2!!!

]]>Thanks! I didn’t go into mathematics to make money, and in fact I’m a lot more well-off than I ever expected to be. Given all that, it seems ridiculous to charge money for my books… especially since these books never made much money in the first place, and illegal Russian websites offer most scientific books for free anyway.

But anyway, I’m glad you’re enjoying all the material on that website.

]]>I was really inspired and overwhelmed by the website you and your colleagues run (http://math.ucr.edu/home/baez/README.html). I think that it is really noble of you to just let the general public access the.pdf files of your books for free and to offer your valuable advice at no cost, when you could have charged them for it. I wish you and your colleagues all the best, and once again, thank you so much for taking your valuable time to maintain and update this blog and the above website. ]]>

I don’t see how to subdivide the cube’s surface into polygons, one for each of its 24 rotational symmetries, such that the symmetries corresponding to the polygons touching the polygon for the identity act to map any other polygon to one that touches it. You seem to be hinting that you can do it.

I _can_ do it for rotational/reflectional symmetries of the cube.

Some relevant bits of topology:

The rotation group of Euclidean 3-space is SO(3), which is topologically the projective 3-space The rotation/reflection group of Euclidean 3-space is O(3), which is topologically two disjoint copies of the projective 3-space

Jim Stasheff proved some very general theorems saying stuff like: there’s no way to make any space that’s homotopy equivalent to into a topological group.

]]>I just meant that it’s continuous in your “funny sense”, and rotations seem more amenable to becoming a Lie group than reflections.

]]>I don’t really get what you mean, Mike. Of course the rotational symmetries of anything 3-dimensional form a subgroup of SO(3), and yes, a rotational symmetry of a cube leaves an antipodal pair of these triangle corners fixed. But I don’t know what you mean by “the rotation group is also continuous”.

]]>The rotation subgroup is also continuous, since a pair of reflections leaves one vertex in common. It’s also a subgroup of SO(3).

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