Uniform harmonic observation in the weighted squarefree complex

1. Problem and contribution

The simplicial complex of squarefree integers up to a height has a known connection to the Mertens function through its Euler characteristic [1]. This paper concerns a different quantity: how much of the harmonic space at an outer height $x$ is visible on faces at an inner height $y$. We use augmented chains throughout, with exterior degree equal to the number of prime factors; the empty face therefore has degree zero.

Let $C_u$ be the real Hilbert space with orthonormal basis $e_n$ indexed by squarefree integers $n\leq u$. On the exterior algebra of prime coordinates, give prime $p$ a fixed positive weight $a_p$, and let $d_u$ be weighted deletion, restricted to $C_u$:

$$ d_u(e_{p_1}\wedge\cdots\wedge e_{p_k}) =\sum_{j=1}^k(-1)^{j-1}a_{p_j} e_{p_1}\wedge\cdots\wedge\widehat e_{p_j}\wedge \cdots\wedge e_{p_k}. $$

All primes up to the outer height are included. Put

$$ D_u=d_u+d_u^*,\qquad \Pi_u=\operatorname{proj}\ker D_u,\qquad S(t)=\sum_{p\leq t}a_p^2. $$

For $y\leq x$, $Q_y:C_x\to C_y$ is the coordinate projection, also regarded as an orthogonal projection on $C_x$.

Theorem 1 (exact observation). There is a compatible homogeneous orthonormal cycle basis $z_n$, indexed by odd squarefree integers, such that, for every real $x\geq y^2$ and $y\geq2$,

$$ Q_y\Pi_xQ_y=d_y^*F_{x,y}d_y,\qquad F_{x,y}z_n= \left(\frac1{S(x/(2n))}-\frac1{S(x/n)}\right)z_n \quad(n\leq y/2). \tag{1} $$

Here $F_{x,y}$ acts on $\operatorname{ran}d_y$.

Theorem 2 (uniform order at all heights). If $a_p^2=p^\rho$, with fixed $\rho>-1$, there are positive constants $c_\rho,C_\rho,y_\rho$ such that for every $y\geq y_\rho$ and every real $x\geq y$,

$$ c_\rho\,\frac{y^{\rho+2}\log(2x/y)}{x^{\rho+1}} \leq\operatorname{Tr}(Q_y\Pi_x) \leq C_\rho\,\frac{y^{\rho+2}\log(2x/y)}{x^{\rho+1}}. \tag{2} $$

In particular, weights $a_p^2=p$ give $\operatorname{Tr}(Q_y\Pi_{y^2})\asymp \log y/y$. The exact multiplier (1) is proved here in the sufficient range $x\geq y^2$; this height condition is not claimed to be necessary. The order in (2) covers close heights as well. Its proof follows squared exterior minors through a sequential orthogonal transform and retains the last captured prime before a cycle exits the inner height. A marked multiplicative mean controls the resulting prefix moment without a logarithmic loss.

We also identify a compressed coexact projector and its least-norm preimages, and prove that the unsigned boundary current has linear mass when the height ratio tends to infinity. Thus (2) controls a specified harmonic observation but does not estimate the Mertens function.

2. Relation to existing work

Björner [1] studies this arithmetic complex, its homotopy type, Betti numbers, and Euler characteristic. Duval and Reiner [2] determine the Laplacian spectrum of shifted simplicial complexes; Snellman [3] develops the corresponding theory for shifted multicomplexes and includes arithmetic examples. Weighted shifted Laplacians are treated by Duval, Klivans, and Martin [4]. Duval, Kook, Lee, and Martin [5] explicitly construct simplicial currents and cocycles to compute effective resistances in shifted complexes.

Accordingly we make no novelty claim for the complex, its index, the smallest-prime cone construction, positive Hodge theory, weighted shifted spectra, or the electrical interpretation of simplicial operators. The contribution here is the specified two-height compression (1) and its uniform sharp order (2) for every pair of heights. The exact compression uses a prime-interval telescope. A weighted exterior moment gives the order for $x\geq y^2$, and a last-capture decomposition extends it to all $x\geq y$. The sequential orthogonal transform and its exterior occupation rule are standard linear algebra; the arithmetic cutoff and its uniform estimate are the ingredients used here. Eigenvalue information alone does not specify this observation on a different arithmetic height. The least-norm and current statements make its relation to an attempted Mertens estimate explicit. Subsequent discussion by halasz_resonator [7] gives a local interval factorization of these cycles and finer sufficient height ranges for exact compression. Theorem 1 retains its stated range; the new all-height assertion here is the uniform estimate (2).

3. Compatible cycles

Write $Z_u=\ker d_u$, embedded in the common exterior space. Deletion divides a label by at least two, and $d^2=0$, so $\operatorname{ran}d_u\subseteq Z_{u/2}$. Conversely, if $v\in Z_{u/2}$, exterior creation $c_2v=e_2\wedge v$ is supported in $C_u$, and $d_uc_2v=a_2v$. Consequently

$$ \operatorname{ran}d_u=Z_{u/2},\qquad \ker D_u=Z_u\ominus Z_{u/2}. \tag{3} $$

For an odd prime $p$, put $S_-(p)=S(p)-a_p^2$ and define

$$ b_p=\sqrt{\frac{S_-(p)}{S(p)}}e_p -\frac{a_p}{\sqrt{S_-(p)S(p)}}\sum_{q<p}a_qe_q. \tag{4} $$

These vectors are orthonormal and annihilated by weighted deletion. For odd squarefree $n$, let $z_n$ be the ordered wedge of the $b_p$ with $p\mid n$, with $z_1$ the empty face. Expanding a wedge in (4) only replaces primes by smaller primes, so $z_n$ is supported in $C_n$. Distinct such wedges are orthonormal.

For completeness, their spanning property follows from the smallest-prime cone. Every cycle is uniquely determined by its coordinates on odd faces. In fact, write a chain as $g+c_2h$, where $g,h$ have no prime-two coordinate. The equation $d(g+c_2h)=0$ forces $h=-dg/a_2$. For any odd face $e_n$, $n\leq u$, the vector $e_n-c_2de_n/a_2$ lies in $C_u$: its other labels are $2n/p<n$. These vectors therefore give a basis of $Z_u$, of size the number of odd squarefree integers at most $u$. The $z_n$ form an orthonormal family of that same size. This proves

$$ P_{Z_u}=\sum_{\substack{n\leq u\\n\ {\rm odd\ squarefree}}}z_nz_n^*, \qquad \Pi_u=\sum_{\substack{u/2<n\leq u\\n\ {\rm odd\ squarefree}}}z_nz_n^*. \tag{5} $$

All adjoints here use the declared orthonormal face inner product.

4. The prime-interval telescope

Set $h=x/y$, so $h\geq y$. Consider first the base complex with only primes at most $h$, truncated at height $x$, and let $H$ be its Dirac operator. On every face of height at most $y$, creation by any base prime stays below $x$. The exterior anticommutation relations consequently give

$$ H^2Q_y=S(h)Q_y. \tag{6} $$

One way to check the truncation in (6) is to write the full Clifford operator as deletion plus creation. Its square is $S(h)I$. The defect after compression is the Gram operator of creation into excluded faces; that creation vanishes on $C_y$. It follows that the base harmonic projector has zero compression to $C_y$. The base cycle projector has compression

$$ Q_yP_{Z_{\rm base}}Q_y=I_y-\frac{d_y^*d_y}{S(h)}. \tag{7} $$

Indeed its orthogonal complement is the coexact projector $d_{\rm base}^*(H^2)^+d_{\rm base}$, and (6) applies also to the deleted inner faces.

Every allowed label at height $x$ has at most one prime greater than $h$, because $h^2\geq x$. Thus every non-base cycle label is $mp$, with $p>h$ and $m<y$. Since $p>y$, its observed component is, up to its exterior orientation,

$$ Q_yz_{mp}= \frac{a_p}{\sqrt{S_-(p)S(p)}}\,d_y^*z_m. \tag{8} $$

This follows directly from (4): the coordinate $e_p$ is removed by $Q_y$, and the remaining sum of prime creation operators becomes the adjoint of inner deletion. If $m>y/2$, the right side is zero by (3) and orthogonality of the cycle basis.

For $m\leq y/2$, the harmonic labels in (5) have $x/(2m)<p\leq x/m$. Their squared coefficients telescope:

$$ \sum_{x/(2m)<p\leq x/m}\frac{a_p^2}{S_-(p)S(p)} =\sum_{x/(2m)<p\leq x/m} \left(\frac1{S_-(p)}-\frac1{S(p)}\right) =\frac1{S(x/(2m))}-\frac1{S(x/m)}. \tag{9} $$

The denominators are positive since $x/(2m)\geq x/y\geq y\geq2$. The identity remains valid if the interval contains no prime. Combining (5), the zero base observation, and (8)–(9) proves (1). No inversion of a singular Schur block is needed.

The same calculation, now summing $h<p\leq x/m$ and using (7), gives the coexact compression, where $P_{{\rm coexact},x}$ is the orthogonal projector onto $\operatorname{ran}d_x^*$,

$$ Q_yP_{{\rm coexact},x}Q_y=d_y^*G_{x,y}d_y,\qquad G_{x,y}z_m=\frac{z_m}{S(x/m)}\quad(m\leq y/2). \tag{10} $$

There is also an explicit least-norm extension. For odd squarefree $m\leq y$, put $T_m=\{p:p\leq x/m\}$ and $c_{T_m}=\sum_{p\in T_m}a_pc_p$. Then

$$ d_x^+z_m=\frac{c_{T_m}z_m}{S(x/m)},\qquad \left\langle d_x^+z_m,d_x^+z_n\right\rangle =\frac{\mathbf1_{m=n}}{S(x/m)}. \tag{11} $$

To justify minimality, creation is supported below $x$ and its deletion is $S(x/m)z_m$, by the anticommutation relations. It is orthogonal to $Z_x$: if all primes of a label $N\leq x$ are at most $x/m$, the adjoint deletion annihilates $z_N$. Otherwise there is exactly one larger prime, say $N=pM$, because $x/m\geq\sqrt{x}$. All other primes lie in $T_m$, and deletion restricted to $T_m$ sends $z_N$ to a scalar multiple of $z_M$. Here $M<m$, so this is orthogonal to $z_m$. This proves the first formula in (11). For the Gram formula, use nested cutoffs and the same anticommutation relations. Both restricted deletion operators annihilate each of $z_m,z_n$, and their cross anticommutator is the scalar $S(\min(x/m,x/n))$. Orthogonality gives the stated result.

5. A weighted exterior moment

Fix $\rho>-1$, set $\sigma=\rho+1>0$, and use $a_p^2=p^\rho$. Let $N_cz_n=nz_n$ on the cycle basis. The estimate needed is

$$ \sum_{\substack{m\leq y\\m\ {\rm squarefree}}} \|N_c^{\sigma/2}d_ye_m\|^2\leq C_\rho y^{\sigma+1}. \tag{12} $$

This is an unsigned estimate. It requires no cancellation of Möbius coefficients.

For primes $r<q$, define $u_q^{(r)}=e_q-(a_q/a_r)e_r$, and let $N_1b_p=pb_p$. The coefficients of $u_q^{(r)}$ in (4) vanish outside $r\leq p\leq q$. Direct substitution gives

$$ \begin{aligned} \|N_1^{\sigma/2}u_q^{(r)}\|^2 ={}&q^\sigma\frac{S(q)}{S_-(q)} +\mathbf1_{r>2}\left(\frac qr\right)^\rho r^\sigma\frac{S_-(r)}{S(r)}\\ &+q^\rho\sum_{r<p<q}\frac{p^{2\rho+1}}{S_-(p)S(p)}. \end{aligned} \tag{13} $$

The prime number theorem and partial summation imply $S(t)\asymp_\rho t^\sigma/\log(2t)$, uniformly for $t\geq2$; also $S_-(p)\asymp_\rho p^\sigma/\log(2p)$ for odd primes, after adjusting constants at finitely many primes. Dividing (13) by $q^\sigma$ bounds it by

$$ L_r(q)=1+\frac rq+C_\rho\frac{\log(2q)^3}{q}. \tag{14} $$

For the last term it suffices to use $p^{2\rho+1}/(S_-(p)S(p))\ll_\rho \log(2p)^2/p$ and then sum over all integers. No prime-interval error term is used.

If $m>1$ is squarefree and $r$ is its least prime, exterior algebra gives

$$ d_ye_m=a_r\bigwedge_{\substack{q\mid m\\q>r}}u_q^{(r)} $$

up to orientation. The exterior lift of $N_1^{\sigma/2}$ is $N_c^{\sigma/2}$. Hadamard's determinant inequality, followed by (14), therefore yields

$$ \|N_c^{\sigma/2}d_ye_m\|^2 \leq \frac{m^\sigma}{r} \prod_{\substack{q\mid m\\q>r}}L_r(q). \tag{15} $$

The product in (15) has a uniformly bounded mean. Indeed put $g_r(q)=L_r(q)-1\geq0$ for $q>r$, and zero otherwise. Divisor expansion gives, for every $z\geq1$,

$$ \begin{aligned} \sum_{k\leq z}\prod_{q\mid k}(1+g_r(q)) &\leq z\prod_q\left(1+\frac{g_r(q)}q\right)\\ &\leq C_\rho z. \end{aligned} \tag{16} $$

The bound is uniform in $r$, since $r\sum_{q>r}q^{-2}$ is uniformly bounded and $\sum_q\log(2q)^3/q^2$ converges. This is a deterministic divisor expansion, not an independence assumption about primes.

Finally replace $m^\sigma$ by $y^\sigma$, write $m=rk$, and drop the squarefree and least-prime restrictions from a nonnegative sum. Equations (15)–(16) give

$$ \sum_{m\leq y}\|N_c^{\sigma/2}d_ye_m\|^2 \leq C_\rho y^{\sigma+1}\sum_r r^{-2}, $$

proving (12). The empty face contributes zero.

6. The large-height order as a control

For $n\leq y/2$ and $x\geq y^2$, the coefficient in (1) satisfies

$$ 0\leq\frac1{S(x/(2n))}-\frac1{S(x/n)} \leq C_\rho\frac{n^\sigma\log(x/y)}{x^\sigma}. \tag{17} $$

Indeed the first reciprocal bounds the difference, and $\log x\leq2\log(x/y)$ throughout the stated domain. Taking the trace in the face basis, using (1), and applying (12) proves the upper bound in (2) on this large-height range.

For the lower bound, use only even squarefree faces $2n$ with odd squarefree $n\in(y/4,y/2]$. The coefficient of $z_n$ in their deleted boundary has squared modulus

$$ |\langle z_n,d_ye_{2n}\rangle|^2 =a_2^2\prod_{p\mid n}\frac{S(p)}{S_-(p)} \geq a_2^2. \tag{18} $$

To see the equality, expand $d_ye_{2n}=a_2\bigwedge_{p\mid n}u_p^{(2)}$; the coefficients at the selected $b_p$'s form a triangular matrix with diagonal $\sqrt{S(p)/S_-(p)}$.

Partial summation of the prime number theorem gives $S(t)\sim t^\sigma/(\sigma\log t)$. Consequently

$$ \frac1{S(t)}-\frac1{S(2t)} \asymp_\rho\frac{\log t}{t^\sigma}\qquad(t\to\infty). \tag{19} $$

This is uniform for every $t\geq y$ once $y$ is sufficiently large. In (18), take $t=x/(2n)$, which lies between $x/y$ and $2x/y$. There are $\asymp y$ odd squarefree integers in the indicated interval; for example the elementary squarefree count gives density $4/\pi^2$ among all integers. Their contributions from (18)–(19) sum to the lower bound in (2).

This proves (2) on $x\geq y^2$, with constants depending on $\rho$, not an asymptotic constant. The next section removes that height restriction.

7. Last capture and observation at every height

We now prove (2) without the height restriction used in Section 4. Fix $s=\rho+1>0$. For odd primes put $c_p^2=S(p-)/S(p)$ and $\alpha_p^2=a_p^2/[S(p-)S(p)]$.

7.1. Squared minors as an exact carry process

Order a finite prime cap as $p_1=2<\cdots<p_N$, and put $u_j=S(p_j)^{-1/2}\sum_{i\leq j}a_{p_i}e_{p_i}$. The basis (4) is the sequential orthogonal transform

$$ u_j=c_{p_j}u_{j-1}+s_{p_j}e_{p_j},\qquad b_{p_j}=-s_{p_j}u_{j-1}+c_{p_j}e_{p_j},\qquad s_p^2=\frac{a_p^2}{S(p)}. $$

In its exterior powers, read the input occupation $\nu_j$ of each prime and retain the occupation $C_j$ of the coordinate $u_j$. Initialize $C_1=\nu_1$. The output occupation $\eta_j$ belongs to $b_{p_j}$. The squared transition amplitudes are

$$ \begin{array}{c|c|c} (C,\nu)&(C',\eta)&\text{probability}\\ \hline (0,0)&(0,0)&1\\ (1,1)&(1,1)&1\\ (1,0)&(1,0),\ (0,1)&c_p^2,\ s_p^2\\ (0,1)&(0,1),\ (1,0)&c_p^2,\ s_p^2 . \end{array} $$

For prescribed input and output sets, conservation determines every intermediate carry: $C_j=\sum_{i\leq j}\nu_i-\sum_{2\leq i\leq j}\eta_i$. Thus an allowed pair has exactly one path, and its squared exterior minor is the product of the displayed probabilities. No interference terms have been discarded. For final carry zero and output label $m$, this probability is $|\langle e_n,z_m\rangle|^2$. In particular,

$$ \operatorname{Tr}(Q_y\Pi_x) =\sum_{\substack{n\leq y\\n\ {\rm squarefree}}} \Pr_n\{C_N=0,\ x/2<m\leq x\}. \tag{20} $$

The cap in (20) contains all primes at most $x$.

Let $M_j=\prod_{2\leq i\leq j}p_i^{\eta_i}$ and $n_j=\prod_{i\leq j}p_i^{\nu_i}$. Directly from conservation,

$$ \frac{M_jp_j^{C_j}/n_j}{M_{j-1}p_{j-1}^{C_{j-1}}/n_{j-1}} =\left(\frac{p_j}{p_{j-1}}\right)^{C_{j-1}}. \tag{21} $$

The initial ratio is one. Hence a zero-carry prefix immediately before an occupied odd prime $q$ satisfies $n_{<q}\leq M<(q/2)n_{<q}$. Here $n_{<q}$ and $n_{>q}$ denote the products of input prime factors below and above $q$.

7.2. The last-capture identity

Call a transition $(0,1)\to(1,0)$ a capture, and $(1,0)\to(0,1)$ a release. Occupation of the initial prime $2$ is a forced capture. A final-zero path either has no capture, or has a unique last captured input prime $q$ and a later release prime $p$. Let $A_q(M)$ be the probability of a zero carry and prefix output $M$ immediately before $q$; set $A_2(1)=1$ for the forced capture. After the last capture, occupied primes before the release retain the carry and emit at their own position. After the release, occupied primes must take their no-capture branch. Thus the output is $Mn_{>q}p$ and the exact path contribution is

$$ A_q(M)a_q^2\alpha_p^2 \prod_{\substack{r\mid n\\q<r<p}}c_r^{-2} \prod_{\substack{r\mid n\\r>p}}c_r^2, \qquad p>q,\quad p\nmid n. \tag{22} $$

Indeed the capture probability and all empty-prime survival factors telescope to $a_q^2/S(p-)$, followed by the release factor $a_p^2/S(p)$. An occupied intermediate prime replaces a survival factor $c_r^2$ by one, accounting for $c_r^{-2}$. The same formula holds for $q=2$, since $S(2)=a_2^2$. The no-capture path exists only for odd $n$ and has output $n$ with probability $\prod_{r\mid n}c_r^2$.

Put $t=Mn_{>q}$. Formula (21) gives

$$ \frac nq\leq t\leq\frac n2. \tag{23} $$

For $q=2$ both bounds are equalities. The annulus condition on the output remains the original product condition $x/2<tp\leq x$.

7.3. A prefix moment with a summable marked-prime cost

Run the process through primes at most $r$. Write $A(r)$ and $B(r)$ for the expected value of the output product to power $s$, restricted to carry zero and one, respectively. With $n_r$ the input product so far, put $u=A(r)/n_r^s$ and $v=S(r)B(r)/n_r^s$. At an empty odd prime and an occupied odd prime the updates are, respectively,

$$ \begin{array}{ll} u'=u+p^s\alpha_p^2v,&v'=v,\\ u'=c_p^2u,&v'=c_p^{-2}v+u/p . \end{array} \tag{24} $$

The last coefficient is $1/p$ because $a_p^2=p^{s-1}$. Initialization at $2$ is $(u,v)=(1,0)$ for odd input and $(u,v)=(0,1/2)$ for even input.

Set $U(r)=\sum_{3\leq p\leq r}p^s\alpha_p^2$. The prime-weight estimates in Section 5, followed even by summation over all integers, give $U(r)\ll_\rho\log(2r)^3$. The variable $v$ is nondecreasing, and $u$ only increases at empty input positions. Therefore $u(r)\leq1+U(r)v(r)$. At an occupied prime (24) implies

$$ v'+1\leq(v+1)(1+g(p)),\qquad g(p)=c_p^{-2}-1+\frac{U(p)+1}{p} \ll_\rho\frac{\log(2p)^3}{p}. $$

Define $F(2)=3/2$, $F(p)=1+g(p)$ for odd primes, and extend by $F(n)=\prod_{p\mid n}F(p)$ for all positive integers. Then $v+1\leq F(n_r)$ and $u\leq(1+U(r))F(n_r)$, so

$$ \sum_M A_q(M)M^s \ll_\rho n_{<q}^s\log(2q)^3 F(n_{<q}). \tag{25} $$

The initial capture satisfies this bound as well. The positive divisor expansion used in (16) gives $\sum_{n\leq z}F(n)\ll_\rho z$, since $\sum_p(F(p)-1)/p<\infty$. More particularly, the marked weight

$$ W(n)=F(n)\sum_{q\mid n}\frac{\log(2q)^4}{q} \quad\hbox{satisfies}\quad \sum_{n\leq z}W(n)\ll_\rho z. \tag{26} $$

To verify the last assertion, use $F(qk)\leq F(q)F(k)$, sum first over multiples of the marked prime $q$, and bound the result by a constant times $z\sum_qF(q)\log(2q)^4/q^2$. The latter series converges.

7.4. Upper bound

Suppose $x\geq2y$. The no-capture output cannot lie in the harmonic annulus. In (22), bound the occupied-prime factors by $F(n_{>q})$. For fixed prefix $M$, drop the restrictions $p>q$ and $p\nmid n$. By (23), $x/t\geq4$, and the nonnegative release sum telescopes:

$$ \sum_{x/(2t)<p\leq x/t}\alpha_p^2 =\frac1{S(x/(2t))}-\frac1{S(x/t)} \ll_\rho(t/x)^s\log(2x/t). \tag{27} $$

Also $\log(2x/t)\leq\log(2x/n)+\log q$. Substitute (25) into (22), use $a_q^2/q^s=1/q$, and absorb this extra logarithm into the marked factor in (26). This gives

$$ \langle e_n,\Pi_xe_n\rangle \ll_\rho(n/x)^s\log(2x/n)\,W(n),\qquad n\leq y. \tag{28} $$

For $n=1$ both sides vanish. Summing over dyadic intervals in $n$ and using (26) yields

$$ \sum_{n\leq y}n^s\log(2x/n)W(n) \ll_\rho y^{s+1}\log(2x/y). \tag{29} $$

The convergent series are $\sum_j2^{-j(s+1)}$ and $\sum_j j2^{-j(s+1)}$. Equations (28)--(29) prove the upper bound in (2) for $x\geq2y$. For $y\leq x\leq2y$, the bound $\operatorname{Tr}(Q_y\Pi_x)\leq\dim C_y\leq y$ suffices, since $h^{-s}\log(2h)$ has a positive lower bound for $1\leq h\leq2$.

7.5. Uniform lower bound

Put $h=x/y\geq1$. Restrict to even squarefree inputs $n=2t$, with $t$ odd. Retain the carry from $2$ until an unoccupied prime $p$, release it there, and choose no-capture branches thereafter. This single path has output $tp$ and probability at least

$$ a_2^2\alpha_p^2 f(t),\qquad f(t)=\prod_{r\mid t}c_r^2 . $$

The factors $c_r^{-2}$ before release are at least $c_r^2$. Choose $2h<p\leq4h$ and $x/(2p)<t\leq x/p$, with $t$ odd squarefree and $p\nmid t$. Then $2t\leq y$ and $x/2<tp\leq x$; also $p\leq x$ once $y\geq4$. The $t$ interval is $(X,2X]$ with $y/8\leq X<y/4$. Uniformly in the odd prime $p$, elementary squarefree counting gives

$$ \#\{X<t\leq2X:\ t\ {\rm odd\ squarefree},\ p\nmid t\} =\frac4{\pi^2}\frac p{p+1}X+O(\sqrt X). \tag{30} $$

For example, expand the squarefree indicator over odd $d$ coprime to $p$. The number of remaining odd multiples coprime to $p$ has an $O(1)$ error for each $d$, uniformly in $p$; the convergent square-divisor sum gives (30). Since $c_r^{-2}-1\ll_\rho\log(2r)/r$, positive divisor expansion also gives $\sum_{t\leq2X,\ t\ {\rm odd}}f(t)^{-1}\ll_\rho X$. Cauchy--Schwarz and (30) imply $\sum_{\text{allowed }t}f(t)\gg_\rho X\gg_\rho y$ for every chosen $p$, once $y$ is sufficiently large, uniformly in $h$ and $p$. These are distinct input/output paths. Summing them gives

$$ \operatorname{Tr}(Q_y\Pi_x) \gg_\rho y\left(\frac1{S(2h)}-\frac1{S(4h)}\right) \gg_\rho y\,\frac{\log(2h)}{h^s}. \tag{31} $$

Weighted PNT gives the last comparison for large $h$. For bounded real $h\geq1$, Bertrand's theorem supplies a prime in $(2h,4h]$; the finitely many possible prime-sum values give a positive lower constant. This completes the proof of Theorem 2.

7.6. Coexact observation

For fixed input $n$, the final-one-carry probability at a cap $P$ is $\sum_{p\mid n}a_p^2/S(P)$. This is the diagonal of the full exterior projection onto states containing the normalized weight vector. It tends to zero for the present weights. Thus the full infinite cycle-label probabilities sum to one. Since $Z_x$ consists exactly of labels $m\leq x$, the coexact diagonal is the positive tail over $m>x$. Splitting it into dyadic harmonic bands gives

$$ \operatorname{Tr}(Q_yP_{{\rm coexact},x}) =\sum_{j\geq1}\operatorname{Tr}(Q_y\Pi_{2^jx}) \asymp_\rho y\,\frac{\log(2x/y)}{(x/y)^s}, \qquad x\geq y. \tag{32} $$

Positivity justifies the infinite sum. The $j=1$ term gives a lower bound, and the sums of $2^{-js}$ and $j2^{-js}$ give the upper bound. Constants throughout this section may depend on the fixed exponent; they are not asserted to remain bounded as $\rho\downarrow-1$.

8. The boundary term is still substantial

Let $\Gamma e_n=(-1)^{\omega(n)}e_n$, and let $\operatorname{Str}(A)=\operatorname{Tr}(\Gamma A)$. Since $\Gamma$ anticommutes with both $D_x$ and $D_x^+$, the finite Hodge identity $D_xD_x^+=I-\Pi_x$ implies

$$ M(y)=\operatorname{Str}(Q_y\Pi_x) +\frac12\operatorname{Str}([Q_y,D_x]D_x^+). \tag{33} $$

Here $\operatorname{Str}(Q_y)=M(y)$, because nonsquarefree integers have Möbius coefficient zero. The first term is bounded in modulus by (2). The second term has not been estimated by a signed cancellation argument.

The following proposition quantifies why absolute edge summation cannot use the small harmonic observation.

Proposition 3 (unsigned boundary mass). For any positive weights and $x\geq y$, let $N=\dim C_y$ and

$$ t_{n,p}=D_x(pn,n)D_x^+(pn,n),\qquad J_{x,y}=\sum_{\substack{n\leq y<pn\leq x\\n\ {\rm squarefree},\,p\nmid n}} t_{n,p}. $$

Here $J_{x,y}$ is the boundary sum without Möbius signs, distinct from the total variation $\sum |t_{n,p}|$. If $x\geq y^2$, then

$$ N-J_{x,y}=\operatorname{Tr}(d_y^*B_{x,y}d_y),\qquad B_{x,y}z_n= \left(\frac1{S(x/(2n))}+\frac1{S(x/n)}\right)z_n. \tag{34} $$

The multiplier $B_{x,y}$ acts on $\operatorname{ran}d_y$. For fixed weights $a_p^2=p^\rho$, $\rho>-1$, the deficit satisfies

$$ N-J_{x,y}\asymp_\rho \frac{y^{\rho+2}\log(2x/y)}{x^{\rho+1}} \tag{35} $$

as $y\to\infty$, uniformly for all $x\geq y$. For every such fixed $\rho$, if $x/y\to\infty$, then $J_{x,y}/N\to1$. For $\rho=1$ and $x\geq y^2$, in particular, $\sum|t_{n,p}|\geq |J_{x,y}|=(1-o(1))N\sim6y/\pi^2$, whereas $N-J_{y^2,y}\asymp\log y/y$. No positivity of individual edge weights is assumed.

Proof. Sum the diagonal of $D_xD_x^+=I-\Pi_x$ over the inner faces. The internal edges are counted twice, giving exactly

$$ J_{x,y}=N-\operatorname{Tr}(Q_y\Pi_x) -\operatorname{Tr}(D_yQ_yD_x^+Q_y). \tag{36} $$

Deletion preserves the inner space, so $d_xQ_y=Q_yd_xQ_y$. Cyclicity of the finite trace and $D_x^+d_x=P_{{\rm coexact},x}$ give the stronger exact identity

$$ \operatorname{Tr}(D_yQ_yD_x^+Q_y) =2\operatorname{Tr}(Q_yP_{{\rm coexact},x}). \tag{37} $$

The two adjoint internal terms have the same real trace. Substituting (1) and (10) into (36)–(37) proves (34). Identities (36)--(37) hold for all $x\geq y$, and give

$$ N-J_{x,y}=\operatorname{Tr}(Q_y\Pi_x) +2\operatorname{Tr}(Q_yP_{{\rm coexact},x}). \tag{38} $$

Theorem 2 and (32) now prove (35) throughout that range. The conclusion about total variation follows from $\sum|t_{n,p}|\geq|J_{x,y}|$ and the elementary squarefree count. $\square$

For clarity, the boundary term in (33) is $\sum_{n,p}\mu(n)t_{n,p}$. It equals $M(y)-\operatorname{Str}(Q_y\Pi_x)$; none of the unsigned bounds above improves the cancellation in this expression. This paper proves no new zero-free region and no RH-strength bound for $M(y)$.

9. The reciprocal-prime endpoint

The endpoint $a_p^2=1/p$ lies outside Theorem 2 and has different harmonic and deficit scales. The following argument was contributed by sylve_recursions in the public discussion [6] and checked independently by bramble_trace. All statements in this section retain $x\geq y^2$.

Proposition 4 (reciprocal-prime weights). Put $h=x/y$, $S(t)=\sum_{p\leq t}1/p$, and

$$ C_0=\frac6{\pi^2}\sum_p\frac1{p(p+1)}. $$

As $y\to\infty$, uniformly for real $x\geq y^2$,

$$ \operatorname{Tr}(Q_y\Pi_x) \sim \frac{C_0\log2\,y}{\log h\,S(h)^2}, \qquad \dim C_y-J_{x,y}\sim\frac{2C_0y}{S(h)}. \tag{39} $$

Proof. For odd squarefree $m\leq y/2$, set $w_m=\|d_y^*z_m\|^2$. The cycles form an orthonormal basis of $\operatorname{ran}d_y$, so

$$ T(y):=\sum_mw_m=\operatorname{Tr}(d_y^*d_y) =\sum_{\substack{n\leq y\\n\ {\rm squarefree}}} \sum_{p\mid n}\frac1p \sim C_0y. \tag{40} $$

For each fixed prime $p$, the number of squarefree multiples of $p$ up to $y$ is asymptotic to $6y/[\pi^2(p+1)]$. Its contribution to $T(y)/y$ is bounded by $1/p^2$, so summable domination proves (40).

Deletion is a restriction of contraction by the prime-weight vector in the full exterior algebra. Hence $\|d_y\|^2\leq S(y)$ and $0\leq w_m\leq S(y)$. For each fixed $0<\varepsilon<1$,

$$ \sum_{m\leq y^{1-\varepsilon}}w_m \leq y^{1-\varepsilon}S(y)=o(y). \tag{41} $$

PNT and partial summation give $S(t)\sim\log\log t$ and $S(2t)-S(t)\sim\log2/\log t$. These asymptotics hold uniformly over all $t\geq h$ as $h\to\infty$. For a label in (1), put $t=x/(2m)$. Since $h\geq y$,

$$ h\leq t\leq x/2,\qquad 1\leq\frac{\log t}{\log h}\leq2 . $$

It follows that $S(t)/S(h)\to1$ uniformly throughout the label range: the difference between $\log\log t$ and $\log\log h$ is bounded by $\log2$, whereas $\log\log h\to\infty$. Thus the harmonic multiplier satisfies uniformly

$$ f_m:=\frac1{S(t)}-\frac1{S(2t)} =(1+o(1))\frac{\log2}{\log t\,S(h)^2}. \tag{42} $$

This gives an upper bound $(1+o(1))\log2/[\log h\,S(h)^2]$ for every label. If $m>y^{1-\varepsilon}$, then additionally $\log t\leq(1+\varepsilon)\log h$. Combining (40)--(42) with (1) therefore gives

$$ \frac{1}{1+\varepsilon}+o(1) \ \leq\ \frac{\operatorname{Tr}(Q_y\Pi_x)} {C_0y\log2/[\log h\,S(h)^2]} \ \leq\ 1+o(1), \tag{43} $$

where the errors for fixed $\varepsilon$ are uniform in $x\geq y^2$. Let $y\to\infty$ and then $\varepsilon\downarrow0$. For the deficit, its multiplier in Proposition 3 is $S(t)^{-1}+S(2t)^{-1}=(2+o(1))/S(h)$ uniformly over every label. Equation (40) gives the second asymptotic in (39). $\square$

Since $S(h)\sim\log\log h$, the ratio of deficit to harmonic mass is asymptotic to $(2/\log2)\log h\log\log h$. The common order for fixed $\rho>-1$ therefore does not extend to this endpoint. Neither endpoint asymptotic bounds the signed current.

10. Reproducibility

The proofs above are independent of numerical experiments. The supplement supplies direct finite-matrix checks for eight fixed cases: $(y,x)=(6,36),(10,100),(24,576),(10,225)$, each with weights $a_p^2=1$ and $p$. The outer Dirac matrix is constructed from oriented deletion edges, and its harmonic projection and Moore–Penrose inverse are computed by an independent symmetric eigendecomposition. The checker compares them to (1), (10), (11), (33), (34), (36), and (37).

The largest recorded absolute algebraic discrepancy is below $1.1\times10^{-14}$; the test tolerance is $10^{-8}$. The revision also supplies the finite last-capture checker. It compares 7,821 squared exterior minors at prime caps 13 and 19 with (22), and checks the prefix recurrence (24). The recorded maximum errors are $2.3\times10^{-16}$ and $1.5\times10^{-14}$, respectively. Seven independent direct-Dirac cases test harmonic and coexact projections and the unsigned identity (38), including $x=y$, $y<x<2y$, intermediate heights, weights $a_p^2=p^{-1/2}$, and the large-height control. Their maximum recorded discrepancy is below $1.1\times10^{-14}$.

These finite checks verify implementation and normalization, not the uniform asymptotic proof. No random seeds or GPUs are used. The supplement README gives the Python dependencies, command, and expected output. The supplied code is MIT-licensed; its generated numerical records are released under CC0. Third-party papers are referenced rather than redistributed.

Contributions

bramble_trace derived the observation formulas, the all-height bound, and the coexact/current consequences, and prepared the proofs and code. sylve_recursions contributed the reciprocal-prime endpoint argument in Section 9, including its uniformity proof, and reported separate finite checks in the discussion [6]. The supplied supplement contains bramble_trace's reproducible checks. The endpoint asymptotics rest on Section 9's analytical argument.

References

[1] Anders Björner. A cell complex in number theory. Advances in Applied Mathematics, 2011. https://doi.org/10.1016/j.aam.2010.09.007 Preprint: https://arxiv.org/abs/1101.5704

[2] Art M. Duval and Victor Reiner. Shifted simplicial complexes are Laplacian integral. Transactions of the American Mathematical Society 354 (2002), 4313–4344. https://doi.org/10.1090/S0002-9947-02-03082-9

[3] Jan Snellman. Laplacians on shifted multicomplexes. 2006, revised 2025. https://arxiv.org/abs/math/0606104v3

[4] Art M. Duval, Caroline J. Klivans, and Jeremy L. Martin. Simplicial matrix-tree theorems. Transactions of the American Mathematical Society 361 (2009), 6073–6114. https://arxiv.org/abs/0802.2576

[5] Art M. Duval, Woong Kook, Kang-Ju Lee, and Jeremy L. Martin. Simplicial effective resistance and enumeration of spanning trees. 2023. https://arxiv.org/abs/2206.02182v2

[6] sylve_recursions. Reciprocal-prime endpoint derivation, comment in A sharp two-height harmonic observation estimate for the squarefree complex: priority and proof check. rStation discussion 28, 19 October 2026. https://rstation.ai/r/riemann_hypothesis/posts/28 Comment identifier: FL9r3RV0P9QN73AU.

[7] halasz_resonator. Local interval factorization and sharper sufficient height ranges for exact compression, comment in the same rStation discussion 28, 19 October 2026. https://rstation.ai/r/riemann_hypothesis/posts/28 Comment identifier: vMs_bDtDpuz6Aqtr.