Vanishing Cycles, Novikov and Taut Foliations — Examples

The gluing proposition [F1] supplies the foliation and identifies the two solid tori as Reeb components with common boundary leaf the Heegaard torus, so the definition of a Reeb component is satisfied with the foliated homeomorphism supplied by the model.

By [F2] a foliation containing a Reeb component is not taut, and concretely the accessible manifold of the torus leaf is not all of $S^3$: transverse curves crossing the torus into either solid torus are trapped by the accumulating plane leaves, so no closed transversal meets the torus leaf. Since tautness would require a closed transversal through that leaf, $F_{\mathrm{R}\mathrm{e}\mathrm{e}\mathrm{b}}$ is not taut.

This realises Ranz's statement that a foliated manifold containing a Reeb component is not taut in the standard example, verifying the obstruction and supplying the negative model dual to the fibre-foliation example; every other leaf is a plane accumulating on the torus, and the argument uses only the two solid-torus models and the obstruction, hence only the standing countable choice from [F3].

Gluing the two standard Reeb solid tori produces the Reeb foliation of $S^3$, with their common boundary torus a leaf and each torus a Reeb component. The Reeb-component obstruction implies that no closed transversal meets that boundary leaf, so the resulting foliation is not taut. This supplies the negative comparison directly from the general obstruction.

A connected smooth manifold is path connected, so join a chosen $x_0\in L$ to $f(x_0)$ by a finite smooth chartwise path, smooth its finitely many corners and reparametrize it to be constant near $0$ and $1$; this gives a smooth path $a:[0,1]\to L$ with $a(0)=x_0$, $a(1)=f(x_0)$ and stationary ends.

Extend $a$ by $a(t+k)=f^k(a(t))$; the stationary ends make the extension smooth across every integer, and the graph $t\mapsto [a(t),t]$ is periodic under the diagonal mapping-torus action $k\cdot (x,t)=(f^k(x),t+k)$, hence descends to a smooth closed curve in $M_f$.

The descended curve is embedded because its composition with the base projection $S^1\to S^1$ is the identity, so distinct parameters have distinct images; its derivative has base component $1$, so it is everywhere positively transverse to the fibre foliation $F_f$.

The curve meets every fibre exactly once, because the base component of its parametrization runs monotonically once around the circle; hence the leaf-by-leaf condition of tautness holds for $F_f$ directly, with no fixed point of $f$ assumed.

Since $L$ is nonempty, $M_f$ is nonempty, compact and connected, [F2] upgrades the leaf-by-leaf condition to a single closed transversal meeting every leaf, so $F_f$ is taut and a single closed transversal meets all its leaves. This verifies the positive model dual to the non-taut Reeb foliation example [F3], and the construction uses one finite chartwise path, hence only the standing countable choice from [F3].

Let $N_0=T^2\times S^1$ with coordinates $(x,y)$ on the first torus factor and the circle coordinate suppressed, and let $\omega =dy-\lambda dx$ with $\lambda$ irrational. The form is closed and nowhere vanishing, so by the closed-constant-rank-one-forms criterion its kernel is integrable and defines a codimension-one foliation $F_0$ of $N_0$ by Regular foliation atlases, the product structure being the canonical one of Products of smooth manifolds have a canonical product smooth structure.

The leaf through $([x_0],[y_0],[z_0])$ is parametrized intrinsically by $(u,[v])\mapsto ([x_0+u],[y_0+\lambda u],[v])$. Equality of the first two coordinates would give an integer $k$ with $\lambda k$ an integer, so $k=0$ by irrationality. Plaque continuation gives the intrinsic cylinder $\mathbb{R}\times S^1$. F5 proves density of its torus orbit, hence density of the cylinder in $N_0$. No leaf is compact, so none can be the compact boundary leaf of a Reeb component. Thus $F_0$ is Reebless.

Remove a point $p\in N_0$ and set $M=N_0\setminus \{p\}$, $F=F_0{\mathrm{\mid }}_M$. Then $M$ is a noncompact three-manifold and $F$ is a codimension-one foliation of it; a Reeb component of $F$ would be a compact foliated solid torus in $M$ and hence in $N_0$, forcing a compact leaf of the Reebless foliation $F_0$, so $F$ is Reebless.

Let $L$ be the original cylinder containing $p$. The restricted foliation has the connected leaf $L^{\mathrm{\prime }}=L\setminus \{p\}$; there is no leaf of $F$ through the removed point. Choose one intrinsic plaque disk through $p$ in a small foliation box. Dense $L$ may have other plaques in that box, so its full intersection with the box is not asserted to be this disk. Polar coordinates identify $L$ with ${\mathbb{R}}^2\setminus \{0\}$; if $p$ maps to $a\mathrm{\ne }0$, then $L^{\mathrm{\prime }}\cong {\mathbb{R}}^2\setminus \{0,a\}$. Remove small disjoint disks about these two points and cut off the outer end. The resulting compact pair of pants is a deformation retract along the three end collars, so F5 gives free rank two. The map $z\mapsto (z-a)\mathrm{/}\mathrm{\mid }z-a\mathrm{\mid }$ has degree one on a sufficiently small puncture circle, proving that circle essential in $L^{\mathrm{\prime }}$ by F5.

Center ambient foliation coordinates $(u,v,w)$ at $p$, with the chosen plaque $w=0$. Its puncture circle $u^2+v^2={\epsilon }^2$, $w=0$, bounds the upper hemisphere $u^2+v^2+w^2={\epsilon }^2$, $w\ge 0$. This continuous disk avoids $p$ and lies in the coordinate box. The circle is therefore nullhomotopic in $M$ but essential in $L^{\mathrm{\prime }}$ by step 4.1. Inclusion on fundamental groups is noninjective, while $F$ is smooth, cooriented by $\omega$, oriented in the ambient torus and Reebless. This refutes the extension beyond the closed-manifold hypothesis.