$$\gdef\End{\text{ End}}$$

Let $\gamma: [0,1] \to M$ be an embedded smooth curve. (If you worry about the boundary, think of an embedded curve $(-\epsilon, 1+\epsilon) \to M$.) Let $E \to M$ be a vector bundle, $\nabla$ be a connection. Our goal is to define the isomorphism $$P_\gamma: E_{\gamma(0)} \to E_{\gamma(1)}. $$ Suppose $u_0 \in E_{\gamma(0)}$, we want to find a section $u_t \in E_{\gamma(t)}$, such that $$ \nabla_{\dot \gamma(t)} u_t = 0. $$ Namely, we should have a 'constant' (or flat) section $t \mapsto u_t$, living over the image of $\gamma$.

The above statement is correct morally, however, $\nabla_{\dot \gamma(t)} u_t$ notation is problematic since $u_t$ is only a section living on a line, not on an open set of $M$. There are two ways to make this rigorous.

1. One is to go to a coordinate patch. Say image of $\gamma$ is contained in a trivializing patch $U$ of $E$, and we have $\{e_\alpha\}$ a frame of $E$, and $x^1, \cdots, x^n$ are base coordinate, then we may express a section using the coefficients $$u_t = \sum_{\alpha} u^\alpha(t) e_\alpha(\gamma(t)) \in E_{\gamma(t)}.$$ The collection of coefficients $u^\alpha(t)$ should satisfy a system of ODE $$ \frac{d u^\alpha(t)}{dt} + \Gamma^\alpha_{i \beta}(\gamma(t)) \dot \gamma^i(t) u^{\beta}(t) = 0.$$
2. The second way, is to define the pullback bundle $\gamma^*E$ and the pull-back connection $\gamma^*\nabla$. In fact, this can be defined more generally. Let $(M, E, \nabla)$ be a bundle with connection, and $F : N \to M$ be a smooth map. We can define $F^*E$ the pull-back bundle on $N$, by setting $(F^*E)_p = E_{F(p)}$ for any $p \in N$, and we can define the connection on $F^*E$ by setting $$(F^*\nabla)_{X_p} (F^* s) = (\nabla_{F_*(X_p)} s)|_p $$ where $s$ is a section of $E$ defined near $F(p)$, and $X_p$ is a tangent vector in $T_p N$. see this mathoverflow discussion for why this defines the pullback connection.

Prop 3.3.8 [Ni] We may extend $\nabla: \Omega^0(M, E) \to \Omega^1(M, E)$ to $\nabla: \Omega^k(M, E) \to \Omega^{k+1}(M, E)$, such that it satisfies the Leibniz rule. If $\omega \in \Omega^r(M)$ and $u \in \Omega^s(M, E)$, then $$ d^\nabla( \omega \wedge u) = d(\omega) \wedge u + (-1)^{|\omega|}\omega \wedge \nabla(u). $$

([Ni] uses $d^\nabla$ for this extension, where I still use $\nabla$.)

Prop For any smooth function $f \in C^\infty(M)$ and $\omega \in \Omega^r(M, E)$, we have $$ (\nabla^2) (f \omega ) = f \nabla^2(\omega) $$

Proof: This is a calculation worth doing, $$ \nabla( df \omega + f \nabla(\omega)) = dd(f) \omega - df \nabla(\omega) + df \nabla(\omega) + f \nabla^2(\omega) = f \nabla^2(\omega). $$

Recall that, if a map $a: \Omega^k(M, E) \to \Omega^{k+s}(M, E)$ is a $C^\infty(M)$-linear map, then the action of $a$ is point-wise (no derivative of section of $E$ is neede). In other word, we may view $a \in \Omega^s(M, \End(E))$.

Curvature We define the curvature $F_\nabla = \nabla^2 \in \Omega^2(M, \End(E))$

Suppose we have a local trivialization $\{e_\alpha\}$ over $U \In M$. Then, we have an induced trivial connection $d_U$ on $E|_U$, and we may write $\nabla|_U = d_U + A$, for some $A \in \Omega^1(U, \End(E))$. We then have $$ F_\nabla|_U = (\nabla|_U)^2 = (d_U + A)^2 = d_U A + A \wedge A $$ where we define $\wedge: \Omega^r(M, \End(E)) \times \Omega^s(M, \End(E)) \to \Omega^{r+s}(M, \End(E)) $ by $\wedge$ on the form factor, and compose on the associative algebra factor $\End(E)$ $$ (\eta \ot a) \wedge (\xi \ot b) = (\eta \wedge \xi) \ot (ab), \quad \eta, \xi \in \Omega^*(M), a,b \in\Omega^0(\End(E)). $$

In the following, $\omega$ denote $k$-form, $u$ denote section of $E$. $X, Y, Z$ are vector fields. We have $$ \nabla_X(\omega \ot u) = L_X(\omega) \ot u + \omega \ot \nabla_X(u) $$ $$ i_X \nabla + \nabla i_X = \nabla_X $$ $$ i_X i_Y + i_Y i_X = 0$$ $$\nabla_X i_Y - i_Y \nabla_X = i_{[X,Y]}$$

For example, we test the last formula on $\Omega^1(M, E)$, $\eta \in \Omega^1(M)$, $u$ a section of $E$, $$ (\nabla_X i_Y - i_Y \nabla_X) (\eta \ot u) = \nabla_X( \eta(Y) \ot u) - i_Y (L_X(\eta)\ot u + \eta \nabla_X u) $$ $$= L_X( \eta(Y)) \ot u + \eta(Y) \ot \nabla_X(u) - [Y (\eta(X)) + d\eta(X, Y)] \ot u - \eta(Y) \ot \nabla_X(u) $$ $$ = \iota_{[X,Y]}(\eta) \ot u $$

Some formula about curvature $$ F(X, Y) = [\nabla_X, \nabla_Y] - \nabla_{[X,Y]}$$ In local coordinates $x_i$ on $U$, we have $$F_{ij} = - F_{ji} = [\nabla_i, \nabla_j]$$ where $\nabla_i = \nabla_{\d_i}$.