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Take the following definition: "A parabolic subgroup of a linear algebraic group defined over $\mathbb{C}$ is a subgroup, closed in the Zariski topology, for which the quotient space is a projective algebraic variety."

My questions are:

(i) Why include closed in the definition?

(ii) What is an example of a projective algebraic variety that is the quotient of a linear algebraic group by a non-Zariski-closed subgroup?

(iii) What is an example of a quotient by a parabolic subgroup that is not a flag manifold?

(iv) Elliptic curves cannot be described as quotients of linear algebraic groups. What are other examples of families of varieties that cannot be expressed in this form?

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  • $\begingroup$ Apart from the topological motivation that for Hausdorff topological gp $G$ and subgp $H$ the quotient $G/H$ is Hausdorff precisely when $H$ is closed in $G$, perhaps (i) may sound better in view of the following fact: if $f:H \rightarrow G$ is a homomorphism between finite type group schemes over a field $k$ and $\ker f = 1$ in the scheme-theoretic sense (i.e., $f$ is injective on $R$-valued points for all $k$-algebras $R$, or $f$ is a functorial monomorphism) then $f$ is necessarily a closed immersion. Thus, (ii) is essentially meaningless in any reasonable algebro-geometric sense. $\endgroup$
    – BCnrd
    Commented May 30, 2010 at 23:33

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If $G={\mathbb C}$ is the additive group and $L\subset G$ is a lattice, then $L$ is not Zariski closed, $G/L$ exists (in a sense of analytic geometry) and is projective. But of course $L$ is not a parabolic subgroup of $G$.

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  • $\begingroup$ Great illustration of importance of paying attention to terminology (here: meaning of "quotient")! $\endgroup$
    – Victor Protsak
    Commented May 31, 2010 at 4:24
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i)-ii) If the subgroup isn't closed, there's no reason in general for the quotient space to even be an algebraic variety. So for the definition to even make sense, there has to be some guarantee that $G/P$ is a variety. A standard result says that if $H \subset G$ is a closed subgroup of a linear algebraic group, then the quotient $G/H$ is a quasi-projective variety. You can find these results, for example, in Borel's text on linear algebraic groups. I don't know of an example of a non-closed subgroups whose quotient both exists and is not complete.

(iii) Pretty much by definition, $G/P$ is what's known as a partial flag variety. For the classical groups, you can directly verify that you get flag-like objects in this way, and for more general groups this is taken as the definition. However, if by flag variety you mean the full flag variety, then just pick any parabolic subgroup which is not a Borel subgroup. The simplest example occurs for $G = GL_3$ where you have parabolic subgroups corresponding the variety of lines in $\mathbb{C}^3$ (i.e., $\mathbb{P}^2$) and its dual, the variety of planes in $\mathbb{C}^3.$

(iv) I'm not sure exactly what you're asking for here. Just examples of varieties that aren't quotients of linear algebraic groups?

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  • $\begingroup$ (iii) So all quotients of complex linear algebraic groups by subgroups containing the Borel subgroup are of the form $U(n)/U(k_1) \times \cdots \times U(k_m)$? When the quotient is by the Borel subgroup, is this reflected in some simplification of $U(k_1) \times \cdots \times U(k_m)$? (iv) Hopefully this is not too bad practice, but I'm going to generalise my last question and ask: What are examples (besides elliptic curves) of complex projective varieties that aren't quotients of comapct semi-simple complex Lie groups by compact subgroups? $\endgroup$
    – Jean Delinez
    Commented Feb 8, 2010 at 18:21
  • $\begingroup$ All quotients of parabolics are of the form you list in the case of $G = U(n),$ but in general there is no "flag-like" description of partial flag-varieties for general $G.$ For the classical groups, however, there are flag-like descriptions, and it's fun finding the Borels and parabolics that they correspond to. As to your last question, there should be lots of examples. You mention elliptic curves, and higher genus curves should also fall into this category. Also, as far as I know, the Schubert subvarieties of flag varieties aren't a quotient of linear algebraic groups. $\endgroup$
    – Mike Skirvin
    Commented Feb 8, 2010 at 21:47
  • $\begingroup$ I'm not sure if I understand. $G/P$ isn't a group, so I don't know what is meant by its Lie algebra. Perhaps I misunderstood the question? $\endgroup$
    – Mike Skirvin
    Commented Feb 8, 2010 at 22:15
  • $\begingroup$ Let me try again: For a general semi-simple compact Lie group $G$ with Lie algebra $\mathfrak{g}$, and a parabolic subgroup $P$ with corresponding Lie subalgebra $\mathfrak{p}$, is there a nice description of the algebra of continuous functions of $G$ in terms of $\mathfrak{g}$ or $\mathbb{p}$, or both? $\endgroup$
    – Jean Delinez
    Commented Feb 9, 2010 at 3:38
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Mike Skirvin has sorted out some of the main issues here, but the language used in the original question tends to confuse matters. The framework is the Borel-Chevalley structure theory of linear algebraic groups over an arbitrary algebraically closed field. Nothing special here about $\mathbb{C}$. Starting with $G$ (which can be assumed connected), define a parabolic subgroup to be a closed subgroup $H$ for which the quotient variety $G/H$ is projective. By analogy with some classical examples, this variety is usually referred to as a "partial flag variety". Note that the construction of a quotient variety in this setting involves a morphism $G \rightarrow G/H$ whose fibers are $H$ and its other left cosets. Thus $H$ must be closed in order for $G/H$ to make sense as a variety. The language of manifolds doesn't come in unless this picture is compared with the Lie group picture for $\mathbb{C}$, where fortunately everything translates well for reductive groups.

In the structure theory, it is shown that $H$ is parabolic precisely when it contains some Borel subgroup, then that $H$ is connected and self-normalizing, etc. All of this is most interesting when $G$ is reductive (and has a nontrivial semisimple derived group).

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(iv) Theorem (Borel-Hirzebruch 1958, maybe?) A projective variety in characteristic zero with a transitive action of a connected linear algebraic group is smooth and Fano, i.e., its anticanonical bundle is ample.

This means "most" varieties don't arise as quotients by parabolic subgroups. In particular, curves of positive genus, and smooth hypersurfaces in $\mathbb{P}^n$ of degree at least $n+1$ make good examples.

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  • $\begingroup$ I guess this theorem comes from the two-part paper by Borel and Hirzebruch on homogeneous spaces and characteristic classes published in 1958-59? In any case, smooth projective varieties (or compact complex manifolds) of the form $G/P$ are scarce thanks to the Borel-Chevalley structure theory and the associated root system data. Even for smooth Schubert varieties, I'd expect few other than flag varieties to be of this form but can't quantify this. $\endgroup$
    – Jim Humphreys
    Commented May 30, 2010 at 22:07
  • $\begingroup$ That sounds like a correct reference, although I have never read the paper itself. I'm inclined to agree that the class of Fano varieties is itself far larger than the class of varieties of the form G/P, but I don't know any more adjectives (except "homogeneous" which is cheating) that narrow things down correctly. $\endgroup$
    – S. Carnahan ♦
    Commented May 31, 2010 at 0:20
  • $\begingroup$ Two important properties that $G/P$ homogeneous spaces have are (1) they are defined over $\mathbb{Z}$ and (2) they have good reduction everywhere (possibly, with modulo assumptions on the center of $G$). Already for del Pezzo surfaces obtained by blowing up a few points on $\mathbb{P}^2$, (2) fails. $\endgroup$
    – Victor Protsak
    Commented May 31, 2010 at 4:22
  • $\begingroup$ Victor, nice observation! The coset space $G/P$ always descends to a smooth proper scheme over the entirety of $\mathbf{Z}$, no hypotheses needed on the center. But proving that a scheme over $\mathbf{C}$ doesn't admit a descent to a smooth proper $\mathbf{Z}$-scheme seems to require more than verifying that a particular $\mathbf{Q}$-descent doesn't have everywhere good reduction. So is the rigorous proof of your surface example nonetheless easy? $\endgroup$
    – BCnrd
    Commented Jun 1, 2010 at 15:35

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