The average score for a n-dimensional convex polytope collection is?

Geometry Level pending

A n n -dimensional convex polytope has counts of each entity class: a null set (entity n , 1 {}_{n,-1} ) [always just 1], vertices (entity n , 0 {}_{n,0} ), edges (entity n , 0 {}_{n,0} ), faces (entity n , 2 {}_{n,2} ), etc. up to (entity n , n {}_{n,n} ) [the object itself and again just 1]. So that words do not have to be invented for each (entity n , m {}_{n,m} -dimensional entity type. I will be referring to them as entity n , m {}_{n,m} or entities n m {}_{n\,m} as appropriate. Please, note that the full dimension entity and the null set entity are also counted.

Given some random collection of convex polytopes embedded in an appropriate genus 0 space (that is, it has no holes) of any integer dimensionality greater or equal to 0 of cardinality greater than 0 (in other words, there is at least one convex polytope and all convex polytopes have at least a vertex), what is the average score of the collection?

An individual convex polytope's score, whose index is m m , is j = 1 n ( ( 1 ) j entity m , j ) \sum _{j=-1}^n ((-1)^j \text{entity}_{m,j}) .


The answer is 0.

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1 solution

This is Euler's characteristic in a genus 0 space (as specified in the problem) of a convex polytope (also specified in the problem) with the addition of the null set and the object itself as specified in the score formula, which always 0. Therefore the average is also always 0. I was sabotaged by the system resetting my intended answer of 1 to 0 and one can not correct wrong answers! Therefore, I edited the problem to match the answer I was forced to use, by using an alternate definition of Euler's characteristic..

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