Izvestiya: Mathematics, 2022, Volume 86, Issue 4, Pages 699–714 DOI: https://doi.org/10.1070/IM9124 (Mi im9124)

DOI: https://doi.org/10.1070/IM9124

Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2022, Volume 86, Issue 4, Pages 85–102
DOI: https://doi.org/10.4213/im9124

Bibliographic databases:

§ 1. Introduction

Let $(M, g)$ be a closed oriented three-dimensional Riemannian manifold and let $\mathcal{F}$ be a transversely oriented smooth foliation of codimension one on $M$. Smooth foliations are always assumed to be of class $C^{\infty}$. The transverse orientability guarantees the existence of a unit vector filed $Z$ orthogonal to the foliation $\mathcal{F}$.

We write $T \mathcal{F}$ for the subbundle of the tangent bundle $TM$ which is tangent to $\mathcal{F}$. Let $\Gamma TM$ and $\Gamma T \mathcal{F}$ be the spaces of smooth sections (vector fields) of the corresponding vector bundles. The space of differential $k$-forms on $M$ is denoted by $\Omega^k(M)$.

The tangent volume form $\chi \in \Omega^{2}(M)$ and the orthogonal volume form $\nu\in \Omega^{1}(M)$ of the foliation $\mathcal{F}$ are defined as follows:

We recall that the second quadratic form of the foliation $ \mathcal{F} $ is defined as

The foliation $ \mathcal{F} $ is said to be minimal if $H:=(1/2)\operatorname{trace}W \equiv 0$. Clearly, $H(x)$ is the mean curvature of the leaf $L_x\in \mathcal{F}$ passing through $x\in M$. Therefore the function $H$ is referred to as the mean curvature of the foliation $\mathcal{F}$.

A taut foliation $\mathcal{F}$ on an oriented three-dimensional manifold $M$ is a transversely oriented foliation of codimension 1 such that for every leaf there is a transversal circle intersecting this leaf.

Zoom inZoom out

Figure 1.A Reeb component and its universal covering

A subset of a manifold $M$ endowed with a foliation $\mathcal{F}$ is said to be saturated if it consists of leaves of $\mathcal{F}$. A saturated subset $\mathcal{G}$ of a three-dimensional compact orientable manifold $M$ endowed with a transversely orientable foliation $\mathcal{F}$ of codimension 1 is called a generalized Reeb component if $\mathcal{G}$ is a connected three-dimensional manifold with boundary and every vector field transversal to $\mathcal{F}$ on the boundary $\partial\mathcal{G}$ is directed either everywhere inwards or everywhere outwards $\mathcal{G}$. Clearly, $\partial\mathcal{G}$ is a finite union of compact leaves of $\mathcal{F}$. In particular, every Reeb component $\mathcal{R}$ is a generalized Reeb component. We recall (see [1]) that the Reeb component $ \mathcal{R} $ is homeomorphic to a solid torus $D^2\times S^1$ all of whose leaves (except for the boundary leaf, which is homeomorphic to the two-dimensional torus $T^2$) are homeomorphic to $\mathbb{R}^2$ and are the images of the graphs of the functions $f_c\colon \operatorname{int} {D^2} \to \mathbb{R}$, $c \in \mathbb{R}$, under the covering map $p\colon D^2\times \mathbb{R} \to D^2\times S^1$ (Fig. 1), where

Sullivan proved the following theorem.

We pose the following questions in this paper. Can we guarantee that a foliation of codimension one on a closed three-dimensional Riemannian manifold is taut if the modulus of mean curvature of its leaves is sufficiently small? Is there an upper bound for this ‘smallness’ in terms of global metric invariants of the Riemannian manifold?

We shall prove the following theorem.

§ 2. Preliminaries and results

2.1. Geometric inequalities and foliations

In this section we state some results that give us necessary geometric estimates to be used in the proof of the main result. For completeness, we give proofs of some of them.

The following theorem of Loewner gives an upper bound for the length of the shortest closed geodesic on a Riemannian two-dimensional torus.

2.2. Supporting spheres and curvatures

The following theorem gives lower bounds for the normal curvatures of a sphere in a Riemannian space whose sectional curvature is bounded from above.

2.3. Novikov’s theorem and vanishing cycles

The following well-known theorem of Novikov gives topological obstructions for the existence of taut foliations.

We shall need the following construction underlying the proof of Novikov’s theorem. Suppose that a simple closed regular curve^2[x]2For simplicity of terminology we sometimes identify a curve and its image. $\alpha\subset L$ lies in the leaf ${ L}\in \mathcal{F}$ and represents the kernel of the homomorphism $\pi_1({ L})\to \pi_1(M)$. Then there is an immersion $g\colon D \to M$ of a two-dimensional disc such that $g (\partial D)=\alpha$. A small perturbation of this immersion is in general position. This means that the induced foliation $\mathcal{F}'$ on $D$, which is tangent to the boundary $\partial D$, has finitely many Morse singularities corresponding to the points where the image $g(D)$ is tangent to the foliation. These are either elliptic singular points (centers), near which the foliation looks like the family of concentric circles, or hyperbolic singular points (saddles), near which the foliation looks like the family of curves $x^2-y^2=t$, $t\in (-\varepsilon, +\varepsilon)$, including the singular point $(0,0)$. Moreover, a small perturbation enables us to assume that there is at most one hyperbolic point on each leaf. The resulting foliation outside the singular points on $D$ can be oriented since $D$ is simply connected and the singular points are of the form given above. Hence $D$ can be endowed with a vector field $X$ tangent to $\mathcal{F}'$. The zeros of $X$ are the singular points of $\mathcal{F}'$. We recall that a separatrix curve outgoing from a singular point and incoming it again, along with this singular point, is called a separatrix loop.

We state Novikov’s results in the following theorem (see also [1], § 25, [7], Lemmas 9.2.2, 9.2.4).

Zoom inZoom out

Figure 2.The disc $D'$

The curve $c:= g\circ f_0\colon S^1 \to T$ is called a vanishing cycle.

§ 3. An auxiliary theorem

Let $A\subset X$ be a closed subset of a topological space $X$ such that $X\setminus A$ is an orientable manifold without boundary. Then the pair $(X,A)$ is called a relative manifold (see [8], Ch. 6, § 2).

Proof. $1 \Leftrightarrow 2$. Consider the following part of the exact homology sequence of a pair:

$$ \begin{equation*} H_2(\mathcal{R})\to H_2(\mathcal{R}, T) \xrightarrow{d} H_1(T) \xrightarrow{q} H_1(\mathcal{R}). \end{equation*} \notag $$

The homomorphism $d$ is induced by the boundary operator $\partial$ on chains. Since $\mathcal{R}\sim S^1$, we have $H_2(\mathcal{R})=0$ and, therefore, $d$ is a monomorphism. Hence $0\ne [S,\partial S] \in H_2(\mathcal{R},T) \Rightarrow 0\ne [\partial S] \in H_1(T)$. However, $[\partial S]\in \ker q$. Therefore, if $ 0\ne [\partial S] \in H_1(T)$, then $ [\partial S]=d[S,\partial S] $ and $[S,\partial S]\ne 0$ in $H_2(\mathcal{R}, T)$.

The assertion in part 5 about a homeomorphism between the surface $S\,{\subset}\,\mathcal{R}$ and any connected component $S_i$ of its lift to the universal covering $\widetilde{\mathcal{R}}$ follows since $\partial S$ is by hypothesis contractible to a point inside $\mathcal{R}$ and, therefore, the embedding $i\colon S\,{\to}\,\mathcal{R}$ can be extended to a continuous map of the sphere $S^2$ and represented as a composite , inducing the composite homomorphism

Zoom inZoom out

which is equal to zero since the sphere is simply connected. The resulting triviality of $i_*(\pi_1(S))$ means that the preimage of $S$ under the covering map $p\colon \widetilde{\mathcal{R}} \to \mathcal{R}$ is a disjoint union $\bigsqcup_i S_i$ of copies $S_i\simeq S$.

The rest of the proof, which will be omitted, follows from the Poincaré–Lefschetz duality and the following easy assertion: $S$ either does not separate $\mathcal{R}$ or divides it into two connected components. $\Box$

We have the following corollary of Theorem 7.

§ 4. Proof of Theorem 2

Assume that $\mathcal{F}$ is not taut. By Theorem 1, $\mathcal{F}$ contains a generalized Reeb component.

By Lemma 3, the boundary of the generalized Reeb component is a family of tori. Let $T^2$ be any of these tori. By Theorem 3, we have $l^2 \leqslant (2/\sqrt{3})\operatorname{Area}(T^2)$, where $l$ is the length of the shortest closed geodesic $\alpha$ which is not null-homotopic in $T^2$. It follows from Proposition 1 that

We consider a normal system of coordinates $(\phi_1,\phi_2,r)$ in the ball $\operatorname{int}B(i_o)$ and denote the projection to the coordinate $r$ by $\operatorname{pr}_r$. By Sard’s theorem, the set of critical values of the function

Zoom inZoom out

Figure 3.Illustration of Case 1

Since $[c]=1$ in $\pi_1(\mathcal{R})$ (see Lemma 5), the preimage of the vanishing cycle $c$ under the covering map $p\colon\widetilde{\mathcal{R}} \to \mathcal{R}$ is a disjoint union of its copies $\widetilde c_k$, $k\,{\in}\, \mathbb{Z}$. The cycles $\widetilde c_k$ lie outside $\bigcup_iW^{\mathrm{c}}_i$ since otherwise some $\widetilde c_k$ would lie inside the disc on $\partial\widetilde{\mathcal{R}}$ bounded by one of the circles belonging to $\partial S_i$, $i\in \mathbb{Z}$, and would represent the trivial element of $\pi_1(\partial \widetilde{\mathcal{R}})$, whence $c$ would represent the trivial element of $\pi_1(T)$ contrary to the definition of $c$. Write $p^{-1}A=\bigsqcup_j A_j$, where $A_j$ are the connected components^4[x]4Possibly a single component. of the preimage under the covering $p\colon \widetilde{\mathcal{R}} \to \mathcal{R}$, and $\widetilde c_k\subset A_j$ for some $j,k\in \mathbb{Z}$ while $S_i\subset \partial A_j$ for some $i\in \mathbb{Z}$. Then $(W^{\mathrm{c}}_i\setminus S_i)\cap A_j=\varnothing$. Otherwise $A_j\subset W_i^{\mathrm{c}}$ by the connectedness of $A_j$, and we would have $\widetilde c_k\subset W_i^{\mathrm{c}}$ contrary to what was proved above. It follows that the inner normal^5[x]5A normal to $S_i$ is said to be inner if its image under the differential of the covering map $p\colon \widetilde{\mathcal{R}} \to \mathcal{R}$ is an inner normal to $ S\subset S(r)$. to $S_i$ is directed towards the non-compact component $W_i$ containing $\widetilde c_k$ (Fig. 3).

We now recall that the lifted Reeb foliation on $\widetilde{\mathcal{R}}$ consists of a boundary leaf $\partial {\widetilde{\mathcal{R}}}\simeq S^1\times {\mathbb{R}}$, which is a covering of the torus $T$, and a family of leaves $\{L_t\}$ that are homeomorphic to $\mathbb{R}^2$ (see Fig. 1). In $\widetilde{\mathcal{R}}\setminus \partial {\widetilde{\mathcal{R}}}$, the foliation $\widetilde {\mathcal{F}}$ is homeomorphic to the direct product $L \times \mathbb{R}$, where $L \simeq L_t$, and the leaves ‘approach’ infinity as $t\to +\infty$ (or $t\to-\infty)$. Since $S_i$ is compact, there is a $t_0\in \mathbb{R}$ such that $L_t\cap S_i=\varnothing$ for $t>t_0$ (or $t<t_0$) and $L_{t_0}\cap S_i\ne\varnothing$. Since the leaves $L_t$, $t\in \mathbb{R}$, are connected, closed and non-compact subsets of $\widetilde{\mathcal{R}}$, we have $L_{t_0}\subset W_i$. Hence, at the tangency point $q\in L_{t_0}\cap S_i$, the surface $S_i$ is supporting^6[x]6We say that $S_i$ is supporting if it is supporting with respect to the inner normal. for the leaf $L_{t_0}$ while the sphere $S(r)$ is supporting for the leaf $p(L_{t_0})$ at the point $p(q)\in S\subset S(r)$.

Case 2. The intersection (4.5) is non-empty and contains a circle not null-homotopic in $T$.

As in Case 1, we choose a regular value $r$ of the function (4.4) in such a way that the inequalities (4.3) and (4.6) hold.

The circle in (4.5) which is not null-homotopic admits an isotopy along the torus $T$ whose result lies inside the ball $B(r)$. Note that this result is an unknotted circle in $B(r)$ since the initial circle of the isotopy lies on the sphere $S(r)$. We can span such a circle by a regular embedded disc and repeat all the arguments above for the disc $D$. Therefore, in what follows we assume that $g\colon D\to\operatorname{int}B(r)$ is an embedding and continue the proof in the same notation. We identify $D$ with $g(D)$ for convenience.

Consider a connected domain $G\subset D'\cap \mathcal{R}$ (with orientation induced from $D$; see Fig. 2) bounded by the vanishing cycle $c$ that represents a closed curve not null-homotopic in $T$ and by some family (possibly empty) of closed integral curves or separatrix loops $c_i\subset T$ that are null-homotopic in $T$ (see Theorem 6). Then

Since $S(r)\cap G =\varnothing$, the connected components of $\partial (A\cap S(r))\subset T$ are either homologous to the vanishing cycle $c$ or homologous to zero in $T$. Hence each of them (as well as $c$) is contractible in $\mathcal{R}$ and, therefore, every surface $S\subset A\cap S(r)$, being homeomorphic to a sphere with holes, satisfies the hypotheses of Theorem 7.

Case 2, a). There is a surface $S\subset A\cap S(r) \subset \partial A$ representing $0=[S,\partial S]\in H_2(\mathcal{R}, T)$.

In this case, as in Case 1, we see from Corollary 2 that every surface $S_i\in p^{-1}(S)$ divides $\widetilde{\mathcal{R}}$ into a compact connected component $W^{\mathrm{c}}_i$ and a non-compact connected component $W_i$ (see (4.7)).

Zoom inZoom out

Figure 4.Illustration of Case 2, a)

Suppose that $ G_k\subset A_j$ for some $k,j\in \mathbb{Z}$ and $S_i\subset \partial A_j$ for some $i\in \mathbb{Z}$, where, as above, $A_j\subset p^{-1}A$ is a connected component of the preimage of $A$ under the covering map ${ p\colon \widetilde{\mathcal{R}} \to \mathcal{R}}$. Note that $G_k\cap S_i =\varnothing$ since $S(r)\cap D=\varnothing$. Assume that $G_k\subset W^{\mathrm{c}}_i$. Then the closure of one of the connected components into which $G_k$ separates $\widetilde{\mathcal{R}}$, does not contain $S_i$ and is a subset of the compact set $W^{\mathrm{c}}_i$. Hence it must be compact, which is not the case (see above). It follows that $G_k\cap\, \operatorname{int} {W}^{\mathrm{c}}_i\,{=}\,\varnothing$ and the inner normal to $S_i$ is directed towards the non-compact part $W_i$ (Fig. 4). Hence, as in Case 1, we can find a leaf $L_{t_0}\subset W_i\subset \widetilde{\mathcal{R}}$ that touches the surface $S_i$ at some point $q\in S_i\cap L_{t_0}$ where $S_i$ is supporting for $L_{t_0}$. This means that the leaf $p(L_{t_0})$ is tangent to the sphere $S(r)$, and $S(r)$ is supporting for $p(L_{t_0})$ at the tangency point $p(q) \in S \subset S(r)$.

Case 2, b). The intersection $ A\cap S(r)$ consists of finitely many connected surfaces not separating $\mathcal{R}$.

In this case, $A\cap S(r)$ has more than one connected component. Indeed, otherwise we would have $A=\mathcal{R}$ and $S=\varnothing$ since every surface $S\subset A\cap S(r) \subset \partial A$ does not separate $\mathcal{R}$. Thus $A\cap S(r)$ consists of at least two connected components. We claim that there are exactly two.

Indeed, assume that there are at least three components. Denote them by $S_1$, $S_2$, $S_3$. Let $A_j\in p^{-1}A$, $G_k\subset A_j$, $j,k \in \mathbb{Z}$, be as above. Then $\partial A_j$ contains at least one copy of $S_1$, $S_2$, $S_3$. We denote them by the same letters for simplicity. Identify $\widetilde{\mathcal{R}}$ with $D^2\times \mathbb{R}$. Choose a sufficiently large constant $t_0 \in \mathbb{R}$ such that $D^2\times [-t_0, t_0]$ contains $S_1$, $S_2$, $S_3$. Then, by Theorem 7, each of the surfaces $S_1$, $S_2$, $S_3$ divides $D^2\times \mathbb{R}$ into two non-compact components, each of which contains one of the discs $D^2\times \{-t_0\}$, $D^2\times \{t_0\}$. To see this, just project to $\mathbb{R} $ all the connected components into which the surfaces $S_i$, $i=1,2,3$, divide $\widetilde{\mathcal{R}}$.

Write $W_i$, $i=1,2,3$, for the closures of those connected components of $\widetilde{\mathcal{R}}\setminus S_i$, $i=1,2,3$, that contain $A_j$. Denote the closures of the remaining components by $\widehat W_i$, $i=1,2,3$. Clearly, $ A_j\subset W_1\cap W_2\subset D^2\times [-t_0,t_0]$. It follows that $A_j$ and $W_1\cap W_2$ are compact. Since $A_j \subset W_3$, we see from the connectedness of $\widehat W_i$ that $\widehat W_i\subset W_3$, $i=1,2$. Therefore we conclude that $\widehat W_3\subset W_1\cap W_2$. Hence $\widehat W_3$ is compact as a closed subset of a compact set. But this contradicts Theorem 7. Thus $A$ contains exactly two connected components of $A\cap S(r)$. It follows from the proof that $A_j\cap p^{-1}(A\cap S(r))$ also consists of two connected components. Suppose that these are $S_1$ and $S_2$. In this case, $A_j=W_1\cap W_2 \subset D^2\times [-t_0,t_0]$. Since $A_j$ contains the surface $G_k$, the inner normals to $S_1$ and $S_2$ are directed inwards the compact set $A_j$ (Fig. 5).

Zoom inZoom out

Figure 5.Illustration of Case 2, b)

Assume that the inner leaves $\{L_t,\, t\in \mathbb{R}\}$ of the lifted Reeb foliation on $\widetilde{\mathcal{R}}$ approach infinity by leaving the component $\widehat W_1$. Hence there is a $t_1$ such that $L_t\cap \widehat W_1 =\varnothing$ for all $t>t_1$ (or $t<t_1$) while $L_{t_1} \cap S_1 \ne\varnothing$. Then $L_{t_1}\subset W_1$ and the surface $S_1$ is supporting for $L_{t_1} $ at the tangency point $q_1\in L_{t_1} \cap S_1 $. In this case it is clear that the sphere $S(r)$ is supporting for $p(L_{t_1})$ at the point $p(q_1)$.

But if the inner leaves $\{L_t,\, t\in \mathbb{R}\}$ of the lifted Reeb foliation approach infinity in the opposite direction, then there is a $t_2$ such that $L_t\cap\widehat W_2 =\varnothing$ for all $t>t_2$ (or $t<t_2$) while $L_{t_2} \cap S_2 \ne\varnothing$. Then $L_{t_2}\subset W_2$, the surface $S_2$ is supporting for $L_{t_2}$ at the tangency point $q_2\in L_{t_2} \cap S_2$ and the sphere $S(r)$ is supporting for $p(L_{t_2})$ at the point $p(q_2)$.

Case 3. The intersection (4.5) is empty.

In this case, the Reeb component $\mathcal{R}$ lies completely inside the ball $B(r)$. Shrinking the radius of $B(r)$, we arrive at a moment when the boundary sphere $S(r)$ touches the torus $T=\partial \mathcal{R}$ and becomes supporting for this torus at the tangency point. Note that shrinking of $r$ preserves the inequality (4.6). $\Box$

It follows from Proposition 2 that there is a leaf $F\in \mathcal{F}$ for which the sphere $S(r)$, $r<i_0$, is supporting at some point $q\in F\cap S(r)$. We denote the normal curvatures of $F$ by $k_H^F$. The mean curvature of the sphere $S(r)$ (resp. of the sphere of radius $r$ in the space of constant curvature $\gamma_0$) is denoted by $H_r$ (resp. $H_r^0$). We always assume that the normal at $q$ is inner for the sphere $S(r)$.

In view of Theorem 4, Remark 3 and Lemma 4, for every vector $v \in T_q F$ we have

Consider the case when $\gamma_0 =0$. Using the inequalities (4.6) and (4.9) and the equality $H_r^0=1/r$, we obtain a system

Eliminating $r^2$, we obtain $1/H_0^2< V_0H_0/(2\sqrt{3})$. Hence $H_0>\sqrt[3]{2\sqrt{3}/V_0}$. However, we should not forget the complementary upper bound imposed on $H_0$, namely, $H_0 \leqslant 2\sqrt{3}\,i_0^2/V_0$. Together with the second inequality in (4.10), it guarantees that the condition $r<i_0$ holds. Therefore, if $H_0\leqslant \min\Bigl\{2\sqrt{3}\,i_0^2/V_0,\sqrt[3]{2\sqrt{3}/V_0}\Bigr\}$, then we arrive at a contradiction.

Zoom inZoom out

Figure 6.The sphere $S^3(R)$

We now consider the case when $\gamma_0 >0$. Since $r<i_0\leqslant \pi/(2\sqrt{\gamma_0})$, the mean curvature of the sphere $S(r)\subset S^3(R)$ of radius $r$ in the sphere of radius $R$ (and constant curvature $\gamma_0=1/R^2$) is of the form $H_r^0=\sqrt{\gamma_0}\, \cot(r\sqrt{\gamma_0})$ (Fig. 6). In view of (4.6) and (4.9), we obtain the following system:

Eliminating $r^2$, we obtain

Thus we have proved that a foliation on $M$ with $|H|<H_0$ has no generalized Reeb components and, therefore, it is taut by Theorem 1. Theorem 2 is proved.

The author is grateful to Professor A. A. Borisenko for his interest in this work and useful discussions.

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Citation in format AMSBIB

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\by D.~V.~Bolotov
\paper Foliations on closed three-dimensional Riemannian manifolds with small modulus of mean curvature of the leaves
\jour Izv. RAN. Ser. Mat.
\yr 2022
\vol 86
\issue 4
\pages 85--102
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