Do flat manifolds only admit trivial bundles?

Question

Suppose I have a flat Riemann surface, such as a torus, an infinite strip, or an infinite cylinder. I can cover the entire surface with a single coordinate patch. Does this imply that I can only have trivial fibre bundles over this flat surface?

Answer 1

Given your usename, I'm guessing you're primarily interested in holomorphic line bundles.

Flatness of the base space turns out to be irrelevant when asking about existence of non-trivial fibre bundles. For one thing, every connected $n$-manifold can be decomposed (in many ways) as an open coordinate ball $B$ and the complement $U$ of a smaller closed coordinate ball. If $F$ is a topological space, it may be possible to "attach" the trivial bundles $B \times F$ and $U \times F$ non-trivially over $B \cap U$. The possibility of doing so rests on the $(n - 1)$th homotopy group $\pi_{n-1}$ of the automorphisms of the fibre $F$.

Particularly, if $T$ is a torus (flat or not), if $p$ is a point of $T$, and if $d$ is a non-zero integer, there exists a holomorphic line bundle that is (holomorphically) trivial in a neighborhood $B$ of $p$, and trivial on the complement $U$ of $p$, but not globally trivial, in the sense that the constant section over $U$ "twists $d$ times" relative to the constant section over $B$ around the boundary circle of $B$. Algebraic geometers say this line bundle is associated to the divisor $dp$, and often use the notation $\mathcal{O}(dp)$, suggesting the sheaf of local holomorphic functions vanishing to order at most $d$ at $p$.

If $p$ and $q$ are distinct points of $T$, the line bundle $\mathcal{O}(p - q)$, associated to the divisor $(+1)p + (-1)q$ is topologically trivial (its total space is diffeomorphic to $T \times \mathbf{C}$; there exists a non-vanishing smooth section), but not holomorphically trivial (there is no non-vanishing holomorphic section). The point of this example is, one has to be specific about what "trivial" means.

On an infinite strip, which is contractible (but not necessarily flat, and in fact often assumed to admit a metric of constant negative curvature), every holomorphic line bundle is (holomorphically) trivial, i.e., admits a non-vanishing holomorphic section, thanks to the Cousin problem.

Answer 2

if $X$ is a complete connected riemannian manifold with non-positive curvature (e;g. a flat metric), the exponential map is a covering (Cartan-Hadamard theorem), see https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem.

This does not imply that the tangent bundle is trivial. For instance the Klein Bottle, or the Möbius strip carries a flat complete riemannian metric but their tangent bundle is not trivial. For instance the Möbius strip is the quotient of the euclidian plane by the group generated by the map $(x,y)\to (x+1, -y)$. In some sense, you have only one chart, but the set of changing of charts is the group generated by this transformation.