This is the case where there are four simple zeros over the $2$-torsion points. The remaining two zeros all lie over a $2$-torsion point, which we arrange so that they lie over exactly one of the points in $\{(0,0), (1,0)\}$.
We recall that in $\mathcal{H}(1^6)$, $d_{opt} = 4$ and by the arrangement above, there is a $2$-torsion point in the boundary of each horizontal cylinder that has exactly one simple zero above it.
import re
#This loads all of the 1-cylinder diagrams in H(3,1^3) formatted as python lists
H1t4_cyl_diags = [[[[0,3,7,4,1,2,6,5],[0,3,6,4,1,2,7,5]]], [[[0,7,2,1,6,4,5,3],[0,6,7,5,2,1,4,3]]],
[[[0,7,5,1,3,2,4,6],[0,1,4,7,3,6,5,2]]], [[[0,7,5,2,3,4,6,1],[0,2,4,7,6,5,3,1]]]]
H1_1_cyl_diags = [[[[0,1,2,3],[0,1,2,3]]]]
#This loads all of the functions for processing cylinder diagrams
%run ./ST5_fcns/cyl_diag_fcns.ipynb
H1t4_vertex_data = strat_odd_sc(H1t4_cyl_diags)
H1_1_vertex_data = strat_odd_sc(H1_1_cyl_diags)
Among partitions of $8$ into four odd numbers the minimum of the maximum numbers among all partitions is $\lceil 2d_{opt}/4 \rceil = 2$. Necessarily the number must be $3$ in this case because $2$ is not odd.
For the partition of $8$ into $8$ positive numbers, there is a unique partition.
Solving $8 - 2s_0 - 2(1) \geq 0$ implies that the largest value of $s_0$ is 3.
In summary:
$$s_0 = 3 \text{ and } t_0 = 1.$$
#This loads all of the standard partition functions needed for nearly every case
#It also loads the partition evaluate function
%run ./ST5_fcns/partition_functions.ipynb
if True:
create_sc_partition_file((), part_length = 4, t0_range = [3], d_opt = 4,
filename_root = 'ST5_data//H_1t6//1_branch_point//partitions//H1t2_part')
if True:
create_sc_partition_file((), part_length = 8, t0_range = [1], d_opt = 4,
filename_root = 'ST5_data//H_1t6//1_branch_point//partitions//H_1t4_part')
#Load the partitions
if True:
with open('ST5_data//H_1t6//1_branch_point//partitions//H1t2_part', 'r') as file:
H1t2_part = eval(file.read())
if True:
with open('ST5_data//H_1t6//1_branch_point//partitions//H_1t4_part', 'r') as file:
H_1t4_part = eval(file.read())
H_1t4_part
We conclude because at this point there are no partitions of $8$ into eight positive integers such that exactly four of them are odd.