Volumes of fibered $2$-fold branched covers of $3$-manifolds

We take numbers $0<t_{1}<t_{2}<\dots<t_{k}<1$. Let $\delta_{i}=\gamma_{i}\times\{t_{i}\}$ be a curve lying on the fiber $F_{i}=\Sigma\times\{t_{i}\}$ for $i=1,\dots,k$ of the mapping torus $T_{f}$. Now $L_{k}=\delta_{1}\cup\dots\cup\delta_{k}$ is a link in $T_{f}$.

We first show that $N$ is atoroidal, i.e. $N$ contains no essential embedded tori. Assume that there exists a torus $T$ embedded in $N$ that is incompressible and not peripheral. Since the fundamental group of a thickened surface $\Sigma\times I$, where $I$ is an interval, does not contain free abelian subgroups of rank $2$, the torus $T$ must intersect some of the fibers $F_{i}$ of $T_{f}$, where the curves $\delta_{i}$ lie.

Without loss of generality, we may suppose that $T$ intersects the fiber $F_{1}$, where $\delta_{1}=\gamma_{1}\times\{t_{1}\}$ lies. We identify $\Sigma$ with the $t_{1}$-level $F_{1}=\Sigma\times\{t_{1}\}$ in $T_{f}$. Let $W_{\Sigma}$ denote the manifold obtained by cutting $T_{f}$ open along $\Sigma$. Since we assumed that $0<t_{1}<t_{2}<\dots<t_{k}<1$, in the beginning of the proof, and $\Sigma$ is identified with the $t_{1}$-level $F_{1}=\Sigma\times\{t_{1}\}$, the level surface $\Sigma\times\{1\}=\Sigma\times\{-1\}$ (as a set) is disjoint from $F_{1}$. We can view $W_{\Sigma}$ as the identification space

and $T_{f}$ is obtained from $W_{\Sigma}$ by identifying the two copies of $\Sigma\times\{t_{1}\}$ in $\partial W_{\Sigma}$ by the identity map.

By using the irreducibility of $T_{f}$ and the incompressibility of $\Sigma$, we may isotope the torus $T$ so that all components of $T\setminus\Sigma$ are annuli, and each component $A$ of $T\setminus\Sigma$ is either vertical with respect to the $I$-product (i.e. $A$ runs around the $S^{1}$ factor of $T_{f}$), or there exists an annulus $\widehat{A}$ in one copy of $\Sigma\subset\partial W_{\Sigma}$ such that $A\cup\widehat{A}$ bounds a solid torus in $W_{\Sigma}$ and $\partial A$ lies in the same copy of $\Sigma$, where the annulus $\widehat{A}$ sits. The former and latter annuli are called the vertical and horizontal annuli respectively. There are two types (A1), (A2) for a horizontal annulus $A$.

1. (A1)

   There exist no curves $\delta_{i}$ which is contained in the solid torus bounded by $A\cup\widehat{A}$.

2. (A2)

   There exists a curve $\delta_{i}$ which is contained in the solid torus bounded by $A\cup\widehat{A}$.

If $A$ is a horizontal annulus of type (A2), then the curve $\delta_{i}$ in the condition of (A2) is unique: If $\delta_{i}$ and $\delta_{j}$ ($i\neq j$) are contained in the solid torus bounded by $A\cup\widehat{A}$, then $\delta_{i}$ and $\delta_{j}$ are isotopic in $W_{\Sigma}$, which implies that $O_{f}(\gamma_{i})=O_{f}(\gamma_{j})$. This contradicts the assumption that the orbits of $\gamma_{1},\dots,\gamma_{k}$ under $f$ are distinct. Hence the curve $\delta_{i}$ in (A2) is unique. In particular $\partial A$ consists of two curves which are parallel to $\gamma_{i}\times\{t_{1}\}$ since $\Sigma$ is identified with the $t_{1}$-level $F_{1}=\Sigma\times\{t_{1}\}$.

Notice that each horizontal annulus of type (A1) can be removed by an isotopy of the torus $T$, and hence we may suppose that each component of $T\setminus\Sigma$ is a vertical annulus or a horizontal annulus of type (A2).

If there exists a horizontal annulus of type (A2), then by replacing the fiber $F_{i}$ (containing the curve $\delta_{i}$) with $F_{1}$ if necessary, we have a horizontal annulus $A_{1}$ of $T\setminus\Sigma$ whose components of $\partial A_{1}$ are parallel to $\delta_{1}=\gamma_{1}\times\{t_{1}\}$.

Suppose that there exist no vertical annuli of $T\setminus\Sigma$. Then a horizontal annulus $A_{1}$ can only connect to a horizontal annulus $A_{2}$ with $\partial A_{2}$ running parallel to $f^{\pm 1}(\gamma_{1})\times\{t_{1}\}$. But then $T$ will be boundary parallel (peripheral) in $N$, contrary to our assumption.

From the above discussion, we may suppose that $T\setminus\Sigma$ contains a vertical annulus $A$. Let $P$ denote the component of $\partial A$ on one copy of $\Sigma\times\{t_{1}\}$ on $\partial W_{\Sigma}$. By the construction of $W_{\Sigma}$ in (2.1), the boundary $\partial A$ is disjoint from the level surface in $W_{\Sigma}$ resulting from the identification of $\Sigma\times\{1\}$ to $\Sigma\times\{-1\}$ via ${(x,1)\sim({{f}(x),-1)}}$. The intersection of the annulus $A$ with the later level surface is the curve resulting from $P\times\{1\}\sim f(P)\times\{-1\}$ under above identification of $\Sigma\times\{1\}$ to $\Sigma\times\{-1\}$. Thus the component of $\partial A$ on the second copy of $\Sigma\times\{t_{1}\}\subset\partial W_{\Sigma}$ is $f(P)$. That is $A$ runs from $P$ to $f(P)$ in $W_{\Sigma}$.

We have two cases.

1. (1)

   $T\setminus\Sigma$ contains a horizontal annulus $A_{1}$, or

2. (2)

   all of the components of $T\setminus\Sigma$ are vertical annuli.

We first consider the case (1). As discussed earlier, without loss of generality, we may assume that $\partial A_{1}$ is formed by two curves parallel to $\gamma_{1}\times\{t_{1}\}$.

Since the case that the horizontal annulus $A_{1}$ connects to another horizontal annulus was excluded earlier, we may now assume that $A_{1}$ connects a vertical annulus $A$. Now this vertical annulus $A$ eventually connects to another horizontal annulus $A^{\prime}$ of type (A2) such that the solid torus bounded by $A^{\prime}\cup\widehat{A^{\prime}}$ contains a curve $\delta_{j}$ for some $j\in\{1,\dots,k\}$.

Assume that $j=1$. Then the torus $T$ must have a self-intersection in $N$, and this is a contradiction.

Next, we assume that $j\in\{2,\dots,k\}$. Then $\partial A^{\prime}$ is formed by two curves parallel to $\gamma_{j}\times\{t_{1}\}$. Recall that the curves $P\times\{t_{1}\}$ and $f(P)\times\{t_{1}\}$, viewed on different copies of $\Sigma\times\{t_{1}\}\subset\partial W_{\Sigma}$, form the boundary of the vertical annulus $A$. This implies that on $\Sigma:=\Sigma\times\{t_{1}\}$ we have $P,f(P)\in O_{f}(\gamma_{j})\cap O_{f}(\gamma_{1})\neq\emptyset$. This contradicts the assumption that the orbits of $\gamma_{1},\dots,\gamma_{k}$ under $f$ are distinct.

We turn to the case (2). To form the torus $T$ from vertical annuli, we have $f^{m}(P)=P$ for some $m>0$. Arguing as above, we conclude that the curves $P,f(P),\dots,f^{m-1}(P)$ on $\Sigma\times\{t_{1}\}\subset T_{f}$ lie on the torus $T$. Since $T$ is embedded in $N$ and all of the components of $T\setminus\Sigma$ are vertical annuli, we have $\gamma_{i}\cap f^{j}(P)=\emptyset$ for any $i=1,\dots,k$ and $j=1,\dots,m$. Equivalently we have $f^{-j}(\gamma_{i})\cap P=\emptyset$ for any $i=1,\dots,k$ and $j=1,\dots,m$. For any $n\in{\mathbb{Z}}$, write $n=m\ell-j$ for some $\ell\in{\mathbb{Z}}$ and some $j=1,\dots,m$. For any $i=1,\dots,k$, we obtain

Thus for any $i=1,\dots,k$, the curve $P$ must be disjoint from $O_{f}(\gamma_{i})$. However this contradicts our assumption that the orbits of $\gamma_{1},\dots,\gamma_{k}$ under $f$ fill $\Sigma$. This implies that $N$ is atoroidal.

To finish the proof of Claim 1, it is enough to show that $N$ contains no essential annuli. Suppose that there exists an essential annulus $A$ in $N$. Then $N$ must be a Seifert manifold (see [9, Lemma 1.16]), and the components of $\partial N$ consist of fibers of the Seifert fibration of $N$. In $N$, we can find a copy of the fiber $\Sigma$ of $T_{f}$, say $S$, that is disjoint from the components $T_{1},\dots,T_{k}$ of $\partial N$ that are created by drilling out the curves $\delta_{1},\dots,\delta_{k}$. Then $S$ is a surface that is essential in the Seifert manifold $N$ with non-empty boundary. Since we assumed that $3g-3+m>0$, $S$ is not a torus or an annulus. Thus up to isotopy, we can make $S$ horizontal which means that $S$ must intersect all the fibers of the Seifert fibration of $N$ transversely, see [9, Proposition 1.11]. Since $S$ is disjoint from the components $T_{1},\dots,T_{k}$ of $\partial N$, it cannot become horizontal. This contradiction implies that $N$ contains no essential annuli. Thus by work of Thurston [20], the manifold $N$ is hyperbolic. This completes the proof of Claim 1.

We now prove the claim (a). We denote by $N_{\bm{n}}$, the mapping torus $T_{f_{\bm{n}}}$ of $f_{\bm{n}}=\tau^{n_{k}}_{\gamma_{k}}\dots\tau^{n_{1}}_{\gamma_{1}}f$ for ${\bm{n}}=(n_{1},\dots,n_{k})\in{{\mathbb{Z}}}^{k}$. We use the fact that $N_{{\bm{n}}}$ is obtained from $N$ by the Dehn filling, where the boundary component $T_{i}\subset\partial N$ corresponding to $\delta_{i}$ is filled. Given ${\bm{n}}=(n_{1},\dots,n_{k})\in{{\mathbb{Z}}}^{k}$, let $s_{i}$ denote the Dehn filling slope on $T_{i}\subset\partial N$ to obtain $N_{\bm{n}}$ for $i=1,\dots,k$. Since $N$ is hyperbolic, each torus boundary component of $N$ corresponds to a cusp of $T_{f}\setminus L_{k}$. Taking a maximal disjoint horoball neighborhood about the cusps, each torus $T_{i}$ inherits a Euclidean structure, well-defined up to similarity. The slope $s_{i}$ can then be given a geodesic representative. We define the length of $s_{i}$, denoted by $\ell(s_{i})$, to be the length of this geodesic representative. (Note that when $k>1$, this definition of slope length depends on the choice of maximal horoball neighborhood. See [16].)

The length $\ell(s_{i})$ of the slope $s_{i}$ is an increasing function of $|n_{i}|$. Let $\lambda>0$ denote the minimum length of the slopes, that is

By Thurston’s hyperbolic Dehn surgery theorem [19], there exists $n\in{\mathbb{N}}$ such that for all ${\bm{n}}=(n_{1},\dots,n_{k})\in{{\mathbb{Z}}}^{k}$ with $|n_{i}|>n$ for $i=1,\dots,k$, the resulting manifold $N_{\bm{n}}(=T_{f_{\bm{n}}})$ obtained by filling $N$ is hyperbolic, and hence $f_{\bm{n}}$ is pseudo-Anosov. Thus the claim (a) holds.

We turn to the claim (b). As $|n_{i}|\to\infty$ for all $i=1,\dots,k$, the volumes of the filled manifolds $N_{\bm{n}}$’s approach the volume of the $3$-manifold $T_{f}\setminus L_{k}$ from bellow. To make things more concrete, we use an effective form proved in [7, Theorem 1.1], which states that if $\lambda>2\pi$, then $N_{\bm{n}}$ is hyperbolic and we have

Since $T_{f}\setminus L_{k}$ is a hyperbolic $3$-manifold with at least $k$ cusps, we have

where $v_{3}=1.01494\dots$ is the volume of the ideal regular tetrahedron, see [1, Theorem 7].

On the other hand, by taking ${\bm{n}}=(n_{1},\dots,n_{k})\in{{\mathbb{Z}}}^{k}$ with all $|n_{i}|$ sufficiently larger than $n$, we can assure that

By (2.2), we obtain

We set ${\bm{n}}_{1}={\bm{n}}$ with the above inequality $\frac{1}{2}\ k<{\rm vol}(N_{{\bm{n}}_{1}})$. Suppose that there exists a finite sequence $\{{\bm{n}}_{\ell}\}_{\ell=1}^{m}$ of the $k$-tuple of integers ${\bm{n}}_{\ell}\in{\mathbb{Z}}^{k}$ such that

Now we choose ${\bm{n}}_{m+1}=(n^{\prime}_{1},\dots,n^{\prime}_{k})\in{{\mathbb{Z}}}^{k}$ with all $|n^{\prime}_{i}|$ sufficiently larger than $n$ so that if we let $\lambda=\lambda_{{\bm{n}}_{m+1}}$ be the minimal length of the slopes corresponding to ${\bm{n}}_{m+1}\in{\mathbb{Z}}^{k}$, then we have

Hence $\left(\frac{2\pi}{\lambda}\right)^{2}<1-x_{m}^{2/3}$. This tells us that

Thus

By (2.2), we have

Putting them together, we obtain

and the conclusion follows inductively. This completes the proof of Theorem 5. ∎

By Theorem 8, for any $g\geq g_{0}$ and $j\in\{1,2\}$, there exists an open book decomposition $(\Sigma_{g,j},h_{g,j})$ of $M$, where $h_{g,j}$ is isotopic to a pseudo-Anosov homeomorphism. We set $F_{g,j}=\Sigma_{g,j}$. Then by Lemma 6, we have mutually disjoint, essential simple closed curves $\gamma_{1},\dots,\gamma_{k}$ in $F_{g,j}$ such that the orbits of $\gamma_{1},\dots,\gamma_{k}$ under $h_{g,j}$ are distinct and fill $F_{g,j}$. Moreover $F_{g,j}\setminus\{\gamma_{1},\dots,\gamma_{k}\}$ is connected.

Let $B=\partial F_{g,1}$ when $j=1$, and let $B$ and $B^{\prime}$ be the components of $\partial F_{g,2}$ when $j=2$. When $j=1$, let $\beta_{1},\dots,\beta_{k}$ be properly embedded, mutually disjoint arcs in $F_{g,1}\setminus\{\gamma_{1},\dots,\gamma_{k}\}$ so that one of the endpoints of each $\beta_{i}$ lies on $\gamma_{i}$ and the other endpoint of $\beta_{i}$ lies on $B=\partial F_{g,1}$. Since $F_{g,1}\setminus\{\gamma_{1},\dots,\gamma_{k}\}$ is connected, one can choose those arcs $\beta_{1},\dots,\beta_{k}\subset F_{g,1}\setminus\{\gamma_{1},\dots,\gamma_{k}\}$ so that they are mutually disjoint. When $j=2$, let $\beta_{1},\dots,\beta_{k}$ be properly embedded, mutually disjoint arcs in $F_{g,2}\setminus\{\gamma_{1},\dots,\gamma_{k}\}$ so that one of the endpoints of each $\beta_{i}$ lies on $\gamma_{i}$ and the other endpoint of $\beta_{i}$ lies on $B$ (resp. $B^{\prime}$) if $i=1,\dots,k-1$ (resp. $i=k$). In both cases $j=1,2$, consider a small neighborhood $\mathcal{N}=\mathcal{N}(\gamma_{i}\cup\beta_{i})$ in $F_{g,j}$. We set $\alpha_{i}$ to be a component of $\partial\mathcal{N}\setminus\partial F_{g,j}$ which is not parallel to $\gamma_{i}$. See Figures 2(1), 3(1). Then $\alpha_{1},\dots,\alpha_{k}$ are mutually disjoint, essential arcs in $F_{g,j}$.