Sukru  lead this week’s discussion and worked through the root systems of type $A_n, B_n, C_n, D_n$. Before covering the maths I should first mention that Sukru feels that we should come up with a better title for the study group as he doesnt think that it is very reflective of what we are actually doing.  We are yet to agree on a new one so I have stuck with the old one.

Recall from the first study group post that a root system is a finite spanning subset $\Phi$ of $E=\mathbb{R}^n\backslash\{0\}$  equipped with a positive definite symmetric bilinear form $(.,.)$ such that

• $\Phi$ is invariant under $\sigma_\alpha$ for each $\alpha\in\Phi$. ($\sigma_\alpha$ is the reflection through the hyperplane orthogonal to $\alpha$.)
• given $\alpha\in\Phi$, the only scalar multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$.
• $\frac{2(\alpha,\beta)}{(\alpha,\alpha)}$ is an integer for all $\alpha,\beta\in\Phi$ (crystallographic condition).

We begin with a general construction. Let $B=\{e_1,\ldots,e_n\}$ be a basis for $E=\mathbb{R}^n$. Let I be the $\mathbb{Z}$-lattice of B and let $\Phi$ be the set of all vectors in I which have certain length or lengths (two lengths). This set is finite, spans E and does not contain $0$  (as we don’t allow vectors of length $0$.)  The certain lengths are  chosen so that the only scalar multiples of $\alpha$ in $\Phi$ are $\pm\alpha$ and the squared lengths divide 2. Note that for all $\alpha,\beta\in I$ we have $(\alpha,\beta)\in\mathbb{Z}$. Since each $\sigma_\alpha$ preserves I (by the crystallographic condition) and preserves lengths it follows that $\sigma_\alpha$  preserves $\Phi$. Hence $\Phi$ is a root system.

Type $A_n$: Choose $E\leqslant \mathbb{R}^{n+1}$ to be the subspace orthogonal to $e_1+e_2+\cdots +e_{n+1}$. Let $I'=I\cap E$ and define $\Phi=\{\alpha\in I'\mid (\alpha,\alpha)=2\}$.  Then

$\Delta=\{e_1-e_2,e_2-e_3,\ldots,e_n-e_{n+1}\}\subset\Phi$

and in fact $\Phi=\{\mp(e_i-e_j)\mid i\neq j\}$ and we can take $\Delta$  to be a set of simple roots. (check the first study group post for definition of simple roots).  Let $W=\langle \sigma_\alpha\mid\alpha\in\Phi\rangle=\langle \sigma_\alpha\mid \alpha\in\Delta\rangle$ and set $\alpha_i=e_i-e_{i+1}$. Then

$\displaystyle{\sigma_{\alpha_i}(e_i)=e_i-\frac{2(e_i-e_{i+1},e_i)}{(e_i-e_{i+1},e_i-e_{i+1})}(e_i-e_{i+1})=e_{i+1}}$.

Also, $\sigma_{\alpha_i}(e_{i+1})=e_i$. Thus we can define an isomorphism $\varphi: W\rightarrow S_{n+1}$, $\sigma_{\alpha_i}\mapsto (i,i+1)$.

The Coxeter graph is

Type $B_n$: Let $E=\mathbb{R}^n$ and

$\Phi=\{\alpha\in I\mid (\alpha,\alpha)=1 \text{ or }2 \} =\{\pm e_i\}\cup\{\mp (e_i\pm e_j) \mid i\neq j\}$.

Then a set of simple roots is $\alpha_1=e_1-e_2, \alpha_2=e_2-e_3,\ldots, \alpha_{n-1}=e_{n-1}-e_n, \alpha_n=e_n$.  In particular, we can write $e_i=(e_i-e_{i+1})+\ldots+(e_{n-1}-e_n)-e_n$. Clearly the first $n-1$ are long roots and $\alpha_n$ is a short root.

Let $W=\langle \sigma_\alpha\mid\alpha\in\Phi\rangle=\langle \sigma_\alpha\mid \alpha\in\Delta\rangle$. Then $\sigma_{e_i}=-e_i$ and corresponds to a signed permutation which changes the sign of i and fixes all the other j. In fact $V=\langle\sigma_{e_i}\rangle\vartriangleleft W$.  Moreover, $W/V\cong \langle\sigma_{\alpha_i}\mid\alpha_i=e_i-e_{i+1}\rangle\cong S_n$. Thus $W\cong C_2^n\rtimes S_n$.

The Coxeter graph is

Type $C_n$, $n\geq 3$: Let $E=\mathbb{R}^n$ and

$\Phi=\{\alpha\in I\mid (\alpha,\alpha)=4 \text{ or } 2\}=\{\pm 2 e_i\}\cup\{\mp(e_i\pm e_j) \mid i\neq j\}$.

This is the dual of $B_n$ so we get the same Weyl group. A set of simple roots  now has $n-1$ short roots and 1 long root.

Type $D_n$, $n\geq 4$: Here $E=\mathbb{R}^n$ and

$\Phi=\{\alpha\in I\mid (\alpha,\alpha)=2\}=\{\pm (e_i\pm e_j)\mid i\neq j\}$.

A set of simple roots is $\alpha_1=e_1-e_2,\alpha_2=e_2-e_3,\ldots,\alpha_{n-1}=e_{n-1}-e_n,\alpha_n=e_{n-1}+e_n$.

As seen in type $A_n$, for $i=1,2,\ldots,n-1$, we have $\sigma_{\alpha_i}(e_i)=e_{i+1}$ and $\sigma_{\alpha_i}(e_{i+1})=e_i$. Thus $\langle \sigma_{\alpha_i}\mid 1\leq i\leq n-1\rangle \cong S_n$. Now

$\displaystyle{ \sigma_{\alpha_n}(e_n)=e_n-\frac{2(e_{n-1}+e_n,e_n)}{(e_{n-1}+e_n,e_{n-1}+e_n)}(e_{n-1}+e_n)=-e_{n-1} }$

and

$\displaystyle{ \sigma_{\alpha_n}(e_{n-1})=e_{n-1}-\frac{2(e_{n-1}+e_n,e_{n-1})}{(e_{n-1}+e_n,e_{n-1}+e_n)}(e_{n-1}+e_n)=-e_n }$

Hence $\sigma_{\alpha_n}$ induces the signed permutation of $\{\pm 1,\ldots, \pm n\}$ which interchanges $n$ and $n-1$ and multiplies both by $-1$. Thus $W=\langle \sigma_{\alpha_i}\rangle$ includes all the changes of signs which change an even number of signs. Hence $W\cong C_2^{n-1}\rtimes S_n$.

The Coxeter graph is

In your general construction, ‘the squared lengths divide 2’ is too strong a condition, and is not satified by $C_n$. Instead it should just be ‘such that $\frac{2(\alpha,\beta)}{(\alpha,\alpha)}\in \mathbb{Z}$‘.