The Atanasov--Ranganathan construction on row 10 #
Row 10 is AR's ninth genus-five family: a triangle {0, 1, 2} with one spoke
from each corner, an apex 3 joining two of the spoke ends, and a banana
{6, 7} reached from the other two.
e0 : 0 -> 2 e1 : 2 -> 1 e2 : 1 -> 0 the triangle
e10 : 0 -> 3 e4 : 1 -> 4 e3 : 2 -> 5 the spokes a, b, c
e5 : 3 -> 4 e9 : 5 -> 3 the apex
e6 : 4 -> 6 e11 : 5 -> 7 e7, e8 : 6 == 7 the banana
The displayed divisor has one of its four chips at
an interior point of a spoke, at an offset equal to another spoke's length. The
certificate is therefore a marked script
(Utilities/Subdivision/SplitRampScript.lean), which bends downward
at the chip and lets the chip pay for the kink.
The two displayed scopes are "a = min(a,b,c)" and "b = min(a,b,c)". The
remaining case, c = min, is the image of the second under the core
automorphism sigma = (1 2)(4 5)(6 7). So exactly the two drawn scopes have to
be proved:
GenusFiveRow10ChamberOne--|e10| ≤ |e4|, |e3|, the mark one4;GenusFiveRow10ChamberTwo--|e4| ≤ |e10|, |e3|, the mark one10.
Both decompose the five chip-free vertices as one ConfigurationMarkedTripod
centre plus AR's eleventh picture (ConfigurationEleven) on the remaining
four, in the special position alpha = gamma ≤ beta that the chamber's two
inequalities supply. GenusFiveRow10Symmetry.chamber_covers and
ClosedOrbit.closedConstruction_of_chamber then finish the closed orthant.
Either scope AR draw gives a degree-four Dhar pencil on its own face.
AR's ninth family on row 10. The paper's own divisor -- two triangle vertices, one banana vertex, and one chip inside the marked spoke -- has rank at least one on every nonloopy forest face, all three chambers at once.