“solar” oscillation scale emerges radiatively, through a loop involving the “dark sector” exchange. We now briefly describe the minimal scotogenic seesaw scenario [@Rojas:2018wym]. The new particles and their charges are given in Table \[quantum number of particles\].

                      $\nu_{R}$      $\eta$       $f$
  ------------------ ----------- --------------- -----
     $SU(2)_{L}$          1             2          1
      $U(1)_{Y}$          0       $\frac{1}{2}$    0
   $\mathbb{Z}_{2}$       +             -          -
     Multiplicity         1             1          1

In Table.\[quantum number of particles\] the additional $\mathbb{Z}_2$ symmetry is the “dark parity” which is responsible for the stablity of the dark matter candidate. The full Yukawa sector can be split as $$\begin{aligned}
 \mathcal{L}=\mathcal{L}_{\text{SM}}+\mathcal{L}_{\text{ATM}}+\mathcal{L}_{\text{DM,SOL}}\end{aligned}$$ where $\mathcal{L}_{SM}$ is the Standard Model Lagrangian and $$\begin{aligned}
 \mathcal{L}_{\text{ATM}}=- \sum_{a} Y_{\nu}^{a}\bar{L}^{a}i\sigma_{2}H^{*}\nu_{R}  + 
 \frac{1}{2}  M_{R} \overline{\nu_{R}^{c}}\nu_{R} + h.c,\end{aligned}$$ which induces the tree level neutrino mass (atmospheric neutrino mass scale) after the electroweak symmetry breaking. The Lagrangian responsible for the solar and dark sector is given by $$\begin{aligned}
 \mathcal{L}_{\text{DM,SOL}}=Y_{f}^{a}\bar{L}^{a}i\sigma_{2}\eta^{*}f+\frac{1}{2}M_{f}\overline{f^{c}}f+h.c.\end{aligned}$$ The dark sector consists of one fermion $f$ and one scalar $\eta$, both of which are odd under the dark $\mathbb{Z}_{2}$ parity. All the [[Standard Model ]{}]{}particles and $\nu_{R}$ are even under this dark $\mathbb{Z}_{2}$ parity.

The scalar sector is given by $$\begin{aligned}
 V&=-\mu_{H}^{2}H^{\dagger}H+m_{\eta}^{2}\eta^{\dagger}\eta+\lambda (H^{\dagger}H)^{2}+\lambda_{\eta}(\eta^{\dagger}\eta)^{2}
 +\lambda_{3}(H^{\dagger}H)(\eta^{\dagger}\eta)+\lambda_{4}(H^{\dagger}\eta)(\eta^{\dagger}H)\nonumber \\
 &+\lambda_{5}\left((H^{\dagger}\eta)^{2}+h.c.\right)
 \label{eq:scoto-pot}\end{aligned}$$ As the $\mathbb{Z}_{2}$ symmetry is exactly conserved, the mixing 