The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of B- and C-Series Root Systems

Starting from our setup, we construct root systems corresponding to the B_n (for n\geq2) and C_n (for n\geq3) Dynkin diagrams. First will be the B_n series.

As we did for the D_n series, we start out with an n dimensional space with the lattice I of integer-coefficient vectors. This time, though, we let \Phi be the collection of vectors \alpha\in I of squared-length {2} or {1}: either \langle\alpha,\alpha\rangle=2 or \langle\alpha,\alpha\rangle=1. Explicitly, this is the collection of vectors \pm(\epsilon_i\pm\epsilon_j) for i\neq j (signs chosen independently) from the D_n root system, plus all the vectors \pm\epsilon_i.

Similarly to the A_n series, and exactly as in the D_n series, we define \alpha_i=\epsilon_i-\epsilon_{i+1} for 1\leq i\leq n-1. This time, though, to get vectors whose coefficients don’t sum to zero we can just define \alpha_n=\epsilon_n, which is independent of the other vectors. Since it has n vectors, the independent set \Delta=\{\alpha_i\} is a basis for our vector space.

As in the A_n and D_n cases, any vector \epsilon_i-\epsilon_j with i<j can be written


This time, any of the \epsilon_i can be written


Thus any vector \epsilon_i+\epsilon_j can be written as the sum of two of these vectors. And so \Delta is a base for \Phi.

We calculate the Cartan integers. For i and j less than n, we again have the same calculation as in the A_n case, which gives a simple chain of length n-1 vertices. But when we involve \alpha_n things are a little different.



If 1\leq i<n-1, then both of these are zero. On the other hand, if i=n-1, then the first is -2 and the second is -1. Thus we get a double edge from \alpha_{n-1} to \alpha_n, and \alpha_{n-1} is the longer root. And so we obtain the B_n Dynkin diagram.

Considering the reflections with respect to the \alpha_i, we find that \sigma_{\alpha_i} swaps the coefficients of \epsilon_i and \epsilon_{i+1} for 1\leq i\leq n-1. But what about \alpha_n? We calculate

\displaystyle\begin{aligned}\sigma_{\alpha_n}(v)&=v-\frac{2\langle v,\alpha_n\rangle}{\langle\alpha_n,\alpha_n\rangle}\alpha_n\\&=v-2\langle v,\alpha_n\rangle\alpha_n\\&=v-2v^n\epsilon_n\end{aligned}

which flips the sign of the last coefficient of v. As we did in the D_n case, we can use this to flip the signs of whichever coefficients we want. Since these transformations send the lattice I back into itself, they send \Phi to itself and we do have a root system.

Finally, since we don’t have any restrictions on how many signs we can flip, the Weyl group for B_n is exactly the wreath product S_n\wr\mathbb{Z}_2.

So, what about C_n? This is just the dual root system to B_n! The roots of squared-length {2} are left unchanged, but the roots of squared-length {1} are doubled. The Weyl group is the same — S_n\wr\mathbb{Z}_2 — but now the short root in the base \Delta is the long root, and so we flip the direction of the double arrow in the Dynkin diagram, giving the C_n diagram.

March 4, 2010 Posted by | Geometry, Root Systems | 2 Comments