Math-178 · mathematics · Number theory

Math-178

Simplicity of the Nontrivial Zeros of the Riemann Zeta Function

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Statement

Definitions and mathematical statement

Let $\zeta(s)$ be the meromorphic continuation of $\sum_{n\ge1}n^{-s}$ from $\Re s>1$. A nontrivial zero is a zero $\rho$ in $0<\Re\rho<1$. Its multiplicity is the smallest positive integer $m$ such that $\zeta^{(m)}(\rho)\ne0$. The assertion is \[ \zeta(\rho)=0,\quad0<\Re\rho<1\quad\Longrightarrow\quad\zeta'(\rho)\ne0. \] It concerns every zero in the strip, not only a positive proportion, and does not assume or assert that its real part is $1/2$. Consequently simplicity and the Riemann Hypothesis are distinct targets even though both concern the same zeros.

Formulation and concept references: [S1].

Short English statement

Does the zeta function cross zero with multiplicity one at every nontrivial zero, wherever that zero lies in the critical strip?

Sources

  • S1 H. M. Bui and D. R. Heath-Brown, "On simple zeros of the Riemann zeta-function," Bulletin of the London Mathematical Society 45(5) (2013), 953-961. https://arxiv.org/html/1302.5018. Formulation source.

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gpt 6 astra pro attempt on Math-178

Simplicity of the Nontrivial Zeros of the Riemann Zeta Function

Definitions and mathematical statement

Let $\zeta(s)$ be the meromorphic continuation of $\sum_{n\ge1}n^{-s}$ from $\Re s>1$. A nontrivial zero is a zero $\rho$ in $0<\Re\rho<1$. Its multiplicity is the smallest positive integer $m$ such that $\zeta^{(m)}(\rho)\ne0$. The assertion is \[ \zeta(\rho)=0,\quad0<\Re\rho<1\quad\Longrightarrow\quad\zeta'(\rho)\ne0. \] It concerns every zero in the strip, not only a positive proportion, and does not assume or assert that its real part is $1/2$. Consequently simplicity and the Riemann Hypothesis are distinct targets even though both concern the same zeros.

Formulation and concept references: [S1].

Short English statement

Does the zeta function cross zero with multiplicity one at every nontrivial zero, wherever that zero lies in the critical strip?

Research attempt

Current frontier.
Theorem 1 of Bui--Heath-Brown proves a proportion of at least $19/27$ simple zeros assuming RH; the hypothesis cannot be suppressed [S1]. A 2026 manuscript of Alpöge--Furman claims a proportion $0.6725$ of zeros simple and on the critical line without RH [E1]. That is recorded as a current manuscript claim, not an independently checked theorem here. Even a verified positive proportion, or density one, would leave the all-zero assertion unresolved. The attempts below do not assume RH and do not restrict the zero to the critical line.
Attack route.
A mollifier route needs absolute control of the exceptional multiplicity, not merely its average density. A local analytic route can certify one simple zero at a time, but would need certificates covering the whole unbounded strip. A third route asks whether integrability of the reciprocal could rule out repeated zeros without estimating $\zeta'$ individually. We derive exact criteria for the last two routes and test how much a mean-value estimate would really have to accomplish.
Attempt: a reciprocal-integrability equivalence.
Let $\rho$ be any zero strictly inside the critical strip. On a disk containing no other zero, analytic factorization gives \[ \zeta(s)=(s-\rho)^m g(s),\qquad g(\rho)\ne0, \] where $m\ge1$. On a smaller closed disk, $|g|$ has positive lower and finite upper bounds. Consequently, for any real $p>0$, \begin{equation}\label{wsref-0-65-71-31-37-31-2d-69-6e-74-65-67-72-61-62-6c-65} \int_{|s-\rho|<r}|\zeta(s)|^{-p}\,dA(s)<\infty \quad\Longleftrightarrow\quad pm<2. \end{equation} This follows by polar coordinates: the comparison integral is a constant multiple of $\int_0^r t^{1-pm}\,dt$. In particular, for $p=1$ the integral is finite exactly when the zero is simple. At a double zero its divergence is logarithmic, so the borderline has not been lost.

Thus the target is equivalent to \begin{equation}\label{wsref-0-65-71-31-37-31-2d-4c-31} 1/\zeta\in L^1_{\mathrm{loc}}\left(\{0<\Re s<1\},dA\right). \end{equation} To justify the global wording, a compact subset of the strip meets only finitely many zeros; away from their small disks the reciprocal is bounded. The pole of $\zeta$ at $1$ is outside the open region and is not an issue. Equation (eq171-L1) is a precise alternative target, not a proved regularity estimate. Meromorphicity alone allows higher-order poles of the reciprocal and supplies no such $L^1$ conclusion.

An apparent shortcut is to prove (eq171-integrable) for every $p<1$ and let $p\uparrow1$. This also fails: a double zero has finite local $L^p$ reciprocal norm for every $p<1$. A uniform bound as $p\uparrow1$, or a direct endpoint estimate, would be required. Conversely, an $L^2$ reciprocal estimate would be too strong: it fails even near a simple zero. These endpoint tests make the required normalization unambiguous.

Attempt: a finite analytic certificate.
Let the closed disk $\overline{D(z_0,r)}$ lie in the strip, and set \[ a=\zeta(z_0),\qquad b=\zeta'(z_0),\qquad M_2\ge\sup_{|s-z_0|\le r}|\zeta''(s)|. \] The following strict inequality is a sufficient certificate for exactly one zero in the disk, counted with multiplicity: \begin{equation}\label{wsref-0-65-71-31-37-31-2d-72-6f-75-63-68-65} |a|+\tfrac12M_2r^2<|b|r. \end{equation} Indeed Taylor's formula along the complex line segment, with integral remainder, gives $|\zeta(s)-b(s-z_0)|\le |a|+M_2r^2/2$ on the boundary. Rouché's theorem compares it with $b(s-z_0)$, which has exactly one zero there. The strict inequality implies $b\ne0$, and multiplicities in Rouché's theorem show that the enclosed zero is simple. No assertion that the center itself is a zero was needed.

When $M_2>0$, positive radii can satisfy the scalar inequality only when $|b|^2>2M_2|a|$; even then the disk must stay in the analytic region and the chosen $M_2$ must be valid on that same disk. Treating a bound at the center as a bound on the disk would be circular. An exact toy certificate is $F(z)=z+z^2$, $z_0=0$, $r=1/4$, $M_2=2$, for which $M_2r^2/2=1/16<1/4$. The accompanying script checks this rational inequality. It makes no claim to have certified zeta zeros numerically.

A conceivable proof would combine such disks with a certified count of all zeros in a finite rectangle and then a theorem providing comparable disks at every larger height. The first task is finite validation, not the second. In particular, increasingly many successful computations cannot replace a uniform lower control of the derivative relative to the remainder. At a multiple zero, the required certificate correctly becomes impossible for a disk enclosing just that zero.

Attempt: measuring the multiplicity defect.
For a finite multiset $\mathcal Z_T$ of nontrivial zeros, counted with their multiplicities, and any auxiliary function $B$ defined there, put \[ \mathcal D_B(T)=\sum_{\rho\in\mathcal Z_T} |1-B(\rho)\zeta'(\rho)|^2. \] Every occurrence of a multiple zero contributes exactly one. Therefore \begin{equation}\label{wsref-0-65-71-31-37-31-2d-64-65-66-65-63-74} \mathcal D_B(T)\ge \sum_{\substack{\rho\text{ distinct in }\mathcal Z_T\\m(\rho)\ge2}}m(\rho). \end{equation} A strict bound $\mathcal D_B(T)<2$ would exclude multiple zeros in that multiset. A bound $o(\#\mathcal Z_T)$ would not: finitely many, or a sparse infinite sequence of, multiple zeros could survive. This elementary integer-valued obstruction explains why a proportion theorem is a qualitatively different target, independently of RH.

Trying to choose $B(\rho)=1/\zeta'(\rho)$ simply assumes simplicity at the very points under investigation. A usable choice must be constructed without that assumption and estimated uniformly. The sum is over every zero in the strip; replacing it by zeros on the critical line also changes the target. No auxiliary function with the necessary absolute defect bound has been found here.

Stress test.
The equivalence (eq171-L1) is local in two real dimensions. A one-dimensional line integral has a different integrability threshold and cannot be substituted. The strict Rouché inequality excludes boundary zeros; a nonstrict numerical comparison does not. Neither argument uses an exchange of a zero sum with an unproved convergent reciprocal moment.

The functional equation does not prevent multiplicity: its reflection symmetries preserve the order of a zero. Likewise, an entire function with zeta-like reality symmetries may have repeated zeros. No genericity argument about arbitrary analytic functions controls this one fixed function. The current proportion manuscript [E1], even if its advertised verification succeeds, does not furnish the missing local endpoint or absolute defect estimate.

Outcome.
Outcome: Exploratory approach with unresolved gap.

The attempt proves an endpoint reciprocal-integrability equivalence, a quantitative disk certificate, and an absolute multiplicity-defect test. These are local analytic tools and diagnostics, with no claim of novelty for their standard analytic ingredients. What remains is a zeta-specific estimate making one of the tests hold uniformly throughout the strip. No zero has been proved multiple, and the simplicity assertion for every zero has not been established.

Sources

  • S1 H. M. Bui and D. R. Heath-Brown, "On simple zeros of the Riemann zeta-function," Bulletin of the London Mathematical Society 45$5$ $2013$, 953-961. https://arxiv.org/html/1302.5018. Catalogue formulation source.
  • E1 Levent Alpöge and Ralph Furman, More than two thirds of the zeta zeros are simple and on the critical line, arXiv:2608.13637, 13 August 2026, abstract. https://arxiv.org/abs/2608.13637. Consulted 27 September 2026. Current manuscript claim; its proof and advertised formal verification have not been audited here.

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