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Total critical point lower bound

Statement

Assume ACω. Let M be a closed smooth n-manifold, f a Morse function and F a field. Then #Crit⁡(f)=∑k=0nmk(f) ≥ ∑k=0nbk(M;F)=PM,F(1), and equality holds if and only if f is F-perfect.

Facts & Assumptions

Given: A closed smooth n-manifold M, a Morse function f:M→R, and a field F.

[F1]

The Morse numbers satisfy mk(f)=0 for k∉{0,…,n} and Mf(1)=∑k=0nmk(f)=#Crit⁡(f) (Morse numbers and the Morse polynomial).

[F2]

The F-Betti numbers are bk(M;F)=dim⁡FHk(M;F) and PM,F(1)=∑kbk(M;F) (Poincare polynomial of a space and of a pair over a field).

[F3]

In the situation of the Morse polynomial identity, mk(f)≥bk(M;F) for every k (Weak Morse inequalities).

[F4]

f is F-perfect exactly when mk(f)=bk(M;F) for every k (Perfect Morse function over a field).

Proof

technique · termwise-sum
1.1F1F2F3givenalgebra

Summing the weak inequalities of [F3] over k=0,…,n gives ∑k=0nmk(f)≥∑k=0nbk(M;F), and both sums are finite. By [F1] the left side is Mf(1)=#Crit⁡(f) and by [F2] the right side is PM,F(1).

2.1F3F4step 1.1algebra∎

Each difference mk(f)−bk(M;F) is nonnegative by [F3]; a finite sum of nonnegative integers vanishes exactly when every summand does. Hence equality in step 1.1 holds if and only if mk(f)=bk(M;F) for all k, which by [F4] is precisely F-perfectness of f.

Remarks

  • The bound is the coarsest numerical obstruction supplied by the page: it uses only the total number of critical points and the total Betti number, and it is attained exactly in the perfect case. The Euler characteristic identity refines the alternating version of this count.

Depends on

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Sources