Prove that the curvature of a connection on a line bundle is a global two form For a connection $\nabla$ on a line bundle, in a local trivialisation, the connection looks like a one form $\nabla s=ds + sa$ but this is not a proper one form cause it depnds on the choice of local trivialization.


*

*Why does the curvature $da$ not depend on the choice of local trivialization??

*Why does $da(X,Y)s=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$?

 A: Recall that for a line bundle $L$, we can take local trivialization $\phi:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb C$. Take local frame (one section, since line bundle is of rank 1), say $s_\alpha=\phi^{-1}(\pi(\cdot),1)$, then any section of $L$ can be represented by $s|_{U_\alpha}=f_\alpha s_\alpha$, for some complex function $f_\alpha$. The transition functions are defined on $U_{\alpha\beta}:=U_\alpha\cap U_\beta$, as
$$
s_\alpha=g_{\alpha\beta}s_\beta.
$$
Thus, for the coordinates of $s$:
$$
f_\beta=g_{\alpha\beta}f_\alpha.
$$
Locally, a connection of $L$ can be writen as:
$$
\nabla=d+A_\alpha,
$$
where $A_\alpha$ are complex valued 1-form over $U_\alpha$.
In fact,
$$
[\nabla s]|_{U_\alpha}=\nabla(f_\alpha s_\alpha)=df_\alpha s_\alpha+f_\alpha\nabla s_\alpha,
$$
thus, if we define $\nabla s_\alpha:=A_\alpha s_\alpha$, then we get
$$
[\nabla s]|_{U_\alpha}=(df_\alpha+f_\alpha A_\alpha) s_\alpha,
$$
which is the precise mean when we write $\nabla=d+A_\alpha$.
We need to know the transition relation of $A_\alpha$. This can be compute as follows:
$$
A_\alpha g_{\alpha\beta}s_\beta=A_\alpha s_\alpha=\nabla s_\alpha=\nabla(g_{\alpha\beta} s_\beta)=dg_{\alpha\beta} s_\beta+g_{\alpha\beta}A_\beta s_\beta.
$$
Thus,
$$
A_\alpha g_{\alpha\beta}=dg_{\alpha\beta}+g_{\alpha\beta}A_\beta\implies
A_\beta=g_{\alpha\beta}^{-1}A_\alpha g_{\alpha\beta}-g_{\alpha\beta}^{-1}dg_{\alpha\beta}=A_\alpha-g_{\alpha\beta}^{-1}dg_{\alpha\beta}.
$$
To move on, I will use a equivalent definition of curvature: $F_\nabla=D^2$, where $D$ is the exterior differential. In particular, by the definition of $D$, we know that(locally)
\begin{align*}
D^2s&=D(\nabla s)=D[(d f_\alpha+f_\alpha A_\alpha)s_\alpha]\\
&=d(df_\alpha+A_\alpha f_\alpha)s_\alpha+(-1)^1(df_\alpha+A_\alpha f_\alpha)\wedge\nabla s_\alpha\\
&=[dA_\alpha f_\alpha-A_\alpha\wedge df_\alpha-(df_\alpha+A_\alpha f_\alpha)\wedge A_\alpha]s_\alpha\\
&=[dA_\alpha f_\alpha-A_\alpha\wedge df_\alpha+A_\alpha\wedge(df_\alpha+A_\alpha f_\alpha)]s_\alpha\\
&=[dA_\alpha+A_\alpha\wedge A_\alpha]f_\alpha s_\alpha\\
&=(dA_\alpha+A_\alpha\wedge A_\alpha)s.
\end{align*}
Therefore,
$$
F_\nabla|_{U_\alpha}=dA_\alpha+A_\alpha\wedge A_\alpha.
$$
Note that $A_\alpha$ is just a complex valued 1-form (in general it is a matrix valued 1-form), we have$A_\alpha\wedge A_\alpha=0$ and
$$
F_\nabla|_{U_\alpha}=dA_\alpha.
$$
Now, we are ready to show, $F_\nabla$ is a global defined 2-form. In fact, 
\begin{align*}
F_\nabla|_{U_\beta}&=dA_\beta=d(A_\alpha-g_{\alpha\beta}^{-1}d g_{\alpha\beta})\\
&=dA_\alpha-dg_{\alpha\beta}^{-1}\wedge dg_{\alpha\beta}\\
&=dA_\alpha+g_{\alpha\beta}^{-1}dg_{\alpha\beta}g_{\alpha\beta}^{-1}\wedge dg_{\alpha\beta}\\
&=dA_\alpha+g_{\alpha\beta}^{-1}dg_{\alpha\beta}\wedge g_{\alpha\beta}^{-1}dg_{\alpha\beta}\\
&=dA_\alpha\\
&=F_\nabla|_{U_\alpha}.
\end{align*}
This answers your first question.
For the second, it is standard in Riemaniann geometry and I believe your can find in text book. 
I would like to remark that, the above computation is trivial but maybe you need to do it once again by yourself.
