Alphabeta Math
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14 results · all verified · 9 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Constant Rank, Submersions, Immersions and Regular Level Sets: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

A Euclidean sphere is a regular level set with tangent hyperplanes

Example

For R>0, the sphere SRm1={xRm:x2=R} is the regular level F1(R2) of F(x)=x22. At xSRm1, TxSRm1=x={h:x,h=0}.

Facts & Assumptions

Given: A radius R>0 and F(x)=x,x on Rm, with m1.

[L2]

A regular level is locally a graph, its tangent space is the derivative kernel, and its tangent vectors are exactly its curve velocities (A regular level set is locally a Ck graph of dimension mn, The tangent space to a regular level set, Tangent vectors to a regular level set are exactly its curve velocities).

Verification

technique · direct
1.1

By [L1], F1(R2)=SRm1 and DF(x)h=2x,h.

givenL1
2.1

If x lies on the sphere, then x0 and DF(x)x=2R20, so DF(x):RmR is surjective.

step 1.1algebra
3.1

Thus R2 is a regular value. By [L2], the sphere is locally a graph and TxSRm1=kerDF(x)=x, with the same set realized by curve velocities.

step 2.1L2
4.1

Negative levels are empty and hence regular by the vacuous convention; the zero level is {0} and is critical because DF(0)=0. Neither boundary case is included in the positive-radius claim.

step 1.1
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

A positive-definite quadratic ellipsoid is a regular level set

Example

Let A be a symmetric positive-definite real m×m matrix and put F(x)=Ax,x. The ellipsoid F1(1) is a regular level set, and TxF1(1)=(Ax).

Facts & Assumptions

Given: A symmetric positive-definite matrix A and the quadratic function F.

[L3]

At a regular level point, the level is locally a graph and its tangent space is the derivative kernel (A regular level set is locally a Ck graph of dimension mn, The tangent space to a regular level set).

Verification

technique · direct
1.1

If F(x)=1, then x0 by [L1], and Ax0 because Ax=0 would give F(x)=0.

givenL1
2.1

By [L2], DF(x)(Ax)=2Ax22>0, so DF(x) is a nonzero functional and hence surjective onto R.

step 1.1L2
3.1

Therefore 1 is a regular value, and [L3] gives TxF1(1)=kerDF(x)={h:Ax,h=0}=(Ax).

step 2.1L3
4.1

The calculation also shows that no singular point can occur on the asserted level; positive definiteness and the level value 1 exclude the only possible degeneracy x=0.

step 1.1step 3.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24Open item page →

The graph of a Ck Euclidean map is a regular level set

Example

Let ψ:URmRn be Ck, k1, and define G:U×RnRn by G(x,y)=yψ(x). Then 0 is a regular value, G1(0) is the graph of ψ, and T(x,ψ(x))G1(0)={(v,Dψ(x)v):vRm}.

Facts & Assumptions

Given: The map ψ and the associated map G.

[L1]

Finite sums and scalar multiples of Ck Euclidean maps are Ck, coordinate maps are Ck componentwise, and total-derivative algebra gives DG(x,y)(v,w)=wDψ(x)v (Ck Euclidean maps are closed under componentwise algebra and composition, Ck Euclidean maps and diffeomorphisms, Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives).

[L2]

A regular level is locally a graph and has tangent space equal to the derivative kernel (A regular level set is locally a Ck graph of dimension mn, The tangent space to a regular level set).

Verification

technique · direct
1.1

The equation G(x,y)=0 is equivalent to y=ψ(x), so G1(0) is precisely the graph.

givenalgebra
1.2

By [L1], DG(x,y)(0,w)=w for every wRn, so DG(x,y) is surjective at every point and 0 is a regular value.

givenL1
2.1

Solving DG(x,ψ(x))(v,w)=0 gives w=Dψ(x)v, and [L2] identifies this kernel with the displayed tangent space.

step 1.2L1L2
3.1

The graph conclusion holds on the whole open set U, including when U is empty, in which case both sides are empty.

step 1.1step 2.1
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

The one-sheeted hyperboloid is a regular surface of revolution

Example

The one-sheeted hyperboloid H={(x,y,z)R3:x2+y2z2=1} is a regular level set and the surface obtained by rotating the profile x=1+z2 in the half-plane y=0, x>0, about the z-axis.

Facts & Assumptions

Given: The polynomial F(x,y,z)=x2+y2z2.

[L2]

Every positive real has a unique positive square root (Existence and uniqueness of n-th roots: a unique a1/n0 with (a1/n)n=a); a regular level is locally a graph with tangent space equal to the derivative kernel (A regular level set is locally a Ck graph of dimension mn, The tangent space to a regular level set).

Verification

technique · direct
1.1

On F1(1) the point (x,y,z) cannot be 0, so the coefficient vector (2x,2y,2z) in [L1] is nonzero and DF(x,y,z) is surjective onto R.

givenL1
1.2

The level equation is x2+y2=1+z2. By [L2], for each z its horizontal section is the circle of positive radius 1+z2, exactly the rotation of the stated profile.

givenL2algebra
2.1

Hence 1 is a regular value, and [L2] gives tangent plane {h:xh1+yh2zh3=0} at (x,y,z).

step 1.1L2
3.1

The radius never vanishes, so the rotation has no apex or rank-drop point; steps 1.1 and 1.2 establish both asserted properties.

step 1.1step 1.2
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

The orthogonal group is a regular level set of dimension n(n1)/2

Example

Let n1. Identify Mn(R) with Rn2 entrywise, and identify the symmetric n×n real matrices with Rn(n+1)/2 by listing the entries in the positions (i,j) with ij. Under these identifications let f:Mn(R)Symn(R),f(A)=ATA, so that f is a map between Euclidean spaces of dimensions n2 and n(n+1)/2.

Then f is C, its derivative is Df(A)H=ATH+HTA, and In is a regular value of f. Consequently O(n)={AMn(R):ATA=In}=f1(In) is a regular level set: near each of its points it is a C graph of dimension n2n(n+1)2=n(n1)2, and its tangent space at AO(n) is TAO(n)={AK:KMn(R), KT=K}, of dimension n(n1)/2.

At n=1 the target dimension equals the source dimension, O(1)={1,1}, and the graph dimension is 0: the two points are isolated.

Facts & Assumptions

Given: A natural number n1, the entrywise identifications above, and the map f with components fij(A)=k<nakiakj for ij.

[L2]

If f is totally differentiable at A, then the directional derivative DHf(A) exists for every H and equals Df(A)H (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L3]

A C1 map is a submersion at a point when its derivative there is surjective, and a value is regular when every point of its fibre is a submersion point (Submersions and immersions between Euclidean open sets, Regular and critical points, regular and critical values, and level sets).

[L4]

Near each of its points a regular level set of a Ck map URmRN is a Ck graph of dimension mN, and its tangent space at such a point is the kernel of the derivative (A regular level set is locally a Ck graph of dimension mn, The tangent space to a regular level set).

[L5]

A linear map is injective exactly when its kernel is trivial, and for a linear map on a finite-dimensional space the dimension of the space is the sum of the dimensions of the kernel and the image (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial, Rank-nullity: dimFV=nullityT+rankT).

Verification

technique · direct
1.1

By [L1], f is C and totally differentiable at every A.

givenL1
1.2

Let Af1(In), so ATA=In. If Ax=0 then x=Inx=ATAx=0, so by [L5] the map xAx is injective and therefore, its kernel being trivial, surjective on Rn; hence A is invertible and A1=AT, so also AAT=In.

givenL5algebra
2.1

Fix A,HMn(R). Then (A+tH)T(A+tH)=ATA+t(ATH+HTA)+t2HTH, a polynomial in t with matrix coefficients, so its derivative at t=0 is ATH+HTA. By [L2] this directional derivative is Df(A)H, so Df(A)H=ATH+HTA. This matrix is symmetric, as the target requires.

step 1.1givenL2algebra
3.1

Let S be symmetric and put H=12AS. Then ATH=12ATAS=12S, and HT=12SAT gives HTA=12SATA=12S. By step 2.1, Df(A)H=S, so Df(A) is surjective onto Symn(R).

step 2.1step 1.2algebra
4.1

By [L3], every point of f1(In) is a submersion point, so In is a regular value and f1(In)=O(n) is a regular level set.

step 3.1L3
5.1

By [L4] with m=n2 and N=n(n+1)/2, near each of its points O(n) is a C graph of dimension n2n(n+1)/2=n(n1)/2, and TAO(n)=kerDf(A)={H:ATH+HTA=0}.

step 4.1L4algebra
6.1

If KT=K and H=AK, then by step 1.2 ATH=K and HTA=KTATA=KT=K, so HkerDf(A). Conversely, if ATH+HTA=0, put K=ATH; then KT=HTA=K and AK=AATH=H by step 1.2. Hence TAO(n)={AK:KT=K}.

step 5.1step 1.2algebra
7.1

The map KAK is linear and injective, because A is invertible by step 1.2, so by [L5] its image has the dimension of its domain. A skew-symmetric matrix is determined freely by its entries strictly above the diagonal and has zero diagonal, so the skew-symmetric matrices have dimension n(n1)/2, and dimTAO(n)=n(n1)/2.

step 6.1step 1.2L5algebra
8.1

At n=1 the source and target both have dimension 1, f(a)=a2, and f1(1)={1,1}, on which f(a)=2a0; the graph dimension n(n1)/2 is 0, so each point is isolated, and the skew-symmetric 1×1 matrices are {0}, in agreement with step 7.1.

step 5.1step 7.1algebra
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

The cone x2+y2=z2 has a rank drop at its apex

Statement refuted

A level set of a smooth map need not have constant derivative rank. For F(x,y,z)=x2+y2z2, the zero level is regular away from its apex and has derivative rank 0 at the apex.

Facts & Assumptions

Given: The smooth map F:R3R, F(x,y,z)=x2+y2z2.

[L2]

A point is regular exactly when its derivative is surjective, and the regular-level graph theorem requires that hypothesis (Regular and critical points, regular and critical values, and level sets, A regular level set is locally a Ck graph of dimension mn).

Counterexample

technique · direct
1.1

The equation F=0 is x2+y2=z2, the double cone, and [L1] gives DF(0,0,0)=0. Thus the derivative rank at the apex is 0.

givenL1
1.2

If (x,y,z)0 lies on the cone, then the row (2x,2y,2z) is nonzero, so the derivative has rank 1 and is surjective.

givenL1
2.1

The rank therefore drops at the apex. Moreover the cone contains the rays with directions (1,0,1), (1,0,1), and (0,1,1), which span R3; no single two-dimensional tangent plane at the apex contains all their velocities, so [L2] cannot supply a regular graph there.

step 1.1step 1.2L2algebra
3.1

This explicit smooth map refutes constant rank on its level and isolates the failure at the critical apex.

step 2.1
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

The cusp y2=x3 has a rank drop at the origin

Statement refuted

A polynomial level curve need not be regular everywhere. The zero level of F(x,y)=y2x3 has derivative rank 1 away from the origin and rank 0 at the origin.

Facts & Assumptions

Given: The polynomial F(x,y)=y2x3 and the curve γ(t)=(t2,t3).

[L2]

Regularity means surjectivity of the derivative, and a regular level is locally a C1 graph (Regular and critical points, regular and critical values, and level sets, A regular level set is locally a Ck graph of dimension mn).

Counterexample

technique · direct
1.1

One has F(γ(t))=t6t6=0, so the parametrized cusp lies in the zero level, and [L1] gives DF(0,0)=0 and γ(0)=0.

givenL1algebra
1.2

If (x,y)(0,0) lies on F1(0), then (3x2,2y)(0,0), so DF(x,y) has rank 1 and is surjective onto R.

givenL1algebra
2.1

Hence the derivative rank drops precisely at the cusp point. The two values y=±x3/2 for x>0 prevent a graph y=g(x) there, while the relation x=y2/3 is not differentiable at 0, in accord with the missing hypothesis in [L2].

step 1.1step 1.2L2algebra
3.1

The polynomial level curve therefore supplies the claimed rank-drop counterexample.

step 2.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24Open item page →

The map (x,y)(x,xy) has nonconstant rank on every neighbourhood of the origin

Example

For f(x,y)=(x,xy), the derivative has rank 1 on the vertical axis and rank 2 off it. Thus no neighbourhood of (0,0) has constant rank, although Df is continuous.

Facts & Assumptions

Given: The polynomial map f:R2R2, f(x,y)=(x,xy).

[L2]

A square matrix has rank 2 exactly when its determinant is nonzero, while its nonzero first row gives rank at least 1; every point of an open set has a ball contained in it (A matrix has rank at least r exactly when it has a nonzero r-rowed minor, The rank of a derivative and constant-rank Euclidean maps, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Verification

technique · direct
1.1

By [L1], detJf(x,y)=x. Thus [L2] gives rank 2 when x0 and rank exactly 1 when x=0.

givenL1L2algebra
1.2

Every open ball about the origin contains (0,0) and also (ε,0) for some nonzero sufficiently small ε.

givenL2choose
2.1

Step 1.1 assigns different ranks to those points, so no neighbourhood of the origin has constant rank. The polynomial entries in [L1] are continuous, proving the final assertion.

step 1.1step 1.2
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-24Open item page →

A critical value can have a smooth level set

Statement refuted

A critical value need not have a singular level set. For F(x,y)=x2, the value 0 is critical although F1(0) is the vertical line.

Facts & Assumptions

Counterexample

technique · direct
1.1

The equation F(x,y)=0 is equivalent to x=0, so F1(0)={(0,y):yR}, the graph of the zero function over the y-axis.

givenalgebra
2.1

By [L1], DF(0,y)=0 for every point of this fibre, so it is not surjective and [L2] makes 0 a critical value of F.

step 1.1L1L2
3.1

The same underlying set is a smooth line and, after swapping coordinates, is the graph covered by [L2]. Thus criticality of this defining function does not force singularity of the set.

step 1.1step 2.1L2
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

Lagrange multipliers locate the extrema of a linear functional on a sphere

Example

Let m1, let R>0, and let aRm. On the sphere x2=R, the linear functional f(x)=a,x has maximum Ra2 and minimum Ra2. If a0, they occur uniquely at x=Ra/a2 and x=Ra/a2 respectively; if a=0, every point is both a maximum and a minimum.

Facts & Assumptions

Given: A natural m1, the radius R>0, vector aRm, objective f(x)=a,x, and constraint G(x)=x22R2.

[L1]

The sphere constraint is regular, and the one-constraint multiplier rule gives f(x)=λG(x) at every constrained local extremum (A Euclidean sphere is a regular level set with tangent hyperplanes, For one regular constraint, the objective gradient is a scalar multiple of the constraint gradient).

Verification

technique · direct
1.1

If a0, [L1] gives a=2λx. The constraint forces x=±Ra/a2, and direct substitution gives the values ±Ra2.

givenL1algebra
2.1

By [L2], every constrained point satisfies Ra2f(x)Ra2, with equality only at the two points from step 1.1.

givenL2
3.1

Hence those points are the unique global extrema when a0. When a=0, f is identically zero, so every constrained point is both an extremum.

step 1.1step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

Two constraints on a sphere-plane circle, where one multiplier solution is only a local maximum

Example

Let G(x,y,z)=(x2+y2+z2,z), let c=(2,1), and let M=G1(c), the unit circle in the plane z=1. The derivative DG is surjective at every point of M, so the two-constraint multiplier rule applies there, and its equation f=DGTλ for the objective f(x,y,z)=xy+xz+yz has exactly four solutions on M: P1=(12,12,1),P2=(0,1,1),P3=(1,0,1),P4=(12,12,1).

On M the objective has maximum 12+2, attained only at P1, and minimum 1, attained exactly at P2 and P3. The fourth solution has f(P4)=122, which is neither of those values, and it is nevertheless a strict local maximum of f on M: f(p)<f(P4) for every pM with 0<pP42<22.

So the multiplier equation does not, by itself, separate a global extremum from a merely local one. Every one of its solutions here is a local extremum of f on M, and three of the four are global; deciding which is which took a separate argument.

Facts & Assumptions

Given: The maps f(x,y,z)=xy+xz+yz and G(x,y,z)=(x2+y2+z2,z) on R3, the value c=(2,1), and M=G1(c). Extrema on M are constrained extrema, comparing f only at points of M: a point aM is a local extremum of f on M when for some r>0 either f(p)f(a) for every pM with pa2<r, or f(p)f(a) for every such p; it is a strict local maximum of f on M when for some r>0, f(p)<f(a) for every pM with 0<pa2<r. This is the sense of "local maximum or minimum of f subject to G(x)=c" in [L3]; [L4] is the unconstrained notion, comparing f at every nearby point of the open set on which it is defined.

[L2]

A C1 map is a submersion at a point when its derivative there is surjective, and a matrix has rank at least r exactly when some r-rowed minor is nonzero (Submersions and immersions between Euclidean open sets, A matrix has rank at least r exactly when it has a nonzero r-rowed minor).

[L3]

If a is a local maximum or minimum of a C1 objective f subject to G(x)=c with G of class C1, and DG(a) is surjective, then there is a unique λ with f(a)=DG(a)Tλ; the condition is necessary and not sufficient (Lagrange multipliers for a regular vector-valued level-set constraint).

[L4]

For URm open, aU and f:UR, the point a is a local minimum when some Euclidean neighbourhood V of a satisfies f(a)f(x) for every xUV, and a strict local minimum when the inequality is strict for xa; local and strict local maxima reverse these inequalities (Local and strict local extrema for scalar fields on Euclidean open sets). The Euclidean norm is x2=(k<nxk2)1/2 (The p-norms xp for rational p1, and x).

Verification

technique · direct
1.1

By [L1], f and G are C1, with f(x,y,z)=(y+z,x+z,x+y) and with the two rows of JG(x,y,z) equal to (2x,2y,2z) and (0,0,1).

givenL1
1.2

A point lies in M exactly when z=1 and x2+y2=1. Writing s=x+y on M gives s2=x2+y2+2xy=1+2xy, so xy=(s21)/2 and f=xy+z(x+y)=(s2+2s1)/2; moreover 2xyx2+y2=1 because (xy)20, so s22 and 2s2.

givenalgebra
1.3

For p=(x,y,1)M and P4=(12,12,1), expanding gives pP422=(x+12)2+(y+12)2=(x2+y2)+2(x+y)+1=2+2s. So on M the distance to P4 determines s and increases with it.

givenL4algebra
2.1

At a point of M the two-rowed minor of JG from columns 1,3 is 2x and the minor from columns 2,3 is 2y, and x2+y2=1 forces (x,y)(0,0), so one of them is nonzero and JG has rank 2. By [L2], DG is surjective at every point of M.

step 1.1step 1.2L2algebra
2.2

By step 1.2, on M the value of f at a point with x+y=s is (s2+2s1)/2, and s2+2s12(1)=(s+1)220 with equality exactly when s=1, while (2)2+2212s2+2s12=(2s)(2+s+2)20 for 2s2, with equality exactly when s=2. Hence f1 on M, with equality exactly at the points where s=1, and f12+2 on M, with equality exactly at the points where s=2.

step 1.2algebra
2.3

By step 1.2 and step 1.3, a point of M with 0<pP42<22 has 0<2+2s<22, that is 2<s<1; and s2+2s12(2)2+2(2)12=(s+2)(s+22)2<0 there, because s+2>0 while s+22<12<0.

step 1.2step 1.3algebra
3.1

On M, s=2 forces 2xy=s21=1=x2+y2, hence (xy)2=0 and x=y=12; and s=1 forces 2xy=s21=0, hence {x,y}={0,1}. Both loci are therefore nonempty, so by step 2.2 the bounds are attained: the maximum of f on M is 12+2, only at P1=(12,12,1), and the minimum is 1, exactly at P2=(0,1,1) and P3=(1,0,1).

step 1.2step 2.2algebra
3.2

By step 2.1 and [L3], every local extremum of f on M satisfies f=DGTλ for a unique λ=(λ1,λ2); by step 1.1 this reads y+1=2λ1x, x+1=2λ1y and x+y=2λ1+λ2, the last of which only determines λ2. Subtracting the second equation from the first gives (yx)(1+2λ1)=0.

givenstep 1.1step 2.1L3algebra
3.3

By step 1.2 and step 2.3, f(P4)=(2)2+2(2)12=122, and f(p)<f(P4) for every pM with 0<pP42<22. So P4 is a strict local maximum of f on M, with r=22.

givenstep 1.2step 2.3L4algebra
4.1

If y=x, then x2+y2=1 gives x=y=±12 and s=±2, so the point is P1 or P4, and both satisfy the equations of step 3.2 with λ1=(x+1)/(2x), which is defined because x0. If instead λ1=12, the first equation gives x+y=1, so s=1, and by step 3.1 the points are P2 and P3, which satisfy all three equations with λ1=12. The two cases cannot both hold, since x=y and x+y=1 give x2+y2=121. So P1,P2,P3,P4 are exactly the solutions of the multiplier equation on M.

step 1.2step 3.1step 3.2algebra
5.1

By step 3.1 the maximum and the minimum of f on M are attained at P1, P2 and P3, each of which is among the four solutions of step 4.1, as [L3] requires. By step 3.3 the remaining solution P4 is a strict local maximum whose value 122 is neither the maximum nor the minimum, because 1<122<12+2 follows from 1<2<32. So the multiplier equation is satisfied at a point that is a local but not a global extremum, and satisfying it does not decide which.

step 3.1step 3.3step 4.1L3algebra
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24Open item page →

FALSE: every level set of a smooth map is locally a graph

Statement

Every level set of a smooth Euclidean map is locally a C1 graph.

Facts & Assumptions

Given: The smooth map F(x,y,z)=x2+y2z2.

[L2]

The regular-level graph theorem assumes surjectivity of the derivative at every point of the fibre (Regular and critical points, regular and critical values, and level sets, A regular level set is locally a Ck graph of dimension mn).

Refutation

technique · direct
1.1

The zero level is the double cone x2+y2=z2, and [L1] gives DF(0,0,0)=0, so the apex is critical and [L2] does not apply there.

givenL1L2
1.2

The cone contains rays from the apex in the linearly independent directions (1,0,1), (1,0,1), and (0,1,1). If it were a C1 graph near the apex, all these curve velocities would lie in its single two-dimensional tangent plane, which is impossible.

givenalgebra
2.1

Thus this smooth polynomial has a level set that is not locally a C1 graph at the apex, refuting the statement.

step 1.1step 1.2
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24Open item page →

FALSE: a critical value must have a singular level set

Statement

If c is a critical value of a smooth map, then the level set over c is singular.

Facts & Assumptions

Refutation

technique · direct
1.1

The zero level of F is the vertical line {(0,y):yR}. By [L1], DF vanishes at every point of this fibre, so 0 is a critical value.

givenL1
2.1

The same line is the graph of the zero function over the y-axis and hence is smooth by [L2].

step 1.1L2
3.1

Therefore a critical value can have a smooth level set; criticality records a failure of this defining map, not necessarily a singularity of the underlying subset.

step 1.1step 2.1
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

FALSE: continuity of the derivative implies constant rank

Statement

If a Euclidean map has continuous derivative, then its derivative has locally constant rank.

Facts & Assumptions

Given: The polynomial map f(x,y)=(x,xy).

[L1]

Its derivative has continuous polynomial entries, rank 1 on x=0, and rank 2 on x0; every neighbourhood of the origin meets both loci (The map (x,y)(x,xy) has nonconstant rank on every neighbourhood of the origin).

[L2]

The correct general conclusion is lower semicontinuity: every rank-at-least-r locus is open (Differential rank is lower semicontinuous).

Refutation

technique · direct
1.1

By [L1], Df is continuous but has two different ranks in every neighbourhood of the origin.

givenL1
2.1

Hence continuity of the derivative does not imply locally constant rank.

step 1.1
3.1

This does not contradict [L2]: the rank-2 locus {x0} is open, so rank jumps upward away from the vertical axis exactly as lower semicontinuity permits.

step 1.1L2

Sources