Band clusters of Gaussian edge scores #
The Gaussian edge-score field gives the edge uv of a graph the score
Internal.edgeScore H q R ω s(u, v) = ⟨(x_u + x_v)/‖x_u + x_v‖, ω⟩, where x are the unit
truncated distance-kernel rows (unitKernel) with decay rate q and radius R, and ω is a
standard Gaussian vector indexed by the vertices.
The band graph bandGraph H score c keeps the edges of H whose score lies in the median band
[-c, c). bigBandEvent H q R c K v is the event that the band cluster of v, the set of
vertices reachable from v in the band graph of the Gaussian edge scores, has more than K
vertices.
BandSubcritical q R c is a percolation hypothesis: the band clusters are subcritical uniformly
over simple cubic graphs. For every ε > 0 some cluster size K makes the expected number of
vertices in band clusters of more than K vertices at most ε h on every simple cubic graph
with h vertices. It is an open hypothesis, not a theorem of this library. Numerically it holds
on the infinite cubic tree for c < 0.1826, which appears to be the worst case; the band-jump
path decompositions (Gaussian.Band) and the circuit bounds derived from them take it as an
explicit premise.
The band graph of an edge score: the edges of H whose score lies in [-c, c).
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The event that the band cluster of v has more than K vertices: more than K vertices
are reachable from v along edges whose Gaussian edge score lies in [-c, c).
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Subcritical band clusters (an open percolation hypothesis). For every ε > 0 there is
a cluster size K such that on every simple cubic graph with h vertices, the expected number
of vertices whose band cluster has more than K vertices is at most ε h. The band cluster of
v is its connected component in the graph of the edges whose Gaussian edge score, for the
kernel with decay rate q and radius R, lies in [-c, c).
This is not proved in this library. It says that band percolation of the Gaussian edge-score
field is subcritical uniformly over cubic graphs. Numerically it holds on the infinite cubic
tree for c < 0.1826, which appears to be the worst case.
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