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:



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> 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 $g$?
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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](http://www.math.harvard.edu/~ctm/papers/home/text/papers/alex/alex.pdf) of McMullen, and the books by Kapovich, Calegari, and Candel-Conlon mentioned in [this question](https://mathoverflow.net/questions/85469/expository-accounts-of-the-thurston-norm).) 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.



Regarding the Thurston norm and the ways that a three-manifold can fiber over the circle
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](http://press.princeton.edu/titles/9494.html). 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 $C$.


