# The Unapologetic Mathematician

## 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 can be written as

$\displaystyle\epsilon_i-\epsilon_j=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{j-1}-\epsilon_j)$

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

$\displaystyle\epsilon_i+\epsilon_n=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{j-1}+\epsilon_j)$

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 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

$\displaystyle\frac{2\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle}{\langle\epsilon_i-\epsilon_{i+1},\epsilon_i-\epsilon_{i+1}\rangle}=\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle$

For $i, 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 $S_n\ltimes\mathbb{Z}_2^{n-1}$.