2

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?

Bryan Shih
  • 9,934
  • I have added this answer: https://math.stackexchange.com/a/4539224/601797, which I think is what you are looking for. At least for me it was the explanation I needed... – A. J. Pan-Collantes Sep 26 '22 at 06:07

2 Answers2

1

The Maurer-Cartan form is

$$ \omega: T_gG \rightarrow T_eG\\ \omega = g^{-1}dg\\ \omega^i_j = (g^{-1})^i_a(dg)^a_j $$

So if $V \in T_eG$, then

$$ gV = (gV)^{b}_c \frac{\partial}{\partial g^b_c} \in T_gG $$

Here, $g^i_j$ are just a fast way to write the coordinates of each element $g \in G$. Since we are considering $G = GL(n, \mathbb{R})$, the coordinates will be the matrix components, so there are two indices.

Now, $dg \in T^*_gG$ are dual elements of $T_gG$ so

$$ \omega^i_j(gV) = (g^{-1})^i_a[(gV)^{b}_c\ \delta^a_b\ \delta^c_j] = (g^{-1})^i_b(gV)^b_j = (g^{-1}gV)^i_j = V^{i}_j $$

Therefore, the way to understand $dg$ is to consider it as the differential of each matrix component. If they are functions $f(x^1, ..., x^n)$, then $(dg)^\alpha_\beta = df = (\partial_\mu f) dx^\mu$

Vicky
  • 569
0

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.

Ted Shifrin
  • 125,228