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 $B$

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 $S$

Link $R$

Link $N$(Rolfsen '76, p. 67)

Link $OMG$(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?