Source–Readout Non-Equivalence: Descent and Equivariant Reconstruction Obstructions

Descent of source relations

Jeremy RodgersSection 5 of 15Original PDF ↗Publication DOI ↗

Descent of source relations

Nontriviality of a fiber is a statement about the map $p$ . To convert it into a statement about lost physics, one must exhibit a physically meaningful quantity that varies across a fiber. This section makes that precise.

Relations and descent

Let $V$ be a set of values and let $r\colon S\longrightarrow V\tag{10}$ be a source relation: a map assigning to each reduced source state a value that the theory regards as physically meaningful. Because the domain of $r$ is $S$ rather than $\Omega$ , the relation is automatically invariant under physical equivalence; this is the payoff of having performed the reduction first. Concrete instances that arise in practice include a global transport class, a boundary relation, a correlation functional, a spectral invariant, or a source property required as an input by some further construction. Nothing in what follows depends on which; $r$ is an arbitrary map out of $S$ .

Definition 4.1 (Descent).

The relation $r$ descends through the readout if there exists a map $\bar r\colon T\longrightarrow V \qquad\text{with}\qquad r=\bar r\circ p .\tag{11}$

If $r$ descends, then $r(s)$ is a function of the readout value $p(s)$ alone, and an agent with access only to public data can evaluate $r$ . If $r$ does not descend, no such evaluation is possible, however the agent proceeds. It is only after descent has been established that it is legitimate to call $r$ an observable of the readout; a source relation is not observable at the readout level merely because it is physically meaningful at the source level.

Theorem 4.2 (Fiber–descent theorem).

Let $r\colon S\to V$ . The following are equivalent.

There exists $\bar r\colon T\to V$ with $r=\bar r\circ p$ .
$r$ is constant on every fiber of $p$ ; that is, $p(s)=p(s') \quad\Longrightarrow\quad r(s)=r(s') \qquad (s,s'\in S).\tag{12}$

When these hold, $\bar r$ is unique.

Proof. 1 $\Rightarrow$ 2. Suppose $r=\bar r\circ p$ and $p(s)=p(s')$ . Then $r(s)=\bar r\bigl(p(s)\bigr)=\bar r\bigl(p(s')\bigr)=r(s').$

2 $\Rightarrow$ 1. Suppose $r$ is constant on fibers. Fix $t\in T$ . By surjectivity of $p$ the fiber $F_{t}$ is nonempty, so we may pick $s\in F_{t}$ and set $\bar r(t):=r(s).\tag{13}$ This prescription is independent of the choice of $s\in F_{t}$ : any two elements of $F_{t}$ have the same image under $p$ , namely $t$ , so (12) forces them to have the same $r$ -value. Hence $\bar r$ is a well-defined map $T\to V$ . For arbitrary $s\in S$ , applying (13) at $t=p(s)$ with the representative $s$ itself gives $\bar r(p(s))=r(s)$ , that is, $r=\bar r\circ p$ .

Uniqueness. If $\bar r_{1}\circ p=\bar r_{2}\circ p$ then, since $p$ is surjective, every $t\in T$ is of the form $p(s)$ , and $\bar r_{1}(t)=r(s)=\bar r_{2}(t)$ . Hence $\bar r_{1}=\bar r_{2}$ . ◻

Note that the proof of 2 $\Rightarrow$ 1 requires only a choice of one representative per fiber for the purpose of evaluating a constant, and the constancy hypothesis makes the choice immaterial. No choice principle is needed, and no representative survives into the conclusion. This is in sharp contrast to the selection problems of Section 5, Section 6, where the chosen representative is the output.

Corollary 4.3 (Descent obstruction).

Suppose there exist $s_{0},s_{1}\in S$ with $p(s_{0})=p(s_{1}), \qquad r(s_{0})\neq r(s_{1}).\tag{14}$ Then $r$ does not descend through $p$ : there is no map $\bar r\colon T\to V$ with $r=\bar r\circ p$ . Consequently no function of the readout alone reproduces $r$ on all of $S$ .

Proof. Condition (14) exhibits a fiber on which $r$ is not constant, so 2 of Theorem 4.2 fails, and therefore 1 fails. ◻

The content of Corollary 4.3 should not be mistaken for a restatement of noninjectivity of $p$ . Noninjectivity says that the fibers are not all singletons. Corollary 4.3 says something further and more specific: that a nominated physically meaningful quantity, one the theory actually cares about, is not among the quantities the readout carries. The hypothesis (14) is a statement one must earn by producing the relation $r$ ; it does not follow from coarseness of the readout by itself, since a noninjective readout may perfectly well transmit every relation the theory needs. This is the difference between observing that a map forgets something and identifying what it forgets.

Relative completeness

The preceding suggests grading the readout not absolutely but against a nominated family of relations.

Definition 4.4 (Relative source completeness; separation).

Let $\mathcal{R}$ be a family of source relations on $S$ (with values in possibly different value sets). The readout is complete relative to $\mathcal{R}$ if every $r\in\mathcal{R}$ descends through $p$ .

The family $\mathcal{R}$ separates the points of $S$ if for all $s\neq s'$ in $S$ there exists $r\in\mathcal{R}$ with $r(s)\neq r(s')$ .

Theorem 4.5 (Separating-family criterion).

Let $\mathcal{R}$ separate the points of $S$ . Then the readout is complete relative to $\mathcal{R}$ if and only if $p$ is bijective.

Proof. ( $\Leftarrow$ ) If $p$ is bijective, then for any $r\colon S\to V$ the map $\bar r:=r\circ p^{-1}$ satisfies $\bar r\circ p=r$ . Hence every relation whatsoever descends, in particular every member of $\mathcal{R}$ .

( $\Rightarrow$ ) Suppose every $r\in\mathcal{R}$ descends, and let $s,s'\in S$ satisfy $p(s)=p(s')$ . By Theorem 4.2, each $r\in\mathcal{R}$ is constant on the fibers of $p$ , so $r(s)=r(s')$ for every $r\in\mathcal{R}$ . Since $\mathcal{R}$ separates the points of $S$ , this forces $s=s'$ . Hence $p$ is injective, and it is surjective by (3). ◻

Theorem 4.5 gives relative completeness an operational reading, but the reading is conditional and the condition must be checked. The theorem says that a readout is source-complete exactly when it carries a family of relations that separates source states. Whether that reading is operational depends entirely on whether the separating family $\mathcal{R}$ has been identified as physically meaningful. An arbitrary separating family always exists , take $\mathcal{R}$ to be all maps $S\to\{0,1\}$ , or all indicator functions of points , and its existence tells one nothing, since its members need not correspond to anything a theory recognizes or an experiment can access. The theorem earns its interpretation only when $\mathcal{R}$ is supplied by the physics.