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Tag: quivers

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

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path algebras

Published September 3, 2004 by lievenlb

The previous post can be found [here][1].
Pierre Gabriel invented a lot of new notation (see his book [Representations of finite dimensional algebras][2] for a rather extreme case) and is responsible for calling a directed graph a quiver. For example,

$\xymatrix{\vtx{} \ar@/^/[rr] & & \vtx{} \ar@(u,ur) \ar@(d,dr) \ar@/^/[ll]} $

is a quiver. Note than it is allowed to have multiple arrows between vertices, as well as loops in vertices. For us it will be important that a quiver $Q $ depicts how to compute in a certain non-commutative algebra : the path algebra $\mathbb{C} Q $. If the quiver has $k $ vertices and $l $ arrows (including loops) then the path algebra $\mathbb{C} Q $ is a subalgebra of the full $k \times k $ matrix-algebra over the free algebra in $l $ non-commuting variables

$\mathbb{C} Q \subset M_k(\mathbb{C} \langle x_1,\ldots,x_l \rangle) $

Under this map, a vertex $v_i $ is mapped to the basis $i $-th diagonal matrix-idempotent and an arrow

$\xymatrix{\vtx{v_i} \ar[rr]^{x_a} & & \vtx{v_j}} $

is mapped to the matrix having all its entries zero except the $(j,i) $-entry which is equal to $x_a $. That is, in our main example

$\xymatrix{\vtx{e} \ar@/^/[rr]^a & & \vtx{f} \ar@(u,ur)^x \ar@(d,dr)_y \ar@/^/[ll]^b} $

the corresponding path algebra is the subalgebra of $M_2(\mathbb{C} \langle a,b,x,y \rangle) $ generated by the matrices

$e \mapsto \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $ $ f \mapsto \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $

$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 name \’path algebra\’ comes from the fact that the subspace of $\mathbb{C} Q $ at the $(j,i) $-place is the vectorspace spanned by all paths in the quiver starting at vertex $v_i $ and ending in vertex $v_j $. For an easier and concrete example of a path algebra. consider the quiver

$\xymatrix{\vtx{e} \ar[rr]^a & & \vtx{f} \ar@(ur,dr)^x} $

and verify that in this case, the path algebra is just

$\mathbb{C} Q = \begin{bmatrix} \mathbb{C} & 0 \\ \mathbb{C}[x]a & \mathbb{C}[x] \end{bmatrix} $

Observe that we write and read paths in the quiver from right to left. The reason for this strange convention is that later we will be interested in left-modules rather than right-modules. Right-minder people can go for the more natural left to right convention for writing paths.
Why are path algebras of quivers of interest in non-commutative geometry? Well, to begin they are examples of _formally smooth algebras_ (some say _quasi-free algebras_, I just call them _qurves_). These algebras were introduced and studied by Joachim Cuntz and Daniel Quillen and they are precisely the algebras allowing a good theory of non-commutative differential forms.
So you should think of formally smooth algebras as being non-commutative manifolds and under this analogy path algebras of quivers correspond to _affine spaces_. That is, one expects path algebras of quivers to turn up in two instances : (1) given a non-commutative manifold (aka formally smooth algebra) it must be \’embedded\’ in some non-commutative affine space (aka path algebra of a quiver) and (2) given a non-commutative manifold, the \’tangent spaces\’ should be determined by path algebras of quivers.
The first fact is easy enough to prove, every affine $\mathbb{C} $-algebra is an epimorphic image of a free algebra in say $l $ generators, which is just the path algebra of the _bouquet quiver_ having $l $ loops

$\xymatrix{\vtx{} \ar@(dl,l)^{x_1} \ar@(l,ul)^{x_2} \ar@(ur,r)^{x_i} \ar@(r,dr)^{x_l}} $

The second statement requires more work. For a first attempt to clarify this you can consult my preprint [Qurves and quivers][3] but I\’ll come back to this in another post. For now, just take my word for it : if formally smooth algebras are the non-commutative analogon of manifolds then path algebras of quivers are the non-commutative version of affine spaces!

[1]: http://www.neverendingbooks.org/index.php?p=71
[2]: http://www.booxtra.de/verteiler.asp?site=artikel.asp&wea=1070000&sh=homehome&artikelnummer=000000689724
[3]: http://www.arxiv.org/abs/math.RA/0406618

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projects in noncommutative geometry

Published March 22, 2004 by lievenlb

Tomorrow
I’ll start with the course Projects in non-commutative geometry
in our masterclass. The idea of this course (and its companion
Projects in non-commutative algebra run by Fred Van Oystaeyen) is
that students should make a small (original if possible) work, that may
eventually lead to a publication.
At this moment the students
have seen the following : definition and examples of quasi-free algebras
(aka formally smooth algebras, non-commutative curves), their
representation varieties, their connected component semigroup and the
Euler-form on it. Last week, Markus Reineke used all this in his mini-course
Rational points of varieties associated to quasi-free
algebras. In it, Markus gave a method to compute (at least in
principle) the number of points of the non-commutative Hilbert
scheme and the varieties of simple representations over a
finite field. Here, in principle means that Markus demands a lot of
knowledge in advance : the number of points of all connected components
of all representation schemes of the algebra as well as of its scalar
extensions over finite field extensions, together with the action of the
Galois group on them … Sadly, I do not know too many examples were all
this information is known (apart from path algebras of quivers).
Therefore, it seems like a good idea to run through Markus’
calculations in some specific examples were I think one can get all this
: free products of semi-simple algebras. The motivating examples
being the groupalgebra of the (projective) modular group
PSL(2,Z) = Z(2) * Z(3) and the free matrix-products $M(n,F_q) *
M(m,F_q)$. I will explain how one begins to compute things in these
examples and will also explain how to get the One
quiver to rule them all in these cases. It would be interesting to
compare the calculations we will find with those corresponding to the
path algebra of this one quiver.
As Markus set the good
example of writing out his notes and posting them, I will try to do the
same for my previous two sessions on quasi-free algebras over the next
couple of weeks.

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