Indiana Jones and the Base of Crystals

Start with a quantum group. For each simple root  define  and similarly for the lowering operators f's. We have unique decompositions

for every u in    for that weight.

Define linear mappings M to itself by its action on  as

Let A be the rational functions of q which are regular at q=0. Now we can define the crystal base (L,B).

L is a free A-submodule of M such that the following hold:

B is a basis of L/qL
Both L and B decompose into weight spaces as you would hope.
Closure under the linear mappings said above (include 0 as a possibility on the basis B).

Using the twidled action instead of the original e's or f's allows regularity at q=0. They are off by a factor of  as you can see by comparing the formulas for e and twidle e.

A way to see all this data is with a graph. Assign a vertex to each element of the base and a colored directed edge if you can get from one vertex to the other with the a modified raising operator. The edge is colored by which i you needed to use.

You can tensor two crystal bases together. The new L and B are what you would expect, but the action of the operators is different. For example,

Even though this construction has passed to the q=0 case for the quantum group. This still retains some information about the representation theory like how reducibility, weights and multiplicities.