The Angular Kronecker Constant for {1, 2, 3} Is 1/4
With the set {1, 2, 3}, we can compute the angular Kronecker constant
using a special metric in RxR, with hexagonal unit balls. The
angular Kronecker constant corresponds to the maximum distance in this
metric of any point in the plane from a discrete (additive) subgroup L
with two generators. Our "canonical" generators for this subgroup
are P=(2/3, 1/3) and Q=(1, 0).
Related to this geometry is a "Half Domain" to which we can narrow the
search for points farthest from L. Our "canonical" half domain in
this case is the rectangle with vertices (0, 0), (0, 1/6), (1, 1/6),
and (1, 0).
Related to this half domain is a set T of members of L such that, for
all points in the half domain, there is at least one member of T which
is among the points of L that are closest. Here there are four
such members of L: (0, 0), Q= (1, 0), P = (2/3, 1/3), and
Q - P = (1/3, -1/3).
Finally, the farthest points from L within the half domain form a
vertical line segment from (1/3, 1/12) TO (1/3, 1/6). In the two
linked pictures, N3.jpg and N3.pdf,
this line segment is continued up to (1/3, 1/4) because these points
are also farthest from L but are outside the given half domain (the
rectangular region); they occur in a "reflection" of the half domain
that could also serve as a half domain (specifically, the vertical
reflection reflection around the line y = 1/6).
The "worst point" (1/3, 1/12) is interesting to us: it is
equidistant from three distinct members of L, namely O = (0, 0) ,
P = (2/3, 1/3) and Q - P = (1/3, -1/3). Notice that two of
these three members of L can serve as generators of L. Also
notice that the metric formulas used to realize the distance to (1/3,
1/12) are two horizontal (with opposite orientation) and one vertical.
Note that this "worst point" is NOT the barycenter of the
triangle formed by O, P and Q-P.