# Riemannian homogeneous aspherical iff flat torus

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

• 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, $G< Isom(R^n)$.... Commented Feb 16, 2022 at 0:39
• 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). Commented Feb 17, 2022 at 10:05
• @IanGershonTeixeira: I was assuming that $X$ is homogeneous (otherwise, it's false). I will add an answer next week when I have more time. Commented Feb 17, 2022 at 21:14
• @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 Commented Feb 20, 2022 at 17:10
• 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 $\pi_1(M) \cong \mathbb{Z}^n$ Commented Feb 20, 2022 at 17:11

## 2 Answers

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 $$T^n$$.

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.

• 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. Commented Feb 20, 2022 at 15:44