Let $M$ be a compact n-dimensional Riemannian manifold with non-negative Ricci curvature. Then its universal cover $\tilde{M}$ is isometric to $\mathbb{R}^p\times N$ for some $p\leqslant n$ and $N$ compact. Consider the group $G$ of isometries on $N$ that are split off by the action of $\pi_1(M)$ on $\tilde{M}$.

We know that $\pi_1(M)$ act discretely on $\tilde{M}$, why does G act discretely on N? Hence it's finite.

Denote the isometry group of $\mathbb{R}^p$ by $Iso(\mathbb{R}^p)$. Consider the projection map $$ \pi_1(M) \to Iso(R^p), $$ the kernel is $G$. The image $H$ is a subgroup of $Iso(\mathbb{R}^p)$ acting discretely on $\mathbb{R}^p$. The Bieberbach theorem says that $H$ contains a normal free abelian subgroup $\mathbb{Z}^p$ such that $[H:\mathbb{Z}^p]<C$.

Let us prove this. Since any element $\alpha \in Iso(\mathbb{R}^p)$ is in the form: $\alpha(x)=Ax+a$, where $A\in O(p-1)$ is a rotation, $a\in \mathbb{R}^p$ is a translation. Consider the projection $$ \alpha=(A,a) \in H \to A \in O(p-1). $$ The kernel is a subgroup acting discretely on $\mathbb{R}^p$ by translation, it's $\mathbb{Z}^p$.

Why is the image finite? i.e. Why does the rotation part act discretely on $\mathbb{R}^p$?

  • $\begingroup$ The second question is part of Bieberbach's theorem (not in the way you state it). What more do you want? $\endgroup$ – YCor Apr 13 '17 at 9:54
  • $\begingroup$ If $M$ is a $2$-dimensional sphere endowed with the standard metric, then $M = N$ and the group of isometries won't act discretely on $N$. $\endgroup$ – HYL Apr 13 '17 at 10:01
  • $\begingroup$ @YCor:I try to prove that $H$ is contains an abelian group with finite index. $\endgroup$ – mathmetricgeometry Apr 13 '17 at 10:37
  • $\begingroup$ @HYL: Then the fundamental group is trivial. $\endgroup$ – mathmetricgeometry Apr 13 '17 at 10:40

The group $G$ need not act discretely on $N$, e.g., consider $G=\mathbb Z$ acting on $S^2\times\mathbb R$ acting by an irrational rotation on the first factor and by translation on the second factor.

To learn more about these matters I suggest you read first Cheeger-Gromoll's paper on the soul theorem (which contains the above example of $\mathbb Z$-action on $S^2\times\mathbb R$), and then Wilking's paper [On fundamental groups of manifolds of nonnegative curvature. Differential Geom. Appl. 13 (2000), no. 2, 129–165]. In the latter Wilking shows that any nonnegatively curved metric as you describe can be deformed through nonnegatively curved metrics to the metric for which the $G$ action on $N$ factors through the action of a finite group.


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