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
: 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
smooth variety whenever
scheme was the orbit space of the principal
Brauer-stable representations (see [last time][3]) which form a Zariski
open subset of the smooth variety
in most cases the reverse implication will also hold, that is, if
is smooth then usually A is a smooth order. However, for low n,
there are some counterexamples. Consider the so called quantum plane
can easily prove (using the fact that the smooth order locus of
everything but the origin in the central variety
the singularities of the Brauer-Severi scheme
corresponding to those nilpotent representations
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
representation has a matrix-description
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
when
indeed singular. I conjecture that this is a general feature for
Brauer-Severi schemes of families (depending on the p.i.-degree
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/