Unitriangular factorization of twisted Chevalley groups. (Q2854970)

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.NEWLINENEWLINE 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 NEWLINE\[NEWLINESU_3(R)=\{g\in SL_3(R):x^tJ x^\sigma=J\}NEWLINE\]NEWLINE 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 NEWLINE\[NEWLINEx_+(a,b)=\left[\begin{smallmatrix} 1&a&b\\ 0&1&a^\sigma\\ 0&0&1\end{smallmatrix}\right],\quad x_-(a,b)=\left[\begin{smallmatrix} 1&0&0\\ a^\sigma&1&0\\ b&a&1\end{smallmatrix}\right]NEWLINE\]NEWLINE are unipotent root elements. The unitriangular groups are NEWLINE\[NEWLINEU=\{x_+(a,b):aa^\sigma=b+b^\sigma\}\quad\text{and}\quad U^-=\{x_-(a,b):aa^\sigma=b+b^\sigma\}.NEWLINE\]NEWLINE The proof is a simple calculation.