The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of D-Series Root Systems

Starting from our setup, we construct root systems corresponding to the D_n Dynkin diagrams (for n\geq4).

The construction is similar to that of the A_n series, but instead of starting with a hyperplane in n+1-dimensional space, we just start with n-dimensional space itself with the lattice I of integer-coefficient vectors. We again take \Phi to be the collection of vectors \alpha\in I of squared-length 2: \langle\alpha,\alpha\rangle=2. Explicitly, this is the collection of vectors \pm(\epsilon_i\pm\epsilon_j) for i\neq j, where we can choose the two signs independently.

Similarly to the A_n case, we define \alpha_i=\epsilon_i-\epsilon_{i+1} for 1\leq i\leq n-1, but these can only give vectors whose coefficients sum to {0}. To get other vectors, we throw in \alpha_n=\epsilon_{n-1}+\epsilon_n, which is independent of the others. The linearly independent collection \Delta=\{\alpha_i\} has n vectors, and so must be a basis of the n-dimensional space.

As before, any vector in \phi of the form \epsilon_i-\epsilon_j for i<j can be written as


while vectors of the form \epsilon_i+\epsilon_j are a little more complicated. We can start with


and from this we can always build 2\epsilon_j=(\epsilon_j-\epsilon_n)+(\epsilon_j+\epsilon_n) for 1\leq j\leq n-1. Then if i<j\leq n-1 we can write \epsilon_i+\epsilon_j=(\epsilon_i-\epsilon_j)+2\epsilon_j. This proves that \Delta is a base for \Phi.

Again, we calculate the Cartan integers. The calculation for i and j both less than n is exactly as before, showing that these vectors form a simple chain in the Dynkin diagram of length n-1. However, when we involve \alpha_n we find


For i<n-2, this is automatically {0}; for i=n-2, we get the value -1; and for i=n-1 we again get {0}. This shows that the Dynkin diagram of \Delta is D_n.

Finally, we consider the reflections with respect to the \alpha_i. As in the A_n case, 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?

\displaystyle\begin{aligned}\sigma_{\alpha_n}(v)&=v-\frac{2\langle v,\alpha_n\rangle}{\langle\alpha_n,\alpha_n\rangle}\alpha_n\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}+v^n)(\epsilon_{n-1}+\epsilon_n)\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}\epsilon_{n-1}+v^n\epsilon_n)-(v^n\epsilon_{n-1}+v^{n-1}\epsilon_n)\end{aligned}

This swaps the last two coefficients of v and flips their sign. Clearly, this sends the lattice I back to itself, showing that \Phi is indeed a root system.

Now we can use \sigma_{\alpha_n}\sigma_{\alpha_{n-1}} to flip the signs of coefficients of v, two at a time. We use whatever of the \sigma_{\alpha_i} we need to get the two coefficients we want into the last two slots, hit it with \sigma_{\alpha_n}\sigma_{\alpha_{n-1}} to flip them, and then invert the first permutation to move everything back where it started from. In fact, this is a lot like what we saw way back with the Rubik’s cube, when dealing with the edge group. We can effect whatever permutation we want on the coefficients, and we can flip any even number of them.

The Weyl group of D_n is then the subgroup of the wreath product S_n\wr\mathbb{Z}_2 consisting of those transformations with an even number of flips coming from the \mathbb{Z}_2 components. Explicitly, we can write \mathbb{Z}_2^{n-1} as the subgroup of \mathbb{Z}_2^n with sum zero. Then we can let S_n act on \mathbb{Z}_2^n by permuting the components, and use this to give an action of S_n on \mathbb{Z}_2^{n-1}, and thus form the semidirect product S_n\ltimes\mathbb{Z}_2^{n-1}.

March 3, 2010 Posted by | Geometry, Root Systems | 2 Comments



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