We know that a Lie group G acts on itself by conjugation. That is, $g \rightarrow I_g = R_{g^{-1}}\circ L_g$ and that this action is an automorphism associated to $G$.
We also know that conjugation maps $e$ (identity on $G$) to itself so we would expect the derivative of $I_g$ to map $T_eG=\mathfrak{g}\rightarrow T_eG=\mathfrak{g}$. Now, according to Marsden and Ratiu's $\textit{Introduction to Mechanics and Symmetry}$ (Chapter 9, pg. 311), differentiating $I_g$ at $e$ gives the $\textbf{adjoint representation}$ of $G$ on $\mathfrak{g}$:$$\textrm{Ad}_g:=T_eI_g:T_eG=\mathfrak{g}\rightarrow T_eG=\mathfrak{g}$$ Explicitly, the adjoint action of $G$ on $\mathfrak{g}$ is given by $$\textrm{Ad}:G\times \mathfrak{g}\rightarrow \mathfrak{g},\:\:\:\:\:\textrm{Ad}_g(\xi) = T_e(R_{g^{-1}}\circ L_g)\xi$$
$\textbf{Question}$: What is that map actually giving us? What does taking the derivative actually mean here? (Yes, I know its a tangent space map but, that doesn't tell me much about this particular case) It's taking in a vector in $\mathfrak{g}$ and outputting what exactly? I feel like almost $\textit{every}$ book just gives something like 'differentiating the conjugate gives a map from $\mathfrak{g}$ to $\mathfrak{g}$' and hides what's happening inside. The example from Marsden and Ratiu leads to more questions than answers too.
$\textbf{Example}:$ For $SO(3)$ we have $I_A(B) = ABA^{-1}$ so differentiating with respect to $B$ at $B=\textrm{identity}$ gives $\textrm{Ad}_A \hat{v} = A\hat{v}A^{-1}$. However $$(\textrm{Ad}_A \hat{v})(w) = A\hat(v)A^{-1}w = A(v \times A^{-1}w) = Av \times w$$ so $$(\textrm{Ad}_A\hat{v})=(Av)^{\hat{}}$$ Identifying $\mathfrak{so}(3)$ with $\mathbb{R}^3$ gives $Ad_A v = Av$.
$\textbf{Question:}$ What????? How did we end up evaluating at $w$. How did we end up taking a cross product in there? And again, what did we really end up with as a result of this map? A representation should be a matrix associated to an element of the Lie group, no? So what is our matrix in this example? Is it $A$?
$\textbf{Further context}$: I'm trying very hard to understand this to get to what is meant by the $\textrm{Ad}^*$-equivariant moment map.
Best Answer
Here is something that may help you. Consider $G=GL_n(\Bbb R)$, that is the group of real invertible $n\times n$ matrices. Now show the following statements:
Show that $GL_n(\Bbb R)$ is open in the vector space $M_{n\times n}(\Bbb R)$ of real $n\times n$ matrices.
As such you may identify every tangent space with $M_{n\times n}(\Bbb R)$. Let $g, h\in GL_n(\Bbb R)$, show that under this identification the derivative of left multiplication $L_g: GL_n \to GL_n, a\mapsto g\cdot a$ at $h$ is the map $D_h L_g: M_{n\times n}\to M_{n\times n}, X\mapsto g\cdot X$.
In the same fashion show that $D_h R_g: M_{n\times n}\to M_{n\times n}$ is given by $X\mapsto X\cdot g$.
Now calculate the derivative of $\mathrm{Ad}(g)=R_{g^{-1}}\circ L_g$ at $\Bbb1$. Show that it is equal to the map $M_{n\times n}\to M_{n\times n}, X\mapsto gXg^{-1}$.
Now you have understood the adjoint mapping for the matrix group $GL_n(\Bbb R)$. Consider now $O\subset GL_n(\Bbb R)$ a closed subgroup. Denote with $\mathfrak{o}$ the tangent space $T_\Bbb1 O\subset T_\Bbb1 GL_n(\Bbb R) = M_{n\times n}$. In particular $\mathfrak o$ is a vector space of matrices. We will now see that the derivatives of $L_g$ and $R_g$ for $g\in O$ act on this space simply by matrix multiplication. So:
In other words you may identify the tangent space of any matrix group at $g$ with a subspace of $M_{n\times n}$ that varies with $g$. The derivatives of the left and right multiplications with $g$ are just the left and right multiplication of $g$ acting on this subspace. Close everything with