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This can be generalised, but let me be fairly concrete. Let $X$ be a simply-connected riemannian manifold and let $G$ denote the Lie group of isometries, assumed nontrivial. Let $F < G$ be a finite subgroup acting freely and consider the smooth quotient $X/F$ with the induced riemannian structure.

The normaliser $N(F) < G$ still acts on $X/F$ isometrically with $F < N(F)$ acting trivially. So $X/F$ inherits an isometric action of the group $N(F)/F$.

Now let $E < N(F)/F$ be a finite subgroup acting freely on $X/F$ and consider the quotient $(X/F)/E$. This is a smooth manifold, locally isometric to $X$ and hence isometric to $X/D$ for some freely-acting subgroup $D<G$.


How is $D$ related to $E$ and $F$? I would expect $D$ to be an extension of $E$ by $F$. Is it? And if so, but does it split? And if not, is there a name for this construction?

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It certainly doesn't have to split: Consider the Z/2Z action on $S^1$ followed by the Z/2Z action on the quotient $S^1$. – Jason DeVito Jun 22 '10 at 23:14
up vote 5 down vote accepted

The group $D$ is the preimage of $E$ in $N(F)$, so it is as you expect. The finiteness hypothesis can be weakened, which is important for many applications. Things become clearer if one thinks categorically in terms of the universal properties.

Say an arbitrary group $G$ acts freely and properly discontinuously and isometrically on $X$ and $H$ is a normal subgroup of $G$. Then $G/H$ acts freely and properly discontinuously and isometrically on $X/H$ with $X \rightarrow (X/H)(G/H)$ a $G$-invariant map. The induced map $f:X/G \rightarrow (X/H)(G/H)$ is an isomorphism. Indeed, both sides composed back with the natural map from $X$ satisfy the same universal property, and $f$ respects the maps from $X$, so $f$ is an isomorphism. QED

In fact, with more work this can all be done more generally with $G$ and Lie group and $H$ a closed normal Lie subgroup, under suitable "niceness" hypotheses for the orbit maps (which are satisfied in the above situation): see Proposition 13 in section 1.6 of Chapter III of Bourbaki LIE.

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