# How many ways are there to prove flag variety is a projective variety?

I am looking for references talking about different ways to prove flag variety $G/B$ is projective variety. Now I have some in mind:

1. There is a proof in Humphreys Linear algebraic groups, he first prove $G/S$ is a complete algebraic variety hence a projective variety, where $S$ is a Borel subgroup of $G$ of largest possible dimension. Then he used the Borel Fixed Point Theorem to obtain the assertion.

2. There is a proof in A.L.Onishchik and E.B.Vinberg Lie groups and algebraic groups, they used Chevalley's theorem to give the unique projective variety structure to the coset variety $G/B$.

3. There is a sketch of the proof in Tanisaki's D-modules, Perverse sheaves and Representation theory, he identified $G/B$(in particular case, suppose $G=GL_{n}(k)$) with set of flags in $k^{n}$. Then it is clear that one can give the projective variety structure on the set of flags.

4. One can also prove the flag variety can be embedded to Grassmannian variety, Then using Plucker embedding to give projective variety structure to this Grassmannian.

5. I think one can prove the line bundle on $G/B$ is positive and then use Kodaira embedding to prove it is a projective variety. But I did not find a proper expository reference.

Are there any other interesting proof? Any related comments are welcome. Thanks!

Edit: $G$ is an algebraic group, $B$ is Borel subgroup and $k$ is algebraic closed and $char k=0$

• What are your precise conditions on G, B, and k? Apr 14, 2010 at 3:16
• @Scott:sorry to mislead, $G$ is an algebraic group, $B$ is Borel subgroup and $k$ is algebraic closed field with characteristic 0 Apr 14, 2010 at 3:31
• I imagine you could construct the homogeneous coordinate ring as follows. The Borel subgroup B acts on the vector space V consisting of n by n matrices, extending to an action on Sym(V). Let the nth degree component A_n consist of those a in Sym(V) such that $\sigma b = (\det x)^n b$ for all $\sigma \in B$. Then recover the flag variety as Proj $A$. However, I have not checked the details. Apr 14, 2010 at 3:33
• @Ben: this is from Borel-Weil-Bott,right? Apr 14, 2010 at 4:11
• The "five" methods listed aren't all distinct (and the characteristic of the field shouldn't matter). It's essential here to put the question in the context of what one already knows and what is the overall goal. Most treatments originate in the 1956-58 Chevalley seminar dealing with structure and classification of semisimple groups, followed by some evolution in exposition and a few simplifications. Shortcuts for general or special linear groups may be useful for some purposes, but the general result takes some work even though it's "elementary". Apr 14, 2010 at 12:43

I'm not sure this is an answer, but it got too long to be a comment! The projectivity of $G/B$ seems to me to follow from two facts: a) a homogeneous space $G/H$ is always a quasi-projective variety (which is due to Chevalley), and b) the variety $G/B$ must be complete.

The first fact clearly has nothing to do with Borel subgroups but it would be maybe interesting to ask about how many proofs we have of it. The idea of the construction is clear: find a representation of $G$ which contains a line $L$ which has $H$ as its stabilizer and then take the orbit of $L$ in $\mathbb P(V)$, but to see that you get a categorical quotient this way takes some more care: you need to use some infinitesimal properties, (which you can tidily say using the Lie algebra of course).

The second fact perhaps depends on what you're willing to assume, but it must come down to the Borel fixed point theorem in some form or other, since this tells you that if $G/H$ is complete, then any solvable subgroup will be contained in a conjugate of $H$, thus Borel subgroups are the only solvable subgroups with a chance of having an associated homogeneous space which is complete.

The argument from Humphreys' book uses the strategy of picking first a maximal solvable subgroup of largest dimension so as to show the orbit has to be closed (essentially because the stabilizer is largest so the orbit has smallest, but even then you use the Borel fixed point and the theorem for $GL_n$ if I remember correctly). I don't know how Onishchik and Vinberg get around this (if they do). Then, as the question says, you get the general result from the Borel fixed point theorem.

The proof for $GL_n$ shows that $G/B$ is projective if you take $B$ to be the subgroup of upper triangular matrices, but one still needs to show that this subgroup is a Borel, which again I only know how to do using the Borel fixed point theorem in some form. (And of course you can use the same strategy for other classical groups if you can eyeball a candidate Borel subgroup).

Given that, it's maybe worth pointing out that in all of this you can get away with a weak version of the Borel fixed point theorem: namely if $V$ is a representation of a solvable group $H$, and $X$ is an $H$-stable closed subvariety of $\mathbb P(V)$ then $X$ has an $H$-fixed point. This can be shown just using the Lie-Kolchin theorem (that is, that a representation of a solvable group contains a one-dimensional subrepresentation), which can be proved directly. Since the standard proof of the general Borel fixed point requires you to use something like Zariski's main theorem, this is maybe a noticeable saving.

All of this leads me to wonde how many proofs do we know for i) the fact that for any closed subgroup $H$ the homogeneous space $G/H$ has to be quasi-projective and ii) the Borel fixed point theorem (or some variant)?

• For the Borel fixed point theorem, see a 1977 note by Steinberg giving a streamlined approach (reprinted by AMS in his collected papers): MR0466336 (57 #6216), Steinberg, Robert, On theorems of Lie-Kolchin, Borel, and Lang. Contributions to algebra (collection of papers dedicated to Ellis Kolchin), pp. 349–354. Academic Press, New York, 1977. For homogeneous spaces, I don't know a real shortcut in arbitrary characteristic where the results belong and tangent spaces (but no Lie algebras) occur. P.S. Dan Allcock found shortcuts to rank one theory: arXiv 2007, then J. Algebra (new title). Apr 14, 2010 at 18:06
• Allcock's paper is a nice shortcut in the theory I agree! It's interesting that he makes use of a slightly stronger form of the fixed point theorem: that an orbit of a connected solvable group acting on a complete variety contains no complete subvariety of positive dimension (this means an orbit of minimal dimension has to be a point, hence the standard theorem). It's not much harder to prove than the standard form once you think of it, but it does seem a little more powerful. Apr 14, 2010 at 18:38

Let $V$ be an irrep for a regular dominant weight, and $\vec v$ a high weight vector. Then the $G$-stabilizer of $k\vec v \in {\mathbb P}V$ is $B$, so the $G$-orbit is $G/B$. I guess I need a bit more thought to say why it's closed, but I'm not sure this will really be different enough from your proofs above to bear thinking about.

• Any closed subvariety of its complement would contain a closed orbit with a Borel fixed point, and thus also contain a highest weight vector. But that's impossible. Apr 14, 2010 at 4:04
• It seems to me that one of the quickest ways of proving the existence of an irrep with a highest weight vector is to use the properness of $G/B$. I guess it can be done in other ways (by for instance first constructing it over the Lie algebra and integrate). However, I think that this illustrates that maybe this is not such a good question: One wants to have the properness of $G/B$ as early as possible as it has many important consequences (conjugacy of Borel subgroups for one, but also that representations induced from $B$ are finite dimensional). Apr 14, 2010 at 4:29
• (cont'd) Hence, one should rather be looking for a proof that uses as few as possible of the properties of (semi-simple) algebraic groups as the properness of $G/B$ is used to establish such properties. Apr 14, 2010 at 4:31
• I'm confused as to what Torsten means, an irrep. has to have a highest weight vector simply because it has finitely many weight spaces. Apr 14, 2010 at 15:21
• @Kevin: What we need is the other direction, given a dominant weight there is an irreducible representation with that weight as highest weight. Apr 14, 2010 at 15:40