A cubic PCF polynomial

# A large arboreal Galois representation for a cubic postcritically finite polynomial

## Abstract.

We give a complete description of the arboreal Galois representation of a certain postcritically finite cubic polynomial over a large class of number fields and for a large class of basepoints. This is the first such example that is not conjugate to a power map, Chebyshev polynomial, or Lattès map. The associated Galois action on an infinite ternary rooted tree has Hausdorff dimension bounded strictly between that of the infinite wreath product of cyclic groups and that of the infinite wreath product of symmetric groups. We deduce a zero-density result for prime divisors in an orbit under this polynomial. We also obtain a zero-density result for the set of places of convergence of Newton’s method for a certain cubic polynomial, thus resolving the first nontrivial case of a conjecture of Faber and Voloch.

Benedetto]rlbenedetto@amherst.edu Faber]awfaber@super.org Hutz]hutzba@slu.edu Juul]jamie.l.rahr@gmail.com Yasufuku]yasufuku@math.cst.nihon-u.ac.jp

## 1. Introduction

Let be a field and a polynomial of degree . Consider the Galois groups of polynomials of the form

 fn(z)−x,

where , and is the -th iterate of (with the convention that ). Such groups are called arboreal Galois groups because (under certain hypotheses) they can be made to act on trees.

Let be the graph whose vertex set is

 ⨆0≤i≤nf−i(x),

and where we draw an edge from to if . Let . Clearly, acts faithfully on , so that . Provided there is no critical point of among the points of the above vertex set, the graph is a regular -ary rooted tree with root . For such , is isomorphic to the -fold iterated wreath product of the symmetric group on letters. Odoni and Juul [9, 12] showed that if and the degree are not both , and if is chosen generically (in the Zariski sense), then .

By contrast, for specific choices of and , the corresponding Galois groups may be much smaller. (See [6, §3] for a high-level explanation and [2, 4, 8, 14] for detailed examples.) Consider a polynomial that is postcritically finite, or PCF for short, meaning that all of its critical points have finite orbit under the iteration of . The simplest examples of PCF polynomials are the power maps and the Chebyshev polynomials defined by . These two examples arise from the -power endomorphism of the algebraic group , which gives a foothold on the associated arboreal Galois representation. (A third type of example, Lattès maps, arises from an endomorphism of an elliptic curve; however, Lattès maps are never conjugate to polynomials. See [15, §6.4].)

Jones and Pink [6, Thm. 3.1] have shown that for PCF maps, the Galois groups have unbounded index inside as . However, their proof does not explicitly describe . One can give an upper bound for  inside by realizing it as a specialization of , with transcendental over . The latter group may be embedded in the profinite monodromy group , where is the strict postcritical orbit; this is precisely the tack taken by Pink [14] in the case of quadratic PCF polynomials.

In this paper, we give the first complete calculation of the arboreal Galois group attached to a PCF polynomial over a number field that is not associated with an endomorphism of an algebraic group. More specifically, we describe the Galois groups for the polynomial

 f(z)=−2z3+3z2

over a number field , where is chosen to satisfy a certain local hypothesis at the primes and . In Section 2 we will define groups that fit between and its Sylow 3-subgroup  — the iterated wreath product of cyclic groups of order 3. The groups are somewhat tricky to handle because their action on the tree lacks a certain rigidity property: for , the kernel of the restriction homomorphism is not the direct product of copies of . That is, in contrast to and , the action of on one branch of the tree above is not independent of its action on another branch. Our main result is the following.

{theorem}

Let be a number field, let , and let . Suppose there exist primes and lying over and , respectively, such that , and either or . Then for each ,

1. The polynomial is irreducible over .

2. We have an isomorphism .

Let and be the corresponding inverse limits. Then the Hausdorff dimension of in is

 limn→∞log|En|log|\Aut(Tn)|=1−13 log2log6≈0.871. (1)
{remark}

In this article, we implicitly work in the category of groups with an action on the regular rooted tree . This applies, for example, to the isomorphism between and the Galois group in the theorem.

The Galois group depends a priori on the number field and the basepoint , but Theorem 1 shows that many choices of and give the same isomorphism type. One key reason is that the discriminant of the second iterate is a square:

 For any x, Disc(f2(z)−x)=[216⋅39⋅x2(x−1)2]2. (2)

This observation will be vital for forcing the Galois group of to lie inside . To fill out the entire group , we utilize ramification above the primes 2 and 3. (See the proof of Proposition 3). These two features are the only arithmetic-dynamical inputs to the theorem; the rest is general theory of groups acting on regular rooted trees.

We are also able to deduce that the geometric Galois group has the same structure:

{corollary}

Let . Let be transcendental over . For every , we have

 \Gal(fn(z)−t/¯\QQ(t))≅En.

For a polynomial , there are two profinite monodromy groups:

 Ggeomg =lim←n\Gal(gn(z)−t/¯\QQ(t))(geometric% monodromy) Garithg =lim←n\Gal(gn(z)−t/\QQ(t))(arithmetic % monodromy).

In general, one knows that . Theorem 1 and its corollary imply that for our special cubic PCF polynomial . By contrast, Pink has shown that for all quadratic PCF polynomials over the rationals [14, Thm. 2.8.4, Cor. 3.10.6]. Similar statements hold upon replacing by essentially any other number field.

While is not an iterated wreath product, it does satisfy the following self-similarity property: the action of on the subtree of height stemming from any fixed vertex at level 1 is isomorphic to . This self-similarity is a property of geometric iterated monodromy groups [11, Prop. 6.4.2], and is such a group by Corollary 1.

Odoni [13] has shown that descriptions of iterated Galois groups of this sort give rise to applications on the density of prime divisors in certain dynamically defined sequences. (See also [5, 9, 10].) More precisely, after counting elements of that fix a leaf of the tree , we have the following arithmetic application.

{theorem}

Let be a number field for which there exists an unramified prime above and above . Let , and define the sequence . Then the set of prime ideals such that

 yi≡0 or 1(mod\fP) for some i≥0

has natural density zero. In particular, the set of prime divisors of the sequence has natural density zero.

Our choice of the polynomial was originally motivated by the following conjecture of Faber and Voloch [3].

{conjecture}

[Newton Approximation Fails for 100% of Primes] Let be a polynomial of degree with coefficients in a number field and let . Define the Newton map and, for each , set . Assume the Newton approximation sequence is not eventually periodic. Let be the set of primes of for which converges in the completion to a root of . Then the natural density of the set is zero.

Faber and Voloch showed that Conjecture 1 holds for any polynomial with at most 2 distinct roots. Thus, the first nontrivial case of the conjecture is a separable cubic polynomial. For reasons explained in [3, Cor. 1.2], the simplest such cubic polynomial is , whose associated Newton map turns out to be conjugate to our polynomial . Our results therefore yield a proof of the first nontrivial case of the Faber-Voloch conjecture:

{theorem}

Let be a number field for which there exists an unramified prime over and over . Let . Choose such that the Newton iteration sequence does not encounter a root of . Then the set of primes of for which the Newton sequence converges in to a root of has natural density zero.

The first and third authors, in collaboration with several others, obtained a weak form of Theorem 1 for a wide class of polynomials [1, Thm. 4.6]. More precisely, they showed that the density of primes as in the theorem has natural density strictly less than 1.

The outline of the paper is as follows. In Section 2, we will define and discuss the group and compute the Hausdorff dimension of equation (1). We will then prove the rest of Theorem 1 in Section 3. Next, we consider the case that is the field of rational functions , proving Corollary 1 in Section 4. In Section 5, we compute the proportion of elements of that fix at least one leaf of the tree , and in Section 6, we prove Theorems 1 and 1.

## 2. Tree automorphisms

Let denote the regular ternary rooted tree with levels, as in Figure 1.

Note that has leaves and vertices. Our results and many arguments will depend on an implicit labeling of the vertices of . We will make this labeling explicit now for purposes of rigor, but we will not comment on it again afterward.

• The level of a vertex is its distance from the root.

• A vertex at level  is given a label , where . The root is given the empty label .

• No two vertices at the same level have the same label.

• The unique path from the root to the vertex with label consists of the vertices with labels .

This labeling enables us to identify certain canonical subtrees of . For example, for each , we consider the subtree that is induced by the set of vertices with labels of the form ; then is isomorphic to .

The automorphism group of the regular rooted ternary tree is isomorphic to the -fold iterated wreath product . Indeed, we may decompose as a copy of (vertices of level at most 1) with 3 copies of attached along the leaves of , so that

 \Aut(Tn)≅\Aut(Tn−1)≀\Aut(T1)≅[\fS3]n−1≀\fS3=[\fS3]n. (3)

Our labeling has the effect of fixing an isomorphism . For any elements and any , the element

 ((a1,a2,a3),b)∈\Aut(Tn−1)≀\Aut(T1)≅\Aut(Tn)

acts on the tree by first acting on the 3 copies of via , and , respectively, and then by permuting these ’s via . More precisely, if we label a vertex of by , where is a vertex of and is the vertex at level 1 that lies below , then

 ((a1,a2,a3),b).(x,i)=(ai.x,b.i).

A labeling of the leaves of induces an injection . Changing the labeling corresponds to conjugating by an element of , and so each automorphism of has a well-defined sign attached to it, corresponding to the parity of the number of transpositions needed to represent it. Thus, for each , we have a homomorphism

 \sgn:\Aut(Tn)→{±1}.
{lemma}

Let be an element of for some , where , and for . Then

 \sgn(g)=\sgn(b) 3∏i=1\sgn(ai).
###### Proof.

Partition the leaves of into three disjoint sets so that the elements of lie over leaf  of , for . Note that . With this notation, is the sign of acting as a permutation on the set .

Consider first the case that . Then permutes the elements of each separately; hence, .

Next, consider the case that for arbitrary . We have already proven the desired result if . If is a 2-cycle — say  — then the induced permutation on the leaves of decomposes as a product of disjoint 2-cycles , where and . Therefore,

 \sgn(g)=(−1)3n−1=−1=\sgn(b).

Similarly, if is a 3-cycle, then the induced permutation on the leaves of decomposes as a product of disjoint 3-cycles. Hence,

 \sgn(g)=1=\sgn(b).

Finally, we consider the general case . Define . Then . The previous two paragraphs show that

 3∏i=1\sgn(ai)=\sgn(hg)=\sgn(h) \sgn(g)=\sgn(b−1) \sgn(g).\qed

For any , we have a restriction homomorphism , where is the subtree with levels with the same root vertex as . We write for the composition of restriction followed by the sign map. Define a sequence of subgroups by the following formula:

 En={\Aut(T1)if n=1(En−1≀\Aut(T1))∩ker(\sgn2)if n≥2.

Here we use the embedding

 En−1≀\Aut(T1)↪\Aut(Tn−1)≀\Aut(T1)≅\Aut(Tn)

from equation (3). Thus, for , writing a given automorphism as , we have

 σ=((a1,a2,a3),b)∈Enif and only ifa1,a2,a3∈En−1 and \sgn2(σ)=1. (4)
{proposition}

For , we have .

###### Proof.

Since , the result is clear for . Suppose it holds for some . Let be the composition

 En≀\Aut(T1)↪\Aut(Tn+1)\lx@stackrel\sgn2→{±1}.

By definition, , and is onto because for any transposition of the leaves of . Thus,

 |En+1|=12∣∣En≀\Aut(T1)∣∣=12|\Aut(T1)|⋅|En|3=12⋅6⋅(23n−1⋅33n−12)3=23n⋅33n+1−12.\qed

Our construction of depends on an identification of with the subtree of lying above a vertex at level 1, which in turn depends on the labeling we have assigned to . In other words, is not normal in (for ): a different labeling yields a conjugate subgroup in .

{proposition}

is normal in if and only if or .

###### Proof.

We have , and has index 2 in . It remains to show that is not normal in for . To that end, we first construct some special elements of .

Define inductively for as follows:

 νn={(12)n=1((νn−1,1,1),1)n≥2.

Thus, transposes two leaves at the -th level and acts trivially on the rest of . In particular, . This yields , and by induction, it follows that for . Note further that .

Next, for fixed , define . Then by (4). However,

 νnaν−1n=((νn−1,1,1),1)((1,1,1),(123))((νn−1,1,1),1)=((νn−1,1,νn−1),(123)),

which does not belong to since for . It follows that is not normal in , as desired. ∎

Write for the infinite ternary rooted tree, which has automorphism group

 \Aut(T∞)=lim←−\Aut(Tn),

where the inverse limit is taken with respect to the restriction homomorphisms

 πm:\Aut(Tn)→\Aut(Tm)form≤n.

The recursive definition of implies that we also have restriction homomorphisms for . Passing to the inverse limit gives a subgroup

 E∞=lim←−En

of .

{corollary}

The Hausdorff dimension of in is given by equation (1).

###### Proof.

Using the facts that and that , a simple induction shows that

 |\Aut(Tn)|=63n−12. (5)

Combining this fact with Proposition 2 gives the desired result. ∎

{corollary}

has infinite index in .

Comparing the cardinalities of and , we see that they share a Sylow 3-subgroup. We can describe one such subgroup explicitly as follows. Let be the cyclic 3-subgroup of . Define a sequence of groups by the following formula:

 Hn={C3if n=1Hn−1≀C3if n≥2.

We identify with a subgroup of using the embedding

 Hn=Hn−1≀C3↪\Aut(Tn−1)≀\Aut(T1)≅\Aut(Tn).

Evidently, , the iterated wreath product. By induction, we see that

 |Hn|=33n−12, (6)

so that is a Sylow 3-subgroup of .

{proposition}

For , is a Sylow 3-subgroup of . It is normal in if and only if .

###### Proof.

For , is an index-2 subgroup of and, hence, it is normal. Next, for some , suppose we know that is a subgroup of . By their recursive definitions, to see that it suffices to show that any restricts to an even permutation on . The restriction of to has the form , where each of , and is either trivial or a 3-cycle. By Lemma 2, we conclude that . Hence, , as desired.

Since is the 3-power part of , we have proven the first statement of the proposition. It remains to show that is not normal in for .

For each , define inductively as follows:

 τn={(12)n=1((τn−1,1,1),(12))n≥2. (7)

We claim that for all . This is clear for . Suppose that it holds for some . Then if and only if its restriction to acts by an even permutation. The restriction to is given by , which has positive sign by Lemma 2.

Note that for , we have . Note also that for , we have , since the restriction of to is .

Next, for fixed , define . Then

 τnaτ−1n=((τn−1,1,1),(12))((1,1,1),(123))((1,τ−1n−1,1),(12))=((1,τ−1n−1,τn−1),(132)),

which does not belong to , since . ∎

{proposition}

The Hausdorff dimension of in is

 limn→∞log|Hn|log|\Aut(Tn)|=log3log6≈0.613.
###### Proof.

Immediate from equations (5) and (6). ∎

Since , the preceding proposition and Corollary 2 show that for large , is substantially larger than , but much smaller than .

Finally, we will need a lemma that constructs certain special elements of :

{lemma}

Let and let be any element that acts as follows:

• On the copy of with the same root as , acts by the identity.

• On each copy of rooted at a vertex of of level , acts by an even permutation of the leaves.

Then .

###### Proof.

We proceed by induction on . For , the second condition on implies that , so . Suppose that the lemma holds for , and let satisfy the given conditions. Let be the vertices of at level 1. Write for the copy of inside that is rooted at . Then restricts to an element of that satisfies the two conditions of the lemma. By the induction hypothesis, for . In addition, is the identity and, hence, is even, on . Thus, by the criterion of (4). ∎

## 3. Main Theorem

Let be a field of characteristic zero, and consider the polynomial

 f(z)=−2z3+3z2∈K[z].

The critical points of in are , , and , all of which are fixed by . Hence, is PCF and, in fact, the union of the forward orbits of its critical points is . Choose a point to be the root of our preimage tree. Note that there is no critical point in the backward orbit of  — i.e., the set .

For each , define

 Kn=K(f−n(x))andGn=\Gal(Kn/K). (8)
{lemma}

For any field of characteristic zero and any , the Galois group of (8) is isomorphic to a subgroup of .

###### Proof.

Because there is no critical point in the backward orbit of , the set consists of exactly distinct elements for each . Identify the vertices of the ternary rooted tree with the set , with vertex lying immediately above if and only if . This identification induces a faithful action of on and, hence, may be identified with a subgroup of .

To see that this subgroup lies inside , we proceed by induction on . For , this is clear because .

Fix , and assume we know the lemma holds for . Write . For each , applying the lemma to the field with root point shows that

 \Gal(K(f−(n−1)(yi))/K(yi))

is a subgroup of . (The labeling of allows us to identify the portion of above with .) Since is a subgroup of , it follows that is isomorphic to a subgroup of .

It remains to show that . Direct computation shows that the discriminant of the degree-nine polynomial is given by equation (2). Since this discriminant is a square in , all elements of act as even permutations of the nine points of . Thus, , as desired. ∎

Our goal is to compute the arboreal Galois groups in the case that is a number field and that the basepoint satisfies the following local hypothesis:

 There exist primes \pp and \qq of K lying above 2 and 3, respectively, ($\dagger$) such that v\qq(x)=1, and either v\pp(x)=±1 or v\pp(1−x)=1.

If this hypothesis holds, we will say that the pair satisfies property (($\dagger$)) (relative to and ).

{example}

If , then the pairs and both satisfy property (($\dagger$)). The latter pair will be important for our arithmetic applications.

{lemma}

Suppose that satisfies (($\dagger$)) relative to and . Then is Eisenstein at for all . In particular, is irreducible for all .

###### Proof.

A simple induction shows that and . Since , it follows immediately that is Eisenstein at . ∎

{proposition}

Let be a number field, and let . Suppose that satisfies property (($\dagger$)) relative to primes and . Let , and let . Then:

1. There are primes and of lying above and , respectively, such that

 e(\pp′/\pp)=2nande(\qq′/\qq)=3n.
2. The pair satisfies property (($\dagger$)) relative to and .

###### Proof.

We proceed by induction on . The statement is trivial for . We therefore assume for the rest of the proof that and that the statement holds for .

Given , let . By our inductive hypothesis, there are primes and of lying over and satisfying the desired properties for . The polynomial

 f(z)−y′′=−2z3+3z2−y′′∈K(y′′)[z]

is Eisenstein at . Thus, there is only one prime of lying above , with ramification index , and with . Moreover,

 e(\qq′/\qq)=e(\qq′/\qq′′)⋅e(\qq′′/\qq)=3⋅3n−1=3n.

Meanwhile, by statement (2) for , we have either , , or . We consider these three cases separately.

If , then the Newton polygon of at has a segment of length and height . Thus, there is a prime of lying above , with ramification index , and with . Moreover,

 e(\pp′/\pp)=e(\pp′/\pp′′)⋅e(\pp′′/\pp)=2⋅2n−1=2n.

If , note that is self-conjugate via ; that is, . Thus, , and the previous paragraph applied to gives the desired conclusion.

Finally, if , then because , the Newton polygon of at has a segment of length and height . Thus, there is a prime of lying above , with ramification index , and with . Once again, then, we have . ∎

{corollary}

Let and be as in Proposition 3. Let and let be the splitting field of over . Then

 6n∣∣[Kn:K].
###### Proof.

Pick , and let and be the primes of given by Proposition 3. Since has intermediate extension , the ramification index of some prime of over must be divisible by . Similarly, the ramification index of some prime of over must be divisible by . Thus, . ∎

{proposition}

Let and be as in Proposition 3. Then

 \Gal(K(f−1(x))/K)≅E1≅\fS3% and\Gal(K(f−2(x))/K)≅E2.
###### Proof.

Let and . Then, as a splitting field of a cubic polynomial, is Galois over with isomorphic to a subgroup of . By Corollary 3 with , we have . Hence, .

By Lemma 3, acts on as a subgroup of . It suffices to show that every element of is realized in .

Write , and for each , write

 f−1(ui)={v