7
$\begingroup$

The title and question have been edited in light of Ian Agol's comment. The previous question was stated in terms of the wrong notion of length to discuss deformations:

What is the longest slope $\gamma$ in the complement of Dehn surgery space of a cusped hyperbolic 3-manifold $M$? Here Dehn surgery space is the space of fillings such that the hyperbolic structure on the filling $M(\gamma)$ can be realized as a deformation of the original $M$.

To keep things focused, consider only fillings of one cusped manifolds.

This question is related to Ken Baker’s question:

Hyperbolic exceptional fillings of cusped hyperbolic 3-manifolds

However, Ken’s question is interested in the total number of slopes in this complement. This question is focusing on the longest such slope measured using normalized length, where the normalized length of $\gamma$, $\mathcal{L}(\gamma)$ is as defined in:

Hodgson, Craig D.; Kerckhoff, Steven P., Universal bounds for hyperbolic Dehn surgery, Ann. Math. (2) 162, No. 1, 367-421 (2005). ZBL1087.57011.

Namely $\mathcal{L}(\gamma)=length(\gamma)/\sqrt{Area(\partial T)}$. Here, $length(\gamma)$ the translation length of $\gamma$ in a cusp neighborhood and $Area(\partial T)$ is the area of that torus in cusp neighborhood.

Of course, it is possible that this question as stated does not have a realizable answer, because there is no longest slope.

Here is a more carefully stated version:

What is the largest $\mathcal{L}_{max}$ such that there exist a family of slopes $\gamma_i$ in (1-cusped hyperbolic) manifolds $M_i$ such that $$\lim_{i \to \infty} \mathcal{L}(\gamma_i) = \mathcal{L}_{max}$$ and each $M_i(\gamma_i)$ is a hyperbolic manifold such that the hyperbolic structure cannot be realized as a deformation of the hyperbolic structure of $M_i$?

Hodgson and Kerckhoff give an upper bound of $\mathcal{L}_{max}\leq C\approx 7.515$.

To give context how normalized length affects length, the (3,3,3) pretzel knot has slope of length 6 yielding a torus filling. However, the normalized length of this slope is $\mathcal{L}=\frac{6}{\sqrt{A}}\approx 1.91673$, $A=\frac{8\sqrt{3}}{(1+3\sqrt{57})^{1/3}}+\sqrt{3}(1+3\sqrt{57})^{1/3}$. This example appears in

Adams, Colin; Bennett, Hanna; Davis, Christopher; Jennings, Michael; Kloke, Jennifer; Perry, Nicholas; Schoenfeld, Eric, Totally geodesic Seifert surfaces in hyperbolic knot and link complements. II, J. Differ. Geom. 79, No. 1, 1-23 (2008). ZBL1158.57004.

and as a filling of the manifold defined by Figure 6:

Agol, Ian, Bounds on exceptional Dehn filling, Geom. Topol. 4, 431-449 (2000). ZBL0959.57009.

However, the cusp area appears to grow for (n,n,n)-pretzels, so I chose the (3,3,3)-pretzel to seemingly get the longest slope (in terms of normalized length) in this family.

$\endgroup$
2
  • 2
    $\begingroup$ It's not clear to me that $L\leq 2\pi$ if you require there to be a deformation. You should look at Hodgson-Kerckhoff's universal bounds on Dehn surgery space. $\endgroup$
    – Ian Agol
    Commented Jan 14, 2020 at 7:05
  • $\begingroup$ Thanks Ian. I missed that The 2-$\pi$ Theorem just guarantees a negatively curved metric can be extend to across the surgery solid torus, but this extension is not claimed to have constant curvature nor be a globally defined deformation of the original metric. $\endgroup$ Commented Jan 24, 2020 at 17:04

0

You must log in to answer this question.

Browse other questions tagged .