Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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An index-one moduli locus can be infinite without compactness

Statement refuted

Without a compactness hypothesis, discrete index-one Morse trajectory data are finite.

Witness

Let M=n1Rn and on Rn put

fn(x)=n2(3xx3),pn=1,qn=1.

Choose the locally normalized bounded downward-gradient-like field described by Xn=2uu near pn, Xn=2vv near qn, and Xn=fn/(1+(fn)2)x away from those charts, patched by disjoint nonnegative bump functions. It is complete. Each component has one unparametrized trajectory from the index-one maximum pn to the index-zero minimum qn; hence the global index-one locus is an infinite discrete union.

Facts & Assumptions

Given: The above disjoint-union field, with the local Morse coordinates and bump-function patching specified in the example.

[F1]

A downward gradient-like field has the stated exact local normal forms and strictly decreases f off critical points (Downward gradient-like vector fields for a Morse function).

Counterexample

technique · direct
1.1

Near pn and qn the displayed local fields are the required Morse normal forms; off them every patched summand has dfn(Xn)<0. The coefficients are bounded on fixed supports and the outside coefficient has absolute value at most 1/2, so every Xn is complete.

F1given
2.1

On each line the interval (1,1) is one flow orbit from pn to qn. Thus M(pn,qn) is a singleton and hence is discrete, without any appeal to Morse--Smale transversality.

step 1.1
3.1

Therefore M1(X):=n1M(pn,qn) is an infinite discrete set. It refutes finiteness of the global index-one locus without a compactness condition, not finiteness for one fixed endpoint pair.

step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources