2. Principal results

Consider a monic irreducible noncyclotomic polynomial $P(x)=\prod _{j=1}^d(x-\alpha _j)$ in $\mathbb Z[x]$ of degree $d>1$ and assume that the polynomial $\prod _{j=1}^d(x-\alpha _j^2)\in \mathbb Z[x]$ is irreducible as well. (Otherwise the Mahler measure of $P(x)$ is bounded from below through the measures of irreducible factors of the latter polynomial.) As in [Reference Dimitrov2], Dimitrov’s cyclotomicity criterion together with Kronecker’s rationality criterion and a theorem of Pólya imply that the hedgehog

$$ \begin{align*}K=K(\beta_1,\dots,\beta_n)=\bigcup_{k=1}^n[0,\beta_j] =\bigcup_{j=1}^d[0,\alpha_j^2]\cup\bigcup_{j=1}^d[0,\alpha_j^4], \end{align*} $$

whose spines originate from the origin and end up at $\alpha _j^2,\alpha _j^4$ for $j=1,\dots ,d$ , has (logarithmic) capacity (or transfinite diameter) $t(K)$ at least 1. Then Dubinin’s theorem [Reference Dubinin3] applies, which claims that $t(K)\le 4^{-1/n}\max _j|\beta _j|$ (with equality attained if and only if the hedgehog K is rotationally symmetric), and produces the estimate for since $n\le 2d$ .

When dealing with Lehmer’s problem instead, one becomes interested in estimating the ‘Mahler measure of the hedgehog’, namely the quantity $\prod _{j=1}^n\max \{1,|\beta _j|\}$ , because any nontrivial (bounded away from 1) absolute estimate for it would imply a nontrivial estimate for the Mahler measure of $P(x)$ . In this setting, Dubinin’s theorem only implies the estimate $\prod _{j=1}^n\max \{1,|\beta _j|\}\ge 4^{1/n}$ for a hedgehog of capacity at least 1, which depends on n. The Mahler measure of the rotationally symmetric hedgehog on n spines, which is optimal in Dubinin’s result, is equal to 4 (thus, independent of n), which certainly loses out to the Mahler measure $1.91445008\dots $ of the ‘Lehmer hedgehog’ attached to the polynomial $x^{10}+x^9-x^7-x^6-x^5-x^4-x^3+x+1$ but also to the measure $3.07959562\dots $ of the hedgehog constructed on Smyth’s polynomial $x^3-x-1$ . The following question arises in a natural way.

Notice that answering this question for hedgehogs of capacity exactly $1$ is sufficient, since the capacity satisfies $t(K_1)\le t(K_2)$ for any compact sets $K_1\subset K_2$ in $\mathbb C$ .

In order to approach Question 1 we use a different construction of hedgehogs outlined in Eremenko’s post on the question in [Reference Lev5] with details set out in [Reference Schmidt6]. Any hedgehog $K=K(\beta _1,\dots ,\beta _n)$ of capacity precisely $1$ is in a bijective correspondence (up to rotation!) with the set of points $z_1,\dots ,z_n$ on the unit circle with prescribed positive real weights $r_1,\dots ,r_n$ satisfying $r_1+\dots +r_n=1$ . Namely, the mapping

$$ \begin{align*}F(z)=\prod_{k=1}^n((z-z_k)(z^{-1}-\overline z_k))^{r_k} \end{align*} $$

is a Riemann mapping of the complement of the closed unit disk to the complement $\hat {\mathbb C}\setminus K$ of hedgehog. It is not easy to write down the corresponding $\beta _j$ explicitly, but for their absolute values we get

$$ \begin{align*}|\beta_j|=\max_{z\in[z_{j-1},z_j]}|F(z)| =\max_{z\in[z_{j-1},z_j]}\prod_{k=1}^n|z-z_k|^{2r_k} \quad\text{for}\; j=1,\dots,n, \end{align*} $$

where we conventionally take $z_0=z_n$ and understand $[z_{j-1},z_j]$ as arcs of the unit circle. This means that if $C\ge 1$ is the minimum of

$$ \begin{align*}\prod_{j=1}^n\max\bigg\{1,\max_{z\in[z_{j-1},z_j]}\prod_{k=1}^n|z-z_k|^{r_k}\bigg\} \end{align*} $$

taken over all n and all possible weighted configurations $z_1,\dots ,z_n$ , then $C^2$ is the minimum in Question 1.

Furthermore, in the spirit of [Reference Konyagin and Lev4] observe that from continuity considerations it suffices to compute the required minimum C for rational positive weights $r_1,\dots ,r_n$ . Assuming the latter and writing $r_j=a_j/m$ for positive integers $a_1,\dots ,a_n$ and $m=a_1+\dots +a_n$ , we look for the mth root of the minimum of

$$ \begin{align*}\prod_{j=1}^n\max\bigg\{1,\max_{z\in[z_{j-1},z_j]}\prod_{k=1}^n|z-z_k|^{a_k}\bigg\} =\prod_{j=1}^m\max\bigg\{1,\max_{z\in[z_{j-1}',z_j']}\prod_{k=1}^m|z-z_k'|\bigg\}, \end{align*} $$

where $z_1',z_2',\dots ,z_m'$ is the multi-set

$$ \begin{align*}\underbrace{z_1,\dots,z_1}_{a_1 \;\text{times}}, \; \underbrace{z_2,\dots,z_2}_{a_2 \;\text{times}}, \; \dots, \; \underbrace{z_n,\dots,z_n}_{a_n \;\text{times}} \end{align*} $$

with prescribed weights all equal to 1. This means that it is enough to compute the minimum for the case of equal weights, $r_1=\dots =r_n=1/n$ , and we may give the following alternative formulation of Question 1.

Question 2. What is the minimum $C_n$ of

$$ \begin{align*}\prod_{j=1}^n\max\bigg\{1,\max_{z\in[z_{j-1},z_j]}\prod_{k=1}^n|z-z_k|\bigg\}^{1/n} \end{align*} $$

taken over all configurations of points $z_1,\dots ,z_n$ on the unit circle $|z|=1$ ? The points are not required to be distinct and $[z_{j-1},z_j]$ is understood as the corresponding arc of the circle, $z_0$ is identified with $z_n$ .

Though there is no explicit requirement on the order of precedence, the minimum corresponds to the successive locations of $z_1,\dots ,z_n$ on the circle.

A comparison with Dubinin’s result suggests that good candidates for the minima in Question 2 may originate from configurations in which all factors in the defining product but one are equal to $1$ . In our answer to the question we show that this is essentially the case by computing the related minima $C_n^*$ explicitly.

Theorem 3. For the quantity $C_n$ we have the inequality $C_n\le C_n^*$ , where $C_n^*=(T_n(2^{1/n}))^{1/n}$ and

$$ \begin{align*}T_n(x) =\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}(x^2-1)^kx^{n-2k} \end{align*} $$

denotes the nth Chebyshev polynomial of the first kind.

Theorem 4. For the quantity $C_n^*$ in Theorem 3 we have the asymptotic expansion

$$ \begin{align*}C_n^*=1 + \nu - \frac14\nu^3 + \frac5{96}\nu^5 - \frac1{128}\nu^7 + O(\nu^9) \end{align*} $$

in terms of $\nu =\sqrt {(\log 4)/n}$ , as $n\to \infty $ . In particular, $(C_n^*)^{\sqrt n}\to e^{\sqrt {\log 4}}$ and $C_n^*\to 1$ as $n\to \infty $ .

Thus, our results imply that the minimum in Question 1 is equal to $1$ , meaning that an analogue of Lehmer’s problem in an analytic setting is trivial. This has no consequences for Lehmer’s problem itself, as we are not aware of a recipe to cook up polynomials in $\mathbb Z[x]$ from optimal (or near optimal) configurations of $z_1,\dots ,z_n$ on the unit circle.

3. Proofs

Proof of Theorem 3

We look for a configuration of the points $z_1,\dots ,z_n$ on the unit circle such that the maximum of $|Q(z)|$ , where $Q(z)=(z-z_1)\dotsb (z-z_n)$ , on all the arcs $[z_{j-1},z_j]$ but one is equal to $1$ :

$$ \begin{align*}\max_{z\in[z_{j-1},z_j]}|Q(z)|=|Q(z_j^*)|=1 \quad\text{for}\; z_j^*\in(z_{j-1},z_j), \quad\text{where}\; j=2,\dots,n. \end{align*} $$

At the same time, the kth Chebyshev polynomial $T_k(x)=2^{k-1}x^k+\dotsb $ is known to satisfy $|T_k(x)|\le 1$ on the interval $-1\le x\le 1$ , with all the extrema on the interval being either $-1$ or $1$ . Note that $T_k(x)$ has k distinct real zeros on the open interval $-1<x<1$ and satisfies $T_k(1)=(-1)^kT_k(-1)=1$ . Therefore, for $n=2k$ even,

$$ \begin{align*} Q(z)=z^k\,T_k\bigg(2^{1/k}\bigg(\frac{z+z^{-1}}2-1\bigg)+1\bigg), \end{align*} $$

we get a monic polynomial of degree n with the desired properties; its zeros $z_1,\dots ,z_n$ ordered in pairs, so that $z_{n-j}=\overline z_j=z_j^{-1}$ for $j=1,\dots ,k$ , correspond to the real zeros $2^{1/k}((z_j+z_j^{-1})/2-1)+1$ of the polynomial $T_k(x)$ on the interval $-1<x<1$ . Then

$$ \begin{align*} \max_{z\in[z_n,z_1]}|Q(z)|=\max_{|z|=1}|Q(z)|=|Q(-1)| =|T_k(1-2^{1+1/k})|=T_k(2^{1+1/k}-1) =T_{2k}(2^{1/(2k)}), \end{align*} $$

where the duplication formula $T_k(2x^2-1)=T_{2k}(x)$ was applied.

The duplication formula in fact allows one to write the very same polynomial $Q(z)$ in the form

$$ \begin{align*}Q(z)=\pm (-z)^{n/2}\, T_n(2^{1/n - 1} {\textstyle\sqrt{2 - (z+z^{-1})}}\,), \end{align*} $$

and this formula gives the desired polynomial, monic and of degree n, for n of any parity. If we set $k=\lfloor (n+1)/2\rfloor $ , the zeros $z_1,\dots ,z_n$ of $Q(z)$ pair as before, that is, $z_{n-j}=\overline z_j=z_j^{-1}$ for $j=1,\dots ,k$ , with the two zeros merging into one, $z_{(n+1)/2}=1$ for $j=k$ when n is odd, so that $2^{1/n - 1} \sqrt {2- \smash [b]{(z_j+z_j^{-1})}}$ for $j=1,\dots ,k$ are precisely the k real zeros of the polynomial $T_n(x)$ on the interval $0\le x<1$ . This leads to the estimate

$$ \begin{align*}\max_{z\in[z_n,z_1]}|Q(z)|=\max_{|z|=1}|Q(z)|=|Q(-1)|=T_n(2^{1/n}) \end{align*} $$

for both even and odd values of n.

Finally, we remark that the uniqueness of $Q(z)$ , up to rotation, follows from the extremal properties of the Chebyshev polynomials.

Proof of Theorem 4

For this part we cast the Chebyshev polynomial $T_n(x)$ in the form

$$ \begin{align*}T_n(x) =\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}2 =\frac{x^n}2\cdot \textstyle((1+\sqrt{1-x^{-2}})^n+(1-\sqrt{1-x^{-2}})^n) \end{align*} $$

leading to

$$ \begin{align*}T_n(2^{1/n}) =\textstyle(1+\sqrt{1-e^{-\nu^2}})^n+(1-\sqrt{1-e^{-\nu^2}})^n \end{align*} $$

in the notation $\nu =\sqrt {(\log 4)/n}$ . Since

$$ \begin{align*} {\textstyle\sqrt{1-e^{-\nu^2}}} &=\bigg(\sum_{k=1}^\infty\frac{(-1)^{k-1}\nu^{2k}}{k!}\bigg)^{\! 1/2} =\nu\bigg(1+\sum_{k=2}^\infty\frac{(-1)^{k-1}\nu^{2k-2}}{k!}\bigg)^{\! 1/2} \\ &=\nu\cdot\bigg(1 - \frac14\nu^2 + \frac5{96}\nu^4 - \frac1{128}\nu^6 + \frac{79}{92160}\nu^8 - \frac3{40960}\nu^{10}+O(\nu^{12})\bigg), \end{align*} $$

we conclude that the term $(1-\sqrt {1-e^{-\nu ^2}})^n=O(\varepsilon ^n)$ for any choice of positive $\varepsilon <1$ , hence

$$ \begin{align*} (T_n(2^{1/n}))^{1/n} =(1+\textstyle\sqrt{1-e^{-\nu^2}})\cdot(1+O(\varepsilon^n)), \end{align*} $$

and the required asymptotics follows.

4. Speculations

Dimitrov’s estimate $t(K)\ge 1$ for the capacity of the hedgehog $K=K(\beta _1,\dots ,\beta _n)$ assigned to a polynomial in $\mathbb Z[x]$ is not necessarily sharp, and one would rather expect to have $t(K)\ge t$ for some $t>1$ . By replacing the polynomial in the proof of Theorem 3 with

$$ \begin{align*}Q(z)=\pm (-z)^{n/2}\, T_n(2^{1/n - 1}t {\textstyle\sqrt{2 - (z+z^{-1})}}\,) \end{align*} $$

and assuming (or, better, believing!) that the corresponding minimum in Question 2 is indeed attained in the case when all but one of the factors are equal to 1, we conclude that the minimum is equal to $(T_n(2^{1/n}t))^{\! 1/n}$ . The asymptotics of the Chebyshev polynomials then converts this result into the answer

$$ \begin{align*}\inf_{\substack{n=1,2,\dots\\ K = K(\beta_1, \dots, \beta_n)\\ t(K) \ge t}}\prod_{j=1}^n\max\{1,|\beta_j|\} \ge t+\textstyle\sqrt{t^2-1} \end{align*} $$

to the related version of Question 1. This is slightly better, when $t>1$ , than the trivial estimate of the infimum by t from below.

In another direction, one may try to associate hedgehogs K to polynomials in a different (more involved!) way, to achieve some divisibility properties for the Hankel determinants $A_k$ that appear in the estimation $t(K)\ge \limsup _{k\to \infty }|A_k|^{1/k^2}$ of the capacity on the basis of Pólya’s theorem. Such an approach has the potential to lead to some partial (‘Dobrowolski-type’) resolutions of Lehmer’s problem. Notice, however, that the bound for $t(K)$ in Pólya’s theorem is not sharp: numerically, the Hankel determinants $A_k=\det _{0\le i,j<k}(a_{i+j})$ constructed on (Dimitrov’s) irrational series

for Smyth’s polynomial $x^3-x-1=(x-\alpha _1)(x-\alpha _2)(x-\alpha _3)$ satisfy $|A_k|\le C^k$ for some $C<2.5$ and all $k\le 150$ , so that it is likely that $\limsup _{k\to \infty }|A_k|^{1/k^2}=1$ in this case.