Time–Frequency Analysis • Notes prepared by Dr. Oussa
HRT Conjecture — Three Lectures
Purpose, Conventions, and Learning Outcomes
Goal. Present a self-contained, undergraduate-accessible development of key verified cases of the Heil–Ramanathan–Topiwala (HRT) Conjecture, organized as three lectures with explicit proofs.
Fourier & shift conventions
- $\widehat f(\xi)=\displaystyle\int_{\mathbb R} f(t)\,e^{-2\pi i \xi t}\,dt$
- $T_x f(t)=f(t-x)$, $M_y f(t)=e^{-2\pi i y t}f(t)$
- Time–frequency shift: $(\pi(x,y)f)(t)=e^{-2\pi i y t}f(t-x)$
Standing regularity (when invoked)
- For mixed-integer arguments: $Zf$ and $Z\widehat f$ continuous (e.g., $f$ Schwartz).
- For the $(m,1)$ case: $g=\widehat f\in C_0(\mathbb R)$ (e.g., $f\in L^1\cap L^2$).
Outcomes
- Understand the line and lattice cases via Fourier/Zak tools.
- Handle the mixed-integer relation using phase rigidity (including the corrected rational subcase).
- Execute the $(m,1)$ proof via compact-group dynamics, Birkhoff averages, and Atkinson’s zero-mean recurrence with the phase-offset device.
Lecture 1 — Foundations, Line Case, and Integer Lattice
Foundations Fourier Zak
Recording.
Watch Lecture 1 (Dropbox).
Watch Lecture 1 (Dropbox).
1.1 HRT statement
For nonzero $f\in L^2(\mathbb R)$ and finitely many points $(x_k,y_k)\in\mathbb R^2$, the set $\{\pi(x_k,y_k)f\}$ is linearly independent.
1.2 Line case via Fourier
If all $(x_k,y_k)$ lie on $y=mx$, apply an appropriate unitary (metaplectic) map $U$ sending $\pi(x_k,mx_k)$ to translations $\pi(x’_k,0)$. Taking Fourier transforms gives
\[
\Big(\sum_{k=1}^{n} c_k\,e^{-2\pi i x_k \xi}\Big)\widehat f(\xi)=0.
\]
A nontrivial trigonometric polynomial has a zero set of Lebesgue measure zero; hence $\widehat f=0$ a.e., contradiction.
\[
\Big(\sum_{k=1}^{n} c_k\,e^{-2\pi i x_k \xi}\Big)\widehat f(\xi)=0.
\]
A nontrivial trigonometric polynomial has a zero set of Lebesgue measure zero; hence $\widehat f=0$ a.e., contradiction.
1.3 Zak transform and boundary laws
\[
(Zf)(t,\omega)=\sum_{\ell\in\mathbb Z} f(t-\ell)\,e^{2\pi i\ell\omega},\quad
Zf(t+1,\omega)=e^{2\pi i\omega}Zf(t,\omega),\quad Zf(t,\omega+1)=Zf(t,\omega).
\]
(Zf)(t,\omega)=\sum_{\ell\in\mathbb Z} f(t-\ell)\,e^{2\pi i\ell\omega},\quad
Zf(t+1,\omega)=e^{2\pi i\omega}Zf(t,\omega),\quad Zf(t,\omega+1)=Zf(t,\omega).
\]
1.4 Integer lattice case
For $(x_k,y_k)\in\mathbb Z^2$,
\[
Z(\pi(x_k,y_k)f)(t,\omega)=e^{-2\pi i (y_k t+x_k\omega)}Zf(t,\omega).
\]
Thus
\[
\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}\,Zf(t,\omega)=0
\]
on $[0,1)^2$. Let $p(t,\omega)=\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}$. If $p\not\equiv0$, its zero set has measure zero on the torus; hence $Zf=0$ a.e. and, since $Z$ is unitary onto its quasi-periodic subspace (injective), $f=0$—a contradiction.
\[
Z(\pi(x_k,y_k)f)(t,\omega)=e^{-2\pi i (y_k t+x_k\omega)}Zf(t,\omega).
\]
Thus
\[
\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}\,Zf(t,\omega)=0
\]
on $[0,1)^2$. Let $p(t,\omega)=\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}$. If $p\not\equiv0$, its zero set has measure zero on the torus; hence $Zf=0$ a.e. and, since $Z$ is unitary onto its quasi-periodic subspace (injective), $f=0$—a contradiction.
Check. Verify directly the quasi-periodicity formulae for $Zf$ and the lattice action from the definition.
Lecture 2 — Mixed-Integer Configurations and Phase Rigidity
Mixed-integer Phase Rational identity
2.1 Mixed-integer identity
Suppose
\[
\sum_{k=1}^n c_k\,\pi(x_k,y_k)f=\pi(\alpha,\beta)f,\qquad (\alpha,\beta)\notin\mathbb Z^2.
\]
Under Zak:
\[
p(t,\omega)\,Zf(t,\omega)=e^{-2\pi i\beta t}\,Zf(t-\alpha,\omega+\beta),
\]
with $p(t,\omega)=\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}$.
\[
\sum_{k=1}^n c_k\,\pi(x_k,y_k)f=\pi(\alpha,\beta)f,\qquad (\alpha,\beta)\notin\mathbb Z^2.
\]
Under Zak:
\[
p(t,\omega)\,Zf(t,\omega)=e^{-2\pi i\beta t}\,Zf(t-\alpha,\omega+\beta),
\]
with $p(t,\omega)=\sum_{k=1}^n c_k e^{-2\pi i (y_k t+x_k\omega)}$.
2.2 Continuous $Zf$ has a zero
If $Zf$ is continuous and obeys $Zf(t+1,\omega)=e^{2\pi i\omega}Zf(t,\omega)$, then $Zf$ must vanish somewhere on $[0,1]^2$; otherwise one would obtain a nowhere-vanishing continuous section of a line bundle with nontrivial monodromy.
2.3 Full irrational case
If $\dim_{\mathbb Q}\mathrm{span}_{\mathbb Q}\{1,\alpha,\beta\}=3$, the orbit of $\delta=(\alpha,\beta)$ is dense in the torus. Iteration forces zeros on a dense set; by continuity, $Zf\equiv0$, contradiction.
2.4 Rational case
Choose $N$ with $N\alpha,N\beta\in\mathbb Z$. Iterating $N$ times yields
\[
\prod_{j=0}^{N-1} p(t-j\alpha,\omega+j\beta)
= e^{-2\pi i\beta\,(Nt-\alpha N(N-1)/2)}\,e^{2\pi i N\alpha\,\omega},
\]
a single character on $\mathbb T^2$. The product on the left expands into a sum of distinct characters unless $p$ itself is a single character; thus a nontrivial configuration is impossible.
\[
\prod_{j=0}^{N-1} p(t-j\alpha,\omega+j\beta)
= e^{-2\pi i\beta\,(Nt-\alpha N(N-1)/2)}\,e^{2\pi i N\alpha\,\omega},
\]
a single character on $\mathbb T^2$. The product on the left expands into a sum of distinct characters unless $p$ itself is a single character; thus a nontrivial configuration is impossible.
2.5 Intermediate case $\dim_{\mathbb Q}=2$
Write $Zf=|Zf|e^{2\pi i\Theta}$ and $p=|p|e^{2\pi i\varphi}$. Phase iteration along $(t,\omega)\mapsto(t-\alpha,\omega+\beta)$, together with the Zak boundary laws, yields integer constraints that force $\beta\in\mathbb Z$, then $\alpha\in\mathbb Z$, contradicting the mixed-integer hypothesis.
Try. Derive the $N$-step identity above directly from the mixed-integer relation by induction on $N$.
Lecture 3 — The $(m,1)$ Configuration and Atkinson’s Theorem
Ergodic averages Compact tori Atkinson
3.1 Setup and iteration
Assume
\[
\sum_{k=1}^m c_k\,f(t-x_k)=f(t-a).
\]
Fourier transform: $g(\xi+a)=p(\xi)g(\xi)$, where
\[
p(\xi)=\sum_{k=1}^{m} c_k e^{-2\pi i x_k \xi}.
\]
Iteration:
\[
g(\xi+na)=\Big(\prod_{j=0}^{n-1}p(\xi+ja)\Big)g(\xi),\qquad
g(\xi-na)=\Big(\prod_{j=1}^{n}p(\xi-ja)\Big)^{-1}g(\xi).
\]
\[
\sum_{k=1}^m c_k\,f(t-x_k)=f(t-a).
\]
Fourier transform: $g(\xi+a)=p(\xi)g(\xi)$, where
\[
p(\xi)=\sum_{k=1}^{m} c_k e^{-2\pi i x_k \xi}.
\]
Iteration:
\[
g(\xi+na)=\Big(\prod_{j=0}^{n-1}p(\xi+ja)\Big)g(\xi),\qquad
g(\xi-na)=\Big(\prod_{j=1}^{n}p(\xi-ja)\Big)^{-1}g(\xi).
\]
3.2 Compact-group model and Birkhoff averages
Put $A=\mathrm{diag}(-2\pi i x_1,\dots,-2\pi i x_m)$, $c=(c_1,\dots,c_m)^\top$, $\mathbf 1=(1,\dots,1)^\top$, so $p(\xi)=\langle e^{\xi A}c,\mathbf 1\rangle$.
Let $G=\overline{\{e^{\xi A}\}}$, $H=\overline{\{e^{jaA}\}}$, and $T(h)=e^{aA}h$.
For fixed $\xi$ define
\[
F(h)=\log\big|\langle e^{\xi A} h c,\mathbf 1\rangle\big|\in L^1(H),\qquad
C_\xi=\int_H F\,d\mu_H.
\]
Rotation $T$ on the compact abelian group $H$ is (uniquely) ergodic, hence
\[
\frac{1}{n}\sum_{j=0}^{n-1}F(T^j h)\to C_\xi\quad(\forall h\in H).
\]
Consequently:
Let $G=\overline{\{e^{\xi A}\}}$, $H=\overline{\{e^{jaA}\}}$, and $T(h)=e^{aA}h$.
For fixed $\xi$ define
\[
F(h)=\log\big|\langle e^{\xi A} h c,\mathbf 1\rangle\big|\in L^1(H),\qquad
C_\xi=\int_H F\,d\mu_H.
\]
Rotation $T$ on the compact abelian group $H$ is (uniquely) ergodic, hence
\[
\frac{1}{n}\sum_{j=0}^{n-1}F(T^j h)\to C_\xi\quad(\forall h\in H).
\]
Consequently:
- If $C_\xi>0$ and $g(\xi)\neq 0$, then $|g(\xi+na)|\to\infty$.
- If $C_\xi<0$, the backward iteration gives $|g(\xi-na)|\to\infty$.
Thus $g\in C_0(\mathbb R)$ is only possible in the zero-mean case $C_\xi=0$ on some fiber.
3.3 Atkinson’s zero-mean recurrence (explicit step)
Fix a transversal $T\subset G$ and choose $r\in T$ with
\[
C_r=\int_H \log|\langle r h c,\mathbf 1\rangle|\,d\mu_H=0.
\]
Set $F_r(h)=\log|\langle r h c,\mathbf 1\rangle|$ and $\phi_r=F_r$.
By Atkinson (ergodic, invertible $T$, mean-zero $\phi_r\in L^1$), for $\mu_H$-a.e. $s\in H$ there exists $n_k\to\infty$ with
\[
\sum_{j=0}^{n_k-1}F_r(T^j s)\to 0.
\]
Approximate $rs$ by phases $\xi_m$ via $e^{\xi_m A}=r s_m\to rs$. For each $k$, continuity of $F_r$ away from its discrete zeros yields $m(k)$ so large that
\[
\Big|\sum_{j=0}^{n_k-1}F_r(T^j s_{m(k)})-\sum_{j=0}^{n_k-1}F_r(T^j s)\Big|<\varepsilon_k,\ \varepsilon_k\downarrow0. \] Using the product identity, \[ |g(\xi_{m(k)}+n_k a)|=\exp\!\Big(\sum_{j=0}^{n_k-1}F_r(T^j s_{m(k)})\Big)\,|g(\xi_{m(k)})| \in (1-\varepsilon_k,1+\varepsilon_k)\,|g(\xi_{m(k)})|. \] As $\xi_{m(k)}\to\xi_\ast$ and $g$ is continuous, $|g(\xi_{m(k)}+n_k a)|\to |g(\xi_\ast)|>0$ whenever $g(\xi_\ast)\neq 0$. This contradicts decay at $+\infty$ unless $g\equiv 0$, hence $f\equiv 0$.
\[
C_r=\int_H \log|\langle r h c,\mathbf 1\rangle|\,d\mu_H=0.
\]
Set $F_r(h)=\log|\langle r h c,\mathbf 1\rangle|$ and $\phi_r=F_r$.
By Atkinson (ergodic, invertible $T$, mean-zero $\phi_r\in L^1$), for $\mu_H$-a.e. $s\in H$ there exists $n_k\to\infty$ with
\[
\sum_{j=0}^{n_k-1}F_r(T^j s)\to 0.
\]
Approximate $rs$ by phases $\xi_m$ via $e^{\xi_m A}=r s_m\to rs$. For each $k$, continuity of $F_r$ away from its discrete zeros yields $m(k)$ so large that
\[
\Big|\sum_{j=0}^{n_k-1}F_r(T^j s_{m(k)})-\sum_{j=0}^{n_k-1}F_r(T^j s)\Big|<\varepsilon_k,\ \varepsilon_k\downarrow0. \] Using the product identity, \[ |g(\xi_{m(k)}+n_k a)|=\exp\!\Big(\sum_{j=0}^{n_k-1}F_r(T^j s_{m(k)})\Big)\,|g(\xi_{m(k)})| \in (1-\varepsilon_k,1+\varepsilon_k)\,|g(\xi_{m(k)})|. \] As $\xi_{m(k)}\to\xi_\ast$ and $g$ is continuous, $|g(\xi_{m(k)}+n_k a)|\to |g(\xi_\ast)|>0$ whenever $g(\xi_\ast)\neq 0$. This contradicts decay at $+\infty$ unless $g\equiv 0$, hence $f\equiv 0$.
Integrability note. $F\in L^1(H)$: near a zero the singularity is $\log|z|$, and $\int_0^\rho |\log r|\,r^{d-1}dr<\infty$ on tori.
3.4 Summary of proved cases
- Colinear configurations (Lecture 1).
- Integer lattice $\mathbb Z^2$ via Zak (Lecture 1).
- No mixed-integer relation under continuity hypotheses (Lecture 2).
- $(m,1)$ configuration via dynamics + Atkinson (Lecture 3).