Proving that bi invariant forms are closed

Question

If $\omega$ is a bi invariant form on a lie group, I want to prove that $d\omega=0$. So first I am trying to prove this in the case of bi invariant $1$ form. But $d\omega(X,Y)=\frac{1}{2}(X\omega(Y)-Y\omega(X)-\omega[X,Y])$. Now using left invariance we get $X\omega(Y)=0,Y\omega(X)=0$. Now I am struck as to why $\omega[X,Y]$ is zero. Thanks.

Answer 1

Biinvariant is equivalent to say that the form is invariant by left and right multiplication. This implies that $Ad(g)^*\omega=\omega$. We deduce that $Ad(exp(tX)^*\omega=\omega$. Differentiate this relatively to $t$, you obtain for every $Y$, $\omega([X,Y])=0$.

Answer 2

Spivak's Book "A comprehensive Introduction to Differential Geometry, Volume 1" gives a different proof (page 395):

He first shows that $\psi^*\omega_e = (-1)^k\omega_e$ for $\psi(a) = a^{-1}$. Then he concludes $\psi^* \omega = (-1)^k\omega$, because $\psi^*\omega$ is also left-invariant. And finally because $\mathrm{d}\omega$ is also bi-invariant: $(-1)^{k+1} \mathrm{d}\omega = \psi^*(\mathrm{d}\omega) = \dots = (-1)^k \mathrm{d}\omega$. Hence $\mathrm{d}\omega = 0$.