Given a Cartan matrix, we want to construct a combinatorial graph. If the matrix is , then we start with points. The rows and columns of the matrix correspond to simple roots, and so each of these points corresponds to a simple root as well. Noting this, we will call the points “roots”.
Now given any two simple roots and we can construct the Cartan integers and . Our data about pairs of roots tells us that the product is either , , , or . We’ll connect each pair of roots by this many lines, like so:
The resulting combinatorial graph is called the “Coxeter graph” of This is almost enough information to recover the Cartan matrix. Unfortunately, in the two- and three-edge cases, we know that the two roots have different lengths, and we have no way of telling which is which. To fix this, we will draw an arrow (that looks suspiciously like a “greater-than” sign) pointing from the larger of the two roots to the smaller, like so:
In each case we know one root must be longer than the other, and by a particular ratio. And so with this information we call the graph a “Dynkin diagram”, and it is enough to reconstruct the Cartan matrix. For example, consider the Dynkin diagram called :
We can use this to reconstruct the Cartan matrix:
One nice thing about Dynkin diagrams is that we really don’t care about the order of the roots, like we had to do for the Cartan matrix. They also show graphically a lot of information about the root system. For instance, we may be able to break a Dynkin diagram up into more than one connected component. Any root in one of these components has no edges connecting it to a root in another. This means that , and thus that and are perpendicular. This breaks the base up into a bunch of mutually-perpendicular subsets, which give rise to irreducible components of the root system .
The upshot of this last fact is that irreducibility of a root system corresponds to connectedness of its Dynkin diagram. Thus our project has come down to this question: which connected Dynkin diagrams arise from root systems? At last, we’re almost ready to answer this question.