5
$\begingroup$

We say that a connected manifold $ M $ is aspherical if $$ \pi_n(M) = 0 $$ for all $ n \geq 2 $.

Equip $ M $ with a metric $ g $ such that $ (M,g) $ is Riemannian homogeneous (i.e. the isometry group acts transitively). If $ M $ is a compact Riemannian homogeneous aspherical manifold must $ M $ be a flat torus?

I believe the answer is yes. Here is the proof:

A compact aspherical manifold (indeed any finite CW complex) has torsion free fundamental group. Since $ M $ is compact Riemannian homogeneous then by

Transitive action by compact Lie group implies almost abelian fundamental group

the commutator subgroup of the fundamental group must be finite. But $ \pi_1(M) $ is torsion free so any finite subgroup is trivial. Thus the commutator subgroup is trivial. In other words $ \pi_1(M) $ is abelian. Since $ M $ is compact $ \pi_1(M) $ is finitely generated. So $ \pi_1(M) $ is a finitely generated torsion free abelian group
$$ \pi_1(M) \cong \mathbb{Z}^n $$ Assuming that a compact Riemannian homogeneous $ K(\mathbb{Z}^n,1) $ must be a flat torus that completes the proof. But I'm not quite sure how to show that a compact Riemannian homogeneous $ K(\mathbb{Z}^n,1) $ must be a flat torus.

What about the case where $ M $ is Riemannian homogenous aspherical but not compact? A Riemannian homogeneous manifold is an isometric product of a contractible piece with a Riemannian homogeneous compact piece. See

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

So as long as the compact piece has dimension at least 2 then the above argument goes through and the compact piece is a flat torus so by homogeneity of the metric the whole thing is flat.

But what about if the compact piece is only one dimensional? I think the group $ H(3, \mathbb{R})/ \Gamma $ with its invariant metric (Nil geometry) is a counterexample where flatness is lost. Here

$$H(3, \mathbb{R}) = \left\{\begin{bmatrix} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\end{bmatrix} : x, y, z \in \mathbb{R}\right\}$$

is the three dimensional Heisenberg group, and

$$\Gamma = \left\{\begin{bmatrix} 1 & 0 & c\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} : c \in \mathbb{Z}\right\}$$

is a discrete central subgroup.

Of course if there is no compact piece and $ M $ is contractible (topologically $ \mathbb{R}^n $) Riemannian homogeneous then there are a million different Riemannian homogeneous metrics that aren't flat. Take for example the hyperbolic metric of even the left invariant metric on any contractible Lie group (all simply connected non-abelian solvable Lie groups are good examples)

This is mostly a proof verification question because this seems too general to be true but I think my proof checks out

$\endgroup$
9
  • $\begingroup$ The sentiment "a (cocompact) action of ${\mathbb Z}^n$ by isometries on a contractible manifold $\tilde M$ must be the action of a lattice on a flat ${\mathbb R}^n$" is false even if $n=2$: Just take a non-flat Riemannian metric on $T^n$ and lift it to the universal cover. There are more exotic examples in dimensions $\ge 7$ when $M$ is homeomorphic but not diffeomorphic to $T^n$. However, you also have the homogeneity requirement for $X=\tilde M$ and this will indeed imply that $M$ is flat: $X=G/K$, $G$ contains an abelian lattice $\Gamma$, hence, $G< Isom(R^n)$.... $\endgroup$ Commented Feb 16, 2022 at 0:39
  • $\begingroup$ Just a comment about your claim Of course if there is no compact piece and $M$ is contractible (topologically $\Bbb R^n$): there are contractible 3 manifolds that are not homeomorphic to $\Bbb R^3$ (see Whitehead manifold). $\endgroup$
    – Didier
    Commented Feb 17, 2022 at 10:05
  • 1
    $\begingroup$ @IanGershonTeixeira: I was assuming that $X$ is homogeneous (otherwise, it's false). I will add an answer next week when I have more time. $\endgroup$ Commented Feb 17, 2022 at 21:14
  • 1
    $\begingroup$ @VitaliKapovitch First, thanks so much for your excellent answer! Second, I agree that a virtually abelian torsion free group need not have finite commutator. However if you reread my argument you will see that is not what I am claiming. Rather, I make the following series of claims: (1) $ M $ is compact and Riemannian homogeneous so (2) $ Iso(M) $ is compact and acts transitively and (3) any manifold which admits a transitive action by a compact group must have fundamental group with finite commutator subgroup so (4) since $ M $ is aspherical the fundamental group is torsion free $\endgroup$ Commented Feb 20, 2022 at 17:10
  • 1
    $\begingroup$ therefore (5) any finite subgroup of $ \pi_1(M) $ is trivial thus (6) the commutator subgroup of $ \pi_1(M) $ is trivial so we can conclude (7) that $ \pi_1(M) $ is abelian now (8) any compact manifold has finitely generated $ \pi_1 $ thus (9) $ \pi_1(M) $ is a finitely generated abelian group and so finally we can conclude (10) that $ \pi_1(M) \cong \mathbb{Z}^n $ $\endgroup$ Commented Feb 20, 2022 at 17:11

2 Answers 2

3
$\begingroup$

Claim: Let $M^n$ be an aspherical closed Riemannian homogeneous manifold. Then $M$ is a flat torus $T^n$.

Let's write $M=G/H$ where $G=Isom_0(M)$ is the connected component of the isometry group of $M$ and $H$ is the isotropy group of a point. Both $G$ and $H$ are compact. $G$ admits a left invariant metric which is biinvariant under $H$ and which induces the original Riemannian metric on $M$ (this is true for all Riemannian homogeneous spaces even if $G$ is not compact).

Let $\tilde G$ be the universal cover of $G$ and $\tilde H$ be the preimage of $H$ under the projection $\pi: \tilde G\to G$. Then $M=\tilde G/\tilde H$.

By the theory of compact Lie groups we have that $\tilde G=\mathbb R^k\times \hat G$ where $G$ is compact semisimple (it's a product of several simple factors).

Since all Lie groups have trivial $\pi_2$ it must hold that the identity component $\tilde H_0$ is simply connected since otherwise $M$ would have nontrivial $\pi_2$. So $ \tilde H_0$ is also isomorphic to $\mathbb R^l\times \hat H$ where $\hat H$ is semisimple and simply connected. Then $\hat H\subset \hat G$ is a closed subgroup. The manifold $\hat G/\hat H$ is closed and simply connected. If it's not a point it has nontrivial top homology and hence has a nontrivial $\pi_k$ for some $k>1$. But that would imply that $M$ also has nontrivial $\pi_k$. Therefore $\hat H=\hat G$. Therefore we can "cancel" $\hat G$ in the homogeneous space $M=\tilde G/\tilde H$. This means that the $\mathbb R^k$ factor in $\tilde G$ already acts transitively on $M$.

Now, the punchline is that any left invariant Riemannian metric on $\mathbb R^k$ is flat and any closed subgroup of $\mathbb R^k$ is flat too. This immediately implies that $M=\tilde G/\tilde H$ is flat. It's well known that the only closed Riemannian homogeneous flat manifolds are flat tori so $M$ is a flat $T^n$.

$\endgroup$
0
$\begingroup$

EDIT: This answer is wrong. See instead the excellent answer by Vitali Kapovitch. As pointed out in the comment what this answer really shows is that the manifold admits a flat metric and is diffeomorphic to a standard torus. It does not show that the original metric must have been the flat metric. I have gone back through and put some edits in [brackets] so that I am at least not claiming anything that is wrong anymore.

A compact Riemannian homogeneous manifold must [admit a metric with] nonnegative sectional curvature (the isometry group is a compact Lie group so has a biinvariant nonnegative sectional curvature metric coming from the Killing form, which is semi-definite since the group is compact).

Cheeger-Gromoll splitting theorem (THE SPLITTING THEOREM FOR MANIFOLDS OF NONNEGATIVE RICCI CURVATURE, theorem 3) states that a compact manifold with nonnegative Ricci curvature has universal cover isometric to the Riemannian product of a flat $ \mathbb{R}^k $ with a compact simply connected Riemannian manifold $ C $.

Thus any nonnegatively curved aspherical closed manifold must flat. In particular any Riemannian homogeneous aspherical closed manifold [admit a] flat [metric]. This argument is from Vitali Kapovitch given in the comments here

https://mathoverflow.net/questions/410547/exact-condition-for-smooth-homogeneous-to-imply-riemannian-homogeneous-for-compa

Once we know that $ M $ [admits a] flat [metric] we can appeal to

Flat Riemannian homogeneous manifolds are trivial

(basically the Bieberbach theorem that a compact flat manifold is determined up to diffeomorphism by its fundamental group) to conclude that $ M $ is [diffeomorphic to] a torus.

$\endgroup$
1
  • 2
    $\begingroup$ It's not true that a compact Riemannian homogeneous has nonnegative sectional curvature. What's true is that any such manifold admits a Riemannian homogeneous metric of nonnegative sectional curvature but it may be different from the original metric. So the argument I gave that you are quoting using the splitting theorem only proves that $M^n$ admits a flat metric and hence is diffeo to a torus. It doesn't prove that the original homogeneous metric is flat. It's true also but you need to work harder to show this. $\endgroup$ Commented Feb 20, 2022 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .