Iterated monodromy groups of rational functions and periodic points over finite fields

Abstract

Let q be a prime power and $\phi$ a rational function with coefficients in a finite field ${\mathbb {F}}_q$. For $n \ge 1$, each element of ${\mathbb {P}}^1({\mathbb {F}}_{q^n})$ is either periodic or strictly preperiodic under iteration of $\phi$. Denote by $a_n$ the proportion of periodic elements. Little is known about how $a_n$ changes as n grows, unless $\phi$ is a power map or Chebyshev polynomial. We give the first results on this question for a wider class of rational functions: $a_n$ has lim inf 0 when q is odd and $\phi$ is quadratic and neither Lattès nor conjugate to a one-parameter family of exceptional maps. We also show that $a_n$ has limit 0 when $\phi$ is a non-Chebyshev quadratic polynomial with strictly preperiodic finite critical point and q is an odd square. Our methods yield additional results on periodic points for reductions of post-critically finite (PCF) rational functions defined over number fields. Our arguments are, at their core, complex dynamical. The difficulty of understanding $a_n$ in general is that ${\mathbb {P}}^1({\mathbb {F}}_{q^n})$ is a finite set with no ambient geometry. In fact, $\phi$ can be lifted to a PCF rational map on the Riemann sphere, where we show that $a_n$ can be bounded by counting elements of the iterated monodromy group (IMG) that act with fixed points at all levels of the tree of preimages. Using a martingale convergence theorem, we translate the problem to determining whether certain IMG elements exist. This in turn can be decisively addressed using the expansion of PCF rational maps in the orbifold metric.

Appendix A: Dynamically exceptional rational functions over finite fields

In this appendix we study the exceptions to Theorem 1.6. In particular, we discuss Lattès maps over finite fields and give a characterization of dynamically exceptional quadratic rational functions over an arbitrary field of characteristic different from 2.

Let K be a field with fixed algebraic closure ${\overline{K}}$, and let $\phi \in K(x)$. For $\alpha \in {\mathbb {P}}^1({\overline{K}})$ with $\alpha \ne \infty$ and $\phi (\alpha ) \ne \infty$, the ramification index $e_\phi (\alpha )$ of $\phi$ at $\alpha$ is the multiplicity of $\alpha$ as a root of the numerator of $\phi (x) - \phi (\alpha )$. If $\alpha = \infty$ or $\phi (\alpha ) = \infty$, then $e_\phi (\alpha ) = e_{\mu \circ \phi \circ \mu ^{-1}} (\mu (\alpha ))$, where $\mu$ is a Mobius transformation mapping both $\alpha$ and $\phi (\alpha )$ away from infinity. We call $\alpha$ a critical point for $\phi$ if $e_\phi (\alpha ) > 1$.

Define the ramification portrait of $\phi$ to be the edge-labeled directed graph whose vertex set is the union of the orbits of all critical points of $\phi \in {\mathbb {P}}^1({\overline{K}})$, and where each vertex $\alpha$ has an arrow to $\phi (\alpha )$ with label $e_\phi (\alpha )$. Note that the graph is not vertex-labeled, so we do not record the specific points involved. For instance, if K has characteristic not equal to 2, then $\phi (x) = (x^2 - 2)/x^2$ has critical points 0 and $\infty$, with $0 \rightarrow \infty \rightarrow 1 \rightarrow -1 \rightarrow -1$. This gives ramification portrait $\bullet \xrightarrow {2} \bullet \xrightarrow {2} \bullet \rightarrow \bullet \circlearrowleft$. Because we deal here with quadratic maps, for which every critical point $\alpha$ has $e_\phi (\alpha ) = 2$, we rewrite this as

$$\begin{aligned} \bullet \rightarrow \bullet \rightarrow \circ \rightarrow \circledcirc , \end{aligned}$$

(A.1)

where $\bullet$ denotes a critical point, $\circ$ a non-critical point, and $\circledcirc$ a non-critical fixed point. Denote a critical fixed point by $\odot$. As another example, if $\phi$ is the degree-2 Chebyshev polynomial $x^2 - 2$, then $\phi$ has ramification portrait

$$\begin{aligned} \odot \qquad \bullet \rightarrow \circ \rightarrow \circledcirc . \end{aligned}$$

(A.2)

We note that the ramification portraits in (A.1) and (A.2) uniquely determine $\phi$ up to Mobius conjugation.

Because we work over an arbitrary field of characteristic not equal to 2, we define $\phi \in K[x]$ to be a Lattès map if there exists a function $r: {\mathbb {P}}^1({\overline{K}}) \rightarrow {{\mathbb {Z}}}$ such that

$$\begin{aligned} r(\phi (\alpha )) = e_\phi (\alpha )r(\alpha )\quad \text {and}\quad r(\alpha ) = 1\text { outside of }P_\phi , \end{aligned}$$

(A.3)

where we recall that $P_\phi$ is the post-critical set for $\phi$. When K is a finite field, these are precisely the liftable maps that lift to Lattès maps defined over ${\mathbb {C}}$ (see Sect. 5 for a definition of lifting). This is because over ${\mathbb {C}}$, the existence of the function r is equivalent to the usual definition of Lattès maps as given by a finite quotient of a self-map of an elliptic curve; see [14, Theorem 4.1].

Proposition A.1

Let K be a field of characteristic not equal to 2, and let $\phi \in K(x)$ have degree 2. Then $\phi$ is a Lattès map if and only if the ramification portrait of $\phi$ is the one in (A.1) or one of the following:

$$\begin{aligned} \bullet \rightarrow \circ \rightarrow \circledcirc \; \; \; \bullet \rightarrow \circ \rightarrow \circledcirc , \quad \quad \quad \bullet \rightarrow \circ \rightarrow \circ \rightleftarrows \circ \leftarrow \circ \leftarrow \bullet \end{aligned}$$

(A.4)

Proof

Let $\Delta = \{\alpha \in {\mathbb {P}}^1({\overline{K}}): r(\alpha ) > 1\}$, and recall that we denote the set of critical points of $\phi$ in ${\mathbb {P}}^1(K)$ by $C_\phi$. By the definition of r, we have $\Delta = P_\phi$ and $\phi ^{-1}(\Delta ) = \Delta \cup C_\phi$. Thus

$$\begin{aligned} 2\#\Delta = \!\sum _{\alpha \in \phi ^{-1}(\Delta )}\! e_\phi (\alpha ) \le \#\Delta + 2\#C_\phi , \end{aligned}$$

(A.6)

with equality if and only if $\Delta$ and $C_\phi$ are disjoint. Because K has characteristic not equal to 2, $\#C_\phi = 2$, and we conclude from (A.6) that $\#\Delta \le 4$, with equality if and only if $\Delta \cap C_\phi = \emptyset$.

Suppose that $\#\Delta < 4$, and let $\gamma \in \Delta \cap C_\phi$. Observe that $\phi ^{-1}(\gamma ) \subset C_\phi \cup P_\phi$ and thus if $\phi ^{-1}(\gamma )$ contains no critical points, then $\phi ^{-1}(\gamma )$ consists of two post-critical points. But there is only one critical point of $\phi$ besides $\gamma$, so it is impossible for both points in $\phi ^{-1}(\gamma )$ to be post-critical. Hence $\phi ^{-1}(\gamma )$ consists of a critical point. Now $\gamma$ cannot be periodic, for otherwise $r(\gamma )$ is not well-defined. Hence if $\phi (\gamma )$ is periodic, then it is a fixed point. But then $\phi ^{-1}(\phi (\gamma ))$ contains both $\gamma$ and $\phi (\gamma )$, which is impossible. Hence $\phi (\gamma )$ cannot be periodic. Because $\#\Delta \le 3$, it must be the case that $\phi ^2(\gamma )$ is a fixed point, and we have ramification portrait (A.1).

Suppose now that $\#\Delta = 4$, and thus $P_\phi \cap C_\phi = \emptyset$. Because $\phi$ cannot have a periodic critical point, $P_\phi$ must contain a cycle, and for each $\alpha$ in this cycle, $\phi ^{-1}(\alpha )$ cannot contain a critical point, as otherwise $\phi ^{-1}(\alpha )$ consists only of a critical point, which must then be periodic. It follows that the length of this cycle can be at most 2. If $P_\phi$ contains a 2-cycle, one easily checks that the only possible ramification portrait is the second one in (A.4).

Now a fixed point in $P_\phi$ cannot have a pre-image that is a critical point, and hence $P_\phi$ can contain at most two fixed points. If there are exactly two, then we must have the first ramification portrait in (A.4). If there is only one, then we must have the ramification portrait in (A.5). $\square$

We now describe quadratic Lattès maps over a field of characteristic not equal to 2. We use the normal form $\phi (x) = (x^2 + a)/(x^2 + b), a \ne b$, which exists for every degree-2 rational function except those conjugate over ${\overline{K}}$ to $x^{\pm 2}$, and can be obtained by conjugating a map’s two critical points to 0 and $\infty$, and then conjugating again so $\phi (\infty ) = 1$. We observe that this conjugation is defined over K if and only if the map’s critical points lie in K; otherwise the conjugation is over a quadratic extension of K. The normal form is unique except that if $ab \ne 0$, then conjugation by $x \mapsto a/(bx)$ takes $(x^2 + a)/(x^2 + b)$ to $(x^2 + (a^2/b^3))/(x^2 + (a/b^2))$. This is the normal form found in [16], and is related to the normal form for critically marked quadratic rational functions given in [12, Section 6].

Proposition A.2

If K is a field of characteristic not equal to 2, then every degree-2 Lattès map is conjugate (over ${\overline{K}}$) to one of the following:

$$\begin{aligned} \frac{x^2-2}{x^2}, \qquad \frac{x^2+\alpha _1}{x^2-\alpha _1}, \qquad \frac{x^2+\alpha _2}{x^2-\alpha _2}, \qquad \frac{x^2 + \alpha _3}{x^2 - (\alpha _3 + 2)}, \qquad \frac{x^2 + \frac{1}{\alpha _3}}{x^2 - \frac{1}{\alpha _3+2}}, \end{aligned}$$

(A.7)

where $\alpha _1$ is a root of $y^2 + 1$ (in ${\overline{K}})$, $\alpha _2$ is a root of $y^2 - 2y - 1$, and $\alpha _3$ is a root of $y^2 + 5y + 8$.

Remark

The two maps $\frac{x^2+\alpha _2}{x^2-\alpha _2}$, where $\alpha _2$ is either root of $y^2 - 2y - 1$, are in fact conjugate to each other by $x \mapsto -1/x$. Otherwise, no two maps in (A.7) are conjugate. Hence there are 8 conjugacy classes of Lattès maps (over ${\overline{K}}$) if K has characteristic not equal to 7. If K has characteristic 7, then $y^2 + 5y + 8$ has only one root in ${\overline{K}}$, and hence there are only 6 conjugacy classes of Lattès maps.

Proof

Let $\phi \in K(x)$ be a degree-2 Lattès map. It follows from Proposition A.1 that $\phi$ is not conjugate to $x^{\pm 2}$, and hence we may write $\phi (x) = (x^2 + a)/(x^2 + b)$ for some $a,b \in {\overline{K}}$ with $a \ne b$. Each of the ramification portraits described in Proposition A.1 then gives rise to two polynomial conditions on a and b. For instance, the portrait in (A.5) forces $\phi ^2(0) = \phi ^2(\infty )$, which implies $b = -a$. The same portrait implies $\phi ^4(\infty ) = \phi ^3(\infty )$, which gives $(a^2 + 1)(a^2 - 2a - 1)=0$. The ramification portrait (A.1) leads to the first map in (A.7), and the portraits in (A.4) lead to the fourth and fifth maps in (A.7), respectively. $\square$

We now give our characterization of dynamically exceptional quadratic rational functions.

Proposition A.3

Let K be a field of characteristic $\ne 2$, and let $\phi \in K(x)$ have degree 2. Then $\phi$ is dynamically exceptional if and only if $\phi$ is a Lattès map or conjugate over ${\overline{K}}$ to $(x^2 + a)/(x^2 - (a+2))$ for some $a \in {\overline{K}}$. In particular, the only dynamically exceptional quadratic polynomials are conjugates of the degree-2 Chebyshev polynomial.

Remark

Maps conjugate to the degree-2 Chebyshev polynomial, as well as Lattès maps with ramification portrait (A.4), are conjugate to $(x^2 + a)/(x^2 - (a+2))$ for appropriate $a \in {\overline{K}}$.

Proof

By definition, there is $\Gamma \subset {\mathbb {P}}^1({\overline{K}})$ with $\phi ^{-1}(\Gamma ) {\setminus } C_\phi = \Gamma$. This implies that $\Gamma \subseteq \phi ^{-1}(\Gamma )$ and $\Gamma \cap C_\phi = \emptyset$. Hence

$$\begin{aligned} 2\#\Gamma = \sum _{\alpha \in \phi ^{-1}(\Gamma )} e_\phi (\alpha ) = \#\Gamma + 2\#(\phi ^{-1}(\Gamma ) \cap C_\phi ), \end{aligned}$$

(A.8)

and it follows that $\#\Gamma \in \{2,4\}$, according to whether $\#(\phi ^{-1}(\Gamma ) \cap C_\phi )$ is 1 or 2.

First suppose that $\#\Gamma = 2$ and $\phi ^{-1}(\Gamma )$ contains a single critical point c. Because $\phi (\Gamma ) \subseteq \Gamma$, c cannot be periodic, for then $c \in \Gamma$. Similarly, $\phi (c)$ cannot be periodic, for then its unique preimage c must be periodic as well. Thus $\phi ^2(c)$ is a fixed point for $\phi$, and after conjugation we may assume $c = \infty$, $\phi (c) = 1$, and $\phi ^2(c) = -1$, giving the map $(x^2 + a)/(x^2 - (a+2))$ for some $a \in {\overline{K}}$. We remark that any map with ramification portrait (A.4) or (A.2) is a special case. Moreover, the only polynomial maps in this family are those with ramification portrait (A.2), namely those conjugate to the degree-2 Chebyshev polynomial.

Now suppose that $\#\Gamma = 4$, and $\phi ^{-1}(\Gamma )$ contains both critical points of $\phi$, i.e., $\phi ^{-1}(\Gamma ) = \Gamma \cup C_\phi$. Then we may define a function $r: {\mathbb {P}}^1({\overline{K}}) \rightarrow {{\mathbb {Z}}}$ satisfying (A.3) by taking $r(\alpha ) = 2$ for $\alpha \in \Gamma$ and $r(\alpha ) = 1$ for $\alpha \not \in \Gamma$. Hence $\phi$ is a Lattès map. The last assertion of the Proposition follows by noting that no map with one of the ramification portraits in Proposition A.1 is conjugate to a polynomial. $\square$

In general we expect a Lattès map $\phi$ defined over a finite field ${\mathbb {F}}_q$ to satisfy

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{\#\textrm{Per}(\phi , {\mathbb {P}}^1({\mathbb {F}}_{q^{n}}))}{q^n+1} > 0, \end{aligned}$$

much as happens with Chebyshev polynomials [11]. Using work of Ugolini [20], we prove this happens in a certain case:

Theorem A.4

Let $K = {\mathbb {F}}_p$ with $p \equiv 1 \bmod {4}$, and suppose that $\phi$ is conjugate over K to the Lattès map $\frac{x^2+a}{x^2-a}$, where $a \in K$ and $a^2 + 1 = 0$. Then

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{\#\textrm{Per}(\phi , {\mathbb {P}}^1({\mathbb {F}}_{p^{n}}))}{p^n+1} \ge \frac{1}{8} \end{aligned}$$

(A.9)

Remark

There are $\phi$ that are ${\overline{K}}$-conjugate to $\frac{x^2+a}{x^2-a}$, where $a \in K$ with $a^2 + 1 = 0$, but not K-conjugate to any such map. Indeed, if $\phi$ is ${\overline{K}}$-conjugate to a map of this kind, then it is K-conjugate to such a map if and only if its critical points lie in K.

Proof

Because $\phi$ is conjugate over K to a map whose critical points are defined over K, the critical points of $\phi$ must be defined over K. Applying a conjugacy that moves these critical points to $\pm 1$, we see that $\phi$ is conjugate over K to $\psi (x) = k(x + x^{-1})$, where $k^2 + \frac{1}{4} = 0$. As detailed in [20, Section 3], the map $\psi$ descends from a degree-2 endomorphism on the elliptic curve $y^2 = x^3 + x$ defined over ${\mathbb {F}}_p$, which has endomorphism ring $R:={{\mathbb {Z}}}[i]$. Moreover, because $p \equiv 1 \bmod {4}$, the two degree-2 maps in R, namely $[1 \pm i]$, are both defined over ${\mathbb {F}}_p$, and indeed have the form $(x,y) \mapsto (\psi (x), y \tau (x))$ with $\tau (x) = c(x^2 - 1)/x^2 \in {\mathbb {F}}_p(x)$.

Our analysis of the action of $\psi$ on ${\mathbb {P}}^1({\mathbb {F}}_{p^n})$ begins by partitioning ${\mathbb {P}}^1({\mathbb {F}}_{p^n})$ into two $\psi$-invariant sets which, by the Hasse bound, have approximately equal size when $p^n$ is large. Let S be the three roots of $x^3 + x$, which lie in ${\mathbb {F}}_p$ since $p \equiv 1 \bmod {4}$. Set

$$\begin{aligned} A_n = {\left\{ \begin{array}{ll} \{x \in {\mathbb {F}}_{p^n} : \text {there is }y \in {\mathbb {F}}_{p^n}\text { with }(x,y) \in E({\mathbb {F}}_{p^n})\} \cup \{\infty \} &{} \text {if }\sqrt{2} \in {\mathbb {F}}_{p^n} \\ \{x \in {\mathbb {F}}_{p^n} : \text {there is }y \in {\mathbb {F}}_{p^n}\text { with }(x,y) \in E({\mathbb {F}}_{p^n})\} {\setminus } S &{} \text {if }\sqrt{2} \not \in {\mathbb {F}}_{p^n} \end{array}\right. } \end{aligned}$$

and take $B_n = {\mathbb {P}}^1({\mathbb {F}}_{p^n}) {\setminus } A_n$.

Because endomorphisms of E preserve $E({\mathbb {F}}_{p^n})$, we immediately have $\psi (A_n) \subseteq A_n$ if $\sqrt{2} \in {\mathbb {F}}_{p^n}$. If $\sqrt{2} \not \in {\mathbb {F}}_{p^n}$ and $\alpha \in A_n$, then $\psi (\alpha ) \in A_n$ unless $\psi (\alpha ) \in S$. But $\psi ^{-1}(S) = S \cup \{\pm 1\}$, and $\pm 1 \not \in A_n$ since $\sqrt{2} \not \in {\mathbb {F}}_{p^n}$. Thus $\psi ^{-1}(S) \cap A_n = \emptyset$. Suppose now that $\alpha \in B_n$, and let $\beta$ satisfy $(\alpha , \beta ) \in E({\overline{{\mathbb {F}}}}_{p})$. The y-coordinate of $[1\pm i](\alpha ,\beta )$ has the form $\beta \tau (\alpha )$. But $\tau (\alpha ) \in {\mathbb {F}}_{p^n}$, so $\beta \tau (\alpha ) \in {\mathbb {F}}_{p^n}$ if and only if $\beta \in {\mathbb {F}}_{p^n}$ or $\tau (\alpha ) = 0$ (i.e., $\alpha = \pm 1$). If $\sqrt{2} \in {\mathbb {F}}_{p^n}$, then $\{\pm 1\} \cap B_n = \emptyset$, whence $\psi (B_n) \subseteq B_n$. If $\sqrt{2} \not \in {\mathbb {F}}_{p^n}$, then the entire orbits of $\pm 1$ under $\psi$ are contained in $B_n$, and so again we have $\psi (B_n) \subseteq B_n$.

If we put $f(n) = (\#A_n)/(p^n+1)$ and $g(n) = (\#B_n)/(p^n+1)$, then the Hasse bound implies that both f(n) and g(n) are $1/2 + O(p^{-n/2})$. In particular,

$$\begin{aligned} \lim _{n \rightarrow \infty } \frac{\#A_n}{p^n+1} = \lim _{n \rightarrow \infty } \frac{\#B_n}{p^n+1} = \frac{1}{2}. \end{aligned}$$

(A.10)

Let $\pi _p \in R$ denote the Frobenius endomorphism of E (which is given explicitly by $(r + \sqrt{r^2 - 4p})/2$ where $r = p + 1 - \#E({\mathbb {F}}_p)$), and let ${{\mathfrak {p}}}$ be the ideal $(1+i)$ of R. Theorem 3.5 of [20] implies that each periodic point in $A_n$ (resp. $B_n$) is the root of a complete binary rooted tree whose depth is given by $v_{{\mathfrak {p}}}(\pi _p^n - 1)$ (resp. $v_{{\mathfrak {p}}}(\pi _p^n + 1))$, where $v_{{\mathfrak {p}}}$ denotes the ${{\mathfrak {p}}}$-adic valuation. The only exception is the fixed point at $\infty$, whose tree includes the critical points $\pm 1$ but otherwise is a complete binary tree with depth given as in the previous sentence. We have

$$\begin{aligned} 2 = v_{{\mathfrak {p}}}(2) = v_{{\mathfrak {p}}}((\pi _p + 1) - (\pi _p - 1)) \ge \min \{v_{{\mathfrak {p}}}(\pi _p+1), v_{{\mathfrak {p}}}(\pi _p-1)\}, \end{aligned}$$

and it follows that either $A_n$ or $B_n$ is composed of periodic points for $\psi$, each one mapped to by a binary tree of non-periodic points of depth at most 2. Without loss of generality, say that $A_n$ satisfies this condition. Then

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{\#\textrm{Per}(\psi , A_n)}{\#A_n} \ge 1/4. \end{aligned}$$

(A.11)

Combining (A.10) and (A.11) gives

$$\begin{aligned} \liminf _{n \rightarrow \infty } \frac{\#\textrm{Per}(\psi , {\mathbb {P}}^1({\mathbb {F}}_{p^{n}}))}{p^n+1}&\ge \liminf _{n \rightarrow \infty } \frac{\#\textrm{Per}(\psi , A_n)}{p^n+1} \\&= \liminf _{n \rightarrow \infty } \left( \frac{\#\textrm{Per}(\psi , A_n)}{\#A_n} \cdot \frac{\#A_n}{p^n+1} \right) \\&\ge \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}. \end{aligned}$$

$\square$

To illustrate the results of this section, we give some further discussion of the maps in Table 1 on p. 2. This collection of maps in fact contains complete information on periodic point counts for all quadratic $\phi \in {\mathbb {F}}_3(x)$ with critical points in ${\mathbb {F}}_3$. First note that the periodic points of $\phi (x) = x^2$ are the same as those of $\phi (x) = 1/x^2$. Any other quadratic $\phi \in {\mathbb {F}}_3(x)$ with critical points in ${\mathbb {F}}_3$ is conjugate over ${\mathbb {F}}_3$ to a map of the form $\phi (x) = (x^2-a)/(x^2-b)$ with $a,b \in {\mathbb {F}}_3$ and $a \ne b$. The second map in Table 1, namely $\phi (x) = x^2 - 1$ is ${\mathbb {F}}_3$-conjugate to map given by $(a,b) = (0,1)$, and is not dynamically exceptional. The third map is the Chebyshev polynomial $x^2 - 2$, which is conjugate to the map $(a,b) = (0,2)$. The fourth map in Table 1 is $(a,b) = (1,0)$, which gives a Lattès map with ramification portrait (A.1). The fifth and sixth maps are $(a,b) = (1,2)$ and $(a,b) = (2,0)$, which are not dynamically exceptional. The only other possibility for (a, b) is (2, 1), which gives a map that is ${\mathbb {F}}_3$-conjugate to the case $(a,b) = (1,2)$.