Product probabilities restrict reversible generators
The first technical step of de la Torre et al.’s Theorem 1 is valid: two-sided reversible evolution and product-test positivity force every local matrix slice of its generator to have a scalar diagonal, symmetric time-space entries and an antisymmetric spatial off-diagonal part. Thus the generator belongs to the tensor power of a seven-dimensional matrix space. This is an expanded proof of an established source step, not a new reconstruction result.
C129, Q01 acceptance audit, 2026-09-14. B77 review. Source: B70 original, PDF pp. 3–4, equations (4), (6), (8)–(14). The written argument below fixes an explicit row/column convention and retains the Taylor remainder.
Object and assumptions
Fix a finite number n of local three-dimensional unit balls. Put u(a)=(1,a) and v(a_1,…,a_n)=u(a_1) tensor … tensor u(a_n) in R(4n). A product preparation has coordinate v(a), and a product effect has coordinate 2^(-n)v(b). All unit Bloch vectors in each factor are available independently. The coordinates and probabilities are dimensionless; no volume or continuum limit is taken.
Let X be a real 4^n by 4^n matrix such that exp(tX) is an allowed reversible transformation for every t in a two-sided neighbourhood of zero. Assume
\[p_{b,a}(t)=2^{-n}v(b)^T e^{tX}v(a)\in[0,1].\]
These assumptions follow for a one-parameter subgroup of the source’s matrix Lie group. We need only these local one-parameter curves, not a claim that every element of an arbitrary connected group is a single exponential. If t is assigned physical time, X has inverse-time units; the argument itself makes no physical clock or energy/action identification.
Boundary derivatives, with the remainder retained
The matrix exponential gives the exact expansion
\[p(t)=p(0)+2^{-n}t\,v(b)^TXv(a) +2^{-n}\frac{t^2}{2}v(b)^TX^2v(a)+O(t^3).\]
Take unit vectors and b_1=-a_1, leaving the remaining b_k and a_k independent. Since u(-a_1)^T u(a_1)=0, p(0)=0. A differentiable nonnegative function on a two-sided neighbourhood has p’(0)=0 and p’’(0)>=0. Therefore
\[v(-a_1,b_2,\ldots,b_n)^TXv(a_1,\ldots,a_n)=0,\] \[v(-a_1,b_2,\ldots,b_n)^TX^2v(a_1,\ldots,a_n)\geq0.\]
With b_k=a_k for every k, p(0)=2^(-n) product_k(1+|a_k|^2)=1. The maximum instead gives p’(0)=0 and
\[v(a)^TX^2v(a)\leq0.\]
These are source equations (10)–(12). Equation (9) is used as an asymptotic expansion, not an exact interval bound on a truncated polynomial. The sign of the second derivative is necessary even when it vanishes; it is not a sufficient criterion for positivity at finite t. For a forward semigroup alone a boundary minimum gives only a one-sided first-derivative inequality, so this inference specifically uses reversibility.
From the vanishing derivative to local matrix blocks
Use row = output and column = input. Fix all spectator output and input indices. The first-order identity implies, for the resulting 4 by 4 slice M,
\[u(-a)^T M u(a)=0\qquad (|a|=1).\]
To justify fixing these indices, u(e_i)+u(-e_i)=2e_0 and u(e_i)-u(-e_i)=2e_i show that unit-vector preparations span R^4. Their tensor products span the spectator space, independently for input and output. The bilinear identity therefore vanishes coefficient by coefficient. This is a linear-span argument; the signed combinations need not be physical states.
Write
\[M=\begin{pmatrix}c&r^T\\s&D\end{pmatrix}.\]
Then the identity becomes c+(r-s) dot a-a^T D a=0. Subtraction at a and -a gives r=s. Evaluation at each e_i gives D_ii=c. Evaluation at (e_i+e_j)/sqrt(2), i unequal j, gives D_ij+D_ji=0. Consequently
\[M=cI_4+\begin{pmatrix}0&r^T\\r&K\end{pmatrix}, \qquad K^T=-K.\]
Conversely every such block satisfies the identity. These are exactly the entry constraints (13)–(14), independent of the source’s index-placement convention. With A the three-dimensional space having only an antisymmetric spatial block, B the three-dimensional space with equal time-space blocks, and I the span of I_4, the slice belongs to L=A direct-sum B direct-sum I. Its dimension is 3+3+1=7. Source equation (6) supplies a basis of A and B.
The same proof works at every site because an orthogonal input/effect pair can be chosen in any factor. In End(R4)(tensor n), the constraints for site k put X in End(R4)(tensor(k-1)) tensor L tensor End(R4)(tensor(n-k)). Choose any vector-space complement of L and expand in a basis adapted to L plus that complement. Intersecting these n conditions removes every tensor coefficient with a complement factor. Hence
\[X\in L^{\otimes n}=(\mathcal A\oplus\mathcal B\oplus\mathcal I)^{\otimes n}.\]
This completes the selected step. L^(tensor n) is a necessary ambient linear space, not by itself the Lie algebra of an admissible group. In particular the second-order conditions, normalization and further group constraints still restrict which elements are actual generators.
Dependencies and strategic consequence
The proof imports finite-dimensional matrix exponentiation, the elementary calculus test at an interior extremum, spanning by product tensors, and a basis/complement intersection argument. Local SO(3) containment, compact-group averaging, entangling-gate universality, identical-copy extension and ancilla/discard closure are not needed for this step. Later portions of the source theorem use additional assumptions and arguments.
Accept equations (10)–(14) and their tensor-space consequence under the stated hypotheses. The next named proof dependency is whether local-rotation averaging isolates a nonzero generator as claimed in equations (15)–(17). It remains supporting until STATE selects it. The minimal-composite exclusion still relies on the rest of Theorem 1; neither its full proof nor the physical origin of reversible descent or a positive universal action constant is established here.