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I know about Ehresmann connections on fiber bundles and covariant derivatives as an (equivalent) way to define linear Ehresmann connections on vector bundles. My question is:

Is there any notion of covariant derivative equivalent to Ehresmann connection in the most general setting concerning fiber bundles?

When I say "the most general setting", I am emphasizing that the fiber bundle do not have any further structure than being just a fiber bundle (it may not be a vector bundle nor a principal bundle).

Thanks in advance,

Diego

PS: I'm concerning the case when the fiber bundles are smooth. I don't worry about the non smooth case.

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I think there is an analog of a covariant derivative associated to an Ehresmann connection, but it produces a more complicated object. Consider a fiber bundle $p:E\to M$ with fiber $F$, and a smooth section $s:M\to E$. Then you can take the tangen map $Ts:TM\to TF$ and use the Ehresmann connection to projctet to the vertical subbundle $VF=\ker(Tp)\subset TF$. So for each $x\in M$, this produces a linear map $T_xM\to V_{s(x)}F$ that can be interpreted as a smooth section of $T^*M\otimes s^*VF$, which is a good analog of the standard covariant derivative. Alternatively, you can insert a vector field $\xi$ on $M$ to obtain a section $\nabla_\xi s$ of $s^*VF$.

This exactly reproduces the standard concept in the case that $F$ is a vector bundle, since in that case $VF$ can be naturally identified with $p^*F$ and thus $s^*VF$ is canonically identified with $F$ for any section $s$.

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No although you can still define parallel transport on fiber bundles. Given the extra structure of a principal bundle, then you have the covariant derivative from a principal connection. See Natural Operator in Differential Geometry for more details.

As for the completness in this very book there is a theorem saying that all fiber bundles admits complete connections (i.e. Ereshmann connections), but the proof is wrong. There is a paper in which the author claims to have found a proof, however I am not sure whether he got it right (he could, just I don't know).

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  • $\begingroup$ Thanks a lot! @Baol $\endgroup$ – Diego Oct 7 '18 at 3:12

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