Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 114 / 3

The assertion is true. The standard filtration ${\mathbb{CP}}^0\subset \cdots \subset {\mathbb{CP}}^n$ gives one cell in each dimension $0,2,\dots ,2n$. There are no odd-dimensional cells, so all cellular differentials vanish. The cellular homology theorem and the universal coefficient theorem for cohomology give one copy of $\mathbb{Z}$ in each even degree from zero to $2n$, and zero in every other degree.

Let $x$ be the Poincare dual of a projective hyperplane, with the complex orientation. It has degree two and evaluates to $+1$ on a complex projective line, so it is the positive generator of $H^2$. We use the intersection interpretation of the cup product: the product of duals of oriented submanifolds in transverse position is the dual of their oriented intersection. Distinct transverse complex hyperplanes intersect in ${\mathbb{CP}}^{n-j}$ after $j$ intersections, with positive complex orientation. Thus $x^j$ is the dual of that linear subspace.

Pairing $x^j$ with a transverse linear ${\mathbb{CP}}^j$ gives one positively oriented intersection point. Therefore $x^j$ is a primitive generator of $H^{2j}$, for every $0\le j\le n$. There is no cohomology above dimension $2n$, so $x^{n+1}=0$. These facts show that the surjective graded ring map from $\mathbb{Z}[x]$ has exactly the indicated kernel:This proves the cohomology ring of complex projective space, rather than only its additive groups. For $n=0$ it is $\mathbb{Z}$, with $x=0$.

The assertion is false. Take $m=3$ and $n=2$. For the constant attaching map, the cell attachment gives $S^2\vee S^4$. Its degree-two generator has square zero: restricting the square to either sphere gives zero, and restrictions identify its degree-four cohomology with that of the $S^4$ summand.

For the Hopf fibration $\eta :S^3\to S^2={\mathbb{CP}}^1$, the attachment instead gives ${\mathbb{CP}}^2$. One can verify the attaching map explicitly with the characteristic mapIts interior maps homeomorphically to the complement of ${\mathbb{CP}}^1$; on the boundary it sends $(z_1,z_2)$ to $[z_1:z_2:0]$, exactly the Hopf fibration. By part (a), the degree-two generator of $H^{\ast }({\mathbb{CP}}^2;\mathbb{Z})$ has nonzero square generating degree four. HenceThe additive groups agree, but multiplication distinguishes the attachments.

The dimension condition explains why this example is the relevant one. In general the only positive-degree additive generators lie in degrees $n$ and $m+1$. A potentially nonzero product can only be the square of the degree-$n$ generator, and only when $m+1=2n$. Its coefficient is the Hopf invariant of the attaching map. Here the constant map has invariant zero, whereas the complex Hopf fibration has invariant one.

For nonzero finite-dimensional $U$ and $V$, the intended assertion is true. Write $a=\dim U$, $b=\dim V$, and replace $W$ by $\mathrm{im}\varphi$, of dimension $r$. Injectivity on the stated slices implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Consequently the bilinear map defines a continuous map of Complex projective spacesThe rank $r$ is at least one, since one such nonzero tensor has nonzero image.

Let $L_U,L_V,L_W$ be the respective tautological bundles. Fiberwise, $\varphi$ identifies $L_U\otimes L_V$ with $F^{\ast }{L}_W$. For the positive hyperplane classes $x=c_1(L_U^{\ast })$, $y=c_1(L_V^{\ast })$ and $z=c_1(L_W^{\ast })$ (zero when the projective space is a point), the first Chern class of a tensor product of complex line bundles givesThe Künneth theorem, together with part (a), identifies the product cohomology ring withThere are no Tor terms because the factor groups are free. In particular, its monomials $x^i{y}^j$ with $0\le i<a$, $0\le j<b$ form an integral additive basis. In top degree,If $r\le a+b-2$, the target relation $z^r=0$ would imply $(x+y{)}^r=0$ and hence contradict this nonzero top power. ThusWhen $a=b=1$, the already-established inequality $r\ge 1$ gives the same conclusion. The complex bilinear dimension bound is sharp: multiplication of complex polynomials of degrees less than $a$ and $b$ has target dimension $a+b-1$, is injective in either nonzero fixed factor, and its image spans every monomial in that target.

Literal zero-space qualification. The printed assertion does not explicitly exclude zero vector spaces. If $U=0$, $V={\mathbb{C}}^2$ and $W=0$, its slice-injectivity hypothesis is vacuous, while the claimed inequality would be $0\ge 1$. Thus, with zero spaces permitted, this is a counterexample to the assertion exactly as written; the proof above supplies the usual nonzero finite-dimensional interpretation.