September 2010 - mathfun@mailman.xmission.com

Re: [math-fun] mechanical realization of a 3-torus
by James Propp 13 Sep '10

13 Sep '10

Bill Thurston writes: >The connected components of the more general configuration space X(a,b,c,d) = >{triples of complex numbers {u,v,w} such that a*u + b*v + c*w + d = 0 can be >a sphere, torus, double torus, triple torus or quadruple torus. I assume Bill means triples of UNIT complex numbers (i.e., complex numbers of magnitude 1), since otherwise we'd get 6-2=4 degrees of freedom instead of 3-1=2. But I'm having trouble getting past the end of the first paragraph; I have a hard time picturing "rotating two adjacent bars about the midpoint of the line joining their endpoints" (though I'm sure it's a simple operation and that once I see what it is I'll have a hard time coming up with a better way to describe it!). Are there any pictures (or better yet animations) available on the web? For that matter, is the Thurston-Weeks article available anywhere online? In my previous email, I should have mentioned what happens to the set S_r = {(u,v,w): |u|=|v|=|w|=1, u+v+w=r} when r = 0: it has two S^1 components. And I'm pretty sure that this holds also for S_r when r is sufficiently close to 0. Jim Propp

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Re: [math-fun] mechanical realization of a 3-torus
by James Propp 12 Sep '10

12 Sep '10

I wrote: >I think I know what happens to the configuration >space if you pin down one of the eight points, but it'd be cool >to really feel what happens. Let me be more concrete about this, and phrase this as a puzzle: What is the topology of the set of triples of unit complex numbers (u,v,w) satisfying u+v+w = some fixed complex number z? Clearly all that matters is the magnitude of z, so we might as well assume z is a real positive number; call it r, and let S_r = {(u,v,w): |u|=|v|=|w|=1, u+v+w=r}. If r > 3, S_r is empty; if r = 3, S_r contains just the one point (1,1,1). But what about r < 3? (Quick: Do you think S_r is connected when r = 2.9, or do you think it has two components?) You may wish to imagine a linkage consisting of three struts of unit length joining points A, B, C, and D; A and D are held fixed (r units apart) while B and C are free to move as long as they stay 1 unit apart from each other and from A and D respectively. (Think of A as 0, B as u, C as u+v, and D as u+v+w.) If you hold A and D fixed, what sort of freedom do you have in moving B and C? (Note that this is related to Dan Asimov's observation that if we just want a mechanical model of the 3-torus, we don't need something as complicated as the mechanism described in my original posting.) Jim

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Re: [math-fun] mechanical realization of a 3-torus
by Dan Asimov 12 Sep '10

12 Sep '10

Oops -- that's *unit* complex numbers, not just complex numbers. Nice geometric questions. --Dan ----- Jim wrote: << Let me be more concrete about this, and phrase this as a puzzle: What is the topology of the set of triples of unit complex numbers (u,v,w) satisfying u+v+w = some fixed complex number z? Clearly all that matters is the magnitude of z, so we might as well assume z is a real positive number; call it r, and let S_r = {(u,v,w): |u|=|v|=|w|=1, u+v+w=r}. If r > 3, S_r is empty; if r = 3, S_r contains just the one point (1,1,1). But what about r < 3? (Quick: Do you think S_r is connected when r = 2.9, or do you think it has two components?) >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] mechanical realization of a 3-torus
by Dan Asimov 12 Sep '10

12 Sep '10

Let K be a complex constant. Then isn't X := {(u, v, w) in C^3 | u+v+w = K} isometric to C^2, via say f: C^2 -> X by f(u,v) = sqrt(1/2) (u, v, K-u-v) --Dan ----------------------------------- Jim wrote: << What is the topology of the set of triples of unit complex numbers (u,v,w) satisfying u+v+w = some fixed complex number z? >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Triangular+Triangular = Factorial
by Dan Asimov 12 Sep '10

12 Sep '10

Very interesting, Richard -- and Ed ! This suggests a few questions to me, possible hard ones: 1) Can we prove there are infinitely many solutions to T[x] + T[y] = z! ? How about a probabilistic heuristic? 2) Generally, given reasonably simple functions F,G: Z+ -> Z+ when can we prove there are infinitely many solutions to F(x) + F(y) = G(z) ? Likewise, what about a probabilistic "proof" ? --Dan ---------------------------------------------- Richard wrote: << . . . Solutions of x(x+1)/2 + y(y+1)/2 = z! are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained: (x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), . . . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] mechanical realization of a 3-torus
by James Propp 12 Sep '10

12 Sep '10

Fred Lunnon writes: >I haven't understood how this linkage is intended to function Start with a bar of length 2 free to pivot around its (fixed) center; call the center 0 and the endpoints +u and -u. Now add a bar of length 2 centered at (and free to pivot around) +u and another bar of length 2 centered at (and free to pivot around) -u; call the endpoints of these bars +u+v, +u-v, -u+v, and -u-v. Add a bar of length 2 joining +u+v and -u+v and another bar of length 2 joining -u+v and -u-v, to ensure that the names we've given these points are truly apt. Now add four more bars of length 2 centered at +u+v, +u-v, -u+v, and -u-v; call the endpoints of these bars +u+v+w, +u+v-w, +u-v+w, +u-v-w, -u+v+w, -u+v-w, -u-v+w, and -u-v-w, and add eight more bars of length 2 joining +u+v+w to +u-v+w, etc. >but have you considered a simulation in Mathematica or Maple; >or more satisfactorily, Java / Javaview or GeomView? I hadn't considered this (since I don't have the relevant programming skills), but it's a good idea, and if anyone implements this please let me know! Jim

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Re: [math-fun] mechanical realization of a 3-torus
by Dan Asimov 10 Sep '10

10 Sep '10

Interesting. But I think the clearest mechanical realization of a 3-torus would be three independently rotating rods. The configuration space is then all ordered triples of angles, which is immediately S^1 x S^1 x S^1 = T^3. This still leaves a wide variety of possibilities for how this could be realized in the plane or 3-space. One possibility is to connect the ends of the rods by elastic bands of 3 different colors, so that the configuration space is displayed as all the resulting triangles. The rods could be set to mechanically rotate at three linearly independent speeds, so that the 3 angles trace out a curve in T^3 that's eventually arbitrarily close to every point of it. (There are many ways to do this so the elastic bands don't get tangled up.) --Dan Jim wrote: << Note that one could describe this linkage simply as a mechanical realization of a 3-torus (hence the title of this post). >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] mechanical realization of a 3-torus
by James Propp 10 Sep '10

10 Sep '10

I have an idea for a physical mathematical demo, and I'm wondering if anyone has any thoughts about it (or any interest in actually building it). Spitzer's "random flight theorem" says that a sum of n independent random unit vectors in the plane has a probability of exactly 1/(n+1) of having magnitude less than 1. In particular, the chance that the sum of 3 independent random unit vectors has magnitude less than 1 is exactly 1/4. It so happens that this particular fact can be deduced from the following geometrical fact, which may or may not be new: If u, v, and w are unit vectors, then exactly two of the eight vectors +u+v+w, +u+v-w, +u-v+w, +u-v-w, -u+v+w, -u+v-w, -u-v+w, -u-v-w have magnitude less than 1 (except in the case when some of these vectors have magnitude exactly 1). There's a nice way to physically model this "two out of eight" fact with a linkage consisting of bars of length 2, so that eight points move jointly so as to trace out the configuration space {(+u+v+w, +u+v-w, +u-v+w, +u-v-w, -u+v+w, -u+v-w, -u-v+w, -u-v-w): |u|=|v|=|w|=1}. Now imagine that the bars are made of a transparent material like plexiglass, and that the model has a backing that shows a disk of radius 1 centered at 0. As you move around in configuration space, there are always exactly two points that lie inside the disk, except for the borderline case where four (or more) points lie on the boundary of the disk. I'm wondering if this might make a good object for the Museum of Mathematics, or maybe just a fun object to own and to display in my office. I think I know what happens to the configuration space if you pin down one of the eight points, but it'd be cool to really feel what happens. Note that one could describe this linkage simply as a mechanical realization of a 3-torus (hence the title of this post). Jim

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Re: [math-fun] Ever seen a spacefilled pentagon?
by Bill Gosper 10 Sep '10

10 Sep '10

WFredLunnon>If you're mapping a (filled) square onto a disc or whatever, why is there an extraneous solid black region at the bottom at the bottom of each drawing? Good question. I hate drawing edges as "lines", i.e. strands of darkened pixels, since the only proper drawing of a line segment is invisible. Much better is to present them as the edges of colored regions. (There was once a World Team Tennis league whose courts were multicolored rectangles instead of outlined with stripes.) So I made the point sequences into polygons. But they're all just various samplings of the Peano "curve", which doesn't close, so I needed to artificially close them. But then I needed to detour significantly downward else the closing segment would cut the figure in numerous places making spurious vertices and color reversals. --rwg "How else would you do it?"--J. von Neumann Actually, the Peano images of [1/6,1/2] and [1/6,5/6] *do* close, but they fill weird looking regions which would be painful to map onto a disk. If you can get at a Mathematica, you can easily try this. Just start the Table iterators 1/6 of the way through and stop (roughly) 5/6 of the way. All the code is there.

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[math-fun] Ever seen a spacefilled pentagon?
by Bill Gosper 10 Sep '10

10 Sep '10

Yesterday I told tutor Julian an idea for spacefilling a pentagon, and he said I could just conformally map Peano's filled square onto whatever, but then he got into it and helped make www.tweedledum.com/rwg/pentagonfill.pdf . It self-contacts, but the median curve doesn't. Note the density variations, which become unbounded in the limit. I don't have a fix, but I can turn a slight variant of this one into one of those exact continuous maps of the rational unit interval onto some dense set of algebraic points. This would permit sampling at various frequencies and phases, which has proven interesting in previous cases (www.tweedledum.com/rwg/sampeano.htm, www.tweedledum.com/rwg/sam306090.htm ). --rwg

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