on more than two thirds of the zeta zeros on the critical line

Overview

A new manuscript proves unconditionally that more than two thirds of the nontrivial zeros of the Riemann zeta function are simple and lie on the critical line. In a dyadic window $T<\mathrm{Im}\rho \le 2T$, it first obtains the proportion $2/3$, then optimizes the test function to reach

$$\frac{3}{2}-\frac{1}{\sqrt{2}}\cot \left(\frac{1}{\sqrt{2}}\right)=0.67250\dots .$$

It also proves that at least $0.83625\dots$ of the zeros are distinct (Claude, 2026b; Claude, 2026a).

The argument turns Weil’s explicit-formula pairing into a finite real symmetric matrix. A zero on the critical line contributes a positive rank-one form, while a symmetric pair of zeros off the line contributes a hyperbolic block of signature $(1,1)$. The prime side of the explicit formula determines the trace and Frobenius norm of the full matrix. A rank-trace inequality then forces enough positive rank to come from zeros on the line.

This is not a mollifier argument and does not assume the Riemann hypothesis. Its analytic range is exactly the range in which the relevant Dirichlet polynomial has length at most $T$.

Status and provenance

The source is a manuscript dated August 10, 2026 and authored by Claude. It reports review by the human mathematicians named in its acknowledgments, a Lean 4 formalization, and separate symbolic checks of the main constants. The manuscript also emphasizes that the symbolic checks do not verify the analytic error estimates. This post explains the argument as presented; it is not an independent audit of a new result of this magnitude.

🏷️ Zeta-Function Preliminaries

For $\mathrm{Re}s>1$, the Riemann zeta function is

$$\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^s}=\prod _p\frac{1}{1-p^{-s}}.$$

It extends meromorphically to the complex plane with a simple pole at $s=1$. The completed function

$$\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma \left(\frac{s}{2}\right)\zeta (s)$$

is entire and satisfies

$$\xi (s)=\xi (1-s),\xi (\overline{s})=\overline{\xi (s)}.$$

The nontrivial zeros lie in the critical strip

$$0<\mathrm{Re}s<1.$$

Write a zero as

$$\rho =\beta +i\gamma$$

and denote its multiplicity by $m_{\rho }$. The functional equation and complex conjugation imply the symmetries

$$\rho ⟷1-\rho ,\rho ⟷\overline{\rho }.$$

At positive height, the symmetry most useful here is

$$\rho ⟷1-\overline{\rho }.$$

It preserves the ordinate $\gamma$ and reflects the real part across the critical line

$$\mathrm{Re}s=\frac{1}{2}.$$

The Riemann hypothesis asserts that every nontrivial zero has $\beta =1/2$. The present argument proves only a lower proportion on that line; it does not constrain the remaining zeros.

Counting Conventions

Let $I=(T,2T]$. The relevant counting functions are

notation	zeros counted in $I$
$N(T,2T)$	all zeros, with multiplicity
$N_0(T,2T)$	critical-line zeros, with multiplicity
$N_0^{\ast }(T,2T)$	distinct critical-line zeros
$N_0^s(T,2T)$	simple critical-line zeros
$N_d(T,2T)$	all distinct zeros

Thus

$$N_0^s\le N_0^{\ast }\le N_0\le N,N_0^s\le N_d\le N.$$

The Riemann-von Mangoldt formula gives

$$N(T,2T)=\frac{T}{2\pi }\left(\log \frac{T}{2\pi }+2\log 2-1\right)+O(\log T).$$

It is convenient to write

$$\ell =\log \frac{T}{2\pi },{\ell }_1=\ell +2\log 2-1,$$

so that

$$N(T,2T)=\frac{T{\ell }_1}{2\pi }+O(\ell ).$$

The mean spacing between zero ordinates near height $T$ is therefore approximately

$$\frac{2\pi }{\ell }.$$

🏷️ Main Proportion Theorems

For $0<\lambda \le 1$, define

$$H(\lambda )=2-\frac{1}{\lambda }-\frac{\lambda }{3},$$ $$H_d(\lambda )=\frac{1+H(\lambda )}{2},F(\lambda )=\frac{\lambda }{1+{\lambda }^2/3}.$$

Unconditional dyadic proportions

For each fixed $0<\lambda \le 1$,

$$N_0^{\ast }(T,2T)\ge \left(H(\lambda )-O_{\lambda }\left(\frac{\log \log T}{\log T}\right)\right)N(T,2T),$$ $$N_0^s(T,2T)\ge \left(H(\lambda )-O_{\lambda }\left(\frac{\log \log T}{\log T}\right)\right)N(T,2T),$$

and

$$N_d(T,2T)\ge \left(\max \{H_d(\lambda ),F(\lambda )\}-O_{\lambda }\left(\frac{\log \log T}{\log T}\right)\right)N(T,2T).$$

When $\lambda <1$, the factor $\log \log T$ can be omitted from these errors.

At the endpoint $\lambda =1$,

$$H(1)=\frac{2}{3},H_d(1)=\frac{5}{6},F(1)=\frac{3}{4}.$$

Consequently,

$$\underset{T\to \infty }{\mathrm{liminf}}\frac{N_0^s(T,2T)}{N(T,2T)}\ge \frac{2}{3},$$

and the same lower bound holds for $N_0^{\ast }/N$, while

$$\underset{T\to \infty }{\mathrm{liminf}}\frac{N_d(T,2T)}{N(T,2T)}\ge \frac{5}{6}.$$

Summing over dyadic intervals transfers the conclusions from $(T,2T]$ to $(0,T]$.

Historical Scale

Before this manuscript, the unconditional critical-line record was more than $5/12$, obtained through refinements of Levinson’s mollifier method (Pratt et al., 2020). Montgomery’s pair-correlation argument had long yielded $2/3$ simple zeros under the Riemann hypothesis (Montgomery, 1973).

The new point is not a stronger prime-side moment. The necessary band-limited pair-correlation evaluation was already known unconditionally (Baluyot et al., 2024). The new point is a way to interpret its zero side without assuming that every zero ordinate is real in the shifted coordinate introduced below.

🏷️ Weil’s Hermitian Form

For a nontrivial zero $\rho =\beta +i\gamma$, set

$${\gamma }_{\rho }=\frac{\rho -\frac{1}{2}}{i}=\gamma -i\left(\beta -\frac{1}{2}\right).$$

Then

$${\gamma }_{\rho }\in \mathbb{R}⟺\beta =\frac{1}{2}.$$

Reflection $\rho \mapsto 1-\overline{\rho }$ becomes

$${\gamma }_{\rho }⟼\overline{{\gamma }_{\rho }}.$$

For $f\in C_c^2(\mathbb{R})$, use the Fourier convention

$$h_f(z)=\hat{f}(z)={\int }_{\mathbb{R}}f(u)e^{izu}du.$$

Because $f$ is compactly supported, $h_f$ extends to an entire function. Two integrations by parts give sufficient decay to make the zero sum below absolutely convergent.

Define the Hermitian pairing

$$W(f,g)=\sum _{\rho }{m}_{\rho }{h}_f({\gamma }_{\rho })\overline{h_g(\overline{{\gamma }_{\rho }})}.$$

The symmetry ${\gamma }_{\rho }\leftrightarrow \overline{{\gamma }_{\rho }}$ makes $W$ Hermitian.

If the Riemann hypothesis holds, every ${\gamma }_{\rho }$ is real and

$$W(f,f)=\sum _{\rho }{m}_{\rho }|h_f({\gamma }_{\rho }){|}^2\ge 0.$$

In fact, positivity of the full Weil form is a criterion equivalent to the Riemann hypothesis. The manuscript does not prove positivity. Instead it compresses $W$ to a carefully chosen finite-dimensional subspace and reads the inertia of that compression.

Spectral Form of the Explicit Formula

Suppose

$$\mathrm{supp}f,\mathrm{supp}g\subset \left[-\frac{L}{2},\frac{L}{2}\right],X=e^L.$$

Weil’s explicit formula can be organized as

$$W(f,g)={\int }_{\mathbb{R}}{h}_f(\tau )\overline{h_g(\tau )}{\nu }_X(\tau )d\tau ,$$

where

$${\nu }_X=\mu +{\Pi }_X+P_X.$$

Here

$$\mu (\tau )=\frac{1}{2\pi }\mathrm{Re}\frac{{\Gamma }^{\prime }}{\Gamma }\left(\frac{1}{4}+\frac{i\tau }{2}\right)-\frac{\log \pi }{2\pi }$$

is the archimedean density,

$$P_X(\tau )=-\frac{1}{\pi }\sum _{n\le X}\frac{\Lambda (n)}{\sqrt{n}}\cos (\tau \log n)$$

is the prime-power contribution, and ${\Pi }_X$ is an explicit lower-order term coming from the pole of $\zeta$.

The support restriction is essential. The explicit formula is applied to the convolution $f\ast \tilde{g}$, whose support lies in $[-L,L]$, so only prime powers satisfying

$$\log n\le L,n\le X=e^L$$

appear. Later,

$$L=\lambda \ell ,X={\left(\frac{T}{2\pi }\right)}^{\lambda }.$$

The range $\lambda \le 1$ is therefore the Dirichlet-polynomial range $X\lesssim T$.

🏷️ Finite Gabor Compression

Choose an even, smooth taper

$$0\le \phi \le 1,\mathrm{supp}\phi \subset \left[-\frac{L}{2},\frac{L}{2}\right].$$

For the first, nonoptimized bound, $\phi$ is essentially $1$ on most of this interval and decreases smoothly to $0$ across fixed-width ramps at the endpoints.

Set

$$h=\frac{2\pi }{L},{\tau }_k=T+kh,0\le k<d,$$

with

$$d=\lfloor \frac{LT}{2\pi }\rfloor .$$

Since $N(T,2T)\sim T\ell /(2\pi )$,

$$d\sim \lambda N(T,2T).$$

Define the modulated windows

$$f_k(u)=\phi (u)e^{-i{\tau }_{k}u}.$$

Their transforms are translates:

$$h_{f_k}(z)=\hat{\phi }(z-{\tau }_k).$$

The compression matrix is

$$G_{k\ell }=W(f_k,f_{\ell }).$$

It has two exactly equal descriptions:

$$G_{k\ell }=\sum _{\rho }{m}_{\rho }\hat{\phi }({\gamma }_{\rho }-{\tau }_k)\hat{\phi }({\gamma }_{\rho }-{\tau }_{\ell })$$

and

$$G_{k\ell }={\int }_{\mathbb{R}}\hat{\phi }(\tau -{\tau }_k)\hat{\phi }(\tau -{\tau }_{\ell }){\nu }_X(\tau )d\tau .$$

The first is the zero side; the second is the prime side. Indeed, the second factor in $W(f_k,f_{\ell })$ initially appears as $\overline{\hat{\phi }(\overline{{\gamma }_{\rho }}-{\tau }_{\ell })}$. Since $\phi$ is real, this equals $\hat{\phi }({\gamma }_{\rho }-{\tau }_{\ell })$.

Critical Sampling Identity

Let

$$a=\frac{1}{L}{\int }_{\mathbb{R}}\phi (u{)}^{2}du.$$

Poisson summation at spacing $2\pi /L$ gives

$$\sum _{k\in \mathbb{Z}}\hat{\phi }(\tau -{\tau }_k)\hat{\phi }({\tau }^{\prime }-{\tau }_k)=L\widehat{{\phi }^2}(\tau -{\tau }^{\prime }).$$

In particular,

$$\sum _{k\in \mathbb{Z}}\hat{\phi }(\tau -{\tau }_k{)}^2=a{L}^2.$$

This identity fixes the useful normalization

$$\hat{G}=\frac{G}{a{L}^2}.$$

In these units, an isolated simple zero on the line, observed through the full sampling grid, contributes trace $1$.

Time-frequency interpretation

The functions $f_k$ form a critically sampled Gabor-type family. The proof uses the sampling identity, not orthogonality. Indeed, orthonormalizing the family would obscure the simple rank-one form contributed by each zero.

🏷️ The Zero-Side Block Structure

Enlarge the height window slightly:

$$I^{\prime }=\left(T-\sqrt{T},2T+\sqrt{T}\right].$$

Split the zeros in $I^{\prime }$ into three classes:

• $S_1$: simple zeros on the critical line, with $s_1=|S_1|$;
• $S_2$: distinct multiple zeros on the critical line, with $s_2=|S_2|$;
• $\mathcal{P}$: unordered off-line pairs $\{\rho ,1-\overline{\rho }\}$, with $p=|\mathcal{P}|$.

Counting multiplicity gives

$$N(I^{\prime })\ge s_1+2s_2+2p.$$

On-Line Zeros

For a zero on the line, ${\gamma }_{\rho }$ is real. Its contribution to the quadratic form of $G$ is

$$m_{\rho }{\left|\sum _k{c}_k\hat{\phi }({\gamma }_{\rho }-{\tau }_k)\right|}^2.$$

This is a nonnegative rank-one form.

Thus the total contribution $P$ from distinct on-line zeros satisfies

$$P⪰0,\mathrm{rank}P\le s_1+s_2.$$

After normalization,

$$\mathrm{tr}P\le \sum _{\rho \in S_1\cup S_2}{m}_{\rho }.$$

Off-Line Pairs

Let

$${\rho }^{\ast }=1-\overline{\rho }.$$

Then

$${\gamma }_{{\rho }^{\ast }}=\overline{{\gamma }_{\rho }}.$$

If

$$x=\sum _k{c}_k\hat{\phi }({\gamma }_{\rho }-{\tau }_k),y=\sum _k{c}_k\hat{\phi }(\overline{{\gamma }_{\rho }}-{\tau }_k),$$

the paired contribution has the form

$$2m_{\rho }\mathrm{Re}(x\overline{y}).$$

Before pullback through the evaluation map, this is represented by

$$m_{\rho }\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right).$$

Its eigenvalues are $m_{\rho }$ and $-m_{\rho }$, so its signature is

$$(1,1).$$

A pullback cannot increase the positive index. Therefore, if $Q$ is the sum of all off-line pair contributions,

$$n_{+}(Q)\le p.$$

No independence among the evaluation vectors is required. This remains valid when a pair lies extremely close to the critical line or when several zeros share the same ordinate.

The essential replacement for positivity

Under the Riemann hypothesis there is no $Q$: every zero contributes a positive square. Unconditionally, $Q$ can be indefinite. The proof retains only the robust information that each off-line pair contributes at most one positive direction.

🏷️ A Rank-Trace Inequality

The central linear-algebra lemma converts trace data and inertia data into a lower bound for positive rank.

Rank-trace inequality

Let $P,Q$ be Hermitian $d\times d$ matrices. Suppose

$$P⪰0,\mathrm{rank}P\le r,n_{+}(Q)\le b.$$

Then, for every $c>0$,

$$\Vert P+Q{\Vert }_{\mathrm{F}}^2\ge c\mathrm{tr}P-\frac{c^2}{4}r+2c\mathrm{tr}Q-c^{2}b.$$

At $c=2$,

$$r\ge 2\mathrm{tr}P+4\mathrm{tr}Q-4b-\Vert P+Q{\Vert }_{\mathrm{F}}^2.$$

Proof: Negative Directions and von Neumann's Inequality

Decompose

$$Q=Q_{+}-Q_{-},Q_{+},Q_{-}⪰0,Q_{+}{Q}_{-}=0.$$

Since $Q$ has at most $b$ positive eigenvalues,

$$\mathrm{rank}{Q}_{+}\le b.$$

Expanding the square gives

$$\Vert P+Q{\Vert }_{\mathrm{F}}^2=\Vert P{\Vert }_{\mathrm{F}}^2+\Vert Q_{+}{\Vert }_{\mathrm{F}}^2+\Vert Q_{-}{\Vert }_{\mathrm{F}}^2+2\mathrm{tr}(P{Q}_{+})-2\mathrm{tr}(P{Q}_{-}).$$

The term $\mathrm{tr}(P{Q}_{+})$ is nonnegative. Let $p_i$ be the eigenvalues of $P$ and $n_i$ those of $Q_{-}$, both in decreasing order. Von Neumann’s trace inequality gives

$$\mathrm{tr}(P{Q}_{-})\le \sum _i{p}_i{n}_i.$$

Since $p_i=0$ for $i>r$,

$$\Vert P{\Vert }_{\mathrm{F}}^2-2\mathrm{tr}(P{Q}_{-})+\Vert Q_{-}{\Vert }_{\mathrm{F}}^2\ge \sum _{i\le r}(p_i-n_i{)}^2+\sum _{i>r}{n}_i^2.$$

Apply

$$x^2\ge cx-\frac{c^2}{4}$$

to the first $r$ terms. Since every $n_i\ge 0$,

$$-c\sum _{i\le r}{n}_i\ge -2c\sum _i{n}_i=-2c\mathrm{tr}{Q}_{-}.$$

The remaining squares $n_i^2$ are nonnegative, so this part is at least $c\mathrm{tr}P-c^{2}r/4-2c\mathrm{tr}{Q}_{-}$. For the positive part, if $q_j>0$ are the nonzero eigenvalues of $Q_{+}$, then

$$\sum _j{q}_j^2\ge \sum _j(2c{q}_j-c^2)\ge 2c\mathrm{tr}{Q}_{+}-c^{2}b.$$

Combining the estimates and using

$$\mathrm{tr}Q=\mathrm{tr}{Q}_{+}-\mathrm{tr}{Q}_{-}$$

proves the claim.

For $c=2$, the scalar inequality is

$$x^2\ge 2x-1.$$

This is the matrix replacement for the classical multiplicity inequality

$$m^2\ge 2m-1.$$

By placing only the simple on-line zeros in $P$, the argument also captures

$$m^2\ge 3m-2,$$

equivalently

$$(m-1)(m-2)\ge 0.$$

This second integrality level is what yields the same lower bound for simple zeros as for all distinct on-line zeros.

🏷️ From Block Structure to Counting Inequalities

Let $A$ be the normalized contribution from zeros in $I^{\prime }$, so the full normalized matrix is

$$\hat{G}=A+E,$$

where $E$ is the contribution from zeros outside $I^{\prime }$.

Regroup

$$A=P_1+Q^{\prime },$$

where $P_1$ contains only simple on-line zeros. Then

$$P_1⪰0,\mathrm{rank}{P}_1\le s_1,\mathrm{tr}{P}_1\le s_1,$$

while

$$n_{+}(Q^{\prime })\le s_2+p.$$

Applying the rank-trace inequality at $c=2$ gives

$$3s_1+4s_2+4p\ge 4\mathrm{tr}A-\Vert A{\Vert }_{\mathrm{F}}^2.$$

Set

$$R(A)=4\mathrm{tr}A-\Vert A{\Vert }_{\mathrm{F}}^2.$$

Since

$$s_1+2s_2+2p\le N(I^{\prime }),$$

we have

$$4s_2+4p\le 2N(I^{\prime })-2s_1.$$

Therefore

$$R(A)\le s_1+2N(I^{\prime }),$$

and hence

$$s_1\ge R(A)-2N(I^{\prime }).$$

For the distinct-zero count, observe that

$$3s_1+4s_2+4p=2(s_1+s_2+p)+(s_1+2s_2+2p).$$

Consequently,

$$s_1+s_2+p\ge \frac{1}{2}\left(R(A)-N(I^{\prime })\right).$$

The actual number of distinct zeros is $s_1+s_2+2p$, so this is a valid lower bound for $N_d$.

These inequalities are purely linear algebra plus the functional-equation symmetry. The arithmetic enters only in the estimates for the trace and Frobenius norm.

🏷️ The Prime-Side Moment Calculation

The explicit formula expresses $G$ using the real density ${\nu }_X$. The critical sampling identity then converts the first two matrix moments into integrals of a single convolution kernel.

After normalization, the manuscript proves

$$\mathrm{tr}\hat{G}=N(T,2T)+o(N(T,2T))$$

and

$$\Vert \hat{G}{\Vert }_{\mathrm{F}}^2=\left(\frac{1}{\lambda }+\frac{\lambda }{3}+o(1)\right)N(T,2T).$$

There are two main pieces in the second moment.

Archimedean Contribution

Stirling’s formula gives, uniformly for $T\le \tau \le 2T$,

$$\mu (\tau )=\frac{1}{2\pi }\log \frac{\tau }{2\pi }+O(T^{-2}).$$

Thus

$${\int }_T^{2T}\mu (\tau )d\tau =N(T,2T)+O(\ell )$$

and

$${\int }_T^{2T}\mu (\tau {)}^{2}d\tau =\frac{T{\ell }_1^2}{4{\pi }^2}\left(1+O({\ell }^{-2})\right).$$

After the window and matrix normalizations, this produces the term

$$\frac{1}{\lambda }N(T,2T)$$

in the squared Frobenius norm.

Prime Diagonal

The prime-power term is a Dirichlet polynomial with frequencies $\log n$:

$$P_X(\tau )=-\frac{1}{\pi }\sum _{n\le X}\frac{\Lambda (n)}{\sqrt{n}}\cos (\tau \log n).$$

When the square is integrated over an interval of length $T$, the diagonal $n=m$ supplies the second main term. For the nearly flat window, partial summation and the prime number theorem give the model asymptotic

$$\sum _{n\le X}\frac{\Lambda (n{)}^2}{n}\left(L-\log n\right)\sim \frac{L^3}{6}.$$

After normalization, this becomes

$$\frac{\lambda }{3}N(T,2T).$$

The off-diagonal terms are controlled by the Montgomery-Vaughan generalized Hilbert inequality applied to the separated frequencies $\log n$. The mixed terms involving $\mu$, $P_X$, and the pole contribution are lower order.

Why $\lambda \le 1$

Since

$$X=e^L={\left(\frac{T}{2\pi }\right)}^{\lambda },$$

the condition $\lambda \le 1$ keeps the Dirichlet polynomial at length at most $T$. Beyond this range the off-diagonal prime sums are no longer negligible. Evaluating them would require Hardy-Littlewood-strength information about prime pairs.

The ratio form of the moment estimate is

$$\frac{(\mathrm{tr}G{)}^2}{\mathrm{tr}(G^2)}=\left(F(\lambda )+o(1)\right)N(T,2T),$$

where

$$F(\lambda )=\frac{\lambda }{1+{\lambda }^2/3}.$$

This is the thresholded Cauchy-Schwarz branch used for one part of the distinct-zero theorem.

🏷️ Control of Zeros Outside the Window

The Fourier transform of a sharp cutoff decays only as $1/|r|$, which is too slow to discard remote zeros. This is why the window is tapered.

For the smooth $\phi$, two integrations by parts give

$$|\hat{\phi }(r-iy)|\lesssim X^{1/4}\Vert {\phi }^{\prime \prime }{\Vert }_1|r{|}^{-2},|y|<\frac{1}{2}.$$

Using the classical local zero count

$$N(t+1)-N(t)\ll \log (t+3),$$

the contribution $E$ of zeros outside

$$\left(T-\sqrt{T},2T+\sqrt{T}\right]$$

has operator norm tending to zero after normalization. Its trace-norm contribution and the zeros in the two boundary strips together produce $o(N)$.

At $\lambda =1$, the final relative error is $O(\log \log T/\log T)$. For fixed $\lambda <1$, the prime off-diagonal component has the power saving $T^{\lambda -1}$, while the fixed-width taper still leaves an overall $O_{\lambda }(1/\log T)$ relative error.

🏷️ Assembly of the $2/3$ Bound

Let

$$N=N(T,2T).$$

The zero-side counting inequality gives

$$N_0^s(T,2T)\ge 4\mathrm{tr}\hat{G}-\Vert \hat{G}{\Vert }_{\mathrm{F}}^2-2N-o(N).$$

Insert the prime-side moments:

$$\mathrm{tr}\hat{G}=N+o(N)$$

and

$$\Vert \hat{G}{\Vert }_{\mathrm{F}}^2=\left(\frac{1}{\lambda }+\frac{\lambda }{3}\right)N+o(N).$$

Then

$$N_0^s(T,2T)\ge \left(4-\frac{1}{\lambda }-\frac{\lambda }{3}-2-o(1)\right)N.$$

Therefore

$$N_0^s(T,2T)\ge \left(H(\lambda )-o(1)\right)N.$$

At $\lambda =1$,

$$4-\left(1+\frac{1}{3}\right)-2=\frac{2}{3}.$$

The same calculation gives the distinct-zero bound:

$$N_d(T,2T)\ge \frac{1}{2}\left(4-\frac{1}{\lambda }-\frac{\lambda }{3}-1-o(1)\right)N=\left(H_d(\lambda )-o(1)\right)N.$$

At $\lambda =1$, this is

$$\frac{1}{2}\left(3-\frac{4}{3}\right)=\frac{5}{6}.$$

The notable point is that the proof never attempts to show that the full matrix is positive semidefinite. The negative directions created by off-line zeros are allowed; their number is controlled through the hyperbolic block structure.

🏷️ Optimization of the Window

The flat-topped taper gives $2/3$, but it is not optimal. Write

$$\phi (u{)}^2=v\left(\frac{u}{L}\right),-\frac{1}{2}\le \frac{u}{L}\le \frac{1}{2}.$$

For a fixed bandwidth $\lambda$, the effective moment ratio is the scale-free functional

$$c_{\lambda }(v)=\frac{\lambda {\left({\int }_{-1/2}^{1/2}v(s)ds\right)}^2}{{\int }_{-1/2}^{1/2}v(s{)}^{2}ds+{\lambda }^2{\int }_{-1/2}^{1/2}{\int }_{-1/2}^{1/2}|s-s^{\prime }|v(s)v(s^{\prime })dsd{s}^{\prime }}.$$

Maximizing this Rayleigh quotient gives the integral equation

$$v+{\lambda }^2\mathcal{T}v=\text{constant},$$

where

$$(\mathcal{T}v)(s)={\int }_{-1/2}^{1/2}|s-s^{\prime }|v(s^{\prime })d{s}^{\prime }.$$

Differentiating twice yields

$$v^{\prime \prime }+2{\lambda }^{2}v=0.$$

The positive even optimizer is

$$v_{\lambda }^{\ast }(s)=\cos (\sqrt{2}\lambda s).$$

At $\lambda =1$,

$$c_1^{\ast }=\frac{\sqrt{2}\tan (1/\sqrt{2})}{1+(1/\sqrt{2})\tan (1/\sqrt{2})}=0.7532960\dots ,$$

so

$$\frac{1}{c_1^{\ast }}=\frac{1}{2}+\frac{1}{\sqrt{2}}\cot \left(\frac{1}{\sqrt{2}}\right)=1.3274992\dots .$$

The optimized on-line and simple-zero proportions are therefore

$$2-\frac{1}{c_1^{\ast }}=\frac{3}{2}-\frac{1}{\sqrt{2}}\cot \left(\frac{1}{\sqrt{2}}\right)=0.67250\dots .$$

For distinct zeros,

$$\frac{1}{2}\left(3-\frac{1}{c_1^{\ast }}\right)=0.83625\dots .$$

This recovers the Montgomery-Taylor optimizing kernel, now inside an unconditional inertia argument.

🏷️ Scope and Structural Limitations

The proof uses the following analytic inputs:

• Weil’s explicit formula;
• the Riemann-von Mangoldt formula;
• Stirling’s formula for ${\Gamma }^{\prime }/\Gamma$;
• Chebyshev-Mertens estimates for prime-power sums;
• the Montgomery-Vaughan generalized Hilbert inequality;
• the local bound $N(t+1)-N(t)\ll \log (t+3)$.

It does not use a zero-density estimate, a zero-free region, or a mollifier.

The result is not the Riemann hypothesis

The argument certifies a lower bound. It says nothing about the location of the remaining proportion of zeros. The explicit formula and the first two moments are insensitive to a sparse set of off-line zeros, so this mechanism cannot prove that every zero is on the line.

The restriction $\lambda \le 1$ is arithmetic, while the finite dimension

$$d\sim \lambda N$$

is geometric. At $\lambda \le 1/2$, the rank bound alone makes the present certificate nonpositive. At $\lambda >1$, the prime off-diagonal requires unavailable correlation information. Higher moments do not bypass these two barriers in the unconditional range analyzed by the manuscript.

The first two moments and the block constraints are also sharp for the extremal multiplicity pattern. A model configuration with

$$\frac{2}{3}N$$

mutually orthogonal simple on-line contributions and

$$\frac{1}{6}N$$

double contributions has

$$\mathrm{tr}\hat{G}=N,\Vert \hat{G}{\Vert }_{\mathrm{F}}^2=\frac{4}{3}N,$$

and saturates both the $2/3$ simple-zero count and the $5/6$ distinct-zero count. Improving the constants therefore requires additional information, not merely a rearrangement of the same two matrix moments.

Dirichlet $L$-Functions

The manuscript also extends the argument to each fixed primitive Dirichlet $L$-function. The functional equation supplies the same conjugate-pair block structure, while the prime coefficients acquire phases of modulus at most $1$. The same $2/3$, $5/6$, and optimized constants follow with the conductor fixed.

🏷️ Conceptual Summary

The proof can be compressed into one chain:

$$\text{explicit formula}⟹\text{finite Hermitian matrix}⟹\{\begin{array}{ll}\text{on-line zero}& \mapsto \text{positive rank one},\\ \text{off-line pair}& \mapsto \text{signature }(1,1),\end{array}$$ $$\text{prime sums}⟹\mathrm{tr}\hat{G}\sim N,\Vert \hat{G}{\Vert }_{\mathrm{F}}^2\sim \left(\frac{1}{\lambda }+\frac{\lambda }{3}\right)N,$$

and finally

$$\text{rank-trace inequality}⟹N_0^s\ge \left(2-\frac{1}{\lambda }-\frac{\lambda }{3}-o(1)\right)N.$$

The conceptual innovation is the middle line. Classical pair correlation reads the zero side term by term under the Riemann hypothesis. The new argument reads the entire finite compression spectrally: positivity is replaced by a count of positive directions, and the functional equation limits how many such directions off-line pairs can supply.

Links

• on interlacing families and Kadison-Singer — both arguments extract global information from eigenvalue constraints on finite Hermitian matrices and from structured rank-one positive semidefinite contributions.
• on Sylvester’s determinantal identity and Schweinsian expansion — this provides nearby historical linear-algebra context, although the present proof uses Sylvester’s law of inertia rather than Sylvester’s determinantal identity.

🐻  Baluyot, S.A.C., Goldston, D.A., Suriajaya, A.I. & Turnage-Butterbaugh, C.L. 2024. An Unconditional Montgomery Theorem for Pair Correlation of Zeros of the Riemann Zeta-Function. Acta Arithmetica 214, 357–376.

🐻  Claude 2026a. 67% of the Zeroes Are on the Line.

🐻  Claude 2026b. More Than Two Thirds of the Zeros of the Riemann Zeta Function Lie on the Critical Line.

🐻  Montgomery, H.L. 1973. The Pair Correlation of Zeros of the Zeta Function. In Analytic Number Theory, pp. 181–193. Proceedings of Symposia in Pure Mathematics, Providence, RI: American Mathematical Society.

🐻  Pratt, K., Robles, N., Zaharescu, A. & Zeindler, D. 2020. More Than Five-Twelfths of the Zeros of ζ Are on the Critical Line. Research in the Mathematical Sciences 7(1), Paper No. 2, 74 pp.