Formula for the differential of a left-invariant 1-form

Question

I am currently working through the first volume of spivak's differential geometry and am currently on the chapter on Lie groups. After showing that the differential of a left invariant form is itself left invariant, he then does the following

$d\omega(\tilde{X}, \tilde{Y})=\tilde{X}(\omega(\tilde{Y}))-\tilde{Y}(\omega(\tilde{X}))-\omega([\tilde{X},\tilde{Y}])$ $=-\omega([\tilde{X},\tilde{Y}])$

to demonstrate a simplified formula for the differential of a 1-form.

Here, $\omega$ is left-invariant 1-form and $\tilde{X}$, $\tilde{Y}$ are left-invariant vector fields.

While I can understand the first equality (it is simply the coordinate-free formula for a 1-form), I can't understand what motivates the second equality.

Is there a reason the two left terms should be equal or is it that they both go to zero for left-invariant vector fields?

Answer 1

Suppose $X$ and $Y$ are left invariant vector fields, and that $\omega$ is a left invariant $1$-form. Then \begin{align} \forall g \in G,~ X(g) &= g\cdot X(e) \\ Y(g)& = g\cdot Y(e) \\ \omega_g(X(g))&= \omega_e(g^{-1}X(g)) = \omega_e(g^{-1}g\cdot X_e) = \omega_e(X_e)\\ \omega_g(Y(g))&= \omega_e(g^{-1}Y(g)) = \omega_e(g^{-1}g\cdot Y_e) = \omega_e(Y_e) \end{align}

Then, $g \mapsto \omega(X) (g)$ and $g\mapsto \omega(Y)(g)$ are constant functions, and thus, $X\cdot \omega(Y) = Y\cdot\omega(X) = 0$. Hence $$ \mathrm{d}\omega(X,Y) = X\cdot \omega(Y) - Y\cdot \omega(X) - \omega\left([X,Y]\right) = - \omega([X,Y]) $$