This notebook computes the equations of the non - distinguished component V_1

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 A in diagonal form. Matrix B generic, with non - zero entries (2, 1) and (3, 1)

B={{a,b,c},{1,d,e},{1,f,g}};A={{1,0,0},{0,I,0},{0,0,-I}};

Impose equations to have a representation of the group \Gamma

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

b=((1/4)+(I/4)) (1+(a^2)+2 e f);c=((1/4)-(I/4)) (1+(a^2)+2 e f);d=((1/2)+(I/2)) (-1+a);g=((1/2)-(I/2)) (-1+a);

We use as variables (u, v), which are (\bsqrdelta, \bar{\bsqrdelta}) in the paper

Solve[{Tr[B],Tr[Inv[B]]}=={u,v},{a,e}]

a=((1+u)/2);e=((-3-2 u+(u^2)-4 v)/(8 f));

Finally impose the condition on the determinant of B

Solve[Det[B]==1,f]

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

delta=-7-4 (u^3)+(u^2) (((-1+v))^2)+2 v+(v^2)-4 (v^3)-2 u (-1-6 v+(v^2));
f=((((1/4)+(I/4)) (5-u-(u^2)+(u^3)-4 u v-2I sqrdelta))/(1+(u^2)-2 v));

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]];

e=Simplify[OpInv[8f] (-3-2 u+(u^2)-4 v)]

MatrixForm[Op[B]]

Now we find the matrix T that conjugates (A, B) to (A.B, B.A.B). For this we find the eigenvectors of A.B with eigenvalues 1, I, -I

Solve[{{1,0,0},{0,1,0}}.Op[(A.B- IdentityMatrix[3])].{{x},{y},{1}}=={{0},{0}},{x,y}]

Solve[{{1,0,0},{0,1,0}}.Op[(A.B-I IdentityMatrix[3])].{{x},{y},{1}}=={{0},{0}},{x,y}]

Solve[{{1,0,0},{0,1,0}}.Op[(A.B+I IdentityMatrix[3])].{{x},{y},{1}}=={{0},{0}},{x,y}]

The general form of the matrix T is :

T=Transpose[{{((1/4)+(I/4)) ((-2+I)-I sqrdelta+u+(1+I) (u^2)-(1+2 I) v+u v)(1+(u^2)-2 v),-(3+I sqrdelta-3 u-I sqrdelta u+(u^2)+(u^3)-2 v-2 u v+2 (v^2)),(1+(u^2)-2 v) ((-1+I)+u-I v)},{((1/4)+(I/4)) ((1+4 I)-sqrdelta-I u+(1+I) (u^2)-(2+3 I) v-I u v)(1+(u^2)-2 v) ,-((5+2 I)-(1-2 I) sqrdelta-(1-2 I) u-sqrdelta u-(1+2 I) (u^2)+(u^3)+2 I v-(4+2 I) u v-2 I (v^2)),(1+(u^2)-2 v) (-4 I+u-v)},{((1/4)+(I/4)) ((1-2 I)+sqrdelta+I u+(1+I) (u^2)-(2+I) v+I u v)(1+(u^2)-2 v),-((5-2 I)+(1+2 I) sqrdelta-(1+2 I) u+sqrdelta u-(1-2 I) (u^2)+(u^3)-2 I v-(4-2 I) u v+2 I (v^2)),(1+(u^2)-2 v) (2+u+v)}}].{{h,0,0},{0,k,0},{0,0,j}};

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

To compute h, j, k we normalize k = 1, and recall that h and j are polynomials in u, v and sqrdelta

h=h0+ sqrdelta h1;j=j0+ sqrdelta j1; k=1;Solve[Flatten[{Coefficient[Op[T.B-B.A.B.T],sqrdelta,0],Coefficient[Op[T.B-B.A.B.T],sqrdelta,1]}]=={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{h0,h1,j0,j1}]

We substitute and clear denominators

h0=2 I ((1+4 I)-I u+(1+I) (u^2)-(2+3 I) v-I u v)(2+u+v);h1=-2 I(2+u+v);j0=-((-2+5 I)-(2+I) u+(2-I) (u^2)+I (u^3)-2 v+(2-4 I) u v+2 (v^2))((-1+I)+u-I v);j1=I ((1-2 I)+u)((-1+I)+u-I v);k=(1+(u^2)-2 v) ((-1+I)+u-I v)(2+u+v);

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

MatrixForm[Op[T]]

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

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

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 / \bar {y} and \bar {z} / y

Simplify[Op[Op[Tr[T] Tr[Inv[T]]]*OpInv[DetT]]]
Simplify[Op[Op[Op[Tr[T]]^3]*OpInv[DetT]]]
Simplify[Op[Op[Op[Tr[Inv[T]]]^3]*OpInv[DetT]^2]]
Simplify[Op[Tr[A.B.Inv[A].Inv[B]]]]
Simplify[Op[Op[Tr[T .Inv[A] .T .A]]*OpInv[Tr[Inv[T]]]]]
Simplify[Op[Op[Tr[Inv[A].Inv[T].A .Inv[T]] *OpInv[Tr[T]]*OpInv[DetT]]]]