Twisters

[Klitzing]

Just to help for readability:

Wendy's "bi-X-prism" should be what is better known as "(X,X)-douprism" or for that reason simply "X-duoprism".
E.g. a bi-square-prism = square-duoprism = tesseract. And a bi-cube-prism = cube-duoprism = hexeract.

--- rk

[wendy]

I tend to avoid 'duoprism', for the reason that X-X prism is an instance of an X-Y prism, and somehow, X-Y duoprism seems to add un-necessary confusion. If on the other hand, an adjectival form is used, eg 'dodecahedral prism', then the remaining element is a line (or point, in the case of pyramids), ie dodecahedron-line-prism, or pentagon-point-pyramid.

duocylinder is an established term, but my preference here, as with the toratopes etc, is to construct these as eg, "bi-circular prism".

[ICN5D]

In four dimensions, you still have great arrows, but the phase-space is now a six-dimensional thing. It's actually a "bi-glomohedrix prism", or the limit of the toratope ((iii)(iii)), as the outer brackets go to zero

Hey, I know that one! In fact, it's probably the simplest and easiest 6D tigroid to visually conceive. Good ole' 330-tiger, plenty of beautiful symmetry. I believe Marek and I discussed the cut array of that one, here: viewtopic.php?f=24&t=801&start=180#p19966

Hmm. This is interesting, but not quite what I was looking for. I was looking more for an interpretative treatment of the curl operator and what intrinsic underlying symmetry causes it to arise (I vaguely recall reading something like that before, but I don't remember what it was). The problem is that by the time you get to equations dealing with the curl operator, you're already squarely stuck in 3D with no obvious way to get beyond without causing ripples of unexpected (and probably unwanted) side-effects and unforeseen consequences to propagate through the whole system. To truly get something that "fits" well in 4D, we need to find out what causes electromagnetism to manifest itself the way it does in 3D, and then figure out what would happen if it were to manifest itself in 4D.

Welp, not sure where to go about looking for that. The wikipedia stuff seemed like the right direction . Hopefully wendy's insight will lead to some good generalization. I only know the ((III)(III)) tigroid, and fairly well at that.

[wendy]

I put the toratope notation to help you, philip.

But i have not got around to feeling what SO5 and SO6 look like. I have a novel by Conway and Smith (quarterions and octonions), on the rotations in 7 and 8 dimensions, but i tend to disagree with it at this time.

[Klitzing]

When you set the icosahedral symmetry over the octahedral symmetry, then these symmetries match: o3a5e and a3o4e. That is, the ID becomes an octahedron, and the D becomes a tetrahedron or cube. The linkage is actually more facinating than it looks.

Here "a" and "e" just represent equivalent, but Independent instances of either "x" or "o" nodes.
E.g. into o3x5o an fq = (1+sqrt5)/sqrt2 scaled x3o4o can be vertex inscribed.
Or into o3o5x an f = (1+sqrt5)/2 scaled o3o4x can be vertex inscribed.

This in fact is due to the common pyritohedral subsymmetry. (Which sadly has no representation as mere reflectional Group. And therefore cannot be displayed by a Dynkin symbol either.)

Note that o3x5o has 30 vertices, x3o4o only has 6. Fixing a pair of opposite ones, there are no further orientations possible (except of the symmetries of oct itself). Thus we get 30/6 = 5 possible ways to inscribe it. - Or, put it an other way, there is a compound of 5 octs, which has its vertices right in the directions of those of an id. (That one is described as [5{3,4}]2{3,5} and known as small icosiicosahedron.)


Similarily, o3o5x has 20 vertices, o3o4x has only 8. Fixing a pair of opposite ones of the latter within the former, then there will be exactly 2 ways to inscribe it (connecting to either of the non-neighbouring pentagon vertices). Thus we get again 2*20/8 = 5 possible ways to inscribe it. - Or, within that other view, there is a compound of 5 cubes, which has its Vertices right in the directions of those of a doe. And then 2 cubes each will capture the same vertex position. (That one is described as 2{5,3}[5{4,3}] and known as rhombihedron.)


For sure you could consider o3x5x (tid) and x3o4x (sirco) too. And there will be a corresponding compound of 5 sircoes with a global icosahedral symmetry (rhombisnub rhombicosicoahedron). But because of the above mentioned different scaling factors, here the hull will NOT be the uniform tid, in fact not even some variant o3y5z therefrom (y and z to be determined accordingly). Rather you'll get some more general variant x3y5z of grid here for its hull!


So, concluding, "a" and "e" finally are NOT Independent choices of "x" and "o". Rather they seem to be "XOR" ones only. - Or, alternatively, you would have to choose the lengths A to D for o3A5B and C3o4D quite properly, so that these could match.

--- rk

[wendy]

The a and e are actually 'atom nodes'. That means, you should not apply stott addition to þem.,

If you must know, here is the reference to Coxeter's "regular complex polytopes"

a as in a3o4o and o3a5o, means, the symmetry represented by the '2' in 3(6)2, 3(10)2, and 5(6)2, and the 4 (but not the 2) in 4(6)2 and 4(4)3.

e as in o3o4e, and o3o5e, means the symmetry represented by the '3' in 3(3)3, 3(6)2, 3(4)4, 3(4)3, 3(8)2, 3(5)3 and 3(10)2.

So, when 3(10)2 has 3(6)2 as a subgroup, this happens in both a and e simulatly. The order of the first is 6*100, of the second as 6*24, a subgroup of order 5. One notes that the '3' groups here are 3(5)3 and 3(4)3, of the second, f5 and xx, the first is five times the second.

It's not as asseteric as it seems. These are actually presented in the complex euclidean hedrix CE2, which maps without distortion onto E4, and the ordinary scalar rotation x => x cis(wt) is a left clifford rotation. While, eg 3(5)3, and 5(3)5 both have the vertices of {3,3,5}, they are orientated differently, and correspond to the icosahedron-icosahedron and dodecahedron-dodecahedron thing in my earlier speak. In the other case, there is a group f2, which aslo has 120 vertices, is the ID.ID prism. Such wild þings comf easily.

Coxeter is not an easy read. It's laced in the sort of write-only runes mathematicians are want to write in. Of RCP, one can largely ignore nearly all the text and look at the occasional piccie and table. On the other hand, a recent read from the UK was largely spoilt by the author's insistance of using 'fixment' for 'fixing', and the insistance of putting decimal imperial measures (eg 5.15 tons for 5 t 3 c), and worse still, metric measures (eg 5.22 kgm). I suppose a marker would fix this, but it's not my book.

[ICN5D]

But i have not got around to feeling what SO5 and SO6 look like

What are SO5 and SO6? Are they the tigroids, like ((III)(III))?

[quickfur]

But i have not got around to feeling what SO5 and SO6 look like

What are SO5 and SO6? Are they the tigroids, like ((III)(III))?

No, "SO" stands for "special orthogonal group", and the number that follows is the dimension. The SO groups are matrices with determinant 1, which correspond with rotations in n dimensions. The interesting thing about this, is that these matrices can be mapped to higher-dimensional vectors, so the set of all matrices in that group would map to a collection of vectors that form some kind of surface in a higher-dimensional space. That's what wendy was doing -- visualizing the shape of such surfaces.

[ICN5D]

Hmm, sounds kind of familiar in a ways. Do those groups contain toratopes? Or, are they more specific to polytopes? If they're orthogonal, I suspect they'd more likely be open toratopes, if so. Like a duocylinder, perhaps? It has rotations and orthogonal faces.

[wendy]

The term 'orthogonal' refers more to matrices. Specifically, if you have a coordinate system, and rotate it somewhat, the vectors have to remain at right angles.

They only contain spheres, i'm afraid. A toratope has a reduced symmetry, usually something like circle-circle symmetry.