Direct Sum of Crystals

sage.combinat.crystals.direct_sum.DirectSumOfCrystals

Direct sum of crystals.

Given a list of crystals \(B_0, \ldots, B_k\) of the same Cartan type, one can form the direct sum \(B_0 \oplus \cdots \oplus B_k\).

INPUT:

  • crystals – a list of crystals of the same Cartan type
  • keepkey – a boolean

The option keepkey is by default set to False, assuming that the crystals are all distinct. In this case the elements of the direct sum are just represented by the elements in the crystals \(B_i\). If the crystals are not all distinct, one should set the keepkey option to True. In this case, the elements of the direct sum are represented as tuples \((i, b)\) where \(b \in B_i\).

EXAMPLES:

sage: C = crystals.Letters(['A',2])
sage: C1 = crystals.Tableaux(['A',2],shape=[1,1])
sage: B = crystals.DirectSum([C,C1])
sage: B.list()
[1, 2, 3, [[1], [2]], [[1], [3]], [[2], [3]]]
sage: [b.f(1) for b in B]
[2, None, None, None, [[2], [3]], None]
sage: B.module_generators
(1, [[1], [2]])
sage: B = crystals.DirectSum([C,C], keepkey=True)
sage: B.list()
[(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)]
sage: B.module_generators
((0, 1), (1, 1))
sage: b = B( tuple([0,C(1)]) )
sage: b
(0, 1)
sage: b.weight()
(1, 0, 0)

The following is required, because DirectSumOfCrystals takes the same arguments as DisjointUnionEnumeratedSets (which see for details).