MathOverflow is a question and answer site for professional mathematicians. 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


Suppose $G$ is a reductive group over an algebraiclly closed field $k$ (suppose $k$ of char zero if you want at first). Let $X$ be its flag variety.

Question: What is the moduli problem that $X$ represents?

EDIT (to clarify): What is the functor of points of $X$?


share|cite|improve this question
Can you clarify the meaning of "the" moduli problem here? In any case, $X$ can be identified with the set of all Borel subgroups of $G$. – Jim Humphreys Apr 22 '12 at 17:23
up vote 4 down vote accepted

Let $X$ be a space and $H$ be a group such that $X\rightarrow X/H$ is a principal bundle. Then $Hom(Y,X/H)$ is in bijection with $H$ torsors over $Y$ equipped with an equivariant map from their total space to $X$. So maps from $Y$ into the flag variety $G/P$ are in bijection with $P$ torsors on $Y$ equipped with a $P$-equivariant map to $G$. Or equally $P$ "subtorsors" of the trivial torsor $G\times Y$.

For example if we take $G=GL_{n}(\mathbb C)$ the data of a $P$ "subtorsor" of $G$ is equivalent to giving a flag (whose type is determined by $P$) of sub-bundles inside the trivial n-dimensional bundle on $Y$.

If for example $P$ consist of all matrices whose first column is zero everywhere except in the upper left corner, we have $G/P=\mathbb P^{n-1}$. Our description says maps into $\mathbb P^{n-1}$ are the same thing as linebundles inside of $Y\times \mathbb C^n$.

Taking the dual of such a linebundle and restricting the coordinate functions of $\mathbb C^n$ to it gives the usual universal property of $\mathbb P^{n-1}$.

All this should work over any field.

share|cite|improve this answer
I think $Hom(Y,X)$ should be $Hom(Y,X/H)$ in your second sentence. – S. Carnahan Apr 23 '12 at 2:03
Thanks, I fixed it. – Jan Weidner Apr 23 '12 at 19:04

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.