Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 102 / 3

A finite-dimensional Lie algebra $L$ is semisimple when its solvable radical is zero, equivalently when it has no nonzero solvable ideals.

We prove the Weyl complete reducibility theorem. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is perfect. By induction, $W$ is a direct sum of irreducible modules. A Casimir element formed using the Killing form commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. ThusInside $\ker \Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb{C}v$ is the required invariant complement.

This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, letwith the natural Hom representation. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f{|}_W=I_W$. Thenso every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.

For the requested example, embed ${\mathfrak{sl}}_2$ as the upper-left $2\times 2$ block in the Special linear Lie algebra ${\mathfrak{sl}}_3$. Under the restricted Adjoint representation,Here $V_n$ is the irreducible ${\mathfrak{sl}}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.

Because $M$ is a finite-dimensional semisimple module, it has an isotypic decompositionwhere the $S_i$ are pairwise nonisomorphic simple right $R$-modules. By Schur lemma,is a division ring, while ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. ThereforeIf the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, Schur lemma gives $D_i=k$.