where Γ$_{R}$$_{j}$ and Γ$_{L}$$_{j}$ are the decay width at tree level, Although the Yukawa couplings of L are much smaller than those of R [cf. (25)], the lepton asymmetries (27a) and (27b) can arrive at a same magnitude, because the Yukawa couplings of R dominate the loop corrections in the decays of both R and L . For generating the observed baryon asymmetry [1] the Yukawa couplings y$_{N}$ and the masses M should fulfill The 3-body decays of L don’t have significant contribu- tions to the total decay width and the lepton asymmetry, but it is sufficient to produce abundant right-handed neu- trinos for the dark matter relic density. We now clarify this interesting scenario in details. It is easy to read the relic density, where κ ′ $_{L$_{j}$}$( y$_{N}$, M ) is a wash out factor and is the branching ratio. From the baryon asymmetry (26) and the relic density (32), we can perform the following ratio, We should keep in mind that the right-handed neutri- nos mix with their left-handed partners due to the Dirac mass term in Eq. (16) so that they will decay at tree level and loop orders. In absence of the mixing between W ± L and W ± R (cf. (8)), the decay width should be [24, 25] The above breaking scale points to the necessity of the resonant [12, 13] leptogenesis because of the parameter choice (39), i.e. M$_{j}$ ≃ 186 GeV. It also indicates that the gauge interactions of the right-handed neutrinos decouple at a high temperature ∼ ( v$_{R}$ v$_{L}$ ) 4 3 O (MeV) = O (100 GeV) so that the non-thermally [23] produced right-handed neutrinos can successfully explain the dark matter relic density. Accordingly, the left-right symmetry breaking scale can be given by we can obtain the dark matter masses which is true in the weak washout region, and With further assumptions, which immediately yields Here m$_{N}$ ∼ 940 MeV is the nucleon mass. On the other hand, as shown in Eq. (23), the mass matrix of the right- handed neutrinos has a same texture with that of the left-handed neutrinos. For example, let’s consider the case with the degenerate neutrino masses, For a given relic density, we can determine the masses of the right-handed neutrinos by the observed baryon asym- metry, Γ$_{R}$$_{j}$ = Γ$_{R}$$_{j}$$_{→}$$_{l}$$_{L}$$_{φ}$$_{L}$ + Γ$_{R}$$_{j}$$_{→}$$_{l}$ c $_{L}$φ ∗ L = 1 8 π ( y † $_{N}$y$_{N}$ ) jj M$_{j}$ , (28a) Γ$_{L}$$_{j}$ = Γ$_{L}$$_{j}$$_{→}$$_{l}$$_{L}$$_{φ}$$_{L}$ + Γ$_{L}$$_{j}$$_{→}$$_{l}$ c $_{L}$φ ∗ L + Γ$_{L}$$_{j}$$_{→}$$_{χφ}$ ∗ $_{L}$φ$_{L}$ = 1 8 π ( M D † N 1 M y † $_{N}$y$_{N}$ 1 M M D N ) jj M$_{j}$ + λ 2 3 2 $^{12}$π$^{3}$λ 2 2 ( y † $_{N}$y$_{N}$ ) jj M 3 j v 2 R ≃ O [ ( M D N M M N ) 2 ] 1 8 π ( y † $_{N}$y$_{N}$ ) jj M$_{j}$ . (28b) m$_{χ}$$_{i}$ = m$_{N}$ c Σ$_{j}$ ( κ$_{R}$$_{j}$ + κ$_{L}$$_{j}$ ) ε$_{j}$ Σ$_{j}$ κ ′ $_{L$_{j}$}$Br$_{ji}$ Ω$_{χ}$$_{i}$ Ω$_{b}$ (35) m$_{1}$ ≃ m$_{2}$ ≃ m$_{3}$ = 0 . 05 eV , (36) m$_{χ}$$_{1}$ ≃ m$_{χ}$$_{2}$ ≃ m$_{χ}$$_{3}$ = 0 . 05 eV v 2 L v 2 R . (37) ε$_{R}$$_{j}$ ≃ ε$_{L}$$_{j}$ ≃ ε$_{j}$ , (29) κ$_{L}$$_{j}$ ≃ κ$_{R}$$_{j}$ ≃ κ ′ L$_{j}$ = 0 . 1 , (38) η = n$_{b}$ n$_{γ}$ = 7 . 04 × n$_{b}$ s = (6 . 225 ± 0 . 170) × 10 − $^{10}$, (30) λ$_{3}$ λ$_{2}$ = 3 . 3 × 10 − $^{4}$, M M N M D N = 1 . 9 × 10 $^{7}$, M$_{j}$ v$_{R}$ = 2 . 1 × 10 − $^{4}$, (39) m$_{χ}$$_{1}$$_{,}$$_{2}$$_{,}$$_{3}$ ≃ 1 . 3 MeV . (40) c Σ$_{j}$ ( κ$_{R}$$_{j}$ + κ$_{L}$$_{j}$ ) ε$_{j}$ ≃ 0 . 99 × 10 − $^{8}$. (31) v$_{R}$ = v$_{L}$ √ m$_{χ}$$_{i}$ m$_{i}$ ≃ 5 . 1 × 10 $^{3}$v$_{L}$ . (41) IV. DARK MATTER n$_{χ}$$_{i}$ s = Σ$_{j}$ κ ′ $_{L$_{j}$}$Br$_{ji}$ g$_{∗}$ , (32) Br$_{ji}$ = Γ$_{L}$$_{j}$$_{→}$$_{χ}$$_{i}$$_{φ}$ ∗ $_{L}$φ$_{L}$ Γ$_{L}$$_{j}$ = O [ ( M M N M D N ) 2 ] M 2 j v 2 R λ 2 3 2 $^{9}$π$^{2}$λ 2 2 ∣ ( U $^{†}$y$_{N}$ ) ij ∣ ∣$^{2}$ y † $_{N}$y$_{N}$ jj (33) Γ$_{χ}$$_{i}$$_{→}$$_{ννν}$ = G 2 F 384 π 3 sin $^{2}$(2 θ$_{i}$ ) m 5 χ$_{i}$ , (42a) Γ$_{χ}$$_{i}$$_{→}$$_{νe}$ $^{+}$e − = 5 G 2 F 3072 π 3 sin $^{2}$(2 θ$_{i}$ ) m 5 χ$_{i}$ , (42b) Γ$_{χ}$$_{i}$$_{→}$$_{νγ}$ = 9 αG 2 F 1024 π 4 sin $^{2}$(2 θ$_{i}$ ) m 5 χ$_{i}$ . (42c) Σ$_{j}$ κ$_{L}$$_{j}$Br$_{ji}$m$_{χ}$$_{i}$ : c Σ$_{j}$ ( κ$_{R}$$_{j}$ + κ$_{L}$$_{j}$ ) ε$_{j}$m$_{N}$ = Ω$_{χ}$$_{i}$ : Ω$_{b}$ . (34) 4