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A projective Verma flag need not split

Statement refuted

Every finite Verma flag of a projective object of O splits, that is, a projective object carrying a Verma flag is the direct sum of the standard factors of that flag.

Facts & Assumptions

Given: The Axiom of Choice, g=sl2, an integer m≥0, and the regular integral block with labels m and −m−2.

[F1]

The projective cover P(−m−2) carries the two-step Verma flag 0⊆Δ(m)⊆P(−m−2) with quotient Δ(−m−2), and the exact sequence 0→Δ(m)→P(−m−2)→Δ(−m−2)→0 is nonsplit; P(−m−2) is indecomposable with head L(−m−2) (The two projectives in the principal sl2 block).

[F2]

Every projective object of O has a finite Verma flag, and the factors of the flag of a projective cover are its standard factors with multiplicities (P(λ):Δ(μ)); the flag of Δ(m) has the single factor Δ(m) (Projectives in category O have finite Verma flags, Finite Verma flags and their multiplicities).

Counterexample

Assume the Axiom of Choice (The Axiom of Choice).

Proof technique: direct: exhibit the two-step flag of P(−m−2) and rule out a splitting by the head.

1.1F1F2

By [F1] the projective P(−m−2) has the finite Verma flag 0⊆Δ(m)⊆P(−m−2) with quotient Δ(−m−2), and the corresponding sequence is nonsplit. If the flag split, then P(−m−2)≅Δ(m)⊕Δ(−m−2).

2.1F1step 1.1

But Δ(m)=M(m) has head L(m), so the direct sum would have L(m) as a simple quotient, in addition to the quotient L(−m−2) from Δ(−m−2); a projective cover has a unique simple quotient, its head, which is L(−m−2) by [F1], and m≠−m−2. Hence no splitting exists.

3.1F1F2step 2.1∎

For m=0 this is the module P(−2) with head and socle L(−2) and middle composition factor L(0): the flag 0⊆Δ(0)⊆P(−2) with quotient Δ(−2)=L(−2) does not split, so the existence of a finite Verma flag for a projective (from [F2]) is strictly weaker than a direct-sum decomposition into standards.

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