Every invariant subspace of the <Adjoint representation of a Lie algebra> is an ideal. Hence the adjoint representation of a <Simple Lie algebra> is irreducible.

Adjoint irreducibility of a simple Lie algebra

The <Adjoint representation of a Lie algebra> is
$$
\operatorname{ad}:\mathfrak g\longrightarrow\mathfrak{gl}(\mathfrak g),
\qquad
\operatorname{ad}_x(y)=[x,y].
$$
The <Killing form> is the symmetric invariant bilinear form
$$
\boxed{\kappa(x,y)=\operatorname{tr}(\operatorname{ad}_x\operatorname{ad}_y)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 1 / a / Solution

The <Adjoint representation of a Lie algebra> is
$$
\operatorname{ad}:\mathfrak g\to\mathfrak{gl}(\mathfrak g),
\qquad \operatorname{ad}_X(Y)=[X,Y].
$$
It is linear. The <Jacobi identity> gives
$$
[\operatorname{ad}_X,\operatorname{ad}_Y]Z
=[X,[Y,Z]]-[Y,[X,Z]]=[[X,Y],Z]
=\operatorname{ad}_{[X,Y]}Z,
$$
so it is a <Lie algebra representation>.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 3 / a / Solution

Let $W\subseteq\mathfrak g$ be invariant under the <Adjoint representation of a Lie algebra>. Then $[\mathfrak g,W]\subseteq W$, so $W$ is an ideal. If $\mathfrak g$ is a <Simple Lie algebra>, its only ideals are $0$ and $\mathfrak g$. Hence the adjoint representation is irreducible. Compact type is compatible with the anti-Hermitian realization used below, although simplicity alone proves this <adjoint irreducibility of a simple Lie algebra>.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 3 / d / Solution

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{title2=$\operatorname{ad}$}
{wiki=Adjoint_representation}

The adjoint representation is $\operatorname{ad}_x(y)=[x,y]$. For a semisimple <Lie algebra>, its nonzero weights are the roots and its zero-weight space is the <Cartan subalgebra>.

= Adjoint representation
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