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I was watching a video on Riemannian Geometry. The lecturer mentions that given the defining condition for a connection on a Riemannian manifold $M$ i.e. :

$$\nabla_X(Y) : \chi(M) \times \chi(M) \to \chi(M),$$ where $\chi(M)$ is the set of $C^{\infty}$ vector fields on M, the second of the Cartesian product from where Y comes as does the resulting quantity itself can be looked at as sections of a tangent bundle, while X is supposed to be looked at as a vector field.

The metric compatibility condition as well as the linearity condition in X and the derivative rule in Y makes sense in such a case.

However, when we move onto the torsion-free condition $$\nabla_{X} Y - \nabla_Y X = [X,Y]$$ inherent in Levi-Civita connections, both X,Y and the resulting $\nabla_{X} Y$ are all to be seen as vector fields. So the torsion free condition does not lend itself to the section of a bundle approach.

Can anyone please explain what this means? I was not able to fathom for myself. Thanks.

Vishesh
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    I don't understand either. Vector fields are precisely sections of the tangent bundle. – Michael Albanese Jul 30 '14 at 11:56
  • Yeah. This is the link for the video.The speaker is Prof J.W.Morgan. Watch from 8:30 mins onwards. http://www.youtube.com/watch?v=ImIQP9szMGs. A small correction which I have made to the question. I apologise for the earlier mistake. – Vishesh Jul 30 '14 at 12:01

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Having seen the part of the video you are referring to, Professor Morgan is talking about the more general context where you are defining a connection on a vector bundle $E$. That is a map

\begin{align*} \Gamma(TM)\times\Gamma(E) &\to \Gamma(E)\\ (X, s) &\mapsto \nabla_Xs. \end{align*}

In this situation, the expressions on either side of the torsion-free condition aren't defined, they are only defined when $X$ and $Y$ are vector fields. That is, it only makes sense to talk about a torsion-free connection when your vector bundle is $E = TM$.

Michael Albanese
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  • Thanks, is there any reference for this where I can glean more? – Vishesh Jul 30 '14 at 12:23
  • There are plenty. Every text I have seen on Riemannian manifolds covers connections on vector bundles. Having said that, there is not much more to say regarding why the notion of a torsion-free connection doesn't make sense on an arbitrary vector bundle $E$. – Michael Albanese Jul 30 '14 at 12:26
  • AAh, that is what I was looking for actually when I made the comment. Thanks a lot, nevertheless. – Vishesh Jul 30 '14 at 12:31