3
$\begingroup$

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 $\mathbb{R}^p$?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.