## Stott expansions being Minkowski sums

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Stott expansions being Minkowski sums

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.

I'm not 100% sure but it may be possible to construct certain EKF polytopes as Minkowski sums of non-convex polytopes too. (Maybe some of you would like to check this? )

The nice thing about the Minkowski sum is that it's defined for arbitrary polytopes (even non-convex ones). So we can have an "algebra" of polytopes, say we start with the regular polytopes (including subdimensional ones), and then arbitrarily mix-and-match them together with Minkowski sums. Could make for a fun imaginary world of polytopes where "atoms" are polytopes, and "chemical reactions" are Minkowski sums. If the starting basis polytopes are unit-edged, many results would also be unit-edged, and some would be CRF. (You knew I was coming to that. ) Of course, summing a polytope with itself would produce a double-length edged copy of itself, but we could adopt the rule that any basis polytope can only be included once. This would give us 2^n "allowed" sums, given n basis polytopes. If we restrict the basis polytopes to full-dimensioned regular polytopes, this yields the set of uniform polytopes (excluding special cases like the 3D snub cube/dodecahedron or the 4D grand antiprism). If we allow subdimensional basis polytopes into the mix, we get also a bunch of partial Stott expansions and possibly something else too.

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).
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### Re: Stott expansions being Minkowski sums

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.

In fact this exactly was the way how Wendy once proved the geometrical existance of prissi = s3s4o3x - as my then new combinatorical reading of this mix of s-, o-, and x-nodes suggested - i.e. as Stott expansion of s3s4o3o, i.e. as Minkowski sum of s3s4o3o and o3o4o3x.

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.

--- rk
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### Re: Stott expansions being Minkowski sums

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.

Haha, the idea just occurred to me recently and I haven't really worked out all of its consequences yet. But already interesting rationalizations arise, e.g., an elongated triangular pyramid could be construed to be a Minkowski sum of a tetrahedron and a line segment, for example.

OTOH, I can't see any easy way to construct, say, a gyroelongated polyhedron, as a Minkowski sum. So it still doesn't fully encompass all of the Johnson solids (even the cut-n-paste ones -- crown jewels of course are a different story). Though it does cover a fair bit more than partial Stott expansion. I wonder what results it could yield when applied to the 4D polytopes...

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.
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### Re: Stott expansions being Minkowski sums

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?
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### Re: Stott expansions being Minkowski sums

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.

I'm an idiot... of course it's easy to visualize, it's just two (not necessarily uniform) truncated octahedra glued together at a hexagonal face. Not 100% sure whether there should be multiple faces where the adjacent square faces would lie, or it would be suitably distorted so that said multiple faces merge into non-regular hexagons.
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### Re: Stott expansions being Minkowski sums

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.

Looks like being the right path.

So first we have to construct the rectification of tridpy = J12. For that purpose we aren't allowed to just take the edge midpoints, rather we have to use diminishing planes orthogonal to the vertex vectors, running slowly inwards, such that their edge intersections meet at a common point. (Else we would produce bent or folded faces underneath the former vertices!) By symmetry the 3 planes according to the equatorial vertices all are alike, so their pace will be the same. Same holds for the truncation velocity of the 2 polar vertices. For the 3 equatorial truncation planes the depth is easily obtained, as those have to intersect at the midpoints of the former equatorial edges. A vertical plane, errected at 2 of those midpoints, intersects the former lacing edges of J12 at 1/4 of its edge size from the tips. That is, the convex hull of those 3 points close to either tip plus the 3 equatorial edge midpoints do define the rectification of J12. Here the 3 rhombical faces then would have a ratio of their diagonals of sqrt(6):1 (if I got it correctly )

After "xoo" and "oxo" we would have to look for "oox" next. That one could be constructed from the above rectification easily. Just stellate that figure at its 6 irregular triangular faces by extending all other 5 faces until those meet. Alternatively you could errect vertical lines at the edge midpoints of the equatorial edges of J12 and furthermore circumscribe a dual triangle each at the above calculated points of the former lacing edges. This then results in a taller triangular prism with latteral rectangular faces, having the same size ratio sqrt(6):1.

Finally you would have to construct the Minkowski sum of those 3 figures, each one possibly scaled up or down ad libitum. In the Wythoffian cases that individual scaling usually is done in such a way that in the result all final edges will have the same size. In our case we deal with too many different edge sizes, therefore this aim cannot be met, and thus we freely could just use scaling factors 1 as well. The resulting figure thus will have a semiregular Hexagon (possibly different side lengths only) for either polar face. Then 3 rectangles underneath the lacing edges of J12 each. The former triangles of J12 become irregular hexagons, the even sides of which still are parallel to the sides of the former regular triangles, but the odd sides are parallel to the isocele acute triangles of those rectifications of J12. Then we have irregular octagons (with 2 perpendicular symmetry axes only) underneath the former equatorial vertices of J12. And finally there are 3 more rectangles underneath the edge midpoints of the equatorial edges of J12.

(Would be interesting to have all those shapes - i.e. J12 = "xoo", "oxo", "oox", "xox" (~J35), and "xxx", possibly also "xxo" and "oxx" in addition - as pictures, possibly all within the same orientations. Mutual scalings would be arbitrary, just as was mentioned for the scaling coefficients in the Minkowski sum.)

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### Re: Stott expansions being Minkowski sums

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?

Minkowski-adding different-symmetrical objects most likely produces some awkward irregular faces with all different edge lengths. Thus those investigations mosty can be omitted within an CRF research.
For Minkowski sums of both full dimensional objects with the same non-regular symmetry we will have to be carefull that all facets (cells) are aligned in parallel pairs at least (as far as applicable), cf. my already described constructions for J12).
The easiest and most promizing cases thus would be Minkowski sums with subdimensional objects, as that parallelness condition there is being achieved most easily.

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### Re: Stott expansions being Minkowski sums

Well yes, if we're looking for CRFs, then it would make the most sense to perform Minkowski addition between polytopes that at least share some symmetry, appropriately aligned.

But I was thinking outside the realm of CRFs, where you add potentially unrelated polytopes together. Could some of those results possibly turn out to be CRF-able? (e.g., via rescaling or deformation of resulting faces). Or is that too unlikely?
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### Re: Stott expansions being Minkowski sums

Further thoughts along the CRF line: Minkowski addition of polytopes where one polytope's symmetry is a subsymmetry of the other polytope's symmetry, is essentially equivalent to partial Stott expansion. So the point of interest, where we might generate new CRFs, is where the two polytopes share some common subsymmetry, but each has a different supersymmetry on top of that.

What would the result be, if we added a cubee and an icosahedron, with pyritohedral symmetry as the common subsymmetry? Probably something non-CRF... but I wonder if in 4D there could be more cases that are CRF-able. E.g., in 3D, adding a square pyramid and a line segment parallel to one of the diagonals of the square face produces a non-CRF with a spheno complex closed by a non-regular hexagon. 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. So there might be more possibilities in 4D where the analogous construction in 3D would be non-CRF. Might be worth exploring.
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### Re: Stott expansions being Minkowski sums

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.

The "bottom" figure then gets a J15 = esquidpy out of the former oct, while the "top" figure then gets a line segment out of the point (tip). That is, you get here "line || esquidpy", which itself is nothing but an elongated squarepyramid dipyramid = esquippidpy.

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### Re: Stott expansions being Minkowski sums

Yes, that's correct. So in a sense it's not something new, just a unified construction scheme for what we already know.

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?
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### Re: Stott expansions being Minkowski sums

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?

ex within 4D brique subsymmetry is
Code: Select all
`oooxxxfffFFFVooof FxfoFfxFofxooVoof xfFFfoFoxxofooVof fFxfoFoxFofxoooVf&#zx`
, i.e. the hull of the compound of those 17 layers. (Here x=1, f=(1+sqrt(5))/2, F=f+x, V=2f.) Thus your aimed sum thus would be described just by adding x onto each node in each layer, i.e. by
Code: Select all
`xxxuuuFFFAAABxxxF AuFxAFuAxFuxxBxxF uFAAFxAxuuxFxxBxF FAuFxAxuAxFuxxxBF&#z(x(,u?))`
- here u=2x, A=F+x=f+u=f+2x, B=V+x=2f+x. That one then ought to be worked out - and esp. investigated, whether there would be unit edges only and whether all contained faces still are regular ones. - Perhaps you could do that a way more faster by visual Investigation? I.e. by converting all those layers into coordinates, Building the union of that total vertex set, Maybe you've some convex hull algorithm at hand, and then investigating generated pix?

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### Re: Stott expansions being Minkowski sums

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.
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### Re: Stott expansions being Minkowski sums

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.

Okay, but those actually are given wrt. brique symmetry, i.e. those coords can be read off directly. I.e. the vertices within the first hexadecant (of 4D) just are (using the already mentioned sizes) and for obvious mirroring reasons scaled up by 2:
Code: Select all
`(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)`

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### Re: Stott expansions being Minkowski sums

Haha, OK, so you did the hard work for me already.

This will have to wait till next week, though. Busy with work today and will be away for the weekend.
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### Re: Stott expansions being Minkowski sums

Spheration is a Minkowski sum (or prism-sum, to use a term derived from Klitzing), of a sphere and a framework.

The normal rounded spheration, as you see in furniture, is a spheration of a small sphere and the base shape. The sort of spheration that provides knobs etc, supposes that the vector sum in Minkwoski, is different over the surtope-dimension. So a vertex is sphere size 1, an edge 1/2, a hedron 1/3, &c. This gives a recessed flat one is likely to encounter in trays.

Richard Klitzing uses 'tegum-sum' to provide a skin over a number of co-spacial elements, where the tegum itsel is a skin over orthogonal elements. So a prism is a replacement of each point with a perpendicular height, (which can be many dimensions), so i suppose a prism-sum is the replacement of points in the same plane.

I had a look at quickfur's notion about the stott sum. It is true that the stott-sum can be represented as a kind of minkowski-sum, but the whole notion of the stott-sum is that we can use 'position-polytope' along with 'position-vector', that is, (1,0,0) represents a point, a vector (0,0,0)-(1,0,0), and a polytope say x3o4o. Vector algebra is the basis of the spreadsheet used in the lace-prisms etc.

The trouble is that Minkowski sums covers any orientation. In spheration, it does not matter, because one element is a sphere. In the stott-vector thing, it is the explicit orientation of the elements that matters. Adding something like q.Octa + x.Cube in any orientation, is different to qo3oo4ox, where the two are specifically orientated that all 12 edges cross.

But carry on. I put my 3d in.
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