Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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. $\square$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.