SetDirectory["~/KappaLib/"];
<<kappaLib-1.2.m
<<helper.m

Define Metaclass I

kappa=emMatrixToKappa[(({{a1, 0, 0, -b1, 0, 0}, {0, a2, 0, 0, -b2, 0}, {0, 0, a3, 0, 0, -b3}, {b1, 0, 0, a1, 0, 0}, {0, b2, 0, 0, a2, 0}, {0, 0, b3, 0, 0, a3}}))];
kappa=kappa//.{b3->b2,a3->a2};

Let us first observe that the Hodge star operator can be rewritten using the emQMedium routine. This simplifies the expressions as we can specify the inverse of g and keep Sqrt(Abs(Det(g))) as a symbolic variable

Metric=emMatrix["g",4,Structure->"Symmetric"];
hodge=emHodge[Metric];
hodgeAlt=emQMedium[Sqrt[Abs[Det[Metric]]],Inverse[Metric]];

Union[Flatten[hodge-hodgeAlt]]

Case: a1 = a2

Define bivector A and metric g

Abivector=(({{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, -1, 0}}));
Metric=Inverse[DiagonalMatrix[{1,-1,-Psi/b2,-Psi/b2}]];

Check that metric has Lorentz signature

Det[Metric]

Formulate equations that should be satisfied

kappaAlt=C1 emQMedium[SqrtAbsDetG,Inverse[Metric]] + emBiProduct[rho,Abivector, Abivector] +C2 emIdentityKappa[];
eqs=Union[Flatten[FullSimplify[kappa-kappaAlt]]];

Show that the equations are satisfied for a suitable choice of parameters:

sub={
a2->a1,
C1->-1/SqrtAbsDetG  ((b2^2)/Psi) ,
C2->a2,
rho->1/2(b2^2-b1^2)/b1,
Psi->b2^2/b1
};
show[simp[eqs//.sub]]

Case: a1 != a2

Define A and metric g

Abivector=(({{0, SqrtAbsXi, 0, 0}, {-SqrtAbsXi, 0, 0, 0}, {0, 0, 0, ((a1-a2)/(2 rho SqrtAbsXi))}, {0, 0, -((a1-a2)/(2 rho SqrtAbsXi)), 0}}));
Metric=Inverse[DiagonalMatrix[{1,-1,-Psi/b2,-Psi/b2}]];

Formulate equations that should be satisfied

kappaAlt=C1 emQMedium[SqrtAbsDetG,Inverse[Metric]] + emBiProduct[rho,Abivector, Abivector] +C2 emIdentityKappa[];
eqs=Union[Flatten[FullSimplify[kappa-kappaAlt]]];

Show that the equations are satisfied for a suitable choice of parameters:

sub={
C1->-(1/SqrtAbsDetG) b2^2 /Psi,
C2->a2,
rho->SgnXi
};
show[simp[eqs//.sub]]

Since Sgn(b1)=Sgn(Psi) the last two equations are equivalent with

         Xi = 1/2( beta1 - beta2^2/beta1 )
         
         -beta1 + Psi - (alpha1 - alpha2)^2/(2Xi) = 0

The first of these is the definition of Xi

The latter holds since Psi satisfies 

        1/beta2 Psi^2 - D3 Psi + beta2 = 0
        
where D3 is defined as above.