TFPT scientific source excerpt
Original repository path: experiments/theory-contracts/original-admissible-source-closure-20261025/QUANTUM-OCTIC-COMPLETION.txt
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000038 | 2. Why the bare octic does not answer that operator question
000039 | -----------------------------------------------------------
000040 | In the original isotropic Gaussian four the normalized state is
000041 | 
000042 |   F8hat = I8(h_+(-1)) Omega/(96 sqrt140),
000043 |   I8(z) = sum z_j^8 +14 sum_(j<k) z_j^4 z_k^4 +168 product z_j^2.
000044 | 
000045 | It remains a valid neutral even-CAR composite, with the previously proved
000046 | original source dictionary and stress responses. Its noninvariance below
000047 | does not invalidate those results or declare an unproved E8 gauge symmetry
000048 | to be a physical requirement.
000049 | 
000050 | For any E8 root alpha, write b_j=(alpha_(2j)-i alpha_(2j+1))/sqrt2 in the
000051 | standard Gaussian chart. Direct substitution gives I8(b)=1 for every root.
000052 | The root zero-mode action follows from the usual vertex shift formula:
000053 | 
000054 |   e_alpha(0) f(h(-1))Omega
000055 |     = [z^-1] exp(sum_(n>=1) alpha(-n)z^n/n)
000056 |                     f(h(-1)-alpha/z)e^alpha.
000057 | 
000058 | Its oscillator-partition (7) coefficient on F8hat is
000059 | 
000060 |   alpha(-7)e^alpha/(672 sqrt140),
000061 | 
000062 | which is nonzero. Thus equality of the classical degree-eight polynomials
000063 | does not identify the bare F8hat with a whole E8 zero-mode invariant field.
000064 |