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Let $ M $ be a compact connected manifold with $$ M \cong \Gamma \backslash G /H $$ where $G $ is a Lie group, $ H $ a compact subgroup, $\Gamma $ a discrete subgroup, and $ G/H $ is connected and simply connected. Then $ \Gamma $ must be the fundamental group of $M$.

  1. What else can we say about uniqueness of $G, H $ and $ \Gamma$ ?

  2. In the 8 model 3-manifold geometries none of the $ G $ are connected. If $ G $ acts transitively on a connected manifold then so does its identity component, so why not just take $ G $ to be connected?

  3. Are there canonical choices for $ G, H $ with particularly nice properties, at least in some cases? For example, can $ G/H $ be chosen to be a symmetric space?

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    $\begingroup$ 2) if $M$ is not oriented then $G$ can't be chosen connected. The point is not that $G$ acts transitively, but that $G$ includes the $\pi_1(M)$-action on the universal covering of $M$. Even among oriented cases, $G$ connected would miss some cases, e.g., in type $\mathbf{E}^1\times\mathbf{H}^2$. 3) Among the eight 3-dimensional Thurston geometries, the locally symmetric ones are the constant curvature ones, and this are only three among those eight. $\endgroup$
    – YCor
    Commented Mar 10, 2020 at 15:32
  • $\begingroup$ If $G$ is any closed subgroup of isometries of the Euclidean space $\mathbf{R}^n$ containing the group $\mathbf{R}^n$ of translations, and $\Gamma=\mathbf{Z}^n$ is the group of integral translations, and $H$ the stabilizer of $0$ in $G$, then the $n$-torus can be viewed as $\Gamma\backslash G/H$. Thus $G$ is far from unique. $\endgroup$
    – YCor
    Commented May 12, 2021 at 16:46

1 Answer 1

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1). Let $G$ be a simply connected solvable Lie group. Then compact $H = \{ e\}$. In this case for fixed $\Gamma$ there are a lot of nonisomorphic $G$, containing $\Gamma$ as a lattice, but all manifolds $\Gamma \setminus G$ are diffeomorphic. There is a construction of all such $G$ - see Auslander L. AN EXPOSITION OF THE STRUCTURE OF SOLVMANIFOLDS. PART I: ALGEBRAIC THEORY, BULLETIN OF THE AMERICAN MATHEMATICAL SOCIETY, V. 79, N 2, 1973. So in this "solvable" case there is no uniqueness of G of any kind. Moreover, it also does not exist a uniqueness in the case of general Lie groups $G$. There is some uniqueness only for the case of semisimple Lie groups $G$.

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  • $\begingroup$ But are they Thurston geometries, i.e., isn't the maximality likely to fail? what does prevent all of them to be isomorphic to cocompact lattices in the same (non-contractible) Lie group? $\endgroup$
    – YCor
    Commented May 12, 2021 at 12:53
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    $\begingroup$ The question 1 was not about Thurston geometries. But some kind of "universal" solvable (!) group $G$ was constructed by Auslander. $\endgroup$
    – Vladimir47
    Commented May 12, 2021 at 16:19

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