Probability and Statistics

[CraigD]

P.P.S. Sorry if the above seems a little polemical at all - I've had some coffee, and now remember that i forgot to mention (although you've uncovered it) that a Rubik's cube is not in any unitary group, except U(1) is a subspace ...

The states of a 2x2x2 appear to me to be isomorphic to U(24), because they are representable by 24x24 matrixes, and for every such matrix [math]A[/math], [math]A^* A =A A^* = I[/math]. Though I’m so rusty on the subject (haven’t used matrixes for other than the practical uses I mentioned in my previous post) that I can’t even recall having studied them in school almost 30 years ago, I gather from the Mathworld and wikipedia articles on unitary groups that these are the only two requirements.

To be a unitary group, Rubik's cubes need skew-symmetry and they don't have it; ...

I don’t think any skew symmetry is required if no entry of any member of the group has a non-zero imaginary part. The U(24) group of a 2x2x2 Rubik’s cube has entry values of 0 and 1 only.

 

Because all of its members have only real value entries, [math]A^* = A^T[/math]. I don’t think the any entries of the members of a unitary group are required have imaginary parts.

 

I think it’s not special unitary, because some [math]\det(A) = 1[/math], some = -1, while all members [math]u[/math] of a special unitary group must have [math]\det(u) = 1[/math].

Oh yeah - how does all this relate to an involutional transform in unitary Hermitian terms, in a Hilbert space of vectors? The Hadamard is still hovering vaguely on my algebraic horizon here; I think I need to get a handle on the different ways it 'works' in unitary/non-unitary vector spaces.

Alas, this is beyond my ability to quickly comment on – I’d have to learn more about these terms before writing about them, which could takes days, weeks, or even longer.

Does the colored, sliced cube have a norm?

The value of it’s norm, which I believe is the absolute value of the determinant of its members, is 1.

It isn't a cube made out of complex numbers, but fixed colors.

It’s important to understand that colored labels on a cube are aesthetic only. It’s more userful, IMHO, to think of each cubie or cubie face as having a unique label, and, if you have a physical cube puzzle, rather than a computer representation of one, to physically label each face, typically with a numeral.

Still, it's kind of mind-boggling that absolute libraries of math are 'in there'.

 

Even if you don't know about the mathematics, you still go through it when you solve a cube;

I agree.

 

That formal system of such terse collections of initial theorems appear to so closely math nature, on so many scales, is, I think, perhaps the most fundamental wonder there is.

[Boof-head]

Whoops, left out the word "special" in that post: the cubes aren't in a special unitary group. SU's do have skew-symmetry, right?

 

My take on the aesthetics is that quark colors are in the same domain as cube colors. A 'color' can represent whatever you say it does, like a number; you could claim that each color has a real and an imaginary part, the color itself and its potential 'diffusion' over the cube. Or that each color has a different 'color-height', which is imaginary (since you can't see it); color-height is then a term in any contrast function, and each facelet is part of a polynomial (of color), etc.

 

I'd like to know if there are braid groups and some analog (maybe a real rep) of a Jones polynomial. I think there should be, but I'm not brave enough just yet. Jones polynomials are connected somehow to certain quantum states..

Mr Jones is something of a hero, at least at one of my old alumni the Auckland School of Math. I also think you could possibly spend a lifetime discovering 'math' in these kinds of puzzles (going quietly crazy in a padded room somewhere and muttering "where is that cycle group" and other weird stuff)

[Boof-head]

Can you verify that the 2-slice cube, as I call it, is in U(24), as you state?

If it is, then it's a skew-Hermitian (color) matrix, with a Lie group. Is it because there are no fixed 'partitions' at the center of each face?

 

It's interesting how the wiki page has a chart for the available positions, of the total permutation space, for full coverage you need 11 full twists, as they call the rotations of each half - equivalent to rotating each half a quarter-twist together (anti-symmetrically), or conjunction, and after which number (11) you can't get any more; here you can see how, on the right of the 'number pile' for full twists, the one for the same number of quarter-twists individually or disjunction takes 3 more to 'fill the space.

[CraigD]

Can you verify that the 2-slice cube, as I call it, is in U(24), as you state?

I’ve not tried proving it, but checking many n x n matrixes B generated from random n-tuples as described in post #32, find that [math]B B^T = I[/math], so I believe any group generated by multiplication of such matrixes is in some U(n).

If it is, then it's a skew-Hermitian (color) matrix, with a Lie group.

None of them are skew-Hermitian, because [math]B^T \not= -B[/math].

 

Because of this, I believe the possible states of any puzzle like a Rubic’s cube with any number of faces and slices, older and simpler puzzles like the fifteen puzzle, and countless other puzzles that can be represented by ordered n-tuples, including a puzzle where pieces may be moved freely (like a child’s wooden tile puzzle) are in some U(n).

 

The ordinary 3x3x3 cube is in U(54).

 

One may trivially add diagonal rows and column so that the 2x2x2 cube is in U(n) for n>24, a 3x3x3 in U(n), n>54, or, in general, a puzzle with m tile faces in U(m).

 

I strongly suspect that the 2x2x2 cube’s states are in U(n) for n<24, the 3x3x3 in U(n), n<54, and many m-faced puzzle in U(n), n<m, though can’t immediately think of how to make such a representation.

 

Computationally, operating on the n x n matrixes to which the n-tuple representations of various cube-type puzzles can be mapped is much less efficient than operating on tuples using what I called in post #32 “a multiplication operation less conventional than … matrix multiplication”, such as the one implemented in the code in post #16.

[Boof-head]

I believe any group generated by multiplication of such matrixes is in some U(n).

Ok, it's just that I went and tried to uncover evidence that U(n) is part of the quotient space, for n >1; I haven't found explicit evidence so in math-speak 'we have no proof yet'. However the wiki entry on Unitary groups has this:

In mathematics, the unitary group of degree n, denoted U(n), is the group of n×n unitary matrices, with the group operation that of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C).

 

In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with absolute value 1 under multiplication. All the unitary groups contain copies of this group.

 

The unitary group U(n) is a real Lie group of dimension n^2. The Lie algebra of U(n) consists of complex n×n skew-Hermitian matrices, with the Lie bracket given by the commutator.

 

The general unitary group (also called the group of unitary similitudes) consists of all matrices A such that A * A is a nonzero multiple of the identity matrix, and is just the product of the unitary group with the group of all positive multiples of the identity matrix.

I think where a color is concerned, if you say "a color is imaginary" then there are 6 imaginary colors with a real part - the area of each facelet - so each position also has an index, relative to the fixed central facelet on the 3-slice cube.

Or the central color and position (and the 'floating' or missing imaginary center on the 2-slice cube) represent the index for each face, so should get a label like '00', for the position and (s^2 + i.color) for the real area + imaginary color rep.

 

I've also seen a paper by someone called Finkelstein and another author; Finkelstein is a pretty abstract guy - he says the weirdest things about something called "Quantum Relativity" and a Rubik's group representation. Except he gets away with it, being a tenured professor at Austin and all.

 

That chart is bugging me - I think it might be worth trying to 'Godelize' it; after any 11 (antisymmetric) moves, or full-twists, you reach the largest state-space of positions or permutations, the series up to 11 of such moves are factorised by the number of moves or are all multiples. With zero moves there is 1 permutations, with 1 move there are 9, with 2 there are 54 permutations, and so on; the first step might be finding the recurrence. The initial state (1), goes to (9) possible positions after 1 move, then 6x(9) or (54) the next number of available states is 321 (??). I need to find a pattern here...

 

But you can already see a 'phase difference' in the chart; the full-twist algebra realises the largest number of possibilities after 11 turns, the quarter-twist (half a full-twist) algebra gets there after 14, so a scrambled 2-cube is never more than 14 turns from a solution. The difference is a factor of 3 quarter-turns which is also the number of spatial dimensions; the cubes are a complete or affine space; a packaged universe of 'mathematical structure'.

These things can drive you nuts, trying to solve them (manufacturer's warning).

[CraigD]

It's interesting how the wiki page has a chart for the available positions, of the total permutation space, for full coverage you need 11 full twists …

I can’t find this chart on the Rubik’s cube wikipedia article. Searching the article for “11” Can you point me to it – give a more specific link, or some text to find? :QuestionM

 

I think it’s impossible that a 3x3x3 cube can be manipulated into any of its 43,252,003,274,489,856,000 possible states within 11 face or slice moves, because only 27 such moves ((6 faces + 3 center slices)(2 quarter turns +1 half turn)) are available, and [math]\sum_{i=0}^{11} 27^i = 5,772,870,588,346,120 < 43,252,003,274,489,856,000[/math]. The number of cube states possible in 11 moves is less than 5,772,870,588,346,120, because it must exclude repeating the same move (eg: <F, F>, which is the same as <F2>) moves that result in a previously reached state (eg: <F,R,-R>, which is the same as <F>), and similar sequences of moves. At least 14 moves should be required, because [math]\sum_{i=0}^{13} 27^i = 4,208,422,658,904,321,508 < 43,252,003,274,489,856,000[/math] and [math]\sum_{i=0}^{14} 27^i = 113,627,411,790,416,680,700 > 43,252,003,274,489,856,000[/math]. As I’ve not worked out this systematic counting in detail, more than 14 moves may be required.

 

According to the wikipedia article “Optimal solutions for Rubik's Cube”, a proof exists (for which no citation is given) that there exists states needing at least 18 moves to solve, and Tomas Rokicki anounced in 8/2008 that he has proven that no more than 22 moves are required.

 

Because sequences of moves are reversable, if 18 to 22 moves are needed to solve some cube, then 18 to 22 moves are needed reach that state starting with a pristine (solved) cube.

[Boof-head]

I think it’s impossible that a 3x3x3 cube can be manipulated into any of its 43,252,003,274,489,856,000 possible states within 11 face or slice moves

Hang on, I meant the pocket cube, the chart is here:

Pocket Cube - Wikipedia, the free encyclopedia

[Erasmus00]

To comment on some of Boof Head's conjectures

 

The Rubik's cube cannot have a full SU(N) symmetry for any N. If we look at the set of ALL unitary matrices of degree N, we need N^2-1 CONTINUOUS parameters. There is not even one continuous parameter involved in the cube (the symmetry group of the cube is continuous!). Further, unitary matrices imply complex parameters, and the positions of cube pieces should probably be real in any obvious mapping.

 

The obvious symmetries of the cube are permutation groups and discrete symmetries.

 

I also don't understand the particle physics references being discussed, and I do particle physics for a living. Maybe you could provide one of the diagrams you are talking about, so I can see what you are talking about and maybe add some clarification?

[Boof-head]

Putting all the pieces together we get that the cube group is isomorphic to

 

[math] (\mathbb Z_3^7 \times \mathbb Z_2^{11}) \rtimes \,((A_8 \times A_{12}) \rtimes \mathbb Z_2). [/math]

...

The symmetry group of the Rubik's cube obtained by dismembering it and reassembling is slightly larger: namely it is the direct product

 

[math] \mathbb Z_4^6 \times \mathbb Z_3 \wr \mathrm S_8 \times \mathbb Z_2\wr \mathrm S_{12}.[/math]

--Rubik's Cube group - Wikipedia, the free encyclopedia

To put it simply, we conclude that spacetime is like a quantum relativistic Rubik's cubic lattice with an indefinite Hilbert metric. This simile has not yet been used up. S Golomb (1982, see also Rubik, 1987) semifactorized [math] S_3 [/math] in the theory of the Rubik cubic lattice much as we semifactorize [math] S_4 [/math] in our null hypercubical lattice. Both [math]\mathbb N^3 [/math] and [math]\mathbb N^4 [/math] have the suggestive symmetries 2 (twain) and 3 (trine); no other powers of N do.

Gravity comes from lattice translations and torsion from spin. Color is trine, and the z component of weak isospin is twain.

 

--Quantum Relativity, David Finkelstein & Michael Gibbs, International Journal of Theoretical Physics Vol 32, No. 10, 1993

[Erasmus00]

All of the groups above are discrete. Z_n are integer groups under addition modulo n (so n discrete elements) and S_n is the permutation group on n objects.

 

The symmetries Finkelstein talks about are symmetries you get when you discretize space-time and put it on a lattice. Instead of full special relativity symmetry, you get a discrete subset, like the cube. This might be useful in simulations and models, but is unrelated to the SU(N) symmetries of the standard model.

[Boof-head]

This might be useful in simulations and models, but is unrelated to the SU(N) symmetries of the standard model.

That must be why the Rubik's cube isn't used to explain QM, or why the lattice doesn't model much and no-one has bothered with it then?

 

See, again there seem to be some misconceptions about what I've had to say so far.

1) I have not said SU(n) symmetry is part of the cube's quotient space or algebra. I have said "there is a connection".

 

2) The Georgi-Glashow SU(5) symmetry is cubic - the quark colors are planar, as 3x2-degree vertex reps along the 'equators' where the poles are colorless; color is a different algebra with an su(3) structure; su(2) and Pauli operators are a subgroup. This symmetry group has a different quotient and algebras, than the 3-d puzzle. SU(5) is not a physical cube, it's "imaginary".

 

3) Rubik's 3-slice lattice, is a quotient space - each face has 9 partitions; the Euclidean symmetry is expressed as the permutation group S_48.

[Boof-head]

Quaternions are a subgroup of these puzzles; I think this does mean it has a vector space - finite dimensional.

And a Coxeter group.

And the icosahedron (the 'rubicon') is Reimannian.

 

All you have to do is say 'the facelets are, or have complex rather than real colors', and you're quids in.

 

So why are fixed areas of color not complex, or "in C"?

[Boof-head]

And, I believe the slice-groups, Coxeter and Quaternion groups, and a few others, mean the lattices are all modular lattices (possibly orthocomplemented); because of the modular arithmetic you do, when you permute rotations.

 

It also has one of the largest groups, a Mathieu group called [math] M_{12} [/math], and the subgroup [math] M_{11} [/math]; these are used in coding theory and compression algorithms, Cayley graphs, Huffman trees, all that stuff.

 

P.S. To prove that a unitary group - U(24) for the 3-cube - exists, I would say that it's a question of proving that conjugate transposition exists as a rotation group, then that an algebraic group exists. Complexity, and what a "complex number" are, is a part of these proofs.

 

Going in reverse, we know complex numbers are binary, one real and one imaginary part; so in what sense is a square of fixed size and color, a complex (number)?

 

Say the area and its position are the real part, so we need to label each area with a "real position"; this is trivial for the fixed facelets in the center of each face.

Then if the color is 'i'. the coefficient has to be different from the position 'value'. Or each facelet has a real area, and an imaginary color coefficient times a 'position'.

 

There are three things to deal with; the positions are fixed at the start position, the areas are fixed and so are the colors of each facelet; rotations change colors at fixed positions. With 3 imaginary colors i,j,k all unitary roots (of 1) as projected from 3 sides in the projection space, the 3 fixed colors are indices of position and color-matching.

 

So the real part might be "the facelet is/ is not in the correct position", or has a metric (+1,-1). There's the quaternion group. Give each 2x2 square matrix (there are 4 on each face) an index, as for any nxn; the fixed center is the matrix pivot at (00), or <00>; Define edges as "<", and ">", for "inner edge" and "outer edge" respectively.

 

Then define "|" as "an area on a surface", and you have a notation like:

 

<10| := 'upper' edge piece, at the "y" position relative to <00> (or |00|);

<01| := 'right' edge piece, at the "x" position ";

|10> := transpose of <10|;

|01> := " ... <10|

<11|, |11> := the corner square and its transpose.

 

That is, the color matrix: [math] c_i\begin{pmatrix}

10 & 11 \\

00 & 01

\end{pmatrix}\;

 

[/math]; where [math] c_i [/math] is a 'complex color', [math] c \in {a,b,c,d,e,f}; i \in \{1,2,...,6\}.[/math]

(that's better}

I believe the other two requirements have been met already.

[Erasmus00]

]P.S. To prove that a unitary group - U(24) for the 3-cube - exists, I would say that it's a question of proving that conjugate transposition exists as a rotation group, then that an algebraic group exists. Complexity, and what a "complex number" are, is a part of these proofs.

 

It is impossible to have a finite dimensional symmetrygroup (the cube) have an uncountably infinite sub-group (any Lie group, including U(24) or SU(24) or U(1) for that matter).

[Boof-head]

It is impossible to have a finite dimensional symmetrygroup (the cube) have an uncountably infinite sub-group (any Lie group, including U(24) or SU(24) or U(1) for that matter).

Except that U(1) is the circle group; are you saying there are no circles in a Rubik's cube?

Can you tell me what this means?:

The three-dimensional Euclidean space [math] R^3 [/math] with the Lie bracket given by the cross product of vectors becomes a three-dimensional Lie algebra.

A Lie algebra can have a sub-algebra, like a group can have a subgroup. Have you heard of Cayley groups? Have a look at G(2).

 

Just thought I would add that U(1), which has a Lie group in it, the same group that EM theory uses, is countably infinite numbers of rotations around a single axis. This is a subgroup of the cube since you can rotate any slice a countably infinite number of times, which isn't all that interesting. But if this wasn't possible the Rubik's wouldn't be interesting either.

[Erasmus00]

Except that U(1) is the circle group; are you saying there are no circles in a Rubik's cube?

 

Yes, I am. A circle has an uncountable number of angles, right? You can turn a circle 2 radians, or 2.1 radians, or 3.14 radians, etc. The number of allowable rotations is NOT countable. Does that make sense?

 

For a Rubik's cube, the configuration space is very large, but finite. Because it is finite, it is countable. You can't have an uncountable group be a subgroup of a finite group.

 

If I have two oranges, I can't divide this into 1 orange and an uncountable set of oranges.

 

Do you see my objection now?

[Boof-head]

A Rubik's cube has 6 faces that can rotate to any angle; there are an infinite number of partial rotations (in the slice group) for any single one of 6 faces.

 

U(1) is a subgroup of the puzzle as a structure-preserving (diffeomorphism) algebra.