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Euler equality alone does not imply perfectness

Statement refuted

False claim: for a Morse function on a closed smooth manifold the Euler characteristic identity $\sum_{p} (- 1)^{ind (p)} = χ (M)$ forces the function to be perfect over every field, equivalently forces the correction polynomial $Q$ of the Morse polynomial identity to vanish.

Assume ${A C}_{ω}$ . Start with the two-critical-point presentation of $S^{2}$ (one $0$ -handle and one $2$ -handle, induced by the height function) and insert a geometrically cancelling $(0, 1)$ -pair between them. The resulting presentation has handles of indices $0, 0, 1, 2$ , and the corresponding Morse function $f^{'}$ on $S^{2}$ has Morse numbers $m_{0} = 2$ , $m_{1} = 1$ , $m_{2} = 1$ , so $M_{f^{'}} (t) = 2 + t + t^{2}, \sum_{k} (- 1)^{k} m_{k} (f^{'}) = 2 - 1 + 1 = 2 = χ (S^{2}) .$ Over every field, $P_{S^{2}, F} (t) = 1 + t^{2}$ , hence $b_{0} = 1 < 2 = m_{0}$ ; the function is not perfect over any field and its correction polynomial is $Q = 1 \neq 0$ . Thus the Euler characteristic identity, which is an equality of alternating sums, does not by itself force perfectness or the vanishing of $Q$ .

Facts & Assumptions

Given: The two-critical-point presentation of $S^{2}$ , a geometrically cancelling $(0, 1)$ -pair inserted between its handles, the resulting presentation, and a Morse function $f^{'}$ inducing it.

[F1]

The creation theorem inserts a cancelling pair of consecutive indices and produces a diffeomorphism of the modified manifold with the original one relative to the incoming boundary, so the modified presentation presents $S^{2}$ (Creation of a cancelling handle pair).

[F2]

Morse functions inducing handle presentations have Morse numbers equal to the handle counts by index (Morse functions and handle decompositions correspond, Morse numbers and the Morse polynomial).

[F3]

The height function on $S^{2}$ is a Morse function with two critical points of indices $0$ and $2$ and Morse polynomial $1 + t^{2}$ (computed below); over every field $b_{0} (S^{2}; F) = b_{2} (S^{2}; F) = 1$ and $b_{1} (S^{2}; F) = 0$ (Homology of spheres, Poincare polynomial of a space and of a pair over a field).

[F4]

$f^{'}$ is $F$ -perfect exactly when $m_{k} (f^{'}) = b_{k} (S^{2}; F)$ for all $k$ (Perfect Morse function over a field).

[F5]

For every field there is a unique $Q \in Z [t]$ with nonnegative coefficients and $M_{f^{'}} = P_{S^{2}, F} + (1 + t) Q$ , and the Euler characteristic identity $\sum_{k} (- 1)^{k} m_{k} (f^{'}) = χ (S^{2})$ holds (Morse polynomial identity, Morse Euler characteristic identity).

Counterexample

technique · explicit-construction

1.1givenalgebra

For $h (x) = x_{3}$ on $S^{2}$ , a point away from the poles has tangent vector $v = e_{3} - x_{3} x$ with $d h (v) = 1 - x_{3}^{2} > 0$ , so it is not critical. In pole charts $h (u) = \pm \sqrt{1 - ∣ u ∣^{2}}$ has Hessian $\mp I_{2}$ at $u = 0$ ; thus the south and north poles have indices $0, 2$ , and $M_{h} = 1 + t^{2}$ . Sphere homology gives $P_{S^{2}, F} = 1 + t^{2}$ , so the correction polynomial is $Q = 0$ .

1.2F1given

Apply [F1] at the disk stage, whose outgoing circle is nonempty, and transport the original final $2$ -handle attaching map across the supplied boundary diffeomorphism. The inserted $(0, 1)$ -pair is cancelling and the modified presentation still presents $S^{2}$ ; its handles are the original $0$ -handle and $2$ -handle together with the new $0$ -handle and $1$ -handle, so the presentation has handles of indices $0, 0, 1, 2$ .

2.1F2step 1.2

Let $f^{'}$ be a Morse function inducing the modified presentation. By [F2] its Morse numbers equal the handle counts by index, that is $m_{0} (f^{'}) = 2$ , $m_{1} (f^{'}) = 1$ , $m_{2} (f^{'}) = 1$ , and all other Morse numbers vanish; hence $M_{f^{'}} (t) = 2 + t + t^{2}$ .

3.1F3F4step 2.1

Over every field $F$ , [F3] gives $b_{0} (S^{2}; F) = 1$ , $b_{1} (S^{2}; F) = 0$ and $b_{2} (S^{2}; F) = 1$ , so $P_{S^{2}, F} (t) = 1 + t^{2}$ . Comparing with step 2.1, $m_{0} (f^{'}) = 2 > 1 = b_{0} (S^{2}; F)$ , and by [F4] the function $f^{'}$ is not $F$ -perfect, for any field $F$ .

4.1F5step 2.1step 3.1

The correction polynomial is computed by the identity of [F5]: $2 + t + t^{2} = (1 + t^{2}) + (1 + t) \cdot 1$ , so the unique correction polynomial is $Q = 1 \neq 0$ .

5.1F5step 2.1step 3.1step 4.1∎

Finally the Euler equality holds: $\sum_{k} (- 1)^{k} m_{k} (f^{'}) = 2 - 1 + 1 = 2$ , and by the Euler identity of [F5] this equals $χ (S^{2})$ ; the same alternating sum computed from the Betti numbers is $1 - 0 + 1 = 2$ . Thus the Euler characteristic identity is satisfied while perfectness fails and $Q \neq 0$ , refuting the displayed false claim.

Remarks

Why the claim fails. The Euler identity is the value at $t = - 1$ of the Morse polynomial identity; the factor $(1 + t)$ vanishes there, so the correction polynomial is invisible to it. Here $Q = 1$ gives $(1 + t) Q = 1 + t$ , an excess in degrees zero and one which cancels in the alternating sum.
Consistency with the weak inequalities. The failure of perfectness is detected by the weak inequality $m_{0} \geq b_{0}$ , which is strict; deleting the cancelling pair recovers the original presentation and leaves homology unchanged.

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Sources

Michele Audin and Mihai Damian, Morse Theory and Floer Homology, Part I Chapter 4 Section 4.4, printed pp. 88-91 (PDF pp. 98-100) (standard reference, not scraped)
Liviu Nicolaescu, An Invitation to Morse Theory (2nd ed.), Chapter 2 Section 2.3, printed pp. 46-53 (PDF pp. 56-63) (standard reference, not scraped)