How do you Calculate the Size of the Exceptional Irreducible Root Systems?

Question

There are three exceptional irreducible root systems $E_6$, $E_7$, and $E_8$ which correspond to the Dynkin diagrams

$$ E_6\; \begin{aligned} &\>\bullet \\[-1ex] &\,\,\mid \\[-1ex] \bullet-\bullet-&\bullet-\bullet-\bullet \end{aligned} \qquad E_7\; \begin{aligned} &\>\bullet \\[-1ex] &\,\,\mid \\[-1ex] \bullet-\bullet-&\bullet-\bullet-\bullet-\bullet \end{aligned} \qquad E_8\; \begin{aligned} &\>\bullet \\[-1ex] &\,\,\mid \\[-1ex] \bullet-\bullet-&\bullet-\bullet-\bullet-\bullet-\bullet \end{aligned} $$

and have $72$, $126$, and $240$ roots respectively. Wikipedia provides brief explanations of how to calculate the size of these root systems for $E_6$, $E_7$, and $E_8$, basically saying the roots of each system must look like a vector of some form, and then counting the number of such vectors. But that doesn't really say why the roots must have that form, or for the case of $E_6$ and $E_7$ why it's a good idea to think of them as embedded in a higher dimensional space. Is there a cleaner way to calculate the size of these root systems? Maybe it would be a good idea to construct $E_8$ and realize $E_6$ and $E_7$ as subsystems of $E_8$ to count their roots? Ideally I'm looking for a method that would work well as a presentation to a class.

Answer 1

Another way to calculate the cardinality of the set of roots is via the formula $|R|=nh$, where $n$ is the rank and $h$ is the Coxeter number. In the case of a real reflection group, the Coxeter number is the largest degree of a fundamental invariant.

For instance for $E_8$, the Coxeter number is $30$, implying that the number of roots is $$|R|=8 \cdot 30=240.$$

In practice, one might wish to start with the Dynkin diagram and nothing else, and compute $h$ and the number of roots. Here is one quick way to do that: first, play the numbers game on the Dynkin diagram to find the highest root. See e.g. here

https://mathoverflow.net/questions/13074/figure-out-the-roots-from-the-dynkin-diagram

Of course, you could use this algorithm for computing all the roots, but it's much faster to get the highest root. Then obtain $h$ using the formula $$m_1+m_2+\cdots+m_n=h-1, \quad \text{where} \quad \phi=m_1\alpha_1+\cdots+m_n \alpha_n$$ is the highest root and $\alpha_1,\dots,\alpha_n$ are the simple roots.