Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 20I (Algebraic Topology)

Write $n=2m-1$. On ${\mathbb{R}}^{2m}$ define the orthogonal linear mapIt satisfies $J^T=-J$ and $J^{T}J=I$. Hence $⟨x,Jx⟩=0$ and $\Vert Jx\Vert =\Vert x\Vert$. Restricting to the unit sphere givesThis continuous unit tangent field is available exactly because the ambient dimension is even.

Triangulate the sphere by the boundary complex of the cross-polytope with simplicial vertices $\pm e_0,\dots ,\pm e_n$: a face contains at most one simplicial vertex from each antipodal pair. Take its barycentric subdivision. Its simplicial vertices are the nonempty faces $F$ of the original complex, and its simplices are strictly nested chains $F_0\subset \cdots \subset F_k$.

Form a quotient complex whose simplicial vertices are pairs $\{F,-F\}$, and whose simplices are the images of these chains. No chain contains both $F$ and $-F$. Moreover a chain of face-orbits lifts uniquely up to simultaneous negation: after choosing a largest face, every smaller face has at most one representative contained in it, because an original face contains no antipodal simplicial vertices. Thus the quotient is an honest simplicial complex, without two different simplices with the same simplicial vertex set. Its realization is homeomorphic to the sphere modulo its antipodal action. This provides a triangulation of Real projective space ${\mathbb{RP}}^n$. The subdivision avoids the duplicate-simplex problem of naively identifying simplicial vertices in the unsubdivided cross-polytope boundary.

The antipodal map is the restriction to the boundary of the unit ball of $-I$ on ${\mathbb{R}}^{n+1}$. The induced orientation on the sphere is reversed or preserved according to the determinant of that ambient linear map, namely $(-1{)}^{n+1}$. Its degree is therefore $(-1{)}^{n+1}$. With integral coefficients, for $n\ge 1$ the induced top-homology map isAt $n=0$, ordinary $H_0(S^0)={\mathbb{Z}}^2$ instead has its two generators exchanged; multiplication by $-1$ is the statement on reduced $H_0(S^0)\cong \mathbb{Z}$. This distinguishes the small-dimension exception from the usual top-degree formula.

Suppose such an $f$ exists. Orthogonality and unit lengths makea continuous map into the unit sphere, since its squared norm is one. This is a homotopy from the identity to the antipodal map. Their degrees would coincide, but for even $n\ge 2$ they are respectively $1$ and $(-1{)}^{n+1}=-1$, a contradiction. Thus no such continuous map exists for even $n$. At $n=0$ there is not even a unit vector orthogonal to either point of $S^0\subset \mathbb{R}$, so nonexistence holds there as well. This is a sphere form of the Hairy ball theorem.