3 manifolds

3-MANIFOLDS, JAKE RASMUSSEN

KEVIN CARLSON

3-manifolds is a subject that’s had extraordinary progress in the last decade, as withthe geometrization conjecture proof. I won’t be able to prove these theorems, but I wantto get us to the point where we understand what they mean. The topology of manifoldsis kind of like the first line of Anna Karenina. The high-dimensional manifolds are happyand understood, but 3- and 4folds have very much their own unique flavors.

1. Basics

1.1. Foundations.

1.1.1. Surfaces. Here are 3 possible definitions of what it means for a compact metrizablespace S to be a surface.

• (Topological) Every p ∈ S has an open neighborhood Up such that Up is homeo-morphic to the open unit ball B2 ⊆ R2.• (Smooth) Same as topological, but the transition functions ϕp ◦ ϕ−1

q : B2 → B2

are smooth.• (Piecewise Linear/Combinatorial) A quotient of a disjoint union of triangles by

identifications of edges in pairs.

It’s a nontrivial fact that these three definitions are equivalent in the following sense:if S is a topological surface, there exists a smooth surface S′ and a homeomorphismbetween them; if smooth surfaces are homeomorphic, then they’re diffeomorphic; if Sis topological there’s a combinatorial S′ homeomorphic to S, and if two combinatorialsurfaces are homeomorphic, then they’re isomorphic by a piecewise linear map aftersubdivision.

A fact we know better is the classification of surfaces as connected sums of the spherewith tori and real projective planes. A note about defining connected sums: use transi-tion maps on the inverse image of the circle of radius 1/2 around a point in each surface,and to have the connected sum be well defined remember to suppose the surfaces areconnected.

1.1.2. 3-manifolds. The first direction of approach is this: which of the facts aboutsurfaces is going to go through for 3-manifolds? The first two definitions are identicalup to dimension. For the third: suppose Y is obtained by starting with a finite set oftetrahedra and identifying faces in pairs. This won’t necessarily give a 3-manifold.

Definition 1.1. Given v ∈ Y as above, define

lk(v) = {x ∈ Y : x ∈ T, v is a vertex of T, dT (x, v) = 1/10}

Date: 10 Oct 2013.

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2 KEVIN CARLSON

So, by construction, lk(v) is a combinatorial surface, and v has a neighborhood homeo-morphic to the cone on lk(v).

Proposition 1.2. Y is a topological 3-manifold iff the link of v is S2 for every vertexv.

The proof will be on the example sheet. That aside, we can now make a

Definition 1.3. Y is a combinatorial 3-manifold if it’s given by identification of tetrahedra-faces pairwise such that every link is a sphere.

Now that we’ve fixed our definitions, we can state

Theorem 1.4 (Moise, 1952). All three definitions are the same, in the same sense asfor surfaces.

This is very much a non-trivial theorem. We remark that it does not hold for n-manifolds even when n = 4. We can define connected sum in the same way, but theobvious analogue of the classification theorem for surfaces is totally false: the semigroupof 3-manifolds under connected sum is not finitely generated.

1.1.3. Manifolds with boundary. Let Y is obtained by identifying some, but not neces-sarily all, faces of a finite set of tetrahedra in pairs.

Definition 1.5. ∂Y , the union of the non-identified spaces, is a combinatorial surface.If the appropriate condition holds about links, specifically each link is either S2 or D2,Y is called a 3-manifold with boundary.

There aren’t any 3-manifolds without boundary that embed into R3, but there areplenty with boundary. For instance, we can solidify any embeddable surface to get thingslike the double solid torus. This always gives “half” a 3-manifold, as follows:

Definition 1.6. The double D(Y ) = Y ∪∂Y Y of a 3-manifold with boundary Y isalways a closed 3-manifold. e.g. S3 = D(D3), D(S1 ×D2) = S1 × S2. Question: what’sthe double of the solid double torus?

1.1.4. Submanifolds. Suppose Y is a combinatorial (resp smooth) 3-manifold andM ⊆ Yis a 1- or 2-manifold in the subspace topology. We say M is nicely embedded if it’s asubcomplex (smooth submanifold) of Y . If Y is a topological 3-manifold and we haveM ⊆ Y a < 3-manifold, we say M is tamely embedded if there’s a homeomorphismY ' Y ′, Y ′ combinatorial, sending M to a nicely embedded submanifold.

So what’s the point of this definition? There’s a

Theorem 1.7. If M ⊆ Y is a tame submanifold, it has an open regular neighborhoodν(M) ⊆ Y such that ν(M) ' the open disk bundle in the normal bundle of M . (don’tyou have to make Y smooth to talk about the normal bundle?)

Here are some non-tame examples.

Example 1.8. A geometric series of non-closed knots that wraps back around to itself isa wild 1-submanifold of §3. Alexander’s horned sphere is a wild 2-submanifold.

3-MANIFOLDS, JAKE RASMUSSEN 3

2. Examples

2.1. Heegard splittings.

Definition 2.1. Suppose Y1, Y2 are compact 3-manifolds and ∂Yi is nonempty. Takepi ∈ ∂Yi and Ui upper-hemisphereB3 neighborhoods of pi. Then the boundary connectedsum Y1#∂Y2 is Y1 \ U1 ∪D2 Y2 \ U2.

For this to work we need connected boundaries and something else...checking links?

Definition 2.2. Handlebody H0 = D3, H1 = S1 × D2, Hg = #g∂S

1 × D2. If ϕ is ahomeomorphism from Σg = ∂Hg to itself, we can define Yϕ = Hg ∪ϕ Hg the Heegaardsplitting for Yϕ.

Example 2.3. Doubling. If ϕi are self-homeomorphisms of Si which restrict to identityon open Euclidean neighborhoods of pi then ϕ1#ϕ2 is well-defined.

Lemma 2.4. Yϕ1#ϕ2 = Yϕ1#Yϕ2.

Proof. Let Hi be a handlebody bounded by Si and H ′i = Hi \ Vi for Vi half-ball neigh-borhoods of pi in Hi. Then H1#∂H2 =: H is a handlebody bounded by S1#S2.

NowYϕ1#ϕ2 = H ∪ϕ1#ϕ2 H = (H ′1 ∪D2 H ′2) ∪ϕ1#ϕ2 (H ′1 ∪D2 H ′2)

= (H ′1 ∪varphi1 H ′1) ∪D2∪D2 (H ′2 ∪ϕ2 H′2) = (Yϕ1 \B3) ∪S2 (Yϕ2 \B3) = Yϕ1#Yϕ2

�

2.2. Lens spaces. Given A ∈ GL2(Z), call ϕA the induced self-homeomorphism of T 2.On the solid torus take two circles through P , ` = S1×P and m = ∂D2 at P . If p, q ∈ Z

are relatively prime then there exist a, b with ap − bq = −1 so A =

(b pa q

)∈ SL2(Z).

Now we have

Definition 2.5. The lens space L(p, q) = YϕA . We claim L(p, q) does not depend onthe choice of a and b.

Lemma 2.6. pi1(L(p, q)) = Z/p.

Proof. Use Seifert-van Kampen. π1(Hi) = 〈`i〉 ' Z. Let ci : T 2 → Hi identify T 2 withthe boundary of the second S1×D2. Then c1(m) = `p1, c1(`) = `b1, c2(m) = 0, c2(`) = `2,giving π1(H) = 〈`1, `2, `p1 = 1, `b2 = `1〉. �

2.3. Mapping Class Groups.

Definition 2.7. Two homeomorphisms ϕ0, ϕ1 : X → X are isotopic if there’s a mapϕ : X × I → X going from ϕ0 to ϕ1 such that ϕt is a homeomorphism for all t.

We remark that isotopy is an equivalence relation.

Proposition 2.8. If ϕ0 is isotopic to ϕ1, then Yϕ0 ' Yϕ1.

Proof. Observe that Hg ' Hg∪fΣg×I, f : Σg×0→id Σg. So Yϕi = [Hg∪fΣg×I]∪ϕiHg.If Φ is an isotopy from ϕ0 to ϕ1 then F : Yϕ0 → Yϕ1 , given by F |Hg = id, F |Σg×I sending

(x, t) 7→ (Φ−1(ϕ0(x)), t) is a homeomorphism. �

4 KEVIN CARLSON

Let NX ⊆ Homeo(X) be the set of homeomorphisms isotopic to idX . Then NX is anormal subgroup, since if Φ is an isotopy from ϕ0 to idX then ΨΦΨ−1 is an isotopy fromψϕ0ψ

−1 to ψ idX ψ−1 = idX for any ψ ∈ Homeo(X). Then we give

Definition 2.9. The mapping class group MCG(X) = Homeo(X)/NX .

So Yϕ only depends on the image of ϕ in the mapping class group.

Example 2.10. GL2(Z) ⊆ Homeo(T 2) intersects NT 2 trivially since any ϕ ∈ NT 2 actstrivially on π1(T 2). So GL2(Z) ⊆MCG(T 2). In fact this is an equality.

On the other hand, on Thursday we’ll discuss the mapping class group of a higher-genus surface is still a subject for research- it’s much more interesting and much biggerthan GL2(Z). One justification for the existence of many distinct 3-manifolds is exactlythe size of these mapping class groups.

Recall we just saw that linear maps define an embedding GL2(Z)→MCG(T 2).Identify S1 ' [0, 1]/ ∼ and map S1× I to itself via T (x, t) = (x+ t, t). Draw a picture

on the fundamental square of the cylinder. Note that T fixes S1 × ∂I. It is isotopic tothe identity of S1× I, but not through homeomorphisms that fix S1× ∂I. Now supposeS is an oriented surface with a simply closed curve γ and ν(γ) a regular neighborhood.Choose an orientation-preserving homeomorphism ϕ : ν(γ) → S1 × I. Note we neededS to be orientable since otherwise ν(γ) could be a Mobius strip.

Definition 2.11 (Dehn Twists). Define the positive Dehn twist around γ by

τγ : S → S, τγ = ϕ−1Tϕ

Away from ν(γ) extend τγ by the identity.

Proposition 2.12. The class represented by τγ in the mapping class group of S doesnot depend on the choice of ϕ or that of ν(γ).

Example 2.13. Consider S = T 2 = m × `, where the simple closed curves m, ` are theedges of the fundamental polygon. We see τm fixes m. But we claim that it sends `to something like {0} × [0, 1/3] ∪ [2/3, 1] ∪ {(x, 1/3x + 1/3), x ∈ I} when we identifyν(γ) with S1 × [1/3, 2/3]. If we similarly move ν(`) into a vertical strip in the centerit should send m into what looks like a hyperbola on the square. Some more detail:[τm(`)] = [m] + [`] and [τ`(m)] = [m]− [`].

So in summary τm acts on H1(T 2) = Z2 by

(1 10 1

)=: A and τ` by

(1 0−1 1

)=: B.

Observe that A and B generate SL2(Z). In fact, if MCG+(S) is the subgroup generatedby orientation-preserving maps, then MCG+(S) is generated by Dehn twists on curvein S. Can we get all of it if we permit orientation-reversing Dehn twist?

2.4. Back to 3-manifolds.

Example 2.14 (Manifolds that fiber over S1). Let ϕ : S → S be a homeomorphismof a connected surface. Define Zϕ = S × I/ ∼ where (x, 1) ∼ (ϕ(x), 0). The sameargument as for Heegaard splitting shows that Zϕ only depends on the equivalence classof ϕ ∈MCG(S). Observe that we have a map π : Zϕ → S1 mapping (x, t) 7→ t ∈ I/ ∼.Observe that π−1(t) is S. We say that Zϕ fibres over S1 with fibre S and monodromy ϕ.


Since S is connected, the map π∗ : π1(Zϕ)→ π1(S1) is surjective: just take a path (x, 0)up to (x, 1) = (ϕx, 0) and connect ϕx to x without moving in the second coordinate.

Not every 3-manifold fibres over S1. For instance, L(p, q) has fundamental group Z/pso only L(0, 1) = S1 × S2 fibres over S1.

Example 2.15 (Brieskorn Spheres). Let p, q, r ∈ Z+ be relatively prime. Let S ⊆ C3 ={(x, y, z) : xp + yq + zr = 0}. This is a singular algebraic surface-highly singular at theorigin. Define Σ(p, q, r) = S5 ∩ S = {(x, y, z) : |x|2 + |y|2 + |z|2 = 1, xp + yq + zr = 0}.It turns out this gives a homology 3-sphere but not an actual 3-sphere unless p, q, or ris 1. This shows us S is singular: a neighborhood of zero looks like a cone on Σ. If itwere a regular surface I guess this should look like a cone on S3.

Example 2.16 (Surgeries). Let Y be a 3-manifold.

Definition 2.17. A knot in Y is a tame embeding K : S1 → Y . A link is an embeddingof a disjoint union of circles.

One nice visual aspect of 3-topology is that it’s easy to draw knots in S3-being a tameembedding, they’re the same as knots in R3.

Definition 2.18. If K ⊆ Y is a knot in an orientable 3-manifold Y then ν(K) is justS1 × D2. Consider Y \ ν(K). Then ∂(Y − ν(K)) = T 2 = ∂S1 × D2. Then let’s glueS1 ×D2 back in along ϕ ∈ MCG(T 2). The resulting quotient space Y ′ is a 3-manifoldobtained by Dehn surgery on Y .

If L ⊆ Y is a link, I can do Dehn surgery on each component separately.

Now let’s make some statements whose proof will be our next main goal. Everyorientable 3-manifold has a Heegaard splitting. Note how much more generality we getfrom the quotient of a disjoint union, rather than of a cross with interval in the manifoldsfibred over the circle. Every orientable 3-manifold is also a surgery on L ⊆ S3.

3. Handlebodies and Heegaard diagrams

In algebraic topology we’re always looking at cell complexes. These are not verymanifoldy. Handlebodies are a decent analogue for geometric topology.

Definition 3.1. The n-dimensional k-handle Hnk is Dk ×Dn−k. This looks dumb-but

think of Dk as big and Dn−k as small. Then this is an n-manifold’s version of a k-cell.

When n = 2, k = 0, we just have a tiny disk. When k = 1, we have a fat interval.When k = 2, we have a real disk. Then if n = 3, k = 0 gives a little ball, k = 1 a narrowcylinder, k = 2 a pancake or a coin, k = 3 a proper ball.

Some terminology: ∂(Hnk ) = ∂Dk ×Dn−k ∪Sk×Sn−k−1 Dk × ∂(Dn−k) and we call the

LHS ∂A and the RHS ∂B. The core of Hnk = Dk × 0. The cocore is 0 × Dn−k. The

attaching sphere is Sk−1 × 0 and the belt sphere is 0× Sn−k−1.

Definition 3.2. If M is an n-manifold with boundary and ϕ embeds ∂AHnk into ∂M

then we can form a new manifold M ′ = M ∪ϕ Hnk .

Then we say M ′ is obtained be attaching a k-handle to M . And we can compute∂M ′ = ∂M − ϕ(∂AH

nk ) ∪Sk−1 ×Sn−k−1∂BHn

k , which we say is obtained by surgery on

6 KEVIN CARLSON

∂M . The manifold ∂M×I∪ϕHnk is the trace of the surgery and has boundary ∂Mt∂M ′.

In general,

Definition 3.3. A handlebody is a manifold build up by successively attaching n-dimensional handles to the empty set.

Example 3.4. So for (n, k) = (2, 1) a gluing looks like sticking a (flat) suitcase handle orrainbow on the boundary of a 2fold. It’s only flat if we attach the S0 × I in oppositeorientations. Note adding a twisted handle makes the surface non-orientable.

When n = 3, H30 = B3.H3

0 ∪H31 glues a barbell to the 3-ball. The orientable gluing

makes a solid torus and the non-orientable something like a solid Klein bottle, althoughI don’t think the latter exists. Similarly if we glue multiple H3

1 s on disjointly we getwhat we earlier called the “handlebody of genus g”.

How about gluing a 2-handle H32 to the solid torus? If we think of the solid torus as

a solid annular cylinder, then we can cork the center with a 2-handle to get D3. Thecorresponding surgery on the torus itself cuts it and caps each end to make a cylinder orsphere. Incidentally, there’s only one other way to do 2-surgery on a torus, since there’sa self-map of the torus sending any nontrivial simple closed curve to any other. Butdifferent gluings may make many different 3-manifolds.

Sometimes we do get closed manifolds as handlebodies: consider H20 ∪ H2

2 = S2.Again, H2

0 ∪H21 has boundary S1, but is a Mobius strip. Attaching an H2

2 gets RP 2. Ifwe attach two 1-cells to a 0-cell in an interleaved way, we get another circular boundary,and by checking the Euler characteristic we can see gluing a 2-cell gives a torus.

If M is a closed handlebody, I can reverse the handle decomposition by the homeo-morphism Hn

k → Hnn−k. Observe this map exchanges ∂A and ∂B. So the embeddings of

∂A into ∂B reverse direction. Note it’s critical that M be closed!

Corollary 3.5. If M is a closed n-manifold with a handle decomposition for n odd, thenχ(M) = 0.

Proof. M is homotopy equivalent to a cell complex with hkk-cells so χ(M) =∑

(−1)khk =∑(−1)khn−k =

∑(−1)n−khn−k = −

∑(−1)khn−k. �

Theorem 3.6. Any combinatorial 3-manifold admits a handle decomposition.

Proof. Take some tetrahedral picture of M . For 0-handles take a neighborhood of everyvertex. For 1-handles take a neighborhood of every edge, for 2- a neighborhood of eachface and for 3- an interior of each space. More precisely: Y0 = ν(F0), Y1 = ν(F1\Y0), Y2 =ν(F2 \ (Y1∪Y2)), Y3 = ν(F3 \ (Y1∪Y2∪Y0)) and the boundary of each Yi is in two pieces,one of which is embedded in the boundary of Yi−1, giving our decomposition. �

Definition 3.7. Suppose c0, c1 : N →M are embeddings. We say c0 is isotopic to c1 ifthere’s an isotopy Φ of M with Φ0 = idM and Φ1c0 = c1.

Proposition 3.8. If ϕi : ∂AHnk → ∂M are isotopic then the results of surgery along the

ϕi are homeomorphic.

Proof. As for Heegaard splittings. �

It seems more natural to define as isotopy of two embeddings simply as a homotopyof embeddings between them. But this is the wrong idea. Consider the embedding


of R → R3 along the x − axis but with a knot sitting above [−1, 1]. Then we coulddefine J : R × I → R3 by J(t, a) = ai(t/a), a > 0, J(t, 0) = (t, 0, 0). We claim thisis a homotopy of embeddings between the x-axis and the knotted x-axis. Remark fordifferential geometers: this definition wouldn’t have such problems if we needed J to besmooth.

Recall the setup from last time: given ϕ : Sk−1 × Dn−k → ∂Mn, we can attach ahandle to M along ϕ to get M ′ = M ∪ϕ Hn

k . The homeomorphism type of M ′ onlydepends on the isotopy class of ϕ. So surely we’d better think about what these isotopyclasses are! Any such ϕ is istopic to a certain ϕ1, whose image is very close to that ofϕ(Sk−1×{0}). So we might suspect the isotopy class is determined by that of ϕ(Sk−1×0.But this won’t work: for instance the oriented and crossed attachings of the 2, 1 handlehave isotopic ...

It’s a fact that the isotopy class of ϕ is determined by ϕ|Sk−1×{0} and a framing. The

set of framings is an affine space for G := πk−1(O(n−k)), ie G acts freely and transitivelyon it.

The action of G on the set of embeddings is as follows. Given γ : Sk−1 → O(n − k),define ψγ : Sk−1 ×Dn−k → Sk−1 ×Dn−k, (θ, v) 7→ (θ, γ(θ)v). For a simple example, ifk = 1 the set of framings is π0(O(n− 1)) = Z/2 so there are 2 ways to attach a 1-handleto Dn. We get either an S1 ×Dn−1 or an M×Dn−2.

Example 3.9. Suppose we add g 1-handles to Bn with orientable framing. If n > 2 allcollections of ≥ g points in Sn−1 are isotopic so the result is always a connected sum ofg S1 ×Dn−1s. This isn’t true for n = 2.

3.1. Digression on orientations. Let’s give three definitions of orientability, roughlyone for each category of manifolds.

Definition 3.10. (1) (Smooth) M is orientable if we can choose charts such thatall transition functions are orientation-preserving, ie have positive Jacobian.

(2) (Combinatorial) M is orientable if we can choose orientations on each top-dimensional simplex so that the induced orientations on the n − 1-dimensionalfaces are opposite.

(3) (Topological) Hn(Y, ∂Y ) ' Z for Y connected.

To justify the topological definition, consider an orientable combinatorial manifoldand the simplicial chain given by the sum of all the top-dimensional simplices. Thishas boundary zero by supposition on the induced orientations on n− 1-simplices and itdamn well isn’t a boundary itself, nor are any of its multiples.

Definition 3.11. An orientation on M is, respectively, a choice of orientable charts, oforientation on top simplices, or of a generator of Hn(Y, ∂Y ).

If U ⊆ Y is an open submanifold and Y is orientable, then U is orientable as well.Remark: orientations are confusing because our intuitions about surfaces are busted

in the case of higher-dimensional manifolds. Specifically, nice manifolds like Sn and theorientable surface of genus g all have orientation-reversing self-homeomorphisms, say byreflection through a plane. But this gives the wrong impression: most orientable mani-folds of dimension bigger than 2 have no such orientation-reversing homeomorphisms.

8 KEVIN CARLSON

Suppose γ ∈ π1(Y, p) is a loop based at p. How can I understand orientability relativeto this loop? Pick an ordered basis (v1, ..., vn) of TpY . Given Y a Riemannian metricand parallel translate this basis along γ to get a new basis (v′1, ..., v

′n). Whether the latter

has the same or the opposite orientation depends only on the class of γ in π1(Y, p); inparticular ϕ : (v1...vn) 7→ (v′1...v

′n) is a homomorphism π1(Y, p)→ Z/2. This must factor

through H1(Y ) so that the action actually only depends on the homology class of γ.This defines the first Stiefel-Whitney class of TY,w1(Y ) ∈ H1(Y,Z/2).

Definition 3.12. The kernel of ϕ is an index- 1 or 2 subgroup so it defines a 2-1 coveringY → Y and ϕ acts trivially on π1(Y ), thus Y is orientable. It’s called the orientationdouble cover of Y .

I’m pretty sure Y is connected iff Y is non-orientable.If H1(Y ) = 0, Y is orientable.

Corollary 3.13. If Y is a handlebody, it’s orientable if and only if its 1-skeleton is.

Proof. Y is homotopy equivalent to Y the associated cell complex. The inclusion mapof the 1-skeleton Y1 induces a surjection π1(Y1)→ π1(Y ). Thus if ϕ is trivial on Y1, it’strivial on Y . �

Example 3.14. Suppose K ⊆ S3 is a knot. Its regular neighborhood ν(K) is a D2 bundleover K. Since S3 is orientable, so in ν(K) and ν(K) must be S1 ×D2.

Regular neighborhoods of surfaces in 3-manifolds. A regular neighborhood of a surfaceS ⊆ Y is a D1 bundle over S. Observe this gives a map π1(S) → Z/2 depending onorientation switching.

A D1 bundle π : E → B is a space E such that B has an open cover by sets Uisuch that π−1(Ui) is homeomorphic to Ui × D1, with the homeomorphism preservingprojection to Ui. In some sense the only disk bundles are the annulus and the Mobiusstrip. The associated sphere bundle is respectively S1 t S1 or S1. Either way I get adouble cover of the circle, the one associated to Z ⊆ π1(S1), the other to 2Z.

Exercise: show we can recover the disk bundle from the sphere bundle.Any such bundle defines a homomorphism ϕπ : π1(B) → Z/2 by pullback of the

sphere bundle along the circle. This along with the exercise implies that D1-bundles arecompletely classified by ϕπ. We can interpret ϕπ also as a homomorphism on H1B andthis deterines the first Stiefel-Whitney class in H1(B; Z/2).

Recall a manifold M determines a homomorphism ϕM : π1(M)→ Z/2 which measuresorientability.

Proposition 3.15. If π : E → M is a D1 bundle, observe E is homotopye quivalentto M so that their fundamental groups are isomorphic under the projection. Then weclaim ϕE(γ) = ϕπ(γ)ϕM (γ).

Proof. T(p,0)E = TpM⊕T0D′ and parallel transport along γ is given as a map on T(p,0)E

by a block-diagonal matrix of TM , parallel transport in M , and Tπ, parallel transportin D1. Then ϕE(γ) = detTE = detTM detTπ = ϕM (γ)ϕπ(γ) as desired. So that justmeans W1(E) = W1(M) +W1(E, π). �

Corollary 3.16. Any M has a unique orientable D1 bundle π : E → M , namely,W1(E, π) = W1(M).


Corollary 3.17. Suppose S ⊆ Y 3 is an orientable surface. Then ν(Y ) is S ×D1.

Proof. W1(ν(E)) = 0, since ν(S) ⊆ Y is orientable. So since W1(S) = 0,W1(ν(S), π) = 0so ν(S)→ S is the trivial D1 bundle. �

Note it’s easy to find examples of non-orientable surfaces in orientable 3-manifolds,eg Y 3 the total space of the appropriate D1-bundle. Or RP 2 ⊆ RP 3.

Our last scholium about orientations will relate them to connected sums. RecallY1#Y2 = Y1 \B3 ∪ϕ Y2 \B3 where ϕ : S2 → S2. We’ve learned we need to worry aboutthe class of ϕ in MCG(S2) here. At the very least MCG(S2) has Z/2 as a quotientdescribing the orientation effects of ϕ, and as with S1 it happens this is the only differencethat need concern us. So to specify what we mean by Y1#Y2, we should orient Y1 andY2, inducing orientations on the S2s we’ll be gluing. Then we should require that ϕ beorientation-reversing. This will ensure that the sum of orientable manifolds is orientable.

Remark: if ψ : Y → Y is a self-homeomorphism, then X ∪ϕ Y where ϕ : A ⊆ Y → Xis homeomorphic to X ∪ϕψ Y . So if Y2 has an orientation-reversing homeomorphism,then applying it won’t change the directed sum. But otherwise it might.

3.2. Heegaard Splittings. We want to prove that every closed connected orientable3-manifold has a Heegaard splitting.

Proof. Y has a handle decomposition Y = Y0 ∪ Y1 ∪ Y2 ∪ Y3. Cut the manifold intoY0 ∪ Y1 and Y2 ∪ Y3. It suffices to show Y0 ∪ Y1 is an Hg since then by reversing thehandle decomposition we can show similarly Y2 ∪ Y3 ' Hg′ . Then Y = Hg ∪Hg′ alongthe common boundary, showing g = g′ since there aren’t embeddings between orientablesurfaces of distinct genus.

To prove the sufficient claim, consider the graph whose vertices are 0-handles and edges1-handles (look at cores.) Since Y is connected there’s a maximal tree corresponding toY ′1 ⊆ Y1 for this graph. Then Y0 ∪ Y ′1 is a B3. Then gluing edges back into the graph isjust attaching handles to a 3-ball. Since Y is orientable, we make all these attachmentsin an orientable way to get Hg where g is the number of 1-handles removed to get Y ′1 . �

We can record a Heegaard splitting as a Heegaard diagram.

Definition 3.18. A generalized Heegaard diagram is a connected orientable surface Stogether with {α1, ..., αk}, {β1, ..., β`} two sets of pairwise disjoint simple closed curveson S-where αs and βs may well intersect.

Given this data we can construct a 3-manifold with boundary called Y (S, α, β) asY = S × [0, 1] ∪βi×{1} H3

2 ∪αi×{0} H32 and filling in S2 boundary components with B3s.

A clarification about cell complexes and handlebodies. Suppose X is a cell complex:we can assume WLOG that the k-skeleton is attached to the k − 1-skeleton. Similarly,if M is a handlebody, ϕ : ∂AH

nk → ∂M is an embedding, ϕ is isotopic to ϕ : ∂AH

nk →

∂B(Mk). SO without loss of generality we can assume handles are attached in order ofindex.

Thinking about generalized Heegaard diagrams again, let’s consider what the bound-ary of S × I ∪α×0 H

32 ∪β×1 H

32 is. For example, if we have no βs and αs the generating

meridians of the genus-g torus, the inner boundary is the torus cut along these curveswith gaps capped. Remark that if Σg is a surface containing a simple closed curve α then

10 KEVIN CARLSON

surgery along α results in Σg−1 if α is nonzero in homology or Σa tΣb with a+ b = g if[α] = 0.

For example, if I write g curves in Σg such that Σg remains connected when I cutthem all out, the inner boundary becomes a sphere which I fill in to get Hg.

Proof. Reverse the handle decomposition to get a manifold built out of a zero-handleand g 1-handles. It’s orientable since gluing 2 and 3 handles doesn’t affect orientability,so it’s Hg. �

Definition 3.19. In particular, a Heegaard diagram is a triple (Σg, {α1, ..., αg}, {β1...βg})with Σg \∪α and Σg \∪β both connected. Then Y (Σ, α, β) is a gluing of two Hgs alongΣg × {1/2}, that is, a Heegaard splitting of a 3-manifold.

Example 3.20. • Lens spaces: L(p, q) = H1∪ϕH1 has a Heegard diagram (T 2, {α}, {β}).If H1(T 2) = 〈m, `〉, α = `, [β] = [pm+ q`].• Observe this specification only took two parameters: surgery along K ⊆ S3 isS3 \ ν(K) ∪ S1 × D2 = S3 \ ν(K) ∪ H3

2 ∪ H33 so we just have to specify the

homology class of the attaching circle for the 2-handle in the homology of T 2.Up to orientation, a standard basis for that homology is a meridian m and theknot K. More precisely, an equator ` uniquely specified as a curve intersectingm once and being nullhomologous in S3 \ ν(K). To specify the orientation, justmake sure ` comes after m. Then we say p/q surgery on K ⊆ S3 means attachinga 2-cell along pm+ q`.• A generalized Heegaard diagram for S3 \ ν(U2), U2 being two unlinked circles.

Write S3 = B3 ∪B3 = X1 ∪X2along a plane disconnecting both components ofU2. Now S3 \ ν(U2)) ∩X2 = H2. Claim we can get S3 \ ν(U2) by attaching a 2-handle to ∂H2. The latter boundary is just a genus-2 orientable surface where wecan make the handles sit in X1 attached to a sphere that’s the aforementionedplane with the point at infinite. Glue a 2-cell into the latter sphere so as todecompose the genus-2 surface into two T 2 neighborhoods, one of each circle inU2. In summary, S3 \ ν(U2) has a generalized Heegard diagram of four circlesnext to each other on the blackboard with the left two and the right two joinedby cylinders behind the blackboard, or of a two-torus with α a central meridianand the two βs loops around the holes.

As an exercise, think about showing S3 \ ν(H) is a handlebody of genus 2. To do thiscut into balls again and untwist to get a ball with two disjoint arcs.

The procedure to finish this exercise is schematized on paper. In essence, cut anduntwist until the outer disks are just the same as in the unknot case, so only the innercircle gets wonky. We can do the same thing with the trefoil knot, untwisting threetimes. Next we’ll consider L a tripled Hopf link. The technique of starting with a Σg

partitioned into glued twisted tori (partitions in a central sphere) and altering the unlinkdiagram as we untwist the outer link gives a general algorithm for Heegaard diagrammingS3 \ ν(L).

3.3. Finding π1 from a Heegaard diagram. Suppose we have a generalized Heegaarddiagram (Σg, {α1...αg}, {β1...βk}) such that (Σg, {αi}, {}) is a handlebody of genus g.Then π1(Y (Σg, α, β)) has a presentation 〈αi : b1 = ... = bk = 1〉 where to find bi I orient


all αs and βs and walk along βi, recording aj if I pass through αj if the local framing

αj , βi is of positive determinant and a−1j otherwise.

So in the Hopf link diagram, we walk along β crossing each α once in each orientation:a1a2a

−11 b−1

1 = 1. So π1(S3 \ ν(H)) = Z2! Exercise: that’s because this manifold is justT 2 × I. Solution: draw one solid torus of the Hopf link going through infinity. Thecomplement of this is another solid torus which you drill a solid torus out of the interiorof for the second component of the link.

In the trefoil diagram, we again get one relation: a1a2a1a−12 a−1

1 a−12 = 1. An amusing

question is to relate the different presentations gotten from different diagrams in thisway.

So, why does this work? Y (Σg, α, β) = Hg ∪ k(H32 ). Now Hg is the wedge of three

circles and under the retraction each meridian αi maps down to a point on a distinctcircle in the wedge away from the meeting point. So we get a generator from each αiplus a tail to and from the meeting point, and relations for each 2-cell as glued to theαi. Observe that trivial passes over an α are immediately cancelled in the presentation:)

Next example: surgery on S3 \ ν(T ), T the trefoil. We just need to attach some 2-handle along a curve in the boundary T 2 = Σ2 after surgery along β. It’s not too hardto find a meridian, beginning with the externally twisted diagram. ` is maybe not aseasy, but just draw something that only intersects m once. We also need ` trivial inH1(S3 − ν(T )) which is the abelianization of π1, H1 = 〈a1, a2 : a1 + a2 = 0〉 = Z. Nowwe can compute ` in homology as 2a1 − 2a2 which is 4. So I need to loop around m abunch!

One last remark on Heegaard diagrams. Suppose γ ⊆ S bounds a disk in S. Then if Iattach a 2-handle along γ and cut I get a sphere and a surface as the resulting boundaryand if I fill in the sphere I get S × I.

Example 3.21. Suppose I want to do∞-surgery on the unlink in S3, i.e. on one of its tori.There are diagrams with two and with just one β, and this is a predictable phenomenonin surgery on multicomponent links-as it should be.

4. Mapping Class Groups

Let M be orientable and connected. Then one can define α : MCG(M) → Z2 thatmaps orientation-reversing isotopy classes to −1. Then we may define MCG+(M) =ker(α).

Lemma 4.1. Suppose ϕ : Dn → Dn restricts to identity on Sn−1 = ∂Dn. Then ϕ isisotopic to the identity, ϕ ' id in local notation.

Proof. Define ϕ : Rn → Rn by identity outside of Dn. Then I have an isotopy ϕt(v) =tϕ(v/t), ϕ0 = id. Restrict this to Dn to get an isotopy of ϕ to idDn . It’s important forthis to apply to diffeomorphisms that topological 3folds are uniquely smooth. �

Proposition 4.2. MCG+(S1) is trivial.

Proof. Given ϕ : S1 ' S1 let α = argϕ(1). Let ϕt(z) = e−iαtϕ(z). This is an isotopy

to ϕ1 which fixes 1. Define ψ : I → I by cutting S1 at 1, e2πiψ(t) = ϕ1(e2πit). Since

12 KEVIN CARLSON

ϕ is orientation preserving ψ must be increasing, so by the lemma ψ is isotopic to idI ,making ϕ1 isotopic to idS1 . �

Proposition 4.3. MCG+(D2) is trivial. Given Φ : D2 ' D2 let ϕ : S1 ' S1 beΦ|S1. By considering homology we see that if Φ orientation-preserving ϕ is orientationpreserving so there’s an isotopy ϕt between ϕ and idS1. The method from the lemmagets an isotopy Φ ' Ψ where Ψ(r, θ) = (r, ϕ(θ)), r ≥ 1/2, 1/2Φ(2r, θ), r ≤ 1/2. Thenblow up,

Ψt(r, θ) =

{(r, ϕ2t(r−1/2)), r ≥ 1/21/2Φ(2r, θ), r ≤ 1/2

Psi1 is identity on the boundary so Phi is isotopic to identity.

Now suppose S is a connected oriented surface. Let TS ⊆MCG+(S) be the subgroupgenerated by Dehn twists along simple closed curves in S.

Theorem 4.4 (Lickorish). TS = MCG+(S).

Proof.

Definition 4.5. Suppose γ1, γ2 ⊆ S are simple closed curves. Say γ1 ∼ γ2 if there’sϕ ∈ TS with ϕ(γ1) ' γ2. This is clearly an equivalence relation.

Lemma 4.6 (1). If |γ1 ∩ γ2| = 1, then γ1 ∼ γ2.

Proof. A regular neighborhood of γ1 ∪ γ2 is a punctured torus. By drawing the pictureswe see τγ1τγ2γ1 is isotopic to γ2. Note this could reverse orientations! But then τ−1

γ1 τ−1γ2 γ1

will also be isotopic to γ2 without switched orientation. �

Lemma 4.7 (2). Suppose γ, δ are simple closed curves on S. Then there’s γ′ ∼ γ suchthat either γ′∩δ = ∅ or |γ′∩δ| = 2, where still the intersection locus has crossing numberzero, and γ′ ∩ (S \ ν(δ)) ⊆ γ ∩ (S \ ν(δ)). If so we say condition ∗ holds of γ and δ.

Proof. By induction on the size of the intersection, k. If k = 0 we’re done. For a strongerbase apply lemma 1; δ is isotopic to something not intersecting it. If k = 2 and the signsare opposite, we’re done.

In all other cases, we have a segment of δ which γ crosses either with the same twosigns in sequence or with three alternating signs. In both cases it’s not too hard to drawa twist reducing the crossing number to 1. �

Corollary 4.8. Suppose γ, δ1, ..., δk are sccs in S and δi ∩ δj = ∅ for each i 6= j. Then∃γ′ ∼ γ such that condition ∗ holds of γ wrt each δ.

Proof. Induction step: ∗ holds for δi, i < k. Let ∗ hold for γ′, δk. Then |γ′ ∩ δI | ≤|γ′ ∩ δi| ≤ 2. The only case to check is when the new intersection number becomes 1, inwhich case γ′ is equivalent to δi and we win. �

Proposition 4.9 (Missing meridians). Consider a system of curves δi, i ≤ 3, neckingeach of the g handles and αi their meridians. If γ is an scc on S, there’s equivalent γ′

intersecting none of the αs.


Proof. First apply the corollary to get γ′ ∼ γ with ∗ holding pairwise with each α andδ. Then look at each α, δ with intersection number 2 with γ′ and isotope γ′ away. Inparticular start by assuming γ′ doesn’t meet α and draw the only possible picture; thennoting that all’s easy if it meets δ but not α manage to turn a double meeting with δ, αinto one just with δ and win. �

Proposition 4.10 (Dehn twists invert mapping class group elements on αs). Givenϕ ∈MCG+(S), ∃ψ ∈ TS such that ψϕ(αi) = αi, 1 ≤ i ≤ g.

Proof. We’ll get ψk by induction. Given ψk inverting ϕ up through αk, let γ = ψkϕ(αk+1).We need to twist γ to αk+1 without interfering with α1...αk. Observe γ∩αi = ψkϕ(αk+1)∩ψkϕ(αi) = ∅. By the missing meridians proposition, we can find ρ ∈ TS\∪kαi such that

ρ(γ) ∩ αi = ∅, k + 1 ≤ i ≤ g. Then γ′ := ργ can be realized as a curve on S2 \ t2gD2.If γ′ separates the circles labeled αk+1, there’s a curve βk+1 between the two sides

of αk+1 which we claim only intersects γ′ once. This claim is supported by the factthat any simple closed curve on S2 is equivalent to the equator under the action ofMCG(S2). Now by the very first lemma γ′ ∼ βk+1 ∼ αk+1, so I can find σ ∈ TS\tkαiwith σ(γ′) = αk+1 and σ|γ′ orientation-preserving. Then since MCG+(S1) is trivial, wecan isotope σρψkϕ to the identity on αk+1.

Otherwise, γ does not separate the two sides of αk+1. Observe that [αk+1] /∈ 〈[α1]...[αk]〉.Then [γ′] /∈ 〈[α1]...[αk]〉, since the latter are the images of the former under the home-omorphism ψkϕ. Then there exists j > k + 1 such that γ′ separates the sides of αj-

otherwise γ′ would be contractible in S \ tkαi. Then draw βj as before. γ′ ∼ βj ∼β′k+1 ∼ αk+1, where β′k+1 intersects both αk+1 and βj in one point.

�

Now we may assume that we have |varphi ∈MCG+(S), ϕ|αi = idαi . Now cut S alongthe αi to get S = S2 \ t2gD2 and ϕ : S → S restricting to identity on the boundary.

Definition 4.11. MCG+(S, ∂) is defined to be the isotopy classes of homeomorphismsof S where both self-maps and isotopies restrict to identity along the boundary. TS isdefined just as for closed surfaces.

Lemma 4.12. Given ϕ ∈ MCG+(S, ∂) and γ connecting two boundary components inthe simplest way, there’s ψ ∈ TS such that ψϕγ intersects γ just in γ’s endpoints.

Proof. By induction on |ϕ(γ) ∩ γ. Let p be the first intersection point of ϕγ with γ.There are two orientations for the intersection. If it’s positively oriented from γ to ϕγ,twist around a small loop about the boundary component γ starts on to reduce to case2. In case 2, see figure for a reduction of intersection number. �

Lemma 4.13. Given ϕ ∈ MCG+(S, ∂), there’s ψ ∈ TS such that ψϕ|γ = idγ where γagain links two boundary components of S.

Proof. By lemma 1, we may assume ϕγ ∩ γ is the endpoints of γ. Twist around a curvethat follows the composition but loops around the circle of one of the endpoints. �

Proposition 4.14. MCG+(Sn, ∂) = TSn.

14 KEVIN CARLSON

Proof. When n = 1, we know both answers are trivial. In general, given ϕ ∈MCG+(Sn, ∂),pick ψ ∈ TSn with ϕ′ = ψϕ restricting to identity on γ. Then cut along γ to get a good

element of MCG+(Sn−1 for induction. �

Proof of theorem Given ϕ ∈MCG+(S), there’s ψ ∈ TS with ψϕ restricting to idαi forevery i. Cut along the αi to get ϕ′ : S2g → S2g, which by the proposition is compositionof Dehn twists. �

5. Surgery

Recall: Dehn surgery on K ⊆ Y is Y ′ = Y \ ν(K) ∪ϕ S1 × D2, ϕ : ∂(S1 × D2) '∂Y \ ν(K). We saw earlier that the homeomorphism type of Y ′ is determined by theclass of ϕ(1× ∂D2) in H1(∂(Y \ ν(K))). Suppose this homology class is pm+ q`. Thenwe say that Y ′ is p/q Dehn surgery on K and denot Y =: Kp/q =: Yp/q(K).

Remark: the choice of m is canonical once K is oriented: m generates the kernelof ι∗ : H1(∂(ν(K)) → H1(ν(K)). The choice of ` is not canonical in general, since(m, km + `) is a basis whenever (m, `) is. We can also see the choice is non-uniquesince (z, w) 7→ (z, zkw) is a self-homeomorphism of the solid torus mapping (,m`) 7→(m, `+ pm).

BUT if [K] = 0 in H!(Y ), there’s a canonical choice of `, i.e. a generator of the kernelof H1(∂(ν(K)))→ H1(Y \ ν(K)) = H1(Y )⊕ Z.

5.1. Surgery and 4-dimensional handle attachment. Consider W = Y × I ∪H42 ,

which we recall means ι : ∂AH42 = S1 ×D2 → Y × 1. Then

∂W = Y × 0 t [(Y × 1) \ ι(∂AH42 )] ∪ ∂BH4

2

Then K = ι(S1 × 0) is a knot in Y along which we’ve done some surgery. In particular,ι(1× ∂D2) = m and ι(S1 × 1) = λ is some longitude for K.

If we do have a canonical longitude ` = λ + pm for some p, we say K is a framingof the handle attachment. (In general, framings of attachings of H4

2 are classified byπk−1(O(n−k)), here π1(O(2)) = Z.) Now ∂W = Y ×0tY ′, Y ′ = Y \ν(K)∪(D2×S1) =Y \ ν(K) ∪ ∂BH4

2 and ι(∂G2 × 1) = ι(S1 × 1) = pm+ `, so Y ′ = p/1 surgery on K.

5.2. Surgery and Dehn twists. Let S be orientable and γ ⊆ S a simple closed curve,Y = S × I,K = γ × 0 a knot in Y . If γ′ is parallel to γ is S, then we claim ` = γ′ × 0 isa longitude for K: just draw a regular neighborhood of γ containing both.

Proposition 5.1. We have a homeomorphism ϕ : Y1(K)→ S× [−1, 1], where ϕ|S×−1 =id, ϕ|S×1 = τ−1

γ .

Lemma 5.2. Let Y = S1 × I × I. Let K = S1 × 0 × 0 and ` = S1 × 1 × 0. Thenthere’s ϕ : Y → Y1(K) such that ϕ|S1×I×−1∪{−1,1}×I is the identity and ϕ|S1×I×1 is τ−1

γ ,γ = K.

Proof. Y \ ν(K) is homeomorphic to T 2 × I, so Y \ ν(K) ∪ S1 ×D2 is always S1 ×D2

and in particular Y1(K) is S1 × D2. Let m, ` be the standard basis for H1(∂Y ). Som = 1× ∂(I × I). And ker ι∗ : ∂Y → Y1(K) is 〈m+ `〉, since we glue ∂D2 to m+ `.

Notice that ` is still a longitude for Y1 = S1 ×D2. So we can find a homeomorphismϕ : Y → Y1(K) fixing `. Now S1 × (I × −1 ∪ −1, 1 × I) is a regular neighborhood of `


is ∂(Y ), so we may assume that ϕ|A = id. To see what happens for ϕ|S1×(I×1) notice

that ϕ∗([m]) = [m] + [`] in ∂Y1(K). So ϕ acts like τ−1γ on ϕ|S1×(I×1) which means it’s

diffeotopic to an actual τ−1γ there. �

Lemma 5.3. If Y = S1 × [−1, 1]2 and K = S1 × 0 and ` = S1 × 1 × 0 then there’sϕ : Y ' Y1(K) with ϕ|S1×A = id, ϕ|S1×B = τ−1γ. Here A is the lower three sides of thesquare, and B is the top.

Setup: S is orientable, γ is simple closed, γ′ parallel to γ, Y = S × [−1, 1],K =γ × 0, ` = γ′ × 0. Then

Proposition 5.4. There’s ϕ : Y ' Y1(K), ϕ|S×−1 = id, ϕ|S×1 = τ−1γ .

Proof. Let Y ′ = ν(γ) × [−1, 1] ' S1 × [−1, 1]2. By the lemma there’s ϕ′ : Y ′ ' Y ′1(K)which restricts to identity on ∂ν(γ) × I and ν(γ) × −1 and τ−1(γ) on ν(γ) × 1. Thensince Y is just Y ′ ∪ (S \ ν(γ))× I we an extend ϕ′ by identity to all of Y ' Y1(K). �

Recall if ϕ is an endomorphism of Σg, Yϕ = Hg ∪ϕ Hg is the associated Heegaardsplitting.

Lemma 5.5. Yτ±1γ ϕ is obtained by integral surgery on K ⊆ Y.

Proof. Yϕ ' Hg ∪ Σg × I ∪Hg. Let K = γ × 0 ⊆ Σg × I, ` = γ′ × 0. Then (Yϕ)1(K) =Hg ∪Σg × I ∪Hg where upon contracting the inner Σg × I we’re identifying Hg to itselfon the boundary by x ∼ ϕ−1(τ−1

γ (x)), which is exactly Yτγϕ. �

Theorem 5.6. Suppose Y is a closed connected oriented 3-manifold. Then Y can beobtained by integer surgery on L ⊆ S3.

Proof. First let’s observe that #gS3 = S3 = H1 ∪ϕ1 Hg = Hg ∪ϕg Hg. And as we’veshown earlier any cco Y has a Heegaard splitting Y = Yϕ. Since MCG+(Σg) is generatedby Dehn twists, ϕ = τγn ...τγ1ϕg. Let ϕ(i) = τγi ...τγϕg. Then Yϕ(i) is obtained by integral

surgery on K ⊆ Yϕ(i−1). So by induction we see Yϕ is integral surgery on L ⊆ S3. �

Corollary 5.7. Any closed connected oriented 3-manifold Y 3 bounds W 4 with π1(W 4) =0.

Proof. Write Y as integer surgery on L ⊆ S3. Take W 4 = D4 ∪L tH42 . �

Most of the time this isn’t true: eg since χ(CP 2) = 3, which is odd, CP 2 doesn’tbound a 5-manifold.

5.3. Surgery Diagrams. Another way to draw pictures of 3-manifolds is to draw alink L ⊆ S3 and label each component of L with an integer that specifies a surgery. e.g.an unknot with a p gives L(p, 1).

Just as for Heegaard diagrams, the surgery diagram is non-unique. For instance, anunknot U with ±1 surgery is equivalent to a ∓1 full twist on all components passingthrough the unknot, where the framings change: a framing f goes down by ∓ the squareof its linking number with U . This is related to blowing down. As an example of whywe might not always use ±1 surgery: 3-surgery on some knot is the same as 1-surgerywith -1 surgeries on two unknots linked with crossing number 1 or −1.

Why does this work? Consider S3 = H1 ∪ϕ1 H1. Then if γ lies in Σ1 as a longitudeit extends across the outer solid torus as a twist Tγ .

16 KEVIN CARLSON

6. Torsion Invariants

6.1. Cellular chain complex. Y 3 is connected oriented compact, maybe with bound-ary, for the rest of the course.

Let H30 ∪gH3

1 ∪gH32 ∪H3

3 = Y for ∂Y = 0, or Y = H30 ∪gH3

1 ∪kH32 otherwise. By the

opposite of the construction of the handle decomposition from the cell decomposition,this describes a cell complex, specifically, Z←0 Zg ←∂ Zg ←0 Z. We’ll justify the right-hand maps being zero later. But for now ∂ is the interesting map. Let 〈ai〉 = C1, αi thebelt circles, and C2 = 〈bj〉, βj the attaching circles. Now if Σg, {α1...αg}, {β1...βg} is aHeegaard diagram for Y , then

Proposition 6.1. ∂bj =∑αi · βjai, where the intersection number α · β is the sum of

the signed intersections.

Proof. Well, ∂(bj) = [βj ] ∈ H1(Gg, H30 ) = H1(Hg). We already computed the class of

βj ∈ π1(Hg), and so we just need to sum the positive and negative intersection of βj andαi. �

Theorem 6.2 (Poincare duality 1). If Y is closed, Hi(Y ) ' H3−i(Y ). If not, Hi(Y ) 'H3−i(Y, ∂Y ).

Proof. Reverse the handle decomposition. The cellular complex is the same, but now C1

is generated by the bj and C2 by the ai. Furthermore ∂ai =∑βj · αibj =

∑−αiβjbj .

As a g × g matrix ∂ is given by Aij = −(αi · βj). If we transpose we get ∂T bj =∑−(αiβj)ai = −∂bj where the latter ∂ came from the original decomposition. So

we have our isomorphism as multiplication by −1. The proof is similar for Y withboundary. �

We used orientability to be able to define the intersection number. But the (mod 2)intersection number is always defined, so we have Poincare duality of Z/2 for any man-ifold.

By universal coefficients, ifH∗(Y ) = (Z,Za⊕T1,Zb⊕T2,Z), thenH∗(Y ) = (Z,Za,Zb⊕

T1,Z ⊕ T2). Then Poincare duality implies a = b and T2 = 0, so we know the whole ofthe homology if we know T1 and a.

Proposition 6.3. Let c∗ : H1(∂Y ; Q)→ H1(Y ; Q). Then dim ker c∗ = 1/2 dimH1(∂Y ).

Lemma 6.4. Suppose 0→ An →ϕ ...→ A2 → A1 → X →i A1 → A2 → ...→ψ An → 0is an exact sequence of vector spaces. Then dim ker i = 1/2 dimX.

Proof. Induction on n. In general, we have an exact sequence 0 → cokerϕ → An−2 →... → A1 → X → A1 → ... → An−2 → kerψ → 0. But dim cokerϕ = dimAn−1 −dimAn = dim kerψ so we may apply the induction hypothesis. �

Proof. Consider the long exact sequence of the pair (Y, ∂Y ), in particular

0→ H3(Y )→ H3(Y, ∂Y )→ H2(∂Y )→ H2(Y )→ H2(Y, ∂Y )→ H1(∂Y )→

H1(Y )→ H1(Y, ∂Y )→ H0(∂Y )→ H0(Y )→ H0(Y, ∂Y )→ 0

And we’re in the situation of the lemma. �


6.2. Covering Spaces. Suppose π : π1(Y ) → G is surjective. Then there’s a regular

covering space Y →ρπ Y with π1(Y ) = kerπ and G its deck transformations.Suppose Y has a cell or, equivalently, handle decomposition. Multiplying by G on

each cell/handle gives a decomposition of Yπ, since ρ−1π (Hn

k ) is a covering space of Hnk ,

which is contractible-making the inverse a trivial covering G×Hnk .

See figure for a drawing of of S3 \ T . The covering space Y corresponding to themap π1(Y ) → H1(Y ) has deck group Z, so we have to multiply a decomposition Y =h0 ∪ h1

1 ∪ h12 ∪ h2 by Z.

6.3. Torsion of chain complexes. Let k be a field and V a finite-dimensional vectorspace over k. A basis {v1...vn} of V determines a generator v1 ∧ ... ∧ vn =: ∧vi ofΛn(V ) =: ΛtopV . If {v′1...v′n} is another basis, then ∧v′i/ ∧ vi = detA, where A changesbasis. Now suppose (C∗, d) is an acyclic chain complex over k and hki is a basis of Ck.Since H∗(C, d) = 0, we can write Ck = ker dk ⊕Ak, dk : Ak ' ker dk−1. Let bki be a basis

for Ak and ck−1i = dk(b

ki ) so ck−1

i is a basis for ker dk−1. Then ck−1i , bk−1

i is a basis forCk−1.

Definition 6.5. The torsion

T (C, hki ) =∏∧h2i

j / ∧ b2ij ∧ ∧c2ij /∏∧h2i+1

j / ∧ b2i+1j ∧ c2i+1

j

Lemma 6.6. The torsion doesn’t depend on bki .

Proof. Let b′ki be another basis for the Ak and A the change of basis matrix. Then A

also changes basis from ck−1j to c

′(k−1)j , so we always get detA once in the numerator

and once in the denominator of the torsion computation. �

Recall from defining torsion C∗, d is acyclic and hij is a basis for Ci and Ci = ker di⊕Aiand bij is a basis for Ai.

Now let Y be a 3-manifold with handle decomposition and hij the i=handles. Suppose

to begin with that H1(Y ) = Zk. Then π1(Y ) surjects on H1(Y ), and the corresponding

cover Y has deck group Zk. Incidentally, Y is usually called the universal abelian coverof Y . C∗(Y ) is a module over Z[Zk] = Z[t±1

1 , ..., t±1k ] the ring of Laurent polynomials,

generated by hij , a choice of lift of each hij . Set K = Q(t1...tk) and C = C∗(Y )⊗K and

pick as basis hij . Finally define

Definition 6.7. T (Y, h) = T (C∗(Y )⊗K, hij), if the cell complex is acyclic.

Observe that the choice of lifts was only well-defined up to the action of Zk, i.e. actionby ta11 ...t

akk .

Proposition 6.8. The torsion does not depend on choice of handle decomposition.

Explanation postponed.

Example 6.9. Let Y = S1×D2. Then Y has a handle decomposition into Z many 1-cellsand 0-cells so the cell complex is R→t−1 R.

18 KEVIN CARLSON

Now take Y = S2 × S1. Then C∗(Y ) = R →t−1→ R →0→ R →t−1 R, because it’s

C∗(S2) ⊗ C∗(S1). Choose b3 = h3, d(b3) = (t − 1)h2, b1 = h1, d(b1) = (t − 1)h0. Then

T (S1 × S2) = 1/1(1/t− 1)1/1(1/t− 1) = 1(t−1)2

.

Now take Y = S3 \ ν(T ), T the trefoil. We saw

C∗(Y ) = R

(t2 − t+ 1−t2 + t− 1

)→ R2

(t− 1 t− 1

)→ R

Setting b2 = h2, d(b2) =

(t2 − t+ 1−t2 + t− 1

). Then we might take b1 =

(10

)= h1

1, so d(b1) =

(t−1)h0. Then to compute the torsion I need to multiply the ratios given by the hes and

the bs. In degree 2 I have 1/1. In degree 1, the basis 〈d(b2), b1〉 =

(t2 = 1 1

−t2 − t+ 1 0

)=: A

and the element of K corresponding to the wedge is detA = t2− t+ 1. In dimension 0 I

get 1/t− 1, so T (Y ) = t2−t+1t−1 6= T (S1 ×D2). Then y \ ν(T )¬ ' Y \ ν(U) so the trefoil

is not an unknot in S3. (check action, multiplication by t±k.)

Finally let Y = S3 \ L,L = T (2, 4). C∗(Y ) was

RA→ R2 B→ R,A =

(−(s− 1)(ts+ 1)(t− 1)(ts+ 1)

), B =

(t− 1 s− 1

)with R = Z[t±1, s±1]. Now b2 = h2, d(b2) = A, b1 =

(10

), d(b1) = (t − 1)h0. Thus

T (Y ) = 1 det

(−(s− 1)(ts+ 1) 1(t− 1)(ts+ 1) 0

)1t−1 = −(ts+ 1) ∼ ts+ 1.

Now to get out of knots and links, let Y = L(p, q) and Y the universal cover of Y ,

so the deck group is Z/p. Y has one handle per dimension, so C∗(Y ) = R → R →R → R,R = Z[Z/p]. Observe this is the complex of a cell decomposition of S3 with pof each kind of handle. To work out the complex, let’s try L(3, 2). There’s a Heegaarddiagram with α the bottom edge of a fundamental square of the boundary T 2 and β aline at slope 3/2. This lift to a generalized Heegaard diagram for Y = S3: stack threeof these diagrams on top of each other, continuing lines across horizontal boundaries toget three different 2-handles. The 0-handles fill in between the αs and the 3-handlesbetween the βs. Now we can compute that the last boundary map sends h1 to (t− 1)h0,

the middle sends h2 to (∑tp−1 + ... + 1)h1, and the first sends h3 to (tq

′ − 1)h2 andq′ = q−1 (mod p).

Exercise: K ⊆ S3 a knot, T (Y \ν(K)) = ∆K(t)/t−1,∆k(t) ∈ Z[t±1] is the Alexanderpolynomial of K. Note H∗(S

3 \K) is the same for any two knots. ∆K is better, thoughfar from perfect. Eg both knots on the front gates of CMS have the same Alexanderpolynomial 1.

If L is a link in S3 and Y = S3\ν(L) then T (Y ) =: ∆(L) is the multivariate Alexanderpolynomial of L.

19 NovLet Y = L(p, q) and Y → Y the covering by S3 with deck group Z/p. We saw last time

C∗(Y ) = R →tq′−1 R →

∑ti R →t−1 R for R = Z[Z/p]. Consider C as an R −module,


where t the generator of Z/p acts as a primitive root of unity ω. Then the tensoredchain complex has middle map 0, so we can compute its torsion.

The good basis is 1, ωq′ − 1, 1, ω − 1 which makes the torsion [(ωq

′ − 1)(ω − 1)]−1,well-defined up to multiplication by ±ωk, k ∈ Z/p. It seems like this is trouble sincethen T (L(p, q′)) = [(ωq − 1)(ω − 1)]−1 which isn’t in the variation. The problem isthat the homeomorphism H1(L(p, q)) ' H1(L(p, q′)) needn’t map 1 to 1, and to do thecomputation we had to send the respective 1s to ω is C. A more effective notation wouldinclude the character on H1(Y ) as an argument of the torsion.

Lemma 6.10. Suppose p > 7 is prime and (ωa−1)(ω−1) = ±ωk(ωnb−1)(ωn−1), ω =

e2πi/p. That is, the two expressions come from lens spaces with the same torsion. Theneither a ≡ ±b (mod p) or ab ≡ ±1 (mod p).

Proof. After reducing the exponents mod p one gets an 8-term linear relation in powersof ω. Since p is prime the minimal polynomial of ω is ωp−1 + ... + 1 which has morethan 8 terms, so all the terms in my relation must cancel-the sets {a + 1, a, 1, 0} and{k, n+ k, bn+ k, (b+ 1)nk} are “congruent (mod p)”. �

It turns out the same holds if p is composite, but it’s more work to prove.

Theorem 6.11. If p is prime then L(p, q) is L(p, q′) iff q ≡ ±q′ or qq′ ≡ ±1.

Proof. One direction was known, the other being immediate from the lemma. �

This also needs the invariance of torsion under handle decomposition, which we’ll nowproceed to not prove. NB:L(p, q) and L(p, q′) have the same homology and homotopy(for π2 consider the covering by S3) so torsion is kind of awesome.

6.4. Handleslides and invariance of torsion. We’ll phrase this in terms of Heegaarddiagrams, which are basically the same as handle decompositions. The problem: giventwo different Heegaard diagrams of Y , to discover how they’re related.

First let’s find some operations that change a Heegaard diagram but not the resultingmanifold.

(1) Stabilization: let Σ1, α1, β1 be the standard genus 1 Heegaard diagram for S3. IfΣg, α, β) is a Heegaard diagram for Y then Σg, α, β)#(Σ1, α1, β1) is a Heegaarddiagram for Y#S3 ' Y . Observe this isn’t so obviously trivial, since it doesincrement the genus of the diagram.

(2) Handleslide: Suppose β1, β2 ⊆ Σ are β-circles for a diagram of Y . Suppose γ isa path from β1 to β2 disjoint from all the βs except at endpoints. We can slideβ1 over β2 using γ: make β′1 the curve that follows β1, then dodges around γ bylooping around β2 to come back and finish β1. Why should these diagrams givethe same manifold? Visualize it as actual 2-handles sitting on a handlebody-thenthere’s an obvious isotopy. Similarly by reversing the handle decomposition wecan slide αs over each other.

We need but won’t prove the following

Theorem 6.12. Let Y 3 be cco. Any 2 Heegaard diagrams are related by a sequence ofstabilizations, destabilizations, and α− and β− handleslides. In case Y has a boundary,we note the same is true for generalized Heegaard diagrams of the flavor (Σg, α|g1, β|k1).

20 KEVIN CARLSON

The best way to prove this is using Morse theory. Assuming this theorem, we’ll nowbe able to prove torsion doesn’t change under these operations.

21 Nov Recall we’re assuming

Theorem 6.13. Any two Heegaard diagrams of Y are related by a sequence of stabiliza-tions and handleslides.

Suppose Σ′, α′, β′) is a stabilization of (Σ, α, β). Then if h′i are the handles of thecorresponding handle decomposition, h′i = hi∪{h2, h1}, the latter handles correspondingto the new α and β.

If Y → Y is a cover, C ′∗(Y ) = c∗(Y )⊕C where C = R→1 R with the first R generated

by h2 and the second by h1.

Proof. of this claim: Let γ be the core of h′. The [γ] = 1 ∈ π1Y since it’s killed by

h2. So in Y , two feet of h1 end on the same lift of h0 so ∂(h1) = 0. Similarly, β′ pasesthrough α′ but no other α, so it lifts up to a small circle, doing the same thing. SO∂(h2) = h1. Since ∂2 = 0, ∂(h3) must miss h2. �

Corollary 6.14. T (Y, h) = T (Y, h′)

Proof. If bi, dbi is a set of basis elements for C∗(Y ) in the original decomposition then

bi, dbi, h2, d(h2) = h′ is a set of basis elements of C∗(Y ) in the new decomposition. Since

the boundary is an isomorphism of Kh2 to Kh1, this doesn’t change the torsion. �

Definition 6.15. If hi is a basis for V let h′i be given by h′i = hi, i 6= j;h′j = hj+λhk, k 6=j. Then hi and h′i are related by an elementary change of basis.

Lemma 6.16. Let h be a preferred basis for an acyclic chain complex C∗ related to h′

by elementary basis changes. Then T (C∗, h) = T (C∗, h′).

Now suppose that (Σ′, α′, β′) is obtained from Σ, α, β by a β-handleslide. Recall this

means I slid some β1 over β2 along a γ intersecting no βs but at the endpoints. Let h2′1

be a lift of h2′1 , a handle corresponding to β′1. Then the basis h2′

1 , h22, h

23, ..., h

2g where h2

j

is the handle corresponding to βj , is obtained by an elementary change of basis from

h21, h

22, h

23, ..., h

2g, namely h2′

1 = h21 + ρh2

2, where ρ is in the deck group. To clarify why ρis necessary, draw a lift of the handleslide diagram. This will show that ρ correspondsto a lift of γ.

Corollary 6.17. T (Y, h′) = T (Y, h).

So we’ve proved, assuming the theorem on stabilizations and handleslides generatingthe Heegaard group, that

Proposition 6.18. T (Y, h) depends only on Y , not on the handle decomposition h.

6.5. Properties of torsion.

6.5.1. Specialization. Suppose R is an integral domain, eg a group ring of Zk, and K itsfraction field. Suppose ϕ : R → L, L a field. The main example is Zk → Zj surjectivefor j < k inducing Z[Zk]→ Z[Zj ] ⊆ Q(t1...tj). Observe this does not extend to K.

Now suppose C∗ is a free chain complex defined over R and C∗⊗K,C∗⊗L are acyclic.Then if h is a basis for C∗, h⊗ 1 becomes a basis for C∗ ⊗K and C∗ ⊗ L.


Lemma 6.19. T (C∗ ⊗R L) = ϕ(T (C∗ ⊗K)) when the latter is defined.

Proof. We want to say a basis bi, d(bi) of C∗⊗K induces a basis ϕ(bi), d(ϕ(bi)) of C∗⊗L.The latter bases could be smaller than we need-but then C∗⊗RL wouldn’t be acyclic. �

Example 6.20. Suppose ϕ : π1(Y ) → Zk and Y the corresponding cover with deck

group Zk. We define Tϕ(Y ) := T (C∗(Y ) ⊗ L, hij), L = Q(t1...tk). Want to say wecan compute these invariants out of a universal invariant as follows: factor ϕ throughH1(Y )/Tors(H1(Y )). Call the resulting map ϕ, and ϕu the map to the free firsthomology. Then the universal torsion Tu(Y ) := Tϕu(Y ) is in the fraction field ofZ[H1(Y )/torsion] and we can compute Tϕ(Y ) = ϕ(Tu(Y )).

6.5.2. Product formula. Let G be finitely generated and Gf its free part-recalling that tomake the free part natural we have to take the quotient, not a subgroup. Suppose Y is a3=manifold and that H1(Y )f is nontrivial. Then I can look at the cover Y corresponding

to π1(Y )→ H1(Y )f . Then C∗(Y ) is a module over R = Z[Hf1 (Y )], which is just a ring of

Laurent polynomials. IfK is the fraction field of R, then T (Y ) = Tu(Y ) = T (C∗(Y )⊗K).Setup: Let Y = Y1 ∪T 2 Y2, with i1, i2, j : Y1, Y2, T

2 → Y . Then these three descend tomaps on free first homology.

Theorem 6.21. If j∗ 6= 0, then T (Y ) = C1∗(T (Y1))C2∗(T (Y2)).

Example 6.22. In notesbook.

Suppose 0 → A → B → C → 0 is an ses of acyclic chain complexes. Let ha be apreferred basis for A and hb extend ι(ha). Since this is an exact sequence, hc = π(hb) isa basis for C.

Lemma 6.23. T (B, {ι(ha), hb}) = T (A, ha)T (C, hc).

Proof. Write B = ker dB ⊕ B′ and sim for A. ker dA ⊆ ker dB under ι. First choose A′

then extend ι(A′) to pick B′. Then we can write ker dB = ker da⊕K with π(K) = ker dCand B′ = A′ ⊕ L with C ′ ⊕ π(K) = C, π(L) =: C ′. Now pick a basis bA for A′, extendto a basis ι(bA) ∪ bB for B′ so bB is a basis for L. Now shut up and multiply. �

theorem. Consider the covering map Y → Y given by π1(Y ) → H1(Y )f . We can write

Y = Y1∪T 2 Y2 where Yi → Y is the covering map induced by ιi∗ : Hf1 (Yi)→ Hf

1 (Y ). We

have a Mayer-Vietoris sequence 0 → C∗(T2) → C∗(Y1) ⊕ C∗(Y2) → C∗(Y ) → 0. Here

the chain complexes are actually still cellular: we can decompose Y into handles in sucha way as to make it so.

The lemma yields that T (C∗(Y1)⊕C∗(Y2)) = T (C∗(Y ))T (C∗(T2)). Specialization says

the left-hand side is ι1∗(Tu(Y1))ι2∗(Tu(Y2)). So it suffices to prove that T (C∗(T )) = 1

whenever j∗ 6= 0. Then we’ll have what we need if we can show T (C∗(T 2)) = 1 whenever

j∗ 6= 0. Note this implies T 2 is a nontrivial cover.Let thatn T 2 be the universal cover of T 2, whose cellular chains we can present as

R →

(t− 1s− 1

)R2 →−s+1,t−1 R

2. Similarly we can write down the complex for the half-covering by S1 ×R, and we go win yeah. �

22 KEVIN CARLSON

6.6. Manifolds that fiber over S1. Suppose Y fibres over S1 via F and projectionP . So Y = F × I/ ∼,∼ gluing the fibres over 0 and 1, probably nontrivially.

There’s a cover Yp∗ corersponding to the surjection π1(Y ) → π1(S1). Then from

example sheet 3 we’ll see that if F is connected and ∂Y 6= ∅, Tp∗(Y ) = ∆t−1 , ∆ ∈ Z[t±1].

Theorem 6.24. ∆ is a monic polynomial.

So we can compute the Alexander polynomials of various knot complements andsometimes decide whether they fibre. In particular ifK ⊆ S3 is alternating then S3\ν(K)fibres iff its Alexander polynomial is monic.

Proof. ∂Y 6= ∅ ⇒ ∂F 6= ∅ ⇒ F has a handle decomposition with no 2-handles, k 1-handles, and a 0-handle. In particular, H1(F ) = Zk. Then Y has a handle decompositioninto F × [0, 1/2]∪F × [1/2, 1] with the gluing along (x, 1/2) ∼ (x, 1/2), (ϕ(x), 0) ∼ (x, 1)with a 0-handle and k 1-handles on the left and a 1-handle and k 2-handles on the right.

Then C∗(Y ) under this decomposition is Rk → Rk ⊕ R → R. The map on the 1-dimensional Rk is zero, since these 1-handles are stuck to other 1–handles. The map onthe loner 1-dimensional R is t − 1. And how about ∂ : Rk → Rk? If ϕ∗ : Rk → Rk isinduced by the monodromy H1(F )→ H1(F ), ∂(x) = tϕ∗(x)− x.

Justifying remark: the word in π1(F × [0, 1/2] ∪H31 represented by βk the attaching

circle of the kth 2-handle is tϕ∗(αk)t−1αk where αk is the generator corresponding to

the kth 1-handle, t to the special 1-handle.Given C∗, we see that T (C∗(Y )) = det ∂/t − 1 = det(tϕ∗ − I)/(t − 1). The highest

power in t is det(ϕ∗) = ±1 since ϕ is a homeomorphism.Remark: if ∂Y 6= ∅ then wlod may assume ϕ(p) = p for p ∈ F .

�

7. Embedded Surfaces

7.1. Thurston norm.

Proposition 7.1. Suppose Y is a (n oriented compact connected) 3-manifold and x ∈H2(Y, ∂Y ). Then there’s an oriented surface S and an embedding (S, ∂S) → (Y, ∂Y )inducing x.

Proof. Let α ∈ H1(Y ) be PD(x).

Lemma 7.2. Suppose X is a cell complex, α ∈ H1(X), H1(S1) = 〈β〉. Then there existsϕ : X → S1 such that ϕ∗(β) = α.

Proof. Construct ϕ inductively on cells of X. Map each 1-cell γi to α[γi] ∈ Z. Now ϕextends to a 2-cell δ just if ϕψδ has degree 0, where ψδ : S1 → X(1) is the attaching map.By construction deg(ϕψδ) = 〈ϕ∗(β), ψδ∗(1)〉 = 〈j∗(α), ψδ∗(1)〉 = 〈α, j∗ψδ∗(1)〉 = 〈α, 0〉 =0.ϕ extends trivially to all higher skeleta since πi(S

1) is then trivial. �

Now take ϕ as given in the lemma, ϕ∗(β) = α. Choose a regular value p ∈ S1 forϕ. That is, find a smooth map ϕ homotopic to ϕ and p ∈ S1 with dϕ : TxY → TpS

1

surjective for all x ∈ ϕ−1(p). Let S = ϕ−1(P ). Claim that PD([S]) = α.Proof of claim: given γ ∈ H1(Y ), then γ ·S = [ϕ∗(γ)] · [p] = 〈β, ϕ∗(γ)〉 = 〈ϕ∗(β), γ〉 =

〈α, γ〉. �


Theorem 7.3. If x ∈ H2(Y, ∂Y ), x is represented by the inclusion of an oriented surface.

Note S may not be connected. For instance, if [S] = 2 in S1 ×D2, by Mayer-Vietorisalso Y \ ν(S) would be connected, and we could get a contradiction.

In general, x ∈ H2(Y ) is represented by a connected S just if x cannot be written asky, k > 1 ∈ Z. Then x is called primitive.

Problem: Suppose x is primitive. What’s the minimal genus g of S with [S] = x?

Proposition 7.4. Suppose Y fibres over S1 with fibre F . Then any connected surfaceSøY with [S] = [F ] has g(S) ≥ g(F ).

Lemma 7.5. Suppose ϕ : S → F oriented surfaces and ϕ∗ : H2(S)→ H2(F ) is an iso.Then g(S) ≥ g(F ).

Proof. Suppose g(S) < g(F ). Then ϕ∗ : H1(F )→ H1(S) has a kernel, say including α.Then cupping α with something that doesn’t kill it gets our contradiction. �

Lemma 7.6. Suppose as above that F fibres Y over S1 and S embeds in Y with [S] = [F ].

Let Y be the cover associated to π1(Y ) → π1(S1). This is an F -bundle over R, which

must then be trivial. Then ι : S → Y lifts to ι : S → Y splitting ι.

Proof. S lifts iff π1(S) → π1(Y ) → π1(S1) is 0. The map π1(Y ) → π1(S1) sends[γ] ∈ π1(Y ) to γ · F = γ · S ∈ Z = π1(S). If γ is a loop on S, then γ · S = 0 sinceν(S) ' S × [−1, 1] and γ′ = γ × 1 is homologous to γ without intersecting S. Thenπ1(S)→ π1(S1) = 0 so S lifts. �

Proposition. Given S → Y with [S] = [F ], lift to S → Y ' R × F . Composing thisinclusion with the projection onto F , we claim α∗ : H2(S) → H2(F ) is iso. Theng(S) ≥ g(F ) by lemma 1. �

Now to get back to our minimizing genus problem, since so many homology classesare represented by disconnected surfaces, we’d rather consider −χ(S). But including alot of contractible spheres makes −χ arbitrarily small, so we can’t have them. Giventhat,

Definition 7.7. If S is connected, the complexity c(S) is defined by −χ(S) if S is nota sphere or disk and 0 otherwise. If S is disconnected, c(S) is the sum of the complexityof its connected subsurfaces, as long as that sum is defined.

Definition 7.8. If x ∈ H2(Y, ∂Y ) then the Thurston (semi)norm is ||x|| = min[S]=x c(S)where S is a compact embedded surface in Y .

Proposition 7.9. i) ||kx|| = k||x|| and ii) ||x+ y|| ≤ ||x||+ ||y||.

Note it really is a seminorm: for instance the generator of second homology in S1×S2

is represented by a sphere.

Lemma 7.10. Suppose y = kx in H2(Y ) and S represents y. If Si are the components ofS there’s a function α : I → Z/k such that if we define T (j) = ∪α(i)≡jSi, then [Tj ] = x.

Proof. Since Y and the Si are oriented, ν(Si) is oriented. ν(Si) ' SI × [−1, 1]. Letpi = p × 1 for some p ∈ Si. Fix p0 ∈ Y and for each i a path γi from p0 to pi. Define

24 KEVIN CARLSON

α(i) = γi ·S. If γ′i is another such path, then γi−γ′i is a cycle, so [γi−γ′i] ·S = [γi−γ′i] ·kxis divisible by k, so α(i) is well defined in Z/k.

Exercise: [T (j)] = [T (j + 1)] Hint find a subset of Y that gives the homology. So,∑[T (j)] = Y = kx⇒ [T (j)] = x. �

Proof. (i) Let S → Y, [S] = X with c(S) = x. Let KS be K parallel copies of S in ν(S).Immediately [kS] = k[S] = kx and c(kS) = kc(S) so ||kx|| ≤ k||x||. For the oppositeinequality we need the lemma.Given S with [S] = y and c(S) = ||y||, the lemma givesT (j) with [T (j)] = x. Then ||kx|| = c(S) =

∑c(T (j)) ≥ kmin c(T (j)) ≥ k||x||.

Recall x ∈ H2(Y, ∂Y ) with ||x|| = min c(S) over surfaces S representing x. Assume∂Y = ∅ and pick norm-defining surfaces S, T representing x, y. After isotopy, we mayassume S is transverse to T , so that S ∩T is a closed submanifold of S and in particulara union of finitely many simple closed curves on S. Let c be a connected component ofS ∩ T .

Case 1: c bounds a disc in S. wlog assume c is an innermost circle, i.e. no componentsof S ∩ T are contained in the disc it bounds. Then modify T by cutting along c andgluing in two disks to the resulting boundary components. The new surface T ′ hasχ(T ′) = χ(T ) + 2 and T has at most one new S2 component, so c(T ′) ≤ c(T ) and wecan replace T with T ′. Since T ′ is homologous to T we can switch them out.

Case 2: c does not bound in S. Resolve S and T along c in an orientation-respectingway: that means essentially to cut along c and smooth out. This produces S′ withχ(S′) = χ(S) + χ(T ). Since c did not bound, this doesn’t create any new sphere com-ponents, so c(S′) ≤ c(S) + c(T ). Now by induction we may eliminate all components ofS ∩ T to get S′ with [S′] = [S] + [T ] so c(S′) ≤ c(S) + c(T ). �

Extend || · || to H2(Y, ∂Y ; Q) via ||x/n|| = 1/n||x||. This is well defined by (i) of theproposition.

Theorem 7.11 (Thurston). ||x|| extends continuously from H2(Y, ∂Y ; Q) to H2(Y, ∂Y ; R).Moreover there is a finite set BT ⊆ H1(Y ; Z)fr such that ||x|| = maxα∈BT α · x.

The convex hull of BT in H1(Y ; R) is called the dual Thurston ball.

Example 7.12. Y = S3 − ν(L), L = T (2, 6). What is BT ? H1(Y ) = 〈m1,m2〉. A Seifertsurface for L is given by an annulus A, which has c(A) = 0, A ·m1 = 1, A ·m2 = −1. Soif am1 + bm2 ∈ BT , (am1 + bm2) ·A ≤ 0 and am1 + bm2) · (−A) ≤ 0, so a = b and BT lieson the diagonal of the m1,m2 plane. If we untwist L by one of its components we getan unknot m1 surrounding m2 with linking number 3. The punctured disk m1 boundshas χ = −2, so ||[D]|| ≤ 2. The linking number of K1 and K2 is sufficiently large thatwe see actually ||[D]|| = 2. D ·m1 = 1, D ·m2 = 0, so if a(m1 +m2) ·D = a ≤ 2. So BTruns between (−2,−2) and (2, 2).

Suppose Y has b1(Y ) > 1 so T (Y ) = ∆(Y ) ∈ Z[H1(Y )fr]. It’s a fact that ∆(Y ) issymmetric under the antipodal map on H1(Y ), since we can multiply by [g] ∈ H1(Y )fr.

Example 7.13. ∆(S3 \ ν(T (2, 4))) = 1 + st ∼ (st)−1/2 = (st)1/2.

We can write ∆(Y ) as∑agg, g ∈ 1/2H1(Y )f .

Definition 7.14. The Newton polygon B(∆(Y )) = {g ∈ 1/2H1(Y )f : ag 6= 0}.


Theorem 7.15 (McMullen). 2B(∆(Y )) ⊆ BT (Y ).

Example 7.16. ∆(S3 \ T (2, 6)) = (st)−1 + 1 + st.

An addendum, McMullen for b1 = 1 : ∂Y 6= ∅, T (Y ) = ∆(Y )/(t − 1) where ∆(Y ) ∈Z[t±1] is symmetric under t 7→ t−1. ∂Y = ∅ : T (Y ) = ∆(Y )/(t−1)2. Seifert: deg∆(Y ) ≤g(Σ) where [Σ] = 1 ∈ H1(Y )f = Z. For example in Y = S3 − ν(T (2, 3)) then ∆(Y ) =t−1 − 1 + t. Seifert makes g(Σ) ≥ 1. A Seifert surface for the trefoil has characteristic−1.

7.2. Incompressible Surfaces.

Theorem 7.17 (Dehn’s lemma). Suppose γ : S1 → Y is an embedded loop. If thereexists ϕ : D3 → Y with ϕ|∂D2 = γ and ϕ(D2,◦) ∩ imγ = ∅, then there’s ϕ′ : D2 → Y anembedding restricting to γ on the boundary.

Observe the condition on ϕ is much stronger than that γ be contractible. For instancethe trefoil knot doesn’t bound a disk in S3, as we’ve just said. In fact only the unknotbounds an embedding disk.

Set S(ϕ) = {x ∈ D2 : #ϕ−1(ϕ(x)) > 1} and D(ϕ) ⊆ S(ϕ) the double points.

Lemma 7.18. Suppose S(ϕ) = D(ϕ) is a finite disjoint union of circles in D2.

Proof. Pick an innermost circle of double points, cut along it, glue disks into the resultinggaps and ignore the S2 component that comes out. �

In general D(ϕ) could be complicated: consider the cone on the figure 8 knot in S2,which is called the Whitney umbrella.

In general, the best we can say after isotoping the image of ϕ is that S(ϕ) = ¯D(ϕ)and D(ϕ) has finitely many components.

Theorem. Induct on the number of components of D(ϕ). Let Y ′ = ν(ϕ(D2)) ⊆ Y . Thisis a 3fold homotopic to ϕ with ∂Y ′ an oriented surface. Case 1: ∂Y ′ = tS2

i . Thenγ ⊆ S2

j implies γ bounds an embedded disk in S2j . Case 2: H1(∂Y ′) has rank at least 2.

Then the kernel of H1(∂Y ′)→ H1(Y ′) is half-dimensional, so H1(Y ′) has rank at least 1,

so there’s α : π1(Y ′)→ H1(Y ′)→ Z/2. Let Y → Y ′ be the corresponding 2-to-1 cover.

Let ϕ : D2 → Y be a lfit of ϕ. Then the number of components of D(ϕ) is strictly lessthan that of D(ϕ). Why? Because some loop through a double point down in Y ′ gotlifted to an arc...

By induction we may assume ϕ : D2 → Y restricts to ϕ on the boundary. Ourcandidate ϕ′ is π ◦ ϕ. D(ϕ′) = ϕ(D2) ∩ T (ϕ(D2) where T is the deck transformation.Now we’re almost in the case of McMullen’s theorem with two intersecting embeddedsurfaces. It looks worse since the two surfaces are coupled under isotopies, but indeedone can show D(π ◦ ϕ) is a finite disjoint union of circles and apply the lemma. �

Corollary 7.19. Let K ⊆ S3 be a knot. If π1(S3 \K) = Z then K is the unknot.

Proof. Let Y = S3 \ ν(K). The map Z2 = π1(∂Y ) → π1(Y ) = Z has kernel sot here’sγ ∈ π1(∂Y ) an embedded loop with [γ] = 0 ∈ π1(Y ) (since all fundamental groupelements in the torus are realized by embeddings.) Let ϕ : D2 → Y be a disk boundedby ϕ. By adding a T 2 × I to ∂Y and γ × I to ϕ, we get γ ∩ ϕ(D2,◦) = ∅. So there’s

26 KEVIN CARLSON

ϕ′ : D2 → Y restricting to γ. γ is a longitude for K, which implies K bounds a disk inS3 and K is the unknot. �

Other similar theorems:

Theorem 7.20 (Loop theorem). Suppose Y is a connected 3-manifold with boundaryand π1(∂Y ) → π1(Y ) has kernel. Then there’s an embedded γ : S1 → ∂Y that boundsan embedded disk in Y .

Definition 7.21. Suppose S → Y, g(S) > 0.S is called incompressible if π1(S)→ π1(Y )in injective.

The theorem is that compressible surfaces are compressible: there’s a nontrivial loopsomewhere on its boundary which bounds a disk in Y . For instance Heegaard surfaces arecompressible and norm-minimizing surfaces for the Thurston norm are incompressible.For otherwise we could use the loop theorem to find S′ representing the same homologyclass with smaller genus. But there also exist incompressible surfaces S with [S] = 0. Forinstance, in S3\ν(K1)∪T 2S3\ν(K2) for K1,K2 nontrivial knots the T 2 is incompressible.

Theorem 7.22 (Sphere theorem). If π2(Y ) 6= 0 then there exists an embedded S2 ⊆ Ywith [S2] 6= 0.

Definition 7.23. S2 → Y is incompressible if [S2] 6= 0 in π2.

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