Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 114 / 5

For unit vectors, the defining inequality for $Y$ is equivalent to $v\cdot w\ge 0$. WriteThen $t=(1-\Vert u{\Vert }^2{)}^{1/2}$, sois a homeomorphism $D(T{S}^{2n})\to Y$. Replacing $v$ by $-v$ gives the same description of $Y^{\prime }$. Their intersection isthe unit tangent bundle, equivalently a Stiefel manifold. Since $e(T{S}^{2n})$ evaluates to $\chi (S^{2n})=2$, its Gysin sequence gives the cohomology of the unit tangent bundle of an even-dimensional sphere:Every product of positive-degree classes vanishes.

Let $G$ consist of the eight signed permutations of the two factors:It is the group of signed permutation matrices in dimension two, hence isomorphic to the dihedral group $D_8$, and every element preserves the equation $v\cdot w=0$. If $a,b$ generate $H^{2n}(X;\mathbb{Z})$ from the two sphere factors, these maps act by the corresponding signed permutation matrix, because the antipodal map of the even-dimensional sphere has degree $-1$. The eight actions are distinct, so only the identity can be homotopic to $1_X$.

Every element of $G$ acts trivially on $H^{\ast }(Z;\mathbb{Z})$. The degree-$2n$ group is $\mathbb{Z}/2$, so its automorphism is forced to be the identity. On the top class, regard $Z$ as the Stiefel manifold of two-frames. The signed permutations are the right action of $O(2)$. The determinant-one component is connected, while a determinant-minus-one element is homotopic within that component to $(v,w)\mapsto (v,-w)$, the antipodal map on the $S^{2n-1}$ fibers. That map has degree $+1$, so the top class is also fixed.

The subgroup preserving $Y$ iswhere $A(v,w)=(-v,w)$, $B(v,w)=(v,-w)$, and $S(v,w)=(w,v)$. It consists exactly of the elements with ${\epsilon }_1{\epsilon }_2=1$. Under the deformation retraction $Y\simeq S^{2n}$ onto the diagonal, $1$ and $S$ act as the identity, while $AB$ and $ABS$ act as the antipodal map. They therefore act on $H^{2n}(Y;\mathbb{Z})$ by $+1,+1,-1,-1$, respectively.

The Thom isomorphism theorem identifies $H^{\ast }(Y,Z)$ with a degree-$2n$ shift of $H^{\ast }(S^{2n})$. On the Thom class in degree $2n$, the same four elements act by $+1,+1,-1,-1$: the factor swap reverses each normal vector, which preserves the orientation because the normal rank $2n$ is even, while the simultaneous antipodal map reverses the oriented tangent fiber. On the degree-$4n$ relative class every element acts by $+1$, since the base and fiber signs for the last two elements cancel.

When $n=1$, $Z=V_2({\mathbb{R}}^3)\cong SO(3)$. Every signed permutation of $(v,w)$ extends, after sending the third frame vector $v\times w$ to the required sign, to right multiplication by an element of $SO(3)$. Since $SO(3)$ is path-connected, every right translation is homotopic to the identity. Therefore