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Week 214 (John Baez)

  1. Apr 26, 2005 #1


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    Regarding This Week's Finds 214

    ...I tried to post something to the SPR thread but I guess there were too many equations. Anyway, here's a PF thread. After reading TWF, I thought of the following quote:

    Continuous geometries...are a generalization of complex projective geometry somewhat in the way that Hilbert space is a generalization of finite dimensional Euclidean space Halperin (1960)

    This quote appears in the book Orthomodular Lattices by G. Kalmbach, on page 191. Just below this quote is the example of the Fano plane and its associated complete modular ortholattice on 16 points.

    These are the sorts of lattices that quantum logicians like.

    Tony Smith, on the page linked by John Baez, mentions the Golden Ratio in connection with the Fano plane. John has mentioned Fibonacci numbers. I thought it would therefore be interesting to bring up the following observations.

    In the paper

    Geometrical measurements in three dimensional quantum gravity,
    J.W. Barrett, http://xxx.lanl.gov/abs/gr-qc/0203018

    John Barrett discusses the following remarkable Fourier transform for [itex]6j[/itex] symbols

    [tex]\frac{1}{N} \sum_{j_{1} \cdots j_{6}} [ j_{4}, j_{5}, j_{6};
    j_{1}, j_{2}, j_{3} ]^{2} H(j_{1},i_{1}) \cdots H(j_{6},i_{6}) = [
    i_{1}, i_{2}, i_{3}; i_{4}, i_{5}, i_{6} ]^{2}[/tex]

    [itex]N[/itex] is a normalisation constant. Choosing [itex]q =
    e^{\frac{i \pi}{5}}[/itex] as in

    A modular functor which is universal for quantum computation,
    M. Freedman M. Larsen Z. Wang,

    gives allowable spin values [itex]j \in 0, \frac{1}{2}, 1, \frac{3}{2}[/itex]. The kernel function is given by the Hopf link invariant

    [tex]H(j,i) = (-1)^{2i + 2j} \cdot \frac{\sin \frac{\pi}{5}(2j +
    1)(2i + 1)}{\sin \frac{\pi}{5}}[/tex]

    Observe that [itex]H[/itex] only takes values [itex]\pm \phi , \pm 1[/itex] where [itex]\phi = 1.61803399 \cdots[/itex] is the golden ratio

    [tex]\phi = \frac{\sin \frac{2 \pi}{5}}{\sin \frac{\pi}{5}}[/tex]

    The full kernel therefore takes values in powers of [itex]\phi[/itex], or in other words the truncated Fibonacci sequence

    [tex]1 , \phi , \phi + 1 , 2 \phi
    + 1 , 3 \phi + 2 , 5 \phi + 3 , 8 \phi + 5[/tex]

    of 7 terms.

    Kea :smile:
    Last edited: Apr 27, 2005
  2. jcsd
  3. Apr 27, 2005 #2


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    Now over on SPR a Kiwi mentioned that 3-holed surfaces are a bit like tubular tetrahedra. Actually, think of a tetrahedron made of untwisted ribbons joined together at the vertices. This is exactly the same as a four punctured sphere, with a little stretching and squeezing. Now a ribbon triangle is dual to a vertex of 3 lines, which becomes a trouser diagram, ie. 3 tubes meeting at a vertex. This is a three punctured sphere. Putting 4 of these trousers (qubits) together turns the tetrahedron into the 3-holed surface.
  4. Apr 27, 2005 #3


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    Three leafed trees with a root (associativity trees), if thickened, become four punctured spheres too. And associativity arrows as tetrahedra naturally appear as maps joining two 4-holed spheres.
  5. Apr 27, 2005 #4

    This is interesting, Kea. The tetrahedron, or sphere with four holes, has four qubits, or three hole spheres, as vertices. One might wonder if the three hole sphere would have two hole spheres as vertices, but I have been dissapointed trying to visualize that. Instead I see a three hole sphere as made up of three ribbons joined at two vertices, each vertice being another three sphere.

    I do wonder about the idea that a ribbon is equivalent to a tube (as in the rather sudden appearance of trousers above). In fact a ribbon is not exactly equivalent to a line either. Is it justifiable to morph a one dimensional line, as in a tetrahedron, into a two dimensional ribbon, into a three dimensional qubit, as you have shown here? Can we just assume more dimensions at will?

    In the other direction, one might wonder if a five hole sphere should have four hole spheres as vertices. Or, one might wonder, how many holes must a sphere have before the next new hole requires the appearance of at least one four hole sphere as a vertice? I am thinking from the perspective of the cubeoctahedral isomatrix, which would have fourteen holes, and twelve vertices, all twelve vertices being equivalent to four hole spheres.

    What branch of mathematics is this anyway? It seems like some kind of topology.


  6. Apr 27, 2005 #5


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    Not equivalent. Dual. Maybe Kea can explain that some more - I don't feel up to it.

    You are right; this is topology.
  7. Apr 27, 2005 #6


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    NC there was one nice elementary fact in what Kea said that would not be hard for you to understand by drawing a quick picture.

    a charming thing with Kea though is that if you understand an inch she will immediately scamper off a mile or more into the blue horizon. So it is a bit risky to say that you understand the initially offered inch. :smile:

    I believe if you simply draw a tetrahedron frame, not the solid block but just the airy frame as if made of aluminum tubing.

    I think if you will simply draw this tubular tet, maybe with rather fat tubing,
    that you will see that it can be mooshed into a

    1. sphere with 3 handles or a
    2. surface of a donut with 3 holes

    you may see this so quickly that it would be foolish of me to say more at this point
  8. Apr 27, 2005 #7
    Thank you selfAdjoint. I noticed the use of the word Dual, when Kea said that the triangle formed by the ribbons is dual to the three ribbons converging on the vertex. I guess I don't really know what this use really means. I was thinking when I first read it that in the tetrahedron model, the mathematics of the triangular face is related to the mathematics of the three-legged vertice. You can either describe the tet by describing a face and its relationship to the other faces, or by describing the vertex and its relation to the other vertices. So the tet, or other form more generally, has these two descriptions, dual descriptions. One describes the surfaces, the other describes the vertices and edges. Since both maths are describing the same object, the descriptions, however different they may appear in formula, are then dual.

    Did I get this right?

    I hope you are feeling better soon, selfAdjoint. Your contributions have always been most valuable to me.


  9. Apr 27, 2005 #8


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    it is in fact foolish to say more for NC benefit, but in case some other reader looks in and is puzzled maybe you can think of those irritating people who sculpt cute dogs out of sausage-shape balloons at children parties.

    imagine you had something better than one of those cute red balloon dogs, namely you had an inflatable balloon forming a tubular tet.

    then because it is stretchy you could set the thing down on the table, like a 3-side pyramid frame with open sides and open bottom, and you could squash the apex down into the middle of the open bottom

    so then it would be like a flat tubey triangle with another point in the middle, and that centerpoint being joined by tubes to the triangles vertices.

    and if you think what that looks like it is a surface of 3-hole donut. draw a picture to be sure.
  10. Apr 27, 2005 #9

    Thank you Marcus. We will all wear tube-tet hats at the 8piG roast.

  11. Apr 27, 2005 #10


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    I am sure looking forward to that 8piG roast. The idea of dual in 2D is that faces go to points, edges to edges, as in the picture.

    I must confess, I mentioned ribbons in the hope we could start discussing tortile tensor categories....


    Attached Files:

  12. Apr 28, 2005 #11
    Twisted tensors, huh? I do want to know more, but I keep running off in my own odd directions. One of those old Chinese guys says you can't poor more tea into a full cup.

    So I must empty my cup and resolve to listen more and babble less.

    Please tell me more about tortile tensor catagoriess, Kea. I have never heard of them before.

    Thank you,

  13. Apr 28, 2005 #12


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    Hi Richard! Kea is telling one of those 'get it' jokes that aussies are so fond of terrifying us with! Good one Kea!!
  14. Apr 28, 2005 #13
    Hi Chronos

    I have been googling. I don't usually think of macreme as an Australian art, but I suppose it could be useful if one wanted to try to catch a mountain parrot. Who would have thought a thesis on macreme would garner so many citations? I am very impressed by the stringyness. Still, many birds are interested in string. Why not a parrot after all?

    I wonder if I asked the author for a copy of the thesis if I would "get it?"

  15. Apr 28, 2005 #14


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    Mountain parrots are quite adept at cutting ropes, given the opportunity. An old explorer tried to catch one with a wire snare hidden under leaves. The Kea dug under the leaves and flicked the wire out of the way, and took off with the bounty unharmed.

    Think about a knot (a closed loop) embedded in ordinary 3-dimensional space. If one wishes to compute a knot invariant using physics (Chern-Simons path integrals) then one needs to thicken the knot into a tube. The tube may be twisted an integral number of times, but this winding number is more clearly represented by the twisting of a ribbon running along a diameter of the tube.
  16. Apr 29, 2005 #15
    Hi Kea

    When does one wish to compute a knot invariant using physics (Chern-Simons path integrals)? I will try to read up on this.

    Do I assume the ribbon wraps edge to edge, covering the surface of the tube along which it makes its transit? Some number of Pi per length of tube, I guess. Ribbon tiles the tube at some angle dependent on width of ribbon.....as width of ribbon approaches zero, angle gets steeper to 90? Maximum width of ribbon is equal to circumfrence of tube. After that the plane of the ribbon width could be rolled up on the tube like plastic wrap on one of those megaphones I sometimes use to drive Tucker into frantic fits of confusion. Very entertaining. Sorry.

    So angle of wrap could be expressed as a width of the ribbon in ratio to the circumfrence of the tube. If you know the width of the ribbon and the angle, you know the circumfrence. If you know the width and the number of revolutions, you know the length along the tube. All this assuming a staight tube. If the tube is bent, the ribbon has to be thinner in places, wider in others, and if you unwrapped it, you would see its wave form. If the tube was bent sufficiently, the width of the ribbon in places would be forced to zero and the ribbon would become a set of spindle shapes.

    What fun.

    It is interesting how when you twist a rope (maybe you are winding it up in a coil after being frustrated in the great Kea Trap debacle) the free end likes to roll back on itself and form spirol twists with a loop. I have often thought that could be a way to get particals from strings.

    So back to the topic. You take a dog annoying megaphone (cardboard tube) and you wrap a pretty red christmas ribbon around it, starting at one end and going edge to edge to the other end. You then carefully grasp both ends of the ribbon and remove the tube, then stretch the ribbon to some rather not too tight length, several many times longer than the original tube. Now the ribbon is a spirol. You look at the spirol in a sharp single source light and it looks like a string of beads. At each twist, the ribbon goes through angles to the light, at regular phases going from zero width to the full width of the ribbon. A string of beads, rather like the strobe image of a ball moving through space when the velocity of the ball is twice the diameter of the ball per each flash interval. If you pull the ribbon tight, it turns into a tube again, a much narrower much longer tube. There is a relatonship between long tube and the original shorter one....they are both made of the same material. And there would be some math to describe the ends of the ribbon tube also, I should think.

    Well I am sure you have much more compact and precise language to talk about these interesting things. Surely these kinds of images are what we need for geometry in higher dimensions. Now there is an explore I can get into.

    An old explorer, actually a refugee but with a hint of pride, from a culture that was rapidly outgrowing its old clothes, once got tangled up in his nets while trying to mist-trap migrating birds. He hung upside down from the gently swaying branches for some time, his face now a few inches, now a few feet from the litter of the forest floor.

    In the third story canopy of the forest above him, a bright eye watched, amused, but also intrigued by the unusual sight. Now there was a bit of rope that could be interesting. Bright feathers and the rush of wings from there to here.

    So. I was going to tell you about the sound of raven wings on a cold night in midwinter when the silence in the deep woods was a bell, and the whoosh whoosh whoosh of air through feather was the beginning and the end of the universe, terrible tingling fear and the bright cold stars dying behind wide dark wings. Whoosh. A raven. Instead, the above.

    It is spring in Minnesota, the first buds pushing out tips of green on the red dosier dogwood brush that hugs the ditches. So it is Autumn in New Zealand. I hope your harvest is the beauty in the bounty.

    Blessed be,

    Last edited by a moderator: Apr 29, 2005
  17. Apr 29, 2005 #16


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    Hi nightcleaner

    Actually, the ribbon runs along the center of the tube. In other words, for each disc slice of the tube, choose a diameter so that they all meet up smoothly as you run around the tube.

    I was disappointed with the movie. Were you?

  18. Apr 30, 2005 #17
    Hi Kea
    Do the two edges of the ribbon make a double helix?

    I havn't seen the movie, but found the trailer by google a few days ago. The trailer is mysterious and provocative, gives me shivers. Someone from a far universe reporting the end of the world on a failing radio......what is she saying anyway?

  19. Apr 30, 2005 #18


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    Yes if you twist a ribbon as Kea describes, the edges will make a double helix.

    What movie is this you two are discussing?
  20. Apr 30, 2005 #19
    Hi selfAdjoint

    That would be telling.

    Interesting things happen to a twisted ribbon if you pull one edge of the double helix into a straight line, forcing the other edge to wind around it.

  21. Apr 30, 2005 #20


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    As usual, nightcleaner has his own wonderful and illuminating perspective on things. If one traces each end of the diameter ribbon around the tube one gets two curves, just like the boundary of a wrapped ribbon. But the number of windings will be different. Nightcleaner, what is the relationship?

    Actually, I've only seen the first movie - it just came out here. The whooshing story is indeed creepier. Sounds more enjoyable! I'll try to view it from a distant planet.

    And yes, the desiduous trees in the city here are blazing autumn in the low morning sun. Personally I prefer the native evergreens, lush under a weight of snow.

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