I'm currently trying to prove or disprove the following claim. First let me set up some notation.


Let $G$ be a connected reductive group over a field $K$, let $S \leq Z \leq N \leq G$ be respectively a maximal split torus, its centralizer and its normalizer. Let $\Phi$ be the root system of the pair $(G,S)$, and for $\alpha \in \Phi$ let $U\_\alpha \leq G$ be the associated root subgroup (possibly noncommutative, when $G$ is not split). Given $u \in U\_\alpha(K)-\{1\}$, there exist unique elements $m(u) \in N(K)$ and $v,v' \in U\_{-\alpha}(K)$ such that $vuv' = m(u)$. The image of the element $m(u)$ in the Weyl group $W\_0 = N(K)/Z(K)$ is the reflection $s\_\alpha$ associated to the root $\alpha$, where $W\_0$ is naturally identified with the Weyl group of $\Phi$ via the action of $N(K)$ on $X^\ast(S)$.


**Now for the claim**: For any two roots $\alpha, \beta \in \Phi$ with $\alpha,\beta$ not parallel (an obvious necessary condition), and **any** $u \in U\_\alpha(K)-\{1\}$ and $v \in U\_\beta(K)-\{1\}$ the braid relation


$n\_\alpha n\_\beta n\_\alpha \ldots = n\_\beta n\_\alpha n\_\beta \ldots$


in $N(K)$ holds true, where the number of factors on both sides equal the order of $s\_\alpha s\_\beta$, and $n\_\alpha = m(u)$ and $n\_\beta = m(v)$.


I think I have a proof of this in the case when $\Phi$ is reduced. So a counterexample should involve non-reduced root systems. I tried to disprove the claim using the example $G = SU(h)$ of a quasi-split unitary group (example 1.15 in Tits Corvallis article), but unfortunately (or fortunately) the claim seems to hold in this case. (**edit**: Even for unitary groups over division algebras the claim seems to hold true.)


So if you either have a counterexample to the above claim or a proof (with or without reference), it would help me a lot.



Braid relations $n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots $ in arbitrary reductive groups
Despite what I wrote earlier **the claim is not true**, despite Proposition 6.1.8 of Bruhat-Tits (first volume). Said proposition states that



> 
> **Proposition 6.1.8**: Let $\Phi$ be a root system of rank $2$ and let $(T, (U\_a,M\_a)\_{a \in \Phi})$ be a *generative* root datum. Order the set $\Phi^{\text{red}} = \{ a \in \Phi : \frac{1}{2}a \not\in \Phi\}$ of reduced roots in a circular order, i.e. write $\Phi^{\text{red}} = \{a\_1,\ldots{},a\_{2k}\}$ with $\Phi \cap (\mathbb{R}\_{> 0} a\_{i-1} + \mathbb{R}\_{> 0} a\_{i+1}) = \{a\_{i-1},a\_i,a\_{i+1}\}$. Then $\Delta = \{a\_1, a\_k\}$ is a basis of $\Phi^{\text{red}}$ with $\{a\_1,\ldots,a\_k\}$ being the set of positive roots. Let $a = a\_1$ and $b = a\_k$.
> 
> 
> For any $u \in U\_a$, $u' \in U\_b$ with $u,u' \neq 1$ the relation
>  $mm'm\ldots = m'mm'\ldots$
>  holds true, where the number of factors on both sides is $k$, and $m = m(u)$, $m' = m(u')$.
> 
> 
> 


I thought that my claim would follow from this by constructing for two given non-parallel roots $\alpha,\beta$ a rank two root subsystem $\Psi \subseteq \Phi$ for which $\{\alpha, \beta\}$ forms a basis. Although it is possible to do so, one cannot restrict the root datum $(T,(U\_a,M\_a)\_{a \in \Phi})$ defined by the reductive group in question to a root datum $(T, (U\_a,M\_a)\_{a \in \Psi})$ in general, since the condition, that for $a,b \in \Psi$ the commutator subgroup $(U\_a, U\_b)$ lies in the subgroup generated by the $U\_{pa+qb}$ with $p,q \in \mathbb{N}\_{\geq 1}$, might be violated, unless $\Psi = \Phi \cap \mathbb{Z}\Psi$. **Hence the proposition implies the claim only for roots** $\alpha, \beta$ **which are (up to a multiple) part of a root basis**.


**To see that the claim is false in general, consider the following example of a special unitary group over a division algebra** (which I had already considered, but the roots $\alpha, \beta$ I chose formed part of a basis). Let $D$ be a central division algebra over a field $K$ which is equipped with an involution $\sigma: D \rightarrow D$ (i.e. $\sigma(ab) = \sigma(b)\sigma(a)$ and $\sigma^2 = \text{id}$) which is the identity on $K$. Let $n \geq 1$, $I = \{\pm 1,\ldots{},\pm n\}$ and let $V$ be the free right-$D$-module with basis $e\_{-n},\ldots{},e\_n$. Consider the hermitian form $h$ on $V$ given in terms of coordinates by $h(x,y) = \sum\_{i \in I} \sigma(x\_i)y\_{-i} + \sigma(x\_0)y\_0$. Let $G$ be the algebraic group $\mathbf{SU}(V,h)$ considered naturally as a subgroup of $\mathbf{GL}\_D(V)$. Let $S$ be the 'obvious' maximal split subtorus of $G$, i.e. on rational points we have


$S(K) = \{ \text{diag}(d\_{-n},\ldots{},d\_n) : d\_i \in K^\times \text{ for }i \in I \cup \{0\}, d\_{-i}d\_i = d\_0 = 1\text{ for } i \in I \}$


The root system $\Phi = \Phi(G,S)$ is then given as follows. Let $\chi\_1,\ldots,\chi\_n$ be the 'obvious' $\mathbb{Z}$-basis of $X^\ast(S)$, i.e. $\chi\_i(\text{diag}(d\_{-n},\ldots{},d\_n)) = d\_i$. Then we have


$\Phi = \{ \chi\_i - \chi\_j : i,j \in I,\ i \neq \pm j\} \cup \{ \chi\_i, 2\chi\_i : i \in I \}$


(type $\text{BC}\_n$) or


$\Phi = \{ \chi\_i - \chi\_j : i,j \in I,\ i \neq \pm j\} \cup \{ \chi : i \in I \}$


(type $\text{B}\_n$), according to whether or not $D$ contains an element $x$ with $\sigma(x) = -x$. In either case, the root subgroups are given as follows. For $i,j \in I$, $i \neq \pm j$ we have


$U\_{\chi\_i - \chi\_j}(K) = \{ u\_{i,j}(d) : d \in D \}$


Here $u\_{i,j}(d) = 1 + d e\_{i,j} - \sigma(d)e\_{-j,-i}$ as a sum in the matrix ring $\text{End}\_D(V)$ and $e\_{i,j}$ denotes the matrix which has a one in the $i$-th row and $j$-th column, and zeroes everywhere else. For $i \in I$


$U\_{\chi\_i}(K) = \{ u\_i(c,d) : c,d \in D,\ \sigma(c)c + d + \sigma(d) = 0 \}$


where $u\_i(c,d) = 1 + ce\_{0,-i} - \sigma(c)e\_{i,0} + de\_{i,-i}$. The elements $m(u)$ are then given as follows.


$m(u\_{i,j}(d)) = \tau \text{diag}(d\_{-n},\ldots{},d\_n)$


Here $\tau$ is the permutation $(i\ j)(-i\ -j)$ of $I\cup \{0\}$ considered as a monomial matrix in $\text{GL}\_D(V)$, and $d\_i = d$, $d\_{-i} = \sigma(d)^{-1}$, $d\_j = -d^{-1}$, $d\_{-j} = -\sigma(d)$ and $d\_k = 1$ for all other $k$. For the other roots we have


$m(u\_i(c,d)) = \tau \text{diag}(d\_{-n},\ldots{},d\_n)$


where $\tau = (i\ -i)$ and $d\_i = d$, $d\_{-i} = \sigma(d)^{-1}$, $d\_0 = cd^{-1}\sigma(c) + 1$ and $d\_k = 1$ for all other $k$.


**Now let** $i \neq \pm j$, $\alpha = \chi\_i$, $\beta = \chi\_j$ and $n\_\alpha = m(u\_i(c,d))$, $n\_\beta = m(u\_j(\tilde{c},\tilde{d}))$. Since $s\_\alpha s\_\beta$ is of order two, the braid relations reduces to the commutativity $n\_\alpha n\_\beta = n\_\beta n\_\alpha$. But an easy computation shows that $n\_\alpha, n\_\beta$ commute if and only if $cd^{-1}\sigma(c)+1$ and $\tilde{c}\tilde{d}^{-1}\sigma(\tilde{c}) + 1$ commute, which is easily seen to be false in general. For instance take $K = \mathbb{R}$, $D$ the quaternions and let $c = \tilde{c} = 1$, $d = -\frac{1}{2} + \frac{1}{2}i$, $\tilde{d} = -\frac{1}{2} + \frac{1}{2}j$. Then $cd^{-1}\sigma(c) + 1 = -\frac{1}{2}i$ and $\tilde{c}\tilde{d}^{-1}\sigma(\tilde{c}) + 1 = -\frac{1}{2}j$.


