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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    I saw a piccy recently of the Riemann sphere on the projective plane; the sphere was tesselated with an octahedral/cubic graph (these of course, are duals).

    ed:

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    So the space of fractional linear transformations of the modular group (aka the Mobius group), obviously acts on this dual graph on the sphere. What I want to know is if I can rotate half the sphere about the octahedral (blue) lines, thus the projective plane will have one half transformed, the other fixed. Fix the other half and rotate the initially fixed half, and that's equivalent to a rotation of the 'whole' sphere (you don't say?).

    What I need is a formula for a Mobius transform that rotates the sphere, say 180°, one hemisphere at a time. Or more generally, one that rotates the sphere any angle, but one hemisphere at a time.

    Anyhoo, I've already worked out that the cube group has a representation (a partition as equivalence classes), which is a tensor product space, the primary 'object' is the direct product of \( \mathbb Z_2 \) with \( \mathbb Z_3 \).

    The tensor product is very much like what happens in the Temperley Lieb diagram monoids, the graph of the cube group (embedded) in \( \mathbb Z × \mathbb Z \) is just a composition of (some finite number of) \( \mathbb Z_2 \oplus \mathbb Z_3 \).

    I could show some of my working (i.e. reasoning), but I'm not sure if the king is really setting that kind of test here.
     
    Last edited: Jan 22, 2019
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  3. arfa brane call me arf Valued Senior Member

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    This is what \(\mathbb Z_2 \oplus \mathbb Z_3 \), embedded in \( \mathbb Z × \mathbb Z = \mathbb Z \oplus \mathbb Z \) looks like:

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    I've drawn a couple of colored lines, which 'technically' are in \( \mathbb R \oplus \mathbb R = \mathbb R^2 \).
    What I need to do is define the subset of points connected by the lines in terms of a relation. (I think so, anyway).
     
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  5. someguy1 Registered Senior Member

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    726
    \(\mathbb Z_2 \oplus \mathbb Z_3\) can't possibly embed in \(\mathbb Z \oplus \mathbb Z\) for the simple reason that \(\mathbb Z_2 \oplus \mathbb Z_3\) has torsion (elements of finite order) and \(\mathbb Z \oplus \mathbb Z\) doesn't.

    Do you understand that, for example, \(\mathbb Z_2\) is not a subset of \(\mathbb Z\)? Assuming that by \(\mathbb Z_2\) you mean \(\mathbb Z / 2 \mathbb Z\) and not something else.

    You seem to be using technical terms in a way I'm not familiar with and that doesn't correspond with standard and accepted usage.

    Also, without looking it up, can you define what a tensor product is? Just a one or two sentence characterization that shows your understanding.
     
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  7. arfa brane call me arf Valued Senior Member

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    What then, does the set of points in my posted image represent?
    Can I at least embed the connected set (obviously a subset)? What does that make the unconnected set into?
    In the context of a diagram monoid, a tensor product is the addition of points, perhaps with connections between them. The monoidal structure is based on the two ways you can compose the diagrams, vertically and horizontally, one is diagram composition (or "direct product"), the other is a tensor product.

    Suppose you have the identity in TLn as n vertical lines between two sets of n horizontal points, then the identity in TLn+1 is the tensor product of n vertical lines and one vertical line (TL1).
     
    Last edited: Jan 23, 2019
  8. arfa brane call me arf Valued Senior Member

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    Ok, so I'm not 'embedding' all of \( \mathbb Z_2 \oplus \mathbb Z_3 \). What I should say is, given the Cartesian product of {0, 1} with {0, 1, 2} is the set of points {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} in \( \mathbb Z^2 \). the subset {(1, 1), (1, 2)} has no zeros, i.e. is the set { \( (a, b)\; |\; a + b ≠ a, b \) } \(, a \in \mathbb Z_2, b \in \mathbb Z_3 \).

    So in some sense, I can disconnect the lines in the diagram from (0, 0) by removing that point.

    The group \( \mathbb Z_2 \) can be written as the set {0, 1} under the operation of addition modulo 2. The tensor product \( \mathbb Z_2 \otimes \mathbb Z_2 \) is then
    \( \{0, 1\} \otimes \{0, 1\} = \{ 0 + 0, 0 + 1, 1 + 1\} = \{0, 1, 2\} ≡ \mathbb Z_3 \).
    So then \( \mathbb Z_2 \oplus \mathbb Z_3 = \mathbb Z_2 \oplus ( \mathbb Z_2 \otimes \mathbb Z_2) \).

    I want a subset of points that excludes any zeros; if I start with { (1,1), (1,2) }, the tensor product is
    { (1,1), (1,2) } "+" { (1,1), (1,2) } = { (2, 2), (2,3), (2,4) }, a proper subset of \( \mathbb Z_3 \oplus \mathbb Z_5 \).

    So you add the numbers 'freely' and thus add points horizontally--the vertical placement is taken care of.

    Why the elements like (0, 1), (1,3) ≡ (1, 0) etc, are excluded from the arithmetic is that you can't rotate part of a 2 x 2 x 2 Rubik's cube on 0 faces . . .
     
    Last edited: Jan 23, 2019
  9. arfa brane call me arf Valued Senior Member

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    Alright, so if I define my fundamental group as \( Δ = \mathbb Z_2 \oplus \mathbb Z_3 \), then I take tensor products:

    \( Δ^{\otimes^2},\; Δ^{\otimes^3},\;. . ., \; Δ^{\otimes^k} \)
    I get a 'stack' of triangular connections corresponding to the triangular numbers; I get the first triangle, or Δ, then two more, and so on.

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    I get families of parallel lines, something like geodesics, or lines of latitude on a more abstract object. In which case, a single 2 x 2 x 2 'matrix' of unit cubes (say, restricted to a cubic shape, a relation between the cubes), is just a way to compose sections of these lines, which actually aren't really lines but transformations of the Riemann sphere with one hemisphere fixed.
     
    Last edited: Jan 23, 2019
  10. someguy1 Registered Senior Member

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    726
    It might be that the language of graph theory and the language of abstract algebra are so profoundly different as to have entirely incompatible definitions of tensor product. Regardless, the things you say fall into two categories: (1) Things I can't make sense of; and (2) Things I know for a fact are false.

    I readily admit to lack of knowledge of higher geometry. Algebraic geometry and projective geometry for example, two disciplines that seem relevant here. But I wish you'd make more of an effort to bridge the gap between whatever book you're reading, and the knowledge level of someone like me who's taken graduate level abstract algebra (and also enough complex analysis to know what the Riemann sphere and linear fractional transformations are) and still can't make heads or tails of your exposition. There's a combination of your expository style and my lack of algebraic geometry that's frustrating to me.

    Take the above quoted bit for example. Anyone who's survived a semester of undergrad abstract algebra knows that \(\mathbb Z_2 \oplus \mathbb Z_3 = \mathbb Z_6\), as can be readily seen from the fact that the element \((1,1)\) has order 6 and generates the group (or ring or \(\mathbb Z\)-module, whichever you intend to denote).

    Secondly, it's an elementary exercise in tensor products to show that \(\mathbb Z_6 \otimes \mathbb Z_6 = \mathbb Z_6\). This follows from the fact that in general, \(\mathbb Z_n \otimes \mathbb Z_m = \mathbb Z_{\text{gcd}(n,m)}\). See for example
    https://math.stackexchange.com/ques...bz-m-mathbbz-otimes-mathbbz-mathbbz-n-mathbbz. If you know what a tensor product is you should be able to prove this directly using bilinearity. But I suspect that your use of tensor products is profoundly different than their definition in abstract algebra. On the other had that can't be. I'm so curious to know the connection!

    So from what I know of the subject, the iterated tensor products you gave all reduce to \(\mathbb Z_6\).

    I wonder if you can bridge the gap between what I know and what you're trying to explain. I don't doubt that you're saying something interesting. I'm only saying that whenever your exposition intersects my knowledge, your exposition is wrong or misleading.
     
    Last edited: Jan 23, 2019
  11. arfa brane call me arf Valued Senior Member

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    Let's go with the "finding a path" context, a naive algorithm.

    If your stuck at (0,0) the "only way out" is to get to (1,1), from which you can get to (1,2). Then any additional sections of your total path must also not contain zeros, with the exception that you're allowed a single vertical transition (corresponding to a second 90° rotation). In the diagram these are the vertical lines, so you're allowed to do (m,n) + (0,1), in one and only one section at a time.

    Let's also assume the graph is bounded by the exclusion of zeros, and that adding (0,1) somewhere doesn't break this rule. We disconnect the lines from the identity (0,0), so we really only need the first two points {(1,1), (1,2)} and the vertical line between them. This corresponds in the algorithmic context to claiming this line is a solution, the cube is one move from the identity by inspection.

    Well, I'm defining it it terms of the addition of points, but clearly 3 + 3 points, for example, in the arithmetic doesn't equal 6 points, this is sort of because one of the points is labeled with a 0, so the highest value in the arithmetic is 2 + 2 and "adding points to points" in this particular case gets you 5 points, labeled 0 1 2 3 4.
     
    Last edited: Jan 23, 2019
  12. someguy1 Registered Senior Member

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    726
    Yeah, I just don't get it. Must be me. If someone asked me for a one-sentence definition of a tensor product I'd either say, "It's a universal object in the category of bilinear maps between R-modules for a commutative ring R," or else, "It's the free Abelian group (or R-module) generated by the Cartesian product of two groups (or R-modules), modulo the bilinearity relations." Now to be fair, either of those one-sentence wonders encapsulates a lot of sophisticated mathematical concepts. Still, once one slogs through the details, at the end one turns around and sees that it was all very obvious all along, as things usually are after the slogging has been endured.

    "Addition of points" fails to capture the essence of a tensor product, the idea of which is to express the "most general bilinear thing imaginable." There's a huge gap or mismatch here. Tensor products are about bilinearity. I wonder if you can provide a reference to your usage that I can study.

    ps -- The other funny thing about tensor products is that they seem to bear hardly any relation with tensors in engineering or physics. But even there, the concept of bilinearity is the thread that links the particular to the abstract.
     
    Last edited: Jan 23, 2019
  13. arfa brane call me arf Valued Senior Member

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    Ok, is it possible then to rotate one half of the Riemann sphere? I have seen someone map the eight vertices of a 2 x 2 x 2 cube puzzle to the projective plane and then show that the permutation space of the puzzle includes fractional linear transformations, if one vertex is the point at infinity. The tesselation of the Riemann sphere in my OP isn't oriented like that I think.

    I agree, it doesn't seem that useful an heuristic. Nonetheless it's common in topological algebras, like TL(x).

    I'll post some links but, for example from Kauffman and Lomanaco:
    (composition is horizontal and tensor multiplication is vertical here)
    more links: https://arxiv.org/pdf/1811.02551.pdf

    https://arxiv.org/pdf/0910.2737.pdf
     
    Last edited: Jan 23, 2019
  14. someguy1 Registered Senior Member

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    726
    I wonder if you're just quote-mining above your level of understanding. Either that, or your knowledge profoundly exceeds your ability to communicate. I looked at your first link and it didn't move me. It surely didn't define tensor products as "addition of points," which frankly is no way at all to talk about tensor products. You linked advanced papers referencing more elementary (yet still advanced) material that's already assumed.

    I'll quit while I'm behind here. If you want me to believe your knowledge is at the level of the papers you linked, you did not convince me of that fact. But it's none of my business and my small attempts to get to the truth of the matter haven't helped; and clearly I have nothing positive to add. I'll leave it alone.
     
    Last edited: Jan 23, 2019
  15. arfa brane call me arf Valued Senior Member

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    What in the figure, from the first link, I posted doesn't say that the tensor product is vertical composition, of lines with points at both ends?

    My graph is not TL which doesn't have multiple lines through points, but I'm using the same idea; the use of the otimes is meant to be used in the context I've defined. So bloody what?
     
  16. someguy1 Registered Senior Member

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    726
    I truly apologize for starting this up. Like I say I'm quitting while I'm behind. I don't doubt that things are being discussed that are far beyond me.
     
  17. arfa brane call me arf Valued Senior Member

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    But you intimated an understanding of the modular group.
    So can I fix 1/2 the Riemann sphere, and have the freedom to rotate the other 1/2?

    I don't know, it just doesn't sound like a hard question.
     
  18. someguy1 Registered Senior Member

    Messages:
    726
    I said I know what a linear fractional transformation is. I mostly showed up because of your remark about tensor products, but we didn't make any progress with that. I don't know enough geometry to be of any help.

    But what kind of rotation to do you mean? Surely not a discontinuous one, holding (say) the lower half of the sphere constant and twisting the top half. There must be some kind of distortion at the equator. I'm guessing it couldn't be analytic but my knowledge doesn't provide me with sufficient insight to be of any help here.
     
  19. arfa brane call me arf Valued Senior Member

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    Well, I don't think it's a hard thing to divide the Riemann sphere into two halves, even if its "South" pole is sitting on a plane surface. Isn't it just a matter of two sets and a couple relations?

    The projection in the plane of the equatorial blue line is a circle, you want this to be the boundary of two disjoint sets.
     
    Last edited: Jan 24, 2019
  20. someguy1 Registered Senior Member

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    726
    I have an intuition about this. I'll work on the Riemann sphere itself, rather than its projective version, because I don't know anything about projective geometry. Definite hole in my knowledge but far from the only one!

    From complex analysis I know that complex analytic functions are smooth in the sense that they are infinitely differentiable at every point. That puts a lot of restrictions on them. They are extremely well behaved.

    In the plane, an analytic function preserve angles (conformality) but I don't know how that relates to the sphere. A moment's thought tells me that if a function preserves angles in the plane, then even if it fails to preserve angles on the sphere, the angular distortion is a function of the stereographic projection. So it must either be conformal, or "almost" conformal up to some factor given by the projection that does NOT induce any badness in the function.

    Now if you think about holding the southern hemisphere constant, and twisting the northern hemisphere through some angle, what happens at the equator? It must get twisted in some non-differentiable way at some level. That is my own personal intuition. My guess is that there's no analytic function that rotates one hemisphere and keeps the other fixed. Please note this is just a guess based on the qualitative facts I remember about analytic functions, so I could well be wrong.

    Now, an LFT is a function of a complex variable of the form \(f(z) = \frac{az + b}{cz + d} \). It's clear that any such function is analytic except for a singularity at the unique point where the denominator vanishes; and that this must be a removable singularity, ie one that's "fixable." In fact the value of \(f\) at the singularity must be \(\infty\) on the Riemann sphere. Again this is from long-ago memory and I could be making it all up.

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

    The article talks about removable singularities in the plane, where \(\infty\) is not allowed, but I'm sure the idea must extend to the Riemann sphere. The bad singularities are the ones that oscillate, and that's not this. So I think my reasoning is ok.

    The main point is that analytic functions are very nice, and can't twist the sphere. And FLT's are very close to being analytic. At worst they fail to be analytic at only a single point, but it's a well-behaved blowup.

    So my intuition is that no linear fractional transformation could twist the top half of a sphere and keep the lower half constant. But of course that requires proof, which I haven't thought about at all.

    In fact here's another point of view. Working in the real numbers, can we find a smooth function that is zero on the left half of the line and nonzero to the right? Like \(f(x) = 0\) if \(x \leq 0\) and \(f(x) = x\) otherwise.

    You can see that such a function must have a corner at 0, and therefore it's not differentiable. I think something analogous must be going on at the equator.

    [In the reals you can fix this by using \(x^2\) on the right but I don't think you can do the equivalent trick in the complex numbers! Now I'm wondering if I'm talking myself into the opposite intuition].

    Here are the rest of the references.

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

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

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

    pps -- One final thought. Simpler to think about in the plane. The equator on the Riemann sphere is a circle, which can be taken to be the unit circle, on the complex plane. So we have to find an analytic function that keeps the inside of the unit circle constant, rotates the outside of the circle, and does something not too messy on the circle.

    Also I think I'm not clear now whether you're asking for an analytic solution or whether an LFT can do it, or some other condition. That wasn't clear from your initial post. A discontinuous map can certainly do it, you just tear it along the equator.
     
    Last edited: Jan 24, 2019
  21. someguy1 Registered Senior Member

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    726
    Nevermind all that, although I'll leave it up because it's valuable on its own.

    No continuous map (let alone a complex differentiable one) can rotate the outer part of the circle while keeping the inside constant. Consider any point on the circle. Every neighborhood of that point contains points that don't move (the points inside the circle) and points that move through \(\theta\) radians, where \(\theta\) is the angle of rotation.

    So you could never satisfy the \(\epsilon\)-\(\delta\) definition of continuity at that point.

    Any function that rotates the upper half of a sphere and leaves the lower half constant MUST be discontinuous. It must tear the equator.

    Ok! What do you think?

    ps -- Aha. I assumed \(\theta\) is constant. Is that true in your question? Or can the angle of rotation vary? You used the word "rotation" and that has a specific meaning, it's not just a variable kind of twist. Can you be more specific about the nature of your rotation?
     
  22. arfa brane call me arf Valued Senior Member

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    Maybe assume that the projective geometry isn't the whole picture. If instead of having some of the integers embedded in \( \mathbb R^2 \) maybe it's an embedding in \( \mathbb C \).

    The modular group is the Mobius group in the domain of the integers, a, b, c, d are integer valued. Does that make a difference since continuity isn't important (i.e. we can consider the integers to be a completely disconnected graph, with labeled vertices).

    --https://ncatlab.org/nlab/show/Moebius transformation
     
    Last edited: Jan 24, 2019
  23. someguy1 Registered Senior Member

    Messages:
    726
    I just watched a video on FLTs and the cross-product and it did bring back some memories of long-forgotten material.

    I think I have a proof that no FLT (also called LFT's) can do your inside-constant outside-rotated map. An FLT maps circles to circles. That's not hard to prove, at least it was easy for the guy on the video. Now (working on the plane again, which is much more straightforward than working on the sphere or projective sphere), if you take a circle centered on the unit circle, the inside semicircle stays where it is; but the outside gets rotated through an angle. The image isn't a circle anymore. So no fractional linear transformation can rotate just the upper half sphere or the plane outside of the unit circle.

    You lost me there. If we are focussed on your question of what are the limitations on the type of map that could rotate the top half of a sphere while leaving the bottom half fixed, then this problem is much more easily analyzed on the plane. Doing it on the sphere or the projective sphere are unnecessary complications in my opinion.

    The video I saw defined it as the set of all normalized Möbius transformations; that is, the set of all 2x2 matrices with ad - bc = 1. Limiting a,b,c and d to integers probably gives you some subgroup of that. Perhaps there's a difference in terminology between various disciplines.

    Is this question drift? I feel that we're making progress by analyzing the question of rotating the top of a sphere. I don't see any graphs or integers here. Clearly there is in some other part of your questions, but for this one limited question of twisting a sphere, I wonder if it's not simpler to just focus on that one question first. I've convinced myself that no continuous map can do it and no fractional linear transformation can do it.

    I then started wondering about what happens if you let your rotation angle be a function of the radius, so that faraway points get rotated by a given angle, but as you get closer to the unit circle there's less rotation, going to zero. So that perhaps we can preserve continuity, that seems possible. And that's where my original intuition comes in that such a map couldn't be complex analytic, though I'm far from knowing how to prove that.
     
    Last edited: Jan 24, 2019

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