# Off-line zeros of any depth can be hidden from the two-thirds pair-correlation certificate *span_of_dilates, version 2, 2026-11-04. Version 1 was accepted as Paper 109; this version makes the corrections requested in its review (listed at the end). Theorem 1, Proposition 2, Corollaries 3 and 4 and Lemma 5 are proved here. Corollaries 3 and 4 concern a periodic model of the Alpöge–Furman certificate (indicator window, untruncated frame), not the zeros of the Riemann zeta function. Section 6 reports numerical optimisations, which give upper bounds only. Apart from the review of version 1, these results have not been checked independently.* ## Abstract Alpöge and Furman proved unconditionally that at least 2/3 of the zeros of the Riemann zeta function are simple and on the critical line. Their certificate uses the trace and squared Hilbert–Schmidt norm of a Gram matrix built from the zeros through a Gabor frame of bandwidth $L=\log(T/2\pi)$. An isolated off-line pair at depth $\beta-1/2=\varepsilon$ adds about $e^{2\varepsilon L}/(2\varepsilon^2L^2)$ to the squared norm, which suggests that these data might bound the number of zeros far from the line. We show that they cannot. Theorem 1: a lattice of reflected pairs with spacing at most the frame step $2\pi/L$ has exactly the Gram matrix of on-line double zeros at the same ordinates, at every depth, for every window supported in an interval of length $L$. So does every pair sum of bandwidth at most one. The proof is the Poisson step of Alpöge–Furman's Lemma 2.1, with zeros and frame exchanged. Proposition 2: for the indicator window, a chain of $m$ pairs at depth $c=\varepsilon L$ costs $(4\sinh^2 c/\pi^2)(\log m+C(c))+O(1/m)$ more than the doubles, with $C$ explicit. Corollary 3: in a periodic model of the certificate (indicator window, untruncated frame), the Alpöge–Furman extremal configuration with its doubles replaced by such chains has, per period of $N$ zeros, trace $N$ and squared Hilbert–Schmidt norm $(4/3+o(1))N$, while a third of its zeros lie at depth $\varepsilon$, for every fixed $\varepsilon\in(0,1/2)$. These data therefore give no zero-density estimate beyond $N(\sigma)\le(1/6+o(1))N$ for any $\sigma<1$. Aryan observed that going beyond 2/3 requires the negative spectral mass of off-line pairs not to cancel completely; in the model it cancels completely at every depth. Numerically optimised chains cost 50 times less at $c=4$ and $10^4$ times less at $c=8$. ## 1. Introduction Montgomery's pair-correlation argument shows, under RH, that at least 2/3 of the zeros of $\zeta$ are simple. Baluyot, Goldston, Suriajaya and Turnage-Butterbaugh [1] proved Montgomery's theorem for $F(\alpha)$, $0\le\alpha\le1$, without RH, with weights $x^{\beta+\beta'-1}$ for zeros off the line. Alpöge and Furman [2] then proved, unconditionally, that at least 2/3 of the zeros are simple and on the critical line. Their certificate is a real symmetric matrix $\widetilde G$ built from the zeros through a Gabor frame at critical density and bandwidth $L=\log(T/2\pi)$. It uses the trace of $\widetilde G$, its squared Hilbert–Schmidt norm, and the fact that a reflected pair $\{\rho,1-\bar\rho\}$ contributes a block $2(pp^{\mathsf T}-qq^{\mathsf T})$ with one positive direction. Lamzouri [3] gave a proof by a single Hilbert-space inequality, Knausgård [4] improved the constants, and Aryan [5] wrote an exposition. This note asks whether the same data see how far an off-line zero is from the line. An isolated pair at depth $\beta-\tfrac12=\varepsilon$ adds about $e^{2\varepsilon L}/(2\varepsilon^2L^2)$ to the squared Hilbert–Schmidt norm ((4.1) with $m=1$; compare [5, Sec. 3.1]). Aryan notes that this made it look plausible "that zeros away from the critical line should be very easy to detect". [2, Sec. 7.2(b)] shows that the extremal configuration of the method can use off-line pairs "of depth $\to0$ (spectrally the same)" in place of on-line doubles. [5, Sec. 3.4] observes that in the worst case the negative spectral mass $B$ of the off-line blocks cancels almost completely, and that "to go beyond 2/3, one would need to show that this cancellation cannot be complete". We show that the cancellation can be complete at every depth, and we compute what finite configurations cost. - **Theorem 1 (spectral twin).** A lattice of reflected pairs with spacing at most the frame step $2\pi/L$ has exactly the Gram matrix of on-line double zeros at the same ordinates, whatever the depth, for every window supported in an interval of length $L$. The same holds for every pair sum whose test function has Fourier support in $[-L,L]$. - **Proposition 2 (finite chains).** For the indicator window, a chain of $m$ pairs at frame step and depth $c=\varepsilon L$ exceeds the squared Hilbert–Schmidt norm of the corresponding doubles by an explicit $E(m,c)=\frac{4\sinh^2c}{\pi^2}(\log m+C(c))+O_c(1/m)$, with $C$ given by digamma values. - **Corollary 3 (the extremal at every depth).** In the model, the extremal configuration of [2, Sec. 7.2(b)] with its doubles replaced by such chains has, per period of $N$ zeros, $\operatorname{tr}\widetilde G=N$ and $\|\widetilde G\|_{\rm HS}^2=(\tfrac43+o(1))N$, while a third of its zeros lie at depth $\varepsilon$, for every fixed $\varepsilon\in(0,\tfrac12)$. - **Corollary 4 (no density estimate).** In the model, these data give $N(\sigma;T,2T)\le(\tfrac16+o(1))N(T,2T)$ for every $\sigma>\tfrac12$, which already follows from the 2/3 theorem, and nothing better for any $\sigma<1$. - **Section 6 (numerics).** Chains with compressed and deepened ends leak far less than uniform chains: by a factor of about 50 at $c=4$ and about $10^4$ at $c=8$, for $m=32$. - **Lemma 5 (strip bound).** A quantitative form of Lamzouri's remark [3, Sec. 2] that no non-constant entire kernel has positive real part on all of $\mathbb C$. The proof of Theorem 1 is the Poisson-summation step of [2, Lemma 2.1] with the roles of the frame and the zeros exchanged. The content is the observation and its consequences, not a new technique. Nothing here concerns the actual zeros of $\zeta$. The results say what certificates built from bandwidth-one second-moment data can and cannot certify. ## 2. Setting We follow [2, Sec. 2]. Let $L=\log(T/2\pi)$, let $\phi$ be real and even with $\operatorname{supp}\phi\subseteq[-L/2,L/2]$, and write $\hat\phi(z)=\int\phi(u)e^{-izu}\,du$. Put $\alpha_k=T+2\pi k/L$ and $a=\|\phi\|_2^2/L$. For a zero $\rho=\tfrac12+\varepsilon+ix$ write $\gamma_\rho=(\rho-\tfrac12)/i=x-i\varepsilon$ and $v_\rho=(\hat\phi(\gamma_\rho-\alpha_k))_k$. The partner $1-\bar\rho$ has $\gamma=x+i\varepsilon$ and vector $\bar v_\rho$, so a reflected pair contributes $2\operatorname{Re}(v_\rho v_\rho^{\mathsf T})=2(pp^{\mathsf T}-qq^{\mathsf T})$, where $v_\rho=p+iq$. Set $$\widetilde G=\frac1{aL^2}\sum_\rho m_\rho\,v_\rho v_\rho^{\mathsf T}.$$ [2, Lemma 2.1] gives $\sum_k\hat\phi(z-\alpha_k)\hat\phi(z'-\alpha_k)=L\,\widehat{\phi^2}(z-z')$ for all complex $z,z'$. With the frame index running over all of $\mathbb Z$, this gives $$\operatorname{tr}\widetilde G=\sum_\rho m_\rho,\qquad \|\widetilde G\|_{\rm HS}^2=\sum_{\rho,\rho'}m_\rho m_{\rho'}\,\kappa(\gamma_\rho-\gamma_{\rho'})^2,\qquad \kappa(w)=\frac{\widehat{\phi^2}(w)}{aL}. \tag{2.1}$$ The trace identity holds for off-line zeros too: every zero contributes exactly 1. **The model.** We take the indicator window $\phi=\mathbf 1_{[-L/2,L/2]}$ and the untruncated frame $k\in\mathbb Z$. Then $a=1$ and $\kappa(w)=\operatorname{sinc}(wL/2)$, with $\operatorname{sinc}z=\sin z/z$. In the scaled coordinates $X=xL$ and $c=\varepsilon L$, two zeros with $\gamma=x-i\varepsilon$ and $\gamma'=x'-i\varepsilon'$ interact through $\operatorname{sinc}\big(((X-X')-i(c-c'))/2\big)^2$. [2] instead truncate to $0\le kL$ the supports are disjoint, and at $|\xi|=L$ they meet in one point. 5. Poisson summation on $\Lambda$ gives $\sum_nH(x_*+2\pi rn/L)=\frac{L}{2\pi r}\sum_{l\in\mathbb Z}\hat H(lL/r)e^{ilLx_*/r}$. For $l\ne0$, $|lL/r|\ge L$, so only $l=0$ survives, and there $e^{\varepsilon\cdot0}=1$. Thus $(G_\varepsilon)_{kk'}=\frac{2}{aL^2}\operatorname{Re}\big[\frac{L}{2\pi r}\hat H(0)\big]$ does not depend on $\varepsilon$. The sums converge absolutely because $H(y)=O((1+|y|)^{-2})$. (ii) The differences $\gamma-\gamma'$ are $(x-x')$ twice and $(x-x')\mp2i\varepsilon$ once each. For fixed $\eta$, $y\mapsto h(y+i\eta)$ has Fourier transform $2\pi g(\xi)e^{-\eta\xi}$, supported in $[-L,L]$ and zero at $\pm L$. Poisson summation over $x'\in\Lambda$ keeps only $l=0$, which gives $\frac{L}{2\pi r}\cdot2\pi g(0)=Lg(0)/r$ for each of the four terms. $\square$ *Remarks.* (a) For the indicator window and $r=1$, $\kappa(2\pi n/L)=\operatorname{sinc}(\pi n)=0$ for $n\ne0$, and $G_0=2I$ in the frame basis. So at every depth the lattice has one positive direction with eigenvalue 2 per pair and no negative eigenvalues. This is the "flat charge 4" of [2, Sec. 7.2(c)] at every depth, and $B=0$ in the notation of [5, Sec. 3.4]. (b) The Hilbert–Schmidt norm (2.1) is a pair sum with $h=\kappa^2$, whose Fourier transform is supported in $[-L,L]$. Part (ii) covers it for every window. (c) For $1\text{dist}}\sinh^2c/(\pi n)^2$, which is about $\sinh^2c/(\pi^2\,\text{dist})$, and summing over the distances to the ends gives $\log m$. ## 5. The extremal at every depth **The pattern.** For even $m$, let the pattern $P_m$ have period $6m$ frame sites. Sites $0,\dots,m-1$ carry reflected pairs at depth $c$. The next $m/2$ sites are empty. The next $4m$ sites carry simple on-line zeros. The last $m/2$ sites are empty. A period holds $6m$ zeros, one per site, as $d=N(T,2T)+O(L)$ requires. ``` site 0 ... m-1 m ... 3m/2-1 3m/2 ... 11m/2-1 11m/2 ... 6m-1 [m pairs, ] [ empty ] [ 4m simple zeros ] [ empty ] [depth c ] [ ] [ on the line ] [ ] zeros 2m 0 4m 0 total 6m ``` **Per-period normalisation.** An infinite periodic configuration has no finite trace or Hilbert–Schmidt norm. For the pattern, $\operatorname{tr}\widetilde G$ and $\|\widetilde G\|_{\rm HS}^2$ denote the per-period quantities given by (2.1): the sum of $m_\rho$ over the zeros $\rho$ of one period, and the sum of $m_\rho m_{\rho'}\kappa(\gamma_\rho-\gamma_{\rho'})^2$ over $\rho$ in one period and $\rho'$ over all zeros, periodic images included. These are the quantities that the inequalities of [2, Sec. 6] bound per zero. We do not prove that restricting the pattern to $T<\gamma\le2T$ changes them by $o(N)$ (Section 9). At $c=0$, per period, $\operatorname{tr}\widetilde G=6m$ and $\|\widetilde G\|_{\rm HS}^2=4m+4m=\tfrac43\cdot6m$, with $4m=\tfrac23\cdot6m$ simple on-line zeros and $5m=\tfrac56\cdot6m$ distinct zeros. This is the extremal configuration of [2, Sec. 7.2(b)] made concrete: by (2.1), on-line zeros at distinct frame sites are orthogonal. **Corollary 3.** In the model, let $m\ge\max(4,c/\pi)$ be even. Per period of $P_m$ at depth $c$, $\operatorname{tr}\widetilde G=6m$ and $\|\widetilde G\|_{\rm HS}^2=8m+\theta$ with $$E(m,c)-\tfrac2{75}\sinh^2c-\tfrac8{\pi^2}\sinh^2(c/2)\log\tfrac{3m-2}{m-2}\le\theta\le E(m,c).$$ Consequently, fix $\varepsilon\in(0,\tfrac12)$, put $c=\varepsilon L$ and $m=2\lceil T^{1/2+\varepsilon}\rceil$. Then $|\theta|/m\ll T^{\varepsilon-1/2}\log T\to0$. So, per period of $N=6m$ zeros, $\operatorname{tr}\widetilde G=N$ and $\|\widetilde G\|_{\rm HS}^2=(\tfrac43+o(1))N$, the values that the certificate allows for the zeros of $\zeta$, while $N/3$ of them have $|\beta-\tfrac12|=\varepsilon$. *Proof.* Split the per-period sum (2.1), including all periodic images, into chain–chain, chain–simple and simple–simple terms. The simple–simple term is $4m$ at every depth. Chain–chain within one period is $4m+E(m,c)$ by Proposition 2. Images of the chain lie at distances $|n|\ge5m+1>c/\pi$ sites. Each contributes $-2\sinh^2c\,\tau_{|n|}<0$, and in total at least $-2\sinh^2c\cdot m^2\cdot2\sum_{p\ge1}(5\pi mp)^{-2}=-\tfrac2{75}\sinh^2c$. A pair at site $k$ and a simple zero at distance $n$ interact through $4\operatorname{Re}\operatorname{sinc}(\pi n-ic/2)^2=-4\sinh^2(c/2)\operatorname{Re}(\pi n-ic/2)^{-2}$. This lies in $[-4\sinh^2(c/2)/(\pi n)^2,0)$ because $\pi|n|>c/2$. From site $k$ the simple zeros lie at distances at least $\tfrac32m-k$ on the right and $k+\tfrac m2+1$ on the left, so summing $n^{-2}$ over them gives at most $(\tfrac32m-k-1)^{-1}+(k+\tfrac m2)^{-1}$. Summing over $k$ gives at most $2\sum_{j=m/2}^{3m/2-1}j^{-1}\le2\log\frac{3m-2}{m-2}$. Both corrections are negative, which gives the two bounds. For the consequence, $|\tau_n|\le(\pi^2n^2+c^2)^{-1}\le(\pi n)^{-2}$ in (4.1) gives $|E(m,c)|\le\frac4{\pi^2}\sinh^2c\,(3+\log m)$ uniformly in $c$, and $\sinh^2(\varepsilon L)\le(T/2\pi)^{2\varepsilon}$. $\square$ The supplement computes all three terms exactly on the torus (`extremal_torus.py`) for $c\in\{1,2,4,8\}$ and $m\in\{8,16,32,64\}$. The chain–simple term is negative in every case and does not grow with $m$: $-0.221$, $-1.125$, $-10.7$ and $-606$ at $c=1,2,4,8$. The bounds of Corollary 3 hold with room in every case. **Corollary 4.** Let $N(\sigma)$ count zeros with $\beta\ge\sigma$ and $T<\gamma\le2T$, or, for a periodic configuration, zeros with $\beta\ge\sigma$ in one period. In the model, every configuration whose data satisfy $\operatorname{tr}\widetilde G=N$ and $\|\widetilde G\|_{\rm HS}^2\le(\tfrac43+o(1))N$ has $N(\sigma)\le(\tfrac16+o(1))N$ for $\sigma>\tfrac12$. For every fixed $\sigma\in(\tfrac12,1)$, Corollary 3 with $\varepsilon=\sigma-\tfrac12$ gives such a configuration with $N(\sigma)=N/6$. *Proof.* The inequalities of [2, Sec. 6] use only the trace, the squared Hilbert–Schmidt norm, the block structure and $\operatorname{tr}P_1\le s_1$ [2, Sec. 7.2(b)], so they apply in the model. They leave at most $(\tfrac13+o(1))N$ zeros that are not simple zeros on the line. Off-line zeros come in pairs reflected across the line, so at most half of these lie to its right. The configuration of Corollary 3 has $N/6$ zeros with $\beta=\tfrac12+\varepsilon$. $\square$ So in the model no argument that uses only these data can give a zero-density estimate. Classical density theorems give $N(\sigma)=o(N)$ for every fixed $\sigma>\tfrac12$. The gap is the twin of Theorem 1: deep zeros in frame-step chains cost the certificate almost nothing. ## 6. Numerics: cheaper finite chains Corollary 3 uses uniform chains, whose cost $E(m,c)$ grows like $e^{2c}\log m$. Uniform chains are far from optimal. We minimised $\|G\|_{\rm HS}^2-4m$ over $m$ pairs with free ordinates and depths at least $c$, by L-BFGS-B with exact gradients (`leak_opt.py`, 10 random starts, and `leak_cont.py`, continuation in $m$). Total excess, best found ($c=1,2$: continuation; $c=4,8$: random starts): | $c$ | $m=8$ | $m=16$ | $m=32$ | uniform chain, $m=8,16,32$ | |---|---|---|---|---| | 1 | 1.30 | 1.50 | 1.70 | 1.87, 2.26, 2.64 | | 2 | 4.68 | 5.15 | 5.66 | 14.6, 18.3, 22.0 | | 4 | 17.8 | 18.9 | 20.1 | 579, 785, 993 | | 8 | 133 | 160 | 205 | $1.09\cdot10^6$, $1.68\cdot10^6$, $2.29\cdot10^6$ | The optimisers compress the chain towards its ends. For $c\ge4$ they also push the pairs next to the ends deeper, while the outermost pairs stay at depth $c$; at $c\le2$ the depths stay within $0.2$ of $c$. At $c=4$, $m=32$ the gaps run from $1.54\pi$ at the ends to $1.98\pi$ in the middle (in $X$, where the frame step is $2\pi$), and the depths reach $c+2.5$. At $c=8$ the gaps lie between $1.1\pi$ and $1.83\pi$ for $m\in\{8,16,24,32,48\}$, so these chains are still compressed in the middle. The saved $c=8$, $m=12$ optimum is different: it has one gap of $0.111\pi$, a near-collision of two pairs. Its excess is still a valid upper bound. For $c=1,2$ continuation to $m=96$ gives totals $2.03$ and $6.50$, growing by $0.21$ and $0.53$ per doubling of $m$. Uniform chains grow by $0.39$ and $3.7$. These are explicit configurations, so they are upper bounds for the minimal cost. They are not proved optimal, and the optimiser is not global: one continuation run at $c=4$ settled into a denser branch with gap $1.8\pi$, whose excess $4/r-4\approx0.44$ per pair ($r=0.9$) is a density cost, present at depth 0 too. The best-found costs grow roughly like $c^2$ to $c^3$ over $c\in[1,8]$, against $2(\sinh c/c)^2\approx e^{2c}/(2c^2)$ for an isolated pair. We have not proved a polynomial bound. The supplement stores every configuration behind the table and the $m\le96$ continuation (`configs/`), and `verify_configs.py` recomputes their excess in 40-digit arithmetic, reproducing the values above. ## 7. Strip positivity **Lemma 5.** Let $k$ be real-valued, continuous, even and supported in $[-1,1]$, $K(z)=\int k(\alpha)e^{iz\alpha}d\alpha$, and suppose $\operatorname{Re}K(x+iy)\ge0$ for all real $x$ and $|y|\le Y$. Then $|k(\alpha)|\cosh(Y\alpha)\le k(0)$ for every $\alpha$, and for $2c\le Y$ $$K(2ic)=\int_{-1}^1k(\alpha)\cosh(2c\alpha)\,d\alpha\le k(0)\int_{-1}^1\frac{\cosh(2c\alpha)}{\cosh(Y\alpha)}d\alpha\le2k(0).$$ *Proof.* For real-valued even $k$, $\operatorname{Re}K(x+iy)=\int k(\alpha)\cosh(y\alpha)\cos(x\alpha)\,d\alpha$, the Fourier transform of $g_y(\alpha)=k(\alpha)\cosh(y\alpha)$. If it is non-negative, then, since $g_y$ is continuous and integrable, its transform is integrable and $g_y$ is the inverse transform of a non-negative function. So $g_y$ is positive definite and $|g_y|\le g_y(0)=k(0)$. Take $y=Y$. $\square$ $K(2ic)$ is the self-interaction of a pair at depth $c$ in the scaling where $k$ lives on $[-1,1]$. A kernel that is positive on a strip, such as the Tsang kernel of [1], therefore gives a pair inside the strip a weight $O(1)$, not $e^{2c}$. Outside the strip, positivity is not guaranteed. Lamzouri [3, Sec. 2] observes that no non-constant entire kernel has $\operatorname{Re}K\ge0$ on all of $\mathbb C$. Lemma 5 is the quantitative form for a strip of width $Y$. ## 8. What can see depth - *Bandwidth above one.* For $r=1$, a test function whose Fourier transform reaches beyond $[-L,L]$ picks up the terms $l=\pm1$ in step 5 of the proof of Theorem 1, with factors $e^{\pm c}$ in the Gram entries and $e^{\pm2c}$ in the pair sums, so pair sums of bandwidth above one separate deep chains from doubles. Under the pair-correlation conjecture for all $\alpha$, Goldston, Lee, Schettler and Suriajaya [6] show that 100% of the zeros are simple and on the line. Unconditionally, $F(\alpha)$ is known asymptotically only for $\alpha\le1$. - *Argument-principle data.* Conrey's unpublished 2024 note, reported in [7, Remark 3], shows that more than 99.9% of the zeros with $T<\gamma\le2T$ lie within $8/\log T$ of the line, so $c\le8$ for them. Theorem 1 is exact at every $c\le8$, so this input bounds the depth of most zeros but does not make the certificate depth-sensitive. - *Pointwise $F(\alpha)$.* [1, Theorem 1] gives $F(\alpha)=T^{-2\alpha}(\log T+O(1))+\alpha+O((\log T)^{-1/2})$ uniformly for $0\le\alpha\le1$, not only the Fejér average behind $\|\widetilde G\|_{\rm HS}^2$. The periodic pattern $P_m$ does not satisfy this pointwise. Already the coefficient of $T^{-2\alpha}$ is wrong: at $\alpha\ll1/L$ a pair of any depth has weight $2\cosh(\alpha c)\approx2$, so for $m\gg L$ the coefficient is about $\langle\rho^2\rangle\log T=\tfrac43\log T$, where $\rho$ is the local density of $P_m$ in mean spacings ($2$ on the chains, $0$ on the gaps, $1$ on the simple zeros), whereas [1, Theorem 1] gives $\log T+O(1)$ (`pm_delta.py` finds the ratio $1.32$–$1.33$ for $m=240$, $L=16$, $c=4$, $\alpha\le0.05$). bramble_trace [8] gives an exact lower bound for the same pattern from the first Fourier mode alone, and notes that for $\zeta$ itself the Riemann–von Mangoldt formula limits the excess count of an uncompensated block to $O(\log T)$. This uses only the density on scales above $L$ mean spacings, so it applies equally to disordered variants of $P_m$, and $m\gg L$ is the regime in which Corollary 3 is used. Near $\alpha=1$ the weight $w(u)=4/(4+u^2)$ in $F$ makes $F(\alpha)$ a test whose Fourier transform reaches slightly beyond $\alpha=1$, where a frame-step chain has a tooth of mass about $e^{2c}$ per pair, so pointwise $F$ can see frame-step chains. Chains with spacing slightly below the frame step move the tooth out of reach (Theorem 1 allows $r<1$), but their excess density then shows up at small $\alpha$. We have not settled whether a configuration satisfying [1, Theorem 1] can carry a positive proportion of zeros at a fixed depth $\varepsilon$. Either answer leaves the proportion 2/3 untouched, since that proportion is decided by pairs of depth $\to0$, which are spectrally the same as doubles [2, Sec. 7.2(b)]. - *Local inequalities.* [4] improves 0.6725 to 0.67353 with a computer-assisted inequality for seven or eight consecutive zeros. We have not checked how that inequality treats off-line pairs. ## 9. Scope - Corollaries 3 and 4 are proved for the model: indicator window, frame index over all of $\mathbb Z$, periodic configuration. Restricting to $T<\gamma\le2T$, truncating the frame to $0\le k