Degrees of stretch factors and counterexamples to Farb’s conjecture 1footnote 11footnote 1Partially supported by NSF Grants DMS-1006553 and DMS-1607374.

Degrees of stretch factors and counterexamples to Farb's conjecture

Abstract.

Franks and Rykken proved that any pseudo-Anosov map with quadratic stretch factor and orientable invariant foliations, can be obtained by lifting a torus map by a branched or unbranched covering map. Farb conjectured that this generalizes to higher degree stretch factors. We construct counterexamples to this conjecture. The proof relies on studying the Perron-Frobenius degrees of our constructed stretch factors and a careful analysis of their Galois conjugates.

1. Introduction

Let be a closed, orientable surface of genus . The mapping class group of , is the group of isotopy classes of orientation preserving homeomorphisms of . Thurston-Nielsen classification states that every mapping class is periodic, reducible or pseudo-Anosov. Pseudo-Anosov maps are the most important class between these three from the point of view of Teichmuller theory and also hyperbolic 3-manifolds. A map is called pseudo-Anosov if there are a pair of transverse measured (singular) foliations and and a number such that the foliations are invariant under the map and their measures get expanded/contracted by , i.e.,

 f(F+,μ+)=(F+,λμ+),
 f(F−,μ−)=(F−,λ−1μ−).

The foliation (respectively ) is called orientable if there is a consistent choice of transverse orientation for it. If one of the invariant foliations is orientable then both of them are orientable. The number is called the stretch factor or the dilatation. It measures how much the map stretches/shrinks in the two canonical directions. It follows from Thurston’s train-track theory that the stretch factor can be obtained as the spectral radius of the action on the vector space of transverse measures. Therefore the stretch factor is an algebraic integer and it makes sense to talk about the algebraic degree of the stretch factor. This algebraic degree is always greater than or equal to two. Hence from an algebraic point of view, one can say that quadratic stretch factors are the ‘simplest’ possible ones. In 1999, Franks and Rykken proved the following [1] (See also the work of Gutkin-Judge [2]).

Theorem 1.1.

(Franks-Rykken) Let be a closed orientable surface of genus and be a pseudo-Anosov map. Assume that the stretch factor of is a quadratic algebraic integer and the invariant foliations of are orientable. There is an Anosov torus homeomorphism and a (branched or unbranched) covering map such that can be obtained by lifting the homeomorphism by the map .

Informally speaking, it states that if the stretch factor of a pseudo-Anosov map is ‘algebraically simple’, then the map itself can be obtained from a ‘simple construction’. Farb conjectured that the above theorem should generalize to higher degree stretch factors, even without the orientability assumption on the invariant foliations (See [7]). More precisely:

Conjecture 1.2.

(Farb-2004) Given any , there exists so that any pseudo-Anosov map with algebraic degree stretch factor on a closed orientable surface arises by lifting a pseudo-Anosov map on some surface of genus at most by a (branched or unbranched) cover.

In this work, we construct counterexamples to the above conjecture.

Theorem 4.12.

There is a sequence of pseudo-Anosov maps , , all of their stretch factors have algebraic degree six, such that at most finitely many of them can possibly satisfy the conclusion of Farb’s conjecture. Moreover, the invariant foliations for ’s can be chosen to be orientable.

The author acknowledges that Reid-Leininger have independently and simultaneously constructed counterexamples to Farb’s conjecture using different methods [4].

Here is the outline of the proof. We use Penner’s construction of pseudo-Anosov maps for a suitable set of curves and obtain a sequence of maps with stretch factors . The stretch factors have algebraic degrees equal to , i.e., . Define the Perron-Frobenius degree of a stretch factor , , to be the smallest size of a non-negative, integral, aperiodic matrix with spectral radius equal to [3]. By Thurston’s train track theory, is finite. In fact if is a lift of a pseudo-Anosov map on a surface of genus , then there is an upper bound, in terms of , for the Perron-Frobenius degree of .

Corollary 4.5.

Let be lift of a pseudo-Anosov map by a branched or unbranched covering. Let be the stretch factor of . Then the Perron-Frobenius degree of is at most .

Therefore in order to construct counterexamples to Farb’s conjecture, it is enough to show that the sequence of Perron-Frobenius stretch factors is unbounded. In other words we use the following Corollary as our strategy:

Corollary 4.6.

Let be a sequence of pseudo-Anosov maps () and let be the stretch factor of . Assume that the followings hold:
1) There is a constant such that for all ,
2) As , we have .
Then all but finitely many elements in the sequence do not satisfy the conclusion of Farb’s conjecture.

Then we use Lind’s theory together with an analysis of the Galois conjugates of ’s, to prove that the sequence is unbounded. Hence we obtain Theorem 4.12. Here is the quantitative version of Theorem 4.12. To set the notation, for two functions and we write if there is a positive constant such that

 1Cg
Theorem 1.3.

There is a sequence of pseudo-Anosov maps , , with stretch factors such that

 dalg(λg)=6anddPF(λg)≥Θ(g13).

In particular, can not be obtained by lifting a pseudo-Anosov map on a surface of genus less than . Moreover, the invariant foliations for ’s can be chosen to be orientable.

One might wonder, how much the bound can be improved. We believe that even the linear bound should be possible.

Conjecture 1.4.

There are pseudo-Anosov maps with such that is constant but grows linearly in .

In fact it is not clear if the constructed examples in Section 2 give a linear bound or not.

1.1. Acknowledgement

Part of this work has been done during the author’s PhD studies at Princeton University. The author would like to thank his advisor, David Gabai, for his support and encouragement. Sincere thanks to Douglas Lind who generously explained how one can give a lower bound for the Perron-Frobenius degree of a Perron number through the mathoverflow website https://mathoverflow.net/questions/228826/lower-bound-for-perron-frobenius-degree-of-a-perron-number. Sincere thanks to Balázs Strenner for many helpful comments. The author gratefully acknowledges the support by a Glasstone Research Fellowship.

2. Construction of the maps

Consider the following set of curves (red curves) and (blue curves) as in Figure 1 (See [7], page 20). In this Figure, there are red curves that intersect the right blue curve exactly twice. Also there are red curves that intersect the right blue curve exactly once. The union fills the surface, therefore we can use Penner’s construction of pseudo-Anosov maps [6]. For any curve , we use the notation to show the positive Dehn twist around . Let be the composition of positive Dehn twists around blue curves, each of them appearing once. Note that the order in which we compose these Dehn twists does not matter since blue curves are disjoint and hence their Dehn twists commute with each other. Let be the composition of positive Dehn twists around red curves where each of them appear exactly once except for that appears times and that appears times. Here and are positive integers that will be specified later (we will be interested in the case for our application). Define

 f:=τr∘(τb)−1 ,where f % depends on g,k and m.

By Penner’s construction, the map is pseudo-Anosov and an invariant bigon track can be obtained from the union by smoothing the intersection points.

3. Computing the stretch factors

Since this section is computational, there is a brief summary of the results at the end of the section.
Let be the vector space of transverse measures on . The homeomorphism acts on by a linear transformation. This action can be represented by a non-negative, integral, aperiodic matrix and the spectral radius of the action is equal to the stretch factor of . Assume . For each curve , define a measure supported on as follows: the value of is equal to one exactly on the branches that are included in and is equal to zero for the rest of the branches. Let be the convex hull of the measures (and their positive scalar multiples) and the zero measure in . The measures form a positive basis for . In general can be a proper sub-cone of the cone of positive measures in . Since is invariant under the action of and moreover the action of is represented by an aperiodic matrix, therefore the spectral radii for the actions on and are equal. As a result the stretch factor of is equal to the spectral radius for the action on . We compute this action in the basis . Let be the intersection matrix for the union , i.e., if then the entry of is equal to the geometric intersection number of and . After possibly re-ordering the basis elements, the matrix has the following form

 (1) Ω=[0XXT0]

,where is a matrix recording the geometric intersection numbers between the curves in and . The action of the Dehn twist (respectively ) on is represented by the following matrix:

 Qj=I+DjΩ,

where is the identity matrix and is the matrix whose entry is equal to one and whose all other entries are equal to zero (See [6], page 194). The action of on is the product of the corresponding elements in the right order (same as the order that the Dehn twists are composed in the definition of ), call it .

Given the linear map , the vector space can be decomposed as:

 H=ker(Ω)⊕Im(Ω),

and acts by identity on (See [7], Proposition 3.14 of version 4). Hence if we define , then there is an induced map

 ^M:^H⟶^H.

If we show the characteristic polynomials of the maps and with and respectively then (See [7], Proposition 3.19 of version 4)

 χ(M)=(x−1)ker(Ω)χ(^M).

Hence the stretch factor of is equal to the spectral radius of . Note that the dimension of is equal to . We calculate the linear map using an explicit basis for and compute the minimal polynomial for this action.
We have

 X=⎡⎢ ⎢⎣110000…00…0011110…00…0000011…12…2⎤⎥ ⎥⎦

,where the columns correspond to the red curves and the rows correspond to the blue curves respectively. Therefore form a basis for , where:

 v1=e1,v2=e2,v3=e3,

Here , and are the first three columns of and span the remaining colmuns of . Since is spanned by the columns of , then ’s form a basis for . The images of ’s under the linear map are as follows.:

Let be the projection map corresponding to the decomposition . In order to compute the map , we want to write ’s as a linear combination of ’s. Define the vectors as the following:

 w=[0,0,0|1,0,0,0,0|0,…,0|0,…,0]T,
 t=[0,0,0|0,0,1,0,0|0,…,0|0,…,0]T.

Then we can write down ’s as linear combinations of ’s together with and .

 M(v1)=v1+v4+(k−1)w,
 M(v2)=v2+v5+(m−1)t,
 M(v3)=v3+v6,
 M(v4)=2v1+v2+(m−1)t+v5+3v4+(2k−2)w,
 M(v5)=v1+4v2+v3+5v5+v6+(4m−4)t+v4+(k−1)w,
 M(v6)=v2+(6g−16)v3+(6g−15)v6+v5+(m−1)t.

Let us use the notation for . We can write and as a linear combination of ’s. Using the fact that and belong to and direct computation, we obtain:

 p(w)=(24g−6542g−114)^v4−(6g−1642g−114)^v5+(142g−114)^v6,
 p(t)=−(3g−821g−57)^v4+(6g−1621g−57)^v5−(121g−57)^v6.

Putting the equations for , and ’s together, we deduce the matrix representing the map . Note that the entries of might not be integers. Giving this matrix as an input to Mathematica, can be computed (or alternatively one can compute the determinant of this matrix by hand). The result is the following.

Lemma 3.1.

Let , and be as defined previously. The characteristic polynomial of is equal to the following:

 χ(^M)=(1+x6)+(6−6g−k−m)(x+x5)+
 +(−96−9k+km−11m+48g+6gk+6gm)(x2+x4)+
 +(211+40m+69k+14km−96g−18gm−30gk−6gkm)x3.

As a corollary, the stretch factor of is equal to the house (largest absolute value of all roots) of the same polynomial.

Definition 3.2.

Let be the map obtained by setting in the above construction. Define to be equal to the stretch factor of .

Putting and in terms of and simplifying the characteristic polynomial , we obtain the next Lemma.

Lemma 3.3.

Let be as defined above. The stretch factor of is the house of the following polynomial:

 Pg(x)=(1+x6)+(6−18g)(x+x5)+(108g2−72g−96)(x2+x4)+
 +(−216g3+216g2+558g+211)x3.

Since is equal to the house of the polynomial , we are interested in solving the equation

 Pg(x)=0.

The polynomial is reciprocal (this is not a coincidence since the action of a homeomorphism preserves the algebraic intersection pairing between two curves, see Corollary 3.17 of [7] version 4), therefore we can write down the above equation in a simpler form by setting

 y=x+1x

as a new variable. We have:

 Pg(x)x3=(1x3+x3)+(6−18g)(1x2+x2)+(108g2−72g−96)(1x+x)+
 +(−216g3+216g2+558g+211).

Using the identities

 1x3+x3=y3−3yand1x2+x2=y2−2,

the equation is equivalent to the following equation:

 y3+(6−18g)y2+(108g2−72g−99)y+(−216g3+216g2+558g+209)=0.

Set

 αg=6−18g,
 βg=108g2−72g−99and
 γg=−216g3+216g2+558g+209.

Therefore, the previous polynomial equation takes the form:

 Qg(y)=y3+αgy2+βgy+γg=0.

The following Lemmas show the key properties that the coefficients of have. In fact, we put in the beginning to have these properties.

Lemma 3.4.

For , the polynomial has exactly one real root.

Proof.

The number of real roots of a cubic can be determined by the sign of its discriminant. The discriminant for the cubic equation is defined as

 Δ:=18abcd−4b3d+b2c2−4ac3−27a2d2.

Direct computation shows that for . Therefore, the polynomial has exactly one real root. ∎

From now on, we assume that . By the previous Lemma, we can use the notation for the unique real root of . Clearly since . Let and be the other two roots. Hence we have . The roots of can be obtained by solving the equation

 x+1x=ω,

where is a root of . Therefore the roots of can be divided into three pairs. The first one consists of . These two are obtained by solving the equation:

 x+1x=ω1.

The four remaining ones can be partitioned into and , where and have absolute value greater than one. We have since . Here is a diagram:

 x+1x=ω1⟹λg,λ−1g.
 x+1x=ω2⟹δg,δ−1g.
 x+1x=ω3⟹θg,θ−1g.

Notation: For two functions , we write if there is a positive constant such that

 1Cg
Lemma 3.5.

We have the following estimates:

 λg=6g+Θ(g13),|δg|=6g−Θ(g13),
 |δg|λg=1−Θ(g−23),|arg(δg)|=Θ(g−13),
 limg→∞1−Re(δgλg)Im(δgλg)=0,
 tan−1(1−Re(δgλg)Im(δgλg))=±Θ(g−13).
Proof.

We first prove the corresponding results for , and .
1) Set . Substitution in the equation gives:

 A3+6A2−99A+209=36g.

Note that surprisingly there are no terms including both and after simplification. Therefore , which implies part one.

2) We have and . Hence

 |ω2|=(216g3−216g2−558g−209ω1)12=
 =(216g3−216g2−558g−2096g+Θ(g13))12=6g−Θ(g13).

This last equality can be checked directly by computation.

3) This part immediately follows from parts one and two:

 |ω2|ω1=6g−Θ(g13)6g+Θ(g13)=1−Θ(g−23).

4,5) We have and . As a result

 2Re(ω2)=ω2+ω3=18g−6−ω1=12g−Θ(g13).

Therefore

6)

 |Im(ω2)|2=|ω2|2−|Re(ω2)|2=(6g−Θ(g13))2−(6g−Θ(g13))2=Θ(g43).

7)

 1−Re(ω2ω1)Im(ω2ω1)=ω1−Re(ω2)Im(ω2)=(6g+Θ(g13))−(6g−Θ(g13))±Θ(g23)=Θ(g13)±Θ(g23)⟶0.

Now we can deduce similar statement for and . Using the fact that and are the larger (in absolute value) roots of the equation

 x+1x=ω1andx+1x=ω2,

one can see that when , the values of and are uniformly bounded above. This implies the corresponding statements for and .

Lemma 3.6.

The polynomials and are irreducible (with coefficients).

Proof.

First we show that is irreducible. This is equivalent to proving that has no integral root. This is true since has no solutions module 2, i.e., no matter if is even or odd, the quantity is always an odd number:

 Qg(ω1)≡(ω1)3−ω1+1≡1(mod 2).

Now we argue that is irreducible. Recall how the roots of can be obtained from the roots of . We drop the subscript in for the sake of convenience. Assume that , where and have integral coefficients and their degrees are non-zero. The polynomial has no roots module 2 and as a result has no real roots. Hence we can assume that and have degrees greater than 1. By Lemma 3.5 each of the roots was asymptotic to , therefore each of the roots of is asymptotic to either or (three for each). On the other hand

 R(0)S(0)=P(0)=1⟹|R(0)|=|S(0)|=1.

This means that the product of the roots of has absolute value equal to one (and similarly for ). Therefore, each of the polynomials and should have an even number of roots. In other words they have even degrees. Without loss of generality, assume that and have degrees two and four respectively. There are two possibilities:
1) The number is a root of . Since , should be the other root of . This implies that

 ω1=λg+(λg)−1

is an integer. Contradiction to what we proved previously.
2) The number is a root of . We can assume that and are the roots of . Therefore

 ω2=δg+(δg)−1

is an integer. Contradiction to the fact that has exactly one real root .
Each of the two possibilities led to a contradiction, so is irreducible. ∎

3.1. Summary of the section

Let be the map defined in the previous section after setting . Let be the stretch factor of . The number is the house of the polynomial

 Pg(x)=(1+x6)+(6−18g)(x+x5)+(108g2−72g−96)(x2+x4)+
 +(−216g3+216g2+558g+211)x3.

The polynomial is irreducible and has exactly two real roots and . The other four roots can be described as two pairs of numbers and , where the absolute values of and are greater than one. If we set then we have the following:

 limg→∞Re(tg)=1,limg→∞Im(tg)=0,
 limg→∞1−Re(tg)Im(tg)=0,
 andtan−1(1−Re(tg)Im% (tg))=±Θ(g−13).

4. Perron-Frobenius degrees of stretch factors

Definition 4.1.

An algebraic integer is called Perron if is strictly greater than the absolute value of its other Galois conjugates [5].

A non-negative, integral matrix is called aperiodic if there is some such that . In the literature, it has been referred by the name Perron-Frobenius as well.

Theorem 4.2.

(Lind [5]) Let be a Perron number. There exists a non-negative, integral, aperiodic matrix whose spectral radius is equal to .

Definition 4.3.

Let be a Perron number. The Perron-Frobenius degree of , , is the smallest possible size of an aperiodic matrix whose spectral radius is equal to [3]:

In other words, Lind’s theorem states that when is a Perron number, its Perron-Frobenius degree is finite.

Theorem 4.4.

(Thurston) Let be the stretch factor of a pseudo-Anosov map on a surface of Euler characteristic . Then is a Perron number and its Perron-Frobenius degree does not exceed .

Proof.

Thurston’s train track theory implies that has an invariant train track . Moreover is the spectral radius of the action of on the space of transverse measures on and this linear action can be represented by a non-negative, integral, aperiodic matrix. The number of branches of any train track, whose complementary regions have negative Euler characteristic, is at most . ∎

Corollary 4.5.

Let be lift of a pseudo-Anosov map by a branched or unbranched covering map. Let be the stretch factor of . Then the Perron-Frobenius degree of is at most .

Proof.

The stretch factor is invariant under taking branched or unbranched covering maps. Therefore, the result follows from Thurston’s Theorem. ∎

Corollary 4.6.

Let be a sequence of pseudo-Anosov maps () and let be the stretch factor of . Assume that the followings hold:
1) There is a constant such that for all ,
2) As , we have .
Then all but finitely many elements in the sequence do not satisfy the conclusion of Farb’s conjecture.

Proof.

This is immediate. Assume that the sequence satisfy the conclusion of Farb’s conjecture. Therefore if we set

 h:=max{h(1),h(2),…,h(d)},

then each arises by lifting a pseudo-Anosov map on a surface of genus at most by a branched or unbrnached covering map. By the previous Corollary we have:

 dPF(λg)<9(2h−2).

This contradicts the second assumption. ∎

Corollary 4.6 is our strategy for constructing counterexamples to Farb’s conjecture.

Given an algebraic integer of degree over and with minimal polynomial , define its companion matrix as

 B=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣00…cd10…cd−101…cd−2⋮⋮⋮00…c1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.

Note that the characteristic polynomial of is equal to up to sign, however can have negative entries. The next Theorem is the more complete version of Theorem 4.2 [5].

Theorem 4.7.

(Lind) Let be a Perron number. Consider its companion matrix acting on the vector space and let be the eigenvector corresponding to the eigenvalue . Let be the positive half-space containing . There are integral points in such that for each , with , and any irreducible component of the matrix is an aperiodic matrix whose spectral radius is equal to .

The next Theorem, also due to Lind [5], gives a converse to the previous theorem.

Theorem 4.8.

(Lind) Let be a Perron number and let and be as before. If is an aperiodic, non-negative, integral matrix with spectral radius equal to , then there are such that for each we have .

We bring Lind’s proof of Theorem 4.8 since we need some of the details of the proof and the proof is not complicated.

Proof.

Consider . By Perron-Frobenius theory, there is a positive eigenvector corresponding to . By working over the field , we can assume that . Let

 v=(v1,v2,…,vn)T∈Rn

and assume that for :

 vi=zi1+zi2λ+⋯+zidλd−1>0,

where the numbers are integers. Define for :

 zi=(zi1,zi2,…,zid)T∈Zd.

Since is an eigenvector for , we have:

 λvi=(Av)i=∑jaijvj.

Let be the map:

In particular, we have for . Taking from both side of the previous equation gives us:

 Ψ(λvi)=∑jai,jΨ(vj).

Note that multiplication by on has matrix with respect to the basis . Hence, we obtain:

 Bzi=∑jaijzj.

Finally, we need to verify that the points ’s belong to the positive half-space . Note that

 v∗=(1,λ,…,λd−1)∈Rd

is a left eigenvector for the linear map corresponding to the eigenvalue . Let be the one-dimensional invariant subspace of corresponding to and be its invariant complement. Therefore,

 C={x∈Rd|v∗x=0}.

If is the projection map from onto , then should be a multiple of . On the other hand

 v∗zi=zi1+zi2λ+⋯+zidλd−1=vi>0.

Hence replacing each by if necessary, we have for each and the proof is complete. ∎

We need the following Lemma and Propositions in order to prove our main result, Theorem 4.12.

Lemma 4.9.

Let be the two dimensional invariant subspace corresponding to the eigenvalues and . Let be the one dimensional eigenspace corresponding to and set:

 W=Eδ⊕Eλ.

If we define and similarly and set then we have . Define to be the projection map corresponding to this decomposition. Assume that ’s are as in the proof of Theorem 4.8. Then for each , the projection does not entirely lie in the one dimensional subspace .

Proof.

Assume the contrary that for some . Therefore we should have (the orthogonal complement inside ). Define the vectors and as follows:

Then we have:

 w∗zi=t∗zi=0⟹(w∗+it∗)zi=0⟹
 ⟹zi1+zi2δ+⋯+zidδd−1=0.

However, this means that satisfies a polynomial equation with degree less than . Therefore all ’s should be zero for . This contradicts the fact that . ∎

The idea of using the next Proposition has been suggested generously by Douglas Lind in the mathoverflow post https://mathoverflow.net/questions/228826/lower-bound-for-perron-frobenius-degree-of-a-perron-number. In this post, the author had asked for a way of finding a lower bound for the Perron-Frobenius degree of a Perron number. This was the answer that Lind gave:
” If a Perron number has negative trace, then any Perron-Frobenius matrix must have size strictly greater than the algebraic degree of , for example the largest root of . If denotes the companion matrix of the minimal polynomial of (which of course can have negative entries), then splits into the dominant 1-dimensional eigenspace and the direct sum of all the other generalized eigenspace. Although I’ve not worked this out in detail, roughly speaking the smallest size of a Perron-Frobenius matrix for should be at least as large as the smallest number of sides of a polyhedral cone lying on one side of (positive -coordinate) and invariant (mapped into itself) under . This is purely a geometrical condition, and there are likely further arithmetic constraints as well. For example, if has all its other algebraic conjugates of roughly the same absolute value, then acts projectively as nearly a rotation, and this forces any invariant polyhedral cone to have many sides, so the geometric lower bound will be quite large.”
Of course Proposition 4.10 is only one way of manifesting the above idea (unfortunately our notation is different from the above quote since we had to choose between the above and Lind’s paper notation) and it would be nice to weaken the geometric assumptions about the roots or to explore the so called arithmetic constraints that Lind mentions.

Proposition 4.10.

Let be a sequence of linear maps. Assume that the eigenvalues of are , and . Moreover:
1) The eigenvalue is a positive real number. The eigenvalues and are complex numbers with non-zero imaginary parts, and .
2) If we set , then we have:

 limg→∞Re(tg)=1,limg→∞Im(tg)=0,
 andlimg→∞1−Re(tg)% Im(tg)=0.

Define as the postive half-space corresponding to the eigenvalue . Let be the minimum number of sides for an arbitrary non-degenerate polygonal cone that is invariant under the map , i.e.,

 Bg(C)⊂C.

Then goes to infinity as goes to infinity. More precisely

 cg.tan−1(1−Re(tg)Im(t