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
    Sure, you can do whatever you like, but the underlying graph says nothing about geometry, it has nothing to say about physics either.
    You seem to be trying to impose some extra structure on a graph, that's fine, no problem.
     
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  3. arfa brane call me arf Valued Senior Member

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    Back to my little problem with an embedding; maybe I should realise that the objection to embedding \( \mathbb Z_6 \). or any of its isomorphisms in \( \mathbb Z^2 \) is irrelevant if I embed the graph of that group instead! Which is what I did anyway. 'sigh'

    I need a rectangular section of \( \mathbb Z^2 \) because the graph is of a finite group. I can decide to include only those edges which I think are relevant to the solution I'm looking for. Which means I can also ignore any irrelevant objections, which is what I should have done.

    The symmetry group of the 2x2x2 cubic puzzle is isomorphic to the group of reflections of (2,2,2) triangles (i.e. the surfaces of octants) on the sphere. I have a table which is a result of a computer solution to counting the cosets in each equivalence class, equivalence here means "the same distance" from the origin.

    Then the elements in the first set of two vertical points are all distance 1 from the origin; in fact the 6 elements at the coordinate (1,1) are the generators of the cube group.
     
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  5. TheFrogger Banned Valued Senior Member

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    Why don't you look at the Longview instead? You must be careful not to step on the axiomatic representation of the computational calculations spacebar: a computer is only capable of certain calculations. For example, the infinitely small:

    [1]a=0
    [2]b=1
    [3]b=b÷2
    [4]a=a+b
    [5]goto [3]

    I can GUARANTEE you the computer cannot see the infinitely small. Put simply: you have bent over backwards to attempt to prove something that does not need proving. I can tell you that from here. If you want to fight about it, let's go! You are a liar and I'm calling you longface.

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    If you want a fight I will kiss your gardenhose penis 'till it dribbles in my mouth
     
    Last edited: Jan 30, 2019
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  7. arfa brane call me arf Valued Senior Member

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    So it's kind of mind-numbing (at least to me), that my OP contains the word "graph" a few times, then I mention an embedding of \( \mathbb Z_2 \oplus \mathbb Z_2 \) in \( \mathbb Z^2 \), and I should have been more specific about the embedding being of a graph in \( \mathbb Z^2 \), how the fuck did I miss that one?
    Although, I mean, if you're in a graph theory lecture, this wouldn't be an uncommon occurrence, referring to a group when what's meant is the graph of that group.

    But I can only embed points in \( \mathbb Z^2 \), the edges don't embed. Oh well, maybe I need the whole thing to be in \( \mathbb R^2 \), and what do you know, it already is!

    Someone then commenting, in a lecture theatre that the group doesn't embed in \( \mathbb Z^2 \), is correct, but the lecturer might say "Yes, but this is graph theory, not group theory". Unless of course, they're someone who just wants to be right all the time.
     
  8. TheFrogger Banned Valued Senior Member

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    • Flaming/insulting other members is unacceptable. Please do not post pseudoscience to our science subforums.
    What about this for a graph Mr Longalreadyis:

    -2--
    xx/
    x/
    /
    -1-

    Z

    Can you tell me how it is possible to traverse point 1 to point 2? Wipe the answer to me while I stare at your open bumhole.
     
  9. TheFrogger Banned Valued Senior Member

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    I'd rather be correct than wrong and PERSEVERE IN MY WRONGNESS!!!
     
  10. James R Just this guy, you know? Staff Member

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    Moderator note: The Frogger has been warned for (a) using inappropriate language in reference to another member, amounting to insulting and/or flaming; (b) posting nonsense pseudo-mathematics to the Physics & Math subforum.

    Due to accumulated warning points, The Frogger will be taking some time out from the forum.
     
  11. arfa brane call me arf Valued Senior Member

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    7,832
    Reviewing some of the things I've said:

    from the OP:
    The tensor product is very much like what happens in the Temperley Lieb diagram monoids, the graph of the cube group (embedded) in ℤ×ℤ is just a composition of (some finite number of) ℤ2⊕ℤ3.

    And the last sentence should end with the word "graphs". But shortly thereafter, someguy1 steps in with objections about my use of the term "tensor", and embedding a group in--a group.

    But maybe I wasn't being clear enough, I construct a graph of ℤ×ℤ (embedding it in \( \mathbb R × \mathbb R \)), which is fine, ℤ×ℤ is a topological space and my "coordinate system" for each copy of the smaller graph. This I present as:

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    And now I'm going to claim that this graph is a (graph of a) presentation of \( \mathbb Z_2 \oplus \mathbb Z_3 \).
    Algebraically, the two colored edges represent elements of the cyclic group of order 6; call the blue one g, the red one h, then we have \( \langle \{g,h\}\, |\, g^6 = h ^6 = gh = e; g^5 = h; h^5 = g \rangle\; g,h \in \mathbb Z_2 \oplus \mathbb Z_3 \).

    I could now add more colored edges, using the same pair of colors (everything is pair-shaped!), and complete the graph, but locally; i.e. map the local symmetries of the group to itself. But what I want to do is treat this graph as a subgraph of a larger, globally embedded graph, in which the smaller graph tiles the larger one. I need a restricted tiling 'function', and this is given by an extended version of the modular arithmetic present in a single tile.

    The tiles overlap when their number is > 1. The overlap (a kind of graph intersection) "pushes" the global graph into a third dimension, so we actually have (ℤ×ℤ)×ℤ.
     
    Last edited: Jan 31, 2019
  12. arfa brane call me arf Valued Senior Member

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    7,832
    The concept of walking through a graph is quite useful; as mentioned there are different types, or a classification of the ways a 'moving point' can 'visit' the vertices of a graph; in a graph with cycles a graph traversal has loops in it, and so on.
    Given that graph theory supposedly kicked off with the bridges of Konigsberg problem (you can walk across bridges, duh), I guess the idea has persisted.

    The idea in a sense gives a graph a dynamic context, but on inspection there are only relations (weightings) between vertices 'living' on the edges. In one context the vertices are bound to the edges connecting them, so that a free vertex is not related except by being a graph vertex. The free addition of vertices, without also adding edges, gives a graph with disconnected components--all the freely added vertices with no connections. Moreover, these cannot be a part of any traversal.

    So the free points in ℤ×ℤ in the flat embedding, above, are "just coordinates", their weights are the embedding of all the free vertices.
     
  13. arfa brane call me arf Valued Senior Member

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    7,832
    Here is a cool thing I can do with my first tile:

    Since it's a locally symmetric object, I can cut out a rectangle in ℝ×ℝ that bounds this tile, then wrap it around a cylinder.
    I can justify this mathematically because the integers are already disconnected from ℝ except at discrete weightings, and my rectangle contains all the points in the order 6 group (and graph). What happens globally? Abstractly a tile is missing, but the square lattice of integers is freely generated, so all I've done is remove the initial set of edges.
     
  14. arfa brane call me arf Valued Senior Member

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    7,832
    So now I have a rectangle which is a subspace of a larger topological space; roughly the integers and the graph's vertices, don't "know anything about" real numbers, so I can identify a pair of edges of the rectangle and glue them together without them noticing (type of thing). But there are two pairs of edges opposite each other, so two ways to roll the rectangle up into a cylinder.

    If I identify the top and bottom edges of the rectangle (here an edge is an object not in the graph), then I get two columns of three points, each column is glued to one of the two cylinder boundaries (after contracting any trailing lines). So now I conjure up a function, a 1-point shift acting on one column, but fixing the other.

    But this 1-point shift is equivalent to twisting the cylinder, actually what you end up with is the union of two distinct shifts (there are two generators in the presentation), hence the union of two distinct twists--a bit hard to visualise.
    However, the shift function is equivalent algebraically to adding the object (0,1) to the object (1,0), like vector addition, with (0,1) + (1,1) for the 2-point shift.

    So you introduce another generator to the presentation: call it f, then you have \( \langle\{f,g,h\}\, |\, fg = h;\, f^2h = g;\, f^3 = e \rangle \)
     
    Last edited: Jan 31, 2019
  15. arfa brane call me arf Valued Senior Member

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    7,832
    I'll complete the picture of the presentation, by presenting some completed graphs. The completions of g and h are a pair of 6-cycles that visit every vertex so <g> and <h> are also generators of the local group; the completion of f is a pair of distinct 3-cycles each visiting 3 vertices (one of two columns of points).

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    The first blue colored graph is of the element g, the second is of the element h; you can see the graphs are equivalent modulo the direction of the graph traversal. The green-colored graph is made a bit bigger so the labels are visible, another little problem with the plotting software is the semicircular arcs not having a direction on them, but this is implied by the other two arrows.

    Recall that the edges represent the difference between the points (vertices, I should really say). In a cyclic graph you can transform all the edges into vertices, and all the vertices into edges and it's an identity transformation . . .
     
    Last edited: Feb 1, 2019
  16. arfa brane call me arf Valued Senior Member

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    7,832
    Then the graph or, a tile, which I notate as \( T_Δ \) is what one author would probably call a subgraph of the identity graph of \( \mathbb Z_2 \oplus \mathbb Z_3 \cong \mathbb Z_6 \).

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    So that it's hard not to see that h is like a vector sum of f and g.
     
  17. arfa brane call me arf Valued Senior Member

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    7,832
    Ok, everything in the last few posts is about the metric space 'induced by' the partition function of the 3,674,160 permutations of a stack of \( 2^3 \)colored cubes, under rotations. Rotations mean the space is restricted--you can take the puzzle (or a loosely stacked 'cube' of 8 unit cubes) apart and reassemble it however you like, but the permutation you get by doing that may not be in the set of 3,674,160 available under rotation.

    \( T_Δ \) is a representation of this rotation group; although there are three edges, in fact it's a reduced graph of \( \mathbb Z_4 \) in which f and g are identical. That is, \( f = g; f^2 = g^2 = h \).

    What's happened is the permutation metric 'acts on' the graph of \( \mathbb Z_4 \), or the reduced "identity graph", overlaying g and its inverse on the same edge (when in fact they're distinct). Also in \( \mathbb Z_4 \), \( gh = g^{-1}; h = h^{-1} \), and h represents the distinction between g and its inverse.

    The metric induces the equivalence \( g \sim g^{-1} \), so the graph \( T_Δ \) has a pair of points at (1,1), separated by the element h.
    This can be visualised (the symmetry of \( \mathbb Z_4 \) can be recovered from the subgraph of \( \mathbb Z_2 \oplus \mathbb Z_3 \)), using some more graphs.

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    Both graphs are of \( \mathbb Z_4 \). The right hand graph is a reduced graph, two edges have been removed; it corresponds to an identity graph because every non-identity vertex element is connected to the identity. To get the reduced graph \( T_Δ \), rotate the triangular part so the two pink edges overlap and the yellow edge of the triangle looks like it covers a single point.

    This is exactly what 'lives' on each permutation of four octants (via rotations or compositions of reflections), of the cube puzzle/tiled sphere.
     
    Last edited: Feb 2, 2019
  18. arfa brane call me arf Valued Senior Member

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    7,832
    I've been reading back over this thread and trying to understand what someguy1's big problem was.

    I see I tried to explain the difference between n disconnected points and a set with a powerset.
    Let's say you construct a graph with n = 6, so formally you have the graph \( K_{6;0} \) where the second number is the number of egdes.

    What have you got? What you've got is six unlabelled points, but you know how many you have because (without really thinking about it) you counted them!

    This is not an irrelevant detail--by counting them as you draw them, although you might then say they all look exactly the same (and you can try transposing a pair of them to see if that changes anything, of course, it doesn't), the counting effectively does label them. You should label them with 1, 2, 3, 4, 5, 6 (because counting them also does this! surprisingly, or maybe not). So now you know the set has a powerset.

    But that's cheating. You really have to consider "being" one of the points. In that case you can't see a thing--six, or n disconnected points have a global perspective (your view of the graph), locally there are no connections so no way to walk through the graph.

    So counting the points "from a distance", is equivalent to embedding them all on the same line (or topological 1-manifold), even if you don't draw any lines or connections (i.e. edges between points). This is one of the subtle things about graphs which is difficult for some people--graphs have a structure but to discern it you need connections. So a disconnected "set" of points can't even be counted, except by someone (unconsciously, perhaps) connecting them all together by counting them, which amounts to weighting each point.

    As to his skepticism about what graph theorists are doing with mathematics, maybe he should take it up with them. Me? I'm just a black hat.
     
  19. arfa brane call me arf Valued Senior Member

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    • Please do not flame other members.
    Ah yes, the precious "subject". I let this one go, but now I'm going to rip into it. Especially since someguy1 also appeared to be skeptical of coloring a graph, something theorists are quite fond of doing. The coloring also means the six points are "aligned", or glued to a boundary (yes! it really does).

    My post was also directed at trying to remove some of the apparent confusion--the "I don't understand" thing from his corner.

    So it is an argument; it isn't a change of subject; as to a conversation, I'm not sure we ever really had one.

    I don't know what to say about all the objections to embedding graphs in graphs, which objections seem to have been prompted by his reading of my initial posts (the first page in this thread), as being about groups--graphs? what the hell are graphs?

    I hate skepticism, I hate being condescended to, and I hate how people look for a way to justify it. So fuck you asshole.
     
  20. arfa brane call me arf Valued Senior Member

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    What the hell is a graph? I posted one of these in the first post, a Platonic tiling of the Riemann sphere.

    Here's another graph, with colored faces; this one's obviously embedded in a Euclidean domain; the colors are a restriction:

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    If you want to, you can turn it back into a set of eight undecorated points by peeling the colored stickers off, or (re)coloring the graph with one color. Then transposing any pair of points is equivalent to transposing a pair of points in an uncolored graph with no edges.
     
  21. arfa brane call me arf Valued Senior Member

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    The completed graph of the partition of the cube group looks like this:

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    The graph which is the 2x2x2 puzzle is "at" one of these points; each point as mentioned is an equivalence class, so the (state of the) puzzle represents a point. The point (0,0) has one representative state, a choice of an "additive" identity. The largest equivalence class has 841,500 representatives, this is the point (9,11).

    This completed graph (in \( Z_12 × Z_15 \) can be reduced. It depends on which tiling you use on the sphere.

    A (2,2,2) tiling with reflections gives you the freedom to color each of eight octants from a set of eight colors, hence a permutation that rotates the orientation of an octant is excluded, so you have a subset of \( S_8 \) in the completed graph. Restricting rotations to 180° further reduces the graph to the section (0,0) through (4,8), the partition of \( S_4 \).

    To get all 3,674,160 representatives you have to include rotations (by 90°) of a single octant "in place", and the reflection group here doesn't have that freedom, but a (2,3,4) tiling does if you color it the right way . . .

    And there's the might-be-important observation that we go from a (2,2,2) to a (2,3,4) tiling/tesselation with a cubic graph representation. A pair of numbers in the Schlafli symbol go from (2,2) to (3,4) (perhaps by adding (1,2)).
     
    Last edited: Feb 4, 2019
  22. arfa brane call me arf Valued Senior Member

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    Sorry, typo. the above should be: "This completed graph in \( \mathbb Z_{12} × \mathbb Z_{15} \) . . ."
     
  23. arfa brane call me arf Valued Senior Member

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    7,832
    The term "free" in a mathematical context, is defined on "things that generate the elements of a group or a set, or the vertices of a graph, maybe that traverse the graph 'freely'", and so on.

    I said the integers are freely generated, what does that mean? One thing it appears to mean is, I can't "freely" permute the integers, I can't exchange a pair of them and still have the integers. I can't embed (glue or bind to a larger space, like the real plane) the integers and then start swapping them around, they're all stuck so that 1,2,3 are all pairwise adjacent. Say I ignore this and exchange 4 and 5, then 2 + 3 = 4, and 1 + 3 = 5, an incongruence. Moreover, the third prime is in the wrong place, number theory collapses around me . . .

    So there's something that corrects my mistake for me, or stops me doing it, called arithmetic: an algebra.

    Another note: the graph of the "cube group", a convenient label for the free monoidal algebra on six letters, or "face rotations" that generates it, is an abstract derivative; Temperley-Lieb algebras are "freely additive", you glue n points to a boundary and you can always add more.

    TLn(x) contains TLn-1(x), but TL1(x=0) looks like the graph \( K_2 \), except it has an orientation, conventionally a vertical line between two points. You're free to generate the identity of TLn(x) for any n, using a tensor product (not of graphs), which amounts to simply adding together n copies of TL1, not vertically, so horizontally (perpendicular to composition, at least).

    This is exactly what you do when you add n copies of one bit of information together and call it an n-bit register. An n-bit register is a tensor product of 1-bit registers, in a free additive algebra.
     
    Last edited: Feb 5, 2019

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