smooth Brauer-Severis

Around the
same time Michel Van den Bergh introduced his Brauer-Severi schemes,
[Claudio Procesi][1] (extending earlier work of [Bill Schelter][2])
introduced smooth orders as those orders $A$ in a central simple algebra
$\mathrm{\Sigma }$ (of dimension $n^2$) such that their representation variety
${\mathbf{trep}}_{n}A$ 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 $X_A$ is a
smooth variety whenever $A$ is a smooth order. Indeed, the Brauer-Severi
scheme was the orbit space of the principal $G{L}_n$-fibration on the
Brauer-stable representations (see [last time][3]) which form a Zariski
open subset of the smooth variety ${\mathbf{trep}}_{n}A\times k^n$. In fact,
in most cases the reverse implication will also hold, that is, if $X_A$
is smooth then usually A is a smooth order. However, for low n,
there are some counterexamples. Consider the so called quantum plane
$A_q=k_q[x,y]:yx=qxy$ with $q$ an $n$-th root of unity then one
can easily prove (using the fact that the smooth order locus of $A_q$ is
everything but the origin in the central variety ${\mathbb{A}}^2$) that
the singularities of the Brauer-Severi scheme $X_A$ are the orbits
corresponding to those nilpotent representations $\varphi :A\to M_n(k)$ which are at the same time singular points in ${\mathbf{trep}}_{n}A$
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 $x\to \left[\begin{array}{cc}a& b\\ c& -a\end{array}\right]y\to \left[\begin{array}{cc}d& e\\ f& -d\end{array}\right]$ such that
$2ad+bf+ec=0$. That is, ${\mathbf{trep}}_{2}A=\mathbb{V}(2ad+bf+ec)\subset {\mathbb{A}}^6$ 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][4] 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 $n\ge 4$ 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.

[1]: http://venere.mat.uniroma1.it/people/procesi/
[2]: http://www.fact-index.com/b/bi/bill_schelter.html
[3]: https://lievenlb.local/index.php?p=341
[4]: http://kappa.math.unb.ca/~colin/