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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!”

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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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