 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
Math Forum / Mathematics / Research / January 2007rotation groups of polyhedra
As far as I can tell, the rotations that you can do to the platonic solids that leave them looking the same is rotating about either the center of a face or a vertex. For a cube, you can rotate each face by either 1/4 rotation, 1/2 rotation, or 3/4 rotation. If you rotate by 4/4 or 360 degees, that's just the identity. You can rotate around each vertex by either 1/3 rotation or 2/3 rotation. So there are 3 possible rotations for each face, and 2 possible rotations for each vertex, which gives you 6 x 3 = 18 8 x 2 = 16 18 + 16 = 34 If you like, you could add the identity, so that's 35. However, in Felix Klein's book, he describes the number of rotations for each polyhedron as being some number larger than the numbers you get from applying the above calculation to each polyhedron. What other possible rotations are there? David >As far as I can tell, the rotations that you can do to the platonic >solids that leave them looking the same is rotating about either the >center of a face or a vertex. That's not *quite* a description, though it's clear enough what you mean. ... >What other >possible rotations are there? Allowing myself the same degree of slop: rotating "about the center of an edge". Lee Rudolph > As far as I can tell, the rotations that you can do to the platonic > solids that leave them looking the same is rotating about either the > center of a face or a vertex. Rotating about the center of a face you get the same thing as rotation bout the center of the opposite face, so you are counting all of yours twice. Why not also allow rotation about the center of an edge? > For a cube, you can rotate each face by > either 1/4 rotation, 1/2 rotation, or 3/4 rotation. If you rotate by [quoted text clipped - 10 lines] > > If you like, you could add the identity, so that's 35. Actually, that's 17+1=18; plus 12 / 2 = 6 for the edges, so I get a total of 24. Does Klein include reflections also? Then we will have 48. > However, in Felix Klein's book, he describes the number of rotations > for each polyhedron as being some number larger than the numbers you > get from applying the above calculation to each polyhedron. What other > possible rotations are there? > > David  SignatureG. A. Edgar http://www.math.ohio-state.edu/~edgar/ >As far as I can tell, the rotations that you can do to the platonic >solids that leave them looking the same is rotating about either the [quoted text clipped - 20 lines] >David > Perhaps a newsgroup post I made some weeks ago is helpful. The topic is the homomorphism of S4 onto S3, which is explained by means of rotations of the cube. The enumeration is easily done: 1 non-rotation; 3 plus 6 half-turns; 6 quarter-turns; 8 120deg-turns. Voilà! Ciao: Johan E. Mebius ( my post in news:sci.math of November 16th 2007 - Subject: Re: Surprisingly hard group theory problem- ) ============================================================================ In my opinion the most beautiful illustration of the S4 - V4 - S3 issue is given by the group of rotational congruence transformations of the cube. This group is isomorphic to S4 by permuting the body diagonals 1, 2, 3, 4 of the cube. Observe what happens with the line segments connecting the three pairs of opposite centres of faces. Denote them by A, B, C. As the body diagonals are being permuted in all ways, A, B and C are also permuted in all ways, and furthermore, composition of permutations of 1, 2, 3, 4 corresponds to composition of permutations of A, B, C. So there is a homomorphism of S4 onto S3. Half-turns around each of the axes A, B, C reverse the directions of the other two axes. They do not permute the set A, B, C. Observe what happens in the meantime to the body diagonals: they are permuted pair-pairwise: (12)(34), (13)(42), (14)(23). Here one sees Klein's four-group V4 faithfully represented as a normal subgroup of S4. It is the kernel of the homomorphism mentioned previously. By now it is not too difficult to visualize and to list a complete homomorphism S4 -> S3: all three-cycles of 1, 2, 3, 4 correspond to rotations over 120 degrees around the body diagonals. In S3 they corresopond to the three-cycles (ABC) and (ACB). In passing one observes here the other famous homomorphism related to Klein's four-group: A4 -> C3 with V4 as its kernel and C3 as its factor group. Four-cycles of 1, 2, 3, 4 correspond to quarter-turns around A, B, C, and in S3 to the transpositions (AB), (BC), (CA). JEMebius a écrit : > >As far as I can tell, the rotations that you can do to the platonic > >solids that leave them looking the same is rotating about either the [quoted text clipped - 10 lines] > > > >18 + 16 = 34 I have very recently read in the issue 'Scientific American' a very nice way of dealing with this kind of problem; it was intitled: Miraculous Burnside's lemma Alain > >If you like, you could add the identity, so that's 35. > > [quoted text clipped - 47 lines] > Four-cycles of 1, 2, 3, 4 correspond to quarter-turns around A, B, C, > and in S3 to the transpositions (AB), (BC), (CA). |
Reply to this Message |
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2010 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.