Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 1 4 Solution Created 2026-10-03 Updated 2026-10-07
A Lie algebra representation is completely reducible when it is a direct sum of irreducible submodules. In finite dimension this is equivalent to saying that every submodule has an invariant complementary subspace. Indeed, one may split off a minimal nonzero submodule and induct on dimension; conversely, if is a finite sum of simple submodules, choose a maximal sum of these with . If , some simple summand is not contained in . Then by simplicity, so is a larger sum disjoint from , a contradiction. Thus .
We prove the required splitting by Haar averaging produces invariant complements. The structural input is the compact real form of a complex semisimple Lie algebra: there is a real semisimple algebra with , whose simply connected integrating Lie group is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses root vectors normalized so that , with real structure constants compatible with the involution and . The real spanis closed under brackets, complexifies to , and has negative-definite Killing form. The compactness criterion for real semisimple Lie algebras then gives a compact simply connected . Thus this route uses root structure and compact integration, not the theorem being proved.
Restrict to . The integration of a Lie-algebra representation gives a representation , since is simply connected. Choose any positive-definite Hermitian inner product and average using normalized Haar measure:The integral exists by compactness. It is positive-definite because for , and it is -invariant by translation invariance of Haar measure. Equivalently, every with is skew-Hermitian for this inner product.
Let be an -submodule. It is invariant under , hence under the exponentials generating the connected group . For , and ,Consequently the orthogonal complement is -invariant, and differentiation makes it -invariant. It is a complex vector subspace, so it is also invariant under and hence under . Therefore as -modules.
Induction on now expresses as a finite direct sum of irreducible submodules. Every finite-dimensional representation of a finite-dimensional complex semisimple Lie algebra is completely reducible. The compact real form need not be the real form originally used to present ; in particular the averaging argument does not require the given presentation to be unitary.