smooth Brauer-Severis

Around the same time Michel Van den Bergh introduced his Brauer-Severi schemes, Claudio Procesi (extending earlier work of Bill Schelter) introduced smooth orders as those orders $A$ in a central simple algebra $\Sigma$ (of dimension $n^2$) such that their representation variety $\wis{trep}n~A$ is a smooth variety. Many interesting orders are smooth : hereditary orders, trace rings of generic matrices and more generally size n approximations of formally smooth algebras (that is, non-commutative manifolds). As in the commutative case, every order has a Zariski open subset where it is a smooth order. The relevance of this notion to the study of Brauer-Severi varieties is that $XA$ is a smooth variety whenever $A$ is a smooth order. Indeed, the Brauer-Severi scheme was the orbit space of the principal $GLn$-fibration on the Brauer-stable representations (see last time) which form a Zariski open subset of the smooth variety $\wis{trep}n~A \times k^n$. In fact, in most cases the reverse implication will also hold, that is, if $XA$ is smooth then usually A is a smooth order. However, for low n, there are some counterexamples. Consider the so called quantum plane
$Aq=kq[x,y]~:~yx=qxy$ with $~q$ an $n$-th root of unity then one can easily prove (using the fact that the smooth order locus of $Aq$ is everything but the origin in the central variety $~\mathbb{A}^2$) that the singularities of the Brauer-Severi scheme $XA$ are the orbits corresponding to those nilpotent representations $~\phi : A \rightarrow Mn(k)$ which are at the same time singular points in $\wis{trep}n~A$ and have a cyclic vector. As there are singular points among the nilpotent representations, the Brauer-Severi scheme will also be singular except perhaps for small values of $n$. For example, if $~n=2$ the defining relation is $~xy+yx=0$ and any trace preserving representation has a matrix-description $~x \rightarrow \begin{bmatrix} a & b \ c & -a \end{bmatrix}~y \rightarrow \begin{bmatrix} d & e \ f & -d \end{bmatrix}$ such that $~2ad+bf+ec = 0$. That is, $~\wis{trep}2~A = \mathbb{V}(2ad+bf+ec) \subset \mathbb{A}^6$ which is an hypersurface with a unique singular point (the origin). As this point corresponds to the zero-representation (which does not have a cyclic vector) the Brauer-Severi scheme will be smooth in this case. Colin Ingalls extended this calculation to show that the Brauer-Severi scheme is equally smooth when $~n=3$ but has a unique (!) singular point when $~n=4$. So probably all Brauer-Severi schemes for $n \geq 4$ are indeed singular. I conjecture that this is a general feature for Brauer-Severi schemes of families (depending on the p.i.-degree $n$) of non-smooth orders.

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