# Inconsistency in the expression for $d\alpha= d\alpha_k \wedge d\xi^k$ contracted with pairs of vector fields

Classical Dynamics: A Contemporary Approach, Jorge v. Jose and Eugene J. Saletan,(1998) states in page 227 that

Globally $$d\alpha$$ is defined by its contraction with pairs of vector fields: $$i_Xi_Yd\alpha=i_Xd(i_Y\alpha)-i_Yd(i_X\alpha)-i_{[X,Y]}\alpha.$$

where $$d\alpha= d\alpha_k \wedge d\xi^k$$ and $$\alpha=\alpha_kd\xi^k$$. But when tried to derive the expression using the definition of Lie derivative got different result. Don't know what's gone wrong, so I'm just replicating my calculation here. As defined in N.M.J. Woodhouse's Introduction to Analytical Dynamics, page 199:

For a $$k$$-form, the lie derivative along $$v$$ is denoted by $$\mathcal{L}_v \alpha$$ and is defined by $$\mathcal{L}_v\alpha=i_vd\alpha+d(i_v\alpha).$$ It has a key propertie, which hold for any $$u, v \in \mathfrak{X}(M)$$ and for any forms $$\alpha$$ and $$\beta$$: $$\mathcal{L}_u(i_v\alpha)=i_{[u,v]} \alpha+i_v\mathcal{L}_u\alpha$$

So by definition, $$i_Yd\alpha=\mathcal{L}_Y\alpha-d(i_Y\alpha)$$$$\Rightarrow i_Xi_Yd\alpha=i_X(\mathcal{L}_Y\alpha)-i_Xd(i_Y\alpha)=\mathcal{L}_Y(i_X\alpha)-i_{[Y,X]}\alpha-i_Xd(i_Y\alpha)$$$$\Rightarrow i_Xi_Yd\alpha=i_Yd(i_X\alpha)-i_{[Y,X]}\alpha-i_Xd(i_Y\alpha)$$ Which is almost opposite to what was expected. Can anyone help me figure out what I've done wrong?

• Well, I think, my calculation is right according to this note, page 12. – Minotaur 3 hours ago