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10 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 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Distributions Integral Manifolds and the Frobenius Theorem: Examples

1 · Prerequisites

2 · Summary

These examples keep the regular Frobenius theory concrete: coordinate-plane and kernel distributions, regular level-set foliations, product and orbit foliations, dense irrational leaves, the Mobius-band line foliation, the standard contact counterexample, and a singular variable-rank family that sits outside the constant-rank theorem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The coordinate-plane distribution and its affine leaves

Example

On Rn, fix 1kn and let

D=span(x1,,xk).

Its maximal connected integral manifolds, or leaves, are the affine coordinate k-planes with xk+1,,xn constant. Connected open subsets of those planes are also integral manifolds.

Facts & Assumptions

Given: The standard coordinates on Rn and the span of the first k coordinate fields.

[A1]

The remaining coordinates are constant along the displayed planes.

Verification

technique · direct
1.1

The fields x1,,xk are smooth and pointwise [given] independent, so they define a smooth rank-k distribution.

given
1.2

For fixed constants ck+1,,cn, the affine plane [given] {xk+1=ck+1,,xn=cn} has tangent space spanned by those same coordinate fields at every point. Hence it is an integral manifold.

given
1.3

Every connected integral manifold has constant transverse coordinates, so it [given] lies in one of the affine planes from step 1.2. Those full planes are connected and cannot be enlarged while retaining that property. Therefore they are exactly the maximal leaves.

given
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The kernel of a submersion as an integrable distribution

Example

Consider F:R3R,F(x,y,z)=x2+y2+z. Since dF=(2x,2y,1) is never zero, F is a submersion. Therefore kerdF is a rank-2 integrable distribution whose leaves are the connected surfaces x2+y2+z=c.

Facts & Assumptions

Given: The submersion F(x,y,z)=x2+y2+z.

[A1]

Its differential never vanishes.

Verification

technique · direct
1.1

Because the third partial derivative of F is 1, the map is a [given] submersion everywhere.

given
1.2

The kernel distribution is therefore integrable, and its maximal connected [given] integral manifolds are the connected components of the level sets of F. Each level set here is a connected paraboloid.

given
2.1

Thus kerdF is an explicit integrable distribution. [given] ∎

given
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The level-set foliation of a regular function

Example

On the punctured plane R2{0}, the function f(x,y)=x2+y2 has no critical points. Its level sets are the circles of positive radius, and they form a regular foliation of the punctured plane.

Facts & Assumptions

Given: The regular function f(x,y)=x2+y2 on R2{0}.

[A1]

Its differential is df=2xdx+2ydy.

Verification

technique · direct
1.1

On the punctured plane, the differential df is never zero, so f is a [given] submersion to (0,).

given
1.2

Therefore kerdf is an integrable line distribution, and its leaves are [given] the connected components of the level sets x2+y2=c. Those components are exactly the circles of radius c.

given
1.3

The correspondence between integrable distributions and regular foliations [given] turns this family of circles into a regular foliation of R2{0}.

given
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The product foliation

Example

On a product manifold M×N, the connected components of the fibres M×{q} for qN form a regular foliation of leaf dimension dimM.

Facts & Assumptions

Given: The product smooth structure on M×N.

[A1]

Product charts have the form (u,v).

Verification

technique · direct
1.1

In a product chart, the slices v=constant are exactly the local [given] pieces of the fibres M×{q}. Their connected components are therefore the local plaques.

given
1.2

Transition maps preserve the second coordinate up to a change depending only [given] on the old second coordinate, so these charts satisfy the defining condition of a regular foliation atlas.

given
2.1

Hence the connected components of the fibres of the projection [given] M×NN form the product foliation.

given
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

Orbit circles of rotation as a foliation away from the origin

Example

On R2{0}, the rotation field X=yx+xy is nowhere zero. Its flow is rotation about the origin, so its leaves are the circles centered at the origin. The origin is excluded because there the field vanishes and the rank drops.

Facts & Assumptions

Given: The rotation vector field on the punctured plane.

[A1]

Its flow preserves the radius function r2=x2+y2.

Verification

technique · direct
1.1

The field is nowhere zero on R2{0}, so it defines a [given] regular one-dimensional distribution there.

given
1.2

Along the flow, [given] X(x2+y2)=2x(y)+2y(x)=0, so the radius is constant on every orbit. The orbits are therefore contained in circles about the origin, and conversely each such circle is an orbit.

givenalgebra
1.3

Hence the punctured plane is foliated by the rotation circles. At the [given] origin the field vanishes, so the regular rank-one hypothesis fails there.

given
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The irrational linear foliation of the two-torus

Example

Let αRQ. On T2=R2/Z2, the constant vector field induced by (1,α) defines a regular foliation. Its leaves are the images of t[p+(t,αt)](pR2), and each leaf is dense in the torus.

Facts & Assumptions

Given: An irrational number α.

[A1]

Translation by (1,α) on R2 descends to the torus.

[A2]

Equip R2/Z2 with its standard quotient smooth structure, equivalently the product smooth structure on two circles.

Verification

technique · direct
1.1

On the smooth torus of [A2], the constant line field spanned by (1,α) [A2] is smooth and nowhere zero, so it gives a regular one-dimensional distribution on T2.

A2given
1.2

The integral curve through [p] is the projected affine line [given] t[p+(t,αt)]. For p=0, fix a point [(u,v)] of the torus and a neighborhood of it. Because α is irrational, the set of classes {[nα]:nZ} is dense in R/Z. Choose n so that [nα] is arbitrarily close to [vαu], and set t=u+n. Then [(t,αt)]=[(u,αu+nα)] has first coordinate [u] and second coordinate arbitrarily close to [v]. Thus the leaf through [0] is dense. Every other leaf is a torus translate of this one, and translations are homeomorphisms, so every leaf is dense.

givenalgebra
2.1

Therefore the torus carries a regular foliation with dense, nonembedded [given] leaves.

given
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The Mobius-band line foliation

Example

Model the Mobius band as the quotient of [0,1]×R by the identification (0,t)(1,t). The horizontal lines [0,1]×{t} descend to a regular one-dimensional foliation.

The line t=0 becomes the central circle leaf, while every leaf with t0 is a circle: one traversal of the horizontal segment changes the transverse sign, and a second traversal returns to the starting point.

Facts & Assumptions

Given: The strip model of the Mobius band and its horizontal lines.

[A1]

The gluing preserves horizontality.

Verification

technique · direct
1.1

The quotient map identifies horizontal tangent directions with horizontal [given] tangent directions, so the horizontal line field descends to a smooth one-dimensional distribution on the Mobius band.

given
1.2

Its local plaques are the images of small horizontal intervals, and the [given] quotient charts preserve that plaque structure. Hence the descended line field defines a regular foliation.

given
1.3

The line t=0 closes up to the central circle. If t0, then [given] the endpoint (1,t) is identified with (0,t), so one horizontal pass through the quotient changes the transverse sign; a second pass along the line t returns to the original class. Thus the leaf through t0 is also a circle.

given
2.1

This yields the standard line foliation of the Mobius band. [given] ∎

given
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The standard contact plane field is not integrable

Statement refuted

The standard contact rank-2 distribution on R3 is integrable.

Facts & Assumptions

Given: Let X1:=x+yz,X2:=y, and let D:=span(X1,X2).

[A1]

This is the kernel of the 1-form α=dzydx.

[L1]

Every integrable smooth distribution is involutive (Integrable distributions are involutive).

Counterexample

technique · direct
1.1

The two fields X1 and X2 are smooth and pointwise independent, so [given] they define a smooth rank-2 distribution on R3.

given
1.2

Their bracket is [given] [X1,X2]=z, which is not a linear combination of X1 and X2. Therefore the distribution is not involutive, even on this global frame.

givenalgebra
1.3

By [L1], an integrable distribution would have to be involutive. Hence this standard [L1] contact distribution is a counterexample to the claim that it is integrable.

L1
2.1

Therefore the displayed statement is refuted. [given] ∎

given
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

A bracket-closed variable-rank family outside regular Frobenius

Statement refuted

Any variable-rank family of tangent subspaces satisfying a bracket-closure condition is covered by the regular Frobenius theorem.

Facts & Assumptions

Given: On R, let Dx:={{0},x=0,TxR,x0.

[A1]

The rank is 0 at the origin and 1 elsewhere.

Counterexample

technique · direct
1.1

Any smooth vector field tangent to this family must vanish at the origin. [given] The Lie bracket of two such fields also vanishes at the origin, so the tangent fields are closed under bracket in this loose sense.

given
1.2

Nevertheless the family is not a smooth distribution of constant rank, so [given] it does not satisfy the hypotheses of the regular Frobenius theorem.

given
2.1

Hence variable-rank bracket closure does not place a family inside regular [given] Frobenius theory.

given
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Leaves of a Lie subalgebra distribution

Example

Let G=GL2(R) and let h be the one-dimensional Lie subalgebra spanned by E12=(0100). The left translates Dg:=d(Lg)I(h) form a rank-1 distribution on G. Its leaves are the left cosets of the subgroup H={I+tE12:tR}.

Facts & Assumptions

Given: The subgroup H={I+tE12:tR} of GL2(R).

[L1]

Through each point of an integrable distribution there is a unique maximal connected integral manifold, its leaf (Existence and uniqueness of maximal connected integral manifolds).

[A1]

Left translation sends h to tangent lines of left cosets of H.

Verification

technique · direct
1.1

The map tI+tE12 is a one-parameter subgroup because [given] E122=0, so (I+sE12)(I+tE12)=I+(s+t)E12. Hence H is a connected immersed Lie subgroup with tangent space TIH=h.

givenalgebra
1.2

For any gG, the left coset gH has tangent line [given] d(Lg)I(h) at the point g. Thus each coset is an integral manifold of the distribution D.

given
2.1

For each gG, the curve cg(t)=g(I+tE12) has image gH and [given] derivative cg(t)=gE12=d(Lcg(t))I(E12), so it is an integral curve of the nowhere-zero rank-1 distribution D. Hence any connected integral manifold through g is locally an open piece of that same integral curve and is contained in gH. Since step 1.2 shows that gH itself is a connected integral manifold through g, [L1] identifies gH as the leaf through g. Varying g gives all leaves.

L1step 1.2given

Sources