# Zero Genesis Game 20｜Arithmetic Darboux Pullback、Multiplicative-Hilbert Geometry 與 Zero-Preserving Dilation Wavelet

**Series:** SECV / Zero Genesis Experimental Playground  
**Experiment:** Zero Genesis Game 20  
**Version:** v0.1  
**Author:** Neo.K  
**AI Collaboration:** Aletheia (GPT-5.6 Sol)  
**Status:** Exploratory / Arithmetic Positive-Energy and Dilation-Filter Study  
**Important:** This document does **not** claim a proof of RH. It derives exact arithmetic factorization and scale-filter identities from classical Dirichlet-series machinery. The multiplicative Hilbert matrix is a known operator. The RH criteria below are derived reformulations unless explicitly stated otherwise.

---

## 0. Motivation

Game 19 produced the exact Darboux pair

$$
AA^\dagger
=
-\partial_s^2
+
\gamma^2
+
\mathcal J_2(s),
$$

$$
A^\dagger A
=
-\partial_s^2
+
\gamma^2
+
\mathcal K_2(s),
$$

where

$$
\mathcal J_2
=
h^2+h'+2\gamma h,
$$

$$
\mathcal K_2
=
h^2-h'+2\gamma h,
$$

and

$$
h
=
-\frac{\zeta'}{\zeta}.
$$

The zero-selective potential $\mathcal J_2$ has signed arithmetic coefficients, while the direct/Selberg partner has nonnegative coefficients.

The main unresolved question was:

> Can the positive Darboux square be pulled back into a genuinely positive arithmetic energy?

Game 20 answers yes.

It then discovers a second issue:

> The first-order positive square reintroduces the equilibrium pole at $s=1$.

This leads to a different construction:

$$
\boxed{
\text{finite dilation high-pass filtering}.
}
$$

A two-tap dilation filter annihilates the deterministic prime-density mode while preserving every nontrivial zeta-zero mode, and its positive weighted energy is norm-equivalent to the original Game 12 critical prime-error energy.

---

# 1. Arithmetic coefficient operators

Let

$$
a=(a_n)_{n\ge2}
$$

be a finitely supported sequence and define the Dirichlet polynomial

$$
\boxed{
F_a(s)
=
\sum_{n\ge2}
a_nn^{-s}.
}
$$

Define logarithmic multiplication:

$$
\boxed{
(L a)_n
=
(\log n)a_n.
}
$$

Define von-Mangoldt divisor convolution:

$$
\boxed{
(C_\Lambda a)_n
=
\sum_{d\mid n}
\Lambda(d)a_{n/d}.
}
$$

---

# 2. First-order arithmetic factors

Since

$$
-\partial_sF_a
=
\sum_n
(\log n)a_nn^{-s},
$$

and

$$
h(s)F_a(s)
=
\sum_n
(C_\Lambda a)_n n^{-s},
$$

define:

$$
\boxed{
T_\gamma
=
L+C_\Lambda+\gamma I,
}
$$

$$
\boxed{
S_\gamma
=
-L+C_\Lambda+\gamma I.
}
$$

Then in the Euler-product half-plane:

$$
\boxed{
-\partial_sF_a+(h+\gamma)F_a
=
F_{T_\gamma a},
}
$$

and

$$
\boxed{
\partial_sF_a+(h+\gamma)F_a
=
F_{S_\gamma a}.
}
$$

Thus the Darboux first-order factors have exact coefficient-space realizations.

---

# 3. Arithmetic commutator

Compute:

$$
(L C_\Lambda a)_n
=
(\log n)
\sum_{d\mid n}
\Lambda(d)a_{n/d},
$$

while

$$
(C_\Lambda L a)_n
=
\sum_{d\mid n}
\Lambda(d)
\log(n/d)
a_{n/d}.
$$

Therefore:

$$
\boxed{
L C_\Lambda-C_\Lambda L
=
C_{\Lambda\log},
}
$$

where

$$
\boxed{
(\Lambda\log)(n)
=
\Lambda(n)\log n.
}
$$

This is the arithmetic commutator underlying the partner swap.

---

# 4. Exact arithmetic Darboux factorization

Multiply the first-order operators:

$$
S_\gamma T_\gamma
=
(-L+C_\Lambda+\gamma I)
(L+C_\Lambda+\gamma I).
$$

Using the commutator identity:

$$
\boxed{
S_\gamma T_\gamma
=
-L^2
+
C_{j_2}
+
\gamma^2 I,
}
$$

where

$$
\boxed{
j_2
=
\Lambda*\Lambda
-
\Lambda\log
+
2\gamma\Lambda.
}
$$

Similarly:

$$
\boxed{
T_\gamma S_\gamma
=
-L^2
+
C_{k_2}
+
\gamma^2 I,
}
$$

with

$$
\boxed{
k_2
=
\Lambda*\Lambda
+
\Lambda\log
+
2\gamma\Lambda.
}
$$

Thus the continuous Darboux pair of Game 19 exists exactly in divisor-convolution algebra.

---

# 5. Partner commutator

Subtract:

$$
\boxed{
T_\gamma S_\gamma
-
S_\gamma T_\gamma
=
2C_{\Lambda\log}.
}
$$

So the complete arithmetic partner difference is a prime-power self-energy convolution.

This recovers the Game 19 Self-Energy / Interaction Separation directly from first-order operator algebra.

---

# 6. Arithmetic Dirac lift

Define the block operator:

$$
\boxed{
\mathbb D_\gamma
=
\begin{pmatrix}
0&S_\gamma\\
T_\gamma&0
\end{pmatrix}.
}
$$

Then:

$$
\boxed{
\mathbb D_\gamma^2
=
\begin{pmatrix}
-L^2+C_{j_2}+\gamma^2I&0\\
0&-L^2+C_{k_2}+\gamma^2I
\end{pmatrix}.
}
$$

This is a purely arithmetic multiplicative Dirac/Darboux lift.

It closely parallels the square-root / Dirac structures encountered earlier in Zero Genesis Game 04.

---

# 7. Real-half-line Gram geometry

For a real parameter:

$$
\sigma>0,
$$

define on finite Dirichlet polynomials:

$$
\boxed{
\langle a,b\rangle_\sigma
=
\int_\sigma^\infty
F_a(s)
\overline{F_b(s)}
\,ds.
}
$$

Since:

$$
\int_\sigma^\infty
(mn)^{-s}ds
=
\frac{
(mn)^{-\sigma}
}{
\log(mn)
},
$$

we obtain:

$$
\boxed{
\langle a,b\rangle_\sigma
=
\sum_{m,n\ge2}
a_m\overline{b_n}
K_\sigma(m,n),
}
$$

where:

$$
\boxed{
K_\sigma(m,n)
=
\frac{
1
}{
(mn)^\sigma\log(mn)
}.
}
$$

This is a positive Gram kernel.

---

# 8. Multiplicative Hilbert matrix

At:

$$
\boxed{
\sigma=\frac12,
}
$$

the kernel becomes:

$$
\boxed{
K_{1/2}(m,n)
=
\frac{
1
}{
\sqrt{mn}\log(mn)
}.
}
$$

This is the known **multiplicative Hilbert matrix**.

Brevig, Perfekt, Seip, Siskakis and Vukotić proved that the associated operator is bounded, strictly positive, has norm $\pi$, and has continuous spectrum $[0,\pi]$.

Thus the positive geometry appearing naturally in Game 20 is an established Dirichlet-series Hilbert geometry rather than an ad hoc kernel.

---

# 9. Arithmetic positive square in the safe half-plane

For:

$$
\sigma>1,
$$

let:

$$
\boxed{
b
=
T_\gamma a.
}
$$

Explicitly:

$$
\boxed{
b_n
=
(\log n+\gamma)a_n
+
\sum_{d\mid n}
\Lambda(d)a_{n/d}.
}
$$

Define:

$$
\boxed{
\mathcal E_\sigma[a]
=
\langle b,b\rangle_\sigma.
}
$$

Then:

$$
\boxed{
\mathcal E_\sigma[a]
=
\sum_{m,n\ge2}
\frac{
b_m\overline{b_n}
}{
(mn)^\sigma\log(mn)
}
\ge0.
}
$$

This is a completely arithmetic positive quadratic form.

---

# 10. Equality with the Darboux square

In the safe half-plane:

$$
F_b
=
-\partial_sF_a
+
qF_a,
$$

where:

$$
\boxed{
q=h+\gamma.
}
$$

Therefore:

$$
\mathcal E_\sigma[a]
=
\int_\sigma^\infty
|
-F_a'+qF_a
|^2ds.
$$

Expand:

$$
|
-F'+qF
|^2
=
|F'|^2
+
q^2|F|^2
-
q(|F|^2)'.
$$

Integrating by parts and using decay at $+\infty$:

$$
\boxed{
\mathcal E_\sigma[a]
=
q(\sigma)
|F_a(\sigma)|^2
+
\int_\sigma^\infty
\left[
|F_a'|^2
+
(\gamma^2+\mathcal J_2)
|F_a|^2
\right]ds.
}
$$

Thus the zero-selective signed potential is contained in an exact positive arithmetic quadratic form.

---

# 11. Fully expanded integer inequality

For $\sigma>1$:

$$
\boxed{
0
\le
q(\sigma)
\left|
\sum_{n\ge2}a_nn^{-\sigma}
\right|^2
}
$$

$$
\boxed{
+
\sum_{m,n\ge2}
a_m\overline{a_n}
\frac{
\log m\log n+\gamma^2
}{
(mn)^\sigma\log(mn)
}
}
$$

$$
\boxed{
+
\sum_{k,m,n\ge2}
j_2(k)
a_m\overline{a_n}
\frac{
1
}{
(kmn)^\sigma\log(kmn)
}.
}
$$

The final term contains the signed zero-selective almost-prime signal.

Yet the complete form is positive.

We call this the:

$$
\boxed{
\textbf{Arithmetic Darboux Quadratic Form}.
}
$$

---

# 12. Adjoint relation in multiplicative-Hilbert geometry

For finite sequences:

$$
\langle S_\gamma a,b\rangle_\sigma
=
\int_\sigma^\infty
(F_a'+qF_a)
\overline{F_b}
\,ds.
$$

Integration by parts gives:

$$
\boxed{
\langle S_\gamma a,b\rangle_\sigma
=
\langle a,T_\gamma b\rangle_\sigma
-
F_a(\sigma)\overline{F_b(\sigma)}.
}
$$

Thus on the endpoint-zero subspace:

$$
\boxed{
F_a(\sigma)=0,
}
$$

the operators become formal adjoints:

$$
\boxed{
S_\gamma
=
T_\gamma^\sharp.
}
$$

Consequently:

$$
\boxed{
S_\gamma T_\gamma
=
T_\gamma^\sharp T_\gamma
}
$$

on the constrained arithmetic Hilbert space.

---

# 13. Positive arithmetic realization achieved

Games 18–19 showed:

$$
j_2
$$

is coefficient-signed but spectrally zero-selective.

Game 20 now shows:

$$
\boxed{
\text{signed zero-selective arithmetic potential}
\rightarrow
\text{positive multiplicative-Hilbert quadratic form}.
}
$$

So positivity has indeed been transferred to a higher structural level.

---

# 14. The equilibrium pole returns in the positive square

However, the first-order factor contains:

$$
q(s)
=
h(s)+\gamma.
$$

Near:

$$
s=1,
$$

$$
q(s)
=
\frac1{s-1}
+
O(s-1).
$$

If:

$$
F_a(1)\neq0,
$$

then:

$$
-F_a'+qF_a
=
\frac{
F_a(1)
}{
s-1
}
+
O(1).
$$

Therefore:

$$
\boxed{
\mathcal E_\sigma[a]
\sim
\frac{
|F_a(1)|^2
}{
\sigma-1
}
}
$$

as:

$$
\sigma\downarrow1.
$$

The positive square reintroduces the equilibrium background that $\mathcal J_2$ had analytically annihilated.

---

# 15. Positivity–Continuation Barrier

We define:

$$
\boxed{
\textbf{Positivity–Continuation Barrier}.
}
$$

> Passing from the signed zero-selective second-order potential to its positive first-order square restores the first-order equilibrium pole. The safe positive arithmetic energy is therefore naturally confined to the Euler-product side unless the test state itself annihilates the equilibrium channel.

This identifies the exact point at which positivity costs background cancellation.

---

# 16. Equilibrium-annihilating test states

To remove the divergence, require:

$$
\boxed{
F_a(1)=0.
}
$$

If:

$$
F_a(s)
=
c(s-1)
+
O((s-1)^2),
$$

then:

$$
-F_a'(s)+q(s)F_a(s)
$$

has cancellation of the constant singular contribution and is regular at $s=1$.

Thus a suitable finite Dirichlet test function can remove the equilibrium pole before the positive square is formed.

---

# 17. Minimal two-tap annihilator

For any integer:

$$
A\ge2,
$$

define:

$$
\boxed{
F_A(s)
=
A^{-s}
-
A(A^2)^{-s}
=
A^{-s}
(
1-A^{1-s}
).
}
$$

Then:

$$
\boxed{
F_A(1)=0.
}
$$

Its zeros satisfy:

$$
A^{1-s}=1,
$$

hence:

$$
\boxed{
\Re s=1.
}
$$

Therefore:

$$
\boxed{
F_A(\rho)\neq0
}
$$

for every nontrivial zeta zero, since all such zeros satisfy:

$$
0<\Re\rho<1.
$$

This is a minimal finite filter that annihilates the equilibrium pole without annihilating any nontrivial zero mode.

---

# 18. Zero-Preserving Dilation Notch

We call:

$$
\boxed{
F_A(s)
=
A^{-s}(1-A^{1-s})
}
$$

a:

$$
\boxed{
\textbf{Zero-Preserving Dilation Notch}.
}
$$

It has:

- a notch at the deterministic exponent $s=1$;
- no notch anywhere in the open critical strip;
- finite arithmetic support in scale space.

---

# 19. Pure prime dilation signal

Multiply the prime logarithmic derivative by $F_A$.

The arithmetic coefficient signal is:

$$
\boxed{
c_A(n)
=
\mathbf1_{A\mid n}
\Lambda(n/A)
-
A\,
\mathbf1_{A^2\mid n}
\Lambda(n/A^2).
}
$$

Its cumulative function is:

$$
\boxed{
D_A(x)
=
\psi(x/A)
-
A\psi(x/A^2).
}
$$

The deterministic main term cancels exactly:

$$
\frac{x}{A}
-
A\frac{x}{A^2}
=
0.
$$

No continuous $x$ background needs to be subtracted.

---

# 20. Zero modes are preserved

The Mellin transform of $D_A$ in the initial half-plane is proportional to:

$$
\boxed{
F_A(s)
\frac{
-\zeta'(s)
}{
s\zeta(s)
}.
}
$$

At a nontrivial zero $\rho$, the pole residue is multiplied by:

$$
\boxed{
F_A(\rho)\neq0.
}
$$

Thus the filter preserves every nontrivial zero pole.

---

# 21. Pointwise RH criterion

The standard prime-error criterion immediately gives:

$$
\boxed{
\mathrm{RH}
\Rightarrow
D_A(x)
=
O_\varepsilon(
x^{1/2+\varepsilon}
)
}
$$

for every $\varepsilon>0$.

Conversely, a bound:

$$
\boxed{
D_A(x)
=
O_\varepsilon(
x^{1/2+\varepsilon}
)
}
$$

for every $\varepsilon>0$ makes its Mellin transform holomorphic in every half-plane:

$$
\Re s>\frac12+\varepsilon.
$$

Since $F_A$ has no zero in the open critical strip, no zeta zero can lie to the right of the critical line.

Functional symmetry gives RH.

Hence:

$$
\boxed{
\mathrm{RH}
\iff
D_A(x)
=
O_\varepsilon(
x^{1/2+\varepsilon}
)
\quad
\forall\varepsilon>0.
}
$$

This is a derived criterion.

---

# 22. Uniform critical-line gain

On the critical line:

$$
s=\frac12+i\gamma,
$$

$$
F_A(s)
=
A^{-1/2-i\gamma}
(
1-A^{1/2-i\gamma}
).
$$

Therefore:

$$
\boxed{
1-A^{-1/2}
\le
\left|
F_A\left(\frac12+i\gamma\right)
\right|
\le
1+A^{-1/2}.
}
$$

The lower bound is independent of the zero height.

Thus critical-line zeros cannot be hidden by high ordinate in this filter.

---

# 23. Height-Independent Passband Principle

We define:

$$
\boxed{
\textbf{Height-Independent Passband Principle}.
}
$$

> The two-tap dilation notch cancels the equilibrium exponent exactly while retaining a uniformly nonzero gain on the entire critical line.

This is a simple anti-camouflage property.

---

# 24. Positive weighted prime energy

For:

$$
\sigma>\frac12,
$$

define:

$$
\boxed{
\mathcal E_{A,\sigma}
=
\int_1^\infty
|D_A(x)|^2
x^{-2\sigma-1}dx.
}
$$

Under RH the pointwise critical estimate gives:

$$
\boxed{
\mathcal E_{A,\sigma}<\infty
}
$$

for every $\sigma>1/2$.

Conversely, if this energy is finite for every $\sigma>1/2$, Cauchy–Schwarz gives holomorphy of the Mellin transform in every corresponding right half-plane.

Since the filter is zero-free in the critical strip, RH follows.

Therefore:

$$
\boxed{
\mathrm{RH}
\iff
\mathcal E_{A,\sigma}<\infty
\quad
\forall\sigma>\frac12.
}
$$

---

# 25. Background-Free Positive Prime Energy

This is the desired combination:

$$
\boxed{
\text{positive energy}
+
\text{no explicit }x\text{ background}
+
\text{full nontrivial-zero preservation}.
}
$$

The price is that the filtered prime signal itself is a signed scale difference.

The quadratic energy, however, is positive.

---

# 26. Filtered prime-power measure

Define the signed dilation-filtered measure:

$$
\boxed{
\nu_A
=
\sum_{n\ge1}
\Lambda(n)
\left[
\delta_{An}
-
A\delta_{A^2n}
\right].
}
$$

Then:

$$
\boxed{
\nu_A([1,x])
=
D_A(x).
}
$$

---

# 27. Dilation-Wavelet Brownian kernel

Using the cumulative-energy identity of Game 13:

$$
\boxed{
\mathcal E_{A,\sigma}
=
\iint
\frac{
d\nu_A(u)d\nu_A(v)
}{
2\sigma
\max(u,v)^{2\sigma}
}.
}
$$

Thus the Game 13 positive max/Brownian kernel is conjugated by a finite dilation high-pass filter.

Symbolically:

$$
\boxed{
K_\sigma^{(A)}
=
H_A
K_\sigma
H_A^\ast
\succeq0.
}
$$

We call this the:

$$
\boxed{
\textbf{Dilation-Wavelet Brownian Kernel}.
}
$$

---

# 28. Explicit pair kernel

Expanding the two dilation taps gives:

$$
\boxed{
\mathcal E_{A,\sigma}
=
\frac1{2\sigma}
\sum_{m,n}
\Lambda(m)\Lambda(n)
K_{A,\sigma}(m,n),
}
$$

where:

$$
\boxed{
K_{A,\sigma}(m,n)
=
\max(Am,An)^{-2\sigma}
}
$$

$$
\boxed{
-
A\max(Am,A^2n)^{-2\sigma}
-
A\max(A^2m,An)^{-2\sigma}
+
A^2\max(A^2m,A^2n)^{-2\sigma}.
}
$$

The full quadratic form is positive by construction, even though the expanded kernel contains mixed signs.

---

# 29. Mellin spectral representation

In the initial absolute-convergence region:

$$
\boxed{
\mathcal E_{A,\sigma}
=
\frac1{2\pi}
\int_{-\infty}^{\infty}
\left|
F_A(\sigma+i\tau)
\frac{
-\zeta'/\zeta(\sigma+i\tau)
}{
\sigma+i\tau
}
\right|^2d\tau.
}
$$

Thus the dilation filter acts as an explicit multiplicative transfer function.

It removes only the equilibrium pole and leaves the nontrivial zero spectrum untouched.

---

# 30. Dilation filter as coherent-state derivative

Define normalized prime density:

$$
\boxed{
\rho(x)
=
\frac{\psi(x)}x,
}
$$

and:

$$
\boxed{
r(x)
=
\rho(x)-1.
}
$$

Put:

$$
x=A^2y.
$$

Then:

$$
D_A(A^2y)
=
\psi(Ay)-A\psi(y).
$$

Divide by $Ay$:

$$
\boxed{
\frac{
D_A(A^2y)
}{
Ay
}
=
\rho(Ay)-\rho(y)
=
r(Ay)-r(y).
}
$$

So $D_A$ is precisely a finite difference of the Game 16 one-dimensional coherent amplitude.

---

# 31. Scale telescoping

The prime number theorem gives:

$$
\boxed{
r(A^ky)\to0
}
$$

as $k\to\infty$.

Therefore:

$$
r(y)
=
-
\sum_{k=0}^\infty
[
r(A^{k+1}y)-r(A^ky)
].
$$

Using the dilation signal:

$$
\boxed{
r(y)
=
-
\sum_{k=0}^{\infty}
\frac{
D_A(A^{k+2}y)
}{
A^{k+1}y
}.
}
$$

This is an exact scale reconstruction formula.

---

# 32. Pointwise criterion without zeros

Assume:

$$
D_A(x)
=
O_\varepsilon(
x^{1/2+\varepsilon}
).
$$

Then:

$$
\left|
\frac{
D_A(A^{k+2}y)
}{
A^{k+1}y
}
\right|
$$

is bounded by:

$$
C_\varepsilon
y^{-1/2+\varepsilon}
A^{-k(1/2-\varepsilon)}
$$

up to an $A$-dependent constant.

For:

$$
0<\varepsilon<\frac12,
$$

the geometric sum converges.

Hence:

$$
\boxed{
r(y)
=
O_\varepsilon(
y^{-1/2+\varepsilon}
).
}
$$

Equivalently:

$$
\boxed{
\psi(y)-y
=
O_\varepsilon(
y^{1/2+\varepsilon}
).
}
$$

Thus the dilation criterion implies the classical RH prime-error criterion using only PNT plus scale telescoping.

---

# 33. Coherent-State Difference Principle

We define:

$$
\boxed{
\textbf{Coherent-State Difference Principle}.
}
$$

> Once the prime-density state is known to converge to equilibrium, controlling one fixed multiplicative-scale difference at critical size is enough to control the full coherent state at critical size.

This is the scale-domain reason the two-tap filter retains complete RH strength.

---

# 34. Weighted Hilbert-space form

For:

$$
\frac12<\sigma<1,
$$

define:

$$
\boxed{
\|r\|_\sigma^2
=
\int_1^\infty
|r(y)|^2
y^{1-2\sigma}dy.
}
$$

Since:

$$
\psi(y)-y
=
yr(y),
$$

this equals the Game 12 prime-error energy:

$$
\boxed{
\|r\|_\sigma^2
=
\int_1^\infty
|\psi(y)-y|^2
y^{-2\sigma-1}dy.
}
$$

---

# 35. Scale shift is a contraction

Define:

$$
\boxed{
(S_A r)(y)
=
r(Ay).
}
$$

Then:

$$
\|S_A r\|_\sigma^2
=
A^{-2(1-\sigma)}
\int_A^\infty
|r(u)|^2
u^{1-2\sigma}du.
$$

Therefore:

$$
\boxed{
\|S_A\|
\le
A^{-(1-\sigma)}
=
q
<
1.
}
$$

So:

$$
I-S_A
$$

is invertible on the weighted space.

---

# 36. Dilation-Difference Isomorphism

The Neumann series gives:

$$
\boxed{
(I-S_A)^{-1}
=
\sum_{k\ge0}S_A^k.
}
$$

Hence:

$$
\boxed{
(1-q)
\|r\|_\sigma
\le
\|
(S_A-I)r
\|_\sigma
\le
(1+q)
\|r\|_\sigma.
}
$$

Using:

$$
(S_A-I)r(y)
=
\frac{
D_A(A^2y)
}{
Ay
},
$$

the background-free dilation energy is norm-equivalent to the original prime-residual energy.

We call this:

$$
\boxed{
\textbf{Dilation-Difference Isomorphism}.
}
$$

---

# 37. Exact energy scaling

A change of variables gives:

$$
\boxed{
\mathcal E_{A,\sigma}
=
A^{2-4\sigma}
\|
(S_A-I)r
\|_\sigma^2.
}
$$

Therefore:

$$
\boxed{
A^{2-4\sigma}
(1-q)^2
\|r\|_\sigma^2
\le
\mathcal E_{A,\sigma}
}
$$

and:

$$
\boxed{
\mathcal E_{A,\sigma}
\le
A^{2-4\sigma}
(1+q)^2
\|r\|_\sigma^2.
}
$$

So in the critical range:

$$
\frac12<\sigma<1,
$$

the two positive energies contain the same Hilbert-space information.

---

# 38. $A=2$ as an anchored Haar detail

For:

$$
A=2,
$$

$$
\boxed{
D_2(x)
=
\psi(x/2)-2\psi(x/4).
}
$$

Rewrite:

$$
\boxed{
D_2(x)
=
[
\psi(x/2)-\psi(x/4)
]
-
\psi(x/4).
}
$$

The two Lebesgue expected masses are both:

$$
x/4.
$$

So $D_2$ is an anchored Haar-type coefficient of the prime-power measure.

It compares the upper half interval:

$$
(x/4,x/2]
$$

against the lower interval:

$$
[1,x/4].
$$

---

# 39. Prime Haar interpretation

This gives:

$$
\boxed{
\textbf{Prime Dilation-Haar Detail}.
}
$$

The RH statement becomes:

> the zero-mean Haar/detail coefficients of the prime-power measure grow only at the square-root critical scale.

The continuous-scale weighted energy is the corresponding Haar square energy.

This connects directly back to the Haar-scale language developed earlier in the Zero Genesis sequence.

---

# 40. Higher-order finite dilation filters

Define:

$$
\boxed{
F_{A,r}(s)
=
A^{-s}
(
1-A^{1-s}
)^r.
}
$$

Then $F_{A,r}$ has an order-$r$ zero at $s=1$.

Its zeros remain on:

$$
\Re s=1.
$$

So every nontrivial zeta zero remains visible.

The corresponding scale-domain signal is:

$$
\boxed{
D_{A,r}(x)
=
\sum_{j=0}^{r}
(-1)^j
\binom rj
A^j
\psi
\left(
\frac{x}{A^{j+1}}
\right).
}
$$

This is an $r$th-order multiplicative finite difference.

---

# 41. Finite-Dilation Regularization Hierarchy

We define:

$$
\boxed{
\textbf{Finite-Dilation Regularization Hierarchy}.
}
$$

> Deterministic poles at a known scaling exponent can be annihilated by placing finite-difference zeros at that exponent, while spectral poles away from the notch line are preserved.

This provides a scale-domain alternative to differential reciprocal-zeta regularization.

---

# 42. Application to multiplicative $k$-pair amplifiers

The raw $k$-fold prime convolution has transfer:

$$
h(s)^k.
$$

Its equilibrium pole at $s=1$ has order $k$.

Therefore:

$$
\boxed{
F_{A,k}(s)h(s)^k
}
$$

has the full equilibrium pole removed.

But since:

$$
F_{A,k}(\rho)\neq0
$$

for every nontrivial zero, the $k$th-order zero pole survives.

Thus one may background-regularize the raw pole amplifier entirely by finite scale differences.

---

# 43. Comparison with reciprocal-Selberg regularization

There are now two different zero-preserving background-removal mechanisms.

## Reciprocal derivative route

$$
\boxed{
(1/\zeta)^{(k)}/(1/\zeta)
}
$$

Advantages:

- natural zero-pole amplification;
- almost-prime interaction cumulants;
- direct Darboux structure.

Cost:

- signed arithmetic coefficients;
- higher divisor-interaction complexity.

## Finite dilation route

$$
\boxed{
F_{A,k}(s)h(s)^k
}
$$

Advantages:

- finite number of scale taps;
- exact equilibrium annihilation;
- zero pole order preserved;
- simple positive square energies.

Cost:

- signed cross-scale differences;
- does not lower the underlying prime-correlation order $k$.

---

# 44. What Game 20 solves

Game 20 closes the Game 19 question:

$$
\boxed{
\text{Can the signed zero-selective Darboux potential be realized as positive arithmetic energy?}
}
$$

Yes.

It admits an exact positive multiplicative-Hilbert quadratic form.

The game also identifies why this positivity does not immediately prove RH:

$$
\boxed{
\text{the first-order square restores the equilibrium pole}.
}
$$

This is the Positivity–Continuation Barrier.

A finite dilation notch removes that equilibrium mode before energy formation and yields a background-free positive prime energy that remains RH-equivalent.

---

# 45. What remains hard

The new energy:

$$
\boxed{
\mathcal E_{A,\sigma}
=
\int_1^\infty
|
\psi(x/A)
-
A\psi(x/A^2)
|^2
x^{-2\sigma-1}dx
}
$$

is positive, background-free, and zero-preserving.

But proving:

$$
\boxed{
\mathcal E_{A,\sigma}<\infty
\quad
\forall\sigma>\frac12
}
$$

from known prime arithmetic is still RH-hard.

The representation removes deterministic background complexity.

It does not create the missing cancellation law.

---

# 46. New representation fixed point

The coherent RH problem can now be stated without:

- explicit zero locations;
- continuous main-term subtraction;
- Ramanujan local-shape bookkeeping;
- signed semiprime potentials.

It becomes:

$$
\boxed{
\text{Why does one fixed zero-mean dilation difference of prime mass have only square-root critical energy?}
}
$$

That is a substantially cleaner prime-side fixed point.

---

# 47. Game 21 candidate

The most promising next object is a **localized multiplicative shell wavelet**, rather than another cumulative anchored filter.

Define:

$$
\boxed{
W_A(y)
=
[
\psi(A^2y)-\psi(Ay)
]
-
A[
\psi(Ay)-\psi(y)
].
}
$$

Equivalently:

$$
\boxed{
W_A(y)
=
\psi(A^2y)
-
(A+1)\psi(Ay)
+
A\psi(y).
}
$$

This compares two adjacent multiplicative shells after equalizing their expected masses.

Its zero transfer factor is proportional to:

$$
\boxed{
(A^s-1)(A^s-A).
}
$$

The zeros lie on:

$$
\Re s=0
$$

or:

$$
\Re s=1,
$$

so no nontrivial zeta zero is annihilated.

This is a genuinely localized shell wavelet.

Possible Game 21 question:

> Does the localized shell-wavelet energy admit stronger arithmetic control than the anchored cumulative dilation energy while retaining full RH sensitivity?

This would test whether locality in multiplicative scale provides real proof leverage.

---

# 48. Monster status

After Game 20:

$$
\boxed{
\text{Arithmetic First-Order Factors: FOUND}
}
$$

$$
\boxed{
\text{Exact Arithmetic Darboux Pair: FOUND}
}
$$

$$
\boxed{
\text{Arithmetic Dirac Lift: FOUND}
}
$$

$$
\boxed{
\text{Multiplicative-Hilbert Positive Geometry: IDENTIFIED}
}
$$

$$
\boxed{
\text{Arithmetic Darboux Quadratic Form: FOUND}
}
$$

$$
\boxed{
\text{Positivity–Continuation Barrier: IDENTIFIED}
}
$$

$$
\boxed{
\text{Zero-Preserving Dilation Notch: FOUND}
}
$$

$$
\boxed{
\text{Background-Free Positive Prime Energy: FOUND}
}
$$

$$
\boxed{
\text{Dilation-Wavelet Brownian Kernel: FOUND}
}
$$

$$
\boxed{
\text{Coherent-State Scale Telescoping: FOUND}
}
$$

$$
\boxed{
\text{Dilation-Difference Isomorphism: FOUND}
}
$$

$$
\boxed{
\text{Finite-Dilation Regularization Hierarchy: FOUND}
}
$$

The surviving monster is:

$$
\boxed{
\textbf{Critical Dilation-Wavelet Prime Cancellation}.
}
$$

---

# 49. Final diagram

$$
\boxed{
\begin{aligned}
a
&\xrightarrow{
T_\gamma=L+C_\Lambda+\gamma
}
b
\\
\|b\|_{K_\sigma}^2
&\ge0
\\
S_\gamma T_\gamma
&=
-L^2+C_{j_2}+\gamma^2
\\[4pt]
F_A(s)
&=
A^{-s}(1-A^{1-s})
\\
F_A(1)
&=0
\\
F_A(\rho)
&\neq0
\\
D_A(x)
&=
\psi(x/A)-A\psi(x/A^2)
\\
\frac{D_A(A^2y)}{Ay}
&=
r(Ay)-r(y)
\\
\mathcal E_{A,\sigma}
&\asymp
\mathcal E_{\rm prime,\sigma}.
\end{aligned}
}
$$

Thus the original critical residual and the background-free dilation-wavelet residual are the same coherent degree of freedom in different gauges.

---

# 50. Conclusion

Game 20 begins by pulling the Game 19 Darboux factorization all the way back into arithmetic coefficient space.

The first-order factors become:

$$
T_\gamma
=
L+C_\Lambda+\gamma I,
$$

$$
S_\gamma
=
-L+C_\Lambda+\gamma I.
$$

Their products recover exactly the signed zero-selective and positive Selberg partner arithmetic operators.

The natural coefficient Hilbert geometry is generated by:

$$
K_\sigma(m,n)
=
\frac1{(mn)^\sigma\log(mn)}.
$$

At $\sigma=1/2$ this is the known multiplicative Hilbert matrix.

In the Euler-product half-plane, the zero-selective signed potential therefore admits an exact positive arithmetic quadratic realization.

However, the positive first-order square restores the equilibrium pole at $s=1$.

This exposes a Positivity–Continuation Barrier.

The way around the deterministic background is surprisingly elementary.

A two-tap finite dilation filter:

$$
F_A(s)
=
A^{-s}(1-A^{1-s})
$$

annihilates the equilibrium exponent $s=1$ while remaining zero-free throughout the open critical strip.

Its prime-side signal is:

$$
D_A(x)
=
\psi(x/A)-A\psi(x/A^2).
$$

No explicit $x$ main term appears.

All nontrivial zero modes survive.

Its weighted square is a positive dilation-wavelet Brownian energy.

Most importantly, this filtered signal is exactly the finite-scale derivative of the normalized coherent prime-density state:

$$
\frac{D_A(A^2y)}{Ay}
=
\frac{\psi(Ay)}{Ay}
-
\frac{\psi(y)}y.
$$

Using only the prime number theorem, the coherent state can be reconstructed by scale telescoping.

On the weighted critical Hilbert space, the scale-difference operator is an isomorphism.

Therefore the Game 12 prime-error energy and the Game 20 background-free dilation energy are norm-equivalent.

The final problem can now be stated in an unusually compact arithmetic form:

> Why does a fixed zero-mean multiplicative Haar difference of prime mass have only critical square-root energy?

That is the surviving **Critical Dilation-Wavelet Prime Cancellation** problem.

---

# References

1. Brevig, O. F., Perfekt, K.-M., Seip, K., Siskakis, A. G., & Vukotić, D. *The multiplicative Hilbert matrix*. Advances in Mathematics 302 (2016), 410–432; arXiv:1411.7294.
2. Perfekt, K.-M., & Pushnitski, A. *On the spectrum of the multiplicative Hilbert matrix*. arXiv:1705.01959.
3. Classical identity:
   $$
   -\zeta'/\zeta(s)
   =
   \sum_n\Lambda(n)n^{-s}
   $$
   for $\Re s>1$.
4. Classical Mellin transform of the Chebyshev function:
   $$
   \int_1^\infty
   \psi(x)x^{-s-1}dx
   =
   -\frac{\zeta'(s)}{s\zeta(s)}.
   $$
5. Classical prime number theorem and RH prime-error equivalence.
6. Classical Gram representation:
   $$
   \int_\sigma^\infty
   (mn)^{-s}ds
   =
   \frac{(mn)^{-\sigma}}{\log(mn)}.
   $$
7. Classical Neumann-series inversion for a strict contraction.