Expression for the Maurer-Cartan form of a matrix group

I understand the definition of the Maurer-Cartan form on a general Lie group $G$, defined as

$\theta_g = (L_{g^{-1}})_*:T_gG \rightarrow T_eG=\mathfrak{g}$.

What I don't understand is the expression

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

when $G$ is a matrix group. In particular, I'm not sure how I'm supposed to interpret $dg$. It seemed to me that, in this concrete case, I should take a matrix $A\in T_gG$ and a curve $\sigma$ such that $\dot{\sigma}(0)=A$, and compute $\theta_g(A)=(\frac{d}{dt}g^{-1}\sigma(t))\big|_{t=0}=g^{-1}A$ since $g$ is constant. So it looks like $\theta_g$ is just plain old left matrix multiplication by $g^{-1}$. Is this correct? If so, how does it connect to the expression above?

Solution 1:

This notation is akin to writing $d\vec x$ on $\mathbb R^n$. Think of $\vec x\colon\mathbb R^n\to\mathbb R^n$ as the identity map and so $d\vec x = \sum\limits_{j=1}^n \theta^j e_j$ is an expression for the identity map as a tensor of type $(1,1)$ [here $\theta^j$ are the dual basis to the basis $e_j$]. In the Lie group setting, one is thinking of $g\colon G\to G$ as the identity map, and $dg_a\colon T_aG\to T_aG$ is of course the identity. Since $(L_g)_* = L_g$ on matrices (as you observed), for $A\in T_aG$, $(g^{-1}dg)_a(A) = a^{-1}A = L_{a^{-1}*}dg_a(A)\in\frak g$.