---
address:
- 'Rice University, Houston, Texas, 77005-1892'
- 'SUNY at Buffalo, Buffalo,N.Y.'
author:
- 'Tim D. Cochran'
- Joseph Masters
title: 'The Growth Rate of the First Betti Number in Abelian Covers of $3$-Manifolds '
---

[^1]

Abstract {#abstract .unnumbered}
========

We give examples of closed hyperbolic 3-manifolds with first Betti number $2$ and $3$ for which no sequence of finite abelian covering spaces increases the first Betti number. For $3$-manifolds $M$ with first Betti number $2$ we give a characterization in terms of some generalized self-linking numbers of $M$, for there to exist a family of $\mathbb{Z}_n$ covering spaces, $M_n$, in which $\beta _1(M_n)$ increases linearly with $n$. The latter generalizes work of M. Katz and C. Lescop \[KL\], by showing that the non-vanishing of any one of these invariants of $M$ is sufficient to guarantee certain optimal systolic inequalities for $M$ (by work of Ivanov and Katz \[IK\]).

Introduction {#introduction .unnumbered}
============

Motivated by Waldhausen’s work on Haken manifolds, and by W. Thurston’s [**Geometrization Conjecture**]{}, it has been variously conjectured that, if $M$ is an orientable, irreducible closed $3$-manifold with infinite fundamental group, then:

$M$ is finitely covered by a Haken manifold;

Some finite cover of $M$ has positive first Betti number;

Either $\pi_1(M)$ is virtually solvable or $M$ has finite covers with arbitrarily large first Betti number;

$M$ has a finite cover that fibers over the circle.

There are easy implications VIBNC$\Longrightarrow$VPBNC$\Longrightarrow$VHC and VFC$\Longrightarrow$VPBNC $\Longrightarrow$VHC. Each implies, if $M$ is atoroidal, the long-standing conjecture of Thurston that such a manifold admits a geometric structure. It is interesting to note that even if $M$ is [**assumed**]{} to be hyperbolic, the conjectures above are open.

In this paper, we restrict our attention to VIBNC. (We note in passing that the alternative “$\pi_1(M)$ is virtually solvable” is sometimes replaced by the a priori stronger alternative that “$M$ is finitely covered by the 3-torus, a nilmanifold or a solvmanifold.”) One rich source of finite covering spaces is the set of iterated (regular) finite [**abelian**]{} covering spaces. Thus specifically, in this paper we consider the question:

Does there exist an integer $m$, such that, if $M$ is any closed, atoroidal $3$-manifold with $\beta_1(M) \geq m$ then $\b_1(M)$ can be increased in a finite abelian covering space?

Note that some condition on $H_1(M)$ is necessary, for if $H_1(M)=0$, then $M$ admits no non-trivial abelian covering spaces. Counter-examples also exist for many manifolds with $\b_1(M)=1$. For if $M$ is zero-framed surgery on a knot in $S^3$, then it is easy to show that $H_1(\wt M;\BQ)\cong\BQ\op
Q[t,t^{-1}]/\<\Delta_k,t^n-1\>$ where $\wt M$ is the $n$-fold cyclic cover and $\Delta_k$ is the Alexander polynomial of $K$. Thus $\b_1(\wt M)=\b_1(M)=1$ except when $\Delta_k$ has a cyclotomic factor. We begin this paper by observing that counter-examples also exist in the cases $\b_1(M)=2$ and $\b_1(M)=3$.

\[mainthm1\] *There exist closed hyperbolic $3$-manifolds $M$ with $\b_1(M)=2$ (respectively $3$) for which no sequence of finite abelian covers increases the first Betti number. More generally, if a sequence of regular covers of M increases the first Betti number, then one of the covering groups contains a non-trivial perfect subgroup.*

It is noteworthy that Question A is still open.

If $\b_1 > 0$, then there is an epimorphism $\pi_1(M)\to \mathbb{Z}$, and a corresponding sequence of finite cyclic covers of $M$. Our second contribution is, in the case $\b_1(M)=2$, to give necessary and sufficient conditions, of a somewhat geometric flavor, for the Betti number of these covers to increase linearly with the covering degree. This is the content of Section 2.

On abelian covers of hyperbolic 3-manifolds with $\b_1(M)=2$ and $3$ {#failure}
====================================================================

In this section, we observe that, if Question A has an affirmative answer, then the integer $m$ must be at least 4.

\[failure23\] There exist closed hyperbolic $3$-manifolds $M$ with $\b_1(M)=2$ (respectively $3$) such that if $\wt M$ is obtained from $M$ by taking a sequence of finite abelian covering spaces, then $\b_1(\wt M)=\b_1(M)$. More generally, if a sequence of regular covers of M increases the first Betti number, then one of the covering groups contains a non-trivial perfect subgroup.

Begin with a “seed” manifold $N$ whose fundamental group is nilpotent. Recall that the [**Heisenberg manifold**]{} with Euler class $e$ is the circle bundle over the torus with Euler class $e$. The fundamental group of such a $3$-manifold is the nilpotent group $\langle x,y,t
: [x,y]=t^e, [x,t], [y,t]\rangle$, called the [**Heisenberg group**]{} of Euler class $e$. For our seed manifold with $\b_1(N)=2$, we shall take $N$ to be the Heisenberg manifold with Euler class $1$, that can also be described as $0$-framed surgery on a Whitehead link. Thus in this case $\pi_1(N)\cong F/F_3$ where $F$ is the free group of rank 2 and $F_3$ is the third term of the lower-central series of $F$. When $\b_1(N)=3$, we take our seed manifold $N$ to be $S^1\x S^1\x S^1$, the Heisenberg manifold of Euler class $0$. Note that each of the Heisenberg groups of non-zero Euler class has $\beta_1=2$ while the Heisenberg group of Euler class $0$ has $\beta_1=3$.

First we claim that *no* finite cover of $N$ will increase the first Betti number, which follows immediately from the Lemma below (which surely is well-known to experts).

\[nilpotent\] Suppose $A$ is a Heisenberg group with non-zero (respectively zero) Euler class. If $\wt A$ is any finite index subgroup of $A$, then $\wt A$ is a Heisenberg group of non-zero (respectively, zero) Euler class. Hence in all cases $\beta_1(\wt A) = \beta_1(A)$.

The result is obvious for $A=\mathbb{Z}\x \mathbb{Z}\x
\mathbb{Z}$ so we assume that $A$ is a Heisenberg group of non-zero Euler class $e$. Then $A$ is a central extension as shown below. $$1\lra \mathbb{Z}\overset{i}{\lra} A \overset{\pi}{\lra} \mathbb{Z}\times\mathbb{Z}\lra 1$$ Since $\wt A$ is a finite index subgroup of $A$, $\pi(\wt A)$ is a finite index subgroup of $\mathbb{Z}\times\mathbb{Z}$ which is hence isomorphic to $\mathbb{Z}\times\mathbb{Z}$. Moreover the kernel of the map $\pi:\wt A\to \pi(\wt A)$ is a finite index subgroup of kernel($\pi$)$=\mathbb{Z}$ which is contained in the center of $A$. It follows that $\wt A$ is also a central extension of the above form and hence is also a Heisenberg group. We claim that $\wt A$ has non-zero Euler class. Suppose not. Then $\wt A$ is abelian. But $A\cong \langle x,y,t : [x,y]=t^e, [x,t], [y,t]\rangle$ where $e\neq 0$. Consider the elements $\{x,y\}$. There is some positive integer $n$ such that both $x^n$ and $y^n$ lie in the subgroup $\wt A$ where they commute. Thus $[x^n,y^n]=1$ in $A$. However since $[x,y]=t^e$, and $t$ commutes with $x$ and $y$, it is easy to see that $x^ny^n=t^ky^nx^n$ where $k=n^2e$ and so $1=[x^n,y^n]=t^k$. This implies that $t$ is of finite order. However, any Heisenberg group is the fundamental group of a circle bundle over the torus, which is an aspherical 3-manifold. Thus $A$ has geometric dimension $3$ and cannot have torsion, for a contradiction.

Next alter the seed manifold in a subtle way using the following result of A. Kawauchi \[Ka1 p. 450-452 , and Ka2 Corollary 4.3\] (see also Boileau-Wang \[BW section 4\]).

\[hyperbolic\] (Kawauchi) For any closed $3$-manifold $N$, there exists a hyperbolic $3$-manifold $M$ and a degree 1 map $f:M\to N$ that induces an isomorphism on homology groups with local coefficients in $\pi_1(N)$. Equivalently, if $\wt N$ is any covering space of $N$ and $\wt f:\wt M  \to \wt N$ is the pull-back, then $\wt f$ induces isomorphisms on homology groups.

To the best of our knowledge, this result was first established by Kawauchi using his theory of [**almost identical imitations**]{}. We sketch a proof using the approach of Boileau and Wang (which overlaps substantially with Kawauchi’s approach). Recall that any $3$-manifold $N$ contains a knot $J$ whose exterior is hyperbolic. With more work, Boileau and Wang ensure that there exists such a knot $J$ which is “totally null-homotopic”, i.e., bounds a map of a 2-disk, $\phi:D^2\to N$, such that the inclusion map $\pi_1({\operatorname{image}}\phi)\to\pi_1(N)$ is trivial. Let $M_n$ be the result of $1/n$-Dehn surgery on $N$ along $J$. By work of W. Thurston, for almost all $n$, $M_n$ is hyperbolic. Choose such an $M_n$ and denote it by $M$. Since $J$ is null-homotopic there is a degree one map $f:M\to N$ that induces an isomorphism on $H_1$.

Let $\wt N$ be a cover of $N$. Since $J$ is null-homotopic, it lifts to $\wt N$, and there is an induced cover $\wt M$ and an induced map $\tl f:\wt M\to\wt N$. Since $J$ is totally null-homotopic, the pre-images of $J$ bound disjoint Seifert surfaces in $\wt M$, and so $\tl f:\wt M\to\wt N$ is an isomorphism on homology.

For any map $f: M \to N$ satisfying the conclusion of Proposition \[hyperbolic\], $ker(f_*)$ is a perfect group. Indeed, Proposition \[hyperbolic\] states that for [**any**]{} covering space $\wt M$ of $M$ that is “induced” from a cover $\wt N$ of $N$, the induced map $\tl f:\wt M\to\wt N$ is an isomorphism on homology, so $\b_1(\wt M)=\b_1(\wt N)$. Specifically, letting $\wt N$ be the universal cover, $H_1(\wt M)\cong H_1(\wt N)=0$ showing that that $\pi_1(\wt M)$ is a perfect group. But $\pi_1(\wt M)$ is kernel$(f_*)$. (Indeed, the condition that $f:M\to N$ induce an isomorphism on first homology with local coefficients in $\pi_1(N)$ is equivalent to the condition that the kernel of $f_*:\pi_1(M)\to\pi_1(N)$ be a perfect group).

Returning to the proof of our theorem, recall that $N$ is our seed Heisenberg manifold, and let $M$ be the manifold guaranteed by Proposition \[hyperbolic\]. We claim that the manifold $M$ satisfies the conclusion of the theorem. For suppose $\wt M\overset{p}{\lra}M$ is a regular finite covering space of $M$ corresponding to a surjection $\psi:\pi_1(M)\to F$, where $F$ is a finite group that contains no nontrivial perfect subgroup (for example if $F$ is abelian). Then, since the kernel of $f_*:\pi_1(M)\to\pi_1(N)$ is a perfect group $P$, and the perfect subgroup $\psi(P)\subset F$ must be trivial, $\psi$ factors through $f_*:\pi_1(M)\to\pi_1(N)$ via a surjection $\phi:\pi_1(N)\to F$. Therefore there is a finite regular cover $\wt N$ of $N$ and a lift $\tl f:\wt M\to\wt N$. Notice that the only property of $M$ and $N$ needed for this argument is that the kernel of $f_*:\pi_1(M)\to\pi_1(N)$ is a perfect group. Proceeding, by Proposition \[hyperbolic\] $H_1(\wt M)\cong H_1(\wt N)$ and by Lemma \[nilpotent\], $\beta_1(\wt N)=\beta_1(N)$. Since $\beta_1(M)=\beta_1(N)$ we conclude that $\beta_1(\wt M)=\beta_1(M)$. This shows that the first Betti number of $M$ cannot be increased by a *single* regular $F$-cover unless $F$ contains a nontrivial perfect subgroup. In particular, it shows that the first Betti number of $M$ cannot be increased by a single *abelian* cover.

Now suppose that $M_k \to ... \to M_0 = M$ is a sequence of regular covers, with covering groups $F_1, ..., F_k$, where no $F_i$ contains a nontrivial perfect subgroup. In the last paragraph we showed that the cover $M_1 \to M_0$ is the pull-back of a corresponding cover $N_1
\to N_0$. We claim that the kernel, $P_1$, of the lift $(f_1)*:\pi_1(M_1)\to\pi_1(N_1)$ is *equal to* the kernel, $P_0$, of $f_*:\pi_1(M_0)\to\pi_1(N_0)$ (here we view $\pi_1(M_1)$ as a subgroup of $\pi_1(M_0)$). For, obviously $P_1 \subset P_0$ and since $F_0$ contains no perfect subgroups, $P_0\subset P_1$. Thus $P_1$ is a perfect group and thus $\wt f_1$ induces an isomorphism on homology (even with twisted coefficients). Thus we have recovered the inductive hypothesis of the previous paragraph and continuing inductively, we get a sequence of finite covers $N_k \to ... \to N_1$, with $\b_1(M_k) = \b_1(N_k)$. Therefore, to finish the proof we only need to observe that $\b_1(N)$ cannot be increased by any sequence of finite covers, which was shown in Lemma \[nilpotent\].

Linear Growth of Betti Numbers in Cyclic Covering Spaces {#lineargrowth}
========================================================

In this section we ask whether or not it is possible to increase the first Betti number with *linear growth rate* in some *compatible family* of cyclic covering spaces. If $M_\infty$ is a fixed infinite cyclic covering space corresponding to an epimorphism $\psi :\pi_1(M)\to \mathbb{Z}$ then by a *compatible family* we mean the usual family of finite cyclic covers $M_n$ associated to $\pi_1(M)\to \mathbb{Z}\to \mathbb{Z}_n$. By a *linear growth rate* we mean $\varinjlim (\beta_1(M_n)/n)$ is positive. It was already known that linear growth occurs precisely when $H_1(M_\infty)$ has positive rank as a $\mathbb{Z}[t,t^{-1}]$-module [@Lu2 Theorem 0.1][@Lu1 pg.35 Lemma 1.34,pg.453]. Therefore our contribution is to offer a more geometric way of viewing this criterion. We also point out an application to certain optimal systolic inequalities for such $3$-manifolds as have appeared in work of Katz \[IK\]\[KL\].

One should note from the outset that if $\pi_1(M)$ admits an epimorphism to $\mathbb{Z}\ast\mathbb{Z}$, then it is an easy exercise to show that $\beta_1(M)$ can be increased linearly in finite cyclic covers since the same is patently true of the wedge of two circles. Such manifolds arise, for example, as $0$-framed surgery on $2$-component boundary links. This condition is not necessary, however, as we shall see in Example \[example3\] below.

Suppose $M$ is a closed, oriented $3$-manifold with $\b_1(M)=2$. Given any basis $\{x,y\}$ of $H^1(M,\BZ)$ we shall define a sequence of higher-order invariants $\b^n(x,y)$; $n\ge1$ taking values in sets of rational numbers. The invariants can be interpreted as certain Massey products in $M$. The invariant $\b^1(x,y)$ is always defined, is independent of basis, and essentially coincides with the invariant $\la$, an extension of Casson’s invariant, due to Christine Lescop \[Les\]. If $\b^i$ is defined for all $i<n$ and is zero, then $\b^n$ is defined (this is why the invariants are called higher-order). If $H_1(M)$ has no torsion, so that $M$ can be viewed as $0$-framed surgery on a 2-component link in a homology sphere (with Seifert surfaces dual to $\{x,y\}$) then $\b^n$, when defined, is the same as the sequence of link concordance invariants of the same name due to the first author \[C1\]. In this case $\b^1$ was previously known as the Sato-Levine invariant.

After defining the invariants $\b^n(x,y)$, we show that their vanishing is equivalent to the linear growth of Betti numbers in the family corresponding to the infinite cyclic cover associated to $x$.

\[linear\] Let $M$ be a closed oriented $3$-manifold with $\b_1(M)=2$. The following are equivalent.

1.  There exists a compatible family $\{M_n|n\ge1\}$ of finite cyclic covers of $M$ such that $\b_1(M_n)$ grows linearly with $n$.

2.  There exists a primitive class $x\in H^1(M;\BZ)$ such that for $\textbf{any}$ basis $\{x,y\}$ of $H^1(M;\BZ)$, $\b^n(x,y)=0$ for all $n\ge1$.

3.  There exists a primitive class $x\in H^1(M;\BZ)$ such that for $\textbf{some}$ basis $\{x,y\}$ of $H^1(M;\BZ)$, $\b^n(x,y)$ can be defined and contains $0$ for each $n\ge1$.

\[systole\] Let $M$ be a closed oriented $3$-manifold with $\b_1(M)=2$. Let $\wt M$ denote the universal torsion-free abelian ($\textbf{Z}\oplus \textbf{Z}$) cover of $M$. Let $[F]$ denote the class in $H_1(\wt M)$ of a lift of a typical fiber of the Abel-Jacobi map of $M$ (represented by a lift of the circle we called $c(x,y)$ below). If, for **some** $\{x,y\}$, and **some** $n$, $\b^n(x,y)\neq 0$ then $[F]$ is non-zero.

The above Corollary generalizes an (independent) result of A. Marin (see Prop.12.1 of \[KL\]), which dealt with only the case $n=1$. The significance of this Corollary is that it has been previously shown by Ivanov and Katz (\[IK, Theorem 9.2 and Cor.9.3\]) that the conclusion of Corollary \[systole\] is sufficient to guarantee a certain optimal systolic inequality for $M$. The interested reader is referred to those works.

Suppose $c$ and $d$ are disjointly embedded oriented circles in $M$ that are zero in $H_1(M;\BQ)$. Then the [**linking number of $c$ with $d$**]{}, ${\ensuremath{\ell k}}(c,d)\in\BQ$ is defined as follows. Choose an embedded oriented surface $V_d$ whose boundary is “$m$ times $d$” (i.e. a circle in a regular neighborhood $N$ of $d$ that is homotopic in $N$ to $md$) for some positive integer $m$, and set: $${\ensuremath{\ell k}}(c,d) = \f1m(V_d\cd c).$$ Given this, the invariants $\b^n(x,y)$ are defined as follows. Let $\{V_x,V_y\}$ be embedded, oriented connected surfaces that are Poincaré Dual to $\{x,y\}$ and meet transversely in an oriented circle that we call $c(x,y)$ (by the proof of \[C1, Theorem 4.1\] we may assume that $c(x,y)$ is connected). Let $c^+(x,y)$ denote a parallel of $c(x,y)$ in the direction given by $V_y$. Note that $\{V_x,V_y\}$ induce two maps $\psi_x$, $\psi_y$ from $M$ to $S^1$ wherein the surfaces arise as inverse images of a regular value. The product of these maps yields a map $\psi:M\to S^1\x S^1$ that induces an isomorphism on $H_1$/torsion. Since $c(x,y)$ and $c^+(x,y)$ are mapped to points under $\psi$, they represent the zero class in $H_1(M;\BQ)$. Therefore we may define $\b^1(x,y)={\ensuremath{\ell k}}(c(x,y)$, $c^+(x,y))$. In fact, $-\b^1(x,y)\cd|{\operatorname{Tor}}H_1(M;\BZ)|$ is precisely Lescop’s invariant of $M$ \[Les; p.90-94\]. An example is shown in Figure \[satolevine\] of a manifold with $\b^1(x,y)= -k$.

(105,92) (10,10)[![Example of $\b^1(x,y)= -k$[]{data-label="satolevine"}](satolevine.eps "fig:")]{} (11,63)[$0$]{} (92,20)[$0$]{} (76,61)[$k$]{} (39,0)[$M$]{}

The idea of the higher invariants is to iterate this process as long as possible (compare \[C1\]). Since $c^+(x,y)$ is rationally null-homologous, there is a surface $V_{c(x,y)}$ whose boundary is “$k$ times $c^+(x,y)$” (in the sense above). We could then define $c(x,x,y)$ to be $V_x\cap V_{c(x,y)}$, an embedded oriented circle on $V_x$. If $c(x,x,y)$ is rationally null-homologous, then $\b^2(x,y)$ is defined as ${\ensuremath{\ell k}}(c(x,x,y),c^+(x,x,y))$ and we may also continue and define $c(x,x,x,y)$. In general $c(x,x\dots,x,y)=c(x^n,y)$ will be able to be defined using the chosen surfaces if $c(x^{n-1},y)$ is defined and is also rationally null-homologous (but to do so involves one more choice of a bounding surface). Once $c(x^{n},y)$ is defined and is rationally null-homologous, we may define $\b^n(x,y)$. In general, we do not claim that the value of $\b^n(x,y)$ is independent of the choices of surfaces. Therefore the invariants can be thought of as taking values in a set, just like Massey products. This indeterminacy will not concern us here, for we are only interested in the first non-vanishing value (if it exists) and we shall see that this is independent of the surfaces.

Much of the time it is convenient to abbreviate $c(\overbrace{x\dots x}^n,y)$ as $c(n)$ so $c(x,y)=c(1)$.

If $c(n)$ is defined and rationally null-homologous then $\b^n(x,y)$ is defined to be the set of rational numbers ${\ensuremath{\ell k}}(c(n),c^+(n))$, ranging over all possible ways of defining such a $c(n)$. If no such $c(n)$ exists then $\b^n(x,y)$ is undefined.

\[Example3\] Consider the manifold $M$, shown in Figure \[example3\], obtained from $0$-framed surgery on a two component link $\{ L_x,L_y\}$. Use a genus one Seifert surface for $L_y$ obtained from the obvious twice-punctured disk and a tube that goes up to avoid $L_x$. Let $V_y$ be this surface capped-off in $M$. Similarly use the fairly obvious Seifert surface for $L_x$ in the complement of $L_y$. Then $c^+(x,y)$ is shown. Since it has self-linking zero with respect to $V_x$, $\b^1(x,y)=\b^1(y,x)=0$. Furthermore $\b^2(x,y)=-1$ (note the link $\{c^+(x,y),L_x\}$ is very similar to that of Figure \[satolevine\]). This means that $\pi_1(M)$ does **not** admit an epimorphism to $\mathbb{Z}\ast \mathbb{Z}$ since that would imply that $\{ L_x,L_y\}$ were a homology boundary link. But $\b^2(x,y)=-1$ precludes this by [@C2]. Nonetheless, further $c(yy...y,x)$ may be taken to be empty since $c^+(x,y)$ and $L_y$ form a boundary link in the complement of $L_x$. Thus $\b^n(y,x)= 0$ for all $n$, indicating, by Theorem \[linear\], that the first Betti numbers will grow linearly in the family of finite cyclic covers corresponding to the map $\pi_1(M)\to \mathbb{Z}$ that sends a meridian of $L_x$ to zero and a meridian of $L_y$ to one.

(123,73) (10,10)[![Example with linear growth in cyclic covers but no map to $\mathbb{Z}\ast\mathbb{Z}$[]{data-label="example3"}](example3.eps "fig:")]{} (72,39)[$c^+(x,y)$]{} (-5,39)[$L_y$]{} (117,17)[$L_x$]{} (65,0)[$M$]{} (36,46)[$0$]{} (119,68)[$0$]{}

\[Example2\] Consider the family of manifolds $M_k$, shown in Figure \[example2\] and Figure \[example2b\], obtained from $0$-framed surgery on a two component link.

(135,98) (10,10)[![Example with $\b^1(x,y)=0$,$\b^2(x,y)=-k$, $\b^2(y,x)= -1$[]{data-label="example2"}](example2.eps "fig:")]{} (83,93)[$0$]{} (41,15)[$0$]{} (110,61)[$k$]{} (65,0)[$M$]{}

(135,98) (10,10)[![The circle $c(x,y)$[]{data-label="example2b"}](example2b.eps "fig:")]{} (83,93)[$0$]{} (41,15)[$0$]{} (110,61)[$k$]{} (69,61)[$c(x,y)$]{} (65,0)[$M$]{}

If $V_x$ denotes the capped-off Seifert surface (obtained using Seifert’s algorithm) for the link component, $L_x$, on the right-hand side and $V_y$ denotes the capped-off Seifert surface for the link component, $L_y$, on the left-hand side, then the dashed circle in Figure \[example2b\] is $c(x,y)=V_x\cap V_y$. The circle $c(y,x)$ is merely this circle with opposite orientation. Since it lies on an untwisted band of $V_x$, $\b^1(x,y)=0=\b^1(y,x)$. Therefore the Lescop invariant of $M$ vanishes. But the link $\{c(x,y),L_x\}$ is the link of Figure \[satolevine\] so $\b^2(x,y)=-k$, whereas the link $\{c(y,x),L_y\}$ is a Whitehead link so $\b^2(y,x)= -1$. We claim further that, as long as $k\neq 0$, for $\textbf{any}$ basis $\{X,Y\}$ of $H^1(M)$, $\b^2(X,Y)\neq 0$. It will then follow from Theorem \[linear\] that the first Betti number of $M$ will grow sub-linearly in **any** family of finite cyclic covers. A general basis, $\{V_X,V_Y\}$, of $H_2(M)$ can be represented as follows. Represent $V_X$ by $p$ parallel copies of $V_x$ together with $q$ parallel copies $V_y$, and represent $V_Y$ by $r$ parallel copies of $V_x$ together with $s$ parallel copies $V_y$, where $ps-qr=\pm 1$. Thus $c(X,Y)=-c(Y,X)$ is represented by $ps-qr$ parallel copies of $c(x,y)$. It follows that $\b^1(X,Y)=\b^1(x,y)=0$, reinforcing our above claim that $\b^1$ is independent of basis. Hence $V_{c(X,Y)}=\pm V_{c(x,y)}$ so $c(X,X,Y)$ is represented by $\pm pc(x,x,y)\mp qc(y,y,x)$. Since $\b^2(X,Y)$ is the self-linking number of this class, it can be evaluated to be $$p^2\b^2(x,y) + q^2\b^2(y,x) - 2pq{\ensuremath{\ell k}}(c(x,x,y),c(y,y,x))$$ but the latter mixed linking number is easily seen to be zero in this case. Hence $\b^2(X,Y)=-kp^2-q^2$ which is non-zero if $k$ is non-zero.

\[equivalence\] Suppose $c(1),\dots,c(n)$ have been defined as embedded oriented curves on $V_x$ arising as $c(1)=V_x\cap V_y$ and $$c(j) = V_x\cap V_{c(j-1)}\qquad2\le j\le n$$ where $V_{c(j)}$, $1\le j\le n-1$, is an embedded, oriented connected surface whose boundary is a positive multiple $k_j$ of $c^+(j)$ (in the sense above). Then $\b^j$ is defined for $1\le j\le n-1$ and the following are equivalent:

1.  $\b^1,\dots,\b^n$ are defined using the given system of surfaces.

2.  $\b^j$ is defined for $1\le j\le n$ and is [**zero**]{} for $1\le
    j\le\[\f n2\]$

3.  $c(n+1)$ exists

4.  For all $s$, $t$ such that $1\le s\le t$ and $s+t\le n$ , ${\ensuremath{\ell k}}(c(s),c^+(t))=0$.

Assume $1\le j\le
n-1$. The hypotheses imply that a positive multiple of $c^+(j)$ is (homotopic to) the boundary of a surface so $c^+(j)$ and $c(j)$ are rationally null-homologous. Thus their linking number is well-defined, establishing the first claim.

[**B1$\Longleftrightarrow$B3**]{}: $\b^n$ is defined precisely when $[c(n)]=0$ in $H_1(M;\BQ)$ which is precisely the condition under which $c(n+1)$ can be defined.

[**B1$\Longrightarrow$B4**]{}: If $n=1$ the implication is true since B4 is vacuous. Thus assume by induction that the implication is true for $n-1$, that is our inductive assumption is that, for all $s+t<n$, ${\ensuremath{\ell k}}(c(s),c^+(t))=0$. Now consider the case that $s+t=n$. Since $\b^n$ is defined $[c(n)]=0$ in $H_1(M;\BQ)$. We claim this is true precisely when $c(n)\cd c(1)=0$ (here we refer to oriented intersection number on the surface $V_x$). For suppose $\psi_x:M\to S^1$ and $\psi_y:M\to S^1$ are maps such that $\psi^{-1}_x(*)=V_x$ and $\psi^{-1}_y(*)=V_y$. Then $(\psi_x)_*([c(n)])=0$ since $c(n)\subset V_x$; and $(\psi_y)_*([c(n)])=0$ precisely when $c(n)\cd V_y=c(n)\cd c(1)=0$. But the map $\psi_x \times \psi_y$ completely detects $H_1(M)$/Torsion. Therefore, once $c(n)$ exists, $\b^n$ is defined if and only if: $$\begin{aligned}
0 = c(n)\cd c(1)  &= \pm(c(1)\cd V_{c(n-1)})\\
&= \pm k_{n-1}{\ensuremath{\ell k}}(c(1),c^+(n-1))\end{aligned}$$ which establishes B4 in the case $s=1$. But we claim that, if B4 is true for $s+t<n$, then for $s+t=n$ and $s< t$, $$k_{t-1}{\ensuremath{\ell k}}(c(s+1),c^+(t-1))=k_s{\ensuremath{\ell k}}(c(s),c^+(t)).$$ This equality can then be applied, successively decreasing $s$, to establish B4 in generality. This claimed equality is established as follows. $$\begin{aligned}
\pm k_{t-1}{\ensuremath{\ell k}}(c(s+1),c^+(t-1)) &= \pm V_{c(t-1)}\cd c(s+1)\\
&= \pm V_{c(t-1)}\cd(V_x\cap V_{c(s)})\\
&= V_{c(s)}\cd(V_x\cap V_{c(t-1)})\\
&= V_{c(s)}\cd c(t)\\
&= k_s{\ensuremath{\ell k}}(c(t),c^+(s))\\
&=k_s{\ensuremath{\ell k}}(c(s),c^+(t)).\end{aligned}$$ The last step is justified by verifying that $c(s)\cd c(t)=0$ if $s< t$. For $$\begin{aligned}
c(s)\cd c(t)  &= c(s)\cd V_{c(t-1)}\\
&= \pm k_{t-1}{\ensuremath{\ell k}}(c(s),c^+(t-1))\end{aligned}$$ which vanishes by our inductive assumption since $s+(t-1)<n$.

[**B4$\Longrightarrow$B1**]{}: Since $\b^1$ is always defined we may assume $n>1$. It follows from B4 that ${\ensuremath{\ell k}}(c(1),c^+(n-1))=0$ if $n>1$. But we saw in the proof of B1$\Longrightarrow$B4 that once $c(n)$ was defined, this was equivalent to $\b^n$ being defined.

[**B2$\Longrightarrow$B1**]{}: This is obvious.

[**B4$\Longrightarrow$B2**]{}: Since B4$\Longrightarrow$B1, we have $\b^j$ defined for $j\le n$. Now suppose $1\le j\le\[\f n2\]$. Since $\b^j={\ensuremath{\ell k}}(c(j),c^+(j)$ and $2j\le n$, this vanishes by B4.

This completes the proof of Lemma \[equivalence\].

The proof shows slightly more, namely that there is a correspondence between the infinite cyclic cover implicit in part $A$ and the class $x$ in parts $B$ and $C$. Suppose $\{M_n\}$ is a family of $n$-fold cyclic covers of $M$ corresponding to the infinite cyclic cover $M_\infty$. Note that $H_1(M_\infty;\BQ)$ is a finitely generated $\La=\BQ[t,t^{-1}]$ module (this involves a choice of generator of the infinite cyclic group of deck translations of $M_\infty$). Throughout this proof, homology will be taken with rational coefficients unless specified otherwise.

[**Step 1**]{}: $\b_1(M_n)$ grows linearly $\Longleftrightarrow H_1(M_\infty;\BQ)$ has positive rank as a $\La$-module.

As remarked above, this fact was previously known. We present a quick proof for the convenience of the reader. We are indebted to Shelly Harvey for showing us this elementary proof. Since $\La$ is a PID, $$H_1(M_\infty)\cong\La^{r_1}\oplus_j\f\La{\<p_j(t)\>}$$ where $p_j(t)\neq0$. By examining the “Wang sequence” with $\BQ$-coefficients $$H_2(M_\infty)\lra H_2(M_n)\overset{\p_*}{\lra}H_1(M_\infty)
\overset{t^n-1}{\lra}H_1(M_\infty)\overset{\pi}{\lra}H_1(M_n)
\overset{\p_*}{\lra}H_0(M_\infty)\overset{t^n-1}{\lra}$$ it is easily seen that $$\begin{aligned}
H_1(M_n)  &\cong\f{H_1(M_\infty)}{\<t^n-1\>}\op\BQ\\
&\cong\(\f\La{\<t^n-1\>}\)^{r_1}\oplus_j\f\La{\<p_j(t),t^n-1\>}\op\BQ.\end{aligned}$$ The first summand contributes $nr_1$ to $\b_1(M_n)$. The $\BQ$-rank of the second summand is bounded above by the sum of the degrees of the $p_j$, a number that is [**independent**]{} of $n$. Therefore $\b_1(M_n)$ grows linearly with $n$ if $r_1\neq0$ and otherwise is bounded above by a constant (independent of $n$).

[**Step 2**]{}: $H_1(M_\infty)$ has positive $\La$-rank $\Longleftrightarrow H_1(M_\infty)$ has no $(t-1)$-torsion (equivalently $t-1$ acts injectively). To verify Step 2, consider the “Wang sequence” with $\BQ$-coefficients $$H_2(M_\infty)\lra H_2(M)\overset{\p_*}{\lra}H_1(M_\infty)
\overset{t-1}{\lra}H_1(M_\infty)\overset{\pi}{\lra}H_1(M)
\overset{\p_*}{\lra}H_0(M_\infty)\overset{t-1}{\lra}H_0(M_\infty)$$ associated to the exact sequence of chain complexes $$0\lra C_*(M_\infty;\BQ)\overset{t-1}{\lra}
C_*(M_\infty;\BQ)\overset{\pi}{\lra}C_*(M;\BQ)\lra0.$$ Since $H_0(M_\infty)\cong\BQ$, ${\operatorname{image}}\p_*\cong\BQ$ on $H_1(M)$. If $\b_1(M)=2$ then it follows that $\BQ\cong{\operatorname{ker}}\p_*={\operatorname{image}}(\pi)\cong{\operatorname{cokernel}}(t-1)$. It follows that $H_1(M_\infty)$ contains at most one summand of the form $\La/\<(t-1)^m\>$ since each such summand contributes precisely one $\BQ$ to ${\operatorname{cokernel}}(t-1)$. Similarly each $\La$ summand of $H_1(M_\infty)$ contributes one $\BQ$ to the cokernel. Therefore $H_1(M_\infty)$ has positive $\La$ rank if and only if it has no summand of the form $\La/\<(t-1)^m\>$. The latter is equivalent to saying that it has no $(t-1)$-torsion, or that $t-1$ acting on $H_1(M_\infty)$ is injective. This completes Step 2.

[**Step 3**]{}: $(t-1):H_1(M_\infty)\to H_1(M_\infty)$ is injective $\Longleftrightarrow$ For any surface $V_x$, dual to $x$, and for each surface $V_y$ such that $\{[V_y],[V_x]\}$ generates $H_2(M;\BZ)$, the class $[\tl c(x,y)]\in
H_1(M_\infty;\BQ)$ is zero. Moreover the latter statement is equivalent to one where “for each” is replaced by “for some”.

Suppose that $t-1$ is injective. Note that the injectivity of $t-1$ is equivalent to $\p_*:H_2(M)\to H_1(M_\infty)$ being the zero map. Then, for [**any**]{} $[V_y]$ as above, $\p_*([V_y])=0$. But we claim that $\p_*([V_y])$ is represented by $[\tl c(x,y)]$, since $V_x$ is Poincaré Dual to the class $x$ defining $M_\infty$. For if $Y=M-{\operatorname{int}}(V_x\x[-1,1])$ then a copy of $Y$, denoted $\wt Y$, can be viewed as a fundamental domain in $M_\infty$, as shown in Figure \[cover\].

(177,84) (10,10)[![Fundamental Domain of $M_\infty$[]{data-label="cover"}](cover.eps "fig:")]{} (15,62)[$\wt c(x,y)$]{} (80,33)[$\wt V_y$]{} (0,12)[$\wt V_x$]{} (173,12)[$t\wt V_x$]{} (179,59)[$t\wt c(x,y)$]{} (87,-5)[$\wt Y$]{}

Moreover if $\wt V_y$ denotes $p^{-1}(V_y)\cap \wt Y$ then $\wt V_y$ is a compact surface in $\wt M$ whose boundary is $t_*(\tl c(x,y))-\tl c(x,y)$. Thus $\wt V_y$ is a 2-chain in $M_\infty$ such that $\pi_\#(\wt V_y)$ gives the chain representing $[V_y]$. Since $\p\wt V_y$ is $(t-1)\tl c(x,y)$ in $C_*(M_\infty;\BQ)$, it follows from the explicit construction of $\p_*$ in the proof of the Zig-Zag Lemma \[Mu,Section 24\] that $\p_*([V_y])=[\tl
c(x,y)]$.

Conversely, if $\p_*([V_y])=0$ for [**some**]{} $[V_y]$ then $\p_*$ is the zero map (note that since $V_x$ lifts to $M_\infty$, $[V_x]$ lies in the image of $H_2(M_\infty)\lra H_2(M)$ so $\p_*([V_x])=0$ t).

Therefore the injectivity statement implies the “for each” statement which clearly implies the “for some” statement. Conversely, the “for some” statement implies the injectivity statement.

[**Step 4**]{}: The class $[\tl c(x,y)]$ from Step 3 is 0 if and only if it is divisible by $(t-1)^k$ for every positive $k$. In fact it suffices that it be divisible by $(t-1)^N$ where $N$ is the largest nonnegative integer such that $\La/\<(t-1)^N\>$ is a summand of $H_1(M_\infty,\BQ)$.

One implication is immediate, so assume that there exists a class $[V_y]$ as in Step 3 such that $\p_*([V_y])=[\tl c_{1}]=(t-1)^N\b$ for some $\b\in H_1(M_\infty)$. Since $[\tl c_{1}]\in{\operatorname{image}}\p_*$, it is $(t-1)$-torsion so $\b$ is $(t-1)^{N+1}$-torsion. Moreover $\b$ lies in the submodule $A\subset
H_1(M_\infty,\BQ)$ consisting of elements annihilated by some power of $t-1$, so, by choice of $N$, $(t-1)^N\b=0=[\tl c_{1}]=0$ as desired. This completes the verification of Step 4.

[**Step 5**]{}: C$\Rightarrow$A Let $\{x,y\}$ be as in the hypotheses of C and let $M_\infty$ correspond to the class $x$. Let $N$ be the positive integer as above. If $\b^{(N+1}$ can be defined, we know in particular that there exists some system of surfaces $\{V_x,V_y,...,V_{c(N)}\}$ that defines $\{c(j)\}$, $1\leq j\leq (N+1)$. Choose a preferred lift $\wt V_x$, of $V_x$ to $M_\infty$ and a preferred fundamental domain $\wt Y$ as above lying on the positive side of $\wt V_x$. Consider any $m, 1\leq m \leq N$. Since $c(m)$ and $c(m+1)$ lie on $V_x$, they lift to oriented 1-manifolds $\tl c(m)$ and $\tl c(m+1)$ in $\wt V_x$. Similarly $c^+(m)$ lifts to $\tl c^+(m)$, which is a push-off of $\tl c(m)$ lying in $\wt Y$. Recall that $c(m+1)=V_{c(m)}\cap V_x$ where $\p
V_{c(m)}=k_mc^+_{(m)}$ for some positive integer $k_m$. Letting $\wt V_{c(m)}$ be $V_{c(m)}$ cut open along $c(m+1)$ we observe that $\wt V_{c(m)}$ can be lifted to $\wt Y$ and viewed as a 2-chain showing that $k_m[\tl c^+(m)]=(t-1)[\tl c(m+1)]$ in $H_1(M_\infty;\BQ)$, as in Figure \[stepfive\].

(177,84) (10,10)[![[]{data-label="stepfive"}](stepfive.eps "fig:")]{} (15,64)[$\wt c(m+1)$]{} (108,72)[$k_m\wt c^+(m)$]{} (65,34)[$\wt V_{c(m)}$]{} (0,12)[$\wt V_x$]{} (173,12)[$t\wt V_x$]{} (179,59)[$t\wt c(m+1)$]{} (83,-5)[$\wt Y$]{}

Thus $[\tl c^+(1)]=(t-1)(1/k_1)[\tl c(2)]=(t-1)^2(1/k_1)(1/k_2)[\tl c(3)]$, et cetera, showing that $[\tl c(1)]$ is divisible by $(t-1)^N$. By Steps 1 through 4, this implies A of Theorem \[linear\], completing Step 5. [**Step 6**]{}: A$\Rightarrow$B We assume that there is a primitive class $x\in H^1(M;\BZ)$ corresponding to $M_\infty$ and $\{M_n\}$ where $\b_1(M_n)$ grows linearly. By Steps 1, 2, and 3, for any $\{x,y\}$ generating $H^1(M;\BZ)$, $H_1(M_\infty;\BQ)$ has no $(t-1)$-torsion, and for any surfaces dual to ${x,y}$, $[\tl c(1)]=0$. Recall that $c(1)$ and $c(2)$ are always defined. We shall establish inductively that for all $m\ge2$, $c(m)$ is defined and that for *any* system of surfaces used to define $c(m-1)$, $[\tl c(m-1)]=0$ in $H_1(M_\infty;\BQ)$. This has already been shown for $m=2$. Suppose it has been established for $m$ (and all lesser values). We now establish it for $m+1$. Since $c(m)$ and $c(m-1)$ exist, the argument in Step 5 (Figure \[stepfive\]) shows that $k_{m-1}[\tl c(m-1)]=(t-1)[\tl c(m)]$ in $H_1(M_\infty;\BQ)$. But $[\tl c(m-1)]=0$ so $[\tl c(m)]$ is $(t-1)$-torsion. Since there is no non-trivial $(t-1)$-torsion, $[\tl c(m)]=0$ in $H_1(M_\infty;\BQ)$. Hence $[c(m)]=0$ in $H_1(M;\BQ)$ so $c(m+1)$ is defined. Since this holds for any system of defining surfaces, this completes the inductive step.

Since $c(m)$ is defined for all $m\ge1$, by Lemma \[equivalence\], $\b^n(x,y)=0$ for all $n\ge1$. This completes the proof of Step 6.

Since B clearly implies C, this completes the proof of Theorem \[linear\].

Assume some $\b^m(x,y)\neq 0$. If $[F]$ were zero then certainly, for the fixed infinite cyclic cover, $M_\infty$, corresponding to $x$, $[\wt c(x,y)]=0$ in $H_1(M_\infty;\mathbb{Q})$ so by Steps $1-3$ of the above proof, $\b_1(M_n)$ grows linearly with $n$. By Theorem \[linear\], this would imply that $\b^m(x,y)= 0$ for all $m$, a contradiction.

[Str]{}

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[^1]: The first author was partially supported by the National Science Foundation. The second author was supported by a National Science Foundation Postdoctoral Research Fellowship
