Symmetry of the cube

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

  1. someguy1 Registered Senior Member

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    727
    Having already been through this issue with tensor products, I have no doubt that graph theorists make up their own definitions for standard mathematical objects. But if you claim there is more than one cyclic group of order 6, that's not something I have to argue, any more than if you said the sun rises in the west and dared me to prove you wrong. I don't need to even have the conversation. You're just wrong on this.

    Are you honestly claiming that \(\mathbb Z_2 \oplus \mathbb Z_3\) is not isomorphic to \(\mathbb Z_6\)? How can you say that and expect to be taken seriously? You might as well claim that 1 + 1 is 3.
     
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  3. arfa brane call me arf Valued Senior Member

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    They are not the same group in the context of graphs of groups. FFS

    --https://en.wikipedia.org/wiki/Graph_of_groups
     
    Last edited: Jan 25, 2019
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  5. someguy1 Registered Senior Member

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    727
    Which groups are they? What are their respective orders? Are they cyclic?

    Look, it's silly for me to actually engage in argument about this. The burden is on you to explain yourself if you honestly think that there is more than one nonisomorphic cyclic group of order 6. Your claim violates the fundamental theorem on finitely generated Abelian groups.

    https://en.wikipedia.org/wiki/Abelian_group#Classification

    "The cyclic group \(\mathbb Z_{mn}\) of order \(m n\) is isomorphic to the direct sum of \(\mathbb Z_m\) and \(\mathbb Z_n\) if and only if m and n are coprime."
     
    Last edited: Jan 25, 2019
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  7. arfa brane call me arf Valued Senior Member

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    Oops
    No it's an element of <5,9>.

    Once more, the graph of the first object is distinct from the graph of the second. Both are however isomorphic under local symmetry operations, but obviously not isomorphic globally (as their graphs show).

    The graphs each have structure, a consequence of the symmetry being mapped to a globally symmetric object; the graph structure dictates that the two isomorphisms (under local operations) are in fact, distinct structures.
    'sigh'

    Just to try and nail this down: the isomorphisms are not part of the graph structure (how it gets "assembled", say). For example in the fundamental object I notate as <2,3>, the set of points {(1,1),(1,2)} does not contain the rest of the group elements, for the simple reason local group symmetries are excluded--we fix these two points and leave the rest alone, in respect of the rectangular section of \( \mathbb Z^2 \) covered by the whole product \( \mathbb Z_2 \oplus \mathbb Z_3 \). The graph has no edges to these points = the exclusions are a coordinate system for the graph.
     
    Last edited: Jan 25, 2019
  8. someguy1 Registered Senior Member

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    727
    You're equivocating graphs and groups, avoiding the question, and tossing in a lot of jargon that obfuscates the fact that you appear to be denying one of the most basic facts in mathematics.

    Do you accept the fundamental theorem of finitely generated Abelian groups? Or not? It dates from 1870.

    Are \(\mathbb Z_2 \oplus \mathbb Z_3\) and \(\mathbb Z_6\) isomorphic as groups or not? Yes or no? No jargon please, just a yes or no.
     
  9. arfa brane call me arf Valued Senior Member

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    7,692
    Yes, they are isomorphic under the operations of addition and multiplication. OK?

    Now a question in your direction: are the operations of external direct product (\( \oplus \)) and addition (tentatively \( \otimes \)) as I've defined them, isomorphic in either group? What's the external direct product of ℤ6?
    No I'm not. The group and its local symmetries are independent of the graph of that group (as a set of connected points). "My" graph, which isn't really mine, is not a graph of either group in question, it's a graph of subgroups. Are we getting the memo yet?
     
    Last edited: Jan 25, 2019
  10. someguy1 Registered Senior Member

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    727
    Multiplication? Where'd that come from? We are talking about groups. YOU are talking about groups. Do you know what a group is?

    But thanks for the straight answer, anyway, even if you just demonstrated that you're not actually sure what a group is.

    Well I don't understand your definitions and I've asked for clarification several times. For what it's worth I would prefer simple explanations to the lengthy, jargon-filled responses you've been giving, accompanied by links to advanced papers in topological quantum field theory.

    I do not understand your definitions and can not discern the algebraic structures you have in mind.

    No. Substituting snark for clarity doesn't help us understand each other.

    If you can keep it simple, please do. If not, I can't respond. Just as I couldn't respond to your links to advanced papers in topological quantum field theory, which turned out to have nothing to do with the matter at hand.

    Likewise your earlier link to graphs of groups. That was a moderately interesting article, but it didn't bear on the subject we were discussing. I'd prefer a simple and focussed conversation. I don't doubt that graph theorists have interesting theories, but you aren't clearly explaining your terminology and concepts. And in many instances so far, your links don't bear on the question.
     
    Last edited: Jan 25, 2019
  11. arfa brane call me arf Valued Senior Member

    Messages:
    7,692
    In the group ℤ6 under addition, multiplication is defined already since say 1 + 1 = 2*1 . I know what a group is, a set with one binary operation. So, you don't need to define 2*1, even though it's isomorphic to 1 + 1.
    And you don't seem to be sure what a graph is.

    You can't follow my notation, you don't understand the difference between a group (or a ring) and a graph of the group (or ring).
    I seem to be wasting time here. Do you at least also get that impression?

    One last time: graphs of groups can be different, not the same (!). This says absolutely nothing about the group symmetry, but does say something about differences between isomorphisms. Clearly, a set of six points on the same line is different to two points on a line plus three points on another line, graphically, and yet there is a group isomorphism (!) iff the points are labeled the right way, that is arithmetically.

    But I'm just blowing smoke up your ass, or something. I have to say, your denigration and condescension leads me to the conclusion that you have a stick up there.

    Look, the graph is genuine, it's a graph of the 2x2x2 cube puzzle, it's a real result. As I mentioned, lots of information is missing, for instance that there are 3 paths from (0,0) to (1,2), and 6 paths from (0,0) to (1,1). Oh yeah, I mentioned the upper boundary of the graph has a partition of \( S_4 \) in it. So what, huh?
     
    Last edited: Jan 25, 2019
  12. someguy1 Registered Senior Member

    Messages:
    727
    I don't understand why you prefer to obfuscate in a way that alters the basic meaning of standard mathematical concepts. What are the addition and multiplication in the group of symmetries of a triangle? You're just making stuff up rather than keeping things simple.

    If you are really talking about rings or \(\mathbb Z\)-modules that's perfectly fine, but just say that.

    You have a tendency to obfuscate, and that's making it hard for me to understand you.


    I've already agreed that I know nothing of the subject. That doesn't relieve you of the obligation to be clear and correct about basic math.

    If you have some structure based on groups that doesn't respect group isomorphism, that might well be the case. To be perfectly clear, I don't say that you're wrong. Only that your exposition is not understandable to me.

    Anyway this no longer seems productive. I'm still interested in the graph-theoretic terminology of tensor products and I'm looking into that. But I don't think the growing frustration on both sides is helpful.
     
  13. arfa brane call me arf Valued Senior Member

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    7,692
    No, the group isomorphisms are independent of the graphs. The graphs do not have operations defined on them which can convert a group into an isomorphism or automorphism. This is because, once again, the group operation is not part of the structure of the graph (although it is obviously part of the group itself, hence 'local').

    I had a lecturer in graph theory once explain a graphical isomorphism of \( S_3 \). All I remember now was it was a bipartite graph and the isomorphism was an incidence matrix.
     
  14. someguy1 Registered Senior Member

    Messages:
    727
    Word salad. I can't parse it.
     
  15. arfa brane call me arf Valued Senior Member

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    7,692
    • Please do not flame other members.
    This is like trying to pull teeth with a pair of pliers.

    You have a finite group, it has one operation acting on a finite set. If you ignore the group operation, say you take another group, maybe a copy of the first group, and combine them 'globally', without invoking the group operation, such that a structure, a graph, appears. What do you have if you don't invoke the group operation on this structure? What does the graph "do"?

    Answer--nothing, it's a graph you dipshit.

    Ok, well how do I convert a graph of ℤ6 into a graph of an isomorphism of ℤ6?
    Answer--it's much easier to start with the G-set of ℤ6, or of the isomorphism which may be a different graph, then invoke valid "moves" that can add and remove points to see if one graph can be morphed into the other. If not the graphs are distinct, but the group isomorphisms are still isomorphisms (maybe because they're local symmetries?). Go figure.

    Maybe see if you can answer this question: given the group ℤ6, how many isomorphisms are there, and how many of the isomorphisms have a graph (a set of points, possibly with edges between points), does the graph need to have numbers labeling each point?
     
    Last edited: Jan 25, 2019
  16. someguy1 Registered Senior Member

    Messages:
    727
    LOL.
     
  17. arfa brane call me arf Valued Senior Member

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    7,692
    So suppose for the G-set of ℤ6, you draw 6 disconnected points and label them with 0,1,...,6. What does this say about the group ℤ6?
    Until you connect the points, it says nothing about the internal symmetry. ℤ6 is cyclic so a cyclic set of edges will do, but the full group symmetry is only "there" when there is an edge from each point to each point, i.e. the graph is completed.

    This reflects in that case, the fact that any element of ℤ6 can be added to any other element. With just two connections, the group symmetry is suppressed, you're expressing two of the additive relations, not all of them. Since addition is commutative and invertible, going from point to point in a forward and backward direction captures this.

    So then, the part of the graph posted, is not a graph of ℤ6 or multiple copies thereof. It's a graph of the symmetry group of a Platonic tiling of the sphere, under the operation of reflection through an edge. Clearly if you choose the set of green+white triangles inside an octant (or face of the octahedral tesselation) on the upper hemisphere, and reflect all 6 of them through say, the right-hand edge, repeating this operation on each of 4 of the upper octants, that's equivalent to a rotation of the upper half of the sphere by π/2. The graph is a graph of this reflection -> rotation group.

    In that, the rotation (composition of 4 reflections through a rh or lh edge) takes (0,0) to (1,1), in this context the numbers stand for "one move, one rotation of π/2". Perhaps the (m,n) coordinates could be rewritten as (m, nπ/2). And here, a rotation in either direction (reflection of 4 octants lying on a slicing plane through the left- or right-hand edge) is (1,1) (after all -1 and 1 are congruent "locally").

    Alrighty?
     
  18. someguy1 Registered Senior Member

    Messages:
    727
    I'll respond to just the first para. I'll just tell you what goes through my mind as I read it, make some comments, and ask some questions.

    Ok. I know what a G-set is. Let me summarize . A group G is said to act on the set X if there is a function \(\varphi : G \times X \to X\) that "respects the group structure." Details are here.

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

    In this case we call X a G-set. Note that there can be many G-sets. G can act on a lot of different kinds of sets.

    Now my picky little brain complains to me about the phrase THE G-set of \(\mathbb Z_6\). There are many possible G-sets. So I'm confused by your use of "the."

    So two things are going on right now:

    * I am bothered that you think the G-set is obvious, when in fact you haven't specified it at all, and there are MANY candidates for a set that some group could act on.

    * But nevermind that, maybe I'm just a grammar fanatic. The "the" isn't important. But here's what is. My ability to follow your post is STUCK right here on this point. If \(\mathbb Z_6\) is acting on something ... what is it acting on? This is obvious to you and totally opaque to me!

    When I was taught this material, we saw examples of groups acting on themselves by left or right multiplication, or by conjugation, or acting on their own powerset by left or right multiplication or conjugation. I haven't seen much in the way of geometric applications other than the various kinds of rotation groups of space; and none in graph theory. So if you intend to educate me, it would be helpful for you to be really explicit about some of these things.

    Haha 7 points gotcha! Just kidding. Ok the set X being acted on is a set of undifferentiated points. Points on a plane I think you mean, not points on a line or n-space for n > 2. Right? Ok. Some random set of points in the plane. I can work with that. So \(\mathbb Z\) acts on this set of points, which I'm perfectly happy to stipulate are called nodes. I'm not completely ignorant, I know a little about trees and Boolean algebra, just not graphs in general.

    Nothing at all, till you tell me what the group action is.

    Ah! I must tell you that I was greatly heartened by this. If you agree that this doesn't say anything (about either the group or the set) then we must be on the same page. We have a group and a set of points but we haven't related them yet so we don't know anything. And I believe we are in agreement on this point.


    Ok now I am lost. Completely over the cliff, no place to hold on to.

    Remember so far we have a finite set of points, 6 in fact, sitting there in the plane. Now you refer to a "cyclic set of edges" and I don't know what that is. Can you tell me? All I know is that there are 6 points in the plane. Walk me through the rest.

    Next, you say that a cyclic set of edges (whatever that is) "will do." What will it do? I have no idea what it does or what you're trying to do.

    I can neither parse nor understand this, since my connection to your exposition has been lost at in the previous sentence.

    So, I've done my best to explain what I know and what I don't know, and which parts of your exposition I can follow and where I'm lost. If you care to fill in the details I'd be grateful. There is a lot of context that you think is obvious that I think isn't obvious at all.

    tl;dr: We have a group action of \(\mathbb Z_6\) on \(\{0,1,2,3,4,5\}\). You think the nature of the action is obvious; and I regard it as completely unknown.

    ps -- Hmmm. Do you mean that you just pick a bijection (any bijection will do since the labeling of the points is arbitrary) and use that to induce the structure of \(\mathbb Z_6\) on the set? I'd certainly believe that, if that's what you mean.

    pps -- One more point. If I define a group action and I replace the group with some other isomorphic representation of it, nothing changes. So I still don't see how the particular representation of a group can EVER make any difference to anything, UNLESS we care a lot about the particular elements of the group as sets. It's really not possible for anything important to change by swapping one version of a group with another. I am a long way from believing the graph theorists have an exception (though anything is possible).
     
    Last edited: Jan 26, 2019
  19. arfa brane call me arf Valued Senior Member

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    7,692
    Yes there are. How then should I choose a particular G-set? How about this one:

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    This one

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    Sorry, should have said a cyclic graph; doing that, connecting the edges in any cycle connects all the elements together, although it doesn't show everything.

    An edge represents a transition from a point to a point by adding the number that gets you there, type of thing. So the edge captures the group action. So the complete map is every vertex connected to every vertex, in which case a graph on 6 vertices has 5 connections at each vertex (are degree five).

    Sorry, in a rush to do something.
     

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    Last edited: Jan 26, 2019
  20. someguy1 Registered Senior Member

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    727
    Even after our conversation I don't understand your notation.

    Why don't you tell me what \(\varphi\) is? For each pair \((n, m)\) where \(n \in \mathbb Z_6\) and \(m \in \{0,1,2,3,4,5\}\), what does it map to?

    You say you have a group action, tell me what it is. Then I'll look at your diagram and probably get it instantly. As it is, I really don't understand your diagrams.

    I have no idea!!!! There is a proper class of sets. From all the sets in the universe of sets; in the category SET; in God's Platonic mathematical heaven where Gottlob Frege and Bertrand Russell tend their collection of sets, occasionally squabbling over what belongs there; you want to pick out SOME set that you regard as the canonical target for the action of the cyclic group of order 6.

    I wish you'd tell me what your criterion is for picking out such a G-set out of all all the sets in the universe of sets. What are you trying to do?

    I think if I could understand this I'd be enlightened. It's opaque at the moment. I'll think about how I'd try to represent the complete graph as a group action.
     
    Last edited: Jan 26, 2019
  21. someguy1 Registered Senior Member

    Messages:
    727
    ps to the last part.

    "An edge represents a transition from a point to a point by adding the number that gets you there, type of thing."

    I get that an edge is an unordered pair of elements that represents a connection or edge between those elements. There are clearly n Choose 2 possible connections or edges in a graph with n nodes. Ok I get that.

    Now ... adding a number? How does that work? Ok I can see that in the complete graph we can get from each node to each of its partners via some mod-n arithmetic. I'll believe that. But then "type of thing," don't know what that refers to. I still don't understand the group action. You're picking out 2-element subsets of a set. Don't see the group action.

    Did you agree with me earlier that by G-set you mean the cyclic 6-group is acting on a set of six points? What's the action? I would be more likely to believe that you just bijected them and induced a circular order on the nodes. I'd believe that much but I don't see the rest of it.
     
  22. arfa brane call me arf Valued Senior Member

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    Are you implying I have some kind of freedom to choose a G-set for any group?

    Nope. I just want at least one target. At a time.
    I thought my group product, isomorphic and all as it is, would get by with a 2x3 section of ZxZ. But what do I know? Is there some criterion that's obviously eluded me?
    Maybe you could show me what it looks like, I'll see if I recognise it . . .
     
  23. arfa brane call me arf Valued Senior Member

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    7,692
    My naive erm, algorithm.

    Without thinking about how you've achieved it, draw a couple of points as little circles.
    Now label one of the points with a number--identify it, a natural choice is the number 1 for the "first" point.

    Now decide, for no real reason other than you can, that you want modular arithmetic to be a player. The first point has an odd number for a label, maybe 2 is going to work? Yes, modulo 2, 2 = 0, so we're away.

    But how to represent addition, in the two fixed points? How do you get from 0 to 1?
    In graph theory, and in category theory, the points are objects, edges (connections) between points are relations.

    In the graph with two numbered points, we have weighted vertices. add an edge to denote a path from 1 t0 2, the group acts on the numbers (hence the vertex set), and on the edges, you need as many as there are transitions, and of course it's a directed graph; each edge is now weighted by the number that represents the difference modulo 2 between two vertices.

    In a more abstract sense, two sets of numbers act on a vertex set and an edge set.

    More abstract, because in the graphs of groups section of graph (and presumably group) theory, they would say \( \mathbb Z_2 \) is the group in both cases.
    The idea is to find a pair of groups that act on a vertex set, and on an edge set. But as we know, nothing is in the way of choosing the pair to be the same group.
     
    Last edited: Jan 26, 2019

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