Suppose we fix a genus $g$ closed surface $S$. Let $f, g \in Map(S)$ be conjugate, for $Map(S)$ the mapping class group of $S$. Then I know that $M_f$ (the mapping torus of $M$ with monodromy $f$) is homeomorphic to $M_g$.

Now, let $M$ is a mapping torus with monodromy $f$, and $N$ is a mapping torus with monodromy $g$. Suppose we know that $M_f$ is abstractly homeomorphic to $N_g$ (not homeomorphic as surface bundles over $S^1$, just homeomorphic as 3-manifolds). If $M_f$ and $N_g$ fiber over $S^1$ with homeomorphic fibers, then need $f, g$ be conjugate?

I think the answer is "no", but I'm not sure why, and I'm not sure how to construct a counterexample. Any help would be greatly appreciated!


3 Answers 3


Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$ and with pseudo-Anosov monodromy, and with 2nd homology of rank $\ge 2$, gives counterexamples.

The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_2(M;\mathbb{Z})$ of $H_2(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_2(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

We are supposing that $M$ fibers and that $\text{rank}(M) \ge 2$, and so there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_2(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But $M$ is a closed hyperbolic 3-manifold and hence its group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$. So at most $A$ homology classes can represent conjugate mapping classes.

  • $\begingroup$ When you write that 'the group of homeomorphic modulo isotopy ... is finite', I suppose you're thinking of $M$ as hyperbolic? $\endgroup$
    – HJRW
    Jun 11, 2016 at 5:57
  • 1
    $\begingroup$ Yes, I added that explicitly to the answer. $\endgroup$
    – Lee Mosher
    Jun 11, 2016 at 13:14

McMullen and Taubes 4-manifolds with inequivalent symplectic forms and 3-manifolds with inequivalent fibrations constructs 3-manifolds $N$ with different fibrations, whose Euler classes do not lie in the same $Diff(N)$-orbit.

The idea of the proof is that two fibrations can not be in the same $Diff(N)$-orbit if the Poincaré duals of their fibers belong to combinatorially inequivalent faces of the Thurston norm ball in $H^1(N;{\mathbb R})$.

They have examples of knot complements as well as a closed manifold. The closed manifold is constructed as follows. Let $L$ be the 4-component link in the 3-torus given by some homology basis $L_0,L_1,L_2$ together with $L_3=L_0+L_1+L_2$. Let $N\to T^3$ be the 2-fold cover branched over $L$ which is trivial over the homology basis but nontrivial over $L_3$. Every fibration of $T^3$ transverse to $L$ lifts to a fibration of $N$.

They show that the branched covering $N\to T^3$ yields an isomorphism of first cohomology, and they identify the Thurston norm ball of $N$ with the norm ball of the norm $\parallel\phi\parallel_L=\mid \phi(L_1)\mid+\mid\phi(L_2)\mid+\mid\phi(L_3)\mid+\mid\phi(L_1+L_2+L_3)\mid$ on $H^1(T^3;{\mathbb R})$. They compute the norm ball of the latter norm and obtain that it has combinatorially inequivalent faces: some are triangles, some are quadrangles. So the same is true for the Thurston norm ball of $N$.

Moreover, every $\phi\in H^1(T^3;{\mathbb R})$ corresponds to a linear fibration of $T^3$ and if $\phi(L_i)\not=0$ for all $i$ then it lifts to a fibration of $N$. So the top-dimensional faces of the Thurston norm ball are all fibered. Picking fiberings in combinatorially inequivalent faces of the Thurston norm ball yields inequivalent fibrations.


To add an explicit counter-example, consider the curves $a$, $b$ and $c$ on the twice punctured torus shown below.

enter image description here

The mapping classes $f = T_a T_a T_b T_c^{-1}$ and $g = T_a T_a T_b^{-1} T_c^{-1}$ are not conjugate (they don't even have the same dilatation). However their mapping tori $M_f$ and $M_g$ are homeomorphic. You can check this by using the latest versions of flipper, SnapPy and the Python code:

>>> import flipper
>>> S = flipper.load('S_1_2')
>>> f = S.mapping_class('aabC')
>>> g = S.mapping_class('aaBC')
>>> f.is_conjugate_to(g)
>>> f.dilatation(), g.dilatation()
(2.618033?, 3.254263?)
>>> import snappy
>>> M_f = snappy.Manifold(f.bundle())
>>> M_g = snappy.Manifold(g.bundle())
>>> M_f.is_isometric_to(M_g)

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