I have to prove the following statement for a 1-form $\alpha\in\Omega^{1}(\mathcal{M})$ and two vector fields $X,Y\in\mathfrak{X}(\mathcal{M})$:
$$\mathrm{d}\alpha (X,Y) = X(\alpha(Y))-Y(X(\alpha))-\alpha([X,Y])$$
For this proof I have to use the general formula
$$\mathcal{L}_{u}(\omega(v_{1},\dots,v_{n})) = (\mathcal{L}_{u}\omega)(v_{1},\dots,v_{n}) + \omega(\mathcal{L}_{u}v_{1},v_{2},\dots,v_{n}) + \omega(v_{1},\mathcal{L}_{u}v_{2},v_{3},\dots,v_{n})+\dots +\omega(v_{1},\dots,\mathcal{L}_{u}\dots,v_{n})$$
for a k-form $\omega\in\Omega^{k}(\mathcal{M})$ and vector fields $u,v_{1},\dots,v_{n}\in\mathfrak{X}(\mathcal{M})$.
Using this formula for my special case of a 1-form $\alpha$ I received
$$\mathcal{L}_{X}(\alpha(Y))=(\mathcal{L}_{X}\alpha)(Y)+\alpha(\mathcal{L}_{X}Y)$$
Using:
(1) $\alpha(\mathcal{L}_{X}Y) = \alpha([X,Y])$
(2) $\alpha(X)\in\mathcal{C}^{\infty}(\mathcal{M})$ $\Rightarrow$ $\mathcal{L}_{X}(\alpha(Y))=X(\alpha(Y))$
(3) $(\mathcal{L}_{X}\alpha)(Y) = (\mathrm{d}i_{X} + i_{X}\mathrm{d})\alpha(Y) = \mathrm{d}\alpha(X,Y) + i_{X}\mathrm{d}\alpha(Y)$
I get the formula:
$$\mathrm{d}\alpha(X,Y) = X(\alpha(Y)) - i_{X}\mathrm{d}\alpha(Y) -\alpha([X,Y]) $$
So to finish the proof, I have to show that $i_{X}\mathrm{d}\alpha(Y) = Y(X(\alpha))$....But I can`t see why this is the case...
Can someone explain it to me?
Thank you!
First you should check the formula you are trying to prove. I think it is $$\mathrm{d}\alpha (X,Y) = X(\alpha(Y))-\underline{Y(\alpha(X))}-\alpha([X,Y]),$$ since $X(\alpha)$ doesn't make sense.
Next as Ted Shifrin said, you should check your computation for (3). You should do $$(\mathcal L_X \alpha)(Y) = [(\mathrm d i_X + i_X \mathrm d)\alpha] (Y) = [\mathrm d (i_X\alpha)](Y) + (i_X\mathrm d\alpha)(Y).$$ I put extra brackets to help clarify the order of operations.
For the first term, $i_X\alpha = \alpha(X)$ is a smooth function. Then using $\mathrm d f(X) = Xf$ for any function $f$, $$[\mathrm d (i_X\alpha)](Y) = [\mathrm d(\alpha(X)](Y)=Y\alpha(X).$$ For the second term, $\mathrm d\alpha$ is a 2-form. Using $i_X\omega(Y) = \omega(X,Y)$ for any 2-form $\omega$, $$(i_X\mathrm d\alpha)(Y) = \mathrm d\alpha(X,Y).$$