Hermitian symmetric spaces of constant curvature have the property that the potential for their Kähler metric can be expresed as some function of the geodesic distance. Does anyone know if there are any general results concerning Kähler manifolds with this property?

$\begingroup$ Could you elaborate on this question? How can a metric fail to be a function of geodesic distance? $\endgroup$– Igor RivinJul 3, 2012 at 14:58

$\begingroup$ @Igor: He means that a potential for the metric is of the form $f(x)=h(d(p,x))$ where $p$ is a fixed point and $h$ is some function of one variable. $\endgroup$– MishaJul 3, 2012 at 21:52

$\begingroup$ For a hermitian symmetric space of constant curvature, this holds for any point $p$ (as defined in Misha's comment). It should not be difficult to give an example where it holds for only one particular choice of $p$ but not for others. So presumably the question is about manifolds where the potential can be written as a function of distance from a point $p$, for any choice of $p$? $\endgroup$– Deane YangJul 3, 2012 at 22:13

$\begingroup$ Also, I am under the impression that the statement is not literally true for a compact manifold, since the Hessian of the distance function has to degenerate at the cut locus. It seems to me that the question needs to be stated more carefully. $\endgroup$– Deane YangJul 3, 2012 at 22:14

$\begingroup$ @Misha: Correct Misha. The FubiniStudy, Begman, and the Euclidean metrics all have potentials of this form. @Yang: I'm speaking locally here; obviously the cut locus will need to be avoided. However, the base point $p$ is arbitrary; you simply avoid the cut locus of $p$. I've thought about this some more and I realized that the potential is also a function of $z^2=\sum_iz_i^2$. Here $(z_1,\cdots, z_n)$ are local coordinates. Rotationally symmetric metrics have this property. $\endgroup$– Oliver JonesJul 4, 2012 at 0:02
2 Answers
Update: I have had a little time to think about this further and have been able to show that if a smooth Kähler metric $g$ on a complex $n$manifold $M$ has a potential $f$ that is a function of the distance from a point $p\in M$, then the metric must be locally rotationally symmetric about $p$, i.e., the group of local $g$isometries that fix $p$ is isomorphic to $\mathrm{U}(n)$. In particular, this shows that, if it has this property with respect to every point $p\in M$, then it must have constant holomorphic sectional curvature. I'll describe the steps in my argument, but I won't put in the details (which are somewhat long) unless there is interest. I'm leaving my original answer below, for notational purposes, even though that preliminary analysis was inconclusive.
Fix notation as in the original answer: There is an open $p$neighborhood $U\subset M$ on which there exists a Kähler potential $f$ for $g$ that is a function of the $g$distance $r:U\to\mathbb{R}$ from $p$, i.e., $f = h\bigl(r^2\bigr)$ for some function $h$ and $\frac{i}{2}\partial\bar\partial f = \Omega$, where $\Omega$ is the Kähler form of $g$. (Clearly, one can assume that $h(0)=0$, and it is not hard to show that $h$ must be smooth on $[0,\epsilon)$ for some $\epsilon>0$ and must satisfy $h'(0) = 1$.) Because $r$ satisfies $\nabla r^2 = 1$, it follows that $\nabla f^2 = 4 r^2 h'(r^2)^2 = \phi(f)$ for some function $\phi$ that satisfies $\phi(0)=0$ and $\phi'(0)=4$. Elementary identities then show that $f$ must satisfy the equation $$ \partial f \wedge \bar\partial f \wedge \bigl(\partial\bar\partial f\bigr)^{n1} = \frac{1}{4n}\phi(f)\ \bigl(\partial\bar\partial f\bigr)^{n}. $$ Conversely, if $f$ is any solution of this equation on a neighborhood of $p\in M$ that satisfies $f(p) =0$, $df_p=0$, and $\frac{i}2\partial\bar\partial f >0$ at $p$ (in the sense of $(1,1)$forms, then the Kähler form $\frac{i}2\partial\bar\partial f$ deinfes a Kähler metric $g$ such that $\nabla f^2 = \phi(f)$ and, from this it easily follows that $f$ is a function of the $g$distance from $p$. It remains now to show that such an $f$ must be rotationally symmetric in the above sense with respect to some $\mathrm{U}(n)$action on a neighborhood of $p$.
This problem can be simplified in the sense that, if such an $f$ exists, it can be shown that there is a function $\psi$ defined on a neighborhood of $0$ such that $\psi(0)=0$ and $\psi'(0)=1$ and such that $s = \psi(f)$ will satisfy $$ \partial s \wedge \bar\partial s \wedge \bigl(\partial\bar\partial s\bigr)^{n1} = \frac{1}{n}s\ \bigl(\partial\bar\partial s\bigr)^{n}. \tag1 $$ In other words, one can reduce to the case $\phi(t) = 4t$ (which is the same as saying that one reduces to the case in which $f = r^2$). Thus, we assume that $s$ satisfies (1) from now on.
The next step is to show, using a straightforward Taylor series argument, that, for any $k\ge2$, there exists a local, $p$centered holomorphic coordinate chart $z_k:U_k\to\mathbb{C}^n$ in which $s  z_k^2$ vanishes to order $k$. This shows that the Kähler metric associated to $s$ is flat to infinite order at $p$, but, since we do not (yet) know that $g$ is real analytic, this does not show that $s$ is flat everywhere. (Also, one does not (yet) know that the sequence of coordinate charts $z_k$ has a convergent subsequence.)
Finally, one uses the fact that each geodesic through $p$ lies in a totally $g$geodesic complex curve passing through $p$ to show that the $(2,0)$ part of the Hessian of $s$ restricts to each such curve to be generated by pullback of a holomorphic map from the complex curve to the Hermitian symmetric space $\mathrm{Sp}(n,\mathbb{R})/\mathrm{U}(n)$. Since this map must have its differential vanish to infinite order at $p$ (by the argument in the previous paragraph), it follows that it must be a constant map, which implies that the $(2,0)$ part of the Hessian of $s$ must vanish identically, i.e., that $\mathrm{Hess}(s) = g$. From this, it is elementary to conclude that $g$ must be flat and hence $s$ must be invariant under the $\mathrm{U}(n)$rotations. Since the original $f$ is a function $s$, it, too, must be invariant under this $\mathrm{U}(n)$action, and, in particular, must be of the form $f = h\bigl(z^2\bigr)$ for some function $h$, as desired.
Original Answer: This is only a preliminary answer, but it's too long to go into a comment, so I'm putting it here. I'll add to it when I have the time to figure out more about it.
It seems that there are two questions here, a pointwise one and a local one: First, let's say that a (smooth) Kähler metric $g$ on a complex manifold $M$ is polar at $p\in M$ if there is an open $p$neighborhood $U\subset M$ on which there exists a Kähler potential $f$ for $g$ that is a function of the $g$distance $r_p:U\to\mathbb{R}$ from $p$, i.e., $f = h\bigl({r_p}^2\bigr)$ for some function $h$ and $\frac{i}{2}\partial\bar\partial f = \Omega$, where $\Omega$ is the Kähler form of $g$. (Clearly, one can assume that $h(0)=0$, and it is not hard to show that $h$ must be smooth on $[0,\epsilon)$ for some $\epsilon>0$ and must satisfy $h'(0) = 1$.) Let's say that such an $f$ is a polar potential for $(g,\Omega)$ at $p$. If a polar potential at $p$ exists on some $\epsilon$ball about $p$, it is unique on that ball.
Now, a polar potential satisfies a natural differential equation, namely, the differential equation that states that the $g$gradient lines of $f$ are $g$geodesics. This is, typically, an overdetermined equation for a pseudoconvex potential $f$, so one can hope to get some information by using this. In fact, calculation shows that this implies that each of the geodesics emanating from $p$ lies in a unique, totally geodesic complex curve passing through $p$. Moreover, the induced metrics on all of these complex curves (one tangent to each complex line in $T_pM$) are isometric and, moreover, they are each possess an isometric rotation about $p$ within the complex curve.
In complex dimension $1$, this implies that the metric is rotationally symmetric about $p$, so the polar potentials are the rotationally symmetric ones. In particular, in dimension $1$, a Kähler metric that is polar at every point has constant curvature and hence is locally symmetric.
In higher dimension, since the curves are totally geodesic and isometric, they all have the same curvature at $p$ and hence it follows that the ambient metric has constant holomorphic sectional curvature at $p$.
In higher dimensions, it appears that, while rotationally symmetric (i.e., $\mathrm{U}(n)$invariant) pseudoconvex potentials are polar with respect to the center of rotation, these are not the only solutions of this equation. Already in dimension $2$, it appears, from the structure equations, that there are (local) pseudoconvex polar potentials that are not rotationally symmetric. I haven't had time to integrate the structure equations, though, so I can't yet say what these solutions look like.
However, by the above argument, a Kähler metric that is polar at every point must have constant holomorphic sectional curvature, and hence the only metrics that have this property at every point are the complex space forms.

$\begingroup$ 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? $\endgroup$ Jul 10, 2012 at 3:36

$\begingroup$ @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. $\endgroup$ Jul 10, 2012 at 12:38

$\begingroup$ @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 $S^2xS^2$. $\endgroup$ Jul 10, 2012 at 19:48

$\begingroup$ @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. $\endgroup$ Jul 10, 2012 at 22:08

$\begingroup$ @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. $\endgroup$ Jul 11, 2012 at 5:16
@Robert: This was too long for the comment field and so I'm putting it here.
I have another example which backs up what you're saying. Hermitian symmetric spaces of compact type admit a K\"{a}hler embedding into complex projective space. If we consider two points $x$ and $y$ on the embedded submanifold $M$ then there are two distances; the geodesic distance $d_M(x,y)$ in the submanifold and the FubiniStudy distance $d_{FS}(x,y)$ in the projective space. An implication of your result is that it is not possible to express $d_{FS}(x,y)$ as a function of $d_M(x,y)$ (except for complex space forms). I looked at the case of the complex Grassmannian for which there are explicit formulas for geodesic distance in terms of stationary angles. It's clear that you are correct; a simple calculation shows that it's not possible to express the FubiniStudy distance solely as a function of the geodesic distance. Thanks very much for your answer.

$\begingroup$ @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 $d_{M}$? $\endgroup$ Jul 12, 2012 at 11:59

$\begingroup$ @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. $\endgroup$ Jul 12, 2012 at 22:00