Spherical symmetry

Discussion in 'Physics & Math' started by arfa brane, Sep 22, 2013.

  1. arfa brane call me arf Valued Senior Member

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    On the 2-sphere, a pair of geodesics defines two antipodal points (separated by 180[sup]o[/sup]), so there are 4 geodesic sections, or paths, from one point to its antipode. So an Euler circuit (visiting each section exactly once) would visit the two points twice, and have 4! possible permutations.
    What happens on the 3-sphere?

    Geodesics on the 2-sphere are lines traced out by points, on the 3-sphere these would be surfaces traced out by lines, so would that mean visiting four 'places' in sequence, and only one permutation for an Euler circuit?
     
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  3. Fednis48 Registered Senior Member

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    I'm not an expert in differential geometry, but I think geodesics on a 3-sphere are still lines. From Wikipedia:

    Anything with higher dimensionality than a line does not have a unique tangent vector, so by this definition it cannot be a geodesic.

    As for how the Euler circuit problem changes, the main difference is that two geodesics don't necessarily define a pair of antipodal points on a 3-sphere. Consider the 3-sphere defined by

    \(x^2+y^2+z^2+t^2=1\)

    Two possible geodesics are {\(x=y=0\), \(z^2+t^2=1\)} and {\(x^2+y^2=1\), \(z=t=0\)}. These curves don't intersect. So in order to generalize the Euler path problem to hyperspheres, you'll probably have to define it more precisely.
     
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  5. arfa brane call me arf Valued Senior Member

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    Ok, thanks Fednis48. I found this:
    --http://www.google.co.nz/url?sa=t&rc...DHxtelmRULz1Qo4rz2iaFYg&bvm=bv.52434380,d.aGc

    Another thing: on the 2-sphere you can transport vectors lying in the same plane, from pole to pole. On the 3-sphere you can transport vectors lying on the same 3-plane?
    To rotate a vector by 90[sup]o[/sup] in the 2-dimensional case, you go from say the pole, to the equator, along the equator and back to the pole. That's three rotations of 90[sup]o[/sup]. Again, what "happens" on the 3-sphere? Are there fewer rotations needed, or what?
     
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  7. Fednis48 Registered Senior Member

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    Looks like a good source you found; far more precise than anything I would have come up with. In general, I think things don't actually change much when you go from the 2-sphere to the 3-sphere. Since you have another variable to work with, you have more options in a lot of situations (like for the transport of vectors in the same plane problem, there's a wider range of planes to pick from). The underlying math often stays the same, though, because at the end of the day you're dealing with the same dimensionality of objects. The shortest path between two points is still 1-D. Planes containing vectors are still 2-D. A 3-sphere probably has some richer properties that the 2-sphere doesn't have if you look at how it relates to 3-and-higher dimensional objects. But if you just work with the 1- and 2-D objects associated with 2-spheres and generalize them up, I think the properties stay mostly unaltered.
     
  8. arfa brane call me arf Valued Senior Member

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    The problem I have seems to have evolved into finding a way to partition S[sub]4[/sub] into four sets. This partitioning is based on the number of ways to travel along a geodesic section in three dimensions. Initially there are 3 choices, then 2 (since the first choice uses one dimension), so the second partition has 6 choices.
    The 'theory' appears to involve the fact that S[sub]4[/sub] is a subgroup of S[sub]8[/sub], and there is a map from one to the other that preserves the orientation of pairs of characters, so since these are abstract edges, each edge or 'invariant pair' in S[sub]8[/sub], has an equivalent single character in S[sub]4[/sub].

    You start with (a,b,c,d) mapped to (ab,cd,ef,gh), say.
    Sorry if this is a bit vague.
     
  9. Fednis48 Registered Senior Member

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    725
    Yeah, you lost me here. Sorry. Let's start at the top: what do you mean by "This partitioning is based on the number of ways to travel along a geodesic section in three dimensions?"
     
  10. arfa brane call me arf Valued Senior Member

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    It's probably easier to forget about geodesics and 3-spheres, all you need is a way to take S[sub]4[/sub] as a subset of S[sub]8[/sub].
    This Cayley table of S[sub]4[/sub] is helpful:

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    Now map the identity permutation (abcd) in S[sub]4[/sub] to (ab,cd,ef,gh) in S[sub]8[/sub]. That means "expanding" a 4x4 matrix to 8x8 (if you use matrices), as if you're mapping points in S[sub]4[/sub] to pairs of points in S[sub]8[/sub] (i.e. "edges"), and you want all the 8x8 matrices that preserve the pairings (i.e. those that leave "edgewise" relations unchanged).

    I actually have the answer, or at least the sizes of each partition, and I'm trying to figure out why the answer is what it is. The "answer" I have leaves out a lot of detail; it's a computer generated solution to a counting problem, and contains only numbers in a table.
     
  11. Fednis48 Registered Senior Member

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    725
    Ok, I'm getting a clearer picture now. But it seems to me like the answer would be trivial: take all matrices in S4, replace each zero by the 2x2 zero matrix, and replace each 1 by the 2x2 identity matrix. What am I missing?
     
  12. arfa brane call me arf Valued Senior Member

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    Nothing much, that's exactly what you do. But the replacement operation requires something called the Kronecker product of matrices. This is also a tensor product, isn't it?
     
  13. Fednis48 Registered Senior Member

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    725
    Gotcha. I think there's some subtle distinction where the tensor product acts on vector spaces while the Kronecker product acts on matrices, but from a physicist's perspective they're the same thing.
     
  14. arfa brane call me arf Valued Senior Member

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    One detail about replacing 1s in a 4x4 permutation matrix with the identity is that reversing this is easy, but that isn't the case in general.
    And one of the things you're missing is that, yes, you can tensor all the 4x4 matrices in S[sub]4[/sub] with I[sub]2[/sub], but that doesn't tell you how they're partitioned. I'm looking at partitions of symmetric groups with a specific example.

    I found another interesting paper which I admit is mostly beyond me, but it does seem to cover some of the ideas I'm pursuing:
    http://www.google.co.nz/url?sa=t&rc...MMMht9VbtBaRsG793r6IcMw&bvm=bv.53217764,d.dGI
     
  15. Fednis48 Registered Senior Member

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    Hot damn! For serious math/science papers, my eyes usually glaze over after about 5 pages. All I can say at this point is it sounds like an interesting problem, and I'm glad someone is pursuing it with enthusiasm. Out of curiosity, though, what brought up this question in the first place?
     
  16. arfa brane call me arf Valued Senior Member

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    Hard to say. I guess I'm just throwing ideas around. The thing is I have this 'concrete' result, and I know it's really a general one (a set of numbers), so I'm trying to formulate meaningful questions about it. None of the questions seem to be trivial at least, for instance, asking why some number appears in two different places leads to a lot more questions.

    I've worked out that this result has to do with directions being transported around the sphere, and that Rubik's cubes in some sense 'unfold' their interiors. Fairly obviously the motions involved take the cube into a fourth dimension and when three spatial dimensions are suppressed, what's left is some kind of map to this fourth dimension--i.e. the result is a map that gives this dimension (whatever it is) 'extra structure'. So maybe it's a nice way to visualise "algorithmic" time, or somesuch.
     
  17. arfa brane call me arf Valued Senior Member

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    Here are some of the obvious, and maybe not so obvious, things about Rubik's cubes:

    1. Regardless of the number of other pieces, or elements, the eight corner 'cubies' are permuted 1/2 at a time. The position and orientation of these 'outer' elements is distinct from the number of 'edgies' (or from any central elements that don't change position).

    2. The identity permutation leaves the subsets of position and orientation unchanged. That means the identity is really two identities, or plays a dual role, and is equivalent to any (hence every) group element that leaves both unaltered. The "bare cube" has a certain mathematical dynamic, since doing nothing to it is equivalent to doing everything (anything at once), as every word is cyclic.

    3. Solving a cube doesn't involve exploring much of this cyclic structure, although it is a fundamental property.

    4. The cube will 'accept' any word, of any length, and 'halt' so it's positioned a finite distance from the initial state.

    5. The only word that has no cyclic structure (is not a cycle in G) is the infinite word.
     
  18. AlphaNumeric Fully ionized Registered Senior Member

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    As has been said, geodesics are ALWAYS lines, regardless of dimension. There are higher dimensional generalisations, such as surfaces of minimal curvature etc, but in ALL (non-pathological examples where geodesics exist) differentiable manifolds geodesics are lines.

    A geodesic is the formalisation of standing at a point in a space M, pointing in some direction \(\mathbf{v}\), and then walking in that direction. The example space you picked, \(S^{2}\), is actually a nice example to work with since it illustrates a number of non-trivial properties which you don't see if you worked with \(\mathbb{R}^{N}\). Despite what \(\mathbb{R}^{N}\) might have you believe locations and vectors are not equivalent. Locations define errr... locations, while vectors define directions of motion within the space of points.

    The 2-sphere is a good example. Stand at the North pole and point in some direction parallel to the ground where you are. If you walked off in that direction you'll end up at the equator eventually, yet if you'd held a laser pen and moved along its original path you'd not, the laser (ignoring gravity) would have been shining out into space, since your arm is not pointing directly at the equator, it's in a plane tangent to the Earth's surface. It's only because the 2-sphere is topologically non-trivial that there's this issue.

    No clearly you cannot say "I walked along that vector!", rather that "The vector gave me the initial direction". If you held your arm out as you walked you'd be changing the direction it is pointing in because each new place you walk to you're now pointing in a different tangent space to the Earth's surface. Moving changes the tangent space.

    More formally you picked a location \(p \in M\) and then defined its tangent space \(T_{p}M\), picked a vector in that space \(\mathbf{v} \in T_{p}M\) and then walked along it. The geodesic which results is swept out by the path length s and at any location you're at \(\exp(s\mathbf{v}) \in M\). Exponentiation maps tangent vectors to locations.

    There's a bit more you can do if you want to consider \(S^{3}\) because of a fortuitous property that the 3-sphere is actually a Lie group (it just so happens I've had my head in precisely this system for the last week....). If you're familiar with rotations in 3 dimensions, ie the SO(3) group, then you have experience with this Lie group. You can view the SO(3) group as a rotation about some axis. The axis is defined by a point on \(S^{2}\) and then you rotate about it, which is an SO(2) rotation. \(SO(3)/SO(2) \sim S^{2}\) and via the Hopf fibration you end up with SO(3) ~ \(S^{3}\) (modulo a few technicalities). I'll now flick to talking about SO(3) instead...

    Suppose you have 2 points (ie a rotation matrix) in your manifold, p and q. You want to know what geodesic links p to q (in that order!). Well you know the definition of the geodesic is that \(q = \exp(d(p,q)\mathbf{v})p\) where d(p,q) is the distance between p and q (ie the geodesic length). Well let's just assume you can do what we might first guess, rearrange to get the vector, \(d(p,q)\mathbf{v} = \log(qp^{-1})\). Since p,q are elements in a non-commuting Lie group the order in the log is necessary. \(\log(qp^{-1})\) is therefore an element of the tangent space, specifically an element of the Lie algebra of G defined at p. Given a basis for your tangent space, \(\mathbf{e}_{j}\) and the resultant metric (ie the Killing form for your Lie algebra) you can then extract the coefficients of the basis expansion, normalise and get \(\mathbf{v}\).

    This way you can construct the geodesic generating vectors which get you from some point p to another point q. You can also do it for general differentiable manifolds but it is less pleasant due to the lack of a binary product for combining locations into other locations (ie the group action on the Lie group doesn't generalise directly).

    Anyway, I'm probably off on a tangent, I thought I'd just monologue seeing as I've had something relevant in my head all week. All notational mistakes are intentional and not because its a Sunday morning

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  19. Pete It's not rocket surgery Registered Senior Member

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    *giggle*
     
  20. arfa brane call me arf Valued Senior Member

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    Code:
         FTM	
     	0	1	2	3	4	5	6	7	8	9	10	11	Total
    QTM 0	1												1
        1		6											6
        2		3	24										27
        3			24	96									120
        4			6	144	384								534
        5				72	768	1,416							2,256
        6				9	564	3,608	4,788						8,969
        7					126	3,600	15,072	14,260					33,058
        8					5	1,248	18,996	57,120	36,780				114,149
        9						120	9,480	85,548	195,400	69,960			360,508
        10							1,728	55,548	341,604	487,724	43,984		930,588
        11							72	14,016	238,920	841,500	256,248	96	1,350,852
        12								1,032	56,440	455,588	268,532	944	782,536
        13								12	928	32,968	54,876	1,496	90,280
        14										8	160	108	276
    
    Total	1      9      54     321     1,847    9,992   50,136  227,536 870,072 1,887,748 623,800 2,644  3,674,160

    This table was produced by a computer program. It's a table of coset data (of the 2x2x2 Rubik's cube).
    The FTM and QTM correspond to "face turn" and "quarter turn" permutation metrics. So the table is also a representatation of where these intersect. The FTM includes all possible moves on a single "face", and QTM is restricted to 90[sup]o[/sup] rotations.

    Anyway, the rubric is "look for patterns". One thing you can see is that the numbers along the lower "diagonal" sum to 24, from FTM = 0 to FTM = 4. Is this a partition of S[sub]4[/sub]? How do you "prove" it?
    The answer to that question appears to involve finding permutations in S[sub]4[/sub] that map to the rotation subgroup of D[sub]4[/sub] initially. You need to 'coordinatize' the (vertices of the) cube in a way that leads naturally to a permutation algebra, say.

    Also notice the number 24 in two places. Also the numbers 6 and 72. And that the sequence along the 'upper' diagonal (where QTM = FTM) increases by a factor of 4, from (1,1) to (4,4).

    But I'm wingin' it here.
     
  21. arfa brane call me arf Valued Senior Member

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    So my conjecture is that, armed with the above table (that is, the information in it) and a bit of linear algebra, you can visualise a way to get from two to three dimensions. As the authors of the paper I linked to in #11 say, once you have a well-defined model, a cubical matrix, you can get to spaces with higher dimensions.

    However, physically permuting the corner elements of Rubik's cubes requires the existence of another dimension, or degree of freedom. In two dimensions reflections of regular polygons about a central axis of symmetry can be said to "require" a third dimension since the reflection is equivalent to a rotation about the (chosen) reflection axis.
    Now instead base a reflection on a non-central axis, i.e. an edge. You get a tiling of the plane which is a tesselation if it covers the whole plane (is a regular tiling). Have I got that right?
     
  22. arfa brane call me arf Valued Senior Member

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    Since the rotation group of a cube is isomorphic to S[sub]4[/sub], and S[sub]4[/sub] has a representation as 24 4x4 matrices, these matrices obviously act on 4 elements of the cube at a time. The cube has 4 diagonals, but also three sets of 4 parallel edges.

    The permutation group that acts on the 8 vertices, 4 at a time, is a "partial permutation" of a solid cube (for which the vertex positions don't change); for instance, rotating one half of the cube through 180[sup]o[/sup], then rotating the other half 180[sup]o[/sup] is a full 180[sup]o[/sup] rotation of the "solid" cube. But if the outer rotation group is factored out, this becomes an identity permutation.

    Now 4 edges contain 8 vertices, so each permutation is in S[sub]8[/sub], and has to contain I[sub]4[/sub] somewhere. So it seems I need elements from S[sub]4[/sub] that contain I[sub]2[/sub].
     
  23. arfa brane call me arf Valued Senior Member

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    So if this particular section of the total permutation space of the 2x2x2 'cubic matrix' is a subset of S[sub]8[/sub] such that I[sub]2[/sub] can be 'contracted to a point' for each element in the subset, and the result is S[sub]4[/sub], I should look a bit harder at partitions of the symmetric groups.

    Apparently you partition n first. Now n = 4 can be partitioned into only one non-decreasing sequence with length 4: (1,1,1,1). If you interpret the numbers as vertices with that degree, then you have a graph which contains two disjoint edges connecting pairs of vertices. When n = 8, there are five partitions of n with length 4: (2,2,2,2), (3,2,2,1), (3,3,1,1), (4,2,1,1), (5,1,1,1).
    The graphs of these partitions are all connected, but three of them have multiple edges, only two are simple (the first two).

    But according to this textbook I have, the conjugacy classes of S[sub]n[/sub] can be parameterized by the partitions of n. The association is the form of cycle decomposition for a permutation ο = (a[sub]1[/sub], a[sub]2[/sub], ..., a[sub]λ[sub]1[/sub][/sub]), (b[sub]1[/sub], b[sub]2[/sub], ..., b[sub]λ[sub]2[/sub][/sub]), ..., (c[sub]1[/sub], c[sub]2[/sub], ..., c[sub]λ[sub]k[/sub][/sub]), where (λ[sub]1[/sub], λ[sub]2[/sub], ..., λ[sub]k[/sub]) is a partition of n
     

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