Which mapping tori are Seifert manifolds?

According to Orlik's lecture notes on Seifert manifolds (and the Wikipedia page on Seifert fiber spaces), a mapping torus over a 2-torus is a Seifert manifold if and only if it is the mapping torus of a mapping with trace less than or equal to 2 in absolute value (under the usual identification $\text{MCG}(T^2) \cong \text{SL}(2,\mathbb{Z})$). For the trace $+2$ ones, the Seifert invariants are simply $\{b; (o_1, 1); \}$ for $b \in \mathbb{Z}$.

Also, it is known that mapping tori arising from finite order maps are all Seifert manifolds.

So, the question is: Is there a general way of determining whether or not a given mapping torus over a genus $g$ surface is a Seifert manifold?

Edit: It seems that if one doesn't mind the geometrization conjecture, it is possible to show that mapping tori of pseudo-Anosov homeomorphisms will give mapping tori that aren't Seifert fibered. By the Nielsen--Thurston classification, this leaves us with reducible mapping classes. Can anything be said in general about mapping tori of these?

• Thanks for the edit. I posted the question in general-topology, since there is nothing really differential about it (even if general-topology is slightly misleading as well). Maybe someone with high enough reputation could add a geometric-topology tag or/and low-dimensional-topology? Aug 15, 2011 at 19:49
• A remark: the result you mention in your edit (that a bundle over the circle with pseudo-Anosov monodromy admits a hyperbolic structure) is easier than the full Thurston conjecture. It is actually one of the parts Thurston himself was able to prove. A proof can be found in Otal's Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3 and probably in Kapovich's Hyperbolic Manifolds and Discrete Groups. Aug 16, 2011 at 12:55
• There is a way, using the fundamental groups. Your mapping torus is a Seifert fibred space if its $\pi_1$ contains an infinite cyclic normal subgroup.
– user641
Mar 29, 2013 at 19:45
• The answer is: If and only if the mapping class has finite order. Mar 15, 2014 at 3:26
• @studiosus: No, that fails for Dehn twists on tori; see the comments below. Mar 18, 2014 at 11:30

Let $f: S\to S$ be a homeomorphism of a compact connected surface (possibly with boundary).

Theorem.

1. If $\chi(S)<0$ then the mapping torus $M=M_f$ of $f$ is a Seifert manifold if and only if the mapping class of $f$ is periodic, i.e., $f$ is isotopic to a periodic homeomorphism.

2. If $\chi(S)\ge 0$ then $M$ is Seifert unless $S$ is the torus and eigenvalues of the action of $f$ on $H_1(S, {\mathbb R})$ are not roots of unity.

Proof. I will prove only Part 1 since the proof is already way too long. Recall that the surface $S$ admits Thurston-Nielsen decomposition with respect to $f$, i.e., a finite collection (possibly empty) of simple loops $L_i\subset S$ which are pairwise disjoint, pairwise non-isotopic and not isotopic to the boundary of $S$ (if there is any), such that:

a. $f$ preserves the multiloop $L=\cup_i L_i$. In particular, there exists $N$ such that $f^N$ preserves each loop $L_i$ and each component $S_j$ of $S\setminus L$.

b. The homeomorphism $f_j^N= f^N|S_j$ is homotopic to a homeomorphism $g_j: S_j\to S_j$ which is either periodic (by taking larger $N$ we can assume that such $g_j$ is the identity) or is pseudo-Anosov.

A homeomorphism $f$ is called reducible if $L$ is nonempty.

A very nice proof of this fundamental theorem can be found for instance in the book by Casson and Bleiler "Homeomorphisms of surfaces after Nielsen and Thurston."

Since a manifold is Seifert if and only if it has finite cover which is Seifert, we can assume that $S$ is oriented and $f=f^N$ (replacing $f$ with $f^N$ amounts to passing to a finite cover of $M$).

We now define tori $T_i$ which are mapping tori of the restrictions $f|L_i$. Then, by construction, the manifold $M=M_f$ is the union of submanifolds with boundary $M_j=M_{f|S_j}$, which meet along the tori $T_i$. As it was observed in the other answer, if $f|S_j$ is pseudo-Anosov, then $M_j$ is hyperbolic and, hence, $M$ cannot be a Seifert manifold in this case. (For instance, this can be seen from the fact that $\pi_1(M_j)$ has trivial center, while $\pi_1$ of a Seifert manifold always has nontrivial center, after passing to a finite cover, unless this cover is the 3-sphere.)

It remains to analyze the case when each $f|S_j$ is isotopic to the identity. Let $A_i=\eta(L_i)$ denote a small annular neighborhood of $L_i$ in $S$. Let $S_j'$ denote the complement in $S_j$ to all the annuli $A_j$ which it meets. By isotopying $f$ further, we can assume that $f_j=f|S_j'$ is the identity while the restriction $f|A_i$ is an iterated Dehn twist $D^{n_i}$ along the loop $L_i$. Then the mapping torus of $f_j$ is the product $S_j'\times S^1$. If some $n_i$ equals zero, then $f|A_i=Id$ and we can eliminate the loop $L_i$ from $L$ without changing topology of $M$ and periodicity (or lack of thereof) of $f$.

Each annulus $A_i$ has two boundary circles which I will denote $A_i^+, A_i^-$; accordingly, each manifold $M_{f|A_i}\cong A_i\times S^1$ has two boundary tori $T_{i}^+, T_i^-$ which are the mapping tori of $f|A_i^\pm$. Each loop $A_i^\pm$ is a boundary loop of a unique component of $$\bigcup_j S_j'$$ which I will denote $S_i^+$ and $S_i^-$ accordingly. (Note that it might happen that $S_i^+=S_i^-$, for instance, if $L$ is a single loop which does not separate $S$.) Accordingly, the (mapping) torus $T_{i}^\pm=M_{f|A_i^\pm}$ is a boundary component of the product manifold $M_i^\pm=M_{f|S_i^\pm}$, the mapping torus of the homeomorphism $f$ restricted to $S_i^+$ or $S_i^-$. The manifold $M_i\pm$ has canonical product structure (since $f$ restricts to the identity on $S_i^\pm$). Therefore, for each torus $T_{i}^+$ we obtain a canonical system of generators of the homology group: "Horizontal'' loop $a_{i}^+$ corresponding to the loop $A_i^+\subset S$ and the "vertical'' loop $b_i^+$ corresponding to the $S^1$-factor of the decomposition $M_i^+=S_i^+\times S^1$. The same applies to the torus $T_{i}^-$, where we switch all pluses to minuses.

Now, consider these loops $a_i^\pm, b_i^\pm$ in the product $M_{f|A_i}=T^2\times [0,1]$. The loops $a_i^+, a_i^-$ are, of course, isotopic to each other in this product manifold, since they correspond to isotopic curves $A_i^+, A_i^-$ on the surface $S$. I will denote by $[a_i]$ their common homology class in this product manifold.

However, this is not the case for for the loops $b^+_i, b^-_i$: From the definition of the Dehn twist we obtain that $$[b_i^-]= [b_i^+]\pm n_i [a_i]$$ the plus or minus depend on which boundary circle of $A_i$ we marked with $+$ and which with $-$; the number $n_i$ here is the power of the Dehn twist that we use.

The bottom line is that when under the gluing $M_i^+$ and $M_i^-$ along $A_i\times S^1$, the fibers of the (Seifert) fibration of $M_i^+$ do not match (up to isotopy) fibers of the Seifert fibration of $M_i^-$.

It is important to note here that the manifolds $M_i^\pm$ here admit unique, up to isotopy, Seifert fibrations, namely, the ones coming from their product decompositions $S_i^\pm \times S^1$, since each component of the surface $S\setminus L$ has negative Euler characteristic. (Here we are using the fact that $S$ is not the torus: cutting torus along a loop results in the annulus $A$ and the product $A\times S^1$ admits infinitely many, up to isotopy, circle fibrations.) This uniqueness theorem should be in Hempel's book on 3-manifolds and in Orlik's book on Seifert manifolds. In particular, up to isotopy, we can talk about the Seifert fibration of the manifold $M_i^\pm$.

The product regions $A_i\times S^1$ of course, admit infinitely many Seifert fibrations; to remedy this, we adjoin each $A_i\times S^1$ to the product manifold $S_i^+\times S^1$. As the result, $M$ is obtained by gluing product manifolds $M_j\cong S_j\times S^1$ along their boundary tori in such a fashion that all the gluing maps do not preserve the (up to isotopy) fibers of the circle fibrations of the manifolds $M_j$.

Now, we can finish the proof of Part 1 of the theorem with the following lemma which, I remember seeing in the book by Jaco and Shalen "Seifert fibered spaces in 3-manifolds", Memoirs of Amer. Math. Soc. 220 (1979).

Lemma. Suppose that $M$ is a 3-dimensional manifold obtained by gluing oriented Seifert manifolds $M_j$ along their incompressible boundary tori, where each $M_j$ admits a unique Seifert fibration. Then $M$ is Seifert if and only if all gluing maps preserve (up to isotopy) fibers of the Seifert fibrations.

Proof. I will skip the proof of one direction of this lemma since it is not needed and assume that one of the gluing maps does not preserve circle fibers. Let $T\subset M$ denote the incompressible torus corresponding to this gluing. Suppose, that $M$ is Seifert fibered; by looking at its fundamental group, it is clear that the base of this fibration has to be of hyperbolic type (i.e., it is a hyperbolic orbifold). It is a standard fact of the theory of Seifert manifolds (I am sure, it is in Jaco and Shalen) that every incompressible torus in such a fibration is "vertical'', i.e., isotopic to a torus foliated by Seifert fibers. To see the algebraic side of this statement, consider the short exact sequence of fundamental groups induced by Seifert fibration: $$1\to {\mathbb Z}\to \pi_1(M)\to B=\pi_1(O)\to 1,$$ where $O$ is the base-orbifold of the fibration and the fundamental group of $O$ is understood in the orbifold sense. Since the torus $T$ is incompressible, its fundamental group yields a subgroup $Z^2$ of $\pi_1(M)$. Projection of this group to $B$ has to be abelian, and, by hyperbolicity assumption on $O$, infinite cyclic. Therefore, the intersection of $Z^2$ with the normal subgroup ${\mathbb Z}$ of $\pi_1(M)$ is a free (abelian) factor of $Z^2$. In particular, $T$ admits a foliation by circles where each fiber is homotopic to the generic fiber of the Seifert fibration of $M$. By working more, one promotes this to an isotopy of $T^2$.

Instead of isotopying the torus we can isotope Seifert fibration itself. Therefore, splitting $M$ along $T$ results in one or two Seifert manifolds and the gluing map preserves Seifert fibrations. Continuing inductively with respect to all the boundary tori of the manifolds $M_i$, we obtain that each $M_i$ admits a Seifert fibration and gluing maps preserve these fibrations. Since Seifert fibration of each $M_i$ was unique (up to isotopy), we are done. QED

• Are you assuming that your homeomorphism $f: S \to S$ is orientation preserving? Or is this true even for the mapping torus of an orientation reversing mapping class? For example the mapping torus of $T^2$ for the mapping class $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$. Is that manifold Seifert fibered? If so, is it obvious which closed nonorientable 3 manifold that is/ what the Seifert fibration symbols are? Apr 8 at 18:56
• @IanGershonTeixeira: This argument is general. With a minor modification it goes through even if $S$ is non-orientable. As for your example, it is a Sol-manifold (hence, is not Seifert): The square of your matrix is hyperbolic. Apr 8 at 22:48

So you have a fiber bundle $M \to S^1$ whose fiber is $F$. Alternatively, $F$ is the quotient of $F \times S^1$ by the equivalence relation $(x,t+1) = (f(x),t)$, where the monodromy $f : F \to F$ is defined up to isotopy.

By Nielsen-Thurston classification of surface diffeomorphisms, $f$ falls into one of 3 cases :

• $f$ can be chosen periodic. In that case, the foliation $\{x\} \times \mathbb R$ of $F \times \mathbb R$ descends into a foliation by circles of $M$, which is therefore Seifert-fibered. (There is probably a slicker proof, but I'm not really a specialist of Seifert fibrations).
• $f$ can be chosen pseudo-Anosov. Then Thurston's celebrated theorem states that $M$ admits a hyperbolic structure. So it cannot be Seifert-fibered. There are probably different ways to show that, but one tool seems best suited to the end of the discussion: Gromov's simplicial volume. This is an invariant $\|M\|$ defined in Volume and Bounded Cohomology which satisfies desirable properties: it is 0 for Seifert-fibered manifolds, it is (up to a universal multiplicative constant) the hyperbolic volume for a hyperbolic manifold, it is additive under connected sum and gluing along incompressible tori, and multiplicative under finite coverings.
• $f$ is reducible. In that case, we take the cyclic covering corresponding to the replacement of $f$ by its power which leaves the pieces of $F$ globally invariant, and one studies these pieces. If one of the pieces is reducible, we play the same game with it and so on, recursively. So now the question boils down to: will we find at some point a pseudo-Anosov map? If yes, we'll have found a piece (delimited by incompressible tori) of a finite covering of $M$ with a hyperbolic structure, so $\|M\| \geq \frac{\text{hyperbolic volume of that piece}}{\text{degree of that covering map}} > 0$ and $M$ isn't Seifert-fibered. In the remaining case, one obtains that, cut along some tori, $M$ is just a bunch of Seifert-fibered space, that is: $M$ is a graph manifold.

So $M$ is a graph manifold if and and only if the monodromy is reducible to a bunch of periodic diffeomorphisms. (Essentially, we have proven that in our case, the manifold is either a graph manifold or contains a hyperbolic piece, which is true in all generality if you believe in geometrization). So what's left is to understand if the implication “graph manifold and surface bundle $\Rightarrow$ Seifert-fibered is true (or how wrong it is). I bet the answer is somehow in Waldhausen's Eine Klasse von 3-dimensionalen Mannigfatigkeiten but I don't have access to it at the moment.

• Thanks for the reply. There is something bugging me about this though. Say $F$ is a torus $T^2$, and the monodromy $f$ is the Dehn twist about the meridian. Then as noted in the original post, $M$ is a Seifert manifold, but $f$ is not isotopic to a periodic diffeomorphism. I suppose this happens since we can cut the torus open along a curve parallel to the meridian to kill off the Dehn twist action, but that would also be the case for higher genus surfaces. See also my comment on Reddit. Aug 16, 2011 at 14:01
• You are absolutely right (I felt like I was losing it on that particular part of the argument). I will edit my answer accordingly. Aug 17, 2011 at 6:59
• On the implication in the edited post: I haven't been able to find it in Waldhausen's paper myself; if it's there, it's probably in the last section. It's an interesting question though. Aug 17, 2011 at 16:15