Given a connection $\nabla$ on a vector bundle $E$ over a smooth manifold $M$, we know there is a unique extension of $\nabla$ to all tensor bundles of $E$ that satisfies Leibniz rule and contraction. I am going to prove this.
We can first define the connection on the dual space of $E$ using the formula forced by the axioms above. Then the rest is essentially just proving one lemma:
Given two bundles $E$, $F$ on $M$ and connections on $E$ and $F$, both denoted by $\nabla$, then there is a unique connection $\nabla$ on $E\otimes F$ such that
$$\nabla_X(s_E\otimes s_F)=\nabla_X s_E\otimes s_F+s_E\otimes \nabla_X s_F$$
I tried to invoke the universal property of tensor product as usual, to use bilinearity to prove well-definedness. However this tensor product of sections is not the strict tensor product in linear algebra tense (it takes tensor product pointwise, and it cannot be understood as a tensor product of two $\mathbb R$-vector spaces. It can be tensor product of $C^\infty(M)$ modules though, but $\nabla_X$ is not $C^\infty$ linear). So I get in problem here, and I ask for a conceptual way to show why $\nabla_X$ is a well defined map from the space of sections of $E\otimes F$ to itself.
Best Answer
It seems to me that people usually include the $\mathbb{R}$-linearity as part of the definition (correct me if I'm wrong); i.e. define $\nabla$ on the simple tensors by the product rule above, and then extend it linearly over $\mathbb{R}$ to the whole space $C^{\infty}(E\otimes F)$.
Actually, due to the fact that $C^{\infty}(E\otimes F)\simeq C^{\infty}(E)\otimes_{C^{\infty}(M)}C^{\infty}(F)$ which OP has mentioned in the comment above, it follows that additivity (i.e. \begin{align} \nabla_X(s\otimes t+s'\otimes t')=\nabla_X(s\otimes t)+\nabla_X(s'\otimes t') \end{align} which is weaker than $\mathbb{R}$-linearity) is sufficient.
With this in mind, here is an alternative proof, which is essentially a direct computation, though perhaps it may look less elegant.
Let $A\in C^{\infty}(E\otimes F)$. Such $A$ can be expressed as a finite sum of simple tensors, though not uniquely in general. So we want to show that if it can be written in the following two ways \begin{align} A=\sum_is_i\otimes t_i=\sum_j\tilde{s}_j\otimes\tilde{t}_j \end{align} (the number of summands in these two expressions may be different in general), then \begin{align} \nabla_X\left(\sum_is_i\otimes t_i\right) =\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) & & (*) \end{align} Let $(e_{\alpha})$ and $(\epsilon_{\beta})$ be smooth local frames for $E$ and $F$ respectively. Then one can write \begin{align} s_i=s_i^{\alpha}e_{\alpha},\qquad t_j=t_j^{\beta}\epsilon_{\beta},\qquad \tilde{s}_i=\tilde{s}_i^{\alpha}e_{\alpha},\qquad \tilde{t}_j=\tilde{t}_j^{\beta}\epsilon_{\beta} \end{align} (Einstein summation convention is assumed.) Then we will have \begin{align} \sum_is_i\otimes t_i =\left(\sum_is_i^{\alpha}t_i^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta}, \qquad \sum_j\tilde{s}_j\otimes\tilde{t}_j =\left(\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta} \end{align} Since $\{e_{\alpha}\otimes\epsilon_{\beta}\}$ is a smooth local frame for $E\otimes F$, by uniqueness of local components we must have \begin{align} \sum_is_i^{\alpha}t_i^{\beta}=\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta} & & (1) \end{align} Taking exterior derivative we then have \begin{align} \sum_i\left(t_i^{\beta}ds_i^{\alpha}+s_i^{\alpha}dt_i^{\beta}\right) =\sum_j\left(\tilde{t}_j^{\beta}d\tilde{s}_j^{\alpha} +\tilde{s}_j^{\alpha}d\tilde{t}_j^{\beta}\right) & & (2) \end{align} Now let $\omega_{\alpha}^{\beta}$ and $\theta_{\alpha}^{\beta}$ be the connection 1-forms of $\nabla^E$ and $\nabla^F$ respectively, so that e.g. \begin{align} \nabla_X s_i=\left[Xs_i^{\alpha}+s_i^{\beta}\omega_{\beta}^{\alpha} (X)\right]e_{\alpha} & & (3) \end{align} and similar identities hold for $\nabla_Xt_i$, $\nabla_X\tilde{s}_j$ and $\nabla_X\tilde{t}_j$. Then we can compute \begin{align} &\nabla_X\left(\sum_is_i\otimes t_i\right) \\ %%% &=\sum_i\left(\nabla_Xs_i\otimes t_i+s_i\otimes\nabla_Xt_i\right) \\ %%% &=\underbrace{\left(\sum_i\left((Xs_i^{\alpha})t_i^{\beta} +s_i^{\alpha}(Xt_i^{\beta})\right)\right)}_{=:I}e_{\alpha}\otimes\epsilon_{\beta} +\underbrace{\left(\sum_is_i^{\gamma}t_i^{\beta}\right)}_{=:II} \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta} +\underbrace{\left(\sum_is_i^{\alpha}t_i^{\gamma}\right)}_{=:III} \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta} \end{align} where the first step follows by additivity and product rule, while the second step is obtained by substitution of (3) and rearranging the terms.
Now $\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)$ will have the same expression, except that all of the $s_i,t_i$ are replaced by $\tilde{s}_j,\tilde{t}_j$; says \begin{align} \nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) =\tilde{I}\cdot e_{\alpha}\otimes\epsilon_{\beta} +\tilde{II}\cdot \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta} +\tilde{III}\cdot \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta} \end{align} Then:
Hence, (*) holds as desired.