Talk:5-cell

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 Field: Geometry

If I'm not mistaken, this is the only figure in either three or four dimensional space in which you can have five equidistant points around a central vertex. This should be noted If it is correct.

This property is true of all in the simplex family - (n+1)-points being equidistant in n-space. Tom Ruen 00:26, 16 November 2006 (UTC)

The symmetry group is listed as A4. I don't think this can be true and should be A5 instead. In general, I believe that the symmetry group of the n-dimenstional simplex is An+1.HannsEwald 01:51, 13 November 2007 (UTC)

A4 actually refers to the Coxeter group A4 not the alternating group. The Coxeter group A4 is isomorphic to the symmetric group S5 and the subgroup of proper rotations is indeed isomorphic to the alternating group A5 as you say. Yes, the notation is unfortunate. -- Fropuff 04:42, 13 November 2007 (UTC)
Thanks for the quick explanation. I suppose then we've got to deal with some inconsistencies: in Tetrahedron the symmetry group is named as Td, while in the Tetrahedral symmetry it is referred to as A4, which in this context must refer to the alternating group.
While I have to admit unfamiliarity with Coexter groups, it seems to me that we would not lose anything if we stayed with the symmetric group when discussing symmetry groups of the n-simplex, especially if my generalization above is correct.HannsEwald 11:36, 13 November 2007 (UTC)
The advantage of using Coxeter groups is that the symmetry group of every regular polytope is a finite Coxeter group. While the symmetric and alternating groups are sufficient to discuss the rotational symmetries of the 3-dimensional regular polytopes the same is not true in higher dimensions, particularly 4. Also, the Coxeter groups have a direct geometric interpretation since the finite ones are usually defined as groups generated by reflections in Euclidean space.
The symmetry group of the n-simplex is the Coxeter group An for all n. This group is isomorphic to Sn+1 as you say, but I find it more helpful to think of the group as a Coxeter group rather than a symmetric group. -- Fropuff 16:43, 13 November 2007 (UTC)

Impossible?[edit]

Forgive my ignorance, but is this shape impossible in 3 dimensions? By the diagrams given, it appears so. If it is impossible, I think that would be useful to note somewhere in the article, particularly in the lead. -kotra (talk) 22:48, 18 October 2008 (UTC)

Simple answer yes - I'll add to the intro. Going down a dimension you can ask if a tetrahedron can exist in a plane. The answer is YES, if you make it degenerate, like a pyramid with zeo height, but faces will overlap. Tom Ruen (talk) 00:42, 20 October 2008 (UTC)
Thanks for your explanation. What you added to the lead is helpful. -kotra (talk) 18:16, 20 October 2008 (UTC)
Is this how it was decided that the object was 4th dimensional? See my post. — Preceding unsigned comment added by MrJosiahT (talkcontribs) 22:45, 22 August 2011 (UTC)

Hypervolume formula?[edit]

Is there a formula for the hypervolume of one of these things? Obviously it would be a function of the volume of a tetrahedral cell multiplied by the minimum distance from that cell to the opposing vertex, and then divided by God knows what (probably 4), but there must be something simpler. —Preceding unsigned comment added by 99.48.55.129 (talk) 09:58, 4 August 2009 (UTC)

According to Regular Polytopes the hypervolume of a regular pentachoron of unit edge is √5/90. —Tamfang (talk) 18:04, 5 August 2009 (UTC)

4th dimension!?[edit]

Sorry if this comment is ignorant, since I haven't actually read the page, but I couldn't make it past the first sentence- how is the shape 4th dimensional when the 4th dimension is time!?

Four geometric dimensions, see the link Four-dimensional space. SockPuppetForTomruen (talk) 23:12, 22 August 2011 (UTC)

Do the cartesian coordinates need to be centered at origin?[edit]

It would be easier to understand if you generate coordinates for each node progressively, like so: For any set of points (A, B, C, etc) all equidistant (say 6ft) from each other, we generate coordinates by setting point A at origin, so Ax=0, as does Ay and so forth.

pointB has x=6 since it is 6 ft from A (By is left at zero). pointC Cx=3 can be determined from law of cosines. cosine for triangle ABC is 0.5*6=3. Cy = 5.196, which can be deduced from pythagorean theorem. squareroot of (6^2-3^2). point Dx=3, point Dy =(((AD^2+AC^2-CD^2)/2)-(Cx*Dx))/Cy which is 1.73. Dz is deduced from pythagorean theorem sqrt(6^2-3^2-1.73^2) = sqrt24 = 4.89 so now we got

x y z
A 0 0 0
B 6 0 0
C 3 5.196 0
D 3 1.73 4.89
- x y z w
E 3 1.73 1.22 4.74

point Ex=3, Ey=1.73, Ez=(((AE^2+AD^2-DE^2)/2)-(Ex*Dx)-(Ey*Dy)/Dz =1.22 and Ew (4th dimension coordinate) can be deduced from pyth. theorem and will be 4.74 One can use this same algorithm for any combination of distances to generate valid coordinatesQdiderot (talk) 08:11, 12 July 2012 (UTC)