Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 16 / 4

Take ${\mathbb{CP}}^3$. It is a closed smooth manifold and is simply connected: its CW complex structure has no one-cells, and attaching cells of dimension at least two to a point cannot create a fundamental group. Its cohomology ring is $\mathbb{Z}[x]/(x^4)$ with $|x|=2$, so $x^3\ne 0$.

The theorem used is the Lusternik–Schnirelmann critical point theorem: every smooth real-valued function on a closed connected smooth manifold has at least its unnormalized Lusternik–Schnirelmann category many distinct critical points, without a nondegeneracy assumption. Here category is the minimum number of open subsets covering the manifold whose inclusions are null-homotopic. The cup length bound gives category at least four. To justify that bound, a cover by three categorical sets would make each positive-degree class vanish on its assigned set. The long exact sequence of each pair lifts three copies of $x$ to relative classes, and their relative cup product would lie in $H^6(M,M;\mathbb{Z})=0$, contradicting $x^3\ne 0$. Thus every smooth function has at least four critical points, including functions with degenerate critical points. The critical-point theorem in this formulation is described in Conley pairs in geometry—Lusternik–Schnirelmann theory and more.

Under the intended convention that the finite group is nontrivial, take ${\mathbb{CP}}^2$. It is simply connected by its CW complex structure. The Lefschetz fixed-point theorem says that a continuous self-map of a compact triangulable space with nonzero rational Lefschetz number has a fixed point.

For any continuous $f:{\mathbb{CP}}^2\to {\mathbb{CP}}^2$, write $f^{\ast }x=dx$ for the integral degree-two generator, with $d\in \mathbb{Z}$. Preservation of the cup product gives $f^{\ast }(x^2)=d^2{x}^2$. Only degrees zero, two and four contribute to the rational Lefschetz number, soConsequently every such map has a fixed point. In a free group action, each nonidentity group element must act by a fixed-point-free homeomorphism. This is impossible here, so no nontrivial finite group can act freely. This is the fixed-point property of even-dimensional complex projective space for complex dimension two. If the trivial group is included, the literal request has no example, since the trivial group acts freely on every space.

Take the Möbius line bundle over $S^1$,A continuous section of a vector bundle corresponds to a continuous real-valued function $s:[0,1]\to \mathbb{R}$ with $s(1)=-s(0)$. If $s(0)=0$, it already vanishes. Otherwise its endpoint values have opposite signs, so the intermediate value theorem gives $s(t_0)=0$ for some $t_0\in (0,1)$. Thus this vector bundle has no nowhere-zero section. Every continuous section must vanish.

Take the Hopf map $\eta :S^3\to {\mathbb{CP}}^1\cong S^2$, given by $\eta (z_0,z_1)=[z_0:z_1]$. The domain has positive-dimensional homology only in degree three, while $H_3(S^2;\mathbb{Z})=0$. In degree two the source group is zero, and all remaining positive-degree source groups are zero too. Hence the induced map on homology is zero in every positive degree.

For nontriviality, use the Hopf fibration $S^1\to S^3\stackrel{\eta }{\to }{S}^2$. It is a fiber bundle: phase multiplication gives each fibre, and choosing a normalized representative on either standard projective chart gives a local trivialization. The theorem invoked is the long exact sequence of homotopy groups of a fibration. Since the universal cover of $S^1$ is contractible, ${\pi }_2(S^1)={\pi }_3(S^1)=0$. Its exact sequence consequently gives an isomorphismThe identity class of $S^3$ is nonzero, since an identity map cannot be null-homotopic while inducing the identity on $H_3(S^3;\mathbb{Z})$. Its image under ${\eta }_{\ast }$ is the class of $\eta$, so that class is nonzero. The Hopf map is not homotopic to a constant despite inducing zero on all positive-degree homology.