Classification of Surfaces
Richard Koch
November 20, 2005

1

Introduction

We are going to prove the following theorem:
Theorem 1 Let S be a compact connected 2-dimensional manifold, formed from a polygon
in the plane by gluing corresponding sides of the boundary together. Then S is homeomorphic to exactly one of the following:• T 2 # . . . #T 2 , that is, a sphere or g-holed torus• T 2 # . . . #T 2 #RP 2 , that is, a connected sum of a g-holed torus and a projective space• T 2 # . . . #T 2 #K, that is, a a connected sum of a g-holed torus and a Klein bottle

It can be proved that any compact connected 2-dimensional manifold can be obtained inthis way, so we are actually classifying all compact connected 2-dimensional manifolds.
But our comments about the general case involve some hand wringing, while our proof of
the above theorem is completely rigorous.
The truth of the above theorem is far from obvious. For example, consider the following
two-dimensional objects. It is not immediately obvious that these have the above form.
However, it is easy to show that each of these objects can be obtained as a quotient objectof a polygon by gluing corresponding sides together. Namely, cut the object into pieces,and mark the boundaries to explain how they are to be glued back together. Then arrange
these pieces in the plane, deforming the pieces if necessary to make them fit, and marking
the remaining unconnected edges along the boundary of this region.

1

Our proof of the theorem comes from Introduction to Topology by Solomon Lefschetz, andfrom Lehrbuch der Topologie by H. Seifert and W. Threlfall. Lefschetz was one of thetwentieth centurie’s greatest topologists, although the Princeton graduate students liked
to say “Lefschetz never stated an incorrect theorem, nor gave a correct proof.” Lefschetz
is one of the characters in the book and movie A Beautiful Mind. Seifert is also a very
important topologist, and Seifert and Threlfall’s book is often quoted. I highly recommendit.

2

Preliminaries

Suppose we have a polygon with sides labeled a, b, and so forth. We can read off thesides clockwise along the boundary, obtaining a string called the symbol of the diagram.
This symbol can be replaced with a cyclic permutation without changing the polygon. For
example, the diagram below has symbol abd−1 bda−1 and also symbol d−1 bda−1 ab.

2We wish to glue corresponding sides together. To get a manifold, each side must occur
exactly twice along the boundary, as the following pictures show. There might also be a
condition at vertices to insure that the quotient space is locally Euclidean at the vertices,
but the proof will show that no such vertex condition is needed.

3

The Standard Models

Theorem 2 The g-holed torus T 2 # . . . #T 2 can be represented by a polygon with 4g sides−1−1 −1
and symbol a1 b1 a−1
1 b1 . . . ag bg ag bg .
Proof: The following picture of T 2 has symbol aba−1 b−1 .

3Inductively assume the above theorem is true for a g-holed torus; we will prove it for theconnected sum of this torus with one additional T 2 . The picture below shows each object
with a disk removed. The boundary of this disk is labeled d. To form their connected sum,
we must glue the two objects together along d. This is shown in the second picture below,−1which produces the same sort of symbol with one additional ag+1 bg+1 a−1
g+1 bg+1 .

Theorem 3 The connected sum of g copies of projective space RP 2 # . . . #RP 2 can be
represented by a polygon with 2g sides and symbol a1 a1 a2 a2 . . . ag ag .
Proof: We prove this by induction; it is certainly true when g = 1 because the object below
describes our standard method of forming projective space.

Assume the theorem is true for a connected sum of g copies of projective space. We
4will show it true when we add one additional copy of projective space. To form this lastconnected sum, we must remove a disk from each object and glue them together along theboundary of these disks. This process is shown below.

Theorem 4 The following objects are homeomorphic: RP 2 #RP 2 #RP 2 ≈ K#RP 2 ≈T 2 #RP 2 .
Proof: The following picture shows that K ≈ RP 2 #RP 2 , and thus proves the first half of
the assertion.

5

The symbol for RP 2 #RP 2 #RP 2 is aabbcc by theorem three. The following picture showsthat the symbol for T 2 #RP 2 is aba−1 b−1 cc. We must prove that these symbols give isomorphic objects.

Lemma 1 If x represents a side and P and Q represent sequences of sides, then xxP −1 Q ≈x1 P x1 Q for an appropriate side x1 .
Proof: See the picture below.

Remark: Using this result, we can prove aba−1 b−1 cc ≈ aabbcc algebraically. Each line of
the argument below is an application of the above lemma. To make the argument clearer,
we place parentheses around P and Q. Sometimes we will cyclically permute the symbols;in that case the original placement is at the end of one line and the permutation is at the
beginning of the next line. We start with aba−1 b−1 cc = cc(ab)(a−1 b−1 ).

6

cc(ab)(a−1 b−1 ) → c1 (ab)−1 c1 (a−1 b−1 ) = c1 b−1 a−1 c1 a−1 b−1
b−1 (a−1 c1 a−1 )b−1 (c1 ) → b1 b1 (a−1 c1 a−1 )−1 (c1 ) = b1 b1 ac−1
1 ac1
a(c−1
1 )a(c1 b1 b1 ) → a1 a1 (c1 )(c1 b1 b1 ) = a1 a1 c1 c1 b1 b1
Theorem 5 The space RP 2 # . . . #RP 2 is homeomorphic to T 2 # . . . #T 2 #RP 2 if thenumber of copies of RP 2 is odd, and to T 2 # . . . #T 2 #K if the number of copies of RP 2
is even.
Proof: Write RP 2 # . . . #RP 2 = (RP 2 #RP 2 )#(RP 2 #RP 2 )# . . . #(RP 2 #RP 2 )#M where
M is either one copy of RP 2 or a connected sum of two copies of RP 2 . Using theorem 4,
we can replace of pair RP 2 #RP 2 by T 2 since there is at least one additional RP 2 at theend. The remaining M is either RP 2 or RP 2 #RP 2 ≈ K.
Remark: The proof of our main theorem thus reduces to showing that any symbol of a poly−1−1 −1
gon can be converted to one of the two forms a1 b1 a−1
1 b1 . . . ag bg ag bg or a1 a1 a2 a2 . . . ag ag
and that the manifolds corresponding to different such symbols are not homeomorphic.

4First Reduction

When the sides of a polygon represented by a symbol are identified, it may or may nothappen that all vertices glue to the same point. For example, the vertices of the polygon
abcabc glue to three distinct points, but the vertices of aba−1 b−1 glue to the same point,
as the pictures below show.

7

Theorem 6 A polygon can always be replaced with an equivalent polygon in which allvertices glue to the same point.
Proof: Suppose that one of the vertices after gluing is P but it is not the only vertex.
We will produce a new diagram with one fewer P -vertex and one additional Q-vertex for
Q 6= P . Start by tracing the boundary of the polygon until an edge joins a P -vertex to a
Q-vertex for Q 6= P . Let a be the previous edge; in the picture R may or may not be a
P -vertex. Cut as indicated and glue to the matching a edge as in the picture below. Noticethat in the new polygon the number of P -vertices has been reduced by one.

Repeat this process until there is only one P -vertex. Then the two edges joining this vertex
must have arrows pointing in to P , or else arrows pointing out from P because otherwise
there would be another P -vertex at the end of one of these edges. We can join the edges
as indicated below, completely eliminating the final P -vertex.

8

5

Reduction Techniques

The remaining reductions depend on two simple observations. The first has already beenpresented, but we list it here again:
Lemma 1 If x represents a side and P and Q represent sequences of sides, then for an
appropriate side x1 :xxP −1 Q ≈ x1 P x1 Q

Lemma 2 If x represents a side and P, Q, and R represent sequences of sides, then for
an appropriate side x1 :xP Qx−1 R ≈ x1 QP x−1
1 R

Remark: Notice that we can tack on an extra S at the front of these symbol sequences,since the sequences can be symmetrically permuted. For instance, SxxP −1 Q = xxP −1 QS ≈x1 P x1 QS = Sx1 P x1 Q.
Each side a in a symbol sequence appears twice, either in the form aP aQ or in the formaP a−1 Q. The lemmas say that any sequence P of sides inside an a-a pair can be thrownoutside if inverted, and any pair P Q of sequences of sides inside an a-a−1 pair can becommuted.

96

Second Reduction

Theorem 7 We can always reduce a symbol sequence to a form in which all a-a pairs
appear side by side.
Proof: If the sequence has the form SxP xQ, replace it with the sequence Sx1 x1 P −1 Q.
This process will never destroy an already existing adjacent pair because both members of
the pair can be assumed to belong to S or P or Q and the pair will still be adjacent in S
or P −1 or Q. Continue until all a-a pairs are adjacent.
Remark: When the author of our book discusses connected sums, he leaves unproved acentral fact: the connected sum M #N only depends on M and N , and not on the choicesof the disks D1 and D2 to be removed from M and N or the gluing homeomorphism ϕfrom the boundary of D1 to the boundary of D2 . The omission of this fact leaves proofs
which depend on connected sums somewhat shaky.
−1−1 −2
Consequently, in the reduction of surface symbols to the forms a1 b1 a−1
1 b1 . . . ag bg ag bg
or a1 a1 . . . ag ag , we are not going to use connected sums. We only use connected sums toidentify the ultimate canonical forms with familiar objects. This step could be replaced
by a direct construction of these two symbols from standard models of our canonical objects.

If we were willing to use connected sums in the reduction argument, we could simplify thatargument a little at this moment. Consider the picture below, which shows an adjacentpair a-a. Suppose we cut off this adjacent pair along the dotted edge d. The

picture shows that the piece which has been cut off is homeomorphic to a Mobius band.
Suppose we connect the sides b and c which came just before and just after the a-a pair.
10

The new polygon would represent a simpler compact surface, and the picture shows thatthe original surface could be obtained from this simpler surface by cutting out a disk andgluing in a Mobius band. Said another way, if the simpler surface is M then the original
surface is M #RP 2 .
We could cut out all adjacent a-a pairs, so the resulting simpler polygon would only containa-a−1 pairs. An argument from the following section of these notes would then show thatsuch a polygon represents a g-holed torus. Consequently the original polygon represents a
connected sum of such a torus with several copies of RP 2 .
But we will not do that, so at the moment our polygon contains a-a pairs and also a-a−1
pairs, and all of the a-a pairs are adjacent.

7

Third Reduction

Definition 1 Two sides a and b in a symbol form a crossed pair if up to cyclic permutation
the symbol has the form P aQbRa−1 Sb−1 T .
Theorem 8 A symbol can always be converted to a reduced form in each crossed pairconsists of adjacent sides aba−1 b−1 and each a-a pair consists of adjacent edges aa.
Proof: Consider the symbol P xQyRx−1 Sy −1 T. We can assume that each adjacent pair aa
and each previously simplified adjacent pair aba−1 b−1 occurs in one of P , Q, R, S, or T .
−1
We will simplify the given symbol so x and y are replaced by adjacent sides x1 y1 x−1
1 y1 .
In the process we will not change the order of sides in P , Q, R, S, or T Hence this process
can be continued until all crossed pairs are adjacent.
The reduction will use the xP Qx−1 ≈ x1 QP x−1
1 simplification of lemma 2. To make theprocess easier to read, we surround P and Q with parentheses before applying the lemma.
Occasionally we will cyclically permute symbols; in these cases, the original order is at theend of one line, and the permuted symbol is at the beginning of the next line.
−1
P x(Q)(yR)x−1 Sy −1 T → P x1 (yR)(Q)x−1
1 Sy T−1−1−1
P x1 y(RQ)(x−1
1 S)y T → P x1 y1 (x1 S)(RQ)y1 T−1−1 −1
y1 x−11 (SRQ)(y1 T P )x1 → y1 x2 (y1 T P )(SRQ)x2−1
x1 y1 x−1
2 y1 T P SRQ

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8

Eliminating Uncrossed Pairs−1
At this point, all a-a pairs are adjacent and all crossed pairs a1 b1 a−1
1 b1 are adjacent, butthese groups of adjacent symbols may be separated by sequences of other unclassified sides.
These sides must belong to a-a−1 pairs because all a-a pairs are adjacent. So suppose wehave a symbol representing a polygon of the form below.

Notice that whenever b is an edge in the top dotted sequence of this picture, the matchingb edge must also belong to this top sequence. This certainly holds for z-z edges or crossed
pairs because these sequences are adjacent, but it also holds for unclassified edges because
if b is in the top sequence and b−1 is in the bottom sequence, then a-b represents a crossededge.
Similarly, whenever b is an edge in the bottom dotted sequence of this picture, the matchingb edge must also belong to this bottom sequence.
But then the starting point P of the side a and the ending point Q of the side a represent
difference vertices in the quotient manifold, contradicting the first reduction of section four.
Indeed, the Q’s from both copies of side a are in the top dotted section, and all sides inthis section have matching sides also in this section, so none of these sides can require us
to glue Q to P . A similar remark holds for the bottom.
The unavoidable conclusion is that there are no unclassified sides and thus the complete
symbol consists of adjacent a-a and adjacent aba−1 b−1 sections, one after another, aroundthe boundary of the polygon.

9

The Final Step

−1
−1 −1
If there are no a-a pairs, then the symbol has the form a1 b1 a−1
1 b1 . . . ag bg ag bg and sois one of our canonical forms. If the symbol has no crossed pairs, then it is a1 a1 . . . ag ag
and so is the other of our canonical forms.

12

Suppose the symbol has both crossed pairs and a-a pairs. It then has the form aaP bcb−1 c−1 Q
up to cyclic permutation. Apply lemma 1 to convert this as follows:aa(P bc)(b−1 c−1 Q) → a1 (P bc)−1 a1 (b−1 c−1 Q) = a1 c−1 b−1 P −1 a1 b−1 c−1 Q
This new symbol has two new pairs b−1 . . . b−1 . . . and c−1 . . . c−1 . . . and one less crossed
pair. Reduce using the method of section 6. If there are other crossed pairs, repeat theargument. Ultimately all pairs of sides will have the form aa.
QED.

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UniquenessWe must still prove that if two surfaces correspond to different canonical symbols, thesurfaces are not homeomorphic. This task will be done in the second term of the course.
I’d like to give a very rough sketch of the method now. I run the risk that if you understand
too much of this paragraph, you won’t take the second term because you already know itall.
Our textbook is called Algebraic Topology, but where’s the algebra? It turns out that it
is possible to assign to each topological space X a group G(X). The assignment depends
only on the topology of X, so if X and Y are homeomorphic, then the groups G(X)and G(Y ) are isomorphic. It is much easier to decide if groups are isomorphic than todecide if topological spaces are homeomorphic; if we discover that G(X) and G(Y ) are
not isomorphic, then X and Y are not homeomorphic. Incidentally, if you have neverheard of a group, you can relax, because they will be defined completely from scratch in432/532.
Many different sorts of group have been assigned to topological spaces. We will discuss aparticular assignment, called the fundamental group and written π1 (X). To obtain π1 (X),
we fix a point p ∈ X and let π1 (X) be the set of all paths which begin and end at p. If
one of these paths can be deformed to another path through paths which all start and end
at p, we call the paths homotopic and declare that they define the same element of π1 (X).
Thus π1 (X) consists of homotopy classes of paths in X which begin and end at p.

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For example, the picture on the left below shows X = R2 − {0}. The paths γ1 and γ2 define
elements of π1 (X). The path γ1 cannot be deformed to the path which is constantly equal
to p because it is trapped by the origin. But γ2 can be deformed to a constant path. Thus
γ1 and γ2 define different elements of π. The picture on the right show two paths which
represent different elements of the fundamental group of a torus.

We can “add” two paths by first traversing one of them and after that the second one.
This gives a group structure to π1 (X). Usually this group is not abelian. For example, inthe picture below we can traverse γ1 and then γ2 , but it is not possible to deform this toa new path which traverses γ2 before it traverses γ1 .

What is π1 (S 1 )? The determination of this group is a high point of the second term.
Perhaps we can guess the answer. One possible path is the path which traverses the circle
exactly one counterclockwise. Call this path [1]. Then the sum of this path with itself
traverses the circle twice. Call this element [2]. Etc. We can also traverse the circle once
counterclockwise. Call this path [−1]. We can consider a constant path which just stays
at p without moving; call this path [0]. These examples and related ones suggest that
π1 (S 1 ) = Z.
A central result in the subject is called the Seifert-Van Kampen theorem. Suppose X =U ∪ V where U and V are open, and suppose U, V, and U ∩ V are pathwise connected. We
obtain a diagram14π1 (U)%&

π1 (U ∩ V)

π1 (X)&%
π1 (V )

According to the theorem, the images of π1 (U) and π1 (V) in π1 (X) generate π1 (X). Moreover, according to the theorem all relations among these generators are consequences ofidentifying an element in π1 (U) with an element in π1 (V) if both elements come from acommon element in π1 (U ∩ V).
Let us apply this theorem to a bouquet of circles shown below. I claim that the fundamentalgroup of such a bouquet is just the free group with one generator for each circle. This follows
from the Seifert-Van Kampen theorem and induction. For example, apply the theorem tothe indicated open sets U and V shown below. Then π1 (U) is the free group on a and b by
induction, π1 (V) = Z is the free group on c by the determination of the fundamental groupof a circle, and π1 (U ∩ V) = 0 because U ∩ V can be deformed to a point. The diagram
then implies that π1 (X) is the free group generated by a, b, c.

15

Finally, apply the Seifert-Van Kampen theorem to the identification polygon of a surface
with U and V as indicated below.

Notice that after the boundary of this disk is glued together, this boundary becomes abouquet of circles, with one circle for each pair of boundary edges. Notice also that U canbe deformed to this bouquet by just pushing everything out to the boundary, notice that
V can be deformed to a single point, and notice that U ∩ V can be deformed to a circle bypushing everything to the circle in the middle of this piece. Thus the diagram
π1 (U)%&

π1 (U ∩ V)

π1 (X)&%
π1 (V )

becomes16

According to the Seifert-Van Kampen theorem, the fundamental group of the surface isgenerated by π1 (U) and by π1 (V). The first of these is generated by the edges of the polygon
and the second group is trivial.
The relations are generated by π1 (U ∩ V), which equals the fundamental group of a circle.
There is just one generator of this group, and thus just one relation. This relation is
obtained by reading off these edges as we go around the boundary of the polygon. In theorientable case, π1 is generated by a1 , b1 , . . . , ag , bg with relation−1−1 −1
a1 b1 a−1
1 b1 . . . ag bg ag |bg = 1

In the nonorientable case, π1 is generated by a1 , a2 , . . . , ag with relation
a1 a1 a2 a2 . . . ag ag = 1
Unfortunately, π1 is not abelian. It is very difficult to determine whether non-abeliangroups defined by generators and relations are isomorphic. But there is a way to abelianize
an arbitrary group. Roughly speaking, whenever a and b are elements of the group, wedeclare that aba−1 b−1 = 1, since this implies that ab = ba. Let us abelianize π1 . This newgroup is called the first homology group and often denoted H1 (X).
It follows that H1 of a torus is generated by commuting symbols a1 , b1 , . . . , ag , bg which−1−1 −1satisfy the relation a1 b1 a−1
1 b1 . . . ag bg ag |bg = 1. If the a’s and b’s commute, this relation is automatically true and H1 is an abelian group with 2g generators. This group isthus
Z ⊕ Z ⊕ . . . ⊕ Z = Z 2g
In the nonorientable case, the group H1 is generated by commuting symbols a1 , a2 , . . . , agsatisfying the relation a1 a1 a2 a2 . . . ag ag = 1. Since these elements commute, this relation
can be written (a1 a2 . . . ag )2 = 1. It is easy to see that the elements
A1 = a1 , A2 = a2 , . . . , Ag−1 = ag−1 , Ag = a1 a2 . . . ag
also generate the group. These new elements commute and satisfy the single relation
A2g = 1. The corresponding group is thusZ ⊕ Z ⊕ . . . ⊕ Z ⊕ Z2
By easy group theory, if H1 (X) and H1 (Y ) are groups of the form Z g or Z g−1 ⊕ Z2 , theonly way H1 (X) and H1 (Y ) could be isomorphic is if both have the first form with thesame g, or both have the second form with the same g. The uniqueness of canonical forms
for fundamental polygons of compact surfaces follows immediately from this result.

17