3
$\begingroup$

I am trying to understand the statement of Satz 1 in Über kompakte homogene Kählersche Mannigfaltigkeiten by Borel. Here is the statement in German

Satz I: Jede zusammenhängende kompakte homogene Kählersche Mannigfaltigkeit ist das direkte Produkt aus einem komplexen Torus und einer projektiv-rationalen Mannigfaltigkeit.

I think this states that a compact homogeneous Kähler manifold is a product of a complex torus and a projective rational manifold. Now surely the projective rational manifold should also be homogeneous? Is this somehow implied? If this is the case then does that mean that projective rational manifold is in fact a complex flag manifold?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ @CarloBeenakker, thanks for repairing those typos... :) $\endgroup$
    – paul garrett
    Commented Apr 14 at 18:06
  • $\begingroup$ A priori, why should each factor of a decomposable homogeneous space be homogeneous? There's no obvious-to-me reason the automorphism group should respect the factors. $\endgroup$
    – LSpice
    Commented Apr 14 at 18:31

2 Answers 2

Reset to default
2
$\begingroup$

If you read through the paper, Borel and Remmert briefly describe their proof strategy:

They show that every compact homogeneous complex manifold $V$ can always be fibered in two different ways:

  1. $V \to \operatorname{Alb}(V)$ is a holomorphic fiber bundle;
  2. $V$ is canonically a holomorphic fiber bundle over a projective rational homogeneous manifold with complex-parallelizable fiber.

If $V$ is additionally Kähler, then these relations can be exploited further.

Then, if I understand correctly, the direct product comes from bundle trivialization, where the second factor is the aforementioned projective rational variety, hence homogeneous - they just don't mention that last part in the formulation of Satz I.

Indeed, the relevant step here appears to be Satz 5: "Let $Q$ be a connected projective rational manifold and let $E \to A$ be a fiber bundle over a complex torus $A$ with typical fiber $Q$. Then $E$ is a homogeneous complex manifold if and only if $Q$ is homogeneous and $E \cong A \times Q$ (biholomorphic and as bundles over $A$)".

In other words, they show that $V \cong \operatorname{Alb}(V) \times Q$, with $Q$ being homogeneous implied by Satz 5.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks a lot for the answer! Is it clear that $Q$ is a complex flag manifold. $\endgroup$
    – Bobby-John Wilson
    Commented Apr 15 at 16:12
  • $\begingroup$ @Bobby-JohnWilson: I was only addressing the first part of your question about the projective rational manifold being homogeneous. I don't know much about flag manifolds and reductive groups. However, it doesn't seem to me to be the case without some further assumptions on $V$. Namely, by Bochner-Montgomery (which they also cite), $\operatorname{Aut}(V)$ is a finite-dimensional complex Lie group. The fiber $Q$ is acted upon by $\operatorname{Ker}(\gamma \colon \operatorname{Aut}(V) \to \operatorname{Aut}(\operatorname{Alb}(V))$, which is a closed complex Lie subgroup. (cont.) $\endgroup$
    – M.G.
    Commented Apr 15 at 17:43
  • $\begingroup$ @Bobby-JohnWilson: (cont.) It's not clear to me (due to my ignorance on the subject) why this subgroup should be reductive (and hence $Q$ be a generalized flag manifold with respect to it). Maybe you should post this as a separate question. Generally speaking, it's not a good idea to post two so different questions in one and the same MO post. It's possible that there might be some hint hidden somewhere in their exposition as to whether this is the case or not, though. $\endgroup$
    – M.G.
    Commented Apr 15 at 17:46
  • $\begingroup$ @Bobby-JohnWilson: Actually, have a look at Folgerung 1 at the end of the paper. They say that $\operatorname{Aut}(V)^0$ ($G(V)$ in their notation) is reductive. Maybe this helps further. $\endgroup$
    – M.G.
    Commented Apr 15 at 17:55
  • 1
    $\begingroup$ Thanks a lot for the detailed response. $\endgroup$
    – Bobby-John Wilson
    Commented Apr 16 at 10:53
5
$\begingroup$

An earlier source than Borel-Remmert (1962) is Matsushima (1957), who writes in French:

Tout espace homogène kählérien compact est produit kählérien d'un tore complexe et d'un espace homogène kählérien d'un groupe de Lie compact semi-simple.

Hasegawa cites both works and formulates the theorem in English, as follows:

A compact homogeneous complex Kähler manifold is biholomorphic to the product of a complex torus and a homogeneous rational manifold (which is a compact simply connected algebraic manifold}.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Will this homogeneous rational manifold necessarily be a complex flag manifold? $\endgroup$
    – Bobby-John Wilson
    Commented Apr 14 at 17:23
  • 1
    $\begingroup$ I don't see anything in the German that corresponds to "biholomorphic" in the English translation. In the other direction, "zusammenhängende" in the German has disappeared in the translation. $\endgroup$
    – Andreas Blass
    Commented Apr 14 at 17:47
  • $\begingroup$ @AndreasBlass "zusammenhängend" means connected. I assume that is only there to avoid silly counter examples. The German original states the equivalence simply as 'is', the French says 'is a Kähler product of', this was translated as 'is biholomorphic to'. I don't know enough about the maths to compare these. $\endgroup$
    – quarague
    Commented Apr 15 at 11:01
  • $\begingroup$ @AndreasBlass: in the original paper, "biholomorph" is actually in Satz 5, where they prove the relevant bundle trivialization. $\endgroup$
    – M.G.
    Commented Apr 15 at 15:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .