# 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:

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

2. 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?

• Are you familiar with the differential of a smooth map between manifolds? Commented Sep 17, 2020 at 6:18
• @Kajelad, I didn't know the pushforward of $\phi$ was also written as $d\phi$, I write it as $\phi_*$. That would make sense after all. Thanks. Why don't you write an answer? Commented Sep 17, 2020 at 16:47

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
1. there's no need for $$G$$ to be an algebraic field with a topology and so on to compute derivatives like in real space $$R$$,
2. 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.