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!


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.

  • $\begingroup$ 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? $\endgroup$ – Sigh_at_psi Sep 27 '17 at 20:58
  • $\begingroup$ 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. $\endgroup$ – jdc Sep 28 '17 at 5:10
  • $\begingroup$ 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! $\endgroup$ – Sigh_at_psi Sep 28 '17 at 10:12
  • $\begingroup$ 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 $g([x]) := \tilde g(x)$. $\endgroup$ – jdc Sep 29 '17 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.