The exterior derivative of the inverse of a function Suppose that $g \in \Lambda ^ 0 ( G) $ is a zero form (a scalar function) which maps from some three dimensional ball $B$ to a Lie group manifold $G$, so that
$$ g : B \to G .$$
Let's assume for the moment that the inverse exists at any point on $G$.
I'm trying to prove that the exterior derivative of $g ^ { - 1} $ is given by the wedge product
$$ d( g ^ { -1 } dg )  = g ^ { - 1 } dg \wedge g ^ { -1 } dg  $$
This seems somewhat intuitively true since we know that, if $G$ is a non-Abelian Lie group, that the variation is given by
$$ \delta g ^ { -1} =  - g ^ { - 1 } \delta g g ^ { - 1 } $$
However, I'm struggling to see how the wedge product comes into play when we do the exterior derivative instead. Any suggestions?
 A: First, your notation of $\Lambda^0(G)$ is terrible. Second, the formula is off by a negative sign.
This is not the inverse of a function. We are considering the mapping that sends a group element to its group inverse. This equation makes sense for matrix groups, where you then compute the wedge product of matrix-valued $1$-forms; more generally, you need to use the Lie bracket on the Lie algebra of $G$ to define this product.
I don't know where your $\delta$ notation is coming from; this is some relic of physicists' notation, I suppose. If you have a matrix group $G$, and you consider the map $G\to G$ given by the group inverse, then $d(g^{-1}) = -g^{-1}dg\,g^{-1}$. (If this is not a matrix group, you have to interpret this with more carefully.) Then we use the standard product rule for differentiating $1$-forms: If $f$ is a function and $\eta$ is a $1$-form, then $d(f\eta) = df\wedge\eta + f\,d\eta$. So
\begin{align*}
d(g^{-1}dg) &= d(g^{-1})\wedge dg + g^{-1} d(dg) = (-g^{-1}dg\,g^{-1})\wedge dg \\ &= -(g^{-1}dg)\wedge (g^{-1}dg).
\end{align*}
At the end we use the fact that functions can pass through the wedge product.
A: I think what you are actually looking for is the so-called left logarithmic derivative. For any smooth manifold $M$, this sends a smooth function $f:M\to G$ to a one-forms with values in the Lie algebra $\mathfrak g$ of $G$. Unfortunately, this is usually denoted by $\delta f\in\Omega^1(M,\mathfrak g)$, I hope this does not lead to confusion with the notation for variations you use. Given $x\in M$ and $X\in T_xM$ the definition is $\delta f(x)(X):=T_{f(x)}\lambda_{f(x)^{-1}}(T_xf(X))$. Here $\lambda_g$ denotes the left translation by $g$.
In the case of a matrix group $G$, you can interpret $f$ as a function to $M_n(\mathbb R)$ and hence $df(X)$ as an element of $M_n(\mathbb R)$. Since left translation in a matrix group is the restriction of a linear map, the definition of $\delta f$ in this case boils down to $f(x)^{-1}df(X)$, which explains the connection to the notation you use.
Now the left logarithmic derivative of any function satisfies the Maurer-Cartan equation, which for $\omega\in\Omega^1(M,\mathfrak g)$ is most safely written as $d\omega(X,Y)+[\omega(X),\omega(Y)]=0$, where the bracket is in $\mathfrak g$. (Writing this as a wedge product, one may need a factor $\tfrac12$, i.e. as $d\omega+\frac12\omega\wedge\omega=0$, but that is rather a matter of conventions.) The easiest way to see that this is true is that $\delta f$ actually is the pullback of the so-called left Maurer-Cartan form, for which the Maurer-Cartan equation follows from the definition of the Lie bracket.
