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My question is "inspired" by the uniformization theorem for Riemmannian surfaces and this post.

Suppose that $X$ is connected (finite-dimensional) topological manifold without boundary. Does there exist $n_{i,j}\in \mathbb{N}\times \{1,2,3\}$, $K_i\in\mathbb{N}$, and finite isometry groups $\Gamma_{i,1}$ acting on $S^{n_k}$, $\Gamma_{i,2}$ on $\mathbb{R_{n_{k,2}}}$ and $\Gamma_{n_{k,3}}$ acting on the hyperbolic space $\mathbb{H}^{n_{k,3}}$ $$ X \cong \prod_{k=1}^{K_1} S^{n_{k,1}}/\Gamma_{k,1} \times \prod_{k=1}^{K_2} \mathbb{R}^{n_{k,2}}/\Gamma_{k,2}, \times \prod_{k=1}^{K_3} \mathbb{H}^{n_{k,3}}/\Gamma_{k,3}? \label{1}\tag{1} $$ Here $\cong$ denotes the existence of a homeomorphism.

If not, what can we say about all objects on the right-hand side of \eqref{1}?

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  • $\begingroup$ This would imply that the universal cover is a product of spheres and euclidean spaces, which clearly is false. $\endgroup$
    – user43326
    Commented Jun 1, 2021 at 11:42
  • $\begingroup$ @user43326 Ah fair, which spaces have this property? $\endgroup$
    – ABIM
    Commented Jun 1, 2021 at 11:57
  • $\begingroup$ Take any closed, orientable, simply connected $4$-manifold which is not $S^4$. $\endgroup$
    – Nick L
    Commented Jun 1, 2021 at 12:04
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    $\begingroup$ Note that the OP asks for finite isometry groups, which implies $\pi_1(X)$ finite... Any compact surface of genus $\geq 1$ is already a counter-example. $\endgroup$
    – abx
    Commented Jun 1, 2021 at 12:20
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    $\begingroup$ What is funny is that the answer to this question is unknown in case you remove the assumption on $\Gamma$ to be finite (and act freely). This is a well known question asked by Gromov here: ihes.fr/~gromov/wp-content/uploads/2018/08/… page 12, second paragraph $\endgroup$
    – Dmitri Panov
    Commented Jun 1, 2021 at 13:07

1 Answer 1

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The answer to your first question is “no”. The first counter-examples appear in dimension four - for example the complex projective plane. There are many simply connected manifolds and that are not aspherical that will fit the bill.

Here is a somewhat wilder example. Suppose that $V$ is a genus two handlebody of dimension three. For example, $V$ can be obtained by embedding a “eye-glasses” graph in three-space and taking a small regular neighbourhood. Let $M$ be the double of $V$ across its boundary. That is, we take two copies of $V$ and glue via the identity on the boundary.

The universal cover of $M$ is a copy of the three-sphere, minus a Cantor set.

I don’t see a path to an answer to your second question. Perhaps you would be interested in Thurston’s geometrisation programme. It is sometimes described as a version of uniformisation in dimension three. It sadly does not generalise to dimension four.

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  • $\begingroup$ Does the class of manifolds which we can describe in this way include any "interesting" family? $\endgroup$
    – ABIM
    Commented Jun 1, 2021 at 12:03
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    $\begingroup$ Sure. Connect sums of simply connected four-manifolds... doubles of higher genus handlebodies... And there are more construction than these... for example torus bundles over hyperbolic surfaces will have nice universal covers but their fundamental groups will not “factor” as a product - instead they will be semi-direct products... I am sure that there are as many examples as there are topologists. :) $\endgroup$
    – Sam Nead
    Commented Jun 1, 2021 at 12:14
  • $\begingroup$ cool haha thanks Sam :) $\endgroup$
    – ABIM
    Commented Jun 1, 2021 at 12:15
  • $\begingroup$ @SamNead, isn't $M = S^1\times S^2 \# S^1\times S^2$ in your example? Do you happen to have a reference for the description you give of its universal cover? $\endgroup$
    – Marco Golla
    Commented Jun 1, 2021 at 13:47
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    $\begingroup$ @MarcoGolla, yes, that is another description of $M$. I don't have a reference to hand for the universal cover. So here is a sketch. Let $S$ and $T$ be disjoint two-spheres in $M$ whose union is non-separating. Cut $M$ along these and denote the result by $X$. So $X$ is (homeomorphic to) $S^3$ with four small, open, round three-balls removed (also, they have disjoint closures). Note that $X$ is simply connected and so lifts to the universal cover. In fact, copies of $X$ tile the universal cover in a tree-like fashion; the result follows. $\endgroup$
    – Sam Nead
    Commented Jun 5, 2021 at 2:37

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