The adjoint representaion of $G$ is a homomorphism $ad_{a}:g \rightarrow aga^{-1}$, $a,g \in G$, what is the meaning of this? Now if we identify $T_{e}G$ with $\mathfrak{g}$ we have the adjoint map $Ad:G \times \mathfrak{g} \rightarrow \mathfrak{g}$, what is the meaning of that? (context of connections (one form) on principal bundles)


1 Answer 1


$ad$ is a homomorphism $ad : G \mapsto \mathrm{Aut}(G)$. By definition, $ad$ takes $a \in G$ to the inner automorphism $ad_a \in \mathrm{Aut}(G)$ that is defined by the formula $\mathrm{ad}_a(g) = aga^{-1}$.

Notice that for each $a \in G$ we have $\mathrm{ad}_a(e)=aea^{-1}=e$. Therefore the derivative $D_{ad_a} : T_e G \to T_e G$ is defined, it is linear automorphism of $T_e G = \mathfrak{g}$, and it is an automorphism of the Lie algebra structure on $\mathfrak{g}$.

$Ad$ is a homomorphism $Ad : G \mapsto \mathrm{Aut}(\mathfrak{g})$. By definition, $Ad$ takes $a \in G$ to $Ad_a = D_{ad_a}$ which, as seen just above, is a Lie algebra automoprhism of $\mathfrak{g}$.

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    $\begingroup$ Maybe it's just me, but the question... and this answer... blur possibly-unhelpfully the distinctions between the various "adjoint" representations. Indeed, "Ad" of $G$ on $\mathfrak g$ is the derivative of (what I'd call) the conjugation action of $G$ on itself, at $1_G$, and then lowercase-ad is the further derivative of that. Just a fretful paraphrase... $\endgroup$ Nov 7, 2014 at 23:03
  • $\begingroup$ So it's a derivative or what? Or is it analogous to the theorem that every linear mapping has a matrix representation?? $\endgroup$ Nov 8, 2014 at 12:06

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