Skip to content →

Author: lievenlb

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

irrelevant list

Published September 25, 2004 by lievenlb

As
there is no way to recover from the previous post, allow me a slow
restart by listing some of the a-typical things done this week :

  • Ate more chocolate than during the last five years

  • Drove the car more than during the rest of the year (minus
    vacations)

  • Didn't do any bicycle exercise

  • Only checked email in the morning (at best)

  • Didn't do any math (apart from helping
    PseudonymousDaughter2)

  • Didn't go in to university at
    all

  • Drank even more coffee than usual

  • Regardless, felt exhausted every evening

  • Did far
    less web-surfing (but managed to find
    this     on academic blogging)

  • Cooked fast and way too
    cholestorol-rich meals

  • Ate even more chocolates
    …

Fortunately, the semester (and teaching)
starts tomorrow!

Leave a Comment

curvatures

Published September 16, 2004 by lievenlb

[Last
time][1] we saw that the algebra $(\Omega_V~C Q,Circ)$ of relative
differential forms and equipped with the Fedosov product is again the
path algebra of a quiver $\tilde{Q}$ obtained by doubling up the arrows
of $Q$. In our basic example the algebra map $C \tilde{Q} \rightarrow
\Omega_V~C Q$ is clarified by the following picture of $\tilde{Q}$
$\xymatrix{\vtx{} \ar@/^/[rr]^{a=u+du} \ar@/_/[rr]_{b=u-du} & &
\vtx{} \ar@(u,ur)^{x=v+dv} \ar@(d,dr)_{y=v-dv}} $ (which
generalizes in the obvious way to arbitrary quivers). But what about the
other direction $\Omega_V~C Q \rightarrow C \tilde{Q}$ ? There are two
embeddings $i,j : C Q \rightarrow C \tilde{Q}$ defined by $i : (u,v)
\rightarrow (a,x)$ and $j : (u,v) \rightarrow (b,y)$ giving maps
$\forall a \in C Q~:~p(a) = \frac{1}{2}(i(a)+j(a))~\quad~q(a) =
\frac{1}{2}(i(a)-j(a))$ Using these maps, the isomorphism $\Omega_V~C
Q \rightarrow C \tilde{Q}$ is determined by $ a_0 da_1 \ldots da_n
\rightarrow p(a_0)q(a_1) \ldots q(a_n)$ In particular, $p$ gives the
natural embedding (with the ordinary multiplication on differential
forms) $C Q \rightarrow \Omega_V~C Q$ of functions as degree zero
differential forms. However, $p$ is no longer an algebra map for the
Fedosov product on $\Omega_V~C Q$ as $p(ab) = p(a)Circ p(b) + q(a) Circ
q(b)$. In Cuntz-Quillen terminology, $\omega(a,b) = q(a) Circ q(b)$ is
the _curvature_ of the based linear map $p$. I\’d better define
this a bit more formal for any algebra $A$ and then say what is special
for formally smooth algebras (non-commutative manifolds). If $A,B$ are
$V = C \times \ldots \times C$-algebras, then a $V$-linear map $A
\rightarrow^l B$ is said to be a _based linear map_ if $ l | V = id_V$.
The _curvature_ of $l$ measures the obstruction to $l$ being an algebra
map, that is $\forall a,b \in A~:~\omega(a,b) = l(ab)-l(a)l(b)$ and
the curvature is said to be _nilpotent_ if there is an integer $n$ such
that all possible products $\omega(a_1,b_1)\omega(a_2,b_2) \ldots
\omega(a_n,b_n) = 0$ For any algebra $A$ there is a universal algebra
$R(A)$ turning based linear maps into algebra maps. That is, there is a
fixed based linear map $A \rightarrow^p R(A)$ such that for every based
linear map $A \rightarrow^l B$ there is an algebra map $R(A) \rightarrow
B$ making the diagram commute $\xymatrix{A \ar[r]^l \ar[d]^p & B
\\\ R(A) \ar[ru] &} $ In fact, Cuntz and Quillen show that $R(A)
\simeq (\Omega_V^{ev}~A,Circ)$ the algebra of even differential forms
equipped with the Fedosov product and that $p$ is the natural inclusion
of $A$ as degree zero forms (as above). Recall that $A$ is said to be
_formally smooth_ if every $V$-algebra map $A \rightarrow^f B/I$ where
$I$ is a nilpotent ideal, can be lifted to an algebra morphism $A
\rightarrow B$. We can always lift $f$ as a based linear map, say
$\tilde{f}$ and because $I$ is nilpotent, the curvature of $\tilde{f}$
is also nilpotent. To get a _uniform_ way to construct algebra lifts
modulo nilpotent ideals it would therefore suffice for a formally smooth
algebra to have an _algebra map_ $A \rightarrow \hat{R}(A)$ where
$\hat{R}(A)$ is the $\mathfrak{m}$-adic completion of $R(A)$ for the
ideal $\mathfrak{m}$ which is the kernel of the algebra map $R(A)
\rightarrow A$ corresponding to the based linear map $id_A : A
\rightarrow A$. Indeed, there is an algebra map $R(A) \rightarrow B$
determined by $\tilde{f}$ and hence also an algebra map $\hat{R}(A)
\rightarrow B$ and composing this with the (yet to be constructed)
algebra map $A \rightarrow \hat{R}(A)$ this would give the required lift
$A \rightarrow B$. In order to construct the algebra map $A
\rightarrow \hat{R}(A)$ (say in the case of path algebras of quivers) we
will need the Yang-Mills derivation and its associated flow.

[1]: http://www.matrix.ua.ac.be/index.php?p=354

Leave a Comment

anyone interested

Published September 15, 2004 by lievenlb

I've been here before! I mean, I did try to set up
non-commutative algebra&geometry sites before and sooner or later
they always face the same basic problems :

a :
dyspnoea : one person does not have enough fresh ideas
to keep a mathematical site updated daily so that it continues to be of
interest (at least, I'm not one of those who can).

b :
claustrophobia : the topic of non-commutative algebra
& non-commutative geometry is too wide to be covered (cornered) by
one person. More (and differing) views are needed for balance and
continued interest.

c : paranoia : if one is
not entirely naive one has to exercise some restraint trying to protect
ones research plans (or those of students) so the most interesting ideas
never even get posted!

By definition, I cannot solve problems
a) and b) on my own. All I can hope is that, now that the basic
technological problems (such as including LaTeX-code in posts) are
solved, other people are willing to contribute. For this reason I
'depersonalized' this blog : I changed the title, removed all
personal links in the sidebar and so on. I want to open up this site
(but as I said, I've tried this before without much success) to
anyone working in non-commutative algebra and/or non-commutative
geometry who is willing to contribute posts on at least a monthly basis
(or fortnightly, weekly, daily…) for the foreseeable future. At
the moment the following 'categories' of posts are available
(others can be added on request) :

  • courses : if you want
    to tell about your topic of interest in small daily or weekly pieces.
  • columns : if you want to ventilate an opinion on something
    related (even vaguely) to na&g.
  • nc-algebra : for anything
    on non-commutative algebra not in the previous categories.
  • nc-geometry : for anything on non-commutative geometry not in the
    previous categories.
  • this blog : for suggestions or
    explanations on the technology of this site.

Mind you,
I am not looking for people who seek a forum to post
their questions (such people can still add questions as comments to
related posts) but rather for people active in na&g with a personal
opinion on relevance and future of the topic.
If you are
interested in contributing, please email me and we will work
something out. I'll also post information for authors (such as, how
to include tex, how to set restrictions etc.) in a _sticky_ post
soon.

Now, problem c) : in running sites for our master class
on noncommutative geometry I've noticed that some people are more
willing to post lectures notes etc. if they know that there is some
control on who can download their material. For this reason there will
be viewing restrictions on certain posts. Such posts will get a
padlock-sign next to them in the 'recent posts' sidebar (they
will not show up in your main page, if you are not authorized to see
them). I will add another sticky on all of this soon. For now, if you
would only be willing to contribute if there was this safeguard, rest
assured, it will be there soon. All others can of course already sign-up
or wait whether any of these plans (resp. day-dreams) ever work
out….

update (febr 2007) : still waiting
but the padlock idea is abandoned.

Leave a Comment

differential forms

Published September 14, 2004 by lievenlb

The
previous post in this sequence was [(co)tangent bundles][1]. Let $A$ be
a $V$-algebra where $V = C \times \ldots \times C$ is the subalgebra
generated by a complete set of orthogonal idempotents in $A$ (in case $A
= C Q$ is a path algebra, $V$ will be the subalgebra generated by the
vertex-idempotents, see the post on [path algebras][2] for more
details). With $\overline{A}$ we denote the bimodule quotient
$\overline{A} = A/V$ Then, we can define the _non-commutative
(relative) differential n-forms_ to be $\Omega^n_V~A = A \otimes_V
\overline{A} \otimes_V \ldots \otimes_V \overline{A}$ with $n$ factors
$\overline{A}$. To get the connection with usual differential forms let
us denote the tensor $a_0 \otimes a_1 \otimes \ldots \otimes a_n =
(a_0,a_1,\ldots,a_n) = a_0 da_1 \ldots da_n$ On $\Omega_V~A =
\oplus_n~\Omega^n_V~A$ one defines an algebra structure via the
multiplication $(a_0da_1 \ldots da_n)(a_{n+1}da_{n+2} \ldots da_k)$$=
\sum_{i=1}^n (-1)^{n-i} a_0da_1 \ldots d(a_ia_{i+1}) \ldots da_k$
$\Omega_V~A$ is a _differential graded algebra_ with differential $d :
\Omega^n_V~A \rightarrow \Omega^{n+1}_V~A$ defined by $d(a_0 da_1 \ldots
da_n) = da_0 da_1 \ldots da_n$ This may seem fairly abstract but in
case $A = C Q$ is a path algebra, then the bimodule $\Omega^n_V~A$ has a
$V$-generating set consisting of precisely the elements $p_0 dp_1
\ldots dp_n$ with all $p_i$ non-zero paths in $A$ and such that
$p_0p_1 \ldots p_n$ is also a non-zero path. One can put another
algebra multiplication on $\Omega_V~A$ which Cuntz and Quillen call the
_Fedosov product_ defined for an $n$-form $\omega$ and a form $\mu$ by
$\omega Circ \mu = \omega \mu -(-1)^n d\omega d\mu$ There is an
important relation between the two structures, the degree of a
differential form puts a filtration on $\Omega_V~A$ (with Fedosov
product) such that the _associated graded algebra_ is $\Omega_V~A$ with
the usual product. One can visualize the Fedosov product easily in the
case of path algebras because $\Omega_V~C Q$ with the Fedosov product is
again the path algebra of the quiver obtained by doubling up all the
arrows of $Q$. In our basic example when $Q$ is the quiver
$\xymatrix{\vtx{} \ar[rr]^u & & \vtx{} \ar@(ur,dr)^v} $ the
algebra of non-commutative differential forms equipped with the Fedosov
product is isomorphic to the path algebra of the quiver
$\xymatrix{\vtx{} \ar@/^/[rr]^{a=u+du} \ar@/_/[rr]_{b=u-du} & &
\vtx{} \ar@(u,ur)^{x=v+dv} \ar@(d,dr)_{y=v-dv}} $ with the
indicated identification of arrows with elements from $\Omega_V~C Q$.
Note however that we usually embed the algebra $C Q$ as the degree zero
differential forms in $\Omega_V~C Q$ with the usual multiplication and
that this embedding is no longer an algebra map (but a based linear map)
for the Fedosov product. For this reason, Cuntz and Quillen invent a
Yang-Mills type argument to “flow” this linear map to an algebra
embedding, but to motivate this we will have to say some things about
[curvatures][3].

[1]: http://www.neverendingbooks.org/index.php?p=352
[2]: http://www.neverendingbooks.org/index.php?p=349
[3]: http://www.neverendingbooks.org/index.php?p=353

Leave a Comment

some smaller steps

Published September 13, 2004 by lievenlb

It
always amazes me how much time I have to waste in trying to get
tech-stuff (such as this weblog) working the way I want. You will barely
notice it but again I spend too much time delving in PHP-scripts,
sometimes with minor success, most of the time almost wrecking this
weblog…

An example : it took me a day to figure out why
this page said there was just 1 visitor online whereas log files showed
otherwise. The PHP-script I used checked this by looking at the
IP-address via _REMOTE_ADDR_ which is perfectly OK on an ordinary
Mac OS 10.3 machine, but _not_ on an OS X-Server! For some reason
it gives as the REMOTE_ADDR just the IP address of the Server (that
is, www.matrix.ua.ac.be in this case) so whoever came by this page got
tagged as 143.129.75.209 and so the script thought there was just one
person around… The trivial way around it is changing every
occurence of REMOTE_ADDR by _HTTP_PC_REMOTE_ADDR_.
Easy enough but it took me a while to figure it out.

Another
example : over the week-end this weblog got a stalker! There were over
100 hits from 38.113.198.9, so whoever that is really liked this site
but didn't have time to read a thing… Again, the standard
solution is to ban the IP-address and most weblog-packages have such a
tool on their admin-page. But whathever I tried and Googled WordPress doesn't seem to have it
on board. There were a few hacks and plugins around claiming to do
something about it but none of them worked! So, I tried more drastic
actions such as editing .htaccess files which I thought would solve
everything (again, no problem under 10.3 but _not_ under
10.3-Server!). Once more, a couple of hours lost trying to figure out
how to get the firewall of a Mac-Server do what I needed. The upshot is
that I know now all dark secrets of the _ipfw_ command, so no
more stalking around this site…

In the process of
grounding my stalker, I decided that I needed better site-stats than my
homemade log-file provided. Fortunetely, this time I picked a package
that worked without too much hassle (one more time I had to make the
REMOTE_ADDR substitution but apart from that all went well). You will
see not too much of the power of this stats-package on the page (apart
from the global counter), I feel that such things are best forgotten
until something strange occurs (like stalkers, spammers and other
weirdos). A nice side-effect though was that for the first time I had a
look at _referring pages_, that is the URL leading to this weblog.
Lots of Google searches (some strange ones) but today there were also a
number of referrals from a Chinese blog. I checked it out and it turned
out to be the brand new Math is Math! Life is Life! weblog…

Another time
consuming thing was getting the BBC-news RSS feeds working in the
sidebar, so that you still get _some_ feel for reality while
being trapped here. I am not yet satisfied with the layout under
Explorer, but then everyone should move on to Safari (so I did give up
trying to work out the PHP-script).

But most time I wasted on
something that so far has left no trace whatsoever here. A plugin that
allows specific posts to be read only by registered users (of a certain
'level', that is WordPress can give users a level from 0 to 10
with specific degrees of freedom). But clearly at the same time I wanted
the rest of the world to have at least some indication of what they were
missing (such as a title with a nice padlock next to it) but so far I
didn't get it working. The only trace of a closed posting would be
in the sidebar-listing of the ten last posts but gives an error message
when an unauthorized user clicks on it. So, still a lot of
headache-sensitive work left to do, but it is about time to get back to
mathematics…

update (febr. 2007) : the
padlock-idea is abandoned.

Leave a Comment

explaining symmetry

Published September 10, 2004 by lievenlb


PseudonomousDaughterTwo learned vector-addition at school and
important formulas such as the _Chasles-Moebius_ equation

$\forall A,B,C \in \mathbb{R}^2~:~\vec{AB}+\vec{BC} = \vec{AC} $

Last evening I helped her a bit with her homework and there was one
problem she could not do immediately (but it was a starred exercise so
you didn't have to do it, but…) :

consider a regular pentagon
with center $\vec{0} $. Prove that

$\vec{0A} + \vec{0B} +
\vec{0C} + \vec{0D} + \vec{0E} = \vec{0} $

PD2 : How would
_you_ do this? (with a tone like : I bet even you can't do
it)
Me : Symmetry!
PD2 : Huh?
Me : Rotate the plane
1/5 turn, then $A \mapsto B $, $B \mapsto C $ and so on. So the vector
giving the sum of all five terms must be mapped to itself under this
rotation and the only vector doing this is the zero vector.
PD2 :
That cannot be the solution, you didn't take sums of vectors and all
other exercises did that.
Me : I don't care, it is an elegant
solution, you don't have to compute a thing!

But clearly
she was not convinced and I had to admit there was nothing in her
textbook preparing her for such an argument. I was about to explain that
there was even more symmetry : reflecting along a line through a vertex
giving dihedral symmetry when I saw what the _intended solution_
of the exercise was :

Me : Okay, if you _have_ to do
sums let us try this. Fix a vertex, say A. Then the sum
$\vec{0E}+\vec{0B} $ must lie on the line 0A by the parallellogram-rule
(always good to drop in a word from the textbook to gain some
trust…), similarly the sum $\vec{0C}+\vec{0D} $ must lie on the
line 0A. So you now have to do a sum of three vectors lying on the
line 0A so the result must lie on 0A
PD2 : Yes, and???
Me : But there was nothing special about $A$. I could have started with
B and do the whole argument all over again and then I would get that
the sum is a vector on the line 0B
PD2 : And the only vector
lying on both 0A and 0B is $\vec{0} $
Me : Right! But
all we did now was just redoing the symmetry argument because the line
0A is mapped to 0B
PD2 : Don't you get started on
_that symmetry_ again!

I wonder which of the two
solutions she will sell today as her own. I would love to see the face
of a teacher when a 15yr old says “Clearly that is trivial because
the zero vector is the only one left invariant under
pentagon-symmetry!”

Leave a Comment

cotangent bundles

Published September 9, 2004 by lievenlb

The
previous post in this sequence was [moduli spaces][1]. Why did we spend
time explaining the connection of the quiver
$Q~:~\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar@(ur,dr)^x} $
to moduli spaces of vectorbundles on curves and moduli spaces of linear
control systems? At the start I said we would concentrate on its _double
quiver_ $\tilde{Q}~:~\xymatrix{\vtx{} \ar@/^/[rr]^a && \vtx{}
\ar@(u,ur)^x \ar@(d,dr)_{x^*} \ar@/^/[ll]^{a^*}} $ Clearly,
this already gives away the answer : if the path algebra $C Q$
determines a (non-commutative) manifold $M$, then the path algebra $C
\tilde{Q}$ determines the cotangent bundle of $M$. Recall that for a
commutative manifold $M$, the cotangent bundle is the vectorbundle
having at the point $p \in M$ as fiber the linear dual $(T_p M)^*$ of
the tangent space. So, why do we claim that $C \tilde{Q}$
corresponds to the cotangent bundle of $C Q$? Fix a dimension vector
$\alpha = (m,n)$ then the representation space
$\mathbf{rep}_{\alpha}~Q = M_{n \times m}(C) \oplus M_n(C)$ is just
an affine space so in its point the tangent space is the representation
space itself. To define its linear dual use the non-degeneracy of the
_trace pairings_ $M_{n \times m}(C) \times M_{m \times n}(C)
\rightarrow C~:~(A,B) \mapsto tr(AB)$ $M_n(C) \times M_n(C)
\rightarrow C~:~(C,D) \mapsto tr(CD)$ and therefore the linear dual
$\mathbf{rep}_{\alpha}~Q^* = M_{m \times n}(C) \oplus M_n(C)$ which is
the representation space $\mathbf{rep}_{\alpha}~Q^s$ of the quiver
$Q^s~:~\xymatrix{\vtx{} & & \vtx{} \ar[ll] \ar@(ur,dr)} $
and therefore we have that the cotangent bundle to the representation
space $\mathbf{rep}_{\alpha}~Q$ $T^* \mathbf{rep}_{\alpha}~Q =
\mathbf{rep}_{\alpha}~\tilde{Q}$ Important for us will be that any
cotangent bundle has a natural _symplectic structure_. For a good
introduction to this see the [course notes][2] “Symplectic geometry and
quivers” by [Geert Van de Weyer][3]. As a consequence $C \tilde{Q}$
can be viewed as a non-commutative symplectic manifold with the
symplectic structure determined by the non-commutative 2-form
$\omega = da^* da + dx^* dx$ but before we can define all this we
will have to recall some facts on non-commutative differential forms.
Maybe [next time][4]. For the impatient : have a look at the paper by
Victor Ginzburg [Non-commutative Symplectic Geometry, Quiver varieties,
and Operads][5] or my paper with Raf Bocklandt [Necklace Lie algebras
and noncommutative symplectic geometry][6]. Now that we have a
cotangent bundle of $C Q$ is there also a _tangent bundle_ and does it
again correspond to a new quiver? Well yes, here it is
$\xymatrix{\vtx{} \ar@/^/[rr]^{a+da} \ar@/_/[rr]_{a-da} & & \vtx{}
\ar@(u,ur)^{x+dx} \ar@(d,dr)_{x-dx}} $ and the labeling of the
arrows may help you to work through some sections of the Cuntz-Quillen
paper…

[1]: http://www.neverendingbooks.org/index.php?p=39
[2]: http://www.win.ua.ac.be/~gvdwey/lectures/symplectic_moment.pdf
[3]: http://www.win.ua.ac.be/~gvdwey/
[4]: http://www.neverendingbooks.org/index.php?p=41
[5]: http://www.arxiv.org/abs/math.QA/0005165
[6]: http://www.arxiv.org/abs/math.AG/0010030

2 Comments

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