## Unitriangular factorization of twisted Chevalley groups.(English)Zbl 1283.20056

Let $$F$$ be one of the fields: (i) $$\mathbb C$$ or (ii) a finite field of the form $$\mathbb F_{q^2}$$. Consider the twisted Chevalley group $$SU_{2n+1,F}$$ of type $$^2A_{2n}$$. The main result of the paper asserts that $$SU_{2n+1}(F)=UU^-UU^-$$ or $$UU^-UU^-U$$ according as to whether $$F$$ is as in case (ii) or as in case (i). Here, $$U,U^-$$ are unipotent radicals of a Borel subgroup and its opposite. The author uses an old, beautiful theorem due to Oleg Tavgen’ which has been employed by others also over the years. The induction base is the group $$SU_3$$ for which the group as well as $$U,U^-$$ can be explicitly described as follows.
Let $$R$$ be any commutative ring and $$\sigma\colon R\to R$$ an involution. In the cases (i) and (ii), $$\sigma$$ is complex conjugation and $$x\mapsto x^q$$, respectively. Then $SU_3(R)=\{g\in SL_3(R):x^tJ x^\sigma=J\}$ where $$J=\left[\begin{smallmatrix} 0&0&-1\\ 0&1&0\\ -1&0&0\end{smallmatrix}\right]$$. If $$a,b\in R$$ such that $$aa^\sigma=b+b^\sigma$$, then $x_+(a,b)=\left[\begin{smallmatrix} 1&a&b\\ 0&1&a^\sigma\\ 0&0&1\end{smallmatrix}\right],\quad x_-(a,b)=\left[\begin{smallmatrix} 1&0&0\\ a^\sigma&1&0\\ b&a&1\end{smallmatrix}\right]$ are unipotent root elements. The unitriangular groups are $U=\{x_+(a,b):aa^\sigma=b+b^\sigma\}\quad\text{and}\quad U^-=\{x_-(a,b):aa^\sigma=b+b^\sigma\}.$ The proof is a simple calculation.

### MSC:

 20G35 Linear algebraic groups over adèles and other rings and schemes 20H05 Unimodular groups, congruence subgroups (group-theoretic aspects) 15A23 Factorization of matrices
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### References:

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