Pensieve Header: An attempt on the ribbon property using Γ-calculus; continues RibbonPropertyV1. Success at last!!!

dir=SetDirectory["C:/drorbn/AcademicPensieve/2014-06/"];
<< KnotTheory`
<< MetaCalculi/MetaCalculi-Program.m
Format[(Subscript[α, a_,b_]), StandardForm]:= Interpretation[(Subscript[α, 10a+b]), (Subscript[α, a,b])];
Format[(Subscript[β, a_,b_]), StandardForm]:= Interpretation[(Subscript[β, 10a+b]), (Subscript[β, a,b])];

(Subscript[γ, 1]):  ((Subscript[γ, 0]) is the same, with all orientations going up)

{n=4; γ0=Γ[ω, (∑_(a=0)^n)(Subscript[h, a])(Subscript[σ, a]), (∑_(a=1)^n)(∑_(b=1)^n)(Subscript[t, a])(Subscript[h, b])(Subscript[α, a,b])], γ1= γ0 // dS[2] // dS[4]}

{Ov = Xp[o1,1]Xp[o2,2]Xp[o3,3]Xp[o4,4] // Γ // dm[o1,o2,o] // dm[o,o3,o] // dm[o,o4,o],
t1=Ov**(γ0*Γ[ϵ[o]]), t2=(γ0*Γ[ϵ[o]])**Ov, ocond=Simplify[t1==t2]}

ocond=FullSimplify[ocond /. {(Subscript[T, o])-> 0}]

{U=Xm[1,u1]Xm[2,u2]Xm[3,u3]Xm[4,u4] // Γ // dm[u1,u2,u]//dm[u,u3,u]//dm[u,u4,u],
t1=U**(γ0*Γ[ϵ[u]]), t2=(γ0*Γ[ϵ[u]])**U, ucond=FullSimplify[t1==t2]}

{cert = γ1 // dm[1,2,1] // dm[3,4,2], eqns=(ϵ[1]ϵ[2] // Γ)==(cert /. (Subscript[σ, _])->1) // Simplify}

γ1 // dm[2,1,1] // dm[4,3,3]

alex= FullSimplify[(γ1 // dm[4,3,3] // dm[3,2,2] // dm[2,1,1])@ω]

{FullSimplify[alex, eqns], FullSimplify[cert@ω], FullSimplify[alex, eqns&&ucond&&ocond]}

Simplify[(alex/(cert@ω))]

Unitarity of (Subscript[γ, 0])

Ωc[n_] :=  Table[Which[i<j, 0,i==j, (1/(1-(Subscript[T, i]))),i>j, 1], {i,n}, {j,n}];
Ωc[4] // MatrixForm

Transpose[γ0[A]].Ωc[4].(γ0[A] /. α->β) // MatrixForm

unita1=Simplify[And@@Thread[
Flatten[Transpose[γ0[A]].Ωc[4].(γ0[A] /. α->β)]==Flatten[Ωc[4]]
]]

unita2=Simplify[unita1 /. {(Subscript[T, 1|3])->T, (Subscript[T, 2|4])->1/T}]

unitarule=Thread[
Flatten[Transpose[γ0[A]]] -> Flatten[Simplify[
Ωc[4].Inverse[Ωc[4].(γ0[A] /. α->β)] /. {(Subscript[T, 1|3])->T, (Subscript[T, 2|4])->1/T}
]]
]

ans1=Simplify[((ω/((Subscript[σ, 2]) (Subscript[σ, 4]) ))/(cert@ω /. unitarule)), eqns&&ucond&&ocond /. α->β]

ans2=Simplify[((Subscript[σ, 2]) (Subscript[σ, 4]) )alex/ω, eqns&&ucond&&ocond]

ans3=((Subscript[α, 24])-(Subscript[α, 2,3]))((Subscript[α, 41])-(Subscript[α, 42]))+((Subscript[α, 21])-(Subscript[α, 22])) ((Subscript[α, 43])-(Subscript[α, 44]))

Simplify[{ans1==ans2 /. β->α, ans2==ans3}]