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Oct 16, 2017 at 15:47 vote accept Benjamin
Jan 27, 2017 at 3:07 comment added user21574 Allen was right, see corollary page 5, cambridge.org/core/services/aop-cambridge-core/content/view/…
Jan 25, 2017 at 5:01 comment added Allen Knutson A note about @YuryUstinovskiy's comment: the K\"ahler metric on a coadjoint orbit of a compact group is not the metric restricted from $\mathfrak g^*$. Indeed, for $SU(2)$ where the coadjoint orbits are concentric $2$-spheres, the K\"ahler area of an orbit is proportional to its radius, not the square of its radius.
Jan 25, 2017 at 4:28 comment added user21574 there is a difference between real coadjoint orbit, and complex coadjoint orbit. real coadjoint orbit is affine, but complex coadjoint orbit is projective . see my old question mathoverflow.net/questions/156394/is-g-t-a-projective-variety
Jan 25, 2017 at 4:12 comment added user21574 Note that, I said complex coadjoint orbit, .ie $G^{\mathbb C}/P$, which is projective variety
Jan 25, 2017 at 0:04 comment added Yury Ustinovskiy If your $M$ is a coadjoint orbit as in Hassan's comment, then $M$ is a submanifold of $\mathfrak g^*$ and naturally reductive metric is given by restricting the metric $\kappa$ from $\mathfrak g^*$.
Jan 25, 2017 at 0:02 comment added Yury Ustinovskiy Depends on what you mean by expression. If you have fixed identification $\mathfrak k^\perp\simeq T_pM$, then the naturally reductive metric is given the restriction of the Killing metric, which for SU(n) is just the Frobenius norm.
Jan 24, 2017 at 22:24 comment added Benjamin Is there an expression for the metric?
Jan 24, 2017 at 21:34 history edited user21574 CC BY-SA 3.0
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Jan 24, 2017 at 21:19 comment added YCor OK I thought you referred to the reductive metric on $SU(n)$. Indeed there's no ambiguity on $SU(n)/K$ (if one still refers to the reductive metric).
Jan 24, 2017 at 21:05 comment added Yury Ustinovskiy Sorry, I am not sure I got your point. I am talking about invariant metrics on homogeneous spaces, and they admit only left group action, unless it is some very special case, like $M=SU(n)$
Jan 24, 2017 at 20:50 comment added YCor in your last sentence, you probably mean left-invariant? the bi-invariant one is unique up to scalar multiplication. "invariant" means both left or bi-invariant according to authors.
Jan 24, 2017 at 20:23 history answered Yury Ustinovskiy CC BY-SA 3.0