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I don't understand this at all!Southafrica6 (talk) 22:24, 12 May 2008 (UTC)Reply[reply]

  • The article assumes the reader has a lot of knowledge coming into the article. There should be a simpler introduction. JohnMGarrison (talk) 03:55, 28 May 2008 (UTC)Reply[reply]
  • This article does link to convex regular polychoron, which explains the necessary background information. Since the 120-cell is but one example of a 4-dimensional polytope, it is hardly reasonably to expect that this article would explain everything from ground zero. That's what hyperlinking is for.—Tetracube (talk) 16:55, 28 May 2008 (UTC)Reply[reply]
and you can't ask for much more than having that link in the first line. —Tamfang (talk) 21:40, 28 May 2008 (UTC)Reply[reply]

which cells around each vertex[edit]

I have no idea who or where to get an answer for this, but thought this might be a good place to start. Is there any way to give a designation to the 120 cells, and then describe the 120 combinations of 4 cells each of which meet at a vertex?

I know that the list would be 120 items long, with 4 entries each, and each cell would appear four times, but if cells 1, 2, 3, and 4 meet at a vertex, and 2,3,4,5 meet at a vertex, does that mean 1,2,3,5 meet at a vertex, or does it absolutely exclude that possibility? I'm trying to understand those combinations but have very few fourth dimensional objects to play with. Ryan Stoughton (talk) 02:35, 14 December 2009 (UTC)Reply[reply]

Yup, those 120 labels are the elements of the binary icosahedral group. Schur's notation used cycle notation for the group together with a sign. Alternatively you can think of the explicit elements in the 3-sphere -- they're listed here in the binary icosahedral group page. You can think of these numbers as being the centres of the faces. Rybu (talk) 04:35, 14 December 2009 (UTC)Reply[reply]
Of course you mean the 600 combinations. — As Rybu suggests, it's much easier to think of polychora as tilings of (curved) 3-space; locally you can mostly ignore the curvature. So think of three cells around an edge; and two more closing the volume around each end of that edge. Using your labels, the first three are "2,3,4" and the last two are "1,5". In the pentachoron every set of four cells meets at some vertex, and there may be some other simple figure whose vertices include "1,2,3,4" and "2,3,4,5" and "1,2,3,5" but it would not be regular. —Tamfang (talk) 06:35, 14 December 2009 (UTC)Reply[reply]
I mean 120 combinations, sorry, I was referring to the boundary that is discussed in the second sentence of the article "The boundary of the 120-cell is composed of 120 dodecahedral cells with 4 meeting at each vertex." I'm really just looking for the equation so I can generate the 120 combinations of 4 items each. Ry (talk) 13:50, 14 December 2009 (UTC) (I am Ryan S above)Reply[reply]
The 120 cells (all of which lie on the hypersurface of the 4D body) have 600 vertices, with four cells meeting at each. Where do you get 120? — There is no one natural numbering of the cells; thus there is no "equation" that will spit out the list of their combinations. You could start with a list of the vertices, and a list of the vertices of the 600-cell (which are centers of the 120 cells), whose coordinates are given in the respective articles; for each of the former, find which four of the latter are nearest. —Tamfang (talk) 18:27, 14 December 2009 (UTC)Reply[reply]
I took the liberty of giving this section a better title than "Question". Do you think you're the only one ever to ask a question?Tamfang (talk) 18:27, 14 December 2009 (UTC)Reply[reply]
Thank you, I understand where I went wrong. I was asking for something impossible because I misinterpreted that second line. Ry (talk) 19:06, 14 December 2009 (UTC)Reply[reply]
I recently added a visualization section which gives a heuristic polar mapping of all the cells, though you'll have to go further to specify the individual cells within each layer. Cloudswrest (talk) 19:45, 2 July 2010 (UTC)Reply[reply]

New images for consideration[edit]

120Cell 2D.png 120Cell 3D.png 120Cell 2Da.png Cell120Petrie.svg

What are you depicting in the "2D.png" picture? It's not immediately clear to me. And you're asking people to consider these pictures for what purpose, to replace some other pictures, to supplement pictures in a particular place? How would you caption the pictures? Rybu (talk) 06:28, 29 January 2010 (UTC)Reply[reply]
Excellent questions. The 2D .png (as the description in Commons explains) takes the orthographic (x,y) projection with 1200 (was 912) "nearest (shortest)" edges of length ). The vertex colors are varied based on the number of overlaps. Edge color is based on projected vertex positions. I suspect the current vertex centered 2D contains more than just the 912 nearest (no way to tell - as the method used to generate is missing). The caption on the 2D would be "Vertex Centered with 1200 (was 912) edges of ". The caption on the 3D would be the same - just adding z. As for deciding to replace or augment, I'm not sure. I think it is good to show the different ways geometery/symmetry are visualized. I like the Stella4D reps from Ruen and the source code generated Rocchini Petrie projection (but I am working on a prettier one with my e8Flyer.nb using Mathematica).Jgmoxness (talk) 15:49, 29 January 2010 (UTC)Reply[reply]
If this shows only those edges which are shortest in projection, then why? Otherwise, I don't understand. (All edges are "really" equal, of course, in a regular figure.) —Tamfang (talk) 02:09, 30 January 2010 (UTC)Reply[reply]

These select only the shortest Norm'd 4D edges. Of course, the 2D (or 3D) projected edges vary based on the projection vectors useed (thus the desire to color code them based on that length, or some other dynamic such as number of overlaps- which I can also show). For the 120 cell of 600 vertices, there are 179700 edges =Binomial[600,2]. That creates a "complete graph" (a very blotchy mess of lines Cell120 all.png. Even the graphs shown in the article only select nearest (e.g. on a 3D cube, the diagonal edges are omitted to show the convex shape). The 2D graph should look the same as in the article, but I suspect the 2D image Cell120-4dpolytope.png is wrong. I use the same code to produce accurate projections for similar objects - and some of that code is adapted from *Weisstein, Eric W. "120-Cell". MathWorld.Jgmoxness (talk) 21:53, 30 January 2010 (UTC)Reply[reply]
"vertex centered" determines only a 3D projection; there are two degrees of freedom for 2D projections of this, so naturally they won't all look alike. (Personally I prefer projections that avoid axes/planes/hyperplanes of symmetry, as they can give a better idea of overall form. But I don't know how to choose a 2-plane out of E4 for that criterion.) —Tamfang (talk) 19:40, 31 January 2010 (UTC)Reply[reply]
Please provide links to the images you suspect are inaccurate, as Tetracube is the author of more than one. Also, please sign your comments as this is starting to get difficult to read. Some issues to clarify, what you call "diagonal edges" are technically not edges at all. The 120-cell is regarded as a 4-dimensional solid, not as a graph. So there are no diagonal edges. When you refer to diagonal edges I believe you're talking about the straight lines (edges) that connect two vertices in the 120-cell but for which the interior of the edge is in the interior of the 120-cell, meaning the edge is not on the boundary of the solid. IMO it's not productive to include images of such things as they're simply distracting. Your figures can't be depicting a projection of the "shortest norm 4d edges" since the underlying graph you're depicting isn't vertex-regular, and the 120-cell is a regular solid. So something isn't matching up, this is why your figures are suspect. Rybu (talk) 19:24, 30 January 2010 (UTC)Reply[reply]
There are 1200 edges in the 120-cell and they all have the same length. Are your 912 edges a subset of these, and how are you choosing them? Rybu (talk) 19:33, 30 January 2010 (UTC)Reply[reply]
Ok, I made corrections above (and signed it). I was mistaken on the reference to Tetracube's contribution to the image in question - the 2D projection "vertex centered" was actually a Ruen creation (with conversions by others - can't find the original). Anyway, to the substance of the discussion. It seems the common terms used across geometry, group, and graph theory are getting in the way, so let me simply describe HOW I created the image (which I referred to as a graph). It takes points (vertices) from the group of 600 (4D vectors) generated by the permutations listed in the 120 cell article. These 600 vertices produces 30 unique groups (or sets) of "possible edges" consisting of different Norm'd 4D length (which total 179700, which is referred to in graph theory as a "complete graph"). Selecting the shortest set of Norm'd 4D edges (the 1200 (was 912) edges of length ) typically represents the convex boundary of the polychora object. These vertices and edges are then projected to 2D (or 3D) by taking the dot product of 2 (or 3) 4D vectors (in this orthographic case x=(1,0,0,0),y=(0,1,0,0)). Jgmoxness (talk) 21:53, 30 January 2010 (UTC)Reply[reply]
What length(s) do you get for the 288 next-shortest edge candidates? Perhaps you've found a flaw in our list of vertex coordinates. Perhaps you've failed to recognize some expression equivalent to 3-√5, and your floating-point computation has a small error (as is common). —Tamfang (talk) 19:40, 31 January 2010 (UTC)Reply[reply]
Correcting "912" to "1200", without leaving any overt sign that you did so, makes others' comments nonsensical. —Tamfang (talk) 07:36, 1 February 2010 (UTC)Reply[reply]
I understand - I did comment at the end that I would make corrections. I also modified all of my responses pertaining to that aspect (now corrected), but did not want to edit others comments. It IS a wiki - please correct your own comments as you see fit (considering the new info).Jgmoxness (talk) 13:29, 1 February 2010 (UTC)Reply[reply]

My apologies, I found I had juxtaposed permutation columns, which surprisingly (given that they are permutations) alters the edge length sets. I will re-post images with corrections ASAP. Jgmoxness (talk) 19:08, 31 January 2010 (UTC)Reply[reply]

Notice the added projection which is more like the "vertex centered" in the article. I left the axes lines in to show it is not strictly a unit (x,y) projection.
x=(1/3, -2/3, 1/3, 0)
y=(1/2, 0, -1/2, 0)
Jgmoxness (talk) 22:53, 31 January 2010 (UTC)Reply[reply]

I just have uploaded a gif file in the hope of being displayed in the 120-cell page.

opposites align[edit]

For visualization purposes, it is convenient that the dodecahedron has centrally aligned opposing parallel faces (a trait it shares with the cells of the tesseract).

What does "centrally aligned" mean? —Tamfang (talk) 21:18, 13 May 2010 (UTC)Reply[reply]

I meant a normal line from the centroid of one face hits the centroid of the other face. I'm not sure this is the case for the octahedron, which also has parallel faces. A string of octahedrons has a zig-zag shape. Cloudswrest (talk) 05:00, 14 May 2010 (UTC)Reply[reply]
It's true of any figure with point reflection symmetry, including all uniform polyhedra except those with only tetrahedral symmetry (tetrahedron, truncated tetrahedron, tetrahemihexahedron). —Tamfang (talk) 07:51, 14 May 2010 (UTC)Reply[reply]
Cool. I was going to do some calculations on the octahedron but you saved me the time. I'll adjust the article. Cloudswrest (talk) 16:14, 14 May 2010 (UTC)Reply[reply]
It's obvious if you consider the octahedron as the 3-antiprism. —Tamfang (talk) 18:02, 15 May 2010 (UTC)Reply[reply]
Odd-numbered prisms and even-numbered antiprisms are uniform polyhedra lacking point reflection symmetry. —Tamfang (talk) 23:00, 2 November 2019 (UTC)Reply[reply]
Oops, small correction: it's not true of "any figure with point reflection symmetry", because there are irregular polyhedra with such symmetry where face centers' normals do not meet the body's symmetry center. But the central normal of each face of a uniform polyhedron necessarily meets that center, because each face is a regular polygon inscribed in a small circle on the circumsphere. —Tamfang (talk) 17:45, 15 May 2010 (UTC)Reply[reply]

Table widths[edit]

Is there any reason the table widths should be fixed at a specific pixel value? It looks unforgiveably ugly at my screen resolution and font size (way too narrow for the screen, with fonts too large to fit comfortably in the narrow column).—Tetracube (talk) 00:12, 3 June 2010 (UTC)Reply[reply]

Sorry, I was just adjusting for what was good on a medium resolution browser, and some uniformity. The images are a fixed size, so sometimes I'll adjust the table to match the images. Please tweak if you can improve.
Well, in general I try to stay away from pixel sizes, because it is just impractical to cater to every possible browser/screen size out there. I've always found it best to just let the browser do its job, or prod it in the right direction with, say, percentage/font-relative directives (like width=60%, or width=25em). But that's just IMNSHO.—Tetracube (talk) 04:05, 3 June 2010 (UTC)Reply[reply]

from the inside[edit]

A 3D projection of a 120-cell performing a simple rotation (from the inside).

Does "from the inside" mean stereographic projection? —Tamfang (talk) 10:00, 7 February 2011 (UTC)Reply[reply]

It just means that I moved the camera inside of the projection... it is the exact same projection as the animation above it. Speaking of which, I'm not sure whether my projections are stereographic or not... is it still a stereographic projection if you are not projecting from a point on a hypersphere that also contains the points being projected? I have simply been using (x, y, z) / (w + k), where k is 'whatever distance looks good'. JasonHise (talk) 10:25, 7 February 2011 (UTC)Reply[reply]
No .. if k=0 it's gnomonic, if k=|(w,x,y,z)| it's stereographic, I don't know what to call the general case unless "perspective". —Tamfang (talk) 10:46, 7 February 2011 (UTC)Reply[reply]
Ok, good to know... I've been calling it perspective, but I've seen similar animations that claim to be stereographic so I was never quite sure my terminology was correct. JasonHise (talk) 18:14, 7 February 2011 (UTC)Reply[reply]
Yes, that's what I use. stereographic is viewpoint "on the n-sphere", and perspective is somewhere else. But there's actually 2 perspectives. There's projection perspective from 4-space, and viewer-perspective in 3-space! Tom Ruen (talk) 20:49, 7 February 2011 (UTC)Reply[reply]

Looks cool, but I wish it was just a bit slower. It's a little too fast to mentally lock onto a cell and track it as the image rotates.Cloudswrest (talk) 18:28, 7 February 2011 (UTC)Reply[reply]

I'm considering going back and rendering double the number of frames for all of the animations, making them both slower and smoother. JasonHise (talk) 18:51, 7 February 2011 (UTC)Reply[reply]

Intertwining rings[edit]

What does File:120-cell_rings.jpg depict? The image description says "Two intertwining rings of the 120-cell." My understanding was, that the 120-cell is bounded by 120 regular dodecahedra. Yet the dodecahedra appear distorted in that image. Therefore I guess some kind of perspective projection is used to project these rings, but isn't really clear to me and the image description also says nothing about this. Can someone clarify? Toshio Yamaguchi (talk) 08:37, 13 May 2011 (UTC)Reply[reply]

Some distortion is inevitable, as each cell is in a different hyperplane. —Tamfang (talk) 16:58, 13 May 2011 (UTC)Reply[reply]
Watch this youtube video: Cloudswrest (talk) 21:31, 13 May 2011 (UTC)Reply[reply]

cant see some symbols.[edit]

If I cant see them, at least put a notice that says "Notice: there are characters that contains mathematical symbols. If you don't have proper rendering, you may see boxes, ?'s, or other symbols instead of math symbols." - User:

I can't guess what you are seeing or where. Everything seems good to me. Perhaps your internet connection was slow and some images timed out loading? Tom Ruen (talk) 08:57, 29 May 2017 (UTC)Reply[reply]

A gif image of a rotating 120 cell[edit]

I made a rotating 120-cell.

You can find it here

I would like to donate it to the Wikipedia. Eviruena (talk) 18:43, 27 January 2023 (UTC)Reply[reply]

The Cartesian coordinates for unit radius[edit]

FYI - I think I have the mid-points of the dual snub 24-cell edges (i.e. rectified) and wanted to fill this table in. I think I may have 3D hull coordinate image for the unit radius 120-cell. It is a bit different, being the alternate (J' 120-cell), but very interesting.

T-120-cell with 600 vertices.png

When I confirm it is consistent with the table data, I will fill it in and add a visualization of it.

Jgmoxness (talk) 22:32, 19 February 2023 (UTC)Reply[reply]
FYI - this is the above 3D projection showing the 16 concentric hulls
Jgmoxness (talk) 21:08, 20 February 2023 (UTC)Reply[reply]
Dual snub 24-cell overlay with the convex hull of the 120-cell.svg
ok, I've done some investigation and did confirm that 3D projections of the 480 4D vertices which are mid-edges of the dual snub 24-cell do NOT coincide with 480 of 600 vertices of either J or J' 120-cell configurations (in fact only 24 do). But what I did find is that the dual snub 24-cell does contain 3D vertices on the outer convex hull which are the tetrahedra that create the pentagons of J. See the figure to the right.
I need some references to understand the suggestion that there should be a projection of the 120-cell that works as suggested in the Unit radius coordinates table. The permutation list with 6 entries for my dual snub 24-cell's 480 edges of (Norm, #) Dual snub 24-cell edges.svg is:
480 mid-edge permutations of the Dual Snub 24-Cell.svg
Jgmoxness (talk) 02:23, 22 February 2023 (UTC)Reply[reply]

Dc.samizdat modified the unit radius entry for the 480 vertex permutations. So I created a vertex permutation list for the 600 vertices of the alternate 120-cell, which I referred to as J' 120-cell (shown above). It looks like this may be what is being referenced as a diminished J' 120-cell. My rows 1-3 are the rows above the 480 vertices of the Diminished 120-cell (with 96 of the 192 in my row 3 also showing up as the 4th row of the 480 in the diminished J' 120-cell). What is left are 3 sets of 96 and 32 making up the other 384 vertices of J' (vs. 3 sets of 120 + 24). BTW - J' comes from the quaternion construction of the 120-cell, but using the dual of T vs. T' in the construction.

Diminished 120-cell vertex permutations.svg

So for completeness, here is an image of the Diminished J' 120-cell. It is J' with the first 120 vertices removed. Those first 120 vertices come from I' (vs. I) in J' making up the 600-cell constructed using T vs. T').

Diminished 120-cell overall convex hull.svg

It is the last five convex hulls contribute to the overall hull shape. Diminished 120-cell convex hulls.svg

Just FYI - there is also a diminished main 120-cell (J) as in this article. It is created by removing the first 120 vertices (i.e. the 600-cell) from J of the main article. Yet, the vertices of that are not even close to the unit radius table's diminished 120-cell.

Jgmoxness (talk) 20:46, 22 March 2023 (UTC)Reply[reply]