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Ben Webster ♦
Ben Webster ♦
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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) containing[EDIT: in the original, this had read "containing the maximal torustorus" since I'd forgotten that there are non-Levi subgroups which contain the maximal torus]. This quotient will always be a projective variety. I

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]

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 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) containing 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).

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.

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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Ben Webster ♦
Ben Webster ♦
  • 44.7k
  • 12
  • 126
  • 260

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 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) containing 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).

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.