Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 26I (Differential Geometry)

A critical point of a smooth map is a point where its derivative is not surjective; its image is a critical value. A regular value is a target value all of whose preimages have surjective derivative.

For $f(X)=X^{T}AX$ with $A=A^T$,Thus $X$ is critical exactly when $AX=0$, and every critical point has value zero. Hence zero is the only possible critical value; it is a critical value precisely when $A$ is singular. If $A$ is singular, every point of its kernel is critical, so there can be infinitely many critical points; if $A$ is invertible, only the origin is critical.

Let ${\mathrm{Skew}}_{2n}$ be the skew-symmetric $2n$ by $2n$ matrices and defineThen $S{p}_n(\mathbb{R})=F^{-1}(J)$, andAt a symplectic $A$, put $H=AX$. The derivative becomes $X^{T}J+JX$. Given skew $K$, choosing $X=-\frac{1}{2}JK$ gives $X^{T}J+JX=K$, so the derivative is surjective.

The preimage theorem makes the level set a submanifold. Sinceits dimension is $2n^2+n$.

If $P=P^T=P^2$ and $\mathrm{tr}P=1$, the spectral theorem gives eigenvalues $1,0,0$. Thus $P$ is orthogonal projection onto a line andfor a unit vector $X$; the only choices are $X$ and $-X$.

The smooth map $\pi :S^2\to S_3(\mathbb{R})$, $X\mapsto X{X}^T$, has image $G{r}_{1,3}(\mathbb{R})$ and fibers $\{X,-X\}$. Near $X_0$, restrict to the hemisphere $X\cdot X_0>0$; there $\pi$ is one-to-one and has smooth inverse obtained by choosing the positive local unit eigenvector. Combining this with a chart of $S^2$ supplies two-dimensional local parametrizations. Hence $G{r}_{1,3}(\mathbb{R})$ is a two-dimensional submanifold, diffeomorphic to $\mathbb{R}{P}^2$.

No. The set is closed and bounded in the finite-dimensional space $S_3(\mathbb{R})$, hence compact and nonempty. A global parametrization would be a homeomorphism from a nonempty open subset of ${\mathbb{R}}^2$ onto this compact surface, but no nonempty open subset of ${\mathbb{R}}^2$ is compact.