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In Tu's book on differential geometry he defines a connection on a smooth vector bundle $E \to M$ as $\nabla : \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E)$ so that $\nabla$ is $C^\infty$-linear on $X \in \mathfrak{X}(M)$ and $\Bbb R$-linear on $s \in \Gamma(E)$.

On the other hand wikipedia defines a connection on a vector bundle as $\nabla : \Gamma(E) \to \Gamma(T^*M \otimes E)$ such that it satisfies the Leibniz rule.

This question is more about the properties of $\Gamma$ and $\otimes$ than it is about connections, but I would be interested in understanding how I can go from $\nabla : \Gamma(E) \to \Gamma(T^*M \otimes E)$ to the one given by Tu?

Do I need that $T^*M\otimes E \cong \mathrm{Hom}(TM,E)$ and is there some property that says something about the Hom functor or the tensor product under $\Gamma$?

Jonathan
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2 Answers2

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I am assuming that everything is smooth.

The tensor product is taken fibre wise for $T^* M \otimes E $. Now we have $T_p M^* \otimes E_p = \mathrm{Hom} (T_pM, E_p)$ for every $p\in M$ which gives us a vector bundle isomorphism $T^*M \otimes E = \mathrm{Hom} (TM,E)$.

A section $ s \in \Gamma(\mathrm{Hom} (T M, E) )$ is the same as a vector bundle homomorphism $TM \to E$ (see here). Under this identification $s$ sends $v_p \in TM$ to $s_p (v_p) \in E_p$.

A vector bundle homomorphism $F: TM \to E$ induces a $C^\infty$-linear map $\tilde{F}: \Gamma(TM) \to \Gamma (E)$ by setting $\tilde{F}(\sigma)= \sigma \circ F$. The converse is also true and known as the "Bundle Homomorphism Characterization Lemma" (this is how we can go from Tu's definition to Wikipedias).

So if we apply the Wikipedia definition of the connection $\nabla^\prime$ to a section $s \in \Gamma(E)$ we get an element $\nabla^\prime s \in \Gamma (T^*M\otimes E) = \Gamma(\mathrm{Hom} (T M, E) )$. By the above we receive an induced $C^\infty$-linear map $ \tilde{\nabla^\prime s}: \mathfrak{X}(M) \to \Gamma(E)$.

We can use this map to define a map $\nabla:\mathfrak{X}(M) \times \Gamma (E) \to \Gamma(E) $ by setting $\nabla (X, s) = \tilde{\nabla^\prime s} (X)$ which satisfies all the properties of Tu's definition of a connection (you forgot to mention the Leibnitz rule in his definition i think).

jd27
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Here's my take: Given a section $s \in \Gamma(E)$, its covariant derivative is a bundle map \begin{align*} \nabla s: T_pM &\rightarrow E_p\\ X &\mapsto \nabla_Xs \end{align*} and therefore defines a map \begin{align*} \Gamma(E) &\rightarrow \operatorname{Hom}(T_*M,E) = T^*M\otimes E\\ s &\mapsto \nabla s. \end{align*} This map is a derivation because it is $\mathbb{R}$-linear (due to constant factor and sum rules of differentiation) and for any $f \in C^\infty(M)$, $$ \nabla (fs) = (df)s + f\nabla s, $$ i.e., it satisfies the Leibniz rule.

Since $X\mapsto \nabla_Xs$ is a bundle map, for each $s \in \Gamma(E)$, it defines a $C^\infty(M)$-linear map \begin{align*} \mathfrak{X}(M) &\rightarrow \Gamma(E)\\ X &\mapsto \nabla_Xs. \end{align*} Therefore, it defines a map \begin{align*} \mathfrak{X}(M) \times \Gamma(E) &\rightarrow \Gamma(E)\\ (X,s) &\mapsto \nabla_Xs, \end{align*} where it is $\mathbb{R}$-linear with respect to $s$ and $C^\infty(M)$-linear with respect to $X$. For fixed $X$, it is also a derivation with respect to $s$.

Deane
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