Let $G$ be an algebraic group and let $B$ be a Borel subgroup. What are the open sets of $G/B$? How does one define a topology on $G/B$?
Does the Bruhat decomposition of $G$ somehow induce a collection of open sets on $G/B$?
2026-04-02 07:09:57.1775113797
Open sets of $G/B$
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Bruhat decomposition decomposes it as a union of subvarieties that are open in their closures; they are finitely many, exist in various dimensions, and only one of them is open (the big cell).
As $B$ is a closed subgroup there is no complication in giving a topology to $G/B$; what is important is it has a natural variety structure and is embeddable as a (closed) subvariety of a projective space. For any closed subgroup $G/H$ is embeddable as subvar of a projective space, when $H$ contains a borel subgroup the embedding is a closed embedding, hence it becomes a projective variety.