Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 26I / c

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.