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Usually, one views the connection $\nabla$ on a vector bundle $E \to M$ as a map $\Gamma(M, E) \to \Gamma(M,T^*M) \otimes \Gamma(M,E)) \simeq \Gamma(M,T^*M\otimes E)$. One can extend this to the exterior power $\Lambda(E)$ of $E$ by the formula $$\nabla(s_1 \wedge ... \wedge s_k)=(\nabla(s_1)) \wedge s_2 \wedge ... \wedge s_k+...+s_1 \wedge s_2 \wedge ... \wedge(\nabla(s_k))$$ However, in some lecture notes I found the alternative descrpition, for which I doubt whether is equivalent to the above. The quote from these notes is the following:
`it [a connection] extends to an odd derivation of degree one on $\Omega(M) \otimes \Gamma(E)=\Gamma(\Lambda(T^*M) \otimes E)=:\Omega(M,E)$ with grading inherited from $\Omega(M)$, leaving $\Gamma(M,E)$ trivially graded, so that \begin{equation}(*) \ \ \nabla(\omega \wedge \sigma)=d\omega \wedge \sigma+(-1)^{|\omega|} \omega \wedge \nabla(\sigma) \end{equation} where $\omega \in \Omega(M), \sigma \in \omega(M,E)$'
Question 1. As I understood, the space $\Omega(M,E)$ does not coincide with $\Gamma(M,\Lambda(E))$ since the second has nothing to do with $T^*M$. Therefore, we extend $\nabla$ in such a way that $\Omega(M,E)$ becomes a domain of the extended $\nabla$ or a target space? Could anybody give me the formula?
Question 2 The property $(*)$ is a definition of the formula by which we extend $\nabla$ or the property which follows from the way in which we extended $\nabla$? If the second is true (as I believe), what is the recipe for the formula?
Further author defines $\iota_X$ to be the usual contraction $\iota_X \omega(Y_1,...,Y_k):=\omega(X,Y_1,...,Y_k)$, extends it to $\Omega(M) \otimes \Gamma(M,E)$ via $\iota_X \otimes id$ and "defines operators $\nabla_X$ on $\Omega(M,E)$, of degree zero by defining: $$\nabla_X:=(\iota_X \otimes id) \circ \nabla+\nabla \circ (\iota_X \otimes id)"$$ Question 3 Before we extend $\nabla$, for $s \in \Gamma(M,E)$ the expression $\nabla s$, being an element of $\Gamma(M,T^*M\otimes E)$ may be viewed as the mapping $$\nabla s:\Gamma(M,TM) \ni X \mapsto \nabla_X S \in \Gamma(M,E).$$ So what is the relation between these two $\nabla_X$'s?
And finally Question 4 What is the relation between the extension of $\nabla$ to the exterior power of $E$ and the described above?

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Question 1: Domain and target; the $\nabla$ takes $\Omega(M,E)$ to itself.

Question 2: Property (*) defines the extension on pure wedge products $\omega \wedge \sigma$, but every element of $\Omega(M,E)$ is a sum of such, so extend by linearity to get a definition.

Question 3: The $\nabla s$ is a 1-form valued in $E$, i.e. element of $\Omega^1(M,E) \subset \Omega(M,E)$, obvious inclusion map.

Question 4: You can try both extensions, to sections of $\Omega^* \otimes \Lambda^* E$, no problem.

Question 5: Why didn't you try math.stackexchange?

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  • $\begingroup$ Thank You for Your answer. About question 2: I posted the formula exactly in the form which I have found in the notes. However, typical element of $\Omega(M,E)$ is a linear combination of expressions like $\omega \otimes s$ where $\omega \in \Omega(M)$ and $s \in \Gamma(M,E)$. How do we interpret $\omega \wedge \sigma$ for $\omega \in \Omega(M)$ and $\sigma \in \Omega(M,E)$? $\endgroup$
    – truebaran
    Sep 2, 2014 at 21:16
  • $\begingroup$ Write $\sigma$ as a linear combination of $\eta \otimes s$, say for simplicity $\sigma=\eta \otimes s$, and let $\omega \wedge \sigma = (\omega \wedge \eta) \otimes s$. $\endgroup$
    – Ben McKay
    Sep 4, 2014 at 18:09

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