Exact spatial logical-bit marginalization

The higher-degree production engine integrates selected logical shifts analytically. It still samples the remaining spacetime background with Markov updates. This page specifies its target and physical estimators.

An orbit identity

Let w(phi) be the original positive finite-M weight. Let c1,…,cr be independent binary cocycles acting by addition on phi. For lambda in \(\mathbb F_2^r\), write \(c_\lambda=\sum_j\lambda_jc_j\). Define the weight on redundant representatives

\[\widetilde w(\phi)=2^{-r}\sum_\lambda w(\phi+c_\lambda),\qquad \overline O(\phi)= \frac{\sum_\lambda w(\phi+c_\lambda)O(\phi+c_\lambda)} {\sum_\lambda w(\phi+c_\lambda)}.\]

Each shift permutes the full configuration set, so

\[\sum_\phi\widetilde w(\phi)=\sum_\phi w(\phi),\qquad \frac{\sum_\phi\widetilde w(\phi)\overline O(\phi)} {\sum_\phi\widetilde w(\phi)} =\frac{\sum_\phi w(\phi)O(\phi)}{\sum_\phi w(\phi)}.\]

This is a finite-sum proof. It needs neither an equal-weight physical-sector assumption nor canonical representative selection. The sampled phi remains a redundant representative. Using raw O(phi) with this target would generally measure the wrong expectation.

Coordinate sheets and factorization

On the spatial d-torus there are \(r=\binom dp\) coordinate logical cocycles. For each spatial orientation I, its sheet S_I consists of all spacetime p-cells of orientation I with \(x_i=0\) for every i in I, arbitrary transverse coordinates and every time slice. Shifting that sheet has zero coboundary. Different sheets have different orientations and are disjoint sets of spins.

Let \(q_I=\sum_{e\in S_I}\sigma_e\) and separate the action into

\[S(\phi)=S_{\rm inv}(\phi)+h_s\sum_Iq_I(\phi).\]

The invariant part contains all face terms, all temporal-spin terms and the spatial-spin terms outside the sheets. A sheet shift changes q_I to -q_I. Consequently

\[\widetilde w(\phi)=C\,e^{S_{\rm inv}(\phi)} \prod_I\cosh(h_s q_I),\qquad \lambda_I^{\rm mean}=\tanh(h_s q_I).\]

The final symbol denotes the conditional mean of the sign \((-1)^{\lambda_I}\), not the conditional mean of the bit lambda_I. Conditionally on phi, the signs are independent with probabilities proportional to exp(h_s*q_I*sign). Physical logical shifts need not have equal weights.

The implemented action is \(S_{\rm inv}+\sum_I\log[2\cosh(h_sq_I)]\). It differs from log(tilde_w/C) by the constant r*log(2), which cancels in Metropolis ratios. When reconstructing the normalized full partition weight, subtract that constant. The six sheets in d=4,p=2 give six integer sums; the factorization avoids an explicit 64-configuration sum on every proposal.

Characters and joint estimates

A supported sign observable transforms by a binary logical character chi:

\[O(\phi+c_\lambda)=O(\phi)(-1)^{\chi\cdot\lambda},\qquad \overline O(\phi)=O(\phi)\prod_{I:\chi_I=1}\tanh(h_sq_I).\]

For a spatial support, chi_I is its mod-two overlap with the corresponding sheet on the measurement slice. A deterministic decoded observable has the same character as its raw support, because the syndrome and correction rule are shift invariant.

For A and B, the compiler uses character XOR for the product:

\[a=\overline A,\quad b=\overline B,\quad d=\overline{AB}=A(\phi)B(\phi) \prod_{I:(\chi_A\mathbin\oplus\chi_B)_I=1}\tanh(h_sq_I).\]

It then estimates f=mean(d)-mean(a)*mean(b). At h_s=0, a nontrivial character has conditional mean zero. If A=B is a sign with such a character, a=b=0 but d=1 exactly. Setting d=a*b would replace a known joint operator by the wrong quantity. Likewise \(\mathbb E[\overline O^2]\) generally differs from \(\mathbb E[\overline{O^2}]\). Use the conditional estimator’s variation for Monte Carlo uncertainty and the actual product insertion for physical second moments.

Scope of the marginalization

Only the spatial coordinate logical shifts are integrated. This is not a sum over equally weighted boundary twists, nor a computation of all separate sector partition functions. Temporal/background degrees of freedom remain in the Markov chain. Exact logical averaging does not prove useful finite-time mixing for them.

At zero longitudinal source, finite-volume symmetry can force a noncontractible Wilson mean to vanish even when a different paired/decoded observable is useful for an ordering question. Interpret the chosen observable and limiting procedure before translating a vanishing mean into a phase statement.