quickfur wrote:Recently, I had this realization that Stott expansions and the CD notation we have been using essentially amounts to Minkowski sums of certain basis polytopes.
For example, take a polytope like x4o3x. We could look at it from the symmetry point of view, as a Wythoff mirror construction, or we could look at it as a bunch of Stott expansions, e.g., o4o3o -> x4o3o -> x4o3x. We can also look at it from Wendy's view as an oblique coordinate system, where we can insert arbitrary numbers between the nodes, e.g., f4o3x and so on.
Fundamentally, though, A4B3C can be thought of as a series of 3 essentially independent Stott expansions based on .4.3. symmetry. In particular, if we consider the "basis polytopes" of .4.3. symmetry as x4o3o, o4x3o, o4o3x, respectively, then a symbol like x4o3x is essentially the Minkowski sum of two polytopes, namely x4o3o and o4o3x. A polytope like f4o3x is the Minkowski sum of a f-scaled x4o3o (IOW f4o3o) and a unit o4o3x.
This is all probably quite obvious and already well-known, of course.
But what this view of Stott expansions as Minkowski sums gives us, is a way to create new polytopes that are not necessarily describable with a single CD symbol, or indeed, any CD symbol at all. Partial Stott expansions can be easily described as a Minkowski sum of a (full-dimensioned) polytope with a subdimensional polytope. For example, consider a Johnson solid like J43 (elongated pentagonal gyrobirotunda). There is no CD symbol that could describe J43 (though we could write it as a tower of 2D CD symbols, I suppose), but it can be neatly described as the Minkowski sum of o5x3o and a line segment. Or J28 (square orthobicupola): it can be described as the Minkowski sum of an octahedron and a square.
One specific point of interest (among many!) is the ability to precisely define an omnitruncate for a larger class of polytopes than just the uniforms. For example, if we start with J12 (triangular bipyramid), and consider its axes of symmetry, we could construct the basis set { J12, vertical line segment, dual triangle to J12's equator }, and thus construct its "omnitruncate analog" as the Minkowski sum of all 3 basis polytopes, which produces J35 (elongated triangular orthobicupola).
Klitzing wrote:quickfur wrote:[...]
But what this view of Stott expansions as Minkowski sums gives us, is a way to create new polytopes that are not necessarily describable with a single CD symbol, or indeed, any CD symbol at all. Partial Stott expansions can be easily described as a Minkowski sum of a (full-dimensioned) polytope with a subdimensional polytope. For example, consider a Johnson solid like J43 (elongated pentagonal gyrobirotunda). There is no CD symbol that could describe J43 (though we could write it as a tower of 2D CD symbols, I suppose), but it can be neatly described as the Minkowski sum of o5x3o and a line segment. Or J28 (square orthobicupola): it can be described as the Minkowski sum of an octahedron and a square.
Indeed a good argument! I'm looking forward to this extension of my idea on partial Stott expansions.
One specific point of interest (among many!) is the ability to precisely define an omnitruncate for a larger class of polytopes than just the uniforms. For example, if we start with J12 (triangular bipyramid), and consider its axes of symmetry, we could construct the basis set { J12, vertical line segment, dual triangle to J12's equator }, and thus construct its "omnitruncate analog" as the Minkowski sum of all 3 basis polytopes, which produces J35 (elongated triangular orthobicupola).
That one looks more like the small rhombicosidodecahedron x3o5x in its relation to the icosahedron x3o5o (say), than the omnitruncate thereof, i.e. the great rhombicosidodecahedron x3x5x. For the former still uses triangles and pentagons, while the latter uses hexagons and decagons instead. J35 only has triangles and squares for faces. No triangles have turned into hexagons here - so far.
[...]
quickfur wrote:[...]
Summing J12, a triangular prism (in dual orientation), and a triaugmented triangular prism would yield something with hexagonal faces, I think. It would be non-CRF, but I'm having a bit of trouble visualizing the result.
quickfur wrote:One specific point of interest (among many!) is the ability to precisely define an omnitruncate for a larger class of polytopes than just the uniforms. For example, if we start with J12 (triangular bipyramid), and consider its axes of symmetry, we could construct the basis set { J12, vertical line segment, dual triangle to J12's equator }, and thus construct its "omnitruncate analog" as the Minkowski sum of all 3 basis polytopes, which produces J35 (elongated triangular orthobicupola).
That one looks more like the small rhombicosidodecahedron x3o5x in its relation to the icosahedron x3o5o (say), than the omnitruncate thereof, i.e. the great rhombicosidodecahedron x3x5x. For the former still uses triangles and pentagons, while the latter uses hexagons and decagons instead. J35 only has triangles and squares for faces. No triangles have turned into hexagons here - so far.
[...]
You're right! For hexagons to form, we need to add another polytope to the basis set. It just occurred to me that because the Minkowski sum is associative, the basis set I gave is equivalent to a basis set with only two elements, { J12, triangular prism }, the triangular prism being the sum of the dual triangle and the line segment, whereas what I had in mind was that having 3 elements in the basis set which are symmetrically related to each other would, when summed together, yield an omnitruncate equivalent.
Now, the triangular prism in this set can be considered the dual of J12, so by analogy with the uniform polyhedra basis sets, the third element required would have to correspond to the rectification of J12. I'm not 100% certain exactly what dimensions exactly the elements would take, but an initial guess would be a triaugmented triangular prism. Summing J12, a triangular prism (in dual orientation), and a triaugmented triangular prism would yield something with hexagonal faces, I think. It would be non-CRF, but I'm having a bit of trouble visualizing the result.
quickfur wrote:If we're looking for CRFs, it would appear that the best bet is to stay with basis sets whose members are at least somewhat aligned with each other in some subsymmetry. But I wonder what might happen if we were to sum non-symmetry-aligned polytopes together... would it yield anything interesting? Or just some non-descript intermediate forms?
quickfur wrote:But in 4D, the analogous case of adding an octahedral pyramid and a line segment parallel to one of the octahedron's axes produces a CRF containing an elongated square bipyramid and a bunch of trigonal prisms and tetrahedra.
quickfur wrote:More interesting would be what happens if we add a tesseract to a 600cell. Would the result be CRF, or require some modification to become CRF?
oooxxxfffFFFVooof FxfoFfxFofxooVoof xfFFfoFoxxofooVof fFxfoFoxFofxoooVf&#zx
xxxuuuFFFAAABxxxF AuFxAFuAxFuxxBxxF uFAAFxAxuuxFxxBxF FAuFxAxuAxFuxxxBF&#z(x(,u?))
quickfur wrote:Haha, actually, I don't have a fast way of generating pics from CD symbols. I need to calculate the coordinates first, which is about as much work (maybe more!) as analysing the CD symbols directly.
(x,A,u,F),
(x,u,F,A),
(x,F,A,u),
(u,x,A,F),
(u,A,F,x),
(u,F,x,A),
(F,u,A,x),
(F,A,x,u),
(F,x,u,A),
(A,F,u,x),
(A,u,x,F),
(A,x,F,u),
(B,x,x,x),
(x,B,x,x),
(x,x,B,x),
(x,x,x,B),
(F,F,F,F)
Users browsing this forum: No registered users and 5 guests