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), δ), where

n ∗ S (X; M) = HomZ(Sn(X),M), δ = 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 α ⊗ β, where α: ∆r −→ X and β : ∆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 µ: R ⊗ R −→ R, we can pair α ∈ Sp(X; R) and β ∈ Sq(Y ; R) to

µ ◦ α ⊗ β ∈ 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); α ⊗ β 7−→ α ∪ β. 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 ∈ Hp(X; R) and v ∈ Hq(X; R) then v ∪ u = (−1)pqu ∪ v. Furthermore, if ϕ: R −→ R0 is a ring homomorphism then there is a natural transformation of graded ring valued functors extending ϕ which agrees with ϕ 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 ∈ Hp(X; R), v ∈ 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 ∈ 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 ∈ H (X; R) and v ∈ H (Y ; R) we have ∗ p ∗ q ∗ ∗ prX u ∈ H (X ×Y ; R) and prY v ∈ H (X ×Y ; R), so can form their cup product prX u∪prY v ∈ p+q ∗ ∗ H (X × Y ; R). If H (X; R) and H (Y ; R) are free R-modules with bases {ur} and {vs} say, ∗ ∗ ∗ then H (X × Y ; R) is also free with basis {prX ur ∪ prY vs}. CUP PRODUCTS, COHOMOLOGY RINGS AND THOM ISOMORPHISMS 3

Example 2.6 (Complex and quaternionic projective spaces). For 1 6 n 6 ∞, ( 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 ∈ 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 ∈ H4(HPn; R) is a generator. Example 2.7 (Real projective spaces). Recall that for 1 6 n 6 ∞,  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 ∞, 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 ∞, H∗(RPn; R) = R[z]/(zn+1), where z ∈ H1(RPn; R) =∼ R is a generator. (b) If n 6 ∞ then ( Z[y]/(2y, yn+1) if n = ∞ or n < ∞ is even, H∗(RPn; Z) = n+1 Z[y]/(2y, y ) ⊗ ΛZ(wn) if n < ∞ is odd, 2 n ∼ n n ∼ where y ∈ H (RP ; Z) = Z/2 (and wn ∈ 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∗ : π1(RP ) −→ π1(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 ∈ 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 ∈ 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 ∗ : π1(RP ) −→ π1(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 ∈ 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) = , |g(−x) − g(x)| 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 ∈ H (Cone(f)) and u2n ∈ 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 ←→ s2n 4n 2n =∼ 4n H (S ) −→ H (Cone(f)); s4n ←→ u4n. Then we must have 2 u2n = h(f)u4n, 2n where h(f) ∈ Z depends on the homotopy class of f, i.e., is well deﬁned on π4n−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: π4n−1S −→ Z with the following properties. 2n (a) h: π4n−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 ∈ π4n−1S for which h(w2n) = 2. CUP PRODUCTS, COHOMOLOGY RINGS AND THOM ISOMORPHISMS 5

Properties (a) and (b) follow from the deﬁnition. For (c) here is an explicit construction of such a map w2n. Recall that Sm × Sm has a cell structure of the following form: m m m m 2m S × S = (S ∨ S ) ∪wem S . m m m The fold map ∇: S ∨ S −→ S can be composed with the attaching map wem to give a map 2m−1 m wm = ∇ ◦ wem : S −→ S and there is a map m m ∇e : S × S −→ Cone(wm) extending ∇. It easy to see that H2m(Sm × Sm) = H2m(Sm) ⊕ H2m(Sm) m with generators obtained by pulling back the generator sm under the two projections to S . Then we have ∗ ∇e (sm) = (sm, sm). Also, ∗ 2m m m =∼ 2m ∇e : H (S × S ) −→ H (Cone(wm)). Now take m = 2n. Then an easy calculation in H4n(S2n × S2n) =∼ Z gives e ∗ 2 e ∗ 2 ∇ (s2n) = ∇ (s2n) 2 = (s2n, s2n) 2 = [(s2n, 0) + (0, s2n)]

= 2(s2n, 0) ∪ (0, s2n) = 2 × (generator), since (sm, 0) ∪ (0, sm) = (generator). Using this it follows that h(w2m) = 2. Question 3.2. For which n is there a map f : S4n−1 −→ S2n with Hopf invariant 1, h(f) = 1 ? Here are some examples. For n = 1, CP2 = Cone(η) where η : S3 −→ S2 can be constructed geometrically as the complex Hopf map. Then from Example 2.6, ∗ 2 3 H (CP ) = Z[u2]/(u2), so we have h(η) = 1. This construction depends on existence and properties of the complex numbers. For n = 2, HP2 = Cone(ν) where ν : S7 −→ S4 can be constructed geometrically as the quaternionic Hopf map. Then from 2.6, ∗ 2 3 H (CP ) = Z[u2]/(u2), we have h(η) = 1. This construction depends on existence and properties of the quaternions. If n = 4, there is a Cayley projective plane OP2 = Cone(σ) where σ : S15 −→ S8 is a certain map. Then ∗ 2 3 H (OP ) = Z[u8]/(u8) and we have h(σ) = 1. This question is deﬁnitively answered in the following major result. Theorem 3.3 (Adams’ Hopf invariant 1 theorem). If n 6= 1, 2, 4 there are no maps of Hopf invariant 1. This also implies that the only spheres that are H-spaces (i.e., groups up to homotopy) are S0, S1, S3 and allowing non-associativity, S7. 6 ANDREW BAKER

4. Thom isomorphisms Let ξ ↓ B be vector bundle of dimension n over a connected space B. Let the total space be E(ξ) and view B as a subspace by identifying it with its image under the 0-section B −→ E(ξ). n n Let E0(ξ) = E(ξ) − B which also gives a ﬁbre bundle over B with ﬁbre R0 = R − {0}. Consider the pair (E(ξ), E0(ξ)). If ξ admits a Riemanian structure then we can also consider the unit disc bundle D(ξ) ↓ B and unit sphere bundle S(ξ) ↓ B and the pair (D(ξ), S(ξ)). Then there is an homotopy equivalence of pairs (D(ξ), S(ξ)) −→ (E(ξ), E0(ξ)). The quotient space Mξ = D(ξ)/S(ξ) is the Thom space of ξ. There is a ‘diagonal map’ E(ξ) −→ E(ξ) × E(ξ) −→' B × E(ξ) which induces a map ψ :Mξ −→ (B+) ∧ Mξ. For any commutative ring k, this gives rise to a product map ∗ ∗ ∗ ∗ ψ ∗ H (B; k) ⊗ He (Mξ; k) −→ He ((B+) ∧ Mξ; k) −→ He (Mξ; k), making He ∗(Mξ; k) into a module over H∗(B; k). We write x · y for the product of x ∈ H∗(B; k) and y ∈ He ∗(Mξ; k). It can be seen that if B is a CW complex then so is Mξ with each k-cell in B corresponding to an (k + n)-cell in Mξ. In particular, the bottom cell of Mξ can be taken to be an n-sphere which can be thought of as the sphere obtained by collapsing to a point the boundary of the disc above a point of B. We say that ξ is k-orientable if there is an element u ∈ He n(Mξ; k) which restricts to the n n ∼ natural choice of generator sn ∈ He (S ; k) = k. Such a u is called a k-orientation. A vector bundle ξ ↓ B of dimension n is said to be orientable if its structure group can be taken to be SLn(R) 6 GLn(R), i.e., if there is a principal SLn(R)-bundle P ↓ B for which ξ =∼ P × Rn. SLn(R) Proposition 4.1. Let ξ ↓ B be a vector bundle.

(a) ξ is F2-orientable. (b) ξ is Z-orientable if and only if it is orientable. Theorem 4.2 (Thom Isomorphism Theorem). If ξ ↓ B is k-orientable then He ∗(Mξ; k) is a free H∗(B; k)-module of rank 1, with any orientation u ∈ He n(Mξ; k) as a generator. Thus for each orientation u ∈ He n(Mξ; k) there is an isomorphism of H∗(B; k)-modules ∗ ∗ Φu : H (B; k) −→ He (Mξ; k); x 7−→ x · u. This means that given an orientation u ∈ He n(Mξ; k), every element w ∈ He k+n(Mξ; k) can be uniquely expressed in the form w = v · u for some v ∈ He k(Mξ; k).

n Example 4.3. Recall that there is a canonical line bundle λn ↓ RP , where

(λn)[x] = {tx : t ∈ R}. Then RPn ⊆ RPn+1;[x] ←→ [x, 0] and there is a homeomorphism . n+1 E(λn) = RP − {[0, 1]}; tx ∈ (λn)[x] ←→ [x, t], where we always take |x| = 1. Consequently, there is a homeomorphism . n+1 Mλn = RP . CUP PRODUCTS, COHOMOLOGY RINGS AND THOM ISOMORPHISMS 7

∗ n Hence there is an isomorphism of H (RP ; F2)-modules

∗ n zn+1 ∗ n+1 H (RP ; F2) −−−→ He (RP ; F2).

In particular, we can inductively prove Example 2.7(a),

∗ n n+1 H (RP ; F2) = F2[zn]/(zn ).

There are various properties of Thom isomorphisms that are useful. For example, if ξ1 ↓ B1 and ξ2 ↓ B2 then . M(ξ1 × ξ2) = Mξ1 ∧ Mξ2,

∗ and we can combine orientations ui ∈ He (Mξi; k) to obtain an orientation for ξ1 × ξ2,

∗ (4.1) u1 ∧ u2 ∈ He (Mξ1 ∧ Mξ2; k).

Another important invariant of a k-orientable bundle of dimension n is the Euler class

∗ n eu(ξ) = s0u ∈ H (B; k),

∗ where s0 : B −→ Dξ −→ Mξ is the zero-section. Then the following relation holds in H (Mξ; k):

2 u = eu(ξ) · u.

It follows that if ξ has a non-vanishing section, then eu(ξ) = 0. Thus eu(ξ) is an obstruction to ξ having a non-vanishing section.

5. Characteristic classes defined using the Thom isomorphism Let ξ ↓ B be a vector bundle of dimension n, where B is connected. Then for an orientation n u ∈ He (Mξ; F2) (in fact this is unique) we have a Thom isomorphism

∗ ∗ Φu : H (B; F2) −→ He (Mξ; F2).

Recall the Steenrod operations Sqk for k > 0. We may deﬁne elements

−1 k k wk(ξ) = Φu Sq u ∈ H (B; F2).

k Of course, w0(ξ) = 1. Notice that for k > n we have Sq u = 0 and so wk(ξ) = 0. Notice also n 2 that since Sq u = u , wn(ξ) = eu(ξ), the Euler class. We also have from Equation (4.1) and the Cartan Formula Xk wk(ξ1 × ξ2) = wj(ξ1) × wk−j(ξ2). j=0 This can be internalised to a Cartan Formula for a direct sum of bundles ξ0 ⊕ ξ00,

Xk 0 00 0 00 wk(ξ ⊕ ξ ) = wj(ξ )wk−j(ξ ). j=0

n Using Example 4.3 we can also show that for λn ↓ RP , w1(λn) = zn. Combining all of these we ﬁnd that wk(ξ) agree with the Seifel-Whitney classes deﬁned by other methods. 8 ANDREW BAKER

6. Manifolds and duality Let M be a compact connected manifold of dimension m. Then the tangent bundle TM ↓ M has dimension m. If TM is k-orientable we say that M is k-orientable. So every manifold is F2-orientable, while M is Z-orientable if and only if it is orientable. If M is k-orientable then for any choice of orientation u ∈ He m(MTM; k) there is a k-Euler class m eu(M) = eu(TM) ∈ H (M; k). It also follows that Hm(M; k) =∼ k and under this isomorphism, eu(M) ←→ χu(M) ∈ k, where χu(M) is the associated Euler characteristic. For example, when M is orientable and k = Z, χu(M) is the geometric Euler characteristic (up to sign) and this vanishes of and only if there is a non-vanishing section of TM. Now take k to b a field and let M be k-orientable. Then for each k, k ∼ k H (M; k) = Hom|(Hk(M; k), k),Hk(M; k) = Hom|(H (M; k), k). Then if x ∈ Hk(M; k) we can cosnider x as defining a k-linear mapping Hm−k(M; k) −→ Hm(M; k) =∼ k; y 7−→ x ∪ y and so it corresponds to an element of Hm−k(M; k). For each k, this sets up an isomorphism k ∼ H (M; k) = Hm−k(M; k) called the Poincaréduality isomorphism. A particular case of this occurs when m = 4n and k = 2n. Then we obtain a pairing 2n 2n 4n H (M; k) ⊗| H (M; k) −→ H (M; k) =∼ k; x ⊗ y 7−→ x ∪ y. This is a non-degenerate bilinear form and is an important invaraint of M. When k = Z, it can be used to essentially classify manifolds. For example, when m = 4 and M is simply connected, this is the main topological invariant.