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Let $W$ be the Weyl group corresponding to the maximal torus $diag(t_1, . . . , t_{n}, t^{−1}_n, . . . , t^{−1}_1)$ in a Borel group of $\operatorname{SO}_{2n}(\mathbb C)$.

What are the matrices corresponding to simple reflections of $W$?

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    $\begingroup$ The group of upper triangular matrices in $SO_{2n}(\mathbf{C})$ is not Borel. One needs to take the quadratic form in a special form, for instance $\sum_{i=1}^nx_ix_{2n+1-i}$. If you're asking for an explicit matrix form, you need to be more specific. Also the maximal torus in a Borel is not unique, so I guess you implicity want it to be defined as the subgroup of diagonal matrices therein. $\endgroup$
    – YCor
    Commented Sep 29, 2019 at 17:48
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    $\begingroup$ Are you asking, for each $s_i \in W$, what is the $T$-coset in $\mathrm{SO}_{2n}(\mathbb{C})$ corresponding to $s_i$ in the identification $W \simeq N(T)/T$? $\endgroup$ Commented Sep 29, 2019 at 19:34
  • $\begingroup$ Yes, you are right, I meant for the torus $T$ to be of the form $diag(t_1, . . . , t_{n}, t^{−1}_n, . . . , t^{−1}_1)$ and $B$ a Borel subgroup containing $T$. $\endgroup$
    – Ami
    Commented Sep 29, 2019 at 20:30
  • $\begingroup$ @YCor, just to be clear, you mean that the group of upper-triangular matrices in $\mathrm{SO}_{2n}$ is not Borel unless one takes a particular quadratic form (in which case it is Borel), right? (I suppose one could say equivalently that the group of upper-triangular matrices isn't Borel unless one chooses a basis adapted appropriately to the choice of quadratic form.) $\endgroup$
    – LSpice
    Commented Sep 30, 2019 at 15:55
  • $\begingroup$ @LSpice Yes. Let's say $T_{2m}$ the whole upper triangular group, and write $\mathrm{SO}(q)$: then $T_{2m}\cap\mathrm{SO}(q)$ is not Borel in $\mathrm{SO}(q)$ in general, notably for the quite usual forms $q=x_1^2+\dots+x_{2m}^2$ (which usually defines $\mathrm{SO}_{2m}$) or $q=x_1^2+\dots+x_m^2-x_{m+1}^2-\dots-x_{2m}^2$ and $q=x_1x_{m+1}+\dots+x_mx_{2m}$ (which are both standard forms for the split form); however it is for $q=x_1x_{2m}+\dots+x_mx_{m+1}$. $\endgroup$
    – YCor
    Commented Sep 30, 2019 at 16:10

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I believe an answer to your question is in the Bourbaki volume dedicated to groups generated by reflections, Lie ..., vol. VI...

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