Adjoint representation of Lie Groups In this Wikipedia article it says that given a lie group $G$ with identity $e$, we can define the automorphism $\Psi_g: G\to G$
as: $\Psi_g(h) = g h g^{-1}$. Then the adjoint representation of $g\in G$ is the map
$Ad: T_e G \to T_e G$ obtained as
$$Ad_g = (d \Psi_g)_e. $$
Then in the case when $G \subset GL(n)$ you can find that $Ad_g (X) = g X g^{-1}$.
I gave a bit thought to the proposition that $Ad_g(X) = g X g^{-1}$
and I came to the conclusion that $Ad_g X$ is the tangent vector to the curve
$g \exp(t X) g^{-1}$ at $t=0$. And that definition for the $Ad_g$ is ok for me.
In fact I can also write precisely what $Ad_g X$ is by looking at it's
action over a function $f: G \to R$
$$
(Ad_g X)f = \frac{d}{dt} f(g \exp(t X) g^{-1})
= X ( f\circ \Psi_g) = d (f\circ \Psi_g)_e X.
$$
But the first one given by Wikipedia and also present in Choquet-Bruhat's Analysis Manifolds and Physics page 166, I cannot understand.
The problem is that I can't see how $d$ actually acts on $\Psi_g: G\to G$ for two reasons:

*

*$G$ is not an algebraic field, so how on Earth I can derive a $G$-valued function
when I have not defined summation on $G$?


*If I take a function $f: G \to R$ I get that
the 1-form $(df)_e$ is a map from $T_e G$ to $R$, then why do they claim
$(d\Psi_g )_e$ is a map from $T_e G$ to $T_e G$, shouldn't it be
$d(\Psi_g)_e: T_e G\to G$?
Let's put words into equations: consider $X\in T_e G$, then let's see how $(d\Psi_g)_e$
acts on $X$:
$$
(d\Psi_g)_e X = X \Psi_g = \frac{d}{dt} (\Psi_g \circ \exp(tX)) 
=\frac{d}{dt} (g \exp(tX) g^{-1}), 
$$
that expression would make sense only if $g \exp(tX) g^{-1} \in G$ was a matrix.
So my question is: what does $(d\Psi_g)_e$ really means in differential geometry?
 A: The adjoint representation of element $g\in G$ is defined as the push-forward
of $\Psi_g$ at the identity $e$:
$$
Ad_g = (\Psi_g)_{e}{}_{*}
$$
That said

*

*there's no need for $G$ to be an algebraic field with a topology and so on to compute derivatives like in real space $R$,

*and $Ad_g$ becomes a linear automorphism of $T_e G$,
since $(\Psi_g)_{e}{}_{*}: T_e G \to T_e G$.

Also defining $Ad_g X$ as the tangent vector to $g \exp(t X) g^{-1}$ at
$t=0$, in the question, is correct.
In fact, for $X\in T_e G$ and $f: G\to R$ smooth:
$$
(Ad_g X)(f)
= \frac{d}{dt} f ( g \exp(t X) g^{-1})
 = X(f \circ \Psi_g)
= ((\Psi_g)_e{}_{*} X )(f),
$$
where the last equality is due to the definition of push-forward.
The definition given in the references indicated in the question
where the same but with a different notation. They wrote $(d\Psi_g)_e$
to indicate the push-forward of $\Psi_g$ not the exterior derivative---unless there is a sophisticated definition of exterior derivative, that I am not aware of, that unifies both concepts, see this answer.
Thanks @Kajelad for the hint.
