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The Math Thread

For the stereographic projection, a geodesic circle with offset c will have a radius related to it.

Spherical: radius = sqrt(c2 + 1)
Hyperbolic: radius = sqrt(c2 - 1)

In both cases, going through the origin is a degenerate case resulting from c going to infinity.

In the spherical case, nondegenerate geodesics surround the origin.

In the hyperbolic case, nondegenerate geodesics do not surround the origin, and their centers are outside the projection's boundary circle. At the boundary-circle intersections, the geodesics' tangent direction is the radial direction.

That means that two different geodesics that intersect on the boundary circle will have zero angle between them. That intersection is sometimes called an "ideal point". A polygon with ideal points as its vertices will have total angle sum equal to zero.

The area of a hyperbolic-plane polygon is much like the area of a spherical polygon.
Spherical: (sum of angles) - (n-2)*pi
Hyp plane: (n-2)*pi - (sum of angles)

Thus, an ideal polygon, a polygon with all its vertices ideal points, has area (n-2)*pi. An ideal triangle thus has area pi -- and all ideal triangles are congruent. One can tilt and rotate any one to get any other one.

On the sphere side, a polygon can be a biangle, a lune, with the vertices at each others' antipodes.

Note on terminology:
Stereographic projection of the hyperbolic plane = Poincaré disk model
Gnomonic projection of the hyperbolic plane = Beltrami–Klein model or Cayley–Klein model or Klein disk model
Hyperboloid surface = hyperboloid model or Minkowski model or Lorentz model

Poincaré half-plane = a conformal mapping of the Poincaré disk model:
zdisk = (zhfpl - i)/(zhfpl + i)
zhfpl = i*(1 + zdisk)/(1 - zdisk)

The z's are from rectangular coordinates: z = x + i*y
 
Solution:

Draw a line between the center of the rectangle and the center of the circle. It will divide both the uncut pizza and the circle into two equal parts, and (equal-sized part) - (equal-sized part) = (equal-sized result)

 
lpetrich said:
Solution:

Draw a line between the center of the rectangle and the center of the circle. It will divide both the uncut pizza and the circle into two equal parts, and (equal-sized part) - (equal-sized part) = (equal-sized result)

Click to expand...


Presumably that line needs to be extended to the edge of the rectangle in both directions, otherwise the pizza remains as a single piece.

 
Response to bilby:

Yes, you're right about that. The line must be extended to the edges of the uncut pizza.

 
Returning to the projection problem, I will consider projection of a space onto itself. Since one can do multiple projections, it is evident that projections form a monoid, and if the projections are invertible, they form a group. A continuous set of invertible projections that includes the identity projection is thus a Lie group, and we can find out about it using infinitesimal departures from the identity projection -- the group's Lie algebra.

I will turn ξ into x and x into x', and for small departures from the identity projection, I will set x' = x + ξ. Thus,
\( X^i{}_a = \frac{\partial x'^i}{\partial x^i} \)

Area:
\( \sqrt{ \frac{|g(x')|}{|g(x)|} } |X| = A \)

Conformality:
\( g(x')_{ij} X^i{}_a X^j_{a} = K g(x)_{ab} \)

In the small-departure case, the area becomes
\( \xi^i{}_{;i} = \delta A \)

Conformality becomes
\( \xi_{i;j} + \xi_{j;i} = (\delta K) g_{ij} \)

This gives us the "conformal Killing vectors". For (δK = 0), one gets the plain Killing vectors. The plain ones are also equal-area. Conformal ones can also be equal-area if δK is constant though nonzero. Strictly speaking, equal-area to within some multiplicative factor.

For flat space with constant metric η, the plain ones are
(Translation) ξ = a where a is a constant vector
(Rotation) ξi = ηia wab xb where w is an antisymmetric 2-tensor
The extra conformal ones are
(Dilation) ξ = x
(Special) ξi = 2xi (x.c) - ci (x.x) where c is a constant vector and
(u.v) = (ηab ua vb)

The first three have a combined integral x' = R.x + D
where (RT).η.R = K*η

and the fourth one a rather complicated integral
\( x' = \frac{x - c (x \cdot x)}{1 - 2 (c \cdot x) + (c \cdot c) (x \cdot x)} \)

Inverting c with c -> c/(c.c) gives us
\( x' = \frac{(c \cdot c) x - c (x \cdot x)}{(c \cdot c) - 2 (c \cdot x) + (x \cdot x)} \)

Adding c to x and x' gives
\( x' = \frac{(c \cdot c) x - 2 (c \cdot x) c}{(x \cdot x)} \)

This suggest a conformal transform that seems rather difficult to derive in Lie-algebra fashion. Inversion: x' = x/(x.x)

For 2D, however, with metric = identity-matrix(2),
ξ1 + i*ξ2 = f(x1 + i*x2) where f is an analytic function.

The 2D case thus gives us x'1 + i*x'2 = F(x1 + i*x2) where F is an analytic function.


Hyperspheres have the same sort of Killing vectors that flat spaces do for more than 3 dimensions, though in fewer categories: rotations (plain) and certain distortions (conformal extra). The dimension condition is because of finiteness. Since spheres are conformally flat, one can carry over all the flat-space conformal Killing vectors. But at the transform origin's antipode, all but a few of them will become infinite there. So a sphere has 3 plain Killing vectors and only 3 extra conformal ones.
 
Now for geodesic-preserving. Let us first start with flat to flat in rectangular coordinates. Referring back to post #334, we have
\( x'^h_{,ij} = U_{,i} x'^h_{,j} + U_{,j} x'^h_{,i} ,\ U_{,ij} = U_{,i} U_{,j} \)

The solution is a bit complicated, so I will skip over some steps. We get this intermediate result:
\( (x'^h e^{-U})_{,ij} = 0 ,\ (e^{-U})_{,ij} = 0 \)

and it is easy to see that the solution is bilinear:
\( x'^h = \frac{R^h{}_i x^i + R^h{}_0}{R^0{}_i x^i + R^0{}_0} \)

for constant matrix (n+1)*(n+1)-dimensional matrix R for n dimensions.


The only conformal ones are linear, as are the only equal-area ones. The area factor is
\( A = \frac{|R|}{(R^0{}_i x^i + R^0{}_0)^{n+1}} \)

ETA: the determinant is for the complete matrix R, with the 0 entries.
 
For the definition of Yab in #331, I get the linearized version Yij = ξi;j + ξj;i.

After some manipulation of the equation that Yab;c satisfies, I find
\( \xi_{i;jk} + \xi_a R^a{}_{kij} = g_{ij} U_{,k} + g_{ik} U_{,j} \)

I then take ;l and subtract (k <-> l), giving
\( \xi_{a;j} R^a{}_{ikl} + \xi_{i;a} R^a{}_{jkl} + \xi_{a;l} R^a{}_{kij} - \xi_{a;k} R^a{}_{lij} - R_{ijkl;a} \xi^a \)
\( = g_{ik} U_{;jl} - g_{il} U_{;jk} \)

A linearized version of an equation in #334 is
\( R^a{}_{ijk} \xi^h{}_{;a} - R^h{}_{ajk} \xi^a_{;i} - R^h{}_{iak} \xi^a_{;j} - R^h{}_{ija} \xi^a_{;k} - R^h{}_{ijk;a} \xi^a \)
\( = \delta^h_j U_{;ik} - \delta^h_k U_{;ij} \)

I have verified that those two expressions are equal. One has to do (ijkl) <-> (hijk) and use some of the symmetries of the Riemann tensor.

Contracting the first one on indices i and j gives us
\( (\xi^a{}_{;a})_{;i} = (n + 1) U_{,i} \)

So U is related to the expansion factor ξa;a.

Contracting the third one on indices h and j gives us
\( R_{ak} \xi^a{}_{;i} + R_{ai} \xi^a{}_{;k} + R_{ik;a} \xi^a = - (n - 1) U_{;ik} \)

On i and k gives us
\( - R_{aj} \xi_{h}{}^{;a} + R_{ah} \xi^a{}_{;j} + R_{hajb}(\xi^{a;b} + \xi^{b;a}) + R_{hj;a} \xi^a = - (g_{hj} U_{;a}{}^a - U_{;hj}) \)

Completing the contraction gives us
\( 2R_{ab} \xi^{a;b} + R_{,a} \xi^a = - (n - 1) U_{;a}{}^a \)

Returning to the i and k contraction, it gives us this result:
\( R_{ai} Y^a{}_j - R_{aj} Y^a{}_i = 0 ,\ Y_{ij} = \xi_{i;j} + \xi_{j;i} \)
 
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I've worked out the Killing vectors for the sphere, using polar angle θ and azimuthal angle φ. These correspond to rotations around the three coordinate axes.
\( \{0, 1\} ,\ \{\cos\phi, -\cot\theta\sin\phi\},\ \{\sin\phi, +\cot\theta\cos\phi\} \)

Now the conformal ones:
\( \{-\sin\theta, 0\} ,\ \{\cos\theta\cos\phi, -\csc\theta\sin\phi\},\ \{\cos\theta\sin\phi, +\csc\theta\cos\phi\} \)


For maximal symmetry, Rhijk = K*(ghj*gik - ghk*gij) with K the curvature value, a constant, I find K*Yij = - U;ij. This has solutions

ξ = (plain Killing vectors) - 1/(2K) * (gradient of U)

(But for the sphere, at least, I find no solution for U -- it must vanish. So the geodesic preservers must be the plain Killing vectors, and those will always preserve geodesics.)

ETA: I discovered a mistake, and I fixed it. I now have some solutions:

position vector
\( p = \{ \sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta \} \)

with U = Uab*pa*pb

ETA again: the conformal Killing vectors are also gradients, but this time of V = Va*pa
 
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Sphere-to-sphere geodesic-preserving transform: vector x to vector x' with arbitrary nonsingular matrix W:
\( x' = \frac{W \cdot x}{|W \cdot x|} \)

If W is proportional to an orthonormal matrix, then this transform does rotations and reflections.

For conformal transforms, one can do

(sphere) -> (strereographic) -> (plane) -> (plane conformal) -> (plane) -> (inverse stereographic) -> (sphere)

The inversion transform for the plane has a simple meaning on the sphere: reflection across the projection equator. 0 = transform origin, infinity = its antipode.
 
Now for a 2-dimensional curiosity. The offset ξ can be expressed as
\( \xi^i = g^{ia} V_{,a} + \epsilon^{ia} W_{,a} \)

where ε is the antisymmetric symbol. Plugging it into the area expression gives
\( V^{;a}{}_{a} = \delta A \)

with W not contributing. This means that W can be an arbitrary function. For an equal-area transform of flat space, we get solutions like
\( V = \frac14 (\delta A) (x^2 + y^2) ,\ \xi = \frac12 (\delta A) \{ x, y \} \)

But for spherical space, there is no nonconstant equal-area solution for V, meaning that only W contributes to ξ.

There is also the curious result that (plain Killing vectors) = ε . (conformal Killing vectors) and vice versa.
 
Now back to maximal symmetry and geodesic-preserving transforms. Getting back to the general case (#331) from the infinitesimal, linearized case (#348), I find
\( Y_{ab} = Y_0 (K g_{ab} - W_{ab}) \)
where Y0 is constant. With Y0 = 1/K, it has the right linearized limit.

This gives us
\( U_{;abc} - 2 (U_{;ab} U_{,c} + U_{;ac} U_{,b} + U_{;bc} U_{,a}) + 4 U_{,a} U_{,b} U_{,c} + K (2g_{ab} U_{,c} + g_{ac} U_{,b} + g_{bc} U_{,a}) = 0 \)

For V = e-2U, it becomes
\( V_{;abc} + K (2g_{ab} V_{,c} + g_{ac} V_{,b} + g_{bc} V_{,a}) \)

This is linear in V, but I cannot proceed any further in this way. I specialized to the spherical case, and I found the same solutions for V that I found earlier for U in the linearized case.

I then tried the 3D-surface hyperspherical case, and I found the same kind of solution as for the 2D-surface sphere. The product of two vectors in a 4-space that the hypersphere can be embedded in.

So I returned to the equation just above and I contracted on a and b. This gives us
\( V^{;a}{}_{a} + 2(n+1) K V = 0 \)

Essentially (Laplacian of V) = (some value) * V. For a 2-sphere, that gives us the l = 2 spherical harmonics, and apparently their counterpart for the 3-sphere.
 
Spherical harmonics can be generalized to arbitrary numbers of dimensions. This can be done recursively, as I will show here.

First, the variables.
\( \theta_1 ,\ \theta_2 ,\ \theta_3 ,\ \dots \theta_n \)

The first one is the azimuthal angle and the rest are polar angles. The position of a point:
\( x_1 = \{ \cos\theta_1 ,\ \sin\theta_1 \} ,\ x_n = \sin\theta_n x_{n-1} \ (+)\ \{ \cos\theta_n \} \)

The metric:
\( (ds_1)^2 = (d\theta_1)^2 ,\ (ds_n)^2 = (d\theta_n)^2 + (\sin\theta_n)^2 (ds_{n-1})^2 \)

The equation can be separated:
\( (V_n)^{;a}{}_a + \lambda_n V_n = 0 ,\ V_n = X_n(\theta_n) V_{n-1}\)
where
\( \frac{1}{\sin^{n-1}\theta_n} \frac{d}{d\theta_n} \left( \sin^{n-1}\theta_n \frac{dX_n}{d\theta_n} \right) + \left( - \frac{\lambda_{n-1}}{\sin^2\theta_n} + \lambda_n \right) X_n = 0 \)

This has the solution
\( X_n = \sin^{m'} \theta_n C^{(\alpha)}_k (\cos\theta_n) \)
with parameters
\( m' = m_{n-1} ,\ \alpha = m' + \frac{n-1}{2} ,\ m_n = m' + k ,\ \lambda_n = m_n (m_n + n - 1) \)

where the C's are the Gegenbauer polynomials. Note that this solution gives the starting solution for n = 1 and m0 = 0, with the Gegenbauer polynomials becoming the Chebyshev polynomials there. For n = 2, the solution for the polar angle is the familiar associated Legendre polynomials.

For n dimensions and m = 2, we get the right eigenvalue for my previous post.
 
Doing a recursive solution, I find
\( V_{;ij} = - K g_{ij} V \)
for m = 1

and
\( V_{;ijk} = - K (2g_{ij} V_{,k} + g_{ik} V_{,j} + g_{ij} V_{,k}) \)
for m = 2

For the second one, one can use the first one as a shortcut: V(m=2) = V(m=1)1 * V(m=1)2 + constant

This shortcut is evident from V(m=0) = constant and from
\( V_n(m=1) = \{ \cos\theta_n ,\ \sin\theta_n V_{n-1}(m=1) \} \)
\( V_n(m=2) = \{ \cos^2\theta_n - (1/(n+1)) ,\ \cos\theta_n \sin\theta_n V_{n-1}(m=1) ,\ \sin^2\theta_n V_{n-1}(m=2) \} \)
to within proportionality constants.

It's also evident that V(m=1) is the position of a point in the space that the hypersphere is embedded in.
 
I will return to conformal Killing vectors for a bit. These vectors are small coordinate changes, x -> x + ξ, that preserve angles and shapes. Ordinary ones also make the space look alike after applying the transform. Conformal Killing vectors satisfy
\( \xi_{i;j} + \xi_{j;i} = 2 g_{ij} X \)
or
(Lie derivative of metric g with respect to ξ) = g * X

Ordinary ones have X = 0.

Doing (ij;k) + (ik;j) - (jk;i) gives us
\( \xi_{i;jk} + \xi_a R^a{}_{kij} = g_{ij} X_{,k} + g_{ik} X_{,j} - g_{jk} X_{,i} \)

and (;l) of that - (k <-> l) gives us
\( R_{ajkl} \xi^a_{;i} + R_{iakl} \xi^a_{;j} + R_{ijal} \xi^a_{;k} + R_{ijka} \xi^a_{;l} + R_{ijkl;a} \xi^a = \)
\( 2X R_{ijkl} - (g_{ik} X_{;jl} + g_{jl} X_{;ik} - g_{il} X_{;jk} - g_{jk} X_{;il}) \)
The LHS is (Lie derivative of Riemann tensor R with respect to ξ)

Contracting on j and l gives this Ricci-tensor result:
\( R_{ak} \xi^a_{;i} + R_{ia} \xi^a_{;k} + R_{ik;a} \xi^a = - ((n-2) X_{;ik} + g_{ik} X^{;a}{}_{a}) \)

and on i and k this Ricci-scalar result:
\( R_{,a} \xi^a = - 2XR - 2(n-1) X^{;a}{}_{a} \)

The shifted Ricci tensor mentioned earlier gives us
\( W_{ij} = R_{ij} - \frac{1}{2(n-1)} g_{ij} R ,\ W_{ak} \xi^a_{;i} + W_{ia} \xi^a_{;k} + W_{ik;a} \xi^a = - (n-2) X_{;ik} \)

Thus, the Weyl tensor gives us
\( C_{ajkl} \xi^a_{;i} + C_{iakl} \xi^a_{;j} + C_{ijal} \xi^a_{;k} + C_{ijka} \xi^a_{;l} + C_{ijkl;a} \xi^a = 2X C_{ijkl} \)
Notice that the derivatives of X drop out of this one, just as derivatives of conformal transforms do.

The Cotton-York tensor gives us
\( Y_{ajk} \xi^a_{;i} + Y_{iak} \xi^a_{;j} + Y_{ija} \xi^a_{;k} + Y_{ijk;a} \xi^a = - (n-2) C^a{}_{ijk} X_{,a} \)
Here, as in the previous ones, derivatives of the conformal factor X appear in the same fashion as derivatives of conformal transforms.
 
Specializing to maximal symmetry, one can find an equation for X from the shifted Ricci tensor. For maximal symmetry, it is W = (1/2)*(n-2)*K*g.

The W-ξ equation gives us (n-2)*K*X*gij = - (n-1)*Xij

Thus, X is the V(m=1) solution for the Laplacian eigenvalue equation, and the conformal Killing vectors ξi = - 1/(2K) * X,i in addition to the plain Killing vectors.

I'll now work out what a conformal transform does to a Killing vector. For raised index, ξ will be unchanged.
\( (e^{2\Phi} \xi_i);j + (e^{2\Phi} \xi_j);i - 2e^{2\Phi} (\xi_i \Phi_{,j} + \xi_j \Phi_{,i} - g_{ij} \xi^a \Phi_{,a}) = 2 g_{ij} e^{2\Phi} (X + \Phi_{,a} \xi^a) \)
giving us the same Killing vector but with its expansion factor X changed.

So one can carry over conformal Killing vectors with a conformal transform, like from a flat space to a maximally-symmetric curved one, because the latter sort of space is conformally flat.
 
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I will now look at some notable general-relativity space-time metrics.

The Friedmann-Lemaître-Robertson-Walker cosmology solution is conformally flat. One can define a "conformal time" as integral of 1/(scale factor) over time, and the metric will look like (scale factor)2 (flat metric).

However, it is only maximally symmetric in the empty (vacuum) and the cosmological-constant (lambdavacuum) cases. The latter solution includes the steady-state and inflation models, where (scale factor) = (constant) * exp(H*t) where H is the Hubble "constant" and t is the time.

I also did the general stationary radially-symmetric case, and it has only time shifts and rotations, with no extra conformal ones.

About the general stationary axially-symmetric case, it has only time shifts and axial rotations, with no extra conformal ones.


Now about counting the maximum number of symmetry generators that a n-dimensional space can have. For plain symmetries, it's (1/2)*n*(n+1), while for conformal ones, it's (1/2)*(n+1)(n+2), at least for n > 2.
 
I will now consider something simpler.

Let's say that one wants to take a large power of some integer. What will its last digit be?

Since there are only 10 digits when using base 10, that means that the digits will repeat themselves, and that we can use those repeats to help find the power. Here are the base sequences. These sequences then repeat indefinitely.
0
1
2 4 8 6
3 9 7 1
4 6
5
6
7 9 3 1
8 4 2 6
9 1

So if we want to find the last digit of 320971134235, we find 7113423235 mod 4 = 73 = 3.

One can do this with any other number-base system. I'll try hexadecimal, base 16. a,b,c,d,e,f = 10,11,12,13,14,15. A (0) means infinitely repeating 0's.
0
1
2 4 8 (0)
3 9 b 1
4 (0)
5 9 d 1
6 4 8 (0)
7 1
8 (0)
9 1
a 4 8 (0)
b 9 3 1
c (0)
d 9 5 1
e 4 8 (0)
f 1

Base 8
0
1
2 4 (0)
3 1
4 (0)
5 1
6 4 (0)
7 1

Base 5
0
1
2 4 3 1
3 4 2 1
4 1

Base 4
0
1
2 (0)
3 1

Base 3
0
1
2 1

Base 2
0
1
 
Now for an interesting curiosity related to spherical harmonics. Consider trigonometric functions of multiple arguments.

cos(2*a) = 2*(cos(a))2 - 1
cos(3*a) = 4*(cos(a))3 - 3*cos(a)
cos(4*a) = 8*(cos(a))3 - 8*(cos(a))2 + 1

sin(2*a) = 2*sin(a)*cos(a)
sin(3*a) = sin(a)*(4*(cos(a))2 - 1)
sin(4*a) = sin(a)*(8*(cos(a))3 - 4*cos(a))

Notice that cos(n*a) has form (polynomial in cos(a)), while sin(n*a) has form (polynomial in cos(a))*sin(a). This is evident from the recurrences
cos((n+1)*a) = cos(a)*cos(n*a) - sin(a)*sin(n*a)
sin((n+1)*a) = sin(a)*cos(n*a) + cos(a)*sin(n*a)
or else
(cos,sin)((n+1)*a) - 2*cos(a)*((cos,sin)(n*a)) + (cos,sin)((n-1)*a) = 0

The polynomials are called the Chebyshev polynomials Tn(x) ~ T(n,x) (first kind) and Un(x) ~ U(n,x) (second kind). They are related to the trigonometric multiples by

cos(n*a) = T(n,cos(a))
sin(n*a) = U(n-1,cos(a)) * sin(a)

One can find several identities for the Chebyshev polynomials by using trigonometric identities. Like
T(n+1,x) = x*T(n,x) - (1-x2)*U(n-1,x)
U(n,x) = T(n,x) + x*U(n-1,x)
(T,U)(n+1,x) - 2*x*((T,U)(n,x)) + (T,U)(n-1,x) = 0

One can find their differential equation from the trigonometric one. For T(n,x):
(1-x2)*y'' - x*y' + n2*y = 0

Another solution is sqrt(1-x2) * U(n-1,x), and U(n,x) has a differential equation of its own:
(1 - x2)*y'' - 3*x*y' + n*(n+2)*y = 0

The Chebyshev polynomials are special cases of the Gegenbauer ones: C(a)n(x) = C(n,a,x). They satisfy differential equation

(1-x2)y'' - (2a+1)*y' + n*(n+2a)*y = 0

The Chebyshev one for the first kind is for a = 0, while for the second kind, a = 1. The Legendre one is for a = 1/2. So in the spherical harmonics that I'd mentioned earlier, Gegenbauer polynomials can be extrapolated into the n = 1 case, with only one surface variable.
 
Do any numbers (a,b,c,d....) other than trivial solutions (0, 1, i, etc.) fulfill the requirement:

\(a^2 = ab = \frac{2ac + b^2}{3} =\frac{2ad + 2bc}{4} =\frac{2ae + 2bd + c^2}{5} = \frac{2af + 2be + 2cd}{6} = \, \cdots\)

No... nm.
 
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