Let $D$ be the unit disk, $D=\{z: |z|\leq 1\}$. Let $S^1$ act on $\partial D\times D$ by pointwise multiplication. What is the orbit space of this action?

I think that since $\partial D=S^1$, this factor will vanish in the orbit space. Is it true that $\frac{\partial D\times D}{S^1}=\frac{\partial D}{S^1}\times \frac{D}{S^1}?$ Is the orbit space of $D$ under $S^1$ is just $D$? the orbits can be identified with radiused $0\leq r\leq 1$.

Thanks!

up vote 2 down vote accepted

Define $p : S^1 \times D^2 \to D^2, p(z_1, z_2) = z_2/z_1$. This is a continuous surjective map. Therefore it is an identification map because domain and range are compact. The orbit of $(z_1,z_2) \in S^1 \times D^2$ consists of all $(zz_1,zz_2)$ with $z \in S^1$. Hence $p$ maps each orbit to a single point of $D^2$. Conversely, $p(z_1,z_2) = p(z_1',z_2')$ means $z_2/z_1 = z_2'/z_1'$. With $z = z_1'/z_1 \in S^1$ we obtain $(z_1',z_2') = (zz_1,zz_2)$ so that $(z_1,z_2),(z_1',z_2')$ belong to same orbit. This shows that $p$ induces a unique homeomorphism $\hat{p} : (S^1 \times D^2) / S^1 \to D^2$ such that $\hat{p} \circ \pi = p$, where $\pi : S^1 \times D^2 \to (S^1 \times D^2) / S^1$ denote the quotient map.

Remarks:

(1) This shows that if you have continuous actions of a topological group $G$ on spaces $X, Y$, then in general $(X \times Y)/G \ne X/G \times Y/G$. This is similar to the situation in a field: You cannot expect $\frac{ab}{c} = \frac{a}{c}\frac{b}{c}$.

(2) If you have a continuous action of a topological group $G$ on a space $X$, you get a continuous action of $G$ on $G \times X$. Defining $p : G \times X \to X, p(g,x) = g^{-1} \cdot x$, you can easily show that $p$ induces a homeomorphism $\hat{p} : (G \times X) / G \to X$.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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