Editing Adjoint representation (section)

===Derivative of Ad===
One may always pass from a representation of a Lie group ''G'' to a [[representation of a Lie algebra|representation of its Lie algebra]] by taking the derivative at the identity.

Taking the derivative of the adjoint map
:<math>\mathrm{Ad} : G \to \mathrm{Aut}(\mathfrak g)</math>
at the identity element gives the '''adjoint representation''' of the Lie algebra <math>\mathfrak g = \operatorname{Lie}(G)</math> of ''G'':
:<math>\begin{align}
\mathrm{ad} : & \, \mathfrak g \to \mathrm{Der}(\mathfrak g) \\
& \,x \mapsto \operatorname{ad}_x = d(\operatorname{Ad})_e(x)
\end{align}</math>
where <math>\mathrm{Der}(\mathfrak g) = \operatorname{Lie}(\operatorname{Aut}(\mathfrak{g}))</math> is the Lie algebra of <math>\mathrm{Aut}(\mathfrak g)</math> which may be identified with the [[differential algebra#Lie algebra|derivation algebra]] of <math>\mathfrak g</math>. One can show that
:<math>\mathrm{ad}_x(y) = [x,y]\,</math>
for all <math>x,y \in \mathfrak g</math>, where the right hand side is given (induced) by the [[Lie bracket of vector fields]]. Indeed,<ref>{{harvnb|Kobayashi|Nomizu|1996|loc=page 41}}</ref> recall that, viewing <math>\mathfrak{g}</math> as the Lie algebra of left-invariant vector fields on ''G'', the bracket on <math>\mathfrak g</math> is given as:<ref>{{harvnb|Kobayashi|Nomizu|1996|loc=Proposition 1.9.}}</ref> for left-invariant vector fields ''X'', ''Y'',
:<math>[X, Y] = \lim_{t \to 0} {1 \over t}(d \varphi_{-t}(Y) - Y)</math>
where <math>\varphi_t: G \to G</math> denotes the [[flow (mathematics)|flow]] generated by ''X''. As it turns out, <math>\varphi_t(g) = g\varphi_t(e)</math>, roughly because both sides satisfy the same ODE defining the flow. That is, <math>\varphi_t = R_{\varphi_t(e)}</math> where <math>R_h</math> denotes the right multiplication by <math>h \in G</math>. On the other hand, since <math>\Psi_g = R_{g^{-1}} \circ L_g</math>, by the [[chain rule]],
:<math>\operatorname{Ad}_g(Y) = d (R_{g^{-1}} \circ L_g)(Y) = d R_{g^{-1}} (d L_g(Y)) = d R_{g^{-1}}(Y)</math>
as ''Y'' is left-invariant. Hence,
:<math>[X, Y] = \lim_{t \to 0} {1 \over t}(\operatorname{Ad}_{\varphi_t(e)}(Y) - Y)</math>,
which is what was needed to show.

Thus, <math>\mathrm{ad}_x</math> coincides with the same one defined in {{section link||Adjoint representation of a Lie algebra}} below. Ad and ad are related through the [[exponential map (Lie theory)|exponential map]]: Specifically, Ad<sub>exp(''x'')</sub> = exp(ad<sub>''x''</sub>) for all ''x'' in the Lie algebra.<ref>{{harvnb|Hall|2015}} Proposition 3.35</ref> It is a consequence of the general result relating Lie group and Lie algebra homomorphisms via the exponential map.<ref>{{harvnb|Hall|2015}} Theorem 3.28</ref>

If ''G'' is an immersely linear Lie group, then the above computation simplifies: indeed, as noted early, <math>\operatorname{Ad}_g(Y) = gYg^{-1}</math> and thus with <math>g = e^{tX}</math>,
:<math>\operatorname{Ad}_{e^{tX}}(Y) = e^{tX} Y e^{-tX}</math>.
Taking the derivative of this at <math>t = 0</math>, we have:
:<math>\operatorname{ad}_X Y = XY - YX</math>.
The general case can also be deduced from the linear case: indeed, let <math>G'</math> be an immersely linear Lie group having the same Lie algebra as that of ''G''. Then the derivative of Ad at the identity element for ''G'' and that for ''G{{'}}'' coincide; hence, without loss of generality, ''G'' can be assumed to be ''G{{'}}''.

The upper-case/lower-case notation is used extensively in the literature.  Thus, for example, a vector {{mvar|x}}    in the algebra <math>\mathfrak{g}</math> generates a [[vector field]] {{mvar|X}}    in the group  {{mvar|G}}.  Similarly, the adjoint map {{math|ad<sub>x</sub>y {{=}} [''x'',''y'']}} of vectors in <math>\mathfrak{g}</math> is homomorphic{{clarify|map is homomorphic??|date=December 2018}} to the [[Lie derivative]] {{math|L<sub>''X''</sub>''Y'' {{=}} [''X'',''Y'']}} of vector fields on the group  {{mvar|G}} considered as a [[manifold]].

Further see the [[derivative of the exponential map]].