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A Morse function on a compact manifold has finitely many critical points

Statement

If $M$ is a compact smooth manifold and $f : M \to R$ is a Morse function, then $f$ has only finitely many critical points.

Facts & Assumptions

Given: A compact smooth manifold $M$ and a Morse function $f : M \to R$ .

[F1]

Every critical point of a Morse function is nondegenerate (Morse functions and excellent Morse functions).

[L1]

Every nondegenerate critical point is isolated (Nondegenerate critical points are isolated).

[L2]

Compactness means that every open cover has a finite subcover (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L3]

The vanishing of $d f$ in a chart is equivalent to the vanishing of all coordinate partial derivatives (Coordinate formula for the differential of a function).

Proof

technique · compactness cover

1.1

For each critical point $p$ , [F1] and [L1] give an open neighbourhood $U_{p}$ containing no critical point other than $p$ .

F1L1givenconstruct

1.2

If $q$ is not critical, choose a chart $x : U \to R^{n}$ around $q$ and write $g : = f \circ x^{- 1}$ . By [L3], some partial derivative $\partial g / \partial x^{i}$ is nonzero at $x (q)$ ; continuity keeps it nonzero on a smaller open neighbourhood $V_{q} \subseteq U$ , so $V_{q}$ contains no critical point.

L3givenconstruct

2.1

The family of all $U_{p}$ together with all $V_{q}$ covers $M$ . By [L2], it has a finite subcover. Only finitely many sets of the form $U_{p}$ occur in that subcover, and each such set contains exactly one critical point by step 1.1. Therefore $f$ has finitely many critical points.

L2step 1.1step 1.2

3.1

Hence every Morse function on a compact manifold has finitely many critical points.

step 2.1∎

Depends on

Used by

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Sources

Michele Audin and Mihai Damian, Morse Theory and Floer Homology (standard reference, not scraped)
Liviu I. Nicolaescu, An Invitation to Morse Theory, 2nd ed. (standard reference, not scraped)