Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 12 / 3 / Solution

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.