Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's an exercise given during a course in Differential Geometry that I'm taking.

Let $M$ denote a smooth manifold and let $G$ be a finite group of diffeomorphisms acting on it without fixed points (that is, $g(p)=p$ for some $p\in M$ forces $g$ to be the identity). We then have on the quotient space $M/G$ a differentiable structure. An atlas for it is obtained via the following observation: if $[p]$ is a point in $M/G$ then there exist a representative $p\in M$ and a chart $(U, \phi)$ in $p$ such that the projection $\pi\colon M\to M/G$ is injective on $U$. It then makes sense to define a chart $$(\overline{U}, \phi_{\overline{U}})=\left(\pi(U), \phi\circ\left(\pi|_{U}\right)^{-1}\right).$$ The family of all such charts is an atlas for $M/G$.

Exercise Show that the transition functions $\phi_{\overline{V}}\circ\phi_{\overline{U}}^{-1}$ can be identified with elements of $G$.

I find this question to be somewhat vague. Identified in which sense? The set of those transition functions might well be infinite, while $G$ is not. Also, I cannot see any relationship between the two. Can somebody provide me with some hint? Thank you.

share|cite|improve this question
up vote 1 down vote accepted

After a talk with the maker of the exercise I finally got it. Let us fix some notation: for every $p \in M$ let $[p]$ denote the equivalence class of $p$ in $M/G$ and let $\pi(p)=[p]$.

Fix point $x\in M$ and a diffeomorphism $g \in G$. Let $\tilde{x}=gx$. Since $G$ is finite, we can take charts $(U, \phi)$ and $(\tilde{U}, \tilde{\phi})$ in $x$ and $\tilde{x}$ respectively such that $\left.\pi\right\rvert_U$ and $\left.\pi\right\rvert_{\tilde{U}}$ are injective. This gives us two charts in quotient space: namely, if $V=\pi(U)$ and $\tilde{V}=\pi(\tilde{U})$, we have the charts $(V, \psi=\phi\circ\left(\left.\pi\right\rvert_U\right)^{-1})$ and $(\tilde{V}, \tilde{\psi}=\tilde{\phi}\circ\left(\left.\pi\right\rvert_\tilde{U}\right)^{-1})$.

The transition function relative to those two charts is related to $g$. Indeed, if $\eta\in \psi(V)\cap\tilde{\psi}(\tilde{V})\subset \mathbb{R}^n$, then \begin{align} (\psi\circ\tilde{\psi}^{-1})(\eta)&=\phi\left( \pi\tilde{\phi}^{-1}(\eta)\right) \\ \notag &=\phi\left(\left.\pi\right\rvert_U\right)^{-1}\left(\pi\tilde{\phi}^{-1}(\eta)\right)\\ \notag &=\left(\phi g^{-1}\tilde{\phi}^{-1}\right)(\eta). \end{align} This concludes the exercise.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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