# A vector bundle admits a local covariantly constant section iff it is flat

Let $p:E\rightarrow M$ be a vector bundle over a manifold $M$ and let $\nabla$ be a connection on $E$. I am trying to show that $E$ admits a covariantly constant section $s$ in a neighborhood of each point (i.e. $\nabla s = 0$), if and only if the curvature of $\nabla$ is 0. I think I can show that a parallel section implies 0 curvature using the various symmetries of the curvature tensor.

I am unsure about the converse though. If we can show that parallel transport depends only on the homotopy class of the path, then we can produce a covariantly constant section by parallel transporting some fixed vector to each point. But, I am not sure how to get this as a consequence of 0 curvature. Any suggestions?

-
Your two statements are not equivalent: the existence of a covariantly constant (or parallel) section in a neighborhood of each point does not imply that the curvature is zero. Here's a counterexample: let $M = \mathbb R\times S^2$, equipped with the product Riemannian metric, let $E=TM$, and let $\nabla$ be the Levi-Civita connection. Then the vector field $s = \partial/\partial t$ (where $t$ is the coordinate on the $\mathbb R$ factor) satisfies $\nabla s = 0$ everywhere, but the curvature is not zero. –  Jack Lee Dec 29 '12 at 20:45

Here is a proof: We pick a homotopy $h_s(t) = H(s,t)$ between two paths $h_0$ and $h_1$ in $M$ and consider the parallel transport along each of the curves $h_s$. We can then show that the resulting vector field along $H$ will be parallel, not only in the $t$-direction, but also in the $s$ direction. This step uses the vanishing curvature assumption. From this the claim follows, because a homotopy leaves the endpoints fixed. Therefore the vector field – being parallel in the $s$-direction at $t=1$ – must be constant there, if we only vary $s$. Here are the details.
Let $I= [0,1]$. Suppose we are given a homotopy $$H: I\times I \to M, \quad (t,s)\mapsto H(t,s) = h_s(t)$$ between two paths $h_0 = H(\cdot,0)$ and $h_1 = H(\cdot,1)$. Let $x = h_s(0)$ and $y=h_s(1)$ denote the two endpoints (which are assumed to be fixed during the homotopy). Let $v\in E$ with $p(v) = x$. We denote by $t\mapsto V(s,t)$ the parallel transport of $v$ along $t\mapsto h_s(t)$ for any $s\in I$, i.e. $V$ satisfies $$\nabla_{t} V = 0 \;\; \forall s,t\in I \quad \text{and} \quad V(s,0) = v\;\; \forall s\in I.$$
Vanishing curvature implies that $\nabla_t \nabla_s V = \nabla_s \nabla_t V$ for all $s,t\in I$. Thus, for any fixed $s\in I$, we have that $t\mapsto \nabla_s V(s,t)$ satisfies $\nabla_t \nabla_s V(s,t) = \nabla_s\nabla_t V(s,t) = 0$ and $\nabla_sV(s,0) = \nabla_s v = 0$. By uniqueness of parallel transport, it follows that $\nabla_s V = 0$. In particular, for $t=1$, we obtain that $\nabla_s V(s,1) = 0$. Since $h_s(1) = y$ is the constant path at $y$, it follows that $V(0,1) = V(1,1)$. So the parallel transport of $v$ along $h_0$ and $h_1$ agrees at their common endpoint. $\square$