I'm learning about the Thurston norm and am trying to understand the implications that the existence of fibered faces has for the ways in which a given three-manifold $M$ can fiber over the circle. In particular, I am interested in the following question:

Let $M$ be a compact, oriented three-manifold without boundary, and suppose that $M$ is irreducible and atoroidal, so that the Thurston norm is non-degenerate. Suppose further that $M$ admits a fibering over $S^1$, so that there exists a fibered face for the unit norm ball. Is it then true that $M$ fibers over $S^1$ with fiber $S_g$ the closed surface of genus $g$ for infinitely many $g$?

I believe the answer is yes but I haven't seen a discussion of a result of this type in any of the sources I have consulted (Thurston's paper, this paper of McMullen, and the books by Kapovich, Calegari, and Candel-Conlon mentioned in this question.) The argument I constructed relies on Theorem 6.1 in the aforementioned McMullen paper, which gives conditions under which a class $\phi \in H^1(M, \mathbb{Z})$ is Poincare-dual to a connected norm-minimizing surface $S$. Specifically, for any $\phi$ that is primitive (i.e. maps onto $\mathbb{Z}$) and for which $b_1(M_\phi)$ is finite (where $M_\phi$ is the cyclic covering of $M$ associated to the kernel of $\phi$), a connected, Thurston-norm minimal $S$ dual to $\phi$ exists. In the case where $\phi$ is associated to a fibration $M \to S^1$, it seems to me that the condition on the finiteness of $b_1$ is automatically satisfied: McMullen remarks that in these circumstances we have $b_1(M_\phi) = b_1(F)$ where $F$ is the fiber of the fibration associated to $\phi$.

Now the argument goes as follows: given any primitive $\phi$ in the cone on the fibered face (such $\phi$ exist with arbitrarily large Thurston norm), take the $S$ associated to $\phi$ by the McMullen theorem. By construction, $S$ is dual to $\phi$, which is in turn dual to the fiber $F$ of the associated fibration, so that $S$ and $F$ represent the same homology class. Per a lemma in Calegari's book, a norm-minimizing surface in an irreducible manifold is incompressible. Then, following the remark of Thurston at the beginning of his Section 3, any incompressible surface homologous to a fiber is in fact isotopic to the fiber; in particular they have the same Euler characteristic. Since we can take $|| \phi ||$ to be arbitrarily large, we find that $\chi(S)$, and hence the genus of the fiber, can be arbitrarily large.

My question then is whether the answer to my above question is yes, and if so, is the argument I give correct, and is there a simpler way to show it? Should this be surprising? Is there a simple example of this phenomenon, preferably one where the various fiberings are easy to "see"? I know of examples when $M$ has boundary, namely link complements in the three-torus, but I would like an example with closed fibers.


2 Answers 2


The answer is: yes if the rank of $H_2(M;\mathbb{Z})$ is $\ge 2$; and no if the rank is $1$ because in that case there is up to isotopy a unique connected surface bundle structure on $M$. The proof uses nothing more than what is in Thurston's original article MR0823443, although there are probably multiple other ways to do it; I'll mention another proof due to Fried.

In the case of rank $\ge 2$, consider a fibered face, and let $C \subset H_2(M;\mathbb{Z})$ be the open cone over that face whose fiber has homology class in $C$. Fix one particular fibration over the circle. Since rank$\ge 2$, $C$ contains primitive elements of $H_2(M;\mathbb{Z})$ arbitrarily far from the origin, i.e. having arbitrarily large norm. So it remains to check that the norm evaluated on the homology class of a fiber equals minus the Euler characteristic of that fiber. To put it another way, you just need to check that a fiber of a fibration over the circle is norm minimizing in its homology class. This is a consequence of the theorem in Thurston's original article saying that any compact leaf of a taut transversely orientable foliation is norm minimizing; in more detail, what he proves is that the absolute value of the Euler class of the foliation evaluated on the compact leaf (which equals $|\chi(S)|$), equals the norm evaluated on that leaf.

Another proof I particularly like is Fried's formula for the restriction of the norm to $C$, from his article here. Consider the fiber $F$, consider its pseudo-Anosov monodromy $\phi : F \to F$, let $S_\phi \subset F$ be the set of singularities of $\phi$, and let $C_\phi \subset M$ be the suspension of $S_\phi$, so $C_\phi$ is a collection of oriented circles in $M$ where $M$ is regarded as the mapping torus of $\phi$. Turn $C_\phi$ into a 1-cycle by assigning to each of its components a coefficient $\frac{p}{2}-1$ where $p$ is the number of prongs of the corresponding singularity in $S_\phi$. The conclusion of Fried's theorem is that the norm is equal to the absolute value of the intersection number with $C_\phi$ (which is obviously equal to $|\chi(S)|$ by the Euler-Poincare formula for $\chi(S)$ expressed in terms of the singularities of $\phi$. Fried proves that intersection number with $C_\phi$ is equal to the norm throughout the entire open cone $C$.

  • $\begingroup$ Thank you for your answer. As I think about this more, I realize that the essential point that I'd like to understand better is why the fiber of a fibration associated to a primitive class is connected. I understand how the theorems in Thurston's paper establish the norm-minimality of fibers, but do they address connectedness? $\endgroup$ Aug 17, 2012 at 14:13
  • 1
    $\begingroup$ If a primitive class is represented by a disconnected fiber, say a 2-component fiber $F = F_1 \cup F_2$ then $F_1,F_2$ would be isotopic, because the monodromy map $F \mapsto F$ would have to transpose the two components (in general, with more components, the monodromy map would cyclically permute them). It would follow that $[F]=[F_1]+[F_2]=[F_1]+[F_1]=2[F_1]$. $\endgroup$
    – Lee Mosher
    Aug 17, 2012 at 15:11

See Autumn Kent's answer to this question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.