The last term gives rise to asymmetric torques. Evaluating the trace in the spin space, the torque reads Symmetric contribution (contribution from the E · A δ term) to the first line in the square bracket was calculated in Ref. [10] by use of where n denotes the magnetization direction. ( The second and third terms in the square bracket in Eq. (D3) lead to a correction to the spin Berry phase (proportional to A $^{z}$) and the standard spin transfer torque (proportional to ∇ n $^{α}$), respectively. ) As for the asymmetric contribution (terms proportional to 1 $_{2}$E$_{ν}$ A δ ν in Eq. (D3)), summation over ν cannot taken independent of the electric field, resulting in complicated torques. We here look into only the correction to the β term (the first line in the square bracket), which is obtained as (neglecting irrelevant terms) The result of the β term is thus ( σ$_{B}$ is the Boltzmann conductivity) and δβ/β = − 1 / 5 in the present model. Thus the β term arising from the spin-orbit interaction has asymmetric contribution, in contrast to the 23 where j$_{‖}$ ≡ n · j , ∇ ‖ ≡ n · ∇ , β ≡ 8 e 9 πm $^{2}$σ$_{B}$ n$_{so}$λ$_{so}$ 2 ∑ kk ′ ∑ σσ ′ k $^{4}$( k $^{′}$) $^{2}$| g r k σ | $^{2}$Re g r k , − $_{σ}$Im g r k $^{′}$σ′ , (D7) T ( β ) ,α so ≃ P e [ β n × ( j · ∇ ) n + δβ n × ( j$_{‖}$ ∇ $_{‖}$) n ] , (D6) − 1 2 ∑ µνβγ ϵ$_{αµν}$R$_{µβ}$R$_{νγ}$E$_{ν}$A δ $_{ν}$( ϵ$_{βγδ}$ − ϵ$_{βγz}$δ$_{δz}$ ) ≃ 1 2 ( E · A α − E$_{α}$A α $_{α}$) . (D5) ∑ µνβγ ϵ$_{αµν}$R$_{µβ}$R$_{νγ}$A δ $_{i}$( ϵ$_{βγδ}$ − ϵ$_{βγz}$δ$_{δz}$ ) = 2( A α i + n $^{α}$Az $_{i}$) = ( n × ∇ $_{i}$n ) $^{α}$, (D4) T α so = 8 e 45 πm $_{2}$n$_{so}$ λ$_{so}$ 2 ∑ µνβγ ϵ$_{αµν}$R$_{µβ}$R$_{νγ}$ ( E · A δ − 1 2 E$_{ν}$A δ ν ) ∑ kk ′ k $^{4}$( k $^{′}$) 2 × ∑ σ [ ( ϵ$_{βγδ}$ − ϵ$_{βγz}$δ$_{δz}$ ) | g r k σ | $^{2}$Re g r k , − σ ∑ σ ′ Im g r k $^{′}$σ′ + 2 ϵ$_{βγz}$δ$_{δz}$ | g r k σ | $^{2}$Re( g r k $_{σ}$)Im g r k $^{′}$, − σ +( δ$_{βz}$δ$_{γδ}$ − δ$_{γz}$δ$_{βδ}$ ) σ | g r k σ | $^{2}$Im g r k , − σ ∑ σ ′ Im g r k $^{′}$σ′ ] . (D3)