# Mahler measure of a totally positive, expanding algebraic integer

Consider a degree-$d$ algebraic integer $\alpha$ all of whose conjugates (including itself) are real numbers greater than 1. Its Mahler measure $M(\alpha)$ is simply equal to the norm $N(\alpha)$. Since $N(\alpha-1) \geq 1$, an application of $H\ddot{o}lder's$ inequality gives the lower bound $N(\alpha) \geq 2^d$, with equality achieved only by $\alpha = 2$.

Question. Does there exist $C > 2$ such that $N(\alpha) \geq C^d$ for all $\alpha \neq 2$?

• Interesting question. I assume you're familiar with work of Smyth, Flammang, etc. on lower bounds for the Mahler measure of totally positive algebraic integers? (These give a lower bound of $C=1.722...$ for $M(\alpha)^{1/d}$ with finitely many exceptions, but valid for all totally positive integers $\alpha$, not just those with all conjugates $>1$). Apr 20, 2016 at 21:13
• On the hypothesis that $\alpha$ is algebraic of degree $d$, the value $\alpha=2$ can only arise in the case $d=1$, so the equality $N(\alpha)=2^d$ is achieved only for $d=1$, $\alpha=2$. Apr 20, 2016 at 22:58
• @BobbyGrizzard Is the "with finitely many exceptions" part effective? Apr 21, 2016 at 5:50
• @FanZheng yes. Here is one effective result: ams.org/journals/mcom/1996-65-213/S0025-5718-96-00664-3/… edit: and here is the one I was probably looking at yesterday, due to Wu and Mu (Quanwu Mu, not the OP!): sciencedirect.com/science/article/pii/S0022314X12001989 Apr 21, 2016 at 11:58
• @BobbyGrizzard NOT the OP LOL Apr 22, 2016 at 4:19

Suppose that $\alpha$ is totally real algebraic integer, and that all its conjugates are greater than $1$. Suppose also that $\alpha \ne 2$.

Note the elementary inequality for $x \in (1,\infty) \setminus \{2\}$:

$$\ln|x| \ge \frac{\ln|x-1| + \ln|x-2|}{3} + \ln C,$$

where $C = 2^{1/3} 3^{1/2} = 2.18 \ldots > 2$. (Equality holds precisely at $3 \pm \sqrt{3}$.)

Denote the conjugates of $\alpha$ by $\alpha_i$. Assuming that $\alpha \ne 1,2$, we see that, because $\alpha - 1$ and $\alpha - 2$ are non-zero algebraic integers, we have inequalities

$$\sum_{i=1}^{d} \ln | \alpha_i - 1| = \ln N(\alpha -1) \ge 0,$$ $$\sum_{i=1}^{d} \ln |\alpha_i - 2| = \ln N(\alpha -2) \ge 0.$$

Hence we deduce that $\displaystyle{\ln N(\alpha) = \sum_{i=1}^{d} \ln |\alpha_i| \ge d \cdot \ln C}$, and thus $N(\alpha) \ge C^{d}$ where $C > 2$.

Gypsum's argument is really nice. In the same spirit, we have the following inequality: $$\ln|x| \geq \frac{2\ln|x-1| + \ln|x-2|}{5} + \ln\sqrt{5}$$ which is achieved at $x = \frac{5\pm \sqrt{5}}{2}$. This way we obtain the optimal constant $C = \sqrt{5}$. Perhaps some further tweaks can yield even larger $C$ with finitely many exceptions, as in the work of Smyth and Flammang.

I wonder how far this approach can be extended. Smyth showed that for totally positive algebraic integers (whose conjugates are not necessarily greater than 1), their $M(\alpha)^{\frac{1}{d}}$ are dense beyond 1.73, which is very close to the lower bound cited by @BobbyGrizzard. Do we have a similar situation here? As a first step, it would be good to construct infinitely many $\alpha$ that give upper bound to $C$. For example, consider the $n$-th $Chebyshev$ polynomial $T_n(x)$. Then the monic polynomial $(-x)^nT_n(\frac{2}{x}-1)$ has all its roots greater than 1. In this case $N(\alpha) = 2^{2n-1}$, suggesting $C \leq 4$.

• Welcome to mathoverflow! This is really a great question, but as you are new here, please be informed that the answers are just "answers"; new questions included in answers may not get enough attention. If you want to probe further, please post a separate question; you may include a link back to this question. Apr 22, 2016 at 6:25
• One can usually extend this approach a certain amount, but not as far as the limit. The fact that equality is achieved at $x = (5+\sqrt{5})/2$ with the optimal bound is related to the fact that $x - 1$ and $x - 2$ are both units. But there is no reason this had to happen. BTW, to improve the bound beyond this point (with the exceptions $x = 2$ and $(5 \pm \sqrt{5})/2$, you simply need to add a very small multiple of the term $\log(x^2 - 5 x + 5)$. Apr 22, 2016 at 13:11
• @FanZheng Thanks for your advice :) Apr 22, 2016 at 17:33