I'm new to geometry, and am confused by the difference between hyperbolic plane and hyperbolic model.

I've been taught that a hyperbolic plane is a plane with negative curvature, a visible example is a saddle. And I've seen that hyperbolic planes can be visualized through crocheting. However, I've also read that "Hilbert says that it is impossible to smoothly embed the hyperbolic plane in Euclidean three-space using the usual Euclidean geometry". It makes me really confused. Do the hyperbolic planes look just like those crochet work?

Moreover, there exist several hyperbolic models such as poincaré disk, poincaré half-plane. None of them "seem" to have negative curvature (they do not look like a saddle). So what I think is that the hyperbolic model is like a framework where distance, geodesic, and something other like these can be easily defined and computed, and is easy to visualize, but hyperbolic plane is hard to visualize in our 3D Euclidean space. I don't know whether I understand it correctly. The difference between the model and the plane and their meaning really confused me.

  • $\begingroup$ Three quick notes: 1. In the sense you're asking, hyperbolic model probably refers to the isometrically-embedded top sheet of a two-sheeted hyperboloid in Minkowski $3$-space. 2. In thehyperbolic plane, circles' circumference grows exponentially with radius; the resulting "crinkliness" prevents the hyperbolic plane from immersing in Euclidean $3$-space (though arbitrarily large disks do embed). 3. A search of this site should turn up a number of related explanations and diagrams. :) $\endgroup$ Commented Jun 17, 2021 at 12:22
  • $\begingroup$ @AndrewD.Hwang Thanks for your reply! I do not specifically refer to hyperbolic planes as hyperboloid. My biggest problem is that I cannot figure out the relationship between hyperbolic plane and hyperbolic model. After reading more materials, I have some new understandings. The hyperbolic models we use also describe hyperbolic planes, they are different projections from hyperbolic planes to Euclidean space, except that they are not isometric. Is it correct? $\endgroup$ Commented Jun 17, 2021 at 13:30
  • $\begingroup$ A local portion of the hyperbolic plane can be embedded in Euclidean 3-space, for example visualized by crochet work. But not the whole of the hyperbolic plane. The hyperbolic plane has constant negative curvature. $\endgroup$
    – GEdgar
    Commented Jun 17, 2021 at 14:17
  • $\begingroup$ @GEdgar Thanks! It is now much clearer for me! So those hyperbolic models can be seen as a "distorted" way to embed the hyperbolic plane in Euclidean 3-space right? $\endgroup$ Commented Jun 17, 2021 at 14:37

1 Answer 1


In differential geometry, a hyperbolic plane is a complete, simply-connected Riemannian $2$-manifold equipped with a metric of constant negative (Gaussian) curvature. By scaling the metric, we may assume the curvature is $-1$; often when people say hyperbolic plane they're assuming $K \equiv -1$. The rest of this answer assumes $K \equiv -1$.

It turns out that any two hyperbolic planes of curvature $-1$ are isometric as Riemannian manifolds. In my experience, when people speak of the hyperbolic plane, they refer to the equivalence class of hyperbolic planes in the space of Riemannian manifolds up to isometry.

It's possible that a hyperbolic model in the sense of the question is what I've called a hyperbolic plane above. There are a number of standard models. The first three below naturally sit inside Minkowski three-space, the real Cartesian three-space equipped with the metric $dx^{2} + dy^{2} - dz^{2}$.

  • The hyperboloid model, the upper sheet of a two-sheeted hyperboloid, $x^{2} + y^{2} - z^{2} = -1$, $z > 0$. Geodesics ("hyperbolic lines") in this model are intersections of the hyperboloid with planes through the origin. (This surface has non-constant _positive  curvature with respect to the Euclidean ambient metric, but acquires constant negative curvature from the ambient Minkowski metric.)

  • The conformal disk model (or Poincaré model), the open unit disk in the plane $z = 0$ with the metric $\frac{4(dx^{2} + dy^{2})}{(1 - (x^{2} + y^{2}))^{2}}$ induced by projection from the point $(0, 0, -1)$, see diagram. Because the metric is a scalar multiple of the Euclidean metric $dx^{2} + dy^{2}$, hyperbolic angles coincide with Euclidean angles. Geodesics turn out to be arcs of circles meeting the unit circle at right angles. At first glance this metric appears flat; see below.

Projecting the hyperboloid to the disk

  • The affine disk model (or Klein-Beltrami model), the open unit disk in the plane $z = 1$ with the metric induced by projection from the origin. Geodesics in this model are Euclidean segments. If the open unit disk and projection point are translated one unit up the $z$-axis in the preceding model, we obtain the affine model.

  • The conformal upper half plane model, induced from the conformal disk by a fractional linear transformation. The metric is $\frac{dx^{2} + dy^{2}}{y^{2}}$, and geodesics are vertical lines and semicircles centered on the $x$-axis.

Curvature is a measure of angular defect per unit area in geodesic triangles. Qualitatively, a surface has negative curvature if the total interior angle of a geodesic triangle (enclosing a topological disk, which is automatic in the hyperbolic plane) is smaller than $\pi$. Given that hyperbolic and Euclidean angles coincide in the conformal models, and that geodesics are arcs of circles, it's visually plausible that the conformal disk and half-plane models have negative curvature.

If we represent a region of the hyperbolic plane as a surface in Euclidean three-space, we obtain a surface having "saddle-like" geometry at each point. The best-known examples are surfaces of rotation, especially the pseudosphere, and surfaces with helicoidal symmetry such as Dini's surface.

A hyperbolic circle of hyperbolic radius has hyperbolic circumference $2\pi \sinh r = \pi(e^{r} - e^{-r})$. Particularly, circumference grows exponentially with radius. The theorem of Hilbert mentioned asserts that the entire hyperbolic plane cannot be represented, even allowing self-intersection. Qualitatively (thinking of crocheted models that start like a saddle and grow outward radially), a hyperbolic disk is "floppy", and the larger the radius, the more area must be accommodated in a Euclidean ball whose Euclidean radius grows like the hyperbolic radius of the disk; "Euclidean space just can't keep up".

As noted in the comments, however, a hyperbolic disk of arbitrarily large hyperbolic radius can be isometrically embedded in Euclidean three-space. One method is to use Dini's surface, taking the edge of the disk to lie on the edge of the surface, and taking the center to lie "exponentially far away along the horn", so that the disk is tightly wrapped around the "axis".

Coda: Hyperbolic geometry also overlaps synthetic geometry. Euclid's parallel postulate is replaced by another, equivalent to "Given a line $\ell$ and a point $p$ not on $\ell$, there exist two lines through $p$ that do not meet $\ell$." Pat Ryan's Euclidean and Non-Euclidean Geometry is an accessible treatment, including the hyperboloid model and the necessary geometry of Minkowski space, and requiring little more than one semester of linear algebra.

  • $\begingroup$ Brilliant! It's of great help! Thanks, Andrew! $\endgroup$ Commented Jun 20, 2021 at 6:02
  • $\begingroup$ You're very welcome! <> For posterity, I should clarify: By "The first three below naturally sit inside Minkowski three-space...", I meant as subsets. The hyperboloid is isometrically embedded in Minkowski space (despite Euclidean appearances), but the two disk models are not. $\endgroup$ Commented Jun 20, 2021 at 10:36
  • $\begingroup$ Sorry but I don't understand what "as subsets" means. I now know that the poincaré disk can be induced by stereographic projection from the hyperboloid model. So by saying it "naturally sit inside Minkowski three-space", does it mean that poincaré disk is also an embedding of a hyperbolic plane in Minkowski three-space, but not an isometric embedding? $\endgroup$ Commented Jun 21, 2021 at 11:43
  • $\begingroup$ Yes, that's right: In the diagram, the conformal unit disk is the subset shown, but the metric induced from the ambient Minkowski metric is the "visually apparent" flat Euclidean metric, not the hyperbolic metric. $\endgroup$ Commented Jun 21, 2021 at 12:15

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