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Given a Jordan curve on the $\mathbb{H}^3$ boundary at infinity, there is a minimal surface (topological disk) in $\mathbb{H}^3$ with the curve as its asymptotic boundary (page.mi.fu-berlin.de/polthier/articles/diss/diss.pdf, Theorem 21 on p37).

Consider boundary curves in the upper half space model. If the boundary curve is a circle, the minimal surface is a geodesic plane (a hemisphere orthogonal to the boundary). A hemiellipsoid orthogonal to the boundary (a triaxial ellipsoid with two axes in the plane at infinity, and the third axis vertical) has a boundary curve that is an ellipse, and it seems likely that a hemiellipsoid is the minimal surface for this boundary curve. Why would it be something more complicated?

For the point on the hemiellipsoid through its vertical axis, it is easy to see that one (hyperbolic) principle curvature is positive and the other is negative, by considering the hemisphere with the same vertical axis through that point. But it would be nice to have an argument that the hyperbolic principle curvatures are equally opposite (update: based on comments/answers, we should be able to pick the ellipsoid parameters to make this true at this specific point), and that this is true for all points on the hemiellipsoid.

The motivation for this question comes from $\mathbb{H}^3$ fibrations, specifically about lifts of geodesics from the base $\mathbb{H}^2$ surface. For some fibrations, the lift is a hemiellipsoid in the upper half space model (see plus.google.com/+RoiceNelson/posts/1w3aoQgj61g and comments therein).

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    $\begingroup$ Can you define the hemiellipsoid more precisely? $\endgroup$
    – Mikhail Katz
    Commented Dec 19, 2015 at 16:46
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    $\begingroup$ If the boundary curve is an ellipse, you still have a free parameter left, the third half axis pointing inside the hyperbolic space. Also, I would not trust argument along the lines of "why should it be more complicated". Your argument would become a tiny bit stronger if you had a Moebius trafo that fixes the boundary ellipse and the halfellipsoid, but turns the rest inside out. $\endgroup$
    – Sebastian Goette
    Commented Dec 19, 2015 at 17:01
  • $\begingroup$ @katz, I added a little more description about the hemiellipsoid. Let me know if you still think this is not precise enough. $\endgroup$
    – Roice Nelson
    Commented Dec 19, 2015 at 17:36
  • $\begingroup$ Thanks @SebastianGoette. I see now that the length of the third axis (even if vertical) is a free parameter. Perhaps the length of this axis will be constrained to some value to make the surface minimal. I agree about not trusting the argument "why should it be more complicated?" I should have left that out of this question :) $\endgroup$
    – Roice Nelson
    Commented Dec 19, 2015 at 17:51
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    $\begingroup$ On p. 21 of Polthier's thesis, equation (2.1), you'll find the explicit minimal surface equation in the upper half-space model. I just plugged the formula for an ellipsoid into this equation in Mathematica (I'm too lazy to do the computation), and found that the only solutions are for hemispheres. $\endgroup$
    – Ian Agol
    Commented Dec 29, 2015 at 3:07

3 Answers 3

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On p. 21 of Polthier's thesis, equation (2.1), you'll find the explicit minimal surface equation in the upper half-space model. I plugged the formula for an ellipsoid into this equation in Mathematica (I'm too lazy to do the computation), and found that the only solutions are for hemispheres. I've included a screenshot from the Mathematica notebook: the appropriately scaled form of the differential equation is a quadratic form which must be identically zero. The only solution which makes the coefficients zero is when the ellipsoid is a sphere.

enter image description here

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It is clear that the triaxial hemiellipsoid is not a minimal surface in general. Indeed, consider the special case the vertical principal axis is the same as one of the axes in the plane at infinity (of course I am comparing the Euclidean lengths of the axes). In other words, this is an ellipsoid of revolution. Then at the point of the ellipsoid on the vertical axis, one of the principal directions is a geodesic and therefore the principal curvature vanishes, whereas the other direction is clearly not a geodesic (unless the ellipsoid is a hemisphere). Thus the mean curvature at this point is nonzero.

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  • $\begingroup$ This is a nice observation, and makes me suspect that the minimal surface hemiellipsoid is unique (still assuming it is ellipsoidal). Perhaps my question would have been better phrased as "what is the shape of the minimal surface for a boundary curve that is an ellipse?" $\endgroup$
    – Roice Nelson
    Commented Dec 19, 2015 at 18:04
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An ellipsoid half is not an isocurve, and therefore can not represent an isocurve in a conformal mapping. Since a minimal surface is an isocurve (in any geometry), an ellipsoid (including those at infinity), is not a minimal surface. Only a hemisphere is.

The base projection of this projection is a Möbius plane, where every circle and straight line represents a 'straight line'. The Möbius plane is the geometry of the plane at infinity of both the H3 and E3. It is just that in E3 you don't move far enough to see any appreciable difference.

The curvature of the plane at infinity is zero. But hyperbolic geometry has a negative curvature, which means the euclidean plane (and hence the Möbius plane, which has the geometry of the complete euclidean plane), is a curve.

The centre of this projection is the point opposite (ie the zenith), which is also at infinity. One of the interesting features of the Möbius plane, is that it's a conformal mapping of the Euclidean plane, if you suppose any point U is the point at infinity, and a straight line is a circle passing through U. When U is the zenith point, the conformal projection of the Euclidean plane at the base is the euclidean plane in nature.

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  • $\begingroup$ It is not clear to me how your answer helps to understand the original problem. $\endgroup$
    – Sebastian Goette
    Commented Dec 20, 2015 at 11:46
  • $\begingroup$ @SebastianGoette The base geometry of this H3 projection is not H2. It's a variety of E2, specifically EM2. The question makes assumptions that are not what I see. $\endgroup$
    – wendy.krieger
    Commented Dec 20, 2015 at 11:54
  • $\begingroup$ I am not sure that the $\mathbb H^2$ mentioned in the question is supposed to lie at infinity. I am not sure, but I thought it was a totally geodesic hyperplane inside $\mathbb H^3$. But that was only a motivational remark anyway. $\endgroup$
    – Sebastian Goette
    Commented Dec 20, 2015 at 11:58
  • $\begingroup$ @SebastianGoette If it is a straight euclidean plane, parallel to the line at infinity, then it is not a plane in H3. All planes H3 in the half-space model, are hemispheres, whose centre lies on the plane at infinity, or a plane orthogonal to the plane at infinity. $\endgroup$
    – wendy.krieger
    Commented Dec 20, 2015 at 12:01
  • $\begingroup$ Could you explain what an isocurve is? Wikipedia does not help ... $\endgroup$
    – Sebastian Goette
    Commented Dec 20, 2015 at 14:52

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