# Why is the generalized flag variety a “variety”?

In several places (for example, Chriss & Ginzburg’s book “Representation Theory and Complex Geometry”), the author says that the set $$X$$ of Borel subalgebras of a semi-simple Lie algebra $$\mathfrak g$$ forms a Zariski-closed subset of a Grassmannian on this semi-simple Lie algebra. A Borel subalgebra is a maximal solvable subalgebra, by general theorems, all of them have the same dimension, say $$d$$. Then the set $$X$$ can be identified with a subset of the Grassmannian $$\mathrm{Gr}(d,\mathfrak g)$$. My question is: why is $$X$$ a Zariski-closed subset? How to translate the condition “solvable” to an algebraic condition? I thought of Lie’s criterion on solvability, but it is not an equivalence condition, so it did not work.

• I think this is addressed in Fulton and Harris. Sep 4, 2021 at 17:02
• @BenMcKay Since you seem to know the answer, perhaps you can spell it out as an answer? Sep 4, 2021 at 21:52
• This is completely standard. It's definitely dealt with in Humphreys and Borel, and probably in Springer's book too. (All these books are called Linear Algebraic Groups.) Sep 5, 2021 at 10:10
• The question specifically asks how to show that the set of Borel subalgebras of a Lie algebra is a closed subvariety of the appropriate Grassmannian (of all subspaces of a given dimension). All of the references mentioned in the comments show that G/B is a projective variety, and then show that all Borel subgroups/algebras are conjugate, along with a normalizer theorem to identify the set of Borels with G/B. To me, this route definitively fails to answer the question (which is not directly about showing G/B is projective, or even about G/B at all without some nontrivial theorems). Sep 9, 2021 at 22:39
• @Grant B. - Projective varieties are complete; the image of the map $G/B\rightarrow {\rm Gr}(d, {\mathfrak g})$ is therefore closed. Doesn't that answer the question? Sep 10, 2021 at 15:22

A nice source on this is Procesi's "Lie Groups: an approach through invariants and representations". There he defines parabolic subgroups as precisely those for which $$G/P$$ is a projective variety and shows that this is equivalent to $$P$$ containing a Borel subgroup.

In a later section (called quadratic relations, I think) he shows that you can cut them out with only quadratic polynomials in an appropriate projectivised representation of $$G$$ (a result originally due to Kostant I believe). This is, in effect, a generalisation of the Plucker embedding for the Grassmannian.

Note, you say generalised flag variety in the title which refers to any quotient $$G/P$$ of semisimple $$G$$ by a parabolic subgroup but in the body of the question only refer to the quotient by a Borel $$G/B$$ which I would call a "full" flag variety.

Edit: said Borel when I meant Kostant

I think, the statement "the set of Borel subalgebras of a semi-simple Lie algebra $$\mathfrak g$$ forms a Zariski-closed subset of a Grassmannian" can be understood in more than one way and then the question becomes less trivial than one thinks. Everything is over $$\mathbb C$$, by the way, otherwise things become even more complicated.

First interpretation: This is the one presumably intentend by Chriss-Ginzburg and others. It consists of two statements:

1. The set of Borel subgroups is in bijection with $$G/B$$.
2. The natural morphism $$G/B\to Gr_d(\mathfrak g)$$ is a closed embedding.

The first assertion uses the conjugacy of Borel subgroups and that Borel subgroups are selfnormalizing. The second uses that $$G/B$$ is projective and that $$B$$ is the normalizer of $$\mathfrak b$$ in $$G$$.

Second interpretation: The statement could also mean that the moduli space of Borel subalgebras is precisely the schematic image of $$G/B$$. This is more subtle and is not addressed in any of the standard text, as far as I know. One can see this as follows:

1. Step: A subalgebra $$\mathfrak b\subseteq\mathfrak g$$ is the Lie algebra of a Borel subgroup if and only if $$\mathfrak b$$ is solvable of maximal possible dimension $$d=\dim B$$. Here one has to prove mainly that a maximal dimensional solvable subalgebra is algebraic in the sense that it is the Lie algebra of some algebraic subgroup.

2. Step: There is a closed subscheme $$X\subseteq Gr_d(\mathfrak g)$$ which is a moduli scheme of $$d$$-dimensional solvable subalgebras. For this one has to translate "solvable" into an algebraic condition. This condition comes from that fact that the degree of solvability of any $$\mathfrak b$$ is bounded by say $$\ell$$. Then $$X$$ is defined by the condition that all "multicommutators" $$[\ldots[[\xi_1,\xi_2],[\xi_3,\xi_4]],\ldots]$$ vanish for all $$\xi_1,\ldots\xi_{2^\ell}\in\mathfrak b$$. This is easily seen to be a closed condition.

3. Step: Step 1 implies that $$X$$ is set theoretically the image of $$G/B$$ in $$Gr_d(\mathfrak g)$$.

4. Step: It remains to prove that $$X$$ is smooth, hence reduced and normal. For this we compute the tangent space $$Z$$ of $$X$$ in $$\mathfrak b$$. First the tangent space of $$Gr_d(\mathfrak g)$$ in $$\mathfrak b$$ is $$H:=\mathrm{Hom}(\mathfrak b,\mathfrak g/\mathfrak b)$$. It turns out that $$Z$$ is the space of $$1$$-cocycles (=derivations) in $$H$$. Moreover, the tangent space of $$G/B$$ in $$\mathfrak b$$ is the space of all $$1$$-coboundaries (=inner derivations) $$C\cong\mathfrak g/\mathfrak b$$. So $$Z/C=H^1(\mathfrak b,\mathfrak g/\mathfrak b)$$. The latter is easily seen to be $$0$$ (take a filtration with $$1$$-dimensional quotients and use that $$\mathfrak b$$ and $$\mathfrak g/\mathfrak b$$ have no weights in common). Thus $$Z=C$$ and therefore $$\dim Z=\dim X$$ and we are done.

• Thank you a lot for your answer! The second interpretation is exactly what I expected. Jan 15, 2022 at 23:35