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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?

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    $\begingroup$ Are you familiar with the differential of a smooth map between manifolds? $\endgroup$
    – Kajelad
    Commented Sep 17, 2020 at 6:18
  • $\begingroup$ @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? $\endgroup$
    – Lagrang3
    Commented Sep 17, 2020 at 16:47

1 Answer 1

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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.

Thanks @Kajelad for the hint.

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