This notebook presents the computations of Table 1.
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using Oscar
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Realizing a cubic surface \(X\) as a blowup of the projective plane at 6 points in general position, there 27 lines or (-1)-curves are given by: - \(E_i\), the exceptional divisors above point \(i\) - \(F_{ij}\), the strict transformations of lines through points \(i\) and \(j\) - \(G_i\), the strict transformations of conics through all points but point \(i\)
The Picard group of \(X\) has rank 7 and is generated by the exceptional divisors as well as \(H\), the pullback of the hyperplane class of \(\mathbb{P}^2\). The (-1)-curves have the following descriptions in terms of \(E_i\) and \(H\): - \(E_i\) - \(F_{ij} = 2H-E_i-E_j\) - \(G_i = H-\sum_{j \neq i} E_j\)
The Weyl group \(W(E_6)\) acts on the 27 lines. It is the subgroup of \(S_{27}\) generated by the permutation action on the points \(\{1,\dots,6\}\) and the Cremona transformation, which is represented below as an automorphism on the Picard group.
[1]:
cremona = Matrix{Int}([2 1 1 1 0 0 0; -1 0 -1 -1 0 0 0; -1 -1 0 -1 0 0 0; -1 -1 -1 0 0 0 0; 0 0 0 0 1 0 0; 0 0 0 0 0 1 0; 0 0 0 0 0 0 1])
[1]:
7×7 Matrix{Int64}:
2 1 1 1 0 0 0
-1 0 -1 -1 0 0 0
-1 -1 0 -1 0 0 0
-1 -1 -1 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
For example, we can see this action takes \(G_2\) to itself.
[24]:
cremona * [2,-1,0,-1,-1,-1,-1]
[24]:
7-element Vector{Int64}:
2
-1
0
-1
-1
-1
-1
[ ]:
G = symmetric_group(27);
H, _ = sub(
G,[
cperm([1,2],[8,12],[9,13],[10,14],[11,15],[22,23]), # (12)
cperm([2,3],[7,8],[13,16],[14,17],[15,18],[23,24]), # (23)
cperm([3,4],[8,9],[12,13],[17,19],[18,20],[24,25]), # (34)
cperm([4,5],[9,10],[13,14],[16,17],[20,21],[25,26]), # (45)
cperm([5,6],[10,11],[14,15],[17,18],[19,20],[26,27]), # (56)
cperm([1,12],[2,8],[3,7],[19,27],[20,26],[21,25]) # Cremona
]
)
[10]:
order(H)
[10]:
51840
An Eckardt point is a common intersection point of three lines. We represent it as a set of three numbers. For example, the set \(\{7,16,21\}\) corresponds to the Eckardt point \(F_{12}+F_{34}+F_{56}\). Given a set of Eckardt points, we want to compute the size of its orbit under \(W(E_6)\).
We compute the stabilizer of the set \(\{7,16,21\}\). It has order 1152. By the orbit-stabilizer theorem, its orbit has size \(51840/1152 = 45\). So there are 45 strata of type 1, i.e., strata with at least one Eckardt point.
[12]:
S = stabilizer(H, Set([7, 16, 21]));
order(S[1])
[12]:
1152
Now we consider a stratum with two Eckardt points. We compute the stabilizer of \(\{\{7,16,21\},\{7,17,20\}\}\). Unfortunately Julia does not have a built-in function for this, so we create our own functions.
[5]:
function make_set(v)
m = Int(length(v)/3)
st = Set()
for i in 1:m
push!(st,Set(v[(i-1)*3+1:i*3]))
end
return st
end
function stab_size(v)
stab = 0
for h in H
acted = on_tuples(v, h)
if make_set(v) == make_set(acted)
stab += 1
end
end
return stab
end
[5]:
stab_size (generic function with 1 method)
We summarize our computations in this table.
Number of Eckardt points |
Size of stabilizer |
Size of orbit |
---|---|---|
1 |
1152 |
45 |
2 |
192 |
270 |
3 |
216 |
240 |
4 |
72 |
720 |
6 |
96 |
540 |
9 |
1296 |
40 |
10 |
240 |
216 |
18 |
1296 |
40 |
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# 2 Eckardt points
@show 51840/size_stab([7, 16, 21,7,17,20])
# 3 Eckardt points
@show 51840/stab_size([11,14,16,10,13,18,9,15,17])
# 4 Eckardt points
@show 51840/stab_size([11,14,16,10,13,18,9,15,17,10,15,16])
# 6 Eckardt points
@show 51840/stab_size([11,14,16,10,15,16,7,17,20,7,18,19,8,13,21,9,12,21])
# 9 Eckardt points
@show 51840/stab_size([11,14,16,9,15,17,10,13,18,1,8,24,4,20,27,3,12,23,6,21,26,2,7,22,5,19,25])
# 10 Eckardt points
@show 51840/stab_size([11,14,16,10,13,18,9,15,17,10,15,16,1,9,25,1,11,27,2,13,25,3,18,27,2,14,26,3,17,26])
# 18 Eckardt points
@show 51840/stab_size([9,14,18,9,15,17,10,13,18,10,15,16,11,13,17,11,14,16,1,7,23,1,8,24,2,7,22,2,12,24,3,8,22,3,12,23,4,19,26,4,20,27,5,19,25,5,21,27,6,20,25,6,21,26])
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