The Unapologetic Mathematician

Mathematics for the interested outsider

Coxeter Graphs and Dynkin Diagrams

We’ve taken our root system and turned it into a Cartan matrix. Now we’re going to take our Cartan matrix and turn it into a pictorial form that we can really get our hands on.

Given a Cartan matrix, we want to construct a combinatorial graph. If the matrix is n\times n, then we start with n points. The n rows and columns of the matrix correspond to n 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 \alpha and \beta we can construct the Cartan integers \alpha\rtimes\beta and \beta\rtimes\alpha. Our data about pairs of roots tells us that the product (\alpha\rtimes\beta)(\beta\rtimes\alpha) is either {0}, {1}, {2}, or {3}. We’ll connect each pair of roots by this many lines, like so:

  • (\alpha\rtimes\beta)(\beta\rtimes\alpha)=0:
  • (\alpha\rtimes\beta)(\beta\rtimes\alpha)=1:
  • (\alpha\rtimes\beta)(\beta\rtimes\alpha)=2:
  • (\alpha\rtimes\beta)(\beta\rtimes\alpha)=3:

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 F_4:

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 \alpha in one of these components has no edges connecting it to a root \beta in another. This means that \alpha\rtimes\beta=0, and thus that \alpha and \beta are perpendicular. This breaks the base \Delta up into a bunch of mutually-perpendicular subsets, which give rise to irreducible components of the root system \Phi.

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.


February 18, 2010 Posted by | Geometry, Root Systems | 10 Comments