# Explicit metrics

Every surface admits metrics of constant curvature, but there is usually a disconnect between these metrics, the shapes of ordinary objects, and typical mathematical drawings of surfaces.

Can anyone give an explicit and intuitively meaningful formulas for negatively curved metrics that are related to an embedding of a surface in space?

There is an easy way to do this for an open subset of the plane. If the metric of the plane is scaled by a function that is $\exp$ of a harmonic function, the scaling factor is at least locally the norm of the derivative of a complex analytic function, so the resulting metric is still flat; the converse is true as well. Therefore, the sign of curvature of a conformally modified metric $\exp(g)$ depends only on the sign of the laplacian of the $g$. If the value of $g$ at a point is less than the average value in a disk centered at that point, then the metric $\exp(g) ds_E$ is negatively curved, where $ds_E = \sqrt(dx^2 + dy^2)$ is Euclidean arc length. alt text http://dl.dropbox.com/u/5390048/NegativeMetrics.jpg

For example, in a region $R$, if we impose a limit that speed is not to exceed the distance to the complement of $R$, this defines a non-positively curved metric. (The metric is 1/(distance to boundary)$ds_E$). In this metric, geodesics bend around corners: it doesn't pay to cut too close, it's better to stay closer to the middle. If the domain is simply-connected, you see one and only one image of everything, no matter where you are.

There are a number of other ways to write down explicit formulas for negatively curved or non-positively curved metrics for a subset of the plane, but that's not the question: what about for closed surfaces in space? Any closed surface $M^2$ has at least a total of $4 \pi$ positive curvature, where the surface intersects its convex hull. If $M$ is a double torus, how can this be modified to make it negative? It would be interesting to see even one good example of a negatively curved metric defined in terms of Euclidean geometry rather than an indirect construction. (In particular: it can be done by solving PDE's, but I wwant something more direct than that.)

• This is a great question, and it is indeed somewhat surprising that nobody has figured anything like this out yet. I find it rather difficult to envision a negatively curved metric on the double torus, given what it looks like when embedded in $R^3$. We know that it is possible to change the "visual" metric into a negatively curved metric via a conformal factor, but it is highly unclear how to do this without using heavy analytic machinery. I imagine that Bill and other visually oriented mathematicians have already tried to do this, so I doubt it's an easy question to answer. – Deane Yang Sep 23 '10 at 20:40
• Although it doesn't answer your question, I've been looking at a related thing recently. Given a finite group $G$ acting on a surface $\Sigma$, try to find an embedding of $\Sigma$ in $S^3$ such that the action of $G$ on $\Sigma$ extends to an action of $G$ on $S^3$. So there's all kinds of restrictions on when this is possible, but looking at examples can be quite pleasant. – Ryan Budney Sep 23 '10 at 20:46
• @Ryan: Yes, that's a fun problem. There are many possible subgroups of $O(3)$, and many possible constructions. An interesting extra geometric condition: when are they equivariant minimal surfaces? Sometimes they're not equivariantly compressible, which I believe means they can be made into minimal surfaces. – Bill Thurston Sep 23 '10 at 21:38
• A slightly cheezy modification of Deane Yang's idea would be to remove a large disc from a hyperbolic surface, so large that the complement is a thin regular neighbourhood of a graph in the surface. Embed that regular neighbourhood in Euclidean space much like how one constructs zero Gauss curvature embeddings of cylinders and Moebius bands in Euclidean space. – Ryan Budney Sep 24 '10 at 0:13
• I'll just point out that Bill's proposed metric of nonpositive curvature for open sets in the plane does not, in general yield a $C^2$-metric. For example, for the unit disk in polar coordinates, the metric Bill proposes would be written in the form $g = (dr^2 + r^2\ d\theta^2)/(1-r)^2$, and a short computation shows that $K = -1/r$ for this metric, so, not surprisingly, it's not $C^2$ at the origin. (In general, for a domain in the plane, the distance to the boundary is continuous, but not even $C^1$.) – Robert Bryant Jun 6 '11 at 18:40

Here's an answer to an analogous question, not Bill's original question, but also a question about how to specify (in a simple way) a non-positively curved metric on a compact Riemann surface $C$ of genus $g>1$, in this instance, one that has been specified as an algebraic curve somewhere (as opposed to being given it as a surface in $3$-space).

The construction is easy: If the curve has been specified as an algebraic curve, then, more-or-less by algorithmic means, one can write down a basis for the holomorphic differentials on $C$ (which is a complex vector space of dimension $g$). Now select two of these differentials, say, $\omega$ and $\eta$, that have no common zeroes on $C$. (Again, this can be tested algebraically). Now consider the metric $g = \omega\circ\bar\omega + \eta\circ\bar\eta$. This $g$ will have non-positive curvature. In fact, the curvature will vanish at only a finite number of points and will otherwise be strictly negative. (Of course, you can add more terms. If you take a basis $\omega_1,\ldots,\omega_g$ of the holomorphic differentials on $C$, then the metric $g = \omega_1\circ\bar{\omega_1} +\cdots + \omega_g\circ\bar{\omega_g}$ will have strictly negative curvature except when $C$ is hyperelliptic, in which case, the curvature will vanish at the Weierstrass points of $C$.)

For example, if you take a hyperelliptic curve, say $y^2 = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{2g+2})$ (with the $\lambda_i$ being distinct and, say, nonzero), then a basis for the holomorphic differentials will be given by $\omega_i = x^{i-1}dx/y$ for $i = 1,\ldots, g$. Moreover, $\omega_1$ and $\omega_g$ (for example) have no common zeros. Thus, the smooth metric $g = (1 + |x|^{2(g-1)})|dx|^2/|y|^2$ has negative curvature on this curve except at a finite number of points.

• Is it possible to make sense of how to embed a hyper-elliptic curve of type $z^2=f(w)$ to the 3D Euclidean space? I can construct a metric of constant negative curvature on the hyper-elliptic curve, but I do not know any explicit embedding that may work. Curiously (according to C.N.Yang) it seems this (Thurston's) problem was considered by Chern in 1930s already. – Bombyx mori Jan 7 at 6:22
• @Bombyxmori: One cannot isometrically immerse in a $C^2$ manner a compact surface of nonpositive curvature in Euclidean 3-space because the curvature at a point of maximum distance from the origin will have to be positive. A theorem of Efimov (I think) says that you cannot immerse isometrically in a $C^2$ manner into Euclidean 3-space any complete surface whose curvature is bounded above by a negative constant. Thus, the embeddings of the type you are asking about do not exist if you require that they be $C^2$. By a theorem of Nash, $C^1$ isometric embeddings do exist, but they are wild. – Robert Bryant Jan 7 at 6:57
• Thank you a lot for the comment. – Bombyx mori Jan 7 at 7:30

This is not a complete answer either -- in particular, I can't write down any formulas yet, but I wanted to share some pictures I made to help build my intuition. Zachary Treisman's construction may be related.

As Bill Thurston's illustration of the (1/d)-metric taught me, by drawing disks living on the surface which represent how far I can get in a certain constant time, I can recover a great deal of tactile intuition for a metric on a surface, even if my eyes are showing me a different one.

So let me try that out in a really simple and special case: flattening a torus, that is visualizing a flat metric on an embedded torus.

The embedding I considered is parametrized by coordinates (u,v) which each live in $[0,2\pi]$:
$$x(u,v)=(c+\cos(v))\cos(u),$$ $$y(u,v)=(c+\cos(v))\sin(u),$$ $$z(u,v)=\sin(v),$$

where the radius of a meridian circle is 1 and c is the distance from the center of a meridian circle to the center of the "outer" hole of the torus. If c>1, then the torus will not self-intersect.

I first placed disks in a triangular lattice in a rectangle with aspect ratio $c\sqrt{3}/2$ (so that the triangular lattice consists of $m$ rows and $cm$ 'columns' of disks). In this figure c=2. These are the disks of constant speed in this flat metric on the torus: I rescaled the rectangle so that it had dimensions $[0,2\pi]\times[0,2\pi]$ and plotted the disks on the surface of the torus. Here are the results (the disks seem to be peeling off the torus because I shrunk the torus so that it wouldn't intersect the disks and then didn't do enough fiddling to make it perfect):

c=1.2 c=2 c=5 I found it useful to imagine the disks moving on the embedded torus by isometries (which are just translations in the rectangle picture). You can see how the disks get sheared on the embedding if we take v to v+c so that the meridians of the torus all rotate -- this is an effect of the differing principal curvatures of the embedding. The outer disks move much faster than the inner ones if we take u to u+c; this rotates the embedding about a central vertical axis.

How hard would it be to make these pictures for negatively curved surfaces? I don't know a nice parametrization for an embedded double torus, let alone one that plays nicely with a negatively curved metric. I also haven't been able to pick out the right features to focus on in these pictures in order to imagine what happens in other cases, so any guidance or comments would be appreciated!

Code:
u3[r_, [Theta], u0] := r Cos[[Theta]] + u0
v3[r_, [Theta], v0] := r Sin[[Theta]] + v0
• @jc: Those are nice pictures! good ingenuity in finding a way to show a flat metric. Ideas: (a) Putting patterns on surfaces is called "texture mapping", implemented in hardware and well-suported in software on modern computers. It should be easy to draw an image of a photograph (say) wrapped around the torus. I don't know what tools do this conveniently. In Mma, check MeshShading and Image Processing. There are other options with separate software. (b) For conformal toruses: you can use stereographi projection from $S^3$ (or work out formula from derivatives). $\to$ next comment – Bill Thurston Sep 25 '10 at 10:40
• @jc, cont'd: For hyperbolic surfaces: the universal covers of $(n m p)$ orbifold is a hypergeometric function (for the special case $22\infty$ it's $\cos$), and many finite-sheeted covers of these orbifolds have simple desriptions as algebraic curves. These are also related to congruence subgroups and modular forms. There are well-developed ways to describe minimal surfaces in $\mathbb R^3$ using complex analytic data of this sort. Willmore surfaces are nice. There are very large bodies of mathematics lurking nearby --- the problem is how to get just a little that you want. – Bill Thurston Sep 25 '10 at 10:59
I'm thinking that the trouble with the metric in the positively curved parts of the surface comes from the fact that when building these things in $R^3$ out of a polygon in a Euclidean or hyperbolic plane, we need to do a bit of stretching, because we run out of dimensions for just rolling - in $R^4$ a torus can be flat, yes? So how about if we fix a particular curvature for a curve on the surface, say the one corresponding to the shortest (in the embedded, Euclidean sense) generator of the fundamental group for a basepoint in the middle of the picture (again, in the embedded, Euclidean sense). Then we can scale tangents by the inverse of the curvature in the direction they specify, so that the directions that were the most stretched if we think of the surface as coming from a glued polygon are the easiest to move along. This could be done so that the osculating circles, once scaled, all end up the same size as our particular circle. This would have the effect, for example, of "tightening the belt" around the outside of the positively curved region, so that Deane Yang's flipping of the cylinder would happen.