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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?

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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.

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