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Let's consider free action of finite abelian group $G = G_1 \oplus G_2$ on a manifold $X$. Is it true that $X/G$ is diffeomorphic to $(X/G_1)/G_2$?

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  • $\begingroup$ What do you mean by $X/G$? Do you mean $|X|/|G|$? right part of equality is also confusing. $\endgroup$ – mesel Mar 5 '14 at 9:13
  • $\begingroup$ $X/G$ is a quotient space. The equality means that $X/G$ is diffeomorphic to $(X/G_1)/G_2$ $\endgroup$ – Gleb Mar 5 '14 at 9:26
  • $\begingroup$ Right,Now it is clear. $\endgroup$ – mesel Mar 5 '14 at 9:30
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yes, it is true (more genrally for free actions of compact lie groups): Let $[x]_1$ denote the equivalence class of the action of $G_1$ via $h.x = (h,e_2)x$, where $e_i$ is the identity element of $G_i$.

First the action of $G_2$ on $X/G_1$ via $g.[x]_1 = [(e_1,g).x]_1$ is well defined and free, what is easy to check. Let its equivalence classes be denoted $[\ ]_2$. So the spaces $X/G$ and $(X/G_1)/G_2$ both have a natural smooth structure. A diffeomorphism $f : X/G \to (X/G_1)/G_2$ is simply given by $$[x] \mapsto [[x]_1]_2.$$ This is true in general. However in your case this is very easy to check: since the groups are finite the projections $p : X \to X/G$ and $\pi : X \to (X/G_1)/G_2$ are local diffeomorphisms. Also clearly $f \circ p = \pi$. Let $p \in X$ and $U$ be a neighborhood such that $p$ and $\pi$ restricted to $U$ are diffeomorphisms onto its images. Then $f_{p(U)} \circ p_U = \pi_U$, or $f_{p(U)} = \pi_{U} \circ (p_U)^{-1}$, that is $f$ is a local diffeomorphism. It is easy to check that $f$ is in fact bijective, so it is a diffeomorphism.

In general one can consider the map $f$ in charts obtained from slices of the action to see that it is a diffeomorphism.

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