# Nontrivial relations of the irreducible root systems

For the root system of the type $$A_n$$, the roots are $$\alpha _{i,j}$$, $$1\le i\neq j\le n$$,
we have the nontrivial relations $$(x_{i,j} (t), x_{j,k}(u)) = x_{i,k}(tu)$$ if $$i, j, k$$ are distinct. ($$x_{i,j}$$ are the elements of the root subgroup $$U_{\alpha_{i,j}}$$)
I'm looking for sources that lists similar commutator relation in the diffrent root systems, and also the permutation of the longest Weyl element on the roots.

Specifically, I'm trying to find relations that enables me to move down the root element by hight using the highest root element and the longest Weyl element.
For example in $$A_n$$, $$(\omega_0 x_{i-1,n}(t)\omega_0^{-1},x_{1,n}(u))=(x_{i,1}(t),x_{1,n}(u))=x_{i,n}(tu)$$ ($$\omega_0$$ is the longest Weyl element and $$x_{1,n}$$ the highest root element).

• I think that the $\omega_0$ (I prefer $w_0$) conjugate of $\alpha_{i - 1, n}$ is $\alpha_{n - i + 2, 1}$, not $\alpha_{i, 1}$. Also note that it doesn't make sense to conjugate an element of $G$ by an element of the Weyl group $W$; the latter is defined only up to translation by the maximal torus $T$, and which translate you choose affects the conjugation on root subgroups. – LSpice Aug 14 at 11:20
• Anyway, the commutation relations among root groups are known as "Chevalley's commutator formula". See, for example, §5.2 of Carter's Simple groups of Lie type or Proposition 1.2.3 of Adler's Refined anisotropic $K$-types …. – LSpice Aug 14 at 11:27
• Yes, I took $T$ to be maximal torus of the Borel subgroup(like in the Bruhat decomposition). By this construction $\sigma \in W \simeq S_n$, $p_σ$ should be the matrix with $1$ on the $(σ(i), i)-entry$ for every $1 ≤ i ≤ n − 1$, $(−1)^{n+1}$ on the $(σ(n), n)-entry$ and zero elsewhere and the longest Weyl element "in" $G$ is $p_{σ_{(12..n)}}$. I'm aware of Chevalley's commutator formula, but I was unsuccessful in finding a general way to move down a root element in Chevalley groups. – Ami Aug 14 at 12:09
• If it helps I took the example from this article:[ Meiri - Generating pairs for finite index subgroups of $SL(n,Z)$] page 3 which I'm trying to expand. : arxiv.org/abs/1511.07798 – Ami Aug 14 at 14:21
• I guess you mean $(-1)^{\operatorname{sgn}(\sigma)}$, not $(-1)^{n + 1}$, for the $(\sigma(n), n)$ entry; otherwise your matrix doesn't lie in $\operatorname{SL}_n$ when $n$ is even and $\sigma$ is the identity. I also guess your notation $\sigma_{(1 2 \ldots n)}$ means just $(1\ 2\ \ldots\ n)$, an $n$-cycle; this is not the longest element of the Weyl group but a Coxeter element, which has length $n$. The longest element is $(1\ n)(2\ n - 1)(3\ n - 2)\dotsb$, which has length $n(n - 1)/2$. (Also notice you've mislabelled your root system; it's $A_{n - 1}$, not $A_n$.) – LSpice Aug 14 at 15:51