The transformed generators are
$$
J'_1=-J_1,
\qquad J'_2=J_2,
\qquad J'_3=-J_3.
$$
For example,
$$
[J'_1,J'_2]=[-J_1,J_2]=-iJ_3=iJ'_3.
$$
The other two cyclic commutators work identically, so
$$
[J'_i,J'_j]=i\epsilon_{ijk}J'_k.
$$
Conjugation by $C$ is therefore an <automorphism of a Lie algebra>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 2 / c / Solution

An automorphism of a Lie algebra is an invertible linear map $f:\mathfrak g\to\mathfrak g$ satisfying $f([X,Y])=[f(X),f(Y)]$.

= Lie algebra representations
{c}
{synonym}


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