I was reading about connections on Wikipedia and I found this section about "Motivation":
My question is the following one. If I look at the vector bundle $E$ like a manifold, then locally I have charts and I can think of $X: M \rightarrow E$ like a smooth map between open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ for some $n$ and $m$. Hence, I should naturally have a definition of derivatives. Am I wrong? What is the point of that argument?
Thanks in advance
To make the story below sound plausible, you should first convince yourself that any reasonable notion of differentiation is going to be a map $\nabla:\Gamma(E)\to\Gamma(T^*M\otimes E)$, with some additional properties. That is, given a section $s$, we get $\nabla s$, and we can feed it a vector field $\nabla_Xs$ to "differentiate in the direction of $X$". Furthermore, it should satisfy the Leibniz rule $\nabla(fs)=df\otimes s+f\nabla s$, and it should satisfy $\nabla_{fX}(s_1+s_2)=f(\nabla_Xs_1+\nabla_Xs_2)$. These are all properties satisfied by the usual derivative. More generally, $\nabla$ will be called a connection.
Let $E\to M$ be a vector bundle for which we want to define a connection. and let $U\subset M$ be a trivialising open subset, so we have $E|_U\cong U\times\mathbb{R}^r$ via a diffeomorphism $\Phi:E|_U\to U\times\mathbb{R}^k$ (which is fibrewise linear). Let's also suppose that $U$ is a chart, so we additionally have a diffeomorphism $\varphi:U\to\varphi(U)\subset\mathbb{R}^n$. Let $s\in\Gamma(E)$ and $v\in T_xM$. Then we can say that $\varphi\times\text{id}\circ\Phi\circ s\circ \varphi^{-1}:\varphi(U)\to U\to E|_U\to U\times\mathbb{R}^r\to\varphi(U)\times\mathbb{R}^r$ which can be identified with a function $f:\varphi(U)\to\mathbb{R}^r$, and you are suggesting that we use the linear structure on $\varphi(U)\times\mathbb{R}^r$. As you can tell from the number of arrows I wrote, there are a lot of identifications to be made to try to do this, in the first place, which is a good indication that such a construction is, at any rate, not instrinsic - if it possible at all.
After making these identifications, we can locally define $$ds(v)_x=\lim_{t\to 0}\frac{s(\gamma(t))-s(\gamma(0))}{t}$$ I have neglected to write out the various chart maps and left the identifications implicit. What we've done here, is defining an element $ds\in \Gamma(T^*U\otimes E|_U)$. It takes a tangent vector (or vector field) and gives us a new section of $E|_U$. If you read a bit more about connections, you will find the definition of the local connection $1$-form $A\in\Gamma(U,\text{End}(E|_U))$ which in our case is $0$, as it is the $1$-form such that $\Phi\circ\nabla\circ\Phi^{-1}=d+A$, and you wanted to define a derivative in the "naive" way. Why does such a $1$-form $A$ exist? It exists because the difference between any two connections lies in $\Gamma(T^*M\otimes \text{End}(E))$. You can verify this by hand quite easily, using the Leibniz rule. On the trivialisation, we the naive connection $d$, which you want to define globally (denoted $\nabla$), and so there exists $A\in\Gamma(T^*U\otimes\text{End}(E|_U))$ such that $\Phi\circ\nabla\circ\Phi^{-1}-d=A$, i.e. $\Phi\circ\nabla\circ\Phi^{-1}=d+A$.
In order to really give a connection with the properties that you would expect of a derivative, the local $1$-form must satisfy $$A_V=\Psi^{-1}\circ A_U\circ\Psi+\Psi^{-1}d\Psi$$ for any trivialising $V\subset M$ with $U\cap V\neq\emptyset$, where $\Psi$ is the transition function between $U\times\mathbb{R}^r$ and $V\times\mathbb{R}^r$. Does the naive definition satisfy this rule? The answer is clearly no, because we would have $A_V=0\neq \Psi^{-1}d\Psi$. That is, unless $d\Psi=0$, which generally is not the case.