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Author: lievenlb

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

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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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megaminx

Published September 7, 2004 by lievenlb

In a few
weeks I will give a _geometry 101_ course! It was decided that in
this course I should try to explain what rotations in $\mathbb{R}^3’$
are, so the classification of all finite rotation groups seemed like a
fun topic. Along the way I’ll have to introduce groups so bringing in a
little bit of GAP
may be a good idea. Clearly, the real power of GAP is lost on the
symmetry groups of the Platonic solids so I’ll do the traditional
computation of the transformation group of the Rubik’s cube. But
then I discovered that there is also a version of it on the dodecahedron
which is called megaminx so I couldn’t resist trying to work out the order of its
transformation group. Fortunately Coreyanne Rickwalt did already the
hard work giving a presentation as
a permutation group. So giving the generators to GAP


f1:=(1,3,5,7,9)(2,4,6,8,10)(20,31,42,53,64)(19,30,41,52,63)(18,29,40,51,62);
f2:=(12,14,16,18,20)(13,15,17,19,21)(1,60,73,84,31)(3,62,75,86,23)(2,61,74,85,32);
f3:=(23,25,27,29,31)(24,26,28,30,32)(82,95,42,3,16)(83,96,43,4,17)(84,97,34,5,18);
f4:=(34,36,38,40,42)(35,37,39,41,43)(27,93,106,53,5)(28,94,107,54,6)(29,95,108,45,7);
f5:=(45,47,49,51,53)(46,48,50,52,54)(38,104,117,64,7)(39,105,118,65,8),(40,106,119,56,9);
f6:=(56,58,60,62,64)(57,59,61,63,65)(49,115,75,20,9)(50,116,76,21,10),(51,117,67,12,1);
f7:=(67,69,71,73,75)(68,70,72,74,76)(58,113,126,86,12)(59,114,127,7,13),(60,115,128,78,14);
f8:=(78,80,82,84,86)(79,81,83,85,87)(71,124,97,23,14)(72,125,98,24,15),(73,126,89,25,16);
f9:=(89,91,93,95,97)(90,92,94,96,98)(80,122,108,34,25)(81,123,109,35,26),(82,124,100,36,27);
f10:=(100,102,104,106,108)(101,103,105,107,109)(91,130,119,45,36),(92,131,120,46,37)(93,122,111,47,38);
f11:=(111,113,115,117,119)(112,114,116,118,120)(102,128,67,56,47),(103,129,68,57,48)(104,130,69,58,49);
f12:=(122,124,126,128,130)(123,125,127,129,131)(100,89,78,69,111),(101,90,79,70,112)(102,91,80,71,113);

and defining the
megaminx group by


megaminx:=Group(f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12); Size(megaminx);

and asking for its order I was a bit surprised to get
after a couple of minutes the following awkward number


33447514567245635287940590270451862933763731665690149051478356761508167786224814946834370826
35992490654078818946607045276267204294704060929949240557194825002982480260628480000000000000
000000000000000

or if you prefer it is
$2^{115} 3^{58} 5^{28} 7^{19} 11^{10} 13^9 17^7 19^6 23^5 29^4 31^3
37^3 41^2 43^2 47^2 53^2 59^2 61 .67 .71. 73. 79 .83 .89 .97. 101 .103.
107 .109 .113$

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