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Hom from a projective counts simple composition factors
Statement
Assume the Axiom of Choice (The Axiom of Choice). Let be a weight and let be the projective cover of produced by Category O has enough projectives. For every finite-length object of , the multiplicity of in a composition series of (Composition series and composition factors of an object).
Facts & Assumptions
Given: The Axiom of Choice, a weight , the projective cover of the previous theorem, and a finite-length object .
is projective, is indecomposable with local endomorphism ring, has a unique maximal proper subobject with simple, and the canonical epimorphism onto the head is essential with head . The functor is exact (Category O has enough projectives, Projective covers in O are indecomposable and unique, Projective object characterisations).
For a simple object of one has if and if : a nonzero morphism is an epimorphism, so is the head of and by [F1]; and for every nonzero morphism has kernel a maximal proper subobject, hence equal to by uniqueness, so all morphisms factor through the fixed quotient . Each endomorphism of this highest-weight simple acts by a scalar on its one-dimensional highest line, which generates the module, so . Simple labels are distinct by The simple objects of O. [F1]
Every object of has a finite composition series, and Jordan–Hölder makes its simple multiplicities independent of the series (Composition series and composition factors of an object, Every object of O has finite length, Jordan-Holder theorem in an abelian category). For , concatenate a composition series of with the inverse images of a composition series of : the resulting series of has precisely their combined factors, proving additivity. The empty series of zero has all multiplicities zero.
Proof
Since is projective, the functor is exact; in particular, for a short exact sequence with all terms of finite length if the two outer Hom spaces are finite-dimensional, so is the middle one and and by [F3].
If both sides are zero, and if is simple then for some and by [F2]; this is the base of the induction on the composition length.
Now let have finite length and induct on the length of a composition series . Assume as induction hypothesis that the identity holds for finite-length objects of smaller length. For both sides are zero. For the exact sequence has simple quotient , and steps 1.1 and 1.2 with the induction hypothesis give .
By induction on the length of a composition series, step 2.1 proves for every finite-length .
Depends on
Used by
- BGG reciprocity Theorem
Dependency tree · two levels
27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Lin Chen, lecture notes (Spring 2024), Lecture 8, Corollary 4.9 (standard reference, not scraped)
- Pavel Etingof, Representations of Lie Groups (18.757, Fall 2023), Proposition 16.2(ii) (standard reference, not scraped)