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.


2 Answers 2


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:

  1. Show that $GL_n(\Bbb R)$ is open in the vector space $M_{n\times n}(\Bbb R)$ of real $n\times n$ matrices.

  2. 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 $D_h L_g: M_{n\times n}\to M_{n\times n}, X\mapsto g\cdot X$.

  3. In the same fashion show that $D_h R_g: M_{n\times n}\to M_{n\times n}$ is given by $X\mapsto X\cdot g$.

  4. 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:

  1. Let $g\in O$, show that $T_g(O) = g\cdot \mathfrak{o}$ and that $T_g(O) = \mathfrak{o}\cdot g$.
  2. Let $h\in O$, show that $D_hL_g: h\cdot \mathfrak o \to g\cdot h\cdot \mathfrak o$ is given by the map $X\mapsto g\cdot X$ (use the result 2.).
  3. Show also that $D_h R_g : h\cdot \mathfrak{o}\to h\cdot \mathfrak o \cdot g$, $X\mapsto X\cdot g$.

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

  1. Show that $\mathrm{Ad}(g):\mathfrak o \to\mathfrak o$ is given by $X\mapsto gXg^{-1}$.
  • $\begingroup$ Thanks @s.harp. I'm working through it $\endgroup$ Commented Aug 19, 2019 at 12:44
  • $\begingroup$ Do you have any pointers on why the cross product comes into the example? I've been at this but I still don't see why $\endgroup$ Commented Aug 19, 2019 at 13:58
  • $\begingroup$ The Lie algebra $\mathfrak{so}_3$ is given by anti-symmetric $3\times3$ matrices, that is by matrices of the form $\begin{pmatrix}0 & x_{12} &x_{13}\\ -x_{12} & 0 & x_{23}\\ -x_{13}& -x_{23}&0\end{pmatrix}$. This a $3$-dimensional vector space and the Lie bracket is given by the matrix commutator. Now there is a very special linear isomorphism $F:\mathfrak{so}_3\to\Bbb R^3$ so that $F([x,y])= x\times y$ where $\times$ is the cross product on $\Bbb R^3$ (I never get the signs quite right, so I'll leave you to finding this map). $\endgroup$
    – s.harp
    Commented Aug 19, 2019 at 19:51
  • $\begingroup$ Now let $g\in SO(3)$ and $x\in\mathfrak{so}_3$, the numbers above tell you that $\mathrm{Ad}(g)\,(x) = gxg^{-1}$. But this is in viewing $\mathfrak{so}_3$ as a matrix space, not with our very special identification $\mathfrak{so}_3\cong \Bbb R^3$. What is $F(gxg^{-1})$? If you can find that you will understand what $\mathrm{Ad}(g)$ looks like when $\mathfrak{so}_3$ is viewed as $\Bbb R^3$ with the cross product as Lie-bracket. $\endgroup$
    – s.harp
    Commented Aug 19, 2019 at 19:54

It's taking in a vector in $\mathfrak{g}$ and outputting what exactly?

It is taking a vector in $\mathfrak{g}$ and giving a vector in $\mathfrak{g}$.

Given $g\in G$, we have $Ad(g):G\rightarrow G$ defined as $h\mapsto ghg^{-1}$. This map sends $e\in G$ to $e\in G$. So, if we consider differential of $Ad(g):G\rightarrow G$ at $e\in G$, we get $Ad(g)_{*,e}:T_eG\rightarrow T_eG$. Changing the notations, we have $ad(g):\mathfrak{g}\rightarrow \mathfrak{g}$.

Adjoint representation is a map $G\times \mathfrak{g}\rightarrow \mathfrak{g}$ which takes an element of Lie group $G$, an element of Lie algebra $\mathfrak{g}$, to give an element of Lie algebra $\mathfrak{g}$

  • $\begingroup$ If you understand this, you can then think about example in the question $\endgroup$
    – user537667
    Commented Aug 19, 2019 at 12:16
  • $\begingroup$ Thank you for this. I'm just still unclear about how the cross product comes in though (in the example). That seems to be some trick from linear algebra that I'm not aware of or I forgot. $\endgroup$ Commented Aug 19, 2019 at 12:42
  • $\begingroup$ @DSS You can see s.harp's answer.. that talks about the example part.. try for some time, if you do not get, ask that user or me... $\endgroup$
    – user537667
    Commented Aug 19, 2019 at 12:59

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