Let $M$ be a manifold, let $L$ be the trivial line bundle over $M$. Define a connection on $L$ by $\nabla:\Gamma(L)\rightarrow \Omega^1(L)$, by $\nabla_X(s)=ds(x)$. Define the curvature $K(\nabla)$ of a connection by $K(\nabla)(X,Y)(s):=\nabla_X \nabla_Y(s)-\nabla_Y \nabla_X (s) - \nabla_{[X,Y]}(s)$.
Show that for $\nabla_X(s)=ds(X)$, $F(\nabla)=0$.
If we use the equivalent definition of the curvature of $F(\nabla)=\nabla \circ \nabla$ it is easy to see that our connection is flat as $d\circ d=0$. But I want to know how to compute these concretely. Please tell me where I go wrong.
$$K(\nabla)(X,Y)(s):=\nabla_X \nabla_Y(s)-\nabla_Y \nabla_X (s) - \nabla_{[X,Y]}(s)$$ $$=d_X d_Y(s)-d_Y d_X (s) - d_{[X,Y]}(s)$$ $$=d_X ds(Y)-d_Y ds(X) - ds({[X,Y]})$$ I am unsure about this step. $ds$ is a one-from so it takes in vectors and returns scalars. Or alternatively it takes in a vector field and returns a smooth function. So $ds(X)$ is a smooth function. Since $d_X$ acts on smooth functions we have. $$=d(ds(Y)) X-d(ds(X))(Y) - ds({[X,Y]})$$
I think for some reason we can chance some brackets around and say $d(ds(Y)) X=d ds(Y)(X)=0$ as $d\circ d=0$.
$$=- ds({[X,Y]})$$ I don't think this is true but it looks like the only thing left to do $$=- ds(XY-YX)$$
$$=- ds(XY)+ds(YX)$$
I'm sure what I've done is incorrect in some way. If you could tell me how that would be helpful.