6
$\begingroup$

For a vector bundle $E\to X$ with a given connection $\nabla$. We say that a section $s$ of $E$ is parallel to a vector space $V$ if $\nabla_V s=0$. If $\gamma:[0,1]\to X$ is a smooth path, we say that $s$ is a parallel transport of $v$ along $\gamma$ provided that $s(\gamma(0))=v$ and $\nabla_{\dot{\gamma}}s=0$ (to be precise, one considers an extension of $\dot{\gamma}$ to a local vector field).

When $E$ has a metric and $\nabla$ is compatible with this metric, then the metric compatibility definition implies that parallel transport defines an isometry (abusing notation a bit, metric compatibility gives $d/dt\langle s_1,s_2 \rangle=\langle \nabla_{t}s_1,s_2\rangle + \langle s_1, \nabla_{t}s_2\rangle = 0$ along $\gamma$ for parallel sections).

But if we do not have a metric. Can we prove that parallel transport is an injective map?

$\endgroup$

1 Answer 1

7
$\begingroup$

Since you can transport along the curve in the inverse direction, and the composition of the two transports is the identity map, yes.

This follows at once from the uniqueness of solutions of differential equations with initial conditions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .