{\mathcal{O}}_L^{\times,+}/({\mathcal{O}}_L^\times)^2.$$ We refer the reader to [@dummit] for heuristics and conjectures about the dimension of the totally positive units (in particular, the conjecture on page 4). In the following theorem, we use the notation of [@dummit].

\[thm-rhos\] Let $\rho$, $\rho^+$, and $\rho_\infty$ be defined by $$\rho = \dim_{{\mathbb F}_2} {{\rm Cl}}(L)/2{{\rm Cl}}(L),\ \rho^+ = \dim_{{\mathbb F}_2} {{\rm Cl}}^+(L)/2{{\rm Cl}}^+(L), \text{ and } \rho_\infty = \dim_{{\mathbb F}_2} {\mathcal{O}}_L^{\times,+}/({\mathcal{O}}_L^\times)^2$$ Then,

1.  $\rho_\infty= p-\dim_{{\mathbb F}_2} {{\rm res}}_\infty({\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2)=\dim_{{\mathbb F}_2} \{\pm 1 \}^{p}/{{\rm res}}_\infty({\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2).$

2.  We have $$0 \to \{\pm 1 \}^{p}/{{\rm res}}_\infty({\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2) \to {{\rm Cl}}^+(L) \to {{\rm Cl}}(L)\to 0.$$ In particular, $\max\{\rho,\rho_\infty\}\leq \rho^+\leq \rho_\infty+\rho$, and $\rho^+=\rho_\infty+\rho$ if and only if the exact sequence splits.

3.  [(Armitage-Fröhlich)]{} $\rho^+-\rho \leq (p-1)/2$.

For part (1), note that $\rho_\infty = \dim_{{\mathbb F}_2} {\mathcal{O}}_L^{\times,+}/({\mathcal{O}}_L^\times)^2$ is the dimension of $\ker\left({{\rm res}}_\infty|_{{\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2}\right)$, and the dimension of ${\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2$ is $p$. Thus, the dimension of the image of ${{\rm res}}_\infty|_{{\mathcal{O}}_L^\times/({\mathcal{O}}_L^\times)^2}$ is $p$ minus the dimension of the kernel.

For part (2), see Section 2 of [@dummit], and in particular Equation (2.9). Part (3) is shown in [@armitage], where it is shown that $\rho^+-\rho\leq \lfloor r_1/2 \rfloor$, where $r_1$ is the number of real embeddings of $L$. Here $r_1=p$ is an odd prime, so 