

08292017, 12:14 AM


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Re: Math trivia
Quote:
Originally Posted by But
Quote:
Originally Posted by lpetrich
I'll now consider groups with continuous parameters, groups with elements M(a) where a is some set of real numbers. Such groups are called "Lie groups", after 19th cy. Swedish mathematician Sophus Lie ("Lee"). ...

That looks a lot like angular momentum in quantum mechanics. Interesting.

Yes indeed. Angular momentum = rotation generators.
I'll illustrate with the zcomponent in spherical coordinates: L z =  i*d/(dφ) for azimuthal angle φ.
Consider exp(i*a*L z) f(φ) for some function f.
It equals sum over n from 0 to infinity of 1/n! * a n * (d nf(φ))(dφ n)
That is a Taylor series, and it evaluates to f(φ + a). That is, rotate by azimuthal angle a.
When one works out the possible angularmomentum states, one works out the representations of the rotation group.

08292017, 03:06 PM


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Re: Math trivia
Just for fun:
Lobachevsky Lyrics  Tom Lehrer
Lobachevsky (song)  Wikipedia
Notice how he sings in a mock Russian accent and how he tries to imitate a Russian speaking English, like not using "the" and putting objects first in some sentences.
NonEuclidean geometry  Wikipedia (1792  1856) and János Bolyai  Wikipedia (1802  1860) independently discovered that Euclid's fifth or parallel postulate was independent of his other axioms by constructing hyperbolic geometry, a system that obeys those axioms but not the parallel one. In hyperbolic geometry, one can have an infinite number of lines that do not intersect some other line, and not just one line (Euclid's fifth). A consequence is that the sum of the angles of a triangle is less than 180d, with comparable inequalities for moreangle polygons.
Their successors also worked out elliptic geometry, where there are no nonintersecting lines, and where the sum of the angles of a triangle is greater than 180d. One of them was Bernhard Riemann  Wikipedia, who, in 1854, generalized further to curvature as an arbitrary function of position.
This bit of math was used by Albert Einstein in General Relativity for describing gravity as spacetime curvature.

08292017, 10:24 PM


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Re: Math trivia
Quote:
Originally Posted by lpetrich
If you have ever worked through quantummechanical angular momentum, much of this should be familiar:
Angularmomentum operators = rotation generators
Angularmomentum magnitude = Casimir invariant of rotation
Etc.

Quote:
Originally Posted by But
Quote:
Originally Posted by lpetrich
...

That looks a lot like angular momentum in quantum mechanics. Interesting.


08312017, 05:36 AM


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Re: Math trivia
I'll now consider regular polygons (2D) and their counterparts in greater numbers of dimensions: polyhedra (3D), polychora (4D), and polytopes in general.
In two dimensions, there is an infinite family of regular polygons, starting with equilateral triangles and squares. For n >= 3, they have n vertices and n edges. In the notation that I will use below:
n, n
In three dimensions, there are only the 5 Platonic solids. They are:
Tetrahedron: 4, 6, 4 (triangle)
Cube: 8, 12, 6 (square)
Octahedron: 6, 12, 8 (triangle)
Dodecahedron: 20, 30, 12 (pentagon)
Icosahedron: 12, 30, 20 (triangle)
In order: vertices, edges, faces. They have the following dualities, from reflecting the list of contents:
Tetrahedron (self), like polygons
Cube  Octahedron
Dodecahedron  Icosahedron
In four dimensions, there are 6 of them. They are:
5cell: 5, 10, 10, 5 (triangle, tetrahedron)
Tesseract: 16, 32, 24, 8 (square, cube)
16cell: 8, 24, 32, 16 (triangle, tetrahedron)
24cell: 24, 96, 96, 24 (triangle, octahedron)
120cell: 600, 1200, 720, 120 (pentagon, dodecahedron)
600cell: 120, 720, 1200, 600 (triangle, icosahedron)
Dualities:
5cell (self)
Tesseract  16cell
24cell (self)
120cell  600cell
Four more than four dimensions, there are 3 of them. Instead of a list, I will give a count of kdimensional objects for n dimensions (point = 0, line = 1, ...).
Simplex: (n+1)!/((k+1)!*(nk)!)
Hypercube: 2nk*n!/(k!*(nk)!)
Crosspolytrope or orthoplex: 2k+1*n!/((k+1)!*(nk1)!)
Duality:
Simplex (self)
Hypercube  Orthoplex
Simplex: triangle, tetrahedron, 5cell, ...
Hypercube: square, cube, tesseract, ...
Orthoplex: square, octahedron, 16cell, ...
So in summary, the number of regular polytopes:
infinite, 5, 6, 3, 3, 3, 3, 3, 3, 3, 3, ...

08312017, 07:47 AM


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Re: Math trivia
I'll now consider plane tilings and their generalizations to nspace tilings, all for flat spaces.
2D: triangle, square, hexagon
3D: cube
4D: tesseract (8cell), 16cell, 24cell
5D and more: hypercube
Polytopes can be interpreted as space tilings for convex spaces, and there is a third possibility: hyperbolic spaces. One can distinguish them by how the circumference of an expanding circle behaves.
If it increases as (2pi)*(radius), then the space is flat. If it increases more slowly, then it is convex, while if it increases faster, then it is hyperbolic.
For radius r,  Flat: 2pi * r
 Convex / Spherical: 2pi * R * sin(r/R)
 Fullysymmetric hyperbolic: 2pi * R * sinh(r/R)
Let's see what happens to a growing disk.
In the flat case, it stays flat, as one would expect.
In the convex case, it starts curling toward one side, and its edge eventually shrinks and becomes a point.
In the hyperbolic case, it gets more and more crinkly as one goes.

08312017, 08:05 AM


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Re: Math trivia
I'll show what regular polyhedra and plane tilings are possible. They are a sort of island in a sea of hyperbolic tilings.
Let's start with a regular triangle, an equilateral one. Its vertex angles are 180d/3 = 60d.
3 triangles: 180d  tetrahedron
4 triangles: 240d  octahedron
5 triangles: 300d  icosahedron
6 triangles: 360d  triangular tiling
7 triangles: 420d  hyperbolic
Now a regular quadrangle, a square. Its vertex angles are 360d/4 = 90d
3 squares: 270d  cube
4 squares: 360d  square tiling
5 squares: 450d  hyperbolic
Now a regular pentagon. Its vertex angles are 540d/5 = 108d
3 pentagons: 324d  dodecahedron
4 pentagons: 432d  hyperbolic
Now a regular hexagon. Its vertex angle = 720d/6 = 120d
3 hexagons: 360d  hexagonal tiling
4 hexagons: 480d  hyperbolic
Now a regular heptagon. Its vertex angle = 900d/7 = (900/7)d
3 heptagons: (2700/7)d  hyperbolic
(Numerical value: 385.714d)
This construction can be generalized using "Schläfli symbols".
(none)  point
{}  line
{p}  polygon
{p,q}  polyhedron
{p,q,r}  polychoron
...
For {p,...,q}, one finds a shape where r of the objects meet at each meeting point, and one gets {p,...,q,r}
Thus, a ngon is {n} and the regular polyhedra and tilings are
{3,3}  tetrahedron
{3,4}  octahedron
{3,5}  icosahedron
{3,6}  triangular tiling
{4,3}  cube
{4,4}  square tiling
{5,3}  dodecahedron
{6,3}  hexagonal tiling
These symbols also express stellated polytopes, like star polygons. One uses fractional values p/q, where p is the total number and q is the skip value as one goes around. Thus a pentagram is {5/2}, with total number 5 and skip value 2.
Instead of 1, 2, 3, 4, 5,
a pentagram does 1, 3, 5, 2, 4
Duality is easy to find for Schläfli symbols  reverse the order.
Last edited by lpetrich; 08312017 at 08:17 AM.

08312017, 09:58 AM


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Re: Math trivia
So here's a list of all nonhyperbolic regular polytopes and tilings, using their Schläfli symbols.
1D: {}
2D convex: {p}
2D stellated: {p/q}
3D convex: {3,3}, {3,4}, {4,3}, {3,5}, {5,3}
3D stellated: {3,5/2}, {5/2,3}, {5,5/2}, {5/2, 5}
2D plane: {3,6}, {6,3}, {4,4}
4D convex: {3,3,3}, {3,3,4}, {3,4,3}, {4,3,3}, {3,3,5}, {5,3,3}
4D stellated: {3,3,5/2}, {5/2,3,3}, {3,5,5/2}, {3,5/2,5}, {5/2,3,5}, {5,3,5/2}, {5,5/2,3}, {5/2,5,3}, {5,5/2,5}, {5/2,5,5/2}
3D plane: {4,3,4}
5D convex: {3,3,3,3}, {3,3,3,4}, {4,3,3,3}
(no stellated ones for 5D and higher)
4D plane: {3,3,4,3}, {3,4,3,3}, {4,3,3,4}
nD convex: {3,...,3}, {3,...,3,4}, {4,3,...,3}
nD plane: {4,3,...,3,4}
Counting them:
1D: 0
2D: (infinite), (infinite)
3D: 5, 4, 3
4D: 6, 10, 1
5D: 3, 0, 3
nD: 3, 0, 1
(n>5)
The spacetiling vertices' coordinates can be expressed in forms with some very interesting properties.
The hypercubic {4,3...3,4} tiling exists in all numbers of dimensions >= 2, and its vertices are {integers}
The triangular tiling {3,6} has vertices {1,0}*n1 + (1/2,sqrt(3)/2}*n2 where n1 and n2 are integers.
The hexagonal tiling {6,3} is like the triangular tiling, having that tiling's vertices an offset of those vertices: n1 and n2 = integers + 1/3
The 16cell tiling {3,3,4,3} has vertices {4 integers} with an even sum. This is {4 even}, {4 odd}, and {2 even, 2 odd}. Offsetting, one gets {4 integers} with an odd sum, {3 even, 1 odd}, {1 even, 3 odd}.
The 24cell tiling {3,4,3,3} has vertices offset from the 16cell vertices by the 24cell vertices: {1 (+1), 3 0} and {4 (+ 1/2)}
This is {4 integers} with either an even or an odd sum, and {4 integers} + 1/2.

09162017, 10:39 PM


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Re: Math trivia
Now some more group theory. As I'd mentioned, the rotation groups SO(n) have spinor representations (spinor = spin vector). Those reps are sometimes called groups Spin(n).
Spinors were first developed by physicist Wolfgang Pauli around 1930 to describe the spins of electrons. This including "Pauli matrices" for electrons' spin operators. Someone or other later discovered that he had reinvented quaternions, because Pauli matrices can be used to create a matrix representation of quaternions.
Generalizing to an arbitrary number of dimensions, we find a difference between even and odd numbers. Spin(2n) describes spinors with size 2[sup]n[/usp], but each one splits into two irreducible ones with sizes both 2 n1. Spin(2n+1) describes irreducible spinors with size 2 n.
Can any spinors be described with real matrices? For some numbers of dimensions, they can. For this, I'll introduce whether or not a group representation is real. Consider an irreducible representation (irrep) D(a) for group elements a. Now try to find a matrix Z that can map D(a) onto its complex conjugate:
Z.D(a).Z 1 = D *(a)
If all of the D(a)'s elements are real, then one can set Z = I and one is done. But if of some of the D(a)'s elements are nonreal complex, then it is more challenging. If it is not possible, then the irrep is called "complex", while if it is possible, there are two special cases for Z:
Z.Z * = Z *.Z = +1 or 1
The +1 case means that the irrep is real, while the 1 case means that the irrep is "pseudoreal" or "quaternionic".
The three cases can be abbreviated real: R, pseudoreal: H, complex: C. The H is after William Rowan Hamilton, the discoverer of quaternions.
Let's see how the spinor reps stack up:  8n + 0: RR
 8n + 1: R
 8n + 2: CC
 8n + 3: H
 8n + 4: HH
 8n + 5: H
 8n + 6: CC
 8n + 7: R
This period8 is sometimes called a "Bott periodicity". The CC ones are conjugates of each other, while the RR and HH are two separate selfconjugate ones each.
For Spin(n1,n2), the counterpart of SO(n1,n2), the reality is like for Spin(n2n1). Thus, a massless spin1/2 particle in our SO(3,1) space has CC and thus divides up into two parts that are mirror images of each other: a lefthanded part and a righthanded part.

09162017, 11:09 PM


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Re: Math trivia
I'll now get into the CayleyDickson construction, because it gives a hierarchy of mathematical entities that lose properties as one goes for the first few.
From an algebra, one constructs the next algebra by taking ordered pairs of elements of the original one and applying various rules to it. One usually starts with the real numbers, and I will do so here, though one may use other algebraic fields.
Addition: (a1,b1) + (a2,b2) = (a1+a2,b1+b2)
Scalar multiplication: c*(a,b) = (c*a,c*b)
Conjugation: (a,b)* = (a*, b)
So the first element's sign is unchanged and all the others have reversed signs.
It undoes itself: (a*)* = a
Multiplication: (a1,b1).(a2,b2) = (a1.a2  (b2*).b1, b2.a1 + b1.(a2*))
This can be generalized by multiplying the (b2*).b1 term by something like 1, but I won't do that here.
It is distributive over addition. Also, (a.b)* = (b*).(a*)
Norming: N(a) = a.(a*) = (a*).a = sum of squares of all components of a.
Reciprocal: a1 = (1/N(a))*(a*)
For real numbers, N((1,1)) = 2
For GF(2), addition and multiplication modulo 2, N((1,1)) = 0
So for finite fields, one has to be careful.

09162017, 11:59 PM


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Re: Math trivia
First, a bit of clarification. The multiplicative identity (1) = (1,0,0,...,0)
Norming is a.(a*) = (a*).a = N(a)*(1)
Now for what properties drop out as one advances in the CayleyDickson construction. I'll start with a big list:
Multiplication is
Commutative: a.b = b.a
Associative: (a.b).c = a.(b.c)
Alternative: like associative but with two of the variables equal: (a.a).b = a.(a.b), (a.b).a = a.(b.a), (b.a).a = b.(a.a)
Powerassociative: it does not matter what order one evaluates the powers in it: ap+q = ap.aq for all p and q
Wellnormed: N(a.b) = N(a)*N(b)
Lacking zero divisors: a.b = 0 implies at least one of a = 0 or b = 0
Also,
Selfconjugate: a = a*
The real numbers (size 1) satisfy all these properties: comm, assoc, alt, pwra, norm, nozd, scjg.
The next CD step gives the complex numbers (size 2). They are not selfconjugate, leaving comm, assoc, alt, pwra, norm, nozd.
The next CD step gives the quaternions (size 4). Their multiplication is not commutative, leaving assoc, alt, pwra, norm, nozd.
The next CD step gives the octonions (size 8). Their multiplication is not associative, though it is alternative, leaving alt, pwra, norm, nozd.
The nonassociativity means that it has no matrix representation.
The next CD step gives the sedenions (size 16). Their multiplication is not alternative or wellnormed, and they have some zero divisors. However, it is still powerassociative, leaving pwra.
Zero divisors are counterintuitive, but they can exist in some cases. Taking the square of matrix {{0,1},{0,0}} gives the zero matrix {{0,0},{0,0}}. Likewise, componentbycomponent multiplication of {1,0} and {0,1} gives {0,0}.
All subsequent CD steps have no changes in properties.
Now for powerassociativity. I'll start with
a1 = P1*(1) + Q1*b
a2 = P2*(1) + Q2*b
a12 = P12*(1) = Q12*b
where
b is antiselfconjugate or imaginarylike: b* =  b
Then,
P12 = P1*P2  Q1*Q2*N(b)
Q12 = P1*Q2 + Q1*P2
or
(P12 + i*nb*Q12) = (P1 + i*nb*Q1) * (P2 + i*nb*Q2)
where nb = sqrt(N(b))
To get the nth power of a*(1) + b, one starts with P = a and Q = 1 and finds
(P(n) + i*nb*Q(n)) = (a + i*nb)n

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