(\varepsilon_1,\ldots,\varepsilon_p)$$ where $\tau_1,\ldots,\tau_p$ are the real embeddings of $L$ and $\varepsilon_i$ is the sign of $\tau_i(u)\in {\mathbb R}$. We order our embeddings in the following manner. Let $g_u(x)$ be the minimal polynomial of $u$ over ${\mathbb Q}$, and let $r_1,\ldots,r_p$ be the real roots of $g_u(x)$ ordered so that $r_1<r_2<\cdots <r_p$. Then, $\{r_i\}=\{\tau_j(u)\}$, and we choose $\tau_i$ so that $\tau_i(u)=r_i$ for all $1\leq i\leq p$. With this notation, ${\rm res}_\infty(u) = (-1,-1,-1,\ldots,1,1,1)$, i.e., it consists of a non-negative number of $-1$ signs followed by a non-negative number of $+1$ signs. Recall that $u$ is in the kernel of ${\rm res}_\infty$ if and only if $u$ is a totally positive unit.

Attached to the generator $\sigma\in \mathcal{G}={\operatorname{Gal}}(L/{\mathbb Q})$, there is a permutation $\phi=\phi_\sigma\in S_p$, where $\phi$ is considered here as a permutation of $\{1,2,\ldots,n\}$, such that $\tau_i(\sigma(u))=r_{\phi(i)}$, and therefore $${\rm res}_\infty(\sigma(u)) = (\varepsilon_{\phi(1)}, \ldots, \varepsilon_{\phi(p)}).$$ Since $\tau$ is an embedding (injective), if $\tau_i(\alpha)=r_i$ for some $\alpha\in L^\times$, then $\tau_i(\sigma(\alpha)) = r_{\phi(i)}$. It follows that $\tau_i(\sigma^n(u)) = r_{\phi^n(i)}$, and $${\rm res}_\infty(\sigma^n(u)) = (\varepsilon_{\phi^n(1)}, \ldots, \varepsilon_{\phi^n(p)}),$$ for all $n\geq 1$.

Now, in addition, suppose that $u$ is a unit of norm $1$, and let $\mathcal{G}\cdot u$ be the subgroup of units generated by 