Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 14 / 1

Let $q:{\mathbb{CP}}^8\to Q$ collapse the indicated Complex projective space. Write $x\in H^2({\mathbb{CP}}^8;\mathbb{Z})$ for the positive generator, so the cohomology ring of complex projective space is $\mathbb{Z}[x]/(x^9)$. The inclusion of ${\mathbb{CP}}^3$ sends $x$ to its corresponding generator and is an isomorphism in degrees $0,2,4,6$.

The CW complex structure makes the inclusion a cofibration, so the positive-degree cohomology of $Q$ identifies with the relative cohomology of the pair. Its long exact sequence shows that $q^{\ast }$ is injective in positive degrees, with image the additive subgroup generated by $x^4,x^5,x^6,x^7,x^8$. Equivalently, the surviving cellular cochain complex has one copy of $\mathbb{Z}$ in each of degrees $8,10,12,14,16$ and zero differential. Let $u_i$ denote the class with $q^{\ast }{u}_i=x^i$, for $4\le i\le 8$.

Naturality of the cup product, together with the injectivity of $q^{\ast }$, now determines all multiplication:Every other product of positive-degree basis elements is zero. Indeed, their degrees exceed $16$. In particular, the surviving classes do not form a polynomial algebra on a degree-eight generator: the independent classes in degrees $10,12,14$ must also be retained. This is an instance of the cohomology ring of a collapsed projective subspace.

Write $A=S^2\times S^2$ and let $i:A\to X$ be its inclusion after the attachment. The cohomology ring of a product of two spheres iswhere $a,b$ come from the first and second factors and $ab$ is the chosen orientation class. The diagonal pulls both $a$ and $b$ back to the same generator of $H^2(S^2;\mathbb{Z})$.

There is one new three-cell. In the cellular chain complex, its boundary has coordinates $(1,1)$ in the two-dimensional cells. Thus the relevant differential isand the original four-cell still has zero boundary. This gives $H_2(X;\mathbb{Z})=\mathbb{Z}$, $H_3(X;\mathbb{Z})=0$, $H_4(X;\mathbb{Z})=\mathbb{Z}$, and no other positive homology groups. Equivalently, the relative cohomology sequence of $(X,A)$ gives an injective restriction in degree two with image $\mathbb{Z}(a-b)$, and an isomorphism in degree four.

Choose $u\in H^2(X;\mathbb{Z})$ and $v\in H^4(X;\mathbb{Z})$ by $i^{\ast }u=a-b$, $i^{\ast }v=ab$. Naturality of the cup product givesSince restriction is injective in degree four, $u^2=-2v$. All further positive-degree cup products vanish by dimension. ThereforeChanging the sign of $v$ would instead give $u^2=2v$; the displayed sign uses the product orientation $ab$. The factor $2$ is the characteristic feature of a cup square after a diagonal sphere attachment.

Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.