Adjoint representation of a Lie group ($\mathrm{Ad}$)

The adjoint representation of a Lie group is its action on its Lie algebra by differentiating conjugation. For a matrix Lie group, ${\mathrm{Ad}}_{g}X=gX{g}^{-1}$.

Let $G$ be connected and compact with nonzero irreducible real Adjoint representation. Its Lie algebra has no proper invariant Lie algebra ideal. If it is Abelian, irreducibility forces dimension one, so $G$ is the circle group. Otherwise its centre has zero-dimensional Lie algebra and is finite. A proper closed normal subgroup likewise has zero-dimensional Lie algebra, hence is finite; connectedness of $G$ makes conjugation on that finite subgroup trivial, so the subgroup is central. A Lie group homomorphism onto a nontrivial connected target with surjective differential consequently has finite central group kernel. The qualifications “proper” and “nontrivial” are essential.

With Adjoint representation convention ${\mathrm{ad}}_X(Y)=[X,Y]$, the matrix exponential identity is $e^{-X}Y{e}^X=e^{-{\mathrm{ad}}_X}Y$. Inverse conjugation reverses product order, so it is a right action or antirepresentation, whereas ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$ is an ordinary left group representation. The first-order terms $Y\mp [X,Y]$ detect a sign mismatch immediately. Negating every adjoint generator without reversing the Lie bracket does not give a new Lie-algebra homomorphism.

Conjugation by $SU(2)$ on its three-dimensional real Lie algebra of traceless skew-Hermitian matrices preserves $-\mathrm{tr}(AB)$ and defines a surjectionwith kernel $\{\pm I_2\}$. It is the universal double covering of $SO(3)$.

A SU(2) representation descends to the SO(3) group exactly when $-I$ acts trivially. In ${\mathrm{Sym}}^n({\mathbb{C}}^2)$ it acts by $(-1{)}^n$, so descent holds precisely for even $n$, or integer spin $j=n/2$. This is the inflation of a group representation criterion for the kernel $\{\pm I\}$. Half-integer-spin representations instead remain representations of the covering group.