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Hi,

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 $X$?

Thanks!

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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

1 Answer 1

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.

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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

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