A Partial Digestion of the HRT Counterexample

Originally discussed by Terence Tao on August 6, 2026, regarding recent breakthroughs in time-frequency analysis.

Introduction to Time-Frequency Shifts

In the study of functions of a single variable, we can manipulate a function $f(t)$ using two primary operations:

1. Spatial Translation: Shifting the function in space.
2. Frequency Modulation: Shifting the function in frequency.

When combined, these create a time-frequency shift, denoted as $\pi(x, \omega)$. While this is essentially a piece of the Weyl representation of the Heisenberg group, we can define it simply as:

$\pi(x, \omega) f(t) := e^{2\pi i \omega t} f(t-x)$

Certain functions exhibit linear dependencies among these shifts. For example, consider a sinusoid: f(t) = A sin(kt + phi) This function satisfies the relation: $\pi\left(\frac{\pi}{k}, 0\right) f + f = 0$

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The HRT Conjecture

The Heil-Ramanathan-Topiwala (HRT) conjecture posits that such linear relations are impossible if the function $f$ possesses sufficient decay.

Conjecture 1 (HRT): If $f$ is a non-zero function, there is no linear relation of the form: $\sum_{k=1}^N c_k \pi(x_k, \omega_k) f = 0$ where the points $(x_k, \omega_k)$ are distinct and the coefficients $c_k$ are not all zero.

A particularly challenging version of this conjecture focused on the case where $f$ is a Schwartz function (smooth and rapidly decaying).

Previous Partial Results

Before the recent counterexample, several positive results were established:

• Linnell's Result: The conjecture holds if the points $(x_k, \omega_k)$ reside within a translate of a discrete subgroup of $\mathbb{R}^2$. This covers all cases where $N=3$.
• Bownik and Speegle: The conjecture holds if $f$ exhibits a "super-exponential" decay rate.

The HRT conjecture was long thought to be true for all Schwartz functions.

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The Counterexample

Recently, Faulhuber, Petersen, van Velthoven, and Voigtlaender successfully resolved the conjecture by proving it false, even for the Schwartz class.

Theorem 2: There exists a non-zero Schwartz function $f$ and distinct points $(x_k, \omega_k)$ with coefficients $c_k$ (not all zero) such that: $\sum_{k=1}^N c_k \pi(x_k, \omega_k) f = 0$

Analysis of the Result

The authors utilized AI to assist in their discovery. Tao notes that this was done responsibly: the final proofs were hand-written, the methodology was transparent, and the results were verified via independent numerical checks.

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Technical Breakdown of the Proof

The proof strategy can be visualized as a sequence of transformations to turn a functional equation into a solvable algebraic problem:

1. The Setup

By setting $N=4$ and choosing points in a translate of the discrete subgroup $\mathbb{Z}^2$ by an irrational shift $\alpha$, the problem becomes an eigenvalue problem.

2. The Zak Transform

Normally, one would use the standard Zak transform for a lattice $\mathbb{Z}^2$. However, scalar quasiperiodic functions with mean zero must have zeroes, creating topological obstacles. To bypass this, the authors used a denser lattice $\mathbb{Z}^2 \times \mathbb{Z}^2$, employing a vector-valued Zak transform taking values in $\mathbb{C}^2$.

3. The Vector Cocycle

The problem is transformed into a "vector cocycle problem": v(x + alpha) = M(x)v(x) Where:

• $v$ is a non-zero smooth quasiperiodic vector-valued function.
• $\alpha$ is an irrational shift.
• $M(x)$ is a matrix-valued function based on the chosen points and coefficients.

4. Approximation and Contraction

The authors sought a scenario where $M(x)$ behaves like a rank-one function. While an exact match is impossible, they used AI-assisted guesswork and numerical computation to find parameters where: $\| M(x) - \text{rank-one function} \| \approx 0.99$ Since this operator norm is $< 1$, they could apply a contraction mapping argument to solve a variant of the equation: $\text{Solve: } v(x+\alpha) = M(x)v(x) + \epsilon(x)$

5. The Small Divisor Problem

The final step involves solving a scalar cocycle equation: f(x + alpha) = g(x)f(x) The standard approach is:

1. Take the logarithm.
2. Apply a Fourier transform.
3. Divide by the multiplier associated with the shift.

This introduces the "small divisor" problem, common in dynamical systems like the KAM theorem. If $\alpha$ is too close to a rational vector, the division becomes unstable.

Summary Checklist for the Counterexample

• Non-zero function $f$
• $f$ is in the Schwartz class
• $N=4$ distinct points
• Coefficients $c_k$ are not all zero
• Linear relation $\sum c_k \pi(x_k, \omega_k) f = 0$ is satisfied