This notebook computes the equations of the distinguished component V_0

Inverse of a matrix of determinant 1

S={{0,0,1},{0,-1,0},{1,0,0}}; Inv[A_]:=Transpose[S.Minors[A].S];

Matrix B has trace and trace of its inverse equals to v=\beta. Matrix A generic, with non - zero entries (2, 1) and (3, 1)

A={{a,b,c},{1,d,e},{1,f,g}};MatrixForm[A]
B={{1,0,0},{0,v-1,1},{0,-1,0}};MatrixForm[B]

Impose equations to have a representation of the group \Gamma

Simplify[Solve[{Tr[A]-Tr[Inv[A]],Tr[A]-Tr[A. B],Tr[B]-Tr[B.Inv[A]],Tr[Inv[B].A]-Tr[Inv[B]],Tr[Inv[A]]-Tr[Inv[A.B]]}=={0,0,0,0,0},{b,c,d,e,f}]]

b=-((2 (-1+a) (a-v))/(-2-v+(v^2)));c=-(((-1+a) (a-v) (-1+v))/(-2-v+(v^2)));d=((-a-g (-2+v)+v)/(-2+v));e=-(1/(2 (-2+v)))(-2 a+2 g+2 v+a v-3 g v-(v^2)+g (v^2)+(Sqrt[(-2+v) ((((a+g (-1+v)-v))^2) (-2+v)-4 (g (g (-2+v)-v)+a (-2+g+v)))]));f=-(1/(2 (-2+v)))(2 a-2 g-2 v-a v+3 g v+(v^2)-g (v^2)+(Sqrt[(-2+v) ((((a+g (-1+v)-v))^2) (-2+v)-4 (g (g (-2+v)-v)+a (-2+g+v)))]));

We use as variables (u, v), which are (\alpha, \beta) in the paper

Simplify[Solve[Tr[A]==u,a]]

a=((u (-2+v)-v)/(-3+v));

Finally impose the condition on the determinant of A

Solve[Det[A]==1,g]

We introduce the discriminant delta. Also the variable sqrdelta=sqrt{delta}

delta=-3-4 u-4 v-6 u v+(u^2) (v^2);
g=(27 u+18 (u^2)-6 (u^3)-27 v+18 u v+33 (u^2) v+2 (u^3) v-36 (v^2)-39 u (v^2)-22 (u^2) (v^2)+6 (u^3) (v^2)+12 (v^3)+10 u (v^3)-24 (u^2) (v^3)-2 (u^3) (v^3)+10 (v^4)+27 u (v^4)+12 (u^2) (v^4)-9 (v^5)-12 u (v^5)-(u^2) (v^5)+2 (v^6)+u (v^6)+(u-v) (((-3-2 v+(v^2)))^2) sqrdelta)/(2 (9 (u^2)-18 u v+12 (u^2) v+9 (v^2)-24 u (v^2)-2 (u^2) (v^2)+12 (v^3)+4 u (v^3)-4 (u^2) (v^3)-2 (v^4)+8 u (v^4)+(u^2) (v^4)-4 (v^5)-2 u (v^5)+(v^6)));

Op[TT] teaches Mathematica to understand that delta = sqrdelta^2. OpConj[TT] conjugates Op[TT] (using the negative root). OpInv[TT] calculates the inverse of TT.

Op[TT_]:=(Simplify[Coefficient[TT,sqrdelta,0]+ delta Coefficient[TT,sqrdelta,2]+ delta^2 Coefficient[TT,sqrdelta,4]+delta^3 Coefficient[TT,sqrdelta,6]+delta^4 Coefficient[TT,sqrdelta,8]+delta^5 Coefficient[TT,sqrdelta,10]+delta^6 Coefficient[TT,sqrdelta,12]])+ sqrdelta(Simplify[Coefficient[TT,sqrdelta,1]+ delta Coefficient[TT,sqrdelta,3]+ delta^2 Coefficient[TT,sqrdelta,5]+delta^3 Coefficient[TT,sqrdelta,7]+delta^4 Coefficient[TT,sqrdelta,9]+delta^5Coefficient[TT,sqrdelta,11]]);
OpConj[TT_]:=(Simplify[Coefficient[TT,sqrdelta,0]+ delta Coefficient[TT,sqrdelta,2]+ delta^2 Coefficient[TT,sqrdelta,4]+delta^3 Coefficient[TT,sqrdelta,6]+delta^4 Coefficient[TT,sqrdelta,8]+delta^5 Coefficient[TT,sqrdelta,10]+delta^6 Coefficient[TT,sqrdelta,12]])-sqrdelta(Simplify[Coefficient[TT,sqrdelta,1]+ delta Coefficient[TT,sqrdelta,3]+ delta^2 Coefficient[TT,sqrdelta,5]+delta^3 Coefficient[TT,sqrdelta,7]+delta^4 Coefficient[TT,sqrdelta,9]+delta^5Coefficient[TT,sqrdelta,11]]);
OpNorm[TT_]:=Simplify[Op[Op[TT] * OpConj[TT]]];
OpInv[TT_]:=Simplify[OpConj[TT]/OpNorm[TT]];

Se have to compute e, f again, to write the square roots in terms of sqrdelta

e=e0+sqrdelta e1;f=f0+sqrdelta f1;Simplify[Solve[Flatten[{Coefficient[Op[{Tr[A]-Tr[Inv[A]],Tr[A]-Tr[A. B],Tr[B]-Tr[B.Inv[A]],Tr[Inv[B].A]-Tr[Inv[B]],Tr[Inv[A]]-Tr[Inv[A.B]]}],sqrdelta,0],Coefficient[Op[{Tr[A]-Tr[Inv[A]],Tr[A]-Tr[A. B],Tr[B]-Tr[B.Inv[A]],Tr[Inv[B].A]-Tr[Inv[B]],Tr[Inv[A]]-Tr[Inv[A.B]]}],sqrdelta,1]}]=={0,0,0,0,0,0,0,0,0,0},{e0,e1,f0,f1}]]

e=((2 (u^2) (-1+v)+v (-9-8 v+5 (v^2))-u v (2+5 v-4 (v^2)+(v^3)))/(2 (-3+v) (1+v) (-u+v)))+((-2+v)/(2 (-u+v)))sqrdelta;f=-((9-4 (u^2)+6 v-7 (v^2)+u (6+9 v-4 (v^2)+(v^3)))/(2 (-3+v) (1+v) (-u+v)))+(1/(2 u-2 v))sqrdelta;

Simplify[MatrixForm[A]]

Op[Simplify[{Tr[A]-Tr[Inv[A]],Tr[A]-Tr[A. B],Tr[B]-Tr[B.Inv[A]],Tr[Inv[B].A]-Tr[Inv[B]],Tr[Inv[A]]-Tr[Inv[A.B]]}]]

Now we need the matrix T that solves T.A=A.B.T and T.B=B.A.B.T. This is a linear problem in the coordinates of  T={{t01,t02,t03},{t04,t05,t06},{t07,t08,t09}}+sqrdelta {{t11,t12,t13},{t14,t15,t16},{t17,t18,t19}}. We give a solution and check that it satisfies the equations

T={{-sqrdelta (u-v) (((-3+v))^2) v (((1+v))^2) (-3+2 u (-2+v)-(-2+v) v)+(u-v) (-3+v) (((1+v))^2) (-3+2 u (-2+v)-(-2+v) v) (-6+v+3 (v^2)-u (8-3 (v^2)+(v^3))),(3+u (-2+v)-2 v) (u-v) (1+v) (-(((3+4 u))^2)+(-5+u) (3+4 u) v+10 (-1+(-1+u) u) (v^2)+(1+u) (13+u) (v^3)+(1-2 u (1+2 u)) (v^4)+(-1+u) u (v^5))+sqrdelta (3+u (-2+v)-2 v) (u-v) (-3+v) (1+v) (3+u (4+(-2+v) v (1+v))-v (2+v (2+v))),-sqrdelta (3+u (-2+v)-2 v) (u-v) (-3+v) (1+v) (3+4 u+v+2 u v+(-1+u) (v^2)-(2+u) (v^3)+(v^4))+(3+u (-2+v)-2 v) (u-v) (1+v) (-(((3+4 u))^2)+(12+u-20 (u^2)) v+(26+u (53+18 u)) (v^2)+(-17+(-18+u) u) (v^3)-(3+4 u (2+u)) (v^4)+(3+u (5+u)) (v^5)-u (v^6))},{sqrdelta (((-3+v))^2) (((1+v))^2) (8 u+(-1+v) (-6+v (-5+(-2+v) v)))+(((-3+v))^2) v (((1+v))^2) (9-8 (u^2)+15 v-(v^2) (5+3 v)+u (6+v (23+v (-3+(-3+v) v)))),sqrdelta (-3+v) (1+v) (18-15 v+8 (v^2)+5 (v^3)-16 (v^4)+4 (v^5)+(u^2) (32-28 v-14 (v^2)+21 (v^3)-8 (v^4)+(v^5))+u (48-41 v+(v^2)+30 (v^3)-20 (v^4)+7 (v^5)-(v^6)))+(-3+v) v (1+v) (27+9 (v^2)+28 (v^3)-22 (v^4)+2 (v^5)+(u^3) (16-12 v-14 (v^2)+21 (v^3)-8 (v^4)+(v^5))+u (81-18 v-30 (v^2)+73 (v^3)-37 (v^4)+7 (v^5))-(u^2) (-72+33 v+70 (v^2)-52 (v^3)+23 (v^4)-7 (v^5)+(v^6))),sqrdelta (-3+v) (1+v) (-9+3 v-25 (v^2)+3 (v^3)+15 (v^4)-8 (v^5)+(v^6)-(u^2) (16-36 v-6 (v^2)+21 (v^3)-8 (v^4)+(v^5))+u (-24+31 v-33 (v^2)-18 (v^3)+28 (v^4)-9 (v^5)+(v^6)))+(-3+v) (1+v) (27+9 v-3 (v^2)-25 (v^3)-23 (v^4)+26 (v^5)-7 (v^6)-(u^3) v (32-4 v-22 (v^2)+21 (v^3)-8 (v^4)+(v^5))+u (72+18 v+22 (v^2)-15 (v^3)-73 (v^4)+54 (v^5)-11 (v^6)+(v^7))+(u^2) (48-16 v+39 (v^2)+46 (v^3)-64 (v^4)+31 (v^5)-9 (v^6)+(v^7)))},{sqrdelta (((-3+v))^2) (((1+v))^2) (3-8 v-(v^2)+2 (v^3)-u (-4+10 v+(v^2)-4 (v^3)+(v^4)))+(((-3+v))^2) (((1+v))^2) (9+6 v-5 (v^2)-8 (v^3)+2 (v^4)+u (24+11 v-13 (v^2)-11 (v^3)+5 (v^4))-(u^2) (-16-4 v+10 (v^2)+(v^3)-4 (v^4)+(v^5))),sqrdelta (-3+v) (1+v) (9-39 v+7 (v^2)+22 (v^3)-14 (v^4)+3 (v^5)+(u^2) (28-34 v-5 (v^2)+26 (v^3)-13 (v^4)+2 (v^5))-u (-33+79 v+3 (v^2)-44 (v^3)+25 (v^4)-7 (v^5)+(v^6)))+(-3+v) (1+v) (27+27 v-63 (v^2)-20 (v^3)+40 (v^4)-17 (v^5)+2 (v^6)+(u^3) (16+12 v-34 (v^2)-5 (v^3)+26 (v^4)-13 (v^5)+2 (v^6))+u (81+108 v-138 (v^2)-73 (v^3)+97 (v^4)-37 (v^5)+6 (v^6))-(u^2) (-72-105 v+92 (v^2)+55 (v^3)-79 (v^4)+31 (v^5)-7 (v^6)+(v^7))),sqrdelta (-3+v) (1+v) (-18+v (6+v (1+v (17+v (8+v (-13+3 v)))))+(u^2) (-20+v (18+v (-11+v (-9+v (16+(-7+v) v)))))-u (39+v (-20+v (13+(-3+v) v (1+v (6+(-4+v) v))))))+(-3+v) (1+v) ((u^3) (16+(-3+v) v (-4+v (-2+v (3+(((-2+v))^2) v))))+v (-36+v (-33+v (-13+v (16+v (29+v (-17+2 v))))))+(u^2) (24-v (23+v (-7+(-2+v) v (3+v (-14+v (11+(-5+v) v))))))+u (9+v (-72+v (-43+2 v (-8+v (-3+v (28+v (-17+3 v))))))))}};MatrixForm[T]

Simplify[Op[T.A-A.B.T]]

Simplify[Op[T.B-B.A.B.T]]

Now that we have the matrices (A, B, T) we do computations with their traces

DetT=Simplify[Op[Det[Op[T]]]]

InvDetT=OpInv[DetT]

We find the equations involving y, \bar{y}, z, \bar{z} in the paper. They are y.\bar{y},y^3,\bar{y}^3, eta, z^3,\bar{z}^3,z.\bar{z},y.z,\bar{y}.\bar{z}

Op[Op[Tr[T] Tr[Inv[T]]]*InvDetT]
Op[Op[Op[Tr[T]]^3]*InvDetT]
Op[Op[Op[Tr[Inv[T]]]^3]*InvDetT^2]
Op[Tr[A.B.Inv[A].Inv[B]]]
Op[Op[Op[Tr[T. Inv[A]. T. A]]^3] *InvDetT^2]
Op[Op[Op[Tr[Inv[A].Op[Inv[T]].A .Op[Inv[T]]]]^3]*Op[InvDetT^4]]
Op[Op[Tr[T. Inv[A]. T. A]]*Op[Tr[Inv[A].Op[Inv[T]].A .Op[Inv[T]]]]*InvDetT^2]
Op[Op[Op[Tr[T]]*Op[Tr[T. Inv[A]. T. A]]]*InvDetT]
Op[Op[Op[Tr[Inv[T]]]*Op[Tr[Inv[A].Op[Inv[T]].A .Op[Inv[T]]]]*InvDetT^2]]