Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2007 / B / 8

With the usual second-countability convention, an $n$-manifold is a Hausdorff second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$. A local orientation of a manifold at $x$ is a choice of generator ofExcision and a coordinate ball identify this relative group with $H_n({\mathbb{R}}^n,{\mathbb{R}}^n\setminus \{0\})$. An orientation of a topological manifold is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with orientation-preserving transition maps.

For a connected closed oriented manifold, its fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to every chosen local generator. Here closed means compact without boundary. Poincare duality states that cap product with this class is an isomorphismOver rational coefficients it gives $b_k=b_{n-k}$. ThereforeWhen $n$ is odd this says $\chi =-\chi$, so $\chi (M)=0$. Torsion does not contribute to the Betti numbers or this Euler characteristic.

Yes, for a nonnegative integer $g$: the closed orientable surface of genus $g$ has precisely these groups. A standard polygon cell structure has one vertex, $2g$ one-cells and one two-cell attached by a product of commutators. Its integral cellular boundary maps are zero, because each generator occurs once in each orientation in the attaching word. Thus $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^{2g}$. At genus zero use the sphere.

No. Its top nonzero group would force dimension three. Poincare duality requires $H_2\cong H^1$, and the universal coefficient theorem gives $H^1\cong \mathrm{Hom}(H_1,\mathbb{Z})=\mathbb{Z}$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give Euler characteristic $-1$, impossible for a closed oriented odd-dimensional manifold.

Yes: for $n\ge 1$, the sphere $S^n$ is connected, closed and oriented, with $H_0=H_n=\mathbb{Z}$ and no intermediate homology. These groups do not require the manifold to be homeomorphic to a sphere, but the sphere provides the requested example.

No. The dimension would again be three. The universal coefficient theorem gives $H^1=0$, since both $\mathrm{Hom}(H_1,\mathbb{Z})$ and $\mathrm{Ext}(H_0,\mathbb{Z})$ vanish. Poincare duality then forces $H_2=0$, contradicting the proposed torsion group. In general the second integral homology of a connected closed oriented three-manifold is free abelian, being dual to its first homology.

Yes: real projective three-space ${\mathbb{RP}}^3$ has these groups. Its usual cell structure has one cell in each dimension zero through three, with boundary maps ${\partial }_1=0$, ${\partial }_2=2$ and ${\partial }_3=0$. Hence its integral homology is $\mathbb{Z}$ in degrees zero and three, $\mathbb{Z}/2$ in degree one, and zero in degree two. It is a connected closed oriented manifold: the antipodal map on $S^3$ preserves orientation, so the sphere's orientation descends to the quotient.