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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02
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Positive sectional curvature with no fixed lower bound on a noncompact manifold

Statement refuted

False claim: every complete, noncompact Riemannian manifold of everywhere positive sectional curvature has sectional curvature bounded below by a positive constant, inf⁡K>0.

The claim is false already in the simplest noncompact model surface: the paraboloid P:={(x,y,z)∈R3:z=x2+y2} with the Riemannian metric induced from Euclidean R3 is complete and noncompact, has positive sectional curvature at every point, and its sectional curvature takes the values K=4(1+4r2)2>0,r2=x2+y2, at the point (x,y,x2+y2), so inf⁡PK=0: there is no positive lower bound.

Facts & Assumptions

Given: The paraboloid P⊆R3 with the metric induced by the Euclidean metric, and the inherited ACω of [A1].

[A1]

The countable-choice premise is the inherited ACω (The Axiom of Countable Choice (ACω)), carried by the hypersurface curvature and submanifold-completeness suppliers used below; the computation itself selects nothing.

[F1]

For an embedded Euclidean hypersurface of dimension m≥2 with a smooth unit normal and orthonormal principal directions ei,ej of principal curvatures κi,κj, the sectional curvature of span⁡(ei,ej) is κiκj (Euclidean hypersurface sectional curvature from principal curvatures). The principal curvatures are the eigenvalues of the shape operator S, so in dimension 2 the only sectional curvature is K=det⁡S=κ1κ2 (Shape operator, Principal curvatures, Gaussian curvature, and mean curvature of an oriented hypersurface).

[F2]

For a parametrized surface with first fundamental form g=(gij) and second fundamental form h=(hij), the identity hij=g(SXi,Xj) gives S=g−1h in any coordinate basis, so det⁡S=det⁡h/det⁡g (Induced connection and second fundamental form, Shape operator). For the graph of a smooth f over the (x,y)-plane with unit normal N=(−fx,−fy,1)/W, W=1+∣∇f∣2, one has gij=δij+fifj, det⁡g=1+∣∇f∣2, and hij=fij/W.

[F3]

The graph of a smooth map is an embedded submanifold (The graph of a smooth map is an embedded submanifold), and the graph of a continuous map into a Hausdorff space is closed in the product (The graph of a continuous map into a Hausdorff space is closed in the product).

[F4]

A closed embedded submanifold of a Riemannian manifold whose connected components are complete is complete in the induced Riemannian metric (Closed embedded submanifolds of complete Riemannian manifolds are complete); R3 with the Euclidean metric is a complete metric space (R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R, Complete metric space: every Cauchy sequence converges in the space), and its Riemannian distance is the Euclidean distance (Riemannian distance on a connected manifold).

[F5]

A metric space is compact exactly when every sequence in it has a convergent subsequence (Open cover, subcover, compact metric space, and compact subset of a metric space); a convergent sequence in R3 is bounded. The infimum of a nonempty set of reals bounded below is its greatest lower bound (Greatest lower bound (infimum)).

Counterexample

1.1F3F4given

The paraboloid is a complete, noncompact surface. [F3, F4, given] The map f:R2→R, f(x,y)=x2+y2, is smooth and continuous, so its graph P={(x,y,f(x,y))} is an embedded submanifold of R2×R≅R3 by [F3] and is closed in R3 by [F3]. The Euclidean metric of R3 is complete [F4], and its Riemannian distance is the Euclidean distance; hence [F4] makes P complete in the induced Riemannian metric. For noncompactness consider pt:=(t,0,t2)∈P for t=1,2,…: the Euclidean norms ∣pt∣2=t2+t4 are unbounded, so (pt) has no convergent subsequence and [F5] shows that P is not compact.

2.1F1F2step 1.1

The curvature at each point. [F1, F2, step 1.1] Parametrize P by X(x,y)=(x,y,f(x,y)), so that Xx=(1,0,2x), Xy=(0,1,2y) and g=(1+4x24xy4xy1+4y2),det⁡g=1+4(x2+y2)=1+4r2, where r2=x2+y2. The upward unit normal is N=(1+4r2)−1/2(−2x,−2y,1), and the second fundamental form has matrix h=(1+4r2)−1/2Hess⁡f=(1+4r2)−1/2(2002),det⁡h=41+4r2. By [F2] the shape operator satisfies S=g−1h and det⁡S=det⁡hdet⁡g=4/(1+4r2)1+4r2=4(1+4r2)2. By [F1] the sectional curvature of the tangent plane at X(x,y), the only tangent two-plane of the surface, is K=det⁡S=4/(1+4r2)2.

3.1F5step 2.1∎

The curvature is positive but its infimum is zero. [F5, step 2.1] For every (x,y) the numerator 4 and the denominator (1+4r2)2 are positive, so K=4/(1+4r2)2>0: the paraboloid has positive sectional curvature everywhere. On the other hand, given any δ>0 choose r>0 with (1+4r2)2>4/δ; then the point (r,0,r2)∈P has 0<K<δ. Hence 0 lies below every value of K but no positive number is a lower bound, so the greatest lower bound of the set of values of K is inf⁡PK=0 by [F5]: a uniform lower bound K≥c>0 fails on the complete noncompact manifold P, refuting the displayed claim. The paraboloid and the chosen radii are explicit, so the inherited ACω of [A1] is not drawn on beyond its declaration.

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