Dimensions of relative homology along the long exact sequence of a triple #
Let B ⊆ A ⊆ X be a triple of spaces and V a module over a division ring k, and write
Hᵢ(-) for relative singular homology with coefficients in V. The long exact sequence of the
triple
⋯ ⟶ Hₙ₊₁(X, A) ⟶ Hₙ(A, B) ⟶ Hₙ(X, B) ⟶ Hₙ(X, A) ⟶ ⋯ ⟶ H₀(X, A) ⟶ 0
controls the dimensions of the homology of (X, B) by those of (A, B) and (X, A):
Hₙ(X, B)is finite-dimensional whenHₙ(A, B)andHₙ(X, A)are (TauCeti.TopTriple.finite_singularHomology_totalPair), and thendim Hₙ(X, B) ≤ dim Hₙ(A, B) + dim Hₙ(X, A)(TauCeti.TopTriple.finrank_singularHomology_totalPair_le);- when the homology of
(A, B)and(X, A)is finite-dimensional in every degree, the alternating sumsSₙ(P) = dim Hₙ(P) - dim Hₙ₋₁(P) + ⋯ ± dim H₀(P)satisfySₙ(A, B) - Sₙ(X, B) + Sₙ(X, A) = dim im (Hₙ₊₁(X, A) ⟶ Hₙ(A, B))(TauCeti.TopTriple.sum_range_neg_one_pow_mul_finrank_singularHomology_eq). HenceSₙis subadditive,Sₙ(X, B) ≤ Sₙ(A, B) + Sₙ(X, A)(TauCeti.TopTriple.sum_range_neg_one_pow_mul_finrank_singularHomology_totalPair_le), and when moreoverHₙ₊₁(X, A)vanishes, the alternating sums with signs(-1)ⁱare additive (TauCeti.TopTriple.sum_range_neg_one_pow_mul_finrank_singularHomology_totalPair_eq).
Along a filtration A 0 ⊆ A 1 ⊆ ⋯ ⊆ A (N + 1) whose steps have finite-dimensional relative
homology in degree i, so does (A (N + 1), A 0)
(TauCeti.finite_singularHomology_ofInclusion_of_monotone). When the steps have
finite-dimensional relative homology in every degree, the defect
∑ⱼ Sₙ(A (j + 1), A j) - Sₙ(A (N + 1), A 0) of subadditivity is the sum of the ranks of the
connecting maps Hₙ₊₁(A (j + 2), A (j + 1)) ⟶ Hₙ(A (j + 1), A 0) of the triples
(A (j + 2), A (j + 1), A 0) (TauCeti.sum_range_finrank_range_singularHomologyδ_eq).
Milnor derives the Morse inequalities from these two subadditivity properties, applied to the triples of sublevels of a Morse function. When a compact band is cut at regular values into pieces containing at most one critical value each, every piece satisfies the Morse inequalities with equality, so each coefficient of the polynomial measuring the failure of the Morse inequalities to be equalities is a sum of ranks of connecting maps along the resulting filtration by sublevels. A coarser filtration can hide cancellations within a piece, and then these ranks need not recover the coefficients.
References #
- J. Milnor, Morse Theory, Annals of Mathematics Studies 51, Princeton University Press, 1963, Section 5.
Finiteness along a triple. If Hₙ(A, B) and Hₙ(X, A) are finite-dimensional, so is
Hₙ(X, B).
Subadditivity of dimension along a triple. If Hₙ(A, B) and Hₙ(X, A) are
finite-dimensional, then dim Hₙ(X, B) ≤ dim Hₙ(A, B) + dim Hₙ(X, A).
The dimension count along the long exact sequence of a triple. If the relative homology of
(A, B) and of (X, A) is finite-dimensional in every degree, then the alternating sums through
degree n satisfy Sₙ(A, B) - Sₙ(X, B) + Sₙ(X, A) = dim im (Hₙ₊₁(X, A) ⟶ Hₙ(A, B)), where
Sₙ(P) = dim Hₙ(P) - dim Hₙ₋₁(P) + ⋯ ± dim H₀(P).
Subadditivity of the alternating sums along a triple. If the relative homology of (A, B)
and of (X, A) is finite-dimensional in every degree, then Sₙ(X, B) ≤ Sₙ(A, B) + Sₙ(X, A),
where Sₙ(P) = dim Hₙ(P) - dim Hₙ₋₁(P) + ⋯ ± dim H₀(P).
Additivity of the alternating sums along a triple. If the relative homology of (A, B)
and of (X, A) is finite-dimensional in every degree and Hₙ₊₁(X, A) vanishes, then
∑ᵢ₌₀ⁿ (-1)ⁱ dim Hᵢ(X, B) = ∑ᵢ₌₀ⁿ (-1)ⁱ dim Hᵢ(A, B) + ∑ᵢ₌₀ⁿ (-1)ⁱ dim Hᵢ(X, A).
Finite filtrations #
Finiteness along a filtration. If A 0 ⊆ A 1 ⊆ ⋯ ⊆ A (N + 1) and every step
Hᵢ(A (j + 1), A j), j ≤ N, is finite-dimensional, then so is Hᵢ(A (N + 1), A 0).
The dimension count along a filtration. Let A 0 ⊆ A 1 ⊆ ⋯ ⊆ A (N + 1) be subspaces of
a space such that every step Hᵢ(A (j + 1), A j), j ≤ N, is finite-dimensional in every
degree, and write Sₙ(P) = dim Hₙ(P) - dim Hₙ₋₁(P) + ⋯ ± dim H₀(P). Then the defect of
subadditivity of Sₙ along the filtration,
Sₙ(A 1, A 0) + ⋯ + Sₙ(A (N + 1), A N) - Sₙ(A (N + 1), A 0), is the sum over j < N of the
ranks of the connecting maps Hₙ₊₁(A (j + 2), A (j + 1)) ⟶ Hₙ(A (j + 1), A 0) of the triples
(A (j + 2), A (j + 1), A 0).