# 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?

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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? –  fuglede Aug 15 '11 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. –  PseudoNeo Aug 16 '11 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 at 19:45

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.

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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. –  fuglede Aug 16 '11 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. –  PseudoNeo Aug 17 '11 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. –  fuglede Aug 17 '11 at 16:15