Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a statement on page 149 of James E. Humphreys' Linear Algebraic Groups:

Let $G$ be an algebraic group. $S \subseteq G$ is an arbitrary torus contained in a Borel subgroup $B$. $\mathscr{B}^S$ denotes the the set of Borel subgroups of $G$ containing $S$. Then $\mathscr{B}^S$ can be identified with a closed subgroup of $G/B$.

I don't know why is the set closed.

Let $Y = \cup_{B' \in \mathscr{B}^S} B'$, and $\phi: G \rightarrow G/B$ be the canonical map. Then $\mathscr{B}^S$ can be identified with $\phi(Y)$. As $|\mathscr{B}^S|$ may be finite or infinite, I don't even know whether $Y$ is closed, then why is $\phi(Y)$ closed?

Thank you very much.

share|cite|improve this question
I don't have Humphrey in front of me, but what you are writing confuses me. What is $G/B$? – Alex B. Aug 16 '11 at 8:05
@Alex B.: I am sorry I didn't see your comment. $B$ is a Borel subgroup of $G$, and $G/B$ is the homogeneous space. – ShinyaSakai Oct 18 '11 at 17:54
@Alex B.: The author has proved, for any algebraic group $G$ and its subgroup $H$, there exists a unique variety $Y$ (on which $G$ acts) and a $G$-equivariant morphism $\phi: G \rightarrow H$, whose fibers are precisely the cosets $xH$, and given a variety $X$ and a morphism $\phi: G \rightarrow X$ whose nonempty fibers are unions of cosets $xH$, there exists a unique morphism $\psi: Y \rightarrow X$ such that $\psi \circ \pi = \phi$. $Y$ is called the homogeneous space and $\pi$ the canonical morphism. – ShinyaSakai Oct 18 '11 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.