# 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?

• 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 Commented Aug 2, 2020 at 16:24
• 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. Commented Aug 2, 2020 at 17:01

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.

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.