# The Unapologetic Mathematician

## 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 can be written

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

This time, any of the $\epsilon_i$ can be written

$\epsilon_i=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{n-1}-\epsilon_n)+\epsilon_n$

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.

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

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

If $1\leq i, 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

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