2
$\begingroup$

If a Riemannian metric is flat, and if the manifold it's defined on is contractible, does that mean any tangent vector can be uniquely extended to a parallel vector field on the entire manifold? Intuitively it at least seems possible.

$\endgroup$
4
$\begingroup$

All you should need is flat and simply connected, so yes. Here's a sketch:

At each point $q$, define $V_q$ by parallel transport of your initial vector $V_p$ along an arbitrary smooth curve joining $p$ to $q$. Since simply connected implies any two such curves with common endpoints are smoothly homotopic, the vanishing of the curvature should tell you that the transported vectors they produce are identical, so this is well-defined. Since there are such curves passing through every point in every direction, the resulting field will be parallel. Smoothness should follow from continuous dependence results for ODE.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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