Identifier
Values
[1] => 0
[2] => 0
[1,1] => 0
[3] => 0
[2,1] => 1
[1,1,1] => 0
[4] => 0
[3,1] => 0
[2,2] => 2
[2,1,1] => 0
[1,1,1,1] => 0
[5] => 0
[4,1] => 1
[3,2] => 0
[3,1,1] => 1
[2,2,1] => 0
[2,1,1,1] => 1
[1,1,1,1,1] => 0
[6] => 0
[5,1] => 0
[4,2] => 0
[4,1,1] => 2
[3,3] => 0
[3,2,1] => 1
[3,1,1,1] => 2
[2,2,2] => 0
[2,2,1,1] => 0
[2,1,1,1,1] => 0
[1,1,1,1,1,1] => 0
[7] => 0
[6,1] => 1
[5,2] => 1
[5,1,1] => 0
[4,3] => 1
[4,2,1] => 0
[4,1,1,1] => 3
[3,3,1] => 0
[3,2,2] => 0
[3,2,1,1] => 0
[3,1,1,1,1] => 0
[2,2,2,1] => 1
[2,2,1,1,1] => 1
[2,1,1,1,1,1] => 1
[1,1,1,1,1,1,1] => 0
[8] => 0
[7,1] => 0
[6,2] => 2
[6,1,1] => 0
[5,3] => 2
[5,2,1] => 1
[5,1,1,1] => 0
[4,4] => 2
[4,3,1] => 1
[4,2,2] => 1
[4,2,1,1] => 1
[4,1,1,1,1] => 0
[3,3,2] => 1
[3,3,1,1] => 1
[3,2,2,1] => 1
[3,2,1,1,1] => 1
[3,1,1,1,1,1] => 0
[2,2,2,2] => 2
[2,2,2,1,1] => 2
[2,2,1,1,1,1] => 2
[2,1,1,1,1,1,1] => 0
[1,1,1,1,1,1,1,1] => 0
[9] => 0
[8,1] => 1
[7,2] => 0
[7,1,1] => 1
[6,3] => 3
[6,2,1] => 0
[6,1,1,1] => 1
[5,4] => 3
[5,3,1] => 2
[5,2,2] => 2
[5,2,1,1] => 0
[5,1,1,1,1] => 1
[4,4,1] => 2
[4,3,2] => 2
[4,3,1,1] => 2
[4,2,2,1] => 2
[4,2,1,1,1] => 0
[4,1,1,1,1,1] => 1
[3,3,3] => 2
[3,3,2,1] => 2
[3,3,1,1,1] => 2
[3,2,2,2] => 2
[3,2,2,1,1] => 2
[3,2,1,1,1,1] => 0
[3,1,1,1,1,1,1] => 1
[2,2,2,2,1] => 3
[2,2,2,1,1,1] => 3
[2,2,1,1,1,1,1] => 0
[2,1,1,1,1,1,1,1] => 1
[1,1,1,1,1,1,1,1,1] => 0
[10] => 0
[9,1] => 0
[8,2] => 0
[8,1,1] => 2
[7,3] => 0
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Description
The smallest integer d such that the restriction of the representation corresponding to a partition of n to the symmetric group on n-d letters has a constituent of odd degree.
For example, restricting $S_{(6,3)}$ to $\mathfrak S_8$ yields $$S_{(5,3)}\oplus S_{(6,2)}$$ of degrees (number of standard Young tableaux) 28 and 20, none of which are odd. Restricting to $\mathfrak S_7$ yields $$S_{(4,3)}\oplus 2S_{(5,2)}\oplus S_{(6,1)}$$ of degrees 14, 14 and 6. However, restricting to $\mathfrak S_6$ yields
$$S_{(3,3)}\oplus 3S_{(4,2)}\oplus 3S_{(5,1)}\oplus S_6$$ of degrees 5,9,5 and 1. Therefore, the statistic on the partition $(6,3)$ gives 3.
This is related to $2$-saturations of Welter's game, see [1, Corollary 1.2].
For example, restricting $S_{(6,3)}$ to $\mathfrak S_8$ yields $$S_{(5,3)}\oplus S_{(6,2)}$$ of degrees (number of standard Young tableaux) 28 and 20, none of which are odd. Restricting to $\mathfrak S_7$ yields $$S_{(4,3)}\oplus 2S_{(5,2)}\oplus S_{(6,1)}$$ of degrees 14, 14 and 6. However, restricting to $\mathfrak S_6$ yields
$$S_{(3,3)}\oplus 3S_{(4,2)}\oplus 3S_{(5,1)}\oplus S_6$$ of degrees 5,9,5 and 1. Therefore, the statistic on the partition $(6,3)$ gives 3.
This is related to $2$-saturations of Welter's game, see [1, Corollary 1.2].
References
[1] Irie, Y. p-Saturations of Welter's Game and the Irreducible Representations of Symmetric Groups arXiv:1604.07214
Code
def branching_symmetric_group(la, p):
"""
Return a dictionary from partitions to multiplicities.
"""
la = Partition(la)
l = {la: 1}
for i in range(la.size()-p):
l_new = dict()
for mu in l:
for r, _ in mu.removable_cells():
nu = mu.remove_cell(r)
l_new[nu] = l_new.get(nu, 0) + l[mu]
l = l_new
return l
def statistic(la):
"""Return the largest number such that the restriction of the
irreducible representation corresponding to la has a component
relative prime to 2.
"""
la = Partition(la)
for m in range(la.size(), 0, -1):
if any(gcd(StandardTableaux(mu).cardinality(), 2) == 1
for mu in branching_symmetric_group(la, m)):
return la.size()-m
Created
Apr 05, 2017 at 11:28 by Martin Rubey
Updated
Sep 14, 2018 at 18:56 by Martin Rubey
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