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Hi, everyone. I am interested in the dehn filling and Hyperbolic 3-manifold.

Suppose M be an orientable compact 3-manifold with one torus boundary and int(M) admit a hyperbolic structure. Thurston proved that almost all the dehn fillings produced hyperbolic 3-manifolds.

My question is:

Among all the hyperbolic dehn fillings, is there possible that two different hyperbolic dehn fillings produce same 3-manifold, for some M? Is it true for all the one cusp hyperbolic 3-manifolds?

If the answer to the question is positive, is there an theorem describing this thing?

Any answer and reference are welcome! Thanks!

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$S^1\times \mathbb{R}^2$ admits a complete hyperbolic structure and has many Dehn fillings which are the same, so I think you want a finite-volume hyperbolic structure (complete hyperbolic metric :). –  Ian Agol Mar 31 '13 at 18:12
    
@Agol, yes! Do you know any about it? –  yanqing Apr 1 '13 at 1:06
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2 Answers 2

up vote 13 down vote accepted

This phenomenon is called "cosmetic surgery."

If $K$ is an amphichiral knot in the $3$--sphere with exterior $M_K$, then $M_K(p/q) \cong - M_K(-p/q)$. So if $p/q$ is a hyperbolic filling slope, then this is the example you seek. Note that the homeomorphism is orientation-reversing. If there is an orientation-preserving homeomorphism between $M_k(p/q)$ and $M_K(p'/q')$, the surgeries are called ``truly cosmetic,'' and it is conjectured that cosmetic surgeries on hyperbolic knots are never truly cosmetic.

Take a look at Bleiler, Hodgson, and Weeks, ``Cosmetic surgery on knots,'' in Proceedings of the Kirbyfest, pp. 23-34 , Geometry and Topology Monographs, Vol. 2, Coventry, 1999 as a beginning reference, and the conjecture that cosmetic surgeries on hyperbolic knots are never truly cosmetic.

EDIT: Regarding your second question, I'm pretty sure there are hyperbolic manifolds $M$ that admit no cosmetic surgeries at all. Here's a sketch:

I'll use $\partial M$ to mean a cusp cross section of a $1$-cusped hyperbolic manifold $M$, normalized to be a flat torus of area one.

By a theorem of Nimershiem, the shapes of flat tori appearing as cusp cross sections of 1-cusped $3$-manifolds is dense in the moduli space of tori. (see Nimershiem, ``Isometry classes of flat 2-tori appearing as cusps of hyperbolic 3-manifolds are dense in the moduli space of the torus,'' in Proceedings of Low-Dimensional Topology, 1992, 133-142.)

By Nimershiem's theorem, we may pick an $M$ such that $\partial M$ has a very short curve $\gamma$ and so that the group of isometries of $\partial M$ up to isotopy is trivial. Let us also assume that $M(\gamma)$ has no short geodesics (you should be able to achieve this by some covering space tricks).

Now, suppose that $M$ admits cosmetic fillings $M(\alpha) \cong M(\beta)$ with $\alpha \neq \beta$. If $\alpha \neq \gamma$, then $\alpha$ is really long in $\partial M$, and it follows from sharp versions of the Hyperbolic Dehn Filling Theorem due to Hodgson and Kerckhoff that the core of the filling torus is short in $M(\alpha)$. This means that $\alpha \neq \gamma \neq \beta$. Furthermore, it means that the cores of the filling tori in $M(\alpha)$ and $M(\beta)$ are the unique short curves in these manifolds, respectively. So any homeomorphism $M(\alpha) \to M(\beta)$ must preserve these cores (this part of the argument is in lemma 1 of the Bleiler-Hodgson-Weeks paper). So, the homeomorphism $M(\alpha) \to M(\beta)$ comes from a homeomorphism (and hence an isometry) of $M$, which induces an isometry of $\partial M$ which is not isotopic to the identity (since we assume that $\alpha \neq \beta$). Since we have chosen $\partial M$ to have no such isometries, we have a contradiction, and must conclude that $M$ has no cosmetic fillings.

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@Kent, Excellent answer! Is there a positive example for the question? –  yanqing Apr 1 '13 at 8:48
    
@yanqing The figure eight knot $K_8$ is amphichiral, and for all but finitely many $p/q$, the manifolds $M_{K_8}(p/q)$ are hyperbolic by the hyperbolic dehn filling theorem, so $M_{K_8}(p/q) = - M_{K_8}(-p/q)$ for these $p/q$ are examples. Note that in these amphichiral knot examples, there is always a homeomorphism of $M_K$ taking one slope to the other. Bleiler has conjectured that this is always the case for hyperbolic cosmetic fillings of a hyperbolic manifold (see Conjecture 3 of Bleiler-Hodgson-Weeks). –  Richard Kent Apr 1 '13 at 12:50
    
@Kent, great! Thanks a lot! –  yanqing Apr 1 '13 at 13:24
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To add even more details to Richard Kent's wonderful answer, the question of cosmetic surgeries showed up as Problem 1.81 on Kirby's list.

Although there are no non-trivial knot complements in $S^3$, that admit non-trivial $S^3$ surgeries the Blieler, Hodgson and Weeks paper that Richard mentions gives knot complements in lens spaces that have this property. In earlier work, Gabai and Berge independently found infinitely many examples of hyperbolic knot complements in $S^1 \times D^2$ that admit at least one (and sometimes two) non-trivial $S^1 \times D^2$ fillings.

A good reference for some recent advances on the cosmetic surgery problem is Yi Ni and Zhongtao Wu's work here. For knots in $S^3$, they are able place some restrictions on what pairs of rational slopes can admit, truly cosmetic (sometimes called purely cosmetic) surgeries.

In particular, they have a succinct statement of the cosmetic surgery conjecture that they attribute to the authors of Kirby's problem list. Namely:

If $M \not\cong S^1 \times D^2$, then for any knot $\gamma \subset M$, $M-\gamma$ does not admit any truly cosmetic fillings.

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@Hoffman, great answer! Thanks! –  yanqing Apr 1 '13 at 14:35
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