5 have zero mass, corresponding to the longitudinal components of Z ′ and W ± $_{R}$. Their orthog- onal combinations have mass-squares − µ ( v 2 4 + 4 v 2 $_{6}$) /v 6 and ( f 2 − µ/v 6 )( v 2 4 + 2 v 2 $_{6}$) respectively. Since n$_{R}$ couples to Δ ± $_{R}$, but not to φ ± $_{R}$, the discussion on dark-matter relic abundance from nn annihilation to lepton pairs through Δ ± R exchange in Ref. [1] applies only if v 2 6 << v 2 $_{4}$. This turns out to be exactly what the model requires because m$_{n}$ comes from v 6 and m$_{n}$ of order 200 GeV is needed for dark-matter relic abundance. To be specific, we will assume in fact that m$_{n}$ = 200 GeV. If this value is changed, some details in the following will be changed, but all the qualitative features of this model will remain. The first thing to notice is that for m$_{n}$ = 200 GeV, Fig. 3 of Ref. [1] requires m$_{Δ}$ + R = 220 GeV. From the Yukawa coupling we get m$_{n}$ = √$_{2}$$_{f}$$_{n}$$_{v}$$_{6}$$_{. Since}$ f$_{n}$ = 1 is assumed in computing the relic abundance in Ref. [1], we obtain v 6 = 141 GeV. Let us now assume M$_{Z}$ ′ = 1 TeV for illustration. Then v 4 = 1851 GeV and M$_{W}$$_{R}$ = 832 GeV, where v 2 = 95 GeV and v 3 = 146 GeV have been used to ensure zero Z − Z ′ mixing at tree level (see next section). The physical charged scalar ξ + R is now 99.4% Δ + R and its mass is given by This implies that f 2 − µ/v 6 = 0.014. We now note that the Δ$_{R}$ scalar triplet masses satisfy the important sum rule This means that both m$_{A}$$_{R}$ and m$_{Δ}$ ++ R are bounded from above as a function of λ 3 which should not be larger than about one in magnitude. We plot in Fig. 1 m$_{Δ}$ ++ R versus m$_{A}$$_{R}$ for various values of λ 3 . We now come to a very important conclusion. To satisfy the dark-matter relic density in this model, m$_{n}$ and m$_{ξ}$ + R have to be of order 200 GeV. This in turn implies that m$_{A}$$_{R}$ m 2 A$_{R}$ 1 + 4 v 2 $_{6}$/v 2 4 + m 2 Δ ++ R = 2 m 2 ξ + R 1 + 2 v 2 $_{6}$/v 2 4 − 2 λ 3 v 2 $_{6}$. (20) m 2 ξ + R = ( f 2 − µ/v 6 )( v 2 4 + 2 v 2 $_{6}$) = [220 GeV]$^{2}$. (19) f$_{n}$ √$_{2}$ (Δ 0 $_{R}$n$_{R}$ n$_{R}$ $_{+}$√$_{2}$$_{δ}$ + $_{R}$n$_{R}$ e$_{R}$ + Δ ++ R e$_{r}$e$_{R}$ ) , (18) A$_{R}$ = √$_{2(}$$_{v}$$_{4}$$_{Im}$$_{Δ}$ 0 R − 2 v 6 Imφ 0 $_{R}$) √ v 2 4 + 4 v 2 6 , ξ ± R = ( v 4 Δ ± R − √$_{2}$$_{v}$$_{6}$$_{φ}$ ± $_{R}$) √ v 2 4 + 2 v 2 6