# 1 Homology and Cohomology

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## The Topology of 4 - Manifolds

### Contents

1 Homology and Cohomology 1

2 Intersection Forms 2

3 Homotopy Type 5

4 The “Big” Structure Theorems 6

5 Whitney Disks and the Failure of the h-Cobordism Principle in Dimension Four 7

6 Appendix 9

6.1 Dual Surfaces via Eilenberg–MacLane Spaces . . . 9

### 1Homology and Cohomology

First we recall the Poincare duality isomorphism. Later we will use it to define the intersection form.

Theorem 1.1. Let𝑋be a closed orientable 4-manifold, then we have an isomorphism 𝑃𝐷:𝐻𝑖(𝑋;Z) −−−− 𝐻2−𝑖(𝑋;Z).

Now we denote by𝑃𝐷1(𝑥)the Poincare dual of any homology class[𝑥] ∈𝐻2−𝑖(𝑋;Z).

Next we will show an easy but important fact about the homology and cohomology of simply- connected four manifolds.

Theorem 1.2. Let𝑋be a simply-connected closed oriented 4-manifold, then𝐻2(𝑋 ,Z)is a free abelian group.

Proof. This is a simple computation: We have:

𝐻2(𝑋;Z)𝐻2(𝑋;Z).

Where we use Poincare duality. And also

𝐻1(𝑋;Z)=Ab(𝜋1(𝑋))=0. Thus by the universal coefficient theorem:

𝐻2(𝑋 ,Z)=Ext1Z(𝐻1(𝑋;Z),Z) ⊕Hom(𝐻2(𝑋;Z),Z)

=Hom(𝐻2(𝑋;Z),Z).

𝐻 (𝑋 (𝐻 (𝑋 ),

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Can we see𝐻2(𝑋;Z)𝐻2(𝑋;Z)geometrically?

• For𝛼∈ 𝐻2(𝑋;Z)choose a complex line bundle𝐿s.t. 𝑐1(𝐿)=𝛼.

• Take a generic section𝜎

• We have an embedded surfaceΣ𝛼 =𝜎1(0)

• [Σ𝛼]=𝑃𝐷(𝛼)

Note: Different construction using Eilenberg-MacLean spaces in appendix of notes.

Next we will define an additional structures on𝐻2(𝑋;Z). Namely we will define the intersection form or intersection product. Please note that we do not assume the manifold to be smooth for this.

But first we need to quickly recall relevant notions from algebraic topology.

### 2Intersection Forms

Recall that for any homology class𝑎∈𝐻2(𝑋;Z)and cohomology class𝑏∈ 𝐻2(𝑋;Z)thecap product is a bilinear map

⌢:𝐻𝑝(𝑋;Z) ×𝐻𝑞(𝑋;Z) →𝐻𝑝−𝑞(𝑋;Z).

Note the important fact that the Poincare isomorphism is given by𝛼↦→ [𝑋]⌢ 𝛼for the fundamental class[𝑋] ∈𝐻𝑛(𝑋;Z). We also have the Kronecker pairing

h·,·i:𝐻𝑝(𝑋;𝐺) ×𝐻𝑝(𝑋;𝐺) →𝐺.

It descends from the evaluation of a cochain on a chain. The last thing we have is thecup product which is a bilinear map

⌣:𝐻𝑖(𝑋;Z) ×𝐻𝑗(𝑋;Z) →𝐻𝑗+𝑖(𝑋;Z).

Now we have everything we need to make this definition:

Definition 2.1. Let𝑋be a closed orientedtopological4-manifold. Then the bilinear map 𝑄:𝐻2(𝑋;Z) ×𝐻2(𝑋;Z) −−−−− Z

given by

(𝛼,𝛽) ↦→ h𝛼 ⌣ 𝛽,[𝑋]i is called (cohomology)intersection formof𝑋.

Here we should keep in mind that choosing a fundamental class [𝑋] ∈ 𝐻4(𝑋;Z)is the same as choosing an orientation of𝑋.

This is a very algebraic definition. For a smooth four manifold we can interpret it in a more geometric way:

Theorem 2.1. Let𝑋 be closed oriented simply-connectedsmooth4-manifold. Let 𝛼,𝛽 ∈ 𝐻2(𝑋;Z) and𝛼],[Σ𝛽] ∈𝐻2(𝑋;Z)be their duals. There are closed 2-forms𝜔𝛼 and𝜔𝛽representing𝛼,𝛽such that

𝑄(𝛼,𝛽)=h𝛼 ⌣ 𝛽,[𝑋]i= Σ𝛼·Σ𝛽=

𝑋𝜔𝛼∧𝜔𝛽.

Since 𝐻2(𝑋;Z) is torsion free we can go forth and back between integral and de Rahm cohomology.

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Proof. First we notice:

𝑄(𝛼,𝛽)=h𝛼 ⌣ 𝛽,[𝑋]i=h𝛼,[𝑋]⌢ 𝛽i

=h𝛼, 𝑃𝐷(𝛽)i=h𝛼,[Σ𝛽]i Switching to de Rahm cohomology:

h𝛼,[Σ𝛼]i=

Σ𝛽

𝜔𝛼

Now we have to show: ∫

Σ𝛽

𝜔𝛼= Σ𝛼·Σ𝛽

ChooseΣ𝛼 tΣ𝛽. Then we have a finite number of intersection points. Since𝜔𝛼vanishes away from Σ𝛼 it is enough to compute the integral at the intersection points.

Around any intersection point choose𝔘and oriented local coordinates𝑥1, 𝑥2, 𝑥3, 𝑥4s.t.

𝔘∩Σ𝛼 ={𝑥3=𝑥4=0} 𝔘∩Σ𝛽={𝑥1 =𝑥2 =0} and𝔘∩Σ𝛼is oriented by𝑑𝑥1∧𝑑𝑥2. Then

𝜔𝛼 = 𝑓(𝑥3, 𝑥4)𝑑𝑥3∧𝑑𝑥4 for a bump function 𝑓 :R2 R. Then

𝔘Σ𝛽

𝑓(𝑥3, 𝑥4)𝑑𝑥3∧𝑑𝑥4=±1 depending on orientation.

By summing over all intersection points we get:

Σ𝛽

𝜔𝛼 = Σ𝛼·Σ𝛽.

For the last equality we have:

h𝜔,[𝑁]i=

𝑁 𝜔 [𝜔1∧𝜔2]=[𝜔1]⌣[𝜔2] Giving us

𝑄(𝛼,𝛽)=

𝑋𝜔𝛼∧𝜔𝛽.

Theorem 2.2. Let𝑋 be closed oriented simply-connected 4-manifold. Then𝑄𝑋 is unimodular, i.e. 𝑎 → 𝑄(·, 𝑎)and𝑏→𝑄(𝑏,·)are isomorphisms.

Proof. By the universal coefficient theorem

𝐻2(𝑋;Z) →Hom(𝐻2(𝑋;Z)) 𝛼↦→ h𝛼,·i

is an isomorphism. This suffices, as

𝑄(𝛼,𝛽)=h𝛼, 𝑃𝐷(𝛽)i

and𝑄is symmetric. Here using the cohomology intersection form really comes in handy, cutting

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This simple example might show how useful our geometric interpretation is when it comes to concrete computations:

Example 2.1. Consider

𝑋=𝑆2×𝑆2.

Then 𝐻2(𝑋;Z)=h𝑃𝐷1([{pt} ×𝑆2]), 𝑃𝐷1([{pt} ×𝑆2])i.

And

𝑄 0 1

1 0

What happens if𝐻2(𝑋;Z)is not free?

Let𝛼∈𝐻2(𝑋;Z)s.t. 𝑛·𝛼=0, then

𝑛𝑄(𝛼,𝛽)=𝑄(𝑛·𝛼,𝛽)=𝑄(0,𝛽)=0. So we can define

𝑄˜ :

𝐻2(𝑋;Z) Ext1

Z(𝐻1(𝑋;Z),Z) 2

−−−−−→ Z

and use the arguments there. Sometimes this is also given as the definition of the intersection form.

With this definition we also get unimodularity without the assumption of a simply-connected manifold. Next we look at some invariants of intersection forms:

Parity:

If𝑄(𝛼,𝛼) ∈2Z𝛼∈𝐻2(𝑋;Z)we call𝑄even. Otherwise it is calledodd.

Definiteness:

If𝑄(𝛼,𝛼)>0∀𝛼∈ 𝐻2(𝑋;Z)we call𝑄positive-definite. If𝑄(𝛼,𝛼)<0∀𝛼 ∈𝐻2(𝑋;Z)we call 𝑄negative-definite.Otherwise it is calledindefinite.

Rank:

The second Betti number𝑏2(𝑋)is called therankof𝑄.

Signature:

OverR𝑄has𝑏+

2 positive and𝑏

2 negative eigenvalues. We call sign𝑄=𝑏+

2 −𝑏

2

thesignatureof𝑄.

Theorem 2.3(Hasse-Minkowski). Let𝐻 be a freeZmodule. If 𝑄 : 𝐻×𝐻 → Zis an odd indefinite bilinear form then

𝑄𝑙(1) ⊕𝑚(−1)

with𝑙, 𝑚∈N0. If𝑄:𝐻×𝐻→Zis an even indefinite bilinear form then 𝑄𝑙

0 1 1 0

⊕𝑚𝐸8

with𝑙, 𝑚∈N0.

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𝐸8 =

­

­

­

­

­

­

­

­

­

­

«

2 −1 0 0 0 0 0 0

−1 2 −1 0 0 0 0 0

0 −1 2 −1 0 0 0 0

0 0 −1 2 −1 0 0 0

0 0 0 −1 2 −1 0 −1

0 0 0 0 −1 2 −1 0

0 0 0 0 0 −1 2 0

0 0 0 0 −1 0 0 2

ª

®

®

®

®

®

®

®

®

®

®

• No easy classification

• Many exotic forms

• Number of unique even definite forms of some ranks:

Rank 8 16 24

# 1 2 5

These numbers grow rapidly with rank and contain no structure (known to me).

Warning: Any intersection form is diagonalizable overQbut might not be overZ. Exercise. Show that

0 1 1 0

is not diagonalizable overZ.

### 3Homotopy Type

We will now find a direct link between homotopy type and intersection form of four manifolds.

Theorem 3.1(Milnor (1958)). The oriented homotopy type of a simply-connected closed oriented 4-manifold is determined by its intersection form.

We need to use an oriented homotopy type here because the intersection form depends on orienta- tion up to sign.

The proof is not given here in detail but rather only sketched. There is a reference to the full proof included below.

Proof. Define𝑋0=𝑋\𝐵4. Then

𝐻𝑘(𝑋0;Z)=

(𝐻2(𝑋) 𝑘=2 0 𝑘=1,3,4. By Hurewicz’s theorem:

𝑓 :𝑆2∨...∨𝑆2 →𝑋0 represents𝜋2(𝑋)𝐻2(𝑋0;Z). This induces an isomorphism 𝐻 (𝑆2 ... 𝑆2 𝐻 (𝑋0

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for every𝑘. Thus

𝑋' (𝑆2∨...∨𝑆2) ∪ 𝑒4

with[ℎ] ∈𝜋3(𝑆2∨...∨𝑆2). Left to show:[ℎ]depends only on𝑄. Complete proof can be found in:

[1, p.141ff] Sketch:

• [𝑋] ∈𝐻4(𝑋;Z)corresponds to[𝑒4] ∈𝐻4((𝑆2∨...∨𝑆2) ∪ 𝑒4;Z)

• 𝑆2∨ · · · ∨𝑆2 =CP1∨ · · · ∨CP1 CP× · · · ×CP

• Long exact sequence on relative homotopy groups:

𝜋4𝑚CP) −−−−− 𝜋4𝑚CP,𝑚𝑆2) −−−−− 𝜋3(∨𝑚𝑆2)

−−−−−→ 𝜋3𝑚CP)

• CPis𝐾(Z,2) =⇒

𝜋3(∨𝑚𝑆2)𝜋4𝑚CP,𝑚𝑆2)𝐻4𝑚CP,𝑚𝑆2)

• 𝜋3(∨𝑚𝑆2)𝜋4𝑚CP,𝑚𝑆2)𝐻4𝑚CP,𝑚𝑆2)

• Oriented manifold:[ℎ]is determined by𝛼𝑘(ℎ([𝑒4]))for𝛼1, ...,𝛼𝑙basis of𝐻4𝑚CP).

• Basis is given by cupping𝑃𝐷1([𝑆2𝑖]). Since

𝐻2𝑚CP)𝐻2(∨𝑚𝑆2)𝐻2(𝑋0;Z)=𝐻2(𝑋;Z) these classes can be seen in𝑋.

• We are done sinceh𝑃𝐷1([𝑆2𝑖])⌣𝑃𝐷1([𝑆2𝑗]), ℎ([𝑒4])i=h𝜔𝑖,𝜔𝑗,[𝑋]i=𝑄(𝜔𝑖,𝜔𝑗)

Exercise. Fill in the gaps in the proof sketch.

### 4The “Big” Structure Theorems

Theorem 4.1(Freedman). Let𝑄be an quadratic (i.e. unimodular symmetric bilinear) form overZ, then there exists atopological4-manifold𝑀 s.t. 𝑄is (up to isomorphism) the intersection form of𝑀. If𝑄is even, then𝑀is unique.

Theorem 4.2(Rohlin). Let𝑋be a simply-connected closed orientedsmooth4-manifold with𝑤2(𝑋)=0.

Then

sign𝑄𝑋∈16Z.

The original proof by Rohlin is very involved. Simpler proof due to Atiyah and Singer using the Atiyah-Singer index theorem. Reference: [2, Theorem 29.9] Such a manifold is also called aspin manifold and it is not by accident that the above proof can be found in lecture notes on “spin geometry”. The spin condition is actually equivalent to the intersection form being even so that we get this corollary:

Corollary 4.2.1. Let 𝑋 be a simply-connected closed orientedsmooth4-manifold with even intersection form𝑄𝑋. Then

sign𝑄𝑋∈16Z.

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Another exciting result we get directly is the existence of four manifolds without any smooth structures.

Corollary 4.2.2. There exists a simply-connected closed 4-manifold𝐸8with intersection form𝐸8that has no smooth structure.

Proof. 𝐸8is a negative definite even form with signature -8. The existence is given by Freedman’s

theorem.

Now we have arrived at the theorem that is the holy grail we are chasing in this seminar:

Theorem 4.3(Donaldson). Let𝑋be a simply-connected closedsmooth4-manifold. If𝑄is definite,𝑄is diagonalizable overZ.

Corollary 4.3.1. Let𝑋be a simply-connected closedsmooth4-manifold. If𝑄is positive-definite then 𝑋#𝑘CP2

as topological manifolds.

### 5Whitney Disks and the Failure of the h-Cobordism Principle inDimension Four

In this last section I will briefly go over the intuition of why dimension four is so different from dimension five and up.

Definition 5.1. Let𝑀and𝑁be closed simply-connected manifolds and𝑊be a cobordism between them (i.e. 𝜕𝑊 =𝑀∪ ¯𝑁). If the inclusions𝑀 →𝑊and𝑁 →𝑊 are homotopy equivalences, then 𝑀and𝑁are calledh-cobordant.

Theorem 5.1(Wall).Two simply-connected four-manifolds with isomorphic intersection form are h-cobordant.

Theorem 5.2(Smale (1961)). Let𝑀and𝑁be cobordant smooth n-manifolds with𝑛 >4. Then𝑀and𝑁 are diffeomorphic.

Warning: This theorem only holds for𝑛 ≥5.

Why does the (smooth) h-cobordism principle fail in dimension four?

The statement

2+2<4 is optimistic but sadly wrong!

• That is not so helpful, we are looking for a longer answer.

Strategy of the proof in higher dimensions:

• Goal: Show that𝑊 𝑀× [0,1]

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Figure 1: Cancelling an index 0 critical point with an index 1 critical point. From [3]

Figure 2: Analogy in dimension three showing𝑆+and𝑆intersection transversely and the resulting flow line. From [3]

• Choose a Morse function 𝑓 :𝑊 → [0,1]with 𝑓(𝑀)=0 and𝑓(𝑁)=1

• If 𝑓 has no critical values we are done!

• Idea: Modify 𝑓 s.t. all critical values disappear

Removing critical points of index 0,1,4,5 works in dimension four. But: Canceling critical points of index 3 and 2 does not work (with this method).

• Suppose 𝑓 has two critical points: 𝑝of index 2 and𝑞of index 3

• Let𝑝and𝑞be separated by𝑍1/2= 𝑓1(1

2)

• Fact: 𝑝and𝑞can be canceled if there is exactly one flow line from𝑝to𝑞 We define

𝑆+={𝑥 ∈𝑍1/2 |𝑥flows to𝑝as𝑡→ ∞}

𝑆={𝑥 ∈𝑍1/2 |𝑥flows to𝑞as𝑡→ −∞}.

These are embedded spheres. If𝑆t𝑆+is a single point we can glue the flow lines and are done.

The algebraic intersection number is 1 because𝑊 is h-cobordism. Problem: The geometric inter- section number might not agree! We need an isotopy to correct this.

Usual procedure:

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Figure 3: Removing intersection points in pairs. From [3]

• Choose intersection points with opposite signs, e.g. 𝑥and𝑦

• Find path𝛼⊂𝑆+and𝛽⊂𝑆joining them

• 𝑊simply-connected =⇒ 𝛼∪𝛽inessential

• There is a disk𝐷⊂𝑊 with𝜕𝐷=𝛼∪𝛽

• If the disk lies outside𝑆+and𝑆we get an isotopy removing the intersection points In dimension𝑛≥5:

• 𝐷is generically embedded

• 𝐷generically does not intersect𝑆+and𝑆in any interior points

In dimension four on the other hand both is not true! The intersection form makes this clear.

With the existence of non-smooth manifolds one the one hand and the failure of the h-cobordism on the other hand, we see that the topology and geometry of four-manifolds is quite unique.

### 6.1Dual Surfaces via Eilenberg–MacLane Spaces

The concept from algebraic topology we need is that of an Eilenberg–MacLane space.

Definition 6.1. Let𝑛 >0 and𝐺be an abelian group then a space𝐾(𝐺, 𝑛)is called Eilenberg-McLane space if it is a CW-complex and

𝜋𝑘(𝐾(𝐺, 𝑛))=

(𝐺 if𝑘=𝑛 0 otherwise

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holds.

Remark. The space𝐾(𝐺, 𝑛)is unique up to homotopy.

Now we can make an important connection between the cohomology groups of a manifold (any CW-complex will work actually) and the homotopy classes of maps from the space into an Eilenberg- MacLane space. Indeed we will see that for the correct Eilenberg-McLane space they are isomorphic.

And it is an isomorphism that is concrete enough so we can actually understand it. One important thing to remember in order to define this is that a continuous map 𝑓 :𝑋→𝑌induces a homomor- phism 𝑓 :𝐻(𝑌;𝐺) →𝐻(𝑋;𝐺)on cohomology. First we notice the following connection:

𝐻𝑛(𝐾(𝐺, 𝑛);𝐺)Ext1Z(𝐻𝑛−1(𝐾(𝐺, 𝑛)), 𝐺) ⊕Hom(𝐻𝑛(𝐾(𝐺, 𝑛)), 𝐺)Hom(𝐺, 𝐺).

So we can choose an element in𝐻𝑛(𝐾(𝐺, 𝑛);𝐺)representing id∈Hom(𝐺, 𝐺). This element is called thecanonical cohomology classand denoted𝜄𝑛. Now we can see this more powerful connection:

Theorem 6.1. Let𝑋be a topological space and𝐺an abelian group. Then define a map [𝑋:𝐾(𝐺, 𝑘)] −−−−−→ 𝐻𝑘(𝑋;𝐺)

by [𝑤] ↦→𝑤(𝜄𝑘).

This map is well-defined and an isomorphism if𝑋is a CW-complex.

Using this isomorphism we can find embedded surfaces𝑆𝛼dual to cohomology class𝛼∈𝐻2(𝑋;Z). In the following let𝑋be a smooth closed oriented 4-manifold.

Strategy for𝛼∈𝐻2(𝑋;Z):

• Observe

𝐻2(𝑋;Z)[𝑋:𝐾(Z,2)][𝑋:CP][𝑋:CP2].

• Identify𝑃𝐷1([CP1]) ∈𝐻2(CP2;Z)(henceforth just[CP1]) with𝜄2 ∈𝐻2(𝐾(Z,2);Z).

• Choose generic 𝑓𝛼 tCP1s.t.[𝑓𝛼]=𝛼i.e.

𝑓

𝛼([CP1])= 𝑓𝛼(𝜄2)=𝛼.

• Define𝑆𝛼= 𝑓1

𝛼 (CP1). This is an embedded surface by transversality.

• With this construction[𝑆𝛼]=𝑃𝐷(𝛼) ∈𝐻2(𝑋;Z).

### References

[1] Alexandru Scorpan. The wild world of 4-manifolds. Providence, R.I., 2005.

[2] Thomas Walpuski. MTH 993 Spring 2018: Spin Geometry. https://walpu.ski/Teaching/

SpinGeometry.pdf.

[3] Simon K Donaldson and Peter B Kronheimer.The geometry of four-manifolds. Oxford mathematical monographs. Oxford, 1. publ. in pbk. edition, 1997.

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