the conjugates of $u$, i.e., $$\mathcal{G}\cdot u = \langle u, \sigma(u), \sigma^2(u),\ldots, \sigma^{p-1}(u)\rangle \subseteq {\mathcal{O}}_L^\times$$ generate a subgroup of ${\mathcal{O}}_L^\times$. Note that the product $\prod_{n=0}^{p-1} \sigma^n(u)=1$, so $$\mathcal{G}\cdot u = \langle u, \sigma(u), \sigma^2(u),\ldots, \sigma^{p-2}(u)\rangle.$$ Then, $${\rm res}_\infty\left(\mathcal{G}\cdot u\right) = \langle 
(\varepsilon_{\phi^n(1)}, \ldots, \varepsilon_{\phi^n(p)}) : 0\leq n\leq p-2\rangle\subseteq V_\infty,$$ where we have defined $V_\infty = \{ \pm 1\}^p$. If we fix an isomorphism $\psi:\{\pm 1\}\cong {\mathbb F}_2$, and write $f_i = \psi(\varepsilon_i)$, then the map ${\rm res}_\infty: \mathcal{G}\cdot u \to V_\infty$ can be written in ${\mathbb F}_2$-coordinates, and the corresponding $p\times (p-1)$ matrix over ${\mathbb F}_2$ is given by $$M_{\infty,u} = \left( f_{\phi^j(i)}\right)_{\substack{1\leq i\leq p\\ 0\leq j \leq p-2}} = \left(\begin{array}{cccc} 
f_1 & f_{\phi(1)} & \cdots & f_{\phi^{p-2}(1)} \\ 
f_2 & f_{\phi(2)} & \cdots & f_{\phi^{p-2}(2)} \\
\vdots & \vdots & \ddots & \vdots\\
f_p & f_{\phi(p)} & \cdots & f_{\phi^{p-2}(p)} \\
\end{array} \right).$$

\[lemma-m\_infty\] Let $u$ be a unit of norm $1$, and let $d_{\infty,u}$ be the dimension of the column space of $M_{\infty,u}$ or, equivalently, the dimension of ${\rm res}_\infty(\mathcal{G}\cdot u)$. Then, $\rho_\infty \leq (p-1) - d_{\infty,u}$. In particular, if $d_{\infty,u} = p-1$, then $\rho_\infty =0$.

If $u$ is of norm $1$, then $-1\not\in \mathcal{G}\cdot u$, because the 