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Jul 12, 2012 at 22:00 comment added Oliver Jones @Robert: yes, good point. When formulating the example above I had in mind Calabi's diastasis potential. This has a nice property with respect to Kahler submanifolds. However, the diastasis is constructed from an arbitrary potential and I don't think it would be difficult to show that if there existed a polar potential then the diastasis must also be polar. I'll check.
Jul 12, 2012 at 11:59 comment added Robert Bryant @Oliver: Well, since the Kähler isometric embedding of a compact type Hermitian symmetric space into a complex projective space that you speak of is not totally geodesic in the higher rank case, the argument I was mentioning above doesn't really apply. I don't see how you are drawing your nonexistence conclusion. I can see that this says that there is a Kähler potential that is expressed in terms of $d_{FS}$ and it's clear that $d_{FS}$ is not a function of $d_{M}$, but how does that prove that there's not a different Kähler potential that is expressible in terms of $d_{M}$?
Jul 12, 2012 at 2:11 history edited Oliver Jones CC BY-SA 3.0
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Jul 12, 2012 at 0:36 history answered Oliver Jones CC BY-SA 3.0