Slice

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For [math]n\in{\Bbb Z}_+[/math] and [math](a,b,c)[/math] in the triangular grid

[math]\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c = n \},[/math]

the slice [math]\Gamma_{a,b,c}[/math] is the set of all strings in [math][3]^n[/math] with [math]a[/math] occurences of [math]1[/math], [math]b[/math] occurences of [math]2[/math], and [math]c[/math] occurences of [math]3[/math]. Thus, for instance,

[math]\Gamma_{1,2,0} = \{ 011, 101, 101 \}.[/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].

The slice [math]\Gamma_{a,b,c}[/math] is incident to [math]2^a+2^b+2^c-3[/math] combinatorial lines.

The equal-slices measure on [math][3]^n[/math] gives each slice a total measure of [math]1[/math].

The [math]k=2[/math] analogue of a slice, sometimes called a layer, plays a role in the proof of Sperner's theorem.

A combinatorial line must touch three slices [math]\Gamma_{a+r,b,c}, \Gamma_{a,b+r,c}, \Gamma_{a,b,c+r}[/math] corresponding to an equilateral triangle in [math]\Delta_n[/math]. Conversely, one might hope that any sufficiently "rich" subsets of three slices [math]\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]\Gamma_{a+r,b,c}[/math] of strings whose first digit is [math]1[/math], and the subsets of [math]\Gamma_{a,b+r,c}, \Gamma_{a,b,c+r}[/math] of strings whose first digit is [math]2[/math].