The Math Thread

[Kharakov]

Every complex number has a direction except for 0, which will have indeterminate direction. Arg(z) isn't analytic, so analytic continuation might be a bit of a challenge.

What if you did it for only the positive quadrant (analytic continuation of the atan function)? What if you interpreted 0 as having orthogonal direction to the complex plane (only in the positive (or negative) direction (orthogonally))?

Is there a 3d (filled) Riemann sphere? I'm thinking of one in which j (z) is constrained to positives.

Stereographic projection will still work for the n-sphere in (n+1)-dimensional space. You could also view the Riemann sphere as the complex projective line and then generalize to the complex projective plane (more generally, complex projective space).

Ok, I'm thinking of doing a pseudo-Riemannian projection, ending the complex space at 2 (instead of infinity), performing some manipulations, projecting back to complex space, doing some manipulations... and wanted to avoid the legwork. lol. funny. So now I have to write it out, instead of looking at a webpage. I know someone who did it (I think), but I am not positive they were doing what I'm thinking of doing.


I'm thinking about writing pseudo-Riemannian projection code so that it's generalized and usable with multiple formulas.

 

[beero1000]

What if you did it for only the positive quadrant (analytic continuation of the atan function)? What if you interpreted 0 as having orthogonal direction to the complex plane (only in the positive (or negative) direction (orthogonally))?

Nope. Real-valued analytic functions are constant.

Is there a 3d (filled) Riemann sphere? I'm thinking of one in which j (z) is constrained to positives.

Stereographic projection will still work for the n-sphere in (n+1)-dimensional space. You could also view the Riemann sphere as the complex projective line and then generalize to the complex projective plane (more generally, complex projective space).

Ok, I'm thinking of doing a pseudo-Riemannian projection, ending the complex space at 2 (instead of infinity), performing some manipulations, projecting back to complex space, doing some manipulations... and wanted to avoid the legwork. lol. funny. So now I have to write it out, instead of looking at a webpage. I know someone who did it (I think), but I am not positive they were doing what I'm thinking of doing.


I'm thinking about writing pseudo-Riemannian projection code so that it's generalized and usable with multiple formulas.

Have you seen Henry Segerman's 3D-printed stereographic projections? It's crazy how rendered they look.

 

[Kharakov]

I was thinking about projecting to pseudoRiemannian space (I think Knighty did this for certain sphere packings, but it's not code/math that I familiarized myself with), shrinking and shifting, project back, do manipulations, project to pR space...


I've heard of Segerman, he does stuff after Paul Nylander: http://bugman123.com/Math/index.html

 

[lpetrich]

Those projections remind me of a curiosity of sphere-to-plane projection. A projection can have any of these three properties:

• Area-preserving: equal area
• Shape-preserving: conformal
• Geodesic-preserving: gnomonic

but not more than one of them.

Here are some radial or azimuthal examples. The sphere has radius 1 and is given by
\( x = \sin\theta\cos\phi ,\ y = \sin\theta\sin\phi ,\ z = \cos\theta \)
with polar angle θ being 0 at the coordinate origin of the plane. The azimuthal angle is φ.

The plane's coordinates are x' and y' with distance r' from the origin.

Area-preserving: equal area:
\(r' = 2 \sin \frac{\theta}{2} ,\ \{x', y'\} = \sqrt{ \frac{2}{1+z} } \{x, y\} \)

Shape-preserving: conformal:
\(r' = 2 \tan \frac{\theta}{2} ,\ \{x', y'\} = \frac{2}{1+z} \{x, y\} \)

Geodesic-preserving: gnomonic:
\(r' = \tan \theta ,\ \{x', y'\} = \frac{1}{z} \{x, y\} \)

 

[lpetrich]

Here are some general formulas for what these constraints satisfy. I will take a space with coordinates ξ and metric γ and got to a space with coordinates x and metric g. I will define for convenience
\( X^i{}_a = \frac{\partial x^i}{\partial \xi^a} \)
since x = x(ξ).

The area-preserving or equal-area case is easy. It is the Jacobian for going between spaces:
\( \frac{\sqrt{|g|}}{\sqrt{|\gamma|}} |X| \)
where || means taking the determinant.

The shape-preserving or conformal case is more complicated.
\( X^T \cdot g \cdot X = K \gamma \)
where K can vary. If one decomposes g and γ into basis vectors or vielbeins:
\( g = E^T \cdot \eta \cdot E ,\ \gamma = {\bar E}^T \cdot {\bar \eta} \cdot {\bar E} \)
one gets
\( Y^T \cdot \eta \cdot Y = K {\bar \eta} \)
where
\( Y = E \cdot X \cdot {\bar E}^{-1} \)
One can then use one's knowledge of orthogonal and pseudo-orthogonal matrices to proceed further.

For two dimensions, and also η and η-bar = the identity matrix, we get a simple result:
\( Y^1{}_1 = Y^2{}_2 ,\ Y^1{}_2 = - Y^2{}_1 \)
For flat space to flat space, one gets the familiar conformal expressions.

Fianlly, here is the geodesic-preserving or gnomonic case. Start with
\( C^k_{ab} = \frac{\partial^2 x^k}{\partial \xi^a \partial \xi^b} + \Gamma(g)^k_{ij} X^i{}_a X^j{}_b - X^k{}_{c} \Gamma(\gamma)^c_{ab} \)
using differential-geometry connection coefficients Γ. If one can find D and E that satisfy
\( C^k_{ab} = D_a X^k{}_b + D_b X^k{}_a + E^k \gamma_{ab} \)
then one has a solution.

 

[lpetrich]

More on map projections.

The orthographic projection is the view-from-a-distance projection. It violates all three properties, and as a polar projection, it has r = sin(θ).

The stereographic projection is the polar conformal one, with r = 2*tan(θ/2).

The equal-area one has r = 2*sin(θ/2), and the gnomonic one has r = tan(θ).

Viewing from distance R gives sphere-to-sphere, with angle from pole χ given by
\( \sin\chi = \frac{\sin\theta}{\sqrt{R^2 + 1 - 2R\cos\theta}} \)
For very close, with altitude h = R - 1 very small, we get
\( \sin\chi = \frac{\theta}{\sqrt{h^2 + \theta^2}} \)
what one gets for a flat surface.

-

In addition to polar projections, cylindrical and pseudo-cylindrical ones are often used.

A cylindrical one can be expressed in rectangular coordinates as
x = φ (the azimuthal angle)
y = h(θ)

The cylindrical equal-area one has h = cos(θ), and the cylindrical conformal one has h = - log(tan(θ/2)).

The latter one is the Mercator projection. It makes Greenland look as big as Africa, though the two regions have the right shape.

A pseudo-cylindrical one can be expressed in rectangular coordinates as
x = w(θ)*φ (the azimuthal angle)
y = h(θ)

For equal area and elliptical meridians (constant-longitude lines), one gets the Mollweide projection:
\( w = \cos \beta ,\ h = \sin \beta ,\ \beta + \sin\beta \cos\beta = \frac{\pi}{2} \cos\theta \)
While Greenland and Africa have appropriate sizes, anything away from the equator and the map's central meridian looks distorted.

 

[lpetrich]

Now for a nice property of conformal projections. One can get another one by taking the flat-surface rectangular coordinates x,y and doing the following to get new ones x',y':

x' + i*y' = f(x + i*y)

where f is "analytic" -- its argument's complex conjugate nowhere enters into it.

We can start with the Mercator projection: x + i*y = log(tan(θ/2)) + i*φ.

Then, (stereographic) = exp(Mercator), where

An interesting variation of the stereographic projection is the Lambert conformal conic one, sometimes used to make conformal projections for midlatitude regions. For power p, it goes

(LCC) = (stereographic)^p

There is a weird variant called the  Peirce quincuncial projection. It projects a map onto a rectangle, one that can be tiled in both polar and azimuthal directions, and not just in the azimuthal direction, as with cylindrical projections.

sn(PQ,m) = sin(-i*Mercator)

where sn is a Jacobi elliptic function with modulus m (using the squared convention). For a hemisphere mapping onto a square, m = 1/2.

-

Finally, I must mention another polar projection, r = θ, what is on the United Nations flag.

 

[lpetrich]

This brings to mind a conformal mapping from a circle to a polygon:
\( K \int \frac{dz}{\prod_i (1 - z/z_i)^{\alpha_i} \)
where the z_i have magnitude 1 and determine the vertex positions, and α_i = 1 - θ_i/π, with the θ's being the polygon's interior angles. If the variable z is on the unit circle, then the integral is on the resulting polygon. This is related to the Schwartz-Christoffel one, which uses a half plane instead of a unit circle.

Circle zc <-> rectangle zr, angle from real axis to vertex directions: a

The solution uses Jacobi elliptic functions and integrals with modulus m = (cos(a))². Not surprisingly, Mathematica has them, and I worked out the solution with it.
\( zr = F(\arcsin(\frac{2 zc}{1 + zc^2}), m) \)
\( zc = \frac{ \text{sn}(zr,m) }{ 1 + \text{cn}(zr,m) } \)

 

[lpetrich]

Back to map projections, one can define a distortion matrix, a matrix that carries an offset in one space into an offset in another space. This matrix can, in turn, be used to quantify the amount of distortion.

Here it is. D is the Y that I'd mentioned earlier:
\( D = E \cdot X \cdot (\{bar E})^{-1} \)

Equal-area:
\( \det D = A \)
where A is a constant.

Conformal:
\( D^T \cdot \eta \cdot D = S \{\bar \eta} \)
where S is a scalar that may vary.

For 2D and both eta's equaling the identity matrix, D22 = D11 and D21 = - D12.

For a polar projection, and for multiplying the azimuthal angle by p, as in LCC,
\( D = \{ \frac{dr}{d\theta} \ 0 \ ; \ 0 \ p \frac{r}{\sin\theta} \} \)

For a cylindrical projection,
\( D = \{ \frac{dh}{d\theta} \ 0 \ ; \ 0 \ \frac{1}{\sin\theta} \} \)

For a pseudo-cylindrical projection,
\( D = \{ \frac{dh}{d\theta} \ 0 \ ; \ \phi \frac{dw}{d\theta} \ \frac{w}{\sin\theta} \} \)
Note the skew term in it: the D21 term, the term with the φ in it.


For no distortion, D = D0*I, a multiple of the identity matrix.

For low distortion nearby, d(D)/d(coordinates) = 0 there, and if possible, also for higher derivatives.


I will now find low-distortion polar equal-area and conformal transforms. These are sometimes used for midlatitude regions.

First, equal-area. The general solution is
\( r = r_0 \sqrt{1 - c \cos \theta} \)

For low distortion around θ = θ₀, I find
\( p = \cos\theta_0 ,\ c = \frac{2p}{1+p^2} \)

Now, conformal. The general solution is
\( r = r_0 ( \tan (\theta/2) )^p \)

with
\( p = \cos\theta_0 \)


I note that the cylindrical equal-area projection has h = cos(θ)
Compare the cylindrical conformal one (Mercator), with h = - log(tan(θ/2))

 

[lpetrich]

For a geodesic-preserving map projection, I had found
\( C^k_{ab} = U_a X^k{}_b + U_b X^k{}_a + V^k \gamma_{ab} \)
It turns out to only be valid for null geodesics in the source space:
\( \gamma_{ab} \frac{d\xi^a}{d\tau} \frac{d\xi^b}{d\tau} = 0 \)

This constraint has the curious consequence that the Mercator projection is geodesic-preserving, when no cylindrical projection is geodesic-preserving.

So in general, one has a stronger constraint:
\( C^k_{ab} = U_a X^k{}_b + U_b X^k{}_a \)

 

[lpetrich]

Here are some more results for projections. Set up a distortion factor
\( Y_{ab} = g_{ij} X^i{}_a X^j{}_b \)

Then relative area A satisfies
\( \frac{|Y|}{|\gamma|} = A^2 \)

conformal invariance for scale factor S means
\( Y_{ab} = S \gamma_{ab} \)

and geodesic preserving means
\( Y_{ab;c} = 2Y_{ab} U_c + Y_{ac} U_b + Y_{bc} U_a \)

Conformal invariance means that the area satisfies
\( A \propto S^{n/2} \)
for n dimensions.

Geodesic preserving and conformal invariance together means
\( \gamma_{ab} (\log S)_{,c} = 2\gamma_{ab} U_c + \gamma_{ac} U_b + \gamma_{bc} U_a \)

For more than one dimension, S is constant and U = 0.

For the area, we have
\( (\log (|Y|/|\gamma|))_{,c} = 2(n+1) U_c \)

So equal-area means that U = 0.

So also being conformal or equal-area mean that Y is "covariantly constant", that Y_ab;c = 0. It also means that C^k_ab = 0.

Taking source-space derivatives and commuting gives us
\( R^l{}_{ijk}(g) X^i{}_a X^j{}_b X^k{}_c - R^d{}_{abc}(\gamma) X^l{}_d = 0 \)
which gives
\( R_{lijk}(g) X^l_{d} X^i{}_a X^j{}_b X^k{}_c - R^e{}_{abc}(\gamma) Y_{de} = 0 \)

Since Y is covariantly constant,
\(Y_{eb} R^e{}_{acd}(\gamma) + Y_{ae} R^e{}_{bcd}(\gamma) = 0 \)
That is satisfied for Y being some constant multiple of the source-space metric γ, and possibly other values.


A consequence is that when one goes from a curved space to a flat space, a geodesic-preserving projection will be neither equal-area nor conformal.

 

[lpetrich]

Being geodesic-preserving seems like a rather strong constraint. Let's see where it get us. I'll do mapping onto a flat space, and use rectangular coordinates for simplicity. Using covariant-derivative notation,
\( X^i{}_a = x^i{}_{,a} ,\ C^i{}_{ab} = X^i{}_{a;b} \)
For geodesic-preserving, we found
\( X^i{}_{a;b} = X^i{}_a U_b + X^i{}_b U_a \)
Taking a second covariant derivative and antisymmetrizing, this gives us
\( X^i{}_{h} R^h{}_{abc} = X^i{}_a U_{[b;c]} + X^i{}_b (U_{a;c} - U_a U_c) - X^i{}_c (U_{a;b} - U_a U_b) \)
It's easy to factor out X, since for a nonsingular mapping, it will be nonsingular.
\( R^h_{abc} = \delta^h_a W_{[bc]} + \delta^h_b W_{ac} - \delta^h_c W_{ab} \)
where
\( W_{ab} = U_{a;b} - U_a U_b \)
Contracting over h and b gives
\( R^{ac} = n W_{ac} - W_{ca} = (n - 1) W_{ac} \)
since W must be symmetric. This means that U must be a gradient, at least locally. We get
\(R_{habc} = \gamma_{hb} W_{ac} - \gamma_{hc} W_{ab} \)
Since the Riemann tensor is symmetric over (ha) <-> (bc), we find
\( W_{ab} = W \gamma_{ab} \)
where no-index W is a scalar.

For more than two dimensions, this means that the source space is maximally symmetric, that it is a hypersphere or some hyperbolic analytic continuation of one. Is it also that for two dimensions?

 

[lpetrich]

With this W, the Riemann tensor becomes
\( R^h{}_{abc} = W (\delta^h_b \gamma_{ac} - \delta^h_c \gamma_{ab}) \)

Let's take the covariant derivative of tensor W and antisymmetrize it. First its definition,
\( W_{a[b;c]} = U_h R^h{}_{abc} - (W_{ac} +U_a U_c) U_b + (W_{ab} + U_a U_b) U_c = 0 \)
Then its value,
\( W_{a[b;c]} = \gamma_{ab} W_{,c} - \gamma_{ac} W_{,b} \)
This means that W is a constant for two dimensions also, and that the source space is maximally symmetric there also: a sphere or a hyperbolic surface.

Thus, only a limited set of spaces can be projected onto a flat space while still preserving geodesics: maximally-symmetric ones, flat ones, hyperspherical ones, or hyperbolic continuations of them.

The equal-area and conformal constraints are much less strict, however.

 

[lpetrich]

Now the case of the destination space being curved. We can express the equation for the destination coordinates as
\( X^h{}_{a;b} = X^h{}_a U_b + X^h{}_b U_a - \Gamma(g)^h_{ij} X^i{}_a X^j{}_b \)

Using the manpulations done earlier, we find
\( X^h{}_f R(\gamma)^f_{abc} = X^h{}_b W_{[b;c]} + X^h{}_b W_{ac} - X^h{}_c W_{ab} + R(g)^h_{ijk} X^i{}_a X^j{}_b X^k{}_c \)
Multiplying by (X^-1)^b_h and contracting gives
\( R(\gamma)_{ac} = n W_{ac} - W_{ca} + R(g)_{ik} X^i{}_a X^k{}_c \)
This has the consequence that W is symmetric, as before, and that U is a gradient, as before.

Thus,
\( X^h{}_f R(\gamma)^f_{abc} = X^h{}_b W_{ac} - X^h{}_c W_{ab} + R(g)^h_{ijk} X^i{}_a X^j{}_b X^k{}_c \)
and
\( R(\gamma)_{ac} = (n-1) W_{ac} + R(g)_{ik} X^i{}_a X^k{}_c \)

The trivial case is a change of coordinates, and in that case, U = W = 0.

It is difficult for me to proceed further.

In the case of 2D or both spaces maximally symmetric,
\( {\hat R(\gamma)}\gamma_{ab} = W_{ab} + {\hat R(g)} Y_{ab} \)
where
\( R(g)^h_{ijk} = {\hat R(g)} ( \delta^h_j g_{ik} - \delta^h_k g_{ij} ) \)

 

[lpetrich]

That is as far as I can go with geodesic-preserving projections. It's also hard to do much with equal-area ones. So I'll consider conformal ones.

Let the source and the destination metrics be conformally related. Each point has its own expansion or contraction factor. Select the coordinates so that it is manifestly the case:
\( \gamma_{ij} = e^{2\Phi} g_{ij} \)

The problem becomes that of a conformal mapping from the destination space to itself, including the trivial case of (destination coordinates) = (source coordinates).

If the destination space is flat, the source space must be "conformally flat" to be conformally related. For the sphere space, that is indeed true. Consider the metric
\( ds^2 = \left( 1 + \frac14 (x^2 + y^2) \right)^{-2} (dx^2 + dy^2) \)
where
\( x = 2 \tan \frac{\theta}{2} \cos\phi ,\ y = 2 \tan \frac{\theta}{2} \sin\phi \)

Looking at x and y, they are our old friend the stereographic projection, but in rectangular coordinates.

The theory of flat-space conformal mappings is well-developed, and one can go from coordinates x,y to coordinates x',y' by making complex numbers out of them and using an analytic function, as it is called. For a complex argument, the argument's conjugate must appear nowhere in the function. Thus,

z = x + i*y
z' = x' + i*y'
z' = f(z)

For the stereographic transform,
\( z = x + i y = 2 \tan \frac{\theta}{2} e^{i\phi} \)

So,
(Lambert conformal conic) = (stereographic)^p for some power p
(Mercator) = log(stereographic)
(quincunx) = (elliptic functions)(Mercator)

 

[lpetrich]

I'll now show how to tell whether some space is conformally flat.

Start with a conformal transformation of a space's metric. Multiply it by a scaling factor that can vary across the space:
\( g'_{ij} = e^{2\Phi} g_{ij} \)

The connection coefficients become
\( \Gamma'^h_{ij} = \Gamma^h_{ij} + (\delta^h_j \Phi_{,i} + \delta^h_i \Phi_{,j} - g_{ij} g^{hl} \Phi_{,l}) \)

and the Riemann curvature tensor becomes
\( R'_{hijk} = e^{2\Phi} ( R_{hijk} + (g_{hj} F_{ik} + g_{ik} F_{hj} - g_{hk} F_{ij} - g_{ij} F_{hk}) ) \)
where
\( F_{ij} = - \Phi_{;ij} + \Phi_{,i} \Phi_{,j} - \frac12 g_{ij} g^{ab} \Phi_{,a} \Phi_{,b} \)

Contracting over h and j, that is, multiplying the Riemann tensor by g^hj and summing over h and j, gives us the Ricci tensor:
\( R'_{ik} = R_{ik} + (n-2)(- \Phi_{;ik} + \Phi_{,i} \Phi_{,k} - g_{ij} g^{ab} \Phi_{,a} \Phi_{,b}) - g_{ij} g^{ab} \Phi_{;ab} \)

Contracting further gives us the Ricci scalar:
\( R' = e^{-2\Phi} (R - 2(n-1) g^{ab} \Phi_{;ab} - (n-1)(n-2) g^{ab} \Phi_{,a} \Phi_{,b} ) \)

Let us now compose a sort of shifted Riemann tensor:
\( W_{ij} = R_{ij} - \frac{1}{2(n-1)} g_{ij} R \)

Then, W' = W + F -- very simple.

-

Define the "Weyl tensor", C, such that
\( R_{hijk} = C_{hijk} + (g_{hj} W_{ik} + g_{ik} W_{hj} - g_{hk} W_{ij} - g_{ij} W_{hk}) \)

Contracting on any set of indices of C will give zero. After a conformal transform, it becomes
\( C'_{hijk} = e^{2\Phi} C_{hijk} \)
The Riemann tensor's extra terms drop out of it.

Also define the "Cotton-York tensor", Y, such that
\( Y_{ijk} = W_{ij;k} - W_{ik;j} \)

Then,
\( Y'_{ijk} = Y_{ijk} - (n-2) \Phi_{,h} g^{hl} C_{lijk} \)

We now have what we need to find out whether or not a space is conformally flat.

 

[lpetrich]

At first sight, we have what we need. For a flat space, C = 0 and Y = 0, and therefore, if the two vanish, the space must be conformally flat. Or so it seems. For three dimensions, C automatically vanishes, and for two dimensions, both C and Y automatically vanish. So one must be more careful.

For two dimensions, the Riemann tensor is completely specified by the Ricci scalar:
\( R_{hijk} = (\frac12 R) \epsilon_{hi} \epsilon_{jk} = (\frac12 R) (g_{hj} g_{ik} - g_{hk} g_{ij}) \)

A conformal transform on a flat space gives this equation for the conformal scaling Φ:
\( R = - 2 e^{-2\Phi} g^{ab} \Phi_{;ab} \)

One can solve for Φ in it, and thus, every 2D space is conformally flat.

For three dimensions, the Riemann tensor is completely specified by the Ricci tensor:
\( R_{hijk} = - G^{ab} \epsilon_{ahi} \epsilon_{bjk} \)

where G, the "Einstein tensor", is
\( G_{ij} = R_{ij} - \frac12 g_{ij} R ,\ G^{ij} = g^{ia} g^{jb} G_{ab} \)
where upper-index g is the inverse of lower-index g.

That means that C automatically vanishes and that we must use Y to test for conformal flatness. There is a nice twist for 3 dimensions. One can construct a 2-index version:
\( Y_{ij} = \frac12 Y_{iab} \epsilon_j{}^{ab} \)
It is symmetric and traceless.

-

Of special-case spaces, an important one is maximal symmetry, where the space has as much symmetry as possible. This has the consequence that the Riemann tensor is given by
R_hijk = K * (g_hj * g_ik - g_hk * g_ij)

where K is a constant. When K = 0, the space is flat. Both the Weyl and the Cotton-York tensors vanish for K nonzero as well as zero, and thus, this sort of space is conformally flat.

Turning to familiar cases, both the 2D plane and the 2D-surface sphere are conformally flat, as are their generalizations to more dimensions.

 

[lpetrich]

Back to map projections, some of them can be expressed very simply in rectangular coordinates.

For a sphere, x = cos(θ)*cos(φ), y = cos(θ)*sin(φ), z = sin(θ)

Rotations around the polar axis (xy) by angle a: {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0, 0, 1}}
Sideways rotations or tilts (xz) by angle a: {{cos(a), 0, -sin(a)}, {0, 1, 0}, {sin(a), 0, cos(a)}}

Vector magnitude: x² + y² + z² = 1

Orthographic: {x,y}
Equal-area: (ortho) / sqrt(1 + z)
Stereographic: (ortho) / (1 + z)
Gnomonic: (ortho) / z

These projections also work for the hyperbolic plane, one sheet of a two-sheet circular hyperboloid.

For it, x = cosh(u)*cos(φ), y = cosh(u)*sin(φ), z = sinh(u)

Rotations around the polar axis (xy) by angle a: {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0, 0, 1}}
Sideways rotations or tilts (xz) by pseudo-angle a: {{cosh(a), 0, -sinh(a)}, {0, 1, 0}, {-sinh(a), 0, cosh(a)}}

Vector magnitude: - x² - y² + z² = 1

Spherical trigonometry carries over, but with hyperbolic functions for side lengths instead of trigonometric ones. To be precise, one must do analytic continuation: (side angle) gets multiplied by i.

A geodesic (shortest-distance curve, straight-line equivalent) is a circle for the sphere and a hyperbola for the hyperbolic plane. A geodesic passing through the pole(s):
Sphere: {sin(s), 0, cos(s)} -- a "great circle", a circle whose center is at the sphere's center
Hyperbolic plane: {sinh(s), 0, cosh(s)}

There is an interesting symmetry between whether the projections are finite or infinite in extent:

Projection	Sphere	Hyp Plane
Orthographic	Finite	Infinite
Equal-Area	Finite	Infinite
Stereographic	Infinite	Finite
Gnomonic	Infinite	Finite

The stereographic projection has a very interesting property. For both the spherical and the hyperbolic cases, circles and geodesics map onto circles.

For the spherical case, the projection center is inside every geodesic circle, while for the hyperbolic case, only some of a geodesic circle is inside the boundary circle.

The other projections do not have such convenient projections -- the projections are usually ellipses and hyperbolas and straight lines. In particular, the gnomonic projection turns geodesics into straight lines.

 

[lpetrich]

There is a very weird kind of regular polygon that a hyperbolic plane can have: a regular "apeirogon", a polygon with an infinite number of sides.

To understand it, consider that a regular polygon's vertices fall upon a circle, and that they can be placed with evenly-spaced angles form the circle's center.

So to make an apeirogon, one needs an infinite-radius circle. While that does not seem very easy to visualize, one can do so with the help of an infinite tilt. By sending its radius and tilt to infinity together in appropriate fashion, one can make some of the apeirogon have a finite distance from the center.

Mathematically, the apeirogon's circle becomes {sinh(w) - (1/2)*s², e^w*s, cosh(w) + (1/2)*s²} with the apeirogon's vertices at evenly spaced values of s. The w is a sort of residual tilt.

A stereographic projection makes it a circle that touches the projection's boundary circle, and the apeirogon's vertices get closer and closer as one approaches the boundary circle.

 

[lpetrich]

Now for tiling a plane with regular polygons. One can find all the regular polyhedra by interpreting them as regular-polygon tilings of spheres, and one can also find regular-polygon tilings for the hyperbolic plane.

To do so, first find the angle sum of a general polygon with n vertices. It can be divided into (n-2) triangles, giving
Angle sum >=< (n-2) * 180d

For the sphere, the relation is >, for the plane, =, and for the hyperbolic plane, <.

For a regular polygon, each angle's size is >=< (n-2)/n * 180d. Making m polygons meet at each vertex makes the sum of the angles 360d, giving
360d >=< m*(n-2)/n * 180d

or
1/n + 1/m >=< 1/2

Both n and m must be at least 3.

The only possible plane ones are (n,m) = (3,6), (4,4), (6,3)
The only possible spherical ones are (n,m) = (3,3), (3,4), (3,5), (4,3), (5,3)
All the others are hyperbolic, like (3,7), (4,5), (4,6), (5,4), (6,4), (7,3), ...

One can even tile with apeirogons, n = infinity.

 Uniform tilings in hyperbolic plane contains some depictions.

Intuitively, if m is too small, then there are not enough polygons to make a plane, and the shape curves in on itself. But if m is too large, then there are too many polygons, and the shape must get crinkly as one goes outward. This is evident from the metrics for all these shapes:

Sphere: ds² = da² + sin(a)² dφ²
Plane: ds² = da² + a² dφ²
Hyp Plane: ds² = da² + sinh(a)² dφ²

A way of illustrating the hyperbolic plane: crocheting a disk that gets crinkly at its edges.
The Institute For Figuring // Online Exhibit: Hyperbolic Space
CABINET // Crocheting the Hyperbolic Plane: An Interview with David Henderson and Daina Taimina
Crocheting the Hyperbolic Plane
Ravelry: Hyperbolic Plane pattern by Daina Taimina