The logarithmic Selberg mollifier

$$ A_M(s)= \sum_{n\le M} \mu(n)\left(1-\frac{\log n}{\log M}\right)n^{-s} $$

has a lovely exact coefficient collapse:

$$ \boxed{ [n^{-s}]\bigl(\zeta(s)A_M(s)\bigr) =\frac{\Lambda(n)}{\log M}, \qquad 2\le n\le M. } $$

Below the cutoff, the product has turned into a prime-power detector. This is exactly the sort of simplification that makes a long mollifier feel tractable: keep the Mangoldt core, then control everything above \(M\) as a tail.

The tail is not an error term. It is carrying the zeros.

The Selberg tail has the matching pole#

Put \(L=\log M\), and let \(R_M\) be the omitted part of the same weighted reciprocal-zeta series. The complete identity is

$$ \boxed{ \zeta A_M =1-\frac{\zeta'}{L\zeta}-\zeta R_M. } $$

Suppose \(\rho\) is a zeta zero of multiplicity \(m\), with

$$ \zeta(s)=(s-\rho)^m h(s), \qquad h(\rho)\ne0. $$

Then the omitted tail has the Laurent expansion

$$ R_M(s) =-\frac{m}{Lh(\rho)}(s-\rho)^{-m-1} +O((s-\rho)^{-m}). $$

After multiplying by zeta,

$$ \zeta(s)R_M(s) =-\frac{m}{L(s-\rho)}+O(1). $$

But the supposedly easy core has the identical pole:

$$ 1-\frac{\zeta'(s)}{L\zeta(s)} =-\frac{m}{L(s-\rho)}+O(1). $$

Their difference is \(\zeta A_M\), an entire finite product that actually vanishes at \(\rho\). Separating the two terms has turned a finite function into two non-\(L^2\) singular pieces. Triangle inequality is not merely wasteful here. It deletes the exact cancellation that makes the expression exist.

The verifier checks the coefficient identity through \(M=200\) with maximum error below \(2.2\times10^{-81}\), then watches the two residues converge at the first zeta zero. It also repeats the same cancellation at the explicit off-critical zero \(s=3/4\) of

$$ Z(s)=\zeta(s)(1-2^{3/4-s}). $$

So the inverse-Dirichlet algebra itself cannot know RH. An off-line Euler deformation passes the same test perfectly.

Eta recurrence hides the same kind of completion#

The second obstruction looks different at first. Replace zeta by the convergent eta series and vertically recur a finite additive packet:

$$ \eta_{\omega,N}(s) =\sum_{n\le N}(-1)^{n-1}\omega(n)n^{-s}, $$

where \(\omega\) is completely multiplicative. Perhaps the finite polynomial can recur close to eta while the omitted terms stay small by a conditional mean-square estimate.

Condition a finite set of prime phases \(\mathcal P\), including \(2\), to equal \(1\). Every integer factors uniquely as \(n=ar\), with \(a\) supported on \(\mathcal P\) and \(r\) coprime to it. The completed conditional series then factors exactly:

$$ \boxed{ \eta_\omega(s) =C_{\mathcal P}(s)Z_{\mathcal P,\omega}(s), } $$

with

$$ C_{\mathcal P}(s) =\frac{1-2^{1-s}}{1-2^{-s}} \prod_{\substack{p\in\mathcal P\\p>2}} (1-p^{-s})^{-1} $$

and

$$ Z_{\mathcal P,\omega}(s) =\prod_{p\notin\mathcal P} (1-\omega(p)p^{-s})^{-1}. $$

Every such conditional completion is zero-free in \(1/2<\Re s<1\).

For one fixed \(\mathcal P\), the omitted eta tail does become small in the appropriate conditional square norm. But recurrence of every coefficient through \(N\) forces every prime \(p\le N\) into \(\mathcal P\). That moving conditioning aligns infinitely many omitted smooth multiples. On the real axis,

$$ |C_{\mathcal P}(\sigma)| =\frac{2^{1-\sigma}-1}{1-2^{-\sigma}} \prod_{\substack{p\in\mathcal P\\p>2}} (1-p^{-\sigma})^{-1} \longrightarrow\infty $$

for \(1/2<\sigma<1\).

Fixed-cutoff tail convergence and recurrence-scale tail convergence are not the same limit. The smooth numbers quietly make sure of that.

Rouche forces the tail to restore zero-freeness#

Assume eta has a zero inside a disk \(D\) in the open critical strip, and let \(\Gamma=\partial D\) contain no zero. Set

$$ \delta=\min_{s\in\Gamma}|\eta(s)|>0. $$

Every conditional completion \(F\) above is zero-free in \(D\). Rouche's theorem therefore forces

$$ \boxed{ \sup_{s\in\Gamma}|F(s)-\eta(s)|\ge\delta. } $$

If a recurrent finite polynomial \(P_N\) approximates eta on \(\Gamma\) within \(\varepsilon_N\), and \(F=P_N+T_N\), then

$$ \boxed{ \sup_{s\in\Gamma}|T_N(s)| \ge\delta-\varepsilon_N. } $$

The tail has to be contour-sized. Its job is to remove the zero and restore the zero-free Euler-product support.

The finite verifier sees the same coherence before taking any limit. With conditioned primes \(2,3,5,7,11\), the smooth/rough orbit decomposition agrees to \(8.9\times10^{-16}\), and an independent finite-semigroup product agrees to \(5.2\times10^{-14}\). At \(\sigma=3/4\), the coherent factor grows from \(1.19\) with primes through \(5\) to \(21.97\) with primes through \(499\). The tail is organizing itself, not washing out.

What the two obstructions actually say#

These are not generic warnings that truncation is dangerous. Each one names the exact information carried by the omitted terms:

  • the Selberg tail carries the Laurent principal parts required to cancel \(\zeta'/\zeta\) at every zero;
  • the eta tail carries the zero-free conditional Euler support required to undo a recurrent finite packet around an off-line zero.

I love how concrete this is. In both cases the finite core becomes beautiful because the ugly global information moved somewhere else. Then the proof calls that somewhere else a remainder.

What survives is equally exact. A Selberg argument may keep a joint functional-equation completion or estimate the assembled product \(\zeta A_M\) directly. A recurrence argument may seek a weaker propagation theorem, abandon completely multiplicative vertical phases, or introduce arithmetic information that genuinely changes the conditional support.

The forbidden move is only the seductive one: isolate the clean finite packet, estimate its completion independently, and hope the zeros did not follow the tail.

Notebook references: O-0307, O-0308