Unitary Steinberg group is centrally closed (Q2849201)

This paper shows that the unitary Steinberg group of a form ring is centrally closed. This was proved by Bak who used matrix entries; the current proof, using roots to classify Steinberg relations among elementary unitary transvections, is shorter and somewhat simpler.NEWLINENEWLINELet \(R\) be an associative ring with involution. Recall that \(\Lambda\) is called a form parameter with respect to a central unitary element \(\lambda\) if \(\{\alpha - \lambda \bar{\alpha} : \alpha \in R\} \subseteq \Lambda \subseteq \{\alpha : \bar{\alpha} = -\lambda \bar{\alpha}\}\) and \(\alpha \Lambda \bar{\alpha} \subseteq \Lambda\) for all \(\alpha \in R\). The form ring \((R,\Lambda)\) and an integer \(n \geq 5\) are fixed from now on. The unitary group \(\text{U} = \text{U}(2n,R,\Lambda)\) is composed of the invertible \(2n\times 2n\) matrices over \(R\) that preserve a standard even \(\lambda\)-Hermitian form \(f \!:\! R^{2n} \times R^{2n} \rightarrow R\) and a standard \(\Lambda\)-quadratic form \(q \!:\! R^{2n} \rightarrow R/\Lambda\).NEWLINENEWLINEAnalogously to Milnor's \(K_2\) theory of a ring, there are elementary unitary transvections in \(\text{U}\), generating the elementary unitary group \(\text{EU} = \text{EU}(2n,R,\Lambda)\). The unitary Steinberg group \(\text{StU} = \text{StU}(2n,R,\Lambda)\) is a generic cover: generated by corresponding formal generators, and subject to relations which hold in any ring. Then, \({K_2U} = {K_2U}(2n,R,\Lambda)\) is the kernel of the natural projection \(\text{StU} \rightarrow \text{EU}\).NEWLINENEWLINEA group is centrally closed if it is perfect and every central extension splits. Showing that \(\text{StU}\) is perfect is not hard. To prove that a central extension splits, the author finds an explicit splitting map, which is verified in a series of lemmas.NEWLINENEWLINEIt then follows that if \(\text{StU}\) is a central extension of \(\text{EU}\), then \(\text{StU}\) is in fact a universal central extension. Moreover, in case where \( {K_2U}\) is central in \(\text{StU}\), then \({K_2U}\) is equal to the Schur multiplier of \(\text{EU}\), namely \({K_2U} = H_2(\text{EU},\mathbb{Z})\).