What is the Cartesian product of n polygons?How do we calculate the Cartesian product between n polygons? Is it always a polytope? If so, can we say anything about its faces?For example, n squares: { S 1 : [ a 1 , a 2 ] × [ b 1 , b 2 ] , … , S n : [ a k , a k + 1 ] × [ b k , b k + 1 ] } so, S 1 × … × S n ∈ R n forms a n − dimensional box.

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What is the Cartesian product of n polygons?How do we calculate the Cartesian product between n polygons? Is it always a polytope? If so, can we say anything about its faces?For example, n squares:   {      S    1    :  [      a    1    ,      a    2    ]  ×  [      b    1    ,      b    2    ]  ,  …  ,      S    n    :  [      a    k    ,      a          k      +      1        ]  ×  [      b          k        ,      b          k      +      1        ]  } so,       S    1    ×  …  ×      S    n    ∈            R        n   forms a   n  − dimensional box.

Haylie Park · Accepted Answer

Step 1Just to add some relevant terms. The cartesian product of polytopes also is known as the prism product.In fact, a polytope P multiplied by some single vertex point v results in   P  ×  v, which is P again (Identity). Multiplication by a single edge e results in   P  ×  e, which is the usual prism with bases being P and lacing edges e. If the dimension of Q also is greater than 1, the product   P  ×  Q is known to be the (P,Q)-duoprism, cf., and using even more factors, you&#039;ll get multiprisms (like (P,Q,R)-triprisms   P  ×  Q  ×  R, then (P,Q,R,S)-quadprisms   P  ×  Q  ×  R  ×  S, etc.)Step 2The total dimension is calculated according to   d  (            ×            P    k    )  =  ∑      d    (          P      k        )  Step 2This applies even for any elements thereof. Thus the vertices of             ×            P    k   are given by the respective products             ×            v    k  , the cartesian products of the vertices each. In order to result in an edge of             ×            P    k  , all components have to be vertices, just a single factor is an edge within the respective polytopal subspace. Similarily 2-faces of             ×            P    k   occure from all vertices and one polygonal factor, or from all vertices and 2 factors being edges, which then results in the correspondingly oriented rectangle.

Nico Patterson · Accepted Answer

Step 1Cartesian products of polytopes P,Q are indeed polytopes again. The vertices are pairs of vertices (p,q) where p is a vertex of P, and q is a vertex of Q.Step 2In fact, if you have two polytopes, say   P  ⊂            R        n   and   Q  ⊂            R        m  , then all faces of   P  ×  Q are just the Cartesian products of faces of P and Q. You can see this by using the hyperplane descriptions: the hyperplanes used to define   P  ×  Q are the same as those for P and Q, but now they are using disjoint sets of variables. So, if   F  ⊂  P  ×  Q is a face, then the first n coordinates of every point in F must satisfy equalities for some face of P and the last m coordinates of every point in F must satisfy equalities for some face of Q.

How do we calculate the Cartesian product between n polygons? Is it always a polytope? If so, can we say anything about its faces? For example, n squares: {S_1:[a_1, a_2] times [b_1, b_2],…,S_n:[a_k, a_{k+1}] times [b_k, b_{k+1}]} so, S_1 times … times S_n in R_n forms a n-dimensional box.

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