Observation boundaries for destructive quantum readout and external geometry

We specify an observation boundary for a finite linearly typed quantum interface with destructive computational-basis readout and primitive classical geometric labels. Geometric rules inspect declared labels; coupling rules inspect declared matter observables; control and resource checks are part of the interface. Under spectator-blind control, arbitrary extensions with the same accessible reduced state preserve geometric traces and accessible matter states. A restricted coherent implementation of destructive readout agrees with immediate readout on their common record–matter output.

For a channel $T$, coarse effects $E_w$, and aggregated fine effects ${\Pi }_w$, record compatibility is equivalent to $T^{\ast }(E_w)={\Pi }_w$. The exact unassisted worst-case total-variation error is a maximum of operator norms over outcome events. Exact record compatibility extends to untouched entangled references, but does not determine residual instruments. For sharp measurements and a coarse Lüders instrument, full compatibility is characterized exactly by the removal of fine coherence within each coarse outcome. A compositional criterion identifies when continuations preserve agreement on a declared observable algebra.

One complete four-face adjacency exchange illustrates the interface. It is induced by a prescribed slicing of a fixed conical spacetime and has explicit charts, edge laws, a compact collar, a transition stratum, an exact guard, location transport, and a decoder-correct finite reversible implementation. The general interface results are independent of this example. A separate dictionary proves three linear observation rules for distributional comparisons without assuming any particular renormalized composite.

Figure fig:split-VII-E records one view of the same separation. Figure fig:chain-VII-E records one view of the same separation. Figure fig:pair-VII-E records one view of the same separation. Figure fig:gate-VII-E records one view of the same separation. The drawings add no hypothesis.

Figure fig:scope-VII-E records that separation. The drawing adds no hypothesis.

Destructive measurement, a coherent record carrier, and a measurement instrument with a surviving quantum output are different interfaces. A geometric extension of a quantum language must say which of these interfaces it can observe. Linear ownership of quantum names alone does not answer that question.

We take destructive computational-basis readout as a given primitive, with type

Its measured quantum interface is consumed. We use its Born semantics, as in the corresponding finite circuit fragment of the Enter calculus Cheung2026Enter. No characterization or uniqueness proof for the measurement law is repeated here.

The semantic additions are fixed throughout: an external classical geometric configuration, locations for active names, declared coupling algebras, and an explicit control and resource interface. The question is then precise:

Which spectator extensions, deferred implementations, and scale/readout comparisons preserve the outputs that the interface declares accessible?

Our answers concern equality of observable outputs, not a mechanism that constructs geometry from quantum syntax.

The paper proves six interface statements.

1. On a prescribed substrate, block channels have an exact finite-step backwards support bound. Retiming a fixed labelled gate partial order preserves the total channel.

2. Under spectator-blind control and resources, any admissible spectator extension with the same accessible reduction preserves the geometric trace and that reduction at every step.

3. Immediate destructive readout agrees with a restricted coherent-then-readout implementation after both outputs are restricted to the same record–matter interface.

4. Record laws across evolution and label aggregation agree exactly when their effects obey a pullback identity. Their unassisted worst-case total-variation error has an exact event-norm formula.

5. Record agreement does not imply residual-instrument agreement. For sharp measurements with a coarse Lüders instrument, an exact criterion isolates the remaining intra-outcome coherence.

6. Agreement on an input observable algebra is preserved as agreement on an output observable algebra exactly when the adjoint continuation maps the latter into the former. A corresponding statement is given for classical block-diagonal state spaces.

Finite-dimensional channel conventions follow Watrous Watrous2018. The circuit principle of deferred measurement has a precise treatment in Gurevich and Blass GurevichBlass2022. Our emphasis is the additional observation boundary: geometry, control, resource tests, and accessible residual observables are all named.

The geometric certificate uses a fixed conical spacetime and the polygonal-slicing setting of DeserJackiwHooft1984,tHooft1992,tHooft1993,Kadar2005. The proofs of the interface theorems do not depend on that certificate. The certificate shows that one nontrivial change of cell incidence can meet the declared interface; it does not supply the assumptions of the interface theorems.

Within the proposed model, a future gravitational update may inspect only its declared observable interface. Destroyed names, inaccessible carriers, and resource bookkeeping are not undeclared geometric source variables. This is an operational requirement for a later quantum–geometric composite, not a derivation of its dynamics.

In particular, the paper proves no gravitational field equation, continuum limit, curvature-convergence theorem, physical causal dominance, or interacting-product theorem. It does not evaluate a singular triangle graph or establish a distributional finite part for one. The short distributional dictionary in sec:dictionary contains conditional linear inference rules only.

The distinction is structural:

Likewise, a scheduled circuit past is not a physical light cone. A computational implementation requires an additional physical model before such identifications can be tested.

For a finite-dimensional Hilbert space $\mathcal{H}$, write $L(\mathcal{H})$ for its operators and $D(\mathcal{H})$ for its density operators. A channel is a completely positive, trace-preserving linear map. Its adjoint satisfies

The adjoint of a channel is unital and completely positive. A Kraus representation is

These are the standard finite-dimensional conventions Watrous2018.

A classical register has a distinguished orthonormal basis. Admissible joint states are block diagonal in that basis, including their correlations with other registers. A diagonal classical marginal alone is insufficient. Operations declared to preserve classicality preserve this joint block-diagonal state space.

A unital observable algebra means a unital ${}^{\ast }$-subalgebra $\mathcal{A}\subseteq L(\mathcal{H})$. Its accessible state functional is

Agreement on $\mathcal{A}$ means equality of these functionals.

Our norm and distance conventions are

The operator norm is $\Vert \cdot {\Vert }_{\infty }$. No ancillary reference is included in an optimization unless explicitly stated.

Wire types are generated by

A judgement $f:A\to B$ consumes the complete input interface and returns the declared output interface. Composition and tensoring obey

Identities and wire permutations are available. Circuit constants used below include quantum unitaries, preparation of $|0⟩$, and destructive computational-basis readout. The finite classical-code implementation also uses reversible NOT, CNOT, and Toffoli operations on classical interfaces. Their quantum-carrier realizations are the corresponding permutation unitaries.

Linearity is an ownership condition. It permits a sequence of gates to consume and return the same physical tensor factor. It provides neither duplication of an arbitrary quantum wire nor an implicit discard operation.

The semantic matter-rule class below permits declared channels. We do not claim that every such channel has an exact implementation in the displayed finite gate library. Exact gate realizability is proved only for the finite reversible event code in sec:event-code.

Let $S$ have computational basis $\{|z⟩:z\in Z\}$, and let $B$ be its unmeasured complement. Put

The interpretation of ${\mathsf{Enter}}_S$ is the channel

The record $R$ is classical. The factor $S$ is absent from the output.

A fresh quantum carrier $A$, with basis indexed by $Z$, is introduced by the isometry

The corresponding channel is ${\mathsf{Copy}}_S(\rho )=V\rho V^{†}$. It returns both $S$ and $A$.

The represented computational-basis instrument is

Its residual output is on $S\otimes B$. It is not the destructive primitive.

Coherent recording is not cloning. For a qubit,

The three interfaces must therefore remain distinct both in theorem statements and in geometric couplings.

An external geometric interface supplies:

1. a set $\mathcal{Q}$ of classical geometric and declared resource labels;

2. an admissible subset

   $${\mathcal{Q}}_{\mathrm{adm}}={\mathcal{Q}}_{\mathrm{reg}}\cup {\mathcal{Q}}_{\mathrm{tr}};$$

3. a finite cell-label set $\mathrm{Cell}(g)$ for each admissible $g$;

4. support, boundary, and resource certificates for permitted updates;

5. an explicitly supplied gauge equivalence, when one is used.

Regular configurations satisfy the model's nondegeneracy conditions. A transition configuration is admitted only with an event certificate specifying its realization and its permitted one-sided continuations.

In the example, positive-length edges belong to regular cellulations. At the event, a vanishing edge slot and coincident endpoint labels may remain in a germ description. Their realization is a single junction; they are not regular embedded one-cells.

A configuration is

where $g\in {\mathcal{Q}}_{\mathrm{adm}}$, $\Gamma$ is a finite labelled matter interface, $\rho \in D({\mathcal{H}}_{\Gamma })$, and $h$ is a finite control state. Classical names satisfy the joint classicality convention. The geometric projection is ${\pi }_g(C)=g$.

A location is an abstract cell label, not an undeclared developed position or a stress-energy assignment. Input and output matter interfaces may change. For example, destructive readout replaces measured names with fresh classical record names.

A finite execution uses three kinds of certified rule.

1. A geometric rule, or $G$-rule, changes supported geometric labels as a specified function of geometric and declared resource labels. It does not inspect the matter density operator or matter-name cardinality. It returns a complete admissible configuration of geometric labels and a location transport map.

   A transition rule includes its common event realization and both one-sided continuations.

2. A matter rule applies a declared channel

   $${\Phi }_g:L({\mathcal{H}}_{{\Gamma }_0})⟶L({\mathcal{H}}_{{{\Gamma }^{\prime }}_0})$$

   on a declared interface, extended by the identity on unacted-on names. It may depend on the current geometric labels but does not change them. Its output-name locations are declared.

3. A coupling rule declares an accessible unital algebra ${\mathcal{A}}_v$ and updates geometry by

   $$g^{\prime }=F_v(g,{\omega }_{\rho }{|}_{{\mathcal{A}}_v}).$$

   It leaves the matter density operator unchanged and supplies a complete admissibility certificate and location transport.

A coupling update must return all affected labels needed for admissibility. Changing a particle parameter while silently freezing incompatible gluing data is not a certified update.

The control state, guards, and resource checks determine which rule is selected. Our executions are nonselective density-operator executions. A selectively normalized measurement branch is not silently treated as a trace-preserving matter rule. Classical control can instead be represented inside a channel that retains its records.

For the compared executions, equal geometric labels, equal finite control states, and equal declared accessible state functionals select the same:

1. rule and matter channel;

2. applicability and capacity-check result;

3. support and resource certificate;

4. transport of original names and successor control state.

Adding inaccessible spectator names does not change these selections. Spectator storage and required transport history are reserved separately from the geometric-update budget. Random external control is compared under a common seed or an explicitly coupled control law.

This is a hypothesis about the complete controller, not merely about the written coupling formula.

The map in eq:coupling need not be affine in $\rho$. Consequently the complete coupled input–output map need not be a quantum channel. The proofs below compare equal accessible inputs step by step. They do not establish general no-signalling or a physical consistency theorem for arbitrary nonlinear backreaction.

A gauge equivalence is an explicitly supplied equivalence relation on descriptions. An admissible geometric observable is constant on the specified classes. Chart changes and retriangulations may be declared gauge when they represent the same metric and marked slice.

Equality of an ambient spacetime does not imply equality of every observable of different marked slices. Likewise, equality of total channels under a prescribed circuit retiming is not automatically a gauge symmetry of coupled geometric evolution.

All geometry, partitions, and channel lists in this section are prescribed. No state-dependent controller changes them.

Let $(X,d)$ be a finite metric space, with site Hilbert spaces ${\mathcal{H}}_x$. For $Y\subseteq X$, set

At tick $t$, let ${\mathcal{B}}_t$ be a partition of $X$, and suppose

where every block map is a channel. If ticks have duration $\tau >0$, the resulting scalar schedule bound is $\Delta /\tau$.

Define the backwards sets recursively by

For a block channel $\Phi ={⨂}_{B\in \mathcal{B}}{\Phi }_B$,

Proof. The complement of the displayed union is a union of blocks on which the input observable is the identity. Each corresponding adjoint maps the identity to itself. The other blocks can enlarge support only within their own union.

For $A\in \mathcal{A}(Y)$ and $0\le s\le t$,

and

The output reduction on $Y$ depends on the state at tick $s$ only through its reduction to $N_{s,t}(Y)$.

Proof. Repeated application of lem:one-layer in reverse tick order proves eq:support-bound.

If $x\in N_{s,t}(Y)$, its block meets $N_{s+1,t}(Y)$ at some $y$. Hence

Induction proves eq:metric-bound.

Finally, every output observable on $Y$ pulls back to an observable on $N_{s,t}(Y)$. Equal input reductions there give equal expectations for all output observables, and therefore equal output reductions.

A prescribed readout on $Y$ at tick $t$ depends only on the initial reduction to $N_{0,t}(Y)$. Replacing a prescribed block channel outside its backwards dependency region does not change its law, provided the supports and schedule remain fixed.

Proof. Apply thm:locality to the readout effects. A replacement outside the backwards region acts on an identity factor in the pulled-back effect, and every channel adjoint preserves that identity.

Readout can be represented on fixed site spaces by reserving output record registers. An inactive placeholder after destructive measurement is not a surviving accessible measured factor.

The backwards region is an exact finite-step upper bound, not necessarily the smallest support for a particular list of channels. Joint record laws require the backwards region of the complete record-retaining network.

Let $R\ge 1$ be an integer. On $\mathbb{Z}$, with $d(i,j)=a|i-j|$, use the blocks

Swap the endpoint factors of every block at each tick. A factor initially at $0$ is at $tR$ after $t$ ticks, attaining speed $Ra/\tau$. In units $R=a=\tau =1$, the attained speed is $1$.

Proof. At each tick the blocks are disjoint consecutive intervals of diameter $Ra$. At tick $t$, the transported factor is at the left endpoint $(t-1)R$ of $B_{t,0}$. The endpoint swap moves it to $tR$, which is the left endpoint of $B_{t+1,0}$. Induction proves the claim.

For any finite horizon, retain a finite interval containing all blocks used by this relay. Keep complete blocks inside that interval and complete cut boundary fragments with singleton blocks. No infinite tensor-product state is needed.

The transported factor may be entangled with an untouched reference; the argument is an identity of wire permutations.

A finite collection of fixed labelled channels has declared supports $S_g\subseteq X$. A strict partial order is admissible if channels with intersecting supports are comparable. A legal execution is a linear extension of this order.

All legal executions of an admissibly ordered labelled channel collection have the same total channel.

Proof. Incomparable channels have disjoint supports, so their extensions by identities commute. Any two linear extensions are connected by adjacent swaps of incomparable elements.

For completeness, compare two extensions at their first disagreement. Move the desired element leftwards in the first extension. Each intervening element is incomparable with it: either order relation would contradict one of the two extensions. Repeating this procedure makes the extensions equal. Every swap preserves the composed channel.

Records are included in gate supports. The theorem does not move a gate past another gate that reads its record. Nor does it change a geometric parameter while treating the resulting channel as the same labelled gate.

In a square-lattice dependency model permitting at most one cardinal step of length $a$ per tick, or a wait, the asymptotic velocity set is

An isotropic physical speed $a/\tau$ is not dominated in every direction by this set.

Proof. After $n$ ticks, $|\Delta x|+|\Delta y|\le na$. Conversely, suitable proportions of cardinal steps and waits approximate every velocity in the displayed diamond. But

has Euclidean norm $a/\tau$ and lies outside that diamond.

This proposition concerns the stated cardinal-hop model, not every block schedule. It illustrates why a directional inclusion theorem cannot be replaced by a comparison of two scalar speeds. More general periodic-graph velocity polytopes are studied in Fritz2013.

A spectator extension of ${\rho }_{\Gamma }$ to a disjoint interface $N$ is any admissible state satisfying

The extension may be entangled. Existing classical registers must remain classical in the full joint state.

For a channel $\Phi :L({\mathcal{H}}_{\Gamma })\to L({\mathcal{H}}_{{\Gamma }^{\prime }})$,

Proof. For every $A\in L({\mathcal{H}}_{{\Gamma }^{\prime }})$,

Equality against every output observable proves the result.

Consider a finite certified execution

and an admissible initial extension satisfying eq:extension, with the same geometry, finite control state, and original-name locations. Assume:

1. each original matter map in the extended execution is tensored with ${\mathrm{id}}_N$;

2. each coupling algebra is embedded as ${\mathcal{A}}_v\otimes I_N$, or its input is a specified function only of the original reduced state;

3. spectator-blind control and resources, ass:blind, hold;

4. spectator locations have the required certified transport, and no spectator outcome is used during the comparison.

Then a paired extended execution is legal and, at every step,

Original-name locations also agree.

Proof. The assertions hold initially. Assume they hold at step $k$. By ass:blind, both executions select the same rule, guard result, support, and resource certificate.

For a geometric rule, the geometric and resource inputs agree. Its output geometry and original-name transport therefore agree. The matter states are unchanged.

For a matter rule, the geometric parameter is common. By lem:partial-trace,

For a coupling rule, every $A\in {\mathcal{A}}_v$ has equal expectation:

The rule receives identical state functionals and returns identical certified geometry and original-name transport. The same conclusion holds when its input is a specified function of the original reduction.

The common control selection gives the same successor control state. Admissibility and the transport certificates establish legality of the next paired configurations. Induction completes the proof.

Under the hypotheses of thm:spectators, inserting any finite product spectator state ${\rho }_0\otimes {\varphi }_N$ leaves the geometric trace and original accessible matter reductions unchanged.

Proof. A product insertion is a special case of eq:extension.

Let $g\in \{0,1\}$, and let a geometric rule request $g\mapsto 1$. Suppose its capacity guard is

At $|\Gamma |=K$, the update is permitted. Adding a single product spectator blocks it, even when every coupling algebra is $\mathbb{C}I$ and no matter operation touches the spectator.

The example violates ass:blind; it does not contradict thm:spectators. Similar leaks can occur through queue priority, repartitioning, support selection, or a spectator-dependent choice of successor location.

Within the theorem's rule class, direct inspection of spectator state or cardinality is excluded as well. It would be inaccurate to infer that every possible leak is a capacity leak. The point is that coupling blindness alone is insufficient: all control and resource observations must also be declared.

A spectator may be returned or consumed by terminal $\mathsf{Enter}$. The partial trace in eq:spectator-conclusion describes the accessible observation; it is not a term-level weakening rule.

Nonselective terminal readout of an entangled spectator preserves the original reduced state. Its conditional states need not be preserved. For

readout of the second factor leaves the first marginal $I/2$ when the outcome is ignored, but conditioning on the outcome gives $|0⟩⟨0|$ or $|1⟩⟨1|$.

Thus the theorem neither identifies the enlarged joint output with the original output nor permits a later geometric rule to read the spectator outcome without changing the comparison.

Immediate ${\mathsf{Enter}}_S$ removes $S$, whereas coherent recording retains it. Their outputs can be compared only after specifying a common accessible interface.

For every state on $S\otimes B$,

Proof. Write

Then

Tracing out $S$ removes precisely the terms with $z\ne z^{\prime }$.

Let ${\mathcal{N}}_A$ denote destructive computational-basis readout of the carrier $A$, returning a classical record $R$.

For every channel $\Lambda :L({\mathcal{H}}_B)\to L({\mathcal{H}}_{B^{\prime }})$,

Proof. A channel acting only on $B$ commutes with restriction over $S$ and with carrier readout. Apply lem:recording-restriction and identify the carrier's diagonal basis with the classical record basis.

Start from geometry $g_0$ and a state $\rho$ on $S\otimes B$. Compare:

1. immediate ${\mathsf{Enter}}_S$, followed by a finite continuation;

2. coherent recording into a fresh $A$, the corresponding continuation, and terminal ${\mathsf{Enter}}_A$.

During the continuation assume:

1. matter commands act only on the current complement $B_k$, never on the retained $S$, carrier $A$, or immediate record $R$;

2. coupling inputs and state-dependent guards depend only on the reduced state on $B_k$;

3. geometric rules inspect only their declared geometric labels;

4. control, support, and resources are blind to the excluded registers, as in ass:blind.

Align executions by their common continuation events. Their geometric traces and complement reductions coincide at every aligned event. After terminal carrier readout, restricting the deferred output away from the retained factor $S$ gives the immediate output on $R\otimes B_T$.

Proof. Immediately after the two initial operations,

Thus the complement reductions agree.

Suppose they agree at a continuation event, together with geometry and control. Conditions (D2)–(D4) and ass:blind give the same guard result, rule, resource certificate, and geometric update. Condition (D1) gives the same matter channel on the complement. By lem:partial-trace, its application preserves equality of complement reductions. Induction establishes the common trace.

For this fixed pair of inputs, compose the matter channels selected along the common trace into a channel $\Lambda$ on the continuation interface. Equation eq:continuation gives equality of the final outputs on $R\otimes B_T$.

The terminal operation has already consumed the carrier $A$. The extra quantum output is the retained original factor $S$. It is $S$, not $A$, that is omitted from the common observation interface.

The selected $\Lambda$ can depend on the input through the coupled geometric trace. The theorem is pairwise and does not assert that the complete coupled map is linear.

The deferred circuit must still return $S$ or consume it in a well-typed continuation. Observational inaccessibility does not create a discard rule. Allowing intermediate access to $R$, $S$, or $A$ requires a different theorem. Standard deferred-measurement constructions can treat more general record-dependent circuits GurevichBlass2022, but their hypotheses are not silently imported into the present geometric comparison.

This section concerns genuine finite-dimensional channels. It does not regard the nonlinear coupling in eq:coupling as a channel.

Let

be a channel. It may include prescribed evolution followed by a coarse-graining channel. Let $\{F_z:z\in Z\}$ be a fine POVM, $\{E_w:w\in W\}$ a coarse POVM, and $q:Z\to W$ a label map. Define

Empty sums are zero. The two record laws are

The two record laws agree for every state if and only if

Proof. Their difference at outcome $w$ is

The bracket is Hermitian. If its expectation vanishes in every state, testing its eigenvectors shows that every eigenvalue is zero. The reverse implication is the adjoint identity.

For example, if $T={\mathrm{Tr}}_{\mathrm{r}}$ and ${\Pi }_w=E_w\otimes I_{\mathrm{r}}$, compatibility is exact because $T^{\ast }(E)=E\otimes I_{\mathrm{r}}$. Readout at another time must first be pulled back through the appropriate evolution; a label aggregation is not an evolution map.

Set

These operators are Hermitian and satisfy ${\sum }_w{D}_w=0$. Define

With the conventions in eq:distance-conventions,

The maximum is attained by a pure input state.

Proof. For real numbers $d_w$ with zero sum,

Use $d_w=\mathrm{Tr}(\rho D_w)$. Since the event set is finite,

For a Hermitian operator $H$, the latter supremum is $\Vert H{\Vert }_{\infty }$, attained at an eigenvector whose eigenvalue has maximal absolute value. This proves the equality and attainment. The bound follows from $|\mathrm{Tr}(\rho D_w)|\le \Vert D_w{\Vert }_{\infty }$.

On a one-dimensional input space, take the two four-outcome POVMs

Their total-variation distance is $1$, whereas the largest individual-outcome discrepancy is $1/2$. The event $\{1,2\}$ attains the correct value. Thus ${\max }_w\Vert D_w{\Vert }_{\infty }$ is not the general formula. For two outcomes, it does suffice because $D_1=-D_0$.

Let $T(\rho )=H\rho H$, where $H$ is the Hadamard unitary. Use computational-basis measurements before and after $T$, with identity label aggregation. Then

The state $|0⟩⟨0|$ gives error $1/2$, so it is not a worst-case input.

The optimization in eq:record-error is unassisted. Equation eq:event-norm is not a diamond-norm formula.

Let ${\mathcal{M}}_E$ and ${\mathcal{M}}_{\Pi }$ denote the record-only measurement channels associated with $E_w$ and ${\Pi }_w$.

Under eq:pullback, for every finite reference system $N$,

In particular, the joint record–reference outputs agree for entangled inputs.

Proof. For every operator $X$,

Equal linear maps remain equal after tensoring with an identity.

An untouched reference is not the residual output produced by an instrument. Equal record channels determine the former comparison, not the latter. Also, an approximate unassisted bound does not become an ancillary-assisted bound merely because exact equality tensors.

For a unital residual algebra $\mathcal{B}\subseteq L({\mathcal{H}}_{\mathrm{c}})$, define

The record register is classical. Thus ${\mathcal{O}}_{\mathcal{B}}$ is the named record–residual observation algebra.

For channels $\Theta ,\Xi$ with a common output space and a unital output algebra $\mathcal{O}$, define

The channels agree on $\mathcal{O}$ for every input if and only if

Moreover,

Proof. For Hermitian $A$, equality of all input expectations is equality of the Hermitian adjoint images. General algebra elements are complex linear combinations of Hermitian elements. For the distance, interchange the two suprema and optimize the Hermitian expectation over states.

For ${\mathcal{O}}_{\mathbb{C}I}$, this is record-law comparison. For ${\mathcal{O}}_{L({\mathcal{H}}_{\mathrm{c}})}$, it compares complete classical-record instrument outputs. Intermediate algebras specify intermediate observation boundaries.

Now let the fine measurement be a PVM $\{P_z\}$. Define

For a coarse instrument with outcome maps ${\mathcal{I}}_w$, put

These channels have the same record–residual output interface.

Record compatibility does not imply agreement of $\mathcal{F}$ and $\mathcal{G}$ on ${\mathcal{O}}_{L({\mathcal{H}}_{\mathrm{c}})}$, even when $T=\mathrm{id}$.

Proof. Take one qubit, computational projectors $P_0,P_1$, and a single coarse outcome $\ast$. Let the coarse instrument be the identity. Then

so the record error is zero. For $\rho =|+⟩⟨+|$,

Their difference has residual block $X/2$, with eigenvalues $\pm 1/2$. Therefore

The full residual algebra distinguishes the outputs.

If two implementations approximate these targets on this input with trace-distance errors at most ${\epsilon }_{\mathrm{f}}$ and ${\epsilon }_{\mathrm{c}}$, their mutual trace distance is at least

This follows from the triangle inequality. More accurate implementation cannot eliminate an incompatibility of the ideal targets.

Assume now that $\{E_w\}$ is also a PVM and the coarse instrument is the Lüders instrument

The aggregated ${\Pi }_w={\sum }_{q(z)=w}{P}_z$ are mutually orthogonal projections.

If $T^{\ast }(E_w)={\Pi }_w$, then every Kraus operator $K_{\alpha }$ in any Kraus representation of $T$ satisfies

Consequently,

Proof. For every vector $v$,

Every summand is nonnegative. Hence $(I-E_w)K_{\alpha }{\Pi }_w=0$ for every $\alpha$. Similarly,

so $E_w{K}_{\alpha }(I-{\Pi }_w)=0$. These two identities give

Substitution into the Kraus representation proves eq:corner-identity.

Define the two dephasing channels

Assume that the fine and coarse measurements are PVMs, the coarse instrument is Lüders, and $T^{\ast }(E_w)={\Pi }_w$ for every $w$. Then the following statements are equivalent:

1. $\mathcal{F}=\mathcal{G}$, equivalently they agree on ${\mathcal{O}}_{L({\mathcal{H}}_{\mathrm{c}})}$ for every input;

2. $$T\circ {\mathcal{D}}_{\Pi }=T\circ {\mathcal{D}}_P;$$

3. for every $X$ and every distinct $z,z^{\prime }$ with $q(z)=q(z^{\prime })$,

   $$T(P_{z}X{P}_{z^{\prime }})=0.$$

Proof. By lem:intertwining, the difference of outcome maps is

Each such difference is supported in the $E_w$-corner. Its sum over $w$ is

Since the $E_w$ are orthogonal, this sum vanishes if and only if every corner vanishes: multiply the sum on both sides by $E_w$. This proves the equivalence of (i) and (ii).

Condition (iii) makes every summand in eq:block-difference zero, proving (i). Conversely, assume (ii) and take $X=P_{z}Y{P}_{z^{\prime }}$, with $z\ne z^{\prime }$ in the same coarse outcome. Then ${\mathcal{D}}_{\Pi }(X)=X$ and ${\mathcal{D}}_P(X)=0$, so $T(X)=0$. This proves (iii).

Under the hypotheses of thm:luders:

1. if $q$ is injective, exact record compatibility already implies full Lüders compatibility;

2. if there is one coarse outcome, full compatibility is equivalent to $T=T\circ {\mathcal{D}}_P$.

Proof. For injective $q$, the two dephasings in eq:dephasings coincide. For one coarse outcome, ${\Pi }_{\ast }=I$, so ${\mathcal{D}}_{\Pi }=\mathrm{id}$. Apply thm:luders.

Thus $T=T\circ {\mathcal{D}}_P$ is sufficient in general but stronger than necessary. Coherence between distinct coarse outcomes need not be destroyed by $T$, because the coarse Lüders instrument removes it. The obstruction is precisely fine coherence within a single coarse outcome.

The preceding comparisons concern specified output channels. For a continuation, a further question is whether input indistinguishability remains indistinguishability.

Let $\mathcal{A}\subseteq L({\mathcal{H}}_{\mathrm{in}})$ and $\mathcal{B}\subseteq L({\mathcal{H}}_{\mathrm{out}})$ be unital ${}^{\ast }$-subalgebras, and let $\Theta$ be a channel. On the unrestricted matrix-state spaces, the following are equivalent:

1. for every pair of states,

   $${\omega }_{\rho }{|}_{\mathcal{A}}={\omega }_{\sigma }{|}_{\mathcal{A}}⟹{\omega }_{\Theta (\rho )}{|}_{\mathcal{B}}={\omega }_{\Theta (\sigma )}{|}_{\mathcal{B}};$$
2. $${\Theta }^{\ast }(\mathcal{B})\subseteq \mathcal{A}.$$

Proof. Condition (ii) implies (i) by pulling each output observable back.

Conversely, assume (i). Let $H=H^{†}$ satisfy

Since $I\in \mathcal{A}$, $\mathrm{Tr}H=0$. With $d=\dim {\mathcal{H}}_{\mathrm{in}}$, choose $\epsilon >0$ small enough that

are positive. They are states agreeing on $\mathcal{A}$. For every Hermitian $B\in \mathcal{B}$, assumption (i) gives

Thus ${\Theta }^{\ast }(B)$ is orthogonal, in the real Hilbert space of Hermitian matrices, to the orthogonal complement of the Hermitian part of $\mathcal{A}$. Finite-dimensional double orthogonal complementation puts it in that Hermitian part. Complex linearity proves ${\Theta }^{\ast }(\mathcal{B})\subseteq \mathcal{A}$.

If successive continuations obey ${\Theta }_k^{\ast }({\mathcal{A}}_{k+1})\subseteq {\mathcal{A}}_k$, their composition preserves the initial observation boundary. This follows by composing the adjoint inclusions.

Let

be a prescribed block-diagonal state algebra, and let

be its trace-orthogonal pinching projection. Admissible input states are the density operators in $\mathcal{C}$. The accessible input algebra $\mathcal{A}$ need not lie inside $\mathcal{C}$.

For all admissible input pairs, agreement on $\mathcal{A}$ is preserved as agreement on $\mathcal{B}$ if and only if

Equivalently,

where ${\mathcal{C}}^{\perp }$ is the Hilbert–Schmidt orthogonal complement.

Proof. An admissible state sees $A$ only through ${\mathcal{E}}_{\mathcal{C}}(A)$. Therefore the displayed inclusion is sufficient.

For necessity, repeat the perturbation proof of thm:composition with Hermitian $H\in \mathcal{C}$ annihilating ${\mathcal{E}}_{\mathcal{C}}(\mathcal{A})$. The states $I/d\pm \epsilon H$ are admissible. The resulting orthogonality puts ${\mathcal{E}}_{\mathcal{C}}{\Theta }^{\ast }(B)$ in ${\mathcal{E}}_{\mathcal{C}}(\mathcal{A})$ for every Hermitian $B$, and then for every $B$. Finally, $\ker {\mathcal{E}}_{\mathcal{C}}={\mathcal{C}}^{\perp }$, proving the equivalent formulation.

The image ${\mathcal{E}}_{\mathcal{C}}(\mathcal{A})$ need not itself be an algebra. The statement uses it as the actual observable linear space on the restricted states.

This section gives three linear inference rules for a separate distributional problem. It assumes no particular graph kernel and proves no finite-part existence theorem for a graph.

Let $X,Y$ be smooth manifolds, with a fixed test-function convention. Write $\mathcal{D}(X)={C_c}^{\infty }(X)$ and ${\mathcal{D}}^{\prime }(X)$ for complex distributions, using the complex-bilinear distribution–test pairing. Convergence below means distributional convergence.

The names VII-C and VII-D refer only to the separate Wick/graph track. Nothing from that track is needed to state or prove the results here.

In finite dimensions, the inclusion ${\Theta }^{\ast }(\mathcal{B})\subseteq \mathcal{A}$ is dual to

Indeed, both say that $\mathrm{Tr}(B\Theta (H))=\mathrm{Tr}({\Theta }^{\ast }(B)H)=0$ whenever $H$ annihilates $\mathcal{A}$. The following quotient criterion has the same formal role.

Let $P:{\mathcal{D}}^{\prime }(X)\to {\mathcal{D}}^{\prime }(Y)$ be continuous and linear, and let $\mathfrak{C}\subseteq {\mathcal{D}}^{\prime }(X)$ and ${\mathfrak{C}}_P\subseteq {\mathcal{D}}^{\prime }(Y)$ be declared real linear subspaces. The assignment

is well defined on the real vector-space quotients if and only if

With quotient topologies, the induced map is continuous.

Proof. If $u-v\in \mathfrak{C}$, then $Pu-Pv=P(u-v)$. Condition eq:descent therefore makes the output class independent of the representative. Conversely, well-definedness applied to $[c]=[0]$, $c\in \mathfrak{C}$, gives $Pc\in {\mathfrak{C}}_P$. Continuity follows from the defining property of the quotient topology.

For $\mathfrak{C}=\mathbb{R}r$, this is a real quotient, even though the ambient distributions are complex. If source and target use the same nonzero residue $r$, descent requires $Pr\in \mathbb{R}r$. Failure of this inclusion is failure of descent for that specified quotient; it is not a proof that the original finite part fails to exist.

No multiplication is defined on these quotients.

Let $r\ne 0$ and $L$ belong to ${\mathcal{D}}^{\prime }(X)$. Choose $\Psi \in \mathcal{D}(X)$ with $⟨r,\Psi ⟩=1$, and set

Then

Proof. Nonzero $r$ permits a test with nonzero pairing, which can be rescaled to obtain $\Psi$. By construction, $⟨r,{\Phi }^{\circ }⟩=0$.

If $L=cr$, $c\in \mathbb{R}$, both conditions follow. Conversely, if they hold, then

This holds for every compact test, so $L=cr$.

For real distributions and real tests, the reality condition is automatic. Over complex tests it must not be omitted: tests annihilating $r$ alone cannot distinguish real from nonreal multiples of $r$.

Suppose a complete regulator difference has already been shown to converge,

Then a single compact test ${\Phi }^{\circ }$ with

obstructs removal of $L_{ab}$ by an allowed finite counterterm in $\mathbb{R}r$. These tests are diagnostics. Existence of a finite part still means convergence against all compact tests, not only against $\ker r$.

For an incidence representation, suppose a pushforward is defined and

For example, ${\gamma }_{ab}$ may be smooth and $\psi$ proper on the relevant supports. The correct condition is

Nonconstancy of ${\gamma }_{ab}$ by itself is not an obstruction unless the needed injectivity or separation property of the pushforward has also been proved.

Fix $\mu >0$, $\epsilon >0$, and

Let $P:{\mathcal{D}}^{\prime }(X)\to {\mathcal{D}}^{\prime }(Y)$ be continuous for distributional convergence.

Suppose the two complete finite parts exist: