on September 28, 2004 by lieven in geometry, Comments (1)
smooth Brauer-Severis
non-commutative geometry
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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.
Hi, I'm a professor of mathematics at the University of Antwerp, in Belgium. My research areas are non-commutative algebra and non-commutative geometry. Sometimes, you can also find me here :
hyper-resolutions | neverendingbooks
January 12, 2008 @ 4:08 pm
[...] Last time we saw that for $A$ a smooth order with center $R$ the Brauer-Severi variety $XA$ is a smooth variety and we have a projective morphism $XA rightarrow mathbf{max}~R$ This situation is very similar to that of a desingularization $~X rightarrow mathbf{max}~R$ of the (possibly singular) variety $~mathbf{max}~R$. The top variety $~X$ is a smooth variety and there is a Zariski open subset of $~mathbf{max}~R$ where the fibers of this map consist of just one point, or in more bombastic language a $~mathbb{P}^0$. The only difference in the case of the Brauer-Severi fibration is that we have a Zariski open subset of $~mathbf{max}~R$ (the Azumaya locus of A) where the fibers of the fibration are isomorphic to $~mathbb{P}^{n-1}$. In this way one might view the Brauer-Severi fibration of a smooth order as a non-commutative or hyper-desingularization of the central variety. This might provide a way to attack the old problem of construction desingularizations of quiver-quotients. If $~Q$ is a quiver and $alpha$ is an indivisible dimension vector (that is, the component dimensions are coprime) then it is well known (a result due to Alastair King) that for a generic stability structure $theta$ the moduli space $~M^{theta}(Q,alpha)$ classifying $theta$-semistable $alpha$-dimensional representations will be a smooth variety (as all $theta$-semistables are actually $theta$-stable) and the fibration $~M^{theta}(Q,alpha) rightarrow mathbf{iss}{alpha}~Q$ is a desingularization of the quotient-variety $~mathbf{iss}{alpha}~Q$ classifying isomorphism classes of $alpha$-dimensional semi-simple representations. However, if $alpha$ is not indivisible nobody has the faintest clue as to how to construct a natural desingularization of $~mathbf{iss}{alpha}~Q$. Still, we have a perfectly reasonable hyper-desingularization $~X{A(Q,alpha)} rightarrow mathbf{iss}{alpha}~Q$ where $~A(Q,alpha)$ is the corresponding quiver order, the generic fibers of which are all projective spaces in case $alpha$ is the dimension vector of a simple representation of $~Q$. I conjecture (meaning : I hope) that this Brauer-Severi fibration contains already a lot of information on a genuine desingularization of $~mathbf{iss}{alpha}~Q$. One obvious test for this seemingly crazy conjecture is to study the flat locus of the Brauer-Severi fibration. If it would contain info about desingularizations one would expect that the fibration can never be flat in a central singularity! In other words, we would like that the flat locus of the fibration is contained in the smooth central locus. This is indeed the case and is a more or less straightforward application of the proof (due to Geert Van de Weyer) of the Popov-conjecture for quiver-quotients (see for example his Ph.D. thesis Nullcones of quiver representations). However, it is in general not true that the flat-locus and central smooth locus coincide. Sometimes this is because the Brauer-Severi scheme is a blow-up of the Brauer-Severi of a nicer order. The following example was worked out together with Colin Ingalls : Consider the order $~A = begin{bmatrix} C[x,y] & C[x,y] \ (x,y) & C[x,y] end{bmatrix}$ which is the quiver order of the quiver setting $~(Q,alpha)$ then the Brauer-Severi fibration $~XA rightarrow mathbf{iss}{alpha}~Q$ is flat everywhere except over the zero representation where the fiber is $~mathbb{P}^1 times mathbb{P}^2$. On the other hand, for the order $~B = begin{bmatrix} C[x,y] & C[x,y] \ C[x,y] & C[x,y] end{bmatrix}$ the Brauer-Severi fibration is flat and $~XB simeq mathbb{A}^2 times mathbb{P}^1$. It turns out that $~XA$ is a blow-up of $~X_B$ at a point in the fiber over the zero-representation. Sphere: Related Content [...]