Essential Laminations in Non-Haken 3-manifolds Mark Brittenham*

In this paper we show that an essential lamination L in a non-Haken 3-manifold M is ‘tightly wrapped’, in the sense that any two leaves of L have intersecting closures; L therefore contains a unique minimal sublamination. We also show that these properties are inherited by any lift of L to a finite cover of M.

Essential laminations are natural generalizations of incompressible surfaces in a 3manifold, and give topologists an object which can be found in many more 3-manifolds than an incompressible surface can (see, e.g., [De] or [Na]). Yet they retain enough properties in common with incompressible surfaces so that they can be used to prove some of the powerful results about 3-manifolds that incompressible surfaces have been used to prove (see, e.g., [G-O] or [G-K]).

This note was motivated by asking the question ‘What \underline{more} can be said about an essential lamination if we know that the 3-manifold M containing it is non-Haken, i.e., M does \underline{not} contain a (2-sided) incompressible surface?’ What we find is the following ‘structure theorem’:

Theorem 1: Let L be an essential lamination in a compact, non-Haken 3-manifold. Then L contains a unique non-empty minimal sublamination L_{0}, i.e., a lamination L_{0} ⊆ L

* Supported in part by NSF grant #DMS-8610730 with the property that L_{0}⊆\overline{L} for every leaf L of L. L_{0} can therefore be defined (a posteriori)

as L_{0} = ⋂_{L⊆L}\overline{L} ≠ ∅.

Since a codimension-1 foliation without Reeb components is an example of an essential lamination, we have as an immediate consequence:

Corollary 2: A codimension-1 foliation F without Reeb components in a compact non-Haken 3-manifold M contains a unique minimal set.

This corollary is of independent interest; it is also an interesting example of a foliationtheoretic result which (seemingly) requires a lamination-theoretic proof. Previously (see [La] or [Co]) it was known that F has only a finite number of distinct minimal sets (that result does apply much more generally, however).

Thus (in a sense), in a non-Haken 3-manifold essential laminations must be ‘tightlywrapped’. This restriction on their structure adds a potentially useful new tool to the further study of essential laminations. In particular, since Haken manifolds are already fairly well-understood, this additional structure appears in exactly those 3-manifolds in which essential laminations will be the most useful: those which do not already contain an incompressible surface. It can therefore turn what might be perceived as a liability (the lack of an compact leaf, to which more classical techniques could apply) into an asset (extra information about the ‘shape’ of the lamination); see, e.g., [Br1]. This usefulness will have its limitations, however: in the second section we demonstrate its lack of application to finding Haken finite coverings of non-Haken 3-manifolds, by showing that the same tightness property is inherited by any lift of L to a finite cover of M.

1. Proof of Theorem 1:

The reader is referred to [G-O] for the basic concepts regarding essential laminations. We will assume throughout that the 3-manifold M in question is compact (hence closed, because M is non-Haken), connected, and orientable.

Proposition 3: If M is an irreducible 3-manifold with non-empty boundary, then either M contains a closed (2-sided) incompressible surface or M is a handlebody.

Proof: Choose a ∂-component F of M. If it is incompressible, we are done. Otherwise there is a compressing disk D for F, and we begin building a \underline{compression} \underline{body} (see [Bo]) for M along F by writing M= M_{0}∪C_{0}, where C_{0} = N(F∪D) (i.e., it is N(F) with a 2-handle attached), and M_{0} is the rest of M. ∂M_{0} has boundary component F_{0} =∂N(F∪D)\F, and we again ask if this is incompressible in M_{0}. Continuing inductively, if at any stage F_{i} is incompressible in M_{i}, then it is incompressible in M, because F_{i} is also incompressible in C_{i} (turning it upside-down (i.e., turning its handlebody structure upside-down), C_{i} is basically N(F_{i}) with a bunch of 1-handles attached, so F_{i} π_{1}-injects). Otherwise, we can decompose F down to a collection of sphere ∂-components of some M_{n}. It is easy to see that M_{n} is irreducible (inductively), and so M_{n} then consists of a union of 3-balls = 0handles; Turning C_{n} upside-down then demonstrates that M is a union of 0-handles with 1-handles attached, i.e., is a handlebody.

Lemma 4: Every lamination L contains a minimal sublamination, i.e., a non-empty lamination L_{0}⊆L which contains no proper (non-empty) sublamination.

Proof: Look at the collection of non-empty sublaminations of L, ordered by inclusion. Because M is compact, any sequence L_{0} ⊇ L_{1} ⊇ … has a lower bound ⋂_{i} L_{i} ≠ ∅ (any finite intersection is non-empty, so the full intersection is; it is a lamination because any leaf of L which has a point in the intersection is entirely contained in each L_{i}, so is entirely contained in their intersection). Therefore, by Zorn’s Lemma, this collection contains minimal elements, i.e., non-empty sublaminations properly containing no others.

Lemma 5: If L′ is an essential lamination obtained from the essential lamination L by splitting L open (see [G-O]) along some finite collection of leaves, and if L′ contains a unique minimal sublamination, then L contains a unique minimal sublamination.

Proof: From the definition of splitting it follows that there is a continuous surjection π : M → M with π(L′)=L which carries leaves of L′ onto leaves of L (just crush the I-fibers of the splitting regions to points).

If L_{0} is a non-empty minimal sublamination of L, then L′_{0}= π^{−1}(L_{0})∩L′ is a sublamination of L′. By assumption, L′ contains a unique minimal sublamination C′, so because there is a minimal sublamination of L′ in L′_{0} (there is a minimal sublamination \underline{for} L′_{0}, thought of on its own, which must then be minimal for L′, as well), it follows that C′ ⊆L′_{0}.

But then π(C′) ⊆ π(L′_{0} )=L_{0} is closed in M (C′ is closed hence compact in M, so π(C′) is compact hence closed in M) and saturated (leaves go to leaves under π), so is a non-empty sublamination of L contained in L_{0}; therefore, L_{0}= π(C′) by the minimality of L_{0}. Therefore \underline{every} minimal sublamination of L is equal to π(C′), i.e., there is only one.

Now let L ⊆M be an essential lamination carried with full support by an essential branched surface B⊆M, with M compact.

Proposition 6: If M is non-Haken, then M\int(N(B)) consists of handlebodies; in particular, B is connected.

Proof: By the above proposition, if some component M_{0} is not a handlebody, then it contains a closed incompressible surface F⊆ M_{0}. The branched surface B∪F then carries the lamination L∪F with full support, and it is easy to verify that B∪F is essential in M. The surface F misses the branch locus of B∪F, so B∪F has no disks of contact or monogons because B doesn’t; ∂_{h}N(B∪F) is incompressible by the choice of F, and has no disk or sphere components because F is not a disk or sphere; and M\N(B∪F) is irreducible, again by choice of F. Finally, B∪F has no Reeb branched surface: since B doesn’t have one, any such would have to include F. But since any lamination carried by F must contain a surface homeomorphic to F (the first leaf you meet falling in along an I-fiber must be a 1-fold cover of a component of ∂N(F)=F∪F), F must be a torus, and the Reeb component must consist of a surface parallel to F together with planar leaves limiting on it from parts of B. But this is impossible, because this would require branch curves along F, which do not exist.

Therefore by [G-O] every leaf of L∪F, and F in particular, is π_{1}-injective in M, making M Haken (it is already irreducible by [G-O]). But this contradicts our hypothesis, so every component M_{0} is a handlebody. It follows that B is connected; if it weren’t, then some component of M\int(N(B)) would have more than one ∂-component.

Essential laminations carried by branched surfaces with the above property are studied in [H-O], where they are called ‘full’; this result therefore says that every essential lamination in a non-Haken 3-manifold is full.

Proposition 7: M and L as above. Then for any two leaves L_{1}, L_{2} of L, the intersection of their closures, \overline{L}_{1} ∩ \overline{L}_{2} is non-empty.

Proof: Consider the essential lamination L_{0}= \overline{L}_{1} ∪ \overline{L}_{2}, carried (see [G-O]) with full support by some essential branched surface B_{0}. Split L_{0} open along L_{1} and L_{2}; call the resulting lamination L′. L′ is still carried by B_{0}, and there is a canonical projection π:M→M with π(L′)=L_{0} which takes an I-fiber of B_{0} to itself. Let A_{k}, for k=1,2, be the inverse image, in L′, of the leaf L_{k}, under π. Each consists of one or two leaves, bounding an I-bundle component of M|L′, and each leaf maps onto L_{k} under π. The essential lamination L′ meets the I-fibers of B_{0} in nowhere-dense sets; this is because L_{1} ∪ L_{2} is dense in L_{0}, and (in an I-fiber) these points have been replaced by intervals meeting L′ only in their endpoints.

By [G-O], the branched surface B_{0} is \underline{infinitely} \underline{splittable} to L′; there is a sequence B_{0}, B_{1}, B_{2}, … of essential branched surfaces carrying L′ (with full support), with L′⊆ N(B_{i}) ⊆ N(B_{i−1}), and ∩N(B_{i}) =L′. Applying Proposition 6 to B_{i}, we can then conclude that it is connected. Since for each of the leaves A_{k}, k=1,2, the support (in B_{i}) of the sublamination \overline{A_{k}} is a (closed) sub-branched surface of B_{i}, it follows that these supports are not disjoint (otherwise they exhibit B_{i} as the union of two disjoint closed subsets). So there is an I-fiber α_{i} of N(B_{i}) which meets both \overline{A_{1}} and \overline{A_{2}}. In particular, α_{i} meets A_{1} and A_{2}, in points x_{i} and y_{i}, respectively.

Consider the set of points x_{i} ⊆ A_{1}; this sequence has a limit point x in L⊆ \overline{A_{1}}. But because L′ met transverse arcs in nowhere-dense sets, it follows that the I-fibers of the N(B_{i}) must be becoming uniformly short (the I-fibers of the N(B_{i}) are \underline{nested} in the Ifibers of N(B_{i−1}), so their lengths must tend to zero, because L′ contains no transverse arcs). Therefore, the distance between x_{i} and y_{i} must be tending to zero; but then since the sequence x_{i} tends to x, it follows that the sequence y_{i} also tends to x. But the y_{i} are all in A_{2}, and so any limit point they have lies in \overline{A_{2}}, and therefore x lies in \overline{A_{2}}, i.e., x ∈ \overline{A_{1}} ∩ \overline{A_{2}}, which is therefore non-empty. But because π(A_{k})=L_{k} (so A_{k}⊆π^{−1}(L_{k})), it then follows that ∅ ≠ \overline{A_{1}} ∩ \overline{A_{2}}⊆π^{−1}(\overline{L}_{1})∩(π^{−1}(\overline{L}_{2}))=π^{−1}(\overline{L}_{1}∩\overline{L}_{2}), so \overline{L_{1}} ∩ \overline{L_{2}} ≠ ∅ in L_{0}.

This now allows us to finish the proofs of the theorem and corollary. First, to unify them, if L is actually a foliation F without Reeb components, split it open along a (finite) collection of leaves to make it an essential lamination L (i.e., so that it is carried by a branched surface). Now by Proposition 7, the closure of any two leaves intersect. Let L_{0} be a minimal sublamination for L, and write it as L_{0}= \overline{L_{0}} for L_{0} a leaf of L_{0} (\overline{L_{0}} is a sublamination of L_{0}, and so equals L_{0} by minimality). Then for any leaf L of L, \overline{L} ∩ \overline{L_{0}} = \overline{L}∩L_{0} is a non-empty sublamination of L_{0} (it is easy to see that it is closed and saturated), and therefore by minimality it equals L_{0}, i.e., L_{0}⊆ \overline{L} for \underline{every} leaf L of L. But if L_{1} = \overline{L_{1}} is any other minimal sublamination for L, then the fact that \overline{L_{0}} ∩ \overline{L_{1}} is non-empty implies that L_{0}= \overline{L_{0}} ∩ \overline{L_{1}} = L_{1}, so there is only one non-empty minimal sublamination. Therefore, by Lemma 5, our original lamination (or foliation) contains a unique minimal sublamination.

Finally, we finish with a slight improvement on the result L_{0}⊆ \overline{L}:

Corollary 8: L, M as above, with unique minimal sublamination L_{0}. Let L be a leaf of L having an end ϵ (see [Ni]). Then the limit set of the end, lim_{ϵ}(L), contains L_{0}.

Proof: lim_{ϵ}(L) is a closed, non-empty, saturated subset of L, i.e., a sublamination. It therefore contains L_{0}.

In other words, not only does every leaf of L limit on L_{0}, but every \underline{end} of every leaf does, as well.

2. Haken coverings of non-Haken 3-manifolds

The structure theorem above shows that the leaves of essential laminations in non-Haken 3-manifolds behave in a qualitatively different fashion, in general, from those of essential laminations in Haken manifolds. It therefore gives, in principle, a way to distinguish Haken manifolds from non-Haken ones: if a manifold contains an essential lamination which fails to have the property stated in the theorem, then the manifold must be Haken.

This distinction is potentially useful in determining when a laminar 3-manifold is virtually Haken. If M is non-Haken and contains an essential lamination L, which therefore has the property that the closures of any two leaves intersect, and has a finite covering π:M̃→M such that the inverse image L′ of L, which is an essential lamination in M̃, has two leaves with disjoint closures, then M̃ must be Haken, and so M is virtually Haken. This is a nice image; however, that is all it is:

Theorem 9: If π:M̃→M is a finite covering, with M non-Haken and L⊆M an essential lamination with L_{0}⊆L its unique minimal sublamination, then L′_{0}=π^{−1}(L_{0})⊆π^{−1}(L)=L′ is the unique minimal sublamination of L′; consequently, any two leaves of L′ have intersecting closures.

This theorem therefore says that an essential lamination in a non-Haken 3-manifold is really tightly wrapped; it can’t be unwrapped by passing to finite covers of M.

The only ingredient of the proof which we do not already have is the following result: Proposition 10: If π, M̃, M, L_{0}, and L′_{0} are as above, and if L′_{0}, L′ are leaves of L′_{0}

with \overline{L′_{0}}⊆\overline{L′}, then \overline{L′_{0}}=\overline{L′}.

Proof: Suppose that \overline{L′_{0}}≠\overline{L′}, so that \overline{L′_{0}} is properly contained in \overline{L′}; in particular, L′∩\overline{L′_{0}}=∅. We will show that π(\overline{L′_{0}})≠L_{0}, which is a contradiction, because π|\overline{L}′_{0}:L′_{0}→L_{0}, where L_{0} is the leaf of L_{0} which L′_{0} maps to, is a covering map, hence onto, so L_{0}=\overline{L}_{0}=\overline{π(L′_{0})}⊆π(\overline{L′_{0}})⊆L_{0}.

To do this, let N(B) be a fibered neighborhood of a branched surface carrying L_{0}, and let B′=π^{−1}(B) and N(B′)=π^{−1}(N(B)) (with I-fibers being carried to I-fibers under π). Pick a fiber X of N(B), and consider π^{−1}(X)=X_{1}∪…∪X_{n}. Pick a point x∈L_{0}∩X, and let \{x_{1},…,x_{k}\}=\overline{L′_{0}}∩π^{−1}(x), with x_{i}∈X_{i}, for all i. Because \overline{L′_{0}} is a closed set, for each x_{i}∈L′_{0}\\overline{L′_{0}} (i.e., each i=k+1,…,n) there is an open neighborhood O_{i} of x_{i} in X_{i} which misses \overline{L′_{0}}.

Consider O=π(O_{1}) ∩ … ∩ π(O_{k})⊆X. Because π maps the fibers X_{i} homeomorphically to X, this is an open subset of X containing x. Consider π^{−1}(O)⊆π^{−1}(X); note that π^{−1}(O)∩X_{i}⊆O_{i} for k+1≤i≤n.

Now look at π^{−1}(O)∩X_{k}=O′ (so π(O′)⊆O). This is an open neighborhood of x_{k} in X_{k}. Because \overline{L′} contains (\overline{L′_{0}} and hence) x_{k}, there are points of L′ in X_{k} which pass arbitrarily close to x_{k} and hence are contained in O′. Choose one; call it y′_{k}. Then y′_{k}∉\overline{L′_{0}}, and π^{−1}(π(y′_{k}))∩X_{i}=π^{−1}(y′) ∩ X_{i} ∈O_{i} are not in \overline{L′_{0}} for k+1≤i≤n. So by choosing a different point y′ of O we have increased the number of points, in the inverse image of a point of L_{0}∩X, which are not in \overline{L′_{0}}. Continuing, we can therefore find a point y of L_{0}∩X (in fact, in L_{0}∩O) with |π^{−1}(y)∩(L′_{0}\L′_{0})|=n, i.e., π^{−1}(y)⊆L′_{0}\L′_{0}. But this means that y∉ π(\overline{L′_{0}}), a contradiction. So \overline{L′_{0}}=\overline{L′}.

The above result is a special case of a more general result on the structure of pullbacks of laminations under non-zero degree maps; see [Br1],[Br2].

Proof of Theorem 9: For technical reasons, we must first (as in Proposition 7) split L_{0} open along a leaf, to insure that L_{0} (and hence L′_{0}) meets I-fibers of some (hence every) branched surface neighborhood in nowhere-dense sets. We will first prove Theorem 9 for this (possibly different) collection of laminations.

Proposition 10 implies that for any pair of leaves L, L′ of L′_{0}, either \overline{L}∩\overline{L′}=∅ or \overline{L}=\overline{L′}. Consequently, L′_{0} is the union of a finite number of disjoint minimal sublaminations, \overline{L}_{1},…,\overline{L}_{n}. This is because the closure of any leaf is a minimal sublamination, and every minimal sublamination of L′_{0} maps onto L_{0} under π. Thus for a given leaf L of L_{0}, every minimal sublamination of L′_{0} contains an inverse image leaf of L, of which there are only finitely many.

Claim: n=1.

Otherwise, there is a branched surface closely approximating L′_{0} which is not connected. To see this, just choose one whose fibered neighborhood has fibers with length less than δ = half of the distance between two components of L′_{0}; this can be done because (as in Proposition 7) L′_{0} meets I-fibers in nowhere-dense sets). But this also means that there is an (essential) branched surface B carrying L_{0} with full support whose inverse image under π is not connected; choose one with fibers of the same short length δ.

But this situation is absurd; B has complement consisting entirely of handlebodies, hence so does π^{−1}(B) (the only thing that finitely covers a handlebody is a handlebody), so every component of M̃\N(π^{−1}(B)) has connected boundary, so π^{−1}(B) is connected.

Therefore L′_{0} is a minimal sublamination of L′. But then since for every leaf L′ of L′, \overline{π(L′)}⊆π(\overline{L′}), and since L_{0}⊆\overline{L}=\overline{π(L′)} for π(L′)⊆L⊆L (since π(L′)=L), we have that π(\overline{L′})∩L_{0}≠ ∅, so \overline{L′}∩π^{−1}(L_{0})=\overline{L′}∩L′_{0}≠ ∅. Therefore L′_{0}⊆\overline{L′}. So any two leaves of L′ have closures whose intersection contains L′_{0}; it is therefore the unique minimal sublamination of L′.

But now if we collapse our split open leaves of L′_{0} back again, Lemma 5 then insures that L′ contains a unique minimal sublamination; in particular, it is the image under collapsing of the minimal sublamination L′_{0}, i.e., it is the original L′_{0}.

As a final note, we should point out that not only is this true for π^{−1}(L)=L′, but also for any sublamination L′_{1} of L′; the closure of any leaf of L′_{1} is the same whether we think we are in L′_{1} or L′, so it contains L′_{0}. In particular, any essential branched surface carrying L′_{1} has complement consisting of handlebodies. For otherwise (by the proof of Proposition 6) there would be an incompressible surface in M̃ missing L′_{1}, hence missing L′_{0}, hence contained in the handlebody complement of some essential branched surface carrying L′_{0}, a contradiction.

References:

[Bo] F. Bonahon, Cobordism of automorphisms of surfaces, Ann. Sci. Ecol. Norm. Sup., ser. 4 no. 16 (1983) 237-270

[Br1] M. Brittenham, Essential laminations and deformations of homotopy equivalences, I: from essential pullback to homeomorphism, preprint

[Br2] M. Brittenham, Essential laminations and deformations of homotopy equivalences, II: the structure of pullbacks, in preparation

[Co] L. Conlon, Foliations of codimension-one, notes from lectures at Washington University, Fall 1989

[De] C. Delman, Essential laminations in knot complements, Thesis, Cornell University, 1992

[G-K] D. Gabai and W. Kazez, 3-manifolds with essential laminations are covered by solid

tori, preprint [G-O] D. Gabai and U. Oertel, Essential laminations in 3-manifolds, Annals of Math. 130

(1989) 41-73 [H-O] A. Hatcher and U. Oertel, Full laminations in 3-manifolds, preprint

[La] C. Lamoreux, Feuilles fermées et captage; applications, C.R. Acad. Sci. Paris 277 (10-73), 579-581

[Na] R. Naimi, Constructing essential laminations in some 3-manifolds, Thesis, Caltech, 1992 [Ni] T. Nishimori, Ends of leaves of codimension-1 foliations, Tôhoku Math. J. 31 (1979)

1-22

Department of Mathematics University of Texas at Austin Austin, TX 78712

AMS Subject Classifications: 57M10, 57M99 Keywords: essential lamination, non-Haken, virtually Haken