## Harmonic function with gradient of constant norm in hyperbolic 3-space

I investigate certain stationary charged perfect fluid solutions to the Einstein-Maxwell equations in general relativity. The classification of these solutions has led me to the following question:

Does hyperbolic 3-space admit a non-constant harmonic function (i.e., with vanishing Laplacian) that has a gradient of constant norm?

A non-constant affine function (i.e., with vanishing Hessian) would fit this description. However, hyperbolic space does not admit such a function.

T. Sakai: "On Riemannian manifolds admitting a function whose gradient is of constant norm", Kodai Math. J. 19 (1996), 39-51.

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This nonexistence follows from an old result of É. Cartan, who shows that the only isoparametric hypersurfaces in hyperbolic space are the trivial ones. – Robert Bryant Oct 26 at 12:21
@ Robert: can you expand your comment to an answer with a reference and explanation of how isoparametric hypersurfaces are related to harmonic functions? – Agol Oct 26 at 15:38
@Agol: OK. I don't have the papers here handy with me, but I'll do my best with what I can find on the net. – Robert Bryant Oct 27 at 0:21

The main paper you want to consult is

É. Cartan, Familles de surfaces isoparamétriques dans les espaces à courbure constante, Annali di Mat. 17 (1938), 177–191.

In this paper, Cartan considers the problem of studying the functions $f$ defined on an open set in a space $M$ of constant curvature that satisfy two equations of the form $|\nabla f|^2 = A(f)$ and $\Delta f = B(f)$ for some functions, $A>0$ and $B$, of one variable. He shows that the (hypersurface) level sets of $f$ are what he called isoparametric, i.e., they have all of their principal curvatures constant.

This had been considered somewhat earlier earlier by Levi-Civita for the case $n=3$ in flat space and by Segre for all $n$ (again, in flat space). They showed that, up to rigid motion and homothety, a connected isoparametric hypersurface must be an open subset of $S^k\times \mathbb{R}^{n-k-1}$ for some $0\le k\le n{-}1$.

Cartan showed that, when the ambient sectional curvature is negative, i.e., in hyperbolic $n$-space, then, again, a connected isoparametric hypersurface must either be totally umbilic (i.e., an open subset of either a totally geodesic hyperplane, an equidistant hypersurface, a horosphere, or a hypersphere) or be an open subset of a hypersurface of the form $S^k\times H^{n-k-1}$, i.e., it must be a totally geodesic hypersurface or horosphere or else the set of points of constant distance from a totally geodesic submanifold of lower dimension. Thus, if you did have a solution of $|\nabla f|^2=1$ and $\Delta f = 0$ (even a local one in a neighborhood of $p$) in hyperbolic $n$-space, then $f-f(p)$ would be the oriented distance from the level set $f=f(p)$, and it is easy to calculate that no such function is harmonic, so you have a contradiction.

By the way, there is a much greater variety of isoparametric hypersurfaces in spaces of constant positive curvature. In fact, even though Cartan found many nontrivial examples, and there has been a lot of work since then constructing more examples and classifying them, the classification of these is still not complete.

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 @Robert, do you know any result regarding the equal-distance flow of hypersurface in Riemannian manifold, which is somehow related to this question. – J. GE Oct 27 at 9:33 @GB: Your question is a bit vague. Could you say more about what you seek? Are you looking for examples of parallel foliations by hypersurfaces, conditions under which such things exist globally (they always exist locally), characterizations of when the distance function from a hypersurface satisfies some 'natural' second order PDE, or what? – Robert Bryant Oct 27 at 11:14 @Robert, The question I ask is the following: Suppose $\Sigma$ is a hypersurface in $M^n$, consider the distance function $f(x)=dist(x, \Sigma)$. Then the level set of $f$ can be considered as the flow of $\Sigma$ along the gradient curve of $\nabla f/ |\nabla f|^2$. This is not mean curvature flow or gauss curvature flow, it is only of first order. I just curious is there any analytic work has been done on the study singularities of this kind of flow? Of course one can add some extra conditions on curvature or $\Sigma$ when needed. – J. GE Oct 27 at 13:04 @GB: There's not really much in the way of 'analytic' work to be done on this flow, which is just the construction of parallel hypersurfaces to a given hypersurface in an ambient manifold. If the ambient manifold has constant sectional curvature, then one knows how to tell from the second fundamental form when the family of parallel hypersurfaces develop singularities and what kind they have. There are global results, such as the fact that for flat $n$-space, the only hypersurfaces for which the 'flow' does not develop singularities are the hyperplanes. Is that what you had in mind? – Robert Bryant Oct 27 at 23:32 @Robert, thanks. I was trying to find an upper bound for the time when singularity developed, given a lower bound of the second fundermantal form. – J. GE Oct 28 at 11:11
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Such a harmonic function does not exist. Here's a sketch of a proof. Consider a level set $\Sigma_c$ of $f=c$. The gradient $\nabla f$ is a constant length vector perpendicular to $\Sigma_c$ at each point of $\Sigma_c$ (let's assume $|\nabla f|=1$). Also, the vector field $\nabla f$ is divergence-free, since $f$ is harmonic, and thus the flow by $\nabla f$ is volume preserving. Now, flow by the vector field $\nabla f$ for time $t$ takes $\Sigma_c$ into $\Sigma_{c+t}$. Since the vectors are constant length, this gives a local orthogonal coordinate system about $\Sigma_c$, which therefore is Fermi coordinates. So the flow lines of $\nabla f$ are geodesics by the Gauss lemma. Since the flow is volume preserving and $|\nabla f|=1$, it also preserves the area of $\Sigma_{c+t}$. This implies that $\Sigma_c$ is a minimal surface, since the derivative of the variation of area under orthogonal deformation is the trace of the second fundamental form. One then computes that the principal curvatures of $\Sigma_c$ at each point are $\pm 1$. This is because under the orthogonal flow for time $t$, the second fundamental form at time $t$ is determined uniquely by the second fundamental form at time $0$. The only way that it can remain trace $0$ is if the principal curvatures are $\pm 1$. Thus, by Gauss' equation, $\Sigma_c$ is isometric to $\mathbb{H}^2_{-2}$. However, by a result of Doug Moore (generalizing a classic result of Hilbert), there is no isometric immersion of $\mathbb{H}^2_{-2}$ into $\mathbb{H}^3_{-1}$.

Addendum: Given GB's comment, I should add some explanation of why the principal curvatures are $\pm 1$. If we take a region $\Omega \subset \Sigma_c$, and let $A_t$ denote the area of the image of $\Omega$ at time $t$ under the flow, then $\partial A_t/\partial t = \int_\Omega H dA$, where $H$ is the mean curvature. Since $\nabla f$ preserves the area of $\Omega$, we see that $H=0$, so $\Sigma_c$ is a minimal surface.

The second variation of area is given by $\partial^2 A_t/\partial t^2 = \int_\Omega (2 Det(B)-Ric(\nabla f)) dA =0$, where $B$ is the second fundamental form, and $Ric$ is the Ricci curvature. Since $\Sigma_c$ is minimal, $Det(B)=-p^2$, where $p$ is a principal curvature. Also, $Ric(\nabla f)=-2$, since $|\nabla f|=1$. So we get that $p=1$, and the principal curvatures are $\pm 1$.

One can also see more directly that the principal curvatures of $\Sigma_t$ evolve like the curvature of plane hyperbolic curves (this can be seen by considering osculating spheres). The only way that the principal curvatures can remain opposite is if they are $\pm 1$.

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 @Agol "This is because under the orthogonal flow for time t, the second fundamental form at time t is determined uniquely by the second fundamental form at time 0." Is this statement trivial? – J. GE Oct 25 at 21:47 @GB: it's not trivial. I'll try to give a more detailed argument later using 2nd variation. But notice the 2nd fund. form depends only on the 2nd derivative at a point. If two surfaces agree to 2nd order, then so will the time t evolute. So the curvature depends only on the 2nd fund. form. – Agol Oct 26 at 2:33 @Agol, Thanks for the details. – J. GE Oct 27 at 9:31