)[g]{}\_D. Writing the charged scalar interaction as $${\cal L}_{\rm charged}=\eta_U{\bar u}_R  g_U  d_L  P^+ + \eta_D{\bar d}_R g_D u_L P^- +{\rm h.c.}$$ we find that, \_U=[\_Hv\_S- \_S v\_H \_Hv\_H+\_S v\_S]{}, \_D=[’\_Hv\_S- ’\_S v\_H ’\_Hv\_H+’\_S v\_S]{}. One way to get $\eta_U$ small while $\eta_D$ is large is to have $v_H \gg v_S$ so that $\eta_U \sim  -\epsilon_S/\epsilon_H$ and $\eta_D \sim \epsilon'_H/\epsilon'_S$ and take the corresponding ratios of $\epsilon$’s to be small and large respectively. For their product to be small we also need, $\epsilon_S \epsilon_H'/\epsilon_S' \epsilon_H$ to be small. This is clearly possible, however there doesn’t appear to be any symmetry reason behind these choices.

Note that there is an interchange symmetry where $H \leftrightarrow S$ that forces both $\eta_U=\eta_D=0$. But in the limit of that symmetry, $P^+=(h^+-S^+)/\sqrt{2}$, and the symmetry’s action on $P^+$ is $P^+ \rightarrow -P^+$. Hence there is a stable charged scalar.

The Glashow-Weinberg model [@Glashow:1976nt] where $H$ couples to the $u$-type quarks and $S$ couples to the $d$-type quarks has, $\epsilon'_H=\epsilon_S=0$ and so $\eta_U=v_S/v_H$ and $\eta_D=-v_H/v_S$. In this model $\eta_D \eta_U=-1$, so when $\eta_U$ is small $\eta_D$ is large, but their product cannot be made small even with a tuning of parameters.

In this section we construct 