Let $S$ be a hyperbolic surface. Suppose $\mathcal{T}$ denotes the Teichmuller space of $S$ and $Mod(S)$ denotes the mapping class group of $S$. Given any pseudo-Anosov element $f\in Mod(S)$, suppose $\lambda_f$ denote the stretch factor (or dilatation) of $f$.

Q) Does there exist $f,g\in Mod(S)$ such that $f\neq g$ and $\lambda_f=\lambda_g.$

  • 1
    $\begingroup$ Do you allow orientation reversing homeomorphisms into your mapping class group? Because that changes the answer, somewhat. Consider the case of the once-punctured torus... $\endgroup$
    – Sam Nead
    Nov 12, 2016 at 2:52

2 Answers 2


Yes. If two pseudo-Anosov mapping classes are conjugate then they must have the same dilatation. So take any pseudo-Anosov $f$ and any mapping class $h$ not in the centraliser of $f$ and let $g = h f h^{-1}$. Then $f$ and $g$ are distinct, both pseudo-Anosov and have the same dilatation.

However, even if you require that $f$ and $g$ are not conjugate the answer is still yes. For example on the twice-punctured torus, let $f = T_aT_bT_c^{-1}$ and $g = T_aT_b^{-1}T_c^{-1}$ where these are Dehn twists about the curves $a$, $b$, $c$ shown below.

Twice-punctured torus

These both have dilatation the largest real root of

$$ 1 - 2x -2x^3 + x^4, $$

which is $\approx 2.2966$, but are not conjugate. You can check this by using flipper and the Python code:

>>> import flipper
>>> S = flipper.load('S_1_2')
>>> f = S.mapping_class('abC')
>>> g = S.mapping_class('aBC')
>>> f.dilatation() == g.dilatation()
>>> f.is_conjugate_to(g)


The following code was used to find this example. It lists all the different mapping classes with word length at most three. It filters out the pseudo-Anosov ones and sorts them by dilatation. We group these by dilatation and then look for things in a group that are not conjugate to the first one.

>>> X = set(S.all_mapping_classes(length=3, letters=list('abcABC')))
>>> Y = sorted([x for x in X if x.is_pseudo_anosov()], key=lambda x: x.dilatation())
>>> Z = [list(b) for _, b in groupby(Y, key=lambda y: y.dilatation())]
>>> Z
[[a.b.C, a.C.b, a.C.B, a.B.C], [a.c.B, a.B.c]]

>>> for P in Z: print([p.is_conjugate_to(P[0]) for p in P])
[True, True, False, False]
[True, True]

It's not too hard to see why this example works. Namely $f$ is actually conjugate to $g^{-1}$ so they must have the same dilatation. But (the stable lamination of $f$) is lacking the symmetry needed for $f$ to be conjugate to $f^{-1}$, so $f \not \equiv f^{-1} \equiv g$:

>>> f.is_conjugate_to(g**-1)
  • 1
    $\begingroup$ How did you find this example, Mark? Is there any deeper theory behind it? $\endgroup$ Oct 8, 2016 at 13:00
  • 1
    $\begingroup$ It was found using (brute force searching with) flipper. But given the mapping classes I think it's easy to convince yourself why they should work. $\endgroup$
    – Mark Bell
    Oct 8, 2016 at 14:37
  • 2
    $\begingroup$ Of course, if you go to length five you'll discover f = "a.B.c.B.c" and g = "a.B.c.c.B". These have the same dilatation, are not conjugate, and $f$ is not conjugate to $g^{-1}$ either. I don't know any reason why these two maps should have the same dilatation. $\endgroup$
    – Mark Bell
    Oct 9, 2016 at 8:47
  • $\begingroup$ Both f and g lie in the stratum [0,0] - that is, they are secretly Anosov maps of the torus. Now there are several ways to win. Perhaps, after capping off the punctures, f and g become inverses. Or perhaps they become identical and f and g were just choosing "different pairs" of fixed points to puncture. Or perhaps they now differ by the hyper-elliptic involution... To figure this out, it should suffice to cap off just once - can flipper do that? $\endgroup$
    – Sam Nead
    Nov 12, 2016 at 2:46
  • $\begingroup$ Excellent suggestion! Unfortunately flipper can't do the capping (but you're welcome to submit a patch). However if I fill in one puncture then the a and c curves become equivalent. On the once punctured torus, flipper then tells me that the maps f = aBaBa and g = aBaaB are indeed conjugate (although not equal). $\endgroup$
    – Mark Bell
    Nov 13, 2016 at 7:50

Here is a general source of examples. Suppose $f$ is a pseudo-Anosov map whose centralizer is not cyclic. Say $\rho$ is finite order and is in the centralizer of $f$. Note that $f$ and $\rho f$ have the same dilatation, because they have a common power. However, $\rho$ can change the way the singularities (and the separatrices) are permuted, and this permutation data is also a conjugacy invariant.

To be concrete, suppose that $f$ is any pseudo-Anosov map on the twice-punctured torus and that $\tau$ is the hyperelliptic element. Then $f$ and $\tau f$ are not conjugate. With a bit of care you can play similar games in hyper-elliptic strata in any genus.

There are more subtle obstructions as well -- see the paper Polynomial invariants of pseudo-Anosov maps by Birman, Brinkmann, and Kawamuro. Near the end of the paper they say: "It seems to be an open question to describe all the ways to construct all pA maps having a fixed dilatation."

[Edit] Here is another construction. Suppose that $f$ is an Anosov map on the torus. Take a power to ensure that $f$ has at least two fixed points $x$ and $y$. Now form $S_g$ by taking a $g$--fold branched cover over $x$ and $y$. So $S_g$ is a surface of genus $g$. We can take a further power of $f$ to ensure that it lifts to a map on $S_g$. Note that $f$, acting on $S_g$, has exactly two singularities. Ok. Now form the surface $T$ by taking a unbranched double cover of $S_2$. So $T$ has genus three and is homeomorphic to $S_3$. There is again a power of $f$ that lifts to $T$, and this lift is a pA map with four singularities. Taking the correct powers of $f$ equips $S_3$ and $T$ with pA maps of the same dilatation, but with different numbers of singularities.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.