# Isotopy classes on the disk and mapping tori

Is the following true?

"The conjugacy classes of two homeomorphisms of the n-times punctured disk have isotopic representatives iff the associated mapping tori are homeomorphic."

By conjugacy class of a homeomorphism is meant the homeomorphism up to (orientation preserving) topological changes of coordinates. The isotopies are not required to keep the boundary pointwise fixed. Thus the question may be rephrased as:

Is it true that, up to changes of coordinates and isotopies, two homeomorphisms of the punctured disk are the same iff their mapping tori are homeomorphic? The "only if" is true; it is the "if" that puzzles me.

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I believe this is answered by results of Gabai and Berge. Namely, there are 1-bridge braids in $D^2\times S^1$ which have surgeries which are 1-bridge braids. Some of these are equivalent braids, but I think Berge has shown that most are not. ams.org/mathscinet-getitem?mr=1093862, ams.org/mathscinet-getitem?mr=1082933 – Ian Agol Apr 26 '13 at 5:26
You must also allow inversion, because there exists an inverse pair of conjugacy classes which are not conjugate to each other, but their mapping tori are homeomorphic. – Lee Mosher Apr 26 '13 at 12:46
Very good point, Lee. Thank you. Thanks to you too, Ian: will look up the references. – André de Carvalho Apr 26 '13 at 15:56
Adding to Ian's comment, based on the relative Thurston norm this just sounds false. Given a compact planar surface homeomorphism $f:S \to S$ with mapping torus $M$, we have $rank(H_2(M,\partial M)) \ge 2$, equal to the number of $f$ orbits of components of $\partial S$. The Thurston norm unit ball has a fibered face, the cone on that face contains an interior lattice point corresponding to $f$, and there's no reason to expect that every surface homeomorphism corresponding to other lattice points is either nonplanar or conjugate to $f$. – Lee Mosher Apr 26 '13 at 19:18
@ Lee: agreed, but the number of fiberings in a given face of Thurston norm ball which are planar surfaces of the same number of punctures is finite (if the manifold is hyperbolic), and there's a chance that all of these fiberings are related by isometries of the manifold, so give the same braid. In fact, this happens for the unique hyperbolic 1-bridge braid in $D^2\times S^1$ which has two Dehn fillings which are $D^2\times S^1$, since the symmetry group permutes the 3 fillings. But I think Berge shows which pairs of 1-bridge braids have the same complement, if I interpret his paper right. – Ian Agol Apr 26 '13 at 20:43

As I said in the comment above, knots in $D^2\times S^1$ which have a Dehn filling giving $D^2\times S^1$ were classified by Gabai and Berge. Gabai proved that knots in $D^2\times S^1$ giving back $D^2\times S^1$ are either cables or 1-bridge braids (note that the original statement was given before the knot complement problem, so Gabai states that the knot could also lie in a ball, but now this is known to not occur). I didn't check, but I believe the cable case will not work: even though there are many solid torus Dehn fillings, the link complement has an infinite image of the mapping class group in the mapping class group of the boundary, so I believe these all give equivalent cablings (although I didn't check this).
To be explicit, I took the braid from Figure 8 of Gabai's paper, which is a braid on 10 strands. I input this into SnapPy, which shows that the braid is hyperbolic, and $(63,1)$ Dehn filling on cusp $0$ gives a manifold with fundamental group $\mathbb{Z}$, which therefore must be $D^2\times S^1$. Moreover, the symmetry group is $\mathbb{Z}/2^2$, and there is a $\mathbb{Z}/2$ subgroup which preserves the cusps, and acts as an elliptic involution, therefore preserving slopes. So the two 1-bridge braids must be inequivalent, but have the same complement.