Covariant derivative

Hi, Is it true that covariant derivative on any vector bundle over a manifold X comes from some connection on the bundle of linear frames of that vector bundle? This statement is true for the case of tangent bundle, but I am not sure if it is true in general or not. If it is true then can someone please suggest some reference for its proof.

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If you know how each vector gets transported along every curve, then you certainly know how a set of basis vector gets transported right? So I believe the answer is yes. – John Jiang Jan 5 '12 at 5:30
This is the kind of thing you really should try to work out yourself. It's OK to consult references to get a rough idea of what's going on, but you should try to fill in all the details yourself. There's no trickery or ingenuity needed at all. – Deane Yang Jan 5 '12 at 11:04

If $E \to X$ is a (finite-dimensional) vector bundle and $P$ is the principal $GL(n)$ bundle of frames, then there is a one-to-one correspondence between covariant derivatives on $E$ and principal connections on $P$. A good reference for details is proposition 4.4 of Lawson and Michelson's Spin Geometry" (unfortunately the relevant part is not on google books).

If $U$ is an open set of $X$ on which $E$ is trivial then relative to some local frame over $U$ we have the connection one-forms in $\Omega^1(U; \mathfrak{gl}(n))$. You can pull these back to $U \times GL(n)$, which then determines a $\mathfrak{gl}(n)$-valued one-form on $P\vert_U$ since the frame gives an isomorphism $P\vert_U \simeq U \times GL(n)$. Then one can show that these locally defined forms on $P$ piece together to form a global connection form.

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thanks for the answer and for the reference. – dushya Jan 5 '12 at 6:33

I don't know much about infinite dimensional things. I am not sure about the right answer but may be following may be useful... Once i saw the following book and statement:

See Page 4 of the book "Lectures on closed geodesics"- W. Klingenberg. Where he says:

"Whereas for Euclidean vector bundles over Euclidean manifolds such a map (Covariant derivative) $\nabla$ always defines a connection $K$, in our more general situation (That is Loop space: Hilbert Manifold) this need not always be true; see [FK] for further details. See also [El 3] for a more general setting."\

EL3: Eliasson, H.: On the geometry of manifolds of maps. J. Diff. Geom. 1, 165 -194 (1967).

So as far as I know in infinite dimension we can define co variant derivative which doesn't come from any so called connection.

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thanks for the references. – dushya Jan 5 '12 at 6:30