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Let $ (M,g) $ be a flat and Riemannian homogeneous (isometry group acts transitively) manifold. Then is $ M $ just diffeomorphic to a product of circles and lines? This is true in dimension 2 (and 1).

Specifically, is every flat compact Riemannian homogeneous manifold a flat torus? I know by theorem of Bieberbach that every flat compact manifold is covered by a flat torus. So the idea would be that any nontrivial covering breaks the symmetry.

Ian Gershon Teixeira
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    What do you mean by a "flat circle" and "flat line"? All lines equipped with Riemannian metrics are isometric to each other and for a circle the length is the complete isometry invariant. And when you say "Cartesian product", do you mean "Riemannian Cartesian product"? That would be false. The Cartesian product as a differentiable manifold is true and not hard to prove. As a hint: Classify all discrete subgroups of isometries of ${\mathbb R}^n$ which commute with all translations. What do you get? – Moishe Kohan Jan 21 '22 at 17:30
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    You only get discrete subgroups of $ \mathbb{R}^n $ i.e. lattices. And any quotient by such a lattice has the trivial topology I described. And the quotient is compact if and only if the lattice is full rank in which case the quotient is a torus. Thanks! – Ian Gershon Teixeira Jan 21 '22 at 19:46

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A Riemannian homogeneous manifold is diffeomorphic to the product of a Euclidean space with a compact Riemannian homogeneous manifold. See

https://mathoverflow.net/questions/410334/noncompact-riemannian-homogeneous-is-trivial-vector-bundle-over-compact-homogene

So it is enough to show this for the compact case.

Let $ M $ be a flat compact Riemannian homogeneous manifold. Then the fundamental group of $ M $ is a torsion free crystallographic group (also known as a Bieberbach group). Since $ M $ is compact and Riemannian homogeneous then by

Transitive action by compact Lie group implies almost abelian fundamental group

the commutator subgroup of the fundamental group is finite. But the fundamental group is torsion free so every finite subgroup is trivial. Thus the commutator subgroup must be trivial. In other words $ \pi_1(M) $ is abelian. So it must be $ \mathbb{Z}^n $. So $ M $ must be isometric to a flat torus.

Ian Gershon Teixeira
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