Symmetry of the cube

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

  1. arfa brane call me arf Valued Senior Member

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    7,832
    So in a sense, it's "just a way" to think about what you're really doing when you play around with a Rubik's cube or similar kind of puzzle.

    Or rather, what your brain is doing, which you simplify--a feedback thing--with an external kind of language you haul out of some level of awareness, of what you're looking at, and how it changes when you carry out abstract graph-moves, but with a physical object, real energy and its consequential dissipation as noise, mostly.

    Where do you take it (the construction), and your awareness?
    Your brain is actually "doing" a lot of mathematical stuff, and the internal language you invent ('I need to move this bit out of the way so I can turn this bit like so . . .') you're mapping complex roots of unity to an external (freely additive) algebra which all those neurons are parsing . . . you "become" in some sense, an FSA in a switching network, connecting incoming requests to outgoing, abstract terminals. These terminals are graphs your brain is "walking" through, in a somewhat larger graph as you focus (it's your brain . . .), on a local region.
     
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  3. arfa brane call me arf Valued Senior Member

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    Hell, nearly forgot. What about an eversion of the 2-sphere? The ultimate topological grail?

    Well, suppose before I embed a permuted version of the square projected-tile (which is a graph of course), I reflect it with s, or one of its rotations in \( Dih_4\)? Does the reflection invert the tile, or is it a tile inversion since recovering the sphere will show an inverted permutation (it will be inside out)?
     
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  5. arfa brane call me arf Valued Senior Member

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    7,832
    So, there is always more. This so far is only at n = 3, so contains n = 2 in the n x n x n cube puzzle family. And with the "edges" back in--the sections I removed from the tiling by colouring them the same--I have a lot more than 3.7 million points-as-graphs.
    But these are covered by the first graph as I pointed out earlier. In principle, n is a free iterant, so the size of larger n permutation spaces must be, almost completely unimaginable. But they (will) all fit because the space is infinitely projected.

    Inversion and the dihedral group over this tile:

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    Does it give me an inversion method? I have vertical, horizontal, and diagonal reflections, along with the rule that two different reflections is a rotation. A rotation of the plane is an isometry, so no inversion method there.
     
    Last edited: Feb 22, 2019
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  7. arfa brane call me arf Valued Senior Member

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    7,832
    And of course, I can always do a sphere inversion first, embed the tile in the inverted orange square (or any square) then invert the sphere again (two inversions or two reflections is the identity), and the square tile is also reflected about one of the diagonals, or vertical or horizontal lines of reflection symmetry (in the dihedral group of the tile). There can be only one, because two is a rotation . . .

    And we're done, but we cheated. Sort of.
     
  8. arfa brane call me arf Valued Senior Member

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    7,832
    Ok so with the recognition that complex numbers take the construction out of the plane (minimal surface with two dimensions) and into -- some other thing with one complex dimension.

    Ok well, the thing about "a" complex number is that the unit circle isn't the only map of its territory. No siree, a complex number is also a point on the 2-torus because you have 'cylindrical' coordinates. The circle that goes around a central axis of symmetry of the 2-torus is the central axis of a cylinder when a torus is cut open--hence the cylinder and the torus have different topologies because cutting a torus open (along a half plane glued to the central axis) is not an ambient isotopy. You have to bend a cylinder and stretch it too, to bring the ends together.

    When I put complex roots as scalars into the vector basis (that I'm using here), I see that since I have a partitioned basis (a bipartite relation, like a graph, or graphlike object), I can use one root on the set under *-closure, and one for each partition which is three roots of unity.

    That is, for \( \omega^{11} = u^5 = v^3 = 1 \), I have \( \omega(ue_i,\,v\bar e_j)^* \). Of course, I have the freedom to choose any roots (of 1), including ones that have very large cycles.

    I know that \( (ue_i,\,v\bar e_j)^* \) is a 15-cycle if I write it as an ordered tuple (or a walk in a directed graph at some latitude).

    So the first *, including the omega "parameter" is a 15 x 11 cycle. I can walk along those 11 points in (the edge set of) \(\mathfrak G \) in 15-cycles (3x5 cycles). Thanks to the vertices in the graph I'm holding (or trying to solve, whatever that means) having 3 orientations, locally.

    The triangles in the two images already posted, of the 3 x 3 x 3 Rubik's cube and its projection, are (supposedly) mapped to this (yup, I think I can see some 3-coloured triangles):

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    In which there is a 5-fold rotational symmetry.
     
    Last edited: Feb 24, 2019
  9. arfa brane call me arf Valued Senior Member

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    7,832
    Mkay? And the ideas you get about how many dimensions there are, of course you know this already, are along the, ah, . . . lines of thinking about how to use them iteratively. That is to say, one dimension at a time--physics (!).

    The puzzle has a rotation over each face--turning circle (oops almost wrote turing there), iterate a singleton, you get a 4-cycle corresponding to the first primitive roots; \( \{\sqrt {-1},\,-\sqrt {-1}\} \) and their squares. This squaring of roots thing is what lets you take the unit circle to the complex plane.

    It also lets you take the union of the projective plane to the complex plane, by identifying an equator on the projected sphere.
    This object (which the real plane and sphere of course, don't know anything about), is where I can put all 3.7 million or so permutations of just the 2 x 2 x 2 cube, because there is only one identity permutation.

    So I first of all put all the points on a unit circle, identify a permutation as the identity and send it to the north pole (project its local quotient in the plane to "almost" infinity). If I then choose a non-identity permutation, I have one of nine points on a line of latitude below the north pole, somewhere between it and the equator.

    To send my first choice to "nowhere" too, I choose another point to move to in a big graph, by transforming the small one I'm holding. I rotate it (completely) and then rotate a second turning circle in my walking graph. Hell, I've done that sitting down.
     
    Last edited: Feb 24, 2019
  10. arfa brane call me arf Valued Senior Member

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    7,832
    And, if you haven't yet, I've recognised that getting the projective plane, or Euclidean plane, or complex plane together is already done with a partition of a vector basis. Simply because a complex number has a representation as a 2 x 2 matrix with negative and positive elements.


    I have \( (e_1)_{ij} + (e_2)_{ij} = \delta_{ij} \) and \( (\bar e_2)_{ij} - (e_1)_{ij} = \epsilon_{ij} \).

    So that the complex number which has a real homomorphism as the real pair \((1,1)\, \rightarrow\, (1,i)\), is represented by 2 x 2 matrices as:

    \( e_1 + e_2 + \bar e_2 - \bar e_1 \).​
     
    Last edited: Feb 24, 2019
  11. arfa brane call me arf Valued Senior Member

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    7,832
    So this means I have to introduce a negative basis element, in fact both of the off-diagonal partial matrices need to be one negative the other positive, and along the way the diagonal elements will, because of the square of i being -1 and that as a 2 x 2 matrix is \( -I_2 \).

    I might have to explain this image of a projection too, so I will:

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    The question I ask is, what happened to the parallel lines at infinity, or the perpendicular ones that go there?

    Well. Projective space is where parallel lines meet at infinity, and perpendicular lines are 180° apart.
    What the projection above needs is a dilation: lift the sphere up to some height, since it has none. The thick black lines get thicker and longer as this continuous transform is made.

    In the above, you have to remember it's a perspective, and you need enough of the whole thing so information is conserved and the projection recovers the sphere. What the lines diverging at the corners look like as they get to infinity is a pair of lines 180° apart (so they're parallel), meeting the vertices of a square at 0°, the square has infinite sides. The parallel lines meet at the vertices too (they are the same lines after all).
     
    Last edited: Feb 24, 2019
  12. arfa brane call me arf Valued Senior Member

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    7,832
    So now my 1st year calculus tells me the point at infinity is at ∞, and at -∞, because if I project two lines in opposite directions (180° apart), they are perpendicular, or meet at a corner of an infinite square region of the plane "out there" beyond a local patch. In every direction.

    Direction is a conserved "quantity" locally, and this holds for a puzzle which represents a sliced 3-ball, with colours.

    One parting shot: a question you can ask is, where is the sphere? The answer is, it's over the plane, it doesn't matter what "where" is supposed to mean. It's localised and it can project at least one 2-pointed object (a kind of graph, meh) over the plane anywhere, so in a sense it's also everywhere one sphere at a time.

    Taking the graph from the sphere in the very first image I posted, to the Euclidean plane is a matter of drawing a cubic graph with eight vertices of equal degree (3), no intersections and one face the global part of the plane not a local face; the dual graph has six vertices and is the octahedral dual, with one vertex not 'localised', but somewhere in the plane.

    So abstractly an infinite plane --minus a small patch -- becomes a vertex in a graph
     
    Last edited: Feb 24, 2019
  13. arfa brane call me arf Valued Senior Member

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    7,832
    One last little firework.

    The projective plane has one sphere in it. What are the consequences of having a pair of them?
    One projects a tiling, as in the first image and subsequent images I've posted (thanks Magic Tile, which I can't get to run on Ubuntu),
    . . . in particular the one I can shrink and put back on the same sphere which projects it, somehow . . . some kind of cutting and gluing operation I can carry out by taking two spheres, one above the other to a common axis.

    Maybe. But why can't I if I only need one Bloch sphere for the complex latitudes for each subgroup of each "cube" group in the n x n x n family? The sphere rotates in the projective plane, but is glued to its 1-sphere boundary, the equator, in the complex plane. In other words I foliate the Bloch sphere as a pair of projective planes, one over the other, Or Something.
     
  14. arfa brane call me arf Valued Senior Member

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    Iterate the logic.

    . . . or Something I think, has to look like this:

    Start with the first tiling of the projective plane, next to the, well, second tiling and consider what's different (in terms of, what kinds of boundaries am I looking at, what can I do to either of the tilings, in terms of starting to colour faces or vertices, to make them start to look the same, somewhere?

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    =

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    ??

    I've started on the right hand one

    In other words, I'm looking at a copy of \( K_2\), the complete graph on two vertices, in which each vertex is one of these graphs on the sphere, and the single edge is some graph moves, or colourings. So I need to find a minimal set of moves that take enough structure between them that I can say they have identical structure = a set of abstract moves that colours each of them.

    The problem of embedding the right hand square back into the "blank" orange square on the sphere is solved by rotations and dilations, then a reflection through four edges. Here, you'll need a reflection that isn't through any axis of central symmetry, but four edges, as horizontal and vertical pairs. You will need to show that it transposes the interior and exterior.

    On the other hand if you do reflect everything through a horizontal or vertical line passing through the centre of the square on the right, respectively a diagonal, then you reflect the plane through a line, but only invert the square (a partial 'eversion' that isn't). Two reflections is a rotation of the square or the plane.
     
    Last edited: Feb 24, 2019
  15. arfa brane call me arf Valued Senior Member

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    7,832
    Then aha!, like that lecturer in the Winter school says, you see that the square with the tile in it, and the orange square with the rest of the sphere outside it, map a pair of spheres to each other, and look like the square graph above has been reflected

    Not a simple reflection through one side, but in the context of having been completely exchanged, in terms of inside and outside, whereas a simple reflection copies the interior--there are two images in each case however.

    A planar reflection through any side over a square tiles the plane, but with two spheres you have a reflection equivalent to two rotations, which exchanges the face of a square graph with its only other face. The dual of this is a graph with two vertices, one at the centre of the infinite plane, exchanged with one at the centre of the face of its dual.
     
  16. arfa brane call me arf Valued Senior Member

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    7,832
    So notice, it's about pairs of objects, abstract vectors or just numbers with one dimension. Or pairs of graphs.

    There's a natural duality in graph theory, over those graphs which embed in the plane and on the sphere--it's the duality of vertex and face "geometry", or just connectedness, which graphs can capture when edges and vertices get weights.

    I'm going to show that the basis elements, the \( e^i_j \), which are partitioned by the relation R ("is the row transpose of") into equivalence classes, is equivalent to a relation C ("is the column transpose of"), because it embeds vertices into a set E (somewhere in a graph with at least one pair of edges), like so:

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    When I take the union of the sets in each relation (which give the same partition), I get the relation "is the row and column transpose of", which is a rotation of the elements of a general 2 x 2 matrix. I had to use the Boolean symbol for the complements, but that's science.

    But union is just a set, something else has to happen, before it "outputs" the leftmost graph (relational map).

    What I do is change the notation, instead of set union I use the fact that R and C are either true or false independently. This is apparent because they have empty intersection.

    So I now have a Boolean algebra over the sets R and C.

    So two sets, two truth values (go figure), each set has two elements which are sets we can write as ordered tuples, because the relations are symmetric. That means the order in the "ordered" tuples vanishes.

    The sets are \( R =\{ (e_1, \bar e_1), (e_2, \bar e_2) \}; C = \{ (e_1, \bar e_2), (e_2, \bar e_1)\}\), and I want \(\{ (e_1, e_2), (\bar e_1, \bar e_2)\}\)

    I see that if I define \( (a,b)(c,d) = (ac,bd) \) as \( (a ∨ b)∧(c ∨ d) = (a∧b) ∨ (c∧d)\), this appears to resolve it. I need meet and join over the sets R and C, in a way that gives me another kind of set product. one that I can see a Boolean algebra in.

    That is, I have \( R∧C \), where \( R ∩ C \) is empty. And I have an algebra that translates to a pair of graph moves, in which two edges and two vertices are removed, leaving two edges and four vertices. Effectively a pair of \( K_2 \) subgraphs are rotated from vertical to horizontal, with some topological moves that leaves them in (the pair of edges in) E.

    The two graphs above, on the right should have an equivalence sign between them then. The union of R with C is factored by the equivalence of R "and" C, which is the logical intersection.

    So I really need this:

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    There is a set of three relations, R, C, and R∧C; the "∧" acting on R and C together, partitions the set of relations into two, shown either side of "≈".
    Note that composition of elements without compositions of sets is the set containing a single pair of zeros.
     
    Last edited: Feb 25, 2019
  17. arfa brane call me arf Valued Senior Member

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    7,832
    Just going over that agin, more closely.

    You have a basis of 4 x (2 x 2) 0-1 matrices. You can partition it with a single relation (which you show has a dual but equivalent relation), so the two sets:

    \( \{e_1, e_2\},\, \{\bar e_1, \bar e_2\} \) can be drawn as a pair of graphs. What about distance? If it was Euclidean geometry, the distance between two points on the same line is a difference, like |a - b| for a and b. For \( |e_1 - e_2|\), the difference is the identity 2 x 2 matrix.

    So that's the distance between the two upper points in each graph. On the lower line it's \(|\bar e_1 - \bar e_2|\) a 'complementary' matrix. This is a consequence of the row-transpose relation.

    Then the left hand unit-counit graph (a relation on a set), is what's left when you start 'gluing' vertices together, i.e. by the numbers. The "inner product" vanishes where \( e_1e_2 \) meet, etc, you then just have to pull an edge up to a boundary.
     
    Last edited: Feb 26, 2019
  18. arfa brane call me arf Valued Senior Member

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    7,832
    But as they say, it isn't over 'til the fat lady sings, or the thickened surface inverts itself:

    You see there is a bit of a problem with just gluing points to points with the same label; you do it twice and get two edges 'upstairs'.
    It seems you need a special move on an edge that transforms each vertex on it, so it goes downstairs, a unit loop (or other half of the one upstairs) . . .

    This move involves multiplying each 'vertex weight' with a unique basis element so the weightings will do the job, under gluing. This is the only way to get a pair of half-loops on distinct boundaries. Essentially you have a real loop upstairs and a complex one downstairs. If you start composing the three graphs vertically the algebra on each vertex should conserve the two sets of relations, R and C.

    I would say the reason for the extra graph move (really you just do a row transpose on each vertex), is because the real loop product "goes twice" otherwise. And it says something about the lower two points and their domain. The crossed over relation between the two boundaries is ambient isotopic, you can move them around each other because they are non-intersecting (relations).
     
  19. arfa brane call me arf Valued Senior Member

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  20. arfa brane call me arf Valued Senior Member

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    But then you realise that all the internal multiplying can be avoided, just by using the lower metric (the distance between points), on one of the edges in C.
    So this means you should be able to use either, the other one will satisfy the metric along the upper boundary.

    But the lower metric is a root of the upper one. You also use it to satisfy the two relations, where R is the lower metric acting on vertices on the left and C is satisfied when it acts on the right, row- and column-transposing elements of a semigroup, the basis under multiplication. The "relation" graphs also fit into other kinds of algebras (no shit?).
     
  21. arfa brane call me arf Valued Senior Member

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    7,832
    The relational graphs I've drawn are topological objects. That means I can redraw them. I can consider what transposing the vertex set on the upper or the lower boundary "means", for instance. Does it make the metrics negative? That doesn't seem to make sense because direction isn't defined; it's just a set of relations as graphs.

    So with that assumption, you see that the rightmost graph has four equivalent graphs, and so does the middle one, each preserving the relations defined on them. So you take the union of sets in each case and say the set of four graphs is an equivalence class for each distinct relation: R and C.

    The loop graph R "and" C is invariant too, and the loops don't change under a transpose, whereas the other two have crossed or uncrossed edges--they each have two planar and two not-planar graphs. So the "graph union" of R and C is a planar graph in a set with one element, invariant under transpose on either boundary.

    Because of those equivalences, you can construct the loop graph in more than one way, in fact there are 16 ways.
     
  22. arfa brane call me arf Valued Senior Member

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    7,832
    Well ain't that a bust. Now I have three graphs with a composition rule. The leftmost graph is obtained with a "R-transform" of one edge in the C-graph, before you take the union that glues R and C together; you do this to one edge and leave the other alone--you need to break a relation in other words, which is a symmetric relation.
    So it doesn't matter which edge is transformed, or which is transformed in the RC graph "product" to get R and C back again.

    But you can now see that the RC graph represents a kind of blocked element, the other two are like switched elements. If you want to use them as actual switches though, an orientation is required; one of R has to be off or on, so then C has to be in a complementary state. So since you have a choice of which graph to use, you essentially have four representations of the same switching network, under vertical and horizontal composition.

    So you can construct more planar diagrams, and say, embed two copies of \(S_4\) on the same plane, but have them actually on two perpendicular planes, just by colouring two rectangular \(S_4\) regions-of-switches (i.e. compositions of R and C as graphs). Each region needs 4 input and 4 output lines to (points on) the boundaries of the rectangles.

    In that case, embedding \( S_4 \) twice, using the graphs as switching elements means the inputs are tuples of 4 letters, and the outputs are their permutations. Since these two switching "tiles" or fabrics, can be embedded in a larger fabric in as many planes as you want, clearly you have then (at least) a pair of slices with \(S_4\)-symmetry (conserved recall, in \( \mathfrak G \)) which are in \( \mathbb R^2 × \mathbb R^2\).
    One more rectangle, perpendicular to the first two, gives you a fabric in \( \mathbb R^2 × \mathbb R^2 × \mathbb R^2\).

    Direction is defined as an operation that takes inputs to outputs--the inputs (where the metric is a unity or, where the action isn't) are acted on by the square root of unity on 'lower boundaries'.
     
    Last edited: Feb 26, 2019
  23. arfa brane call me arf Valued Senior Member

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    7,832
    So just checking the graphs-as-switches idea flies: Each of the R and C graphs, however you choose them, has a pair of inputs and a pair of outputs.

    The outputs are on the lower boundary, which generates loops on both boundaries when the lower metric acts (i.e. we have a unit as 2 x 2 matrix, and one of its roots, partitioned by relations, all of which have the strictly positive unit on the upper, and the (so far, positive) root on the lower boundary).

    The loops are needed so an output can be defined: this is a permutation of the inputs \( e_1, e_2 \). So you need to transform a pair of lower elements into a loop which is then embedded in the space with the positive-only metric. That is, you connect the outputs together and block the last switch; but you need to leave the inputs open (or do you?), so here the output is another special operation.

    So you then have the relation on pairs of outputs that says \( (e_1, e_2)\) is a permutation of \( (e_2, e_1)\) using the composition of relations R and C. You just compose, vertically, the graphs R and C and apply transforms where needed, which then takes pairs of input "letters" to pairs of outputs. Start with one bank of switches, then upgrade to 2, then 4, and so on. Eventually you have a permutation algebra, as some switching/graph move 'calculus' on 2n letters.
     

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