Maurer-Cartan 1-form

Can anyone help me with the following? Let $\rho$ be the right-invariant Maurer-Cartan 1-form

$$\rho = dg\ g^{-1}$$

I want to show that the MC equation

$$d\rho - \rho \wedge\rho = 0$$

holds.

So I compute

$$d\rho = -dg \wedge d(g^{-1}) = dg\wedge g^{-1}dg\ g^{-1}$$

Why am I allowed to take the $g^{-1}$ through the wedge to get the right result? Naively it seems this should be wrong, because the wedge is essentially a commutator of matrices. Or is my notation too simplistic.

I'm aware that I can get this result more generally by considering the structure equation for right-invariant forms, but ideally I'd like to make this direct computation rigorous, if possible!

Many thanks in advance!

Solution 1:

The question comes down to prove that:

$$ \omega g\wedge \phi = \omega\wedge g\phi ,$$

where $\omega$ and $\phi$ are Lie algebra-valued differential forms, and g is a group operation.

If (Einstein summation convention) $\omega=\omega_i\,dx^i$ and $\phi=\phi_i\,dx^i$, where $\omega_i, \phi_i$ are non-commuting "matrices" (Lie algebra elements), then:

$$ \omega g\wedge \phi = (\omega_i\,g\,\phi_j)\,dx^i\wedge dx^j = \omega\wedge g\phi .$$

Inside the parentheses above there is a "matrix" product. As you see, the components of the forms, which are in the Lie algebra, are not interchanged. The $dx^i$ are the same as for usual differential forms.