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The cube <math>[3]^n</math> has <math>\frac{(n+1)(n+2)}{2}</math> slices, and each slice <math>\Gamma_{a,b,c}</math> has cardinality <math>\frac{n!}{a!b!c!}</math>. | The cube <math>[3]^n</math> has <math>\frac{(n+1)(n+2)}{2}</math> slices, and each slice <math>\Gamma_{a,b,c}</math> has cardinality <math>\frac{n!}{a!b!c!}</math>. | ||
Each slice <math>\Gamma_{a,b,c}</math> is incident to <math>2^a+2^b+2^c-3</math> combinatorial lines. | Each slice <math>\Gamma_{a,b,c}</math> is incident to <math>2^a+2^b+2^c-3</math> [[line|combinatorial lines]]. | ||
The [[equal-slices measure]] gives each slice a total measure of 1. | The [[equal-slices measure]] gives each slice a total measure of 1. |
Revision as of 01:48, 18 February 2009
A slice of the cube [math]\displaystyle{ [3]^n }[/math] is a set of the form [math]\displaystyle{ \Gamma_{a,b,c} }[/math] for some (a,b,c) in the triangular grid
- [math]\displaystyle{ \Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c = n \}, }[/math]
where [math]\displaystyle{ \Gamma_{a,b,c} }[/math] is the set of all strings with a 1s, b 2s, and c 3s. Thus for instance
- [math]\displaystyle{ \Gamma_{1,2,0} = \{ 011, 101, 101 \}. }[/math]
The cube [math]\displaystyle{ [3]^n }[/math] has [math]\displaystyle{ \frac{(n+1)(n+2)}{2} }[/math] slices, and each slice [math]\displaystyle{ \Gamma_{a,b,c} }[/math] has cardinality [math]\displaystyle{ \frac{n!}{a!b!c!} }[/math].
Each slice [math]\displaystyle{ \Gamma_{a,b,c} }[/math] is incident to [math]\displaystyle{ 2^a+2^b+2^c-3 }[/math] combinatorial lines.
The equal-slices measure gives each slice a total measure of 1.
The k=2 analogue of a slice is sometimes known as a layer, and plays a role in the proof of Sperner's theorem.
A combinatorial line must touch three slices [math]\displaystyle{ \Gamma_{a+r,b,c}, \Gamma_{a,b+r,c}, \Gamma_{a,b,c+r} }[/math] in an equilateral triangle. Conversely, one might hope that any sufficiently "rich" subsets of three slices [math]\displaystyle{ \Gamma_{a+r,b,c}, \Gamma_{a,b+r,c}, \Gamma_{a,b,c+r} }[/math] in an equilateral triangle will have many combinatorial lines between them. However there appear to be quite a lot of obstructions to this hope, consider e.g. the subset of [math]\displaystyle{ \Gamma_{a+r,b,c} }[/math] of strings whose first digit is 1, and the subsets of [math]\displaystyle{ \Gamma_{a,b+r,c}, \Gamma_{a,b,c+r} }[/math] of strings whose first digit is 2.