Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 26H (Differential Geometry)

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.

Choose $Q\in SO(n+1)$ with $Q{e}_{n+1}=v$. Conjugation by $Q$ identifieswithIt follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to $SO(n)$, andAt $R\in S_v$, its tangent space is

For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,Identify $\mathfrak{so}(n+1)$ with the exterior square ${\Lambda }^2{\mathbb{R}}^{n+1}$. The orthogonal complement of $T_I{S}_v={\Lambda }^2(v^{\perp })$ is $v\wedge v^{\perp }$. If $v$ and $w$ are linearly independent, the nonzero bivector $v\wedge w$ belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of $T_{I}SO(n+1)$, so the intersection is not transverse. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequentlyas summarized by distinct unit-vector stabilizers are not transverse.