1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – John Jiang
    Jan 5, 2012 at 5:30
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Jan 5, 2012 at 11:04

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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