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The Möbius band's central leaf has reflection holonomy

Example

Assume Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $M o b = (R \times (- 1, 1)) / Z$ be the smooth Möbius band, with the action $k \cdot (t, x) = (t + k, (- 1)^{k} x)$ , and let $F$ be the foliation whose leaves are the images of the horizontal segments $R \times {x}$ ; the projection $[t, x] \mapsto [t]$ is a bundle over the circle with fibre the open interval $(- 1, 1)$ , to which $F$ is transverse. Then the middle leaf $L_{m i d} = π (R \times {0})$ is diffeomorphic to $S^{1}$ and its holonomy group at every point is $Z / 2$ : the generator of $π_{1} (L_{m i d}) ≅ Z$ has holonomy germ the reflection $x \mapsto - x$ of the transversal, and its square has trivial germ. Every other leaf is diffeomorphic to $S^{1}$ (it wraps twice around the band) and has trivial holonomy group.

Facts & Assumptions

Given: The quotient $M o b = (R \times (- 1, 1)) / Z$ under $k \cdot (t, x) = (t + k, (- 1)^{k} x)$ , the foliation $F$ by images of the horizontal segments, the projection $π (t, x) = [t, x]$ and the circle quotient map $t \mapsto [t]$ of $R / Z$ (The circle as $S^{1} = R / Z$ with basepoint $[0]$ ).

[F1]

The band is the suspension of the representation $ρ : Z \to Diff ((- 1, 1))$ , $ρ (k) (x) = (- 1)^{k} x$ , over $B = S^{1}$ : the diagonal action $k \cdot (t, x) = (t + k, ρ (k) x)$ is the displayed action, so the quotient and its descended foliation are those of the suspension construction (The suspension foliation of a representation of the fundamental group, Leaves of a regular foliation).

[F2]

In a suspension, the leaf through $y$ is $\tilde{B} / K_{y}$ with $K_{y} = {γ : ρ (γ) y = y}$ , the isomorphism $π_{1} (L_{y}) ≅ K_{y}$ holds, and the holonomy representation is $γ \mapsto {germ}_{y} (ρ (γ))$ ; forward-path holonomy is its inverse (Suspension holonomy is the germ of the represented monodromy action).

[F3]

The holonomy group of a leaf is the image of the holonomy representation, i.e. the isotropy of the holonomy groupoid at a point of the leaf (The isotropy of the holonomy groupoid is the leaf holonomy group, The holonomy representation and the holonomy group of a leaf).

[F4]

The diffeomorphisms $x \mapsto x$ and $x \mapsto - x$ of $(- 1, 1)$ generate a group isomorphic to $Z / 2$ , and $(- 1)^{k}$ equals $1$ for even $k$ and $- 1$ for odd $k$ (The holonomy representation and the holonomy group of a leaf).

Verification

technique · direct

1.1F1F2F3F4

The middle leaf. By [F1] the band is the suspension of $ρ$ over $S^{1}$ with the displayed action. Let $y = 0$ . Then $ρ (k) (0) = (- 1)^{k} 0 = 0$ for every $k$ , so $K_{0} = Z$ and by [F2] the leaf through $[0, 0]$ is $R / Z ≅ S^{1}$ with $π_{1} (L_{m i d}) ≅ Z$ . Its holonomy representation is $k \mapsto {germ}_{0} (ρ (k))$ . The generator $k = 1$ accordingly has holonomy germ the reflection $x \mapsto - x$ : the map $ρ (1)$ is not the identity near $0$ (it sends $x$ to $- x$ ), so its germ is a nonidentity involution, and $k = 2$ gives $ρ (2) = i d$ . Hence the holonomy group is the two-element group generated by this reflection, isomorphic to $Z / 2$ by [F4].

1.2F2F3F4

The other leaves. Let $y \neq 0$ . Then $ρ (k) y = (- 1)^{k} y = y$ holds exactly when $k$ is even, so $K_{y} = 2 Z$ and by [F2] the leaf through $[0, y]$ is $R / 2 Z ≅ S^{1}$ , the leaf wrapping twice around the band. Its holonomy representation sends $k \in 2 Z$ to the identity germ of $ρ (k) = i d$ , so the holonomy group is trivial.

2.1step 1.1step 1.2∎

Conclusion. The middle leaf is a circle whose holonomy group is $Z / 2$ , generated by the reflection germ $x \mapsto - x$ , while every other leaf is a circle with trivial holonomy group; the projection to the base circle exhibits the band as an interval bundle over the circle transverse to $F$ .

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Sources

Leiden NCG seminar, Noncommutative Geometry of Foliations (2023 seminar notes) (standard reference, not scraped)
Eckhard Meinrenken, Lie Groupoids and Lie Algebroids, lecture notes (University of Toronto MAT1341, Fall 2017) (standard reference, not scraped)