# Symmetry of the cube

Discussion in 'Physics & Math' started by arfa brane, Jan 22, 2019.

1. ### arfa branecall me arfValued Senior Member

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A closer look at my set of basis elements:

$e_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, e_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, e_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.

$e_i, i \in \{1,2,3,4\}$ is a standard way to notate basis vectors. But here I can use the fact that the set of matrices has some symmetry: the unique 1 in each matrix has a unique row/column address.

I might, if I choose to, form two subsets such that each 2-element subset is related. An obvious relation is this: if you transpose the rows of $e_1$, you have $e_3$, likewise for $e_2$, it's the row-transpose of $e_4$.

In other words $e_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}e_1$.

But $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = e_2 + e_3$.

I want to change the index set so it looks simpler than this. What to do?
What I can choose to do (because I decided I have that freedom), is use an extra symbol to denote which of the $e_j$ is the row-transpose of $e_i$ so I write the basis set like this:

$e_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \bar {e_1} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, {e_2} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \bar {e_2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.

$\begin{array} {| c | c |} * & e_1 & \bar {e_1} & e_2 & \bar {e_2}\\ \hline e_1 & e_1 &0 & 0 & \bar {e_2}\\ \bar {e_1}& \bar {e_1 }& 0&0 & e_2\\ e_2 &0& \bar {e_1}& e_2&0\\ \bar{e_2}&0&e_1&\bar{e_2}&0\\ \hline \end{array}$

From the multiplication table, we see that $e_ie_j = \begin{cases} e_j, & i=j \\ 0, & i \ne j \end{cases}$.

But $\bar {e_i}\bar{e_j} = \begin{cases} 0, & i = j \\{e_j}, & i \ne j \end{cases}$.

Last edited: Feb 11, 2019

3. ### arfa branecall me arfValued Senior Member

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So what I'm doing is all justified mathematically because I have a result, a solution to a counting problem which is a table indexed by the number of group elements 'living' on each tiled section of the graph.

I know that there's a category of monoids, from which I can choose the free category, so there's a monoidal structure for everything including an abstract vector basis, so I can use this idea of iterating over the operation of concatenation, for any G-sets I identify.

For instance the "*" in the FSA context means an action that is the union of all finite length "strings in the alphabet". It must include the empty string everywhere because the monoid is free to generate it.
Take that to the concept of walking freely through a graph, iterating a string with algorithmic structure.

But then the tile, I'll abbreviate it to the object T, also has this "*" defined on it, so I can always find $T^0$ the identity tiling 'operation" or empty string in the monoid.

Clearly this is just a square lattice with coordinates. The graph (we're then in) is $K_{\infty; 0}$, the infinitely disconnected graph.

Last edited: Feb 11, 2019

5. ### arfa branecall me arfValued Senior Member

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Or, in another sense, T acts on K, identifying the centre, then generates a pair of connections to its boundary in T (having acted, will act), so the whole graph shows up because all the edges do, "along with" the action that maps points to points. The integers go for a ride.

7. ### arfa branecall me arfValued Senior Member

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If you've noticed that relation between elements of the basis above, the relation "is the row transpose of" can be written using a couple of well known symbols, the Kronecker delta and the permutation (Levi-Civita) symbol.

As follows: $\{e_j,\,\bar {e_j} \} |_{j \in \{1,2\}} = \delta_{ij}e_ie_j ∪ \delta_{ij}\bar{e_i}e_j$, and
$\{e_j,\,\bar {e_j}, -e_j, -\bar {e_j}\} |_{j \in \{1,2\}} = \epsilon_{ij}\bar{e_i}\bar{e_j} ∪ \epsilon_{ij}e_i\bar{e_j}$, except for the minus sign.

The permutation symbol almost works, but you have elements with a negative sign.
This little problem can be fixed by using a complementary Kronecker delta: $\bar \delta_{ij} = \begin{cases} 0, & i=j\\ 1, & i \ne j\end{cases}$, which is then equivalent to $\epsilon_{ij}$ modulo sign.

Also well-known is that the Kronecker delta defines the elements of a square matrix; when $i,j \in \{1,2\}$, we have the elements of the matrix $I_2 = e_1 + e_2$.
So then $\bar \delta_{ij} = \bar {e_1} + \bar{e_2}$

Last edited: Feb 11, 2019
8. ### arfa branecall me arfValued Senior Member

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So all the stuff in this thread is an outline of a larger idea, which is that everything in the domain--points or vertices in a graph, subgraphs like the object <2,3>, or <p,q> are also like abstract objects with connections (morphisma) between them; there's a path from any tile to any other tile, a metagraph.

All is in an abstraction called a category; category theory is actually an attempt to simplify mathematical objects (such as graphs, or algebras defined on graphs, or sets of permutation matrices and other G-sets), into a classification schema based on two simple ideas: objects in the domain are points, and arrows between points are morphisms.

The whole graph (not its 2-d projection or shadow) is itself one of these abstract objects, with at least one morphism connecting it to a graph of a larger permutation group (the 2x2x2 puzzle expands to a 3x3x3, etc). It just keeps going; categories are a way to keep all the objects you can define (or "find" in the graph and its algebraic structure) in the context of say, data types, so you don't make mistakes.

A mistake one might make is thinking there's an implied summation (Einstein summation) in the "algebra of indices" in the last post, no there isn't.

The objects $g_i, g_j$ composed together, are piecewise linear elements, in a string: that is we have objects like $g_i g_j g_k g_l . . .$. Besides there would be an overloading if I introduce $\sum_{j=1}^2 e_j$, fairly obviously. Wrong data type--compile-time error!

Graphs are very common in IT, when you write software they're almost a requirement. There is a "theory of graphs", which says I can take this object:

And represent it with another graph, or as many graphs as I need with as many coloured edges or vertices as I need in it, and I can find a map from this object and the set of graphs I construct to say, this one:

simply because both objects are graphs already . . .
Defining a set of rotations or reflections gives me a permutation graph of each object.

In the large graph, I can embed each of the above graphs (with their rotation/reflection algebras) in a point--(0,0). In which sense I shrink a graph with |V| vertices and |E| edges (the faces are then defined because the graphs have cycles), which is k-coloured, to a point in the larger structure. The vertices in the large graph are 3-dimensional objects, the edges therefore have four, and the graph embeds in three more Euclidean dimensions (is a tiling of the 3-plane). How many dimensions is that, 3 + 3, or 4 + 3?

Last edited: Feb 12, 2019
9. ### arfa branecall me arfValued Senior Member

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Alrighty, so WTF is all this "really" about?

One of the things it's about is that the human body has symmetries--bilateral, mostly. In particular we have two hands, like mirror images, and we know "when" we hold our right hand palm outwards up to a real mirror, it looks like a left hand. Ok, well it's actually our brains doing that, so maybe I should call it a brain move, which we do twice for a pair of mirrors, suitably aligned.
Reflections can be composed (use another mirror), so they look like rotations, the real image is like the mathematical image of a function, a rotation.

So our hands, attached to the rest, can hold this puzzle, or set of puzzles and rotate parts of them, then we have say, a pair of mirror images corresponding to a reflected rotation, which is still a rotation. I.e. if rotations form another mathematical class of object than reflections alone, then rotations 'absorb' reflections. What we need to show is first of all a group symmetry.

But first, the topological manifold. Hold one of these

...and check out what you can do with it, say, holding it in the open palm of one hand.
One thing you can do is rotate your palm twice while keeping the palm up, topologically wrapping a path for a cube with a locally fixed orientation around the circle twice, this is just like the so-called belt trick or twisted ribbon. Along the path from a fixed point of view, the cube rotates with one face hidden. This can be described by a fixed length string of changes in the vertex/edge/face set, based on one face pointing up. The string has to "start" with a choice of a fixed orientation, but proceeds over time afterwards in steps, then ends.

You can throw it like a cricket ball, it will rotate around three axes in normal 3-space. This is being freely rotated by a monoidal object, the set of all strings as compositions of even more abstract "changes of coordinates", from a global perspective. So, mathematics.

For a 2x2x2 sliced up 3-ball, fixing half and rotating the other is invariant if you switch it around, also if you rotate each half simultaneously--the parallel version.
And so on.

But "when" does the graph exist? It isn't meaningful, not really a question, a graph $\mathfrak G$ exists, it will exist and has. The existence of time is a mathematical convenience.

However, the graph on $\mathfrak R$ the Riemann sphere is a foliation of the 3-sphere, along three orthogonal spacelike surfaces intersecting the centre, so then $\mathfrak G$ is the partition along timelike surfaces, and each edge has 1 + 1 dimensions, being one of 3 + 1, when time is embedded in it.

Last edited: Feb 14, 2019
10. ### arfa branecall me arfValued Senior Member

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In this video, think about your perspective of the direction the handle points, and divide it by four points around the orbit it makes with the cup's centre.

The cup is stuck to his hand. It follows a path in which his shoulder, elbow and wrist rotate. There are points you can mark in an abstract graph.

11. ### arfa branecall me arfValued Senior Member

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So now we have a model in which a physical "graph", the puzzle, or the cup rotating, is like a particle of sorts.

With the 2x2x2 permutation cube, you have a model of a controlled descent into a much larger space, or of a free descent like a particle "bouncing around" a container, a kind of box. So two cubes permuted freely is a pair of particles, and n cubes is a model of n particles.

So we can ask, how many particles does it take to fill up the container, or how long does it take a free particle to visit each node and bounce off or reflect from, each boundary? Here, freely permuted is like free "fall" within a bounded graph.

There are 11 green coloured vertices in a coloured graph (a projection onto $\mathbb E^2$), $\mathfrak G$, which is an incomplete graph $K_{54;|E|}$, if you remove the point at (0,0), there are 53 vertices . So there are 42 red ones. 53 and 11 are both prime, hence coprime; 42 factors as 2x3x7.

And it's well-known, even by non-mathematicians among which I count myself, that as the n in n x n x n increases, the puzzles get harder to solve. This is the domain of cryptography--code breaking. In which context you crack a code by solving a permuted state, or "bringing it closer" to the chosen identity permutation of an abstract string of letters, the letters are vertices in a graph.

Also not quite as well known is that the positive octant of complex projective space $\mathbb C \mathbb P^n$ is the space of pure states of spin n/2 (the projections onto the surface), that is, the Riemann sphere mapped to a positive octant of the 3-sphere via a bundle of flat tori.

But the graph $\mathfrak G$ says the physical puzzle is what's left when you factor out the complex plane, or when time is embedded (you go to the real plane and the real 3-ball).

Last edited: Feb 14, 2019
12. ### TheFroggerValued Senior Member

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What's the deal? A cube contains six-wrecked-angles.

13. ### arfa branecall me arfValued Senior Member

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What's the deal? I have no idea really.
But I know there's a connection to cryptography (I have links to papers that also say this), and to something called a symplectic vector space.

The standard basis for such a mathematical object or space is explained here
So obviously in $\mathbb R^{2n}$, when n = 1, we have $\omega = \begin{bmatrix} 0 & 1\\-1 & 0 \end{bmatrix} = \epsilon_{ij}|_{i,j \in \{1,2\}}$, the Levi-Civita symbol with 2 indices. More generally we have $\omega = \epsilon_{ij} \otimes I_{n}$ in $\mathbb R^{2n}$.

For reasons unspecified, all the negatives vanish--a permutation metric is positive everywhere, Euclidean distances can't be negative. But perhaps the time dimension can be mapped to negative integers, maybe I can think about reflecting $\mathfrak G$ through a line or a point. The question is then, why do that?

Last edited: Feb 14, 2019
14. ### arfa branecall me arfValued Senior Member

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What about the fact that 1 and -1 are congruent modulo 2? It means the point (1,1) in the graph has a point at (-1,-1) congruent to it.

But every point along the subgraph y = x in the graph, or compositions of walks along it, is congruent modulo (p,p), to (-1,-1). The point of congruence is what happens when I reflect points, modulo the algebra, through the origin.

Points lying along vertical columns are distributed when we go to modulo (p,q), but again (1,2) is congruent to (-1,-1), mod(2,3). (1,1) reflects as (-1,-2) mod(2,3)

Last edited: Feb 15, 2019 at 3:29 AM
15. ### arfa branecall me arfValued Senior Member

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On the foliation of the 3-sphere. The 3-sphere as the interior of a 2-sphere, is sliced by cutting planes into eight distinct topological spheres.

Looking at the physics, the physical model says doing this means permutations involve a pair of surfaces interacting--there is some noise, a bit of friction, some shit happens between particles on each surface, then it's over and the noise vanishes, your input, some amount of energy, also vanishes (because you, erm, chose that to be true). Now you've got a derivative of sorts, but it's discrete because you can do it really slowly, so the "noise derivative" diminishes. That's really inefficient in terms of processing power though.

But you can see that rotating half a 2 x 2 x 2 puzzle in effect changes a connected graph into two graphs which are disconnected, but still a graph, then you reconnect some edges and it's the same again except four edges have been reconnected after a cyclic permutation in the two disconnected components (the timelike graph).

Doing this freely--the random graph walk context--is like turning the 3-ball inside out or dissecting it by permuting the orientation and position of each octant; the graph, being a solution the the problem of counting the number of permutations, has a few group theorems hovering around it, one essential is the Orbit-Stabiliser Theorem.

16. ### arfa branecall me arfValued Senior Member

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I realised some time ago that I needed to define some relations on permutation matrices.

First of all, a permutation matrix is a 0-1 matrix, such that the sum of element in each row is 1, and the sum of elements in each column is 1.
So they're a subset of symmetric matrices with 0s and 1s in them, i.e. of the set $M_n(\mathbb Z_2)$.

A matrix $A_{ij}$ is a permutation matrix if the sum over row i equals the sum over row i+1, up to i = n, and if the sum over column j equals the sum over column j+1, up to j = n.

So all the $e_j$, where j is a column index, have sums over all rows and columns equal to 1, but a permutation matrix sums over all rows and columns to n, since sums over individual rows/columns equals 1.
So if I bring back the row index for the set $\{e_j, \,\bar {e_j}\}$, I can use this summing over rows/columns to filter out the matrices in $M_n(\mathbb Z_2)$, which don't have the right sums.

If I restrict addition to modulo 2 addition, then addition is closed over the $\{e_j, \,\bar {e_j}\}$ when the "is a permutation matrix" relation is included. Then the basis set is a vector space under this restricted addition, with multiplicative scalars $\{0,1\}$.

17. ### arfa branecall me arfValued Senior Member

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Well, my graph theory of the cube says something about an eversion of the 3-ball in piecewise linear fashion, I think.

It's obviously discrete, but a continuous eversion of say, the 2-sphere has a halfway point, in which half the exterior and half the interior 2-d surface intersect in $\mathbb R^3$. But what happens to the 3-ball, the solid interior region isomorphic to $S^3$ in an eversion of the 2-sphere?

Can the eversion have a discrete tiling embedded, a dihedral group, say, such that octants or other sections can be permuted under reflections/rotations?

Last edited: Feb 16, 2019 at 7:28 AM
18. ### arfa branecall me arfValued Senior Member

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And on permutation matrices: a permutation matrix is incomplete in the sense not all elements are nonzero.

If you have a square 2 x 2 matrix of four different elements, a,b,c,d, and if none are zero, the matrix is complete in the same sense a graph is complete.

Thus a permutation matrix isn't complete because it has "disconnected" elements, however the basis elements $\{ e^j_k \}$ of a complete 2 x 2 matrix aren't permutations, but partial permutations (since you can add pairs together and have a permutation matrix), so are partial incomplete 0-1 matrices.

19. ### arfa branecall me arfValued Senior Member

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Pictures can say a lot. You might have come across the hypercube permutation puzzle--the sliced 4-ball.
This is a 'natural' extension of the 3-ball puzzles, but doesn't have a physical model.

The reason it has to live on a computer speaks to something about storing information (preserving it) in n dimensions.

The reason people build models of Rubik's cubes on computers (build virtual machines), is efficiency. Perhaps the exploration of higher dimensions this way wasn't really a flier until efficient electronic computers arrived.

Anyway, from the community pursuing this approach (whence the table of results and my derived graphs), a few more piccys, which should speak for themselves.

20. ### arfa branecall me arfValued Senior Member

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One idea that I believe is useful, which I've seen elsewhere is a kind of resurrection of Maxwell's demon.
In which a small demon at a centre (let's make it the point (0,0), in my case), can see a surface changing around them--our map or graph on $S^2$ which we're permuting, in a controlled or free way.

If the surface is being everted as well, what does the demon see? We can give our demon as many resources as needed, say a large but finite amount of memory, the ability to store and compare two discrete states. We can have a discrete version and a continuous version.

21. ### arfa branecall me arfValued Senior Member

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One more thing about topological demons. My abstract subgraph-as-tile, can tile the graph, under translations (restricted by the modular algebra of addition, I multiply the number of points in each line of constant x, along y).

I can also do this if my tile is a cylinder, rolled up in one of two ways (by gluing together one pair of opposite sides of the rectangle); because I can draw my subgraph on a cylinder both ways, I do, then tile the graph both ways by rolling the cylinder(s) and rotating them in a third dimension to translate them.

So where is the graph-tiling demon, inside a cylinder?

22. ### arfa branecall me arfValued Senior Member

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--https://golem.ph.utexas.edu/category/2011/03

Last edited: Feb 16, 2019 at 8:26 PM
23. ### arfa branecall me arfValued Senior Member

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What does colouring the vertices of a graph (resp, colouring the faces of its dual, if it has one), mean we can do?
It means we can de-colour them, and investigate what the consequences are.

So with this idea, a colouring algorithm which is very simple, we can quotient out faces of a planar graph (a projection), and see what happens to the embedding on the sphere, torus, Klein bottle, or even more exotic kinds of surfaces.

Just by de-colouring each copy of a graph which is initially on the sphere, we recover the "big" graph I've been posting (a little graph, in other words, a kind of vehicle we can walk through the big one).

This algebraic looking object:

remove the colours from the sections between pairs of parallel slices, you have a 2x2x2 graph.

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