Go Back   Freethought Forum > The Marketplace > The Sciences

Reply
 
Thread Tools Display Modes
  #76  
Old 08-28-2017, 11:14 PM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default Re: Math trivia

Quote:
Originally Posted by But View Post
Quote:
Originally Posted by lpetrich View Post
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 z-component in spherical coordinates: Lz = - i*d/(dφ) for azimuthal angle φ.

Consider exp(i*a*Lz) f(φ) for some function f.
It equals sum over n from 0 to infinity of 1/n! * an * (dnf(φ))(dφn)

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 angular-momentum states, one works out the representations of the rotation group.
Reply With Quote
Thanks, from:
But (08-29-2017)
  #77  
Old 08-29-2017, 02:06 PM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default 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.

Non-Euclidean 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 more-angle polygons.

Their successors also worked out elliptic geometry, where there are no non-intersecting 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 space-time curvature.
Reply With Quote
Thanks, from:
JoeP (08-29-2017)
  #78  
Old 08-29-2017, 09:24 PM
But's Avatar
But But is offline
This is the title that appears beneath your name on your posts.
 
Join Date: Jun 2005
Gender: Male
Posts: MVCXLVIII
Default Re: Math trivia

Quote:
Originally Posted by lpetrich View Post
If you have ever worked through quantum-mechanical angular momentum, much of this should be familiar:

Angular-momentum operators = rotation generators
Angular-momentum magnitude = Casimir invariant of rotation
Etc.
Quote:
Originally Posted by But View Post
Quote:
Originally Posted by lpetrich View Post
...
That looks a lot like angular momentum in quantum mechanics. Interesting.
:blush:
Reply With Quote
Thanks, from:
lpetrich (08-31-2017)
  #79  
Old 08-31-2017, 04:36 AM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default 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:
5-cell: 5, 10, 10, 5 (triangle, tetrahedron)
Tesseract: 16, 32, 24, 8 (square, cube)
16-cell: 8, 24, 32, 16 (triangle, tetrahedron)
24-cell: 24, 96, 96, 24 (triangle, octahedron)
120-cell: 600, 1200, 720, 120 (pentagon, dodecahedron)
600-cell: 120, 720, 1200, 600 (triangle, icosahedron)

Dualities:
5-cell (self)
Tesseract - 16-cell
24-cell (self)
120-cell - 600-cell

Four more than four dimensions, there are 3 of them. Instead of a list, I will give a count of k-dimensional objects for n dimensions (point = 0, line = 1, ...).
Simplex: (n+1)!/((k+1)!*(n-k)!)
Hypercube: 2n-k*n!/(k!*(n-k)!)
Cross-polytrope or orthoplex: 2k+1*n!/((k+1)!*(n-k-1)!)

Duality:
Simplex (self)
Hypercube - Orthoplex

Simplex: triangle, tetrahedron, 5-cell, ...
Hypercube: square, cube, tesseract, ...
Orthoplex: square, octahedron, 16-cell, ...

So in summary, the number of regular polytopes:
infinite, 5, 6, 3, 3, 3, 3, 3, 3, 3, 3, ...
Reply With Quote
  #80  
Old 08-31-2017, 06:47 AM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default Re: Math trivia

I'll now consider plane tilings and their generalizations to n-space tilings, all for flat spaces.

2D: triangle, square, hexagon
3D: cube
4D: tesseract (8-cell), 16-cell, 24-cell
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)
  • Fully-symmetric 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.
Reply With Quote
  #81  
Old 08-31-2017, 07:05 AM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default 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 n-gon 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; 08-31-2017 at 07:17 AM.
Reply With Quote
  #82  
Old 08-31-2017, 08:58 AM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default Re: Math trivia

So here's a list of all non-hyperbolic 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}

n-D convex: {3,...,3}, {3,...,3,4}, {4,3,...,3}
n-D 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 space-tiling 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 16-cell 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 24-cell tiling {3,4,3,3} has vertices offset from the 16-cell vertices by the 24-cell 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.
Reply With Quote
  #83  
Old 09-16-2017, 09:39 PM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default 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 2n-1. Spin(2n+1) describes irreducible spinors with size 2n.

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 non-real 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 "pseudo-real" or "quaternionic".

The three cases can be abbreviated real: R, pseudo-real: 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 period-8 is sometimes called a "Bott periodicity". The CC ones are conjugates of each other, while the RR and HH are two separate self-conjugate ones each.

For Spin(n1,n2), the counterpart of SO(n1,n2), the reality is like for Spin(|n2-n1|). Thus, a massless spin-1/2 particle in our SO(3,1) space has CC and thus divides up into two parts that are mirror images of each other: a left-handed part and a right-handed part.
Reply With Quote
  #84  
Old 09-16-2017, 10:09 PM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default Re: Math trivia

I'll now get into the Cayley-Dickson 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: a-1 = (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.
Reply With Quote
  #85  
Old 09-16-2017, 10:59 PM
lpetrich's Avatar
lpetrich lpetrich is offline
Member
 
Join Date: Jul 2004
Location: Lebanon, OR, USA
Gender: Male
Posts: CDXCVI
Default 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 Cayley-Dickson 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)
Power-associative: it does not matter what order one evaluates the powers in it: ap+q = ap.aq for all p and q
Well-normed: N(a.b) = N(a)*N(b)
Lacking zero divisors: a.b = 0 implies at least one of a = 0 or b = 0

Also,
Self-conjugate: a = a*

The real numbers (size 1) satisfy all these properties: comm, assoc, alt, pwra, norm, nozd, scjg.

The next C-D step gives the complex numbers (size 2). They are not self-conjugate, leaving comm, assoc, alt, pwra, norm, nozd.

The next C-D step gives the quaternions (size 4). Their multiplication is not commutative, leaving assoc, alt, pwra, norm, nozd.

The next C-D step gives the octonions (size 8). Their multiplication is not associative, though it is alternative, leaving alt, pwra, norm, nozd.

The non-associativity means that it has no matrix representation.

The next C-D step gives the sedenions (size 16). Their multiplication is not alternative or well-normed, and they have some zero divisors. However, it is still power-associative, 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, component-by-component multiplication of {1,0} and {0,1} gives {0,0}.

All subsequent C-D steps have no changes in properties.


Now for power-associativity. I'll start with
a1 = P1*(1) + Q1*b
a2 = P2*(1) + Q2*b
a12 = P12*(1) = Q12*b
where
b is anti-self-conjugate or imaginary-like: 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
Reply With Quote
Reply

  Freethought Forum > The Marketplace > The Sciences


Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
 
Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump

 

All times are GMT +1. The time now is 07:47 PM.


Powered by vBulletin® Version 3.8.2
Copyright ©2000 - 2017, Jelsoft Enterprises Ltd.
Page generated in 0.19602 seconds with 14 queries