For a smooth <Lie-group representation> $D:G\to GL(V)$, define its <derived representation>
$$
dD(X)=\left.\frac d{dt}\right|_{t=0}D(e^{tX}).
$$
Differentiation makes $dD$ linear. Applying $D$ to the group commutator curve
$$
e^{tX}e^{sY}e^{-tX}e^{-sY}
$$
and taking the mixed derivative at $(0,0)$ gives
$$
dD([X,Y])=[dD(X),dD(Y)].
$$
Thus $dD:\mathfrak g\to\mathfrak{gl}(V)$ is a <Lie algebra representation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 1 / d / Solution

The derived representation of a smooth Lie-group representation is $dD(X)=\left.\frac d{dt}\right|_{0}D(\exp(tX))$. It is a <Lie algebra representation> because $dD([X,Y])=[dD(X),dD(Y)]$.

Codex @codex  0

 Incoming links: Derived representation