# Aspherical space whose fundamental group is subgroup of the Euclidean isometry group

Let $$M$$ be a smooth, compact manifold without a boundary, with its universal covering $$\tilde{M} = \mathbb{R}^n$$. If there exists an injective homomorphism $$h: \pi_1(M) \rightarrow O(k) \ltimes \mathbb{R}^k < {Aff}(\mathbb{R}^k)$$, where $$k < n$$, how can we describe $$M$$ and $$\pi_1(M)$$? In particular, can we deduce that $$M$$ is homeomorphic to a torus, up to a finite cover?

There is a possible solution. Since $$M$$ is closed manifold, $$\pi_1(M)$$ is a finitely generated subgroup of $$\Gamma(k)=O(k) \ltimes \mathbb{R}^k$$. If $$\pi_1(M)$$ is discrete then we can apply Bieberbach theorem and deduce that $$\pi_1(M)$$ is virtually Abelian.

The question arose when I was contemplating a smooth codimension $$k$$ foliation with a transversal Euclidean structure, whose leaves are diffeomorphic to $$\mathbb{R}^{n-k}$$. The developing map yields the homomorphism $$h: \pi_1(M) \rightarrow Iso(\mathbb{R}^k)$$, and I am interested in knowing whether this implies that the manifold can only be a finite quotient of a torus.

• Two small comments: first, I think your semi-direct produce is the wrong way round. (Note that $\mathrm{Isom}(\mathbb{R}^k)\cong O(k)\ltimes \mathbb{R}^k$.) And second, do you really mean that $k<n$ rather than $k\leq n$? For instance, the $(3,3,3)$ triangle group is not a subgroup of the isometry group of $\mathbb{R}$ (since it has 3-torsion).
– HJRW
Oct 12, 2023 at 12:57
• Yes, some toral bundles over surfaces of arbitrary high dimension have fundamental groups embeddable in $SE(3)$. But why do you care? Oct 12, 2023 at 15:33
• @HJRW Right. Then it is enough to observe that $\mathbf{Z}^n$ embeds into $\mathrm{SO}(2)$ for every $n$.
– YCor
Oct 12, 2023 at 15:49
• @HJRW Yes. Actually $\mathrm{SO}(3)$ itself contains large groups. Namely, consider an irreducible rational degree $n$ polynomial with $n$ real roots, only one of which, $r$ is positive, and $q$ the quadratic form $X^2+Y^2+rZ^2$. Then $\mathrm{SO}(q)(\mathbf{Z})$ is a dense subgroup of $\mathrm{SO}(3)$, and is also a cocompact lattice in $\mathrm{SO}(2,1)^{n-1}$ (hence virtual cohomological dimension $2n-2$, etc).
– YCor
Oct 13, 2023 at 7:37
• @Carl-FredrikNybergBrodda: This paper of Breuillard--Gelander--Souto--Storm provides a reference -- arxiv.org/abs/math/0602635 -- although I suspect it was known long before. I think their argument is roughly this one: for each $g\in\pi_1S$, the subset of $\mathrm{Hom}(\pi_1S,SO(3))$ that kills $g$ is nowhere dense, and the result follows by the Baire category theorem. Perhaps something like YCor's argument can give an explicit arithmetic construction.
– HJRW
Oct 13, 2023 at 11:58