## The conformal modulus of a mapping class

(Warning — this post concerns math I don’t know well and is all questions, no answers.)

Suppose you have a holomorphic map from C^* to M_g,n, the moduli space of curves.  Then you get a map on fundamental groups from $\pi_1(\mathbf{C}^*)$ (otherwise known as Z) to $\pi_1(\mathcal{M}_{g,n})$ (otherwise known as the mapping class group) — in other words, you get a mapping class.

But not just any mapping class;  this one, which we’ll call u, is the monodromy of a holomorphic family of marked curves around a degenerate point.  So, for example, the image of u on homology has to be potentially unipotent.  I’m not sure (but I presume others know) which mapping classes u can arise in this way; does some power of u have to be a product of commuting Dehn twists, or is that too much to ask?

In any event, there are lots of mapping classes which you are not going to see.  Let m be your favorite one.  Now you can still represent m by a smooth loop in M_g,n.  And you can deform this loop to be a real-analytic function

$f: \{z: |z| = 1\} \rightarrow \mathcal{M}_{g,n}$

Finally — while you can’t extend f to all of C^*, you can extend it to some annulus with outer radius R and inner radius r.

Definition:  The conformal modulus of a mapping class x is the supremum, over all such f and all annuli, of (1/2 pi) log(R/r).

So you can think of this as some kind of measurement of “how complicated of a path do you have to draw on M_{g,n} in order to represent x?”  The modulus is infinite exactly when the mapping class is represented by a holomorphic degeneration.  In particular, I imagine that a pseudo-Anosov mapping class must have finite conformal modulus.  That is:  positive entropy (aka dilatation) implies finite conformal modulus.   Which leads Jöricke to ask:  what is the relation more generally between conformal modulus and (log of) dilatation?  When (g,n) = (0,3) she has shown that the two are inverse to each other.  In this case, the group is more or less PSL_2(Z) so it’s not so surprising that any two measures of complexity are tightly bound together.

Actually, I should be honest and say that Jöricke raised this only for g = 0, so maybe there’s some reason it’s a bad idea to go beyond braids; but the question still seems to me to make sense.  For that matter, one could even ask the same question with M_g replaced by A_g, right?  What is the conformal modulus of a symplectic matrix which is not potentially unipotent?  Is it always tightly related to the size of the largest eigenvalue?

## 2 thoughts on “The conformal modulus of a mapping class”

1. JSE says:

One of my topologist readers who doesn’t like commenting tells me that the monodromy of a map from C^* has to be have the property that some finite power is a product of disjoint Dehn twists.

2. chris says:

I think Royden’s Theorem (Teichmueller metric = Kobayashi metric) answers a couple of these questions. My constants here are probably off…

A holomorphic map of an annulus (with finite modulus) lifts to a periodic
holomorphic map of the hyperbolic plane H^2 into Teichmueller space. The
action of Z on H^2 is generated by a hyperbolic isometry f, while the
action on Teichmuller space is generated by the mapping class u determined by the annulus.

According to Royden’s Theorem, the map of H^2 (with curvature -4) is a
contraction, and so the translation length L(f) on H^2 is at least the
translation length L(u) on Teichmueller space. On the other hand, the
modulus of the annulus H^2/ is exactly pi/(2L(f)) (maybe the 2 is in
the wrong place… I get confused with curvature -4). Therefore, the
modulus pi/(2L(f)) is at most pi/(2L(u)).

Now, Bers’ proof of Thurston’s classification theorem says that L(u) = 0
if and only if u has a power which is a product of commuting Dehn twists.
Therefore infinite modulus implies L(u) = 0, and hence u is a product of
commuting Dehn twists.

On the other hand, if u is pseudo-Anosov, then L(u) = ent(u), the entropy
of u. So the modulus is at most pi/(2 ent(u)). The Teichmueller disk
defined by the axis of u gives an isometric periodic embedding of H^2 into
Teichmueller space, making the inequalities into equalities, so that
modulus = pi/(2ent(u)).

If u is reducible, it’s not clear to me whether you can find a sequence of
annuli for which the supremum of the associated moduli is exactly
pi/(2L(u)).