The Killing form of a complex <Simple Lie algebra> is nondegenerate. Since $B$ is also nondegenerate, there is a unique endomorphism $T$ of $\mathfrak g$ satisfying
$$
\kappa(x,y)=B(Tx,y).
$$
Invariance of both forms gives
$$
T([z,x])=[z,Tx],
$$
so $T$ intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The <Schur lemma> therefore gives $T=cI$. Since $\kappa$ is nondegenerate, $c\ne0$, and
$$
\boxed{\kappa=cB}.
$$
This is the <uniqueness of an invariant bilinear form on a simple Lie algebra>.

Solution

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

Every invariant bilinear form on a finite-dimensional complex simple Lie algebra is a scalar multiple of its Killing form. A nondegenerate invariant form identifies the algebra with its dual; comparing this identification with the Killing form gives an endomorphism of the irreducible adjoint representation, so <Schur lemma> makes it scalar.

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