TITLE: Conceptual definition of the extension of a connection to 1-forms
QUESTION [25 upvotes]: I have a question that arose while reading Milnor's "Characteristic Classes".  I will use the notation from that book.
Let $M$ be a smooth manifold and let $\zeta$ be a complex vector bundle on $M$.  Milnor defines a connection on $M$ to be a map $\nabla\colon C^{\infty}(\zeta) \rightarrow C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying the Leibniz identity, where $\tau_{\mathbb{C}}$ is the complexified tangent bundle of $M$.
In Lemma 4 of Appendix C, he proves that such a connection can be extended to a map $\hat{\nabla}\colon C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta) \rightarrow C^{\infty}(\wedge^2 \tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying an appropriate Leibniz rule.  However, his proof is just a definition in local coordinates, with the details left to the reader.  I verified these details, though they were a bit of a pain.
However, I feel like there must be a more conceptual definition of $\hat{\nabla}$ that makes no reference to local coordinates.  Does anyone know one?

REPLY [3 votes]: Although the answer is already given, it may be helpful to point out how the coordinate definition contains the multilinear definition, in a single step of work. From the definition
$$(d^\nabla s)_{ij}^\alpha=\frac{\partial s_j^\alpha}{\partial x^i}-\frac{\partial s_i^\alpha}{\partial x^j}+\Gamma_{i\beta}^\alpha s_j^\beta-\Gamma_{j\beta}^\alpha s_i^\beta$$
you can contract with $X^iY^j$ to get (using product rule from calculus twice)
$$(d^\nabla s)(X,Y)^\alpha=\Big(\frac{\partial (s(Y))^\alpha}{\partial x^i}-s_j^\alpha\frac{\partial Y^j}{\partial x^i}\Big)X^i-\Big(\frac{\partial (s(X))^\alpha}{\partial x^j}-s_i^\alpha\frac{\partial X^i}{\partial x^j}\Big)Y^j+\Gamma_{i\beta}^\alpha s(Y)^\beta X^i-\Gamma_{j\beta}^\alpha s(X)^\beta Y^j.$$
The first and fifth terms form the definition of $\nabla_X(s(Y))$, the third and sixth terms form the definition of $\nabla_Y(s(X))$, and the second and fourth terms define $-s([X,Y]).$ So you have
$$(d^\nabla s)(X,Y)=\nabla_X(s(Y))-\nabla_Y(s(X))-s([X,Y]).$$