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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    7,692
    Ok, so a graph can represent the symmetries of a group in a total or a partial way, no problems tend to show up as long as there's support.

    The support for the structure of the graph, in the first rectangular 2x3 section, is either the direct product of two well-known groups of co-prime order, or its what you get when you rotate part of a Rubik's cube. It sounds kinda dumb, it's just a puzzle.

    Yes, but the graph's complexity increases dramatically even with low numbers in the vertex elements (m,n). The simplest part of the graph is the upper section mentioned, where the rotation group of a solid cube gets partitioned by--a partition function (?)
    This pops out when you restrict rotations in the physical model to 180°. You can keep it in one copy of \( S_4 \) by restricting again, to a pair of adjacent faces, and their opposite faces (hemispheres, I should say). So each of three copies of \( S_4 \) is wrapped this way, into a set of "face moves" around the same centre (acting, so to speak, on the sides of a box, not the top or bottom).

    This gives you the partition, and a way to weight the vertex subset along the boundary with numbers of elements in equivalence classes. The 2x2x2 puzzle has I think about 3.7 million permutations, and that's only the rotation subgroup.

    The graph is the structure of a table or matrix of numbers of these elements, the coset space with a partial order.

    K?
     
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  3. arfa brane call me arf Valued Senior Member

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    7,692
    On the topic of the isomorphisms of my fundamental object. \( \mathbb Z_6 \) and all that.

    The symmetries of this group, or of the isomorphism I'm using, are not relevant in the graph. I could add them anytime as weighted edges. I can add a redundant loop to each of the 6 points to represent adding 0, or (0,0). And so on.

    To reiterate, local group symmetries are suppressed--they exist but you use a minimal set. Both the elements (1,1) and (1,2) are order 6, thus "visit" every vertex, they capture enough local symmetry to do the job, and anyway, are already presented by the partition of the graph.

    If there's a partial order, you probably know there are objects, subgraphs, called chains and antichains, and yes I've already counted them all.
    Now I know how many holes it takes . . .
     
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  5. arfa brane call me arf Valued Senior Member

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    7,692
    And about the compositional symmetry of the graph.

    It breaks down at the coordinate (4,8), because of the modular arithmetic, in the context of my "adding a column of points" operation.

    It has to do with the way the <2,3> object acts on itself, or rather, how the objects (1,1) and (1,2) act on themselves. It remains a bit obscure but I'm convinced the heuristic (the physical model) has a plateau between (4,8) and (5,9) that can be explained by modular arithmetic, but the explanation requires me to formulate a way for the local symmetry of <2,3> to act "across" the sublattice of five points, so it "breaks"--the boundary isn't smooth--when you add the next column.

    What I'm talking about can be demonstrated easily enough.

    Add (1,2) to itself by multiplying by 1,2, ..., take the 'direct' sum to mean a point in the column to the right (the maximum), but also write the sum modulo (2,3).

    1(1,2) = (1,2) ≡ (1,2)
    2(1,2) = (2,4) ≡ (0,1)
    3(1,2) = (3,6) ≡ (1,0)
    4(1,2) = (4,8) ≡ (0,2)
    5(1,2) = (5,10) ≡ (1,1)
    6(1,2) = (6,12) ≡ (0,0)
    . . .
     
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  7. someguy1 Registered Senior Member

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    727
    Of course. I imply it, I state it, I shout it from the highest mountain. I tweet it. I proclaim it on cable tv news. I editorialize it on the op-ed page of the Sunday New York Times. I say it in every possible way it is humanly possible to say anything.

    Given a group G there are MANY possible G-sets. Unless you are using the term G-set in a nonstandard way.

    When a group G acts on a set X, we call X a G-set. That's the definition.

    The definition of a group acting on a set is here:

    https://en.wikipedia.org/wiki/Group_action

    Given any group there are many possible sets it could act on. Standard examples are G acting on itself by left multiplication; G acting on itself by conjugation; G acting on its powerset by left multiplication; G acting on its powerset by conjugation.

    There are also of course many geometric examples. The group of planar rotations acts on the plane. The group of isometries in 3-space acts on 3-space. Physics uses a lot of these kinds of spacial group actions.

    So you use the term G-set, but you think that given a set G there is only one possible G-set. That's false. You use the terminology of group actions, but you give no group action. That's murky and indicates either a flaw in your thinking or a flaw in your understanding or a flaw in your exposition. Those are not mutually exclusive.

    [Or perhaps as with tensor products, the graph theorists are making up their own words again. That's certainly possible].

    But let's go back to your idea that \(\mathbb Z_6\) induces, via group action, the complete graph on six points. I have been thinking about this and I just don't see it.

    Say we have a set of points labelled {a, b, c, d, e, f}. The labels don't matter. Now what is the complete graph? It's the following set of pairs, or connections, or edges.

    {a,b}
    {a,c}
    {a,d}
    {a,e}
    {a, f}
    {b,c} (We needn't include {b,a} because this graph is not directed)
    {b,d}
    {b,e}
    {b,f}
    {c,d}
    {c,d}
    {c,f}
    {d,e}
    {d,f}
    {e,f}

    All together I get 15 edges. That's 6 Choose 2, the number of 2-element subsets of a 6-element set. I assume we're agreed on all this so far.

    Now please, show me a group action, in the sense of the actual definition of what a group action is, whereby \(\mathbb Z_6\) somehow generates this collection of 15 edges or 2-element subsets. I just can't figure it out myself.

    On the other hand, I CAN visualize the following. Suppose we take any bijection from \(\mathbb Z_6\) to {a,b,c,d,e,f}. This would induce an ordered, circular graph structure on six points, for example

    c -> e -> b -> d -> a -> f -> c ...

    Even then, we have used \(\mathbb Z_6\) to induce a circular directed graph structure on six points via bijection; but still I don't see a group action or a G-set as you claim. Of course there are 6! = 720 such bijections possible. Are any of them group actions? The collection of all of them? I am just not seeing any G-sets.

    Until we nail down this terminological problem we can not move forward.

    What is your definition of a G-set?

    [Note: I did not read any of your four intervening posts. Till we nail down what you mean by G-set, no progress can be made. That's because I know what a G-set is, and your claim that given G there is only one possible G-set flies in the face of standard terminology of what a G-set is. And nobody likes flies in their face!]
     
    Last edited: Jan 26, 2019
  8. arfa brane call me arf Valued Senior Member

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    7,692
    Yes it is and I can't for the life of me understand where you got the idea I think that given a set G there is only one possible G-set.

    The given set of points, labeled with coordinates in \( \mathbb Z^2\) is a G-set. I didn't make this graph up out of ideas, the Rubik's puzzle, under computer analysis, is the basis ultimately.

    If I have this freedom to choose (which of course I already know I have), then surely the graph is just a choice? But what kind of graph is it?
    The thing with that question is I already know what kind of graph it is, yet here we are.
    Well, it might be helpful to start weighting the vertices with, you know, numbers and try adding numbers to numbers.

    Then you can say you have a set of numbers 'acting' on vertices, that gives you an additive relation between vertices of your 6-point graph.

    But wait, your graph has enough structure that you can weight the vertices by ordering them, so you have a point in a "first place", another in a "second place", based on their relative positions. Numbers.
     
  9. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    My definition of a G-set, courtesy of the Comp Sci & Mathematics School at Waikato:

    Group action on a set.


    Let X be a set and G a group. An action of G on X is a map * : G × X → X, such that ∀x ∈ X:

    1. e*x = x
    2. (gh)*x = g(h*x)​

    ∀g,h ∈ G.

    X is a called a left G-set and G is said to act on the left.
     
  10. someguy1 Registered Senior Member

    Messages:
    727
    Ok. We agree.

    So:

    1) When you use \(\mathbb Z_6\) to represent or induce or generate the complete graph on six points ... what is the group action? I have been trying to figure it out and can't.

    2) Why are you surprised that a given group G may have many different G-sets? After all every group acts on itself by left multiplication and also by conjugation. That's at least two reasonable G-sets for any group, and there are others. Well ok to be fair that's only one G-set with two different actions. But then G acts on its powerset by conjugation as well. So that's another possible G-set.

    \(\mathbb Z_4\) acts on itself by addition. But it also acts on the complex plane via multiplication by \(i^n\), which is the same as rotating the plane by quarter turns. So there are two obvious potential G-sets for the same group.
     
  11. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    The graph represents the symmetry group. Suppose you have six points in general position, a graph which is equivalent to six disconnected graphs of one point.

    Graphs are topological spaces, you can redraw a six point graph with the points anywhere, so why not the vertices of a hexagon?
    If you want arithmetic, even modular arithmetic, numbers are your Huckleberry.
    You don't strictly need them, any symbols will do, but the graph has to have weighted vertices. Weighting is a group action on vertices.

    If you use non-numeric symbols arithmetic is trickier, more tedious, but do-able. You have the choice though.

    K, so decorate your hexagonal disconnected graph with the following tuples, in a cycle: (0,0),(0,1), (0,2), (1,0), (1,1), (1,2). Or with single numerals, if you choose to.

    Now add an edge from (0,0) to (1,2), and weight it with (1,2). The edge represents adding the vertex weights together, it's the difference between them. To "get to" the vertex (1,2) when at (0,0), move to it by adding the number on the edge, to go back add the inverse.
    Keep adding (1,2) at each vertex you go to in the "additive cycle subgraph", and add the edge (so that the operation of addition generates this edge and weights it).

    Each vertex in the subgraph is degree 2. Ok add some more edges between vertices that give you the edges of the hexagon and degree-4 vertices; there's a pattern in the edge weights, a modular kind of pattern.

    I'll see If I can do this with the geometry package I have . . ., but it shouldn't be too difficult an exercise
     
    Last edited: Jan 26, 2019
  12. someguy1 Registered Senior Member

    Messages:
    727
    The symmetry group of six points is \(S_6\), a non-Abelian group of order 720. It's the group of all possible bijections of a 6-element set to itself, with function composition as the operation. (Abelian is always capitalized after Neils Henrik Abel, who first proved the unsolvability of the quintic (before Galois) and who died tragically at 27.

    https://en.wikipedia.org/wiki/Symmetric_group

    However that's different than the symmetry group of a regular hexagon, which is \(D_6\), a dihedral group of order 12.

    http://www.cs.uleth.ca/~holzmann/notes/dihedral/dihedral.html

    You are claiming a relationship with \(\mathbb Z_6\), an Abelian group of order 6.

    Please clarify your meaning.

    Better if we focus on six points and your claim that they are a "G-set" of \(\mathbb Z_6\) where you won't tell me what the group action is.

    I'm being insistent because I'm trying to find points of reference between what I know and what you are trying to express. I don't doubt that you have some geometrical insight in mind. I don't think you're being very clear in your mathematical exposition.

    The symmetry group of 6 points and the symmetry group of a regular hexagon are profoundly different; and each are different from the cyclic group of order 6. Those are three different groups.
     
    Last edited: Jan 26, 2019
  13. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    Aren't you forgetting that the points in a graph aren't fixed? You can exchange any two points in a disconnected graph of n points, and you have a disconnected graph of n points.

    You can permute the vertices arbitrarily, change their relative positions, and all you have still, is a disconnected n-point graph. That is to say, unless you distinguish the points somehow, you got nothin'. You have a topological set of six points that you can arrange by say, sprinkling them on the surface of a sphere, there is nothing to permute.

    An example. I can draw six points at the edges of a hexagon, say "this isn't an object with geometric symmetries, there are no reflection axes in this frame".
    Then I draw the exact same graph as a cyclic graph that looks like a T, nothing at all like a hexagon; in graph theory the shape of a graph is not relevant, in fact geometry means nada.

    'Yeesh'
     
    Last edited: Jan 26, 2019
  14. someguy1 Registered Senior Member

    Messages:
    727
    I stand by what I wrote. I don't understand what you are saying. If you don't think the symmetry group of a six element set is \(S_6\) and the symmetry group of a regular hexagon is \(D_6\), consult any undergrad text on abstract algebra, or just read the links I supplied.

    Correct. That's one bijection. There are 719 others. That's 6!, the number of bijections from a 6-element set to itself. Those are morphisms in the category of sets if you like.
     
    Last edited: Jan 27, 2019
  15. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    I'll try again since you have admitted graph theory isn't one of your strong points.

    Ok, draw six points on a sheet of paper. It doesn't matter where they are = they can be anywhere, they don't have a position, they are unrelated (the edge set of this graph is empty).
    Now find the powerset of the six points, without changing the graph. You can count the points, but the result is always the same: 6, no matter what you do. Maybe you can count them twice, but don't ask me what that has to do with graph theory.
    I know what you're saying about those groups is true though (seriously, I do), but six identical points in a graph, has neither symmetry.

    Which is not to say, however, that you aren't allowed to arrange the six points in a hexagonal pattern (or however you choose) and say it's the same graph. However, adding an edge somewhere changes the graph into a different graph. Coloring the six points means now you can transpose the positions of two points, because that's two more different graphs.

    Sorry, just read this more closely.
    Ok, you are saying a graph of six disconnected points is a set with six elements in it? Well, kind of. There is no order in the set so far. The points in the graph can be anywhere = the points in the graph are anywhere.
    So they are, in a certain context, "wandering around", although there is a superficial appearance of being fixed, on inspection it isn't real.

    An analogy
    Suppose the points are six identical chickens, roaming freely about the countryside. You need a relation between these identical (i.e. indistinguishable) chickens, and that relation is a way to distinguish one from the other. That is, if your aim is to count the permutations or something. There does seem to be a few more problems to address though. They're all still free, not bound to anything that will let you define what a permutation is.
     
    Last edited: Jan 27, 2019
  16. someguy1 Registered Senior Member

    Messages:
    727
    Ok, the powerset has \(2^6 = 64\) elements. When you say find it, I don't understand that because powersets don't live anywhere in particular, they just have mathematical existence. If you draw six points in the plane you could certainly draw 63 distinct closed curves that enclose all the various nonempty subsets, plus the empty set makes 64. And you definitely can't "find" the empty set anywhere! That's a matter of philosophy. In math it exists. As it relates to a set of physical objects or points drawn on paper, I have my doubts.



    Yeah ok. I don't think anyone would disagree.

    What are identical points? Are they different from "different" points? All points are exactly alike, except for their location in some space. If two points have the same location, they're the same point. There's only one point at the origin for example. I think you're wandering into philosophical backwaters best left alone for now. But whether the points are identical or not, if there are six of them then there are 720 possible bijections of them. I don't follow the point of your exposition.

    Ok. Still not following the exposition but I'll certainly agree you can rearrange six points (or locations). I'm suddenly not sure what you even mean by a point, whether your points have locations, whether some of them can be "identical" and some "different." This post seems unmoored from anything I can understand.

    Yeah sure, but what about the group actions? Don't think you can avoid coming to terms with the fact that you're using terminology like G-sets, and when challenged, you are changing the subject.

    Now we are COLORING the points? I'm perfectly willing to allow that this can be done mathematically, but it just seems like yet another attempt at avoiding the very specific questions I've asked you.

    That doesn't even make sense. Assigning colors to the nodes of a graph doesn't transpose the nodes. It doesn't do that at all. It's transpositions that transpose points. Undergrad algebra again.

    https://en.wikipedia.org/wiki/Cyclic_permutation#Transpositions

    And what happened to the powerset? You brought it up and then forgot about it.

    I'm going to take a break from this. If you know enough abstract algebra to understand my posts then you must see that you are avoiding all my questions and failing to back up your own claims. I'm still waiting for you to tell me about "the" G-set of \(\mathbb Z_6\), which you were very adamant about a while ago and now seem to have abandoned.
     
  17. arfa brane call me arf Valued Senior Member

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    7,692
    And you repeat the same error, in the very next post!

    Since a GRAPH of six points with no edges, doesn't have this "closed curve" thing anywhere in it, you need a different graph, which is what you blithely go on to do with your "set of 6 points" and "63 closed curves enclosing subsets".

    You step almost unconsciously out of the context of a graph, straight into something graph theory actually has nothing to say about, i.e. geometry.
    And therefore I'm assuming the remainder of your post, including the "questions" I'm avoiding, has gone there too.
     
  18. someguy1 Registered Senior Member

    Messages:
    727
    The nodes of a graph form a set. That's basic graph theory, which you claim to know. Now I really am taking a break.
     
  19. arfa brane call me arf Valued Senior Member

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    7,692
    I'll keep going with my "why a graph has no symmetries" thing.

    A "set" of 6 free points is a set of points none of which are glued to a boundary. Until you fix, somehow, all 6 points on some manifold (because you need the notion of a distance between points), there isn't any algebra, no arithmetic.

    Coloring the vertices of a graph is standard practice. Now the graph is different to a set of 6 indistinguishable things; if the uncolored graph is a set, it's a set containing six copies of the same thing, so it's a multiset. Take the intersection of six points with one point and you have one point. Take the union of one point with six points, you have . . . one point.

    Color the set with six different colors--make the points into six different things, and bingo, algebra, maybe some numbers.
     
    Last edited: Jan 27, 2019
  20. someguy1 Registered Senior Member

    Messages:
    727
    Oh manifolds. Now you're name-dropping differential geometry. Come on, man. Anyway:

    "In the most common sense of the term,[1] a graph is an ordered pair G = (V, E) comprising a set V of vertices or nodes or points ..." (my emphasis).

    https://en.wikipedia.org/wiki/Graph_theory#Graph

    The nodes are taken from a set. A set has subsets and bijections. You're correct that a six element set has no arithmetic or algebra. It does have set-ness. It obeys the laws of set theory. It's an object in the category of sets. It has bijections, subsets, and a powerset.

    Ok I am taking a break. This is going nowhere. Manifolds. Right.
     
    Last edited: Jan 27, 2019
  21. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    But six points, all by themself, don't. You just admitted you have to draw closed curves around subsets to get the powerset relation. You embedded, perhaps unconsciously, the points on a manifold.

    The argument we're having isn't about whether a graph is a set of vertices and a set of edges, it's about whether you can draw six unconnected points and derive their powerset.
    I did say "different graph".
    . . .
    Draw a graph of six points, what the hell make it a row of six points. Now color them with six colors. Ok, now transpose two adjacent points, by drawing another graph, different to the first "coloring operation", with the transposed colors.
     
    Last edited: Jan 27, 2019
  22. someguy1 Registered Senior Member

    Messages:
    727
    No, I said that in response to your asking where I can "find" the powerset.

    But we know that from basic graph theory, the nodes form a set. And from set theory, a set has a powerset. You don't have to find it, and in the infinite case you CAN'T find all the subsets. They simply exist by virtue of the powerset axiom.


    Not at all. You gave me a set, the set of nodes. Given a set I can posit the existence of its powerset and the set of all possible bijections (injective and surjective maps) from the set to itself. That's set theory. No algebra, no space. Given a pure set we can talk about the functions on that set, some of which are bijections. We can talk about the subsets of that set; and we can even form the powerset, the set consisting of all the subsets. This is basic high-school set theory. No algebra, no distance, and definitely no manifold, which is a far more sophisticated mathematical object.


    We're no longer having an argument. Or a conversation. Every time you misuse a technical term and I explain the proper meaning, you change the subject to something else. That's no longer fun for me. I've explained a lot of math to you and you simply keep changing the subject. What's the point?

    That still answers none of the many direct questions I've put to you. You lost me at G-set and you have NEVER justified your use of the term. You even posted the correct definition, yet you haven't tried to apply it to your ideas.
     
  23. arfa brane call me arf Valued Senior Member

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    7,692
    You have a set of six colors, you also have a set of six points. Coloring the points is the set of colors acting on the six points.
    You can color the edges and faces of graphs too.

    But now, you have a way to permute the positions = change the colors. You now have an algebra, where you had six things just kind of floating around.
     
    Last edited: Jan 27, 2019

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