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Let $ G $ be a Lie group with finitely many connected components, $ K $ a maximal compact subgroup, and $ \Gamma $ a torsion free cocompact lattice. Then $$ \Gamma \backslash G/K $$ is an aspherical closed manifold with fundamental group $ \Gamma $.

This construction gives, for example, all closed manifolds which are locally symmetric spaces of noncompact type or of Euclidean type.

In general this construction gives all aspherical locally Riemannian homogeneous manifolds. This includes all compact solvmanifolds as well as all the 3-manifolds with the six aspherical Thurston geometries ($ S^3 $ and $ S^2 \times E^1 $ geometries are excluded since they are not aspherical).

In dimensions $ n \leq 2 $ every aspherical closed manifold is a locally symmetric space of noncompact type or of Euclidean type.

However for $ n=3 $ this is no longer true since the closed manifolds of the form $$ \Gamma \backslash G/K $$ given above are exactly the manifolds admitting a Thurston geometry. The only nontrivial connected sum with a geometric structure is $ \mathbb{R}P^3 \# \mathbb{R}P^3 $. In particular that means any connected sum of two aspherical closed 3-manifolds gives an example of a 3-manifold which cannot be expressed as $$ \Gamma \backslash G/K $$ Is it true for all dimensions $ n \geq 3 $ that a connected sum of two aspherical closed manifolds is never locally Riemannian homogeneous? In other words there is no way to write $ M_1 \# M_2 $ as $$ \Gamma \backslash G/K $$ where $ G $ is a Lie group with finitely many connected components, $ K $ a maximal compact subgroup, and $ \Gamma $ a torsion free cocompact lattice.

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    $\begingroup$ I am not sure if this answers your question, but mathoverflow.net/questions/46874/… might be relevant. $\endgroup$ Commented Feb 17, 2022 at 22:11
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    $\begingroup$ @OtisChodosh that link does answer the question. The fundamental group of $M_1\#M_2$ is a free product and has infinitely many ends. Hence so does the universal cover of $M_1\#M_2$. On the other hand the universal cover of a locally homogeneous space is homogeneous and hence has at most two ends. $\endgroup$ Commented Feb 17, 2022 at 22:39
  • $\begingroup$ @VitaliKapovitch interesting, thanks. Can you explain why a homogeneous space has at most two ends (I assume this is a very basic question)? $\endgroup$ Commented Feb 17, 2022 at 22:41
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    $\begingroup$ @OtisChodosh Sorry, I didn't think it through very carefully but what I had in mind was this. A connected lie group $G$ topologically is $\mathbb R^n\times K$ where $K$ is maximal compact subgroup. This has 1 end if $n>1$ and two ends if $n=1$. If $K$ is a compact subgroup of $G$ then $G/K$ will have no more than 2 ends. I have not thought through the case if $K$ is not compact (it should be true too) but it does not arise for Riemannian homogeneous spaces which I think is what the OP is interested in. $\endgroup$ Commented Feb 17, 2022 at 23:05
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    $\begingroup$ sorry, I should have used a different letter for the subgroup in the second part of the above comment. So, $G$ is a connected Lie group and $H$ is a compact subgroup (not necessarily maximal compact $K$). Then still $G/H$ has at most two ends by the argument I gave. $\endgroup$ Commented Feb 17, 2022 at 23:18

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