Reduction of the structure group of $\mathbb{R}^n$-fiber bundles to a special subgroup of $\mathrm{Homeo}(\mathbb{R}^n)$

Question

Let $G$ be the group of all self-homeomorphisms $f$ of $\mathbb{R}^n$ which satisfy $$f(x+m)=f(x)+m,\quad \forall m\in \mathbb{Z}^n.$$

In other words, $G$ is the group of all equivariant self-homeomorphisms of the Euclidean $n$-space with respect to the standard action of $\mathbb{Z}^n$ on $\mathbb{R}^n$.

Is there a fiber bundle whose fibers are homeomorphic to $\mathbb{R}^n$ but the structure group of the bundle cannot be reduced to $G$? Is there a manifold $M$ such that the structure group of $TM\to M$, which is counted as a fiber bundle, cannot be reduced to $G$?

Answer 1

Your group $G$ consists of homeomorphisms of $\Bbb R^n$ which induce homeomorphisms on $T^n$ with induced map $f_*: \pi_1(T^n) \to \pi_1(T^n)$ equal to the identity. (If $f$ satisfied $f(x+m) = f(x)+g(m)$, for some homomorphism $g: \pi_1(T^n) \to \pi_1(T^n)$, then the induced map $f_*$ would be $g$.) Because the cohomology ring $H^*(T^n)$ is generated by $H^1$, and the induced map on $H^1$ is the identity, we see that the induced map on $H^n$ is the identity; therefore, $f$ is an oriented homeomorphism.

Write $G' \subset \text{Homeo}(T^n)$ for the subgroup of homeomorphisms inducing the identity on $\pi_1(T^n)$. The group $G'$ is a union of connected components. The relation to your group $G$ is that there is a surjective homomorphism $\pi: G \to G'$ with kernel the translations $\Bbb Z^n$; that is to say, every homeomorphism $g: T^n \to T^n$ lifts to a $\Bbb Z^n$-equivariant homeomorphism $\tilde g: \Bbb R^n \to \Bbb R^n$. $\require{AMScd}$

From here forward I assume $n \leq 3$. In these dimensions, we have a group homomorphism which is a homotopy equivalence $GL_n(\Bbb Z) \ltimes T^n \to \text{Homeo}(T^n)$, and in particular there is a group homomorphism $T^n \to G'$ which is a homotopy equivalence. Because we can make a diagram \begin{CD} \Bbb Z^n @>>> \Bbb R^n @>>> T^n \\ @| @VVV @V\simeq VV \\ \Bbb Z^n @>>> G @>>> G' \end{CD}

where the rows are fiber sequences and the outer two vertical arrows are homotopy equivalences, the middle arrow is a group homomorphism which is a homotopy equivalence, and hence $G$ is contractible. Thus if you can reduce the structure group of a fiber bundle to $G$, you can trivialize the fiber bundle: the only $\Bbb R^n$-fiber bundles which induce $T^n$-bundles, for $n \leq 3$, are trivial. Furthermore, for $n \leq 3$, one has the equivalence $\text{Homeo}(\Bbb R^n) \simeq O(n)$, and so in this range $\Bbb R^n$-fiber bundles up to isomorphism are the same as vector bundles up to isomorphism.

In particular $T\Sigma$ is a vector bundle which you cannot give structure group $G$ for any surface $\Sigma$, so long as $\Sigma \neq T^2$. The same is true for any non-oriented 3-manifold, but for an oriented 3-manifold $TM$ is trivial.

For bundles of higher rank $n$, the group $G$ is still a covering space of $G'$, but the latter group is more complicated: once $n \geq 5$, we have $\pi_0(G') = \Bbb Z_2^\infty$ (Wikipedia), and at a quick pass I can't find anything known about even the higher homotopy groups of $G'$, so it's not clear to me what you can say.