Start with a quantum group. For each simple root

define
![e^{(n)}_i=e^{n}_i/[n]_{q_i}!](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ugA_suHQm8_bG4QdYrLWF4rf3yFO2JmlcHmYlK151HAHEn8LrQFQecSCyt95Mv95rqSsG33j5i9dWD9JWBtbl8rJqGWPrteMHK4kRAxp3o0g_UTw-ZUtByG_-PmkEz-SlhXvU8DJUqEhQFgl9zU_sINtTshQ6CrQ=s0-d)
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
![[n+1]_q](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uSrlqznCv8SQpcO7OeIu9mxH1NFs2g-sQ2GKWaZFo5IDXyTEH0IXIPKhf1UgNxzKuCbH49Ihs2G_G5wejftPK0MXfB5p1E5fnU-qcxBvOODeM=s0-d)
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.