorem $SO(3)/SO(2)\rightarrow S^2$ is automatically continuous from the fact that the map $SO(3)\rightarrow S^2$ is continuous? – Sigh_at_psi Sep 27 '17 at 20:58
• I believe this is just the universal property of the quotient topology. Let me know if you agree. If $X \overset f\to Y$ is a quotient map a composition $X \overset f\to Y \overset g\to Z$ is continuous, and $U \subseteq Z$ is open, then $f^{-1}(g^{-1}(U))$ is also open. But a set in $Y$ is defined to be open just if its preimage in $X$ is open, so $g^{-1}(U)$ is also open. Since $U$ was arbitrary, this means $g$ is continuous. – jdc Sep 28 '17 at 5:10
• I think we need $Y=\{(g\circ f)^{-1}(z)|z\in Z\}$ because other wise the existence of $g$ from the map $X\rightarrow Z$ is not guaranteed. Although I don't know if the orbit-stabilizer theorem yields the automatic map. Confusing! – Sigh_at_psi Sep 28 '17 at 10:12
• I am not sure I understand your reasoning. Any function $\tilde g\colon X \to Z$ that respects a equivalence relation $\sim$ descends uniquely to a well-defined function $g\colon Y = {X/\sim} \to Z$ via the formula $g([x]) := \tilde g(x)$. – jdc Sep 29 '17 at 1:33