Kähler potentials that depend only on geodesic distance - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:11:26Z http://mathoverflow.net/feeds/question/101196 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101196/kahler-potentials-that-depend-only-on-geodesic-distance Kähler potentials that depend only on geodesic distance Oliver Jones 2012-07-03T01:41:13Z 2012-08-12T14:11:58Z <p>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? </p> http://mathoverflow.net/questions/101196/kahler-potentials-that-depend-only-on-geodesic-distance/101766#101766 Answer by Robert Bryant for Kähler potentials that depend only on geodesic distance Robert Bryant 2012-07-09T12:27:31Z 2012-08-12T14:11:58Z <p><strong>Update:</strong> 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.</p> <p>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 $$\partial f \wedge \bar\partial f \wedge \bigl(\partial\bar\partial f\bigr)^{n-1} = \frac{1}{4n}\phi(f)\ \bigl(\partial\bar\partial f\bigr)^{n}.$$ 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 $p$.</p> <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 $$\partial s \wedge \bar\partial s \wedge \bigl(\partial\bar\partial s\bigr)^{n-1} = \frac{1}{n}s\ \bigl(\partial\bar\partial s\bigr)^{n}. \tag1$$ 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.</p> <p>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.) </p> <p>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. </p> <p><strong>Original Answer:</strong> 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.</p> <p>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 <em>polar</em> 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 <em>polar potential</em> for $(g,\Omega)$ at $p$. If a polar potential at $p$ exists on some $\epsilon$-ball about $p$, it is unique on that ball.</p> <p>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 $p$ <em>within the complex curve</em>. </p> <p>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. </p> <p>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 $p$.</p> <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.</p> <p>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.</p> http://mathoverflow.net/questions/101196/kahler-potentials-that-depend-only-on-geodesic-distance/102007#102007 Answer by Oliver Jones for Kähler potentials that depend only on geodesic distance Oliver Jones 2012-07-12T00:36:14Z 2012-07-12T02:11:22Z <p>@Robert: This was too long for the comment field and so I'm putting it here. </p> <p>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 Fubini-Study 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 Fubini-Study distance solely as a function of the geodesic distance. Thanks very much for your answer.</p>