Minimal pseudoAnosov stretch factors on nonorientable surfaces
Abstract.
We determine the smallest pseudoAnosov stretch factor (restricting to pseudoAnosov maps with an orientable invariant foliation) on the closed nonorientable surfaces of genus 4, 5, 6, 7, 8, 10, 12, 14, 16, 18 and 20. We also determine the smallest stretch factor of an orientationreversing pseudoAnosov map (again restricting to orientable invariant foliations) on the closed orientable surfaces of genus 1, 3, 5, 7, 9 and 11.
1. Introduction
Let be a surface of finite type. A homeomorphism of is pseudoAnosov if there are transverse singular measured foliations and and a real number such that and [Thu88]. The number is called the stretch factor of .
On any surface , the set of pseudoAnosov stretch factors forms a discrete set [AY81, Iva88]. In particular, there is a minimal stretch factor. For an orientable surface , we denote by the minimal stretch factor of orientationpreserving pseudoAnosov maps. One motivation for studying the number is that the shortest geodesic on the moduli space of algebraic curves homeomorphic to has length .
Another motivation for studying small stretch factors comes from 3manifold theory. The mapping torus of a pseudoAnosov map is a hyperbolic 3manifold and the stretch factor of is related to the hyperbolic volume of ([KKT09, KM18]). This relates small volume hyperbolic manifolds ([Ago02, AST07, GMM09, Mil09]) to small stretch factor pseudoAnosov maps.
The asymptotic behavior of is wellunderstood: Penner [Pen91] showed that , where denotes the closed orientable surface of genus . For other constructions of small stretch factors and asymptotics for different sequences of surfaces, see [Bau92, McM00, Min06, HK06, Tsa09, Val12, Yaz18].
Finding the exact values of minimal stretch factors turns out to be much more difficult. The value of for hyperbolic is only known when [CH08]. This value is approximately 1.72208, the largest root of .
More is known for pseudoAnosov maps whose invariant foliations and are transversely orientable. We denote the minimal stretch factor of orientationpreserving pseudoAnosov maps on with orientable invariant foliations by . The known values are summarized in the table below.
Minimal polynomial of  

1  2.61803  
2  1.72208  
3  1.40127  
4  1.28064  
5  1.17628  
7  1.11548  
8  1.12876 
Initially, the pseudoAnosov maps realizing the stretch factors in the table were constructed in different ways. The construction is due to Zhirov [Zhi95] for , Lanneau & Thiffeault [LT11b] for , Leininger [Lei04] for , Kin & Takasawa [KT13] and Aaber & Dunfield [AD10] for and Hironaka [Hir10] for . Hironaka [Hir10] then showed that all of the examples above except the example arise from the fibration of a single hyperbolic 3manifold, the mapping torus of the “simplest hyperbolic braid”. The fact that the values in the table are indeed the minimal stretch factors were shown by Lanneau and Thiffeault [LT11b] by a systematic way of narrowing down the set of possible minimal polynomials of the minimal stretch factors. For the larger values of , their proof is computerassisted.
Main results
In this paper, we study the analogous problem for pseudoAnosov maps whose mapping torus is a nonorientable 3manifold. There are two types of such pseudoAnosov maps: maps on nonorientable surfaces and orientationreversing maps on orientable surfaces.
First we state our theorem for closed nonorientable surfaces. Denote by the closed nonorientable surface of genus (the connected sum of projective planes) and by the minimal stretch factor among pseudoAnosov homeomorphisms of with an orientable invariant foliation. (Only one of the foliations can be orientable, otherwise the surface would have to be orientable as well.)
Theorem 1.1.
The values and minimal polynomials of for , , , , , , , , , and are as follows:
Minimal polynomial of  singularity type  

4  1.83929  (6)  
5  1.51288  (4,4,4)  
6  1.42911  (10)  
7  1.42198  (4,4,4,4,4)  
8  1.28845  (14)  
10  1.21728  (18)  
12  1.17429  (22)  
14  1.14551  (26)  
16  1.12488  (30)  
18  1.10938  (34)  
20  1.09730  (38) 
The table also contains the singularity type of the minimizing pseudoAnosov map. For example, (4,4,4) means that the pseudoAnosov map has three 4pronged singularities.
Next, we state our result for orientationreversing maps. Let us denote by the minimal stretch factor among orientationreversing pseudoAnosov homeomorphisms of with orientable invariant foliations.
Theorem 1.2.
The values and minimal polynomials of for , , , , and are as follows:
Minimal polynomial of  singularity type  

1  1.61803  no singularities  
3  1.25207  (4,4,4,4)  
5  1.15973  (6,6,6,6)  
7  1.11707  (8,8,8,8)  
9  1.09244  (10,10,10,10)  
11  1.07638  (12,12,12,12) 
Moreover, we have
for and .
Based on these results, we make the following conjectures.
Conjecture 1.3.
For all , is the largest root of
Conjecture 1.4.
For all , is the largest root of
Since the larger root of the polynomial is , these conjectures would imply the following conjectures on the limits of the normalized minimal stretch factors.
Conjecture 1.5.
We have
Conjecture 1.6.
We have
We expect the limits to be different for other genus sequences. For example, in a forthcoming paper [LS18b] we show that if denotes the minimal stretch factor among pseudoAnosov mapping classes on obtained from Penner’s construction, then the sequence has exactly two accumulation points as . One accumulation point, , is the limit for the sequence restricted to even . The other accumulation point, strictly greater than , is the limit for the sequence restricted to odd . We expect this dichotomy to be indicative how the sequence behaves for odd and even genus sequences, respectively, since so far no pseudoAnosov mapping class of a nonorientable surface is known to not have a power arising from Penner’s construction (compare with Question 1.10 below).
In order to compare Conjectures 1.6 and 1.5 to the orientationpreserving case, we recall that Hironaka asked in [Hir10, Question 1.12] whether
(1.1) 
Since any pseudoAnosov map on can be lifted to a pseudoAnosov map on with the same stretch factor, it is natural that the limit in Conjecture 1.5 is larger than the limit in (1.1). The fact that the limit in Conjecture 1.6 is smaller than the limit in (1.1) is consistent with the fact that nonorientable hyperbolic 3manifolds can have smaller volume than orientable ones. For example, the smallest volume noncompact hyperbolic 3manifold is the Gieseking manifold, a nonorientable manifold [Ada87]. Since the stretch factor is related to the volume of the mapping torus [KM18], on a fixed surface one can expect to find orientationreversing pseudoAnosov maps with smaller stretch factor than orientationpreserving ones.
We also have a conjecture about the minimal stretch factors on the genus 9 and 11 nonorientable surfaces. We will discuss evidence for this conjecture in Section 5.4.
Conjecture 1.7.
The approximate values and minimal polynomials of for are as follows:
Minimal polynomial of  singularity type  

9  1.35680  (16)  
11  1.22262  (8,8,8) 
In the orientationpreserving case, the minimal stretch factor does not monotonically decrease with the genus, since (see Table 1). Theorem 1.2 shows that fails to be strictly decreasing at every other step. We conjecture that in fact the value of strictly increases in every other step. We will discuss evidence for this after Proposition 5.6.
Conjecture 1.8.
For all , we have
Comparison with the orientationpreserving case
In the orientationpreserving case, the concrete descriptions of the examples are all very different. For , Zhirov describes the example by the induced homomorphism . Lanneau and Thiffeault [LT11b, Appendix C] describe the same example as a product of the Humphries generators. For , Lanneau and Thiffeault use Rauzy–Veech induction, and for , Leininger uses Thurston’s construction. While Hironaka gives a unified construction in [Hir10] using fibered face theory, her work does not an give explicit description of the maps.
In contrast, the descriptions of our examples are explicit and uniform: all of our examples are constructed as a composition of a Dehn twist and a finite order mapping class. As we will explain shortly, such constructions cannot work in the classical setting.
We also remark that it is also possible to construct the examples in Theorem 1.1 and Theorem 1.2 by studying fibrations of certain small volume nonorientable hyperbolic 3manifolds, although we will not discuss this construction in this paper.
Galois conjugates and Penner’s construction
All of our examples have a power that arises from Penner’s construction of pseudoAnosov mapping classes. In sharp contrast, none of the classical minimal stretch factor examples have a power that arises from Penner’s construction. This is because these stretch factors have Galois conjugates on the unit circle. However, Shin and the second author showed in [SS15] that examples with this property do not have a power arising from Penner’s construction.
One may wonder what the reason of this discrepancy is. A heuristic reason for why Galois conjugates of small stretch factors should lie on the unit circle is that every pseudoAnosov stretch factor is a biPerron algebraic unit: a real number larger than 1 whose Galois conjugates of lie in the annulus . If is close to 1, this annulus is a thin neighborhood of the unit circle, so it seems natural for the Galois conjugates to lie on the unit circle.
However, in Section 4 we will prove the following theorem that explains why the nonorientable cases are different.
Theorem 1.9.
If is a pseudoAnosov map on a nonorientable surface or an orientationreversing pseudoAnosov map on an orientable surface, then the stretch factor of does not have Galois conjugates on the unit circle.
Penner’s conjecture on nonorientable surfaces
Penner asked in [Pen88] whether every pseudoAnosov map has a power that arises from his construction.^{1}^{1}1The conjecture that this is true is known colloquially as Penner’s conjecture. However, from the writing in [Pen88, p. 195], it is unclear whether Penner intended to pose this as a question or a conjecture, or even whether he conjectured the opposite. This was answered in the negative by Shin an the second author in [SS15] by providing the obstruction mentioned earlier: if the stretch factor has a Galois conjugate on the unit circle, the pseudoAnosov map cannot have a power arising from Penner’s construction.
However, Theorem 1.9 demonstrates that this obstruction is vacuous for nonorientable surfaces and for orientationpreserving maps. Since there are no other known obstructions, it is possible that the answer to Penner’s question is in fact “yes” in these settings. Some evidence for this is provided by the fact that all the minimal stretch factor examples we give in Theorems 1.2 and 1.1 have a power arising from Penner’s construction. Some evidence against is provided by failure of the second author in [Str17a, Section 7] to construct certain pseudoAnosov maps on nonorientable surfaces using Penner’s construction. We pose this problem as a question.
Question 1.10.
Does every pseudoAnosov map on a nonorientable surface have a power arising from Penner’s construction?
Question 1.11.
Does every orientationreversing pseudoAnosov map on an orientable surface have a power arising from Penner’s construction?
Outline of the paper
In Sections 3 and 2, we construct the examples for Theorems 1.2 and 1.1. In Sections 3 and 2, this is done by a generalization of the construction we gave for the Arnoux–Yoccoz pseudoAnosov maps in [LS18a].
To show that our examples have minimal stretch factor, we follow Lanneau and Thiffeault’s approach for orientable surfaces [LT11b, LT11a]: we run a bruteforce search for integral polynomials whose largest root is smaller than our candidate for the minimal stretch factor and hope that we do not find any. Aside from some low genus cases, this search is computerassisted. Our code can be found at https://github.com/b5strbal/polynomialfiltering.
In Section 4, we give various properties the polynomials have to satisfy. Some of these properties are analogous to Lanneau and Thiffeault’s, but there are also some that are unique to the nonorientable case. We also give the proof of Theorem 1.9 here.
In Section 5, we describe the polynomial elimination process and prove Theorem 1.1 without computer assistance in the case . This elimination process ends up being significantly cleaner for us than it was for Lanneau and Thiffeault. In their case, the restrictions on the polynomials alone are not sufficient to rule out all polynomials, so they were left with a few polynomials that needed to be ruled out by studying the possible singularity structures of the pseudoAnosov maps and by using Lefschetz number arguments. For us, no arguments like these are necessary.
Acknowledgements
2. Construction of pseudoAnosov maps on nonorientable surfaces
In this section, we use Penner’s construction to construct pseudoAnosov mapping classes on nonorientable surfaces. We briefly recall Penner’s construction on nonorientable surfaces below. For more details, see [Pen88, Section 4] or [Str17a, Section 2].
In Penner’s construction, we have a collection of twosided simple closed curves that fill the surface (the complement of the curves is a union of disks and oncepunctured disks) and that are marked inconsistently. This means that there is a small regular neighborhood for each curve and an orientation of each annulus such that the orientation of and are different at each intersection whenever . Penner showed that any product of the Dehn twists is pseudoAnosov assuming that

each twist is righthanded according to the orientation of ,

each twist is used in the product only with positive powers,

each twist is used in the product at least once.
We will present the construction as follows. First we define the rotationally symmetric graphs that will be the intersection graphs of the collections of curves. Then we describe the rotationally symmetric surfaces and curves on these surfaces whose intersection matrices realize the given graphs. Finally, we define our mapping classes as a composition of a Dehn twist and a rotation.
2.1. The graphs
Let and be integers of different parity such that and . Let be the graph whose vertices are the vertices of a regular gon and every vertex is connected to the vertices that are the farthest away from in the cyclic order of the vertices.
2.2. The surfaces
For each , we will construct a nonorientable surface that contains a collection of curves with intersection graph . To construct , start with a disk with one crosscap. By this, we mean that we cut a smaller disk out of the disk and identify the antipodal points of the boundary of the small disk. We indicate this identification with a cross inside the small disk, see Figure 2. The resulting surface is homeomorphic to the Möbius strip.
Next, we consider disjoint intervals on the boundary of the disk and label the intervals with integers from 1 to so that each label is used exactly twice. In the cyclic order, the labels are where and all labels are understood modulo .
For each label, the corresponding two intervals are connected by a twisted strip, as on Figure 2.
Lemma 2.1.
The Euler characteristic of is .
Proof.
The disk with a crosscap has zero Euler characteristic (it is homeomorphic to a Möbius strip), and each attached twisted strip has contribution 1. ∎
Lemma 2.2.
The number of boundary components of is .
Proof.
We will show that the number of boundary components of is the same as the number of orbits of the dynamical system in the group . The number of such orbits is , since is odd.
To prove our claim, we identify with the intervals in the cyclic order. We claim that the right endpoint of the interval at position lies on the same boundary component as the right endpoint of the interval at position . One can see this by induction. In the case , the cyclic order of labels is , so the twisted strips identify the right endpoint of every interval with the right endpoint of the next interval. When , the cyclic order is , in which case every third right endpoint is on the same boundary component, and so on. ∎
Proposition 2.3.
The surface is homeomorphic to the nonorientable surface of genus with boundary components.
Proof.
The Euler characteristic of the nonorientable surface of genus with boundary components is . By Lemmas 2.2 and 2.1, we obtain the equation . Rearranging, we obtain . ∎
2.3. The curves
We construct a twosided curve for each label as follows. Each curve consists of two parts. One part of each curve is the core of the strip corresponding to the label. The other part is an arc inside the disk that passes through the crosscap and connects the corresponding two intervals. The curve is shown on Figure 2.
Note that every pair of curves intersects either once or not at all. The curves and are disjoint if and only if the two labels and the two labels link in the cyclic order. In order words, if the two labels separate the two labels.
Lemma 2.4.
The intersection graph of the curves on is .
Proof.
We proof the lemma by induction. If , then , so the cyclic order is . Since the no two labels link, all pairs of curves intersect and the intersection graphs the complete graph .
Now suppose is decreased by 2. Then is decreased by 1, and we obtain the cyclic order . As a consequence, 1 becomes linked with 2 and . Hence the intersection graph is indeed .
It is easy to see that every time is decreased by two each label is linked with two more labels, hence the intersection graph is always . ∎
Lemma 2.5.
The curves can be marked so that all intersections are inconsistent.
Proof.
Choose markings for the which are invariant under the rotational symmetry. See Figure 3. The marking of the curves is indicated by the coloring as follows. Consider the orientable surface obtained by removing the crosscap and cutting the strips attached to the disk in the middle. Choose an orientation of this surface. Then color the arcs composing the curves using red and blue depending on whether the orientation of the neighborhood of the curve matches the orientation of the surface or not. Note that the color of a curve changes when it goes through the crosscap or the middle of a strip.
Since blue and red meets at every intersection, the marking is inconsistent. ∎
2.4. The mapping classes
Denote by the rotation of by one click in the clockwise direction. Define the mapping class
where is a Dehn twist about the curve . (There are two possible directions for the Dehn twist, but either choice works for our purposes.) Note that
so arises from Penner’s construction. In particular, is pseudoAnosov and so is .
We remark that for , the mapping class coincides with the nonorientable ArnouxYoccoz mapping class , described as a product of a Dehn twist and a finite order mapping class by the authors in [LS18a].
Proposition 2.6.
The stretch factor of is the largest root of where .
Proof.
To compute the stretch factor, we use Penner’s approach in the section titled “An upper bound by example” in [Pen91]. Penner constructed an invariant bigon track by smoothing out the intersections of the curves . Each defines a characteristic measure on this bigon track, defined by assigning 1 to the branches traversed by and zero to the rest. The cone generated by the is invariant under both and , hence it contains the unstable foliation, and the stretch factor is given by the largest eigenvalue of the action of on this cone. The rotation acts by a permutation matrix and the matrix corresponding to is the sum of the identity matrix and matrix obtained by the intersection matrix by zeroing out all rows except the first row. The product of these two matrices takes the following form:
This particular matrix belongs to .
This matrix is the companion matrix of the polynomial in the statement of the proposition. Hence the characteristic polynomial of this matrix is indeed that polynomial. ∎
Our next goal is to determine the singularity structure of the mapping classes . For this, first we need a lemma.
Consider the complementary regions of the curves . There are two types of regions depending on whether a region contains a boundary component of (type 1) or not (type 2). A region of type 1 is an annulus that is bounded by a boundary component of on one side and by a polygonal path consisting of arcs of the curves on the other side. The shaded region on Figure 3 illustrates a region of type 1.
Lemma 2.7.
The length of these polygonal paths is .
Proof.
This follows from the observation that every point in the orbit in corresponding to the boundary component (see the proof of Lemma 2.2) has two associated arcs. Since the number of orbits is , the length of each orbit is , and hence the length of the polygonal path with twice this quantity. ∎
Proposition 2.8.
The pseudoAnosov mapping class has singularities, one for each boundary component. The number of prongs of each singularity is .
Proof.
Each complementary region of the curves contains either one singularity or none. The number of prongs of a singularity equals the number of cusps of the bigon track obtained by the smoothing process that are contained in the same region as the singularity. If the number of cusps is 2, then the region does not contain a singularity. If the number of cusps is , then it contains a pronged singularity.
Regions of type 2 are rectangles (bounded by four subarcs of the curves ), and hence contain two cusps. So they do not correspond to singularities.
As a corollary of Propositions 2.3, 2.8 and 2.6, we have the following.
Corollary 2.9.
There exist pseudoAnosov mapping classes with orientable invariant foliations on the surfaces with the data below. All of these examples belong to the family for the and shown in the table.
minimal polynomial  singularity type  

4*  3  2  1.83929  (6)  
5*  6  3  1.51288  (4,4,4)  
6*  5  2  1.42911  (10)  
7*  10  5  1.42198  (4,4,4,4,4)  
8*  7  2  1.28845  (14)  
9  8  3  1.35680  (16)  
10*  9  2  1.21728  (18)  
11  12  3  1.22262  (8,8,8)  
12*  11  2  1.17429  (22)  
13  22  11  1.27635  (4)  
14*  13  2  1.14551  (26)  
15  14  3  1.18750  (28)  
16*  15  2  1.12488  (30)  
17  18  3  1.14259  (12,12,12)  
18*  17  2  1.10938  (34)  
19  18  5  1.20514  (36)  
20*  19  2  1.09730  (38) 
(4 means that there are 11 singularities with 4 prongs.)
In each genus, the family contains several examples. In the table above, we have listed only the example with the smallest stretch factor. In the starred cases, we will be able to certify that the given stretch factors are not only minimal in the family but among all pseudoAnosov maps with orientable invariant foliations.
3. Orientationreversing pseudoAnosov mapping classes on odd genus surfaces
In this section, we construct an orientationreversing pseudoAnosov mapping class with small stretch factor on every odd genus surface. The construction is analogous to the construction in the previous section, but simpler. As in the previous section, we separate the construction of the surfaces, the curves and finally the mapping classes.
3.1. The surfaces
For every , consider the surface obtained by chaining together annuli in a cycle as on Figure 4.
Proposition 3.1.
The number of boundary components of is 4 if is even and 2 if is odd.
Proof.
The boundary of is composed of arcs, 4 arcs for each annulus. Our goal is to determine which of belong to the same boundary component.
Denote be the rotation of by one click. By tracing the boundary, one can see that every boundary point lies on the same boundary component as . Moreover, the path between and traverses each of the 4 types of arcs exactly once. Therefore it suffices to pick any boundary point and determine how many equivalence class the set falls apart. The number of such equivalence classes is 4 if is even and 2 if is odd. ∎
Proposition 3.2.
The surface is homeomorphic to if is even and if is odd.
Proof.
We have . From the equation , where is the genus and is the number of boundary components, it follows that . The statement now follows from Proposition 3.1. ∎
As a consequence, the construction only produces odd genus examples.
3.2. The curves
From now on, suppose that is even. Consider the set of core curves of the annuli. Our numbering will differ from the standard cyclic numbering; we will explain this shortly. As in Section 2.3, any rotationally symmetric marking of the curves is an inconsistent marking.
The intersection graph of is a cycle of length . We draw this cycle as on Figure 5: the vertices are the vertices of a regular polygon and every vertex is connected to the two vertices that are the second furthest in the cyclic order. We number the curves according to the cyclic orientation induced by this picture.
3.3. The mapping classes
Denote by the rotation of (see Figure 4) by one click in the clockwise direction. Since and intersect for all , the rotation induces a rotation of the cycle on Figure 5 by clicks. So rotates the cycle by clicks, which is congruent to 1 modulo if is even. Therefore induces rotating the cycle on Figure 5 by one click (in the clockwise direction, assuming that we have chosen the numbering of the curves accordingly). In particular, we have .
We are now ready to define our the mapping class:
Note that
so arises from Penner’s construction. In particular, is pseudoAnosov and so is .
Proposition 3.3.
The stretch factor of is the largest root of .
Proof.
The proof is similar to the proof of Proposition 2.6. We have
where is the matrix of the action of on the cone of measures (the product of a permutation matrix and the sum of the identity matrix and the first row of ). The matrices above illustrate the case . The matrix is the companion matrix of the polynomial on the proposition. ∎
Proposition 3.4.
The pseudoAnosov mapping class has four pronged singularities.
Proof.
By Proposition 3.1 and its proof, each of the four boundary components of consists of arcs if is even. There is a prong for every second corner of the boundary path, therefore there are prongs for each singularity. ∎
Corollary 3.5.
There exist orientationreversing pseudoAnosov mapping classes with orientable invariant foliations on the surfaces with the data below. All of these examples are belong to the family for the shown in the table.
minimal polynomial  singularity type  

1  2  1.61803  no singularities  
3  4  1.25207  (4,4,4,4)  
5  6  1.15973  (6,6,6,6)  
7  8  1.11707  (8,8,8,8)  
9  10  1.09244  (10,10,10,10)  
11  12  1.07638  (12,12,12,12) 
Proof.
The statement follows from Propositions 3.2, 3.4 and 3.3. The reason we have no singularities in the genus 1 case is that by Proposition 3.4 the “singularities” have two prongs, so they are not actually singularities. ∎
We remark that in the genus 1, there is an alternative, simpler construction that yields the same stretch factor. Consider the matrix
with determinant . The corresponding linear map maps to , hence it descends to an Anosov diffeomorphism of the torus . Its stretch factor is the largest root of the , the characteristic polynomial of .
4. Restrictions on polynomials
PseudoAnosov stretch factors are roots of integral polynomials. The properties of these integral polynomials are similar, but slightly different depending on whether a pseudoAnosov mapping class is an orientationpreserving or orientationreversing mapping class on an orientable surface or a mapping class on a nonorientable surface. In this section, we discuss these properties for nonorientable surfaces and orientationreversing mapping classes.
A polynomial of degree is called reciprocal if , in other words, when its coefficients are the same in reverse order up to sign. Analogously, we define to be antireciprocal if .
Proposition 4.1.
Let be a pseudoAnosov map with a transversely orientable invariant foliation on the closed nonorientable surface of genus . Then its stretch factor is a root of a (not necessarily irreducible) polynomial with the following properties:


is monic and its constant coefficient is

The absolute values of the roots of other than lie in the open interval . In particular, is not reciprocal or antireciprocal.

is reciprocal mod 2.
Proof.
Note that exactly one of the stable and unstable foliations is transversely orientable (otherwise the surface itself would be orientable). We will assume that it is the stable foliation, otherwise we replace by its inverse.
Consider the action defined by pullback on cohomology with real coefficients. Since the stable foliation is orientable, it is represented by a closed real 1form, that is, an element of . The stable foliation is the one whose leaves are contracting and hence the surface is expanding in the transverse direction. Hence the measure of transverse arc in the pullback is times its measure in . Hence is an eigenvector of the map with eigenvalue or .
Let be the characteristic polynomial of . Note that , hence . This proves (1).
The polynomial has integral coefficients, since restricts to an action . This restriction is invertible, since the action of on is a group representation, so the determinant of is . Therefore the constant coefficient of is . Also, as a characteristic polynomial, is monic. This proves (2).
Let be the orientationpreserving lift of to the orientable double cover of . Applying [McM03, Theorem 5.3 (1)] for , we obtain that any root of other than satisfies . Applying the same theorem for , we conclude that any root of other than satisfies . Therefore absolute values of the roots of other than and possibly lie in the open interval . However, it was shown in the proof of [Str17b, Proposition 2.3] that if or is a root of , then and cannot be roots of , hence mentioning the edge case in the previous sentence is not necessary.
If was reciprocal, then and or and would have to be roots. If it was antireciprocal, then and or and would have to be roots. As we have just shown, these scenarios are impossible, because is not a root of . This proves (3).
Finally, notice that we have not guaranteed that is a root of —we have only shown that either or is a root. If it is , then the polynomial or satisfies all the required properties. ∎
We call a matrix antisymplectic if it the corresponding linear transformation sends the standard symplectic form on to its negative. Formally, this can be written as
where and is the identity matrix.
Proposition 4.2.
The characteristic polynomial of a antisymplectic matrix is antireciprocal.
Proof.
Let be an antisymplectic matrix. Since is symplectic, we have and . Since , we have
hence is antireciprocal. ∎
The proof above is a straightforward modification of the standard proof of the fact that the characteristic polynomials of symplectic matrices are reciprocal.
Proposition 4.3.
Let be an orientationreversing pseudoAnosov map with transversely orientable invariant foliations. Then its stretch factor is a root of a (not necessarily irreducible) polynomial with the following properties:


is monic and its constant coefficient is .

.

.

The absolute values of the roots of other than and lie in the open interval .
Proof.
Let be the characteristic polynomial of . Clearly, (1) holds. Applying [McM03, Theorem 5.3] for the square of , we obtain (5) as well.
An orientationreversing homeomorphism sends the intersection form on to its negative. Proposition 4.2 implies that . To decide which sign is right, we only need to compute the sign of the constant coefficient of . If the constant coefficient , then the sign is positive. If the constant coefficient is , then the sign is negative. To put this in another way, we have
(4.1) 
For orientationpreserving homeomorphisms, the action on homology is symplectic, hence its determinant is . It follows that, for fixed , the determinant is either for all orientationreversing homeomorphisms of or for all orientationreversing homeomorphisms of . It is sufficient to check only one homeomorphism to decide which one. For example, consider the reflection about the plane containing the curves on Figure 6.
The curves form a homology basis. We have and for all . So the matrix of is a diagonal matrix whose diagonal contains s and s, of each. Hence the determinant is and we have shown (2).
We are now ready to prove Theorem 1.9. We emphasize that, unlike in the previous propositions, in this theorem we are not assuming that the surface is closed or that the pseudoAnosov mapping class has a transversely orientable invariant foliation.
Proof of Theorem 1.9.
An irreducible polynomial with a (complex) root on the unit circle is reciprocal. This is because then is also a root of , therefore is a root of the polynomial , where is the degree of . But the minimal polynomial is unique up constant factor, so . So is indeed reciprocal. So if a stretch factor has a Galois conjugate on the unit circle, then the minimal polynomial of is reciprocal and is also a root of the minimal polynomial.
However, by [Str17b, Proposition 2.3], and are not Galois conjugates if is a stretch factor of a pseudoAnosov map (possibly with no orientable invariant foliations) on a nonorientable surface (possibly with punctures). This completes the proof in the case when the pseudoAnosov map is supported on a nonorientable surface.
We now prove the orientationreversing case. If our surface is closed, then, by Proposition 4.3, and are not Galois conjugates if is a stretch factor of an orientationreversing pseudoAnosov map with orientable invariant foliations. If the foliations are not orientable, we can lift the map to the orientation double cover of the foliations to obtain an orientationreversing pseudoAnosov map with orientable invariant foliations and with the same stretch factor. Therefore and are not Galois conjugates in this case, either.
If our surface has punctures, then we can fill in the punctures after making the foliations orientable to obtain a pseudoAnosov map with the same stretch factor on a closed surface, reducing to the closed case discussed in the previous paragraph. This completes the proof in the case when the pseudoAnosov map is orientationreversing. ∎
5. Elimination of polynomials
In this section we first prove bounds on the sum of th powers of roots of a polynomial when the absolute values of the roots are bounded by some . These bounds are improved versions of Lemma A.1. of [LT11b], using the special properties of the polynomials in Propositions 4.3 and 4.1.
Then we describe how we use this lemma and Propositions 4.3 and 4.1 in order to systematically narrow down the set of possible minimal polynomials of the minimal stretch factors.
5.1. Power sum bounds
We begin by proving two elementary lemmas.
Lemma 5.1.
Suppose and are positive real numbers such that . Then
Proof.
The function is increasing on the interval . So if there are so that , then we can increase the sum by moving and away from each other by keeping their product unchanged, until at least one of them is or . After every such operation, the number of that are equal to or increases. So eventually we get to a point where at most one is not or . When , no such can exist, and exactly half of the equal , the other half , otherwise their product would not be 1. When , exactly one such exists, it equals 1 and exactly half of the remaining equal , the other half . ∎
Lemma 5.2.
Suppose and are positive real numbers such that and and . Then
Moreover, the inequalities are strict if .
Proof.
Similarly to the proof of Lemma 5.1, our approach is to change the numbers to increase as much as possible while keeping the hypotheses true. Whenever there are such that , we push and apart until at least one of them equals or . The end result is the same as before, so we have