1 Homology and Cohomology

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The Topology of 4 - Manifolds

Adrian Dawid January 12, 2021

1 Homology 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:𝐻^𝑖(𝑋;Z⁾ ⁻⁻⁻⁻⁻^→ ^𝐻2−𝑖(𝑋;Z⁾^.

Now we denote by𝑃𝐷⁻¹(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⁾⁼Ext¹_Z(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 surfaceΣ_𝛼 =𝜎⁻¹(0)

• [Σ_𝛼]=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.

2 Intersection Forms

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

⌢:𝐻_𝑝(𝑋;Z^{) ×}^𝐻^𝑞^(𝑋;Z^{) →}^𝐻^𝑝−𝑞^(𝑋;Z^).

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

h·,·i:𝐻^𝑝(𝑋;𝐺) ×𝐻_𝑝(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

⌣:𝐻^𝑖(𝑋;Z^{) ×}^𝐻^𝑗^(𝑋;Z^{) →}^𝐻^𝑗+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⁾ and [Σ_𝛼],[Σ_𝛽] ∈𝐻₂(𝑋;Z⁾be their duals. There are closed 2-forms𝜔𝛼 and𝜔𝛽representing𝛼,𝛽such that

𝑄(𝛼,𝛽)=h𝛼 ⌣ 𝛽,[𝑋]i= Σ_𝛼·Σ_𝛽=

∫

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,[Σ_𝛽]i Switching to de Rahm cohomology:

h𝛼,[Σ_𝛼]i=

∫

Σ_𝛽

𝜔𝛼

Now we have to show: ∫

Σ_𝛽

𝜔𝛼= Σ_𝛼·Σ_𝛽

ChooseΣ_𝛼 t^Σ𝛽. Then we have a finite number of intersection points. Since𝜔𝛼vanishes away from Σ_𝛼 it is enough to compute the integral at the intersection points.

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

𝔘∩Σ_𝛼 ={𝑥₃=𝑥₄=0} 𝔘∩Σ_𝛽={𝑥₁ =𝑥₂ =0} and𝔘∩Σ_𝛼is oriented by𝑑𝑥₁∧𝑑𝑥₂. Then

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

∫

𝔘∩Σ_𝛽

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

By summing over all intersection points we get:

∫

Σ_𝛽

𝜔𝛼 = Σ_𝛼·Σ_𝛽.

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. Then𝑄_𝑋 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⁾⁼^h𝑃𝐷⁻¹^([{pt} ×𝑆²]), 𝑃𝐷⁻¹([{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) ∈2Z^∀𝛼∈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:

OverR^𝑄has𝑏⁺

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

𝑄^𝑙(¹^{) ⊕}^𝑚(−¹⁾

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

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

®

¬ What about definite forms?

• 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.

3 Homotopy 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 orientation 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

𝐻_𝑘(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) ∪_ℎ 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) ∪_ℎ 𝑒⁴;Z⁾

• 𝑆²∨ · · · ∨𝑆² =CP¹^{∨ · · · ∨}CP¹ ^⊂CP^∞^{× · · · ×}CP^∞

• Long exact sequence on relative homotopy groups:

𝜋4(×_𝑚CP^∞^{) −}⁻⁻⁻⁻^→ 𝜋4(×_𝑚CP^∞^,^∨^𝑚^𝑆²^{) −}⁻⁻⁻⁻^→ 𝜋3(∨_𝑚𝑆²)

−−−−−→ 𝜋3(×_𝑚CP^∞⁾

• CP^∞is𝐾(Z^,2) =⇒

𝜋3(∨_𝑚𝑆²)𝜋4(×_𝑚CP^∞^,^∨^𝑚^𝑆²⁾^𝐻4(×_𝑚CP^∞^,^∨^𝑚^𝑆²⁾

• 𝜋3(∨_𝑚𝑆²)𝜋4(×_𝑚CP^∞^,^∨^𝑚^𝑆²⁾^𝐻4(×_𝑚CP^∞^,^∨^𝑚^𝑆²⁾

• Oriented manifold:[ℎ]is determined by𝛼^𝑘(ℎ_∗([𝑒⁴]))for𝛼1, ...,𝛼^𝑙basis of𝐻⁴(×_𝑚CP^∞⁾.

• Basis is given by cupping𝑃𝐷⁻¹([𝑆²_𝑖]). Since

𝐻²(×_𝑚CP^∞⁾^𝐻²^(∨^𝑚^𝑆²⁾^𝐻²^(𝑋⁰;Z⁾⁼^𝐻²^(𝑋;Z⁾ these classes can be seen in𝑋.

• We are done sinceh𝑃𝐷⁻¹([𝑆²_𝑖])⌣𝑃𝐷⁻¹([𝑆²_𝑗]), ℎ_∗([𝑒⁴])i=h𝜔^𝑖,𝜔^𝑗,[𝑋]i=𝑄(𝜔^𝑖,𝜔^𝑗)

Exercise. Fill in the gaps in the proof sketch.

4 The “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

sign𝑄_𝑋∈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 form𝑄_𝑋. Then

sign𝑄_𝑋∈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.

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

• Short answer:

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₊and𝑆₋intersection 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= 𝑓⁻¹(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. If𝑆₋t^𝑆⁺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 intersection 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⊂𝑆₋joining them

• 𝑊simply-connected =⇒ 𝛼∪𝛽inessential

• There is a disk𝐷⊂𝑊 with𝜕𝐷=𝛼∪𝛽

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

• 𝐷is generically embedded

• 𝐷generically does not intersect𝑆₊and𝑆₋in 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 Appendix

6.1 Dual 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

𝜋^𝑘(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:

𝐻^𝑛(𝐾(𝐺, 𝑛);𝐺)^Ext¹_Z^(𝐻^𝑛−1(𝐾(𝐺, 𝑛)), 𝐺) ⊕Hom(𝐻_𝑛(𝐾(𝐺, 𝑛)), 𝐺)^Hom^{(𝐺, 𝐺).}

So we can choose an element in𝐻^𝑛(𝐾(𝐺, 𝑛);𝐺)representing id∈Hom(𝐺, 𝐺). This element is called thecanonical cohomology classand denoted𝜄^𝑛. 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)] −−−−−→ 𝐻^𝑘(𝑋;𝐺)

by [𝑤] ↦→𝑤^∗(𝜄^𝑘).

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

Using this isomorphism we can find embedded surfaces𝑆_𝛼dual 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^].