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1. Cohomology Let X Be a Space and S CUP PRODUCTS, COHOMOLOGY RINGS OF SPACES AND THOM ISOMORPHISMS [04/02/2003] ANDREW BAKER 1. Cohomology Let X be a space and S¤(X) the graded abelian group of singular chains on X, so Sn(X) is the free abelian group generated by the singular n-simplices ∆n ¡! X. With the usual boundary homomorphism d,(S¤(X); d) is a chain complex and H¤(X) = H¤(S¤(X); d): Dually, for any abelian group M we can from a cochain complex (S¤(X; M); ±), where n ¤ S (X; M) = HomZ(Sn(X);M); ± = d : The cohomology of X with coefficients in M is then the cohomology ¤ ¤ H (X; M) = H¤(S (X; M); d): This is a covariant homotopy invariant functor in X. For each X, H¤(X; M) is also a covariant functor in M and given a short exact sequence of abelian groups 0 ! L ¡! M ¡! N ! 0 there is an induced long exact sequence ¢ ¢ ¢ ¡! Hn¡1(X; N) ¡! Hn(X; L) ¡! Hn(X; M) ¡! Hn(X; N) ¡! Hn+1(X; L) ¡! ¢ ¢ ¢ which is natural with respect to such short exact sequences. 2. External products and the Eilenberg-Zilber Theorem Now suppose that we have two spaces X; Y and want to calculate H¤(X £ Y ). As an abelian n group, S¤(X £ Y ) has a basis consisting of the singular simplices ∆ ¡! X £ Y , but each of these is equivalent to a pair of singular simplices ∆n ¡! X and ∆n ¡! Y . On the other hand we can also consider the tensor product S¤(X) ­ S¤(Y ) where M [S¤(X) ­ S¤(Y )]n = Sk(X) ­ Sn¡k(Y ); k and a suitable boundary homomorphism is defined. This is much bigger than S¤(X £ Y ) since it has a basis consisting of elements of the form ® ­ ¯, where ®: ∆r ¡! X and ¯ : ∆s ¡! Y are singular simplices. Theorem 2.1 (Eilenberg-Zilber Theorem). There are inverse chain homotopy equivalences Φ / S¤(X) ­ S¤(Y ) o S¤(X £ Y ) Ψ Moreover these are unique up to chain homotopy and natural in X and Y . 1 2 ANDREW BAKER Given this, we can now form an external multiplication of S¤(X) and S¤(Y ) into S¤(X £ Y ). Dually, given any commutative ring R with multiplication ¹: R ­ R ¡! R, we can pair ® 2 Sp(X; R) and ¯ 2 Sq(Y ; R) to ¹ ± ® ­ ¯ 2 HomZ(Sp(X)­Sq(Y );R): Z Extending up to a graded homomorphism we obtain an external product ¤ ¤ S (X; R)­S (Y ; R) ¡! HomZ(S¤(X)­S¤(Y );R): Z Z Then we can precompose this with Ψ to form the external product ¤ ¤ t: S (X; R)­S (Y ; R) ¡! HomZ(S¤(X £ Y );R): Z These are both cochain maps. Now specialising to the case Y = X we can use the diagonal ∆: X ¡! X £X and its induced map ∆¤ to form the cup product ¤ ¤ ¤ [ = t ± ∆ : S (X; R)­S (X; R) ¡! HomZ(S¤(X);R); ® ­ ¯ 7¡! ® [ ¯: Z Again, this is a cochain map. Theorem 2.2. If R is a commutative ring then the cup product turns H¤(; R) into a graded commutative ring valued contravariant functor on the homotopy category of spaces, i.e., if u 2 Hp(X; R) and v 2 Hq(X; R) then v [ u = (¡1)pqu [ v: Furthermore, if ': R ¡! R0 is a ring homomorphism then there is a natural transformation of graded ring valued functors extending ' which agrees with ' when evaluated on a point ¤. We usually just write uv for u [ v. Remark 2.3. In fact, H¤(; R) is really an R-algebra valued functor, where for each space X, the projection to a point X ¡! ¤ induces the unit R =» H¤(¤; R) ¡! H¤(X; R): Notice that if R has characteristic 2 (i.e., 2 = 0 in R) then H¤(X; R) is commutative in the usual sense. In general, if p; q are odd and u 2 Hp(X; R), v 2 Hq(X; R), then vu + uv = 0; 2u2 = 0: In particular, if Hp(X; R) has no 2-torsion, u2 = 0: Example 2.4. Let Sn be the unit sphere in Rn+1 where n > 1. Then ¤ n H (S ; R) = ΛR(sn); n n the exterior algebra over R on a generator sn 2 H (S ; R). Proof. Recall that ( R if k = 0; n; Hk(Sn; R) =» ¤ 0 otherwise: Example 2.5. Suppose that X and Y are two spaces and that prX : X £ Y ¡! X and p q prY : X £ Y ¡! Y are the product maps. Then for u 2 H (X; R) and v 2 H (Y ; R) we have ¤ p ¤ q ¤ ¤ prX u 2 H (X £Y ; R) and prY v 2 H (X £Y ; R), so can form their cup product prX u[prY v 2 p+q ¤ ¤ H (X £ Y ; R). If H (X; R) and H (Y ; R) are free R-modules with bases furg and fvsg say, ¤ ¤ ¤ then H (X £ Y ; R) is also free with basis fprX ur [ prY vsg. CUP PRODUCTS, COHOMOLOGY RINGS AND THOM ISOMORPHISMS 3 Example 2.6 (Complex and quaternionic projective spaces). For 1 6 n 6 1, ( Z if k 6 n is even; Hk(CPn) = Hk(CPn; Z) = 0 otherwise: Then for any commutative ring R, H¤(CPn; R) = R[x]=(xn+1); where x 2 H2(CPn; R) is a generator. Similarly, ( Z if k 6 n is divisible by 4; Hk(HPn) = Hk(HPn; Z) = 0 otherwise: Then for any commutative ring R, H¤(HPn; R) = R[y]=(yn+1); where y 2 H4(HPn; R) is a generator. Example 2.7 (Real projective spaces). Recall that for 1 6 n 6 1, 8 >Z if k = 0; > > <>Z=2 if n and k are even and k 6 n; Hk(RPn) = Hk(RPn; Z) = Z=2 if n is odd or 1, k is even and k < n; > > >Z if n is odd and k = n; > :0 otherwise: (a) If R = F2 or any F2-algebra, then for n 6 1, H¤(RPn; R) = R[z]=(zn+1); where z 2 H1(RPn; R) =» R is a generator. (b) If n 6 1 then ( Z[y]=(2y; yn+1) if n = 1 or n < 1 is even; H¤(RPn; Z) = n+1 Z[y]=(2y; y ) ­ ΛZ(wn) if n < 1 is odd; 2 n » n n » where y 2 H (RP ; Z) = Z=2 (and wn 2 H (RP ; Z) = Z if n is odd) are generators. m n m n Application 2.8. Let f : RP ¡! RP be a map such that f¤ : ¼1(RP ) ¡! ¼1(RP ) is non-trivial . Then m 6 n. m n Proof. The Hurewicz Theorem implies that f¤ : H1(RP ; Z=2) ¡! H1(RP ; Z=2) is also non- ¤ 1 n 1 m 1 k zero and therefore so is f : H (RP ; F2) ¡! H (RP ; F2). Writing zk 2 H (RP ; F2) for the ¤ generator, we have f zn = zm and so ¤ m m f zn = zm 6= 0: m If m > n then zn = 0, giving a contradiction, so we must have m 6 n. ¤ A map f : Sm ¡! Sn is called antipodal if for all x 2 Sm, f(¡x) = ¡f(x): Such a map induces a map f : RPm ¡! RPn fitting into a commutative diagram f Sm ¡¡¡¡! Sn ? ? ? ? y y f RPm ¡¡¡¡! RPn 4 ANDREW BAKER m n and it can be shown that f ¤ : ¼1(RP ) ¡! ¼1(RP ) is non-trivial. Hence we have Application 2.9. Let f : Sm ¡! Sn be an antipodal map. Then m 6 n. Application 2.10 (Borsuk-Ulam). If n > 1, for every map g : Sn ¡! Rn, there is an x 2 Sn for which g(¡x) = g(x). Proof. If not then g(¡x) 6= g(x) for every x. The map g(¡x) ¡ g(x) f : Sn ¡! Sn¡1; f(x) = ; jg(¡x) ¡ g(x)j is an antipodal map, contradicting 2.9. ¤ Calculating cup products is not very straightforward in general and tends to require geometric input rather than just homotopy theoretic input. In the next section we discuss some important examples of the latter type. 3. The Hopf invariant Let f : S4n¡1 ¡! S2n be a map (here n > 1). The mapping cone Cone(f) is a 2-cell complex, 2n 4n Cone(f) = S [f e ; whose cohomology has the form ( Z if k = 0; 2n; 4n; Hk(Cone(f)) = 0 otherwise: 2n 4n Choose generators u2n 2 H (Cone(f)) and u2n 2 H (Cone(f)). We can ‘normalise’ these by requiring that in the long exact sequence associated with the cofibration sequence S2n ¡! Cone(f) ¡! S2n, we have 2n =» 2n 2n H (Cone(f)) ¡! H (S ); u2n Ã! s2n 4n 2n =» 4n H (S ) ¡! H (Cone(f)); s4n Ã! u4n: Then we must have 2 u2n = h(f)u4n; 2n where h(f) 2 Z depends on the homotopy class of f, i.e., is well defined on ¼4n¡1S . h(f) is the Hopf invariant of f. Now recall that for a map `: Sm ¡! Sm, the degree of ` is defined to be the integer deg ` for which ¤ m m m ¤ ` : H (S ) ¡! Hm(S ); ` (sm) = (deg `)sm: We could equally well have used homology here. 2n Proposition 3.1. For n > 1, the Hopf invariant gives a function h: ¼4n¡1S ¡! Z with the following properties. 2n (a) h: ¼4n¡1S ¡! Z is a group homomorphism. (b) If g : S2n ¡! S2n and k : S4n¡1 ¡! S4n¡1 are maps, then h(g ± f ± k) = (deg g)2(deg k)h(f): 2n (c) There is an element w2n 2 ¼4n¡1S for which h(w2n) = 2.
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