username5243 wrote:One other thing - in 5D and higher, you might get ursatopes with a uniform base, if it has triangular faces only. For instance, oox3xfo3oox3ooo&#xt could be called the rap ursateron (rap = rectified pentachoron) with 1 rap, 5 tet-ursachora, 5 oct-ursachora, 10 triangular prism pyramids, and 1 srip (small rhombated pentachoron, x3o3x3o). The only other polychora with only triangular faces are sadi (snub icositetrachoron) and rox (rectified hexacosichoron) - but I get the feeling neither of these will work. The sadi case would be a diminishing of the ex (hexacosichoron) ursateron, so if that one doesn't work, the sadi case won't either. Rox probably won't work either, especially if the ex case is degenerate...
With regards to the expansions, in 5D for the regular based ones, there are several types. The first kind is of the type xfo3oox3ooo3xxx&#xt and has teddi-prisms connecting the ursachoral facets. The other kind is of type xfo3oox3xxx3ooo&#xt and replaces the ursachoral cells with expanded-ursachoral cells. There is also a combined expansion xfo3oox3xxx3xxx&#xt with expanded-ursachoron cells with teddi-prisms connecting them.
quickfur wrote:username5243 wrote:One other thing - in 5D and higher, you might get ursatopes with a uniform base, if it has triangular faces only. For instance, oox3xfo3oox3ooo&#xt could be called the rap ursateron (rap = rectified pentachoron) with 1 rap, 5 tet-ursachora, 5 oct-ursachora, 10 triangular prism pyramids, and 1 srip (small rhombated pentachoron, x3o3x3o). The only other polychora with only triangular faces are sadi (snub icositetrachoron) and rox (rectified hexacosichoron) - but I get the feeling neither of these will work. The sadi case would be a diminishing of the ex (hexacosichoron) ursateron, so if that one doesn't work, the sadi case won't either. Rox probably won't work either, especially if the ex case is degenerate...
Rap ursateron sounds like it might actually exist! That's very exciting. I've never thought of it that way, that the only requirement is that the base facet have only triangular 2-faces. (Well, besides the decagon radius restriction as Wendy explained. (And also the expanded form must also be CRF, since that's what ends up being the base facet.))
I'm pretty sure the sadi case won't work, since as you said it's a diminishing of the ex ursateron, which I think somebody has already proven is either flat or hyperbolic.
The rox case is unlikely to work, but it's hard to tell for sure: the "bad" thing is that the dichoral angles in rox are even wider than in ex, so it leaves even less room for the lacing teddy cells to fit in spherical curvature. However, the 600 octahedral teddies lacing the octahedral cells do have a narrower dichoral angle, so perhaps they might compensate just enough for it to be CRF? It's highly unlikely, but might be worth a more careful investigation before we rule it out.With regards to the expansions, in 5D for the regular based ones, there are several types. The first kind is of the type xfo3oox3ooo3xxx&#xt and has teddi-prisms connecting the ursachoral facets. The other kind is of type xfo3oox3xxx3ooo&#xt and replaces the ursachoral cells with expanded-ursachoral cells. There is also a combined expansion xfo3oox3xxx3xxx&#xt with expanded-ursachoron cells with teddi-prisms connecting them.
Cool. I think these should exist, as per Klitzing's note on Stott expansions. But I'd like to see proof before I commit to it.
Mercurial, the Spectre wrote:It appears that the series of ursatopes is infinite and orbiform, although only ones based on the simplex and the orthoplex.
To consider, look at them as vertex figures.
...
Conclusion:
s[x]'s verf happens to be the (x-1)-dimensional ursachoron. Its infinitude (and orbiformity) is due to it being represented as a verf of s[x], which can be derived from t[x], from which t[x]'s verf can always be deformed to equal lengths! Though its circumradius does not hold, it keeps on increasing so that at 4D, you'd have an ursatope with a circumradius equal to its edge length, and in 5D, it is more than 1, so it can only be represented as a verf of a hyperbolic honeycomb. Same applies to the hyperoctahedral ursatopes.
I think all of you missed a simple trick, haha!
But that's ok , I'm contributing to your efforts.
Cheers,
Mercurial
Klitzing wrote:Mercurial, the Spectre wrote:It appears that the series of ursatopes is infinite and orbiform, although only ones based on the simplex and the orthoplex.
To consider, look at them as vertex figures.
...
Conclusion:
s[x]'s verf happens to be the (x-1)-dimensional ursachoron. Its infinitude (and orbiformity) is due to it being represented as a verf of s[x], which can be derived from t[x], from which t[x]'s verf can always be deformed to equal lengths! Though its circumradius does not hold, it keeps on increasing so that at 4D, you'd have an ursatope with a circumradius equal to its edge length, and in 5D, it is more than 1, so it can only be represented as a verf of a hyperbolic honeycomb. Same applies to the hyperoctahedral ursatopes.
I think all of you missed a simple trick, haha!
But that's ok , I'm contributing to your efforts.
Cheers,
Mercurial
Thus you are saying that these ursachora do exist just because they are vertex figures of according (possibly hyperbolic) hyperhoneycombs.
Moreover you relate xfo3oox(3ooo)*&#xt (N+2 nodes) to s3s4o(3o)* (N+3 nodes) and furthermore xfo4oox(3ooo)*&#xt ((N+1)+2 nodes) to s3s4o(3o)*4o (N+4 nodes).
But by mere formal analogy then xfo5oox(3ooo)*&#xt ((N+1)+2 nodes) ought occur as vertex figure of s3s4o(3o)*5o (N+4 nodes).
So the stated non-existance here of the formers (for any N>0) thus then should imply the non-existance of the latter - is that correct?
How could one understand that?
--- rk
wendy wrote:Of course, the rap ursulatope exists.
For the uniform polytopes, the base can consist of a singly marked dynkyn-graph, with the x-node connected only by '3' branches.
You then get by the Conway-Hart rule, that the rectate consists of unmarking the marked node, and marking the adjacent nodes. This will turn a figure to a rectate, or a n-rectate into an n-cantelate (eg o3x3o3o -> x3o3x3o. ). From this, one makes the ursulatope between these two sections.
There are an awful lot. All of the rectate simplexes up to something like 19 dimensions are in the list, and a good few of the rectates of the cross-polytope. The half-cube is in there, as well as many of the Elte-Gosset polyhedra.
ofx3xoo3ooo3ooo *c3ooo3ooo3ooo3ooo
ooo3ooo3xoo3ofx *c3ooo3ooo3ooo3ooo
ooo3ooo3ooo3ooo *c3ooo3xoo3ofx3xoo
ooo3ooo3ooo3ooo *c3ooo3ooo3xoo3ofx
xoo3ofx3xoo *b3xoo&#xt = ofx3xoo4ooo3ooo&#xt
wendy wrote:1: only 1 marked node (produces no squares via x..x)
2: only '3' branches join the marked node.
3: the circumradius is less than f.
o3x3o x
o3u3o is bistratic cap of 3
x3x3x o3x3o
xux3oox3ooo&#xt = x3x3o3o
xux3oox3xxx&#xt = bistratic x3o3x-cap of x3x3o3x
xux3oox3ooo3ooo&#xt = x3x3o3o3o
xux3oox3xxx3ooo&#xt = bistratic x3o3x3o-cap of x3x3o3x3o
xux3oox3ooo3xxx&#xt = bistratic x3o3o3x-cap of x3x3o3o3x
xux3oox3xxx3xxx&#xt = bistratic x3o3x3x-cap of x3x3o3x3x
wendy wrote:I'm pretty sure too, since all the vertices move in the same angle according to rays from the centre. That's the effect of stott expansion.
Klitzing wrote:Thought of replacing teddies by tuts. We already had that somewhere in the past. Those ursatope like friends already were named by someone. I don't recall how, by whom, nor when.
There you will have to stack a quasiregular, a u-scaled copy, and the truncate of that quasiregular (i.e. ringing the directly neighbouring nodes in addition).
quickfur wrote:Klitzing wrote:Thought of replacing teddies by tuts. We already had that somewhere in the past. Those ursatope like friends already were named by someone. I don't recall how, by whom, nor when.
There you will have to stack a quasiregular, a u-scaled copy, and the truncate of that quasiregular (i.e. ringing the directly neighbouring nodes in addition).
At least for the case where the base cell is an n-simplex, replacing the teddies (n-teddies) with tuts produces nothing other than the truncation of the n-simplex x3x3o3...3o.
In 4D, if you start with an octahedral base cell, then you get the bisected truncated 16-cell (i.e., half of o4o3x3x, with the equatorial octs cut into square pyramids). I'm not 100% certain, but I think starting with an n-cross base cell and replacing the (n-1)-teddies with (n-1)-simplex truncates would also give you a bisected n-cross.
The icosahedral case in 4D gives you a bistratic cap of the truncated 600-cell o5o3x3x, which I think we named a "rotunda" in the past (IIRC there was some debate over whether "rotunda" should be reserved for hemi-n-spherical polytopes and polystratic caps should just be called caps).
I think I've considered these cases before... and came to the conclusion that you would always get some kind of section of a uniform polytope, i.e., nothing new unlike the ursatope family. But I don't have a formal proof of this, so it's possible I missed some cases that might be more interesting.
Klitzing wrote:Moreover it then seems to happen that these then just are the bistratic quasiregular first caps of that truncate polytope, which adds to that single marked node of that quasiregular a further leg marked 3, and that end node will be ringed as well.
E.g.
- Code: Select all
o3x3o x
o3u3o is bistratic cap of 3
x3x3x o3x3o
Klitzing wrote:1: only 1 marked node (produces no squares via x..x)
2: only '3' branches join the marked node.
3: the circumradius is less than x.
Klitzing wrote:What I'm not fully clear here is whether the Stott expansions of across-symmetry would translate into those of the truncate.
E.g. would it be correct that we obtain from the true sentence "oox3xux3oox3ooo&#xt is a bistratic cap of o3x3x *b3o3o" a further true sentence - via correspondingly applied Stott expansion on either side - i.e. something along the lines "oox3xux3oox3xxx&#xt is(?) a bistratic cap of o3x3x *b3o3x"?
Klitzing wrote:Klitzing wrote:What I'm not fully clear here is whether the Stott expansions of across-symmetry would translate into those of the truncate.
E.g. would it be correct that we obtain from the true sentence "oox3xux3oox3ooo&#xt is a bistratic cap of o3x3x *b3o3o" a further true sentence - via correspondingly applied Stott expansion on either side - i.e. something along the lines "oox3xux3oox3xxx&#xt is(?) a bistratic cap of o3x3x *b3o3x"?
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