Symmetries of the regular tetrahedron...?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Symmetries of the regular tetrahedron...?

Postby Paul » Fri Oct 13, 2006 11:47 pm

Hello all,

I posted this Symmetries of the regular tetrahedron...? and perhaps something more... to the Math Message board...

It occurred to me that perhaps it's more likely that someone here might be able to find the missing 6 reflections...?

Does anyone have any ideas?
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Postby wendy » Sat Oct 14, 2006 7:49 am

Every edge of a tetrahedron is a mirror-edge, that is, there is a mirror that reflects one end to the other. It only has six mirrors, because there are always pairs of mirrors that form an angle 180/n degrees, n odd.

The 12 may have come in, because the perpendiculars to these six mirrors form the 12 vertices of a cuboctahedron.
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Postby Paul » Sat Oct 14, 2006 3:14 pm

Hello Wendy,

Thanks for your reply.

Doesn't there need to be 12 singular reflections simply because the group of symmetries of 4 objects (vertices here), S<sub>4</sub>, is equal to 4! = 24, and the rotational symmetries of the tetrahedron take up 12 of those, A<sub>4</sub>, which leaves 12 elements of S<sub>4</sub> which must be singular reflections...?
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Postby thigle » Sat Oct 14, 2006 6:02 pm

in your post that you provided link to, you stated (at the end):
Reference [3] describes the remaining 6 reflection symmetries of the regular tetrahedron:

"reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to -x): the rotations correspond to those of the cube about face-to-face axes" (Bold added)


you can image those like this:

take any 2 faces of tetrahedron.
omit their common edge, which leaves you with a closed loop of 4 edges.
all their midpoints lay in the plane that the above quote talks about, the plane that is bisective & perpendicular to a segment connecting 2 opposite edges of tetrahedron.
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Postby Paul » Sat Oct 14, 2006 8:12 pm

Hello Thigle,

Yes... I believe that's it. Let me know if these two figures do, in fact, depict what it is that you're speaking of:

Image

Image

Thanks Thigle, I believe you've found the 6 missing singular reflections of the regular tetrahedron!
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Postby thigle » Sun Oct 15, 2006 12:01 pm

well, i didnt, just few years ago, i was asking my father's cousin (who's a dean here in bratislava at the faculty of geometry) about symmetry groups, and he used tetrahedron to illustrate the idea, showing me all the possible transformations along the way. actually he explained it on a phonecall, so i visualised those through his descriptions.

you wrote:
I may have some additional questions relating to all this if we can find the missing 6 reflection symmetries...


what are those ?
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Postby Paul » Mon Oct 16, 2006 10:47 pm

Hello Thigle and all,

If any of you have access to John Stillwell's new book The Four Pillars of Geometry, pp. 163-166, that'd be better. However, I'll do my best to paraphrase what I believe John Stillwell's saying...

Basically, John Stillwell defines a mapping between the rotational symmetries of the tetrahedron and the 24-cell using quaternions of the form:

Image

For this, he positions a tetrahedron inside a cube with each vertex of the tetrahedron coinciding with a vertex of the cube:

Image

The red axis represents a tetrahedron rotation of pi through opposite edges. These correspond to +/- i,j,k. The red axis depicted in the figure above is the rotational axis for k. Obviously, the axes corresponding to the other two pairs of edges goes through the i and j axes.

Notice that the quaternion gets pi/2, since θ here equals pi. Also notice we get 6 vertices of the 24-cell for 3 rotations of the tetrahedron because of the +/-. Further, the identity rotation maps to +/-1. (I think... with -1...?)

The blue axis corresponds to the tetrahedron rotations of 2pi/3 through a vertex and its opposing face, with coordinates (1/sqrt(3)*(+/-i +/-j +/-k). (1/sqrt(3) is a scaling factor to give the quaternions absolute value 1) These 8 tetrahedron rotations, along with the identity, are again doubled by +/- to correspond to the 16 vertices, as 8 pairs of opposites, of the 24-cell +/-(1/2), +/-(i/2), +/-(j/2), +/-(k/2). Again, the quaternions here have an angle of 2pi/6 since θ equals 2pi/3.


First of all, did I depict all that in the figure correctly?


My questions here now are... Can a similar mapping be made between the 3 rotational symmetries of the triangle, and say, the vertices of the octahedron in 3-space?

Further, is there always some similar mapping between the n-simplex and some (n+1)-d polytope? When is there, and when is there not such a mapping between the n-simplex and some (n+1)-d polytope?


Also... is a reflection in dimension n always going to be expressible as a rotation in dimension n+1?
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Postby wendy » Tue Oct 17, 2006 7:14 am

The number of mirrors in a simplex group correspond to an exchange of pairs of vertices. So for the tetrahedron, which has four vertices, there are 4*3/2 = 6 such edges, and a like number of mirrors.
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Postby thigle » Sat Oct 21, 2006 1:56 pm

hi paul. as for the last question, answer is yes. one can find explanation of this in for ex.: 'Manning: Fourth Dimension Simply Explained', in the introduction.

as for the other questions, i don't have a clue. right now i am trying to understand how the quaternions map to the 4-space, through analogies of how complex numbers map onto plane and that onto rieman's sphere.
there should be a similar construction from quaternions to 4-space to 4-sphere. the issue is discussed here: http://tetraspace.alkaline.org/forum/viewtopic.php?t=829
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Postby Paul » Sun Oct 22, 2006 6:56 pm

Hello Thigle,

Thanks for your response.

Yes, I do have that book... although I think those essays may still be posted on the Internet.

I check out the Introduction.
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