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SURFACE GROUPS

R. ANDREW KREEK

1. Introduction

Definition 1.1. We define a surface group to be the group given the by presentation

g = a1, b1, a2, b2 . . . ag, bg|[a1, b1] . . . [ag, bg] = 1.

Question 1.2. For what integers g and h is h a subgroup of g?

As a first step, we observe the connection between the surface group g and the closed,orientable genus-g surface g.

Computation 1.3. The fundamental group 1(g) of the closed, orientable genus-g surfaceg is the surface group g.

Proof. First, recall Proposition 1.26 from Hatchers Algebraic Topology regarding applicationsof van Kampens Theorem to cell complexes: If we attach 2-cells to a path connected space Xvia maps , making a space Y, and N 1(X, x0) is the normal subgroup generated by allloops

1

, then the inclusion X Y induces a surjection 1(X, x0) 1(Y, x0) whosekernel is N. Thus 1(Y) 1(X)/N.

Consider the wedge-sum of 2g circles, labeled a1, a2, . . . , ag, and b1, b2, . . . , bg. This will bethe 1-skeleton of a cell-decomposition of the genus-g surface. By gluing a single 2-cell along theword a1b1a

1

1b11

. . . agbga1

g b1

g , we obtain the closed, orientable genus-g surface. This can bevisualized, as in Figure 1 below, by the familiar identification space of a genus-g surface as a4g-gon with pairs of edges, and all of the vertices, identified.

a

a

b

b

c

c

d

d

-1

-1

-1

-1

Figure 1. Genus-2 surface as a quotient of an octogon

By van Kampens theorem, 1(2gi=1 S

1

i ) of 2g circles is the free group on 2g generators, and

by Proposition 1.26, 1(g) is the quotient of 1(2g

i=1 S1

i ) by the normal subgroup generated

Date: August 17, 2007.

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by the word [a1, b1] . . . [ag, bg], which is precisely the surface group g. This can be explainedby the observation that the loop [a1, b1] . . . [ag, bg] (or any of its conjugates) is nullhomotopicas a result of having added the 2-cell.

2. Preliminaries

Proposition 2.1. h is a covering space of g if and only if (g)|(h).

Proof. It is well known, and easy to verify, that (g) = 2 2g.Suppose (g)|(h), then for some n N, n(2 2g) = 2 2h, so h = n(g 1) + 1, and

we see by cutting along the loops L, as in figure 2, and identifying them, that h is a coveringspace of g.

L

L

L

L

L

h

g

p

Figure 2. 5 covering 2 with 4-fold symmetry

Now suppose that h is an n-sheeted covering space of g. Recall that ifTg is a triangulationof the surface g, and V, E, and F are the number of vertices, edges, and faces ofT respectively,

then we can calculate the Euler characteristic of g by (g) = V E + F. Let Tg be atriangulation of g. Then Tg lifts to a triangulation Th of h. Note, by Proposition 1.33,

suppose given a covering space p : ( X, x0) (X0, x0) and a map f : (Y, y0) (X, x0) withY path-connected and locally path-connected. Then a lift f : (Y, y0) ( X, x0) of f exists ifff(1(Y, y0)) p(1( X, x0)). Any contractible space has trivial fundamental group, henceany map from a contractible space to g lifts. Then, for each vertex e0 in Tg, there are npre-images of e0 in Th. Hence, if V is the number of vertices in Tg, nV is the number ofvertices in Th. Similarly, since the edges and the faces of the triangulation are contractible,each one lifts to n distinct pre-image edges and and faces. Thus, ifTg has E edges and F faces,Th has nE edges and nF faces. Thus (g) = V E+ F, and (h) = nV nE + nF =n(V E+ F) = n(g).

If the covering space

g of g is an infinite genus surface, then the statement (

g)|(g)

is not well defined. In general, for an infinite-sheeted cover, it may not be true that (g)|(g).But this does not create any problems because the fundamental group of an infinite genus sur-face is not a surface group.

Proposition 2.2. The fundamental group of a non-compact surface is free.

We shall use this fact without proof.2

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Figure 3. Example of an infinite-genus surface

3. The Answer

Theorem 3.1. Given integers g and h, h < g if and only if g 1|h 1.

Proof. This proof hinges on the Fundamental Theorem of Covering Spaces, quoted here fromfrom Hatcher:

Let X be path-connected, and semilocally simply-connected. Then there is a bijection be-tween the set of basepoint-preserving isomorphism classes of path-connected covering spaces

p : (

X,x0) (X0, x0) and the set of subgroups of 1(X, x0), obtained by associationg the

subgroup p(1( X, x0)) to the covering space ( X, x0). If basepoints are ignored, this corre-spondences gives a bijection between isomorphism classes of path-connected covering spaces

p : X X and conjugacy classes of subgroups of 1(X, x0).Now, suppose g 1|h 1. Then, 2 2g|2 2h, so (g)|(h) for the genus-g and -

h surfaces g and h. Hence, by Proposition 2.1, h is a covering space of g, so h =1(h) < 1(g) = g.

Next, suppose h < g. By the Fundamental Theorem of Covering Spaces, there is a

covering space g of g such that 1(g) = h. By proposition 2.2, g is not an infinite-sheeted cover since its fundamental group is h. Therefore, it must be some closed, orientablesurface with finite genus. However, since only the surface h has the fundamental group h,g = h. Thus, h is a covering space of g, hence g 1|h 1.

References[1] A. Hatcher. Algebraic Topology. Cambridge University Press. 2002.

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