Let $e_i$ be the standard unit basis vector in $\mathbb{R}^{l+1}$
I'm trying to show that the $R = \{ \pm(e_i-e_j) : 1 \leq i < j \leq l+1\}$ is a root system for $E = spanR = \{ \Sigma \alpha_i e_i : \Sigma \alpha_i = 0\}$
I'm struggling to show that given $\alpha \in R$, we have that the reflection about the hyperplane normal to $\alpha$ permutes the elements of $R$. The reflection operator of $\alpha$ is denoted $s_\alpha$.
We have that $s_\alpha(\gamma) = \gamma - 2\frac{(\gamma,\alpha)}{(\alpha,\alpha)}\alpha$.
My calculations have not been working out. Can somebody help me with this? Thank you
Let $v_{ij} = e_i - e_j$. We find that $$ s_{v_{ij}}\left(\sum_k \alpha_k e_k \right) = \alpha_j e_i + \alpha_i e_j + \sum_{k \notin\{i,j\}} \alpha_k e_k. $$ From there, I think it should be clear that $s_{v_{ij}}$ indeed permutes the elements of $R$.
With that established, we find the following: if $\{i,j\},\{p,q\}$ are disjoint, $$ s_{v_{ij}} v_{pq} = v_{pq} \in R. $$ If $p = i$ and $q \notin \{i,j\}$, we have $$ s_{v_{ij}} v_{pq} = s_{v_{ij}} v_{iq} = v_{jq} \in R. $$ Since $s_{v_{ij}} = s_{v_{ji}}$ and $v_{pq} = -v_{qp}$, all cases where there is one common index are similar to this one. Finally, if $\{i,j\} = \{p,q\}$, we find that $s_{v_{ij}}v_{pq} = -v_{pq} \in R$.
In each case, $s_{v_{ij}}$ takes elements of $R$ to $R$. Because $s_{v_{ij}}$ is invertible, the map induced on $R$ must be bijective.