False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Zero mean curvature implies a submanifold is totally geodesic

Statement refuted

False claim: if a positive-dimensional Riemannian submanifold has zero mean-curvature vector, then it is totally geodesic.

Assume ${A C}_{ω}$ . Mean curvature is only the trace of the second fundamental form, so nonzero trace-free extrinsic curvature can remain.

Facts & Assumptions

Given: Countable choice and the open parameter domain $U = (- π, π) \times R$ .

[F1]

For a positive-dimensional immersion, the averaged mean-curvature vector is $H = \frac{1}{m} {tr}_{g} I I$ . Mean curvature vector.

[F2]

An embedded Riemannian submanifold is totally geodesic exactly when its normal-valued second fundamental form vanishes identically. Totally geodesic submanifold.

[F3]

The second fundamental form is the normal component of the ambient covariant derivative, and the Euclidean Levi–Civita symbols vanish in Cartesian coordinates. Induced connection and second fundamental form, Christoffel formula for the levi civita connection.

[F4]

The derivatives and identities $(\cosh v)^{'} = \sinh v$ , $(\sinh v)^{'} = \cosh v$ , $\cosh^{2} v - \sinh^{2} v = 1$ , $(\sin u)^{'} = \cos u$ , $(\cos u)^{'} = - \sin u$ , and $\sin^{2} u + \cos^{2} u = 1$ hold. Addition formulas, identities, parity, and derivatives of the hyperbolic functions, The derivatives of sine and cosine are cosine and minus sine, Pythagorean and parity identities for all six trigonometric functions on their natural domains.

Refutation

technique · direct counterexample

1.1

Define $X : U \to R^{3}$ by $X (u, v) = (\cosh v \cos u, \cosh v \sin u, v)$ . Using [F4], $X_{u} = (- \cosh v \sin u, \cosh v \cos u, 0)$ and $X_{v} = (\sinh v \cos u, \sinh v \sin u, 1)$ , so $⟨ X_{u}, X_{u} ⟩ = ⟨ X_{v}, X_{v} ⟩ = \cosh^{2} v$ and $⟨ X_{u}, X_{v} ⟩ = 0$ . Thus $X$ is an immersion with induced metric $\cosh^{2} v (d u^{2} + d v^{2})$ . The coordinate $v$ is recovered from the third component, and $u \in (- π, π)$ is recovered from the seam-free circle coordinate, so this chart is an embedding onto its image.

F4algebraconstruct

2.1

The cross product from step 1.1 has length $\cosh^{2} v$ , yielding the smooth unit normal $N = (sech v \cos u, sech v \sin u, - \tanh v)$ . The second derivatives are $X_{u u} = (- \cosh v \cos u, - \cosh v \sin u, 0)$ , $X_{u v} = (- \sinh v \sin u, \sinh v \cos u, 0)$ , and $X_{v v} = (\cosh v \cos u, \cosh v \sin u, 0)$ . Their inner products with $N$ are respectively $- 1, 0, 1$ , so [F3] gives $I I (X_{u}, X_{u}) = - N$ , $I I (X_{u}, X_{v}) = 0$ , and $I I (X_{v}, X_{v}) = N$ .

F3F4step 1.1algebra

3.1

The vectors $e_{1} = X_{u} / \cosh v$ and $e_{2} = X_{v} / \cosh v$ are orthonormal by step 1.1. Step 2.1 therefore gives $I I (e_{1}, e_{1}) = - {sech}^{2} v N$ and $I I (e_{2}, e_{2}) = {sech}^{2} v N$ . By [F1], $H = \frac{1}{2} (I I (e_{1}, e_{1}) + I I (e_{2}, e_{2})) = 0$ everywhere.

F1step 1.1step 2.1algebra

4.1

Nevertheless step 2.1 gives $I I (X_{u}, X_{u}) = - N \neq 0$ at every point (at $(u, v) = (0, 0)$ it is the explicit vector $(- 1, 0, 0)$ ). By [F2] the catenoid chart is not totally geodesic, which refutes the claim.

F2step 1.1step 2.1step 3.1

5.1

The witness is nonempty, boundaryless, two-dimensional, and its induced conformal factor $\cosh^{2} v$ is strictly positive. The open angular interval excludes both seam endpoints; no limiting assertion is made. Empty and zero-dimensional cases are outside the positive-dimensional claim, while in dimension one [F1] makes $H = I I (e, e)$ , so the implication happens to hold there. Countable choice is inherited exactly through [F1]–[F3]; the parametrization, frame, and normal are explicit and add no choice. No biconditional is asserted.

F1F2F3F4step 1.1step 2.1step 3.1step 4.1∎

Depends on

Used by

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Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Danny Calegari, Minimal Surfaces (standard reference, not scraped)
John M. Lee, Riemannian Manifolds: An Introduction to Curvature (standard reference, not scraped)