What is the equation for the saddle-like 2d surface (embeded in 3d Euclidean space with cartesian coordinates x, y and z) with constant negative curvature frequently used to illustrate open universe (for example in the following image is taken from Wikipedia)?

enter image description here

Flat universe (zero curvature) is illustrated by a plane, with equation say

$z = 0$

Closed universe (constant positive curvature) is illustrated by a surface of sphere, say

$x^2 + y^2 + z^2 = r$


2 Answers 2


One shouldn't expect to have a "good" formula for the local isometric embeddings of a constant negative curvature surface in Euclidean $\mathbb{R}^3$. This is due to a little theorem proved by David Hilbert around 1901:

Theorem There does not exist a smooth immersion of the hyperbolic plane into Euclidean 3 space.

The theorem has been further studied in the years following. In 1961 Efimov showed that any complete surface with curvature strictly bounded above (that is to say, if there exists a negative number $K_0 < 0$ such that the Gaussian curvature is always strictly less than $K_0$) cannot admit a smooth (twice continuously differentiable) isometric immersion into Euclidean three space.

That is to say, if you try to "extend" any surface in Euclidean 3 space that satisfies constant negative curvature, you are guaranteed to hit a singularity. In particular, you cannnot expect the surface to be described by $F(x,y,z) = 0$ where $F(x,y,z)$ has a nice algebraic expression (say, polynomial) and has smooth level sets.

Typically the image one usually use to illustrate the notion of negative (but not constant) curvature is the graph $$ z = x^2 - y^2 $$ which produces a classical saddle, or the catenoid whose Gaussian curvature, while everywhere negative, is not constant. (Though it has constant [in fact everywhere vanishing] mean curvature.)

Lastly, however, despite the above, it is possible to embed "patches" of hyperbolic plane into Euclidean 3 space. There are many ways of doing so (one can search for the term pseudosphere; though some people use the same term for the hyperboloid/de Sitter spaces embedded in higher dimensional Minkowski space), but one of the more well-known is the tractricoid. (See Wiki entry here.) Parametrically in cylindrical coordinates $(z,r,\theta)$ the surface can be described by:

$$ \mathbb{R}_+\times\mathbb{S}^1 \ni (t,\omega) \mapsto \left(z=\frac{1}{\cosh t},r=t-\tanh t , \theta = \omega\right) $$

and has constant negative curvature.

  • $\begingroup$ +1 - thanks, this is something I've always sort of known without ever understanding why! $\endgroup$ Oct 30, 2012 at 17:18
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    $\begingroup$ As an aside, the saddle is a very poor mental image for a surface of constant Gaussian curvature. Remember that Gauss curvature is the product of the principal curvatures. To have negative curvature one needs to curve "up (positive $z$)" in the $x$ direction and "down" in the $y$ direction. To get constancy the amount of $x$ curvature must be inversely proportional to the amount of $y$ curvature. So once you fixed the ridge of the saddle to be roughly parabolic, you need the front and back end of the saddle to pinch tightly compared to the middle of the saddle to ensure constant curvature. $\endgroup$ Oct 30, 2012 at 17:21
  • $\begingroup$ @WillieWong Thanks for the detailed question. I have read the article about tractricoid - its surface has finite area (because it cannot be extended without singularity), but I assume that 3d hyperbolic space with constant negative curvature has infinite volume? $\endgroup$
    – Leos Ondra
    Oct 31, 2012 at 14:31
  • $\begingroup$ @Leos: yes. The simply connected hyperbolic space $\mathbb{H}^n$ in any dimension $n \geq 2$ has infinite (hyper)volume. In fact the volume enclosed in each geodesic ball is strictly greater than the volume enclosed in the Euclidean ball of the same radius. $\endgroup$ Oct 31, 2012 at 15:31
  • $\begingroup$ @WillieWong I have found a popular article "Experiencing Meanings in Geometry" (math.cornell.edu/~dwh/papers/Aesthetics/Aesthetics.pdf) by David W. Henderson and Daina Taimina where they write that it is possible to construct complete hyperbolic surface with constant curvature in 3d Euclidean space. $\endgroup$
    – Leos Ondra
    Nov 3, 2012 at 21:15

I don't think you can embed a surface of constant negative curvature in Euclidean space. However you can embed it in Minkowski space. See http://en.wikipedia.org/wiki/De_Sitter_space for details. If you have Minkowski space defined by:

$$ ds^2 = -dx_0^2 + dx_1^2 + dx_2^2 + dx_3^2 $$

then the corresponding surface is:

$$ -x_0^2 + x_1^2 + x_2^2 + x_3^2 = a^2$$

  • $\begingroup$ @WillieWong: apologies, I deleted then completely rewrote my answer after you (quite correctly!) commented. Actually I'm not 100% sure the rewrite is correct :-) $\endgroup$ Oct 30, 2012 at 16:54
  • $\begingroup$ The edit looks good. I'll comment a bit more on the Euclidean case. $\endgroup$ Oct 30, 2012 at 16:56
  • $\begingroup$ @JohnRennie "However you can embed it in Minkowski space". What exactly makes the difference? The fact that the space is Minkowskian (with the minus sign at the first coordinate) or that we have four coordinates in total? $\endgroup$
    – Leos Ondra
    Oct 31, 2012 at 14:26
  • $\begingroup$ @LeosOndra: is Minkowskian a word? :-) Anyhow it is the signature of the metric and not the number of dimensions. The example I gave is for 4 dimensions, but it applies to any number of dimensions. Please don't ask me to prove it though: I know it's the case, but not how it's proved. $\endgroup$ Oct 31, 2012 at 14:39
  • $\begingroup$ @JohnRennie So is it possible to embed a complete and smooth 2d surface with negative constant curvature in Minkowski space 2+1 defined by ds^2 = -dt^2 + dx^2 + dy^2 ? $\endgroup$
    – Leos Ondra
    Oct 31, 2012 at 14:53

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