The Mapping Class Group

Last time, we touched on the mapping class group, so now, we’re going to dig in. Now, we’re not going to dig too deeply, there’s a LOT here (see the wonderful book by Farb and Margalit for a hint at what’s there) and for now, my primary reference is Minsky’s set of lectures from the PCMI summer institute in 2011 on moduli of Riemann surfaces.

Let $S$ be a topological surface.  Then it has an astoundingly large group $\mathrm{Homeo}(S)$ of homeomorphisms.  This is WAY too big, and has infinitely many connected components.  So first, we restrict to the homeomorphisms that preserve orienations, $\mathrm{Homeo}^+(S)$, and then we mod out by isotopy, that is, continuous families of homeomorphisms.  So we’re looking at the set of connected components of $\mathrm{Homeo}^+(S)$.  We will denote this group equivalently by $\mathrm{MCG}(S)$, by $\mathrm{Mod}(S)$ and by $\Gamma_{g,n}$ where $g$ is the genus of $S$ and $n$ is the number of punctures.

For a moment, let’s just notice that this is an extremely robust notion.  Though there’s work required, as long as we restrict to punctured, oriented surfaces, we can replace homeomorphisms with diffeomorphisms and isotopies with homotopies and we get the same group.  However, there are two changes that happen occasionally that do change the group, albeit in small ways:

1. If we forget about the orientation and allow it to be reversed, then we get an index two extension compared to the above group
2. If we allow boundary discs instead of punctures, things get a little bit more complicated.  If we restrict to homeomorphisms and isotopies that fix the boundary pointwise, then we end up with an abelian extension of the above group by Dehn twists (mentioned last time, but more on them below) around the boundaries.

For the very simplest surfaces, we can actually compute the mapping class group very explicitly.  For the sphere with at most three punctures, as well as the torus with no punctures, we can get really explicit (though the proofs are long and technical, and thus serve little point in a blog post!)

Note that $\mathbb{R}^2$ is the sphere with one puncture.  Both it and the sphere have trivial mapping class groups.  This is essentially because we can take a circle, apply a homeomorphism, and then isotope so that the circle is back to where it started.  A hint in the direction of the proof is to use the Jordan Curve Theorem and Alexander’s Trick.

The twice punctured sphere is the annulus, and it turns out to have a single nontrivial element in its mapping class group, consisting of imagining the annulus as a right cylinder and rotating it 180 degrees around a diameter of the circle.  Essentially, what this does is it swaps the two punctures.  A similar story happens with the pair of pants, which is the three times punctured sphere.  It turns out that $\Gamma_{0,3}\cong S_3$, the permutations of the punctures.  For $n\geq 4$, things get more complex, and this essentially reflects the fact that (as we will see) the moduli space $M_{0,n}$ for $n\geq 4$ is positive dimensional.

Now, before we talk a bit about the specific types of elements of mapping class groups in general (next time! This post is already getting a bit longer than intended), we’ll talk about the mapping class group of the torus.

In general, we get a natural map $\Gamma_{g,n}\to \mathrm{Aut}(H_1(S_{g,n};\mathbb{Z}))$ because isotopically trivial homeomorphisms act trivially on homology.  This preserves the intersection form, so when there are no punctures, we get $\Gamma_{g,0}\to \mathrm{Sp}(2g,\mathbb{Z})$.  If $g=1$, then we have an isomorphism $\mathrm{Sp}(2,\mathbb{Z})\cong \mathrm{SL}_2(\mathbb{Z})$. (And a side note, personally, this map is essentially the exact same thing as the Torelli map taking a curve to its Jacobian, and so is closely related to the Schottky problem, on which I wrote my thesis)

So, we want to analyze this map to compute $\Gamma_{1,0}$, and it will turn out to be an isomorphism.  Surjectivity follows because any element of $\mathrm{SL}_2(\mathbb{Z})$ acts on $\mathbb{R}^2$, but preserves $\mathbb{Z}^2$, and so acts on the torus, and because it’s invertible, this is a homeomorphism.  Injectivity isn’t really much harder, it can be done by lifting the question to $\mathbb{R}^2$ and showing that any map that is the identity on homology must be isotopic to the identity.

So now that we can describe the mapping class group of a torus, there are three types of elements to look at:

• Elliptic: An elliptic element is any element of finite order.  These are very rare, only corresponding to rotation of the square lattice, the hexagonal lattice, the powers of these maps, and their conjugates.  These have trace of -1, 0 or 1.
• Parabolic: These elements are also called twist maps, and generalize to Dehn twists that keep coming up.  These maps preserve (up to isotopy) exactly one simple closed curve on the torus, and they have trace 2 or -2.
• Hyperbolic: The hyperbolic elements, also called Anosov, have trace outside of $[-2,2]$, and this is where things get interesting, which means we won’t go too deep.  They have two foliations by straight lines (well, on the plane covering it), and they’re transverse invariant foliations, one of which has lengths along it expanded by the element, and one of which is contracted. Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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2 Responses to The Mapping Class Group

1. prof dr mircea orasanu says:

this subject is very interesting with great implications stated prof dr mircea orasanu and prof drd horia orasanu for Riemann surfaces and analytical complex functions

2. prof dr mircea orasanu says:

in thus cases must to precised that MAPPING GROUP THEORY represented as a consequence of conformal transform introduced by Riemann and other and here are stated with prof dr mircea orasanu and prof drd horia orasanu and followed that appear in many studies as to say in LAGRANGIAN OPERATORS and for thus the proportionality constant for the drag force and it has units of mass/time, , where m is the charged particle’s mass and is the effective time between collisions. The full equation can be solved as follows. First assume that the electric field is zero. We are then left with the equation,
which can be easily solved to find
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where is an unknown constant. Now we know that the electric field will cause a steady state drift, thus, we can assume that the velocity is of the form,
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The differences between these types of materials can be understood from solid state theory. [ ]
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