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For various reasons, I'm interested in working with complex projective varieties that are also principal bundles. I began by looking at projective spaces themselves $\mathbb{CP}^n = SU(n+1)/U(n)$, then generalised to Grassmannian spaces $Gr(n,d) = U(n)/U(d) \times U(n-d)$, and finally to flag manifolds $U(n)/U(k_1) \times \cdots \times U(k_m)$. I am now looking for other examples of projective varieties that are also principal bundles. Does anyone know of any?

EDIT: Sorry for asking such an ill-posed question, I didn't realize I had made so many tacit assumptions. What I am looking for are principal $G$ bundles $\pi:P \to X$ such that $G$ and $P$ are both compact Lie groups (ideally matrix groups), and $X$ is a projective variety. As far as I understand it is not always possible to construct such a bundle for a projective variety, one obstruction being the absence of a continuous symmetry group.

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It looks like you mean "homogeneous space" (space with a transitive group action) rather than "principal bundle" (morphism with a free transitive group action on fibers). –  Anton Geraschenko Feb 6 '10 at 6:32
But isn't every homogeneous space automatically a principal bundle? As far as I can see, for any homogeneous $G$-space $X$, with stabilizer subgroup $H$ of some point $x \in X$, we have $X = G/H$. The projection $\pi:G \to G/H$, gives $G$ a fibred structure and the action of $H$ on the fibres is free and transitive, making $\pi:G \to X$ a principal $H$-bundle. Yes, all the examples I mentioned are homogeneous spaces, but I would be interested in examples that only principal bundles. –  Jean Delinez Feb 6 '10 at 8:38
I am not sure I understand this question. What in the principal fibration is supposed to be the complex projective variety? The base? The total space? In the examples you have provided, it seems that it is the base of the fibration. If so, then just simply take your favourite variety and stick a principal bundle on it. Clearly this is too trivial an answer, so more likely than not I'm misunderstanding the question. –  José Figueroa-O'Farrill Feb 6 '10 at 10:12
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2 Answers

You probably mean that you are interested in principal bundles on a projective variety $X$ with fiber being a compact Lie group $G$. If you are interested in the topological classification of such bundles, then this is a classical problem, and the bundles are classified by homotopy classes of mappings from $X$ to the classifying space $BG$ (the structure of a projective variety on $X$ is not important here, only its topology). A lot is known about the classification of such mappings; in particular, this problem gives rise to the theory of characteristic classes, and (in the case of $G=U(n)$, i.e. complex vector bundles) to topological $K$-theory, which describe invariants of such bundles. Wikipedia contains lots of information and references on this.

Another problem is to look at bundles with an additional structure, usually a connection. For example, if $X$ is a curve, it is interesting to look at bundles with a flat connection. These bundles form an interesting moduli space, which is the same as the space of stable holomorphic bundles with fiber being the complexification $G_{\Bbb C}$ of $G$, for simple simply connected $G$ (the Narasimhan-Seshadri theorem). If $G=U(1)$, this moduli space is just the Jacobian of $X$. There is an analog of the Narasimhan-Seshadri theorem to higher dimensions, which is called the Donaldson-Uhlenbeck-Yau theorem.

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He seems to like principal bundles whose total space is a projective variety. One needs compact base and a compact group for that, at least, but that is probably not sufficient. Maybe he is asking for more examples of this situation? –  Mariano Suárez-Alvarez Feb 6 '10 at 13:54
Well, in the examples given in the question it is the base that is projective, not the total space (which is $U(n)$). –  Pavel Etingof Feb 6 '10 at 14:10
Urgh. You are right. I do need coffee... –  Mariano Suárez-Alvarez Feb 6 '10 at 14:32
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As I interpret your question, you're looking for principal bundles where the base is projective, and both the fiber and total space are both compact groups.

There's an obvious class of these, which is taking any compact group $P$ and modding out by a Levi subgroup $G$ (I'm trying to match the notation in your question. I apologize to the Lie theorists in the audience for using the most confusing notation ever) [EDIT: in the original, this had read "containing the maximal torus" since I'd forgotten that there are non-Levi subgroups which contain the maximal torus]. This quotient will always be a projective variety.

I believe that one can prove that this is the only way of getting a quotient which is projective (note that the complexification of $P$ acts on the quotient, since it is compact, and the complexification of the Lie algebra acts; now use the Borel fixed point theorem). [EDIT: it seems that this isn't true. For example, elliptic curves exist.]

It's possible that there's some strange way of making $G$ act on $P$ that's not a subgroup, but I think the above are the right class of examples generalizing the Grassmannian. [EDIT: this paragraph at least is vague enough to not be false, but as pointed out in comments, there are other examples]

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Using biquotient constructions (which are natural generalizations of homogenous actions), one can find an actions of $T^2$ on $S^3\times S^3$ whose quotient spaces are$\mathbb{C}P^2$#$\pm \mathbb{C}P^2$ . One still gets a principal fiber bundle $H\rightarrow G\rightarrow G/H$ just as in the homogeneous action case. That said, I don't know enough algebraic geometry to know whether either of these two spaces are projective varieties. –  Jason DeVito Feb 6 '10 at 17:34
I think the statement that this is the only way to get a projective quotient is not quite right (although easy to correct). Take P=2-torus, G=1. Perhaps the irreducible projective varieties admitting the representation P/G are products of an abelian variety and a coadjoint orbit of a compact group as explained above. –  Pavel Etingof Feb 6 '10 at 21:36
Also I think "G containing the maximal torus of P" is not sufficient: G has to be a Levi subgroup of P (i.e. $P/G$ is a coadjoint orbit of $P$). E.g. $SO(5)/SO(4)=S^4$ is not a complex manifold. –  Pavel Etingof Feb 6 '10 at 21:40
Good points all. This is why I should never try to phrase things in terms of compact groups. Complex ones are so much nicer. –  Ben Webster Feb 6 '10 at 21:50
The "compact" way of saying "Levi" is: centraliser of a torus. At the complex level these are the generalised flag manifolds $H/Q$ for $H$ complex semisimple and $Q$ parabolic. These are the only projective homogeneous spaces with $H$ semisimple (theorem of Wang from the 1950's). –  Fran Burstall Jun 5 '12 at 21:06
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