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The height function on a sphere is perfect

Example

Assume ACω. For n≥1 let h:Sn→R,h(x)=xn+1, be the height function on the unit sphere (Euclidean spheres and closed balls as subspaces of Rn). Its only critical points are the two poles ±en+1, nondegenerate of indices 0 and n, so Mh(t)=1+tn,#Crit⁡(h)=2. Over every field F, the homology of spheres gives b0(Sn;F)=bn(Sn;F)=1 and bk(Sn;F)=0 otherwise, hence PSn,F(t)=1+tn=Mh(t): the height function is F-perfect for every F, with correction polynomial Q=0, and every weak inequality is an equality. For n=1 this is the circle with one minimum and one maximum.

Facts & Assumptions

Given: An integer n≥1, the height function h(x)=xn+1 on the unit sphere Sn, and a field F.

[F2]

The Morse numbers are mk(h)=#{p∈Crit⁡(h):ind⁡(p)=k} and Mh(t)=∑kmk(h)tk (Morse numbers and the Morse polynomial).

[F3]

PX,F(t)=∑kdim⁡FHk(X;F)tk is the Poincare polynomial over F and bk(X;F)=dim⁡FHk(X;F) the F-Betti numbers (Poincare polynomial of a space and of a pair over a field).

[L1]

For n≥1, H~k(Sn;G) is G for k=n and 0 otherwise; in particular H0(Sn;G)≅G. (Homology of spheres).

[F4]

There is a unique Q∈Z[t] with nonnegative coefficients and Mh(t)=PSn,F(t)+(1+t)Q(t) (Morse polynomial identity).

[F5]

h is F-perfect when mk(h)=bk(Sn;F) for all k, equivalently when Mh=PSn,F (Perfect Morse function over a field).

Verification

technique · direct-local-model
1.1F1givenalgebra

If x∈Sn is not a pole, put v:=en+1−xn+1x. Then x⋅v=xn+1−xn+1(x⋅x)=0, so v∈TxSn, and dhx(v)=vn+1=1−xn+12≠0 because xn+12<1 off the poles. Hence only the poles can be critical points of h.

2.1F1F2step 1.1

Near the north pole write the upper hemisphere as u↦(u,1−∥u∥2), so that h(u)=1−∥u∥2=1−12∥u∥2+O(∥u∥4); the Hessian at u=0 is −In, hence the north pole is a nondegenerate critical point of index n. Near the south pole write u↦(u,−1−∥u∥2), so that h(u)=−1−∥u∥2=−1+12∥u∥2+O(∥u∥4); the Hessian at u=0 is In and the south pole has index 0. Therefore Crit⁡(h)={±en+1} and, by [F2], Mh(t)=1+tn,#Crit⁡(h)=2.

3.1L1F3step 2.1

By [L1] read in unreduced form, b0(Sn;F)=bn(Sn;F)=1 and bk(Sn;F)=0 for k∉{0,n}; hence by [F3] PSn,F(t)=1+tn=Mh(t).

4.1F4F5step 2.1step 3.1∎

The correction polynomial of the Morse polynomial identity is unique [F4]; since Q=0 satisfies Mh=PSn,F+(1+t)Q, the actual correction polynomial is Q=0 and mk(h)=bk(Sn;F) for every k. By [F5] the height function is F-perfect, for every field F, and every weak inequality mk≥bk is an equality.

Remarks

  • Endpoint indices. The example realizes the extreme indices 0 and n and shows that the equality case of every inequality occurs simultaneously; it is the simplest perfect Morse function.
  • The case n=1. For the circle the two poles are a minimum and a maximum of indices 0 and 1, and Mh(t)=1+t=PS1,F(t) over every field.

Depends on

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Sources