Math 528: Algebraic Topology Class Notes
Dr. Denis Sjerve
Term 2, Spring 2005 2 Contents
1January4 7
1.1ARoughDefinitionofAlgebraicTopology...... 7
2January6 13
2.1TheMayer-VietorissequenceinHomology...... 13
2.2Example:twospaceswithidenticalhomology...... 17
2.3Aseminar...... 19
3 January 11 21
3.1Hatcher’sWebPage...... 21
3.2CWComplexes...... 21
3.3CellularHomology...... 24
3.4Apreviewofthecohomologyring...... 25
3.5 Boundary Operators in Cellular Homology ...... 26
3.6Atopologyvisitor...... 26
4 January 13 27
3 4.1 Homology of RP n ...... 27
4.2Duality...... 29
5 January 18 31
5.1Assignment1 ...... 31
5.2Tworeadingcourses...... 32
5.3HomologywithCoefficients...... 32
5.4Application:LefschetzFixedPointTheorem...... 34
6 January 20 37
6.1TensorProducts...... 37
6.2LefschetzFixedPointTheorem...... 39
6.3Cohomology...... 41
7 January 25 43
7.1Examples:LefschetzFixedPointFormula...... 43
7.2ApplyingCohomology...... 46
7.3History:theHopfinvariant1problem...... 47
7.4AxiomaticdescriptionofCohomology...... 48
7.5MyProject...... 48
8 January 27 49
8.1AdifferencebetweenHomologyandCohomology...... 49
8.2Axiomsforunreducedcohomology...... 49
4 8.3Eilenberg-Steenrodaxioms...... 50
8.4Aconstructionofacohomologytheory...... 51
8.5Universalcoefficienttheoremincohomology...... 54
9February1 57
9.1CommentsontheAssignment...... 57
9.2 Proof of the UCT incohomology...... 58
9.3 Properties of Ext(A, G)...... 60
10 February 3 63
10.1NaturalityintheUCT...... 63
10.2ProofoftheUCT...... 64
10.3Somehomologicalalgebra...... 65
10.4Grouphomologyandcohomology...... 66
10.5MilnorConstruction...... 68
11 February 8 71
11.1ASeminar...... 71
11.2Products...... 72
5 6 Chapter 1
January 4
Math 528 notes, jan4
Instructor: Denis Sjerve (Pronounced: Sure’ vee)
Office: Math 107 [email protected]
Send an email to the professor, with the following information:
• computer languages that I know (c++, java, html, tex, maple, etc.) •
Marks: homework, term projects.
Text: Algebraic Topology, Allen Hatcher.
1.1 A Rough Definition of Algebraic Topology
Algebraic topology sort of encompasses all functorial relationships between the worlds of algebra and topology: world of topological problems →F world of algebraic problems
7 Examples
1. The retraction problem: X is a topological space, A ⊆ X is a subspace. Does there exist a continouous map r : X → A such that r(a)=a for all a ∈ A? r is called a retraction and A is called a retract of X.
A −−−i → X −−−r → A
where r ◦ i = idA. If r exists, then F (i) F (r) F (A) −−−→ F (X) −−−→ F (A) or F (i) F (r) F (A) ←−−− F (X) ←−−− F (A)
where the composites must be idF (A). Consider a retraction of the n-disk onto its boundary:
Sn−1 = ∂(Dn) −−−i → Dn −−−r → ∂(Dn01) = Sn−1 Suppose, in this case, that the functor here is “take the nth homology group”.
n n n Hn−1(∂(D )) −−−→ Hn−1(D ) −−−→ Hn−1(∂D ) Z −−−→ 0 −−−→ Z (assuming n>1). This is not possible, so ∂Dn is not a retract of Dn
2. When does a self map f : X → X haveafixedpoint?I.e.istherex ∈ X such that f(x)=x? From now on, “map” means a continuous map. Let us consider X = Dn,f : X → X is any map. Assume that f(x) =6 x for all x ∈ D. Project x through f(x)onto∂Dn, as follows: r(x)
x
f(x)
Dn
Then r : Dn → ∂Dn is continuous and r(x)=x if x ∈ ∂Dn. This is a contradiction, since f must have a fixed point.
8 3. What finite groups G admit fixed point free actions on some sphere Sn?Thatis,there exists a map G × Sn → Sn,(g,x) 7→ g · x, such that h · (g · x)=(hg) · x, id · x = x; furthermore, for any g =6 id, g · x =6 x for all x ∈ Sn. This is “still” unsolved (although some of the ideas involved in the supposed proof of the poincare conjecture would do it for dimension 3). However, lots is known about it. For example, any cyclic group admits a fixed-point free action on any odd-dimensional sphere. Say, G = Zn,and X 2k−1 k S = {(z1,...,zk) ⊆ C | ziz¯i =1}.
A generator for G is T : S1 → S1, T (x)=ξx,whereξ = e2πi/n. The action of G on 2k−1 S is T (z1,...,zk)=(ξz1,...,ξzk). 4. Suppose M n is a smooth manifold of dimension n. What is the span of M? i.e. whit is the largtest number k such that there exists a k-plane varing continouously with respect to x?Thatis,ateachpointx ∈ M we have k linearly independent vectors v1(x),...,vk(x)inTxM, varying continously with respect to x.
$x$ Tx(M)
Definition: if k = n then we say that M is parallelizable. In all cases, of course, k S3 is also parallizable. Consider the unit quaternions: X { | 2 } q0 + q1i + q2j + q3k qn =1 n 9 These form a group: i2 = j2 + k2 = −1,ij = kk = −ji, etc. Pick three linearly inde- pendent vectors at some fixed point in S3. Then use the group structure to translate this frame to all of s3. 5. The homeomorphism problem. When is X homeomorphic to Y ? f X −−−→ Y F y F y F (f) F (x) −−−→ F (Y ) 6. The homotopy equivalence problem. f p 7. The lifting problem. Given X → B and E → b, can we find a f˜ : X → E such that pf˜ ' f? 8. The embedding problem for manifolds. What is the smallest k such that the n- dimensional manifold M can be embedded into Rn+k? Suppose Sn ⊆ Rn +1. RP n =real projective space of dimension n,whichisSn in which antipodal points are identified: RP n def= Sn/x ∼−x. Alternatively, RP n is the space of lines through the origin in Rn+1. Unsolved problem: what is the smallest k such that RP n ⊆ Rn+k? 9. Immersion problem: What is the least k such that RP n immerses into Rn+k? embedding $S^1$ R^2 immersion 10. The computation of homotopy groups of spheres. Πk(X) = the set of homotopy classes k n of maps f : S → X. What is Πk(S )? The answer depends upon k − n, essentially ; this is the Freudenthal suspension theorem. 10 2 3 4 n We know that Π3(S )=Z;Π4(S ) ≈ Π5(S ) ≈···≈Z2.Andπn(S ) ≈ Z,andsoon. 11 12 Chapter 2 January 6 Notes for Math 528, Algebraic Topology II, jan6 2.1 The Mayer-Vietoris sequence in Homology Recall the van Kampen Theorem: Suppose X isaspacewithabasepointx0,andX1 and X2 are subspaces such that x0 ∈ X1 ∩ X2,andX = X1 ∪ X2. X ∩ X −−−i1→ X 1 2 1 i2y j1y j2 X2 −−−→ X Apply the fundamental group ‘functor’ Π1 to this diagram: i Π (X ∩ X ) −−−1#→ Π (X ) 1 1 2 1 1 i2#y j1#y j2# Π1(X2) −−−→ Π1(X) Question: How do we compute Π1(X) frim this data? We will need some mild assumptions: X1,X2,X1 ∩ X2 are path connected, etc. 13 There exists a group homomorphism from the free product Π1(X1) ∗ Π1(X2)intoΠ1(X), given by c1 · c2 7→ j1#(c1) · j2#(c2). Fact: This map is onto Π1(X). However, there exists a kernel coming from Π1(X1 ∩ X2). In −1 fact, i1#(α) · i2#(α ), for every α ∈ Π1(X1 ∩ X2), is in the kernel because j1#i1# = j2#i2#. Theorem: (van Kampen): X1,X2,X1 ∩X2 contain the base point x0 ∈ X = X1 ∪X2,every space is path connected, etc. Then Π1(X) ≈ Π1(X1) ∗ Π2(X2)/K where k is the normal subgroup containing all elements of the form −1 i1#(α) · i2#(α ) where α ∈ Π2(X1 ∩ X2). Definition: Let X be a space with a base point x0 ∈ X.Thenth homotopoy group is the n n set of all homotopy classes of maps f :(I ,∂I ) → (X, X0). Here, n I = {(t1,...,tn)|0 ≤ ti ≤ 1}; ∂In = the boundary of In = {(t1,...,tn)|0 ≤ ti ≤ 1, some ti =0or1} Notation: Πn(X, x0)=Πn(X)=thenth homotopy group. n n n Fact: I /∂I ≈ S . Therefore, Πn(X) consists of the homotopy classes of maps n f :(S , ∗) → (X, x0) . Question: Is there a van Kampen theorem for Πn? Answer: NO. But there is an analogue of the van Kampen Theorem in Homology: it is the Meyer–Vietoris sequence. Here is the setup: ∩ −−−i1→ X1 X2 X1 i2y j1y j2 X2 −−−→ X = X1 ∪ X2 14 Question: What is the relationship amongst H∗(X1 ∩ X2), H∗(X1), H∗(X2)andH∗(X)? Theorem: (Mayer-Vietoris) Assume some mild hypotheses on X1,X2,X (they are, in some sense, “nice”. We will see what niceness properties they have later) Then there exists a long exact sequence: α∗ β∗ δ α∗ ···→Hn(X1 ∩ X2) → Hn(X1) ⊕ Hn(X2) → Hn(x) → Hn−1(X1 ∩ X2) →···→H0(X) → 0. The maps α∗ and β∗ come from the structure on X: β∗ : Hn(X1) ⊕ Hn(X2) → Hn(x) β∗(c1,c2)=j1∗(c1)+j2∗(c2) α∗ : Hn(X1 ∩ X2) → Hn(X1) ⊕ Hn(X2) c → (i1#(c), −i2#(c)) The minus sign gets included in α∗ for the purpose of making things exact (so that β∗α∗ =0). One could have included it in the definition of β∗ instead and still be essentially correct. Proof: There exists a short exact sequence of chain complexes α β 0 −−−→ C∗(X1 ∩ X0) −−−→ C∗(X1) ⊕ C∗(X2) −−−→ C∗(X1 + X2) −−−→ 0 Cn(X1 + X2) is the group of chains of the form c1 + c2,wherec1 comes from X1 and c2 comes from X2. The ‘mild hypotheses’ imply that C∗(X1 + X2) ≤ C∗(X) is a chain equivalence. Lemma: If β 0 → C00 →α C0 → C → 0 is an exact sequence of chain complexes, then there exists a long exact sequence 00 α∗ 0 β∗ δ 00 α∗ ···→Hn(C ) → Hn(C ) → Hn(C) → Hn−1(C ) →··· To prove this, one uses the “snake lemma” which may be found in Hatcher, or probably in most homological algebra references. Remarks: There exists a Mayer-Vietoris sequence for reduced homology, as well: α∗ β∗ δ α∗ ···→H˜n(X1 ∩ X2) → H˜n(X1) ⊕ H˜n(X2) → H˜n(x) → H˜n−1(X1 ∩ X2) →··· Here, the reduced homology group ( def Hn(X)ifn =06 , H˜n = Hn(X, ∗)= H˜0(X) ⊕ H0(x0)ifn =0. 15 That is, the reduced homology groups are the same as the usual ones except in dimension 0, where they differ by a direct-sum copy of the integers. Often, the reduced homology groups eliminate distracting exceptional cases in dimension 0. Examples. X is a space, the unreduced suspension of X is SX := X × [0, 1]/(x × 0=p, x × 1=q) single point q t =1 I t =0 single point p We also have the reduced suspension, which is ΣX = SX/(X × [0, 1]) q x0 x0 p Fact: If W is a nice space, and A ⊆ W is a contractible nice subspace, then W → W/A is a homotopy equivalence. ≈ Corollary: H˜n(SX) → H˜n−1(X). Proof: 16 cone C+ 1 SX = X(t = 2 ) cone C− So, in the exact sequence, we have ···→ ˜ ∩ →α∗ ˜ ⊕ ˜ →β∗ ˜ →δ ˜ ∩ →···α∗ Hn(X1 X2) |Hn{z(C+}) H| n{z(C−}) Hn(x) Hn−1(X1 X2) =0 =0 k k−1 Cor H˜n(S ) ∼ H˜n−1(S ). Pf. S(Sk−1)=Sk. 2.2 Example: two spaces with identical homology Recall that real projective n-space is RP n = Sn/(x ∼−x), the n-sphere with antipodal points identified. Let us write S2(RP 2) for S(S(RP 2)). Define def X = RP 2 ∨ S2(RP 2) def Y = RP 4, where A ∨ B is the one point union of A, B.Now, Z if i =0 H (RP 4)= Z if i =1, 3 i 2 0 otherwise. Note that it is a corollary of the Mayer-Vietoris sequence that H˜i(A ∨ B) ≈ H˜i(A) ⊕ H˜i(B). 17 So we can compute that Z if i =0 H (RP 4)= Z if i =1, 3 i 2 0 otherwise, which is the same homology as X. IsitthecasethatX and Y are “the same” in some sense? Perhaps “same” means “homeo- morphic”. But Y = RP 4 is a 4–dimensional manifold, whereas X = RP 2 ∨ S2(RP 2)isnot a manifold. So X is not homeomorphic to Y . Can “same” mean “homotopy equivalent?” Still no. The universal covering space of Y is S4 The universal covering space of X is: copy of S2(RP 2) S2 =universal covering space of RP 2 copy of S2(RP 2) If X and Y were homotopically equivalent then their universal covering spaces would also ˜ ˜ ˜ be homotopically equivalent. Let X and Y be the universal covering space. H2(X)=Z, but H2(Y˜ )=0. Question: Does there exist a map f : X → Y (or g : Y → X) such that f∗ (resp, g∗)isan isomorphism in homology? Again, the answer is no, and we shall see why next week. 18 2.3 A seminar There is a seminar in Topology, 11:00 on Monday, in the PIMS seminar room, by a visiting speaker of Ailana Fraser. 19 20 Chapter 3 January 11 Notes for Math 528, Algebraic Topology II, jan11 3.1 Hatcher’s Web Page Hatcher’s web page is: http://www.math.cornell.edu/∼hatcher There, you can find an electronic copy of the text. 3.2 CW Complexes Attaching an n-cell en toaspaceA: Suppose we have φ : §n − 1 → A.WewanttoextendthistoamapDn →Φ X,where X = A ] Dn/φ(x) ∼ x for all x ∈ ∂Dn = §n − 1. 21 en φ A Dn Sn−1 Definition: ofaCWcomplex:X = X0 ∪ X1 ∪··· 0 0 0 X is a discrete set of points {x1,x2,...}. Now attach 1-cell(s) via maps φα : S → X , where α ∈ A, the same index set. X1 is the result of attaching 1-cells. 0 eα x3 x4 x1 x2 Suppose we have constructed Xn−1,then − 1-skeleton. Then Xn is the result of attaching n−1 n−1 n-cells by maps φα : S → X .Then ] n n−1 n ∈ n → ∼ X := X Dα x ∂Dα x φα(x). α∈A Examples. − 1. X = Sn =pt ∪ en = e0 ∪ en. A =pt=e0. φ : Sn 1 → A is the constant map. Thus, X = A ] Dn x ∈ ∂Dn,x∼ e0 = Sn. 2. Sn = ∂(∆n+1), where ∆n+1 is the standard n + 1-simplex. For example, consider ∆3, the surface of the tetrahedron. ∂(∆3)=S2. To describe S2 this way (as a simplicial complex) we need: 4 vertices, 6 edges and 4 faces. This is far less efficient than the CW-complex. 3. RP n := the space of lines through the origin in Rn+1 = Sn x−˜ x. 22 n C+ = D+ −x x n C− = D− Note that we have the double covering φ : Sn → RP n−1,whereφ(x)=(x, −x). So RP n = RP n−1 ∪ en; the attaching map is φ. Therefore RP n = e0 ∪ e1 ∪···∪en. 4. CP n, complex projective n-space = the space of one-dimensional compex subvector spaces of Cn+1 = S2n+1 X ∼ ζx, for all complex unit numbers ζ. (We can think of ζ ∈ S1. Another way to say this is: there exists an action of S1 on S2n+1.LetusthinkofS2n+1 in the following way: 2n+1 S = {(z1,...,zn +1)| z1z¯1 + ···+ zn+1z¯n+1 =1}. 1 2n+1 Then S acts on S by ζ · (z1,...,zn+1)=(ζz1,...,ζzn+1). Exercie: φ : S2n+1 →§2n+!/S1 = CP n.VerifythatCP n+1 = CP n ∪ e2n with attaching map φ. Thus CP n = e0 ∪ e2 ∪ e4 ∪···∪e2n. Remark: Suppose X = A ∪ en,φ : Sn − 1 → A.Wehavei : A ⊆ X (recall that X = A ] Dn xφ˜(x))). So X A = Dn ∂Dn = Sn . n n−1 n Now suppose X is a CW complex. What is X X ? Suppose the n-cells are eα, 23 where α ∈ A, the same index set. Then _ n n−1 n X X = Sα. α∈A Let X be a CW complex. We have a commutative diagram: dn+1 dn n+1 n n n−1 n−1 n−2 ... Hn+1(X ,X ) Hn(X ,X ) Hn − 1(X ,X ) i∗ δi∗ δ n 0 n−1 Hn(X ) Hn−1(X ) This diagram serves to define the maps dn.Observethatdndn+1 =0,sowehaveachain n n−1 complex. C∗(X):Cn(X)=Hn(X ,X ). CW Definition: H∗(C∗(X)) = H∗ (X). CW AbigTheorem: H∗ (X) is naturally isomorphic to H∗(X). Examples. 1. Sn = e0 ∪ en X0 = pt, Xn = Sn.(SoX1,X2, ··· ,Xn−1 are empty) If n>1, then ( Zi=0,n Ci(x)= . 0 otherwise Therefore, all dx =0so ( Zi=0,n HCW (Sn)= . i 0 otherwise n 0 2 2 2. CP = e ∪ e ∪···∪e n. Here, all dk are 0, so ( Zi=0, 2, 4, ···,n HCW (CP n)= . i 0 otherwise 24 3. X = CP n,Y = S2 ∨ S4 ∨···S2n and all attaching maps are trivial. (i.e. constant). It is easy to check that ( Zi=0, 2, 4, ···, 2n Hi(Y )= 0 otherwise That is, H∗(X) ≈ H∗(Y ). But X is not homeomorphic to Y if n>1. X is a compact wn-dimensional manifopld with no boundary; Y is not even a manifold if n>1. Exercies: CP 1 ≈ S1. Question: if n>1isthereamapf : X → Y inducing an isomorphism in H∗? Answer: No, but one needs cohomology in order to see it. 3.4 A preview of the cohomology ring The cohomology groups of a space X (whatever they are) actually form a graded ring. The cohomology ring structure of X = CP n is: H∗(X) ≈ Z[x] (Xn+1) x ∈ H2(CP n) Here, we have 1 ∈ H0(X) ≈ Z, x ∈ H2(X) ≈ Z, x2 ∈ H4(X) ≈ Z, ..., xn ∈ H2n(X) ≈ Z. Moreover, Hi(X)=0ifi is odd. The answer for Y is: ( Z i =0, 2,...,2n Hi(Y )= . 0 otherwise This is the same information, on the group theoretic level, as homology. However, all “prod- ucts” turn out to be zero in H∗(Y ). So the cohomology rings of these two spaces are nonisomorphic. 25 3.5 Boundary Operators in Cellular Homology n n−1 X aCW-complex,Cn(X)=Hn(X ,X ). We have n n−1 n n−1 Hn(X ,X ) ≈ Hn(X X , ∗). (This theorem is known as “excision”, among other names) More generally, H∗(X, A) ≈ H∗(X/A, ∗). Therefore, we have _ n n−1 ≈ ˜ n Hn(X ,X ) Hn( Sα), α∈A n a free abelian group with one generator for each n-cell eα. Therefore, we can identify the n generators with the n-cells eα. n n−1 n−1 n−2 Let us define dn : Hn(X ,X ) → Hn−1(X ,X )by: X n n−1 dn(eα)= dαβeβ β∈B n ∈ − n−1 ∈ Here, the n-cells are eα,α A,andthen 1–cells are eβ for β B. n n−1 → n−1 eα has an attaching map φα : S X .Wehavemaps W n−1 −−−φα→ n−1 −−−c → n−1 n−2 ≈ n−1 −−−c → n−1 S X X X β∈B Sβ Sβ n−1 n−1 Consider the composite of these maps, fαβ : S → S . f has a degree, dαβ. m m m m Definition: :iff : S → S then f∗ : Hm(S ) → Hm(S ) is multiplication by deg f (Z → Z,k 7→ (deg f)k). 3.6 A topology visitor Dylan Thurston (son of Fields medalist William Thurston) is applying for a job at UBC. Seminar on Thursday, 3:00-4:00 in PIMS 216. Colloquium Friday, 3:00-4:00 in MATX 1100 Meeting with graduate and undergraduate students, Thursday 4:00-5:00 PIMS 110 26 Chapter 4 January 13 Notes for Math 528, Algebraic Topology II, jan13 4.1 Homology of RP n n n Today’s goal will be to compyut H∗(RP ) using cellular homology. Recall that RP = |e0 {z∪ e}1 ∪e2 ∪···∪en. | S1{z } RP 2 ∞ n ∞ We define RP = limn→∞ RP = S /x ∼−x. n n−1 Cellular Homology: Cn = Hn(X ,X ) = a free abelian group; its generators are in 1-1 correspondence with the n-cells in X. dn+1 dn ··· −−−→ Cn+1 −−−→ Cn −−−→ Cn − 1 −−−→ ··· −−−→ C0 −−−→ 0 InthecaseofRP n, it turns out that we will have 0 ×2 0 ··· −−−→ 0 −−−→ Cn −−−→ ··· −−−→ C1 −−−→ C0 −−−→ 0 Corollary: Zi=0 H (RP n)= Z i =1, 3, 5, ···, 2k 27 and ( n 0 n even Hn(RP )= Z n odd Description of dk : Ck → Ck−1 Sk−1 RP k−1 RP k−1/RP k−2 = Sk−1 Note that RP k−2 comes from the equator of the sphere, and is collapsed to a point: Sk−1 Sk−1 Sk−2 id on top sphere collapse equator a pm bottom sphere where a : k−1 k−1 S → S is the antipodal map. Therefore, dk : Ck → Ck−1 is deg(id+a) Z −−−−−→ Z where ( 2 k even deg(id + a)=1+(−1)k = 0 k odd 28 4.2 Duality I Y is the space of mappings I =[0, 1] → Y .IfY has a base point y0 then we can define the ω loop space ΩY = the space of maps I → Y such that ω(0) = ω(1) = y0. We have Maps(ΣX, Y ) ≈ Maps(X, ΩY ) f(x, t)=f 0(x)(t) This correspondence is valid at the level of homotopy: [ΣX, Y ] ≈ [X, ΩY ]. ΩY has extra structure: it is an H–space. H probably stands for either “homotopy” or “Heinz Hopf”, depending on wheter you are Heinz Hopf or not. ΩY is a group up to homotopy: The group multiplication µ is concatenation of paths: ( ω(2t), 0 ≤ t ≤ 1 µ :ΩY × ΩY → ΩY (ω,η) 7→ µ(ω,η)=ω ∗ η = 2 − 1 ≤ ≤ η(2t 1), 2 t 1 Exercises: Let be the constant map at y0. ∗ ω ' ω ∗ ' ω. Inverses: the invers of ω is the path ω−1(t)=ω(1 − t). That is, ω ∗ ω−1 ' ' ω−1 ∗ ω.The operation is associative because composition of functions is associative. Therefore, [X, ΩY ] is an actual group, so [ΣX, Y ] is also a group. Here, the group operation is defined thus: f :ΣX → Y and g :ΣX → Y .Then ( f(x, 2t)0≤ t ≤ 1 f · g(x, t)= 2 − 1 ≤ ≤ g(x, 2t 1) 2 t 1 Examples. 1. Let f : S1 7→ S1 be f(z)=zk.Thendegf = k. 2. g : Sn → Sn. We wish to define a map g with deg g = k. Definition: Suppose f : X → Y is base point preserving. Then we can define Σf :ΣX → ΣY ,Σf(x, t)=(f(x),t). Now, suspending f : z 7→ zk n − 1timesgivesamapΣn−1f : Sn 7→ Sn of degree k. We’ll go into this more next time. 29 30 Chapter 5 January 18 Notes for Math 528, Algebraic Topology II, jan18 5.1 Assignment 1 p.155 #1,3,5,7,9(ab); p156#12, p.257 #19 Here are some comments: (1) paraphrased is: Show that f : Dn → Dn has a fixed point. this is the Brouwer Fixed point theorem and we have proved it already. The question says to prove it in a special way, though: Apply degree theory to the map that sends both northern and southern hemispheres to the southern hemisphere by f. (3) No comment. n (5) Two reflections r1,r2 of S through hyperplanes. Show that r1 ' r2 through a homo- topy consisting of reflections through hyperplanes. There is a hint. (7) Let f : Rn → Rn be an invertible linear transformation.We get an induced map on reduced homology: ˜ Rn n −{ } −−−f∗→ ˜ Rn n −{ } Hn( ,R 0 ) Hn( ,R 0 ) ≈y ≈y n f∗ n Z ≈ H˜n(R −{0}) −−−→ H˜n(R −{0}) ≈ Z 31 This is an invertible map Z → Z. Show that f∗ = id if and only if detf > 0. Use linear algebra, gaussian reduction, etc. (9ab) This one is different from the online one. The original question had all 4 parts. Compute the homology groups of: (a) S2 north pole = south pole. Hint earlier in the book involving CW complexes. (b) S1 × (S1 ∨ S2) – two tori. Again, Cellular homology probably the best thing. (12) Show that the map S1 × S1 → S1 × S1 S1 ∧ S1 ≈ S2 is not homotopic to a point. hint: look a what happens in Z ≈ H2(S1 × S1) → 2 H2(S ) ≈ Z. Compute the degree of this map; if it were null-homotopic, the degree would be zero. n m (19) Compute the homology of truncated projective space; H∗(RP RP ), m The due date is Feb. 1 5.2 Two reading courses Dr. Sjerve is offering reading courses in Riemann Surfaces and also in 3-manifold topology (a book by Hempel, Annals of Math. Studies). 5.3 Homology with Coefficients Example: f : X → RP ∞ = S∞/x ∼−x. Pick a model for S∞: it is the set of sequences (x1,x2,x3,...) P ∈ R ∞ 2 with xi such that i=1 xi =1andxk = 0 for k 0(means:k sufficiently large). 32 1 The homotopy class of f is a unique element [f] ∈ H (X; Z2) ≈ Hom (H1(X; Z),Z2). Which leads us to the question: how does one put coefficients into a homology theory? There is a kind of natural way to do it. X =some space; C∗(X) = some chain complex. A typical n-chain is X niσi,ni ∈ Z. i Z Z Q R C LetP G be some abelian group, like n, , , , . Now, a typical chain in Cn(X; G)is ∈ i niσi,ni G. We can define boundary operators in the same way as before, since the coefficients are fairly “inert” and acted on the simplices themselves. One can describe this in terms of tensor products as follows: Cn(X; G) ≈ Cn(X) ⊗ G. The boundary operators are then just ∂⊗ id. One can then define homology in the obvious way: H∗(X; G)=H∗(C∗(X; G)). Essentially everything goes through verbatim in this setting. One thing that does change is the following: Theorem: (The dimension axiom) H0(pt, Z) ≈ G (instead of Z). Example: RP n = e0 ∪ e1 ∪···∪en. The cellular chain complex: δ δ δ δ δ δ δ ··· −−−→ Ck+1 −−−→ Ck −−−→ ··· −−−→ C2 −−−→ C1 −−−→ C0 −−−→ 0 × ··· −−−→ Z −−−→ Z −−−→ ··· −−−→ Z −−−→ Z −−−2→ Z −−−0→ 0 n With Z2 coefficients, C∗ ⊗ Z2 = C∗(RP ; Z2), all maps become 0: 0 0 0 0 0 0 0 ··· −−−→ Z2 −−−→ Z2 −−−→ ··· −−−→ Z2 −−−→ Z2 −−−→ Z2 −−−→ 0 Therefore, ( n Z2 i =0, 1,...,n Hi(RP ; Z2)= 0 otherwise. 33 5.4 Application: Lefschetz Fixed Point Theorem Definition: Let A be an abelian group (say finitely generated). Suppose that φ : A → A is a group homomorphism. Then, A = F ⊕ T where F is free abelian of rank r<∞ and T is a finite group. φ induces a group homomorphism φ : A/T → A/T . One can represent this map by an n × n matrix. The trace of φ is the trace of this n × n matrix over Z (the sum of the diagonal entries). The choice of basis is not important, since the trace is invariant under conjugation. Definition: Suppose X is a finite complex, and f : X → X is a map. f induces the map f∗ : Hn(X) → Hn(X), which is a map of finitely generated abelian groups. The Lefschetz Number X def n λ(f) = (−1) trace(f∗(Hn(X; Q) → (Hn(X; Q)). n The purpose of doing things over the rational numbers Q is to kill off the torsion part. Theorem: (Lefschetz) If f is fixed point free, then λ(f)=0. Example: If f ' id, then we are computing the traces of identity matrices, and λ(f)= χ(X), the Euler characteristic of the space X. Let βn be the rank of the finite dimensional vector space Hn(X; Q). βn is called the nth Betti number. Then the Euler characteristic of X was defined to be X n χ(X)= (−1) βn. n Let γn be the the number of n-dimensional cells in X. P n Theorem: (essentially using only algebra) χ(X)= (−1) γn. Let X = §2n.Thenχ(X)=2,since ( Z i =0, 2n Hi(X)= 0 otherwise Suppose that X has a nonvanishing vector field V 34 x V (x) So x ∈ Rn+1 is a vector with norm 1; V (x) ∈ Rn+1 is perpendicular to x (x · V (x)=0). Suppose V is a nonzero vector field on S2n,thatis,x 7→ V (X) is continuous in x, V (x) =06 for all x,andx · V (x) = 0 for all x. Without loss of generality, assum that |V (x)| = 1 for all x. There exists a unique geodesic on S2n going through x in the direction of V (x). Let f : S2n → S2n be the map that takes x to that point f(x) whose distance from x along this geodesic is 1. x V (x) f(x) f is clearly homotopic to the identity map: take the homotopy which parameterizes the geodesics by arclength.. Therefore λ(f)=λ(id)=χ(S2n)=2.Butf has no fixed points (λ(f) = 0). This is a contradiction. We can define geodesics on any manifold, so we have the following: 35 Theorem: If M n is a comopact n-manifold with Euler characteristic χ(n)=neq0thenM does not have a nonzero vector field. Corollary: CP n = e0∪e2∪···∪e2n = S2n+1/x ∼ λx, |λ| =1.Notethatχ(CP n)=n+1 =0,6 so CP n does not have a nonzero vector field. P 2n+1 ∈ 2n+1 2 Example: S has a nonzero vector field. Let x =(x1,...,x2n+2) S , xi =1. We would like to define V (x) ∈ R2n+2 such that x · V (x) = 0. And this is easy enough: we pair the coordinates off and swap them, thus: V (x)=(−x2,x1, −x4,x3, ··· , −x2n+2,x2n+1). Now x · V (x)=x1 · (−x2)+x2 · x1 + ···=0. Example. Let G be a topological group which is also a smooth manifold. Then χ(G)=0. Examples: S1, S3 ≈ unit quaternions, SO(n)=n × n orthogonal matrices. The Euler characteristic of SO(n)is0. Here is how to construct the nonvanishing field: Let e ∈ G be the identity element. Take v ∈ Te(G),v=0.6 Then we construct a nonzero vector field V by V (g)=(dRg)(v)whereRg : G → G is right multiplication by g,andd is differentiation. v Rg e g dRg 36 Chapter 6 January 20 Notes for Math 528, Algebraic Topology II, jan20 6.1 Tensor Products A, B are abelian groups. We define A ⊗ B = F (A × B)/N where F (A × B) is the free abelian group on pairs (A, B), and N is the subgroup which is generated by all elements of the form (a1 + a2,b) − (a1,b) − (a2,b), (a, b1 + b2) − (a, b1) − (a, b2) That is, we are forcing linear relations in both coordinates. Notation: a ⊗ b deontes the class of (a, b)inA ⊗ B. Zero element: a ⊗ 0=0⊗ a =0 Inverses: −(a ⊗ b)=(−a) ⊗ b = a ⊗ (−b). Properties: • A ⊗ Z ≈ A. a ⊗ n 7→ na. Also, the inverse map A → A ⊗ Z, a 7→ a ⊗ 1. • A ⊗ Zn ≈ A/nA, φ(a)=a ⊗ 1. φ induces a map A/nA → A ⊗ Zm since φ(na)= na ⊗ 1=a ⊗ n · 1=0. 37 • A ⊗ B = B ⊗ A. •⊗is associative. • Suppose A is finite. Then A ⊗ Q = 0. To see why: r 1 a ⊗ = r a ⊗ s s = 0 for some r. • A ⊗ (B ⊕ C)=(A ⊗ B) ⊕ (A ⊗ C) • Zm ⊗ Zn ≈ Zd,whered =gcd(m, n). • A ⊗ B is functorial in both A and B. For example, if φ : A → A0 is a group homomor- phism, then there exists a group homomorphism φ ⊗ 1:A ⊗ B → A0 ⊗ B. Question: What is the relationship between Hn(X; G)andHn(X) ⊗ G? C∗(X; G)=C∗(X) ⊗ G,andH∗(X; G)=H∗(C∗(X) ⊗ G). Take c ∈ Cn(X), and suppose that c is a cycle. Let [c] be the homology class, [c] ∈ Hn(X). Now, c ⊗g ∈ Cn(X) ⊗G is a cycle, since ∂(c ⊗g)=∂(c) ⊗g. Therefore, [c ⊗g] is a homology class in Hn(X; G). Definition: µ : Hn(x) ⊗ G → Hn(X; G) [c] ⊗ g 7→ [c ⊗ g] µ is not an isomorphism, in general. However, we have the following theorem: Theorem: (Universal Coefficient theorem for Homology): There exists a natural short exact sequence µ 0 −−−→ Hn(X) ⊗ G −−−→ Hn(X; G) −−−→ Hn−1(X) ∗ G −−−→ 0 We denote Hn−1(X) ∗ G by Tor(Hn−1(X),G). 38 6.2 Lefschetz Fixed Point Theorem Lemma: Suppose we have a commutative diagram with exact rows: 0 −−−→ A −−−→ A0 −−−→ A00 −−−→ 0 y φy φ0y φ00y 0 −−−→ A −−−→ A0 −−−→ A00 −−−→ 0 where A, A0,A00 are finitely generated abelian groups.. Then Trφ0 =Trφ +Trφ00 Lemma: (Hopf Trace Formula) Suppose C∗ is a chain complex of abelian groups such that: • Each Cn is a finitely generated abelian group. • Cn =6 0 for only finitely many n. Suppose that φ : C∗ → C∗ is a chain map. Then X X n n (−1) Tr(φ : Cn → Cn)= (−1) Tr(φ∗ : Hn(C) → Hn(C) n n Proof: ∂n+1 ∂n ··· −−−→ Cn+1 −−−→ Cn −−−→ Cn−1 −−−→ ··· Let Zn := ker ∂n, Bn := ∂n+1(Cn+1), so Hn := Zn/Bn. Consider 0 −−−→ B −−−→ Z −−−→ H −−−→ 0 n n n 0 y φy φ y φ∗y 0 −−−→ Bn −−−→ Zn −−−→ Hn −−−→ 0 and −−−→ −−−→ −−−∂n→ −−−→ 0 Zn Cn Bn−1 0 y φ0y φy φy ∂n 0 −−−→ Zn −−−→ Cn −−−→ Bn−1 −−−→ 0 39 Now, 0 00 Tr(φ∗ : Hn → Hn)=Tr(φ : Zn → Zn) − Tr(φ : Bn → Bn) 0 00 Tr(φ : C → C )=Tr(φ : Z → Z )+Tr(φ : B − → B − ) X n n X n n n 1 n 1 n n =⇒ (−1) Tr(φ∗ : Hn → Hn)= (−1) Tr(φ : Cn → Cn) Proof: of the Lefschetz fixed point formula: Assume that X is a finite complex, and f : X → X is a map with no fixed points. Assume for simplicity that X is a simplicial complex and f is a simplicial map. Suppose f(x) =6 x for all x ∈ X. Then it is not necessarily true that f(σ)∩σ = for all simplices σ in X. We shall subdivide to make it true. σ Repeat until the simplices are small enough so that f(σ) ∩ σ = ∅. We can do this due to compactnes; in a more general setting, we should use the simplicial approximation theorem. P n Then (−1) Tr(f : Cn → Cn)=0,sincef is defined on simplices and there is nothing on the diagonal of f.Soλ =0. Example: f : RP 2k+1 → RP 2k+1, λ(f)=1− deg f. f∗ H (RP 2k+1) −−−→ H (RP 2k+1) 2k+1 2k+1 ≈y ≈y Z −×−−−deg→f Z If f is fixed point free, then deg f = 1 (e.g. a homoemorphism that preserves orientation). f(x1,x2,...,x2k+1,x2k+2)=(x2, −x1,...,x2k+2, −x2k+1) f is a homeomorphism. Moreover, f preserves orientation, because as a linear map, f has 40 matrix 01 −10 .. . 01 −10 6.3 Cohomology Start with a chain comoplex C∗(x). ∂1 ∂2 0 ←−−− C0 ←−−− C1 ←−−− C2 ←−−− ··· ←−−− Cn ←−−− Cn+1 ←−−− ∗ Make the cochain complex: C := Hom(C∗(X),G) 0 1 n 0 −−−→ C0 −−−δ → C1 −−−δ → C2 −−−→ ··· −−−→ Cn −−−δ → Cn+1 n where δ :Hom(Cn(X),G) → Hom(Cn+1(X),G)mapsα : Cn(X) → G to α ◦ δn +1: δn+1 α Cn+1 −−−→ Cn −−−→ G H∗(X; G) =the homology of this chain complex. Cohomology is contravariant – i.e. if f : X → Y , f ∗ : Hn(Y ) → Hn(X). ∗ Question: What is the relationship between H (X; G)andHom(Hn(X),G)? Cohomology has more structure than homology. There exists a product Hp(X) × Hq(X) → Hp+q(X × X). Let d : X → X × X be the diagonal: d(x)=(x, x). Thereforeore, there exists a product, called the cup product: Hp(X) ⊗ Hq(X) → Hp+q(X) u ⊗ v 7→ u ∪ v H∗(X) is a graded ring. Examples: • H∗(CP n; Z) is a polynomial algebra; it is isomorphic to Z[x]/(xn+1), x ∈ H2(CP n; Z). ∗ n n+1 1 n 2 2 • H (RP ; Z2) ≈ Z2[x]/(x ), x ∈ H (RP ; Z2) ≈ Z2; x = x ∪ x ∈ H . 41 42 Chapter 7 January 25 Notes for Math 528, Algebraic Topology II, jan25 Today: the Lefschetz Fixed point Formula (examples), and cohomology. 7.1 Examples: Lefschetz Fixed Point Formula Basic notion: A is a finitely generated abelian group, and φ : A → A is an endomorphism. The fundamental theorem of finitely generated abelian groups says that A ≈ F ⊕ T ; F ≈ Zr,T finite group. To get a trace for φ we do one of the following: • Take the trace of the linear transformation φ ⊗ 1: φ ⊗ 1 A ⊗ Q - A ⊗ Q ≈ ≈ ? ? Qr - Qr linear 43 • Take the trace of φ¯ : φ¯ A/T - A/T ≈ ≈ ? φ¯ ? Zr - Zr Lefschetz Fixed Point Theorem. If X is finite and f : X → X is a map wiht no fixed points, then λ(f) = 0. Recall: X n λ(f)= −1 Tr(f∗ : Hn(X; Q) → Hn(X; Q)) Question. Is the converse of the Lefschetz Fixed Point Theorem true? Answer: NO. Remark: The Lefschetz Fixed Point Theorem is true for coeffecients in any field F . Example. f : X → X, for a finite complex X. Consider the suspension: Sf : SX → SX. P Q Note that Sf has two fixed points: P and Q. Similarily, the k-fold suspension Skf : SkX → SkX has a sphere of fixed points. Recall: 44 ≈ - H˜i(X) H˜i+k(X) k f∗ (S f)∗ ? ≈ ? - H˜i(X) H˜i+k(X) Therefore, we can relate λ(f)toλ(Skf): i 01 2... k k+1 ... Hi(X) Z H1 H2 ... Hk Hk+1 ... k Hi(S X) Z 0000 H1 ... k k Assume that X is connected. Then λ(f) − 1=(−1) (λ(S f) − 1) ( f∗ : H0(X) → H0(X)is identity on Z) The converse of the Lefschetz Fixed Point Theorem would say that λ(g)=0⇒? g has no fixed points. We shall choos f and k such that λ(Skf) = 0 but Skf haslotsoffixedpoints. For example, start with a map f : X → X such that λ(f) = 0 and assume that k is even. Then λ(Skf)=0. This suggests the question: let f : X → X,whereX is a finite complex. What are the possible sequences of integeres k0,k1,k2,... that realize the traces f∗ : Hi → Hi? Assume 1 1 that X is connected, so that l0 = 1. Recall: there exists a map of degree l, φ : S → S such that z 7→ zl,wherez ∈ C, |z| = 1. This map has a fixed point (the complex number 1) so it is base point preserving. The (k − 1)fold suspension of φ, Skφ : Sk → Sk has degree l, and it is also base point preserving. n1 n2 nr Suppose that X = S ∨X = S ∨···∨X = S where n1 ≤ n2 ≤···≤nr. Oneachsphere nk S ,chooseaselfmapofdegreekj. Therefore, we can realize any sequence (1,k1,k2,...)by choosing a one point union of spheres. This is a kind of “cheap” example, since all of the attaching maps are trivial. 45 Now choose X = S2 ∨ S4 ∨ S6 ∨···∨S2nY = CP n = e2 ∨ e4 ∨ e6 ∨···∨e2n Here, H∗(X, Z) ≈ H∗(Y ; Q). We can realize any sequence (1,k1,k2,...,kn) for traces of self maps f : X → X. This is not the case in Y . i 01 23 4... 2n − 12n Hi(Y ) Z 0 Z 0 Z ... 0 Z Trf∗ 10k1 0 k2 0 kn 2 3 The values of k are determined completely by k1: k2 = k1, k3 = k1, and so forth. To prove this, we need to use cohomology: we need a ring structure. 7.2 Applying Cohomology For any space W and a ring R, we can put a natural graded ring structure on the cohomology groups H∗(W ; R). That is, if u ∈ Hp(W ; R)andv ∈ Hq(W ; R), the their product, u ∪ v ∈ Hp+q(W ; R). If f : W → Z is a map, then the (contravariantly) induced map f ∗ : H∗(Z, R) → H∗(W ; R) is a ring homomorphism, then there exists an identity 1 ∈ H0(X; R). Fact: H∗(CP n; Z) ≈ Z[x]/(xn+1), where x ∈ H2(CPn; Z) ≈ Z is a generator. Take any self map f : Z → X. Then we can compute the traces using cohomology. Consider the map f∗: f ∗ H2(CP n) - H2(CP n) ≈ ≈ ? ? Z - Z x - kx Here, f ∗(xl)=f ∗(x)l =(kx)l = klxl,sincef ∗ is a ring homomorphism. Therefore, the sequence of trances is (1,k,k2,...,kn). Moreover, we can realize any k. 46 7.3 History: the Hopf invariant 1 problem 3 2 3 0 2 Suppose f : S → S .Thenf∗ : H˜∗(S ) → H˜∗(S ). Question: Does f induce the 0 map in n 3 2 homotopy? Here, Πn(X) := the group of homotopy classes of maps S → X. f : S → S 3 2 induces f# :Πn(S ) → Πn(S ), [g] 7→ [f ◦ g]. In the 1930s, Heinz Hopf constructed the following map f : S3 → S2: z1 (z1,z2) 7→ z2 where z1,z2 ∈ C,z1z¯1 + z2z¯2 = 1 and the image lies in the one point compactification C ∪∞ of C. Pick x, y ∈ S2,x=6 y.Thenf −1(x),f−1(y) are both circles. Facts: 1. f# is an isomorphism for n>3. 2 2. π3S ≈ Z. 2n−1 n n 2 Let f : S → S (n ≥ 2). Then X = S ∪f e n.Then ( Zi=0,n,2n Hi(X) ≈ 0 otherwise Let y ∈ H2n(X) ≈ Z be a generator. Then x2 = x ∪ x = ky. What is k? In particular, when is there a map f : S2n−1 → Sn with k =1? k is called the Hopf invariant of f 47 7.4 Axiomatic description of Cohomology Definition: A cohomology theory is a sequence of contravariant functors from the category of CW complexes to the category of abelian groups (say h˜n,n ∈ Z) together with natural transformations h˜n(A) →∂ h˜n+1(X/Z) for every CW pair (X, A) satisfying the following axioms: 1. (homotopy axiom) If f ≈ g : X → y then f ∗ = g∗ : h˜n(Y ) → h˜n(X). 2. The following is a long exact sequence: c∗ i∗ ∂ ... - h˜n(X/A) - h˜n(X) - h˜n(A) - h˜n+1(X/A) ... To say that the transformations ∂ are natural means that if f :(X, A) → (Y,B)isa map of CW complexes, then the following diagram is commutative: c∗ i∗ ∂ ... - h˜n(X/A) - h˜n(X) - h˜n(A) - h˜n+1(X/A) - ... 6 6 6 6 f ∗ f ∗ f ∗ f ∗ c∗ i∗ ∂ ... - h˜n(Y/B) - h˜n(Y ) - h˜n(B) - h˜n+1(Y/B) - ... W → 3. Suppose X = α∈A Xα,whereA is the same index set. Let iα : Xα X be the ∗ ˜n → ˜n ∈ inclusion maps. Therefore, iα : h (X) h (Xα), α A. These maps induce an ≈ Q ˜n → ˜n isomorphism h (X) α∈A h (Xα). The first axiomatization of cohomology was done by Eilenberg and Steenrod, in the book “Algebraic Topology”. 7.5 My Project Start to read Milnor’s book: Characteristic Classes. 48 Chapter 8 January 27 Notes for Math 528, Algebraic Topology II, jan27 : 8.1 A difference between Homology and Cohomology Example. Let G be a free abelian groups on generators eα, α ∈ A. P ∈ Z Elements of G: a∈A nαeα,wherenα iandthenα are finitely nonzero. ∗ Let G ≈⊕a∈AZα.ThenG =Hom(G, Z). f : G → Z, xα = f(eα), for α ∈ A. Q Z Z Z Hom (G, )= α∈A α. A typical element in πα∈A α is a sequence (xα)α∈A, not necessarily finitely nonzero. 8.2 Axioms for unreduced cohomology Last time we gave axioms for reduced cohomology: a sequence of contravariant functors h˜n. 49 Unreduced cohomology: hn contravariant functors. hn(X, A)=h˜n(X/A) hn(X)=hn(X, ∅) = h˜n(X/∅) where X/∅ = X+ = X] point. Then there are axioms for the unreduced cohomology theory. See text. 8.3 Eilenberg-Steenrod axioms Eilenberg-Steenrod (c. 1950) in Foundations of Algebraic Topology, gave the first axiomatic treatment, including the following: ( Z n =0 Dimension Axiom: hn(pt)= . 0 otherwise The above axioms and the dimension axiom give a unique theory. Remark: there exist homology (cohomology) theories which do not satisfy the dimension axiom. (for a trivial example, homology with coefficients) An uninteresting example: Suppose h∗ is a cohomology theory satisfying the dimension axiom. Define another cohomology theory by kn = hn−m for a fixed integer m. 1960s: Atiyah, Bott, Hirzebruch constructed another cohomology theory (complex K-theory) which does not satisfy the dimension axiom: ( Z n =0, 2, 4,... Kn(pt)= 0 otherwise This is a very powerful theory and is part of the reason that Atiyah got a fields medal. Definition: The coefficients of a cohomology theory are h∗(pt). Another example of a sophisticated theory is stable cohomotopy theory; the groups are stable cohomotopy groups of spheres. 50 8.4 A construction of a cohomology theory Remark: We can develop cohomology from the axioms. But we need a construction of a cohomology theory. Start with a chain complex C∗. ∗ Definition: the cochain group C := Hom (C∗,G)whereG is an abelian group, called the coefficients. n A typical element in C := Hom (Cn,G) is a group homomorphism α : Cn → G. - ∂n+1 - ∂n - - ··· Cn+1 Cn Cn−1 ··· α ◦ ∂ α n +1 - ? G n n n+1 n Therefore, there exists a coboundary map δ : C → C , δ (α):=α ◦ ∂n+1. It is an elementary fact that δ ◦ δ =0. Definition: Hn(C; G)=kerδ/imageδ. ∗ ∗ Example: H (X; G)=H (Hom (C∗(X),G))) We can do it for a pair also: C(X) H∗(X, A; G):=H∗ Hom ,G C(A) Then the axioms follow algebraically (although there is some work to be done in order to check that this is so!) Example: Long exact sequence in homology. Suppose E is the short exact sequence of abelian groups. i j 0 - A - B - C - 0 51 Let G be an abelian group. Then there exists an exact sequence i∗ j∗ Hom (A, G) Hom (B,G) Hom (C, G) 0 with i∗(β)=β ◦ i for a homomorphism β : B → G. i j 0 - A - B - C - 0 i ∗ (β )= β β ◦ - i ? G Moreover, if E is a split exact seqnece, then i∗ :Hom(B,G) → Hom (A, G)isanepimor- phism, i.e. i∗ j∗ 0 Hom (A, G) Hom (B,G) Hom (C, G) 0 is exact. Proof: Exactness at Hom (C, G):j∗ is 1-1: Let γ : C → G be in ker j∗, i.e. γ ◦ j =0.But j : B → C is an epimorphism, so γ =0. The following is clear: imj∗ ⊆ ker i∗. Next, we show that ker i∗ ⊆ im j∗.Takeanelement β ∈ ker i∗: i j 0 - A - B - C - 0 β ◦ i β =0 γ - ? G Define γ : C → G by γ(c)=β(b), where b is any element such that j(b)=c. 52 Now assume that E is split exact: i j 0 - A - B - C - 0 That is, there exists p : C → B such that j ◦ p =idC , or equivalently, there exists q : B → A such that q ◦ i =idA. Exercise: In this case, B ≈ A ⊕ C. Then i∗ 0 Hom (A, G) Hom (B,G) is exact: q 0 - A - B - ... i α β ? G Comments. 1. If C is a free abelian group, then E is split exact. 2. Consider the short exact sequence of chain complexes: - - - - 0 C∗(A) C∗(X) C∗(C, A) 0 These chain groups are free, and therefore the sequence is split. Therefore, we have a short exact sequence of cochain complexes, ∗ ∗ ∗ 0 C w(A) C w(X) C (X,w A) 0 w w w w w w w w w w w w 0 Hom (C∗(A),G) Hom (C∗(X),G) Hom (C∗(X, A),G) 0 By general nonsense, this gives the long exact sequencne in cohomology. 53 8.5 Universal coefficient theorem in cohomology ∗ Question: what is the relationship between H (X, G)andHom(H∗(X),G)? In the first group, we dualize the chain complex and then apply homology; in the second, we apply homology to the chain complex first, and then dualize. The two groups are not isomorphic; however, There exists a group homomorphism ∗ h- H (X; G) Hom (H∗(X),G) which is onto. n Definition: of h:leth ∈ H (X; G) be a cohomology class, represented by α : Cn(X) → G: - - - - ... Cn+1(X) Cn(X) Cn−1(X) ... α ◦ δ n α +1 β =0 - ? G h(u) is to be a homomorphism Hn(X) → G. α is not unique, α + β∂n will also do. 0 Restrict α to α :ker∂n = Zn(X) → G. 0 But α (Bn) = 0 and therefore there exists a homomorphism Hn(X) → G. Define h(u)tobe this homomorphism. Exercise: h is onto. h is not always an isomorphism. Example: - - ∗ n h- n - 0 ker h H (RP ) Hom (H∗(RP ), Z) 0 n We will compute H∗(RP ) cellularly: ... - C - C - C - C - 0 w3 w2 w1 w0 w w w w w w w w w w w w w w w w ×2 0 ×2 0 ... - Z - Z - Z - Z 54 and therefore Z if i =0 Z2 if i is odd; 1 ≤ i ≤ n R n ≤ ≤ Hi( P )=0ifi is even; 2 i n 0ifi = n is even Z if i = n is odd ∗ Dualize the chain complex C∗: C =Hom(C∗, Z) 0 Cn ··· C2 C1 C0 0 w w w w w w w w w w w w w w w w w w w w Z Z Z Z Therefore, Z if i =0 Hi(RP n)= 0ifi odd, 1 ≤ i ≤ n Z2 if i even, 1 ≤ i ≤ n Note that these are very different from the homology groups! So, i n h- n - H (RP ) Hom (Hi(RP ), Z) 0 is the zero map for 0 ≤ i ≤ n. Therefore Hi(RP n) ≈ ker h in these dimensions. 55 56 Chapter 9 February 1 Notes for Math 528, Algebraic Topology II, feb1 9.1 Comments on the Assignment . #2 p.184: Use the Lefschetz Fixed Point Theorem to show that a map f : §n →§n has a fixed point unless deg f = the degree of the antipodal map. Fairly elementary: just look at what the LF P number for f is λ(f)=1+(−1)n deg f, and the theorem relates this to fixed points. #4 p.184: Suppose that X is a finite simplicial complex, f : X → X is a simplicial homeo. Let F = {x ∈ X|f(x)=x). Show that λ(f)=X(F ). We may assume that F is a subcomplex of X (if not, one can subdivide). #13 p.206: hX, Y i is the set of base point preserving homotopy classes of maps f : X → Y .We are to show that hX, K(G, 1)i→H1(X, G) is an isomorphism, where G is some abelian group. A space Q is a K(Π, 1) if Π1(W )= Π and the universal covering space W˜ is contractible (⇐⇒ Πn(W˜ )=1). 1 map hX, K(G, 1)i→H (X; G), (f : X → K(G, 1) induces 7→ f∗(H1(x) → H1(K(G, 1)). H1(Y ) ≈ Π1(Y )/[Π1(Y ), Π1(Y )], and H1(K(G, 1)) ≈ G/[G, G]=G. 57 The UCT in Cohomology says: - - n h- - 0 ker h H (Y ; G) Hom (Hn(y),G) 0 1 ker h =0ifn = 1. Therefore H (Y ; G) ≈ Hom (H1(Y ),G)). #3(a,b), p.229 Use cup products to show that there is no map f : RP n → RP m,wheren>m, ∗ 1 m 1 n ∗ k inducing a nontrivial map f : H (RP , Z2) → H (RP , Z2). Here, H (RP ; Z2)= k+1 1 k ∗ I m Z2[x]/(x )wherex ∈ H (RP ; Z2). (If this were to exist, f : H (RP ; Z2) → ∗ n H (RP ; Z2) is a ring homomorphism...) What is the corresponding results for map n m CP → CP ?(wouldhavetobeH2... (b) says: Prove the Borsuk-Ulam theorem: if f : §n → Rn then f(−x)=f(x) for some x. Suppose not. Then f(−x) =6 f(x). Define g : Sn →§n−1, f(x) − f(−x) g(x)= |f(x) − f(−x)| .Theng(−x)=−g(x),so g induces a map RP n → RP n−1. Can we find a lift h˜ in the diagram below? g Sn - Sn−1 - ˜ p h p ? ? h RP n - RP n−1 #4 p.229 f : CP n → CP n. Compute λ(f)usingH∗. ∗ ∞ 2 1 2 #5 p.229 H (RP ; Zm) ≈ Zm[α, β]/(2α, 2β,α )whereα ∈ H ,β ∈ H and m>2. There is a corresponding statement in the book for m =2. #12 p.229 X = S1 × CP ∞/S1 × pt. Y = S3 × CP ∞. Show that H∗(X; Z) ≈ H∗(Y ; Z). actually true for any coefficients. One can detect the difference between these using the Steenrod algebra. This problem will require some reading; should use the Kunneth formula. 9.2 Proof of the UCT in cohomology There exists a short exact sequence - - n h- - 0 ker h H (X; G) Hom(Hn(X),G)) 0 ? Ext(Hn−1(X),G) 58 Today’s order of business: define Ext, and identify it with the kernel. Definition: (the functor Ext) Let A,G be abelian groups. Choose free abelian groups F1,F0 such that there exists a presentation - ∂ - - - 0 F1 F0 A 0 i.e. this is an exact sequence, called a free presentation of A. Apply the functor Hom (·,G): δ∗ 0 coker ∂∗ Hom (F ,G) Hom (F ,G) Hom (A, G) 0 w 1 0 def Ext(A, G) Remark: This defninition of Ext(A, G) suggests that it does not depend on the choice of the resolution. One needs the following lemma to se this: Lemma: Consider 2 resolutions and a group homomorphism φ : A → A0: - ∂ - - - 0 F1 F0 A 0 φ 0 0 ? - 0 ∂ - 0 - - 0 F1 F0 A 0 0 Then there exists a chain map α : F∗ → F∗ and moreover, any two such are chain homotopic. Proof: : Diagram chasing: - ∂ - - - 0 F1 F0 A 0 α1 α0 φ ? 0 ? 0 ? - 0 ∂ - 0 - - 0 F1 F0 A 0 We must find α0, α1 making this diagram commute. To define α0, pick a generator x0 ∈ F0. 0 ∈ 0 Look at φ(X0). There exits x0 F0 (not necessarily unique) such that 0 0 φ(x0)= (x0). 0 → 0 Define α(x0)=x0. Extend linearly to a map α : F0 F0. 59 0 0 Now choose a generator x1 ∈ F1. α0∂(x1)=0(since α0∂ = φ∂ =0).Thusα0∂(x1) ∈ 0 0 ∈ 0 0 0 0 ker , so there exists a unique x1 F1 such that ∂ (x1)=α0∂(x1). Define α1(x1)=x1,and extend linearly. 0 Now, suppose that α, β : F∗ → F∗ are chain maps; we would like to say that α, β are chain homotopic. - ∂ - - - 0 F1 F0 A 0 0 D α1 − β1 α0 − β0 φ ? 0 ? 0 ? - 0 ∂ - 0 - - 0 F1 F0 A 0 → 0 Diagram chasing (exercise): There exists D0 : F0 F1 such that 0 ∂ D0 = α0 − β0,D0∂ = α1 − β1. D0 is the chain homotopy from α to β.Thisimpliesthatα∗ = β∗ on homology. Exercise: Ext(A, G) is well defined. Universal Coefficient Theorem: There exists a short exact sequece - - n h- - 0 Ext(Hn−1(X),G) H (X; G) Hom(Hn(X),G)) 0 Moreover, this exact sequence is functorial in X and G, and is split (but the splitting is not functorial in X). n Thus H (X; G) ≈ Hom (Hn(X),G) ⊕ Ext(Hn(X),G). This tells us how to compute coho- mology. 9.3 Properties of Ext(A, G) • If A is free abelian then Ext(A, G) = 0: choose a resolution - ∂ - A - - 0 F1 F0 A 0 such that F0 = A, = idA,F1 =0. 60 • Ext(A1 ⊕ A2,G)=Ext(A1,G) ⊕ Ext(A1,G) (splice 2 resolutions). • Ext(Zn,G) ≈ G/nG: 0 - Z×n - Z - Z - 0 w w n F1 F1 is a resolution of Zn. So Ext is defined by the exact sequence 0 Ext(Zn,G) Hom (Z, G) Hom (Z, G) Hom (Zn,G) 0 ?≈ ?≈ ?≈ o G/nG G G • If A = Zn ⊕ T ,whereT is a finite group, then Ext(A, G) ≈ Ext(T,G). ≈ ⊕···⊕ | | | To see why, decompose T Zn1 Znk , n1 n2 ... nk. Therefore, Ext(A, G) ≈ G/n1G ⊕···⊕G/nkG. Example: n ≈ Z ⊕ Z H (X; Z) Hom(Hn(X), ) ExtHn−1(X), ) (9.1) ≈ (free part of Hn(X) ⊕ torsion part of Hn−1(X) (9.2) 61 62 Chapter 10 February 3 Notes for Math 528, Algebraic Topology II, feb3 10.1 Naturality in the UCT Because the Universal Coefficient Theorem is natural, A map f : X → Y induces maps as follows: - - i h- - 0 Ext(Hi−1(X),G) H (X; G) Hom (Hi(X),G) 0 6 6 6 ∗ Ext(f∗,id) f Hom (f∗,id) - - i h- - 0 Ext(Hi−1(Y ),G) H (Y ; G) Hom (Hi(Y ),G) 0 There also exist splittings i i sX :Hom(Hi(X),G) → H (X; G) sX :Hom(Hi(X),G) → H (X; G) but the splittings are not natural inX. Example: #11, p.205. A Moore space X = M(Zm,n), i.e. ( Zm i = m H˜i(X) ≈ 0 otherwise 63 n n+1 n n X = S ∪f e , where the attaching map f : S → S has degree m =0.6 The cellular chain complex is: ------0 Cn+1 Cn 0 ... 0 C0 0 0 - Z - Z - 0 - ... - 0 - Z - 0 ×m So ( Zm i = n H˜i(X)= 0 otherwise. 0 C ∗ +1 C∗ 0 ... 0 C∗ 0 wn wn w0 Z Z Z ×m So ( Z i = n +1 H˜ i(X)= 0 otherwise Let c : X → Sn+1 be the map that collapses the n-skeleton. n+1 ∗ i n+1 i c∗ : H˜i(X) → H˜i(S ) is always 0, but c : H˜ (S ) → H˜ (X) is not 0. n+1 Put X = M(Zm,n) in the diagram, Y = S , f = c, i = n +1. 10.2 Proof of the UCT Consider the short exact sequence of chain complexes: - - - - 0 Z∗ C∗ B¯∗ 0 where in degree n this is - i - ∂ - - 0 Zn Cn B¯n−1 0 • The short exact sequence of chain complexes is split, because all groups are free abelian. 64 • The boundary operators in Z∗ and B¯∗ are 0. Now take Hom (·,G): 0 Hom (Z∗,G) Hom (C∗,G) Hom (B¯∗,G) 0 This is a short exact sequence of cochain complexes, so we get the long exact sequence: - ∂- - n - ∂- - ... Hom (Zn−1,G) Hom (Bn−1,G) H (C; G) Hom (Zn,G) Hom (Bn,G) ... n the image H (C; G) → Hom (Zn,G)=the{α : Zn → G : α|Bn =0} =Hom(Hn(C),G). Now let us find Coker Hom (Zn−1,G) → Hom (Bn−1,G). Consider the exact sequence - - - - 0 Bn−1 Zn−1 Hn−1(C) 0 Apply Hom (·,G): 0 Ext(Hn−w1(C),G) Hom (Bn−1,G) Hom (Zn−1,G) ... coker Exercises: Check the details. 10.3 Some homological algebra Let R be a ring with 1, not necessarily commutative. Definition: a left R–module is an abelian group M together with scalar multiplication R × M → M,(r, m) 7→ rm, satisfying: (1) (r1 + r2)m = r1m + r2m (2) r(m1 + m2)=rm1 + rm2 (3) (r1r2)m = r1(r2m)) 65 (4) 1m = m Ditto for right R-modules. Examples: (1) If M is a left R-module, then it becomes a right R-module by the defnition mr = r−1m. (2) R = Z:AleftR-module = an abelian group = a right R-module. (3) R = F , a field. Then modules are vector spaces over F . (4) The integral group ring of a group G. R := Z[G] = the set of all integral linear combinations of elements of G: n1g1 + n2g2 + ···+ nrgr. (5) The group algebras: k[G]. (6) Given left R-modules A, B, we can consider the abelian group Hom R(A, B)ofR- module homomorphisms φ : A → B.Exercise:IsHomR(A, B)anR-module? (it turns out not to be) (7) Let A be a left R-module, and B be a right R module. Then B ⊗R A is the free abelian group on pairs (b, a) (for any b ∈ B,a ∈ A), modulo the relations: • (b1 + b2,a)=(b1,a)+(b2,a) • (b, a1 + a2)=(b, a1)+(b, a2) (so far, this is the tensor product over the integers) • (br, a)=(b, ra). Notation: b ⊗ a is the class of (b, a)inB ⊗R A. Remark: B ⊗R A = B ⊗Z A/(relations br ⊗Z a = b ⊗Z ra). 10.4 Group homology and cohomology Let R = Z[G], and A be a left R-module. We will define H∗(G; A). Take a free R-resolution of Z. That is, find free R-modules Fi (i ≥ 0) and an exact sequence; ∂2 - ∂1 - - - ... F1 F0 Z 0 (We say that is an augmentation). 66 Remark: the kernel of may not be a free R-module: choose some F1 and ∂1 : F1 → ker , and continue. The complex may be infinite. P P Example: Suppose F0 = R = Z[G]. Define : F0 → Z by ( ngi)= ni.Thenker =the augmentation ideal I[G]. Choose F1 and ∂1 such that F1 is a free R-module, and ∂1 : F1 → F0 has image I[G]. Etc. Now F∗ is a chain complex with all homology groups equal to 0 (in positive dimension s). ∗ ∗ Define H (G; A)=H (Hom R(F∗,A)). Exercise: Show that this definition does not depend on the choice of free resolution F∗.The argument is similar to the argument about chain homotopies earlier. Suppose that B is a right R-module. Then H∗(G; B):=H∗(B ⊗R F∗). Example. Suppose that p : X˜ → X is a regular covering such that: • X˜ is contractible. • π1(X)=G. X is called a K(G, 1) space. We will relate C∗(X˜)toaresolutionofZ. Let us use simplicial homology, and consider Cn(X). n n Let σ =some n-simplex (i.e. a generator of Cn(X)). So σ :∆ → X,where∆ is the standard n-simplex. Then there exists a liftσ ˜ :∆n → X˜, making the following diagram commute: X˜ - σ p ? ∆n - X σ Choose some particular liftσ ˜.ThenG acts on X˜ by covering transformations. Therefore, the set of lifts of σ is {g ◦ σ˜|g ∈ G}. Let σ1,...,σk be the n-simplices in X. Make a choice of a liftσ ˜i for each σi.Now,atypical ˜ element of CnX is an integral linear combination of the generators gσ˜i,1≤ i ≤ r, g ∈ G. So, Cn(X˜ is a free Z[G]-module of rank k. 67 Since X˜ is contractible, we have a free Z[G]-resolution of Z: - ∂2- ∂1- - - ... C2(X˜) C1(X˜) C0(X˜) Z 0 ∗ ∗ Therefore, H∗(G; B)=H∗(X˜; B), and H (G; A)=H (Hom (C∗(X˜),Z)). 10.5 Milnor Construction G = a discrete group. Take the join G ∗ G. Definition: The join of two spaces X, Y is (x, y, 0) = (x0,y,0) X ∗ Y ∗ I (x, y0, 1) = (x, y0, 1) Take X, Y ⊆ Rhuge, X ∩ Y = ∅. Then the picture is X x y y0 y line tx +(1− t)y G ×G = E2(G). There exists a diagonal action of G on E2(G) (Extend linearly to the lines). ∗ ∗ ∗ Iterate this construction: Ee(G)=(G G) G,etc.ThusP En+1(G)=(En(G)) G). A typical point in EN (G)ist1g1 + ···+ tngn,whereti ≥ 0and ti =1. BG = EG/G = K(G, 1). Example: G = Z2. 68 1 1 2 E2G = S E3G = S ∗ G = S n−1 ∞ ∞ EnG = S , E∞G = S is contractible. Then BG = RP . Theorem: EG is contractible. 69 70 Chapter 11 February 8 Notes for Math 528, Algebraic Topology II, feb8 11.1 A Seminar Joseph Maher. Questions: 1. Which finite groups G admit a fixed-point free action on S3? (still open) 2. Which finite groups G admit a fixed-point free linear action on S3? i.e. when there are representations ρ : G → SO(4) such that ρ(g) does not have +1 as an eigenvalue for g ∈ G, g =6 e? Recall that SO(4) is the group of 4 × 4 orthogonal matrices of determinant 1. It is the group of rotations of the 4 sphere. Eigenvalue +1 means that ρ(g) · v = v for some eigenvector v. This question has been answered, 60 years ago. 3 3. Suppose that G admits a fixed-point free linear action on S (e.g. G ≈ Zn). Suppose there exists some topological action of G on S3, without fixed points. Is the topological action conjugate to a linear one? 71 linear S3 - S3 6 h ≈ h−1 ≈ ? top. action S3 - S3 Livesay: circa 1950. Yes, for G = Z2. Maher: Yes for G = Z3. (this is the subject of the seminar, likely) Perelman: Announced a proof of the Thurston Geometrization Conjecture. This im- plies the Poincare conjecture, and yes to (3). 11.2 Products There exist three types of products. 1. Cup product. Hi(X; R) × Hj(X; R) - Hi+j(X; R) (u, v) - u ∪ v It will be bilineaer in u, v: u ⊗ v 7→ u ∪ v. 2. Cross product. Hi(X; R) × Hj(X; R) - Hi+j(X × Y ; R) (u, v) - u × v 3. The Cap product. i - Hi(X; R) × H (X; R) Hi−j(X; R) (u, v) - u ∩ v 72 Motivation. Suppose we try to find a “cup product” in homology: Hi(X; R) × Hi(X; R) → Hi+j(X; R). Indeed, there exists a cross product in homology, Hi(X; R) × Hi(X; R) → Hi+j(X × Y ; R). i j i × j × Let eα be an i-cell in X and eβ beaj-cell in Y .Theneα eβ is an i + j cell in X Y .The theory says that this passes to a product in homology. Now specialize to X = Y . - i+j ?- Hi(X; R) × Hj(X; R) H (X × X; R) Hi+j(X; R) There’s no really good map from X × X to X, however, which doesn’t throw away a lot of information. In cohomology: × ∆∗ Hi(X; R) × Hj(X; R) - Hi+j(X × X; R) - Hi+j(X; R) where ∆ : X → X × X is the diagonal. Definition of the cup product: Hi(X; R) ⊗ Hj(X; lR) → Hi+j(X; R). u ∈ Hi(X; R),v ∈ Hj(X; R). These are cohomology classes, and can be realized by cocycle maps: u is realized by α : Ci(X) → R v is realized by β : Cj(X) → R Then α ∪ β will be the homomorphism α ∪ β : Ci+j(X) → R defined by what it does on singular simplices: i+j i+j Let σ :∆ be a singular i + j simplex. δ will have vertices v0,v1,...,vi+j.Then ∪ | · α β(σ)=α(σ [v0,...,vi]) β(σ[vi,...,vi+j ]) i where the product is in R. Here, [v0,...,vi]=δ ,[vi,...,vi+1,...,vi+j]. Exercise: Let δ denote a coboundary operator δ : Cn → Cn−1.Then δ(α ∪ β)=δ(α) ∪ β +(−1)iα ∪ δ(β). From this it follows that the cup product of cochains yields a cup product in cohomology, Hi(X; R) ⊗ Hj(X; R) - Hi+j(X; R) (u, v) - u ∪ v 73 Properties of the cup product: 1. u ∪ v is bilinear in u and v. 0 2. There exists an identity 1 ∈ H (X; R), namely the class that takes the value idR on every point x ∈ X.HereC0(X) is the free abelian group on the points of X,and 0 C (X)=Hom(X0(X),R). 3. If u ∈ Hi(X; R), v ∈ Hj(X; R), and f : Y → X,thenf ∗(u ∪ v)=f ∗(u) ∪ f ∗(v). 4. u ∈ Hi(X; R),v∈ Hj(X; R). Then u ∪ v =(−1)ijv ∪ u 5. We can make cohomology into a graded–commutative associative ring with identity: ∗ ⊗∞ i H (X; R)= i=0H (X; R). The ring structure is the cup product. (Graded–comutative means precisely that prop- erty (4) holds) 6. Moreover, if f : U → X is a map, then f ∗ : H∗(X; R) → H∗(Y ; R) is a ring homomor- phism. 7. There exist relative versions: Hi(X, A; R) ⊗ Hj(X, B; R) - Hi+j(X, A ∪ B; R) 2n−1 n n 2n Example: The Hopf invariant one problem: f : S → S , X = S ∪f e , n>1. Then ( Zi=0,n,2n Hi(X; Z)= 0 otherwise Let u ∈ Hn(X; Z),v ∈ H2n(X;Z) be generators. Then u ∪ u ∈ H2n(X; Z). Therefore, u ∪ u = kv,wherek is called the Hopf invariant of f. k = H(f). k is unique up to sign; it depends on v but not u. Theorem: H(f)=0ifn is odd. Proof: u ∪ u =(−1)n·nu ∪ u = −u ∪ u,sou ∪ u =0(Z has no 2-torsion). Hopf: There exist elements of Hopf invariant 1 for n =2, 4, 8. 74 Example (Thursday) We apply the mod 2 Steenrod algebra to the Hopf Invariant one prob- lem. The conclusion is: Theorem: If f : S2n−1 → Sn has odd Hopf invariant, then n is a power of 2. Idea: u 7→ u2 = u ∪ u Hn(X) - H2n(X) - - H∗(X)=0 n n 2n i n Sq : H (X; Z2) → H (X; Z2) is the cup square. There exist other squares: Sq : H → n+i H ,coefficientsinZ2.Ifn isnotapwoerof2,then Xn−1 n j n−k Sq = ajSq · Sq . j=1 Definition of the Cross product: Hi(X; R) ⊗ Hi(Y ; R) - Hi+j(X × Y ; R) u ⊗ v - u × v p1 : X × Y → X is projection onto X. p2 : X × Y → Y is projection onto Y . ∗ ∈ i × ∗ ∈ j × Here p1(U) H (X Y ; R), p2(U) H (X Y ; R). Then × ∗ ∪ ∗ u v := p1(u) pw(v). j Definition of the cap product: Hi(X; R)⊗H (X; R) → Hi−j(X; R). Take a singular i-simpex σ ∈ C (X), and a singular cochain α : C (X) → R.Thenσ ∩ α = α(σ| ) · σ| . i j | [{zv0,...,vj ]} | [v{zj,...,vi}] ∈R ∈Ci−j (X) j This passes to a cap product Hi ⊗ H → Hi−j. 75