Affine dynamics and traces

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These are checked reformulations and obstructions, not a resolution. They identify where three proposed dynamical arguments require the unproved bounded-quotient first-kill estimate. Write $B$ for the delayed excluded union and $q(n)$ for the first killing modulus of $n\in B$. The case of modulus $1$ is trivial; the operator discussion assumes all moduli are at least $2$.

Physical dilation has an exact, non-negligible-looking defect

Suppose an integer $p\ge2$ is coprime to every modulus. For $0\le b<p$, define a branch system, with the same distinct moduli, by

$$a_{q,b}\equiv p^{-1}(a_q-b)\pmod q, \qquad0\le a_{q,b}<q.$$

Let $B_b$ denote its delayed union. Then

$$B=\bigsqcup_{b=0}^{p-1}(pB_b+b)\ \sqcup\ D_p, \qquad D_p=\{n\in B:n<pq(n)\} =\bigcup_q(E_q\cap[q,pq)). \tag{1}$$

Indeed, for $n=pt+b$, membership in its branch image means that some original matching modulus satisfies $q\le t=\lfloor n/p\rfloor$. This holds precisely when the least matching modulus satisfies that inequality. If the branch condition holds, activation in the original system certainly holds as well.

Let $\beta$ be the full Haar union measure. All branches have the same $\beta$, since multiplication by $p$ is an automorphism on the profinite factor generated by these moduli. For $\Delta_F(X)=H_{B_F}(X)-\beta H_{\lfloor X\rfloor}$, (1) gives

$$\Delta_F(X)=\frac1p\sum_{b=0}^{p-1}\Delta_{F_b}(X/p) +H_{D_p}(X)+O_p(1). \tag{2}$$

The weighting error is bounded because $\sum_{t\ge1}|(pt+b)^{-1}-(pt)^{-1}|<\infty$; the cutoff and harmonic-number errors are also $O_p(1)$. For $p=2$, $D_p$ is exactly the first-endpoint contribution. More generally, replacing activation $n\ge q$ by $n\ge cq$ gives the difference $\bigcup_q(E_q\cap[q,cq))$ for every fixed $c\ge1$. It is not legitimate to assume these changes logarithmically negligible.

Measurable affine harmonicity alone would still be insufficient

On an infinite profinite factor on which $p$ is invertible, put

$$Pf(z)=p^{-1}\sum_{b=0}^{p-1}f((z-b)/p).$$

Its iterates mix continuous functions uniformly to their Haar means: for functions on a finite quotient, the $p^k$ branch shifts become uniformly distributed, and continuous functions admit uniform finite-quotient approximation. The conclusion is false pointwise for bounded measurable functions.

In fact choose a Haar-null comeager set $C$, and let $\Gamma$ be the countable group generated by integer translations and multiplication and division by $p$. Then

$$E=\bigcap_{\gamma\in\Gamma}\gamma C$$

is still null and comeager, is invariant under $\Gamma$, and satisfies $P\mathbf1_E=\mathbf1_E$. A null comeager $C$ exists by intersecting dense open sets of arbitrarily small Haar measure. Thus affine harmonicity, almost-everywhere regularity, and Baire-category information do not by themselves force pointwise regularity. This is an abstract obstruction, not an exceptional congruence family.

Operator covariance retains the endpoint obstruction

On $\ell^2(\mathbb N)$ let

$$V_{q,a}e_t=e_{qt+a},\qquad D e_n=n^{-1}e_n.$$

The range of $V_{q,a}$ is precisely the delayed class. Compress to the quotient set corresponding to the actual first-kill set $E_q$ and transport back to $E_q$. The positive error between $D/q$ and $V_{q,a}^*DV_{q,a}$ has the diagonal value

$$C(n)=\frac1{qt}-\frac1n=\frac{a/q}{nt}, \qquad n=qt+a\in E_q. \tag{3}$$

Each individual error is trace class with trace $O(1/q)$, since its quotient entries are at most $1/(qt^2)$. Their orthogonal sum need not be trace class, even for actual first kills.

For every prime $p>3$, choose $a_p=(p-1)/2$. The endpoint $r_p=(3p-1)/2$ is a first kill: an earlier selected prime dividing $2r_p+1=3p$ is impossible. All earlier congruences are active there. Therefore the aggregate trace is at least

$$\sum_{p>3}\left(\frac1p-\frac2{3p-1}\right)=\infty.$$

The original prime-modulus system nevertheless has natural survivor density zero by the established coprime case.

A trace-class equivalent reference makes the remaining issue exact. Set $D_-e_n=(n-\tfrac12)^{-1}e_n$; then $D_--D$ is trace class. The analogous aggregate has diagonal

$$C_-(n)=\frac1{q(t-\tfrac12)}-\frac1{n-\tfrac12} =\frac{a+(q-1)/2}{q(t-\tfrac12)(n-\tfrac12)}.$$

For $q\ge2$, $t\ge1$, and $0\le a<q$,

$$\frac1{4nt}\le C_-(n)\le\frac4{nt}. \tag{4}$$

For the lower bound, the numerator is at least $q/4$ and the denominator at most $qtn$. For the upper bound, use numerator at most $3q/2$, $t-1/2\ge t/2$, and $n-1/2\ge3n/4$. Hence

$$\frac1{\log X}\operatorname{Tr}(C_-\mathbf1_{[1,X]})\to0$$

if and only if every set $R_T=\{qt+a\in E_q:t\le T\}$ has logarithmic density zero. One direction follows from the lower bound $H_{R_T}(X)/(4T)$. For the other, (4) bounds the part with $t>T$ by $4H_{\lfloor X\rfloor}/(T+1)$, after which $T\to\infty$. Thus this proposed aggregate covariance estimate contains exactly the unresolved endpoint assertion; it is not an available trace-ideal lemma.

Individual finite-rank equivalence loses countable activation data

Choose $q_i=2^{i+2}$ and $a_i=i$ for $i\ge1$. Let $F_i$ project onto the full class and $P_i$ onto its delayed version. Then

$$F_i-P_i=|e_i\rangle\langle e_i|.$$

The full classes cover every positive integer via these roots, whereas the delayed union has natural density at most $\sum_i1/q_i=1/4$. Thus their countable joins are different even though every pair of generators agrees modulo finite rank. Passing to a compact-operator quotient before taking the join cannot justify the desired conclusion.

Also, distinct moduli and orthogonality alone do not imply negligible aggregate errors: compress $V_{q,q-1}$ to its first endpoint $2q-1$ for every $q\ge2$. These ranges are orthogonal and their errors sum to $(1/2)\log X+O(1)$ below $X$. They are not actual first kills: the class $1\pmod2$ already covers all those endpoints.

A forward flow and its precise limitation

Define $T(n)=n+q(n)$ on $B$. Then $q(T(n))\le q(n)$. For a fixed target $m$, its predecessor labels form a primitive set: if $p<q$, $p\mid q$, and $m-p,m-q$ were predecessors with these labels, then $m-q\in[a_p]_p$ and $m-q\ge q>p$, contradicting its first-kill label.

Every orbit reaches a class disjoint from all earlier full classes. Indeed, while its label is $q$, compatibility with an earlier modulus $p$ forces a drop within at most $p/(p,q)$ steps along the progression of step $q$. A nonempty first-kill cell cannot have $p/(p,q)=1$. Thus a nonterminal label costs less than $q^2$ in forward displacement. Labels strictly decrease at transitions, so a terminal class is reached before position

$$n+\sum_{j\le q(n)}j^2=n+O(q(n)^3).$$

The terminal cylinders are pairwise disjoint; each carries one absorbing arithmetic progression. A globally private endpoint $r_q\in[q,2q)$ has no predecessor: privacy would force predecessor label $q$, but then its predecessor would be $r_q-q<q$ and inactive.

These properties do not control the required mass transport. For a large integer $M$, take every $M/3<q<M/2$, with

$$C_q=[M]_q,\qquad r_q=M-q.$$

All moduli are distinct, $q<r_q<2q$, and the endpoints are globally private because $|r_q-r_p|=|q-p|<\min(p,q)$. Yet

$$T(r_q)=M\quad\hbox{for every selected }q.$$

Thus $\asymp M$ genuine private sources immediately merge into the terminal class of the least modulus. Their harmonic mass tends to $\log(4/3)$, whereas that terminal cylinder has measure $\asymp1/M$. This rules out a uniform charge by terminal measure or a uniformly bounded merger cost. It has only bounded harmonic mass and is not an infinite counterexample. A further flow argument would have to use global progression consistency, not merely nonincreasing labels, primitive predecessor sets, and polynomial arrival bounds.

Affine-invariant logarithmic means retain the countable-union gap

The final continuation also considered a finitely additive formulation. It gives an exact reformulation, not a proof of the needed continuity.

Fix a free ultrafilter $\mathcal U$ on positive integer cutoffs and define, for every $E\subseteq\mathbb N$,

$$\nu_{\mathcal U}(E)= \lim_{N\to\mathcal U}\frac{H_E(N)}{H_N}. \tag{5}$$

This is a finitely additive probability measure, and agrees with logarithmic density whenever that density exists. For every fixed integer $q\ge1$ and $a\ge0$,

$$\nu_{\mathcal U}(qE+a)=\frac1q\nu_{\mathcal U}(E). \tag{6}$$

For dilation, use the exact identity

$$H_{qE}(N)=q^{-1}H_E(\lfloor N/q\rfloor)$$

and

$$0\le H_E(N)-H_E(\lfloor N/q\rfloor) \le H_N-H_{\lfloor N/q\rfloor}=O_q(1).$$

Translation by $a$ changes the harmonic sum by at most $H_a$, as in the translation-dynamics note. These errors vanish after division by $H_N$, proving (6).

For the actual increasing-modulus first kills, write

$$E_i=a_i+q_iS_i,\qquad d(S_i)=q_ie_i,$$

where $S_i$ is the finite periodic quotient-survivor set from the basic reductions and its quotient variable starts at $1$. Activation is therefore unchanged. Equation (6) gives $\nu_{\mathcal U}(E_i)=e_i$ for every $i$. Finite additivity recovers only

$$\nu_{\mathcal U}(B)\ge\sum_i e_i=\beta.$$

The missing conclusion is exactly

$$\nu_{\mathcal U}\!\left(\bigcup_{i>k}E_i\right)\longrightarrow0, \tag{7}$$

for every such logarithmic mean. Assuming countable additivity would be invalid: every singleton has measure zero under (5), whereas $\nu_{\mathcal U}(\mathbb N)=1$.

Neither this formulation nor its affine covariance uses the singleton-distinctness restriction. No argument deriving (7) from that restriction was found. Likewise, the cumulative spectral innovation budget still does not control high-conductor point evaluation, and the constant disjoint first-kill entropy estimate remains unproved. The continuation supplies no new transfer theorem for either route.