Tag: representations

why nag? (2)

Published March 28, 2005 by lievenlb

Now, can
we assign such an non-commutative tangent space, that is a $\mathbf{rep}~Q$ for some quiver Q, to $\mathbf{rep}~\Gamma$ ? As $\Gamma = \mathbb{Z}_2 \ast \mathbb{Z}_3$ we may
restrict any solution $V=(X,Y)$
in $\mathbf{rep}~\Gamma$ to the finite subgroups $\mathbb{Z}_2$ and $\mathbb{Z}_3$ . Now, representations of finite cyclic groups are
decomposed into eigen-spaces. For example

$V \downarrow_{\mathbb{Z}_2} = V_+ \oplus V_-$

where $V_{\pm} = \{ v \in V~|~g.v = \pm v \}$ with g the
generator of $\mathbb{Z}_2$ . Similarly,

$V \downarrow_{\mathbb{Z}_3} = V_1 \oplus V_{\rho} \oplus V_{\rho^2}$

where $\rho$ is a
primitive 3-rd root of unity. That is, to any solution $V \in \mathbf{rep}~\Gamma$ we have found 5 vector spaces $V_+,V_-,V_1,V_{\rho}$ and $V_{\rho^2}$ so we would like them to correspond to the vertices
of our conjectured quiver Q.

What are the arrows of Q, or
equivalently, is there a natural linear map between the vertex-vector
spaces? Clearly, as

$V_+ \oplus V_- = V = V_1 \oplus V_{\rho} \oplus V_{\rho^2}$

any choice of two bases of V (one
compatible with the left-side decomposition, the other with the
right-side decomposition) are related by a basechange matrix B which we
can decompose into six blocks (corresponding to the two decompositions
in 2 resp. 3 subspaces

$B = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ B_{31} & B_{32} \end{bmatrix}$

which gives us 6 linear maps between the
vertex-vector spaces. Hence, to $V \in \mathbf{rep}~\Gamma$ does correspond in a natural way a
representation of dimension vector $\alpha=(a_1,a_2,b_1,b_2,b_3)$ (where $dim(V_+)=a_1,\ldots,dim(V_{\rho^2})=b_3$ ) of the quiver Q which
is of the form

$\xymatrix{ & & & & \vtx{b_1} \\ \vtx{a_1} \ar[rrrru]^(.3){B_{11}} \ar[rrrrd]^(.3){B_{21}} \ar[rrrrddd]_(.2){B_{31}} & & & & \\ & & & & \vtx{b_2} \\ \vtx{a_2} \ar[rrrruuu]_(.7){B_{12}} \ar[rrrru]_(.7){B_{22}} \ar[rrrrd]_(.7){B_{23}} & & & & \\ & & & & \vtx{b_3}}$

Clearly, not every representation of $\mathbf{rep}~Q$ is obtained in this way. For starters, the
eigen-space decompositions force the numerical restriction

$a_1+a_2 = dim(V) = b_1+b_2+b_3$

on the
dimension vector and the square matrix constructed from the arrow-linear
maps must be invertible. However, if both these conditions are
satisfied, we can reconstruct the (isomorphism class) of the solution in
$\mathbf{rep}~\Gamma$ from this quiver representation by taking

$X = B^{-1} \begin{bmatrix} 1_{b_1} & 0 & 0 \\ 0 & \rho^2 1_{b_2} & 0 \\ 0 & 0 & \rho 1_{b_3} \end{bmatrix} B \begin{bmatrix} 1_{a_1} & 0 \\ 0 & -1_{a_2} \end{bmatrix}$

$Y = \begin{bmatrix} 1_{a_1} & 0 \\ 0 & -1_{a_2} \end{bmatrix} B^{-1} \begin{bmatrix} 1_{b_1} & 0 & 0 \\ 0 & \rho^2 1_{b_2} & 0 \\ 0 & 0 & \rho 1_{b_3} \end{bmatrix} B$

Hence, it makes sense to
view $\mathbf{rep}~Q$ as a linearization of, or as a tangent space to,
$\mathbf{rep}~\Gamma$ . However, though we reduced the study of
solutions of the polynomial system of equations to linear algebra, we
have not reduced the isomorphism problem in size. In fact, if we start
of with a matrix-solution $V=(X,Y)$
of size n we end up with a quiver-representation of total dimension 2n.
So, can we construct some sort of non-commutative normal space to the
isomorphism classes? That is, is there another quiver Q whose
representations can be interpreted as normal-spaces to orbits in certain
points?

Leave a Comment

B for bricks

Published January 28, 2005 by lievenlb

Last time we
argued that a noncommutative variety might be an _aggregate_
which locally is of the form $\mathbf{rep}~A$ for some affine (possibly
non-commutative) $C$-algebra $A$. However, we didn't specify what we
meant by 'locally' as we didn't define a topology on
$\mathbf{rep}~A$, let alone on an arbitrary aggregate. Today we will start
the construction of a truly _non-commutative topology_ on
$\mathbf{rep}~A$.
Here is the basic idea : we start with a thick
subset of finite dimensional representations on which we have a natural
(ordinary) topology and then we extend this to a non-commutativce
topology on the whole of $\mathbf{rep}~A$ using extensions. The impatient
can have a look at my old note A noncommutative
topology on rep A but note that we will modify the construction here
in two essential ways.
In that note we took $\mathbf{simp}~A$, the
set of all fnite dimensional simple representations, as thick subset
equipped with the induced Zariski topology on the prime spectrum
$\mathbf{spec}~A$. However, this topology doesn't behave well with
respect to the gluings we have in mind so we will extend $\mathbf{simp}~A$
substantially.

Leave a Comment

A for aggregates

Published January 9, 2005 by lievenlb

Let us
begin with a simple enough question : what are the points of a
non-commutative variety? Anyone? Probably you\’d say something like :
standard algebra-geometry yoga tells us that we should associate to a
non-commutative algebra $A$ on object, say $X_A$ and an arbitrary
variety is then build from \’gluing\’ such things together. Ok, but what
is $X_A$? Commutative tradition whispers $X_A=\mathbf{spec}~A$ the
[prime spectrum][1] of $A$, that is, the set of all twosided prime
ideals $P$ (that is, if $aAb \subset P$ then either $a \in P$ or $b \in
P$) and \’points\’ of $\mathbf{spec}~A$ would then correspond to
_maximal_ twosided ideals. The good news is that in this set-up, the
point-set comes equipped with a natural topology, the [Zariski
topology][2]. The bad news is that the prime spectrum is rarely
functorial in the noncommutative world. That is, if $\phi~:~A
\rightarrow B$ is an algebra morphism then $\phi^{-1}(P)$ for $P \in
\mathbf{spec}~B$ is not always a prime ideal of $A$. For example, take
$\phi$ the inclusion map $\begin{bmatrix} C[x] & C[x] \\ (x) & C[x]
\end{bmatrix} \subset \begin{bmatrix} C[x] & C[x] \\ C[x] & C[x]
\end{bmatrix}$ and $P$ the prime ideal $\begin{bmatrix} (x) & (x) \\ (x)
& (x) \end{bmatrix}$ then $P Cap \begin{bmatrix} C[x] & C[x] \\ (x) &
C[x] \end{bmatrix} = P$ but the corresponding quotient is
$\begin{bmatrix} C & C \\ 0 & C \end{bmatrix}$ which is not a prime
algebra so $\phi^{-1}(P)$ is not a prime ideal of the smaller algebra.
Failing this, let us take for $X_A$ something which obviously is
functorial and worry about topologies later. Take $X_A = \mathbf{rep}~A$
the set of all finite dimensional representations of $A$, that is
$\mathbf{rep}~A = \bigsqcup_n \mathbf{rep}_n~A$ where $\mathbf{rep}_n~A
= \{ Chi~:~A \rightarrow M_n(C)~\}$ with $Chi$ an algebra morphism. Now,
for any algebra morphism $\phi~:~A \rightarrow B$ there is an obvious
map $\mathbf{rep}~B \rightarrow \mathbf{rep}~A$ sending $Chi \mapsto Chi
Circ \phi$. Alernatively, $\mathbf{rep}_n~A$ is the set of all
$n$-dimensional left $A$-modules $M_{Chi} = C^n_{Chi}$ with $a.m =
Chi(m)m$. As such, $\mathbf{rep}~A$ is not merely a set but a
$C$-_category_, that is, all objects are $C$-vectorspaces and all
morphisms $Hom(M,N)$ are $C$-vectorspaces (the left $A$-module
morphisms). Moreover, it is an _additive_ category, that is if
$Chi,\psi$ are representations then we also have a direct sum
representation $Chi \oplus \psi$ defined by $a \mapsto \begin{bmatrix}
Chi(a) & 0 \\ 0 & \psi(a) \end{bmatrix}$. Returning at the task at
hand let us declare a _non-commutative variety_ $X$ to be (1) _an
additive_ $C$-_category_ which \’locally\’ looks like $\mathbf{rep}~A$
for some non-commutative algebra $A$ (even if we do not know at the
momemt what we mean by locally as we do not have defined a topology,
yet). Let is call objects of teh category $X$ the \’points\’ of our
variety and $X$ being additive allows us to speak of _indecomposable_
points (that is, those objects that cannot be written as a direct sum of
non-zero objects). By the local description of $X$ an indecomposable
point corresponds to an indecomposable representation of a
non-commutative algebra and as such has a local endomorphism algebra
(that is, all non-invertible endomorphisms form a twosided ideal). But
if we have this property for all indecomposable points,our category $X$
will be a Krull-Schmidt category so it is natural to impose also the
condition (2) : every point of $X$ can be decomposed uniquely into a
finite direct sum of indecomposable points. Further, as the space of
left $A$-module morphisms between two finite dimensional modules is
clearly finite dimensional we have also the following strong finiteness
condition (3) : For all points $x,y \in X$ the space of morphisms
$Hom(x,y)$ is a finite dimensional $C$-vectorspace. In their book
[Representations of finite-dimensional algebras][3], Peter Gabriel and
Andrei V. Roiter call an additive category such that all endomorphism
algebras of indecomposable objects are local algebras and such that all
morphism spaces are finite dimensional an _aggregate_. So, we have a
first tentative answer to our question **the points of a
non-commutative variety are the objects of an aggregate** Clearly, as
$\mathbf{rep}~A$ has stronger properties like being an _Abelian
category_ (that is, morphisms allow kernels and cokernels) it might also
be natural to replace \’aggregate\’ by \’Abelian Krull-Schmidt category
with finite dimensional homs\’ but if Mr. Abelian Category himself finds
the generalization to aggregates useful I\’m not going to argue about
this. Are all aggregates of the form $\mathbf{rep}~A$ or are there
other interesting examples? A motivating commutative example is : the
category of all coherent modules $Coh(Y)$ on a _projective_ variety $Y$
form an aggegate giving us a mental picture of what we might expect of a
non-commutative variety. Clearly, the above tentative answer cannot be
the full story as we haven\’t included the topological condition of
being locally of the form $\mathbf{rep}~A$ yet, but we will do that in
the next episode _B for Bricks_. [1]:
http://planetmath.org/encyclopedia/PrimeSpectrum.html [2]:
http://planetmath.org/encyclopedia/ZariskiTopology.html [3]:

1/ref=sr_1_8_1/026-3923724-4530018

2 Comments

From Galois to NOG

Published December 22, 2004 by lievenlb

Evariste Galois (1811-1832) must rank pretty high on the all-time
list of moving last words. Galois was mortally wounded in a duel he
fought with Perscheux d\’Herbinville on May 30th 1832, the reason for
the duel not being clear but certainly linked to a girl called
Stephanie, whose name appears several times as a marginal note in
Galois\’ manuscripts (see illustration). When he died in the arms of his
younger brother Alfred he reportedly said “Ne pleure pas, j\’ai besoin
de tout mon courage pour mourir ‚àö‚Ä† 20 ans”. In this series I\’ll
start with a pretty concrete problem in Galois theory and explain its
elegant solution by Aidan Schofield and Michel Van den Bergh.
Next, I\’ll rephrase the problem in non-commutative geometry lingo,
generalise it to absurd levels and finally I\’ll introduce a coalgebra
(yes, a co-algebra…) that explains it all. But, it will take some time
to get there. Start with your favourite basefield $k$ of
characteristic zero (take $k = \mathbb{Q}$ if you have no strong
preference of your own). Take three elements $a,b,c$ none of which
squares, then what conditions (if any) must be imposed on $a,b,c$ and $n
\in \mathbb{N}$ to construct a central simple algebra $\Sigma$ of
dimension $n^2$ over the function field of an algebraic $k$-variety such
that the three quadratic fieldextensions $k\sqrt{a}, k\sqrt{b}$ and
$k\sqrt{c}$ embed into $\Sigma$? Aidan and Michel show in \’Division
algebra coproducts of index $n$\’ (Trans. Amer. Math. Soc. 341 (1994),
505-517) that the only condition needed is that $n$ is an even number.
In fact, they work a lot harder to prove that one can even take $\Sigma$
to be a division algebra. They start with the algebra free
product $A = k\sqrt{a} \ast k\sqrt{b} \ast k\sqrt{c}$ which is a pretty
monstrous algebra. Take three letters $x,y,z$ and consider all
non-commutative words in $x,y$ and $z$ without repetition (that is, no
two consecutive $x,y$ or $z$\’s). These words form a $k$-basis for $A$
and the multiplication is induced by concatenation of words subject to
the simplifying relations $x.x=a,y.y=b$ and $z.z=c$.

Next, they look
at the affine $k$-varieties $\mathbf{rep}(n) A$ of $n$-dimensional
$k$-representations of $A$ and their irreducible components. In the
parlance of $\mathbf{geometry@n}$, these irreducible components correspond
to the minimal primes of the level $n$-approximation algebra $\int(n) A$.
Aidan and Michel worry a bit about reducedness of these components but
nowadays we know that $A$ is an example of a non-commutative manifold (a
la Cuntz-Quillen or Kontsevich-Rosenberg) and hence all representation
varieties $\mathbf{rep}n A$ are smooth varieties (whence reduced) though
they may have several connected components. To determine the number of
irreducible (which in this case, is the same as connected) components
they use _Galois descent, that is, they consider the algebra $A
\otimes_k \overline{k}$ where $\overline{k}$ is the algebraic closure of
$k$. The algebra $A \otimes_k \overline{k}$ is the group-algebra of the
group free product $\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}
\ast \mathbb{Z}/2\mathbb{Z}$. (to be continued…) A digression : I
cannot resist the temptation to mention the tetrahedral snake problem
in relation to such groups. If one would have started with $4$ quadratic
fieldextensions one would get the free product $G =
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}$. Take a supply of
tetrahedra and glue them together along common faces so that any
tertrahedron is glued to maximum two others. In this way one forms a
tetrahedral-snake and the problem asks whether it is possible to make
such a snake having the property that the orientation of the
\’tail-tetrahedron\’ in $\mathbb{R}^3$ is exactly the same as the
orientation of the \’head-tetrahedron\’. This is not possible and the
proof of it uses the fact that there are no non-trivial relations
between the four generators $x,y,z,u$ of $\mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z}$ which correspond to reflections wrt. a face of
the tetrahedron (in fact, there are no relations between these
reflections other than each has order two, so the subgroup generated by
these four reflections is the group $G$). More details can be found in
Stan Wagon\’s excellent book The Banach-tarski paradox, p.68-71.

Leave a Comment

double Poisson algebras

Published October 26, 2004 by lievenlb

This morning,
Michel Van den Bergh posted an interesting paper on the arXiv
entitled Double
Poisson Algebras. His main motivation was the construction of a
natural Poisson structure on quotient varieties of representations of
deformed multiplicative preprojective algebras (introduced by
Crawley-Boevey and Shaw in Multiplicative
preprojective algebras, middle convolution and the Deligne-Simpson
problem) which he achieves by extending his double Poisson structure
on the path algebra of the quiver to the 'obvious' universal
localization, that is the one by inverting all $1+aa^{\star} $ for $a $ an
arrow and $a^{\star} $ its double (the one in the other direction).
For me the more interesting fact of this paper is that his double
bracket on the path algebra of a double quiver gives finer information
than the _necklace Lie algebra_ as defined in my (old) paper with Raf
Bocklandt Necklace
Lie algebras and noncommutative symplectic geometry. I will
certainly come back to this later when I have more energy but just to
wet your appetite let me point out that Michel calls a _double bracket_
on an algebra $A $ a bilinear map
$\{ \{ -,- \} \}~:~A \times A
\rightarrow A \otimes A $
which is a derivation in the _second_
argument (for the outer bimodulke structure on $A $) and satisfies
$\{ \{ a,b \} \} = – \{ \{ b,a \} \}^o $ with $~(u \otimes v)^0 = v
\otimes u $
Given such a double bracket one can define an ordinary
bracket (using standard Hopf-algebra notation)
$\{ a,b \} = \sum
\{ \{ a,b \} \}_{(1)} \{ \{ a,b \} \}_{(2)} $
which makes $A $ into
a Loday
algebra and induces a Lie algebra structure on $A/[A,A] $. He then
goes on to define such a double bracket on the path algebra of a double
quiver in such a way that the associated Lie structure above is the
necklace Lie algebra.

2 Comments

hyper-resolutions

Published September 30, 2004 by lievenlb

[Last time][1] we saw that for $A$ a smooth order with center $R$ the
Brauer-Severi variety $X_A$ is a smooth variety and we have a projective
morphism $X_A \rightarrow \mathbf{max}~R$ 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][2])
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][3]) of the Popov-conjecture for quiver-quotients (see for
example his Ph.D. thesis [Nullcones of quiver representations][4]).
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][5] : 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
$~(Q,\alpha)$ $\xymatrix{\vtx{1} \ar@/^2ex/[rr] \ar@/^1ex/[rr]
& & \vtx{1} \ar@/^2ex/[ll]} $ then the Brauer-Severi fibration
$~X_A \rightarrow \mathbf{iss}_{\alpha}~Q$ 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 $~X_B \simeq \mathbb{A}^2 \times
\mathbb{P}^1$. It turns out that $~X_A$ is a blow-up of $~X_B$ at a
point in the fiber over the zero-representation.

[1]: http://www.neverendingbooks.org/index.php?p=342
[2]: http://www.maths.bath.ac.uk/~masadk/
[3]: http://www.win.ua.ac.be/~gvdwey/
[4]: http://www.win.ua.ac.be/~gvdwey/papers/thesis.pdf
[5]: http://kappa.math.unb.ca/~colin/

Leave a Comment

smooth Brauer-Severis

Published September 28, 2004 by lievenlb

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
$\Sigma$ (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 $GL_n$-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 $~\phi : A \rightarrow
M_n(k)$ 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 $~x \rightarrow
\begin{bmatrix} a & b \\ c & -a \end{bmatrix}~y \rightarrow
\begin{bmatrix} d & e \\ f & -d \end{bmatrix}$ such that
$~2ad+bf+ec = 0$. That is, $~\mathbf{trep}_2~A = \mathbb{V}(2ad+bf+ec)
\subset \mathbb{A}^6$ 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 $n \geq 4$ 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.

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

One Comment

Brauer-Severi varieties

Published September 27, 2004 by lievenlb

![][1]
Classical Brauer-Severi varieties can be described either as twisted
forms of projective space (Severi\’s way) or as varieties containing
splitting information about central simple algebras (Brauer\’s way). If
$K$ is a field with separable closure $\overline{K}$, the first approach
asks for projective varieties $X$ defined over $K$ such that over the
separable closure $X(\overline{K}) \simeq
\mathbb{P}^{n-1}_{\overline{K}}$ they are just projective space. In
the second approach let $\Sigma$ be a central simple $K$-algebra and
define a variety $X_{\Sigma}$ whose points over a field extension $L$
are precisely the left ideals of $\Sigma \otimes_K L$ of dimension $n$.
This variety is defined over $K$ and is a closed subvariety of the
Grassmannian $Gr(n,n^2)$. In the special case that $\Sigma = M_n(K)$ one
can use the matrix-idempotents to show that the left ideals of dimension
$n$ correspond to the points of $\mathbb{P}^{n-1}_K$. As for any central
simple $K$-algebra $\Sigma$ we have that $\Sigma \otimes_K \overline{K}
\simeq M_n(\overline{K})$ it follows that the varieties $X_{\Sigma}$ are
among those of the first approach. In fact, there is a natural bijection
between those of the first approach (twisted forms) and of the second as
both are classified by the Galois cohomology pointed set
$H^1(Gal(\overline{K}/K),PGL_n(\overline{K}))$ because
$PGL_n(\overline{K})$ is the automorphism group of
$\mathbb{P}^{n-1}_{\overline{K}}$ as well as of $M_n(\overline{K})$. The
ringtheoretic relevance of the Brauer-Severi variety $X_{\Sigma}$ is
that for any field extension $L$ it has $L$-rational points if and only
if $L$ is a _splitting field_ for $\Sigma$, that is, $\Sigma \otimes_K L
\simeq M_n(\Sigma)$. To give one concrete example, If $\Sigma$ is the
quaternion-algebra $(a,b)_K$, then the Brauer-Severi variety is a conic
$X_{\Sigma} = \mathbb{V}(x_0^2-ax_1^2-bx_2^2) \subset \mathbb{P}^2_K$
Whenever one has something working for central simple algebras, one can
_sheafify_ the construction to Azumaya algebras. For if $A$ is an
Azumaya algebra with center $R$ then for every maximal ideal
$\mathfrak{m}$ of $R$, the quotient $A/\mathfrak{m}A$ is a central
simple $R/\mathfrak{m}$-algebra. This was noted by the
sheafification-guru [Alexander Grothendieck][2] and he extended the
notion to Brauer-Severi schemes of Azumaya algebras which are projective
bundles $X_A \rightarrow \mathbf{max}~R$ all of which fibers are
projective spaces (in case $R$ is an affine algebra over an
algebraically closed field). But the real fun started when [Mike
Artin][3] and [David Mumford][4] extended the construction to suitably
_ramified_ algebras. In good cases one has that the Brauer-Severi
fibration is flat with fibers over ramified points certain degenerations
of projective space. For example in the case considered by Artin and
Mumford of suitably ramified orders in quaternion algebras, the smooth
conics over Azumaya points degenerate to a pair of lines over ramified
points. A major application of their construction were examples of
unirational non-rational varieties. To date still one of the nicest
applications of non-commutative algebra to more mainstream mathematics.
The final step in generalizing Brauer-Severi fibrations to arbitrary
orders was achieved by [Michel Van den Bergh][5] in 1986. Let $R$ be an
affine algebra over an algebraically closed field (say of characteristic
zero) $k$ and let $A$ be an $R$-order is a central simple algebra
$\Sigma$ of dimension $n^2$. Let $\mathbf{trep}_n~A$ be teh affine variety
of _trace preserving_ $n$-dimensional representations, then there is a
natural action of $GL_n$ on this variety by basechange (conjugation).
Moreover, $GL_n$ acts by left multiplication on column vectors $k^n$.
One then considers the open subset in $\mathbf{trep}_n~A \times k^n$
consisting of _Brauer-Stable representations_, that is those pairs
$(\phi,v)$ such that $\phi(A).v = k^n$ on which $GL_n$ acts freely. The
corresponding orbit space is then called the Brauer-Severio scheme $X_A$
of $A$ and there is a fibration $X_A \rightarrow \mathbf{max}~R$ again
having as fibers projective spaces over Azumaya points but this time the
fibration is allowed to be far from flat in general. Two months ago I
outlined in Warwick an idea to apply this Brauer-Severi scheme to get a
hold on desingularizations of quiver quotient singularities. More on
this next time.

[1]: http://www.neverendingbooks.org/DATA/brauer.jpg
[2]: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Grothendieck.html
[3]: http://www.cirs-tm.org/researchers/researchers.php?id=235
[4]: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Mumford.html
[5]: http://alpha.luc.ac.be/Research/Algebra/Members/michel_id.html

Leave a Comment

moduli spaces

Published September 9, 2004 by lievenlb

In [the previous part][1] we saw that moduli spaces of suitable representations
of the quiver $\xymatrix{\vtx{} \ar[rr] & & \vtx{}
\ar@(ur,dr)} $ locally determine the moduli spaces of
vectorbundles over smooth projective curves. There is yet another
classical problem related to this quiver (which also illustrates the
idea of looking at families of moduli spaces rather than individual
ones) : _linear control systems_. Such a system with an $n$ dimensional
_state space_ and $m$ _controls_ (or inputs) is determined by the
following system of linear differential equations $ \frac{d x}{d t}
= A.x + B.u$ where $x(t) \in \mathbb{C}^n$ is the state of the system at
time $t$, $u(t) \in \mathbb{C}^m$ is the control-vector at time $t$ and $A \in
M_n(\mathbb{C}), B \in M_{n \times m}(\mathbb{C})$ are the matrices describing the
evolution of the system $\Sigma$ (after fixing bases in the state- and
control-space). That is, $\Sigma$ determines a representation of the
above quiver of dimension-vector $\alpha = (m,n)$
$\xymatrix{\vtx{m} \ar[rr]^B & & \vtx{n} \ar@(ur,dr)^A} $
Whereas in control theory (see for example Allen Tannenbaum\’s Lecture
Notes in Mathematics 845 for a mathematical introduction) it is natural
to call two systems equivalent when they only differ up to base change
in the state-space, one usually fixes the control knobs so it is not
natural to allow for base change in the control-space. So, at first
sight the control theoretic problem of classifying equivalent systems is
not the same problem as classifying representations of the quiver up to
isomorphism. Fortunately, there is an elegant way round this which is
called _deframing_. That is, for a fixed number $m$ of controls one
considers the quiver $Q_f$ having precisely $m$ arrows from the first to
the second vertex $\xymatrix{\vtx{1} \ar@/^4ex/[rr]^{B_1}
\ar@/^/[rr]^{B_2} \ar@/_3ex/[rr]_{B_m} & & \vtx{n} \ar@(ur,dr)^A} $
and the system $\Sigma$ does determine a representation of this new
quiver of dimension vector $\beta=(1,n)$ by assigning to the arrows the
different columns of the matrix $B$. Isomorphism classes of these
quiver-representations do correspond precisely to equivalence classes of
linear control systems. In [part 4][1] we introduced stable and
semi-stable representations. In this framed-quiver setting call a
representation $(A,B_1,\ldots,B_m)$ stable if there is no proper
subrepresentation of dimension vector $(1,p)$ for some $p \lneq n$.
Perhaps remarkable this algebraic notion has a counterpart in
system-theory : the systems corresponding to stable
quiver-representations are precisely the completely controllable
systems. That is, those which can be brought to any wanted state by
varying the controls. Hence, the moduli space
$M^s_{(1,n)}(Q_f,\theta)$ classifying stable representations is
exactly the moduli space of completely controllable linear systems
studied in control theory. For an excellent account of this moduli space
one can read the paper [Introduction to moduli spaces associated to
quivers by [Christof Geiss][2]. Fixing the number $m$ of controls but
varying the dimensions of teh state-spaces one would like to take all
the moduli spaces $ \bigsqcup_n~M^s_{(1,n)}(Q_f,\theta)$
together as they are all determined by the same formally smooth algebra
$\mathbb{C} Q_f$. This was done in a joint paper with [Markus Reineke][3] called
[Canonical systems and non-commutative geometry][4] in which we prove
that this disjoint union can be identified with the _infinite
Grassmannian_ $ \bigsqcup_n~M^s_{(1,n)}(Q_f,\theta) =
\mathbf{Gras}_m(\infty)$ of $m$-dimensional subspaces of an
infinite dimensional space. This result can be seen as a baby-version of
George Wilson\’s result relating the disjoint union of Calogero-Moser
spaces to the _adelic_ Grassmannian. But why do we stress this
particular quiver so much? This will be partly explained [next time][5].

[1]: http://www.neverendingbooks.org/index.php?p=350
[2]: http://www.matem.unam.mx/~christof/
[3]: http://wmaz1.math.uni-wuppertal.de/reineke/
[4]: http://www.arxiv.org/abs/math.AG/0303304
[5]: http://www.neverendingbooks.org/index.php?p=352

Leave a Comment

quiver representations

Published September 6, 2004 by lievenlb

In what
way is a formally smooth algebra a _machine_ producing families of
manifolds? Consider the special case of the path algebra $\mathbb{C} Q$ of a
quiver and recall that an $n$-dimensional representation is an algebra
map $\mathbb{C} Q \rightarrow^{\phi} M_n(\mathbb{C})$ or, equivalently, an
$n$-dimensional left $\mathbb{C} Q$-module $\mathbb{C}^n_{\phi}$ with the action
determined by the rule $a.v = \phi(a) v~\forall v \in \mathbb{C}^n_{\phi},
\forall a \in \mathbb{C} Q$ If the $e_i~1 \leq i \leq k$ are the idempotents
in $\mathbb{C} Q$ corresponding to the vertices (see this [post][1]) then we get
a direct sum decomposition $\mathbb{C}^n_{\phi} = \phi(e_1)\mathbb{C}^n_{\phi} \oplus
\ldots \oplus \phi(e_k)\mathbb{C}^n_{\phi}$ and so every $n$-dimensional
representation does determine a _dimension vector_ $\alpha =
(a_1,\ldots,a_k)~\text{with}~a_i = dim_{\mathbb{C}} V_i = dim_{\mathbb{C}}
\phi(e_i)\mathbb{C}^n_{\phi}$ with $ | \alpha | = \sum_i a_i = n$. Further,
for every arrow $\xymatrix{\vtx{e_i} \ar[rr]^a & &
\vtx{e_j}} $ we have (because $e_j.a.e_i = a$ that $\phi(a)$
defines a linear map $\phi(a)~:~V_i \rightarrow V_j$ (that was the
whole point of writing paths in the quiver from right to left so that a
representation is determined by its _vertex spaces_ $V_i$ and as many
linear maps between them as there are arrows). Fixing vectorspace bases
in the vertex-spaces one observes that the space of all
$\alpha$-dimensional representations of the quiver is just an affine
space $\mathbf{rep}_{\alpha}~Q = \oplus_a~M_{a_j \times a_i}(\mathbb{C})$ and
base-change in the vertex-spaces does determine the action of the
_base-change group_ $GL(\alpha) = GL_{a_1} \times \ldots \times
GL_{a_k}$ on this space. Finally, as all this started out with fixing
a bases in $\mathbb{C}^n_{\phi}$ we get the affine variety of all
$n$-dimensional representations by bringing in the base-change
$GL_n$-action (by conjugation) and have $\mathbf{rep}_n~\mathbb{C} Q =
\bigsqcup_{| \alpha | = n} GL_n \times^{GL(\alpha)}
\mathbf{rep}_{\alpha}~Q$ and in this decomposition the connected
components are no longer just affine spaces with a groupaction but
essentially equal to them as there is a natural one-to-one
correspondence between $GL_n$-orbits in the fiber-bundle $GL_n
\times^{GL(\alpha)} \mathbf{rep}_{\alpha}~Q$ and $GL(\alpha)$-orbits in the
affine space $\mathbf{rep}_{\alpha}~Q$. In our main example
$\xymatrix{\vtx{e} \ar@/^/[rr]^a & & \vtx{f} \ar@(u,ur)^x
\ar@(d,dr)_y \ar@/^/[ll]^b} $ an $n$-dimensional representation
determines vertex-spaces $V = \phi(e) \mathbb{C}^n_{\phi}$ and $W = \phi(f)
\mathbb{C}^n_{\phi}$ of dimensions $p,q~\text{with}~p+q = n$. The arrows
determine linear maps between these spaces $\xymatrix{V
\ar@/^/[rr]^{\phi(a)} & & W \ar@(u,ur)^{\phi(x)} \ar@(d,dr)_{\phi(y)}
\ar@/^/[ll]^{\phi(b)}} $ and if we fix a set of bases in these two
vertex-spaces, we can represent these maps by matrices
$\xymatrix{\mathbb{C}^p \ar@/^/[rr]^{A} & & \mathbb{C}^q \ar@(u,ur)^{X}
\ar@(d,dr)_{Y} \ar@/^/[ll]^{B}} $ which can be considered as block
$n \times n$ matrices $a \mapsto \begin{bmatrix} 0 & 0 \\ A & 0
\end{bmatrix}~b \mapsto \begin{bmatrix} 0 & B \\ 0 & 0 \end{bmatrix}$
$x \mapsto \begin{bmatrix} 0 & 0 \\ 0 & X \end{bmatrix}~y \mapsto
\begin{bmatrix} 0 & 0 \\ 0 & Y \end{bmatrix}$ The basechange group
$GL(\alpha) = GL_p \times GL_q$ is the diagonal subgroup of $GL_n$
$GL(\alpha) = \begin{bmatrix} GL_p & 0 \\ 0 & GL_q \end{bmatrix}$ and
acts on the representation space $\mathbf{rep}_{\alpha}~Q = M_{q \times
p}(\mathbb{C}) \oplus M_{p \times q}(\mathbb{C}) \oplus M_q(\mathbb{C}) \oplus M_q(\mathbb{C})$
(embedded as block-matrices in $M_n(\mathbb{C})^{\oplus 4}$ as above) by
simultaneous conjugation. More generally, if $A$ is a formally smooth
algebra, then all its representation varieties $\mathbf{rep}_n~A$ are
affine smooth varieties equipped with a $GL_n$-action. This follows more
or less immediately from the definition and [Grothendieck][2]\’s
characterization of commutative regular algebras. For the record, an
algebra $A$ is said to be _formally smooth_ if for every algebra map $A
\rightarrow B/I$ with $I$ a nilpotent ideal of $B$ there exists a lift
$A \rightarrow B$. The path algebra of a quiver is formally smooth
because for every map $\phi~:~\mathbb{C} Q \rightarrow B/I$ the images of the
vertex-idempotents form an orthogonal set of idempotents which is known
to lift modulo nilpotent ideals and call this lift $\psi$. But then also
every arrow lifts as we can send it to an arbitrary element of
$\psi(e_j)\pi^{-1}(\phi(a))\psi(e_i)$. In case $A$ is commutative and
$B$ is allowed to run over all commutative algebras, then by
Grothendieck\’s criterium $A$ is a commutative regular algebra. This
also clarifies why so few commutative regular algebras are formally
smooth : being formally smooth is a vastly more restrictive property as
the lifting property extends to all algebras $B$ and whenever the
dimension of the commutative variety is at least two one can think of
maps from its coordinate ring to the commutative quotient of a
non-commutative ring by a nilpotent ideal which do not lift (for an
example, see for example [this preprint][3]). The aim of
non-commutative algebraic geometry is to study _families_ of manifolds
$\mathbf{rep}_n~A$ associated to the formally-smooth algebra $A$. [1]:
http://www.matrix.ua.ac.be/wp-trackback.php/10 [2]:
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Grothendieck.
html [3]: http://www.arxiv.org/abs/math.AG/9904171

One Comment