Is the principal $\Bbb{Z}^n$-bundle $\Bbb{R}^n \to \Bbb{R}^n/ \Bbb{Z}^n$ a trivial bundle?

Question

consider the translation group action $\Bbb{Z}^n$ on $\Bbb{R}^n$, which is principal action, therefore it induce principal $\Bbb{Z}^n$-bundle $\Bbb{R}^n \to \Bbb{R}^n/ \Bbb{Z}^n$.

we know $\Bbb{R}^n/ \Bbb{Z}^n$ has flat metric, and the tangent bundle is trivial, it looks very like $\Bbb{R}^n$.

This principal bundle is not a trivial G-bundle correct? the intuition is there is no global section, as you wrap around the circle on the tori, it should come back to the starting point, which is not possible for a section like this.

Answer 1

It is not a trivial bundle, and your heuristic picture is correct. One simple proof is that the (total space of the) trivial $\mathbb{Z}^n$ bundle is disconnected, but $\mathbb{R}^n$ is connected.

Answer 2

It is in fact as far from trivial as possible: because the total space is contractible, this is the universal principal $\mathbb{Z}^n$-bundle, meaning that any principal $\mathbb{Z}^n$-bundle on any other (reasonable) space $X$ is the pullback of this one along some map $X \to \mathbb{R}^n/\mathbb{Z}^n$, which is unique up to homotopy. This says that the space $\mathbb{R}^n/\mathbb{Z}^n$ is the classifying space $B \mathbb{Z}^n$.