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 :math:`\mathbb F_2^r`,
write :math:`c_\lambda=\sum_j\lambda_jc_j`. Define the weight on redundant
representatives

.. math::

   \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

.. math::

   \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 :math:`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 :math:`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 :math:`q_I=\sum_{e\in S_I}\sigma_e` and separate the action into

.. math::

   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

.. math::

   \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**
:math:`(-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
:math:`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:

.. math::

   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:

.. math::

   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 :math:`\mathbb E[\overline O^2]` generally differs from
:math:`\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.