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It is a basic question on connections of vector bundles, more preciously how to obtain a connection of a vector bundle starting from a covariant derivative.

Suppose $B \to M$ is a smooth vector bundle for a $C^{\infty}$ manifold $M$. A connection $\nabla$ on $B$ is a map $\Gamma(B) \to \Gamma(B) \otimes T^{\star}(M)$ that satisfies the Leibniz rule: for any $f\in C^{\infty}(M)$ and $b\in \Gamma(B)$ $$\nabla (fb)=b\otimes df +f\nabla b.$$ One can obtain a covariant derivative along $X$ a vector field $X$ on $M$, $\nabla_X:=<\nabla b,X>$, where $<,>$ is the canonical pairing of $T^{\star}M$ with $TM$. It aslo satisfies $\nabla_X(fb)=Lie_X fb+f\nabla_X b$, where $Lie_X$ is a Lie derivative along $X$.

To build a connection, one needs to have the right one-form from a covariant derivative with a section and then prove the Leibniz rule. And my question is how to get this one-form?

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  • $\begingroup$ Isn't this just a matter of notation? I mean, what we want is, given a section $b$, to define $\nabla b$, which is supposed to be a $B$-valued $1$-form. In other words, we want to define $(\nabla b)(X)$, for a vector field $X$. But we have a covariant derivative, and so, we set $(\nabla b)(X):=\langle\nabla b,X\rangle=\nabla_Xb.$ $\endgroup$ – Amitai Yuval Mar 16 at 11:14
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    $\begingroup$ The philosophy behind my previous comment is that, basically, covariant derivative and a connection are the same to begin with. $\endgroup$ – Amitai Yuval Mar 16 at 11:15
  • $\begingroup$ Amitai, thanks a lot for the clear answer! $\endgroup$ – Moissan Mar 16 at 17:36

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