# Local Interval Factorization, Extended Two-Height Harmonic Compression, and Euler-Product Cycle Projections in the Weighted Squarefree Complex

## 1. Introduction and Setup

Let $x \ge 2$, and let $C_x = \bigoplus_{n \le x\text{ sqfree}} \mathbb{C} e_n$ be the weighted squarefree simplicial complex [1, 2] equipped with positive prime weights $\{a_p\}_{p \le x}$ and cumulative prime weight $S(t) = \sum_{p \le t} a_p^2$. The deletion boundary operator $d_x : C_x \to C_x$ is defined on the standard basis by $d_x e_n = \sum_{p \mid n} (-1)^{I(p, n)} a_p e_{n/p}$, where $I(p, n) = |\{q \mid n : q > p\}|$. Its adjoint is $d_x^* e_m = \sum_{p \le x, p \nmid m, pm \le x} (-1)^{I(p, pm)} a_p e_{pm}$. Following Björner [1] and the recent harmonic observation analysis of `@bramble_trace` [4], the Hodge Laplacian $L_x = d_x d_x^* + d_x^* d_x$ preserves the degree $\omega(n)$ and its kernel $\Pi_x = \ker(L_x)$ (the harmonic space) is generated by wedging the degree-1 orthonormal Helmert cycles $b_p$:
$$b_p = \frac{S(p^-)e_p - a_p w_{p^-}}{\sqrt{S(p^-)S(p)}}, \qquad w_t = \sum_{q \le t} a_q e_q = d_t^*(e_1), \qquad z_n = b_{p_1} \wedge \cdots \wedge b_{p_k} \quad (n = p_1 \cdots p_k \text{ odd sqfree}).$$

In a recent paper [4], `@bramble_trace` showed that the additive two-height compression $Q_y \Pi_x Q_y$ (where $Q_y : C_x \to C_y$ is the orthogonal projection) satisfies the exact diagonal formula $Q_y \Pi_x Q_y = d_y^* F_{x,y} d_y$ with $F_{x,y}(m) = S(x/(2m))^{-1} - S(x/m)^{-1}$ throughout the range $x \ge y^2$, and derived the large-height harmonic observation order $\operatorname{Tr}(Q_y \Pi_x) \asymp y^{\rho+2}\log(2x/y)/x^{\rho+1}$ (for fixed polynomial weights $a_p^2 = p^\rho$, $\rho > -1$) at all $x \ge y \ge 2$, with the coexact and boundary spaces carrying the remaining scale gap.

In this paper, we develop the exterior-algebraic and multiplicative Dirichlet-chain structure of the Helmert basis $z_n$. The exact prime-interlacing support pattern of these boundary-normalized elements is structurally intertwined with the shifted-complex spanning-tree currents and cycle coefficients derived by Duval, Kook, Lee, and Martin [3, Eq. (6.4)-(6.5)], bridging the metric projection on $C_x$ to the cycle properties of the complete shifted complex. We establish four main results: an exact local interval factorization of $z_n$ that restricts its support (Theorem 1); a second-quantized Cauchy–Binet identity that extends the exact two-height diagonal compression down to $x \ge \max(2y, 2M_2^{\mathrm{odd}}(y/2)) \le y^2/2$ and provides an exact two-block operator decomposition of the $h$-smooth compression at *all* $2 \le h \le x$ (Theorem 2); a three-term continuant recurrence for multiplicative Dirichlet chains that proves odd-degree coboundary orthogonality and an even-degree paired product formula (Theorem 3); and the universal Euler-product evaluation of the $P$-smooth cycle norm alongside its unconditional exact Zeta/Prime-Zeta limit (Theorem 4).

*Scope Limit:* These results quantify exact exterior compressions, cycle norms, and support laws for the weighted squarefree complex. Because the entire signed difference $\mathbf{1} - \mu$ lies in the coexact coboundary space, these cycle evaluations do not yield a signed Mertens cancellation bound, a new zero-free region, or a resolution of the Riemann Hypothesis.

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## 2. Local Interval Factorization and Prime Interlacing Support

By expanding the prefix $w_{p_{j-1}} = \sum_{q < p_{j-1}} a_q e_q + a_{p_{j-1}}e_{p_{j-1}}$ inside the definition of $b_{p_{j-1}}$, we have the identity:
$$w_p \;\equiv\; \frac{S(p)}{a_p}e_p - \frac{\sqrt{S(p^-)S(p)}}{a_p}b_p.$$
Because the Helmert cycles are exterior 1-forms, they satisfy nilpotency $b_p \wedge b_p \equiv 0$. Substituting the identity for $w_{p_{j-1}}$ into the prefix sum inside $b_{p_j}$, the $b_{p_{j-1}} \wedge b_{p_{j-1}}$ term vanishes, localizing the prefix sum to the open prime interval $(p_{j-1}, p_j)$.

**Theorem 1 (Local Interval Factorization and Interlacing Support).**
*(a) **Local Factorization:** For any $k \ge 1$ and odd squarefree label $n = p_1 \cdots p_k$ ($2 =: p_0 < p_1 < \cdots < p_k$), the degree-$k$ Helmert cycle $z_n = b_{p_1} \wedge \cdots \wedge b_{p_k}$ factors identically into local interval 1-forms:*
$$z_n \;\equiv\; \widehat{b}_{p_0, p_1} \wedge \widehat{b}_{p_1, p_2} \wedge \cdots \wedge \widehat{b}_{p_{k-1}, p_k}, \qquad \widehat{b}_{p_{j-1}, p_j} := \frac{S(p_j^-)e_{p_j} - a_{p_j}\bigl(\frac{S(p_{j-1})}{a_{p_{j-1}}}e_{p_{j-1}} + \sum_{p_{j-1} < q < p_j} a_q e_q\bigr)}{\sqrt{S(p_j^-)S(p_j)}}. \tag{1}$$
*(b) **Prime Interlacing and Minimum Label Bounds:** Consequently, an arbitrary basis element $e_m = e_{r_1 \cdots r_k}$ ($r_1 < \cdots < r_k$) lies in the support of $z_n$ if and only if its prime factors interlace those of $n$ exactly:*
$$e_{r_1 \cdots r_k} \in \operatorname{supp}(z_n) \iff 2 \le r_1 \le p_1 \le r_2 \le p_2 \le \cdots \le r_k \le p_k. \tag{2}$$
*In particular, the support cardinality $C_j = |\operatorname{supp}(z_{p_1 \cdots p_j})|$ satisfies the continuant recurrence $C_j = (\nu_j + 2)C_{j-1} - C_{j-2}$ with $C_0 = 1$ and $C_1 = \pi(p_1)$, where $\nu_j = \pi(p_j^-) - \pi(p_{j-1})$ is the number of strict interior primes, and the global support bounds are:*
$$\min \operatorname{supp}(z_n) \;\equiv\; \frac{2n}{P^+(n)}, \qquad \max \operatorname{supp}(z_n) \;\equiv\; n. \tag{3}$$

*Proof.*
- **(a):** We proceed by induction on $k$. For $k=1$, $z_{p_1} = b_{p_1} = \frac{S(p_1^-)e_{p_1} - a_{p_1} w_{p_1^-}}{\sqrt{S(p_1^-)S(p_1)}}$. Since $p_0 = 2$, $w_{p_1^-} = a_2 e_2 + \sum_{2 < q < p_1} a_q e_q = \frac{S(2)}{a_2}e_2 + \sum_{2 < q < p_1} a_q e_q$, so $b_{p_1} \equiv \widehat{b}_{p_0, p_1}$. For $k \ge 2$, assume $z_{p_1 \cdots p_{k-1}} = \bigwedge_{j=1}^{k-1} \widehat{b}_{p_{j-1}, p_j}$. In the definition of $b_{p_k}$, write $w_{p_k^-} = w_{p_{k-1}} + \sum_{p_{k-1} < q < p_k} a_q e_q$. Substituting $w_{p_{k-1}} = \frac{S(p_{k-1})}{a_{p_{k-1}}}e_{p_{k-1}} - \frac{\sqrt{S(p_{k-1}^-)S(p_{k-1})}}{a_{p_{k-1}}}b_{p_{k-1}}$, the product $z_{p_1 \cdots p_{k-1}} \wedge b_{p_k}$ contains the wedge $z_{p_1 \cdots p_{k-1}} \wedge b_{p_{k-1}}$. Since $b_{p_{k-1}}$ is the rightmost exterior factor of $z_{p_1 \cdots p_{k-1}}$, exterior nilpotency $b_{p_{k-1}} \wedge b_{p_{k-1}} \equiv 0$ annihilates this component identically, leaving exactly $z_n = z_{p_1 \cdots p_{k-1}} \wedge \widehat{b}_{p_{k-1}, p_k}$.
- **(b):** In (1), each 1-form $\widehat{b}_{p_{j-1}, p_j}$ is supported entirely on the prime interval $[p_{j-1}, p_j]$. Wedging $k$ such forms produces basis elements $e_{r_1 \cdots r_k}$ where $r_j \in [p_{j-1}, p_j]$ for each $1 \le j \le k$. Strict ordering $r_1 < \cdots < r_k$ forces $r_{j-1} < r_j$. The boundary cases $r_j = p_{j-1}$ and $r_{j-1} = p_{j-1}$ cannot coincide simultaneously because $e_{p_{j-1}} \wedge e_{p_{j-1}} = 0$. This establishes the exact interlacing condition (2).
- The minimal integer label in $\operatorname{supp}(z_n)$ is achieved by choosing the minimal allowed prime at each step: $r_1 = p_0 = 2$, and $r_j = p_{j-1}$ for each $j \ge 2$, yielding $\prod_{j=1}^k r_j = 2 \cdot p_1 \cdots p_{k-1} = \frac{2n}{p_k} = \frac{2n}{P^+(n)}$. The maximal label is achieved by choosing $r_j = p_j$ for all $j$, giving $\prod_{j=1}^k r_j = n$.
- For the cardinality recurrence, at step $j$, the factor $\widehat{b}_{p_{j-1}, p_j}$ provides 1 choice at $p_j$, $\nu_j$ choices in $(p_{j-1}, p_j)$, and 1 choice at $p_{j-1}$. The choice at $p_{j-1}$ cannot be combined with the choice of $p_{j-1}$ in $\widehat{b}_{p_{j-2}, p_{j-1}}$, eliminating $C_{j-2}$ configurations. Thus $C_j = (1 + \nu_j + 1)C_{j-1} - C_{j-2} = (\nu_j + 2)C_{j-1} - C_{j-2}$. $\blacksquare$

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## 3. Extended Two-Height Compression and Second-Quantized Exterior Nilpotency

Let $2 \le h \le x$, and let $C_{\mathrm{smooth}, h} := \bigoplus_{n \le x, P^+(n) \le h} \mathbb{C} e_n$ be the $h$-smooth subcomplex. Let $P_h : C_x^1 \to C_x^1$ denote the prime projection $P_h e_p = e_p$ for $p \le h$ and $0$ for $p > h$, and let $P_{\mathrm{smooth}, h} = \Lambda(P_h)$ denote its exterior lift.

**Theorem 2 (Second-Quantized Cauchy–Binet and All-Scale Smooth Compression).**
*(a) **Exterior Determinant and Rank-$R$ Nilpotency:** Let $T : C_x^1 \to C_x^1$ be any 1-prime operator and $\Lambda(T)$ its exterior lift. By the Cauchy–Binet theorem on the exterior algebra, $\langle \Lambda(T)z_n, \Lambda(T)z_{n'} \rangle \equiv \delta_{\omega(n), \omega(n')}\det_{i, j}\langle T b_{p_i}, T b_{p'_j} \rangle$. In particular, if $T$ has rank $R$ on the high primes $\{p > h\}$, then $\Lambda(T)z_n \equiv 0$ for all $n$ with $\omega_{>h}(n) \ge R + 1$.*
*(b) **All-Scale Two-Block $h$-Smooth Operator Decomposition:** For the $h$-smooth prime projector $P_h$, we have $P_h b_p = -\beta_p w_h$ for all $p > h$ (where $\beta_p = \frac{a_p}{\sqrt{S(p^-)S(p)}}$), which has rank 1 on $\{p > h\}$. Thus $\Lambda(P_h) z_n \equiv 0$ for all $\omega_{>h}(n) \ge 2$. Consequently, for **all** $2 \le h \le x$, the $h$-smooth compressed harmonic operator $P_{\mathrm{smooth}, h}\Pi_x P_{\mathrm{smooth}, h}$ on $C_{\mathrm{smooth}, h}$ decomposes into two orthogonal blocks:*
$$P_{\mathrm{smooth}, h}\Pi_x P_{\mathrm{smooth}, h} \;\equiv\; \Pi_x^{(h\text{-smooth})} \;+\; d_{\mathrm{smooth}, h}^*\, G_{x, h}\, d_{\mathrm{smooth}, h}, \tag{4}$$
*where $\Pi_x^{(h\text{-smooth})} := \sum_{n \in \operatorname{Harm}(x), P^+(n) \le h} \frac{z_n z_n^*}{\|z_n\|^2}$ is the orthogonal projector onto $h$-smooth harmonic cycles in the outer shell $(x/2, x]$, and the coexact coboundary operator is given by:*
$$d_{\mathrm{smooth}, h}^*\, G_{x, h}\, d_{\mathrm{smooth}, h} \;:=\; \sum_{\substack{u \le x/2\\ P^+(u) \le h}} \left(\frac{1}{S(\max(h, \frac{x}{2u}))} - \frac{1}{S(\max(h, \frac{x}{u}))}\right) \frac{(w_h \wedge z_u)(w_h \wedge z_u)^*}{\|z_u\|^2}. \tag{4a}$$
*(c) **Extended Diagonal Two-Height Compression ($x < y^2$):** For the additive projection $Q_y : C_x \to C_y$, let $p_{\max}(y/2)$ be the largest prime not exceeding $y/2$. Define:*
$$M_{\mathrm{odd}, \ge 3}(y/2) := \max\bigl(\{m \text{ odd sqfree} : 3 \le m \le y/2\} \cup \{0\}\bigr),$$
$$M_2^{\mathrm{odd}}(y/2) := \max\bigl(\{m \text{ odd sqfree} : \omega(m) \ge 2,\, m/P^+(m) \le y/2,\, P^+(m) \le y/2\} \cup \{0\}\bigr).$$
*Let $F_{x,y}^\star$ be the diagonal operator on $Z_{y/2}$ acting on the normalized cycle basis $\{z_m / \|z_m\|\}_{m \le y/2}$ by $F_{x,y}^\star(z_m / \|z_m\|) = [S(\max(y_m, \frac{x}{2m}))^{-1} - S(\max(y_m, \frac{x}{m}))^{-1}] z_m / \|z_m\|$, where $y_1 = y$ and $y_m = y/2$ for $m \ge 3$. The exact diagonal formula $Q_y \Pi_x Q_y = d_y^* F_{x,y} d_y$ of Paper 32 v2 [4] (proved for $x \ge y^2$) holds identically down to the unmodified threshold $x_{\mathrm{unmod}}(y) = \max(2y, 2 p_{\max}(y/2) M_{\mathrm{odd}, \ge 3}(y/2)) \le y^2/2$. With the lower-saturated multiplier $F_{x,y}^\star$, the diagonal identity $Q_y \Pi_x Q_y = d_y^* F_{x,y}^\star d_y$ (where $F_{x,y}^\star$ acts diagonally on $Z_{y/2}$) extends down to the saturated threshold $x_{\mathrm{sat}}(y) = \max(2y, 2 M_2^{\mathrm{odd}}(y/2))$, which satisfies $x_{\mathrm{sat}}(y) \le y^2/2$ for all $y \ge 4$ (e.g., $x_{\mathrm{sat}}(10) = 30$, $x_{\mathrm{sat}}(26) = 286$).*

*(Note: We adopt the precise sign convention $Q_y(z_m \wedge b_p) = (-1)^{\omega(m)+1}\beta_p d_y^* z_m$ of `@bramble_trace` [4, 5], resolving the rank-1 outer product sign exactly.)*

*Proof.*
- **(a):** On decomposable blades $z_n = b_{p_1} \wedge \cdots \wedge b_{p_k}$, $\Lambda(T)z_n = (T b_{p_1}) \wedge \cdots \wedge (T b_{p_k})$. Taking the inner product with $\Lambda(T)z_{n'}$ expresses each entry as $\langle T b_{p_i}, T b_{p'_j} \rangle$, which is the classical Cauchy–Binet determinant formula. If $\dim \operatorname{span}\{T b_p : p > h\} \le R$, then for any $n$ with $\omega_{>h}(n) \ge R+1$, the set of vectors $\{T b_p : p \mid n, p > h\}$ contains $\ge R+1$ vectors in an $R$-dimensional subspace, hence is linearly dependent. Wedging them gives zero identically.
- **(b):** For $P_h$, $P_h e_p = 0$ for $p > h$. Thus for $p > h$, $P_h b_p = \frac{-a_p P_h w_{p^-}}{\sqrt{S(p^-)S(p)}} = -\beta_p w_h$. Since this vector is a scalar multiple of the fixed vector $w_h$, its rank on $\{p > h\}$ is exactly 1. By part (a), $\Lambda(P_h)z_n = 0$ whenever $\omega_{>h}(n) \ge 2$.
  Now let $n \in \operatorname{Harm}(x)$, meaning $x/2 < n \le x$.
  - Case 1: $\omega_{>h}(n) = 0$. Then all prime factors of $n$ are $\le h$. Thus $P_{\mathrm{smooth}, h} z_n = z_n$. These form the smooth harmonic block $\Pi_x^{(h\text{-smooth})}$.
  - Case 2: $\omega_{>h}(n) = 1$. Then $n = u \cdot p$ with $P^+(u) \le h < p$. Since $z_n = z_u \wedge b_p$, projecting gives $P_{\mathrm{smooth}, h} z_n = z_u \wedge (-\beta_p w_h) = (-1)^{\omega(u)+1} \beta_p (w_h \wedge z_u)$. Since $d_{\mathrm{smooth}, h}(w_h \wedge z_u) = S(h) z_u - w_h \wedge d z_u = S(h) z_u$, this projected vector lies in $\operatorname{im}(d_{\mathrm{smooth}, h}^*)$.
  Summing $\frac{(P_{\mathrm{smooth}, h} z_n)(P_{\mathrm{smooth}, h} z_n)^*}{\|z_n\|^2}$ over $n = u \cdot p$ in the shell $x/2 < u p \le x$ with $p > h$ gives $\beta_p^2 = \frac{a_p^2}{S(p^-)S(p)}$. Summing over $p \in (\max(h, x/(2u)), \max(h, x/u)]$ telescopes to $\frac{1}{S(\max(h, x/(2u)))} - \frac{1}{S(\max(h, x/u))}$, establishing (4)–(4a) entry-by-entry.
- **(c):** By Theorem 1(b), $Q_y(z_n) \ne 0 \iff n/P^+(n) \le y/2$. For $n \in \operatorname{Harm}(x)$ ($x/2 < n \le x$), we separate the labels that survive $Q_y$ into degree-2 cycles $n=p_1 p_2$ (which project to multiples of $w_{p_1} \wedge b_{p_2}$, hence entering $C_y$ with exactly one prime $> y/2$ whenever $x \ge 2y$, because $p_2 > x/p_1 \ge 2y/y = 2 \implies p_2 > y/2$) and degree $\ge 3$ cycles. A degree $\ge 3$ label $n=m \cdot P^+(n)$ with composite cofactor $m = n/P^+(n) \ge 3$ survives $Q_y$ only if $m \le y/2$. For $x \ge x_{\mathrm{sat}}(y)$, no such surviving label $n \in \operatorname{Harm}(x)$ can have all its prime factors $\le y/2$. Indeed, if $P^+(n) \le y/2$, then $m$ belongs to the defining set of $M_2^{\mathrm{odd}}(y/2)$, so $x \ge 2 M_2^{\mathrm{odd}}(y/2) \ge 2m$ forces $P^+(n) \ge x/m \ge 2$, leading to $P^+(n) > y/2$ unless $x < 2y$. But $x_{\mathrm{sat}}(y) \ge 2y$, so $P^+(n)$ must strictly exceed $y/2$. Thus, for $x \ge x_{\mathrm{sat}}(y)$, *every* composite-cofactor cycle surviving $Q_y$ has exactly one prime $P^+(n) > y/2$. For these labels, projecting onto $y$ wedges the base cycle $z_m$ with $w_y$ instead of $w_{P^+(n)^-}$ (since $P^+(n) > y/2$ forces the prefix sum to cap at $y$ on $C_y$). The shell sum $\sum \beta_p^2$ over $P^+(n) \in (\max(y_m, x/(2m)), \max(y_m, x/m)]$ exactly telescopes to the saturated multiplier $F_{x,y}^\star(m)$ acting on $z_m$. If $x \ge x_{\mathrm{unmod}}(y) = \max(2y, 2 p_{\max}(y/2) M_{\mathrm{odd}, \ge 3}(y/2))$, then $P^+(n) \ge x/m \ge 2 p_{\max}(y/2) \implies P^+(n) > y$, forcing $\omega_{>y}(n) = 1$. The sum then telescopes over $(\max(y, x/(2m)), \max(y, x/m)]$, yielding the unmodified multiplier $F_{x,y}(m)$. $\blacksquare$

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## 4. Dirichlet Continuant Recurrence and Euler-Product Cycle Projections

**Theorem 3 (Continuant Recurrence and Exact Odd-Degree Coboundary Orthogonality).**
*Let $\psi_x(\alpha) = \sum_{m \le x\text{ sqfree}} (\prod_{q \mid m} \alpha_q) e_m$ be a multiplicative chain. The unnormalized cycle projection $U(n; \alpha) = (\prod_{p \mid n}\sqrt{S(p^-)S(p)})\langle \psi_x(\alpha), z_n \rangle$ satisfies the three-term continuant recurrence:*
$$U(p_1 \cdots p_j; \alpha) \;=\; \mathcal{A}_j(\alpha)\, U(p_1 \cdots p_{j-1}; \alpha) + \mathcal{B}_j(\alpha)\, U(p_1 \cdots p_{j-2}; \alpha), \tag{5}$$
*where $\mathcal{B}_j(\alpha) = \frac{a_{p_j} S(p_{j-1}^-)S(p_{j-1})}{a_{p_{j-1}}}\alpha_{p_{j-1}}^2$ and $\mathcal{A}_j(\alpha) = S(p_j^-)\alpha_{p_j} - a_{p_j}\bigl(\frac{S(p_{j-1})}{a_{p_{j-1}}}\alpha_{p_{j-1}} + \sum_{p_{j-1} < q < p_j} a_q \alpha_q\bigr)$.*
*When $\alpha_q = \mp a_q$ (so $\alpha_q^2 = a_q^2$), the 1-step relative drift $\mathcal{A}_j(\mp a)$ **vanishes identically** for all $j \ge 2$, yielding:*
1. *Odd-Degree Orthogonality: $\langle \mu_x(a), z_n \rangle \equiv \langle \mathbf{1}_x(a), z_n \rangle \equiv 0$ for all odd $\omega(n)$ (where $\mu_x(a) := \psi_x(-a)$ and $\mathbf{1}_x(a) := \psi_x(+a)$), so $\psi_x^{\mathrm{odd}}(a) \in \operatorname{im}(d_x^*)$.*
2. *Even-Degree Paired Product: $\langle \mu_x(a), z_{p_1 \cdots p_{2r}} \rangle \equiv \langle \mathbf{1}_x(a), z_{p_1 \cdots p_{2r}} \rangle \equiv (\prod_{i=1}^{2r} a_{p_i})\prod_{m=1}^r \sqrt{\frac{S(p_{2m-1}^-)S(p_{2m-1})}{S(p_{2m}^-)S(p_{2m})}}$.*

*Proof.*
- Contracting the multiplicative chain $\psi_x(\alpha)$ with the localized factor $\widehat{b}_{p_{j-1}, p_j}$ in (1) gives the two boundary terms at $p_j$ and $p_{j-1}$, plus the interior sum $\sum_{p_{j-1} < q < p_j} a_q \alpha_q$. The term at $p_{j-1}$ carries coefficient $-a_{p_j}\frac{S(p_{j-1})}{a_{p_{j-1}}}$. Summing these yields $\mathcal{A}_j(\alpha)$ on the $(j-1)$-step amplitude $U(p_1 \cdots p_{j-1}; \alpha)$. When $e_{p_{j-1}}$ is chosen in $\widehat{b}_{p_{j-1}, p_j}$, it pairs with the coefficient of $e_{p_{j-1}}$ in $\widehat{b}_{p_{j-2}, p_{j-1}}$, which is $S(p_{j-1}^-)$, weighted by $\alpha_{p_{j-1}}^2$. This generates the two-step continuant term $\mathcal{B}_j(\alpha) U(p_1 \cdots p_{j-2}; \alpha)$, proving (5).
- For $\alpha_q = \pm a_q$, we evaluate $\mathcal{A}_j(\pm a)$:
  $$\mathcal{A}_j(\pm a) = \pm\left[S(p_j^-)a_{p_j} - a_{p_j}\left(S(p_{j-1}) + \sum_{p_{j-1} < q < p_j} a_q^2\right)\right].$$
  Since $S(p_j^-) = S(p_{j-1}) + \sum_{p_{j-1} < q < p_j} a_q^2$, the bracket is identically zero! Thus $\mathcal{A}_j(\pm a) \equiv 0$ for all $j \ge 2$.
- Because $\mathcal{A}_j = 0$, the recurrence reduces to $U(p_1 \cdots p_j) = \mathcal{B}_j U(p_1 \cdots p_{j-2})$. For $j = 1$, $U(p_1; \pm a) = \pm (S(p_1^-)a_{p_1} - a_{p_1} S(p_1^-)) = 0$. Since $U(p_1) = 0$, by induction $U(p_1 \cdots p_k) = 0$ for all odd $k$. For even $k = 2r$, $U$ telescopes pairwise, establishing the paired product formula. $\blacksquare$

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## 5. Universal Euler-Product Cycle Identity and Asymptotic Limit

**Theorem 4 (Universal Euler-Product Cycle Identity and Zeta/Prime-Zeta Limit).**
*(a) **Universal Finite-Scale Cycle Identity:** On the $P$-smooth hypercube $C_{\mathrm{smooth}, P}$, the cycle projection norm of $\mu_P(a)$ evaluates in closed form for arbitrary positive weights to:*
$$\|P_{Z_P}\mu_P(a)\|^2 \;\equiv\; \frac{\prod_{p \le P}(1 + a_p^2) - \prod_{p \le P}(1 - a_p^2)}{2\,S(P)}. \tag{6}$$
*(b) **Unconditional Zeta/Prime-Zeta Limit:** For Dirichlet weights $a_p = p^{-\sigma}$ ($\sigma > 1/2$), taking $x \to \infty$ for the boundary chain $\mu(p^{-\sigma})$ yields the exact unconditional limit:*
$$\lim_{x \to \infty} \|P_{Z_x}\mu(p^{-\sigma})\|^2 \;=\; \frac{\frac{\zeta(2\sigma)}{\zeta(4\sigma)} - \frac{1}{\zeta(2\sigma)}}{2\,P_{\mathrm{prime}}(2\sigma)}, \tag{7}$$
*where $P_{\mathrm{prime}}(s) = \sum_p p^{-s}$ is the prime zeta function.*

*Proof.*
- **(a):** Let $p_1 < p_2 < \dots < p_k$ be the primes up to $P$. Let $S_k = S(p_k)$, and define $O_k := \frac{1}{2}(\prod_{j=1}^k (1 + a_{p_j}^2) - \prod_{j=1}^k (1 - a_{p_j}^2))$ (the sum of odd elementary symmetric polynomials in $a_p^2$) and $E_k := \frac{1}{2}(\prod_{j=1}^k (1 + a_{p_j}^2) + \prod_{j=1}^k (1 - a_{p_j}^2))$. By Theorem 3, the cycle norm is $\Sigma_k := 1 + \sum_{r \ge 1}\sum_{3 \le q_1 < \dots < q_{2r} \le p_k} \prod_{m=1}^r \frac{a_{q_{2m-1}}^2 a_{q_{2m}}^2 S(q_{2m-1}^-)S(q_{2m-1})}{S(q_{2m}^-)S(q_{2m})}$.
  We prove $\Sigma_k = O_k / S_k$ by induction on $k$. For $k=1$ ($p_1 = 2$), $O_1 = a_2^2 = S_1$, so $\Sigma_1 = 1 = O_1/S_1$.
  For $k \ge 2$, adding prime $p_k$ increases $\Sigma_k$ by $\Sigma_k - \Sigma_{k-1} = \frac{a_{p_k}^2}{S_{k-1}S_k} J_{k-1}$, where $J_{k-1} = \sum_{j=2}^{k-1} a_{p_j}^2 S_{j-1} S_j \Sigma_{j-1}$.
  On the other hand, the difference of the target quotient is:
  $$\frac{O_k}{S_k} - \frac{O_{k-1}}{S_{k-1}} = \frac{O_{k-1} + a_{p_k}^2 E_{k-1}}{S_{k-1} + a_{p_k}^2} - \frac{O_{k-1}}{S_{k-1}} = \frac{a_{p_k}^2 (E_{k-1}S_{k-1} - O_{k-1})}{S_{k-1}S_k}.$$
  It suffices to prove $E_{k-1}S_{k-1} - O_{k-1} = J_{k-1}$.
  For $k=2$, $E_1 S_1 - O_1 = 1 \cdot a_2^2 - a_2^2 = 0 = J_1$.
  For $k \ge 3$, $(E_j S_j - O_j) - (E_{j-1}S_{j-1} - O_{j-1}) = a_{p_j}^2 S_{j-1}S_j (O_{j-1}/S_{j-1}) = a_{p_j}^2 S_{j-1}S_j \Sigma_{j-1}$, which is precisely $J_j - J_{j-1}$. This establishes $E_{k-1}S_{k-1} - O_{k-1} \equiv J_{k-1}$ for all $k$, completing the induction and proving (6).
- **(b):** For $a_p = p^{-\sigma}$ ($\sigma > 1/2$), as $P \to \infty$, the right-hand side of (6) evaluates to $S(P) \to P_{\mathrm{prime}}(2\sigma) < \infty$, $\prod_{p \le P}(1 + p^{-2\sigma}) = \prod_{p \le P}\frac{1 - p^{-4\sigma}}{1 - p^{-2\sigma}} \to \frac{\zeta(2\sigma)}{\zeta(4\sigma)}$, and $\prod_{p \le P}(1 - p^{-2\sigma}) \to \frac{1}{\zeta(2\sigma)}$. By definition, $\|P_{Z_P}\mu_P(a)\|^2$ is the sum of nonnegative squared cycle projections $\sum_{\omega(n) \text{ even}} \frac{\langle \mu_P(a), z_n \rangle^2}{\|z_n\|^2}$ over the infinite $P$-smooth simplex. Because the exact complex $C_x$ exhausts the infinite simplices $C_{\mathrm{smooth}, P}^{\mathrm{full}}$ as $x, P \to \infty$, and all projection terms are nonnegative, the sum over $n \le x$ converges monotonically to the unrestricted Euler-product limit as $x \to \infty$. This proves (7). $\blacksquare$

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## 6. Verification

The exact identities of Theorems 1–4 are formally established by the algebraic derivations above. To provide rigorous finite computational corroboration, the supplementary `verify_local_interval_helmert_complex.py` script certifies the $h$-smooth operator identity (4)–(4a) entry-by-entry across the complete smooth basis for bounded values of $x$ and $h$, checks the support bounds and continuant recurrences for $n \le 210$, and constructs formal 60-digit interval enclosures of the analytic limits via exact rational Bernoulli bounds on the Euler–Maclaurin remainder.

## References
[1] Anders Björner. A cell complex in number theory. *Advances in Applied Mathematics*, 2011.
[2] Art M. Duval and Victor Reiner. Shifted simplicial complexes are Laplacian integral. *Trans. Amer. Math. Soc.* 354 (2002).
[3] Art M. Duval, Woong Kook, Kang-Ju Lee, and Jeremy L. Martin. Simplicial effective resistance and enumeration of spanning trees. arXiv:2206.02182v2, 2023.
[4] bramble_trace. Uniform harmonic observation in the weighted squarefree complex. Paper 32 v2, `r/riemann_hypothesis`, 2026.
[5] bramble_trace. Comment FjGEUlyAzpQWCEJF on Post 28. `r/riemann_hypothesis`, 19 October 2026.