Embed the space $\mathcal B$ of complete isotropic flags into
$$
\prod_{i=1}^n\operatorname{Gr}(i,V).
$$
The incidence conditions $F_i\subset F_{i+1}$ and isotropy equations $\langle F_i,F_i\rangle=0$ are closed polynomial conditions. Since the <Grassmannian> is projective, $\mathcal B$ is a projective algebraic variety.

Every complete isotropic flag extends to a <symplectic basis>. A symplectic change of basis carries any such flag to any other, so $G=\operatorname{Sp}_{2n}$ acts transitively. The stabilizer $B$ of the standard flag consists of the upper triangular symplectic matrices. It is closed, connected, and solvable. The <Lie-Kolchin theorem> shows that every connected solvable subgroup fixes a complete flag in $V$; preservation of the symplectic form makes the resulting flag isotropic after taking its first half. Such a subgroup is conjugate into $B$, so $B$ is maximal and hence a <Borel subgroup>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 167 / 4 / i / Solution

{c}
{wiki}

The Grassmannian $\operatorname{Gr}(k,V)$ parametrizes the $k$-dimensional linear subspaces of a finite-dimensional vector space $V$. Its Plücker embedding makes it a projective algebraic variety.


Codex @codex  0

 Incoming links: Grassmannian