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This article is cited in 1 scientific paper (total in 1 paper) Foliations on closed three-dimensional Riemannian manifolds with small modulus of mean curvature of the leaves D. V. Bolotov B. Verkin Institute for Low Temperature Physics and Engineering, National Academy of Sciences of Ukraine, Khar'kov
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2022, Volume 86, Issue 4, DOI: https://doi.org/10.4213/im9124 
Bibliographic databases:
    § 1. IntroductionLet $(M, g)$ 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:
$$
\begin{equation*}
\begin{gathered} \, \chi(X,Y):= (i(Z)\mu)(X,Y)=\mu(Z,X,Y)=\mu (X, Y, Z)\quad \forall\, X, Y \in \Gamma TM, \\ \nu (X):=g(X, Z) \quad \forall\, X \in \Gamma TM, \end{gathered}
\end{equation*}
\notag
$$
where $\mu \in \Omega^3(M)$ is the volume form on $M$ and $i(Z)\colon \Omega^k(M)\to\Omega^{k-1}(M)$ is the interior product with the vector field $Z\in\Gamma TM$. It sends every $k$-form $\omega$ to the following $(k-1)$-form $i(Z)\omega$:
$$
\begin{equation*}
(i(Z)\omega)(X_1,\dots,X_{k-1}):= \omega(Z,X_1,\dots,X_{k-1})\quad \forall\, X_i \in \Gamma TM, \quad i=1,\dots, k-1.
\end{equation*}
\notag
$$
Remark 1. In the local basis $\{\theta^1, \theta^2, \theta^3\}$ which is dual to a local field $\{e_1, e_2, e_3\}$ of orthonormal frames with $e_3 \equiv Z$ and $\mu=\theta^1 \wedge \theta^2 \wedge \theta^3$, we have $\nu=\theta^3$ and $\chi=\theta^1 \wedge \theta^2$. Clearly, $\mu=\chi \wedge \nu$. We recall that the second quadratic form of the foliation $ \mathcal{F} $ is defined as
$$
\begin{equation*}
B(X)=g (\nabla_XX, Z)=g(W(X), X) \quad \forall\, X \in \Gamma T \mathcal{F},
\end{equation*}
\notag
$$
where $ \nabla $ is the Levi-Civita connection on $(M, g)$ and $W (X)=-\nabla_XZ$ is the Weingarten operator of $\mathcal{F}$. 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. 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. Theorem 1 (see [2]). Let $M$ be a closed oriented three-dimensional Riemannian manifold and let $\mathcal{F}$ be a smooth transversely orientable foliation of codimension one on $M$. Then the following assertions are equivalent. 1) $\mathcal{F}$ is taut. 2) $\mathcal{F}$ contains no generalized Reeb components. 3) $\mathcal{F}$ is a minimal foliation for some Riemannian metric on $M$. 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. Theorem 2. Let $V_0>0$, $i_0>0$, $\gamma_0\geqslant 0$ be fixed constants and let $M$ be a closed oriented three-dimensional manifold with the following properties. 1) The volume $\operatorname{Vol}(M)\leqslant V_0$. 2) The sectional curvature $\gamma$ of $M$ satisfies the inequality $\gamma \leqslant \gamma_0$. 3) $\min \{\operatorname{inj}(M),\pi/(2\sqrt{\gamma_0})\} \geqslant i_0 $, where $\operatorname{inj}(M)$ is the injectivity radius of $M$. Put where $x_0$ is a root of the equation Then every smooth transversely orientable foliation $\mathcal{F}$ of codimension one on $M$ whose modulus of mean curvature of the leaves satisfies the inequality $|H|<H_0$, must be taut. In particular, it must be minimal for some Riemannian metric on $M$.§ 2. Preliminaries and results2.1. Geometric inequalities and foliationsIn 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. Lemma 1 (Reeb). Let $(M,g)$ be a closed oriented three-dimensional Riemannian manifold endowed with a smooth transversely orientable foliation $\mathcal{F}$ of codimension one. Then Proof. Let $\{e_1, e_2, e_3 \}$ and $\{\theta^1, \theta^2, \theta^3\}$ be as in Remark 1. We recall that the $1$-forms of the Levi-Civita connection are defined in the following way:
$$
\begin{equation*}
\omega_i^j (X):= g(\nabla_{X}e_i,e_j), \qquad i,j =1,2,3.
\end{equation*}
\notag
$$
Note that $\omega_i^j =-\omega_j^i $ since $g(\nabla_{X}e_i,e_j)+ g(e_i,\nabla_{X}e_j)=Xg(e_i,e_j)=0$. Using the structure equations
$$
\begin{equation*}
d\theta^i=\theta^j\wedge \omega_j^i, \qquad i,j =1,2,3,
\end{equation*}
\notag
$$
we obtain
$$
\begin{equation}
\begin{aligned} \, d\chi &=d(\theta^1\wedge \theta^2)= (\theta^j\wedge \omega_j^1 \wedge \theta^2- \theta^1\wedge \theta^j\wedge \omega_j^2)= (\theta^3\wedge \omega_3^1 \wedge \theta^2-\theta^1\wedge \theta^3\wedge \omega_3^2) \nonumber \\ &=\bigl(\theta^3\wedge g(W(e_1),e_1)\theta^1 \wedge \theta^2-\theta^1\wedge \theta^3\wedge g(W(e_2),e_2)\theta^2\bigr) \nonumber \\ &=\bigl(g(W(e_1),e_1)+g(W(e_2),e_2)\bigr)\mu=2H\mu. \qquad\Box \end{aligned}
\end{equation}
\tag{2.1}
$$
Lemma 2. Let $(M,g)$ be a three-dimensional closed orientable manifold endowed with a smooth transversely orientable foliation $\mathcal{F}$ of codimension one. Let $A$ be a generalized Reeb component of $\mathcal{F}$. Then Proof.
$$
\begin{equation*}
2\int_{A} H\mu=\int_A d\chi=\int_{\partial A}\chi=\pm \operatorname{Area}(\partial A),
\end{equation*}
\notag
$$
where the first equality follows from Lemma 1 and the second equality follows from the Stokes theorem. $\Box$ Remark 2. 1) The boundary $\partial A$ need not be connected. 2) The sign in (2.2) depends on whether the vector field $Z$ on the boundary $\partial A$ is directed inwards or outwards $A$. Corollary 1 (see [2]). Taut foliations contain no generalized Reeb components. Proof. If the foliation $ \mathcal{F} $ is taut, then it is minimal for some Riemannian metric by Theorem 1. Therefore, if $\mathcal{F}$ contains a generalized Reeb component $A$, then Lemma 2 yields that We arrive at a contradiction. $\Box$ Proposition 1. Let $(M,g)$ be a closed oriented three-dimensional Riemannian manifold endowed with a transversely oriented smooth foliation $\mathcal{F}$ of codimension one. Suppose that $\mathcal{F}$ contains a generalized Reeb component $A$ and there is an upper bound $|H|<H_0$ for the modulus of mean curvature of $\mathcal{F}$. Then $\operatorname{Area}(\partial A)<H_0 \operatorname{Vol}(M)$. Proof. Put $B=\overline{M\setminus A}$. Clearly, $B$ is a generalized Reeb component and $\partial A=\partial B$. Then we have
$$
\begin{equation*}
\begin{aligned} \, 2 \operatorname{Area} (\partial A) &= \biggl| \int_{\partial A} \chi\biggr| + \biggl|\int_{\partial B} \chi\biggr| = \biggl|\int_{A} d\chi \biggr|+ \biggl|\int_{B} d\chi \biggr| \\ &=2\biggl|\int_AH\mu\biggr|+2\biggl|\int_BH\mu\biggr|< 2H_0\biggl|\int_A\mu\biggr| +2H_0\biggl|\int_B\mu\biggr| \\ &=2H_0(\operatorname{Vol}(A)+\operatorname{Vol}(B))=2H_0 \operatorname{Vol}(M). \end{aligned}
\end{equation*}
\notag
$$
Here the second equality follows from the Stokes theorem and the third equality follows from Lemma 1. $\Box$. Lemma 3 (see [1], Ch. 6, § 25). Let $M$ be a closed orientable three-dimensional manifold endowed with a smooth transversely orientable foliation $\mathcal{F}$ of codimension one. Suppose that $A$ is a generalized Reeb component of $\mathcal{F}$. Then $\partial A$ is a family of tori. Proof. Since $M$ is orientable and the foliation is transversely orientable, the leaves are orientable manifolds. Since $\mathcal{F}$ contains a generalized Reeb component, it has no sphere $S^2$ as a leaf. Otherwise $\mathcal{F}$ would be an $S^2$-bundle over a circle by the Reeb stability theorem (see, for example, [1], § 21) and hence it would contain no generalized Reeb components. By the definition of a generalized Reeb component, we can choose a smooth vector field $\xi$ which is transversal to $\mathcal{F}$ (for example, orthogonal to $\mathcal{F}$ if $M$ is Riemannian) and directed inwards $A$ on the boundary. Let $\phi_t$ be the one-parameter group of diffeomorphisms generated by $\xi$. We note that$\phi_t (A)\subset A$ for all $t \geqslant 0$. Since $\xi$ is non-degenerate, there is a small $t_0>0$ such that the diffeomorphism $\phi_{t_0}$ has no fixed points. By the Lefschetz fixed point theorem, the algebraic number of fixed points of $\phi_{t_0}$ is equal to the Lefschetz number where $k$ is an arbitrary field. The family $\phi_{t \cdot t_0}$, $t\in [0,1]$, is a homotopy between $\phi_{t_0}$ and the identity map. Therefore the traces in the Lefschetz formula coincide with the Betti numbers, and the Lefschetz number coincides with the Euler characteristic $\chi(A)$. Since $\phi_{t_0}$ has no fixed points, the Euler characteristic $\chi(A)=0$. Replacing $\xi$ by $-\xi$ and repeating the argument, we obtain $\chi(B)=0$, where $B=\overline {M \setminus A}$. Since $M=A \cup B$ is a closed $3$-manifold, we have $\chi(A \cup B)=0$ and the formula $\chi(A \cup B)=\chi(A)+\chi(B)-\chi(A \cap B)$ yields that $\chi(\partial A=A \cap B)=0$. Since $\mathcal{F}$ contains no leaves homeomorphic to $S^2$, we have where $\{T_i\}$, $i=1, \dots, n$, are the compact leaves comprising $\partial A$. We conclude that $\chi(T_i)=0$ for every $i=1, \dots, n$. Hence all the sets $T_i$, $i= 1, \dots, n$, are homeomorphic to the two-dimensional torus. $\Box$ The following theorem of Loewner gives an upper bound for the length of the shortest closed geodesic on a Riemannian two-dimensional torus. Theorem 3 (Loewner; see [4]). Let $(T^2, g)$ be a two-dimensional torus endowed with an arbitrary Riemannian metric. Write $l$ for the length of the shortest closed non-contractible geodesic on $T^2$. Then 2.2. Supporting spheres and curvaturesThe following theorem gives lower bounds for the normal curvatures of a sphere in a Riemannian space whose sectional curvature is bounded from above. Theorem 4 (see [5], § 22.3.2). Suppose that $p\in M$, $\beta\colon [0,r]\to M$ is a normal geodesic which is a radius of the ball $B(p,r)$, and the point $\beta(r)$ is not conjugate to $p$ along $\beta$. Let the radius $r$ be such that the space of constant curvature $\gamma_0$ also contains no conjugate points at distance smaller than $r$. If, at every point $\beta (t)$, the sectional curvatures $\gamma$ of $M$ do not exceed $\gamma_0$, then the normal curvatures $k^S_H$ of the sphere $S(p,r)$ at the point $\beta(r)$ with respect to the normal $-\beta'$ are greater than or equal to the similar curvature $k_H^0$ of yje sphere of radius $r$ in the space of constant curvature $\gamma_0$. Remark 3. As an immediate corollary of this theorem, we see that if the hypotheses of Theorem 2 hold for $M$, then all normal curvatures of the sphere $S(r)\subset M$ are positive provided that $r<i_0$ and the normal to $S(r)$ is directed inwards the ball $B(r)$ bounded by it.1 Such a normal is said to be inner. Definition 1. We say that a hypersurface $S\subset M$ in a Riemannian manifold $M$ is supporting for a subset $A\subset M^n$ at a point $p\in \partial A\cap S$ with respect to a normal $n_p \perp T_pS$ if $S$ divides some ball $B_p$ centered at $p$ into two components (we include $S\cap B_p$ in both) and $A\cap B_p$ is contained in the component to which $n_p$ is directed. Definition 2. We say that a sphere $S(r)\subset M$, $r< i_0$, is supporting for a set $A\subset M$ at a point $q\in A\cap S(r)$ if it is supporting for $A$ at $q$ with respect to the inner normal. Lemma 4. Suppose that a surface $F\subset M$ is tangent to the sphere $S(r_0)$, $r_0< i_0$, at a point $q$ and $S(r_0)$ is supporting for $F$ at $q$. Then $k^S_H(v) \leqslant k_H^F(v)$ for every $v\in T_qS(r_0)$. Proof. We endow the ball $B(r_0)$ with polar coordinates $(r,\phi,\psi)$ in such a way that the coordinate $\phi$ corresponds to the angular coordinate of the polar coordinate system $(\rho, \phi)$ on the sphere $S(r_0)$ centered at a point $q\in S(r_0)$. For every tangent vector $v\in T_q(S(r_0))$ we have a surface $\Psi\colon \phi =c_1=\mathrm{const}$, on which the surfaces $S(r_0)$ and $F$ cut out the curves $\gamma_1\colon r=r_0$ and $\gamma_2\colon r=g(\psi)$ passing through $q$ and having tangent vector $v$ at $q$. The curvatures at $q$ of the curves $\gamma_1$, $\gamma_2$ on the surface $\Psi$ coincide with the normal curvatures $k^S_H(v)$ and $k_H^F(v)$ at $q$ since the direction along the $r$-coordinate is collinear to the direction of the normal, to $\gamma_1$, $\gamma_2$ at the point $q\in \Psi$ and to the direction of the common normal (which we choose to be inner for the sphere $S(r_0)$) to the surfaces $S(r_0)$ and $F$ at $q$.
Now, it suffices to calculate the curvatures at $q$ of the curves on $\Psi$. Since the $r$-coordinate lines are geodesic in the ambient space, they are geodesic on $\Psi$ and, therefore, the metric on $\Psi $ is of the form We endow the curves with an arclength parameter $t$ in such a way that $q=\gamma_1(0)=\gamma_2(0)$ and $\dot{\gamma_1}|_{t=0}=\dot{\gamma_2}|_{t=0}$. Taking the direction of the normal into account, we obtain an expression for the curvature of the first curve at $q$:
$$
\begin{equation*}
-\nabla^1_{\dot{\gamma_1}}\dot{\gamma_1}|_{t=0} =-\ddot{\gamma}^1_1|_{t=0}-\Gamma^1_{ij}\dot{\gamma^i_1}\dot{\gamma^j_1}|_{t=0},
\end{equation*}
\notag
$$
while the curvature of the second curve is Since the curve $\gamma_1$ lies on the sphere $S(r_0)$, its first coordinate $\gamma^1_1$ is equal to a constant and, therefore, $\dot{\gamma}^1_1=\ddot{\gamma}^1_1 \equiv 0$. Moreover, $\dot{\gamma_1}|_{t=0}=\dot{\gamma_2}|_{t=0}$. Hence we have The last inequality holds because the sphere $S(r_0)$ is supporting for $F$ at $q$ and, therefore, the curve $\gamma_2$ lies below the horizontal line $\gamma_1\colon r=r_0$ in some neighborhood of zero and has a local maximum at $t=0$. $\Box$ 2.3. Novikov’s theorem and vanishing cyclesThe following well-known theorem of Novikov gives topological obstructions for the existence of taut foliations. Theorem 5 (see [6]). Let $M$ be a closed three-dimensional manifold endowed with a smooth foliation $\mathcal{F}$ of codimension one. Suppose that any of the following conditions holds: 1) $\pi_1(M)$ is finite; 2) there is a leaf $ L \in \mathcal{F}$ such that the homomorphism $i_*\colon \pi_1({ L})\to \pi_1(M)$ induced by the embedding $i\colon L\to M$ has a non-trivial kernel. Then the foliation contains a Reeb component. We shall need the following construction underlying the proof of Novikov’s theorem. Suppose that a simple closed regular curve2 $\alpha\subset L$ 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). Theorem 6 (see [6]). 1) The interior of $D$ always contains either a closed integral curve of $X$ or a separatrix loop bounding a disc $D'\subset D$ such that all integral curves inside $D'$ are either closed curves or separatrix loops. Moreover, the curve $g(\partial D')$ is not null-homotopic on its leaf while the $g$-image of any closed integral curve or separatrix loop inside $D'$ is null-homotopic on its leaf. 2) There is a continuous one-parameter family of closed curves $f_t\colon S^1 \to D'$, $t\in [0,1]$, with the following properties:
The curve $c:= g\circ f_0\colon S^1 \to T$ is called a vanishing cycle. Lemma 5. The vanishing cycle $c$ is a non-trivial element of the kernel of the homomorphism $\pi_1(T)\to \pi_1(\mathcal{R})$. Proof. Since $0\ne [g( \partial D')]\in \pi_1(T) $ by definition and the $g$-images of the closures of all other trajectories of $X$ inside $D'$ are closed curves contractible along their leaves, we have a homotopy $g\circ f_t, \ t\in [0,1]$, such that the vanishing cycle $c=g\circ f_0$ represents a non-trivial element of the fundamental group $\pi_1(T)$ and, for every $t$ with $ 0<t \leqslant 1$, the loop $g\circ f_t\colon S^1\to L_t\subset \mathcal{R}$ is null-homotopic in the leaf $L_t \in \mathcal{F}$, whence $c$ represents the zero element of the fundamental group $\pi_1(\mathcal{R})$. $\Box$
§ 3. An auxiliary theoremLet $A\subset X$ 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). Theorem 7. Let $(S,\partial S) \subset (\mathcal{R}, T)$ be a connected relative two-dimensional compact submanifold (a relative surface) of the solid torus $\mathcal{R}$ with $\partial \mathcal{R} =T$. Suppose that it is homeomorphic to the sphere $S^2$ with finitely many open disjoint discs deleted, and the boundary $\partial S$ is a set of circles (not necessarily disjoint) contractible inside $\mathcal{R}$. Endowing $S$ with a triangulation and an orientation (for example, induced from an orientation of the sphere), we obtain a singular relative cycle $[S,\partial S] \in H_2(\mathcal{R},T)$. The following assertions are equivalent: 1) $0\ne [S,\partial S] \in H_2(\mathcal{R},T)$; 2) $0\ne [\partial S] \in H_1(T)$; 3) $S$ does not separate $\mathcal{R}$; 4) there is a closed smooth curve in $\operatorname{int}\mathcal{R}$ intersecting $S$ transversally at a single point; 5) every connected component of the lift of $S$ to the universal covering $\widetilde{\mathcal{R}}$ is homeomorphic to $S$ and divides $\widetilde{\mathcal{R}}$ into two non-compact components. Proof. $1 \Leftrightarrow 2$. Consider the following part of the exact homology sequence of a pair: 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 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. Corollary 2. If the relative surface $(S,\partial S) \subset (\mathcal{R}, T)$ satisfies the hypotheses of Theorem 7 and one of the conditions 1)–5) of the theorem does not hold, then any connected component $S_j$ of the lift of $S$ to the universal covering $\widetilde{\mathcal{R}}$ divides $\widetilde{\mathcal{R}}$ into two connected components, one of which is compact. Proof. By Theorem 7, if one of the conditions 1)–5) does not hold, then neither of them holds. Assuming that $S_i$ does not divide $\widetilde{\mathcal{R}}$, we can easily find a closed smooth curve $\gamma \in \widetilde{\mathcal{R}} \setminus \partial \widetilde{\mathcal{R}}$ that intersects $S_i$ transversally at a single point $s\in \widetilde{\mathcal{R}}$. We identify $\widetilde{\mathcal{R}} \setminus \partial \widetilde{\mathcal{R}}$ with $\mathbb{R}^3$. The class dual to $\gamma \subset \mathbb{R}^3$ is the Thom class $\Phi_1$ of the normal bundle of the curve $\gamma$. This class can be localized in an arbitrary tubular neighborhood $U_1$ of $\gamma$. Hence it is compactly supported and represents an element of $H^2_c(\mathbb{R}^3)$ (see [9], Ch. 1, § 6). The Thom class $\Phi_2$ of the normal bundle of the submanifold $S_i\setminus \partial S_i \subset \mathbb{R}^3$ represents an element of $H^1(\mathbb{R}^3)$ and can be localized in an arbitrary tubular neighborhood $U_2$ of the manifold $S_i\setminus \partial S_i $. Then $\Phi_3 =\Phi_2\wedge \Phi_1$ is the Thom class dual to $s\in \mathbb{R}^3$. It is represented by a bell-like form in the neighborhood $U_1\cap U_2$ of the point $s$ and, therefore, represents a non-trivial element of $H^3_c(\mathbb{R}^3)$ (see [9], Ch. 1, § 6). Since $H^2_c(\mathbb{R}^3)=H^1(\mathbb{R}^3)=0$, this contradicts the existence of a non-degenerate pairing (see [9], Ch. 1, § 5) in view of the following formula:
Hence $S_i$ divides $\widetilde{\mathcal{R}}$ into two components. One of them is compact because condition 5) of Theorem 7 does not hold. Since the universal covering $\widetilde{\mathcal{R}}$ is non-compact, the second component cannot be compact. $\Box$ § 4. Proof of Theorem 2Assume 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 By the hypothesis of Theorem 2, whence $l^2 <(2i_0)^2$. Therefore $\alpha$ is contained in the open ball $\operatorname{int}{B}(r)$ centered at $o \in \alpha$ of radius $r$, where the exponential map $\exp|_{B(r)}\colon T_oM \to M$ is one-to-one and, therefore, $\alpha$ is contractible inside $\operatorname{int}{B}(r)$. We span the geodesic $\alpha$ by a disc $g\colon D\to \operatorname{int}B(r)$ as in Theorem 6 above and find a vanishing cycle $c\subset D' \subset D$ belonging to the torus $T$ that bounds the Reeb component $\mathcal{R}$.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
$$
\begin{equation}
\operatorname{pr}_r|_ {T\,{\cap}\operatorname{int}B(i_o)}\colon T\cap \operatorname{int}B(i_o) \to \mathbb{R}
\end{equation}
\tag{4.4}
$$
is of Lebesgue measure zero. Hence we can choose $r$ in (4.3) in such a way that the intersection where $S(r):= \partial B(r)$, is either the empty set or the preimage of a regular value of (4.4) and, therefore, a disjoint union of circles. Remark 4. The regular value of $r$ in (4.3) can be chosen arbitrarily close to $l/2$. In particular, in view of (4.1), it can be chosen in such a way that Then it follows automatically from (4.6) and (4.2) that $r<i_0$. Proposition 2. There is a value of $r$ satisfying (4.6) such that the sphere $S(r)$ is supporting for a leaf of the Reeb component $\mathcal{R}$. Proof. Case 1. The intersection (4.5) is non-empty and all the circles occurring in it are null-homotopic in $T$.
Choose a regular value $r$ of the function (4.4) such that (4.3) and (4.6) hold. Consider the connected component $A$ of the set $\mathcal{R} \cap B(r)$ containing the vanishing cycle $c$. The set $ A\cap S(r) \subset \partial A$ is a finite disjoint union of surfaces, each of which is homeomorphic to a sphere with some number of holes. Let $S$ be one of these surfaces. Its boundary $\partial S$ is a subset of the family of circles (4.5), each of which is contractible in $T$ by assumption. Hence the hypotheses of Theorem 7 hold. By this theorem, the preimage of $S$ under the covering map $p\colon \widetilde{\mathcal{R}}\to\mathcal{R}$ is a disjoint union $\bigsqcup_iS_i$ of copies $S\simeq S_i$. The boundary $\partial S_i\subset \partial\widetilde{\mathcal{R}}$ clearly consists of circles contractible along $\partial\widetilde{\mathcal{R}}$, each of which bounds a two-dimensional disc in $\partial\widetilde{\mathcal{R}}$ by Jordan’s theorem. We obtain from Corollary 2 that every surface $S_i$ cuts $\widetilde{\mathcal{R}}$ into two connected pieces: where $W^{\mathrm{c}}_i$ (resp. $W_i$) is the compact (resp. non-compact) connected component.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 components4 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 normal5 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 supporting6 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 in $H_1(T)\cong \pi_1(T)\cong \mathbb{Z}^2$. We easily see that $G$ is a relative surface homeomorphic to a sphere with finitely many open discs removed. Since $c_i$ and $c$ are null-homotopic in $\mathcal{R}$, Theorem 7 yields that the preimage of $G \subset \mathcal{R}$ under the covering map $p\colon \widetilde{\mathcal{R}} \to \mathcal{R}$ consists of a countable family of copies $G_k\cong G$, $k\in \mathbb{Z}$, each of which divides $\widetilde{\mathcal{R}}$ into two non-compact parts in view of (4.8).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)). 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). 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 Since all the normal curvatures $k^0_H$ of a sphere of radius $r<i_0$ in the sphere of constant curvature $\gamma_0$ are constant and positive, we have $0<k^0_H\leqslant \min_v k^S_H(v) \leqslant \min_v k_H^F(v)$ and $k^0_H\leqslant \max_v k^S_H(v) \leqslant \max_v k_H^F(v)$. Since the mean curvature is half the sum of the (minimal and maximal) principal curvatures, we have the following inequalities at the point $q$: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
$$
\begin{equation}
\begin{cases} \dfrac1{r} <H_0, \\ 2\sqrt{3}\, r^2 <V_0H_0. \end{cases}
\end{equation}
\tag{4.10}
$$
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. 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:
$$
\begin{equation}
\begin{cases} \sqrt{\gamma_0}\,\cot(r\sqrt{\gamma_0}) <H_0, \\ 2\sqrt{3}\, r^2 <V_0H_0. \end{cases}
\end{equation}
\tag{4.11}
$$
Eliminating $r^2$, we obtain
$$
\begin{equation*}
\frac{1}{\gamma_0}\operatorname{arccot}^2\frac{H_0}{\sqrt{\gamma_0}}-\frac{V_0}{2\sqrt{3}}H_0< 0.
\end{equation*}
\notag
$$
Note that the left-hand side of the inequality is a decreasing function of $H_0$. This function is positive near zero and negative at infinity. Hence there is a unique positive root $x_0$ of the equation
$$
\begin{equation*}
\frac{1}{\gamma_0}\operatorname{arccot}^2\frac{x}{\sqrt{\gamma_0}}-\frac{V_0}{2\sqrt{3}}\,x=0.
\end{equation*}
\notag
$$
In the domain $0< H_0 \leqslant\min \{2\sqrt{3}\,i_0^2/V_0, x_0\}$ we clearly obtain a contradiction. 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. Corollary 3. Let $(M, g)$ be a three-dimensional closed orientable Riemannian manifold satisfying the hypotheses of Theorem 2. If $\pi_1(M)\,{<}\,\infty$, then $(M, g)$ admits no transversely orientable foliation of codimension one such that the mean curvature $H$ of its leaves satisfies the inequality $|H|< H_0$. The author is grateful to Professor A. A. Borisenko for his interest in this work and useful discussions. 
Citation in format AMSBIB
\Bibitem{Bol22} |
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