A partial digestion of the HRT counterexample

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摘要

Terry Tao discusses a recent counterexample to the Heil-Ramanathan-Topiwala (HRT) conjecture in harmonic analysis, explaining the structure of time-frequency shifts and the conjecture's status.

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# A partial digestion of the HRT counterexample Source: [https://terrytao.wordpress.com/2026/08/06/a-partial-digestion-of-the-hrt-counterexample/](https://terrytao.wordpress.com/2026/08/06/a-partial-digestion-of-the-hrt-counterexample/) A function![{f(t) \in L^2({\bf R})}](https://s0.wp.com/latex.php?latex=%7Bf%28t%29+%5Cin+L%5E2%28%7B%5Cbf+R%7D%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)of one variable can be translated in space by a spatial shift![{x}](https://s0.wp.com/latex.php?latex=%7Bx%7D&bg=ffffff&fg=000000&s=0&c=20201002)to obtain a new function ![\displaystyle \pi(x,0) f(t) := f(t-x),](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cpi%28x%2C0%29+f%28t%29+%3A%3D+f%28t-x%29%2C&bg=ffffff&fg=000000&s=0&c=20201002) and also modulated in frequency by a frequency shift![{\omega}](https://s0.wp.com/latex.php?latex=%7B%5Comega%7D&bg=ffffff&fg=000000&s=0&c=20201002)to obtain a new function ![\displaystyle \pi(0,\omega) f(t) := e^{2\pi i \omega t} f(t).](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cpi%280%2C%5Comega%29+f%28t%29+%3A%3D+e%5E%7B2%5Cpi+i+%5Comega+t%7D+f%28t%29.&bg=ffffff&fg=000000&s=0&c=20201002) One can compose these two operations to obtain a time\-frequency shift: ![\displaystyle \pi(x,\omega) f(t) := e^{2\pi i \omega t} f(t-x).](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cpi%28x%2C%5Comega%29+f%28t%29+%3A%3D+e%5E%7B2%5Cpi+i+%5Comega+t%7D+f%28t-x%29.&bg=ffffff&fg=000000&s=0&c=20201002) \(This can be viewed of as a portion of the[Weyl representation of the Heisenberg group](https://en.wikipedia.org/wiki/Heisenberg_group), but we will not adopt a representation\-theoretic perspective here\.\) Some functions obey finite linear relations between their time\-frequency shifts\. For instance, a sinusoid![{f(t) = A \sin(k t + \phi)}](https://s0.wp.com/latex.php?latex=%7Bf%28t%29+%3D+A+%5Csin%28k+t+%2B+%5Cphi%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)obeys the relation ![\displaystyle \pi(\frac{\pi}{k},0) f + f = 0.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cpi%28%5Cfrac%7B%5Cpi%7D%7Bk%7D%2C0%29+f+%2B+f+%3D+0.&bg=ffffff&fg=000000&s=0&c=20201002) However, the Heil\-Ramanathan\-Topiwala \(HRT\) conjecture states that once one imposes some reasonable decay condition on![{f}](https://s0.wp.com/latex.php?latex=%7Bf%7D&bg=ffffff&fg=000000&s=0&c=20201002), no such relations exist: > **Conjecture 1 \(HRT conjecture\)**If![{f \in L^2({\bf R})}](https://s0.wp.com/latex.php?latex=%7Bf+%5Cin+L%5E2%28%7B%5Cbf+R%7D%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)is non\-zero, then there is no relation of the form![\displaystyle c_1 \pi(z_1) f + \dots + c_n \pi(z_n) f = 0](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++c_1+%5Cpi%28z_1%29+f+%2B+%5Cdots+%2B+c_n+%5Cpi%28z_n%29+f+%3D+0&bg=ffffff&fg=000000&s=0&c=20201002) for some distinct points![{z_1,\dots,z_n \in {\bf R}^2}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_n+%5Cin+%7B%5Cbf+R%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)and some coefficients![{c_1,\dots,c_n \in {\bf C}}](https://s0.wp.com/latex.php?latex=%7Bc_1%2C%5Cdots%2Cc_n+%5Cin+%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), not all zero\. A special case of the HRT conjecture, which was also open, makes the additional assumption that![{f}](https://s0.wp.com/latex.php?latex=%7Bf%7D&bg=ffffff&fg=000000&s=0&c=20201002)was Schwartz\. Many positive results towards this conjecture were known\. I will mention only a few here\. There is a[result of Linnell](https://arxiv.org/abs/math/9807057)that the conjecture is true if![{z_1,\dots,z_n}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_n%7D&bg=ffffff&fg=000000&s=0&c=20201002)lie in a translate of a discrete subgroup of![{{\bf R}^2}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+R%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002); this \(together with an argument handling the collinear case\) establishes all cases where![{n \leq 3}](https://s0.wp.com/latex.php?latex=%7Bn+%5Cleq+3%7D&bg=ffffff&fg=000000&s=0&c=20201002), and several partial results involving the![{n=4}](https://s0.wp.com/latex.php?latex=%7Bn%3D4%7D&bg=ffffff&fg=000000&s=0&c=20201002)cases are also known\. The conjecture is also known if![{f}](https://s0.wp.com/latex.php?latex=%7Bf%7D&bg=ffffff&fg=000000&s=0&c=20201002)is decays at a suitably super\-exponential rate, by[work of Bownik and Speegle](https://math.umd.edu/~rvbalan/TEACHING/RIT2023/Bownick_Speegel2012.pdf)\. I was aware of this conjecture through various talks and conversations with colleagues, and even briefly tried my hand at it for a while, though not with particularly serious effort \(or progress\)\. It was thus a nice surprise to see that it has[just been resolved by Faulhuber, Petersen, van Velthoven, and Voigtlaender](https://arxiv.org/abs/2608.05044), even in the Schwartz case: > **Theorem 2**There exist complex numbers![{c_1,\dots,c_{12}\in {\bf C}}](https://s0.wp.com/latex.php?latex=%7Bc_1%2C%5Cdots%2Cc_%7B12%7D%5Cin+%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), not all zero, distinct points![{z_1,\dots,z_{12} \in {\bf R}^2}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_%7B12%7D+%5Cin+%7B%5Cbf+R%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002), and a non\-zero Schwartz function![{f_* \in \mathcal{S}({\bf R})}](https://s0.wp.com/latex.php?latex=%7Bf_%2A+%5Cin+%5Cmathcal%7BS%7D%28%7B%5Cbf+R%7D%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)such that![\displaystyle c_1 \pi(z_1) f_* + \dots + c_{12} \pi(z_{12}) f_* = 0.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++c_1+%5Cpi%28z_1%29+f_%2A+%2B+%5Cdots+%2B+c_%7B12%7D+%5Cpi%28z_%7B12%7D%29+f_%2A+%3D+0.&bg=ffffff&fg=000000&s=0&c=20201002) It is perhaps unsurprising in this current era that this result is AI\-assisted\. However, I think the authors have disclosed their AI use responsibly, with the final arguments written by hand with a readable overview of the argument, as well as proper discussion of methods, relation to past literature, and other independent numerical checks on the result\. The negative result lies only a little beyond the positive results:![{n}](https://s0.wp.com/latex.php?latex=%7Bn%7D&bg=ffffff&fg=000000&s=0&c=20201002)is now increased to![{12}](https://s0.wp.com/latex.php?latex=%7B12%7D&bg=ffffff&fg=000000&s=0&c=20201002), and all but one of the points![{z_1,\dots,z_{12}}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_%7B12%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)lie in \(a translate of\) a discrete subgroup of![{{\bf R}^2}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+R%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)\(in fact the explicit subgroup![{{\bf Z} \times \frac{1}{2}{\bf Z}}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+Z%7D+%5Ctimes+%5Cfrac%7B1%7D%7B2%7D%7B%5Cbf+Z%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)is used\)\. The functions constructed are smooth and rapidly decaying, but not analytic or super\-exponentially decaying, which would start being in conflict with the known positive results\. In addition to AI being used to come up with the initial proof strategy, a more traditional numerical computation was used to verify one step of the argument\. I have not had the time to do a full digestion of the result, but \(after reading the introduction, and[using a little AI assistance of my own](https://chatgpt.com/share/6a74dfc3-c614-83e8-b36b-9b07bf8d55fa)\) I was able to understand the main ideas at a high level\. The first few reductions are relatively standard\. Setting![{z_{12} = 0}](https://s0.wp.com/latex.php?latex=%7Bz_%7B12%7D+%3D+0%7D&bg=ffffff&fg=000000&s=0&c=20201002)and![{c_{12} = -c_*}](https://s0.wp.com/latex.php?latex=%7Bc_%7B12%7D+%3D+-c_%2A%7D&bg=ffffff&fg=000000&s=0&c=20201002), one can view the problem as one of solving an eigenvalue problem ![\displaystyle (c_1 \pi(z_1) + \dots + c_{11} \pi(z_{11})) f_* = c_* f_*.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%28c_1+%5Cpi%28z_1%29+%2B+%5Cdots+%2B+c_%7B11%7D+%5Cpi%28z_%7B11%7D%29%29+f_%2A+%3D+c_%2A+f_%2A.&bg=ffffff&fg=000000&s=0&c=20201002) The time\-frequency shifts![{z_1,\dots,z_{11}}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_%7B11%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)are chosen to lie in a translate of the discrete subgroup![{{\bf Z} \times \frac{1}{2}{\bf Z}}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+Z%7D+%5Ctimes+%5Cfrac%7B1%7D%7B2%7D%7B%5Cbf+Z%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)by a certain irrational shift![{(\alpha, \beta/2)}](https://s0.wp.com/latex.php?latex=%7B%28%5Calpha%2C+%5Cbeta%2F2%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. If one was working in the shifts of a standard lattice![{{\bf Z} \times {\bf Z}}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+Z%7D+%5Ctimes+%7B%5Cbf+Z%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), it would be natural to work with the[Zak transform](https://en.wikipedia.org/wiki/Zak_transform)of![{f_*}](https://s0.wp.com/latex.php?latex=%7Bf_%2A%7D&bg=ffffff&fg=000000&s=0&c=20201002), but it turns out that the approach does not quite work when doing this for topological reasons \(relating to the fact that scalar quasiperiodic functions of mean zero are forced to have zeroes\), and so the authors used the slightly denser lattice instead![{{\bf Z} \times \frac{1}{2}{\bf Z}}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+Z%7D+%5Ctimes+%5Cfrac%7B1%7D%7B2%7D%7B%5Cbf+Z%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), which relates to a vector\-valued version of the Zak transform taking values in![{{\bf C}^2}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+C%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)rather than![{{\bf C}}](https://s0.wp.com/latex.php?latex=%7B%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. Applying this transform, the eigenvalue problem can be transformed by standard calculations to a “vector cocycle problem” ![\displaystyle B_*(z) F_*(z - \tau) = c_* F_*(z), \ \ \ \ \ (1)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++B_%2A%28z%29+F_%2A%28z+-+%5Ctau%29+%3D+c_%2A+F_%2A%28z%29%2C+%5C+%5C+%5C+%5C+%5C+%281%29&bg=ffffff&fg=000000&s=0&c=20201002)where![{F_* : {\bf R}^2 \rightarrow {\bf C}^2}](https://s0.wp.com/latex.php?latex=%7BF_%2A+%3A+%7B%5Cbf+R%7D%5E2+%5Crightarrow+%7B%5Cbf+C%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)is a non\-zero smooth quasiperiodic vector\-valued function,![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)is an irrational shift![{\tau= (\alpha,\beta)}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%3D+%28%5Calpha%2C%5Cbeta%29%7D&bg=ffffff&fg=000000&s=0&c=20201002), and![{B_*(z)}](https://s0.wp.com/latex.php?latex=%7BB_%2A%28z%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)is a certain explicit![{2 \times 2}](https://s0.wp.com/latex.php?latex=%7B2+%5Ctimes+2%7D&bg=ffffff&fg=000000&s=0&c=20201002)matrix\-valued function depending on the choices of![{c_1,\dots,c_{11}}](https://s0.wp.com/latex.php?latex=%7Bc_1%2C%5Cdots%2Cc_%7B11%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002),![{z_1,\dots,z_{11}}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_%7B11%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), and![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. How to solve this equation? The motivating scenario here is if the matrix function![{B_*(z)}](https://s0.wp.com/latex.php?latex=%7BB_%2A%28z%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)was replaced by a rank one function ![\displaystyle B_0(z) = \chi(z) \chi(z-\tau)^*](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++B_0%28z%29+%3D+%5Cchi%28z%29+%5Cchi%28z-%5Ctau%29%5E%2A&bg=ffffff&fg=000000&s=0&c=20201002) for some smooth vector\-valued function![{\chi : {\bf R}^2 \rightarrow {\bf C}}](https://s0.wp.com/latex.php?latex=%7B%5Cchi+%3A+%7B%5Cbf+R%7D%5E2+%5Crightarrow+%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)of unit magnitude\. Then one could solve the equation by taking![{F_*(z) = \chi(z)}](https://s0.wp.com/latex.php?latex=%7BF_%2A%28z%29+%3D+%5Cchi%28z%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)and![{c_* = 1}](https://s0.wp.com/latex.php?latex=%7Bc_%2A+%3D+1%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. It is not possible to make the function![{B_*}](https://s0.wp.com/latex.php?latex=%7BB_%2A%7D&bg=ffffff&fg=000000&s=0&c=20201002)exactly of this form, but through some numerical computation and clever AI\-assisted guesswork, the authors were able to find a choice of![{c_1,\dots,c_{11}}](https://s0.wp.com/latex.php?latex=%7Bc_1%2C%5Cdots%2Cc_%7B11%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)and![{z_1,\dots,z_{11}}](https://s0.wp.com/latex.php?latex=%7Bz_1%2C%5Cdots%2Cz_%7B11%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), and![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)that made![{B_*}](https://s0.wp.com/latex.php?latex=%7BB_%2A%7D&bg=ffffff&fg=000000&s=0&c=20201002)*approximately equal*to a rank one function![{B_0}](https://s0.wp.com/latex.php?latex=%7BB_0%7D&bg=ffffff&fg=000000&s=0&c=20201002)of this form, in fact getting a uniform estimate ![\displaystyle \sup_{z \in {\bf R}^2} \| B_*(z) - B_0(z) \|_{op} < \frac{1}{3}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Csup_%7Bz+%5Cin+%7B%5Cbf+R%7D%5E2%7D+%5C%7C+B_%2A%28z%29+-+B_0%28z%29+%5C%7C_%7Bop%7D+%3C+%5Cfrac%7B1%7D%7B3%7D.&bg=ffffff&fg=000000&s=0&c=20201002) As it turns out, such an approximation is sufficient to run a contraction mapping argument to find a solution to a variant of[\(1\)](https://terrytao.wordpress.com/2026/08/06/a-partial-digestion-of-the-hrt-counterexample/#bzt), namely ![\displaystyle B_*(z) v_*(z - \tau) = q_*(z) v_*(z)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++B_%2A%28z%29+v_%2A%28z+-+%5Ctau%29+%3D+q_%2A%28z%29+v_%2A%28z%29&bg=ffffff&fg=000000&s=0&c=20201002) for some smooth![{v_* : {\bf R}^2 \rightarrow {\bf C}^2}](https://s0.wp.com/latex.php?latex=%7Bv_%2A+%3A+%7B%5Cbf+R%7D%5E2+%5Crightarrow+%7B%5Cbf+C%7D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)and![{q_* : {\bf R}^2 \rightarrow {\bf C}}](https://s0.wp.com/latex.php?latex=%7Bq_%2A+%3A+%7B%5Cbf+R%7D%5E2+%5Crightarrow+%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. \(Here it was important to get the operator norm bound below![{\frac{1}{3}}](https://s0.wp.com/latex.php?latex=%7B%5Cfrac%7B1%7D%7B3%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002); they are barely able to do this, with a numerically obtained bound of![{0.333032}](https://s0.wp.com/latex.php?latex=%7B0.333032%7D&bg=ffffff&fg=000000&s=0&c=20201002), though this bound might not be optimal\.\) The main remaining obstacle is that the “eigenvalue function”![{q_*(z)}](https://s0.wp.com/latex.php?latex=%7Bq_%2A%28z%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)is varying in the parameter![{z}](https://s0.wp.com/latex.php?latex=%7Bz%7D&bg=ffffff&fg=000000&s=0&c=20201002)rather than constant\. \(This issue was, by the way, anticipated to some extent in[previous work of Demeter](https://arxiv.org/abs/1006.0732), who observed that eigenfunctions of the almost Matthieu discrete Schrödinger operator gave a near\-miss counterexample to the HRT conjecture, but with an eigenvalue that depended on an auxiliary phase shift parameter rather than constant\.\) However, if one was able to solve the scalar cocycle equation ![\displaystyle q_*(z) h(z-\tau) = c_* h(z) \ \ \ \ \ (2)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++q_%2A%28z%29+h%28z-%5Ctau%29+%3D+c_%2A+h%28z%29+%5C+%5C+%5C+%5C+%5C+%282%29&bg=ffffff&fg=000000&s=0&c=20201002) for some smooth![{h : {\bf R}^2 \rightarrow {\bf C}}](https://s0.wp.com/latex.php?latex=%7Bh+%3A+%7B%5Cbf+R%7D%5E2+%5Crightarrow+%7B%5Cbf+C%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002), then one could solve the equation[\(1\)](https://terrytao.wordpress.com/2026/08/06/a-partial-digestion-of-the-hrt-counterexample/#bzt)by setting![{F_*(z) = h(z) v_*(z)}](https://s0.wp.com/latex.php?latex=%7BF_%2A%28z%29+%3D+h%28z%29+v_%2A%28z%29%7D&bg=ffffff&fg=000000&s=0&c=20201002)\. The approach to solve[\(2\)](https://terrytao.wordpress.com/2026/08/06/a-partial-digestion-of-the-hrt-counterexample/#qzh)is standard: take logarithms, apply a Fourier transform, and then divide out by the multiplier associated to the![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)shift\. This can cause a well\-known “small divisor” problem \(which arises in various dynamical contexts, such as in the[KAM theorem](https://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Arnold%E2%80%93Moser_theorem)\) if![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)behaves too much like a rational vector, but the standard resolution to this is to select a shift![{\tau}](https://s0.wp.com/latex.php?latex=%7B%5Ctau%7D&bg=ffffff&fg=000000&s=0&c=20201002)that obeys good Diophantine approximation properties\. For the purposes of numerics the authors selected an extremely concrete shift, namely ![\displaystyle \tau = (2^{1/3} - 1, 2^{2/3} - 1)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Ctau+%3D+%282%5E%7B1%2F3%7D+-+1%2C+2%5E%7B2%2F3%7D+-+1%29&bg=ffffff&fg=000000&s=0&c=20201002) but I get the impression that the exact choice here was not crucial for the argument, and that many other irrational algebraic numbers could have worked here\.

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