Recall that Pisot numbers are algebraic integers greater than $1$, whose other Galois conjugates have modulus $<1$. The set of Pisot numbers is usually denoted $S$. It is known that $S$ is denumerable and closed (Salem). Alors, the sequence of derived sets $S',S'', ...$ does not terminate. The smallest accumulation point of $S$ is the golden ratio $\phi$.

Let $\lambda$ be an accumulation point in $S$. Therefore it is the limit of a sequence of Pisot numbers $\mu_n\ne\lambda$. It is clear that the degree of $\mu_n$ tends to $+\infty$, whereas its norm remains bounded as $n\rightarrow+\infty$. My question is about the Galois conjugates of $\mu_n$; most of them must be close to the unit circle, because their product equals $N(\mu_n)/\mu_n$. Given a Galois conjugate $\tau$ of $\lambda$, does there exist a Galois conjugate $\tau_n$ of $\mu_n$ such that $\tau_n\rightarrow\tau$ ? What is the statistics of the Galois conjugates of $\mu_n$ as $n\rightarrow+\infty$ ?

Example: the multinacci number of degree $d$ is the root $a_d>1$ of the polynomial $X^d-X^{d-1}-\cdots-X-1$. When $d\rightarrow+\infty$, one has $a_d\rightarrow2$. Then I have a proof that the empirical measure $$m_d:=\frac1d\sum_{a\sim a_d}\delta_a$$ where the sum runs over all the Galois conjugates of $a_d$ (including itself) converges vaguely towards the uniform measure over the unit circle.

Edit. The same property holds true in the situation depicted by Dufresnoy and Pisot. Let $P\in{\mathbb Z}[X]$ be the minimal polynomial of $\lambda$, unitary. Let $A\in{\mathbb Z}[X]$ be such that $|A(z)|\le|P(z)|$ on $\mathbb T$, the equality arising only at finitely many points. Define the polynomials $P_n^\pm(X)=X^nP(X)\mp A(X)$. For infinitely many pairs $(n,\pm)$, $P_n^\pm$ has only one root $\lambda_n^\pm$ away from the unit disk, and none over $\mathbb T$. This Pisot number tends to $\lambda$, and the empirical measure defined as above tends to $\frac1{2\pi}d\theta$. It is thus tempting to conjecture that in every situation, the empirical measure tends to the uniform measure.

  • $\begingroup$ What is the norm of a Pisot number? $\endgroup$ Mar 19, 2012 at 17:11
  • $\begingroup$ @Anthony. This is the norm of an algebraic integer, the product of the number with its Galois conjugates; therefore a rational integer. $\endgroup$ Mar 19, 2012 at 17:36
  • $\begingroup$ When you say "Alors" up there, you clearly do not mean "therefore". Perhaps you mean "indeed", since what follows is quite a strong assertion. $\endgroup$ Mar 19, 2012 at 17:52

1 Answer 1


The answer to the first question is "certainly not". Consider the polynomials $P_d(X)=X^d-4X^{d-1}-X^{d-2}-1$. They have $d-1$ roots in the unit disk by Rouche, so their positive real roots are Pisot numbers. Also, the positive real roots of $P_d$ tend to the larger root of $X^2-4X-1$, which is a Pisot number. However, any disk or radius $r<1$ is eventually free of roots of $P_d$, so nothing tends to the conjugate.

  • $\begingroup$ I accept your response, because it sets the first question. I should like however that somebody give an answer about the second question (statistics). $\endgroup$ Mar 21, 2012 at 10:19
  • $\begingroup$ Sure. If I could answer both, I would answer both. I just didn't want to keep you waiting, so I posted the "half-answer" I had. Feel free to unaccept if somebody figures out the distribution problem :). $\endgroup$
    – fedja
    Mar 21, 2012 at 10:32
  • $\begingroup$ For what it's worth, the second question is equivalent to asking if the coefficient $a_1$ in every Pisot polynomial $X^d+a_1 X^{d-1}+\dots+a_d$ is much less than $d$ when $d\to\infty$ and the corresponding Pisot number stays bounded. Since all your examples involve polynomials with small coefficients ($a_1$ can be comparable to $d$ only if $\max|a_j|$ is exponential in $d$), they do not really tell much about the general case... $\endgroup$
    – fedja
    Mar 21, 2012 at 17:32

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