Let \(T\) be a CartanType with index set \(I\), and \(W\) be a realization of the type \(T\) weight lattice.
A type \(T\) crystal \(C\) is a colored oriented graph equipped with a weight function from the nodes to some realization of the type \(T\) weight lattice such that:
Each edge is colored with a label in \(i \in I\).
For each \(i\in I\), each node \(x\) has:
Furthermore, when they exist,
This crystal actually models a representation of a Lie algebra if it satisfies some further local conditions due to Stembridge [St2003].
REFERENCES:
[St2003] | J. Stembridge, A local characterization of simply-laced crystals, Trans. Amer. Math. Soc. 355 (2003), no. 12, 4807-4823. |
EXAMPLES:
We construct the type \(A_5\) crystal on letters (or in representation theoretic terms, the highest weight crystal of type \(A_5\) corresponding to the highest weight \(\Lambda_1\)):
sage: C = CrystalOfLetters(['A',5]); C
The crystal of letters for type ['A', 5]
It has a single highest weight element:
sage: C.highest_weight_vectors()
[1]
A crystal is an enumerated set (see EnumeratedSets); and we can count and list its elements in the usual way:
sage: C.cardinality()
6
sage: C.list()
[1, 2, 3, 4, 5, 6]
as well as use it in for loops:
sage: [x for x in C]
[1, 2, 3, 4, 5, 6]
Here are some more elaborate crystals (see their respective documentations):
sage: Tens = TensorProductOfCrystals(C, C)
sage: Spin = CrystalOfSpins(['B', 3])
sage: Tab = CrystalOfTableaux(['A', 3], shape = [2,1,1])
sage: Fast = FastCrystal(['B', 2], shape = [3/2, 1/2])
sage: KR = KirillovReshetikhinCrystal(['A',2,1],1,1)
One can get (currently) crude plotting via:
sage: Tab.plot()
If dot2tex is installed, one can obtain nice latex pictures via:
sage: K = KirillovReshetikhinCrystal(['A',3,1], 1,1)
sage: view(K, pdflatex=True, tightpage=True) #optional - dot2tex graphviz
or with colored edges:
sage: K = KirillovReshetikhinCrystal(['A',3,1], 1,1)
sage: G = K.digraph()
sage: G.set_latex_options(color_by_label = {0:"black", 1:"red", 2:"blue", 3:"green"}) #optional - dot2tex graphviz
sage: view(G, pdflatex=True, tightpage=True) #optional - dot2tex graphviz
For rank two crystals, there is an alternative method of getting metapost pictures. For more information see C.metapost?
See also the categories Crystals, ClassicalCrystals, FiniteCrystals, HighestWeightCrystals.
Todo
Most of the above features (except Littelmann/alcove paths) are in MuPAD-Combinat (see lib/COMBINAT/crystals.mu), which could provide inspiration.
Bases: sage.combinat.backtrack.GenericBacktracker
Time complexity: \(O(nF)\) amortized for each produced element, where \(n\) is the size of the index set, and \(F\) is the cost of computing \(e\) and \(f\) operators.
Memory complexity: \(O(D)\) where \(D\) is the depth of the crystal.
Principle of the algorithm:
Let \(C\) be a classical crystal. It’s an acyclic graph where all connected component has a unique element without predecessors (the highest weight element for this component). Let’s assume for simplicity that \(C\) is irreducible (i.e. connected) with highest weight element \(u\).
One can define a natural spanning tree of \(C\) by taking \(u\) as the root of the tree, and for any other element \(y\) taking as ancestor the element \(x\) such that there is an \(i\)-arrow from \(x\) to \(y\) with \(i\) minimal. Then, a path from \(u\) to \(y\) describes the lexicographically smallest sequence \(i_1,\dots,i_k\) such that \((f_{i_k} \circ f_{i_1})(u)=y\).
Morally, the iterator implemented below just does a depth first search walk through this spanning tree. In practice, this can be achieved recursively as follow: take an element \(x\), and consider in turn each successor \(y = f_i(x)\), ignoring those such that \(y = f_j(x^{\prime})\) for some \(x^{\prime}\) and \(j<i\) (this can be tested by computing \(e_j(y)\) for \(j<i\)).
EXAMPLES:
sage: from sage.combinat.crystals.crystals import CrystalBacktracker
sage: C = CrystalOfTableaux(['B',3],shape=[3,2,1])
sage: CB = CrystalBacktracker(C)
sage: len(list(CB))
1617
sage: CB = CrystalBacktracker(C, [1,2])
sage: len(list(CB))
8