# Connection from a covariant derivative

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?

• 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 $(\nabla b)(X):=\langle\nabla b,X\rangle=\nabla_Xb.$ – Amitai Yuval Mar 16 at 11:14
• The philosophy behind my previous comment is that, basically, covariant derivative and a connection are the same to begin with. – Amitai Yuval Mar 16 at 11:15
• Amitai, thanks a lot for the clear answer! – Moissan Mar 16 at 17:36