The MO answer https://mathoverflow.net/a/98176/11926 notes the following: Let $\gamma$ be an algebraic number that is not a root of unity. Then Baker's theorem implies that there is a constant $C(\gamma)>0$ such that for all $n\ge2$ we have $\gamma^n1\ge n^{C(\gamma)}$. An alternative way to phrase this is that $\gamma$ cannot be too close to a root of unity. Thus if we let $\boldsymbol{\mu}_n$ denote the $n$'th roots of unity, then the result says that $\min_{\zeta\in\boldsymbol{\mu}_n}\gamma\zeta\ge n^{C(\gamma)}$. My question is whether a higher dimensional analogue of this is known. Let $F(X_1,\ldots,X_k)\in\overline{\mathbb{Q}}[X_1,\ldots,X_k]$ be a polynomial with algebraic coefficients. Does there exist a constant $C=C(F)>0$ such that $$ \min_{\substack{\zeta_1,\ldots,\zeta_k\in\boldsymbol{\mu}_n\\ F(\zeta_1,\ldots,\zeta_k)\ne0\\}} \biglF(\zeta_1,\ldots,\zeta_k)\bigr \ge n^{C} \quad\hbox{for all $n\ge2$?} $$ Or if this is not known, for the application that I have in mind it would suffice to have a lower bound which implies that $$ \liminf_{n\to\infty} \frac{1}{n} \min_{\substack{\zeta_1,\ldots,\zeta_k\in\boldsymbol{\mu}_n\\ F(\zeta_1,\ldots,\zeta_k)\ne0\\}} \log\biglF(\zeta_1,\ldots,\zeta_k)\bigr \ge 0. $$
Since a monomial in roots of unity is again a root of unity, you can assume your polynomial is linear. But unfortunately very little is known about this question. It has been discussed in MO before:

$\begingroup$ Good point about reducing to a linear polynomial. Thanks. However, the MO question you cite is only for sums of roots of unity, which would not include general linear forms with algebraic coefficients. Of course, quite likely the proof of a nontrivial lower bound for the former could be adapted to give one for the latter. (Nontrivial = better than exponential) $\endgroup$ – Joe Silverman Aug 20 '14 at 14:20

$\begingroup$ @Joe Can't you reduce algebraic coefficients to rational by multiplying conjugates? Integer coefficients reduce to sums by $3\zeta = \zeta + \zeta +\zeta$. But the point is we can't even deal with sums. Maybe something more can be done on the liminf question. $\endgroup$ – Felipe Voloch Aug 20 '14 at 14:31

$\begingroup$ Very clever reduction (although if one wants reasonable dependence on $F$, this might not be the way to go). The liminf question is exactly asking if one can improve on the trivial exponential bound. I would guess that if any such were known, then it would have been mentioned in the answers to Terry's question that you cited. $\endgroup$ – Joe Silverman Aug 20 '14 at 14:37
Here is a related, but different, problem, as mentioned by @Andreas. Let $f$ be a Laurent polynomial in $n$ variables with integer coefficients. Define $\log_0(t)$ to be $\log t$ if $t>0$ and 0 if $t=0$. Consider finite subgroups $F$ of the multiplicative $n$torus and the Riemann sums of $\log_0 f$ over $F$ as $F$ becomes more and more uniformly distributed. Do these always converge to the integral of $\log_0 f$ over the torus with respect to Lebesgue measure (which is the log of the Mahler measure of $f$)? This question is motivated by a dynamical question about periodic points.
Concrete case: Let $f(x,y) = 3  x  x^{1}  y  y^{1}$. The zero set of $f$ on the 2torus is an oval containing exactly four roots on unity (the intersections of the oval with the coordinate axes), so there are no subtleties about how various finite $F$s can intersect this oval. And yet, as far as I know, whether or not there is convergence of the Riemann sums is open even in this simple case.

$\begingroup$ Thanks, Doug. Actually, the "related, but different" problem is the one I'm interested in! But I thought that the question that I posed was likely to be easier, and that it would be an essential step in proving the convergence to log Mahler measure. That's interesting that you say the question is motivated by a dynamical question about periodic points. Now that you say that, it seems clear, but I came at it from the viewpoint of torsion points on algebraic groups. Do you know any other references or discussion of the problem of $\hbox{Avg}\log_0f\to\log M(f)$? $\endgroup$ – Joe Silverman Aug 21 '14 at 15:10

$\begingroup$ In arxiv.org/abs/1108.4989 we proved that the averages converge provided that the zero set of $f$ on the $n$torus has dimension at most $n2$ (this is a successor to the paper @Andreas mentions). The argument uses homoclinic points for the associated algebraic action, which we can create with the assumption on the zero set by finding another polynomial, not a multiple of $f$, whose zero set on the torus contains the zero set of $f$ on the torus. This is "atonality" as introduced by Agler, McCarthy, and Stankus in their work on function theory on polydisks. $\endgroup$ – Douglas Lind Aug 21 '14 at 15:39

$\begingroup$ Thanks for the reference. I was looking at the earlier paper, where you'd assumed that the intersection is finite (which already looks quite intricate). I'll take a look at the 1108.4989 paper. You're talking about real dimension, right? So $f=0$ on $(\mathbb{C}^*)^d$ has real codimension 2, so generically it should intersect $(S^1)^d$ in a set of real codimension 2. This would seem to indicate that your result applies to "most polynomials" in some appropriate sense. Or am I confused about codimension here? $\endgroup$ – Joe Silverman Aug 21 '14 at 15:54

$\begingroup$ Yes, real dimension as a subset of the $d$torus, which has real dimension $d$. And yes, this covers "most" polynomials. The concrete case above is where all our techniques fail. $\endgroup$ – Douglas Lind Aug 22 '14 at 0:26
Baker's theorem applied to an integer Laurent polynomial $f$ can be used to understand the entropy of a certain algebraic action of $\mathbb Z$ (which associated with $f$, it is the Pontryagin dual of the $\mathbb Z$module $\mathbb Z[\mathbb Z]/(f \cdot \mathbb Z[\mathbb Z])$ seen as a discrete abelian group) in terms of the growth rate of periodic points. It is wellknown that a higherdimensional analogue of Baker's theorem would imply a similar description of the entropy also for $\mathbb Z^n$actions  which is currently not known to hold.
However, for certain algebraic $\mathbb Z^n$actions associated with socalled atoral Laurent polynomials with integer coefficients, different methods can be found to relate the growth rate of periodic points to the entropy.
(A Laurent polynomial in $n$ variables is called atoral if the intersection of its zeroset with the $n$torus has real codimension at least $2$.)
As a consequence the itegral of the logarithm of the absolute value of the atoral polynomial over the $n$torus can be computed as some sort of Riemann integral by summing over the values at roots of unity and taking a limit. Moreover, this implies that some weak form of Baker's theorem has to hold in high dimensions  at least if the polynomial is assumed to be atoral. For a precise statement see the reference below.
All this was proved and explained in detail in:
D. Lind, K. Schmidt, E. Verbitskiy, Entropy and growth rate of periodic points of algebraic $\mathbb Z^d$actions, Numbers and Dynamics. Contemporary Mathematics, Amer. Math. Society, Volume 532 (December, 2010). [arXiv:0912.5169]

$\begingroup$ Thanks. As Doug Lind pointed out in his answer, the paper you cite has the result if the unitary set of $f$ is finite. The result for codimension 2 or greater unitary sets is in the paper: [MR3082539] Lind, Douglas; Schmidt, Klaus; Verbitskiy, Evgeny; Homoclinic points, atoral polynomials, and periodic points of algebraic $\mathbb{Z}^d$actions. Ergodic Theory Dynam. Systems 33 (2013), no. 4, 1060–1081. $\endgroup$ – Joe Silverman Aug 21 '14 at 19:08

$\begingroup$ Sorry, you are right, I got the reference wrong. $\endgroup$ – Andreas Thom Aug 21 '14 at 19:49
As in Baker's theorem $\prod_i \alpha_i^{n_i}1 \gg n^{C}$, $n := \max_i{n_i}$, this would be another instance where the trivial Liouville lower bound (exponential in $n$) should actually be replaced by a polynomial bound in $1/n$. [In fact the two questions seem not unrelated. I believe Baker's inequality is expected to extend to sums of a fixed number of roots of $S$units: $\prod_i \alpha_i^{n_i}$ with rational exponents $n_i$ of height $\leq n$. However, Waldschmidt's book Diophantine Approximation on Linear Algebraic Groups indicates that such a conjectural generalization of Baker's theorem is wide open. ]
Anyway, the problem posed here concerns a lower bound on the absolute value of a linear form in roots of unity. As in the question that Terry Tao asked four years ago, let us restrict to rational coefficients; the generalization to algebraic coefficients is straightforward as Felipe Voloch notes. It is then convenient to reformulate the problem thus:
Given $L < \infty$, prove that there is a $C(L) < \infty$ such that the following holds for all $n \gg_L 0$. If $f \in \mathbb{Z}[x]$ is an integer polynomial of length bounded by $L$, then either $f(\zeta_n) = 0$ or $f(\zeta_n) > n^{C(L)}$.
This is quite a coincidence: yesterday I spotted Tao's earlier question, and was also thinking about the same problem today. Once the problem is phrased this way, I think it might be possible to apply, to the roots of the polynomial $f$ and with test function an appropriate smoothed compactly supported cutoff of the logarithmic distance to $\zeta_n$, the quantitative form of Bilu's equidistribution theorem due to Favre and RiveraLetelier (cf. the statement on page 3 in this paper of Fili and Miner: http://arxiv.org/pdf/1210.7887.pdf). For the roots of $f$ have very small height: bounded by $\log{L}/\deg{f}$, which we may arrange to be $O(1/n)$ as we may take $f$ to have degree $\asymp n$. But for this to work, we will also need to know that not one of these roots $\alpha_i$ can get too close to $\zeta_n$, and here is where the difficulty lies. See my second answer and the comments below it.
By the way, we could also pose the problem in semiabelian varieties, which would be an archimedean counterpart to the TateVoloch conjecture:
If $X \subset A$ is a subvariety of a semiabelian variety over $\bar{\mathbb{Q}}$, and under a fixed complex embedding of $\bar{\mathbb{Q}}$, show that a point of order $n$ not lying on $X$ is of distance at least $\gg n^{C(X)}$ apart from $X$.
There is even an analogous question one could ask about CM points near subvarieties of Shimura varieties; Habegger in the $p$adic case has some results on the analog of TateVoloch for powers of the modular curve.

$\begingroup$ Interesting. I don't immediately see why one can reformulate the problem using onevariable polynomials of length bounded by $L$, unless the $\gg$ constant in $f(\zeta_n)\gg n^{C(L)}$ is supposed to be independent of $f$. Also, in your statement of TateVoloch, don't you need $A$ and $X$ to be defined over $\overline{\mathbb{Q}}$? I think that it is not true, for example, if we take $A=\mathbb{G}_m(\mathbb{C})$ and if we're allowed to take $X=\{\beta\}$ with $\beta\in\mathbb{C}^*$ transcendental. $\endgroup$ – Joe Silverman Aug 20 '14 at 15:05

$\begingroup$ @JoeSilverman: Yes, I meant the implied constant in $\gg$ to depend only on $L$ (or the inequality $> n^{C(L)}$ to hold for $n \gg_L 0$). Sorry for this omission, I will edit. And thanks for correcting the TateVoloch statement. $\endgroup$ – Vesselin Dimitrov Aug 20 '14 at 15:08

$\begingroup$ @JoeSilverman: Also, by the quantitative Bilu theorem I had in mind the result of Favre and RiveraLetelier, given on page 3 in this paper by Fili and Miner: arxiv.org/pdf/1210.7887.pdf . I thought we could try to apply this theorem to the roots of $f$, which we know to have height only $\leq (\log{L})/n$. $\endgroup$ – Vesselin Dimitrov Aug 20 '14 at 16:06

$\begingroup$ @JoeSilverman: It did not work as I hoped, my apology. The real difficulty is that $\zeta_n$ could get too close to a root of $f$, and my hope that it cannot get as close as $o(1/n)$ (when $\deg{f} < n$ and the length $k$ of $f$ is fixed), remains unproved. The trivial lower bound of $C^{n}$ is of no use. Oh well; at least this led to another interesting problem. (How small can the distance be between two distinct roots of an integer polynomial of degree $n$ and length bounded by $L$? I had guessed not smaller than $O_L(1/n)$.) $\endgroup$ – Vesselin Dimitrov Aug 21 '14 at 11:46
After a couple of failed attempts at a proof, I have come to appreciate the difficulty of even the subexponential bound $e^{o(n)}$. The lemma I had asserted does not follow from the Theorem below if $\delta$ is too small. The difficulty is related to the following problem, which I think is the crux of the matter in the question:
Fix a $L < \infty$. How small can the infimum be, as a function of $m$, of the difference $\alpha  \beta$ of a pair of distinct roots of an integer polynomial of degree $m$ having length bounded by $L$?
My guess was that this difference cannot get smaller than $O_L(1/m)$, but I was wrong that this follows from an effective equidistribution theorem. I do not know about any literature on this question. I only needed this with $\beta = \zeta_n$ a root of unity of order $n \asymp m$, but the available lower bounds on $\alpha^n  1$ do not seem to be of any help for this. If my hope that $\alpha  \beta > c(L)/m$ were correct, the subexponential bound would follow by the argument left below.
My apology for the two failed attempts. I will still preserve the text that I had written below, as it shows that if my guess in the previous paragraph was anywhere near the truth, we would have as a consequence the subexponential bound in the original problem.
To recall the statement we want: $$ \liminf_{n\to\infty} \frac{1}{n} \min_{\substack{\zeta_1,\ldots,\zeta_k\in\boldsymbol{\mu}_n\\ F(\zeta_1,\ldots,\zeta_k)\ne0\\}} \log\biglF(\zeta_1,\ldots,\zeta_k)\bigr \ge 0. $$ As noted by Felipe Voloch, we are through if (for any fixed $k$) we show that a nonvanishing sum of $k$ roots of unity is at least $e^{o(n)}$ in absolute value. As I note in my first answer, evey such sum is a polynomial $f(\zeta_n)$ in a primitive $n$th root of unity $\zeta_n$ with nonnegative integer coefficients whose sum is at most $k$. We may assume that $d < n$, and also that $d \asymp n$ if we want (as we are free in our choice of primitive root $\zeta_n$). We may also assume that $f$ has no multiple roots.
The following is a variant of the quantitative equidistribution theorem of Favre and RiveraLetelier, taken from page 3 of this paper (On totally real numbers and equidistribution) by Fili and Miner: http://arxiv.org/pdf/1210.7887.pdf . The only modification I make to the statement is that the polynomial $P$ is not required to be irreducible; but irreducibility is irrelevant in such statements.
Theorem. There is an absolute constant $c > 0$ such that the following is true. Let $P \in \mathbb{Z}[x]$ have Mahler measure $M = M(P)$ and complex zeros $\beta_1,\ldots,\beta_m$ (with multiplicities accounted). Then, all $C^1$ functions $\lambda$ on $\mathbb{P}^1(\mathbb{C})$ having Lipschitz constant $\mathrm{Lip}(\lambda)$ satisfy:
$$\Big \frac{1}{m} \sum_{i=1}^m \lambda(\beta_i)  \int_{S^1} \lambda(z) \frac{d\theta}{2\pi} \Big$$ is at most $$\leq \frac{\mathrm{Lip}(\lambda)}{m} + {\Big( \frac{\log{M(P)} + c\log{m}}{m} \Big)}^{1/2} \cdot { \Big( \int_{\mathbb{P}^1(\mathbb{C})} d\lambda \wedge d^c \lambda \Big)}^{1/2}$$.
I had thought this had the following as consequence. It certainly holds for $\delta$ bounded away from zero, but my intended application had to have $\delta$ as small as $O(1/m)$; then the first term on the righthand side of the estimate destroys all hope, as Terry Tao noted in the comments. The Laplacian under the integral poses no problem, but I had overlooked the Lipschitz constant.
Hypothesis. For any $P \in \mathbb{Z}[x]$ with degree $m$ and Mahler measure $< B$, and all $\delta > 0$, the neighborhood $\beta_i  z < \delta$ of any one of the roots contains fewer than $O_B(\delta m)$ other roots $\beta_j$.
While such a strong hypothesis is probably unlikely to be true, here is how the subexponential would follow as a consequence (just to preserve the argument I had written).
Claim. Assume the Hypothesis. Then, for any $k$ and $\varepsilon > 0$, as soon as $n \gg_{k,\varepsilon} 0$ is sufficiently large with respect to them, all nonvanishing sums of $k$ roots of unity of order dividing $n$ have absolute value at least $e^{\varepsilon n}$.
Proof. Consider the complex zeros $\alpha_1,\ldots,\alpha_d$ of the polynomial $f \in \mathbb{Z}[x]$; as remarked above, we lose no generality in assuming that $d \asymp n$ and that these roots are pairwise distinct. As $M(f)$ is bounded by $k$, we may use the equidistribution theorem above. On the other hand, denoting $D(n,C)$ the disk $\zeta_n  z \leq C/n$ with $C$ a constant on our disposal, the Hypothesis applied to the polynomial $P(z) := f(z)(z^n1)$ (of degree $< 2n$) reduces our claim to showing that there is a $C < \infty$ making $\sum_{\alpha_i \notin D(n,C)} \log{\zeta_n  \alpha_i} >  \varepsilon n/2$ as soon as $n \gg_C 0$.
Note that, applying the triangle inequality to the equation $f(\alpha_i) = 0$ (whose length $k$ is fixed), all roots $\alpha_i$ lie in the fixed compact subset $K = \{z \leq k\} \subset \mathbb{C}$. There is a smooth function $\lambda \in C_c^{\infty}(\mathbb{C}^{\times})$ supported in $2K$ which coincides with $\log{z\zeta_n}$ on $K \setminus D(n,C)$ and satisfies the following conditions:
 it has a Lipschitz constant bounded by $10n/C$;
 it is bounded in magnitude by $10\log{(n/C)}$ in $D(n,C)$, and by a constant independent of $n$ in $\mathbb{C} \setminus K$;
 its Laplacian is bounded in magnitude by $10(n/C)^2$ in $D(n,C)$ and by a constant independent of $n$ in $\mathbb{C} \setminus K$.
Applying the above Theorem, now with $P$ our polynomial of degree $d \asymp n$ and the function $\lambda$ just constructed, we conclude that $\Big \frac{1}{n} \sum_{\alpha_i \notin D(n,C)} \log{\zeta_n  \alpha_i} \Big = O(1/C)$. As the constant $C$ was left on our disposal and can be arbitrarily large, this is enough to conclude by the previous paragraph.

$\begingroup$ Why is the Mahler measure of f bounded by k? I thought it was the logarithmic height that was controlled by k, which would make the Mahler measure exponential in k and n. $\endgroup$ – Terry Tao Aug 21 '14 at 4:44

$\begingroup$ @TerryTao: The Mahler measure is bounded by the length. This is just an application of the triangle inequality under the integral in $\log{M(f)} = \int_{S^1} \log{f(z)} \frac{d\theta}{2\pi}$. Then the absolute logarithmic height is $(\log{M}(f)) / \deg{f}$. So, in our situation, the logarithmic height is bounded by $(\log{k})/n$, and hence goes to zero. $\endgroup$ – Vesselin Dimitrov Aug 21 '14 at 4:48

$\begingroup$ OK, got it. (I got confused and thought $f$ was the minimal polynomial for the sums of roots of unity.) I'm not seeing how the lemma follows from the theorem though  in the limit as $\delta \to 0$, wouldn't one need a very large Lipschitz constant for $\lambda$? $\endgroup$ – Terry Tao Aug 21 '14 at 5:00

$\begingroup$ @TerryTao: I think you are right. The smallest admissible choice for $\delta$ is $O(1/m)$ (else the statement in the lemma is vacuous), but even this seems too small to have the lemma as written. On the other hand a much weaker statement would suffice for the desired application. Let me see if I can get the required lemma correct. $\endgroup$ – Vesselin Dimitrov Aug 21 '14 at 5:23

$\begingroup$ @TerryTao: Indeed it may not be possible to preclude in this way that no $\alpha_i$ is really close to $\zeta_n$; if this is so, it would ruin the argument. Certainly we have a crude lower bound $\zeta_n  \alpha_i > C^{n}$ (where any $C > 2$ will do for $n \gg_C 0$), but this of course is inadequate. I will try to see later if we could still say something. $\endgroup$ – Vesselin Dimitrov Aug 21 '14 at 5:36