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

A finite marked-carrier occupation model

Jeremy RodgersSection 8 of 15Original PDF ↗Publication DOI ↗

A finite marked-carrier occupation model

We now realize the equivariant obstruction in a completely explicit finite setting.

The model

Fix integers $N\geq 2$ and $K\geq 2$ . Consider $N$ individually marked carriers distributed among $K$ cells. The source state space is the set of assignments $S=\{1,\ldots,K\}^{\{1,\ldots,N\}},\tag{25}$ so that a source state $s\in S$ is a function assigning to each marked carrier $i\in\{1,\ldots,N\}$ the cell $s(i)\in\{1,\ldots,K\}$ that it occupies. Then $|S|=K^{N}$ .

The physical status of the marks is a modeling declaration, and here it is made explicitly: the carrier identities are physically meaningful source distinctions. Two assignments that differ by which marked carrier sits in which cell are different source states, not two descriptions of one source state. Accordingly, the physical equivalence $\simeq_{\mathrm{phys}}$ on the underlying realization space is trivial as far as the marks are concerned, and (25) already is the reduced source space $S$ in the sense of Definition 2.1. Remark 7.4 below examines what happens if this declaration is reversed.

The readout records only how many carriers occupy each cell: $p(s)=\bigl(n_{1}(s),\ldots,n_{K}(s)\bigr), \qquad n_{j}(s):=\#\{\, i : s(i)=j \,\},\tag{26}$ with values in $T=\Bigl\{\, (n_{1},\ldots,n_{K})\in\mathbb{N}_{0}^{K} : \sum_{j=1}^{K}n_{j}=N \,\Bigr\}.\tag{27}$ Every such vector is realized by some assignment, so (3) holds with the $T$ of (27).

Fibers

Proposition 7.1 (Fiber cardinality).

For $n=(n_{1},\ldots,n_{K})\in T$ , $|F_{n}|=\binom{N}{n_{1},\ldots,n_{K}}=\frac{N!}{n_{1}!\,\cdots\,n_{K}!}.\tag{28}$

Proof. An element of $F_{n}$ is precisely an ordered partition of the set $\{1,\ldots,N\}$ of marks into blocks $B_{j}=s^{-1}(j)$ with $|B_{j}|=n_{j}$ for $j=1,\ldots,K$ ; conversely any such ordered partition determines a unique $s$ with $p(s)=n$ . Counting these partitions: choose $B_{1}$ in $\binom{N}{n_{1}}$ ways, then $B_{2}$ from the remaining $N-n_{1}$ marks in $\binom{N-n_{1}}{n_{2}}$ ways, and so on. The product telescopes to the multinomial coefficient (28). ◻

The symmetry and the obstruction

Let $G=S_{N}$ , the symmetric group on the carrier marks, acting on $S$ by $(g\cdot s)(i):=s(g^{-1}i) \qquad (g\in S_{N},\ s\in S,\ i\in\{1,\ldots,N\}).\tag{29}$ This is a left action: $(g\cdot(h\cdot s))(i)=(h\cdot s)(g^{-1}i)=s(h^{-1}g^{-1}i)=((gh)\cdot s)(i)$ . Physically it permutes which marked carrier occupies which cell, while the marks themselves retain their identities.

Proposition 7.2 (Invariance and transitivity).

For every $g\in S_{N}$ and $s\in S$ , $p(g\cdot s)=p(s).\tag{30}$ Consequently $S_{N}$ acts trivially on $T$ , the map $p$ is equivariant for this pair of actions, and $G_{n}=S_{N}$ for every $n\in T$ . Moreover $S_{N}$ acts transitively on each fiber $F_{n}$ .

Proof. Invariance: for each cell $j$ , $n_{j}(g\cdot s)=\#\{\,i : s(g^{-1}i)=j\,\} =\#\{\,i' : s(i')=j\,\} =n_{j}(s),$ where the middle equality is the substitution $i'=g^{-1}i$ , a bijection of $\{1,\ldots,N\}$ . This proves (30). If $S_{N}$ is given the trivial action on $T$ , then $g\cdot p(s)=p(s)=p(g\cdot s)$ , which is exactly the equivariance condition (17); and every $g$ fixes every $n$ , so $G_{n}=S_{N}$ .

Transitivity: let $s,s'\in F_{n}$ . Then for each $j$ the sets $s^{-1}(j)$ and $s'^{-1}(j)$ both have $n_{j}$ elements, so there is a bijection $\beta_{j}\colon s'^{-1}(j)\to s^{-1}(j)$ . The blocks $s'^{-1}(1),\ldots,s'^{-1}(K)$ partition $\{1,\ldots,N\}$ , as do the blocks $s^{-1}(1),\ldots,s^{-1}(K)$ , so the $\beta_{j}$ assemble into a permutation $g\in S_{N}$ with $g\bigl(s'^{-1}(j)\bigr)=s^{-1}(j)$ for all $j$ . Then $s(g\,i')=j$ exactly when $i'\in s'^{-1}(j)$ , i.e. $s\circ g=s'$ , i.e. $s'(i)=s(g i)$ . Writing $h=g^{-1}$ gives $(h\cdot s)(i)=s(h^{-1}i)=s(gi)=s'(i)$ , so $h\cdot s=s'$ . ◻

The situation is thus the one isolated by Corollary 6.7 in its most extreme form: the symmetry group is entirely invisible to the readout, so every stabilizer is the whole group, and that whole group acts transitively on every fiber.

Example 7.3 (A fiber with no invariant representative).

Take $n_{0}=(1,\,N-1,\,0,\ldots,0)\in T,\tag{31}$ which is a legitimate element of (27) since $1+(N-1)=N$ and $K\geq 2$ . By Proposition 7.1, $|F_{n_{0}}|=\frac{N!}{1!\,(N-1)!\,0!\cdots 0!}=N>1,\tag{32}$ using $N\geq 2$ : the fiber consists of the $N$ assignments in which exactly one marked carrier occupies cell $1$ and the remaining $N-1$ occupy cell $2$ , indexed by which carrier is the solitary one.

By Proposition 7.2, $G_{n_{0}}=S_{N}$ acts transitively on $F_{n_{0}}$ , and $|F_{n_{0}}|>1$ . Corollary 6.7 therefore gives $\mathop{\mathrm{Fix}}_{S_{N}}(F_{n_{0}})=\varnothing,\tag{33}$ and no $S_{N}$ -equivariant deterministic section $T\to S$ exists. That is: no permutation-equivariant rule assigns to occupation data a specific marked-carrier assignment. Any rule that names the carrier sitting alone in cell $1$ must name a particular mark, and conjugating the rule by the transposition exchanging that mark with another produces a different answer from identical readout data.

Directly: if $\sigma$ were equivariant, then $\sigma(n_{0})\in F_{n_{0}}$ would be some assignment in which carrier $i_{\ast}$ is alone in cell $1$ ; taking $g$ to be a transposition moving $i_{\ast}$ , we get $g\cdot\sigma(n_{0})\neq\sigma(n_{0})$ , while $\sigma(g\cdot n_{0})=\sigma(n_{0})$ because $S_{N}$ acts trivially on $T$ . This contradicts (21).

Note that ordinary sections abound here: $T$ is finite, and one may simply declare that the solitary carrier is always mark $1$ . What fails is not selection but symmetric selection. The convention “mark $1$ ” is a choice with no warrant in the data, and the symmetry of the problem exposes it as such.

Remark 23 (The modeling boundary).

The obstruction in Example 7.3 depends entirely on the declaration made after (25), and it is worth stating exactly where the line falls.

Suppose instead that the carrier marks were declared physically redundant , that permuting which carrier is which is a relabeling and not a physical operation. Then the permutations (29) belong to $\simeq_{\mathrm{phys}}$ , and the reduction of Definition 2.1 must be performed with respect to them. The reduced source space becomes $S'=\{1,\ldots,K\}^{\{1,\ldots,N\}}\big/ S_{N},$ whose elements are exactly the occupation vectors: two assignments have the same $S_{N}$ -orbit if and only if they induce the same $(n_{1},\ldots,n_{K})$ , by the transitivity established in Proposition 7.2. The induced readout $p'\colon S'\to T$ is then a bijection, every reduced fiber is a singleton, and by Proposition 3.2 the readout is source-equivalent. The obstruction has vanished , and correctly so, because on this declaration there was never anything to lose.

The two models are mathematically adjacent and physically opposite. The theorems of Section 4, Section 6 detect lost source structure only after the theory has independently specified which distinctions are physical. They do not, and cannot, make that specification on the theory’s behalf. In particular, in Example 7.3 the group $S_{N}$ is not a gauge group: it acts on physically distinct states and its orbits are not points of the reduced source space. Treating it as gauge would be a different model, the one just described, with a different and trivial conclusion.