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 SRm−1={x∈Rm:∥x∥2=R} is the regular level F−1(R2) of F(x)=∥x∥22. At x∈SRm−1, TxSRm−1=x⊥={h:⟨x,h⟩=0}.

Facts & Assumptions

Given: A radius R>0 and F(x)=⟨x,x⟩ on Rm, with m≥1.

[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 m−n, The tangent space to a regular level set, Tangent vectors to a regular level set are exactly its curve velocities).

Verification

technique · direct
1.1givenL1

By [L1], F−1(R2)=SRm−1 and DF(x)h=2⟨x,h⟩.

2.1step 1.1algebra

If x lies on the sphere, then x≠0 and DF(x)x=2R2≠0, so DF(x):Rm→R is surjective.

3.1step 2.1L2

Thus R2 is a regular value. By [L2], the sphere is locally a graph and TxSRm−1=ker⁡DF(x)=x⊥, with the same set realized by curve velocities.

4.1step 1.1∎

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.

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 F−1(1) is a regular level set, and TxF−1(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 m−n, The tangent space to a regular level set).

Verification

technique · direct
1.1givenL1

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

2.1step 1.1L2

By [L2], DF(x)(Ax)=2∥Ax∥22>0, so DF(x) is a nonzero functional and hence surjective onto R.

3.1step 2.1L3

Therefore 1 is a regular value, and [L3] gives TxF−1(1)=ker⁡DF(x)={h:⟨Ax,h⟩=0}=(Ax)⊥.

4.1step 1.1step 3.1∎

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.

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:U⊆Rm→Rn be Ck, k≥1, and define G:U×Rn→Rn by G(x,y)=y−ψ(x). Then 0 is a regular value, G−1(0) is the graph of ψ, and T(x,ψ(x))G−1(0)={(v,Dψ(x)v):v∈Rm}.

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)=w−Dψ(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 m−n, The tangent space to a regular level set).

Verification

technique · direct
1.1givenalgebra

The equation G(x,y)=0 is equivalent to y=ψ(x), so G−1(0) is precisely the graph.

1.2givenL1

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

2.1step 1.2L1L2

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

3.1step 1.1step 2.1∎

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

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+y2−z2=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+y2−z2.

[L2]

Every positive real has a unique positive square root (Existence and uniqueness of n-th roots: a unique a1/n≥0 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 m−n, The tangent space to a regular level set).

Verification

technique · direct
1.1givenL1

On F−1(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.

1.2givenL2algebra

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.

2.1step 1.1L2

Hence 1 is a regular value, and [L2] gives tangent plane {h:xh1+yh2−zh3=0} at (x,y,z).

3.1step 1.1step 1.2∎

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

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(n−1)/2

Example

Let n≥1. 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 i≤j. Under these identifications let f:Mn(R)→Sym⁡n(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)={A∈Mn(R):ATA=In}=f−1(In) is a regular level set: near each of its points it is a C∞ graph of dimension n2−n(n+1)2=n(n−1)2, and its tangent space at A∈O(n) is TAO(n)={AK:K∈Mn(R), KT=−K}, of dimension n(n−1)/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 n≥1, the entrywise identifications above, and the map f with components fij(A)=∑k<nakiakj for i≤j.

[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 U⊆Rm→RN is a Ck graph of dimension m−N, 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 m−n, 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: dim⁡FV=nullity⁡T+rank⁡T).

Verification

technique · direct
1.1givenL1

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

1.2givenL5algebra

Let A∈f−1(In), so ATA=In. If Ax=0 then x=Inx=ATAx=0, so by [L5] the map x↦Ax is injective and therefore, its kernel being trivial, surjective on Rn; hence A is invertible and A−1=AT, so also AAT=In.

2.1step 1.1givenL2algebra

Fix A,H∈Mn(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.

3.1step 2.1step 1.2algebra

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 Sym⁡n(R).

4.1step 3.1L3

By [L3], every point of f−1(In) is a submersion point, so In is a regular value and f−1(In)=O(n) is a regular level set.

5.1step 4.1L4algebra

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

6.1step 5.1step 1.2algebra

If KT=−K and H=AK, then by step 1.2 ATH=K and HTA=KTATA=KT=−K, so H∈ker⁡Df(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}.

7.1step 6.1step 1.2L5algebra

The map K↦AK 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(n−1)/2, and dim⁡TAO(n)=n(n−1)/2.

8.1step 5.1step 7.1algebra∎

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

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+y2−z2, the zero level is regular away from its apex and has derivative rank 0 at the apex.

Facts & Assumptions

Given: The smooth map F:R3→R, F(x,y,z)=x2+y2−z2.

[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 m−n).

Counterexample

technique · direct
1.1givenL1

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.

1.2givenL1

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.

2.1step 1.1step 1.2L2algebra

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.

3.1step 2.1∎

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

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)=y2−x3 has derivative rank 1 away from the origin and rank 0 at the origin.

Facts & Assumptions

Given: The polynomial F(x,y)=y2−x3 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 m−n).

Counterexample

technique · direct
1.1givenL1algebra

One has F(γ(t))=t6−t6=0, so the parametrized cusp lies in the zero level, and [L1] gives DF(0,0)=0 and γ′(0)=0.

1.2givenL1algebra

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

2.1step 1.1step 1.2L2algebra

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=∣y∣2/3 is not differentiable at 0, in accord with the missing hypothesis in [L2].

3.1step 2.1∎

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

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:R2→R2, 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
1.1givenL1L2algebra

By [L1], det⁡Jf(x,y)=x. Thus [L2] gives rank 2 when x≠0 and rank exactly 1 when x=0.

1.2givenL2choose

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

2.1step 1.1step 1.2∎

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.

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 F−1(0) is the vertical line.

Facts & Assumptions

Counterexample

technique · direct
1.1givenalgebra

The equation F(x,y)=0 is equivalent to x=0, so F−1(0)={(0,y):y∈R}, the graph of the zero function over the y-axis.

2.1step 1.1L1L2

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.

3.1step 1.1step 2.1L2∎

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.

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 m≥1, let R>0, and let a∈Rm. On the sphere ∥x∥2=R, the linear functional f(x)=⟨a,x⟩ has maximum R∥a∥2 and minimum −R∥a∥2. If a≠0, they occur uniquely at x=Ra/∥a∥2 and x=−Ra/∥a∥2 respectively; if a=0, every point is both a maximum and a minimum.

Facts & Assumptions

Given: A natural m≥1, the radius R>0, vector a∈Rm, objective f(x)=⟨a,x⟩, and constraint G(x)=∥x∥22−R2.

[L1]

The sphere constraint is regular, and the one-constraint multiplier rule gives ∇f(x)=λ∇G(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
1.1givenL1algebra

If a≠0, [L1] gives a=2λx. The constraint forces x=±Ra/∥a∥2, and direct substitution gives the values ±R∥a∥2.

2.1givenL2

By [L2], every constrained point satisfies −R∥a∥2≤f(x)≤R∥a∥2, with equality only at the two points from step 1.1.

3.1step 1.1step 2.1∎

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

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=G−1(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)=12−2, which is neither of those values, and it is nevertheless a strict local maximum of f on M: f(p)<f(P4) for every p∈M with 0<∥p−P4∥2<2−2.

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=G−1(c). Extrema on M are constrained extrema, comparing f only at points of M: a point a∈M is a local extremum of f on M when for some r>0 either f(p)≤f(a) for every p∈M with ∥p−a∥2<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 p∈M with 0<∥p−a∥2<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 U⊆Rm open, a∈U and f:U→R, the point a is a local minimum when some Euclidean neighbourhood V of a satisfies f(a)≤f(x) for every x∈U∩V, and a strict local minimum when the inequality is strict for x≠a; local and strict local maxima reverse these inequalities (Local and strict local extrema for scalar fields on Euclidean open sets). The Euclidean norm is ∥x∥2=(∑k<n∣xk∣2)1/2 (The p-norms ∥x∥p for rational p≥1, and ∥x∥∞).

Verification

technique · direct
1.1givenL1

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).

1.2givenalgebra

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=(s2−1)/2 and f=xy+z(x+y)=(s2+2s−1)/2; moreover 2xy≤x2+y2=1 because (x−y)2≥0, so s2≤2 and −2≤s≤2.

1.3givenL4algebra

For p=(x,y,1)∈M and P4=(−12,−12,1), expanding gives ∥p−P4∥22=(x+12)2+(y+12)2=(x2+y2)+2(x+y)+1=2+2 s. So on M the distance to P4 determines s and increases with it.

2.1step 1.1step 1.2L2algebra

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.

2.2step 1.2algebra

By step 1.2, on M the value of f at a point with x+y=s is (s2+2s−1)/2, and s2+2s−12−(−1)=(s+1)22≥0 with equality exactly when s=−1, while (2)2+22−12−s2+2s−12=(2−s)(2+s+2)2≥0 for −2≤s≤2, with equality exactly when s=2. Hence f≥−1 on M, with equality exactly at the points where s=−1, and f≤12+2 on M, with equality exactly at the points where s=2.

2.3step 1.2step 1.3algebra

By step 1.2 and step 1.3, a point of M with 0<∥p−P4∥2<2−2 has 0<2+2 s<2−2, that is −2<s<−1; and s2+2s−12−(−2)2+2(−2)−12=(s+2)(s+2−2)2<0 there, because s+2>0 while s+2−2<1−2<0.

3.1step 1.2step 2.2algebra

On M, s=2 forces 2xy=s2−1=1=x2+y2, hence (x−y)2=0 and x=y=12; and s=−1 forces 2xy=s2−1=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).

3.2givenstep 1.1step 2.1L3algebra

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 (y−x)(1+2λ1)=0.

3.3givenstep 1.2step 2.3L4algebra

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

4.1step 1.2step 3.1step 3.2algebra

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 x≠0. 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=12≠1. So P1,P2,P3,P4 are exactly the solutions of the multiplier equation on M.

5.1step 3.1step 3.3step 4.1L3algebra∎

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 12−2 is neither the maximum nor the minimum, because −1<12−2<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.

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+y2−z2.

[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 m−n).

Refutation

technique · direct
1.1givenL1L2

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.

1.2givenalgebra

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.

2.1step 1.1step 1.2∎

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

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
1.1givenL1

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

2.1step 1.1L2

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

3.1step 1.1step 2.1∎

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.

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 x≠0; 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
1.1givenL1

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

2.1step 1.1

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

3.1step 1.1L2∎

This does not contradict [L2]: the rank-2 locus {x≠0} is open, so rank jumps upward away from the vertical axis exactly as lower semicontinuity permits.

Sources