Brauer character basis theorem 2026-10-05
Over a splitting field for finite group representations of characteristic , the irreducible Brauer characters form a complex basis of the class functions on the p-regular elements. Independence is the character form of the Brauer–Nesbitt theorem. To obtain spanning, extend such a class function by zero on the p-singular classes. Ordinary irreducible characters form a basis of all class functions by character orthogonality. Restricting them to p-regular elements yields Brauer characters of reductions of an integral form of a group representation in a compatible splitting p-modular system; if needed, first extend scalars, which does not change the simple-module list under the splitting hypothesis. Each restriction is a nonnegative integral sum of simple Brauer characters by exact-sequence additivity and the Jordan–Hölder theorem. Thus these restrictions span, proving the assertion. Consequently the number of simple modules equals the number of p-regular conjugacy classes, and evaluation identifies the complexified modular representation ring with the product of one copy of for each such class.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 138 4 i Solution Created 2026-10-03 Updated 2026-10-05
An integral form of a group representation is a -stable finite free -submodule with . Such a form exists: take a basis lattice and replace it by , which is finite and torsion-free and therefore free over the discrete valuation ring .
Choose ordinary simple modules with forms , and modular simple modules with projective covers . With a uniformizer, put . The decomposition matrix and Cartan matrix of a group algebra have entriesThe first numbers do not depend on the integral form: the Brauer character of its reduction is the ordinary character restricted to p-regular elements, and part 2(a) determines all composition multiplicities from this restriction.
For the assertion in (i), set . It is a submodule of the finite free module , hence is finite free over . There is a natural injective mapFor any homomorphism in the target, clearing the finitely many denominators of its values on an -basis of gives . Thus , and the map is surjective. Therefore , regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.
Let be choices of integral form of a group representation over a complete p-modular system. Then is finite free, and : clearing denominators proves the spanning assertion. Also . If is a projective module, applying the Hom functor to yieldsConsequently these ordinary and modular Hom spaces have the same dimension of a vector space. Projectivity is essential: for over , the trivial and sign lattices have zero Hom between them, whereas their reductions in characteristic coincide and have a one-dimensional Hom space.