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The Borel-Weil theorem

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let λ∈X∗(T). If λ is not dominant integral then H0(X,Lλ)=0. If λ is dominant integral then H0(X,Lλ)≅L(λ)∗ as g-modules, and Hi(X,Lλ)=0 for every i>0.

Facts & Assumptions

Given: The Axiom of Choice, the group G, its Borel B, the flag variety X=G/B of dimension N=∣Φ+∣, a weight λ∈X∗(T) and the equivariant line bundle Lλ.

[F1]

The function space Fλ={f∈O(G):f(gb)=λ(b)f(g)} is identified with H0(X,Lλ); it is a finite-dimensional g-module for the derived left translation action (x⋅f)(g)=ddt∣0f(exp⁡(−tx)g), and annihilation by every x∈n− is equivalent to invariance under left translation by the whole group U− (Sections of an associated line bundle as equivariant functions, The cohomology of a Borel-character line bundle is a rational G-module, A U−-invariant section is determined on the big cell).

[F2]

The space of U−-invariant functions in Fλ has dimension at most 1 and, when nonzero, consists of the weight-(−λ) line for the torus action; moreover f is determined by f(1) on this space (A U−-invariant section is determined on the big cell).

[F3]

For every dominant integral μ the subspace of n−-invariants of the finite-dimensional irreducible module L(μ) equals its lowest weight space L(μ)w0μ and is one-dimensional (The lowest weight space is the nilradical-invariant line).

[F4]

Every finite-dimensional g-module is completely reducible, and the finite-dimensional irreducible modules are exactly the L(μ) with μ dominant integral; the dual L(λ)∗ of L(λ) is irreducible of highest weight −w0λ (Weyl's complete reducibility theorem, Highest-weight classification, Highest weight of the dual representation).

[F5]

If λ is dominant integral, then there is a regular function v∈O(G) with v(gb)=λ(b)v(g) and v(1)=1; in particular Fλ≠0 (The dominant Borel-Weil section extends from the big cell).

[F6]

For μ=λ+ρ dominant and regular, there is a reduced expression w0=siN⋯si1 whose partial products wj=sij⋯si1 satisfy N(wjμ)=j and end at the strictly antidominant weight w0μ; consequently, with λj=wj⋅λ, the simple reflection used at step j satisfies ⟨λj,αij+1∨⟩≥0 (A regular weight has a unique dominant dot translate).

[F7]

If ⟨ν,α∨⟩≥−1 for a simple root α, then Hi(X,Lν)≅Hi+1(X,Lsα⋅ν) for all i≥0 (Rank-one cohomology shifts across a simple wall).

[F8]

X is smooth projective of pure dimension N=∣Φ+∣ and Lλ is locally free, so Hq(X,Lλ)=0 for all q>N, and each Hq is finite-dimensional (Serre duality for locally free sheaves on a smooth projective variety, A semisimple flag variety is smooth and projective, The cohomology of a Borel-character line bundle is a rational G-module).

[F9]

A dominant weight pairs nonnegatively with every positive root. A weight ν is dominant if and only if −w0ν is dominant: if ν is dominant then w0ν is antidominant, so −w0ν is dominant, and conversely if −w0ν is dominant then w0(−w0ν)=−ν is antidominant, so ν is dominant (Integral, dominant, and strictly dominant weights, Dot-Weyl facets and single-wall translation data, Weyl length equals inversion number).

Proof

1.1F1F2F3F4F9givenalgebra

By [F1], Fλ is a finite-dimensional g-module. If it is nonzero, complete reducibility [F4] gives Fλ≅⨁j=1kL(μj) with each μj dominant integral. By [F3] its n−-invariants have dimension k, while [F1]–[F2] identify them with the at-most-one-dimensional U−-invariant space. Thus k=1, and comparison of the invariant weights gives w0μ1=−λ. The root-sign property of w0 gives w0−1=w0 and makes −w0 preserve dominant integral weights by [F9]; therefore λ=−w0μ1 is dominant integral. Now [F4] applies to L(λ)∗, identifying it with L(−w0λ)=L(μ1)≅Fλ.

2.1F5step 1.1algebra

Suppose λ is dominant integral. By [F5], Fλ≠0, so step 1.1 gives Fλ≅L(λ)∗; this proves the second clause for H0. Conversely, if Fλ≠0 for an arbitrary weight λ, step 1.1 shows that λ is dominant integral, so for non-dominant λ one has H0(X,Lλ)=Fλ=0.

3.1F6F7F8step 2.1algebra∎

It remains to prove Hi(X,Lλ)=0 for i>0 when λ is dominant integral, which is the case in which step 2.1 has settled H0. Put μ=λ+ρ, which is dominant and regular, and use the reduced expression w0=siN⋯si1 and partial products wj of [F6], with λj=wj⋅λ, so that λj+ρ=wjμ. At the step passing from j to j+1 the construction of [F6] chooses the simple reflection sij+1 with ⟨wjμ,αij+1∨⟩>0 (the reflection increases the count N by one), so ⟨λj,αij+1∨⟩=⟨wjμ,αij+1∨⟩−1≥0≥−1 by [F6]; hence [F7] gives Hi(X,Lλj)≅Hi+1(X,Lλj+1) for all i≥0. Composing the N isomorphisms gives Hi(X,Lλ)≅Hi+N(X,Lw0⋅λ) for all i≥0. For i>0 one has i+N>N=dim⁡X, so Hi+N(X,Lw0⋅λ)=0 by [F8]. Therefore Hi(X,Lλ)=0 for every i>0, completing the proof of the second clause.

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