# 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!

• Wedge does nothing to $g$, one should only care about the differentials. After all one can think of $dg\,g^{-1}$ as being a set of ordinary differential 1-forms. – Start wearing purple Jun 6 '13 at 11:46
• But my understanding was that $(dg \wedge g^{-1}dg\ g^{-1})(X,Y) = [dg(X), g^{-1}dg(Y)g^{-1}] \neq [dg(X)g^{-1},dg(Y)g^{-1}]$. What is the flaw in my reasoning there? And I don't quite see how one can think of them as a set of ordinary 1-forms, because they are $\mathfrak{g}$-valued! – Edward Hughes Jun 6 '13 at 11:52
• No, there's no bracket here. If you write $[\omega,\omega]$, then you use the bracket. Yes, $dg\cdot g^{-1}(A)= dg(A)g^{-1}=Ag^{-1}$ for $A\in T_gGL(n)\cong \mathfrak{gl}(n)$. – Ted Shifrin Jun 6 '13 at 14:07
• Let me elaborate. The wedge product is giving you the commutator, but "scalars" commute with wedge. – Ted Shifrin Jun 6 '13 at 14:28
• Ah thanks Ted - that second comment made it completely clear. I hadn't been thinking of the $g$ as a scalar but of course it is in this context if you look at the components! – Edward Hughes Jun 6 '13 at 19:09

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