Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 102 5 c Solution Created 2026-10-03 Updated 2026-10-05
We first prove the useful lemma that dominant root-lattice highest weights have zero weight. Write for the root lattice, and let a dominant integral weight be the highest weight of a finite-dimensional irreducible representation.
First, is a nonnegative integral combination of simple roots. Indeed, write , separating its positive and negative coefficients in the root basis. The two parts have disjoint supports, and distinct simple roots have nonpositive inner product, so . If , thenOn the other hand, dominance gives for each simple root, hence . This contradiction proves .
Now suppose a nonzero weight has all . The identityshows that some satisfies and . We use the standard sl2 Lie algebra fact that its lowering operator is injective on any positive eigenspace in a finite-dimensional representation. This follows from the classification of finite-dimensional sl2 representations: in each irreducible , the only weight killed by the lowering operator is the lowest weight .
Consequently a nonzero vector of weight lowers to a nonzero vector of weight . Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at , repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.
For a minuscule representation, every weight belongs to , so the zero weight just obtained lies in this orbit. Every Weyl group element is invertible, and forces . This proves that minuscule weights in the root lattice are zero. Under the assumption , every possible highest weight lies in , henceHere is the one-dimensional trivial Lie algebra representation, by the classification of finite-dimensional irreducible highest-weight representations.