False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-05

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Every leaf of a regular foliation is an embedded submanifold

Statement

Every leaf of a regular foliation is an embedded submanifold.

Facts & Assumptions

Given: An irrational $α$ and, on the standard torus $T^{2} = R^{2} / Z^{2}$ , the constant distribution $D$ spanned by the image of $(1, α)$ .

[L1]

The quotient $R^{2} / Z^{2}$ is the two-torus, with the product topology and its standard product smooth structure (The two-dimensional torus $T^{2} = (R / Z)^{2}$ , Products of smooth manifolds have a canonical product smooth structure).

[L2]

An integrable rank-one distribution determines a regular foliation whose leaves are its maximal connected integral manifolds (Regular foliations and integrable distributions correspond).

[L3]

For an integrable distribution, the leaf through a point is its tangent-curve reachability class and carries the unique maximal connected integral-manifold structure (The leaf equivalence relation of an integrable distribution, Existence and uniqueness of maximal connected integral manifolds).

[L4]

Every real number has an integer part $⌊ x ⌋$ satisfying $⌊ x ⌋ \leq x < ⌊ x ⌋ + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[L5]

The real numbers are Archimedean (Every complete ordered field is Archimedean).

[L6]

Among $N + 1$ objects placed in $N$ classes, two lie in the same class (The pigeonhole principle on $N$ ).

[L7]

A one-dimensional embedded submanifold of a two-manifold is locally an ambient coordinate line (Embedded submanifolds and slice charts).

Refutation

technique · direct

1.1

On every lifted quotient chart, the linear coordinate $r = y - α x$ is constant in the direction $(1, α)$ , so $D$ is locally the span of a coordinate vector field and is integrable. By [L2] it determines a regular foliation. Its integral curve through $[0]$ is $γ (t) = [(t, α t)]$ . Conversely, a piecewise smooth tangent curve lifts locally with derivative $h (t) (1, α)$ , so each lifted segment has displacement parallel to $(1, α)$ ; summing the segments shows that every endpoint reachable from $[0]$ lies in $γ (R)$ . Thus [L3] identifies $γ (R)$ with the leaf through $[0]$ .

L1L2L3givenalgebra

1.2

The derivative of the lifted curve is the nonzero vector $(1, α)$ , so $γ$ is an immersion. If $γ (s) = γ (t)$ , then $t - s$ and $α (t - s)$ are integers. Irrationality of $α$ forces $t - s = 0$ , so $γ$ is injective.

givenalgebra

1.3

Fix $ε > 0$ . By [L5], choose a positive integer $N$ with $1 / N < ε$ . Put $r_{m} = m α - ⌊ m α ⌋ \in [0, 1)$ for $0 \leq m \leq N$ . Partition $[0, 1)$ into the $N$ half-open intervals of length $1 / N$ . By [L6], two distinct $r_{m}, r_{l}$ lie in one interval; after interchanging them if necessary, $0 < δ : = r_{l} - r_{m} < 1 / N$ , where positivity follows because equality would make $(l - m) α$ an integer. For the nonzero integer $q = l - m$ , one has $q α \equiv δ (m o d Z)$ .

L4L5L6givenchoosealgebra

2.1

Fix $[(u, v)] \in T^{2}$ and let $w \in [0, 1)$ represent the class of $v - α u$ . With $j = ⌊ w / δ ⌋$ , [L4] gives $0 \leq w - j δ < δ$ . Set $n = j q$ and $t = u + n$ . Then $[(t, α t)] = [(u, α u + j δ)]$ , which is within $ε$ of $[(u, v)]$ in a product quotient chart. Since the point and $ε$ were arbitrary, the origin leaf is dense in $T^{2}$ .

L1L4step 1.3algebra

3.1

This leaf has dimension $1$ in the $2$ -manifold $T^{2}$ . If it were embedded, [L7] would give an ambient chart in which its intersection with the chart domain is a coordinate line, which is not dense in that domain. But step 2.1 makes the leaf's intersection with every nonempty open chart domain dense there, a contradiction. Hence the leaf is not embedded.

L1L7step 2.1

4.1

Thus a regular foliation has a nonembedded leaf, so the universal statement is false.

step 1.1step 3.1∎

Depends on

Used by

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Sources

Will J. Merry, Differential Geometry (standard reference, not scraped)