# Braid relations $n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots$ in arbitrary reductive groups

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.

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 $\tilde{c}\tilde{d}^{-1}\sigma(\tilde{c}) + 1 = -\frac{1}{2}j$.