Timeline for Kähler potentials that depend only on geodesic distance
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2012 at 14:11 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Completed the answer to the original question
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| Jul 12, 2012 at 0:36 | vote | accept | Oliver Jones | ||
| Jul 11, 2012 at 11:57 | comment | added | Robert Bryant | @Oliver: Yes, he slipped on this. Consider the Poincaré disk metric in polar coordinates: $g=dr^2+\sinh^2rd\theta^2$ and the area form is $\Omega=\sinh rdr\wedge d\theta$. The Kähler potential is $f=8\log(\cosh(\frac12r))$. On the product of two copies, $g=dr_1^2+\sinh^2r_1\ d\theta_1^2+dr_2^2+\sinh^2r_2\ d\theta_2^2$ and $\Omega=\sinh r_1dr_1\wedge d\theta_1+\sinh r_2dr_2\wedge d\theta_2$ while $r = \sqrt(r_1^2+r_2^2)$ is the geodesic distance from the point $(r_1,r_2)=(0,0)$. However, there is no function $h$ of 1 variable so that $f=h(r)$ is a Kähler potential for this metric. | |
| Jul 11, 2012 at 6:19 | comment | added | Oliver Jones | @Robert: OK, so you're saying Gromov is wrong. I'll need time to study your comments. | |
| Jul 11, 2012 at 5:16 | comment | added | Robert Bryant | @Oliver: I looked at Gromov's paper; the example you quote is on the second page. He clearly intended the polar claim for all Hermitian symmetric spaces, just not the claim of boundedness for $df$ (which is what he really cared about) if it has an Euclidean factor. In spite of his claim, Hermitian symmetric spaces of rank bigger than one do not have the polar property (whether it has an Euclidean factor or not). The argument that I gave above for $S^2xS^2$ works for the product of two Poincaré disks as well, which shows that this product, which has no Euclidean factor, isn't polar either. | |
| Jul 10, 2012 at 22:08 | comment | added | Oliver Jones | @Robert: Gromov had the proviso that the Hermitian symmetric space has no Euclidean factor. That takes care of your product example. Sorry for the omission. His paper is available online or I can send it to you if you're interested. | |
| Jul 10, 2012 at 19:48 | comment | added | Robert Bryant | @Oliver: Here's a way to see that the irreducible higher rank Hermitian symmetric spaces can't be polar (at any point): The polar property is clearly inherited by any totally geodesic complex submanifold (with the induced Kähler structure). So, for example, if the real Grassmannians $Gr(2,n)=SO(n)/(SO(2)SO(n))$ were polar for $n>3$, then $Gr(2,4)=S^2xS^2$ would be polar, but it's not. The Kähler potential for $S^2$ is $f(r) = 8\log(\sec(r/2))$ and the distance function for $S^2xS^2$ is $r = \sqrt(r_1^2+r_2^2)$, but, clearly, no function of this $r$ can be a Kähler potential for $S^2xS^2$. | |
| Jul 10, 2012 at 12:38 | comment | added | Robert Bryant | @Oliver: Yes, I chose polar arbitrarily so that I'd have something to call the property; I was thinking of 'pole of rotation', nothing deeper than that. I'm surprised that Gromov would claim this; it's not true for the product of the Poincaré disk and the complex plane, which is an Hermitian symmetric space, albeit a reducible one. Perhaps Gromov meant to write 'rank 1 Hermitian symmetric space' and inadvertently omitted the rank 1 part? I can put in a sketch of the argument if you are interested. | |
| Jul 10, 2012 at 3:36 | comment | added | Oliver Jones | Robert, is the term \textit{polar} your terminology? If so, what motivated it? I found a paper by Gromov entitled "K\"{a}hler Hyperbolicity and $L_2$-Hodge Theory" in which he states that for Hermitian symmetric spaces the K\"{a}hler potential is expressible as a function of geodesic distance. I didn't see any restrictions on curvature. Have I missed something? | |
| Jul 9, 2012 at 16:04 | history | edited | Robert Bryant | CC BY-SA 3.0 |
fixed typos
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| Jul 9, 2012 at 12:27 | history | answered | Robert Bryant | CC BY-SA 3.0 |