2
$\begingroup$

Let $(M,L) $ be a polarized projective variety and is Chow stable, then under which condition it is K-stable in sense of Donaldson's definition?

Here is a good referrence for the definitions of Chow stability and k-stability

Added: I know that the coadjoint orbits have balanced metrics and so they are Chow stable, but I wanted to see that "are the coadjoint orbits, K-stable?". So this is the motivation of this question.

$\endgroup$

2 Answers 2

3
$\begingroup$

Yes, coadjoint orbits polarized by the anticanonical bundle are $K$-stable.

Let $O$ be an orbit of a vector $v$ in $\mathfrak{k^*}$ under the coadjoint action of a compact lie group $K$ with complexification $G$. Then the moment map $$ μ:G/P\rightarrow \mathfrak{k^*} $$ is a diffeomorphism between $O$ and the flag variety $G/P$ where $P$ is the complexification of the stabilizer of $v$. The coadjoint orbit has the structure of a projective variety induced by this diffeomorphism, so $μ$ is trivially an isomorphism.

Anticanonically polarized flag varieties are Kähler-Einstein Fano manifolds, so they are $K$-stable by the stability part of Yau-Tian-Donaldson correspondance.

Berman, Robert J. "K-polystability of Q-Fano varieties admitting Kahler-Einstein metrics." arXiv preprint arXiv:1205.6214 (2012).

D. V. Alekseevsky and A. M. Perelomov: Invariant K\"ahler-Einstein metrics on compact homogeneous spaces, Funct. Anal. Appl. 20 (3) (1986) 171--182.

What is the Explicit Relationship between Coadjoint Orbits and Flag Manifolds?

Edit: Since $Aut(G/P)$ is certainly not discrete, other Kähler classes do not seem to be handled by these general theorems, contrary what I originally wrote.

$\endgroup$
2
  • $\begingroup$ Coadjoint orbits are subvarieties of the affine space $\mathfrak{g}^*$, they cannot be isomorphic to a flag variety which is projective. $\endgroup$
    – abx
    Commented Feb 11, 2015 at 7:46
  • $\begingroup$ I have clarified your point in my answer. $\endgroup$ Commented Feb 11, 2015 at 16:34
2
$\begingroup$

I guess you are asking about asymptotic Chow stability (as defined in your reference) rather than Chow stability. There is no real link between Chow stability and K-stability.

It is an open problem to show that for a smooth polarised variety (with discrete automorphism group), K-stability implies asymptotic Chow stability. This seems to have been noted first by Ross-Thomas in "A study of the Hilbert-Mumford criterion for the stability of projective varieties", after Theorem 3.9.

If you allow automorphisms, Ono-Sano-Yotsutani have a counterexample in " An example of asymptotically Chow unstable manifolds with constant scalar curvature". Note that their example is Kaehler-Einstein, so Berman's paper "K-polystability of Q-Fano varieties admitting Kahler-Einstein metrics" implies K-polystability.

If you relax the smoothness condition, Odaka has examples of K-stable orbifolds which are not asymptotically Chow stable. See "The Calabi Conjecture and K-stability", Section 3.

Note that in Odaka's examples asymptotic Chow stability involves embeddings into projective space. If one uses embeddings to weighted projective space as in Ross-Thomas's "Weighted projective embeddings, stability of orbifolds and constant scalar curvature Kähler metrics", I believe it is expected that orbifolds are asymptotically Chow stable in that sense.

As you have probably noticed in your reference, it is true that asymptotic Chow semistability implies K-semistability. This is essentially because the Donaldson-Futaki invariant appears as the leading order term in the polynomial governing asymptotic Chow semistability. Since this polynomial is non-negative, its leading term must be non-negative.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer. I almost knew your comments. In fact, I know that the coadjoint orbits have balanced metrics and so they are Chow stable, but I wanted to see that "are the coadjoint orbits, K-stable?". So thats why I asked this question $\endgroup$
    – user21574
    Commented Jan 14, 2015 at 16:35

You must log in to answer this question.