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

such that the dilatation of g is the Perron-Frobenius eigenvalue of .

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

can’t have a subsequence converging to x, since

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 is an integral matrix, so all the Galois conjugates of must be among its eigenvalues. In other words, is an algebraic number none of whose Galois conjugates lie on the unit circle. But there are lots of pseudo-Anosovs whose dilatations *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.

Cool!

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?

### Like this:

Like Loading...