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With the D$_{2}$ symmetry, the relationship between S and N$_{s}$ can be further derived. In fact, we can rewritten S = ( A ↓ + A $^{↑}$) /[∫ dEF ( ε ) ( a ↑ + a $^{↓}$)] . The D$_{2}$ sym- metry gives Δ T$_{23}$ = − Δ T$_{21}$ . Noticing T$_{3}$$_{↑}$$_{,}$$_{1}$ = T$_{3}$$_{↓}$$_{,}$$_{1}$ , the spin Nernst coefficient can be simplified as N$_{s}$ = ( A ↓ − A $^{↑}$)/ (2 πk$_{B}$ T ). Here ε = E − E$_{F}$ and F ( ε ) = f (1 − f ), a ↑ denotes the spin up term: a ↑ = T$_{2}$$_{↑}$$_{,}$$_{1}$ + T$_{3}$$_{↑}$$_{,}$$_{1}$ , and a ↓ the spin down term a ↓ = T$_{2}$$_{↓}$$_{,}$$_{1}$ + T$_{3}$$_{↓}$$_{,}$$_{1}$ , we also use the notation of the integral term A ↑ = ∫ dEεF ( ε ) a $^{↑}$/ T and A ↓ = ∫ dEεF ( ε ) a $^{↓}$/ T . Because of the symmetry, only leads-1,2,3 are used in the simplified expression of S and N$_{s}$ , we only need the upper half of the sample for our investigation. In fact, a ↑ ( a $^{↓}$) and A ↑ ( A $^{↓}$) reflects transport properties of spin-up (spin-down) electrons in the upper half of the sample. Roughly speaking, S can be seen as the sum of spin-up and spin-down terms, while N$_{s}$ as the difference of them. 

In the numerical calculations, t = ℏ $^{2}$/ (2 m $^{∗}$a $^{2}$) is set as the energy unit. If taking the effective electron mass m ∗ = 0 . 05 m$_{e}$ and the lattice constant a = 12 . 5 nm , t is about 5 meV . Temperature is fixed by k$_{B}$ T = 0 . 01 t , which is about 1 K . The size of center region is L = 19 a , about 237 nm . In a reasonable experimental range thus far V$_{R}$ ∈ [0 , 0 . 1] $^{12}$. However, in order to thoroughly study the relationship between the spin Nernst coefficient N$_{s}$ and the seebeck coefficient S , we extend the range of V$_{R}$ up to [0 , 1] in our calculation. 

Fig.2 shows the spin Nernst coefficient N$_{s}$ and the see- beck coefficient S versus the Fermi Energy E$_{F}$ in the clean system ( W = 0). It is clearly seen that the see- 

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beck coefficient S peaks at the positions where there are step-changes of transmission function T$_{1}$$_{,}$$_{2}$ + T$_{1}$$_{,}$$_{4}$ . These peaks can be explained by a simple model only with a 2-lead system without the Rashba SOI, shown in Fig.3(a). The transmission coefficient T$_{1}$$_{,}$$_{3}$ is a step func- tion (solid-black curve). The reason is as follows. The sample can be considered as a multi-channel system at a low temperature (here T ∼ 1 K ). When fermi energy increases, more channels in the lead are used to trans- port current. Thus, Δ V of two leads as well as the cur- rent increase with increasing of fermi energy (red-dashed curve). However, the S (blue-dotted curve) can not ac- cumulates while E$_{F}$ increases, it only peaks while the channel number changes and S is close to zero with a fixed channel number. This can be seen from the LB formula (2), if lead-p and lead-q have different tempera- tures, f$_{p}$ ( E,T +Δ T ) − f$_{q}$ ( E,T − Δ T ) is an antisymmetry function of E − E$_{F}$ (see plot in small box of Fig.3(a)): when E < E$_{F}$ , f$_{p}$ < f$_{q}$ , current flows from lower tem- perature lead to higher temperature one; when E > E$_{F}$ , current flows in the opposite direction. Only when the two flows are not equal, i.e. T$_{pσ,q}$ has an antisymmetry part, we can have a nonzero current. Thus for Fig.3(a), only when T$_{1}$$_{,}$$_{3}$ is at the step-change point, it has anti- symmetry part and can give a non-zero S . 

This conclusion can also be used to analyst spin- involved quantities. From Fig.2, we can see that N$_{s}$ (red solid line) shows an oscillatory structure. Besides the peaks at V$_{R}$ = 0 ( N$_{s}$ is zero at this point), the mag- nitude of N$_{s}$ oscillation is also large at the peaks of S ; but at the exact maximum point of S , where V$_{R}$ is quite small ( V$_{R}$ ≲ 0 . 1), N$_{s}$ is generally close to zero. This is because the spin transmission coefficient Δ T$_{2}$$_{,}$$_{3}$ generally has an extreme value when the transmission coefficient 

