\\
\frac{1}{7 \nu^8} & \frac{1}{7 \nu^7} & \frac{1}{7 \nu^6} & \frac{1}{7 \nu^5} &
    \frac{1}{7 \nu^4} & \frac{1}{7 \nu^3} & \frac{1}{7 \nu^2} & -\frac{1}{\nu}
\end{pmatrix}.$$ Clearly, the first $2\times 2$ minor of $B$ in this case is non-zero.

It is shown in [@ShSo Theorem 2.6] that if $\nu < 0.6684$ that $A_\nu$ is disconnected, and hence totally disconnected, whence $\nu \not \in {{\mathcal{Z}}}_J$. Here we have that if $\nu > 0.8315$ then $\nu \in {{\mathcal{Z}}}_J$. Where exactly this dividing line is between these two conditions is still unclear. For that matter, it is not even clear if ${{\mathcal{Z}}}_J$ is a connected set, so the term “dividing line” might not be an accurate description of the boundary.

The complex case
----------------

Theorem \[thm:tool2\] does not seem to be applicable here, so we use a different method. Notice that this method works for the other three cases as well (and even higher-dimensional ones – see [@HS-multi]) but gives worse bounds.

If $A_{{\kappa}^2}$ is connected, then $A_{\kappa}$ has non-empty interior. In particular, this is the case if $|\kappa| \ge 2^{-1/4}$.

Note first that if $|\kappa^2| \ge 1/\sqrt{2}$ then $A_{\kappa^2}$ is connected – see [@BH Proposition 1]. Moreover, by the Hahn-Mazurkiewicz theorem, $A_{{\kappa}^2}$ is path 