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My question is as follows.

Does every hyperbolic Brunnian link have rectangular cusp shapes on all its components?

Here is what I mean:

The Borromean rings form a famous link $B$ (a smooth closed 1-manifold in the three-sphere):

Link $B$

alt text

This link has the property that removing a single component yields an unlink. This property is called the Brunnian property. Links with the Brunnian property are Brunnian links.

Here are some other Brunnian links:

Link $S$

alt text


Link $R$

alt text


Link $N$

alt text (Rolfsen '76, p. 67)


Link $OMG$

alt text (Rolfsen '76, p. 67)


The complement $S^3 - L = K_L$ of any link $L$ (Brunnian or not) may admit a complete hyperbolic structure (cf. Thurston '97, pp. 131–132, p. 147) with finite volume (cf. SnapPy). By Mostow-Prasad rigidity, this structure (if it exists) is a topological invariant. We may then write $K_L$ as the image of a quotient $D: \mathbb{H}^3 \to K_L$ of hyperbolic space by a freely acting discrete group $G \simeq \pi_1(K_L)$ of isometries. Each component $C$ of the link admits a regular neighborhood $N(C)$ that is the image under $q$ of an open horoball $\tilde{N}$. In fact it admits many such. The maximal such neighborhood is the maximal cusp neighborhood $m_C$.

The lifts of $m_C$ to the universal cover are horoballs. Pick one such lift $M_C$. The closures of all the lifts abut the (horospherical) boundary $\partial M_C$ at a discrete set $\Lambda$ of points. Let $\Gamma$ be the subgroup of $G$ that preserves $\partial M_C$, a peripheral subgroup of $G$. This subgroup acts transitively and freely on this set of points, so we may identify $\Gamma$ with $\Lambda$ by picking one point $o \in \Lambda$ to represent the identity of $\Gamma$.

As it turns out, in the induced metric, the horosphere $\partial M_C$ is isometric to the Euclidean plane, and $\Gamma$ acts on this by isometries. So $\Gamma$ is a discrete, torsion-free, freely acting group of isometries of the Euclidean plane. Therefore it is either $0,$ $\mathbb{Z}$, or $\mathbb{Z} \oplus \mathbb{Z}$.

If $\Gamma \simeq 0$ or $\mathbb{Z}$, then the quotient of the maximal cusp neighborhood by $\Gamma$ (and therefore by $G$) would have infinite volume. Conversely, assuming that $\Gamma \simeq \mathbb{Z} \oplus \mathbb{Z}$, one can show easily that the quotient of the maximal cusp neighborhood has finite volume.

Assume therefore that $\Gamma \simeq \mathbb{Z} \oplus \mathbb{Z}$. Fix an isometry $\phi$ of $\partial M_B$ with $\mathbb{C}$ sending $o$ to $0$. The elements $\gamma_m,$ $\gamma_\ell$ of $\Gamma$ corresponding to the meridian and the longitude of the link component $C$ generate $\Gamma$, so we might as well also ensure that our choice of isometry sends one of them to a positive real number. SnapPy's convention is to send the longitude to a positive real number, apparently, so let's use that convention. Then we say the cusp shape of the link component $C$ is the ordered pair of complex numbers $(\phi(\gamma_m), \phi(\gamma_\ell))$.

Here are SnapPy's computations for the cusp shapes of the components of the above Brunnian links (I have rounded the decimals):

  • Link $B$: all of shape $(5\cdot 10^{-17}+0.569 i,\\,1.140)$ (by symmetry)
  • Link $S$: all of shape $(1\cdot 10^{-16}+0.550 i,\\,1.181)$ (by symmetry)
  • Link $N$: no hyperbolic structure
  • Link $OMG$:
    • cusp 0: $(1\cdot 10^{-16}+0.258 i,\\,2.514)$
    • cusp 1: $(-9\cdot 10^{-16}+0.262 i,\\,2.471)$
    • cusp 2: $(1\cdot 10^{-16}+0.316 i,\\,2.051)$
    • cusp 3: $(1\cdot 10^{-15}+0.431 i,\\,1.506)$
    • cusp 4: $(2\cdot 10^{-17}+0.5848 i,\\,1.111)$
    • cusp 5: $(-5\cdot 10^{-16}+0.232 i,\\,2.796)$

That is to say, each of the above hyperbolic Brunnian links (almost certainly) has a rectangular cusp shape on every one of its components.

Does every hyperbolic Brunnian link have rectangular cusp shapes on all its components?

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2 Answers 2

up vote 7 down vote accepted

Any two-component two-bridge link is Brunnian, but most of them do not have rectangular cusp shapes. For a description of the geometry of 2-bridge link complements, you may consult this appendix by David Futer.

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Thank you very much! –  Robert Haraway Jan 23 '13 at 15:23
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The rectangular shapes that you find are probably due to the particular symmetries these links have. When there is a symmetry that fixes a component $C$ of a link which fixes both the meridian and the longitude but inverts the orientation to only one of the two curves, then the shape is rectangular.

This holds because the symmetry induces an isometry on the flat cusp section, which must preserve the unsigned angle between the (geodesic representatives of the) meridian and the longitude, and the only possibility is that they stay at right angles.

Sometimes we find such symmetries when the link is very symmetric and the component $C$ is an unknot: just try to do a $\pi$-rotation that inverts $C$ and see if this gives a symmetry of the whole link. This works for instance for the Borromean rings and the second link you draw.

Added: no, the $\pi$-rotation inverts both the meridian and the longitude... to invert only one of them, mirror the diagram and see if you get the same link up to isotopy, possibly after permuting the components other than $C$. This works for the links $B$ and $N$ drawn above, and maybe on others.

So the reason for rectangular shapes in the pictures above should be "unknotted components and many symmetries".

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