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
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
Best Answer
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 $T^n$.