The Schwartz One-Rogue Critical Case for the HRT Conjecture
A self-contained proof for the critical lattice-plus-one-point configuration, written for readers who want the mathematical argument first, the connection with Oussa’s prior HRT work second, and the formal audit trail third.
Purpose, Scope, and Navigation
The goal is to isolate one precise HRT configuration and prove it completely in the Schwartz class. The background lattice has critical covolume, the configuration contains finitely many lattice points, and there is one additional point outside the lattice. After the standard symplectic normalization, the lattice is $\mathbb Z^2$ and the rogue point is $(\alpha,\beta)$.
If $\Gamma\subset\mathbb Z^2$ is finite, $(\alpha,\beta)\notin\mathbb Z^2$, and $0\ne f\in\mathcal S(\mathbb R)$, then
$$
\{M_nT_m f:(m,n)\in\Gamma\}\cup\{M_\beta T_\alpha f\}
$$
is linearly independent.
Author’s related work
Zak-transform reduction
Arithmetic trichotomy
Case $r=1$
Holonomy and winding guide
Mixed arithmetic case $r=2$
Dense case $r=3$
Certification links
References
The proof below is written so that a reader can follow the argument without running Lean or any other software. Formal links are provided for auditability, not as a substitute for the proof.
Author’s Related Work on the HRT Conjecture
This article should be read as part of a longer research program on the HRT conjecture, time-frequency shifts, and the role of the Zak transform. The present proof is intentionally self-contained, but it is not isolated from that earlier work. The relevant prior papers provide the representation-theoretic viewpoint, the one-lattice-plus-one-point viewpoint, the zero-set viewpoint, and the mixed-integer trichotomy that motivate the proof below.
| Paper | What it contributes | Connection with this article |
|---|---|---|
| Currey–Oussa, Heisenberg-group formulation Canad. Math. Bull. 63(4), 2020 |
Places the HRT conjecture inside a finite-translate problem for the Heisenberg group and makes explicit the bridge between time-frequency shifts and group translates. | Explains why a geometric Gabor-system question can be attacked through algebraic and representation-theoretic structure. DOI · arXiv |
| Okoudjou–Oussa, special configurations J. Fourier Anal. Appl. 31(4), 2025; arXiv:2110.04053 |
Studies special HRT configurations, including the case where many points lie in an integer lattice and one point is off the lattice, using ergodic ideas. | This is the closest published antecedent to the one-rogue viewpoint: lattice structure plus one exceptional point. arXiv |
| Oussa, zero set of the Zak transform arXiv:2305.12299, 2023 |
Focuses on how a linear dependence forces invariance properties of the zero set of the Zak transform under torus translations. | Supplies the conceptual reason that a single Zak zero may propagate along a rogue orbit, which is exactly what happens in the dense case below. arXiv |
| Oussa, mixed-integer trichotomy arXiv:2508.04613v2, 2026 |
Develops the finite/dense/infinite-proper orbit trichotomy for mixed-integer HRT configurations and extracts cocycle rigidity constraints from the Zak transform. | Provides the dynamical language used here: the rogue point generates a torus translation, and arithmetic determines the orbit closure. arXiv |
| Oussa, the $(m,1)$-configuration arXiv:2508.18529v3, 2025 |
Proves a one-rogue linear-independence principle for smooth vectors in step-two square-integrable-mod-center representation settings, with an HRT specialization. | Shows that one-rogue phenomena are not accidental: they also appear naturally in nilpotent representation-theoretic models. arXiv |
The Normalized Theorem
Operators and sign convention. For $a,b\in\mathbb R$, write
$$
T_a f(t)=f(t-a),\qquad M_b f(t)=e^{-2\pi i bt}f(t),\qquad \pi(a,b)=M_bT_a.
$$
This is the convention used throughout the certification note: integer lattice shifts become multiplication by torus characters, and the rogue point shifts the Zak variables by $(x,\omega)\mapsto(x-\alpha,\omega+\beta)$.
Theorem. Let $\Gamma\subset\mathbb Z^2$ be finite and let $(\alpha,\beta)\in\mathbb R^2\setminus\mathbb Z^2$. Put
$$
\Lambda=\Gamma\cup\{(\alpha,\beta)\}.
$$
If $0\ne f\in\mathcal S(\mathbb R)$, then the finite Gabor system
$$
\mathcal G(f,\Lambda)=\{\pi(a,b)f:(a,b)\in\Lambda\}
$$
is linearly independent in $L^2(\mathbb R)$.
Illustration 1. Critical lattice plus one rogue point
The proof converts a geometric statement in the time-frequency plane into a dynamical statement on the Zak torus.
The Zak-Transform Reduction
For $f\in\mathcal S(\mathbb R)$, define
$$
Zf(x,\omega)=\sum_{k\in\mathbb Z}f(x+k)e^{-2\pi i k\omega}.
$$
The series converges rapidly and defines a continuous function satisfying
$$
Zf(x,\omega+1)=Zf(x,\omega),\qquad
Zf(x+1,\omega)=e^{2\pi i\omega}Zf(x,\omega).
$$
The second identity is the topological source of the unavoidable zero used later.
Reduction lemma. Assume that a nontrivial dependence relation exists:
$$
\sum_{(a,b)\in\Gamma}c_{a,b}M_bT_af+c_*M_\beta T_\alpha f=0.
$$
The coefficient $c_*$ must be nonzero; otherwise the relation would involve only the integer-lattice subsystem, which is already linearly independent.
With the sign convention $M_\xi f(t)=e^{-2\pi i\xi t}f(t)$, the Zak identities are
$$
Z(M_bT_af)(x,\omega)=e^{-2\pi i(a\omega+bx)}F(x,\omega),\qquad (a,b)\in\mathbb Z^2,
$$
and
$$
Z(M_\beta T_\alpha f)(x,\omega)=e^{-2\pi i\beta x}F(x-\alpha,\omega+\beta),
$$
where $F=Zf$.
Define the integer-lattice trigonometric polynomial
$$
P_\Gamma(x,\omega)=\sum_{(a,b)\in\Gamma}c_{a,b}e^{-2\pi i(a\omega+bx)}.
$$
Applying the Zak transform gives
$$
P_\Gamma(x,\omega)F(x,\omega)+c_*e^{-2\pi i\beta x}F(x-\alpha,\omega+\beta)=0.
$$
Solving for the rogue translate gives
$$
F(x-\alpha,\omega+\beta)=R(x,\omega)F(x,\omega), \tag{1}
$$
where
$$
R(x,\omega)=-c_*^{-1}e^{2\pi i\beta x}P_\Gamma(x,\omega).
$$
Thus the one-step multiplier is $R$, not $Q$.
The Arithmetic Trichotomy
The orbit of $(x,\omega)$ under the rogue translation
$$
T_{\alpha,\beta}(x,\omega)=(x-\alpha,\omega+\beta)\pmod{\mathbb Z^2}
$$
is controlled by
$$
r=\dim_{\mathbb Q}\operatorname{span}_{\mathbb Q}\{1,\alpha,\beta\}.
$$
| Arithmetic type | Meaning | Orbit behavior | Mechanism |
|---|---|---|---|
| $r=1$ | $\alpha,\beta\in\mathbb Q$ | finite | finite lattice refinement |
| $r=2$ | one rational relation, but not both coordinates rational | infinite non-dense | mixed arithmetic holonomy obstruction |
| $r=3$ | $1,\alpha,\beta$ rationally independent | dense in $\mathbb T^2$ | zero propagation |
Illustration 2. Finite, one-dimensional, and dense orbit closures
The difficult middle case is neither finite nor fully dense. It requires a one-dimensional obstruction: holonomy.
Case $r=1$: Rational Rogue Coordinates
Claim. If $r=1$, the theorem follows from the classical lattice-contained theorem.
Indeed, $r=1$ means $\alpha,\beta\in\mathbb Q$. Choose $N\in\mathbb N$ such that $N\alpha,N\beta\in\mathbb Z$. Then
$$
(\alpha,\beta)\in\frac1N\mathbb Z^2,
\qquad
\Gamma\subset\mathbb Z^2\subset\frac1N\mathbb Z^2.
$$
Hence $\Lambda\subset(1/N)\mathbb Z^2$, a full-rank lattice. The classical discrete-subgroup result of Linnell gives linear independence for every finite Gabor subsystem contained in such a discrete subgroup. This eliminates the rational case.
Example. If $(\alpha,\beta)=(1/2,1/3)$, then $\Lambda\subset(1/6)\mathbb Z^2$. The point is visually off the original integer lattice, but it is not off all lattices.
Holonomy and Winding: The Reader’s Guide to the Difficult Step
The mixed arithmetic case is the subtle one. The orbit is too large for a finite-lattice argument and too small for the dense-torus zero-propagation argument. What remains is a one-dimensional obstruction. The obstruction is easiest to understand through three related notions: winding number, logarithm, and holonomy.
1. Winding number tells us whether a loop admits a periodic logarithm
Let $B:S^1\to\mathbb C^\times$ be a continuous nonvanishing loop. Its winding number $\operatorname{wind}(B)$ counts the net number of times $B(x)$ travels around the origin as $x$ goes once around the circle. The key fact used here is:
$$
\operatorname{wind}(B)=0
\quad\Longleftrightarrow\quad
B(x)=e^{\ell(x)}\text{ for a continuous periodic }\ell:S^1\to\mathbb C.
$$
Thus winding zero is the permission slip that allows us to take a logarithm without tearing the circle.
2. Holonomy is the exponential average of that logarithm
When $\operatorname{wind}(B)=0$, choose a periodic logarithm $\ell$ with $e^\ell=B$ and define
$$
\operatorname{Hol}(B)=\exp\left(\int_0^1\ell(x)\,dx\right). \tag{2}
$$
This is well-defined: another periodic logarithm differs from $\ell$ by $2\pi i k$, and exponentiating the integral removes that ambiguity.
3. Why the quotient $K(x-\theta)/K(x)$ is not always holonomically trivial
Suppose $K:S^1\to\mathbb C^\times$ has winding $n$. Choose a lift $k$ on $[0,1]$ such that $K(x)=e^{k(x)}$ and
$$
k(x+1)=k(x)+2\pi i n.
$$
Then
$$
\int_0^1\bigl(k(x-\theta)-k(x)\bigr)\,dx=-2\pi i n\theta.
$$
Therefore
$$
\operatorname{Hol}\left(\frac{K(x-\theta)}{K(x)}\right)=e^{-2\pi i n\theta}. \tag{3}
$$
The quotient has winding zero, but it can still carry a nontrivial holonomy phase. This distinction is the heart of the mixed case.
4. Why this creates a contradiction
The dynamics force a holonomy of the form
$$
C e^{-2\pi i\theta s}, \qquad \theta\notin\mathbb Q.
$$
The algebraic origin of the cocycle forces the same holonomy to be algebraic over Laurent polynomials in $e^{2\pi is}$. An irrational exponential character cannot satisfy such a finite Laurent-polynomial relation on an interval. The two descriptions are incompatible.
Illustration 3. Winding, logarithm, and holonomy
Winding zero permits a logarithm. The average of the logarithm gives the holonomy used in the contradiction.
Case $r=2$: The Mixed Arithmetic Holonomy Obstruction
Theorem. Suppose $r=2$. Then no nontrivial dependence relation can exist.
Step 1. Mixed normal form
Since $r=2$, there is exactly one rational relation among $1,\alpha,\beta$. A lattice automorphism of $\mathbb Z^2$ preserves the normalized critical problem and sends the rogue vector to a form in which one coordinate is rational and the other is irrational. We use the form
$$
\beta=\frac pm\in\mathbb Q,\qquad \gcd(p,m)=1,
\qquad \alpha\notin\mathbb Q.
$$
Set
$$
\theta=m\alpha.
$$
Then $\theta\notin\mathbb Q$.
Step 2. Close the rational coordinate
Start from the correct one-step Zak equation
$$
F(x-\alpha,\omega+\beta)=R(x,\omega)F(x,\omega). \tag{1}
$$
Iterating $m$ times closes the second coordinate because $m\beta=p\in\mathbb Z$ and $F$ is periodic in the second Zak variable. Thus, for each fixed $s$, one obtains
$$
F(x-\theta,s)=B_s(x)F(x,s), \tag{4}
$$
where
$$
B_s(x)=R(x-(m-1)\alpha,s+(m-1)\beta)\cdots R(x-\alpha,s+\beta)R(x,s).
$$
Because $R$ is built from finitely many torus characters, the $m$-fold product is a Laurent-polynomial loop:
$$
B_s(x)=L(e^{2\pi is},e^{2\pi ix})
$$
for a Laurent polynomial $L(w,z)\in\mathbb C[w^{\pm1},z^{\pm1}]$.
Step 3. Nonzero fibers are nowhere zero
Put $H_s(x)=F(x,s)$. If $H_s\not\equiv0$ and $H_s(x_0)=0$, then (4) gives
$$
H_s(x_0-n\theta)=0\qquad(n\ge0).
$$
Since $\theta$ is irrational, the orbit $\{x_0-n\theta:n\ge0\}$ is dense in $\mathbb T$. Since $H_s$ is continuous, $H_s$ must vanish on all of $\mathbb T$, contradicting $H_s\not\equiv0$. Therefore every nonzero fiber is nowhere zero. It follows from (4) that $B_s$ is nowhere zero on every such fiber.
Step 4. There is an interval of admissible fibers
Since $F\not\equiv0$, choose $(x_*,s_*)$ with $F(x_*,s_*)\ne0$. By continuity there is an open interval $I$ around $s_*$ such that $H_s\not\equiv0$ for all $s\in I$. By Step 3, each $H_s$ is nowhere zero on the circle. Thus, for $s\in I$, both $H_s$ and $B_s$ define nonvanishing loops.
Step 5. Gauge away the Zak quasi-periodicity
The Zak fiber satisfies
$$
H_s(x+1)=e^{2\pi is}H_s(x).
$$
Define the periodic gauge
$$
K_s(x)=e^{-2\pi isx}H_s(x).
$$
Then $K_s(x+1)=K_s(x)$ and (4) becomes
$$
K_s(x-\theta)=e^{2\pi is\theta}B_s(x)K_s(x),
$$
or equivalently
$$
e^{2\pi is\theta}B_s(x)=\frac{K_s(x-\theta)}{K_s(x)}. \tag{5}
$$
The quotient on the right is a multiplicative coboundary. Consequently $B_s$ has winding zero, and its holonomy $J(e^{2\pi is})=\operatorname{Hol}(B_s)$ is defined by the exponential average of a periodic logarithm.
Step 6. Why the dynamics force a specific holonomy
Fix $s\in I$. Since $K_s$ is a nonvanishing loop on the circle, it has an integer winding number
$$
n_s=\operatorname{wind}(K_s).
$$
Winding number is locally constant under continuous deformation through nonvanishing loops. After shrinking $I$ if necessary, write this constant winding number as $n$.
$$
K_s(x)=e^{k_s(x)},\qquad k_s(x+1)=k_s(x)+2\pi i n. \tag{6a}
$$
The integer $n$ records how many times $K_s$ winds around the origin during one trip around the circle.
Solving (5) for $B_s$ gives
$$
B_s(x)=e^{-2\pi is\theta}\frac{K_s(x-\theta)}{K_s(x)}.
$$
A periodic logarithm of $B_s$ is therefore
$$
b_s(x)=-2\pi is\theta+k_s(x-\theta)-k_s(x). \tag{6b}
$$
To integrate the last two terms, write
$$
k_s(x)=p_s(x)+2\pi i n x,
$$
with $p_s$ $1$-periodic. Then
$$
\begin{aligned}
\int_0^1[k_s(x-\theta)-k_s(x)]\,dx
&=\int_0^1[p_s(x-\theta)-p_s(x)]\,dx-2\pi i n\theta\\
&=-2\pi i n\theta.
\end{aligned}
$$
The periodic part integrates to zero because translation does not change the average of a periodic function.
Hence
$$
\begin{aligned}
\operatorname{Hol}(B_s)
&=\exp\left(\int_0^1 b_s(x)\,dx\right)\\
&=\exp(-2\pi is\theta-2\pi i n\theta)\\
&=e^{-2\pi i\theta(s+n)}.
\end{aligned}
$$
With $C=e^{-2\pi i n\theta}$, this is
$$
J(e^{2\pi is})=C e^{-2\pi i\theta s}. \tag{6}
$$
Step 7. The root-count and holonomy computation, with every contribution visible
$$
J(w)=a_d(w)\,\prod_{|r_\nu(w)|>1}(-r_\nu(w)). \tag{7c}
$$
The main point is not merely that this is true, but why it is true. The Laurent denominator $z^{-M}$ is cancelled, in holonomy, by exactly $M$ interior roots. The only factors that remain visible to holonomy are the exterior roots.
Step 7.1. Turn the Laurent loop into an ordinary polynomial in $z$
For fixed $s$, write
$$
w=e^{2\pi is},\qquad z=e^{2\pi ix},\qquad B_s(x)=L(w,z).
$$
The expression $L(w,z)$ is a Laurent polynomial. It may contain negative powers of $z$, for example $z^{-1}$ or $z^{-2}$. Choose $M\ge0$ large enough so that
$$
Q(w,z):=z^M L(w,z)
$$
is an ordinary polynomial in $z$:
$$
Q(w,z)=a_d(w)z^d+a_{d-1}(w)z^{d-1}+\cdots+a_0(w).
$$
On a small arc of $w$-values, the leading coefficient $a_d(w)$ stays nonzero, so the degree $d$ is fixed. Since the loop $L(w,e^{2\pi ix})$ is nonzero on the unit circle, $Q(w,\cdot)$ has no root on $|z|=1$. Therefore, at each such $w$, we may write
$$
Q(w,z)=a_d(w)\prod_{\nu=1}^{d}(z-r_\nu(w)), \tag{7a}
$$
with roots counted with multiplicity.
Graph 1. What clearing the Laurent denominator does
The proof never studies an arbitrary loop. It studies a loop coming from a finite polynomial in the circle variable $z$.
Step 7.2. Winding number is a factor-by-factor root count
Since $L(w,z)=z^{-M}Q(w,z)$, formula (7a) gives
$$
L(w,z)=a_d(w)z^{-M}\prod_{\nu=1}^{d}(z-r_\nu(w)). \tag{7b-pre}
$$
Now let $z$ go once around the unit circle. The winding number of a product is the sum of the winding numbers of its factors.
| Factor | Path as $z$ moves on $|z|=1$ | Winding contribution |
|---|---|---|
| $a_d(w)$ | A nonzero constant. | $0$ |
| $z^{-M}$ | $z^{-1}$ travels once clockwise; $M$ copies travel $M$ times clockwise. | $-M$ |
| $z-r_\nu(w)$ with $|r_\nu(w)|<1$ | A circle of radius $1$ centered at $-r_\nu(w)$ that contains the origin. | $+1$ |
| $z-r_\nu(w)$ with $|r_\nu(w)|>1$ | A circle of radius $1$ centered at $-r_\nu(w)$ that does not contain the origin. | $0$ |
Therefore
$$
\operatorname{wind}_z L(w,z)
=\#\{\nu:|r_\nu(w)|<1\}-M. \tag{7b}
$$
Step 5 says this winding number is zero. Hence
$$
\#\{\nu:|r_\nu(w)|<1\}=M.
$$
In words: the number of interior roots is exactly the number of negative powers we had to clear.
Graph 2. Why an interior root contributes one winding and an exterior root contributes none
This is the argument-principle computation in its most elementary form: count whether the translated circle encloses the origin.
Step 7.3. The only logarithmic average lemma we need
The following elementary fact replaces the black-box use of Jensen’s formula.
In plain language: a factor of the form $1-$small rotating term does not contribute to holonomy. Its logarithm oscillates, but its average over a full period is zero.
Step 7.4. Pair the $M$ interior roots with the $M$ copies of $z^{-1}$
Let
$$
I(w)=\{\nu:|r_\nu(w)|<1\},\qquad E(w)=\{\nu:|r_\nu(w)|>1\}.
$$
From Step 7.2, $|I(w)|=M$. Therefore the $M$ copies of $z^{-1}$ in $z^{-M}$ can be paired with the $M$ interior-root factors.
Multiplying over all interior roots gives
$$
\operatorname{Hol}\left(z^{-M}\prod_{\nu\in I(w)}(z-r_\nu(w))\right)=1. \tag{7c-int}
$$
This is the precise meaning of the phrase the interior roots cancel the $z^{-M}$ factor. They cancel in holonomy because each paired factor has logarithmic average zero.
Step 7.5. Each exterior root contributes exactly $-r_\nu(w)$
If $|r_\nu|>1$, then
$$
z-r_\nu=-r_\nu\left(1-\frac{z}{r_\nu}\right).
$$
Since $|z/r_\nu|<1$ on the unit circle, the factor $1-z/r_\nu$ has holonomy $1$ by the same logarithmic-average lemma. The constant factor $-r_\nu$ remains. Therefore
$$
\operatorname{Hol}(z-r_\nu)=-r_\nu,
\qquad |r_\nu|>1. \tag{7c-ext}
$$
Graph 3. Holonomy contribution ledger
The formula (7c) is a bookkeeping statement: after all invisible holonomy-one factors are removed, only the leading coefficient and exterior constants remain.
Step 7.6. Multiply the contributions: proof of (7c)
We now return to the complete factorization
$$
L(w,z)=a_d(w)z^{-M}\prod_{\nu=1}^{d}(z-r_\nu(w)).
$$
The holonomy of a product is the product of the holonomies, because logarithms add and exponential averages multiply.
| Part of the product | Computation | Holonomy contribution |
|---|---|---|
| constant $a_d(w)$ | It does not depend on $x$. | $a_d(w)$ |
| inside $z^{-M}\prod_{\nu\in I}(z-r_\nu)$ | Pair each interior root with one $z^{-1}$. | $1$ |
| outside $\prod_{\nu\in E}(z-r_\nu)$ | Each factor is $-r_\nu(1-z/r_\nu)$. | $\prod_{\nu\in E}(-r_\nu)$ |
J(w)=a_d(w)\prod_{\nu\in E(w)}(-r_\nu(w))
=a_d(w)\prod_{|r_\nu(w)|>1}(-r_\nu(w)). \tag{7c}
$$
This completes the proof of (7c). Notice that the usual Jensen formula is not being used as a mysterious black box here. We have only used the power series for $\log(1-u)$ when $|u|<1$, together with elementary factorization.
Example 1. One interior root and one exterior root.
Let
$$
L(z)=z^{-1}(z-1/2)(z-2).
$$
Here $M=1$. The root $1/2$ is inside the unit disk and the root $2$ is outside. Pair the interior root with $z^{-1}$:
$$
z^{-1}(z-1/2)=1-\frac12 z^{-1}.
$$
This factor has holonomy $1$. The exterior factor is
$$
z-2=-2\left(1-\frac z2\right),
$$
and $1-z/2$ has holonomy $1$. Therefore
$$
J=-2.
$$
Formula (7c) gives exactly the same answer: the only exterior root is $2$, so $J=(-2)$.
Example 2. Two interior roots and two exterior roots.
Let
$$
L(z)=z^{-2}(z-1/4)(z+1/2)(z-3)(z+2).
$$
Here $M=2$. The interior roots are $1/4$ and $-1/2$. The exterior roots are $3$ and $-2$. The two interior roots pair with $z^{-2}$ and contribute $1$. The exterior roots contribute
$$
(-3)(-(-2))=(-3)(2)=-6.
$$
Hence
$$
J=-6.
$$
This example shows why the sign matters: every exterior root contributes the factor $-r$.
Step 7.7. From exterior-root product to Laurent-polynomial certificate
Formula (7c) says that $J(w)$ is built from a product of exterior roots. The exterior roots form a subset of size $d-M$ among the $d$ roots of $Q(w,z)$. Even if we do not want to label the exterior roots globally, there are only finitely many possible subsets of size $d-M$.
Define a polynomial in a new variable $T$ by multiplying over all such subsets:
$$
\Psi(w,T)=\prod_{|S|=d-M}\left(T-a_d(w)\prod_{\nu\in S}(-r_\nu(w))\right).
$$
The actual exterior-root set is one of these subsets. Therefore
$$
\Psi(w,J(w))=0.
$$
The coefficients of $\Psi$ are symmetric in the roots. By Vieta’s formulas, symmetric expressions in the roots can be rewritten in terms of the coefficients of $Q(w,z)$. Since the coefficients of $Q$ are Laurent polynomials in $w$, after clearing harmless denominators we obtain
$$
A_0(w)+A_1(w)J(w)+\cdots+A_D(w)J(w)^D=0, \tag{7}
$$
where the $A_j(w)$ are Laurent polynomials and not all of them vanish.
Example 3. The quadratic certificate computed completely.
Suppose
$$
Q(w,z)=z^2-(3+w)z+\frac12,
\qquad L(w,z)=z^{-1}Q(w,z).
$$
Then $M=1$, so one root is inside and one root is outside. Let the roots be $r_1,r_2$. The possible exterior-root products are $-r_1$ and $-r_2$. Therefore
$$
\Psi(w,T)=(T+r_1)(T+r_2)=T^2+(r_1+r_2)T+r_1r_2.
$$
Vieta gives
$$
r_1+r_2=3+w,
\qquad r_1r_2=\frac12.
$$
Hence
$$
\Psi(w,T)=T^2+(3+w)T+\frac12.
$$
Since $J(w)$ is one of $-r_1$ or $-r_2$, we get
$$
J(w)^2+(3+w)J(w)+\frac12=0.
$$
This is the Laurent-polynomial certificate in the simplest nontrivial case.
Step 8. The contradiction: irrational frequency versus finite algebra
$$
J(e^{2\pi is})=C e^{-2\pi i\theta s},\qquad \theta\notin\mathbb Q. \tag{6}
$$
Step 7 says
$$
A_0(w)+A_1(w)J(w)+\cdots+A_D(w)J(w)^D=0. \tag{7}
$$
We substitute (6) into (7) and show that the resulting finite exponential sum cannot vanish on an interval.
Start with the Laurent relation
Put $w=e^{2\pi is}$
Use $J=Ce^{-2\pi i\theta s}$
Collect frequencies $k-j\theta$
Irrationality forces contradiction
Step 8.1. Substitute the dynamic holonomy into the algebraic certificate
Substitute $w=e^{2\pi is}$ and $J(e^{2\pi is})=C e^{-2\pi i\theta s}$ into (7):
$$
\sum_{j=0}^{D}A_j(e^{2\pi is})\bigl(Ce^{-2\pi i\theta s}\bigr)^j=0. \tag{8a}
$$
Since $A_j$ is a Laurent polynomial,
$$
A_j(w)=\sum_k a_{j,k}w^k
$$
with only finitely many nonzero coefficients. Therefore
$$
A_j(e^{2\pi is})=\sum_k a_{j,k}e^{2\pi iks}.
$$
Hence (8a) becomes
$$
\sum_{j,k}a_{j,k}C^j e^{2\pi i(k-j\theta)s}=0. \tag{8b}
$$
Step 8.2. Why the frequencies cannot collide
The frequencies in (8b) are
$$
k-j\theta.
$$
Suppose two of them were equal:
$$
k-j\theta=k’-j’\theta.
$$
Then
$$
k-k’=(j-j’)\theta.
$$
The left side is an integer. If $j\ne j’$, then this would force $\theta$ to be rational. But in the mixed case $\theta\notin\mathbb Q$. Therefore $j=j’$, and then $k=k’$. So all the frequencies are distinct.
Graph 4. Frequency separation after substitution
After substitution, the algebraic relation becomes a finite sum of exponentials with distinct real frequencies.
Step 8.3. Distinct exponentials cannot cancel on an interval
We use the elementary linear independence fact: if $\lambda_1,\ldots,\lambda_N$ are distinct, then
$$
c_1e^{\lambda_1s}+\cdots+c_Ne^{\lambda_Ns}=0
$$
on a nonempty interval implies $c_1=\cdots=c_N=0$.
$$
\sum_{q=1}^N c_q\lambda_q^m e^{\lambda_qs_0}=0,
\qquad m=0,1,\ldots,N-1.
$$
The matrix $[\lambda_q^m]$ is a Vandermonde matrix. Its determinant is nonzero because the $\lambda_q$ are distinct. Therefore every $c_q=0$.
Applying this to (8b), every coefficient $a_{j,k}C^j$ is zero. Since $C\ne0$, every coefficient $a_{j,k}$ is zero. Hence every Laurent polynomial $A_j$ is identically zero.
This contradicts Step 7, where the Laurent-polynomial relation was constructed to be nontrivial. Therefore the assumed dependence relation cannot exist in the mixed arithmetic case.
Example 4. A concrete frequency contradiction.
Suppose for illustration that Step 7 produced
$$
A_0(w)+A_1(w)J(w)+A_2(w)J(w)^2=0,
$$
with
$$
A_0(w)=1-w,
\qquad A_1(w)=2+w^{-1},
\qquad A_2(w)=3w.
$$
If $J(e^{2\pi is})=Ce^{-2\pi i\theta s}$, substitution gives the frequencies
$$
0,\\ 1,\\ -\theta,\\ -1-\theta,\\ 1-2\theta.
$$
When $\theta$ is irrational, these are all distinct. The term with frequency $0$ has coefficient $1$, so the exponential sum cannot vanish identically. This is the small-scale version of the general contradiction.
Graph 5. The mixed arithmetic contradiction
The contradiction is not hidden in formalism: it is the incompatibility between irrational frequency dynamics and finite Laurent-polynomial algebra.
Case $r=3$: Dense Torus Dynamics and Zero Propagation
Theorem. Suppose $r=3$, equivalently $1,\alpha,\beta$ are linearly independent over $\mathbb Q$. Then no nontrivial dependence relation can exist.
Step 1. The Zak transform must vanish somewhere
Suppose $F=Zf$ were nowhere zero. For fixed $x$, the loop $\omega\mapsto F(x,\omega)$ is a nonvanishing loop in $\mathbb C^\times$, so its winding number $N(x)$ is defined. The function $x\mapsto N(x)$ is continuous and integer-valued, hence locally constant. But
$$
F(x+1,\omega)=e^{2\pi i\omega}F(x,\omega)
$$
increases the winding number by one. Thus $N(x+1)=N(x)+1$, contradicting periodic consistency. Hence there exists $(x_0,\omega_0)$ with
$$
F(x_0,\omega_0)=0.
$$
Step 2. The zero propagates densely
From (1), if $F(x,\omega)=0$, then
$$
F(x-\alpha,\omega+\beta)=0.
$$
Iterating gives
$$
F(x_0-n\alpha,\omega_0+n\beta)=0\qquad(n\ge0).
$$
Since $1,\alpha,\beta$ are rationally independent, this forward orbit is dense in $\mathbb T^2$. Therefore the zero set of $F$ is dense. It is also closed because $F$ is continuous. Hence $F\equiv0$.
Step 3. Injectivity finishes the contradiction
The Zak transform is injective on $L^2(\mathbb R)$. Thus $F\equiv0$ implies $f=0$, contradicting the hypothesis $f\ne0$. Hence the dense case is impossible.
Assembly of the Proof
The arithmetic trichotomy is exhaustive. In the rational case, a finer lattice reduces the problem to the classical discrete-subgroup theorem. In the mixed arithmetic case, holonomy produces the contradiction. In the maximally irrational case, dense zero propagation gives the contradiction. Therefore the normalized Schwartz one-rogue critical theorem follows.
For every finite $\Gamma\subset\mathbb Z^2$, every $(\alpha,\beta)\notin\mathbb Z^2$, and every nonzero $f\in\mathcal S(\mathbb R)$,
$$
\{M_nT_m f:(m,n)\in\Gamma\}\cup\{M_\beta T_\alpha f\}
$$
is linearly independent.
Concrete Calculations and Diagnostic Examples
The proof above is intentionally abstract because it must cover every finite lattice set $\Gamma$ and every rogue point $(\alpha,\beta)\notin\mathbb Z^2$. The following calculations show exactly how the mechanism behaves in representative cases. The notation is aligned with the certification note: $R$ is the one-step Zak multiplier, $L(w,z)$ is the $m$-step Laurent cocycle, and $Q(w,z)=z^M L(w,z)$ is the polynomial used in the local holonomy certificate.
$$
F(x-\alpha,\omega+\beta)=R(x,\omega)F(x,\omega).
$$
The symbol $Q$ is reserved for
$$
Q(w,z)=z^M L(w,z),
$$
the ordinary polynomial obtained from the Laurent polynomial $L$ by clearing negative powers of $z$.
$$
r=\dim_{\mathbb Q}\operatorname{span}_{\mathbb Q}\{1,\alpha,\beta\}.
$$
Then use the corresponding line of the proof: finite lattice refinement for $r=1$, holonomy for $r=2$, and dense zero propagation for $r=3$.
| Rogue point | Rank calculation | Orbit behavior | Proof mechanism |
|---|---|---|---|
| $(1/2,1/3)$ | $r=1$ | finite modulo $\mathbb Z^2$ | all points lie in a finer lattice $(1/6)\mathbb Z^2$ |
| $(\sqrt2,1/3)$ | $r=2$ | infinite but non-dense: $\mathbb T\times\{0,1/3,2/3\}$ | iterate three times, then use holonomy |
| $(\sqrt2,\sqrt3)$ | $r=3$ | dense in $\mathbb T^2$ | one Zak zero propagates densely |
Example A. Rational rogue point: why $r=1$ is a finite-lattice problem.
Take
$$
(\alpha,\beta)=\left(\frac12,\frac13\right).
$$
Then
$$
\operatorname{span}_{\mathbb Q}\left\{1,\frac12,\frac13\right\}=\mathbb Q,
$$
so $r=1$. With common denominator $6$,
$$
\frac12=\frac36,\qquad \frac13=\frac26.
$$
Therefore
$$
(\alpha,\beta)=\left(\frac36,\frac26\right)\in (1/6)\mathbb Z^2.
$$
Since $\Gamma\subset\mathbb Z^2\subset(1/6)\mathbb Z^2$, the whole set $\Gamma\cup\{(1/2,1/3)\}$ lies in the single discrete lattice $(1/6)\mathbb Z^2$.
$$
(x,\omega)\mapsto\left(x-\frac12,\omega+\frac13\right).
$$
After $6$ steps,
$$
(x,\omega)\mapsto(x-3,\omega+2)\equiv(x,\omega)\pmod{\mathbb Z^2}.
$$
Hence every orbit has at most $6$ points.
Example B. The mixed orbit $\alpha=\sqrt2$, $\beta=1/3$.
Let
$$
\alpha=\sqrt2,\qquad \beta=\frac13.
$$
Then
$$
\operatorname{span}_{\mathbb Q}\{1,\sqrt2,1/3\}=\operatorname{span}_{\mathbb Q}\{1,\sqrt2\},
$$
so $r=2$. The cyclic subgroup generated by the rogue motion is
$$
\mathcal O(\sqrt2,1/3)=\{n(-\sqrt2,1/3)\pmod{\mathbb Z^2}:n\in\mathbb Z\}.
$$
The second coordinate assumes only three values:
$$
0,\quad \frac13,\quad \frac23\pmod1.
$$
If $n=3k+r$, where $r\in\{0,1,2\}$, then the first coordinate is
$$
-n\sqrt2\equiv-r\sqrt2-3k\sqrt2\pmod1.
$$
Since $3\sqrt2\notin\mathbb Q$, the set $\{-r\sqrt2-3k\sqrt2:k\in\mathbb Z\}$ is dense in $\mathbb T$. Therefore
$$
\overline{\mathcal O(\sqrt2,1/3)}
=\mathbb T\times\left\{0,\frac13,\frac23\right\}.
$$
This orbit is infinite, but it is not dense in $\mathbb T^2$. This is the precise mixed arithmetic situation where the holonomy obstruction is needed.
$$
\theta=m\alpha=3\sqrt2\notin\mathbb Q.
$$
Example C. The three-step Zak calculation for $\alpha=\sqrt2$, $\beta=1/3$.
The correct one-step equation is
$$
F(x-\sqrt2,s+1/3)=R(x,s)F(x,s).
$$
Iterating once more gives
$$
F(x-2\sqrt2,s+2/3)=R(x-\sqrt2,s+1/3)R(x,s)F(x,s).
$$
A third iteration gives
$$
F(x-3\sqrt2,s+1)=R(x-2\sqrt2,s+2/3)R(x-\sqrt2,s+1/3)R(x,s)F(x,s).
$$
Since $F(x,s+1)=F(x,s)$, this becomes
$$
F(x-3\sqrt2,s)=B_s(x)F(x,s),
$$
where
$$
B_s(x)=R(x-2\sqrt2,s+2/3)R(x-\sqrt2,s+1/3)R(x,s).
$$
This is the concrete form of the $m$-step cocycle in the mixed case.
Example D. The correct role of $Q(w,z)$ in the local holonomy certificate.
Suppose the $m$-step cocycle has the form
$$
B_s(x)=L(w,z),\qquad w=e^{2\pi is},\quad z=e^{2\pi ix},
$$
with $L\in\mathbb C[w^{\pm1},z^{\pm1}]$. If $L$ contains negative powers of $z$, choose $M\ge0$ so that
$$
Q(w,z)=z^M L(w,z)
$$
is an ordinary polynomial in $z$:
$$
Q(w,z)=a_d(w)\prod_{\nu=1}^d(z-r_\nu(w)).
$$
Since $L=z^{-M}Q$, the winding number is
$$
\operatorname{wind}_z L(w,z)=\#\{\nu:|r_\nu(w)|<1\}-M.
$$
Therefore winding zero means that exactly $M$ roots of $Q(w,\cdot)$ lie inside the unit disk. The holonomy is then
$$
J(w)=a_d(w)\prod_{|r_\nu(w)|>1}(-r_\nu(w)).
$$
Example E. A quadratic local holonomy certificate.
Let
$$
L(w,z)=z^{-1}\left(z^2-(3+w)z+\frac12\right),\qquad |w|=1.
$$
Here $M=1$, and the ordinary polynomial is
$$
Q(w,z)=zL(w,z)=z^2-(3+w)z+\frac12.
$$
If $|z|=1$ and $Q(w,z)=0$, then after dividing by $z$ we get
$$
z+\frac{1}{2z}=3+w.
$$
But
$$
\left|z+\frac{1}{2z}\right|\le1+\frac12=\frac32,
\qquad |3+w|\ge3-|w|=2,
$$
a contradiction. Thus no root lies on the unit circle.
The roots $r_+(w),r_-(w)$ satisfy
$$
r_+(w)+r_-(w)=3+w,
\qquad r_+(w)r_-(w)=\frac12.
$$
There is one interior root and one exterior root. Denote the exterior root by $r_+(w)$. Since $a_d(w)=1$, the holonomy is
$$
J(w)=-r_+(w).
$$
Substituting $r_+(w)=-J(w)$ into the equation
$$
r_+(w)^2-(3+w)r_+(w)+\frac12=0
$$
gives
$$
J(w)^2+(3+w)J(w)+\frac12=0.
$$
Thus the explicit Laurent certificate is
$$
A_0(w)=\frac12,\qquad A_1(w)=3+w,
\qquad A_2(w)=1.
$$
Example F. A cubic local holonomy certificate.
Let
$$
L(w,z)=z^{-1}\left(z^3-(4+w)z+\frac18\right),\qquad |w|=1,
$$
so
$$
Q(w,z)=zL(w,z)=z^3-(4+w)z+\frac18.
$$
On $|z|=1$,
$$
Q(w,z)=-(4+w)z+\left(z^3+\frac18\right).
$$
Since
$$
|(4+w)z|=|4+w|\ge3,
\qquad \left|z^3+\frac18\right|\le1+\frac18=\frac98,
$$
Rouch\’e’s theorem shows that $Q(w,\cdot)$ has the same number of zeros inside the unit disk as $-(4+w)z$, namely one. Since $M=1$, the winding of $L$ is zero.
Let $r_0(w)$ be the interior root and let $r_1(w),r_2(w)$ be the exterior roots. The holonomy is
$$
J(w)=(-r_1(w))(-r_2(w))=r_1(w)r_2(w).
$$
Vieta’s formulas for
$$
z^3-(4+w)z+\frac18
$$
give
$$
r_0+r_1+r_2=0,
\qquad r_0r_1+r_0r_2+r_1r_2=-(4+w),
\qquad r_0r_1r_2=-\frac18.
$$
Since $J=r_1r_2$, the last identity gives
$$
r_0J=-\frac18,
\qquad r_0=-\frac{1}{8J}.
$$
Also $r_1+r_2=-r_0$, so
$$
-r_0^2+J=-(4+w).
$$
Hence
$$
J+(4+w)=\frac{1}{64J^2},
$$
and therefore
$$
J(w)^3+(4+w)J(w)^2-\frac1{64}=0.
$$
Thus an explicit local certificate is
$$
A_0(w)=-\frac1{64},\qquad A_1(w)=0,
\qquad A_2(w)=4+w,
\qquad A_3(w)=1.
$$
Example G. The final contradiction as a finite list of frequencies.
In the mixed example, $w=e^{2\pi is}$ and the dynamic calculation forces
$$
J(w)=C e^{-2\pi i\theta s},
\qquad \theta=3\sqrt2.
$$
Suppose, only to illustrate the algebraic relation, that Step 7 produced a relation of degree $2$:
$$
A_0(w)+A_1(w)J(w)+A_2(w)J(w)^2=0,
$$
with
$$
A_0(w)=1-w,
\qquad A_1(w)=2+w^{-1},
\qquad A_2(w)=3w.
$$
Substitution gives
$$
\begin{aligned}
0={}&(1-e^{2\pi is})
+(2+e^{-2\pi is})Ce^{-2\pi i\theta s}
+3e^{2\pi is}C^2e^{-4\pi i\theta s}\\
={}&1-e^{2\pi is}+2C e^{-2\pi i\theta s}
+C e^{2\pi i(-1-\theta)s}
+3C^2 e^{2\pi i(1-2\theta)s}.
\end{aligned}
$$
The frequencies are
$$
0,\quad 1,
\quad -\theta,
\quad -1-\theta,
\quad 1-2\theta.
$$
Since $\theta=3\sqrt2$ is irrational, these frequencies are all distinct. Therefore the finite exponential sum cannot vanish on an interval unless all coefficients vanish. In this illustrative relation, the coefficient of the frequency $0$ is $1$, so the relation is impossible.
Example H. Dense zero propagation in the maximally irrational case.
Take
$$
(\alpha,\beta)=(\sqrt2,\sqrt3).
$$
Then $1,\sqrt2,\sqrt3$ are rationally independent, so $r=3$. If the Zak transform has a zero at
$$
(x_0,\omega_0)=(0.10,0.20),
$$
then the one-step equation propagates zeros to
$$
(x_n,\omega_n)=(x_0-n\sqrt2,\omega_0+n\sqrt3)\pmod{1}.
$$
The first few approximate points are:
| $n$ | $x_n$ mod $1$ | $\omega_n$ mod $1$ |
|---|---|---|
| 0 | 0.100 | 0.200 |
| 1 | 0.686 | 0.932 |
| 2 | 0.272 | 0.664 |
| 3 | 0.857 | 0.396 |
| 4 | 0.443 | 0.128 |
| 5 | 0.029 | 0.860 |
The table is only a numerical glimpse. The theorem used in the proof says that the full orbit is dense in the torus. Since the zero set of a continuous function is closed, a dense set of zeros forces every point to be a zero. Thus $Zf\equiv0$, and injectivity of the Zak transform gives $f=0$, contradicting the nonzero hypothesis.
Lean Certification Links and Formal Audit Trail
The proof above is intended to stand on its own. The following links are included so that readers can audit the formal artifacts that support the one-rogue classification, rational refinement, and especially the mixed arithmetic holonomy endpoint.
Primary Lean certification archive for the difficult mixed arithmetic endpoint.
The Dropbox archive HRT-infinite-not-dense.zip is the main external Lean package for the $r=2$ infinite non-dense Zak/holonomy contradiction. In the language of this article, it is the formal audit trail for the step where periodic logarithms, holonomy, and the winding-number obstruction are converted into a contradiction.
| Proof component | Role in this article | Formal status language | Access |
|---|---|---|---|
| One-rogue taxonomy | Identifies the normalized lattice-plus-one-point model and the three arithmetic classes. | formal infrastructure | direct archive · shared archive folder |
| Rational refinement | Supports the $r=1$ reduction to a finer lattice, relative to Linnell’s theorem. | formal reduction, classical external input | direct archive · lattice benchmark archive |
| Primary Lean archive: mixed arithmetic endpoint | Supports the $r=2$ infinite non-dense Zak/holonomy contradiction. This is the formal companion to the holonomy/winding-number part of the proof. | Lean endpoint package | HRT-infinite-not-dense.zip |
| Endpoint theorem name | The named final result for the mixed arithmetic formal package. | one_rogue_mixed_integer_HRT_infinite_nondense_certified |
local summary |
| Dense zero-propagation proof | The $r=3$ proof is written explicitly in this article using continuity, unavoidable Zak zeros, and dense orbits. | human-readable proof included here | jump to proof |
Reader Checks
Check 1. Show that $r=1$ implies $(\alpha,\beta)\in(1/N)\mathbb Z^2$ for some $N$.
Check 2. In the mixed form $\beta=p/m$, verify that $m$ iterations of the rogue translation close the second coordinate and leave an irrational rotation in the first coordinate.
Check 3. Prove that if a continuous nonzero loop $B:S^1\to\mathbb C^\times$ has winding zero, then it admits a continuous periodic logarithm.
Check 4. If $k(x+1)=k(x)+2\pi i n$, verify directly that $\int_0^1[k(x-\theta)-k(x)]\,dx=-2\pi i n\theta$.
Check 5. For $L(w,z)=z^{-q}a(w)\prod_{\nu}(z-r_\nu(w))$, explain why winding zero means that exactly $q$ roots lie inside the unit disk.
Check 6. Use the Vandermonde argument to prove that finitely many exponentials with distinct frequencies are linearly independent on any nonempty interval.
Check 7. In the dense case, explain why one zero of $Zf$ implies $Zf\equiv0$.
References and Attribution
- Author’s related HRT papers.
- B. Currey and V. Oussa, “Translates of Functions on the Heisenberg Group and the HRT Conjecture,” Canadian Mathematical Bulletin 63(4) (2020), 871–881. DOI · arXiv.
- K. A. Okoudjou and V. Oussa, “The HRT conjecture for two classes of special configurations,” Journal of Fourier Analysis and Applications 31(4) (2025), Paper No. 48. arXiv.
- V. Oussa, “Zero set of Zak transform and the HRT Conjecture,” arXiv:2305.12299 (2023). arXiv.
- V. Oussa, “A Trichotomy and Rigidity Constraints for the HRT Conjecture in the Mixed-Integer Case,” arXiv:2508.04613v2 (2026). arXiv.
- V. Oussa, “On the $(m,1)$-configuration for the HRT Conjecture,” arXiv:2508.18529v3 (2025). arXiv.
- HRT conjecture. C. Heil, J. Ramanathan, and P. Topiwala, “Linear independence of time-frequency translates,” Proceedings of the American Mathematical Society 124 (1996), 2787–2795. Public discussion and references: Heil–Speegle survey PDF.
- Zak transform. K. Gröchenig, “Zak Transform Methods,” in Foundations of Time-Frequency Analysis, Birkhäuser. DOI page: 10.1007/978-1-4612-0003-1_9. Quick reference: Zak transform.
- Lattice-contained systems. P. Linnell, “Von Neumann algebras and linear independence of translates,” Proceedings of the American Mathematical Society 127 (1999), 3269–3277. arXiv record.
- Winding and holonomy background. See winding number, contour winding number, and holonomy.
- Jensen formula and logarithmic averages. See Jensen’s formula. In this article the needed use is the elementary polynomial-root version explained in Step 7.
- Finite exponential independence. A finite family of exponential functions with distinct exponents is linearly independent; see A. Eremenko’s note Linear dependence of exponentials for a reference-level discussion.
- Torus dynamics. See irrational rotation, linear flow on the torus, and Kronecker’s theorem.