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Frankel (section $18.1$a) calculates the Maurer-Cartan form $\theta$ of the Lie group $\operatorname{SO}(2)$ by using the formula $\theta=g^{-1}dg$, where $g$ is viewed as the identity on $\operatorname{SO}(2)$,$$dg=\begin{pmatrix}d(g_{11})&d(g_{12})\\d(g_{21})&d(g_{22})\end{pmatrix}\in \operatorname{Mat}(2\times 2)\otimes\Omega^1(\operatorname{SO}(2))$$ is a matrix of $1$-forms and $\theta=g^{-1}dg$ is simply the product of the two matrices, i.e.$$\theta_{ij}=\sum_k (g^{-1})_{ik}(dg)_{kj}\in\Omega^1(G).$$ In principle, this makes sense for any matrix group, i.e. if $G\subseteq \operatorname{GL}(n)$ is a Lie subgroup, then we can use the last equation to define the Maurer-Cartan form. Of course the question is whether this is equivalent to the standard definition using the pushforward of the left translation. One issue is obvious:

The definition described above yields an element$$\theta\in\operatorname{Mat}(n\times n)\otimes\Omega^1(G)$$and it is not clear that $\theta$ is even an element of the subset $\mathfrak g\otimes\Omega^1(G)$, i.e. $$\theta\in\mathfrak g\otimes\Omega^1(G)\subseteq \operatorname{Mat}(n\times n)\otimes\Omega^1(G).$$ In the example discussed in the beginning this can be checked explicitly, but is there a more general argument?

Filippo
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  • Yes, this is the Mauer-Cartan form. It is equivalent to the standard definition using pushforward by left translation, because left translation for a matrix Lie group is just left multiplication. – Max Feb 14 '25 at 21:35
  • @Max Thank you for the comment. Interesting, sounds like this question is strongly related to my former question, which asks for a clarification of the equation $(L_g)_*=L_g$. If you can have a look at my other question and perhaps resolve my confusion, that would be much appreciated – Filippo Feb 14 '25 at 21:42
  • Yes, $(L_g)_\ast = L_g$ for matrix Lie groups. Here we always take advantage of the embedding of a matrix Lie group in $n \times n$ matrices, which we consider a vector space. – Max Feb 14 '25 at 22:26
  • @Max Thank you for the hint, now finally everything makes sense! – Filippo Feb 15 '25 at 19:50

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Yes, $g^{-1}dg$ can be interpreted as the product of two matrices, but there is also another equivalent interpretation which implies the first interpretation.

First of all, $g$ should be viewed as the canonical embedding $G\to M_{n\times n}$. Given $p\in G$, let $\ell_p\in\mathrm{End}(M_{n\times n})$ be left-multiplication, then $dg_{pq}\circ d(L_p)_q=\ell_p\circ dg_q$ (see this answer). Also, we view $\theta\in\Omega^1(G)\otimes T_eG$ as a matrix by setting $\tilde\theta_p:=dg_e\circ \theta_p\in L(T_pG,M_{n\times n})$. That being said,$$\tilde\theta_p=dg_e\circ d(L_{p^{-1}})_p=dg_{p^{-1}p}\circ d(L_{p^{-1}})_p=\ell_{p^{-1}}\circ dg_p$$ or $\tilde\theta_p=p^{-1}dg_p$ for short. In addition, $$\forall X\in T_pG:(\ell_{p^{-1}}\circ dg_p)(X)_{ij}=(p^{-1}dg_p(X))_{ij}=\sum_k(p^{-1})_{ik}d(g_{kj})(X)$$ as desired.

Filippo
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