# Geodesics and Diffeomorphisms

1. Nov 23, 2007

### WWGD

Hi, everyone:

I was just going over some work on Hyperbolic geometry, and noticed that

the geodesics in the disk model are the same as the geodesics in the upper-

half plane, i.e, half-circles or line segments, both perpendicular to the boundary.

Now, I know the two domains are diffeomorphic: the Mobius map

M(z)=(z-i)/(z+i) takes H diffeomorphically into D, the open unit disk..

Is this last the explanation for why both have the same geodesics,

i.e, do diffeomorphisms preserve geodesics ? Is there some other

relation between the two domains that explains that they have

the same geodesics?.

Thanks.

2. Nov 23, 2007

### Chris Hillman

Poincare's UHP versus disk models of the hyperbolic plane

Careful what you ask for, I might toss a small book at you!

You are correct that the Moebius transformation
$$w \mapsto \frac{w+i}{w-i}$$
takes $0, \, i, \, \infty$ to $-1, \, 0, \, 1$ and thus maps the UHP conformally to the interior of the unit disk. If you express this as a real transformation, we can regard it as a coordinate transformation from Poincare's upper half plane model of the hyperbolic plane, which in the UHP chart has the line element
$$ds^2 = \frac{dx^2+dy^2}{y^2}, \; -\infty < x < \infty, \; 0 < y < \infty$$
to Poincare's unit disk model of the hyperbolic plane, which in the polar stereographic chart has the line element
$$ds^2 = \frac{4 \, \left( d\rho^2 + \rho^2 \, d\phi^2 \right)}{1-\rho^2}, \; 0 < \rho < 1, \; -\pi < \phi < \pi$$
This chart is readily transformed to the more familar cartesian stereographic chart with the line element
$$ds^2 = \frac{4 \, \left( d\overline{x}^2+d\overline{y}^2 \right)} {\left( 1-\overline{x}^2-\overline{y}^2 \right)^2}, \; \overline{x}^2+\overline{y}^2 < 1$$
which can indeed be obtained by stereographic projection of the hyperboloid of one sheet considered as the variety $-x_1^2 + x_2^2 + x_3^2 = -1$ in $E^{1,2}$ from the point $(x_1,x_2,x_3)=(-1,0,0)$ to the plane $x_1 = 0$.

Poincare himself recalled, in a famous passage in one of his popular books written for a general audience, that he noticed these facts at the exact moment in which he set his foot down to get onto a bus!

To see this, consider the inverse transformation
$$z \mapsto \frac{i \, (1-z)}{1+z}$$
Writing $z = \rho \, \exp(i \, \phi)$, the transformation becomes
$$x = \frac{2 \rho \, \sin(\phi)}{\rho^2 + 2 \rho \, \cos(\phi) + 1}, \; y = \frac{1-\rho^2}{\rho^2 + 2 \rho \, \cos(\phi) + 1}$$
with inverse transformation
$$\rho = \sqrt{ \frac{x^2+(y-1)^2}{x^2+(y+1)^2}}, \; \phi = \arctan \frac{2 x}{1-x^2-y^2}$$
Plugging into the first line element above, we obtain the second.

Let's study $H^2$ in the UHP chart.

You can readily solve the Killing equations to find three generators of the Lie algebra of Killing vectors, namely
$$\partial_x, \; x \, \partial_x + y \, \partial_y, \; \frac{x^2-y^2}{2} \, \partial_x + x y \, \partial_y$$
which I hope you recognize as three infinitesimal Moebius transformations!

To obtain the uniparameteric subgroups of transformations corrresponding to each generator, we proceed as follows:

To find the integral curves of the vector field $\partial_x$, we solve
$$\dot{x} = 1, \; \dot{y} = 0, \; x(0)=x_0, \; y(0) = y_0$$
which gives $x(s) = x_0 + s, \; y(s) = y_0$, or better yet, $(x,y) \mapsto (x + \lambda, y)$, i.e. horizontal translation .

To find the integral curves of the vector field $x \, \partial_x + y \, \partial_y$, we solve
$$\dot{x} = x, \; \dot{y} = y, \; x(0)=x_0, \; y(0) = y_0$$
which gives $x(s) = x_0 \, \exp(s), \; y(s) = y_0 \, \exp(s)$ ("additive" group parameterized by s), or better yet, $(x,y) \mapsto (x \, \lambda, y \, \lambda)$ ("multiplicative" group parameterized by $\lambda > 0$). That is, dilation from the origin in euclidean terms, or "vertical" translation in hyperbolic terms.

Exercise: find the integral curves of the vector field $\frac{x^2-y^2}{2} \, \partial_x + x y \, \partial_y$; the resulting one-parameter group of transformations should consist of the hyperbolic rotations about the point $(x,y)=(0,1)$, with orbits which are coordinate circles with various centers which in hyperbolic terms are concentric circles around this point.

Thus, we have a three dimensional Lie group of self-isometries of the hyperbolic plane, which is fully analogous to the three dimensional Lie group of self-isometries of the euclidean plane.

The geodesic equations are readily obtained from the geodesic Lagrangian; the result is
$$\ddot{x} - \frac{2 \dot{x} \dot{y}}{y} = 0, \; \ddot{y} + \frac{\dot{x}^2-\dot{y}^2}{y} = 0$$
From this we readily obtain the first integrals
$$\dot{x} = A \, y^2, \; \dot{y} = y \, \sqrt{(1- A^2 \, y^2}$$
From this we obtain
$$\frac{dx}{dy} = \frac{\dot{x}}{\dot{y}} = \frac{A y}{\sqrt{1-A^2 y^2}}$$
which gives $(x-x_0-\sqrt{y_m-y_0^2})^2 + y^2 = y_m^2$. This is evidently, in euclidean terms, a semicircular arc passing through $(x,y) = (x_0, y_0)$, orthogonal to $y=0$, and with maximal height $y=y_m$.

From our knowledge of conformal mappings in general and Moebius transformations in particular, we can immediately infer that in the unit disk model, the geodesics appear (in euclidean terms) as semicircular arcs orthogonal to the unit circle (the image of the real line under our diffeomorphic mapping giving the change of coordinate charts as above). We could also confirm this directly by studying the geodesics in the polar stereographic chart. Readers should draw a sketch at this point to verify that given a line L (geodesic) in $H^2$ and a point P off L, there are infinitely many lines through P which do not intersect L!

PF readers will no doubt be familiar with formulas from analytic geometry giving the equation of the line L through two given points P1, P2, aka the straightedge equation, and also the equation of a circle of radius A with center P, aka the compass equation.

Exercise: find a formula giving the equation of the hyperbolic geodesic through two points P1, P2 in the unit disk. Find a formula giving the equation of the circle of hyperbolic radius A and hyperbolic center P.

Once one has these in hand, note that most euclidean constructions with straightedge and compass do not depend upon the parallel postulate, so we can simply substitute our hyperbolic versions. In this way we can confirm that if we assume that euclidean geometry is self-consistent, we must conclude that hyperbolic geometry is also self-consistent. In this way Gauss and independently Lobachevski and Janos Bolyai built up the synthetic theory of hyperbolic geometry.

There is no need to use analytic formulas here; what GLB actually did was to use euclidean constructions with straightedge and compass to model the required hyperbolic straighedge and compass. See Chaim Goodman-Strauss, "Compass and Straightedge in the Poincare Disk", American Mathematical Monthly 108 (2001): 38-49, for a detailed discussion.

Actually, Poincare's flash of inspiration covered much more ground than we can cover here; he also saw how doubly periodic functions encountered in complex analysis (these arise naturally for example in solving various ODEs) are connected with tilings of the hyperbolic plane and with their symmetry groups. See for example Jones and Silverman, Complex Functions for doubly periodic groups and see Armstrong, Groups and Symmetry for an elementary enumeration of the symmetry groups.

I feel that we ought to have a sticky listing some examples of structure at various levels in manifold theory and Riemannian (or Lorentzian) geometry. As an example of global structure, we see that $H^2$ is aptly named, for it is indeed diffeomorphic (and thus homeomorphic) to the euclidean plane, but because ${R\left[H^2\right]}_{1212} = -1[/tex] while [itex]{R\left[E^2\right]}_{1212} = 0$, it is certainly not isometric to the euclidean plane! Furthermore, from considering "ideal endpoints" (on the unit circle) of geodesics in the disk model, you can see that the space of lines (geodesics) is diffeomorphic to a Moebius band, i.e. real projective plane with a disk removed.

There are many other charts on $H^2$ which are often encountered, including the radial chart
$$ds^2 = \frac{dr^2}{1+r^2} + r^2 \, d\phi^2, \; 0 < r < \infty, \; -\pi < \phi < \pi$$
the cylindrical conformal chart (analog of Mercator chart, in which loxodromes appear as coordinate lines, which gives an angle-preserving but not area-preserving euclidean representation)
$$ds^2 = \frac{d\zeta^2 + d\phi^2}{\sinh(\zeta)^2}, \; 0 < \zeta < \infty, \; -\pi < \phi < \pi$$
the cylindrical axial projection chart (area preserving)
$$ds^2 = \frac{dz^2}{z^2-1} + \left( z^2-1 \right) \, d\phi^2, \; -1 < z < 1, \; -\pi < \phi < \pi$$
$$ds^2 = d\theta^2 + \sinh(\theta)^2 \, d\phi^2, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
and the planar central projection chart
$$ds^2 = \frac{dv^2}{\left( 1-v^2 \right)^2} + \frac{v^2}{1-v^2} \, d\phi^2 , \; 0 < v < 1, \; -\pi < \phi < \pi$$
which is geodesic-preserving; the geodesics appear as coordinate line segments in the unit disk, which gives Klein's unit disk model of $H^2$.

Exercise: find explicit coordinate transformations between these, and determine which properties each chart has: angle-preserving, area-preserving, or geodesic-preserving? Some of these are best considered to "live" in part of the plane, and some, in part of a cylinder. Which are which? (Some hints provided above.)

Exercise: study Flanders, Differential Forms and their Physical Applications. For each chart, read off a natural coframe field. If neither unit vector gives a geodesic vector field, rotate by a suitable amount at each point to obtain a frame field which is geodesic. Express the Laplace-Beltrami operator in terms of each chart. Express the three Killing vectors in terms of each chart.

Exercise: study Olver, Applications of Lie Groups to Differential Equations. Explain why the heat equation in the hyperbolic plane becomes (in the UHP chart)
$$\frac{u_t}{y^2} = u_{xx} + u_{yy}$$
Determine its point symmetry group and find the fundamental solution using the methods of Lie.

I suggested above considering diffeomorphisms as coordinate transformations. From this point of view, the fact that our transformation arises from a conformal mapping in the sense of complex variables ensures that since our first chart was "conformal", our second chart (obtained by applying this mapping as a coordinate transformation) must also be "conformal", and it is.

As suggested above, we can compare any coordinate chart on any Riemannian two-manifold with a locally euclidean model (above I used some cylindrical and some planar models), and we can decide whether or not our chart gives a euclidean representation of the intrinsic geometry on our Riemannian two-manifold which preserves angles, geodesics, or area. Gauss himself showed that no representation can preserve all three if the Gaussian curvature is nonzero. But as we saw above we can find representations which preserve either angles or geodesics or area!

Exercise: for the sphere $S^2$, find analogous charts and work out exercises analogous to the ones above. The same, for $H^{1,1}$.

Last edited: Nov 24, 2007
3. Nov 23, 2007

### WWGD

Thanks: will get back to you in 2010, after I finish digesting the answer :).

Another one (maybe you should copyright it before posting it):

I read of lines in the saddle and the hyperboloid as models for Hyperbolic

geometry. Specifically (Borwein and Borowski, Dict. of Math-great book)

"Hyperbolic Geometry....can be modeled as the geometry of lines in the saddle.."

I don't know if I am too far off here, but, from what I understand, no

hyperbolic surface ( I guess this would mean constant curvature=-1)

can be embedded in IR^n . Isn't the saddle embedded in IR^3.?

Thanks Again

4. Nov 23, 2007

### Chris Hillman

Hi, WWGD,

Your post is a bit hard to read--- see this PF page for some formatting ideas.

Yes, I should have said that by $H^2$ I mean the hyperbolic plane with constant curvature $R_{1212} = -1$. This can be embedded in $E^{1,2}$, flat space with the indefinite but nondegenerate quadratic form $Q(\vec{x}) = -x_1^2 + x_2^2 + x_3^2$. As your book states, this cannot be globally embedded in $R^3$, although it can be locally embedded as various surfaces of revolution. Indeed, Beltrami's original model was the surface of revolution obtained from the tractix; see for example Dirk J. Struik, Lectures on Classical Differential Geometry, Dover reprint.

The saddle surface has negative curvature but not constant negative curvature, and yes, it can be embedded in $R^3$. In general, whether or not a surface has constant curvature, in regions of negative curvature, initially parallel geodesics will diverge whereas in regions of positive curvature, initially parallel geodesics will converge. That was the point of mentioning the saddle surface.

5. Nov 24, 2007

### Chris Hillman

A Very Quick Overview of Stereographic Projection of H^2

Here is some more detail about stereographic projection of H^2.

We take the H^2 to be the lower component of the hyperboloid of two sheets $-y_1^2+y_2^2+y_3^2 = -1$ embedded in $E^{1,2}$ and define the projection by drawing a straightline in the flat embedding space from (1,0,0) to $(y_1,y_2,y_3)$ and taking the project to map to the intersection (0,x,y) of this line with the plane R=0. Then
$$(y_1,y_2,y_3) - (1,0,0) = t \, \left( (0,x,y) - (1,0,0) \right)$$
from which
$$\left[ \begin{array}{c} y_1\\ y_2 \\ y_3 \end{array} \right] = \left[ \begin{array}{c} -\frac{1+x^2+y^2}{1-x^2-y^2} \\ \frac{2x}{1-x^2-y^2} \\ \frac{2y}{1-x^2-y^2} \end{array} \right], \; x^2 + y^2 < 1$$
This is a parameterization of $H^2$ by parameters which we can identify with the coordinates of the cartesian stereographic chart. Taking partials wrt x,y and taking the $E^{1,2}$ inner product of these two three-vector fields (which span the tangent spaces in the embedded manifold), we obtain the line element (basically, the metric tensor, or the first fundamental form in the language of surface theory):
$$ds^2 = \frac{dx^2 + dy^2}{ \left( \frac{1-x^2-y^2}{2} \right)^2 }, \; x^2 + y^2 < 1$$
Incidently, the cartesian stereographic chart has the advantage over the polar stereographic chart that it is well defined at the origin, but the latter chart has geodesic equations which are easier to solve!

We can prove using analytic geometry (compare the less elementary differential equation plus conformal mapping approach in my Post #2) the fact that in the unit disk model, geodesics appear as euclidean semicircular arcs lying inside and orthogonal to the unit circle.

Recalling the scale factor t above, if we consider the chordal distance between two points on $H^2$ as a parameterized surface in $E^{1,2}$--- that is the flat embedding space distance between the two points, which are spacelike separated!--- then we can prove the useful fact that the ratio of the euclidean distance between the images of the two points in the unit disk model to the chordal distance is the geometric mean of the two scale factors! That is,
$$\frac{\sqrt{(x-x^\prime)^2 + (y-y^\prime)^2}} {\sqrt{-(y_1-y_1^\prime)^2+(y_2-y_2^\prime)^2+(y_3-y_3^\prime)^2}} = \sqrt{t \, t^\prime}$$
Then a bit of trignometry gives the distance formula
$$d\left( (x,y), (x^\prime, y^\prime) \right) = 2 \, \operatorname{arcsinh} \, \frac{\sqrt{(x-x^\prime)^2 + (y-y^\prime)^2}} {\sqrt{1-x^2-y^2} \, \sqrt{1-{x^\prime}^2-{y^\prime}^2}}$$
This formula gives the hyperbolic distance measured along a hyperbolic geodesic between any two points in the hyperbolic plane. It defined a metric in the sense of metric topology (see for example Munkres, Topology).

Students are often confused by how the notion of a metric space relates to the notion of a (Riemannian) metric tensor. The answer is very simple! If we expand the two variable Taylor series of our distance formula in terms of (dx,dy), we obtain
$$\begin{array}{rcl} d\left( (x,y), (x+dx, y+dy) \right) & = & 2 \, \operatorname{arcsinh} \, \frac{\sqrt{dx^2 + dy^2}} {\sqrt{1-x^2-y^2} \, \sqrt{1-(x+dx)^2-(y+dy)^2}} \\ & = & \frac{2 \, \sqrt{dx^2+dy^2}}{1-x^2-y^2} \; \left( 1 + \frac{x \, dx + y \, dy}{1-x^2-y^2} \right) + O(dx^3, dy^3) \end{array}$$
Here, the first term is simply the square root of the now familiar expression
$$ds^2 = \frac{dx^2+dy^2}{\left( \frac{1-x^2-y^2}{2} \right)^2}$$
the next term gives a quadratic correction, and the remaining terms (not shown) give higher order corrections.

In other words, to find the distance between two points using the line element, we need to find the proper-distance-parameterized geodesic arc between them and then to integrate ds along this arc. Our distance formula allows us to circumvent that messy process! Clearly, we can only expect to find a nice distance formula in the case of particularly simple manifolds (here we are dealing with a Riemannian two-manifold of constant Gaussian curvature). OTH, it is always true that the distance between two nearby points in any Riemannian manifold is givento first order in the separation by the line element.

Exercise: try to fill in the missing details above. Consider $S^2$ as the ordinary round sphere $y_1^2+y_2^2+y_3^2 = 1$ and work out an analgous distance formula for the stereographic chart on the sphere. Similarly for $H^{1,1}$ (see [thread=195445]this recent PF thread[/thread]).

Last edited: Nov 24, 2007
6. Nov 25, 2007

### HallsofIvy

Staff Emeritus
I'm not nearly as knowlegdgeable about this as Chris Hillman but I might point out that the disk and half-plane models have the same geodesics because the are basically the same model.

If you take a small portion of the disk, next to the boundary circle, and enlarge it until the circle becomes indistinguishable from a line, you have the half-plane model.

7. Nov 25, 2007

### Chris Hillman

Hi, Halls, I highly recommend the excellent textbook on the philosophy of space and time by Sklar, Space, Time, and Spacetime which offers among other things perhaps the clearest nontechnical discussion I've seen of Riemannian geometry, including comparing the Poincare upper half place and Klein unit disk model.

If you then study (if you haven't already) the readable UG textbook by Struik, Lectures on Classical Differential Geometry, or an equivalent textbook, for classical theory of surfaces and some of the classical theory of plane and space curves as developed by mathematicians like Poncelet, Gauss, Darboux, Klein and Lie (collaboratively), Frenet & Serret (independently), and Beltrami in the nineteenth century (and greatly improved by Elie Cartan c. 1905), and apply this to these two models you will see that in classical terms
• if we consider the UHP model as defined by the particular parameterization and UHP chart as in my posts above, then this UHP chart is a conformal, non-geodesic, non-area-presvering chart,
• if we consider Klein's disk model as defined by the particular parameterization and central project chart as in my posts above, then this central projection chart is a orthogonal but non-conformal, geodesic, non-area-preserving chart.
Compare the discussion in Sklar's book.

As I pointed out above, it is probably best for our purposes to consider a local diffeomorphism on M to define a change of local coordinate chart. Then we can change freely between any of the charts I discussed above (and infinitely many more, of course). In particular, the Poincare unit disk model, the one beloved by Coxeter and frequently illustrated by Escher in his woodcuts, when considered as the parameterization and chart I gave above, is like the UHP chart a conformal, non-geodesic, non-area-presvering chart, and the "conformal" part is immediate from the fact that the diffeomorphism to compare these two charts happens to be the real number incarnation of the conformal mapping mentioned by WWGD.

If it is not clear that these three charts give different "maps" (in the sense of cartography) of the hyperbolic plane, I advise drawing to scale, in all three charts, a fixed array of small disks and a fixed configuration geodesics. You can then easily "see" the differences! In particular, all the properties I mentioned will be visually apparent.

I think you are focusing on the application of conformal mapping as per complex variables, which is relevant for the reason just stated, but I think that WWGD was asking about the Riemannian geometry of the hyperbolic plane. However, I grant you that it is no coincidence that further Moebius transformations showed up with we computed the three-dimensional Lie algebra of Killing vector fields! But here I think that's a bit of distraction.

Part of the point here is that in differential geometry, there are often various distinct notions of "equivalence" which can sometimes be a bit difficult to disentangle, often because of the many "levels of geometric structure" which are often implicitly or explicitly present in the discussion. In particular, I chose to suppress mentioning self-isometries, apart from the discussion of the three-dimensional real Lie group generated by the Lie algebra of Killing vector fields, which is of course a subgroup of the six-dimensional Lorentz group.

Last edited: Nov 25, 2007
8. Dec 9, 2007

### funinthesun

Why does (z-i)/(z+i) have to be on the unit disk? I guess the question can be rephrased, why is it a complex number with complex abs value < 1?

9. Dec 9, 2007

### Chris Hillman

The claim is that we have a pair of inverse holmorphisms (in fact, Moebius transformations) which map the upper half plane to the interior of the unit disk and vice versa, and which map the real line to the unit circle and vice versa. To understand this, you need to know a bit about complex variables and Moebius transformations. One very readable advanced undergraduate level textbook is Boas, Invitation to Complex Analysis.