## When is a three-manifold deck transformation group solvable?

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Suppose that $\pi:Y \to Y'$ is a regular covering of closed, connected, orientable three-manifolds and let $G$ be the deck transformation group. Furthermore, suppose that $Y$ is a rational homology sphere (I don't know how much this condition matters). It's not always the case that $G$ is solvable, since one can take $Y$ to be $S^3$ and $Y'$ to be the Poincare homology sphere. Are there other examples where $G$ is not solvable? Are such examples classified? If $Y$ has elliptic geometry, this is the only example (I think), but I have no clue in general. This seems like it could be related to residual finiteness of three-manifold groups/RFRS/LERF/other four-letter acronyms I don't understand.

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## 3 Answers

Cooper and Long showed you can realize any finite group acting on a rational homology sphere.

There's no general sort of classification that I know of. Maybe what you're asking for is, given $Y'$ a rational homology sphere, and a homomorphism $\varphi: \pi_1(Y')\to K$, where $K$ is a finite group, when is the cover $Y\to Y'$ corresponding to $ker(\varphi)$ again a rational homology sphere? One perspective is that this amounts to computing $H_1(Y; \mathbb{Z}[K])$, homology with twisted coefficients. Dunfield and Thurston took advantage of this to compute covers of manifolds in the Snappea census with positive first betti number, but such computations can get rapidly complicated, so I don't know of any general pattern when $K$ is non-solvable.

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Not exactly an answer to the question, but a classification of solvable groups which are fundamental groups of compact 3-manifolds is given the following celebrated paper:

Thomas, C. B. On 3-manifolds with finite solvable fundamental group. Invent. Math. 52 (1979), no. 2, 187–197.

It is very easy to check which of the manifolds produced are rational homology spheres...

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You are asking for free actions of finite, nonsolvable groups on $Y$. If you truly don't care that $Y$ is a rational homology sphere, for any finite group $G$ there exists a closed, connected, orientable 2-manifold $X$ and a free action of $G$ on $X$, so $G$ also acts freely on $Y = X \times S^1$.

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Yes, that is a good observation.  I really want to restrict to rational homology spheres then.   – Tye Lidman May 19 2012 at 22:05
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