Surface Topology

Tara E. Brendle July 22, 2011

Contents

1Lecture1:Whatisasurface? 3 1.1 Definitionofasurface ...... 3 1.2 Surfaceswithboundary ...... 4 1.3 Closedsets ...... 4 1.4 Compactness ...... 4 1.5 Connectedness,path-connectedness ...... 6 1.6 Wherewe’reheading...... 7

2Lecture2:Quotients 8 2.1 Quotientsassetswithoutmetrics ...... 8 2.2 Whathappenstothemetric? ...... 8 2.3 Pathification ...... 8 2.4 Examples ...... 9

3Grouptheory 11 3.1 Groupsandhomomorphisms ...... 11 3.2 Groupactions...... 12 3.3 Forming quotients of spaces via group actions ...... 13

4Lecture4:Eulercharacteristic 14 4.1 Euler characteristic: definition/examples ...... 14 4.2 Eulercharacteristicandgenus...... 15

5Lecture5:FundamentalGroups 18 5.1 Homotopy...... 18 5.2 FundamentalGroups...... 18 5.3 Basepoints and functoriality ...... 19

6Lecture6:Functoriality,CoveringSpacesandDeckTransformations 21 6.1 Functoriality ...... 21 6.2 First examples: Convex sets in Rn; the circle S1 ...... 21 6.3 Coveringspaces...... 22 6.4 Decktransformations...... 22 6.5 Liftingpathsandhomotopies ...... 23 6.6 Finishing the proof of the Isomorphism Theorem ...... 24

1 6.7 TheTorus,Again...... 25

7Lecture7:MoreonFundamentalGroupsandCoveringSpaces 26 7.1 HomotopyEquivalence...... 26 7.2 Actionofthedeckgrouponthecover ...... 27 7.3 Fundamental groups of surfaces given as plane models ...... 28

8Lecture8:JordanCurveTheorem 29 8.1 GraphTheoryCheatSheet ...... 29 8.2 PolygonalJordanCurveTheorem...... 30 8.3 GeneralJordanCurveTheorem...... 31

9Lecture9:TriangulabilityofSurfaces 33 9.1 FinishingupJordanCurveTheorem ...... 33 9.2 Corollaries of Jordan-Schoenflies ...... 34 9.3 Proof of Triangulability, assuming Jordan-Schoenflies...... 35 9.4 Well-definednessofconnectedsum ...... 36 9.5 Finishing Classification ...... 36

10 Lecture 10: Proof of Triangulability of Surfaces 37 10.1 Well-definednessofconnectedsum ...... 37 10.2 Finishing Classification ...... 37 10.3 Classification of simple closed curves on surfaces ...... 38

11 Lecture 11: Basics of curves on surfaces 40 11.1EssentialCurves ...... 40 11.2 Algebraic intersection ...... 40 11.3 Geodesicrepresentatives ...... 40 11.4Torusexample ...... 41 11.5Bigoncriterion ...... 41

12 Lecture 12: Mapping Class Groups I 43 12.1 PuncturesversusBoundaryComponents ...... 43 12.2Finiteorderexamples ...... 43 12.3Somewarmupexamples ...... 43

13 Lecture 13: Mapping Class Groups II 45 13.1Dehntwists...... 45 13.2Torus,again...... 46 13.3Exactsequences...... 46

14 Lecture 14: Mapping Class Groups III 47 14.1Generatingsets...... 47 14.2Thecurvecomplex ...... 47 14.3 Variationsonthecurvecomplex ...... 47 14.4 Generation of Mod(S)...... 48

2 Acknowledgement: This lecture follows two excellent treatments of the basics of surface topology: Armstrong’s Basic Topology and Goodman’s Beginning Topology.

1Lecture1:Whatisasurface?

Note that all “spaces” are metric spaces. Those with a background in topology can think of these as topological spaces, with the topology generated by the open sets induced by the metric.

1.1 Definition of a surface The OED’s first definition of a surface is: “The outermost boundary (or one of the boundaries) of any material body, immediately adjacent to the air or empty space, or to another body.” Compare with the mathematical definition: A surface S is a metric space which is locally homeomorphic to a disk. In other words, for every point p ∈ S, there exists an open set U ⊆ S such that p ∈ U and such that U is homeomorphic to the open disk {(x, y) | x2 + y2 < 1}. A topological space X is called Hausdorff, if for every a, b ∈ X with a #= b, there exist disjoint open sets U(a),U(b) containing a, b, respectively. In a topology course, a surface is usually defined with the word “Hausdorff” replacing “metric”, which gives a slightly more general notion of a surface. Note that every metric space is automatically Hausdorff, since if ! d(a, b)=!,thenwecantakeU(a),U(b) each to be balls of radius less than 2 centered at a, b, respectively. Note to those with topology background: a compact Hausdorffspace admits a metric if and only if it’s second countable. Also, a locally metrizable space is metrizable iffit is Hausdorff and paracompact.

Examples.

• The plane R2 with the standard metric. • The sphere S2: : by symmetry it’s enough to find a neighborhood of (0, 0, 1) homeomor- phic to a disk. Then take, say, all points with z>0 and show that projection to xy-plane is a homeomorphism. • The torus T 2, version one: surface of revolution in R2 • The torus T 2, version two: S1 × S1 in R4

Nonexample.

• To see why being locally homeomorphic to a disk is not a sufficient condition: Start with the plane. Remove (0, 0) and replace it with two “copies”.

3 1.2 Surfaces with boundary A surface with boundary S is a metric space such that for every point p ∈ S, there exists an open set U ⊆ S such that p ∈ U and such that U is homeomorphic to an open disk or to the half disk {(x, y) | x2 + y2 < 1andx ≥ 0}.Ifp has a neighborhood of the first type it is an interior point, otherwise p is a boundary point.

Examples of surfaces with boundary. • The disk D2 = {(x, y) | x2 + y2 ≤ 1} • The annulus A = {(x, y) | 1 ≤ x2 + y2 ≤ 2} Exercise 1.1. Which of the following are surfaces? Surfaces with boundary? 1. {(x, y, z) ∈ R3 | x2 + y2 + z2 =1and z ≥ 0} 2. {(x, y, z) ∈ R3 | 4x2 + y2 +9z2 =1} 3. {(x, y, z) ∈ R3 | x =0or z =0}

1.3 Closed sets A subset of a space is closed if its complement is open. Exercise 1.2. Think of an example of a subset of R2 with the usual metric which is both open and closed, and one which is neither open nor closed. Let S be a subset of a space X.Thenp ∈ X is a limit point of S (sometimes known as a point of accumulation) if every open set which contains p also contains at least one point of S −{p}. Exercise 1.3. Determine the limit points of [0, 1) in R.

1 Exercise 1.4. Determine the limit points of the set of points of the form n in R,forn ∈ N. The union of a set S and its limit points is called the closure of S. Exercise 1.5. Asetisclosedifandonlyifitcontainsallitslimitpoints. Therefore a set is closed if and only if it is equal to its closure.

1.4 Compactness

A space is compact if every open cover has a finite subcover. That is, if X = ∪α∈AVα,where n Vα is open for all α in some (possibly infinite) index set A,thenX = ∪i=1Vαi for some finite subset of the Vα’s.

Note about terminology. Just to ensure constant confusion, mathematicians often use the term “surface” to refer to both surfaces and surfaces with boundary as defined above, and use the term “closed surface” to mean “compact surface without boundary”, i.e., as opposed to a set that’s closed in the sense that its complement is open.

4 Theorem 1.6 (Heine-Borel for R). Aclosedintervalofthereallineiscompact. Proof. Let C be an open cover of [a, b]. Consider the set X of all x ∈ [a, b]suchthat[a, x] is contained in a finite subcover of C. First note that X has a least upper bound s since it is nonempty and bounded above. We will show that s ∈ X and that s = b.LetU ∈ C be an open set containing s.Ifscompact space X is compact. Proof. Let F be an open cover of C.ThenF ∪ (X − C) is an open cover of X,andhencehasa finite subcover F".ThenF" (or possibly F" − (X − C)) is the finite subcover of C we are looking for. Theorem 1.8. Prove that if f : X → Y is continuous, and then f(A) is compact if A is compact. Proof. Without loss of generality, we may assume that X is compact and that f is surjective. Let F be an open cover of Y . Then since f is continuous, {f −1(U) | U ∈ F} is an open −1 −1 cover of X and has a finite subcover of the form f (U1),...,f (Un). Since f is surjective, −1 f(f (U)) = U for all U,hence{U1,...,Un} is the finite subcover of Y we require. Theorem 1.9 (Heine-Borel for Rn). AsubsetofRn is compact if and only if it is closed and bounded. We’ll leave the “forward” direction as a series of exercises. To prove the converse, we’ll need the following proposition:

Proposition 1.10. AfiniteproductofclosedintervalsinRn is compact. Proof of Proposition. We’ll show that S =[a, b] × [c, d] is compact and leave the general case as an exercise (it’s virtually identical). Let F be an arbitrary open cover of S. Without loss of generality, we may assume F is consists of sets of the form U × V ,whereU and V are both open intervals. (Exercise: explain why this is true.) Now, let (x, y) be a point in S. Since F also covers {x}×[c, d], and since {x}×[c, d] is homeomorphic to [c, d], an interval, there must be a finite subcover of {x}×[c, d] consisting of x x x elements of the form Ui × Vi ,fori =1,...,m(x). Let U denote the intersection of all the x Ui .

5 But then the union of all the U x, over all x ∈ [a, b], is an open cover of [a, b], which must have a finite subcover U xj ,forj =1,...,n.Nowwehave

[a, b] × [c, d] ⊂ (U x1 × [c, d]) ∪···∪(U xn × [c, d]) ⊂ ((U x1 × V x1 ) ∪···∪(U x1 × V x1 )) ∪···∪((U xn × V xn ) ∪···∪(U xn × V xn )) 1 1 m(x1) m(x1) 1 1 m(xn) m(xn) and we’re done. Closed and bounded ⇒ compact. Now, suppose we have a closed and bounded subset S of Rn. Since S is bounded it must be contained in the n-fold product [−a, a] × [−a, a] ×···×[−a, a] for some a ∈ R. (We can take this as the definition of a bounded set in Rn, if we like.) By the preceding proposition, this set is compact. Hence S is a closed subset of a compact space, and by the theorem above, we’re done. Exercise 1.11. Finishing the proof of Heine-Borel for Rn:

1. Recalling that metric spaces satisfy the Hausdorffproperty, prove that a compact subset K of a metric space X is closed. (HINT: Show that if x ∈ X − K,thenthereexistdisjoint open sets containing x and K,respectively.) 2. Complete the proof of the general Heine-Borel Theorem by considering the open cover of Rn by open balls of radius n ∈ Z,centeredattheorigin.

1.5 Connectedness, path-connectedness A space is disconnected if it is the disjoint union of (at least) two nonempty open sets. Otherwise, it is connected. Though the definition is easy to state, there are two caveats: first, it can be very difficult to prove that a space is connected, and second, our intuition about connectedness doesn’t always serve as good guide (see the example of the topologist’s sine curve).

Exercise 1.12. Prove each of the following statements. 1. The continuous image of a connected space is connected. 2. (Finite) products of connected spaces are connected. Some examples without proof.

• Rn is connected for all n ≥ 1. • Intervals of all kinds (open, closed, half-open) in R are connected. • Sn for all n ≥ 1 is connected (S0 is disconnected). • T 2 is connected. • 1 Topologist’s sine curve {(x, sin( x )) | 0

6 The last example above (the topologist’s sine curve) seems to defy our sense of what “con- nected” should mean, perhaps, and if you have never done so, it is well worth carefully working through the proof that it satisfies the above definition of connectedness in order to fully appre- ciate what this definition is trying to capture. For our purposes, however, we will mostly use what is a more natural notion of connectivity (at least to Prof. Brendle’s way of thinking): [Recall: Prof. Leininger defined a path at the beginning of the previous lecture.] A space is path-connected if for any two points x, y, there is a path from x to y.Notethat the topologist’s sine curve fails to be path-connected. Prof. Leininger will say more about this in his next lecture (I think!), and we’ll revisit this notion probably in Week2.

1.6 Where we’re heading A simple closed curve in a space X is a map f : S1 → X such that the restriction f : S1 → f(S1) is a homeomorphism.

1. Jordan Curve Theorem: The complement of any simple closed curve in R2 has two connected components. 2. Classification of Surfaces: A complete list of all closed orientable surfaces. 3. Mapping class groups of surfaces: The appropriate notion of a group of “symmetries” or “automorphisms” of a surface.

Note that there other versions of the Jordan Curve Theorem appearing in the literature. We will end up using a stronger statement, known as the Jordan-Schoenflies Theorem.

7 2Lecture2:Quotients

Acknowledgement: Lecture 2 will follow the treatment of quotient spaces given in Schwartz’s wonderful new book “Mostly Surfaces”.

2.1 Quotients as sets without metrics An equivalence relation on a set X is binary relation ∼ which is reflexive, symmetric, and transitive. If ∼ is an equivalence relation on a set X, and if x ∈ X,thentheequivalence class [x]ofx under ∼ is defined to be the set of all y ∈ X such that y ∼ x.Thequotient of X by ∼,denoted X/ ∼, is the set of all equivalence classes (which form a partition of X). A priori, X/ ∼ is just a set, but in all the cases we care about, X/ ∼ will also inherit a metric from X. Exercise 2.1. Explain what is wrong with the following “proof” that the reflexive property is redundant in the above definition: By symmetry, we have x ∼ y ⇒ y ∼ x.Thus,bytransitivity, x ∼ x.Henceanyrelationthatissymmetricandtransitiveisalsoreflexive. Exercise 2.2. For every possible choice of two out of the above three properties, give an example of a set, and a relation on that set, which satisifies those two properties but not the third.

2.2 What happens to the metric? Let X be a space with metric d". One natural way to try to put a metric on the quotient space X/ ∼ is as follows. Let S1,S2 be two equivalence classes in X/ ∼.Defineδ(S1,S2)to " be the infimum of the set {d (s1,s2) | s1 ∈ S1 and s2 ∈ S2}. The intuition here is that the distance between the sets S1 and S2 is taken to be the ”shortest” distance between individual elements in the two sets. Since the construction involves infimums of a metric, we can see that δ satisfies the axioms of nonnegativity and of symmetry. But it’s no longer necessarily the case that δ(S1,S2) = 0 if and only if S1 = S2, nor that the triangle inequality is satisfied. There’s a general construction for taking such a function and creating a new one which retains most of the information encoded in and also satisfies the triangle inequality, which we discuss next.

2.3 Pathification The idea here is to start with a function that’s a little bit like a metric, and build slowly towards something that is a metric. Pathification is one step in this process. Let Y be a set (think Y = X/ ∼)and let δ : Y × Y → R be a function satisfying symmetry and nonnegativity, but not necessarily the triangle inequality, and also not necessarily requiring that δ(x, y) = 0 if and only if x = y. Now, we’re going to add in the triangle inequality (so we still won’t quite have a metric). If we have two points x, y ∈ Y ,thenachain from x to y is a finite sequence of points in Y : x = x0,x1,...,xn = y.IfCx,y is such a chain, then define δ(Cx,y)=δ(x0,x1)+δ(x1,x2)+ ···+ δ(xn−1,xn). Define

d(x, y) = infCx,y δ(Cx,y) Here we’re taking the infimum over all possible chains from x to y. The function d is sometimes called the pathification of δ. It is clear from the definition of d that it satisfies nonnegativity and symmetry again.

8 Exercise 2.3. Show that d satisfies the triangle inequality. Metrics for Quotients. Back to quotient spaces: we now take the pathification d of the function δ we defined on the quotient X/ ∼ using the original metric d" on X. The only way d can fail to be a metric now is that we might have d(x, y)=0forx #= y. We will see an example of this soon. If, however, d satisfies d(x, y) = 0 if and only if x = y,thend is a metric, which we will refer to as the quotient metric for X/ ∼ derived from the original metric d" on X,orwesaythat X/ ∼ “inherits” the metric d from the metric d". In this case, we will also say X/ ∼ is a good quotient. Otherwise, X/ ∼ is a bad quotient and inherits a metric from X in no “natural” way.

2.4 Examples Examples of good quotients.

1. The standard example of a torus. This generalizes to the standard 4g-gon “plane model” of a surface of genus g (see definition of genus below). 2. Mobius band 3. Any “plane model” of a surface or surface with boundary. 4. Cell complexes, simplicial complexes, δ-complexes. 5. Connected sum (we will prove that this is independent of choices involved later). Exercise 2.4. Prove that the standard representation of a torus by identifying opposite sides of 3 3 the unit square is a good quotient. Calculate the distance from (0, 0) to ( 4 , 4 ) in this quotient. Exercise 2.5. Prove that the Mobius band is a good quotient. Exercise 2.6. Prove that the hexagon with sides abca−1b−1c−1 is homeomorphic to a torus. (A cutting and pasting argument is sufficient at this point.) Generalize this example to find a 4g +2-gon plane model of a connected sum of g copies of the torus.

2 Exercise 2.7. Forming S as a quotient of two disks: Let D1 and D2 be two copies of the 2 standard unit disk, each with the usual metric inherited from R .LetX = D1 ∪ D2,with d(x, y)=2if x, y are in different disks, or d(x, y) is the standard distance in the disk if x and y live in the same copy. Now, define x ∼ y if and only if x = y or if x, y are the “same” point on the boundaries of D1 and D2.ProvethatX/ ∼ is a good quotient which is homeomorphic to S2.Inthisexercise,whyisitimportantthatthedistancebetween points in the two different disks in X is 2? What goes wrong if it’s, say, 1 instead? Exercise 2.8. The projective plane is the quotient S2/ ∼,wherex ∼ y if x, y are antipodal points (i.e., x = −y.Provethattheprojectiveplaneisasurface. Orientability and genus. A surface is non-orientable if it contains a copy of the Mobius band. Otherwise it is orientable. If a closed orientable surface S is homeomorphic to a connected sum of g tori, we say that it has genus g. The classification of surfaces tells us the genus of an orientable surface is a well-defined invariant. (We can also talk about genus of a non-orientable surface, but won’t do that here.)

9 Exercise 2.9. A Bad Quotient. Consider R with the usual metric, with x ∼ y if and only if x − y ∈ Q.Provethatnotonlydoes this quotient fail to be good, but it fails in a fairly spectacular way, in that the pathification of the inherited ”pre-distance” function on R/ ∼ is actually the zero map.

10 3Grouptheory

There are deep connections between understanding the structure of surfaces and group theory, so we’ll start with a review of some basics.

3.1 Groups and homomorphisms A group is a pair (G, *), where G is a set and ∗ : G × G → G is a binary operation satisfying: • (G1: Associativity) The operation ∗ is associative.

• (G2: Identity) There exists an element eG ∈ G such that e ∗ g = g ∗ e = g for all g ∈ G. • (G3: Inverses) For every g ∈ G, there exists an element h ∈ G such that g ∗ h = h ∗ g = eG. An additional axiom (G0: Closure) is often included in the definition. “Closure” means that g ∗ h ∈ G for all g,h ∈ G. I haven’t included it above since this is covered by our definition of ∗ as a binary operation from G × G into G, but I’m mentioning it anyway because students often forget to check it. Notation: The element h in (G3) is the inverse of g in G and is usually denoted g−1 org ¯.

Exercise 3.1. Prove that the identity element eG is unique. Prove that inverses are unique.

Exercise 3.2. Let g ∈ G be an arbitrary but fixed element. Prove that the function µg : G → G defined by µg(x)=g ∗ x is a bijection. Examples: • Z, under addition • R∗ = R −{0}, operation is the usual multiplication of real numbers.

• Sn, the group of bijections of a set with n elements (operation is function composition)

• GLn(R) (operation is matrix multiplication) A function f : G → G" satisfying f(gh)=f(g)f(h) for all g,h ∈ G is called a homomor- phism. A bijective homomorphism is an isomorphism. An isomorphism from a group G to itself is an automorphism.

Examples:

∗ • The determinant homomorphism from GLn(R)toR is surjective but not injective.

−1 • If G is any group, then for a fixed g ∈ G,themapig : G → G given by ig(x)=gxg is an automorphism of G. ab • The set of all nonzero matrices of the form where a and b are real numbers ! −ba" forms a group under matrix multiplication. The map from C∗ to this matrix group given ab by a + bi .→ is an isomorphism. ! −ba"

11 Exercise 3.3. Prove that the set of all automorphisms of G is itself a group under function composition.

The kernel of a homomorphism f : G → H is the set of all g ∈ G such that f(g)=eH . Exercise 3.4. Prove that the kernel is a subgroup of G,i.e.,agroupinitsownrightunder the restriction of the multiplication in G. Exercise 3.5. Prove that a group homomorphism is injective if and only if itskernelistrivial.

3.2 Group actions We mainly care about groups because they act on sets. Let X be a set and G agroup.A(left) action of G on X is a function · : G × X → X such that for all x, y ∈ X: 1. e · x = x,and 2. g · (h · x)=(gh) · x. We sometimes say in this case that X is a G-set. Examples:

• Sn acts on the set {1, 2, 3, ..., n}. n n • GLn(Z)actsonthesetZ and on the vector space R (note that for the purposes of this example, we only need that Rn is a set and can forget that it also has the extra structure of a vector space). • The multiplicative group R∗ also acts on Rn. In fact, if F is the scalar field for a vector space V ,thenF ∗ = F −{0} acts on V . (This is built into the axioms of a vector space, in fact.) • Every group acts on itself by left multiplication, (also by right multiplication)

Let x ∈ X.WedefineGx = {g ∈ G | g · x = x}. This is known as the stabilizer of x in G, or the isotropy subgroup of x.

Exercise 3.6. Prove that for any x ∈ X,withX a G-set, then Gx is indeed a subgroup of G. Let x ∼ y if there exists a g ∈ G such that g · x = y. Exercise 3.7. Prove that ∼ is an equivalence relation on G and hence partitions X into equivalence classes. Each equivalence class is known as an orbit in X under the action of G. If X has just one orbit under the action of G,thenwesaythatG acts transitively on X.

Exercise 3.8. Prove that Sn acts transitively on {1, 2,...,n} Exercise 3.9. Prove that Z acts on R by the operation of addition. Describe the orbits in R under this action. Exercise 3.10. Prove that Z2 acts on R2 by the operation of addition. Describe the orbits in R under this action.

We say that G acts faithfully on X if g · x = x for all x ∈ X implies that g = eG. Exercise 3.11. Which of the examples of group actions we’ve described in thislectureare faithful, if any?

12 3.3 Forming quotients of spaces via group actions A group action on a space gives us a natural way to form a quotient of that space in the manner described in the previous lecture, and we can talk about “good” and “bad” quotients if the metric on the original space descends to a metric on the quotient, as before.

Examples:

• Z on R. What is the quotient space? Is it a good quotient? What is the metric? • Q on R. What is the quotient space? Is it a good quotient? What is the metric? • Z2 on R2. What is the quotient space? Is it a good quotient? What is the metric? • Now consider the action of 2Z × 2Z on R2 by addition. What is the quotient space? Is it a good quotient? What is the metric? Note that 2Z × 2Z is isomorphic to Z × Z as a group. Describe this example instead as the group Z × Z acting on R2. What is the key difference between this example and the previous example? Exercise 3.12. Generalize the previous result to describe all possible different metrics that one could obtain on T 2 in this way.

NOTE: In fact all higher genus orientable surfaces arise this way. We’ll see more about this later. Some more terminology: if a group G acts on a space X, we usually insist that G acts by homeomorphisms (in other words for every fixed g,themaponX defined by x .→ g · x is a homeomorphism). If the result is a compact quotient space, we say that G acts cocompactly on X. So we see from the above examples that Z2 acts cocompactly (in infinitely many ways) on R2. Another adjective that is frequently attached to group actions is “properly discontinuous”. The foundation of everything we’ve done so far is the notion of an open set, where open sets are defined with reference to some metric. However, we could cheat and take any given set S and just say that every subset of S is open (and hence every subset is closed). In this case we are “giving S the discrete topology”. We often want to do this in the case that S is a group, since many groups do not come equipped with any obvious metric of their own. But in some cases, for example the group R, we already have a more natural metric at hand for defining our open sets, which does not result in every subset being both open and closed. However, it may be the case for some subset of this space that the “usual” metric actually induces the discrete topology on the subset. For example, the subgroup Z in R inherits the discrete topology, but Q does not. (Prove!) If a group is being considered with the discrete topology, whether it inherits this in some natural way or not, we call it a discrete group. We say that a discrete group G acting on a (locally compact Hausdorff) space X acts properly discontinuously if for all compact K ⊂ X, we have that (g · K) ∩ K #= ∅ for at most finitely many g ∈ G. This condition might seem somewhat strange at first, but the take-home point is that it guarantees nice properties in the quotient space, which you’ll hear more about from Prof. Leininger. Back to examples, note that any surface is a locally compact Hausdorffspace. Again, we see that Z2 acts properly discontinuously on R2.

13 4Lecture4:Eulercharacteristic

We begin by giving a complete statement of the classification of surfaces. Theorem 4.1 (Classification of Surfaces). Every closed, connected, orientable surface is homeomorphic to the connected sum of g tori for precisely one g ∈ Z≥0.

Recall from Lecture 2 notes that a surface S has genus g if S is homeomorphic to the con- nected sum of g tori. A priori, a surface could have more than one genus, but the classification theorem tells us that genus is a complete invariant of closed, connected, orientable surfaces. Also, at this point we haven’t proved well-definedness of connected sum. So right now, “genus” is not necessarily a meaningful concept. It is important to realize that the classification of surfaces is really a multi-part theorem, so to make this point crystal clear, and to also keep track of where we use each part, we will restate the theorem as follows: Theorem 4.2 (Classification of Surfaces). Let S be a closed, connected orientable surface. Then: 1. The surface S is homeomorphic to a surface of genus g. 2. Any surface of genus g is homeomorphic to any other surface of genus g.Inotherwords, the operation of connected sum is well defined. 3. If T is a surface of genus g" #= g,thenT is not homeomorphic to S. In this lecture we will define an incredibly useful tool for studying surfaces known as the Euler characteristic and prove its basic properties. Let S be a surface. A cell decomposition of S is a CW-complex C together with a homeomor- phism h : C → S. (Think: plane model for a surface.) Note that S admits a cell decomposition if and only if S is triangulable (which means replacing “cell complex” with “simplicial 2-complex” in the above definition). Theorem 4.3 (Kerekjarto, Rado). Every compact surface is triangulable. The triangulability of surfaces is a deep result which we will use heavily in what follows. We will be proving this carefully next week. This is also a key ingredient in proving Part I of the Classification Theorem.

4.1 Euler characteristic: definition/examples Let S be a surface equipped with a cell decomposition C with v vertices, e edges, and f faces. We define the Euler characteristic of S with respect to C by

χC (S)=v − e + f.

In fact, we will see that in fact the Euler characteristic does not depend on the particular cell decomposition of S, so we will usually just write χ.

Examples.

14 • The disk D2, with cell decomposition given by a single 2-simplex, has Euler characteristic equal to 3 − 3+1=1. • The sphere S2, with cell decomposition given by two 2-simplices glued along their common boundaries, has Euler characteristic equal to 3 − 3+2=2. • The torus T 2, with cell decomposition given by the usual identification of the sides of the unit square, has Euler characteristic equal to 1 − 2+1=0. Exercise 4.4. Find a different cell decomposition for each of the above and check explicitly that you get the same result. Exercise 4.5. Show that each of the following 3 “moves” (and their inverses)onagivencell decomposition leaves the Euler characteristic invariant: 1. Subdividing an edge by adding a vertex. 2. Subdividing a face by connecting two vertices with a new edge. 3. Introducing a new vertex in the interior of a face and a new edge connecting that vertex to an existing vertex adjacent to that face. Exercise 4.6. Nontrivial Fact: Any two cell decompositions on the same surface overlap in just a finite number of points, possibly after perturbing them slightly. We will see a technique for proving this later on. For now, assume the fact, and use it to prove that Euler characteristic is independent of the choiceoftriangulation,byshowingthata finite number of the 3 types of moves above are sufficient to “transform” any cell decomposition into any other.

4.2 Euler characteristic and genus The next formula is one of the most important in all of topology. Remember that at this stage, “having genus g” means “homeomorphic to some connected sum of g tori”, a construction which aprioridepends on choices of disks and gluing maps, although we will later see that these choices don’t end up mattering. Theorem 4.7 (Euler’s Formula). If S is a closed surface of genus g with respect to a cell decomposition C,thenχC (S)=2− 2g. Proof. WLOG we can consider the case where S is given as a plane model. (Here we are using triangulability in a big way.) This is because the process of “forgetting” an edge in a cell decomposition doesn’t change the Euler characteristic. (See exercise above, Type 2 move: You’re exchanging 2 faces for 1 face, while eliminating one edge, so the net change in f − e + v is 0. ) Recall that a plane model has exactly 1 face, but the edges are paired up in some way which makes it slightly unclear how many distinct vertices there are, and we also want to clarify the relationship between the number of edges and the genus of the surface. Begin by (cyclically) reading the boundary word for the plane model for S. If it contains a subword of the form ...a...a... then it contains a Mobius band and is not orientable. So if we see a in the boundary word it can only be followed by an a−1. If it (cyclically) contains a subword of the form aa−1, then we can glue these together and delete the corresponding edge and vertex, leaving Euler characteristic unchanged. Repeat this until no such subwords exist. Now, if the new boundary word contains a pattern of the form ...a...b...a−1 ...b−1 ... we say that the boundary word contains a cross.

15 Lemma 4.8. If the boundary word contains a cross, then S is not a sphere. Proof. Think of the unit square with the usual toroidal identification of sides, with small bit of each corner deleted. So we see that our surface S has a decomposition as the connected sum of a torus and some other surface. But then S is not homeomorphic to S2 since there exists a simple closed curve C in the connected sum of a torus and another surface such that S −C is still connected and path-connected. However, by the Jordan Curve Theorem (which we will prove independently next week), any simple closed curve separates S2 into two connected components. Exercise 4.9. Prove the converse: if the boundary word does not contain a cross, then S is a sphere. Lemma 4.10. The Euler characteristic of S2 is 2. Proof. The proof is by induction on the number of edges in our plane model. Our base case is a plane model with boundary word aa−1. (Exercise: check that this really is a “good” quotient homeomorphic to S2.) Now, by the previous lemma (or really, its converse, which was given as an exercise immediately following the lemma), we know that our boundary word does not contain a cross. So if we see something of the form a...b...a−1, it must actually be of the form a...b...b−1 ...a−1. Now the “distance” from b to b−1 is shorter than from a to a−1,so we keep going in this way until we must eventually arrive at a subword oftheformcc−1.We make the usual identification here, which reduces the number of edges in our plane model by 2 without changing the Euler characteristic. By induction, we are donenow. Now, to finish the proof of the theorem, we induct on the number of edges in the plane model. If S is not a sphere, then by a previous lemma, its plane model has a cross.(Youcan prove this similarly to the above.) Let’s reproduce the construction we did in the proof of that lemma, so that we can reinterpret our plane model. Let’s reintroduce enough edges in our cell decomposition so that we can “cut out” the torus and replace it with a disk. The resulting plane model (after forgetting some interior edges again) has 4 fewer edges, identified in pairs, which means our total edge count for Euler characteristic has gone down by 2. But the plane model has the same set of vertices, glued together in the same way. Scholium 4.11 (Classification Part I). Every surface is homeomorphic to some connected sum of some number of tori (possibly zero). Unfortunately, the construction in the previous proof depended heavily on an initial, specific choice of triangulation, not just on the fact that the surface was triangulable. If we knew that χ is an invariant of the underlying space (see exercise above which proves this, assuming a Nontrivial Fact), then we would also have Classification Part II. Similarly, we could prove that χ is independent of the choice of triangulation by using Classification Part II. The exercise above gives you an idea of one approach to the first direction, and we will later encounter some machinery that would equip us for doing this. We will also learn some different machinery that will enable us to prove Classification Part II without knowing that χ is an invariant of the underlying space. So basically, we’ll be covered. Exercise 4.12. Assuming the full classification of surfaces, identify the surface given by a plane model with boundary word dbec−1e−1cb−1a−1d−1a.

16 Exercise 4.13. Prove that a connected compact surface of genus g with b boundary components has Euler characteristic equal to 2 − 2g − b.(FeelfreetoassumethatEulercharacteristicisan invariant of the underlying space here – just be aware that you’re using it!) Exercise 4.14. It turns out that surfaces in each of the three cases χ>0,χ=0,χ<0 share certain fundamental properties. Assuming the full classification theorem of surfaces, write down acompletelistofcompactconnectedsurfaceswitheachofthese properties. Also, once you know that Euler characteristic is independent of the choice of cell decom- position of a surface, it’s easy to see how Euler characteristic behaves under connected sum. Suppose we have two surfaces S1,S2, with

Proposition 4.15. χ(S1%S2)=χ(S1)+χ(S2) − 2

Proof. Suppose D1 and D2 are the two disks you’re going to use for the connect sum. Choose cell compositions for S1,S2, for which D1,D2 are 2-cells, with their respective boundaries each decomposed into n edges and n vertices. Then the connected sum has two fewer faces (corresponding to D1 and D2)thanS1. Exercise 4.16. Let P denote the nonorientable surface given by the plane model with boundary word aa.Thissurfaceiscalledtheprojective plane.(Wehaveseenadifferentdescription previously.) Prove that P%P%P =∼ P%T2. Exercise 4.17. Redo everything in this lecture for nonorientable surfaces.Inotherwords, formulate a reasonable version of the Classification Theoremfornonorientablesurfaces.Include areasonablenotionof“genus”andaversionofEuler’sformula.

17 5Lecture5:FundamentalGroups

Throughout this lecture, “map” means “continuous map”.

5.1 Homotopy

Let X, Y be spaces, and let I = [0, 1]. We say that two (continuous) maps f0,f1 : X → Y are homotopic if there exists a third (continuous) map F : X × I → Y such that:

F (x, 0) = f0(x) for all x ∈ X

F (x, 1) = f1(x) for all x ∈ X

We often denote F (x, t)byft(x). The map F is called a homotopy from f0 to f1, and if such an F exists, we write f0 ∼ f1. Proposition 5.1. Homotopy gives an equivalence relation on the set of all continuous functions from X to Y (for a given X, Y ). Proof. The map F (x, t)=f(x) is a homotopy from f to itself, so ∼ is reflexive. Also, if F is a homotopy from f0 to f1,thenG(x, t)=F (x, 1 − t) is a homotopy from f1 to f0. Exercise 5.2. Finish the proof by showing that ∼ is transitive.

Exercise 5.3. Let X be any space, and let f : X → Rn be a map. Prove that f is homotopic to the zero map defined by ζ(x)=0for all x ∈ X.ConcludethatallmapsfromX into R are homotopic. For our purposes, we will nearly always use I = [0, 1] as the domain X.

5.2 Fundamental Groups

A loop based at x0 in a space S is a path f : [0, 1] → S such that f(0) = f(1) = x0. The point x0 is called the basepoint. Two loops f0,f1 are loop homotopic or homotopic rel x0 if there exists a homotopy F from f0 to f1 such that ft is a loop based at x0 for all 0 ≤ t ≤ 1. Note that this is definitely different from saying that f0 and f1 are homotopic – the difference is that a loop homotopy doesn’t allow the basepoint to move at all during the homotopy. We denote the loop homotopy class of a loop f by [f]. Having a basepoint that doesn’t move can be an annoying restriction sometimes, but it allows us to put a group structure on loop homotopy classes of loops as follows. Let [f], [g]be two homotopy classes of loops based at a point x0 in a space X. Note that this notation means that f,g are actual loops based at x0 representing their respective classes. We define [f] ∗ [g] roughly by the rule “first do f,thendog, only do them each twice as fast so that we still finish in a unit’s worth of time”. More precisely, we define a loop h by

1 f(2t)for0≤ t ≤ 2 h(t)=  1 g(2t − 1) for 2 ≤ t ≤ 1 

18 The problem with this definition is that it depends on a choice of representative of the loop homotopy classes [f], [g]. We need to prove that if f " ∼ f and if g" ∼ g,thenf " ∗ g" ∼ f ∗ g. Let F be the homotopy between f " and f and let G be the homotopy between g" and g.Then we can define a homotopy J as follows:

1 F (2s, t)for0≤ s ≤ 2 J(s, t)=  1 G(2s − 1,t)for2 ≤ s ≤ 1 Basically we’re just doing each individual homotopy twice as fast. So we have a well defined operation on our set of based loops. The fundamental group of a space X based at x0 is the set of all loop homotopy classes of loops based at x0 together with the operation of loop multiplication, and is denoted π1(X, x0). We need to prove that this is actually a group, but this is perhaps more difficult than it first appears. For example, it’s fairly clear that the constant loop defined by e(s)=x0 for all s ∈ [0, 1] should represent the identity element, but the composed map f ◦ e (meaning first do f,thendoe) is not the same loop as the map f, so again, we need to show the existence of a homotopy between them.

2s t+1 f t+1 for 0 ≤ s ≤ 2 F (s, t)= ' (   t+1 x0(= e(s)) for 2 ≤ s ≤ 1  2 Think of this as f being parametrized by s, but multiplied by t+1 ,sothatatt =0,weare doing f twice as fast, and then sitting at x0 for the remaining half-unit of time, but at t =1, we are just doing f at the usual speed. Exercise 5.4. Write an explicit formula for a homotopy between e ◦ f and f. Similarly, it’s intuitively clear that the inverse of a loop should be “do the loop backwards”. −1 −1 So let f be a loop based at x0.Definef (s)=f(1 − s). We need to show that [f] ∗ [f ]= [f −1] ∗ [f]=[e]. To show that [f] ∗ [f −1]=[e], we have the following homotopy:

1−t f(2s)for0≤ s ≤ 2   1−t 1+t F (s, t)= f(1 − t)for2 ≤ s ≤ 2  1+t  f(2 − 2s)for2 ≤ s ≤ 1   Exercise 5.5. Write an explicit formula for a homotopy between f −1 ◦ f and e. Exercise 5.6. Complete the proof that the fundamental group of a surface actually is a group by proving that multiplication of loops is associative by writing down an explicit homotopy.

5.3 Basepoints and functoriality In defining “the” fundamental group of a space Y , we needed to make a choice of basepoints y0. But how much does the fundamental group actually depend on this choice? Obviously the elements themselves will be different – a loop based at y0 is not the same as a loop based at y1 #= y0. But it turns out that in all of the cases we care about, the resulting group will be the

19 same no matter what choice of basepoint we make, at least, up to isomorphism. In fact, one should really think of changing basepoints as being analogous to a change of basis for a vector space.

Theorem 5.7. Let Y be a space with y0,y1 ∈ Y such that there exists a path α from y0 to y1. Then π1(Y,y0) =∼ π1(Y,y1). Corollary 5.8. If Y is path-connected, then we can safely talk about “the” fundamental group of Y and we often denote it just by π1(Y ).

Proof. Let α be a path from y0 to y1. Then we define the following map:

φ : π1(Y,y1) → π1(Y,y0) [f] .→ [αfα−1]

Exercise 5.9. Prove that φ is a well defined isomorphism of groups. (The techniques are very similar to those we used to prove that the fundamental group isagroup,andacompleteproof can be found in any good textbook, e.g., Munkres, Armstrong, Goodman, or Hatcher. To prove bijectivity, it is usually easier to write down an inverse mapratherthanusemethodssuchas proving the kernel is trivial, etc.)

20 6Lecture6:Functoriality,CoveringSpacesandDeck Transformations

Some terminology: A path-connected space X is simply connected if π1(X) is the trivial group.

6.1 Functoriality When dealing with basepoints, it is standard to write the space and the basepoint as a pair (Y,y0) and refer to the pair as a pointed space.

Proposition 6.1. Let (Y,y0), (Z, z0) be two pointed spaces. Let φ : Y → Z be a (continu- ous) map such that φ(y0)=z0 (this is called a basepoint-preserving map). Then there is a homomorphism φ∗ : π1(Y,y0) → π1(Z, z0).

Proof. Let [f] ∈ π1(Y,y0). Then φ(f) is a loop based at z0 in Z,sowedefineφ∗([f]) = [φ(f)]. If [f]=[g], where F is a homotopy from f to g,thenφ ◦ F is a homotopy from φ(f)toφ(g), so this is a well defined function.

Exercise 6.2. Prove that φ∗ is a homomorphism. (This is actually even easier than well- definedness.)

The next theorem tells us that we have constructed a useful algebraic invariant for distin- guishing between topological spaces.

Theorem 6.3. If Y,Z are path-connected homeomorphic spaces, then π1(Y ) =∼ π1(Z). The theorem is proved by the following two (very short) exercises.

Exercise 6.4. Let φ :(X, x0) → (Y,y0) and ψ :(Y,y0) → (Z, z0) be two basepoint-preserving maps. Prove that (ψ ◦ φ)∗ = ψ∗ ◦ φ∗. Exercise 6.5. Prove that if the map φ in the above proposition is actually a homeomorphism, then φ∗ is an isomorphism of groups. When we pass from the continuous map φ of spaces to the induced homomorphism of groups φ∗, we see that we are somehow respecting the natural structures and transformations that exist between spaces and groups. Going from φ to φ∗ is an example of a functor from the category of spaces to the category of groups. We also use the term naturality to describe this kind of process.

6.2 First examples: Convex sets in Rn;thecircleS1 Actually computing fundamental groups can be difficult without some more machinery that we haven’t developed yet. We’ve already proved that EVERY map of EVERY space into R2 is homotopic to the zero map ζ.IfwetakeaconvexsubsetA of Rn, we can use the so-called straight-line homotopy to prove that π1(A) =∼ 1. So if f : [0, 1] → A is a loop based at a0 ∈ A,thenthemap

F (s, t)=t ∗ a0 +(1− t)f(s) gives a homotopy to the constant loop at a0. This works by sliding each point f(s)toa0, which we can do by convexity. In particular, we have shown:

21 n Proposition 6.6. The fundamental group π1(R ) is trivial for all n. The standard first “interesting” example is actually the circle, and indeed, it’s often easier to find non-trivial loops than it is to prove that there aren’t any.

Winding number. Define p : R → S1 by p(x)=(cos(2πx), sin(2πx)). If we think of S1 in 2πix 2 the complex plane C, we can write this as p(x)=e . If we let x0 =(1, 0) ∈ S ,thenwesee −1 that p (x0)=Z. Proposition 6.7. Let f be a loop in S1.Thenthereisauniquemapf˜ : [0, 1] → R such that 1 f˜(0) = 0 and p ◦ f˜ = f on [0, 1].Iff0,f1 are homotopic loops in S ,thenf˜0(1) = f˜1(1) This proposition will follow directly from more general results which will be proved shortly. In light of the proposition, we define the winding number of the homotopy class [f]tobe the integer f˜(1) which is uniquely determined by f. We denote this by w(f). Now consider the map

1 Φ:π1(S ,x0) → Z given by Φ([f]) = w(f). Exercise 6.8. Prove that Φ is an isomorphism of groups. (HINT: Use what you already know about π1(R).)

Exercise 6.9. Let X, Y be path-connected spaces. Prove that π1(X × Y ) =∼ π1(X) × π1(Y ).

2 Exercise 6.10. Calculate π1(T ).

6.3 Covering spaces Let X,X˜ be path connected spaces. Let p : X˜ → X be a continuous map. Then we say that p is a covering map if every point in X has a path-connected neighborhood U such that the −1 preimage p (U) consists of countably many disjoint sets U˜1, U˜2,... such that the restriction of p to each component U˜i is a homeomorphism. In this case, we say that X˜ is a covering space or a cover of X, and we say that the neighborhood U is evenly covered by the U˜i. In the previous lecture we defined a map p : R → S1 by p(x)=(cos(2πx), sin(2πx)). This map is a covering map, and hence R covers S1.

6.4 Deck transformations Let p : X˜ → X be a covering map. A deck transformation (or covering automorphism) is a homeomorphism φ : X˜ → X˜ such that p ◦ φ = p. Proposition 6.11. The set of all deck transformations of a given covering map p : X˜ → X forms a group. Exercise 6.12. Prove the proposition. We call this group the deck group or the automorphism group of the covering map p,and we denote it Aut(X,p˜ )orjustAut(X˜). Let’s revisit the covering map p(x)=(cos(2πx), sin(2πx)). Since cos(2πx), sin(2πx)areboth periodic with period 1, we have that any function φn : R → R where φn(x)=x + n for some

22 n ∈ Z is a deck transformation for this covering space. Conversely, any deck transformation clearly must be of this form. So the deck transformation group for p(x)=(cos(2πx), sin(2πx)) is just Z. We’ve seen the group Z appearing in association with S1 before, namely, as its fundamental group. This is not a coincidence, as the next theorem tells us. Theorem 6.13 (The Isomorphism Theorem). Suppose that p : X˜ → X is a covering map where X,X˜ are path-connected and X˜ is simply connected. Then Aut(X,p˜ ) =∼ π1(X). Proof that there is a homomorphism. First we choose a basepoint x for X. Since X is path- connected, our choice doesn’t really matter. Choose any pointx ˜ ∈ p−1(x). Let ψ ∈ Aut(X,p˜ ). Theny ˜ = ψ(˜x) is another point in p−1(x), and since X˜ is also path-connected there exists some path f˜ fromx ˜ toy ˜.Thenp ◦ f is a loop in X based at x.Sowedefine ˜ Φ:Aut(X,p) → π(X, x) ψ .→ [p ◦ f˜]

In order to prove that Φis well-defined, we need to know that the particular choice of path f˜ doesn’t affect the final outcome. This follows easily from the next exercise. Exercise 6.14. Let Y be a simply connected space, and let x, y ∈ Y .Thenanytwopathsfrom x to y are homotopic. (HINT: If f,g are the two paths, then fg−1 is a loop based at x.)

The next step is to show that Φis a group homomorphism. Let ψ1,ψ2 be two deck trans- formations. Choose paths f˜1, f˜2 fromx ˜ to ψ1(˜x),ψ2(˜x), respectively, and let f1 = p ◦ f˜1 and similarly f2 = p ◦ f˜2. Choose a third pathg ˜ fromx ˜ to (ψ1 ◦ ψ2)(˜x), and let g = p ◦ g˜.Weneed to show that f1f2 is loop homotopic to g. To see this, note that ψ1 ◦ f˜2 is a path from ψ1(˜x)toψ1 ◦ ψ2(˜x). Therefore the product f˜1 ∗ (ψ1 ◦ f˜2) is a path fromx ˜ to (ψ1 ◦ ψ2)(˜x). But since we proved above that Φis independent of choice of path, we have that

Φ(ψ1 ◦ ψ2)=[p ◦ (f˜1 ∗ (ψ1 ◦ f˜2))]

= [(p ◦ f˜1) ∗ (p ◦ ψ1 ◦ f˜2)]

=[f1 ∗ (p ◦ f˜2)]

=[f1 ∗ f2]

=[f1] ∗ [f2]

=Φ(ψ1)Φ(ψ2)

In order to complete the proof of the Isomorphism Theorem, we need a bit more machinery.

6.5 Lifting paths and homotopies Suppose we have a covering map p : X˜ → X.Supposewehaveamaph from a space Y into the base space X. Then we can ask the general question of whether there exists a lifting of h to X˜, that is, a map h˜ : Y → X˜ such that p ◦ h˜ = h. We will be particular interested in the case where Y is either an interval I or a square I × I (the latter because we will be interested

23 in lifting homotopies of paths). So we will use Q to denote either I or I × I, but actually this works for any cube, i.e., finite product of I’s. We will say that v is a vertex of Q if v is an endpoint of I or if v ∈{(0, 0), (0, 1), (1, 0), (1, 1)}. The theorem that we want to prove is the following. Theorem 6.15. Let p : X˜ → X be a covering map. Let f : Q → X be a continuous map. Let v be a vertex of Q,letx = f(v),andletx˜ ∈ p−1(x).Thenthereisauniqueliftf˜ : Q → X˜ such that f˜(v)=˜x. In other words, paths and homotopies can always be lifted, and this lifting is unique as soon as we specify where a vertex goes. In order to prove this theorem, we will need two lemmas.

Lemma 6.16. Consider the case where Q =∼ I × I.ThereexistsarealnumberN such that if " 1 " Q is a subcube of Q with sidelength less than N ,thenf(Q ) is contained in an evenly covered neighborhood of X.

Exercise 6.17. Prove this lemma. (HINT: Consider a sequence {Qn} of subcubes with diameter going to 0 with Qj not contained in an evenly covered neighborhood. Use Bolzano-Weierstrass.) Lemma 6.18. Suppose that f(Q) is contained in an evenly covered neighborhood. Then there is a unique lift f˜ : Q → X˜ such that f˜(v)=˜x.

Proof. Let U be the evenly covered neighborhood containing Q.LetU˜j be the component “upstairs” containingx ˜.Letpj denote the restriction of p to Uj.Thenpj is a homeomorphism −1 ˜ −1 and hence has an inverse pj .Thenf = pj ◦ f is the unique lift of f we are looking for. Now we’re ready to complete the proof of the Lifting Property for Paths and Homotopies (and of maps of cubes in general). Proof. First we partition Q into subcubes which are small enough to invoke the first lemma above which lets us find evenly covered neighborhoods for each of them. Let’s label these small subcubes Q1,...,Qm.WLOGwecanorderthemsothatQk shares at least one vertex vk with some Qj for j

6.6 Finishing the proof of the Isomorphism Theorem We now have the necessary technology to prove injectivity and surjectivity of the homomorphism Φdefinedabove.

Injectivity. Let Φ(ψ) = 1 in π(X, x). Consider a path f˜ joiningx ˜ to ψ(˜x), and let f = p ◦ f˜. Then by definition Φ(ψ)=[f]. But since we assumed ψ is in the kernel of Φ, there is a homotopy F from f to the constant loop. In this case, the homotopy F : Q → X is a continuous map of the square Q = I × I into X, so we can use the lifting theorem to find a unique lift F˜ : Q → X˜ such that F˜(0, 0) =x ˜.

24 Now, one “side” of F is the path f, so by uniqueness of lifts, that “side” of F˜ must be f˜, which joinsx ˜ to ψ(˜x). Similarly, the other “side” of F˜ must be the constant path atx ˜.Further, F (0,t)=F (1,t)=x,andhenceF˜(0,t)andF˜(1,t) are constant maps. But this must mean that in fact ψ(˜x)=˜x. So, we have shown that any deck transformation which is in the kernel of Φmust preserve the pointx ˜. We now wish to show that in fact any such deck transformation mustbethe identity element. Lety ˜ be any other point of X˜,andchooseanypathf˜ fromx ˜ toy ˜. As usual, let x = p(˜x),y = p(˜y), and f = p ◦ f˜,sothatf is a path in X from x to y. Now, f˜ is a lift of f with initial pointx ˜. But similarly, ψ ◦ f˜ is also a lift of f and also has initial pointx ˜. By uniqueness of path lifting, we must have that these two paths are the same; in particular, they have the same terminal point. This means thaty ˜ = ψ ◦ f˜(1) = ψ(˜y).

Surjectivity. Let [f] ∈ π1(X, x). We need to find an element ψ ∈ Aut(X,p˜ ) which maps to [f]. So lety ˜ ∈ X˜, and letα ˜ be a path fromx ˜ toy ˜.Thenα = p ◦ α˜ is a path from x to y = p(˜y) in X.Thatmeansthatf ◦ α is a path from x to y, which we can lift uniquely to a path f ◦˜ α fromx ˜ to some other pointz ˜. So we will define ψ(˜y)=˜z.Weneedtocheckthatthemapψ we have defined is actually a deck transformation and also that it mapsto[f]underΦ. Recall that to compute Φ(ψ), we need to compute ψ(˜x). In this case, we can just take α to be a constant path, so we are just lifting f to the unique lift f˜ with initial pointx ˜.Butthen Φ(ψ)=[p ◦ f˜]=[f]. Exercise 6.19. It remains to show that the map ψ we just defined is actually a deck transfor- mation. Prove first that p ◦ ψ = p,andthenfinishtheproofoftheIsomorphismTheoremby showing that ψ is a homeomorphism of X˜. Exercise 6.20. Prove that any deck transformation is completely determinedbyitsactionon just one point; in particular, the only deck transformation that fixed any point of X˜ is the identity.

6.7 The Torus, Again Compute the deck group of the covering of the torus T 2 by the plane R2 directly, and also by using the Isomorphism Theorem.

25 7Lecture7:MoreonFundamentalGroupsandCovering Spaces

The first part of this lecture follows Hatcher’s Algebraic Topology.

7.1 Homotopy Equivalence Recall that: Let X, Y be spaces, and let I = [0, 1]. We say that two (continuous) maps f0,f1 : X → Y are homotopic if there exists a third (continuous) map F : X × I → Y such that:

F (x, 0) = f0(x) for all x ∈ X

F (x, 1) = f1(x) for all x ∈ X

Amapf : X → Y is a homotopy equivalence if there is another map g : Y → X such that the composition f ◦ g ∼ IdY and g ◦ f ∼ IdX . If a homotopy equivalence exists from X to Y , then we write XY˜ ,andsaythatX and Y are homotopy equivalent and that they have the same homotopy type.

Observation. A homeomorphism from X to Y is a homotopy equivalence. Exercise 7.1. Prove that homotopy equivalence is an equivalence relation on spaces. Deformation retracts. A special case of homotopy equivalence. Let X be a space with subspace A.Thenadeformation retract of X to A is a (continuous) map F : X × I → X such that

• F (x, 0) is the identity map IdX , • F (x, 1) maps X onto A

• For all t ∈ I,themapF (x, t) is equal to IdA when restricted to A. Example: the symbols + and - are homotopy equivalent but not homeomorphic. Exercise 7.2. Prove that a deformation retract gives a homotopy equivalence between spaces. We’ve already seen that two homeomorphic spaces have isomorphic fundamental groups. It turns out that we can extend this to a more general result.

Theorem 7.3. If f : X → Y is a homotopy equivalence, then the induced map f∗ : π1(X, x) → π1(Y,f(x)) is an isomorphism. This theorem is fairly straightforward to prove, with the following lemma/exercise: Exercise 7.4. Suppose that F : X × I → Y is a homotopy from f = F (x, 0) to g = F (x, 1). Let x0 ∈ X,andletα denote the path F (x0,t).Thenwegetinducedmapsf∗ : π1(X, x0) → π1(Y,f(x0)) and g∗ : π1(X, x0) → π1(Y,g(x0)),andwehaveanisomorphismoffundamental groups induced by the path α: φα : π1(Y,f(x0)) → π1(Y,g(x0)).Provethatφα ◦ g∗ = f∗.

26 Proof of Theorem. Let g : Y → X be a homotopy inverse for the homotopy equivalence f : X → Y . Consider the composition f∗ ◦ g∗ ◦ f∗ : π1(X, x0) → π1(X, f ◦ g ◦ f(x0)). The previous exercise implies that the composition of the first two maps: g ◦ f : π1(X, x0) → π1(X, g ◦ f(x0)) is φα for some α. In particular, this implies that f∗ is one-to-one. Similarly, we can see that g∗ is also one-to-one, which in turn implies that f∗ is also surjective.

7.2 Action of the deck group on the cover Notation: if a group G acts on a space X, we often denote the quotient by X/G. Let p : X˜ → X be a covering map where as usual X,X˜ are path-connected and X˜ is simply connected. The deck group Aut(X˜) clearly acts on the space X˜, and it acts by homeomorphisms (see end of Lecture 2 notes). In particular, for G =Aut(X˜), the following condition holds: (*) Each x˜ ∈ X˜ has a neighborhood U such that g · U ∩ U = ∅ for all nontrivial g ∈ G. Exercise 7.5. Prove this. Recall that any deck transformation is completely determined by its action on just one point; in particular, the only deck transformation that fixed any point of X˜ is the identity. We have seen that Aut(X˜) =∼ π1(X). So this means that π1(X) also acts on X˜ by homeo- morphisms. There is a natural way to define this action, as follows. Letx ˜ ∈ X˜, let x = p(˜x), and let f ∈ π1(X, x). Then there is a unique lift of f to a path f˜ with initial pointx ˜.Let˜y be its terminal point, and define f · x˜ =˜y. Exercise 7.6. Prove that this action is compatible with the action of Aut(X˜) under the Iso- morphism Theorem. Some terminology: a covering p : X˜ → X is normal if for every x ∈ X and every pair of pointsx, ˜ x˜" ∈ p−1(x), there exists a deck transformation ψ such that ψ(˜x)=x˜". Theorem 7.7. If an action of a group G on a path-connected space X˜ satisfies the condition above, then the following hold: 1. The map X˜ → X/G˜ defined by x˜ .→ [˜x] is a normal covering map. (This is true even if X˜ is not path connected.) 2. The group G is the deck group for the covering X˜ → X/G˜ .

3. If X˜ is also simply connected, then G =∼ π1(X/G˜ ). Proof. To prove the first statement, let [x] ∈ X/G˜ . Then take the neighborhood U of x in X˜ guaranteed by the condition (*) above. Then p(U) is a neighborhood of [x] which is evenly covered. Clearly G is at least a subgroup of the deck group Aut(X˜), and the covering is normal because each pair of points in the preimage of a single point “downstairs” lives in a g1(U),g2(U), −1 respectively, so g2g1 takes one to the other. Now, to prove that G is the full set of a deck transformations, suppose that ψ is an arbitrary element of Aut(X˜). Let [x] ∈ X/G˜ , with x a point in the preimage p−1([x]). Then by definition, ψ(x) is in the same orbit as x, so there exists an element g ∈ G

27 7.3 Fundamental groups of surfaces given as plane models Recall that a subgroup N is normal in a group G if gng−1 ∈ N for all g ∈ G. This is precisely the condition required to make the multiplication of cosets aN · bN = abN be well defined, hence giving a group structure on G/N, the set of all (left) cosets of N in G. If R1,...,Rm are elements in a group H,thenthenormal closure of R1,...,Rm ∈ H is the smallest normal subgroup of H containing the elements R1,...,Rm. We say that a group G has presentation 2g1,...,gn | R1,...,Rm3 if the group G is generated by the elements g1,...,gn and if G =∼ F/N,whereF is the free group on {g1,...,gn} and N is the normal closure of the elements R1,...,Rm. (Typically Ri is given as a word in the gj .) Theorem 7.8 (Seifert-van Kampen Theorem). Let X = A ∪ B,whereA, B, A ∩ B are all nonempty, open, and path-connected. Let jA,jB denote the respective inclusion maps of A ∩ B into A, B.Chooseabasepointx0 ∈ A ∩ B.Now,supposeweknowthefollowing:

π1(A, x0)=2a1,...,an | R1,...Rp3

π1(B,x0)=2b1,...,bm | S1,...Sq3

π1(A ∩ B,x0)=2c1,...,ck | T1,...Tr3

Then we can write down the following presentation for the fundamental group of X:

π1(X, x0) =∼ 2a1,...,an,b1,...,bm | R1,...Rp,S1,...Sq, (jA)∗(c1)=(jB)∗(c1),...,(jA)∗(ck)=(jB )∗(ck)3

Proving this is nontrivial. Goodman contains a good sketch. Armstrong proves a version using simplicial approximations. Other textbooks like Munkres, Hatcher, and Massey, prove it in greater generality.

Corollary 7.9. If P is a plane model for a surface S with boundary word w(x1,...,xn),then apresentationforπ1(S) is given by:

2x1,...,xn | w =13

Exercise 7.10. Prove the corollary assuming the Seifert-Van Kampen Theorem. Exercise 7.11. Assuming the full Classification Theorem, prove that a closedorientablesurface of genus g has fundamental group with presentation:

2a1,b1,...,ag,bg | [a1,b1] ···[ag,bg]=13

28 8Lecture8:JordanCurveTheorem

Our proof of the Jordan Curve Theorem will show us some deep connections with elementary combinatorics, in the form of graph theory. We will closely follow the treatment of Carsten Thomassen’s excellent (and relatively short) paper The Jordan-Schoenflies Theorem and the Classification of Surfaces.

8.1 Graph Theory Cheat Sheet • An arc is a path which is one-to-one. (Exercise: Prove that a space is path-connected if and only if it is arc-connected.) • A polygonal arc is an arc which is the union of a finite number of (straight) line segments. • A region of an open set in the plane is a maximal path-connected subset. • A graph Γis a pair (V,E,φ)whereV is a FINITE set of vertices, E is a FINITE set of edges,andφ is a function from E to (unordered) pairs of vertices. We usually leave out the φ from the notation and when convenient denote an edge e by vw if φ(e)={v, w}. In this case we say that e is incident to v and w.DOWEALLOWLOOPS?

• If there exist edges e1 #= e2 such that φ(e1)=φ(e2), then we say that Γhas multiple edges.

• A path is a graph (V = {v1,...,vn},E = {v1v2,v2v3,...vn−1vn}, where all the vi are distinct.

• If we add the edge v1vn to a path we get a cycle. We denote both the path and the cycle by listing the vertices and hope that it’s clear from the context what we mean. • If A ⊂ V ,thenΓ− V is the graph obtained by deleting all • A graph is connected if every two vertices are joined by a path. • A graph Γis 2-connected if it is connected and if for every v ∈ V ,Γ−{v} is connected. • If X is a space, we say that Γ= (V,E)canbeembedded in X if there is a one-to-one function from V to X such that each edge e = vw ∈ E can be represented by a simple arc in X joining v to w. • If Γcan be embedded in the plane R2, we say that Γis a planar graph.

• If Γactually is embedded in the plane R2, we say that Γis a plane graph. • A subdivision ofΓis a graph obtained from Γby inserting vertices on edges in the obvious way.

Exercise 8.1. Define an isomorphism of graphs. Write down a precise definition of “subdivi- sion”.

29 8.2 Polygonal Jordan Curve Theorem Lemma 8.2 (Polygonal Approximations of Paths in Open Sets). If A is an open path-connected set in R2,thenanytwopointsinA are joined by a simple polygonal arc in A. Proof. Let p, q ∈ A.Letf be any path in A from p to q. Consider the set

S = {t ∈ [0, 1] | A contains a simple polygonal arc from p to t}. Then S is nonempty since 0 ∈ S, and it’s clearly bounded above by 1. As we’ve seen before, it’s an easy exercise to show that the least upper bound s of S is actually in S,andmoreover that s =1. Lemma 8.3. If Γ is a planar graph, then Γ admits an embedding in R2 where all the edges are simple polygonal arcs. Proof. Let’s abuse notation and think of Γas an actual plane graph. For every vertex p ∈ Γ, we choose a closed disk D(p) which is small enough so that it does not intersect any edges which are not incident to p. We also choose our disks so that D(p) ∩ D(q)=∅ for p #= q.Ifpq is an edge from p to q, we can consider 3 subintervals of pq determined by D(p),D(q). Then we can use the previous lemma (carefully) to replace pq with a simple polygonal arc. Theorem 8.4 (Jordan Curve Theorem: Polygonal Case). If C is a simple closed polygonal curve in R2,thenR2 − C has precisely two regions, each of whose boundaries is C.

2 Proof. Suppose that p1,p2,p3 belong to distinct regions of R − C. Choose any disk D such that D ∩ C is a straight line segment. For each i, we can choose a polygonal arc from pi to D (not intersecting C). Since we’re working with an actual disk with a straight line running through it, we conclude that we can join at least two of the three points by a simple polygonal arc, which is a contradiction. We’ve proved that R2 − C contains at most 2 regions. Now we’ll show it contains at least 2 regions, i.e., that R2 − C is not path-connected. For each point q ∈ R2 − C, we take a (straight) ray R starting at q.ThenR ∩ C is a finite collection of intervals and points. Let Q be an interval (for simplicity, we’ll consider points to be intervals as well). Since we’re in R2 we can talk about “sides” of lines and intervals and we can distinguish in the obvious way between C touching R (entering and leaving Q on the same side of R)andcrossing R (entering and leaving on different sides). Let k be the number of crossing points of R with C. Note that the parity of k is independent of the choice of R and depends only on q and C. Now it is easy to find two points with different parity by finding a ray that intersects C precisely once, and one that does not intersect C at all (which exists since C is a bounded subset of R2.

More Terminology. Now that we understand polygonal closed curves in the plane, we can talk about the interior of a polygonal closed curve C (the bounded component of R2 − C), denoted int(C). and similarly the exterior of a polygonal closed curve C (the unbounded component), denoted ext(C). We will also denote the closure of int(C)=int(C) ∪ C in the usual way: int¯ (C), and similarly for the closure of the exterior... Exercise 8.5. Let C be a simple closed polygonal curve. Let P be a simple polygonal arc properly embedded in int¯ (C),i.e.,suchthatP joins two points p, q on C but the interior of P does not intersect C.LetP1, P2 be the two subarcs of C with endpoints p, q.Provethat

30 2 R − (C ∪ P ) has precisely three regions, and their boundaries are C, P1 ∪ P ,andP2 ∪ P , respectively. (HINT: This exercise is a sort of scholium of the Polygonal JCT.) Conclude that if r, s are two points on the interiors of P1,P2,respectively,thenitisnotpossibletojointhem by a simple polygonal arc in int(C) without intersecting P .

8.3 General Jordan Curve Theorem The work we did in the previous subsection was mainly to prove the following fact, which most of us figured out as kids without actually being motivated to actually prove it. (Or was that just me?)

Theorem 8.6. The complete bipartite graph K3,3,alsoknownastheUtilities Graph,isnot planar. Please note that this is NOT the same as the famous theorem of Kuratowski, which says that a graph is nonplanar if and only if it contains a finite list of subgraphs, one of which is the Utilities Graph. That is harder to prove.

Proof that the Utilities Graph is Nonplanar. There is a cycle C of length 6 in K3,3 and we can label its vertices x1x2x3x4x5x6. We can think of C as a polygonal simple closed curve with two “chords” x1x4,x2x5,x3x6.IfK3,3 were planar, then at least two of the three chords would have to be either both in int(C) or both in ext(C). But this would contradict the Exercise above. Exercise 8.7. Prove that the intersection of any straight line in R2 with a simple closed curve is compact. Proposition 8.8. Let C be a simple closed curve in R2.ThenR2 − C is not path-connected.

Proof. If we project C to the x-axis, we will get a closed subinterval [a1,a2]ofR.LetL1,L2 denote the vertical lines x = a1,x = a2, respectively, and let L3 be a vertical straight line situated somewhere in between them. For i =1, 2, let pi denote the highest point in the intersection Li ∩ C.LetP1,P2 be the two arcs on C with end points p1,p2.ThenPi ∩ L3 are compact and disjoint. So L3 contains an interval L4 joining P1 to P2 only intersecting C in its endpoints. Now, let L5 be an arc in ext(C) joining p1 to p2 consisting of segments of L1,L2,anda horizontal straight line segment above C.IfL4 is in ext¯ (C) then there is a simple polygonal arc in ext¯ (C)fromL4 to L5. But, then C ∪ L4 ∪ L5 ∪ L6 is a plane graph isomorphic to K3,3, which is a contradiction. Hence L4 does not lie in ext¯ (C), so int(C) #= ∅.

We are assuming all our surfaces are closed and connected, in other words, compact with no boundary. Lemma 8.9 (Key Lemma). If G is a 2-connected graph, H is a 2-connected subgraph of G, then G can be obtained from H by successively adding paths such that each of these paths joints two distinct vertices in the current graph and has all other vertices outside the current graph.

31 Proof. We induct on N = the number of edges in the set EG\EH .IfN =0thenG = H and we’re done. Otherwise, if H is not a maximal 2-connected subgraph of G,thenwecanfind H" which is a maximal 2-connected subgraph of G containing H, and then apply the induction hypothesis to go from H to H",andthenfromH" to G. So now assume H is a maximal 2-connected subgraph of G. Since G is connected, there exists an edge x1x2 ∈/ H with x1 ∈ H. Further, since G − x1 is connected, there exists a path x2x3 ...xk where xk ∈ H but x2,...,xk−1 ∈/ H. (We allow the possibility that k = 2.) But then joining the edge x1x2 and this path x2x3 ...xk to H, we get a 2-connected graph, which must therefore be all of G, and the lemma is proved.

32 9Lecture9:TriangulabilityofSurfaces 9.1 Finishing up Jordan Curve Theorem More Terminology. The term faces is also used for the regions of the complement of a plane graph Γin R2. In particular, the unbounded region is often called the outer face. IfΓis 2-connected, then the boundary of the outer face is the outer cycle. Exercise 9.1. If Γ=(V,E) is a plane 2-connected graph with |V |≥3,allofwhoseedgesare represented as simple polygonal arcs, prove that R2 − Γ has |E|−|V | +2 regions, each of which has a cycle of Γ as boundary. Define the union of two plane graphs in the obvious way: subdivide edges such that each edge is actually straight, not just polygonal. If two edges cross, or if a vertex of one graph lies on an edge of another, subdivide further. then the following lemma is clear.

Exercise 9.2. Prove that if Γ1, Γ2 are two plane graphs such that each edge is a simple polygonal arc, then the union of Γ1, Γ2 is a graph Γ3.

Note that if Γ1, Γ2 are 2-connected then their union Γ3 is too, as long as Γ1, Γ2 share at least two points in common.

Lemma 9.3. Let Γ1, Γ2,...,Γk be plane 2-connected graphs all of whose edges are simple polygonal arcs such that Γi has at least two points in common with Γi−1, Γi+1,butshares no points in common with any other Γj .AlsoassumethattheabstractgraphsΓi are pairwise disjoint (i.e., we’re not embedding any subgraphs separately in the plane). Then any point which is in the outer face of Γ1 ∪ Γ2, Γ2 ∪ Γ3,...,Γk−1 ∪ Γk is also in the outer face of Γ1 ∪···∪Γk.

Proof sketch. Take a point p in a bounded face of Γ1 ∪···∪Γk. Find a cycle C which contains p in its interior. Choose C minimally in that it is contained in Γi ∪ Γi+1 ···∪Γj with j − i as small as possible. Argue that j − i =1. Proposition 9.4. If A is a simple arc in the plane, then R2 − A is path-connected. Proof sketch. Let p, q be two points in R2 − A.Choosed such that p, q are both at least a distance of 3d from the arc A.Now,A is the image of a uniformly continuous map of I into 2 R . So we can partition A into segments Pi with end points p = p1,p2,...,pn such that any " point on the path in A from pi to pi+1 is at most distance d away from pi. Then let d be the " minimum distance between all the Pi,Pj , with |i − j|≥2. So d ≤ d. Keep going like this, further subdividing each Pi into segments Pi,j . The idea is to cover A with boxes of a size designed to ensure that graphs formed from the union of the boundaries of the boxes satisfies the hypotheses of the previous lemma, thus guaranteeing that p, q are in the outer face of this union, and the arc A is completely disjoint from this outer face. Thus p, q can be joined by a simple polygonal arc in the complement of P . Theorem 9.5 (General Jordan Curve Theorem for R2). If C is a simple closed curve in R2, then R2 − C has precisely two regions, each of which has C as boundary.

Proof. Suppose that p1,p2,p3 are distinct points contained in three distinct regions R1,R2,R3, 2 respectively, of R −C.LetQ1,Q2,Q3 be three pairwise disjoint segments of C. It follows from the preceding lemma (plus a little thought) that we can find 9 polygonalarcsAij satisfying:

33 • Aij is a a simple polygonal arc from pi to Qj .

• Aij is contained in Ri

• Aij ∩ Aik = pi if j #= k

• Aij ∩ Ai!j! = ∅

But then union of the Aij can be extended to form a plane graph isomorphic to K3,3. this is a contradiction. So by Propositon 8.8, R2 − C has precisely two components. Finally, by Proposition 9.4, we know that C is the boundary of each. Exercise 9.6. State and prove an analogue of the Jordan Curve Theorem for thesphereS2. Theorem 9.7 (Jordan-Schoenflies). If f is a homeomorphism of one simple closed curve in the plane onto another, then f can be extended to a homeomorphism of the whole plane. Prove the following generalizations of Exercises 8.5 and 9.1, respectively, which are useful in proving the Jordan-Schoenflies Theorem. Exercise 9.8. Let C be a simple closed curve and let P be a simple polygonal arc in int(C) such that P joins two points p, q on C but does not otherwise intersect C.LetP1,P2 be the two arcs on C with endpoints at p, q.ThenR2\(C ∪P ) has precisely three regions whose boundaries are C, P1 ∪ P ,andP2 ∪ P ,respectively. Exercise 9.9. If Γ=(V,E) is a plane 2-connected graph containing a cycle C (which is a simple closed curve) such that all edges in Γ\C are simple polygonal arcs in int¯ (C),thenR2\Γ has |E|−|V | +2 regions, each of which has a cycle of Γ as boundary. If we assume the Seifert-van Kampen Theorem, then we can prove the Jordan Curve The- orem as an application of fundamental groups. We sketch this approach as follows. Let C be a simple closed curve in R2. Choose a homeomorphism h : R2 → S2 −{(0, 0, 1}. Choose a point p ∈ C, and then choose a second homeomorphism k : R2 → S2 −{p}.Let L = k−1(C) −{p},sothatL is some kind of infinite “line” in R2. Exercise 9.10. Use L to form an appropriate decomposition of R2 − L such that Seifert-van Kampen will tell us that R2 − L is simply connected. Exercise 9.11. Prove that there is a homeomorphism of R3 taking L to the z-axis. Exercise 9.12. Assuming any results you may need about extending continuousfunctions(e.g., Tietze Extension Theorem; see, e.g., Armstrong), finish the proof of the Jordan Curve Theorem.

9.2 Corollaries of Jordan-Schoenflies Let F be a closed set in the plane. We say that a point p ∈ F is curve accessible if for each point 1 ∈/ F , there is a simple arc A from p to q such that A ∩ F = {p}. Corollary 9.13. If F is a closed set in the plane with at least three curve-accessible points, then R2\F has at most two regions. Exercise 9.14. Prove Corollary 9.13 assuming the Jordan-Schoenflies Theorem. Corollary 9.15. Let Γ, Γ" be 2-connected plane graphs such that g is a homeomorphism and plane-isomorphism of Γ onto Γ".Theng can be extended to a homeomorphism of all of R2. Exercise 9.16. Prove Corollary 9.15. You will need to use the Key Lemma from previous lecture.

34 9.3 Proof of Triangulability, assuming Jordan-Schoenflies. Theorem 9.17. Every surface S is homeomorphic to a triangulated surface. Proof. As we’ve observed before, it is sufficient to prove that S has a cellular structure.

Setup. For each point p on S, choose a disc neighborhood D(p), which we think of as an actual disc in the plane, disjoint from all others. Inside D(p) we draw two quadrilaterals Q1(p) ⊂ Q2(p) containing p (interiors nested). Since S is compact, we can take a finite cover of S by int(Q1(pi). The idea of the proof is basically to modify the choices of the Q1’s so that eventually they give a cell structure on S.

The Painful Step. The main difficulty will be arranging things so that the Q1(pi)sothathave only a finite number of points of intersection on S. The idea is to build up starting with Q1(p1), and show how to choose Q1(p2) to ensure that the two only have a finite number of points of intersection. More precisely, we will assume inductively that Q1(p1),Q1(p2),...Q1(pk−1)have been chosen to achieve finitely many intersection points. Now, look at Q2(pk), the “outer square” for pk.ChooseQ3(pk)tobesomeothersquare nested between Q1(pk)andQ2(pk). We are going to use Q3(pk)tofinda“new”Q1(pk) which will complete the inductive step. Consider intersections of Q1(pj ) with Q2(pk). We say a segment P of Q1(pj ) is bad if it joins two points of Q2(pk) and is otherwise contained in the interior of Q2(pk). We say that such a P is very bad if it also intersects Q3(pk).

CLAIM 1: There are aprioriinfinitely many bad segments in Q2(pk) but only finitely many very bad ones. Exercise 9.18. Prove the claim *carefully* using an argument similar to thatusedtoprove Proposition 1.14 on p. 35 of Hatcher’s Algebraic Topology.

Since there are only finitely many very bad segments, their union together with Q2(pk) itself forms a 2-connected graph which we will call Γ. Using Lemma 8.3, we can replace Γwith a plane-isomorphic graph Γ" whose edges are all simple polygonal arcs. Corollary 9.15 allows us " to extend the isomorphism from Γto Γ to a homeomorphism of the interior of Q2(pk) which is " " fixed on Q2(pk) itself. Let Q1,Q3 be the images of Q1(pk),Q3(pk) under this homeomorphism.

"" CLAIM 2: We can now choose a simple closed polygonal curve Q3 in the interior of Q2(pk) " "" "" such that Q1 ⊆ int(Q3 )andsuchthatQ3 intersects only the very bad segments (i.e., disjoint from any “merely” bad segments). Exercise 9.19. Prove Claim 2. The main thing we’ve achieved now is that the very bad segments are now simple polygonal arcs. " "" Now, Γ ∪ Q3 is a 2-connected plane graph. If we use Corollary 9.15, we can now take "" " "" Q3 to be a quadrilateral whose interior contains Q1.SowecanuseQ3 as our “new” Q1(pk), and we have finitely many intersection points with Q1(p1),Q1(p2),...,Q1(pk−1). Also, our construction ensures that the Q1(pi) still cover S.

Final Step. The union of the Q1(pi)’s now gives a nice graph Γon the surface S.The Jordan-Schoenflies theorem now gives the desired homeomorphism to a cellular surface.

35 More precisely, each Q2(pi) contains only finitely many very bad segments, which are all simply polygonal arcs which form a 2-connected graph. Thus each region of the complement of Γ= ∪Q1(pi) is bounded by a cycle C which lives inside some Q2(pi). And For each such C, draw a new convex polygon C" of, say, unit sidelength, whose vertices are the same as those of C. Taking the union, this gives a new graph Γ" which is plane-isomorphic to Γ. So now we use the Jordan-Schoenflies to extend this to a homeomorphism on the interiors of the C’s.

9.4 Well-definedness of connected sum Let’s redefine the operation of connected sum as follows. We now know that any surface carries a cell structure with a single face F , which we think of as sitting in the plane R2.Taketwo congruent disjoint triangles T1,T2 in the interior of F , and delete their interiors. Each triangle 2 Ti inherits an orientation from R , which just means a cyclical ordering of vertices – we’ll choose, say, a clockwise ordering. (The choice doesn’t matter, but we just have to make one.) We now identify T1 with T2 in such a way that their orientations disagree. This gives us a new surface S". It is an easy exercise to show that we can continuously map the interior of F to itself taking any triangle to any other, so this operation is well defined. Exercise 9.20. Prove that the surface S" which results from this operation is homeomorphic to the result of doing a connected sum of S with a torus T 2. Exercise 9.21. Prove that we can relax this a little bit by allowing the triangles to be taken from different faces.

9.5 Finishing Classification Exercise 9.22. So...where are we? Over the weekend, take stock. Try to write down an outline of the proof of classification theorem, including any missingpiecesifany.Whathaveweproved? What remains to finish offour proof of classification of surfaces, if anything?

36 10 Lecture 10: Proof of Triangulability of Surfaces 10.1 Well-definedness of connected sum Let’s redefine the operation of connected sum as follows. We now know that any surface carries a cell structure with a single face F , which we think of as sitting in the plane R2.Taketwo congruent disjoint triangles T1,T2 in the interior of F , and delete their interiors. Each triangle 2 Ti inherits an orientation from R , which just means a cyclical ordering of vertices – we’ll choose, say, a clockwise ordering. (The choice doesn’t matter, but we just have to make one.) We now identify T1 with T2 in such a way that their orientations disagree. This gives us a new surface S". It is an easy exercise to show that we can continuously map the interior of F to itself taking any triangle to any other, so this operation is well defined. Exercise 10.1. Prove that the surface S" which results from this operation is homeomorphic to the result of doing a connected sum of S with a torus T 2. Exercise 10.2. Prove that we can relax this a little bit by allowing the triangles to be taken from different faces.

10.2 Finishing Classification N.B.: I am purposely not writing a lot of details in this particular section of the lecture notes.

Things we have proved, more or less:

1. Every surface is triangulable, using the Jordan Curve Theorem and the Jordan-Schoenflies Theorem. 2. Every triangulated surface can be represented by a “plane model” or “polygonal model”, that is, a polygon with its sides identified in pairs. 3. Every orientable triangulated surface is homeomorphic to a connected sum of the sphere S2 with some number of tori. This number of tori is the genus of the surface. Apriori, the genus of a surface is not even well-defined for a particular surface, let alone for a surface up to homeomophism. 4. Just a note: our argument for showing that every orientable triangulated surface is home- omorphic to a connected sum of S2 with some number of tori can be modified without much trouble to show how to change any given plane model into the “standard” 4g-gon. (Exercise?)

5. Connected sum is well-defined. In other words, the connected sum of g tori (done in one particular way) is homeomorphic to the connected sum of g tori (done in some other specified way). 6. The Euler characteristic of any triangulation of any plane model representing S2 is 2. 7. The Euler characteristic of any triangulation of any plane model representing a (closed orientable) surface which is homeomorphic to the connected sum of S2 with some nonzero number g of tori is 2 − 2g.

37 8. By a connectedness argument, we know that any connected sum involving tori is not homeomorphic to S2. To sum up: what we have so far is a complete infinite list of possibilities, but we don’t know yet whether it’s possible for a surface to be homeomorphic to more than one thing on the list. For example, how do we know that there is no homeomorphism from theconnectedsumofg tori to the connected sum of h tori for g #= h?

Two ways to finish offthe proof, which we haven’t yet proved:

1. Euler characteristic is an invariant of the choice of triangulation. 2. The abelianization of the fundamental group of a surface of genus g is Z2g (and Zk !Zl for k #= l, but this part isn’t so hard).

If we were going to try to use fundamental groups as an invariant of surfaces, we would have to give a full proof of Seifert - van Kampen. If we don’t care about obtaining the most general possible statement of SVK, then this can be done for surface groups using something called the Simplicial Approximation Theorem and the technique of “barycentric subdivision” of a given triangulation. These techniques are similar at least in spirit to those we have seen developed in Thomassen’s graph theoretic approach and are completely accessible at this point, though the proof is quite long and involved. So, we will proceed by showing the former. Theorem 10.3. Euler characteristic is an invariant of the choice of triangulation and hence invariant under homeomorphism. Proof. Let T,T" be two different triangulations of S. Then we can represent them each (dis- jointly) as polygonal plane graphs Γ, Γ", respectively, with some sort of homeomorphism taking Γ" to the polygon for Γ. Now, the image of Γ" under this map may not be polygonal, but using Lemma 8.3 we can arrange that it is. Thus, we can form the graph G =Γ∪ Γ".Now,eachof Γ, Γ" can be related to G (and hence to each other) by a finite sequence of moves of the type described in Exercise 4.5. So we’re done. Theorem 10.4. The homeomorphism type of a compact orientable surface is completely deter- mined by its genus and the number of boundary components.

10.3 Classification of simple closed curves on surfaces We have seen how important it is to understand just the case of a simple closed curve in the plane, since the entire classification of surfaces rests mainly on the strength of the Jordan Curve Theorem. So we need to understand all the possible ways a simple closed curve can be embedded in an arbitrary surface. Just like with the Jordan Curve Theorem, what we really need to do is describe the complement of the curve. Hence given a simple closed curve C and a surface S (with C ⊂ S), we define the surface SC to be a surface with two boundary components c1,c2 such that the result of gluing c1 to c2 is the surface S, and such that the image of c1,c2 in this quotient surface is the curve C. From here on out, curve means simple closed curve. AcurveC ⊂ S is nonseparating in S if SC is path-connected, otherwise it is separating.

38 Proposition 10.5. Up to homeomorphism, there is a unique nonseparating curve onagiven surface. More precisely, if α,β are two nonseparating curves on a surface S,thenthereexists ahomeomorphismh : S → S such that h(α)=β.

Proof. By Theorem 10.4, there exists a homeomorphism φ : Sα → Sβ.Moreover,wecan choose φ so that it “correctly” matches up boundary components. The map φ induces a homeomorphism of S taking α to β. A similar argument shows that if α,β are two separating curves on S, then there is a homeomorphism of S taking α to β if and only if the corresponding “pieces” of Sα,Sβ are homeomorphic. If S is closed, we often say that the genus of a separating curve C is the minimum of the genus of the components of SC, although this term is not always consistently used, and saying that a curve C has genus g may simply mean that one of the components of SC has genus g.

Exercise 10.6. Use the Jordan Curve Theorem to prove that SC has at most 2 components. Exercise 10.7. Classify all topological types of curves on a closed surface of genus g. If α,β are two curves on a surface S,wedefinethegeometric intersection number of α and β to be |α∩β|. We denote this by i(α,β). Apriori,thesetα∩β could have infinite cardinality. Using something called transversality, it turns out that in fact one can always homotope a finite collection of curves to have finitely many points of intersection which are “transverse”, as opposed to “tangencies”. Unfortunately, we don’t have the time in this course to develop the necessary tools to express this rigorously. Exercise 10.8. Let α be a nonseparating curve on a surface S.Proveeachofthefollowing. 1. If the genus of S is at least 1, then there exists a curve β such that i(α,β)=1 2. If S is genus two, prove that there exists a curve β such that α and β fill S,thatis, S\(α ∪ β) is a disjoint union of disks. 3. Show that there exists a curve α" which is disjoint from α and not isotopic to α so that the union α ∪ α" separates S.(ForthisexerciseyouwillneedtoassumethatS has genus at least 3.

39 11 Lecture 11: Basics of curves on surfaces

Acknowledgement: following the Primer by Farb-Margalit Let S = Sg,b,n denote a surface with genus g, with b boundary components and n punctures (points removed). Recall that throughout, “curve” means “simple closed curve”.

11.1 Essential Curves A curve is essential on S if it is NOT homotopic to a point or a boundary component or a puncture.

11.2 Algebraic intersection We defined geometric intersection of two curves in the previous lecture. To define a notion of algebraic intersection, we require some notion of an orientation of a surface. Roughly speaking, an orientation is a consistent choice of ordered basis in a neighborhood locally homeomorphic to R2 (usually we choose {(1, 0), (0, 1)}). We need the machinery of smooth manifolds to explain precisely what we mean by a “consistent choice”, but the idea is that we need to have a well- defined sense of “left”, “right”, “up” and “down”. Note that we don’t need any of this to talk about an orientation of just a curve, since that can be given by the usual parametrization. Let α,β be two oriented, transverse curves on an oriented surface. Then ˆi(α,β) is the sum of all signed intersections, where we assign the value +1 to an intersection which “agrees” with our orientation and −1 otherwise. Clearly then we have:

ˆi(α,β)=−ˆi(β,α) Exercise 11.1. Prove the following: 1. If α is separating, then for all transverse curves β,wehavei(α,β) ∈ 2Z,andˆi(α,β)=0. 2. In general, algebraic intersection and geometric intersection of α,β agree mod two. 3. Assuming the bigon criterion, for a =[α],b=[β],thenˆi(a, b) is well defined. 4. i(a, a)=ˆi(a, a)=0.

11.3 Geodesic representatives Here S is a hyperbolic surface. Recall that there is a bijective correspondence between the set of all conjugacy classes in the fundamental group of a surface S and the set of all free homotopy classes of oriented curves in S. So we will often abuse notation and use α to denote a curve in S and also to denote an element of π1(S), even though that is only well defined up to conjugacy. A geodesic in a hyperbolic surface S is the image of a geodesic in H2.

Proposition 11.2. Suppose S is a hyperbolic surface (i.e., χ(S) < 0). Let α ∈ π1(S) be a (not necessarily simple) curve in S which is not homotopic to a puncture. Then there exists a unique geodesic γ in the homotopy class of α.Moreover,ifα is simple, then so is γ.

40 Before we prove the proposition, recall that individual elements in the conjugacy class of 2 α #= 0 in π1(S) are in one-to-one correspondence with the set of all lifts to H of α.Thegroup π1(S) acts on the set of lifts, and we can interpret this action as π1(S) acting on cosets of the cyclic group 2α3≤π1(S). The stabilizer of the coset γ2α3 is the subgroup generated by γα0γ where α0 is a primitive root of α. Exercise 11.3. Explain why this correspondence fails to be true for T 2 (with the usual covering by R2). Proof of Proposition. We will not prove the last statement of the proposition here. And we’ll 2 simplify things by assuming here that α is primitive in π1(S). Choose a liftα ˜ ∈ H .Letφ be the generator of the stabilizer ofα ˜.Thenφ is some hyperbolic isometry of H2.LetA be its axis. By the straight-line homotopy, we see that the image of A under the covering map H2 → S is a geodesic in [α]. Now, S1 × I is compact, and hence distances of lifted homotopies in H2 are bounded. So any other homotopic lift has the same endpoints at infinity, so we get the same geodesic A in H2. Hence the geodesic representative of [α] is unique.

Exercise 11.4. Finish the proof of the proposition in the case when α is not primitive.

Exercise 11.5. Let α be a simple closed curve. Then the conjugacy class of α in π1(S) consists of primitive elements.

11.4 Torus example Here is a basic problem: given a surface S, classify all homotopy classes of simple closed curves on S. This is not really possible in general, at least not in any meaningful sense, but on the torus we can give a complete answer. Consider the usual covering map R2 → T 2.Thenwehave

Aut(R2,p) =∼ Z2 =∼ 2(1, 0), (0, 1)3

2 q Then we can represent (p, q) ∈ π(T ) by the projection to the unit square of the line y = p x. Conversely, any lift of an element in π1(S) ends ,at some (p, q), and then we can just use the straight line homotopy. Exercise 11.6. Prove the following. In each case, you might want to try first proving the case (p, q)=(1, 0). • i((p, q), (p",q")) = |pq" − p"q| • ˆi((p, q), (p",q")) = pq" − p"q

11.5 Bigon criterion Lemma 11.7. Let χ(S) < 0.Letα,β be transverse curves in S.Ifα,β form no bigons, then any pair of lifts α˜, β˜ in the universal cover (H2)intersect at most in one point. Proof sketch. Suppose that there are two (or more) points of intersection; thensomesegments ofα ˜, β˜ form a disk upstairs. By an innermost disk argument, this disk must project via the covering map to a disk downstairs, contradicting the assumption that there are no bigons.

41 Theorem 11.8 (Bigon Criterion). Two transverse simple closed curves α,β in a surface S are in minimal position if and only if α,β do not form a bigon. The Bigon Criterion can be proved as follows: Exercise 11.9. Prove that any two lifts α˜, β˜ intersect in at most one point, and that their corresponding axes have distinct endpoints. Conclude that there cannot be any intersection- reducing homotopy of α or β.

42 12 Lecture 12: Mapping Class Groups I 12.1 Punctures versus Boundary Components In Mod(S), punctures are fixed setwise and boundary components are fixed pointwise, not just by the maps but also during isotopies. We can think of a puncture as being obtained by removing a closed disk from a surface, whereas a boundary component is obtained by removing an open disk. Thus we can also think of a puncture as a boundary component that is allowed to be twisted and/or swapped with other boundary components. Finally, it is equivalent to think of punctures as marked points on the surface (rather than deleted points), as long as we make the condition that isotopies don’t allowed non-marked points to wander over the marked points.

12.2 Finite order examples 1. The hyperelliptic involution (kebab embedding, rotated by π) 2. Element of order g (flower embedding)

Exercise 12.1. Find examples of the following finite order elements in the mapping class groups of the following surfaces:

1. An element of order 4g +2,foranySg 2. Another involution besides the one described above. (Makesuretogiveaconvincingreason why it’s different!)

3. An element of order 5 in Mod(S2).

12.3 Some warmup examples

Notation: Sg,b,n has genus g, with b boundary components and n punctures. Proposition 12.2 (Alexander Lemma). The mapping class group of a disk Mod(D2) is trivial.

2 So in the triple-index notation, D = S0,1,0,orjustS0,1.

2 2 Proof. Given φ : D → D with φ = id∂D2 ,wedefine:

x (1 − t)φ( 1−t )for0≤|x|≤1 − t F (x, t)=  x for 1 − t ≤|x|≤1 

2 Exercise 12.3. The mapping class group is also trivial for the following surfaces: S0,1,1,S0,0,1,S , that is, the disk with one puncture, the sphere with one puncture, and the sphere. The next calculation gives some idea of what happens with punctures. Exercise 12.4. If α,β are simple arcs with the same endpoints, then α and β are homotopic rel endpoints.

43 Proposition 12.5. The mapping class group Mod(S0,0,3) =∼ Σ3,whereΣ3 denotes the sym- metric group on three letters. Proof sketch: The map is the obvious one induced by what happens to the puncturesundera mapping class. It’s clearly a homomorphism and clearly surjective. To see that it is injective, consider an element in the kernel. Since endpoints are preserved, it takes any arc to another arc with the same endpoints, which is isotopic to the original arc by the previous exercise. We invoke an isotopy extension theorem and assume WLOG that this arc is in fact fixed. Then we cut open along the arc to obtain a disk with one puncture, which has trivial mapping class group.

Remark: The surface S0,0,3 is sometimes called an “open pair of pants”. A pair of pants is 3 the surface S0,3,0.(ItturnsoutthatMod(S0,3,0 is isomorphic to Z .)

Exercise 12.6. Prove that Mod(S0,0,2) =∼ Z/2Z.

Now consider the annulus A = S0,2,0. Proposition 12.7. The mapping class group of the annulus Mod(A) =∼ Z. Proof sketch. The annulus A is covered in the obvious way by R × [0, 1], which we think of as embedded in the upper-half plane in the obvious way. Let [φ] ∈ Mod(A), and let f be a properly embedded arc in A, say, corresponding to {(r, t) | t =0and1≤ r ≤ 2}. Think of f as a path, then we have a unique lift φ(˜f)toR × [0, 1] based at the origin. The endpoint f˜(1) = (n, 1) for some integer n.Thenwedefineamap

Mod(A) → Z [φ] .→ n

Well definedness follows from homotopy lifting, and it’s also clear that the map is a homo- morphism. If we think of the annulus as those points in R2 in terms of polar coordinates with radius 1 ≤ r ≤ 2, then we can define a map TA by

(r, θ) .→ (r, θ + πr)

n Then TA .→ n as described above, so the map is surjective. We can prove injectivity using the straight-line homotopy.

44 13 Lecture 13: Mapping Class Groups II 13.1 Dehn twists

The map TA defined on an annulus A in the previous section is an example of a Dehn twist. More generally, if α ⊂ S is a simple closed curve, we can take an annular neighborhood A of α, do TA on A, and then extend by the identity to S. We want to call this a “twist about α”. A priori, this depends on the choice of the neighborhood A, and on a particular parametrization of A, but up to isotopy, these choices don’t matter. So if a denotes the isotopy class of α,then we can define Ta to be the Dehn twist about a. Note that this does not require any choice of orientation for a, but instead only relies on an orientation of the surface S.

Exercise 13.1. The Dehn twist Ta is trivial in Mod(S) if and only if a is null-homotopic or parallel to a puncture. Proposition 13.2. Let a, b be isotopy classes of simple closed curves with k ∈ Z.Then

k i(Ta (b),b)=|k|i(a, b)

k Proof. You can draw a quick picture to find a representation of Ta (b)andb which realizes the intersection number given in the proposition. I’ll explain this in lecture Exercise 13.3. Use the Bigon Criterion to show that this picture actually realizes minimal position, hence completing the proof of the proposition.

Exercise 13.4. Verify the proposition in each of the following cases: 1. Any example with i(a, b)=1,andk =1, 2, 3. 2. Any example with a separating and i(a, b)=2,andk =1, 2 (you can also do k =3if you’re really brave). The “calculus” of Dehn twists is very beautiful. Below we list several Facts/Exercises, and you should attempt them all. Some are easy, some are hard, but it’s worth attempting them all to appreciate the difference. Exercise 13.5 (Facts about Dehn twists). Here a, b are isotopy classes of curves and f is an arbitrary mapping class.

1. Ta = Tb if and only if a = b

−1 2. fTaf = Tf(a)

3. f commutes with Ta if and only if f(a)=a

4. If a, b are both nonseparating, then Ta is conjugate to Tb

5. TaTb = TbTa if and only if i(a, b)=0

6. TaTbTa = TbTaTb if and only if i(a, b)=1

7. Ta,Tb generated a free group (of rank two) if and only if i(a, b) ≥ 2.

45 Here are two specific results we’ll need later, so even if you didn’t make through all of the above, make sure to do these:

Exercise 13.6. If i(a, b)=1,thenTaTbTa(b)=a.

2 Exercise 13.7. If i(a, b)=1,thenTbTa TbTa reverses the orientation of a.

13.2 Torus, again Chris proved in his lecture that Mod(T 2) =∼ SL(2, Z). It is known classically that SL(2, Z)can be generated with two elements; these correspond to two Dehn twists about a meridian and a longitude on T 2 (the curves corresponding to the projected images of the x− and y− axes, respectively.

13.3 Exact sequences A sequence of maps between groups is exact if the image of the first is precisely equal to the kernel of the next.

Exercise 13.8. Suppose you have the following short exact sequence of groups:

1 → K → G → Q → 1

In particular, this is one way of saying that Q =∼ G/K.ProvethatG can be generated by a set of generators for K together with a lift of a set of generators for Q.

46 14 Lecture 14: Mapping Class Groups III 14.1 Generating sets 1. Dehn twist generators 2. Finite order generators 3. Bounded generating sets

14.2 The curve complex Let χ(S) < 0. We define a simplicial complex C(S) corresponding to S as follows: 1. Vertices: C0(S) ↔ isotopy classes of essential simple closed curves on S. 2. Edges: C1(S) ↔ disjoint representatives.

k 3. k-simplices: C (S):[a1],...,[ak+1]spanak-simplex if and only if there exist pairwise disjoint representatives. The simplicial complex C(S) is usually called the curve complex (as opposed to “a” curve complex). The condition on k-simplices means that C(S) is completely determined by its 1- skeleton C1(S). In other words, C(S) is a flag complex. Despite the simplicity and elegance of its definition, the curve complex remains an active area of research. CHANGE OF NOTATION: Sg,n = Sg,b,n

Theorem 14.1 (Harvey, Lickorish). If 3g + n ≥ 5,thenC(Sg,n) is connected.

Proof. Given a, b ∈ (S), we must find a sequence c1 = a, c2,...,ck = b such that i(ci,ci+1)=0. We will prove this by induction; we will actually need two base cases: if i(a, b) = 0, the result is obvious. If i(a, b)=1,thentaketheboundaryγ of a regular neighborhood of a ∪ b.Thecurve γ must be essential in S otherwise we violate the assumption that 3g + n ≥ 5, Then a, [γ],b is the desired sequence. Now, we are ready to induct on i(a, b) ≥ 2. Consider transverse (oriented) representatives α,β of a, b, respectively. There are two cases to consider based on whether two consecutive intersections of α with β have the same or opposite signs. In either case we can do some kind of “surgery” on α,β to reduce intersection. [FILL IN YOUR OWN PICTURES!] In the case of the torus, we have shown that no two distinct essential curves can be realized disjointly. In this case, we use the minimum possible intersection to determine where to put edges, which in the case of the torus is 1, and we get what’s known as the Farey complex.

14.3 Variations on the curve complex Let N(S) denote the subcomplex of C(S) spanned by (isotopy classes of) nonseparating curves.

Proposition 14.2. For g ≥ 2,thecomplexN(Sg,n) is connected.

47 Proof. Let a, b ∈ N(S). We can connect a, b in C(S), possibly using some separating curves: a = c1,...,ck = b.Supposeci is separating. WLOG, we can assume that ci−1,ci+1 are both nonseparating, since if, say, ci−1 is also separating, then ci−1,ci cobound a subsurface of genus at least 1, so there are infinitely many choices for a nonseparating curve d that we can insert into the sequence: ...,ci−1,d,ci,.... So, we assume that our separating curve ci has two nonseparating neighbors in the sequence ci−1,ci+1.Thenci has two sides, and ci−1,ci+1 are either both on the same side, or else are on different sides. If they’re on different sides, we can simply delete ci from the sequence. If they’re on the same side, then we choose any nonseparating curve from the other side, insert it into the sequence in place of ci, and we’re done. One more variation: we define a complex Nˆ (S) with the same vertex set as N(S), but we put edges down whenever i(a, b) = 1 instead of i(a, b) = 0. This complex is 1-dimensional.

Exercise 14.3. Use connectedness of N(S) to prove that Nˆ (S) is also connected for g ≥ 2(n = 0).

The result of the exercise also holds for S1,n,whenn ≥ 0. This can be proved by induction using the Farey graph as the base case. Note that Mod(S)actsonNˆ (S), a path-connected space. Variations of the following lemma appear all over the place in Geometric Group Theory. This version has basically been cooked up for this example. Lemma 14.4. Suppose G acts by simplicial automorphisms on a connected 1-dimensional simplicial complex X.SupposeG acts transitively on the vertex set X0 and on ordered pairs of vertices connected by an edge. Let v, w be two vertices which share an edge in X,andleth ∈ G take w to v.Thentheset{h}∪ Stab (v) generates G. Note that apriori, the lemma only guarantees a potentially infinite generating set. Much depends on what we can say about the stabilizer of a vertex. When you see these kinds of proofs, remember that everything is cooked up to work out – the hard part is thinking of the right complex for your group to act on in the first place!

Proof. Let g ∈ G.Connectv to g(v) in X: v = v0,v1,...,vk = g(v). Now G acts transitively 0 on X ,sochoosegi such that gi(v)=vi.(Wechooseg0 = e,andgk = g.) We will use induction to show gi is in the subgroup H generated by the element h and the elements of Stab(v). The base case is g0 = e. −1 −1 Now, if ei is the edge between vi and vi+1,thengi (ei) is an edge between v and gi gi+1(v). Since G acts transitively on ordered pairs of vertices connected by an edge, there is an element −1 r ∈ G such that r · (v, gi gi+1(v)) = (v, w). So in particular, r ∈ Stab(v). But then we have

−1 −1 rgi gi+1(v)=w = h (v)

−1 So hrgi gi+1 ∈ Stab(v), which implies that gi+1 ∈ H (assuming by induction that gi ∈ H).

14.4 Generation of Mod(S)

If S = Sg,n is a surface of genus g with n punctures, then PMod(S) means the version of the mapping class group where the set of punctures is fixed pointwise. The “P” stands for “pure”.

48 Theorem 14.5. The pure mapping class group PMod(S0,n) is finitely generated by Dehn twists (about separating curves). We can use the previous theorem, together with the following short exact sequence, to work out Mod(S0,n)

1 → PMod(S0,n → Mod(S0,n) → Σn → 1 What we’re going to prove more carefully is the following theorem.

Theorem 14.6. The group PMod(Sg,n) is generated by finitely many Dehn twists about non- separating curves for g ≥ 1,n≥ 0. Proof. The proof is by double induction on g and n. Chris provided one base case: Mod(T 2), and actually a similar proof goes through for Mod(S1,1. The key ingredient for this proof is the Birman Exact Sequence (BES), which holds for the values of g,n as above.

1 → π1(Sg,n → PMod(Sg,n+1 → PMod(Sg,n) → 1 The first map is the “Push” map, where you “push a marked point alongacurve”.That this is well defined is not at all clear, and unfortunately the proof uses the “long exact sequence of a fibration” and is beyond the scope of this course. The key pointsforourpurposesare:

1. The group π1(Sg,n) is finitely generated by nonseparating loops.

2. If γ is a simple representative of a homotopy class in π1(S), and if a regular neighborhood −1 of γ has boundary components a, b,thenPush(γ)=TaTb (or the inverse). So we’re done with induction on punctures, and now need to induct on genus. We apply Lemma 14.4 to Mod(S)onNˆ (S). By Exercise 13.6, we can take h in the lemma to be the product TaTbTa, if a, b are the vertices playing the role of v, w.Sowejustneedtounderstand Stab(a)=Mod(S, a). There is another short exact sequence which, together with Exercise 13.7, tells us we can assume that the curve a is fixed not only setwise, but with orientation preserved.

1 → Mod(Sg,-a) → Mod(Sg,a) → Z/2Z → 1

Here Mod(Sg,-a)meansthesubgroupofMod(Sg) consisting of elements which fix the ori- ented curve -a . The final piece is to consider the “Delete map” from Mod(Sg,-a)toPMod(S − α), where α ∈ [a]. Exercise 14.7. Prove that the following sequence is exact, if the “right-hand” map is the Delete map: 1 →2Ta3→Mod(Sg,-a) → PMod(S − α) → 1

49