Poincaré half-plane model

In non-Euclidean geometry, the Poincaré half-plane model is the upper half-plane (denoted below as H), together with a metric, the Poincaré metric, that makes it a model of two-dimensional hyperbolic geometry.

It is named after Henri Poincaré, but originated with Eugenio Beltrami, who used it, along with the Klein model and the Poincaré disk model (due to Riemann), to show that hyperbolic geometry was equiconsistent with Euclidean geometry. The disk model and the half-plane model are isomorphic under a conformal mapping.

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Metric

The metric of the model on the half-plane

\{ \langle x, y \rangle | y > 0 \} \,

is given by

(d s)^2 = \frac{(d x)^2 %2B (d y)^2}{y^2} \,

where s measures length along a possibly curved line. The straight lines in the hyperbolic plane (geodesics for this metric tensor, i.e. curves which minimize the distance) are represented in this model by circular arcs perpendicular to the x-axis (half-circles whose origin is on the x-axis) and straight vertical lines ending on the x-axis. The distance between two points measured in this metric along such a geodesic is

\operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arccosh} \left( 1 %2B \frac{ {(x_2 - x_1)}^2 %2B {(y_2 - y_1)}^2 }{ 2 y_1 y_2 } \right) \,.

This model is conformal which means that the angles measured at a point are the same in the model as they are in the actual hyperbolic plane.

This can be generalized to model an n+1 dimensional hyperbolic space by replacing the real number x by a vector in an n dimensional Euclidean vector space.

Special curves

In addition to the straight lines mentioned above, there are other special curves on the hyperbolic plane which can be modeled in the Euclidean half-plane:

Constructing the curves

Here is how one can use compass and straightedge constructions in the model to achieve the effect of the basic constructions in the hyperbolic plane. For example, how to construct the half-circle in the Euclidean half-plane which models a line on the hyperbolic plane through two given points.

Draw the line segment between the two points. Construct the perpendicular bisector of the line segment. Find its intersection with the x-axis. Draw the circle around the intersection which passes through the given points. Erase the part which is on or below the x-axis.

Or in the special case where the two given points lie on a vertical line, draw that vertical line through the two points and erase the part which is on or below the x-axis.

Draw the radial line (half-circle) between the two given points as in the previous case. Construct a tangent to that line at the non-central point. Drop a perpendicular from the given center point to the x-axis. Find the intersection of these two lines to get the center of the model circle. Draw the model circle around that new center and passing through the given non-central point.

Or if the two given points lie on a vertical line and the given center is above the other given point, then draw a horizontal line through the non-central point. Draw a circle around the intersection of the vertical line and the x-axis which passes through the given central point. Construct the tangent to that circle at its intersection with the horizontal line. The midpoint between the intersection of the tangent with the vertical line and the given non-central point is the center of the model circle. Draw the model circle around that new center and passing through the given non-central point.

Or if the two given points lie on a vertical line and the given center is below the other given point, then draw a circle around the intersection of the vertical line and the x-axis which passes through the given central point. Draw a line tangent to the circle which passes through the given non-central point. Draw a horizontal line through that point of tangency and find its intersection with the vertical line. The midpoint between that intersection and the given non-central point is the center of the model circle. Draw the model circle around that new center and passing through the given non-central point.

Find the intersection of the two given semicircles (or vertical lines).

Find the intersection of the given semicircle (or vertical line) with the given circle.

Find the intersection of the two given circles.

Symmetry groups

The projective linear group PGL(2,C) acts on the Riemann sphere by the Möbius transformations. The subgroup that maps the upper half-plane, H, onto itself is PSL(2,R), the transforms with real coefficients, and these act transitively and isometrically on the upper half-plane, making it a homogeneous space.

There are four closely related Lie groups that act on the upper half-plane by fractional linear transformations and preserve the hyperbolic distance.

The relationship of these groups to the Poincaré model is as follows:

Important subgroups of the isometry group are the Fuchsian groups.

One also frequently sees the modular group SL(2,Z). This group is important in two ways. First, it is a symmetry group of the square 2x2 lattice of points. Thus, functions that are periodic on a square grid, such as modular forms and elliptic functions, will thus inherit an SL(2,Z) symmetry from the grid. Second, SL(2,Z) is of course a subgroup of SL(2,R), and thus has a hyperbolic behavior embedded in it. In particular, SL(2,Z) can be used to tessellate the hyperbolic plane into cells of equal (Poincaré) area.

Isometric symmetry

The group action of the projective special linear group PSL(2,R) on H is defined by

\left(\begin{matrix}a&b\\ c&d\\ \end{matrix}\right) \cdot z = \frac{az%2Bb}{cz%2Bd} = {(ac|z|^2%2Bbd%2B(ad%2Bbc)\Re(z))  %2B i\Im(z)\over|cz%2Bd|^2}.

Note that the action is transitive, in that for any z_1,z_2\in\mathbb{H}, there exists a g\in {\rm PSL}(2,\mathbb{R}) such that gz_1=z_2. It is also faithful, in that if gz=z for all z in H, then g=e.

The stabilizer or isotropy subgroup of an element z in H is the set of g\in{\rm PSL}(2,\mathbb{R}) which leave z unchanged: gz=z. The stabilizer of i is the rotation group

{\rm SO}(2) = \left\{ \left(\begin{matrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\\ \end{matrix}\right)\,:\,\theta\in{\mathbf R}\right\}.

Since any element z in H is mapped to i by an element of PSL(2,R), this means that the isotropy subgroup of any z is isomorphic to SO(2). Thus, H = PSL(2,R)/SO(2). Alternatively, the bundle of unit-length tangent vectors on the upper half-plane, called the unit tangent bundle, is isomorphic to PSL(2,R).

The upper half-plane is tessellated into free regular sets by the modular group SL(2,Z).

Geodesics

The geodesics for this metric tensor are circular arcs perpendicular to the real axis (half-circles whose origin is on the real axis) and straight vertical lines ending on the real axis.

The unit-speed geodesic going up vertically, through the point i is given by

\gamma(t) = \left(\begin{matrix}e^{t/2}&0\\ 
                   0&e^{-t/2}\\ \end{matrix}\right) \cdot i 
   = ie^t.

Because PSL(2,R) acts transitively by isometries of the upper half-plane, this geodesic is mapped into the other geodesics through the action of PSL(2,R). Thus, the general unit-speed geodesic is given by

\gamma(t) = 
\left(\begin{matrix}a&b\\ c&d\\ \end{matrix}\right) \left(\begin{matrix}e^{t/2}&0\\ 
                   0&e^{-t/2}\\ \end{matrix}\right) \cdot i
  = \frac {aie^t %2Bb} {cie^t %2Bd}.

This provides the complete description of the geodesic flow on the unit-length tangent bundle (complex line bundle) on the upper half-plane.

See also

References