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246:, means that the second triangle is instead wound backwards - try testing out what the Schläfli symbol {3/2} really means. Because it is wound backward it is not exactly quasiregular. But it still is a triangle, so Steelpillow, I agree with you that it should be quasiregular. Unfortunately, that is not what many people think, so we cannot put that into the article.
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Then, yes it has regular faces of two types alternating around each vertex. But there are more subtle issues which mean that it is not usually regarded as quasiregular. For example its vertex figure is sometimes written as {3.4.3/2.4}. Personally I think it should be (for an even more subtle reason),
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that would be an immersion but for the vertices (0-skeleton) a "near-immersion". Not a standard term though. Knowledge calls
Steiner's Roman surface "... a self-intersecting mapping of the real projective plane into three-dimensional space, with an unusually high degree of symmetry ..." A bit of a
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I don't see anything wrong with my edit. My open-link edits like this were done over about 30 articles, getting started linking some common terminology, still useful even without linked-article yet. I purposely removed the vertex/edge counts since it includes ALL of them, left the face info since
554:
Away from the vertices, the image in R^3 of the tetrahemihexahedron is also an immersion. Three distant points of the original polyhedron (in the projective plane) are mapped to the centre. But as you move along an axis from the centre towards a vertex, the self-intersection is between closer and
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terms. Doesn't seem to have gotten any uptake, so I agree that it should be left out. But yes, this is an infinite family that is uniform if the number of dimensions is at least 3 (and is in fact semiregular). The vertex figure is the demicross polytope of one less dimension, which is uniform if
528:
Immersion vs. embedding always confuses me, but you are the first person ever to explain to me that the tetrahemihexahedron in R^3 is neither. What then is it? An injection, a projection, or what? I also fail to understand why the vertices are not locally homeomorphic to R^2 - they are after all
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Thank you for the clearest explanation yet. One last piece of the puzzle, if I want to talk about the tetrahemihexahedron as "a xxxxxxx of the projective plane in R^3", what word should I use for "xxxxxxx"? "Injection" or "mapping" are very broad, is there a more specific term? â Cheers,
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embedded in a projective plane (or should that be diffeomorphic not homeomorphic? why mention both in the above, and what's the difference anyway? I do struggle with these long words which always seem to get explained in terms of each other.) â Cheers,
551:. The image self-intersects, so it is not an embedding. But no matter where you are on the original Klein bottle, if you consider a small area around you, the image of that area is non-self-intersecting. So it is a local embedding, AKA an immersion.
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Exactly so (though I am always unclear whether an "immersion" is allowed to self-intersect). All we need is a verifiable reference. Knowledge can't follow the logic through unless somebody else has already done so - and published. â Cheers,
184:
Viewing the HTML page source in my browser options, the graphic box is standards-compliant HTML table markup, set to right-align. I know that IE6 is obsolescent and not standards-compliant, but I wouldn't expect it to be
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Thanks for the clarification. I bet I forget it again though, sigh. These models of the projective plane, and many others, are well documented. What I cannot find is any study of their quasiregularity. â Cheers,
769:. However, itâs interesting to try to interpret that claim. Hereâs my guess, which will hopefully inspire someone else to look into the matter further (and possibly even create a citation for that claim?)
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This article claimed that the tetrahemihexahedron was the three-dimensional âdemicrossâ polytope. Iâve commented out that claim, since I canât find any other source replicating it, which violates
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closer points of the original. At the vertex itself, no matter how small a scale you consider, the self-intersection is always in your neighbourhood. So it fails to be an embedding, even locally.
488:. The immersion cannot have the same characteristics intrinsic to the surface - after all, it is not even an embedding. However, none of this is relevant to quasiregularity. HTH. â Cheers,
140:
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A fun fact is that you can pair up the vertices along three edges and perform an unwinding operation to get a genuine polyhedral immersion of the projective plane, that is very similar to
892:
What a shame. There've been so many kinds of polytopes and properties thereof discovered during the last decade, and it's just unfortunate that none have yet been published. â
288:
I don't consider the hemicuboctahedron as "wound backwards" in any way. I regard it as existing quite normally in the projective plane. The "winding backwards" occurs when we
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bad. Check around. Maybe IE6 really is that bad by modern standards, or you need to reinstall. (I use
Firefox and it's just fine). Sorry I can't help more. --
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in the projective plane). Therefore I believe that while the hemicuboctahedron is not a quasiregular polyhedron, it should still be regarded as quasiregular.
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can self-intersect but that is not the issue, the neighbourhoods of the six vertices are not homeomorphic to R^2. The article draws a good analogy to the
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OK with the broken links for now. It also has the same vertex arrangement as the octahedron, so I have added that in to replace the bit you deleted. --
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we're already in 4D or above, so we can truncate and rectify such polytopes in 4D and up and still get uniform polytopes.
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Thanks anyway. I am coming to suspect that there is no accepted term. Guess that leaves me free to invent one. â Cheers,
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Somebody else has already (although not exactly in the way
Maproom did) shown the relationship between thah and co; see
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They are indeed disclike, or homeomorphic to R^2 as you describe them, as are the six points on the Roman surface
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In the rod-and-ball image, the squares are yellow. You can only see half of a given square at a time. -- Cheers,
303:. Arguably, "hemicuboctahedron" denotes the quasiregular map, and "tetrahemihexahedron" denotes its immersion in
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Seems like an interesting family, itâs a shame that uniform polytopes are still such an obscure field of study.
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Perhaps in the end not relevant, but for the record, that is not correct. The meaning of
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Have a look at the "elco" entry at
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A quasiregular polyhedron is one derived from a regular polyhedron by
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I just removed this from the stellation and quasiregular categories.
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I think I'll try to explain some subtle issues here. 3.4.3.4 is the
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It looks nice, but I don't quite see where the square faces are. --
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list of all the polyhedra that he considers to be quasiregular.
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The squares intersect right through the center of the model.
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Under this assumption, the demicross 2-polytope would be a
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George Hart considers the hemipolyhedra to be quasiregular.
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Sorry, never mind the long words, I'd say the key word is
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Firstly, it is not a stellation of any convex core.
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is not even an immersion, and this is not either. --
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827:. The demicross 5-polytope would consist of 16
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562:. But this destroys "quasiregularity". --
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