Notions and Notations.

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

Re: Four Products, etc

There is another product (or something like a product), involving digons.

For two abstract polytopes A and B, let A'=A\{body(A)} and B'=B\{nulloid(B)}, and take the (disjoint) union C=A'∪B'. Define a partial order on C in terms of the partial orders on A and B, thus: a₁≤a₂ in C if and only if a₁≤a₂ in A'; b₁≤b₂ in C if and only if b₁≤b₂ in B'; and a≤b and not b≤a (where a is anything in A', and b is anything in B'). So, each vertex of B takes the place of the body of A.

The result C is always a polytope, if A and B are greater than (-1)-polytopes. The dimension is dim(C)=dim(A)+dim(B). Every [dim(A)-2, dim(A)+1]-subpolytope in C is a digon.

If B is a 1-polytope, then C is a ditope. If A is a 1-polytope, then C is a hosotope.
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mr_e_man
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Re: Notions and Notations.

Can you write this as an actual constuction of a polytope, and not in some mathematical puree.

It is listed as a product, but you describe it as a sum. A polytope product of A and B is the outer product of some subset of these (ie a set of all members of A, B, of the surface and some additional points), not a union.

The union of two sets (A, B, C) and (C, D, E) is the set (A, B, C, D, E). These five elements do not consist or comprise the outer product (AC, AD, AE, BC, BD, BE, CC, CD, CE). Further more, you have not indicated exactly how these set members would be realised in actual polytopes.
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wendy
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Re: Notions and Notations.

Huh. Somehow I missed that reply.

wendy wrote:The union of two sets (A, B, C) and (C, D, E) is the set (A, B, C, D, E).

The disjoint union of {A, B, C} and {C, D, E} is the set {A1, B1, C1, C2, D2, E2}, where C1 ≠ C2.
It is a kind of sum. Indeed the sizes of the sets are simply added.

I thought of it as a product, because the dimensions add.

Further more, you have not indicated exactly how these set members would be realised in actual polytopes.

The rank (or dimension) of an element can be deduced from the partial order. The incidence relations are also specified. Beyond that, you can build the polytope however you want.

The ranks of elements of the polytope A remain the same in the "product" C. The ranks of elements of B are all increased by dim(A). Everything in A (except its body) is incident with everything in B (except its nulloid).

I wrote that post at the same time as this one. Perhaps I should have linked them both ways, or kept them together in one thread.
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mr_e_man
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Re: Notions and Notations.

mr_e_man wrote:Huh. Somehow I missed that reply.

wendy wrote:The union of two sets (A, B, C) and (C, D, E) is the set (A, B, C, D, E).

The disjoint union of {A, B, C} and {C, D, E} is the set {A1, B1, C1, C2, D2, E2}, where C1 ≠ C2.
It is a kind of sum. Indeed the sizes of the sets are simply added.

[...]

No-one told you to apply the disjoint union. It more is like having within programming languages the List() and Set() distinctions:
in the former you would have contained C as often as you would add it, within the latter it would be kept just once, i.e. like a distinct() applies.
Thus I'd assume that Wendy there used the word "set" more in the common sense usage ...

--- rk
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Re: Notions and Notations.

Klitzing wrote:No-one told you to apply the disjoint union.

I told myself!

viewtopic.php?p=28303#p28303

I defined an operation on two abstract polytopes, in terms of the disjoint union of sets. It's similar to the four products of polytopes, though it's not commutative, and it doesn't involve the Cartesian product of sets. The operation makes generalized hosotopes/ditopes.

(It doesn't make sense to use the non-disjoint union, since, in abstract polytopology, we usually don't care what the elements are, or whether some element is in two different polytopes; we only care about the relations between the elements of one polytope.)

I think Wendy's issue was not the disjointness, but that I called a union a product.
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mr_e_man
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Re: Notions and Notations.

The Hasse diagram of my "product" looks like this:

not product.png (33.31 KiB) Viewed 480 times

Here A is a triangle, B is a pentagon, and C = {3, 2, 5} (Schlafli symbol).
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Re: Notions and Notations.

The actual formula for wythoff nullitopes, is a prism with one side 'r'.eg x5o2r3o. However if you need to resort to nullitopes, it means you are making a lot of extra work fou yourself, since they only appear in a top-down reading of the CD. read from the bottom up with vertex nodes and accounting for s and a symmetries, you never create nullitopes.

With sets, we actually learn them in the sense that all members in a set were unique. This leads in to boolean arithmetic. But people are useing set in a different meaning and blaming the old people for not keeping up. So if the factors of 120 be (2 2 2 3 5), that is a list object, but you can reduce members by making it a set(remove duplicates), or a pariity (toggles members so only), eg py (2,2,3,3,5)= (5), meaning that 180 is a fifth square, while the set of factors of 120 and 180 become (2,3,5) = set of prime divisors..

In relation to abstract polytopes, a number of people told me of various thing they use, such as the nulliod is the empty set.[it is actually different in each set, but the way abstract polytopes are implement, there is only one set on the table], that a supernom or top incidence must exist,[multitopes do not require a top incidence]. From what i can make of it is just a babbleing brook of nonsense, which barely coverthe scope of polytopes.
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wendy
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Re: Notions and Notations.

When the nulloid and body are omitted, my construction becomes very simple: You stack two Hasse diagrams, one on top of the other, and make incidence connections everywhere in between. You can call it a product, or a sum, or something else.

The disjoint union thing was merely a mathematician's formalization of this construction.

I was using "set" with the same old meaning. The union of {A,B,C} and {C,D,E} is {A,B,C,D,E}. The disjoint union is something like {{A,1},{B,1},{C,1},{C,2},{D,2},{E,2}}, enforcing distinction between elements of the two sets. (Since 1≠2, the sets {C,1} and {C,2} are distinct.)
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mr_e_man
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Re: Notions and Notations.

The old meanin og set is taken that the duplicates are removed onvr. There was never a 'didjoint union' It supposes thrat every element has an srparatr identy,which id not true of set.The whole point og setd is c1=c2.

tHe slice of the hasse digram as describrf givr what NWJ call cluster It is not s udeful esy of joinig polytoped, dince it heasvily deprnds on msting incidencrd.
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