Classification of Z\(_2\) spin liquids phases in Ba\(_2\)CuO\(_{3+\delta }\)
1.1 Projective symmetry groups
Under coordinates we choose in Sect. 3, space group symmetries, including translation, parity, and inversion, read
$$\begin{aligned} T_x:(i_x,i_y,i_z)\mapsto & {} (i_x+1,i_y,i_z), \end{aligned}$$
(29)
$$\begin{aligned} T_y:(i_x,i_y,i_z)\mapsto & {} (i_x,i_y+1,i_z), \end{aligned}$$
(30)
$$\begin{aligned} T_z:(i_x,i_y,i_z)\mapsto & {} (i_x,i_y,i_z+1), \end{aligned}$$
(31)
$$\begin{aligned} P_x:(i_x,i_y,i_z)\mapsto & {} (-i_x-i_z,i_y,i_z), \end{aligned}$$
(32)
$$\begin{aligned} P_y:(i_x,i_y,i_z)\mapsto & {} (i_x,-i_y-i_z,i_z), \end{aligned}$$
(33)
$$\begin{aligned} I:(i_x,i_y,i_z)\mapsto & {} (-i_x,-i_y,-i_z), \end{aligned}$$
(34)
and the time-reversal symmetry is
$$\begin{aligned} {\mathcal {T}}:u_{ij}\mapsto -u_{ij}. \end{aligned}$$
(35)
The symmetries above satisfy equalities
$$\begin{aligned} T_x^{-1}T_y^{-1}T_xT_y= & {} 1, \end{aligned}$$
(36)
$$\begin{aligned} T_y^{-1}T_z^{-1}T_yT_z= & {} 1, \end{aligned}$$
(37)
$$\begin{aligned} T_z^{-1}T_x^{-1}T_zT_x= & {} 1, \end{aligned}$$
(38)
$$\begin{aligned} T_xP_xT_xP_x= & {} 1, \end{aligned}$$
(39)
$$\begin{aligned} T_yP_yT_yP_y= & {} 1, \end{aligned}$$
(40)
$$\begin{aligned} T_x^{-1}P_yT_xP_y= & {} 1, \end{aligned}$$
(41)
$$\begin{aligned} T_y^{-1}P_xT_yP_x= & {} 1, \end{aligned}$$
(42)
$$\begin{aligned} P_xT_xT_z^{-1}P_xT_z= & {} 1, \end{aligned}$$
(43)
$$\begin{aligned} P_yT_yT_z^{-1}P_yT_z= & {} 1, \end{aligned}$$
(44)
$$\begin{aligned} P_xP_yP_xP_y= & {} 1, \end{aligned}$$
(45)
$$\begin{aligned} P_x^2=P_y^2= & {} 1, \end{aligned}$$
(46)
$$\begin{aligned} T_xIT_xI= & {} 1, \end{aligned}$$
(47)
$$\begin{aligned} T_yIT_yI= & {} 1, \end{aligned}$$
(48)
$$\begin{aligned} T_zIT_zI= & {} 1, \end{aligned}$$
(49)
$$\begin{aligned} P_xIP_xI= & {} 1, \end{aligned}$$
(50)
$$\begin{aligned} P_yIP_yI= & {} 1, \end{aligned}$$
(51)
$$\begin{aligned} I^2= & {} 1. \end{aligned}$$
(52)
\({\mathcal {T}}\) commutes with all the space group symmetries. Following Ref [20], we can first determine \(G_x\), \(G_y\), \(G_z\), and \(G_T\) through Eqs. (36)–(38) and the commutation relations between \({\mathcal {T}}\) and translations. Unlike the 2D case [20], here we choose the gauge that
$$\begin{aligned} G_z(i)= & {} \tau ^0, \end{aligned}$$
(53)
$$\begin{aligned} G_x(i)= & {} \eta _x^{i_z}\tau ^0, \end{aligned}$$
(54)
$$\begin{aligned} G_y(i)= & {} \eta _y^{i_z}\eta _{xy}^{i_x}\tau ^0, \end{aligned}$$
(55)
$$\begin{aligned} G_T(i)= & {} g_T\eta _{xt}^{i_x}\eta _{yt}^{i_y}\eta _{zt}^{i_z}, \end{aligned}$$
(56)
where two gauge inequivalent choices of \(g_T\) are \(g_T=\tau ^0\) or \(i\tau ^3\), and seven \(\eta \)s are \(\pm 1\).
Then we consider parities \(P_x\) and \(P_y\). The PSG equations given by Eq. (39) to (46) read
$$\begin{aligned} G_x(P_x(i))G^{-1}_{P_x}(i+{\hat{x}})G_x(i+{\hat{x}})G_{P_x}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(57)
$$\begin{aligned} G_y^{-1}(P_x(i))G^{-1}_{P_x}(i)G_y(i)G_{P_x}(i-{\hat{y}})\in & {} {\mathcal {G}}, \end{aligned}$$
(58)
$$\begin{aligned} G_x^{-1}(P_y(i))G^{-1}_{P_y}(i)G_x(i)G_{P_y}(i-{\hat{x}})\in & {} {\mathcal {G}}, \end{aligned}$$
(59)
$$\begin{aligned} G_y(P_y(i))G^{-1}_{P_y}(i+{\hat{y}})G_y(i+{\hat{y}})G_{P_y}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(60)
$$\begin{aligned} G_{P_x}(i)G_x(P_x(i))G^{-1}_z(T_zT_x^{-1}P_x(i))\cdot \nonumber \\ G_{P_x}(T_zT_x^{-1}P_x(i))G_z(T_zT_x^{-1}(i))\in & {} {\mathcal {G}}, \end{aligned}$$
(61)
$$\begin{aligned} G_{P_y}(i)G_y(P_y(i))G^{-1}_z(T_zT_y^{-1}P_y(i))\cdot \nonumber \\ G_{P_y}(T_zT_y^{-1}P_y(i))G_z(T_zT_y^{-1}(i))\in & {} {\mathcal {G}}, \end{aligned}$$
(62)
$$\begin{aligned} G_{P_x}(i)G_{P_y}(P_x(i))G^{-1}_{P_x}(P_y(i))G^{-1}_{P_y}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(63)
$$\begin{aligned} G_{P_x}(i)G_{P_x}(P_x(i))\in & {} {\mathcal {G}}, \end{aligned}$$
(64)
$$\begin{aligned} G_{P_y}(i)G_{P_y}(P_y(i))\in & {} {\mathcal {G}}. \end{aligned}$$
(65)
Since \(P_x\) and \(P_y\) do not change \(i_z\), in our gauge Eq. (57) to (60) reduce to
$$\begin{aligned} G^{-1}_{P_x}(i+{\hat{x}})G_{P_x}(i)= & {} \eta _{xpx}\tau ^0, \end{aligned}$$
(66)
$$\begin{aligned} G^{-1}_{P_x}(i+{\hat{y}})G_{P_x}(i)= & {} \eta _{ypx}\eta _{xy}^{i_z}\tau ^0, \end{aligned}$$
(67)
$$\begin{aligned} G^{-1}_{P_y}(i+{\hat{x}})G_{P_y}(i)= & {} \eta _{xpy}\tau ^0, \end{aligned}$$
(68)
$$\begin{aligned} G^{-1}_{P_y}(i+{\hat{y}})G_{P_y}(i)= & {} \eta _{ypy}\tau ^0, \end{aligned}$$
(69)
which give \(G_{P_x}(i)\) and \(G_{P_y}(i)\) the generic form
$$\begin{aligned} G_{P_x}(i)= & {} g_{P_x}\varTheta _{P_x}(i_z)\eta _{xpx}^{i_x}\eta _{ypx}^{i_y}, \end{aligned}$$
(70)
$$\begin{aligned} G_{P_y}(i)= & {} g_{P_y}\varTheta _{P_y}(i_z)\eta _{xpy}^{i_x}\eta _{ypy}^{i_y}\eta _{xy}^{i_yi_z}, \end{aligned}$$
(71)
where \(\varTheta \)s are (\(\pm 1\))-valued functions of \(i_z\). Eq. (61) to (65) then reduce to
$$\begin{aligned} g_{P_x}^2\varTheta _{P_x}(i_z)\varTheta _{P_x}(i_z+1)\eta _{xpx}^{i_z+1}G_x(i)= & {} \eta _{5px} \tau ^0, \end{aligned}$$
(72)
$$\begin{aligned} g_{P_y}^2\varTheta _{P_y}(i_z)\varTheta _{P_y}(i_z+1)\eta _{xy}^{i_y}G_y(i)= & {} \eta _{5py} \tau ^0, \end{aligned}$$
(73)
$$\begin{aligned} g_{P_x}g_{P_y}g^{-1}_{P_x}g^{-1}_{P_y}(\eta _{xpy}\eta _{ypx})^{i_z}= & {} \pm \tau ^0, \end{aligned}$$
(74)
$$\begin{aligned} g^2_{P_x}\eta _{xpx}^{i_z}= & {} \pm \tau ^0, \end{aligned}$$
(75)
$$\begin{aligned} g^2_{P_y}(\eta _{ypy}\eta _{xy})^{i_z}= & {} \pm \tau ^0, \end{aligned}$$
(76)
for all sites i. Eq. (74) to (76) require that
$$\begin{aligned} \eta _{xpy}=\eta _{ypx},\eta _{xy}=\eta _{xpx}=\eta _{ypy}=1, \end{aligned}$$
(77)
while Eqs. (72) and (73) give two \(\varTheta \)s a specific form. All gauge inequivalent \(\varTheta \)s are
$$\begin{aligned} \varTheta _{P_x}(i_z)= & {} \eta _{5px}^{i_z},\text { for }G_x(i)=\tau ^0; \end{aligned}$$
(78)
$$\begin{aligned} \varTheta _{P_y}(i_z)= & {} \eta _{5py}^{i_z},\text { for }G_y(i)=\tau ^0; \end{aligned}$$
(79)
$$\begin{aligned} \varTheta _{P_x}(i_z)= & {} \eta _{5px}^{i_z}\cdot \sqrt{2}\sin (\frac{\pi }{2}i_z+\frac{\pi }{4}),\nonumber \\ \text {for }G_x(i)= & {} (-)^{i_z}\tau ^0; \end{aligned}$$
(80)
$$\begin{aligned} \varTheta _{P_y}(i_z)= & {} \eta _{5py}^{i_z}\cdot \sqrt{2}\sin (\frac{\pi }{2}i_z+\frac{\pi }{4}),\nonumber \\ \text {for }G_y(i)= & {} (-)^{i_z}\tau ^0. \end{aligned}$$
(81)
Finally we consider inversion I. Equations (47) to (49) induce PSG equations
$$\begin{aligned} G_x(I(i))G_{I}(i+{\hat{x}})G_x(i+{\hat{x}})G_{I}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(82)
$$\begin{aligned} G_y(I(i))G_{I}(i+{\hat{y}})G_y(i+{\hat{y}})G_{I}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(83)
$$\begin{aligned} G_z(I(i))G_{I}(i+{\hat{z}})G_z(i+{\hat{z}})G_{I}(i)\in & {} {\mathcal {G}}. \end{aligned}$$
(84)
Under our gauge, \(G_{I}(i)\) has the generic form
$$\begin{aligned} G_{I}(i)=g_I\eta _{xI}^{i_x}\eta _{yI}^{i_y}\eta _{zI}^{i_z}. \end{aligned}$$
(85)
According to Eqs. (50) and (51),
$$\begin{aligned} G_I(P_x(i))G^{-1}_{P_x}(I(i))G_x(I(i))G_{P_x}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(86)
$$\begin{aligned} G_I(P_y(i))G^{-1}_{P_y}(I(i))G_y(I(i))G_{P_y}(i)\in & {} {\mathcal {G}}, \end{aligned}$$
(87)
we have
$$\begin{aligned} g_Ig^{-1}_{P_x}g_Ig_{P_x}\eta _{xI}^{i_z}= & {} \pm \tau ^0, \end{aligned}$$
(88)
$$\begin{aligned} g_Ig^{-1}_{P_y}g_Ig_{P_y}\eta _{yI}^{i_z}= & {} \pm \tau ^0, \end{aligned}$$
(89)
for all sites i. Therefore, \(\eta _{xI}=\eta _{yI}=1\). From just the same argument, \(\eta _{xt}=\eta _{yt}=1\). Then, Eqs. (56), (70), (71) and (85) reduce to
$$\begin{aligned} G_T(i)= & {} g_T\eta _{t}^{i_z}, \end{aligned}$$
(90)
$$\begin{aligned} G_{P_x}(i)= & {} g_{P_x}\varTheta _{P_x}(i_z)\eta _{p}^{i_y}, \end{aligned}$$
(91)
$$\begin{aligned} G_{P_y}(i)= & {} g_{P_y}\varTheta _{P_y}(i_z)\eta _{p}^{i_x}, \end{aligned}$$
(92)
$$\begin{aligned} G_I(i)= & {} g_I\eta _{I}^{i_z}, \end{aligned}$$
(93)
where \(\eta _{t}\), \(\eta _{p}\) and \(\eta _{I}\) can take value of \(\pm 1\), and \(\varTheta _{P_x}(i_z)\) and \(\varTheta _{P_y}(i_z)\) are determined above. The constraints of gs reduce to
$$\begin{aligned} g_{P_x}^2=g_{P_y}^2=g_I^2= & {} \pm \tau ^0, \end{aligned}$$
(94)
$$\begin{aligned} g_{P_x}g_{P_y}g^{-1}_{P_x}g^{-1}_{P_y}= & {} \pm \tau ^0, \end{aligned}$$
(95)
$$\begin{aligned} g_Ig^{-1}_{P_x}g_Ig_{P_x}= & {} \pm \tau ^0, \end{aligned}$$
(96)
$$\begin{aligned} g_Ig^{-1}_{P_y}g_Ig_{P_y}= & {} \pm \tau ^0, \end{aligned}$$
(97)
$$\begin{aligned} g_Tg^{-1}_{P_x}g_Tg_{P_x}= & {} \pm \tau ^0, \end{aligned}$$
(98)
$$\begin{aligned} g_Tg^{-1}_{P_y}g_Tg_{P_y}= & {} \pm \tau ^0, \end{aligned}$$
(99)
$$\begin{aligned} g_Ig^{-1}_Tg_Ig_T= & {} \pm \tau ^0. \end{aligned}$$
(100)
All gauge inequivalent choices of gs are
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=\tau ^0\quad g_{P_y}=\tau ^0\quad g_I=\tau ^0; \end{aligned}$$
(101)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(102)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=\tau ^0\quad g_I=\tau ^0; \end{aligned}$$
(103)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(104)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^1; \end{aligned}$$
(105)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=\tau ^0\quad g_I=\tau ^0; \end{aligned}$$
(106)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=\tau ^0; \end{aligned}$$
(107)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=\tau ^0\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(108)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=\tau ^0\quad g_I=\tau ^0; \end{aligned}$$
(109)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=\tau ^0; \end{aligned}$$
(110)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(111)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=\tau ^0\quad g_I=\tau ^0; \end{aligned}$$
(112)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^1\quad g_I=\tau ^0; \end{aligned}$$
(113)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^1; \end{aligned}$$
(114)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=\tau ^0; \end{aligned}$$
(115)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(116)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(117)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=\tau ^0; \end{aligned}$$
(118)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(119)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(120)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^1\quad g_I=\tau ^0; \end{aligned}$$
(121)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^0\quad g_I=i\tau ^1; \end{aligned}$$
(122)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^1; \end{aligned}$$
(123)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=\tau ^0; \end{aligned}$$
(124)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^1\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(125)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1; \end{aligned}$$
(126)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=\tau ^0; \end{aligned}$$
(127)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^3; \end{aligned}$$
(128)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1; \end{aligned}$$
(129)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^3\quad g_I=\tau ^0; \end{aligned}$$
(130)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=\tau ^0\quad g_I=i\tau ^1; \end{aligned}$$
(131)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=\tau ^0\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^3; \end{aligned}$$
(132)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1; \end{aligned}$$
(133)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^3; \end{aligned}$$
(134)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(135)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1; \end{aligned}$$
(136)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^3; \end{aligned}$$
(137)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^3; \end{aligned}$$
(138)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^1; \end{aligned}$$
(139)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1; \end{aligned}$$
(140)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^3; \end{aligned}$$
(141)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^2; \end{aligned}$$
(142)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^2\quad g_I=i\tau ^1; \end{aligned}$$
(143)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^2\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^1; \end{aligned}$$
(144)
$$\begin{aligned} g_T= & {} \tau ^0\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^2\quad g_I=i\tau ^3; \end{aligned}$$
(145)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^1\quad g_{P_y}=i\tau ^2\quad g_I=i\tau ^3; \end{aligned}$$
(146)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^3\quad g_{P_y}=i\tau ^1\quad g_I=i\tau ^2; \end{aligned}$$
(147)
$$\begin{aligned} g_T= & {} i\tau ^3\quad g_{P_x}=i\tau ^2\quad g_{P_y}=i\tau ^3\quad g_I=i\tau ^1. \end{aligned}$$
(148)
There are 48 different gauge inequivalent choices of gs. Therefore, the total number of PSGs is \(48\times 2^5\times 4=6144\). However, when \(g_T=\tau ^0\), to acquire non-vanishing ansatzes, \(\eta _T\) must be identical to [20] \(-1\). Therefore, \(15\times 2^4\times 4=960\) PSGs are killed and the total number of PSGs reduces to 5184.
1.2 Ansatzes of the nearest-neighbour spin coupling model
In this section we assume that only \(u_{i,i+x}\), \(u_{i,i+y}\), \(u_{i,i+z}\), \(u_{i,i-x+z}\), \(u_{i,i-y+z}\) and \(u_{i,i-x-y+z}\) are non-vanishing. First, an ansatz \(u_{i,i+m}\) under \(T_xG_x\), \(T_yG_y\) and \(T_zG_z\) reads
$$\begin{aligned} u_{i,i+m}=\eta _x^{i_xm_z}\eta _y^{i_ym_z}u_m^l\tau ^l,\quad l=0,1,2,3, \end{aligned}$$
(149)
where \(u_m^i\), \(i=1,2,3\) is real and \(u_m^0\) is pure imaginary. \({\mathcal {T}}\) and I further give constraints
$$\begin{aligned} \eta _t^{m_z}g_Tu_m^l\tau ^lg_T^{-1}= & {} -u_m^l\tau ^l; \end{aligned}$$
(150)
$$\begin{aligned} \eta _I^{m_z}g_Iu_m^l\tau ^lg_I^{-1}= & {} u_{I(m)}^l\tau ^l. \end{aligned}$$
(151)
Using \(u_{I(m)}=u_{-m}=u_m^{\dagger }\), we can conclude that when
$$\begin{aligned} \eta _t= & {} 1,\eta _I=1,g_T=i\tau ^3,\text { and }g_I=i\tau ^3, \end{aligned}$$
(152)
all \(u_m\) vanish. This kills \(10\times 2^3\times 4=320\) PSGs and the totally number of PSGs is \(5184-320=4864\). When
$$\begin{aligned} \eta _t\eta _I=-1,g_T=i\tau ^3,g_I=i\tau ^{1,2}\text { or }\tau ^0, \end{aligned}$$
(153)
all \(u_m^l\) vanish for odd \(m_z\), namely only \(u_{i,i+x}\) and \(u_{i,i+y}\) remain non-zero. These ansatzes degenerate to describe spin liquids in a rectangular lattice in 2D plane, which is irrelevant to us. There is another similar case. When \(G_x=(-)^{i_z}\tau ^0\), \(G_{P_x}\) will give the constraint
$$\begin{aligned}&\left( \cos \left( \frac{\pi }{2}m_z\right) +(-)^{i_z}\sin \left( \frac{\pi }{2}m_z\right) \right) \cdot \nonumber \\&\quad \eta _p^{m_y}g_{P_x}u_m^l\tau ^lg_{P_x}^{-1}=u_{P_x(m)}^l\tau ^l. \end{aligned}$$
(154)
For odd \(m_z\), l.h.s. is a function of \(i_z\) while r.h.s. is not, which indicates that all the \(u_m^l\) for odd \(m_z\) must vanish to satisfy the equation. As indicated in the previous argument, we do not take consideration of these ansatzes. Therefore, only \(G_x(i)=G_y(i)=\tau ^0\) case will be under consideration. The number of PSGs left is \((4864-(9+14)\times 2^3\times 4)\div 4=1032\).
When \(G_x(i)=G_y(i)=\tau ^0\), \(P_x\) and \(P_y\) give constraints on ansatzes
$$\begin{aligned} g_{P_x}u_x^l\tau ^lg_{P_x}^{-1}= & {} u_x^{l\dagger }\tau ^l; \end{aligned}$$
(155)
$$\begin{aligned} \eta _pg_{P_x}u_y^l\tau ^lg_{P_x}^{-1}= & {} u_y^l\tau ^l; \end{aligned}$$
(156)
$$\begin{aligned} \eta _{5px}g_{P_x}u_z^l\tau ^lg_{P_x}^{-1}= & {} u_{-x+z}^l\tau ^l; \end{aligned}$$
(157)
$$\begin{aligned} \eta _p\eta _{5px}g_{P_x}u_{-y+z}^l\tau ^lg_{P_x}^{-1}= & {} u_{-x-y+z}^l\tau ^l; \end{aligned}$$
(158)
$$\begin{aligned} \eta _pg_{P_y}u_x^l\tau ^lg_{P_y}^{-1}= & {} u_x^l\tau ^l; \end{aligned}$$
(159)
$$\begin{aligned} g_{P_y}u_y^l\tau ^lg_{P_y}^{-1}= & {} u_y^{l\dagger }\tau ^l; \end{aligned}$$
(160)
$$\begin{aligned} \eta _{5py}g_{P_y}u_z^l\tau ^lg_{P_y}^{-1}= & {} u_{-y+z}^l\tau ^l; \end{aligned}$$
(161)
$$\begin{aligned} \eta _p\eta _{5py}g_{P_y}u_{-x+z}^l\tau ^lg_{P_y}^{-1}= & {} u_{-x-y+z}^l\tau ^l. \end{aligned}$$
(162)
These equations determine the constraints of ansatzes in numerical calculation. Two of the constraints are employed. One is the periodic condition, that the periodicity of all the ansatzes is 1, and the other is the sector condition, that the ansatzes satisfy
$$\begin{aligned} u_z=s_x u_{-x+z}=s_y u_{-y+z}=s_{xy} u_{-x-y+z}, \end{aligned}$$
(163)
with \(s_x,s_y,s_{xy}=\pm 1\). According to numerical results, there are at most 311 inequivalent ansatzes.
Numerical method and data for strong coupling case
Differential evolution (DE), originally developed by Storn and Price [42], is a meta-heuristic algorithm that globally optimizes a given objective function in an iterative manner. Usually, the objective function is treated as a black box and no assumptions are needed. For example, unlike traditional gradient descent, conjugate gradient and quasi-Newton methods, derivatives are not needed. Evaluation of derivatives of mean-field energy defined previously is time-consuming for which DE is suitable. Besides, another algorithm, the Nelder-Mead method is also tested but doesn’t perform as well as DE.
DE works with a group (called population) of solution candidates (called agents), which is initialized randomly. In each iterative step, a certain agent is selected and a new agent is generated from this agent and two other randomly selected agents in a random, linear way. If the new agent is better that the old agent, the old agent is replaced by the new one. If not, the trial agent is simply discarded. This procedure continues until some certain accuracy is reached.
In this paper, DE is used to optimize the mean-field energy with respect to ansatzes. Constrained by PSG’s, the number of optimizing variables is restricted to be 12. The number of agents is set to be 120, 10 times the number of variables, with differential weight being 0.9 and cross-over probability being 0.5.
1.1 Fourier transformation of the mean-field Hamiltonian
The mean-field Hamiltonian reads
$$\begin{aligned} H_{{\text {MF}}}=H_{{\text {MF}}}^f-H_{{\text {MF}}}^b \end{aligned}$$
(164)
with
$$\begin{aligned} H_{{\text {MF}}}^f= & {} \frac{3}{4} \sum _{\mathbf {r}} \sum _\alpha J_\alpha [ \psi _{\alpha _1}^\dagger (\mathbf {r})U_\alpha \psi _{\alpha _2}(\mathbf {r}) \nonumber \\&+\psi _{\alpha _2}^\dagger (\mathbf {r})U_\alpha ^\dagger \psi _{\alpha _1}(\mathbf {r}) ] \end{aligned}$$
(165)
and
$$\begin{aligned} H_{{\text {MF}}}^b= & {} \sum _{\mathbf {r}} \sum _\alpha t_\alpha [ b_{\alpha _1}^\dagger (\mathbf {r})U_\alpha b_{\alpha _2}(\mathbf {r})\nonumber \\&+b_{\alpha _2}^\dagger (\mathbf {r})U_\alpha ^\dagger b_{\alpha _1}(\mathbf {r})] \end{aligned}$$
(166)
Here superscripts f and b mean fermion and boson respectively. \(\mathbf {r}\) refers to the coordinate of one certain super-cell. \(\alpha \) refers to the index of one certain bond in a cell. \(\alpha _1\) is the index of the first end of bond \(\alpha \), \(\alpha _2\) is the other end. They are assigned for each bond \(\alpha \) before practical calculation. Order of these two sites, \(\alpha _1\) and \(\alpha _2\), matters, which means they are not symmetric in formulas. \(U_\alpha \) is defined to be \(u_{\alpha _1,\alpha _2}\), so it becomes a bond-dependent quantity. Note that \(u_{\alpha _1,\alpha _2}\) not necessarily equals to \(u_{\alpha _2,\alpha _1}\). In this holon-condensed case, bosons are treated as scalars.
Only derivation of Fourier-transformed form for the fermion Hamiltonian is shown in detail. The Fourier-transformed form of the boson Hamiltonian can be obtained by just replacing \(\psi \) with b since commutation relations are not included in derivation.
Take substitutions
$$\begin{aligned} \psi _{\alpha _1}(\mathbf {r})=\frac{1}{\sqrt{N_\text {cell}}}\sum _{\mathbf {k}}{\psi _{\alpha _1}(\mathbf {k})}e^{\mathbbm {i}\mathbf {k}\cdot (\mathbf {r}+\mathbf {l}_{\alpha _1})} \end{aligned}$$
(167)
and
$$\begin{aligned} \psi _{\alpha _2}(\mathbf {r})=\frac{1}{\sqrt{N_\text {cell}}}\sum _{\mathbf {k}}{\psi _{\alpha _2}(\mathbf {k})}e^{\mathbbm {i}\mathbf {k}\cdot (\mathbf {r}+\mathbf {l}_{\alpha _2})}, \end{aligned}$$
(168)
we further have
$$\begin{aligned} H_{{\text {MF}}}^f= & {} \frac{3}{4} \sum _{\mathbf {k}}\sum _{\alpha } J_\alpha [ \psi _{\alpha _1}^\dagger (\mathbf {k})U_\alpha e^{\mathbbm {i}\mathbf {k}\cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} \psi _{\alpha _2}(\mathbf {k}) \nonumber \\&+\psi _{\alpha _2}^\dagger (\mathbf {k})U_\alpha ^\dagger e^{-\mathbbm {i}\mathbf {k}\cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} \psi _{\alpha _1}(\mathbf {k}) ]. \end{aligned}$$
(169)
It should be noted that this Hamiltonian is block-diagonalized with respect to \(\mathbf {k}\). So we can calculate eigenvalues and eigenvectors of each block-matrix individually to reduce calculation workload.
$$\begin{aligned} {\mathcal {H}}_{{\text {MF}}}^b= & {} \sum _{\mathbf {k}}\sum _{\alpha } t_\alpha [ b_{\alpha _1}^\dagger (\mathbf {k})U_\alpha e^{\mathbbm {i}\mathbf {k}\cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} b_{\alpha _2}(\mathbf {k}) \nonumber \\&+b_{\alpha _2}^\dagger (\mathbf {k})U_\alpha ^\dagger e^{-\mathbbm {i}\mathbf {k}\cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} b_{\alpha _1}(\mathbf {k}) ]. \end{aligned}$$
(170)
These two equations can be rephrased in matrix form:
$$\begin{aligned} H_{{\text {MF}}}^f=\sum _{\mathbf {k}}{\psi ^\dagger (\mathbf {k}) Q_{{\text {MF}}}^f(\mathbf {k}) \psi (\mathbf {k})} \end{aligned}$$
(171)
with
$$\begin{aligned} \psi ({\mathbf {k}})= \begin{pmatrix} \psi _{1,\uparrow }(\mathbf {k})\\ \psi _{2,\uparrow }(\mathbf {k})\\ \vdots \\ \psi _{N_\text {site},\uparrow }(\mathbf {k})\\ \psi _{1,\downarrow }(\mathbf {k}) \\ \vdots \\ \psi _{N_\text {site},\downarrow }(\mathbf {k}) \end{pmatrix}. \end{aligned}$$
(172)
Here \(N_\text {site}\) means the number of sites in one unit cell. And the number of cells is inidicated by \(N_\text {cell}\).
The \(Q_{{\text {MF}}}^f\) can be diagonalized as
$$\begin{aligned} Q_{{\text {MF}}}^f(\mathbf {k})=S^f(\mathbf {k})D^f(\mathbf {k})S^{f \dagger }(\mathbf {k}). \end{aligned}$$
(173)
Denote \(\phi (\mathbf {k})=S^{f \dagger }(\mathbf {k}) \psi (\mathbf {k})\), we further have
$$\begin{aligned} H_{{\text {MF}}}^f=\sum _{\mathbf {k}}\sum _{i=1}^{2N_\text {site}}{\lambda _i^f(\mathbf {k})\phi _i^\dagger (\mathbf {k})\phi _i(\mathbf {k})}. \end{aligned}$$
(174)
To obtain energy of the original Hamiltonian, we need to evaluate the average \(\langle \psi _i^\dagger (\mathbf {k_1})\psi _j(\mathbf {k_2}) \rangle _0\). The subscript 0 means that average is taken in a Gaussian level. These two averages can be expressed as
$$\begin{aligned} \langle \psi _i^\dagger (\mathbf {k_1})\psi _j(\mathbf {k_2}) \rangle _0= \delta _{\mathbf {k_1},\mathbf {k_2}} \sum _{l=1}^{2N_\text {site}}{S_{il}^{f*}S_{jl}^f \langle \phi _l^\dagger (\mathbf {k_1})\phi _l(\mathbf {k_1}) \rangle _0 },\nonumber \\ \end{aligned}$$
(175)
where
$$\begin{aligned} \langle \phi _l^\dagger (\mathbf {k})\phi _l(\mathbf {k}) \rangle _0= {\left\{ \begin{array}{ll} 0, &{}\lambda _l^f(\mathbf {k})>0 \\ 1, &{}\lambda _l^f(\mathbf {k})<0. \end{array}\right. } \end{aligned}$$
(176)
For simplicity, we define several functions:
$$\begin{aligned} n_f(i,s)= & {} \sum _{\mathbf {k}}\langle \psi _{i,s}^\dagger (\mathbf {k}) \psi _{i,s}(\mathbf {k}) \rangle _0, \end{aligned}$$
(177)
$$\begin{aligned} n_b(i,s)= & {} \sum _{\mathbf {k}} \langle b_{i,s}^\dagger (\mathbf {k}) b_{i,s}(\mathbf {k}) \rangle _0, \end{aligned}$$
(178)
$$\begin{aligned} O_f(\alpha ,s_1,s_2)= & {} \sum _{\mathbf {k}}e^{\mathbbm {i}\mathbf {k} \cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} \langle \psi _{\alpha _1,s_1}^\dagger (\mathbf {k}) \psi _{\alpha _2,s_2}(\mathbf {k}) \rangle _0, \nonumber \\ \end{aligned}$$
(179)
$$\begin{aligned} O_b(\alpha ,s_1,s_2)= & {} \sum _{\mathbf {k}}e^{\mathbbm {i}\mathbf {k} \cdot (\mathbf {l}_{\alpha _2}-\mathbf {l}_{\alpha _1})} \langle b_{\alpha _1,s_1}^\dagger (\mathbf {k}) b_{\alpha _2,s_2}(\mathbf {k}) \rangle _0.\nonumber \\ \end{aligned}$$
(180)
Here i is the site index in one super-cell, s, \(s_1\) and \(s_2\) can be \(\uparrow \) or \(\downarrow \).
1.2 Evaluation of the energy
With the assistance with definitions above, the energy of the original mean-field Hamiltonian (Eq. 164) can be expressed as
$$\begin{aligned} \langle H_{\text {MF}} \rangle _0= & {} \langle H_{\text {MF}}^f \rangle _0-\langle H_{\text {MF}}^b \rangle _0, \end{aligned}$$
(181)
$$\begin{aligned} \langle H_{{\text {MF}}}^f \rangle _0= & {} \sum _{\alpha }\frac{3J_\alpha }{4}\left[ \frac{N_\text {cell}}{4}-\frac{1}{4}n_f(\alpha _1,\uparrow )-\frac{1}{4}n_f(\alpha _1,\downarrow ) \right. \nonumber \\&-\frac{1}{4}n_f(\alpha _2,\downarrow )-\frac{1}{4}n_f(\alpha _2,\downarrow ) \nonumber \\&-\frac{1}{2N_\text {cell}}O_f(\alpha ,\downarrow ,\uparrow )O_f(\alpha ,\uparrow ,\downarrow ) \nonumber \\&+\frac{1}{2N_\text {cell}}O_f(\alpha ,\uparrow ,\uparrow )O_f(\alpha ,\downarrow ,\downarrow ) \nonumber \\&-\frac{1}{2N_\text {cell}}O_f^*(\alpha ,\downarrow ,\uparrow )O_f^*(\alpha ,\uparrow ,\downarrow ) \nonumber \\&+\frac{1}{2N_\text {cell}}O_f^*(\alpha ,\uparrow ,\uparrow )O_f^*(\alpha ,\downarrow ,\downarrow ) \nonumber \\&-\frac{1}{4N_\text {cell}}O_f^*(\alpha ,\uparrow ,\uparrow )O_f(\alpha ,\uparrow ,\uparrow ) \nonumber \\&+\frac{1}{4N_\text {cell}}n_f(\alpha _1,\uparrow )n_f(\alpha _2,\uparrow ) \nonumber \\&-\frac{1}{4N_\text {cell}}O_f^*(\alpha ,\uparrow ,\downarrow )O_f(\alpha ,\uparrow ,\downarrow ) \nonumber \\&+\frac{1}{4N_\text {cell}}n_f(\alpha _1,\uparrow )n_f(\alpha _2,\downarrow ) \nonumber \\&-\frac{1}{4N_\text {cell}}O_f^*(\alpha ,\downarrow ,\uparrow )O_f(\alpha ,\downarrow ,\uparrow ) \nonumber \\&+\frac{1}{4N_\text {cell}}n_f(\alpha _1,\downarrow )n_f(\alpha _2,\uparrow ) \nonumber \\&-\frac{1}{4N_\text {cell}}O_f^*(\alpha ,\downarrow ,\downarrow )O_f(\alpha ,\downarrow ,\downarrow ) \nonumber \\&\left. +\frac{1}{4N_\text {cell}}n_f(\alpha _1,\downarrow )n_f(\alpha _2,\downarrow ) \right] , \end{aligned}$$
(182)
and
$$\begin{aligned} \langle H_{{\text {MF}}}^b \rangle _0= & {} \frac{1}{2N_\text {cell}}\sum _{\alpha } t_\alpha [ O_b(\alpha ,\uparrow ,\downarrow )O_f(\alpha ,\downarrow ,\uparrow ) \nonumber \\&+O_b(\alpha ,\downarrow ,\uparrow )O_f(\alpha ,\uparrow ,\downarrow ) \nonumber \\&-O_b(\alpha ,\uparrow ,\uparrow )O_f(\alpha ,\downarrow ,\downarrow ) \nonumber \\&-O_b(\alpha ,\downarrow ,\downarrow )O_f(\alpha ,\uparrow ,\uparrow ) \nonumber \\&+O_b^*(\alpha ,\uparrow ,\downarrow )O_f^*(\alpha ,\downarrow ,\uparrow ) \nonumber \\&+O_b^*(\alpha ,\downarrow ,\uparrow )O_f^*(\alpha ,\uparrow ,\downarrow ) \nonumber \\&-O_b^*(\alpha ,\uparrow ,\uparrow )O_f^*(\alpha ,\downarrow ,\downarrow ) \nonumber \\&-O_b^*(\alpha ,\downarrow ,\downarrow )O_f^*(\alpha ,\uparrow ,\uparrow ) \nonumber \\&+O_b(\alpha ,\uparrow ,\uparrow )O_f^*(\alpha ,\uparrow ,\uparrow ) \nonumber \\&+O_b(\alpha ,\uparrow ,\downarrow )O_f^*(\alpha ,\uparrow ,\downarrow ) \nonumber \\&+O_b(\alpha ,\downarrow ,\uparrow )O_f^*(\alpha ,\downarrow ,\uparrow ) \nonumber \\&+O_b(\alpha ,\downarrow ,\downarrow )O_f^*(\alpha ,\downarrow ,\downarrow ) \nonumber \\&+O_b^*(\alpha ,\uparrow ,\uparrow )O_f(\alpha ,\uparrow ,\uparrow ) \nonumber \\&+O_b^*(\alpha ,\uparrow ,\downarrow )O_f(\alpha ,\uparrow ,\downarrow ) \nonumber \\&+O_b^*(\alpha ,\downarrow ,\uparrow )O_f(\alpha ,\downarrow ,\uparrow ) \nonumber \\&+O_b^*(\alpha ,\downarrow ,\downarrow )O_f(\alpha ,\downarrow ,\downarrow ) ]. \end{aligned}$$
(183)
Construction of the interaction hamiltonian in weak coupling limit
A generic interaction term reads (spin indices neglected) [27]
$$\begin{aligned} H'= & {} \int \prod _a\frac{k_a}{2\pi }\sum _{P_a,i_a}\delta (Q)V({P_a,i_a,k_a})\nonumber \\&\psi _{i_1}^{P_1\dag }(k_1)\psi _{i_2}^{P_2\dag }(k_2)\psi _{i_3}^{P_3}(k_3)\psi _{i_4}^{P_4}(k_4), \end{aligned}$$
(184)
where \(P=1(-1)\) for R(L). The constraint of momentum conservation
$$\begin{aligned} 0=Q=- & {} P_1k_{Fi_1}-P_2k_{Fi_2}+P_3k_{Fi_3}+P_4k_{Fi_4}\nonumber \\&-k_1-k_2+k_3+k_4. \end{aligned}$$
(185)
As the momentum of chiral fermions (\(k_i\)) are much smaller than Fermi vectors (\(k_{Fi}\)), the momentum conservation is reduced to
$$\begin{aligned} -P_1k_{Fi_1}-P_2k_{Fi_2}+P_3k_{Fi_3}+P_4k_{Fi_4}=0. \end{aligned}$$
(186)
Therefore, only two types of interactions are allowed. The first one is intra-band scattering \(\psi _{i}^{R\dag }\psi _{j}^{L\dag }\psi _{j}^{L}\psi _{i}^{R}\) and the second one is inter-band scattering \(\psi _{i}^{R\dag }\psi _{i}^{L\dag }\psi _{3-i}^{L}\psi _{3-i}^{R}\). The purely chiral terms like \(\psi _{i}^{R\dag }\psi _{j}^{R\dag }\psi _{j}^{R}\psi _{i}^{R}\) do not generate renormalization at leading order [27] and thus are neglected. When spin is included, we define charge and spin current
$$\begin{aligned} T_{ij}= & {} \sum _{s}\psi _{i,s}^{\dag }\psi _{i,s}, \end{aligned}$$
(187)
$$\begin{aligned} \mathbf {T}_{ij}= & {} \frac{1}{2}\sum _{s,s'}\psi _{i,s}^{\dag }\mathbf {\sigma }_{ss'}\psi _{i,s'}, \end{aligned}$$
(188)
and the interaction Hamiltonian density in real space can be written as
$$\begin{aligned} \mathcal {H'}= & {} f_{ij}^{\rho }T_{ii}^R T_{jj}^L-f_{ij}^{\sigma }\mathbf {T}_{ii}^R\cdot \mathbf {T}_{jj}^L\nonumber \\&+{f'}_{ii}^{\rho }T_{i,3-i}^RT_{i,3-i}^L-{f'}_{i,i}^{\sigma }\mathbf {T}_{i,3-i}^R\cdot \mathbf {T}_{ij}^L. \end{aligned}$$
(189)
Derivation of RG equations
Define \(z_i=v_i\tau -ix\), where \(v_i\) is the Fermi velocity. Determined by the operator product expansion (OPE) in terms of chiral fermions
$$\begin{aligned} \psi _{i,s}^R(x,\tau )\psi _{j,s'}^{R\dag }(0,0)\sim \frac{\delta _{ij}\delta {ss'}}{2\pi z_i}, \end{aligned}$$
(190)
$$\begin{aligned} \psi _{i,s}^L(x,\tau )\psi _{j,s'}^{L\dag }(0,0)\sim \frac{\delta _{ij}\delta {ss'}}{2\pi z_i^*}, \end{aligned}$$
(191)
the current algebra reads [27]
$$\begin{aligned} T_{ij}^R(x,\tau )T_{lm}^R(0,0)\sim & {} \frac{\delta _{il}}{2\pi z_j}T_{jm}^R-\frac{\delta _{jm}}{2\pi z_i}T_{il}^R, \end{aligned}$$
(192)
$$\begin{aligned} T_{ij}^{Ra}(x,\tau )T_{lm}^{Rb}(0,0)\sim & {} \frac{\delta _{ab}}{4}\left( \frac{\delta _{il}}{2\pi z_j}T_{jm}^R-\frac{\delta _{jm}}{2\pi z_i}T_{il}^R\right) \end{aligned}$$
(193)
$$\begin{aligned}&+\frac{i\epsilon _{abc}}{2}\left( \frac{\delta _{il}}{2\pi z_j}T_{jm}^{Rc}+\frac{\delta _{jm}}{2\pi z_i}T_{il}^{Rc}\right) ,\nonumber \\ T_{ij}^{Ra}(x,\tau )T_{lm}^R(0,0)\sim & {} \frac{\delta _{il}}{2\pi z_j}T_{jm}^{Ra}-\frac{\delta _{jm}}{2\pi z_i}T_{il}^{Ra}, \end{aligned}$$
(194)
where \(T^{Ra}\) is the components of the vector current \(\mathbf {T}^R\). For \(T^L\), the current algebra is the same, except \(z_i\rightarrow z_i^*\). Employing the standard method [39] and using the current algebra above, we obtain the RG equations
$$\begin{aligned} {\dot{f}}_{ii}^{\rho }= & {} -\frac{16({f'}_{ii}^{\rho })^2+3({f'}_{ii}^{\sigma })^2}{32\pi v_i}, \end{aligned}$$
(195)
$$\begin{aligned} {\dot{f}}_{ii}^{\sigma }= & {} -\frac{2(f_{ii}^{\sigma })^2+4{f'}_{ii}^{\rho }{f'}_{ii}^{\sigma }+({f'}_{ii}^{\sigma })^2}{4\pi v_i}, \end{aligned}$$
(196)
$$\begin{aligned} {\dot{f}}_{i,3-i}^{\rho }= & {} \frac{16({f'}_{ii}^{\rho })^2+3({f'}_{ii}^{\sigma })^2}{16\pi (v_i+v_{3-i})}, \end{aligned}$$
(197)
$$\begin{aligned} {\dot{f}}_{i,3-i}^{\sigma }= & {} \frac{2(f_{i,3-i}^{\sigma })^2+4{f'}_{ii}^{\rho }{f'}_{ii}^{\sigma }-({f'}_{ii}^{\sigma })^2}{2\pi (v_i+v_{3-i})}, \end{aligned}$$
(198)
$$\begin{aligned} \dot{f'}_{ii}^{\rho }= & {} \frac{16f_{ij}^{\rho }{f'}_{ii}^{\rho }+3f_{ij}^{\sigma }{f'}_{ii}^{\sigma }}{8\pi (v_i+v_{3-i})}\nonumber \\&-\sum _i \frac{16f_{ii}^{\rho }{f'}_{ii}^{\rho }+3f_{ii}^{\sigma }{f'}_{ii}^{\sigma }}{32\pi v_i}, \end{aligned}$$
(199)
$$\begin{aligned} \dot{f'}_{ii}^{\sigma }= & {} \frac{2f_{ij}^{\rho }{f'}_{ii}^{\sigma }+2f_{ij}^{\sigma }{f'}_{ii}^{\rho }-f_{ij}^{\sigma }{f'}_{ii}^{\sigma }}{\pi (v_i+v_{3-i})}\nonumber \\&-\sum _i\frac{2f_{ii}^{\rho }{f'}_{ii}^{\sigma }+2f_{ii}^{\sigma }{f'}_{ii}^{\rho }+f_{ii}^{\sigma }{f'}_{ii}^{\sigma }}{4\pi v_i}. \end{aligned}$$
(200)
Symmetries of the coupling constants are employed in the derivation of the equations above. The initial value of these coupling constants are
$$\begin{aligned} f_{ij,0}^{\rho }= & {} \frac{1}{4} f_{ij,0}^{\sigma }=J'\left( 1-(-1)^{i+j}\cos \left( \frac{k_{Fi}+k_{Fj}}{2}\right) \right) \nonumber \\&+\frac{1}{2}J(1-\cos (k_{Fi}+k_{Fj}))+\delta _{ij}J_y, \end{aligned}$$
(201)
$$\begin{aligned} {f'}_{ii,0}^{\rho }= & {} \frac{1}{4} {f'}_{ii,0}^{\sigma }=J\sin (k_{F1})\sin (k_{F2})\nonumber \\&-2J'\sin \left( \frac{k_{F1}}{2}\right) \sin \left( \frac{k_{F2}}{2}\right) -J_y. \end{aligned}$$
(202)
The RG flows are calculated numerically.
Bosonization dictionary [39]
The bosonization dictionary [39] reads
$$\begin{aligned} \psi _{i,s}^{R/L}\sim \eta _{i,s}e^{i\sqrt{4\pi }\phi ^{R/L}_{i,s}}, \end{aligned}$$
(203)
where the chiral boson fields satisfy commutation relation [27]
$$\begin{aligned}{}[\phi ^{R}_{i,s}(x),\phi ^{R}_{i',s'}(y)]= & {} -[\phi ^{L}_{i,s}(x),\phi ^{L}_{i',s'}(y)]\nonumber \\= & {} \frac{i}{4}\text {sgn}(x-y)\delta _{ii'}\delta _{ss'}, \end{aligned}$$
(204)
$$\begin{aligned}= & {} \frac{i}{4}\delta _{ii'}\delta _{ss'}, \end{aligned}$$
(205)
and \(\eta \)s are Klein factors satisfying \(\{\eta _{i,s},\eta _{i',s'}\}=2\delta _{ii'}\delta _{ss'}\). To describe spin and charge modes, we further define
$$\begin{aligned} \phi _{i,\rho }= & {} \frac{1}{\sqrt{2}}(\phi ^R_{i,\uparrow }+\phi ^R_{i,\downarrow }+\phi ^L_{i,\uparrow }+\phi ^L_{i,\downarrow }), \end{aligned}$$
(206)
$$\begin{aligned} \theta _{i,\rho }= & {} \frac{1}{\sqrt{2}}(\phi ^R_{i,\uparrow }+\phi ^R_{i,\downarrow }-\phi ^L_{i,\uparrow }-\phi ^L_{i,\downarrow }), \end{aligned}$$
(207)
$$\begin{aligned} \phi _{i,\sigma }= & {} \frac{1}{\sqrt{2}}(\phi ^R_{i,\uparrow }-\phi ^R_{i,\downarrow }+\phi ^L_{i,\uparrow }-\phi ^L_{i,\downarrow }), \end{aligned}$$
(208)
$$\begin{aligned} \theta _{i,\sigma }= & {} \frac{1}{\sqrt{2}}(\phi ^R_{i,\uparrow }-\phi ^R_{i,\downarrow }-\phi ^L_{i,\uparrow }+\phi ^L_{i,\downarrow }), \end{aligned}$$
(209)
where subscript \(\rho \) represents charge mode and \(\sigma \) represents spin mode, respectively.
