## Pseudo-Anosov puzzle 2: homology rank and dilatation

In fact, following on what I wrote about the two Farb-Leininger-Margalit theorems below, one might ask the following.  Is there an absolute constant c such that, if f is a pseudo-Anosov mapping class on a genus g surface, and the f-invariant subspace of H_1(S) has dimension at least d, then

log λ(f) >= c (d+1)  / g?

This would “interpolate” between Penner’s theorem (the case d=0) and the F-L-M theorem about Torelli (the case d=2g).

## 2 thoughts on “Pseudo-Anosov puzzle 2: homology rank and dilatation”

1. ianagol says:

Hey Jordan,
I think I can show that g log(lambda) grows like log(d). This follows from Theorem 6.2 from my paper “Ideal Triangulations of Pseudo-Anosov Mapping Tori”, which gives an alternate proof of Farb-Leininger-Margalit. If one has a pA map with a d-dimensional fixed subspace of homology, then the mapping torus has first betti number at least d+1. If one takes the singular fibers of the pA flow, then one can show that they generate H_1 rationally. So the number of singular fibers is at least d+1. When you drill these fibers, you get a cusped hyperbolic 3-manifold with at least d+1 cusps. One can show that the volume of such a manifold grows at least linearly in d (due to Colin Adams), and therefore the minimal number of tetrahedra in an ideal triangulation grows linearly in d. By my result Theorem 6.2, log(number of tetrahedra) gives a linear lower bound for g log(lambda), so it grows at least like log(d).

Here’s a link to the paper:
http://front.math.ucdavis.edu/1008.1606

Your conjecture sounds plausible, I think the sort of estimates that are made in the paper are far from being sharp.

best,
Ian Agol

2. ianagol says:

This is an addendum to the comment: Actually, the part I said about the singular fibers generating H_1 rationally might not be right – I thought I had a proof of this, but it’s not right. In any case, it’s really easy to see that the number of tetrahedra has to grow linearly with d (volume is irrelevant), since the rank grows linearly with the number of tetrahedra, so the argument should still work.