Talk:5-cell

old comments on symmetries

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 A_n+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 T_d, 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 A_n for all n. This group is isomorphic to S_n+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?

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 (talk • contribs) 22:45, 22 August 2011 (UTC)

Hypervolume formula?

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!?

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?

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)

Yet Another Alternate Name

The 5-cell is also known as a pyrochoron, so maybe, someone should add that in as an alternate name. I don't want to do it myself because I don't have an account, and I've seen these things get inexplicably very contentious.--69.112.209.47 (talk) 03:58, 16 May 2018 (UTC)

There are no references on that website, but I'll link it as an external reference. Tom Ruen (talk) 20:59, 23 May 2018 (UTC)

Yeah, I guess it wouldn't be a good source for an encyclopedia—I'm pretty sure that the people on that website basically made up the names they use themselves even if their names are better than "n-cell"-style names. Thanks for adding it as an external link, though! 69.112.209.47 (talk) 05:03, 31 May 2018 (UTC)

Lexical Ambiguity Alert!

The "Related polytopes and honeycomb" section of the article states that "It [the 5-cell] is similar to three regular polychora: the tesseract {4,3,3}, 600-cell {3,3,5} of Euclidean 4-space, and the order-6 tetrahedral honeycomb {3,3,6} of hyperbolic [4] space. All of these have a tetrahedral cell." However, this is not referring to geometric similarity; instead, it's referring to "similarity" in a non-geometric way! Since this is arguably a geometry article, I think this is worth changing. — Preceding unsigned comment added by 69.112.209.47 (talk) 04:07, 16 May 2018 (UTC)

I reworded it, and it was wrong anyway, listing tesseract rather than 16-cell. Tom Ruen (talk) 20:41, 23 May 2018 (UTC)

Hypertetrahedron listed at Redirects for discussion

An editor has asked for a discussion to address the redirect Hypertetrahedron. Please participate in the redirect discussion if you wish to do so. _signed, Rosguill ^talk 19:37, 26 May 2019 (UTC)

"Pentagonal hyperdisphenoid"

Listed as a name for a lower-symmetry variant of the 5-cell, added by User:Tomruen in 2014. Google gives no relevant results for "hyperdisphenoid" except copies of this very article. Where does it come from? The name i'm familiar with for this shape is "5-2 step prism", which seems to come from Wendy Krieger. --Goomba1729 (talk) 16:05, 26 April 2022 (UTC)

Yes, it was an attempted descriptive dimensional analogue of the tetragonal disphenoid which has a cycle of 4 equal edges, and two equal cross-edges. The 5-cell form has a cycle of 5 equal edges and 5 equal cross-edges. I wish there was better term available to all of the n-simplices in this series. Their symmetries are isomorphic to the dihedral group of the regular polygons, which can be seen in these 2D projections, so the n-gonal description is an accurate generalization. Note: the 4 and 6 edge labels are not lengths, but imply edge lengths from positions for squares and hexagons in the honeycomb. Tom Ruen (talk) 16:48, 26 April 2022 (UTC)

Vertex figures of omnitruncated simplectic honeycomb

n	2-simplex	3-simplex	4-simplex	5-simplex	6-simplex	7-simplex	8-simplex
Order	6	8	10	12	14	16	18
	Equilateral triangle	Tetragonal disphenoid	Pentagonal hyperdisphenoid?

The hyperpyramid coordinates (now moved up with the top paragraph) are in fact related to the unit radius coordinates now just below it by a ${\tfrac {4}{\sqrt {5}}}$ scaling, along with rotating wxyz to xyzw and ordering the vertices for consistency. I think adding a comment about that relationship would help.

More importantly, it looks like we have a significant error in the unit-radius coordinates, which by table 3 in reference , xyz should be ${\tfrac {\sqrt {5}}{4}}$ not ${\sqrt {\tfrac {5}{2}}}$. The norms are not consistent as it entered.

In addition, my change before that regarding the "simplest set of coordinates" circumcenter being translated to the origin is not precisely the same as the hyperpyramid coordinates, even though the circumradius and edge length's are the same.

I suggest changes something like...

The simplest set of Cartesian coordinates is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (φ,φ,φ,φ), with edge length 2√2, where φ is the golden ratio. While these coordinates are not origin-centered, subtracting $(1,1,1,1)/(2-{\tfrac {1}{\varphi }})$ from each translates the 4-polytope's circumcenter to the origin with radius $2(\varphi -1/(2-{\tfrac {1}{\varphi }}))={\sqrt {\tfrac {16}{5}}}\approx 1.7888$, with the following coordinates:

$\left({\tfrac {2}{\varphi }}-3,1,1,1\right)/({\tfrac {1}{\varphi }}-2)$

$\left(1,{\tfrac {2}{\varphi }}-3,1,1\right)/({\tfrac {1}{\varphi }}-2)$

$\left(1,1,{\tfrac {2}{\varphi }}-3,1\right)/({\tfrac {1}{\varphi }}-2)$

$\left(1,1,1,{\tfrac {2}{\varphi }}-3\right)/({\tfrac {1}{\varphi }}-2)$

$\left({\tfrac {2}{\varphi }},{\tfrac {2}{\varphi }},{\tfrac {2}{\varphi }},{\tfrac {2}{\varphi }}\right)/({\tfrac {1}{\varphi }}-2)$

The following set of origin-centered coordinates with the same radius and edge length as above can be seen as a hyperpyramid with a regular tetrahedral base in 3-space:

$\left(1,1,1,{\frac {-1}{\sqrt {5}}}\right)$

$\left(1,-1,-1,{\frac {-1}{\sqrt {5}}}\right)$

$\left(-1,1,-1,{\frac {-1}{\sqrt {5}}}\right)$

$\left(-1,-1,1,{\frac {-1}{\sqrt {5}}}\right)$

$\left(0,0,0,{\frac {4}{\sqrt {5}}}\right)$

Scaling this and the previous coordinates by ${\tfrac {\sqrt {5}}{4}}$ give unit-radius origin-centered regular 5-cells with edge lengths ${\sqrt {\tfrac {5}{2}}}$. The hyperpyramid has coordinates:

$\left({\sqrt {5}},{\sqrt {5}},{\sqrt {5}},-1\right)/4$

$\left({\sqrt {5}},-{\sqrt {5}},-{\sqrt {5}},-1\right)/4$

$\left(-{\sqrt {5}},{\sqrt {5}},-{\sqrt {5}},-1\right)/4$

$\left(-{\sqrt {5}},-{\sqrt {5}},{\sqrt {5}},-1\right)/4$

$\left(0,0,0,1\right)$

Jgmoxness (talk) 00:03, 9 April 2023 (UTC)

I thought it might be helpful to show some overlaying 3D projections of the 4-polytope to compare the differences, with code and text for explanation.

For a MTM notebook for more interaction, see this post [1].

Jgmoxness (talk) 02:05, 9 April 2023 (UTC)

As usual, I got it wrong and you spotted it and corrected me. Thanks! Your proposed changes look great to me -- the illustrations are especially nice. Dc.samizdat (talk) 22:44, 9 April 2023 (UTC)

Thanks.

I committed those edits.

I was looking at the last set of coordinates and wondered if it would be better for consistency to show these scaled to unit-radius, e.g.

$\left({\sqrt {3}},{\sqrt {5}},{\sqrt {10}},\pm {\sqrt {30}}\right)/(4{\sqrt {3}})$

$\left({\sqrt {3}},{\sqrt {5}},{\sqrt {40}},0\right)/(4{\sqrt {3}})$

$\left({\sqrt {3}},{\sqrt {45}},0,0\right)/(4{\sqrt {3}})$

$\left(0,0,0,1\right)$

Opinions?

Jgmoxness (talk) 02:34, 11 April 2023 (UTC)

Yes, those unit-radius coordinates are very interesting, quite illuminating actually, you've simplified them well -- it would be great to give them too, but i'd leave the edge-length-2 coordinates in the article as well, if only because Coxeter's metric tables of the 4-polytopes are all based on edge length 2 -- I find unit-radius coordinates so much more "normal" for nearly all purposes myself, but Coxeter's unit-2-edge-length comparison is canonical, and should always be given also Dc.samizdat (talk) 07:40, 11 April 2023 (UTC)

different hyperplanes

Any five vertices in five different hyperplanes constitute a 5-cell, though not usually a regular 5-cell.

Not loving this language. Any vertex inhabits an infinite number of hyperplanes.

Any five vertices not all in one 3-plane, no four of which are in one 2-plane, no three of which are in a line …

—Tamfang (talk) 04:48, 20 July 2023 (UTC)

Ha, I forgot having written that. Now it says:

A 5-cell is formed by any five points which are not all in the same hyperplane (as a tetrahedron is formed by any four points which are not all in the same plane, and a triangle is formed by any three points which are not all in the same line).

I was about to object that this doesn't specify that every set of four must not be in one plane, and so on down; but then it hit me, if four are in a plane, then the five form a skew pyramid, which is in a 3-plane, so an explicit recursion is not necessary – but could it be mentioned somehow for clarity? —Tamfang (talk) 00:17, 8 November 2023 (UTC)