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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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