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Let $$ f(x) = \sum_{j=1}^n c_j e^{2\pi i\alpha_j x}, g(x) = \sum_{k=1}^m d_k e^{2\pi i\beta_k x}$$ be two (quasi-periodic) trigonometric polynomials, where the coefficients $c_j, d_k$ are complex and the frequencies $\alpha_j,\beta_k$ are real (but are not necessarily commensurable). Suppose that $f,g$ have infinitely many zeroes in common, thus $f(x)=g(x)=0$ for infinitely many reals $x$. Does this imply that $f,g$ have a non-trivial common factor (in the ring of trigonometric polynomials)? In the commensurable case (so that $f,g$ have a common period) this is clear from the factor theorem, but my interest is in the incommensurable case. I would also be interested in the special case when $f,g$ are both real-valued, though I suspect that this restriction does not materially impact the question.

Here is a closely related question. Let $G = \{ (e^{2\pi i \alpha_1 x}, \dots, e^{2\pi i \alpha_k x}): x \in {\bf R} \}$ be a one-parameter subgroup of the torus $(S^1)^k$ for some reals $\alpha_1,\dots,\alpha_k$ that are linearly independent over ${\bf Q}$ (so that $G$ is dense in $(S^1)^k$). Is it true that every codimension two real subvariety of $(S^1)^k$ intersects $G$ in at most finitely many points? Note that if one specialises to the subvarieties $\{ z \in (S^1)^k: P(z)=Q(z)=0\}$ for two polynomials $P,Q$ with no common factor one basically obtains a version of the first question.

I'm somewhat aware of the literature on exponential polynomials (e.g., Ritt's theorem), but I couldn't find quite the appropriate tool; results in transcendental number theory (e.g., the six exponentials theorem) also seem vaguely relevant, but again it's not quite a perfect fit.

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    $\begingroup$ My first thought was to try to deduce it from Schanuel's conjecture, which would at least clarify whether it should be true, but this itself seems quite hard. If we let $x_1,\dots, x_n$ be the roots and mandate that the $\alpha_i x_j$ be $\mathbb Q$-linearly independent, then the $e^{ 2 \pi i \alpha_i x_j}$s generate a field of transcendence degree at least $nk-n-k$ under Schanuel's conjecture. $\endgroup$
    – Will Sawin
    Commented Mar 20, 2021 at 22:26
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    $\begingroup$ Demanding that the $n \times k$ matrix with those entries have rank $< k-1$ is a codimension $n+1-k$ condition, and the same for the matrix with inverse entries, and assuming these are independent that is codimension $2n+2-2k$ in total, so this can't happen if $2n+2-2k > n+k$ which happens for $n> 3k-2$, so indeed $n$ must be bounded under the conjecture and this strong linearly independence assumption. If there are any linear relations, we get additional multiplicative relations on the entries of the matrix for free, and I guess the goal is to prove those don't help... $\endgroup$
    – Will Sawin
    Commented Mar 20, 2021 at 22:28
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    $\begingroup$ This paper seems to have a proof assuming Schanuel's conjecture: "From Schanuel’s Conjecture to Shapiro’s Conjecture" doi.org/10.4171/CMH/328 They attribute this conjecture to H.S. Shapiro, and looking at other citing papers it seems like it is still open. They also say that the case when one of the exponential polynomials has commensurable frequencies is known unconditionally and is due to van der Poorten and Tijdeman. $\endgroup$
    – Gjergji Zaimi
    Commented Mar 20, 2021 at 23:03
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    $\begingroup$ @GjergjiZaimi Wow, my question is virtually identical to the Shapiro conjecture! If you convert your comment to an answer I will gladly accept it. $\endgroup$
    – Terry Tao
    Commented Mar 21, 2021 at 0:09
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    $\begingroup$ Here is an example by Montgomery showing why this is difficult, adapted from the end of the 1975 paper below. Let $f(z)=1-\exp(2\pi i z)$, $g(z)=1-\exp(2\pi i \alpha z)$, where $a\sim 1839103/2097152$. Then $f(z)$ is $0$ at every integer, and $|g(1)|\le 1$, $|g(8)|\le 1/8$, $|g(512)|\le 1/512$. With the right $\alpha$ we can find an infinite sequence where $|g(2^n)|\le 1/2^n$. Then $f$ and $g$ have infinitely many near-zeroes in common but no common factor. $\endgroup$
    – user44143
    Commented Mar 21, 2021 at 4:36

1 Answer 1

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I'm converting the comment above to an answer:

Shapiro made the following conjecture in the paper "The expansion of mean-periodic functions in series of exponentials" Comm. Pure and Appl. Math. 11 (1958) 1–22:

If two exponential polynomials have infinitely many zeros in common they are both multiples of some third (entirely transcendental) exponential polynomial.

In the paper "On common zeros of exponential polynomials" Enseignement Math. (2) 21 (1975), no. 1, 57–67, van der Poorten and Tijdeman prove a special case of Shapiro's conjecture for the case when one of the polynomials has rational frequencies. They observe that in a certain sense this statement is equivalent to the Skolem–Mahler–Lech theorem.

More recently, in the paper "From Schanuel’s Conjecture to Shapiro’s Conjecture", D'Aquino, Macintyre, and Terzo show that the general conjecture can be proven assuming Schanuel's conjecture.

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    $\begingroup$ That 1975 paper ends describing this as an algebraic attack on an apparently analytic problem. After the work of D'Aquino, Macintyre, and Terzo (and noting the appearance of Skolem too), the updated description might be that this is a problem in analysis solved algebraically by logicians! Curious. $\endgroup$
    – user44143
    Commented Mar 21, 2021 at 2:59

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