Two-corner homology bound with incoming and outgoing ranks
ProvedSmooth4Algebra.unsaturated_corner_budgetLet be square-zero in finite dimension over a field. Let and be two split, distinct linear blocks with inclusions and projections that are identities on their own block and zero on the other. Assume the outgoing/incoming ranks of are at most and , and those of are at most and . Then
This is the linear-algebra form of the unsaturated opposite-corner bound. In a bifiltered application, the corner-incidence argument must separately establish the four stated rank bounds.
import Definitions.Def_Smooth4AlgebraHomology set_option autoImplicit false
theorem Smooth4Algebra.unsaturated_corner_budget
{K V E F : Type*} [Field K] [AddCommGroup V] [Module K V]
[FiniteDimensional K V] [AddCommGroup E] [Module K E]
[FiniteDimensional K E] [AddCommGroup F] [Module K F]
[FiniteDimensional K F]
(d : V →ₗ[K] V) (h_square : d.comp d = 0)
(incE : E →ₗ[K] V) (incF : F →ₗ[K] V)
(projE : V →ₗ[K] E) (projF : V →ₗ[K] F)
(hE : projE.comp incE = LinearMap.id)
(hF : projF.comp incF = LinearMap.id)
(hEF : projE.comp incF = 0) (hFE : projF.comp incE = 0)
(rPlus rMinus cPlus cMinus : ℕ)
(hEout : Module.finrank K (LinearMap.range (d.comp incE)) ≤ cMinus)
(hEin : Module.finrank K (LinearMap.range (projE.comp d)) ≤ rPlus)
(hFout : Module.finrank K (LinearMap.range (d.comp incF)) ≤ rMinus)
(hFin : Module.finrank K (LinearMap.range (projF.comp d)) ≤ cPlus) :
(Module.finrank K E - rPlus - cMinus) +
(Module.finrank K F - rMinus - cPlus) ≤
Module.finrank K (Smooth4Algebra.Homology d) := by sorryRead-back
What the Lean code literally says, in plain math · Codex independent blind auditor; exact model identifier unavailable
For every field and every three finite-dimensional -vector spaces , each given as an additive commutative group with a -module structure, let be -linear with , and let , , , and be -linear maps satisfying , , , and . For any natural numbers satisfying , , , and , the inequality holds. The quotient views the intersection inside , and this intersection equals because . Every subtraction in the two summands is natural-number subtraction truncated at zero, so the summands are respectively and with integer expressions inside the maxima. Any of the spaces or bounds may be zero.
Confirmed by the mission captain (proposal self-audit).