If $dX_1 = dX_2$ then curvatures of $\nabla^{X_1}$ and $\nabla^{X_2}$ agree

Question

Let $E \simeq M \times \mathbb C$ be a trivial smooth complex line bundle over the Riemann surface $M$ and let $S \colon M \to E$ be its smooth nowhere vanishing section. Let $\nabla^1$ and $\nabla^2$ be two covariant derivatives on $E$, then $$ \nabla^1 S=S \otimes X^1,\\ \nabla^2 S = S \otimes X^2, $$ where $X^1$, $X^2 \in \Omega^1(M)$ are called the connections' one-forms. It is stated that if $dX^1 = dX^2$ then curvatures of $\nabla^1$ and $\nabla^2$ are the same. I tried to show this by definition but I have a problem. Let $U$, $V \in \Gamma(TM)$, $f \in C^\infty(M)$. Then $$ \nabla^1_U \nabla^1_V (fS) = \nabla^1_U ( (V(f)+fX^1(V))S) \\ = \bigl( U(V(f))+U(f)X^1(V)+fU(X^1(V)) + X^1(U)(V(f)+fX^1(V)) \bigr)S $$ and hence $$ \nabla^1_U \nabla^1_V(fS)-\nabla^1_V \nabla^1_U(fS) \\ = \bigl([U,V](f)+fU(X^1(V))-fV(X^1(U)) \bigr)S, $$ and then $$ R^1(U,V)(fS) = \nabla^1_U \nabla^1_V(fS)-\nabla^1_V\nabla^1_U(fS)-\nabla^1_{[U,V]}(fS) \\ = \bigl( U(X^1(V))-V(X^1(U))- X^1([U,V])\bigr)fS $$ But how to show that this expression depends only on $dX^1$?

Answer 1

There is the following formula for the exterior derivative that you should get to know (it comes up a lot): http://en.wikipedia.org/wiki/Exterior_derivative#Invariant_formula.

In the case of a 1-form $\mu$, it says that $$ d\mu(U,V) = U\cdot \mu(V) - V \cdot \mu(U) - \mu([U,V]). $$ Therefore the last thing you wrote is exactly $dX^1(U,V)$.

This formula can be proved using Cartan's formula: $$ d i_U \mu + i_U d \mu = L_U \mu. $$ Rearranging gives $$ i_U d\mu = L_U \mu - d(\mu(U)). $$ Now apply to a vector field $V$ and get $$ d\mu(U,V) = (L_U \mu)(V) - V\cdot \mu(U). $$ Now we use the fact that the Lie derivative satisfies the product rule: $$ U\cdot (\mu(V)) = L_U (\mu(V)) = (L_U \mu)(V) + \mu(L_U V) = (L_U \mu)(V) + \mu([U,V]). $$ So $(L_U \mu)(V) = U\cdot(\mu(V)) - \mu([U,V])$, as desired.