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Today and Friday, we will follow Nicolascu's note 3.3, and discuss connection on vector bundle.
Suppose we are given a smooth manifold $M$. Let $f$ be a smooth function on $M$. We know two things:
So, why things are different when we talk about a vector bundle $\pi: E \to M$? Let's think about the following thing:
Of course, “yes, we can!”, you may say, since we can find a local trivialization of $E$ over $U$ $$ \pi^{-1}(U) \cong U \times \R^k $$ then a section $\sigma$ over $U$ can be described by an $k$-tuple of functions $(\sigma_1, \cdots, \sigma_k): U \to \R^k$, and we all know how to take derivative of functions. Any objection?
The above approach is problematic. How do you know the answer you get does not depends on the local trivialization? Giving a local trivialiation is like giving a local frame $(e_1, \cdots, e_k)$ of the vector bundle. If we use the above prescription to define 'taking derivatives' of a section, then, we are declaring these local sections $e_1, \cdots, e_p \in \Gamma(U, E)$ are 'constant'. That is a huge bias, a huge choice to make. In other words, the notion of 'constant section' would depends on the choice of the trivialization. Hence, the approach in the previous paragraph is not right (in general).
You may remember that we have encountered the same problem, when we try to take derivative of a tangent vector field $X$, and we found a solution called 'Lie derivative'. Here, we introduce the notion of connection and covariant derivatives.
$\gdef\vect\z{Vect} \gdef\lcal{\mathcal L}$ Let $\pi: E \to M$ be a vector bundle. A connection should satisfies the following
$$ \nabla_X(f \sigma) = \nabla_X(f) \sigma + f \nabla_X(\sigma) = X(f) \sigma + f \nabla_X(\sigma).$$