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The absolute LKB cellular boundary and fraction-field rank

Statement

Let C be the unordered configuration of two points in C∖{p1,…,pn}, where p1<⋯<pn and n≥1. Let C~ be the regular cover determined by Φ(α)=qa(α)tb(α), where a is the sum of the two mobile points' winding numbers around all punctures and b is their mutual half-twist exponent. Write Λ=Z[q±1,t±1] and K=Q(q,t).

In the preceding two-dimensional model, the absolute covering chain groups have bases ai,bi,ci in degree one and Aij,Bir in degree two, with dAij=(q−1)(aj−bi), dBi1=(1−t)ai−ci+qci+1,dBi2=−tai+tbi−ci+ci+1, dBi3=(t−1)bi−qci+ci+1. Ordinary absolute H2(C~;Z) is the kernel of this differential. Its natural map to K⊗ΛH2(C~;Z) is injective, and that vector space has dimension (n2). These claims do not assert integral freeness or injection into end-relative homology.

Facts & Assumptions

Given: n≥1, the real punctures, the cover determined by Φ, and the rings Λ and K.

[F1]

An equivariant two-dimensional model for the LKB configuration space supplies the two-dimensional quotient cell model, its attaching words, and the deck-equivariant lifted homotopy equivalence.

[F2]

Cellular homology computes singular homology identifies the cellular homology of the covering CW complex with ordinary absolute singular homology.

Proof

1.1F1givenconstruct

Choose the directed edges in [F1] to make a positively oriented puncture meridian on returning along its barred edge. The barred grid is contractible, so its lift can be fixed consistently with every barred edge having displacement 1. The resulting ai and bi loops move one mobile point counterclockwise once around pi, with the other outside that small meridian disk. Thus their exponents are (a,b)=(1,0) and their displacement is q. Each ci crosses the diagonal between two real chambers exchanged by coordinate interchange; in the unordered quotient it exchanges the mobile points counterclockwise in their common puncture interval, enclosing no puncture. Its exponents are (0,1) and its displacement is t. Paths in the barred grid provide the basepoint paths, and changing those paths by a homotopy has no effect on these displacements. Both q and t occur, so the deck group is Z2.

2.1F1F2step 1.1construct

Lift every cell after fixing one basepoint lift. For a word, a positive edge contributes its prefix displacement times that edge, and an inverse edge contributes minus the displacement after traversing that inverse edge times the positive edge. Apply this to the four attaching words of [F1]. The word biajbi−1aj−1 contributes bi+qaj−qbi−aj. The word aici+1ai−1ci−1 contributes ai+qci+1−tai−ci. The word ci+1biai−1ci−1 contributes ci+1+tbi−tai−ci. Finally ci+1bici−1bi−1 contributes ci+1+tbi−qci−bi. These are exactly the displayed differential formulas. The lifted model has no 3-cells, so [F1] and [F2] identify H2(C~;Z) with ker⁡d⊂C2. These are absolute chains; no end neighbourhood or relative quotient has entered the construction.

3.1step 2.1algebra

Over K, let W be the span of all Bir. Suppose ∑i(riBi1+siBi2+uiBi3) has zero boundary. Its ai coefficient gives si=(1−t)t−1ri, and its bi coefficient then gives ui=ri. The remaining boundary is (q+t−1)∑iri(ci+1−ci). Since q+t−1≠0, its c1 coefficient forces r1=0, its c2 coefficient then forces r2=0, and successive coefficients force every ri=0. Hence also si=ui=0, and d∣W is injective. Projection onto the (n2) A-coordinates is consequently injective on ker⁡(d⊗K), giving an upper bound (n2) for its dimension.

4.1step 2.1step 3.1algebra

Put S=(t−1)(qt+1) and define Vib=−qtBi1+q(t−1)Bi2+Bi3, Via=Bi1+q(t−1)Bi2−qtBi3, and Vi0=−tBi1+(t−1)Bi2−tBi3. Substitution gives dVib=Sbi−(q−1)(qt+1)ci+1, dVia=−Sai+(q−1)(qt+1)ci, and dVi0=(qt+1)(ci−ci+1). Therefore the integral chains Eij=SAij+(q−1)Vib+(q−1)Vja+∑i<k<j(q−1)2Vk0 are cycles: the aj,bi terms cancel the boundary of SAij, and the c terms telescope. Their A-coordinates are S in coordinate (i,j) and zero elsewhere. Since S≠0, they are independent over K. Together with step 3.1 this proves that the field kernel has dimension (n2) and basis {Eij}. For n=1 there are no such chains, and step 3.1 says the full kernel is zero.

5.1step 2.1step 4.1algebra∎

The ring Λ is a domain: it is the localization of the polynomial domain Z[q,t] by monomials. Its finite free module C2 is torsion-free, and so is its submodule ker⁡d. If an element of ker⁡d maps to zero after localization, a nonzero denominator annihilates it; torsion-freeness makes it zero. Thus the localization map on absolute H2 is injective. Moreover every field cycle becomes an integral cycle after multiplying by a common nonzero denominator of its finitely many cellular coordinates. Hence localization of ker⁡d is exactly ker⁡(d⊗K), not just a subspace thereof. Step 4.1 now gives the asserted dimension. Integral spanning by the Eij was never used or inferred.

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