norm of $-1$ is $-1$, and the norm of every element in $\mathcal{G}\cdot u$ is $1$. In particular, $\langle {{\rm res}}_\infty(-1), {{\rm res}}_\infty(\mathcal{G}\cdot u)\rangle$ is a space of dimension $1+d_{\infty,u}$. Hence, the kernel of ${{\rm res}}_\infty$ is at most of dimension $p-(1+d_{\infty,u})$. It follows that $\rho_\infty\leq (p-1)-d_{\infty,u}$ as desired.

Totally positive units in cyclic extensions of prime degree {#sec-totposprimedegree}
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Let $L$ be a cyclic extension of prime degree $p>2$, and let ${\mathcal{O}}_L^\times$ be the unit group of ${\mathcal{O}}_L$. Let ${\mathcal{O}}_L^{\times, 1}$ be the units of norm $1$, so that ${\mathcal{O}}_L^\times \cong \{ \pm 1 \}\times {\mathcal{O}}_L^{\times,1}$. In this section we show the following result:

\[prop-p3p5totpos\] Let $p\geq 3$ be a prime, let $L$ be a cyclic extension of degree $p$, and suppose that the polynomial $\phi_p(x)=(x^p-1)/(x-1)$ is irreducible over ${\mathbb F}_2$. Then, either $\rho_\infty=0$ (i.e., ${\mathcal{O}}_L^{\times,+} = ({\mathcal{O}}_L^{\times})^2$), or $\rho_\infty = p-1$ in which case every unit in ${\mathcal{O}}_L^{\times,1}$ is totally positive.

If every unit in ${\mathcal{O}}_L^{\times,1}$ is totally positive, then $\rho_\infty=p-1$ since we would have $$\rho_\infty = \dim_{{\mathbb F}_2} {\mathcal{O}}_L^{\times,+}/({\mathcal{O}}_L^\times)^2 = \dim_{{\mathbb F}_2} {\mathcal{O}}_L^{\times,1}/({\mathcal{O}}_L^\times)^2= p-1.$$ Otherwise, there must be a unit $u\in {\mathcal{O}}_L^{\times,1}$ that is not totally positive (in particular, $u$ is not in $\pm 1 \cdot  ({\mathcal{O}}_L^{\times})^2$). 