# Maurer Cartan $1$-form explanation of notation

This is the construction I am given:

Let $$G$$ be a Lie group, $$\mathfrak{g}:=T_eG$$ its Lie algebra. We define $$1$$-form $$w_G \in \Omega^1(G) \otimes \mathfrak{g}$$, i.e. closed $$1$$-forms with value in $$\mathfrak{g}$$ by the rule: At each $$g \in G$$, $$(w_G)_g: X_g \mapsto (L_{g^{-1}})_* X_g$$ where $$X_g$$ is a vector in $$T_gG$$.

1. It doesn't seem to me clear that $$w_G$$ is a smooth form.
2. It is claimed that when $$G=GL_n(\Bbb R)$$, $$w_G = g^{-1}dg$$. What does the $$dg$$ even mean in this case?

My reference: page 103, Definition 7.4.13, Prop 7.4.15

EDIT: I became aware of this post for my second question. Still I am confused, why and how can we regard $$g:G \rightarrow G$$ as the identity map?

Since $$GL_n(\Bbb R)$$ is an open subset of $$\Bbb R^{n^2}$$, you can think of the inclusion map $$g\colon G\to \Bbb R^{n^2}$$ and its derivative $$dg$$ as a vector-valued $$1$$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.

Notice that if $$a\in G$$ is fixed, and $$X_a\in T_aG$$, then $$\omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$$, as desired.

Since $$g^{-1}$$ is a smooth function and the components of $$dg$$ are smooth, the matrix-valued $$1$$-form $$g^{-1}dg$$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $$L_g$$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $$X$$ gives a smooth function.