A connection on a vector bundle $E$ is a map
$ D:\Gamma(E)\rightarrow \Gamma(T^*(M)\otimes E)$ satisfying
1) For any $s_1,s_2\in \Gamma(E)$,
$D(s_1+s_2)=Ds_1+Ds_2$
2)
For $s\in \Gamma(E)$ and $\alpha\in C^{\infty}(M)$,
$D(\alpha s)=d\alpha \otimes s + \alpha Ds$
I am having trouble understanding these conditions.
I don't understand the notation $d\alpha \otimes s$. I have only ever dealt with tensor products of linear functionals. Ie $v^*\otimes w^*(v,w)=v^*(v)w^*(w)$. The quantity $d\alpha \otimes s$ is the tensor product of the differential of $\alpha$ and a smooth section of $E$. How is this quantity defined? Can anyone shed any light here? Thanks in advance.
Let $E \stackrel{\pi}{\to} M$ be the vector bundle, then given a smooth section $s \in \Gamma(E)$, what does the connection output when applied to $s$? We need an element of $\Gamma(T^* M \otimes E)$, which is a section of the tensor product bundle $T^* M \otimes E$. This is given by the following data: for any point $x \in M$ in the base space, $(Ds)(x) = \omega \otimes v$, where $\omega \in T^*_xM$ and $v \in \pi^{-1}(x)$. That is, $(Ds)(x)$ consists of a covector at $x$ and an element in the fibre over $x$.
Furthermore, if $s \in \Gamma(E)$ and $\alpha \in C^{\infty}(M)$, then the product rule says that $D(\alpha s) = d\alpha \otimes s + \alpha Ds$. That is, for any point $x \in M$ in the base space, $$D(\alpha s)(x) = (d \alpha)(x) \otimes s(x) + \alpha(x) (Ds)(x).$$ Does everything live in the appropriate space? Well, $d\alpha$ is a 1-form on $M$ so $(d\alpha)(x) \in T^*_x M$ and $s(x) \in \pi^{-1}(x)$, so we get that $(d\alpha)(x) \otimes s(x) \in T^*_x M \otimes \pi^{-1}(x)$. How about the other term? $\alpha(x) \in \mathbb{R}$ so we are just scaling the tensor product $(Ds)(x) \in T_x^* M \otimes \pi^{-1}(x)$.