Then, as $t_{\ensuremath{\mathbf{w}}}\neq 0$, $$\left| \sum_{(i_1,\dots,i_d) \in {\ensuremath{\mathcal{S}}}}w_1^{i_1+1}\cdots w_d^{i_d+1} \right| = \bigg| (w_1\cdots w_d)S({\ensuremath{\mathbf{w}}})\bigg| = \left|\frac{1}{t_{\ensuremath{\mathbf{w}}}}\right| \geq |{\ensuremath{\mathcal{S}}}|.$$ But $({\ensuremath{\mathbf{w}}},t_{\ensuremath{\mathbf{w}}}) \in D({\mbox{\boldmath$\rho$}})$ implies $|w_j|\leq1$ for each $1\leq j \leq d$. Thus, the above inequality states that the sum of $|{\ensuremath{\mathcal{S}}}|$ complex numbers of modulus at most one has modulus $|{\ensuremath{\mathcal{S}}}|$. The only way this can occur is if each term in the sum has modulus one, and all terms point in the same direction in the complex plane. By symmetry, and the assumption that we take a positive step in each direction, there are two terms of the form $w_2^{i_2+1}\cdots w_d^{i_d+1}$ and $w_1^2w_2^{i_2+1}\cdots w_d^{i_d+1}$ in the sum, so that $w_1^2$ must be 1 in order for them to point in the same direction. This shows $w_1=\pm1$, and the same argument applies to each $w_k$, so there are at most $2^d$ points in ${\ensuremath{\mathcal{V}}}\cap D({\mbox{\boldmath$\rho$}})$.

By Proposition \[prop:critpt\] every such point $({\ensuremath{\mathbf{w}}},t_{\ensuremath{\mathbf{w}}}) \in {\ensuremath{\mathcal{V}}}\cap D({\mbox{\boldmath$\rho$}})$ is critical, and to show it is isolated it is sufficient to prove $S_1^{(k)}({\ensuremath{\mathbf{w}}}_{\hat{k}}) \neq 0$ for all $1 \leq k \leq d$. Indeed, if $S_1^{(k)}({\ensuremath{\mathbf{w}}}_{\hat{k}})=0$ then ${\ensuremath{\mathbf{w}}}\in {\ensuremath{\mathcal{V}}}$ implies $$|t_{\ensuremath{\mathbf{w}}}| =\frac{1}{\left| w_1\cdots w_d S_0^{(k)}({\ensuremath{\mathbf{w}}}_{\hat{k}}) \right|} \geq \frac{1}{\left|S_0^{(k)}({\ensuremath{\mathbf{w}}}_{\hat{k}})\right|} \geq \frac{1}{S_0^{(k)}({\ensuremath{\mathbf{1}}})} > \frac{1}{|{\ensuremath{\mathcal{S}}}|},$$ by our assumption that 