Shin-Strenner: Pseudo-Anosov mapping classes not arising from Penner’s construction

Balazs Strenner, a Ph.D. student of Richard Kent graduating this year, gave a beautiful talk yesterday in our geometry/topology seminar about his recent paper with Hyunshik Shin.  (He’s at the Institute next year but if you’re looking for a postdoc after that…!)

A long time ago, Robert Penner showed how to produce a whole semigroup M in the mapping class group with the property that all but a specified finite list of elements of M were pseudo-Anosov.  So that’s a good cheap way to generate lots of certified pseudo-Anosovs in the mapping class group.  But of course one asks:  do you get all pA’s as part of some Penner semigroup?  This can’t quite be true, because it turns out that the Penner elements can’t permute singularities of the invariant folation, while arbitrary pA’s can.  But there are only finitely many singularities, so some power of a given pA clearly fixes the singularities.

So does every pA have a power that arises from Penner’s construction?  This is what’s known as Penner’s conjecture.  Or was, because Balazs and Hyunshik have shown that it is falsitty false false false.

When I heard the statement I assumed this was going to be some kind of nonconstructive counting argument — but no, they actually give a way of proving explicitly that a given pA is not in a Penner semigroup.  Here’s how.  Penner’s semigroup M is generated by Dehn twists Q_1, … Q_m, which all happen to preserve a common traintrack, so that there’s actually a representation

\rho: M \rightarrow GL_n(\mathbf{R})

such that the dilatation of g is the Perron-Frobenius eigenvalue \lambda of \rho(g).

Now here’s the key observation; there is a quadratic form F on R^n such that F(Q_i x) >= F(x) for all x, with equality only when x is a fixed point of Q_i.  In particular, this shows that if g is an element of M not of the form Q_i^a, and x is an arbitrary vector, then the sequence

x, g x, g^2 x, \ldots

can’t have a subsequence converging to x, since

F(x), F(gx), F(g^2 x), \ldots

is monotone increasing and thus can’t have a subsequence converging to F(x).

This implies in particular:

g cannot have any eigenvalues on the unit circle.

But now we win!  Because \rho(g) is an integral matrix, so all the Galois conjugates of \lambda must be among its eigenvalues.  In other words, \lambda is an algebraic number none of whose Galois conjugates lie on the unit circle.  But there are lots of pseudo-Anosovs whose dilatations \lambda do have Galois conjugates on the unit circle.  In fact, experiments by Dunfield and Tiozzo seem to show that in a random walk on the braid group, the vast majority of pAs have this property!  And these pAs, which Shin and Strenner call coronal, cannot appear in any Penner semigroup.


Anyway, I found the underlying real linear algebra question very appealing.  Two idle questions:

  • If M is a submonoid of GL_n(R) we may say a continuous real-valued function F on R^n is M-monotone if F(mx) >= F(x) for all m in M, x in R^n.  The existence of a monotone function for the Penner monoid is the key to Strenner and Shin’s theorem.  But I have little feeling for how it works in general.  Given a finite set of matrices, what are explicit conditions that guarantee the existence of an M-monotone function?  Nonexistence?  (I have a feeling it is roughly equivalent to M containing no element with an eigenvalue on the unit circle, but I’m not sure, and anyway, this is not a checkable condition on the generating matrices…)
  • What can we say about the eigenvalues of matrices appearing in the Penner subgroup?  Balazs says he’ll show in a later paper that they can actually get arbitrarily close to the unit circle, which is actually not what I had expected.  He asks:  are those eigenvalues actually dense in the complex plane?
Tagged , , , ,

2 thoughts on “Shin-Strenner: Pseudo-Anosov mapping classes not arising from Penner’s construction

  1. Concerning your first idle question, I’m a little confused why one couldn’t set F(x) = 0. On the other hand, if you define the function F(x) = |x| + 1/|x|, then one has the inequality F(m^k x) > F(x) for all non-zero x and any sufficiently large k (depending on x) exactly when m has no eigenvalues of absolute value 1.

  2. JSE says:

    Re the first thing, you really want the inequality to be strict “most of the time” in whatever the right sense is. Maybe an explicit finite set of curves is allowed to be excluded, or something?

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: