In what sense is $(L_g)_*=L_g$ for matrix Lie groups?

Question

Let $G\subseteq \operatorname{GL}(n)\subseteq \operatorname{Mat}(n\times n)$ be a Lie subgroup.

I am currently studying the Maurer Cartan form and I read this answer which explains why it is sometimes denoted by $g^{-1}dg$. Apparently the key observation is that \begin{equation}\tag{1} \forall X\in T_pG:(L_{p^{-1}})_*(X)=L_{p^{-1}}X, \end{equation} and indeed the answer claims that $(L_p)_*=L_p$. Of course the LHS is a function $TG\to TG$ and the RHS is a function $G\to G$, so this equality requires to be interpreted correctly - but I am struggling to do so. I know that $\mathfrak g$ can be identified with a matrix subspace, but this doesn't seem to help here.

Answer 1

To me, the most intuitive way to understand this identification is to follow along Kurt G.'s comment and to just work with the smooth structure of $G$ as a manifold and multiplication as a smooth map.

A matrix Lie group $G\leq \operatorname{GL}(n)$ has the smooth structure of an embedded submanifold of the surrounding Euclidian space $\mathbb{R}^{n^2}$ with its natural smooth structure given by the identity chart. All this means is that maps to and from $G$ are smooth (in the manifold sense) if and only if they are smooth in the Euclidian sense, i.e. when $G$ is seen as a subset of $\mathbb{R}^{n^2}$. Differentiability and the derivative of a map on the subset $G$ are defined by the existence of an extension to an open neighbourhood of $G$.

On the one hand, this allows us to identify tangent vectors as $$T_g G = \left\{\frac{d}{dt}_{|t=0} \gamma : \quad \gamma\colon (-\varepsilon,\varepsilon)\to G, \gamma(0) = g\right\}\subseteq M_{n\times n},$$ where we just derive curves in $\mathbb{R}^{n^2}$ in the usual way.

On the other hand, we can see that matrix multiplication is a bilinear map $$m\colon M_{n\times n} \times M_{n\times n}\to M_{n\times n},\quad m(A,B) = AB.$$

As for any bilinear map, it is then straightforward to show

$$(dm)_{(A,B)}(V,W) =\frac{d}{dt}_{t=0}m(A+tV,B+tW) = m(A,W) + m(V,B) = AW + VB.$$ The multiplication on $G$ has an extension to all of $\mathbb{R}^{n^2}$ - it was defined as a restriction. This implies that the derivative of the restriction $m_{|G}$ to $G$ is given by the restriction $dm_{|TG\times TG}$ of the derivative on $\mathbb{R}^{n^2}\times \mathbb{R}^{n^2}$ to the tangent space.

We obtain the derivative of left-multiplication at $A$ by setting $V = 0$:

$$(dL_A)_B W = dm_{(A,B)}(0,W) = AW = L_AW.$$

Alternatively, you can consider multiplication $L_A\colon G\to G$ with a fixed matrix from the left directly, i.e. $L_A(X) = AX.$ This map is linear, and as for any linear map $$(dL_A)_X W = L_AW = AW,$$ which can be seen directly by applying the definition of the total derivative in Euclidian space and using the linearity of $L_A$ once.

Again, since the smooth structure on $G$ is given by the smooth structure on the surrounding $\mathbb{R}^{n^2}$, this shows $$(L_A)_* = L_A\colon TG\to TG.$$

The "confusion" stems from the fact that the symbol $L_A$ does not contain all the information about the map. It just means "left-multiplication with $A$" and tells us nothing about the domain and codomain. If we evaluate the derivative of left-multiplication with $A$ at any point $X$ applied to a tangent vector $W$, which is also a matrix, we obtain the tangent vector $AW$ at $AX$. It can be expressed as $L_A$, but with different domain and codomain than the original map.

Answer 2

The result is trivial given the hint that one must consider the embedding$$g\colon G\to M_{n\times n}$$in order to identify $T_pG$ with a subspace of $M_{n\times n}$. Indeed, the pushforward $dg_p\in L(T_pG,M_{n\times n})$ is the appropriate function. In addition, it is important to distinguish between $L_p\colon G\to G$ and the extension $\ell _p\colon M_{n\times n}\to M_{n\times n}$.

That being said, by $d(L_p)_q=L_p$ we mean that $dg_{pq}\circ d(L_p)_q=\ell_p\circ dg_q$, which is equivalent to$$d(g\circ L_p)_q=d(\ell_p\circ g)_q$$ (use the chain rule and the fact that $d(\ell_p)_q=\ell_p$). But the last equality is obvious since $g\circ L_p=\ell_p\circ g$.