# Torsionless connection and parallel transport over loops

Let suppose I have a Riemannian manifold and I define a connection on it. Is it true and safe to say that if a connection is torsionless and I parallel transport a vector along a loop made by geodesics I should have the same vector at the end of the loop?

Can anybody clarify this to me in plain english (I will translate in a formal way by myself). Thanks in advance.

Edit. I asked this question because I wanted to have a clearer picture about Torsion. So let's suppose for a while to have a Manifold with zero curvature. Is then safe to say that if the connection is torsionless when I make a loop along geodesics then the angle of the transported vector is the same as the original one?

• By "have the same vector at the end of the loop" what do you mean exactly? E.g. if you have a geodesic loop but not a closed geodesic, the vector field at the end of the loop would have a different angle than when you started. – Kal S. Feb 9 '17 at 10:21
• I think that in order to have the new vector the norm of the old one, the connection it must be compatible with the metric, I'm wondering the relation of torsionless – Dac0 Feb 9 '17 at 10:26
• If it is compatible with the metric and torsion free then the unique such connection is the Levi-Civita. But still you need to clarify what you mean by "the same vector at the end of the loop". – Kal S. Feb 9 '17 at 10:31
• Thank you Test123. I was thinking about a loop made by more than one geodesic. – Dac0 Feb 9 '17 at 10:37
• It doesn't make any difference. If at your initial point you don't have smoothness after the parallel transport you won't get the same angle for the vector. You won't even need to think about holonomy. So what you are asking is not clear at all. – Kal S. Feb 9 '17 at 10:40

This is not true if the curvature is not zero. Every Riemannian connection is torsionless but it is not always flat as shows $S^2$ endowed with the metric inheritated from $R^3$. The Lie algebra of holonomy around loops is charcterized by Ambrose Singer which expressed it with the curvature, and a curvature is not always zero in a Riemannian manifod.
• I think I do not agree with this but I may misunderstood the answer. E.g. in $S^2$ with the induced euclidean metric, if you parallel transport a vector field $X$ along a closed geodesic then $X(0)=X(1)$. – Kal S. Feb 9 '17 at 10:27