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:

1. We know what it means for $f$ to be constant.
2. Given a tangent vector $v_p \in T_p M$, we know how to take directional derivative $v_p(f)$.

So, why things are different when we talk about a vector bundle $\pi: E \to M$? Let's think about the following thing:

1. Given a local section $\sigma \in \Gamma(U, E)$ for an open set $U \In M$, what does it mean when we say $\sigma$ is a 'constant section'? Or can we say it? (recall a section means a smooth map $\sigma: U \to E$ such that $\pi \circ \sigma = id:U \to U$.)
2. Given a point $p \in M$, $\sigma$ a section of $E$ defined in a neighborhood of $p$, and a tangent vector $v_p \in T_p M$, can we take 'directional derivative' of $\sigma$ along $v_p$?

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.

$$ \gdef\vect{\text{Vect}} \gdef\lcal{\mathcal L} $$ Let $\pi: E \to M$ be a vector bundle. A connection should satisfies the following

1. Data: $\nabla: \vect(M) \times C^\infty(M, E) \to C^\infty(M, E)$, $(X, \sigma) \mapsto \nabla_X(\sigma)$.
2. $\C^\infty(M)$-linear: suppose $f \in C^\infty(M)$, we want $\nabla_{fX}(\sigma) = f \nabla_X(\sigma)$. Note that Lie derivative does not satisfy this property: $\lcal_{fX} Y = [fX, Y] = f[X,Y] + [f, Y]X = f \lcal_X Y - Y(f) X$
3. Leiniz rule: suppose $f \in C^\infty(M)$, we want

$$ \nabla_X(f \sigma) = \nabla_X(f) \sigma + f \nabla_X(\sigma) = X(f) \sigma + f \nabla_X(\sigma).$$

One can concisely reformulate the above conditions, using differential one-form. A connection is the following data $$ \nabla: C^\infty(M, E) \to C^\infty(M, T^* M \otimes E) $$ such that $$ \nabla(f \sigma) = df \ot \sigma + f \nabla(\sigma), \quad \forall f \in C^\infty(M) $$ where $T^*M \ot E$ is the tensor product of two vector bundles, and $C^\infty(M, T^* M \otimes E)$ is smooth section of it. Sometimes, we use the following notation: recall $\Omega^0(M)$ is 0-form, ie. smooth function, and $\Omega^1(M)$ is the space of 1-forms. We use $\Omega^0(M, E) = C^\infty(M, E)$, and $\Omega^1(M,E) = C^\infty(M, T^* M \otimes E)$, then $$\nabla: \Omega^0(M, E) \to \Omega^1(M, E). $$