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I want to find a differentiable $n$-dimensional compact manifold $M$ which can be endowed with an affine structure but cannot be endowed with a euclidean structure.

An affine (resp. euclidean) structure is a geometric structure with $X=\Bbb R^n$ and $G$ is the group of affine (resp. euclidean) transformations of $\Bbb R^n$.

I would like to find such a manifold for every possible dimension $n\geq 1$. I know that in dimension $1$ and $2$, such a manifold doesn't exist, since the only affine manifolds in that case are the circle, the torus and the Klein bottle and all of them are euclidean manifolds.

In dimension $3$, I think that $S^1\times S^2$ is an example. It is an affine manifold since it is diffeomorphic to the quotient $$\Bbb R^3-0/x\sim 2x.$$ However I don't really know how to prove that $S^1\times S^2$ has no euclidean structure (maybe we can use some theorem of Thurston about geometries of $3$-manifolds but it seems to be a "big tool")

In dimension $n\geq 4$, maybe $S^1\times S^{n-1}\simeq \Bbb R^n-0/x\sim 2x$ could be an example, but again I don't know how to prove that this manifold doesn't admit a euclidean structure.

Is there an elementary proof that these compact affine manifolds don't have euclidean structures? Are there some better examples?

Thanks in advance.

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Each Euclidean manifold $M$ admits a flat Riemannian metric (obtained by pull-back of the flat Riemannian metric on $E^n$ via coordinate charts of the Euclidean atlas). If $M$ is compact, the metric is complete (Hopf-Rinow theorem) , so is its lift to the universal covering space $X$ of $M$ (see this question). By Cartan-Killing-Hopf theorem, each simply connected complete flat $n$-dimensional manifold (such as $X$) is isometric to the Euclidean space $E^n$. In particular, its higher homotopy groups ($\pi_k(X)$, $k\ge 2$) vanish. Thus, your examples (covered by $S^{n-1}\times {\mathbb R}$, $n\ge 3$) do not admit Euclidean structures since $\pi_{n-1}\cong {\mathbb Z}$). These are the simplest examples. I can give you examples of aspherical affine manifolds not admitting Euclidean structures (say, product of genus 2 surface and the circle), but proofs are more complicated. Instead of higher homotopy groups you would use Bieberbach's theorem that the fundamental group of every complete flat Riemannian manifold is virtually abelian.

In the answer I assumed that you know basic Riemannian geometry and some algebraic topology; if you do not, I suggest do Carmo's book "Riemannian Geometry" for Riemannian geometry and, say, Hatcher, for algebraic topology.

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  • $\begingroup$ Thank you for this detailed answer. I've began studying Riemannian geometry recently but it's fine I get the arguments. In the meantime I read about Bieberbach theorem which says that every compact Euclidean manifold is finitely covered by a flat torus. It solves the problem but it's just a more complicated way to say that the universal cover of a compact Euclidean manifold is $E^n$. $\endgroup$ – Adam Chalumeau Jul 15 at 8:42

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