# The Unapologetic Mathematician

## 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:

$\displaystyle\begin{pmatrix}2&-1&0&0\\-1&2&-2&0\\{0}&-1&2&-1\\{0}&0&-1&2\end{pmatrix}$

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.

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February 18, 2010 - Posted by | Geometry, Root Systems

## 10 Comments »

1. It’s a minor point, but just above the final diagram you talk about the Dynkin diagram being enough to reconstruct the “Coxeter matrix”; I think you mean “Cartan matrix”.

Like many other readers, I expect, the fact that I haven’t commented before doesn’t mean I’m not enjoying the posts. Keep up the good work!

Comment by Jonathan | February 19, 2010 | Reply

2. Thanks, Jonathan. Fixed.

Comment by John Armstrong | February 19, 2010 | Reply

3. [...] up to isomorphism. We’ve shown that for each such root system we can construct a connected Dynkin diagram, which determines a Cartan matrix, which determines the root system itself, up to isomorphism. So [...]

Pingback by The Classification of (Possible) Root Systems « The Unapologetic Mathematician | February 19, 2010 | Reply

4. [...] strategy is to determine which Coxeter graphs can arise from actual root systems, and then see what Dynkin diagrams we can get. Since Coxeter [...]

Pingback by Proving the Classification Theorem I « The Unapologetic Mathematician | February 22, 2010 | Reply

5. [...] proven the classification theorem, we know all about root systems, right? No! All we know is which Dynkin diagrams could possibly arise from root systems. We don’t know whether there actually exists a root [...]

Pingback by Construction of Root Systems (setup) « The Unapologetic Mathematician | March 1, 2010 | Reply

6. [...] Transformations of Dynkin Diagrams Before we continue constructing root systems, we want to stop and observe a couple things about transformations of Dynkin diagrams. [...]

Pingback by Transformations of Dynkin Diagrams « The Unapologetic Mathematician | March 5, 2010 | Reply

7. [...] But we need another short root which will both give a component in the direction of and will give us access to . Further, it should be orthogonal to both and , and should have a Cartan integer of with in either order. For this purpose, we pick , which then gives us the last vertex of the Dynkin diagram. [...]

Pingback by Construction of the F4 Root System « The Unapologetic Mathematician | March 9, 2010 | Reply

8. [...] of . But each shuffles around the roots in , and these roots correspond to the vertices of the Dynkin diagram of the root system. And for to be an automorphism of , it must preserve the Cartan integers, and [...]

Pingback by The Automorphism Group of a Root System « The Unapologetic Mathematician | March 11, 2010 | Reply

9. [...] To classify these, we defined Cartan matrices and verified that we can use it to reconstruct a root system. Then we turned Cartan matrices into Dynkin diagrams. [...]

Pingback by Root Systems Recap « The Unapologetic Mathematician | March 12, 2010 | Reply

10. Isn’t the Cartan matrix the transpose of what it should be? Kac says that $|a_{ij}|>1$ implies the arrow points towards vertex i. It points towards 3, so $a_{32}=-2$, right?

Comment by Alex | February 20, 2011 | Reply