1 Homology and Cohomology

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The Topology of 4 - Manifolds

Adrian Dawid January 12, 2021


1 Homology and Cohomology 1

2 Intersection Forms 2

3 Homotopy Type 5

4 The โ€œBigโ€ 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 Eilenbergโ€“MacLane Spaces . . . 9

1 Homology and Cohomology

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

Theorem 1.1. Let๐‘‹be a closed orientable 4-manifold, then we have an isomorphism ๐‘ƒ๐ท:๐ป๐‘–(๐‘‹;Z) โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ ๐ป2โˆ’๐‘–(๐‘‹;Z).

Now we denote by๐‘ƒ๐ทโˆ’1(๐‘ฅ)the Poincare dual of any homology class[๐‘ฅ] โˆˆ๐ป2โˆ’๐‘–(๐‘‹;Z).

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

Theorem 1.2. Let๐‘‹be a simply-connected closed oriented 4-manifold, then๐ป2(๐‘‹ ,Z)is a free abelian group.

Proof. This is a simple computation: We have:


Where we use Poincare duality. And also

๐ป1(๐‘‹;Z)=Ab(๐œ‹1(๐‘‹))=0. Thus by the universal coefficient theorem:

๐ป2(๐‘‹ ,Z)=Ext1Z(๐ป1(๐‘‹;Z),Z) โŠ•Hom(๐ป2(๐‘‹;Z),Z)


๐ป (๐‘‹ (๐ป (๐‘‹ ),


Can we see๐ป2(๐‘‹;Z)๐ป2(๐‘‹;Z)geometrically?

โ€ข For๐›ผโˆˆ ๐ป2(๐‘‹;Z)choose a complex line bundle๐ฟs.t. ๐‘1(๐ฟ)=๐›ผ.

โ€ข Take a generic section๐œŽ

โ€ข We have an embedded surfaceฮฃ๐›ผ =๐œŽโˆ’1(0)

โ€ข [ฮฃ๐›ผ]=๐‘ƒ๐ท(๐›ผ)

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

Next we will define an additional structures on๐ป2(๐‘‹;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 class๐‘Žโˆˆ๐ป2(๐‘‹;Z)and cohomology class๐‘โˆˆ ๐ป2(๐‘‹;Z)thecap product is a bilinear map

โŒข:๐ป๐‘(๐‘‹;Z) ร—๐ป๐‘ž(๐‘‹;Z) โ†’๐ป๐‘โˆ’๐‘ž(๐‘‹;Z).

Note the important fact that the Poincare isomorphism is given by๐›ผโ†ฆโ†’ [๐‘‹]โŒข ๐›ผfor the fundamental class[๐‘‹] โˆˆ๐ป๐‘›(๐‘‹;Z). We also have the Kronecker pairing

hยท,ยทi:๐ป๐‘(๐‘‹;๐บ) ร—๐ป๐‘(๐‘‹;๐บ) โ†’๐บ.

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

โŒฃ:๐ป๐‘–(๐‘‹;Z) ร—๐ป๐‘—(๐‘‹;Z) โ†’๐ป๐‘—+๐‘–(๐‘‹;Z).

Now we have everything we need to make this definition:

Definition 2.1. Let๐‘‹be a closed orientedtopological4-manifold. Then the bilinear map ๐‘„:๐ป2(๐‘‹;Z) ร—๐ป2(๐‘‹;Z) โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ Z

given by

(๐›ผ,๐›ฝ) โ†ฆโ†’ h๐›ผ โŒฃ ๐›ฝ,[๐‘‹]i is called (cohomology)intersection formof๐‘‹.

Here we should keep in mind that choosing a fundamental class [๐‘‹] โˆˆ ๐ป4(๐‘‹;Z)is the same as choosing an orientation of๐‘‹.

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

Theorem 2.1. Let๐‘‹ be closed oriented simply-connectedsmooth4-manifold. Let ๐›ผ,๐›ฝ โˆˆ ๐ป2(๐‘‹;Z) and [ฮฃ๐›ผ],[ฮฃ๐›ฝ] โˆˆ๐ป2(๐‘‹;Z)be their duals. There are closed 2-forms๐œ”๐›ผ and๐œ”๐›ฝrepresenting๐›ผ,๐›ฝsuch that

๐‘„(๐›ผ,๐›ฝ)=h๐›ผ โŒฃ ๐›ฝ,[๐‘‹]i= ฮฃ๐›ผยทฮฃ๐›ฝ=



Since ๐ป2(๐‘‹;Z) is torsion free we can go forth and back between integral and de Rahm cohomology.


Proof. First we notice:

๐‘„(๐›ผ,๐›ฝ)=h๐›ผ โŒฃ ๐›ฝ,[๐‘‹]i=h๐›ผ,[๐‘‹]โŒข ๐›ฝi

=h๐›ผ, ๐‘ƒ๐ท(๐›ฝ)i=h๐›ผ,[ฮฃ๐›ฝ]i Switching to de Rahm cohomology:





Now we have to show: โˆซ


๐œ”๐›ผ= ฮฃ๐›ผยทฮฃ๐›ฝ

Chooseฮฃ๐›ผ tฮฃ๐›ฝ. Then we have a finite number of intersection points. Since๐œ”๐›ผvanishes away from ฮฃ๐›ผ it is enough to compute the integral at the intersection points.

Around any intersection point choose๐”˜and oriented local coordinates๐‘ฅ1, ๐‘ฅ2, ๐‘ฅ3, ๐‘ฅ4s.t.

๐”˜โˆฉฮฃ๐›ผ ={๐‘ฅ3=๐‘ฅ4=0} ๐”˜โˆฉฮฃ๐›ฝ={๐‘ฅ1 =๐‘ฅ2 =0} and๐”˜โˆฉฮฃ๐›ผis oriented by๐‘‘๐‘ฅ1โˆง๐‘‘๐‘ฅ2. Then

๐œ”๐›ผ = ๐‘“(๐‘ฅ3, ๐‘ฅ4)๐‘‘๐‘ฅ3โˆง๐‘‘๐‘ฅ4 for a bump function ๐‘“ :R2 โ†’R. Then



๐‘“(๐‘ฅ3, ๐‘ฅ4)๐‘‘๐‘ฅ3โˆง๐‘‘๐‘ฅ4=ยฑ1 depending on orientation.

By summing over all intersection points we get:



๐œ”๐›ผ = ฮฃ๐›ผยทฮฃ๐›ฝ.

For the last equality we have:



๐‘ ๐œ” [๐œ”1โˆง๐œ”2]=[๐œ”1]โŒฃ[๐œ”2] Giving us




Theorem 2.2. Let๐‘‹ be closed oriented simply-connected 4-manifold. Then๐‘„๐‘‹ is unimodular, i.e. ๐‘Ž โ†’ ๐‘„(ยท, ๐‘Ž)and๐‘โ†’๐‘„(๐‘,ยท)are isomorphisms.

Proof. By the universal coefficient theorem

๐ป2(๐‘‹;Z) โ†’Hom(๐ป2(๐‘‹;Z)) ๐›ผโ†ฆโ†’ h๐›ผ,ยทi

is an isomorphism. This suffices, as

๐‘„(๐›ผ,๐›ฝ)=h๐›ผ, ๐‘ƒ๐ท(๐›ฝ)i

and๐‘„is symmetric. Here using the cohomology intersection form really comes in handy, cutting


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

Example 2.1. Consider


Then ๐ป2(๐‘‹;Z)=h๐‘ƒ๐ทโˆ’1([{pt} ร—๐‘†2]), ๐‘ƒ๐ทโˆ’1([{pt} ร—๐‘†2])i.


๐‘„ 0 1

1 0

What happens if๐ป2(๐‘‹;Z)is not free?

Let๐›ผโˆˆ๐ป2(๐‘‹;Z)s.t. ๐‘›ยท๐›ผ=0, then

๐‘›๐‘„(๐›ผ,๐›ฝ)=๐‘„(๐‘›ยท๐›ผ,๐›ฝ)=๐‘„(0,๐›ฝ)=0. So we can define

๐‘„หœ :

๐ป2(๐‘‹;Z) Ext1

Z(๐ป1(๐‘‹;Z),Z) 2

โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ 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:

โ€ข Parity:

If๐‘„(๐›ผ,๐›ผ) โˆˆ2Zโˆ€๐›ผโˆˆ๐ป2(๐‘‹;Z)we call๐‘„even. Otherwise it is calledodd.

โ€ข Definiteness:

If๐‘„(๐›ผ,๐›ผ)>0โˆ€๐›ผโˆˆ ๐ป2(๐‘‹;Z)we call๐‘„positive-definite. If๐‘„(๐›ผ,๐›ผ)<0โˆ€๐›ผ โˆˆ๐ป2(๐‘‹;Z)we call ๐‘„negative-definite.Otherwise it is calledindefinite.

โ€ข Rank:

The second Betti number๐‘2(๐‘‹)is called therankof๐‘„.

โ€ข Signature:


2 positive and๐‘โˆ’

2 negative eigenvalues. We call sign๐‘„=๐‘+

2 โˆ’๐‘โˆ’



Theorem 2.3(Hasse-Minkowski). Let๐ป be a freeZmodule. If ๐‘„ : ๐ปร—๐ป โ†’ Zis an odd indefinite bilinear form then

๐‘„๐‘™(1) โŠ•๐‘š(โˆ’1)

with๐‘™, ๐‘šโˆˆN0. If๐‘„:๐ปร—๐ปโ†’Zis an even indefinite bilinear form then ๐‘„๐‘™

0 1 1 0


with๐‘™, ๐‘šโˆˆN0.


๐ธ8 =













2 โˆ’1 0 0 0 0 0 0

โˆ’1 2 โˆ’1 0 0 0 0 0

0 โˆ’1 2 โˆ’1 0 0 0 0

0 0 โˆ’1 2 โˆ’1 0 0 0

0 0 0 โˆ’1 2 โˆ’1 0 โˆ’1

0 0 0 0 โˆ’1 2 โˆ’1 0

0 0 0 0 0 โˆ’1 2 0

0 0 0 0 โˆ’1 0 0 2












ยฌ What about definite forms?

โ€ข No easy classification

โ€ข Many exotic forms

โ€ข 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 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. Define๐‘‹0=๐‘‹\๐ต4. Then


(๐ป2(๐‘‹) ๐‘˜=2 0 ๐‘˜=1,3,4. By Hurewiczโ€™s theorem:

๐‘“ :๐‘†2โˆจ...โˆจ๐‘†2 โ†’๐‘‹0 represents๐œ‹2(๐‘‹)๐ป2(๐‘‹0;Z). This induces an isomorphism ๐ป (๐‘†2 ... ๐‘†2 ๐ป (๐‘‹0


for every๐‘˜. Thus

๐‘‹' (๐‘†2โˆจ...โˆจ๐‘†2) โˆชโ„Ž ๐‘’4

with[โ„Ž] โˆˆ๐œ‹3(๐‘†2โˆจ...โˆจ๐‘†2). Left to show:[โ„Ž]depends only on๐‘„. Complete proof can be found in:

[1, p.141ff] Sketch:

โ€ข [๐‘‹] โˆˆ๐ป4(๐‘‹;Z)corresponds to[๐‘’4] โˆˆ๐ป4((๐‘†2โˆจ...โˆจ๐‘†2) โˆชโ„Ž ๐‘’4;Z)

โ€ข ๐‘†2โˆจ ยท ยท ยท โˆจ๐‘†2 =CP1โˆจ ยท ยท ยท โˆจCP1 โŠ‚CPโˆžร— ยท ยท ยท ร—CPโˆž

โ€ข Long exact sequence on relative homotopy groups:

๐œ‹4(ร—๐‘šCPโˆž) โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ ๐œ‹4(ร—๐‘šCPโˆž,โˆจ๐‘š๐‘†2) โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ ๐œ‹3(โˆจ๐‘š๐‘†2)

โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ ๐œ‹3(ร—๐‘šCPโˆž)

โ€ข CPโˆžis๐พ(Z,2) =โ‡’


โ€ข ๐œ‹3(โˆจ๐‘š๐‘†2)๐œ‹4(ร—๐‘šCPโˆž,โˆจ๐‘š๐‘†2)๐ป4(ร—๐‘šCPโˆž,โˆจ๐‘š๐‘†2)

โ€ข Oriented manifold:[โ„Ž]is determined by๐›ผ๐‘˜(โ„Žโˆ—([๐‘’4]))for๐›ผ1, ...,๐›ผ๐‘™basis of๐ป4(ร—๐‘šCPโˆž).

โ€ข Basis is given by cupping๐‘ƒ๐ทโˆ’1([๐‘†2๐‘–]). Since

๐ป2(ร—๐‘šCPโˆž)๐ป2(โˆจ๐‘š๐‘†2)๐ป2(๐‘‹0;Z)=๐ป2(๐‘‹;Z) these classes can be seen in๐‘‹.

โ€ข We are done sinceh๐‘ƒ๐ทโˆ’1([๐‘†2๐‘–])โŒฃ๐‘ƒ๐ทโˆ’1([๐‘†2๐‘—]), โ„Žโˆ—([๐‘’4])i=h๐œ”๐‘–,๐œ”๐‘—,[๐‘‹]i=๐‘„(๐œ”๐‘–,๐œ”๐‘—)

Exercise. Fill in the gaps in the proof sketch.

4 The โ€œBigโ€ Structure Theorems

Theorem 4.1(Freedman). Let๐‘„be an quadratic (i.e. unimodular symmetric bilinear) form overZ, then there exists atopological4-manifold๐‘€ s.t. ๐‘„is (up to isomorphism) the intersection form of๐‘€. If๐‘„is even, then๐‘€is unique.

Theorem 4.2(Rohlin). Let๐‘‹be a simply-connected closed orientedsmooth4-manifold with๐‘ค2(๐‘‹)=0.



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 โ€œspin geometryโ€. The spin condition is actually equivalent to the intersection form being even so that we get this corollary:

Corollary 4.2.1. Let ๐‘‹ be a simply-connected closed orientedsmooth4-manifold with even intersection form๐‘„๐‘‹. Then



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-manifold๐ธ8with intersection form๐ธ8that has no smooth structure.

Proof. ๐ธ8is a negative definite even form with signature -8. The existence is given by Freedmanโ€™s


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

Theorem 4.3(Donaldson). Let๐‘‹be a simply-connected closedsmooth4-manifold. If๐‘„is definite,๐‘„is diagonalizable overZ.

Corollary 4.3.1. Let๐‘‹be a simply-connected closedsmooth4-manifold. If๐‘„is positive-definite then ๐‘‹#๐‘˜CP2

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. Let๐‘€and๐‘be closed simply-connected manifolds and๐‘Šbe a cobordism between them (i.e. ๐œ•๐‘Š =๐‘€โˆช ยฏ๐‘). If the inclusions๐‘€ โ†’๐‘Šand๐‘ โ†’๐‘Š are homotopy equivalences, then ๐‘€and๐‘are calledh-cobordant.

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

Theorem 5.2(Smale (1961)). Let๐‘€and๐‘be cobordant smooth n-manifolds with๐‘› >4. Then๐‘€and๐‘ are diffeomorphic.

Warning: This theorem only holds for๐‘› โ‰ฅ5.

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

โ€ข Short answer:

The statement

2+2<4 is optimistic but sadly wrong!

โ€ข That is not so helpful, we are looking for a longer answer.

Strategy of the proof in higher dimensions:

โ€ข Goal: Show that๐‘Š ๐‘€ร— [0,1]


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

Figure 2: Analogy in dimension three showing๐‘†+and๐‘†โˆ’intersection transversely and the resulting flow line. From [3]

โ€ข Choose a Morse function ๐‘“ :๐‘Š โ†’ [0,1]with ๐‘“(๐‘€)=0 and๐‘“(๐‘)=1

โ€ข If ๐‘“ has no critical values we are done!

โ€ข Idea: Modify ๐‘“ 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).

โ€ข Suppose ๐‘“ has two critical points: ๐‘of index 2 and๐‘žof index 3

โ€ข Let๐‘and๐‘žbe separated by๐‘1/2= ๐‘“โˆ’1(1


โ€ข Fact: ๐‘and๐‘žcan be canceled if there is exactly one flow line from๐‘to๐‘ž We define

๐‘†+={๐‘ฅ โˆˆ๐‘1/2 |๐‘ฅflows to๐‘as๐‘กโ†’ โˆž}

๐‘†โˆ’={๐‘ฅ โˆˆ๐‘1/2 |๐‘ฅflows to๐‘žas๐‘กโ†’ โˆ’โˆž}.

These are embedded spheres. If๐‘†โˆ’t๐‘†+is a single point we can glue the flow lines and are done.

The algebraic intersection number is 1 because๐‘Š is h-cobordism. Problem: The geometric inter- section number might not agree! We need an isotopy to correct this.

Usual procedure:


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

โ€ข Choose intersection points with opposite signs, e.g. ๐‘ฅand๐‘ฆ

โ€ข Find path๐›ผโŠ‚๐‘†+and๐›ฝโŠ‚๐‘†โˆ’joining them

โ€ข ๐‘Šsimply-connected =โ‡’ ๐›ผโˆช๐›ฝinessential

โ€ข There is a disk๐ทโŠ‚๐‘Š with๐œ•๐ท=๐›ผโˆช๐›ฝ

โ€ข If the disk lies outside๐‘†+and๐‘†โˆ’we get an isotopy removing the intersection points In dimension๐‘›โ‰ฅ5:

โ€ข ๐ทis generically embedded

โ€ข ๐ทgenerically does not intersect๐‘†+and๐‘†โˆ’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 Eilenbergโ€“MacLane Spaces

The concept from algebraic topology we need is that of an Eilenbergโ€“MacLane space.

Definition 6.1. Let๐‘› >0 and๐บbe an abelian group then a space๐พ(๐บ, ๐‘›)is called Eilenberg-McLane space if it is a CW-complex and

๐œ‹๐‘˜(๐พ(๐บ, ๐‘›))=

(๐บ if๐‘˜=๐‘› 0 otherwise



Remark. The space๐พ(๐บ, ๐‘›)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 ๐‘“ :๐‘‹โ†’๐‘Œinduces a homomor- phism ๐‘“โˆ— :๐ปโ€ข(๐‘Œ;๐บ) โ†’๐ปโ€ข(๐‘‹;๐บ)on cohomology. First we notice the following connection:

๐ป๐‘›(๐พ(๐บ, ๐‘›);๐บ)Ext1Z(๐ป๐‘›โˆ’1(๐พ(๐บ, ๐‘›)), ๐บ) โŠ•Hom(๐ป๐‘›(๐พ(๐บ, ๐‘›)), ๐บ)Hom(๐บ, ๐บ).

So we can choose an element in๐ป๐‘›(๐พ(๐บ, ๐‘›);๐บ)representing idโˆˆHom(๐บ, ๐บ). This element is called thecanonical cohomology classand denoted๐œ„๐‘›. Now we can see this more powerful connection:

Theorem 6.1. Let๐‘‹be a topological space and๐บan abelian group. Then define a map [๐‘‹:๐พ(๐บ, ๐‘˜)] โˆ’โˆ’โˆ’โˆ’โˆ’โ†’ ๐ป๐‘˜(๐‘‹;๐บ)

by [๐‘ค] โ†ฆโ†’๐‘คโˆ—(๐œ„๐‘˜).

This map is well-defined and an isomorphism if๐‘‹is a CW-complex.

Using this isomorphism we can find embedded surfaces๐‘†๐›ผdual to cohomology class๐›ผโˆˆ๐ป2(๐‘‹;Z). In the following let๐‘‹be a smooth closed oriented 4-manifold.

Strategy for๐›ผโˆˆ๐ป2(๐‘‹;Z):

โ€ข Observe


โ€ข Identify๐‘ƒ๐ทโˆ’1([CP1]) โˆˆ๐ป2(CP2;Z)(henceforth just[CP1]) with๐œ„2 โˆˆ๐ป2(๐พ(Z,2);Z).

โ€ข Choose generic ๐‘“๐›ผ tCP1s.t.[๐‘“๐›ผ]=๐›ผi.e.


๐›ผ([CP1])= ๐‘“๐›ผโˆ—(๐œ„2)=๐›ผ.

โ€ข Define๐‘†๐›ผ= ๐‘“โˆ’1

๐›ผ (CP1). This is an embedded surface by transversality.

โ€ข With this construction[๐‘†๐›ผ]=๐‘ƒ๐ท(๐›ผ) โˆˆ๐ป2(๐‘‹;Z).


[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/


[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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