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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?

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    $\begingroup$ If $g\in \Lambda ^0 (G)$ then it is a function $g: G \to \mathbb R$? if $g: B\to G$, then $g^{-1} : G\to B$. Then what is the exterior derivative of $g^{-1}$? I guess you can take exterior only to some sort of differential forms $\endgroup$ Commented Aug 2, 2020 at 16:24
  • $\begingroup$ It's not even clear to me if the exterior derivative on the inverse in this case is even defined, but since $G$ is also a manifold I thought that it was okay. $\endgroup$ Commented Aug 2, 2020 at 17:01

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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.

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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.

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