I am a physics undergrad.
The adjoint action of a group on itself is $\operatorname{Ad}: G \times G \to G$ is defined to be $\operatorname{Ad}:(g,h) \to g^{-1}hg$. The adjoint representation of the group is the action of the group on its lie algeba, and is defined as the derivative of this map at the identity element, i.e. $\mathrm{D}\operatorname{Ad}_1=:\operatorname{ad}:G\to\operatorname{GL}(\mathfrak{g})$. How do I prove that for matrix lie groups, this map is actually $\operatorname{ad}(g)(X)=g^{-1} X g $ for $X \in \mathfrak{g}$.
What is the linear map $\operatorname{ad}$ in general?
I am sorry if this question is really trivial, but all the math books are inaccessible to me, and I couldn't find this in any physics book.
Best Answer
Let $c:[0,1]\to G$ be a differentiable curve in $G$ such that $c(0)=1$ (identity matrix).
Then, if $g\in G$, the (group) adjoint action on the points of the curve $c(t)$ is $g^{-1}c(t)g.$ Now every element of the Lie algebra can be written as a tangent vector to some curve $c$ at the identity ($t=0$). If $x$ is such tangent vector, that is: $$ x = \frac{dc(t)}{dt}\bigg|_{t=0}, $$
then the derivative of the group adjoint is: $$ \frac{d}{dt}g^{-1}c(t)g\bigg|_{t=0} = g^{-1}xg, $$
by simple Leibniz rules, which is exactly the adjoint action on the algebra.