January 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Moebius madness on Youtube
by Eugene Salamin 13 Jan '08

13 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Saturday, January 12, 2008 2:21:36 PM Subject: Re: [math-fun] Moebius madness on Youtube On 1/12/08, Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > > > ----- Original Message ---- > From: Fred lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Sent: Friday, January 11, 2008 12:54:07 PM > Subject: Re: [math-fun] Moebius madness on Youtube > > > Nicely done. And it grieves me to have to admit that, although I've > been > concerned with these groups for a good few years, I had no idea that > the > isomorphism of the Moebius group in n-space with the Euclidean group in > (n+1)-space had such a straightforward demonstration! WFL > > On 1/11/08, Henry Baker <hbaker1(a)pipeline.com> wrote: > > Very cool animation of Moebius transformations: > > > > http://youtube.com/watch?v=JX3VmDgiFnY > > > Those two groups are not isomorphic. The Euclidean group has a normal subgroup > (the translations), while the Mobius group is semisimple. > > Gene About ten years ago, I read somewhere (Coolidge?) that they were isomorphic (qua Lie groups, presumably); I remember being very surprised at the time, but never got around to investigating the matter until now. If this claim turns out to be wrong, that extricates me very conveniently from my elephant trap. I'd also have the consolation of company: Coolidge relates with sadistic relish how an unfortunate Ph. D. student by the name of Lohrl expended considerable time and effort in a futile attempt to prove that Moebius n-space and Lie-sphere (n-1)-space groups were isomorphic. [In that case at least, it seems perfectly obvious that they cannot possibly be; since only the latter group has a 1-parameter normal subgroup, to wit the offset transformations.] In the present case --- Moebius n-space and Euclidean (n+1)-space, continuous components say --- we have a transformation (stereographic projection) inducing a smooth map from isometries in (n+1)-space acting on the projecting sphere (though its north pole remains northwards), to conformalities in n-space. This map is a homomorphism, composition mapping to composition (I _think_ this is obvious!). It must surely be invertible, since its codomain contains generators for the whole Moebius group: translations, rotations, (proper) inversions. Therefore the groups are isomorphic ... no? Unless of course, there's a mistake somewhere in the reasoning above. And if there isn't, I'm right back there, still sitting in deep do-dos ... Fred Lunnon _______________________________________________ I should have been more specific. The Euclidean group in 3-space contains the 3-dimensional abelian normal subgroup of translations. The Mobius group on the 2-sphere (thanks, Dan), being semisimple cannot contain a solvable normal subgroup. The Mobius group has a 2-dimensional abelian subgroup (of translations preserving infinity), but it is not normal, since the fixed point, infinity, can be conjugated to any other point. In the YouTube video, translation of the sphere normal to the plane induced the magnification Mobius transformation. But magnification by factor m and translation by t do not commute: m(z+t) /= mz + t. So the YouTube map is not an isomorphism. (Is it of any interest to think about mappings between groups that are not homomorphisms?) Fred, what is a Lie sphere? While Googling for the answer, I discovered the following Indra's Pearls web site. http://klein.math.okstate.edu/IndrasPearls/ Gene ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs

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Re: [math-fun] Moebius madness on Youtube
by Dan Asimov 12 Jan '08

12 Jan '08

<< [The Moebius group on n-space] is also isomorphic to Lorentz group in (n+2)-space, i.e. SO(n+1,1). To demo this isomorphism, just fly through space at relativistic velocities. A Lorentz transformation of your spacecraft induces a mapping of the celestial sphere which is a Moebius transformation. >> I suspect the "Moebius group on n-space" is really SO(n+1,1)/{+-I}rather than SO(n+1,1). As I understand it, the original Moebius group is defined as all conformal bijections of the Riemann sphere = S^2 = C u {oo} with itself (conformal meaning orientation-preserving as well as angle-preserving). This is the group of all linear fractional transformations given by z -> (az+b)/(cz+d), a,b,c,d complex. After eliminating repeats, this can be seen to be the group PSL(2,C) = SL(2,C)/{+-I}. And PSL(2,C) can be identified with the orientation-preserving isometries of hyperbolic 3-space. This group is connected, so it must be SO(3,1)/{+-I}. (So in fact it's not natural to say the original Moebius group acts on 2-space, since many transformations take some point to oo and take oo to a (finite) point.) So a natural definition of the "Moebius group on n-space" would be the orientation-preserving conformal transformations of the n-sphere, or alternatively the orientation-preserving isometries of hyperbolic (n+1)-space. One could also make a good case to ignore the orientation-preserving condition, since that may be seen in the original Moebius group as an artifact of the 2-dimensional sphere that it acts on conformally. With orientation-preserving, I think the "n-dimensional Moebius group" of conformal transformation of S^n can be identified with SO(n+1,1)/{+-I}. Without the orientation-preserving condition, I think it would then be O(n+1,1)/{+-I}. (It seems that most authors define the Lorentz group on (n+2)-space as just O(n+1,1).) --Dan _____________________________________________________________________ "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] Moebius madness on Youtube
by Eugene Salamin 12 Jan '08

12 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Friday, January 11, 2008 12:54:07 PM Subject: Re: [math-fun] Moebius madness on Youtube Nicely done. And it grieves me to have to admit that, although I've been concerned with these groups for a good few years, I had no idea that the isomorphism of the Moebius group in n-space with the Euclidean group in (n+1)-space had such a straightforward demonstration! WFL On 1/11/08, Henry Baker <hbaker1(a)pipeline.com> wrote: > Very cool animation of Moebius transformations: > > http://youtube.com/watch?v=JX3VmDgiFnY Those two groups are not isomorphic. The Euclidean group has a normal subgroup (the translations), while the Mobius group is semisimple. Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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[math-fun] Moebius madness on Youtube
by Henry Baker 12 Jan '08

12 Jan '08

Very cool animation of Moebius transformations: http://youtube.com/watch?v=JX3VmDgiFnY

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Re: [math-fun] Moebius madness on Youtube
by Eugene Salamin 11 Jan '08

11 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Friday, January 11, 2008 12:54:07 PM Subject: Re: [math-fun] Moebius madness on Youtube Nicely done. And it grieves me to have to admit that, although I've been concerned with these groups for a good few years, I had no idea that the isomorphism of the Moebius group in n-space with the Euclidean group in (n+1)-space had such a straightforward demonstration! WFL On 1/11/08, Henry Baker <hbaker1(a)pipeline.com> wrote: > Very cool animation of Moebius transformations: > > http://youtube.com/watch?v=JX3VmDgiFnY And, it is also isomorphic to Lorentz group in (n+2)-space, i.e. SO(n+1,1). To demo this isomorphism, just fly through space at relativistic velocities. A Lorentz transformation of your spacecraft induces a mapping of the celestial sphere which is a Moebius transformation. Gene ____________________________________________________________________________________ Looking for last minute shopping deals? Find them fast with Yahoo! Search. http://tools.search.yahoo.com/newsearch/category.php?category=shopping

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Re: [math-fun] Maple 11
by Eugene Salamin 08 Jan '08

08 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Monday, January 7, 2008 11:47:53 AM Subject: Re: [math-fun] xkcd points out dangers of math fun On 1/7/08, Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > ... > Maple 11 can do these. I will say that Maple 11 is a winner, while Maple 9 is a loser. > ... That's useful to know. Maple 10 was less appallingly flaky than Maple 9, but their @£$% licensing software kept clobbering it. I might try upgrading again, I suppose (groan!) > The denominators in the diagonal terms, R[p,p] seem to fit the pattern 1*3*5*...*(2p -1). As I remarked earlier (attachment Jan 3rd), nobody seems previously to have noticed that u^{kk} = ( 1 + 1/3 + 1/5 + ... + 1/(2k-1) ) 2/\pi --- all the remaining values follow from these and u^{01} = 1/2, via \del^2 u^{kl} = 0. WFL _______________________________________________ Check maplesoftsoft.com. See if you can qualify for the student edition. It's identical to the professional edition, but costs only $160 or so. When I bought mine, I had to provide a school and student ID number, which I could do because I'm a distance student at U of Arizona. Do you have a proof for the value of u^{kk} ? Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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Re: [math-fun] xkcd points out dangers of math fun
by Eugene Salamin 07 Jan '08

07 Jan '08

On a square grid of 1 ohm resistors, the effective resistance between the node at (0,0) and the node at (p,q) is given by the following expression. R[p,q] = (1/(4 pi^2)) int (F(x,y), x=-pi..pi, y=-pi..pi), F(x,y) = [1 - cos(p x + q y)] / [2 - (cos x + cos y)]. Maple 11 can do these. I will say that Maple 11 is a winner, while Maple 9 is a loser. Here are some low-order results. The pi's are always in the denominator. R[1,0] 1/2 R[1,1] 2/pi R[2,0] 2 - 4/pi R[2,1] 4/pi - 1/2 R[2,2] 8/3pi R[3,0] 17/2 - 24/pi R[3,1] 46/3pi - 4 R[3,2] 1/2 + 4/3pi R[3,3] 46/15pi R[4,0] 40-368/3pi R[4,1] 80/pi - 49/2 R[4,2] 6 - 236/15pi R[4,3] 24/5pi - 1/2 R[4,4] 352/105pi R[5,5] 1126/315pi R[6,6] 13016/3465pi R[7,7] 176138/45045pi The denominators in the diagonal terms, R[p,p] seem to fit the pattern 1*3*5*...*(2p -1). Gene ____________________________________________________________________________________ Looking for last minute shopping deals? Find them fast with Yahoo! Search. http://tools.search.yahoo.com/newsearch/category.php?category=shopping

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Re: [math-fun] Fwd: septoku mother lode
by Dan Asimov 06 Jan '08

06 Jan '08

Thane wrote: << [Lots of interesting stuff & links re septoku] >> Um, I wonder if we can change the name to, say, hexoku? Or else heptoku? Before it gets carved in stone? (I find the word septoku to be uncomfortably close to seppuku.) --Dan

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Re: [math-fun] xkcd points out dangers of math fun
by Eugene Salamin 06 Jan '08

06 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Sunday, January 6, 2008 12:50:21 PM Subject: Re: [math-fun] xkcd points out dangers of math fun On 1/6/08, Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > ... > On a square grid of 1 ohm resistors, the effective resistance between the node at (0,0) and the node at (p,q) is given by the following expression. > > R[p,q] = (1/(4 pi^2)) int (F(x,y), x=-pi..pi, y=-pi..pi), > > F(x,y) = [1 - cos(p x + q y)] / [2 - (cos x + cos y)]. > This is pretty much identical to formula (18) in Cserti, for the 2-D case. No doubt one could be convinced of its correctness simply by checking that it satisfied the Kirchhoff-inspired equations I quoted earlier [ my u^{kl} being your R[p,q] ] : [Lat1] u^{00} = 0; [Lat2] u^{01} = 1/2; [Lat3] \del^2 u^{kl} = 0 for (k,l) <> (0,0); where \del^2 u^{kl} == u^{k,l-1} + u^{k,l+1} + u^{k-1,l} + u^{k+1,l} - 4 u_{kl} . But how did you arrive at it; and exactly what assumptions did you make in order to do so? Fred Lunnon _____________________________________________ We have the Kirchhoff equations (1/4)(v[j+1,k] + v[j-1,k] + v[j,k+1] + v[j,k-1]) - v[j,k] = 0, except at the nodes where current enters or exits. If 1 A of current exits at a node, the current is divided equally among the 4 resistors, and the 4 adjacent nodes will be (1/4) V higher. So at that node, the RHS should be (1/4). If 1 A exits at (p,q) and enters at (r,s), the RHS is (1/4)(delta(j,p) delta(k,q) - delta(j,r) delta(k,s)). Now define the Fourier series V(x,y) = sum(v[j,k] exp(i j x + i k y)) summed over all integers j and k. Applying this to the Kirchhoff equations, we can solve for V(x,y). It is a fraction whose numerator is exp(i r x + i s y) - exp(i p x + i q y), and whose denominator is 4 - 2(cos x + cos y). To calculate v[j,k], multiply V(x,y) by exp(- i j x - i k y)/(2 pi)^2 and integrate over (x,y) in [-pi, pi]^2. The effective resistance between (r,s) and (p,q) is equal to the voltage difference v[r,s] - v[p,q]. You can set (p,q) = (0,0), and get my solution, where I replaced (r,s) by (p,q). Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ

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[math-fun] Fwd: septoku mother lode
by Thane Plambeck 06 Jan '08

06 Jan '08

here's some fun analysis of septoku, a sudoku variant invented by bruce oberg that's played on a hexagon-tiled board ---------- Forwarded message ---------- From: Bruce Oberg <bruce(a)oberg.org> Date: Jan 4, 2008 4:20 PM Subject: septoku mother lode To: Thane Plambeck <tplambeck(a)gmail.com>, Nick Baxter <nickb(a)baxterweb.com>, Matt Scott <matt(a)loosecannonstudios.com>, Chris Zimmerman <chrisz(a)suckerpunch.com>, David Nixon <david(a)docnixon.com>, Amit Khetan <amitkhet(a)gmail.com>, David Moulton <moulton(a)idaccr.org> Cc: George Bell <gibell(a)comcast.net> hi y'all george bell has done a bunch of research into septoku, and produced some very interesting results. he's found that there are only 6 unique septoku boards (modulo substitution, rotation, and reflection). with this in hand, he's built some puzzles with only 6 seeds (they are very difficult to solve). i've posted his paper on my website: http://oberg.org/septoku_theorems.pdf with these results, all the questions i posed about septoku have been answered. george will claim $154 for answering questions #1 and #4 first. questions #2 and #3 had already been solved by earlier submissions: http://oberg.org/septoku_centers_proof.pdf (David Nixon) http://oberg.org/septoku_corners_proof.pdf (Amit Khetan and David Moulton) my original paper is here: http://oberg.org/septoku_g4g7.pdf thanks to everyone for spending some time on my little doodle. bruce oberg -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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