Tag: representations

noncommutative curves and their maniflds

Published March 17, 2007 by lievenlb

Last time we have
seen that the noncommutative manifold of a Riemann surface can be viewed
as that Riemann surface together with a loop in each point. The extra
loop-structure tells us that all finite dimensional representations of
the coordinate ring can be found by separating over points and those
living at just one point are classified by the isoclasses of nilpotent
matrices, that is are parametrized by the partitions (corresponding
to the sizes of the Jordan blocks). In addition, these loops tell us
that the Riemann surface locally looks like a Riemann sphere, so an
equivalent mental picture of the local structure of this
noncommutative manifold is given by the picture on teh left, where the surface is part of the Riemann surface
and a sphere is placed at every point. Today we will consider
genuine noncommutative curves and describe their corresponding
noncommutative manifolds.

Here, a mental picture of such a
_noncommutative sphere_ to keep in mind would be something
like the picture on the right. That is, in most points of the sphere we place as before again
a Riemann sphere but in a finite number of points a different phenomen
occurs : we get a cluster of infinitesimally nearby points. We
will explain this picture with an easy example. Consider the
complex plane $\mathbb{C} $, the points of which are just the
one-dimensional representations of the polynomial algebra in one
variable $\mathbb{C}[z] $ (any algebra map $\mathbb{C}[z] \rightarrow \mathbb{C} $ is fully determined by the image of z). On this plane we
have an automorphism of order two sending a complex number z to its
negative -z (so this automorphism can be seen as a point-reflexion
with center the zero element 0). This automorphism extends to
the polynomial algebra, again induced by sending z to -z. That
is, the image of a polynomial $f(z) \in \mathbb{C}[z] $ under this
automorphism is f(-z).

With this data we can form a noncommutative
algebra, the _skew-group algebra_ $\mathbb{C}[z] \ast C_2 $ the
elements of which are either of the form $f(z) \ast e $ or $g(z) \ast g $ where
$C_2 = \langle g : g^2=e \rangle $ is the cyclic group of order two
generated by the automorphism g and f(z),g(z) are arbitrary
polynomials in z.

The multiplication on this algebra is determined by
the following rules

$(g(z) \ast g)(f(z) \ast e) = g(z)f(-z) \astg $ whereas $(f(z) \ast e)(g(z) \ast g) = f(z)g(z) \ast g $

$(f(z) \ast e)(g(z) \ast e) = f(z)g(z) \ast e $ whereas $(f(z) \ast g)(g(z)\ast g) = f(z)g(-z) \ast e $

That is, multiplication in the
$\mathbb{C}[z] $ factor is the usual multiplication, multiplication in
the $C_2 $ factor is the usual group-multiplication but when we want
to get a polynomial from right to left over a group-element we have to
apply the corresponding automorphism to the polynomial (thats why we
call it a _skew_ group-algebra).

Alternatively, remark that as
a $\mathbb{C} $-algebra the skew-group algebra $\mathbb{C}[z] \ast C_2 $ is
an algebra with unit element 1 = 1\aste and is generated by
the elements $X = z \ast e $ and $Y = 1 \ast g $ and that the defining
relations of the multiplication are

$Y^2 = 1 $ and $Y.X =-X.Y $

hence another description would
be

$\mathbb{C}[z] \ast C_2 = \frac{\mathbb{C} \langle X,Y \rangle}{ (Y^2-1,XY+YX) } $

It can be shown that skew-group
algebras over the coordinate ring of smooth curves are _noncommutative
smooth algebras_ whence there is a noncommutative manifold associated
to them. Recall from last time the noncommutative manifold of a
smooth algebra A is a device to classify all finite dimensional
representations of A upto isomorphism Let us therefore try to
determine some of these representations, starting with the
one-dimensional ones, that is, algebra maps from

$\mathbb{C}[z] \ast C_2 = \frac{\mathbb{C} \langle X,Y \rangle}{ (Y^2-1,XY+YX) } \rightarrow \mathbb{C} $

Such a map is determined by the image of X and that of
Y. Now, as $Y^2=1 $ we have just two choices for the image of Y
namely +1 or -1. But then, as the image is a commutative algebra
and as XY+YX=0 we must have that the image of 2XY is zero whence the
image of X must be zero. That is, we have only
two one-dimensional representations, namely $S_+ : X \rightarrow 0, Y \rightarrow 1 $
and $S_- : X \rightarrow 0, Y \rightarrow -1 $

This is odd! Can
it be that our noncommutative manifold has just 2 points? Of course not.
In fact, these two points are the exceptional ones giving us a cluster
of nearby points (see below) whereas most points of our
noncommutative manifold will correspond to 2-dimensional
representations!

So, let’s hunt them down. The
center of $\mathbb{C}[z]\ast C_2 $ (that is, the elements commuting with
all others) consists of all elements of the form $f(z)\ast e $ with f an
_even_ polynomial, that is, f(z)=f(-z) (because it has to commute
with 1\ast g), so is equal to the subalgebra $\mathbb{C}[z^2]\ast e $.

The
manifold corresponding to this subring is again the complex plane
$\mathbb{C} $ of which the points correspond to all one-dimensional
representations of $\mathbb{C}[z^2]\ast e $ (determined by the image of
$z^2\ast e $).

We will now show that to each point of $\mathbb{C} – { 0 } $
corresponds a simple 2-dimensional representation of
$\mathbb{C}[z]\ast C_2 $.

If a is not zero, we will consider the
quotient of the skew-group algebra modulo the twosided ideal generated
by $z^2\ast e-a $. It turns out
that

$\frac{\mathbb{C}[z]\ast C_2}{(z^2\aste-a)} =
\frac{\mathbb{C}[z]}{(z^2-a)} \ast C_2 = (\frac{\mathbb{C}[z]}{(z-\sqrt{a})}
\oplus \frac{\mathbb{C}[z]}{(z+\sqrt{a})}) \ast C_2 = (\mathbb{C}
\oplus \mathbb{C}) \ast C_2 $

where the skew-group algebra on the
right is given by the automorphism g on $\mathbb{C} \oplus \mathbb{C} $ interchanging the two factors. If you want to
become more familiar with working in skew-group algebras work out the
details of the fact that there is an algebra-isomorphism between
$(\mathbb{C} \oplus \mathbb{C}) \ast C_2 $ and the algebra of $2 \times 2 $ matrices $M_2(\mathbb{C}) $. Here is the
identification

$~(1,0)\aste \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $

$~(0,1)\aste \rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $

$~(1,0)\astg \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $

$~(0,1)\astg \rightarrow \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $

so you have to verify that multiplication
on the left hand side (that is in $(\mathbb{C} \oplus \mathbb{C}) \ast
C_2 $) coincides with matrix-multiplication of the associated
matrices.

Okay, this begins to look like what we are after. To
every point of the complex plane minus zero (or to every point of the
Riemann sphere minus the two points ${ 0,\infty } $) we have
associated a two-dimensional simple representation of the skew-group
algebra (btw. simple means that the matrices determined by the images
of X and Y generate the whole matrix-algebra).

In fact, we
now have already classified ‘most’ of the finite dimensional
representations of $\mathbb{C}[z]\ast C_2 $, namely those n-dimensional
representations

$\mathbb{C}[z]\ast C_2 =
\frac{\mathbb{C} \langle X,Y \rangle}{(Y^2-1,XY+YX)} \rightarrow M_n(\mathbb{C}) $

for which the image of X is an invertible $n \times n $ matrix. We can show that such representations only exist when
n is an even number, say n=2m and that any such representation is
again determined by the geometric/combinatorial data we found last time
for a Riemann surface.

That is, It is determined by a finite
number ${ P_1,\dots,P_k } $ of points from $\mathbb{C} – 0 $ where
k is at most m. For each index i we have a positive
number $a_i $ such that $a_1+\dots+a_k=m $ and finally for each i we
also have a partition of $a_i $.

That is our noncommutative
manifold looks like all points of $\mathbb{C}-0 $ with one loop in each
point. However, we have to remember that each point now determines a
simple 2-dimensional representation and that in order to get all
finite dimensional representations with det(X) non-zero we have to
scale up representations of $\mathbb{C}[z^2] $ by a factor two.
The technical term here is that of a Morita equivalence (or that the
noncommutative algebra is an Azumaya algebra over
$\mathbb{C}-0 $).

What about the remaining representations, that
is, those for which Det(X)=0? We have already seen that there are two
1-dimensional representations $S_+ $ and $S_- $ lying over 0, so how
do they fit in our noncommutative manifold? Should we consider them as
two points and draw also a loop in each of them or do we have to do
something different? Rememer that drawing a loop means in our
geometry -> representation dictionary that the representations
living at that point are classified in the same way as nilpotent
matrices.

Hence, drawing a loop in $S_+ $ would mean that we have a
2-dimensional representation of $\mathbb{C}[z]\ast C_2 $ (different from
$S_+ \oplus S_+ $) and any such representation must correspond to
matrices

$X \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $Y \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

But this is not possible as these matrices do
_not_ satisfy the relation XY+YX=0. Hence, there is no loop in $S_+ $
and similarly also no loop in $S_- $.

However, there are non
semi-simple two dimensional representations build out of the simples
$S_+ $ and $S_- $. For, consider the matrices

$X \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $Y \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $

then these
matrices _do_ satisfy XY+YX=0! (and there is another matrix-pair
interchanging $\pm 1 $ in the Y-matrix). In erudite terminology this
says that there is a _nontrivial extension_ between $S_+ $ and $S_- $
and one between $S_- $ and $S_+ $.

In our dictionary we will encode this
information by the picture

$\xymatrix{\vtx{}
\ar@/^2ex/[rr] & & \vtx{} \ar@/^2ex/[ll]} $

where the two
vertices correspond to the points $S_+ $ and $S_- $ and the arrows
represent the observed extensions. In fact, this data suffices to finish
our classification project of finite dimensional representations of
the noncommutative curve $\mathbb{C}[z] \ast C_2 $.

Those with Det(X)=0
are of the form : $R \oplus T $ where R is a representation with
invertible X-matrix (which we classified before) and T is a direct
sum of representations involving only the simple factors $S_+ $ and
$S_- $ and obtained by iterating the 2-dimensional idea. That is, for
each factor the Y-matrix has alternating $\pm 1 $ along the diagonal
and the X-matrix is the full nilpotent Jordan-matrix.

So
here is our picture of the noncommutative manifold of the
noncommutative curve $\mathbb{C}[z]\ast C_2 $ : the points are all points
of $\mathbb{C}-0 $ together with one loop in each of them together
with two points lying over 0 where we draw the above picture of arrows
between them. One should view these two points as lying
infinetesimally close to each other and the gluing
data

$\xymatrix{\vtx{} \ar@/^2ex/[rr] & & \vtx{}
\ar@/^2ex/[ll]} $

contains enough information to determine
that all other points of the noncommutative manifold in the vicinity of
this cluster should be two dimensional simples! The methods used
in this simple minded example are strong enough to determine the
structure of the noncommutative manifold of _any_ noncommutative curve.

So, let us look at a real-life example. Once again, take the
Kleinian quartic In a previous
course-post we recalled that
there is an action by automorphisms on the Klein quartic K by the
finite simple group $PSL_2(\mathbb{F}_7) $ of order 168. Hence, we
can form the noncommutative Klein-quartic $K \ast PSL_2(\mathbb{F}_7) $
(take affine pieces consisting of complements of orbits and do the
skew-group algebra construction on them and then glue these pieces
together again).

We have also seen that the orbits are classified
under a Belyi-map $K \rightarrow \mathbb{P}^1_{\mathbb{C}} $ and that this map
had the property that over any point of $\mathbb{P}^1_{\mathbb{C}}
– { 0,1,\infty } $ there is an orbit consisting of 168 points
whereas over 0 (resp. 1 and $\infty $) there is an orbit
consisting of 56 (resp. 84 and 24 points).

So what is
the noncommutative manifold associated to the noncommutative Kleinian?
Well, it looks like the picture we had at the start of this
post For all but three points of the Riemann sphere
$\mathbb{P}^1 – { 0,1,\infty } $ we have one point and one loop
(corresponding to a simple 168-dimensional representation of $K \ast
PSL_2(\mathbb{F}_7) $) together with clusters of infinitesimally nearby
points lying over 0,1 and $\infty $ (the cluster over 0
is depicted, the two others only indicated).

Over 0 we have
three points connected by the diagram

$\xymatrix{& \vtx{} \ar[ddl] & \\ & & \\ \vtx{} \ar[rr] & & \vtx{} \ar[uul]} $

where each of the vertices corresponds to a
simple 56-dimensional representation. Over 1 we have a cluster of
two points corresponding to 84-dimensional simples and connected by
the picture we had in the $\mathbb{C}[z]\ast C_2 $ example).

Finally,
over $\infty $ we have the most interesting cluster, consisting of the
seven dwarfs (each corresponding to a simple representation of dimension
24) and connected to each other via the
picture

$\xymatrix{& & \vtx{} \ar[dll] & & \\ \vtx{} \ar[d] & & & & \vtx{} \ar[ull] \\ \vtx{} \ar[dr] & & & & \vtx{} \ar[u] \\ & \vtx{} \ar[rr] & & \vtx{} \ar[ur] &} $

Again, this noncommutative manifold gives us
all information needed to give a complete classification of all finite
dimensional $K \ast PSL_2(\mathbb{F}_7) $-representations. One
can prove that all exceptional clusters of points for a noncommutative
curve are connected by a cyclic quiver as the ones above. However, these
examples are still pretty tame (in more than one sense) as these
noncommutative algebras are finite over their centers, are Noetherian
etc. The situation will become a lot wilder when we come to exotic
situations such as the noncommutative manifold of
$SL_2(\mathbb{Z}) $…

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the noncommutative manifold of a Riemann surface

Published March 16, 2007 by lievenlb

The
natural habitat of this lesson is a bit further down the course, but it
was called into existence by a comment/question by
Kea

I don’t yet quite see where the nc
manifolds are, but I guess that’s coming.

As
I’m enjoying telling about all sorts of sources of finite dimensional
representations of $SL_2(\mathbb{Z}) $ (and will carry on doing so for
some time), more people may begin to wonder where I’m heading. For this
reason I’ll do a couple of very elementary posts on simple examples of
noncommutative manifolds.

I realize it is ‘bon ton’ these days
to say that noncommutative manifolds are virtual objects associated to
noncommutative algebras and that the calculation of certain invariants
of these algebras gives insight into the topology and/or geometry of
these non-existent spaces. My own attitude to noncommutative geometry is
different : to me, noncommutative manifolds are genuine sets of points
equipped with a topology and other structures which I can use as a
mnemotechnic device to solve the problem of interest to me which is the
classification of all finite dimensional representations of a smooth
noncommutative algebra.

Hence, when I speak of the
‘noncommutative manifold of $SL_2(\mathbb{Z}) $’ Im after an object
containing enough information to allow me (at least in principle) to
classify the isomorphism classes of all finite dimensional
$SL_2(\mathbb{Z}) $-representations. The whole point of this course is
to show that such an object exists and that we can make explicit
calculations with it. But I’m running far ahead. Let us start with
an elementary question :

Riemann surfaces are examples of
noncommutative manifolds, so what is the noncommutative picture of
them?

I’ve browsed the Google-pictures a bit and a picture
coming close to my mental image of the noncommutative manifold of a
Riemann surface locally looks like the picture on the left. Here, the checkerboard-surface is part of the Riemann surface
and the extra structure consists in putting in each point of the Riemann
surface a sphere, reflecting the local structure of the Riemann surface
near the point. In fact, my picture is slightly different : I want to
draw a loop in each point of the Riemann surface, but Ill explain why
the two pictures are equivalent and why they present a solution to the
problem of classifying all finite dimensional representations of the
Riemann surface. After all why do we draw and study Riemann
surfaces? Because we are interested in the solutions to equations. For
example, the points of the _Kleinian quartic Riemann
surface_ give us all solutions tex \in
\mathbb{C}^3 $ to the equation $X^3Y+Y^3Z+Z^3X=0 $. If (a,b,c) is such
a solution, then so are all scalar multiples $(\lambda a,\lambda
b,\lambda c) $ so we may as well assume that the Z$coordinate is equal
to 1 and are then interested in finding the solutions tex \in
\mathbb{C}^2 $ to the equation $X^3Y+Y^3+X=0 $ which gives us an affine
patch of the Kleinian quartic (in fact, these solutions give us all
points except for two, corresponding to the _points at infinity_ needed
to make the picture compact so that we can hold it in our hand and look
at it from all sides. These points at infinity correspond to the trivial
solutions (1,0,0) and (0,1,0)).

What is the connection
between points on this Riemann surface and representations? Well, if
(a,b) is a solution to the equation $X^3Y+Y^3+X=0 $, then we have a
_one-dimensional representation_ of the affine _coordinate ring_
$\mathbb{C}[X,Y]/(X^3Y+Y^3+X) $, that is, an algebra
morphism

$\mathbb{C}[X,Y]/(X^3Y+Y^3+X) \rightarrow \mathbb{C} $

defined by sending X to a and Y to b.
Conversely, any such one-dimensonal representation gives us a solution
(look at the images of X and Y and these will be the coordinates of
a solution). Thus, commutative algebraic geometry of smooth
curves (that is Riemann surfaces if you look at the ‘real’ picture)
can be seen as the study of one-dimensional representations of their
smooth coordinate algebras. In other words, the classical Riemann
surface gives us already the classifcation of all one-dimensional
representations, so now we are after the ‘other ones’.

In
noncommtative algebra it is not natural to restrict attention to algebra
maps to $\mathbb{C} $, at least we would also like to include algebra
maps to $n \times n $ matrices $M_n(\mathbb{C}) $. An n-dimensional
representation of the coordinate algebra of the Klein quartic is an
algebra map

$\mathbb{C}[X,Y]/(X^3Y+Y^3+X) \rightarrow M_n(\mathbb{C}) $

That is, we want to find all pairs of $n \times n $ matrices A and B satisfying the following
matrix-identities

$A.B=B.A $ and $A^3.B+B^3+A=0_n $

The
first equation tells us that the two matrices must commute (because we
took commuting variables X and Y) and the second equation really is
a set of $n^2 $-equations in the matrix-entries of A and
B.

There is a sneaky way to get lots of such matrix-couples
from a given solution (A,B), namely by _simultaneous conjugation_.
That is, if $C \in GL_n(\mathbb{C}) $ is any invertible $n \times n $
matrix, then also the matrix-couple $~(C^{-1}.A.C,C^{-1}.B.C) $
satisfies all the required equations (write the equations out and notice
that middle terms of the form $C.C^{-1} $ cancel out and check that one
then obtains the matrix-identities

$C^{-1} A B C = C^{-1} BA C $ and $C^{-1}(A^3B+B^3+A)C = 0_n $

which are satisfied because
(A,B) was supposed to be a solution). We then say that these two
n-dimensional representations are _isomorphic_ and naturally we are
only interested in classifying the isomorphism classes of all
representations.

Using classical commutative algebra theory of
Dedekind domains (such as the coordinate ring $\mathbb{C}[X,Y]/(X^3Y+Y^3+X) $)
allows us to give a complete solution to this problem. It says that any
n-dimensional representation is determined up to isomorphism by the
following geometric/combinatorial data

a finite set of points $P_1,P_2,\dots,P_k $ on the Riemann surface with $k \leq n $.
a set of positive integers $a_1,a_2,\dots,a_k $ associated to these pointssatisfying $a_1+a_2+\dots_a_k=n $.
for each $a_i $ a partition of $a_i $ (that is, a decreasing sequence of numbers with total sum
$a_i $).

To encode this classification I’ll use the mental
picture of associating to every point of the Klein quartic a small
loop. $\xymatrix{\vtx{}
\ar@(ul,ur)} $ Don\’t get over-exited about this
noncommutative manifold picture of the Klein quartic, I do not mean to
represent something like closed strings emanating from all points of the
Riemann surface or any other fanshi-wanshi interpretation. Just as
Feynman-diagrams allow the initiated to calculate probabilities of
certain interactions, the noncommutative manifold allows the
initiated to classify finite dimensional representations.

Our
mental picture of the noncommutative manifold of the Klein quartic, that
is : the points of the Klein quartic together with a loop in each point,
will tell the initiated quite a few things, such as : The fact
that there are no arrows between distict points, tells us that the
classification problem splits into local problems in a finite number of
points. Technically, this encodes the fact that there are no nontrivial
extensions between different simples in the commutative case. This will
drastically change if we enter the noncommutative world…

The fact that there is one loop in each point, tells us that
the local classification problem in that point is the same as that of
classifying nilpotent matrices upto conjugation (which, by the Jordan
normal form result, are classified by partitions) Moreover,
the fact that there is one loop in each point tells us that the local
structure of simple representations near that point (that is, the points
on the Kleinian quartic lying nearby) are classified as the simple
representations of the polynmial algebra $\mathbb{C}[x] $ (which are the
points on the complex plane, giving the picture
of the Riemann sphere in each point reflecting the local
neighborhood of the point on the Klein quartic)

In general, the
noncommutative manifold associated to a noncommutative smooth algebra
will be of a similar geometric/combinatorial nature. Typically, it will
consist of a geometric collection of points and arrows and loops between
these points. This data will then allow us to reduce the classification
problem to that of _quiver-representations_ and will allow us to give
local descriptions of our noncommutative manifolds. Next time,
I’ll give the details in the first noncommutative example : the
skew-group algebra of a finite group of automorphisms on a Riemann
surface (such as the simple group $PSL_2(\mathbb{F}_7) $ acting on the
Klein quartic). Already in this case, some new phenomena will
appear…

ADDED : While writing this post
NetNewsWire informed me that over at Noncommutative Geometry they have a
post on a similar topic : What is a noncommutative space.

One Comment

The cartographers’ groups

Published March 13, 2007 by lievenlb

Just as cartographers like
Mercator drew maps of
the then known world, we draw dessins
d ‘enfants to depict the
associated algebraic curve defined over
$\overline{\mathbb{Q}} $.

In order to see that such a dessin
d’enfant determines a permutation representation of one of
Grothendieck’s cartographic groups, $SL_2(\mathbb{Z}),
\Gamma_0(2) $ or $\Gamma(2) $ we need to have realizations of these
groups (as well as their close relatives
$PSL_2(\mathbb{Z}),GL_2(\mathbb{Z}) $ and $PGL_2(\mathbb{Z}) $) in
terms of generators and relations.

As this lesson will be rather
technical I’d better first explain what we will prove (so that you can
skip it if you feel comfortable with the statements) and why we want to
prove it. What we will prove in detail below is that these groups
can be written as free (or amalgamated) group products. We will explain
what this means and will establish that

$PSL_2(\mathbb{Z}) = C_2
\ast C_3, \Gamma_0(2) = C_2 \ast C_{\infty}, \Gamma(2)
= C_{\infty} \ast C_{\infty} $

$SL_2(\mathbb{Z}) =
C_4 \ast_{C_2} C_6, GL_2(\mathbb{Z}) = D_4 \ast_{D_2} D_6,
PGL_2(\mathbb{Z}) = D_2 \ast_{C_2} D_3 $

where $C_n $ resp.
$D_n $ are the cyclic (resp. dihedral) groups. The importance of these
facts it that they will allow us to view the set of (isomorphism classes
of) finite dimensional representations of these groups as
noncommutative manifolds . Looking at the statements above we
see that these arithmetical groups can be build up from the first
examples in any course on finite groups : cyclic and dihedral
groups.

Recall that the cyclic group of order n, $C_n $ is the group of
rotations of a regular n-gon (so is generated by a rotation r with
angle $\frac{2 \pi}{n} $ and has defining relation $r^n = 1 $, where 1
is the identity). However, regular n-gons have more symmetries :
flipping over one of its n lines of symmetry

The dihedral group $D_n $ is the group generated by the n
rotations and by these n flips. If, as before r is a generating
rotation and d is one of the flips, then it is easy to see that the
dihedral group is generated by r and d and satisfied the defining
relations

$r^n=1 $ and $d^2 = 1 = (rd)^2 $

Flipping twice
does nothing and to see the relation $~(rd)^2=1 $ check that doing twice a
rotation followed by a flip brings all vertices back to their original
location. The dihedral group $D_n $ has 2n elements, the n-rotations
$r^i $ and the n flips $dr^i $.

In fact, to get at the cartographic
groups we will only need the groups $D_4, D_6 $ and their
subgroups. Let us start by finding generators of the largest
group $GL_2(\mathbb{Z}) $ which is the group of all invertible $2
\times 2 $ matrices with integer coefficients.

Consider the
elements

$U = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},
V = \begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix}/tex] and $R =
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

and form the
matrices

$X = UV = \begin{bmatrix} 1 & -1 \\ 0 & 1
\end{bmatrix}, Y = VU = \begin{bmatrix} 1 & 0 \\ 1 & 1
\end{bmatrix} $

By induction we prove the following relations in
$GL_2(\mathbb{Z}) $

$X^n \begin{bmatrix} a & b \\ c & d
\end{bmatrix} = \begin{bmatrix} a-nc & b-nd \\ c & d \end{bmatrix} $
and $\begin{bmatrix} a & b \\ c& d \end{bmatrix} X^n =
\begin{bmatrix} a & b-na \\ c & d-nc \end{bmatrix} $

$Y^n \begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} a & b \\ c+na & d+nb \end{bmatrix} $ and
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} Y^n = \begin{bmatrix}
a+nb & b \\ c+nd & d \end{bmatrix} $

The determinant ad-bc of
a matrix in $GL_2(\mathbb{Z}) $ must be $\pm 1 $ whence all rows and
columns of

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in
GL_2(\mathbb{Z}) $

consist of coprime numbers and hence a and
c can be reduced modulo each other by left multiplication by a power
of X or Y until one of them is zero and the other is $\pm 1 $. We
may even assume that $a = \pm 1 $ (if not, left multiply with U).

So,
by left multiplication by powers of X and Y and U we can bring any
element of $GL_2(\mathbb{Z}) $ into the form

$\begin{bmatrix}
\pm 1 & \beta \\ 0 & \pm 1 \end{bmatrix} $

and again by left
multiplication by a power of X we can bring it in one of the four
forms

$\begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{bmatrix}
= { 1,UR,RU,U^2 } $

This proves that $GL_2(\mathbb{Z}) $ is
generated by the elements U,V and R.

Similarly, the group
$SL_2(\mathbb{Z}) $ of all $2 \times 2 $ integer matrices with
determinant 1 is generated by the elements U and V as using the
above method and the restriction on the determinant we will end up with
one of the two matrices

${ \begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix},\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} } =
{ 1,U^2 } $

so we never need the matrix R. As for
relations, there are some obvious relations among the matrices U,V and
R, namely

$U^2=V^3 $ and $1=U^4=R^2=(RU)^2=(RV)^2$ $

The
real problem is to prove that all remaining relations are consequences
of these basic ones. As R clearly has order two and its commutation
relations with U and V are just $RU=U^{-1}R $ and $RV=V^{-1}R $ we can
pull R in any relation to the far right and (possibly after
multiplying on the right with R) are left to prove that the only
relations among U and V are consequences of $U^2=V^3 $ and
$U^4=1=V^6 $.

Because $U^2=V^3 $ this element is central in the
group generated by U and V (which we have seen to be
$SL_2(\mathbb{Z}) $) and if we quotient it out we get the modular
group

$\Gamma = PSL_2(\mathbb{Z}) $

Hence in order to prove our claim
it suffices that

$PSL_2(\mathbb{Z}) = \langle
\overline{U},\overline{V} : \overline{U}^2=\overline{V}^3=1
\rangle $

Phrased differently, we have to show that
$PSL_2(\mathbb{Z}) $ is the free group product of the cyclic groups of
order two and three (those generated by $u = \overline{U} $ and
$v=\overline{V} $) $C_2 \ast C_3 $

Any element of this free group
product is of the form $~(u)v^{a_1}uv^{a_2}u \ldots
uv^{a_k}(u) $ where beginning and trailing u are optional and
all $a_i $ are either 1 or 2.

So we have to show that in
$PSL_2(\mathbb{Z}) $ no such word can give the identity
element. Today, we will first sketch the classical argument based
on the theory of groups acting on trees due to Jean-Pierre
Serre and Hyman Bass. Tomorrow, we will give a short elegant proof due to
Roger Alperin and draw
consequences to the description of the carthographic groups as
amalgamated free products of cyclic and dihedral groups.

Recall
that $GL_2(\mathbb{Z}) $ acts via Moebius
transformations on
the complex plane $\mathbb{C} = \mathbb{R}^2 $ (actually it is an
action on the Riemann sphere $\mathbb{P}^1_{\mathbb{C}} $) given by the
maps

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}.z =
\frac{az+b}{cz+d} $

Note that the action of the
center of $GL_2(\mathbb{Z}) $ (that is of $\pm \begin{bmatrix} 1 & 0
\\ 0 & 1 \end{bmatrix} $) acts trivially, so it is really an action of
$PGL_2(\mathbb{Z}) $.

As R interchanges the upper and lower half-plane
we might as well restrict to the action of $SL_2(\mathbb{Z}) $ on the
upper-halfplane $\mathcal{H} $. It is quite easy to see that a
fundamental domain
for this action is given by the greyed-out area

To see that any $z \in \mathcal{H} $ can be taken into this
region by an element of $PSL_2(\mathbb{Z}) $ note the following two
Moebius transformations

$\begin{bmatrix} 1 & 1 \\ 0 & 1
\end{bmatrix}.z = z+1 $ and $\begin{bmatrix} 0 & 1 \\ -1
& 0 \end{bmatrix}.z = -\frac{1}{z} $

The first
operation takes any z into a strip of length one, for example that
with Re(z) between $-\frac{1}{2} $ and $\frac{1}{2} $ and the second
interchanges points within and outside the unit-circle, so combining the
two we get any z into the greyed-out region. Actually, we could have
taken any of the regions in the above tiling as our fundamental domain
as they are all translates of the greyed-out region by an element of
$PSL_2(\mathbb{Z}) $.

Of course, points on the boundary of the
greyed-out fundamental region need to be identified (in order to get the
identification of $\overline{\mathcal{H}/PSL_2(\mathbb{Z})} $ with the
Riemann sphere $S^2=\mathbb{P}^1_{\mathbb{C}} $). For example, the two
halves of the boundary by the unit circle are interchanged by the action
of the map $z \rightarrow -\frac{1}{z} $ and if we take the translates under
$PSL_2(\mathbb{Z}) $ of the indicated circle-part

we get a connected tree with fundamental domain the circle
part bounded by i and $\rho = \frac{1}{2}+\frac{\sqrt{3}}{2} i $.
Calculating the stabilizer subgroup of i (that is, the subgroup of
elements fixing i) we get that this subgroup
is $\langle u \rangle = C_2 $ whereas the stabilizer subgroup of
$\rho $ is $\langle v \rangle = C_3 $.

Using this facts and the general
results of Jean-Pierre Serres book Trees
one deduces that $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $
and hence that the obvious relations among U,V and R given above do
indeed generate all relations.

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Monsieur Mathieu

Published March 1, 2007 by lievenlb

Even a virtual course needs an opening line, so here it is : Take your favourite $SL_2(\mathbb{Z}) $-representation Here is mine : the permutation presentation of the Mathieu group(s). Emile Leonard Mathieu is remembered especially for his discovery (in 1861 and 1873) of five sporadic simple groups named after him, the Mathieu groups $M_{11},M_{12},M_{22},M_{23} $ and $M_{24} $. These were studied in his thesis on transitive functions. He had a refreshingly direct style
of writing. I’m not sure what Cauchy would have thought (Cauchy died in 1857) about this ‘acknowledgement’ in his 1861-paper in which Mathieu describes $M_{12} $ and claims the construction of $M_{24} $.

Also the opening sentenses of his 1873 paper are nice, something along the lines of “if no expert was able to fill in the details of my claims made twelve years ago, I’d better do it myself”.

However, even after this paper opinions remained divided on the issue whether or not he did really achieve his goal, and the matter was settled decisively by Ernst Witt connecting the Mathieu groups to Steiner systems (if I recall well from Mark Ronan’s book Symmetry and the monster)

As Mathieu observed, the quickest way to describe these groups would be to give generators, but as these groups are generated by two permutations on 12 respectively 24 elements, we need to have a mnemotechnic approach to be able to reconstruct them whenever needed.

Here is a nice approach, due to Gunther Malle in a Luminy talk in 1993 on “Dessins d’enfants” (more about them later). Consider the drawing of “Monsieur Mathieu” on the left. That is, draw the left-handed bandit picture on 6 edges and vertices, divide each edge into two and give numbers to both parts (the actual numbering is up to you, but for definiteness let us choose the one on the left). Then, $M_{12} $ is generated by the order two permutation describing the labeling of both parts of the edges

$s=(1,2)(3,4)(5,8)(7,6)(9,12)(11,10) $

together with the order three permutation obtained from cycling counterclockwise
around a trivalent vertex and calling out the labels one encounters. For example, the three cycle corresponding to the ‘neck vertex’ is $~(1,2,3) $ and the total permutation
is

$t=(1,2,3)(4,5,6)(8,9,10) $

A quick verification using GAP tells that these elements do indeed generate a simple group of order 95040.

Similarly, if you have to reconstruct the largest Mathieu group from scratch, apply the same method to the the picture above or to “ET Mathieu” drawing on the left. This picture I copied from Alexander Zvonkin‘s paper How to draw a group as well as the computational details below.

This is all very nice and well but what do these drawings have to do with Grothendieck’s “dessins d’enfants”? Consider the map from the projective line onto itself

$\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}}$

defined by the rational map

$f(z) = \frac{(z^3-z^2+az+b)^3(z^3+cz^2+dz+e)}{Kz} $

where N. Magot calculated that

$a=\frac{107+7 \sqrt{-11}}{486},
b=-\frac{13}{567}a+\frac{5}{1701}, c=-\frac{17}{9},
d=\frac{23}{7}a+\frac{256}{567},
e=-\frac{1573}{567}a+\frac{605}{1701} $

and finally

$K =
-\frac{16192}{301327047}a+\frac{10880}{903981141} $

One verifies that this map is 12 to 1 everywhere except over the points ${
0,1,\infty } $ (that is, there are precisely 12 points mapping under f to a given point of $\mathbb{P}^1_{\mathbb{C}} – { 0,1,\infty } $. From the expression of f(z) it is clear that over 0 there lie 6 points (3 of which with multiplicity three, the others of multiplicity one). Over $\infty $ there are two points, one with multiplicity 11 and one with multiplicity one. The difficult part is to compute the points lying over 1. The miraculous fact of the given values is that

$f(z)-1 = \frac{-B(z)^2}{Kz} $

where

$B(z)=z^6+\frac{1}{11}(10c-8)z^5+(5a+9d-7c)z^4+(2b+4ac+8e-6d)z^3+$
$(3ad+bc-5e)z^2+2aez-be) $

and hence there are 6 points lying over 1 each with mutiplicity two.

Right, now consider the complex projective line $\mathbb{P}^1_{\mathbb{C}} $ as the Riemann sphere $S^2 $ and mark the six points lying over 1 by a white vertex and the six points lying over 0 with a black vertex (in the source sphere). Now, lift the real interval $[0,1] $ in the target sphere $\mathbb{P}^1_{\mathbb{C}} = S^2 $ to its inverse image on the source sphere. As there are exactly 12 points lying over each real
number $0 \lneq r \lneq 1 $, this inverse image will consist of 12 edges which are noncrossing and each end in one black and one white vertex. The obtained graph will look like the \”Monsieur Mathieu\” drawing above with the vertices corresponding to the black vertices and the three points over 1 of multiplicity three corresponding to the
trivalent vertices, those of multiplicity one to the three end-vertices. The white vertices correspond to mid-points of the six edges, so that we do get a drawing with twelve edges, one corresponding to each number. From the explicit description of f(z) it is clear that this map is defined over $\mathbb{Q}\sqrt{-11} $ which is also the
smallest field containing all character-values of the Mathieu group $M_{12} $. Further, the Galois group of the extension $Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) =
\mathbb{Z}/2\mathbb{Z} $ and is generated by complex conjugation. So, one might wonder what would happen if we replaced in the definition of the rational map f(z) the value of a by $a = \frac{107-\sqrt{-11}}{486} $. It turns out that this modified map has the same properties as $f(z) $ so again one can draw on the source sphere a picture consisting of twelve edges each ending in a white and black vertex.

If we consider the white vertices (which incidentally each lie on two edges as all points lying over 0 are of multiplicity two) as mid-points of longer edges connecting the
black vertices we obtain a drawing on the sphere which looks like \”Monsieur Mathieu\” but this time as a right handed bandit, and applying our mnemotechnic rule we obtain _another_ (non conjugated) embedding of $M_{12} $ in the full symmetric group on 12 vertices.

What is the connection with $SL_2(\mathbb{Z}) $-representations? Well, the permutation generators s and t of $M_{12} $ (or $M_{24} $ for that matter) have orders two and three, whence there is a projection from the free group product $C_2 \star C_3 $ (here $C_n $ is just the cyclic group of order n) onto $M_{12} $ (respectively $M_{24} $). Next
time we will say more about such free group products and show (among other things) that $PSL_2(\mathbb{Z}) \simeq C_2 \star C_3 $ whence the connection with $SL_2(\mathbb{Z}) $. In a following lecture we will extend the Monsieur Mathieu example to
arbitrary dessins d\’enfants which will allow us to assign to curves defined over $\overline{\mathbb{Q}} $ permutation representations of $SL_2(\mathbb{Z}) $ and other _cartographic groups_ such as the congruence subgroups $\Gamma_0(2) $ and
$\Gamma(2) $.

9 Comments

noncommutative geometry : a medieval science?

Published February 28, 2007 by lievenlb

According to a science article in the New York Times, archeologists have discovered “signs of advanced math” in medieval mosaics. An example of a quasi-crystalline Penrose pattern was found at the Darb-i Imam shrine in Isfahan, Iran.

“A new study shows that the Islamic pattern-making process, far more intricate than the laying of one‚Äôs bathroom floor, appears to have involved an advanced math of quasi crystals, which was not understood by modern scientists until three decades ago. Two years ago, Peter J. Lu, a doctoral student in physics at Harvard University, was transfixed by the geometric pattern on a wall in Uzbekistan. It reminded him of what mathematicians call quasi-crystalline designs. These were demonstrated in the early 1970s by Roger Penrose, a mathematician and cosmologist at the University of Oxford. Mr. Lu set about examining pictures of other tile mosaics from Afghanistan, Iran, Iraq and Turkey, working with Paul J. Steinhardt, a Princeton cosmologist who is an authority on quasi crystals and had been Mr. Lu’s undergraduate adviser. ”

Penrose tilings are one of the motivating examples of Alain Connes’ book as there is a $C^* $-algebra associated to it. In fact, the algebra is AF ( a limit of semi-simple finite dimensional algebras) so is even a formally smooth algebra in Kontsevichian noncommutative geometry (it is remarkable how quickly one gets used to silly terminology…). However, the Penrose algebra is simple, so rather useless from the point of view of finite dimensional representations… Still, Connesian noncommutative geometry may be a recent incarnation of the medieval Tehran program (pun intended). Thanks to easwaran for the link (via Technorati).

Added, March 1 : I haven’t looked at the Connes-Marcolli paper A walk in the noncommutative garden for a while but now that I do, I see that they mentioned the above already at the end of their section on Tilings (page 32). They also include clearer pictures.

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master class 2007

Published February 9, 2007 by lievenlb

Next
week our master programme on noncommutative geometry
will start. Here is the list of all international mini-courses (8 hours
each) and firm or tentative dates. For the latest update, it is always
best to check with the Arts seminar
website.

Hans-Juergen Schneider (Munich) “Hopf Galois extensions and
quotient theory of Hopf algebras”. February 20-23 each day from
10h30-12h30.
Markus Reineke
(Wuppertal) “Representations of quivers”. February 27-28, March 1-2
each day from 10h30-12h30.
Arthur Ruuge (Moscow) “Semiclassical approximation of quantum
mechanics”. March 6-9 each day from 10h30-12h30.
Rupert Yu
(Poitiers) in March or April.
Isar Stubbe (Antwerp) in April.
Fred Van Oystayen (Antwerp) in April.
Raf
Bocklandt (Antwerp) in April or May.
Goro Kato (Los Angeles)
in May.
Florin Panaite (Bucharest) in May.
Pjotr
Hajac (Warsaw) in June.

Apart from these mini-courses
there will be four regular courses (approx. 30hrs each) during the whole
semester.

Raf Bocklandt “Knot theory”.
Lieven Le Bruyn “Noncommutative geometry”.
Geert Van
de Weyer “Quantum groups”.
Fred Van Oysyaeyen
“Noncommutative algebra”.

p>Dates and places of all
lectures will be made available through the Arts seminar
site.

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coalgebras and non-geometry 3

Published September 11, 2006 by lievenlb

Last
time we saw that the _coalgebra of distributions_ of a
noncommutative manifold can be described as a coalgebra
Takeuchi-equivalent to the path coalgebra of a huge quiver. This
infinite quiver has as its vertices the isomorphism classes of finite
dimensional simple representations of the qurve A (the coordinate ring
of the noncommutative manifold) and there are as many directed arrows
between the vertices corresponding to the simples S and T as is the
dimension of $Ext^1_A(S,T) $.

The fact that this
coalgebra of distributions is equivalent to the path coalgebra of
_some_ quiver is in the Kontsevich-Soibelman
paper though it would have been nice if they had given reference for
this fact to the paper Wedge Products and
Cotensor Coalgebras in Monoidal Categories by Ardizzoni or to
previous work by P. Jara, D. Llena, L. Merino and D. Stefan,
“Hereditary and formally smooth coalgebras”, Algebr.
Represent. Theory 8 (2005), 363-374. In those papers it is shown that a
coalgebra with coseparable coradical is hereditary if and only if it
is formally smooth if and only if it is a cotensor coalgebra of some
bicomodule.

At first this looks just like the dual version of
the classical result that a finite dimensional hereditary algebra is
Morita equivalent to the path algebra of a quiver (which is indeed what
the proof does) but again the condition that the coradical is
coseparable does not require the coradical to be finite dimensional…
In our case, the coradical is indeed coseparable being the direct sum
over all matrix coalgebras corresponding to the simple representations.
Hence, we can again recover the _points_ of our noncommutative manifold
from the direct summands of the coradical. Fortunately, one can
compute this huge coalgebra of distributions from a small quiver, the
_one quiver to rule them all_, but as I’ve been babbling about all of
this here [numerous
times](http://www.neverendingbooks.org/?s=one+quiver) I’ll let the
interested find out for themselves how you use it (a) to get at the
isoclasses of all simples (hint : morally they are the smooth points of
the quotient varieties of n-dimensional representations and enough tools
have been developed recently to spot some fake simples, that is smooth
proper semi-simple points) and (b) to compute the _ragball_, that is the
huge quiver with vertex set the simples and arows as described
above. Over the years I’ve calculated several one-quivers for a
variety of qurves (such as amalgamated free products of finite groups
and smooth curves). If you are in for a puzzle, try to determine it for
the qurve $~(\mathbb{C}[x] \ast C_2) \ast_{\mathbb{C}
C_2} \mathbb{C} PSL_2(\mathbb{Z}) \ast_{\mathbb{C} C_3}
(\mathbb{C}[x] \ast C_3) $ The answer is a mysterious
hexagon

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coalgebras and non-geometry 2

Published September 7, 2006 by lievenlb

Last time we
have seen that the _coalgebra of distributions_ of an affine smooth
variety is the direct sum (over all points) of the dual to the etale
local algebras which are all of the form $\mathbb{C}[[
x_1,\ldots,x_d ]] $ where $d $ is the dimension of the
variety. Generalizing this to _non-commutative_ manifolds, the first
questions are : “What is the analogon of the power-series algebra?” and
do all ‘points’ of our non-commutative manifold do have such local
algebras? Surely, we no longer expect the variables to commute, so a
non-commutative version of the power series algebra should be
$\mathbb{C} \langle \langle x_1,\ldots,x_d \rangle \rangle $,
the ring of formal power series in non-commuting variables. However,
there is still another way to add non-commutativity and that is to go
from an algebra to matrices over the algebra. So, in all we would expect
to be our _local algebras_ at points of our non-commutative manifold to
be isomorphic to $M_n(\mathbb{C} \langle \langle x_1,\ldots,x_d
\rangle \rangle) $ As to the second question : _qurves_ (that is,
the coordinate rings of non-commutative manifolds) do have such algebras
as local rings provided we take as the ‘points’ of the non-commutative
variety the set of all _simple_ finite dimensional representations of
the qurve. This is a consequence of the _tubular neighborhood theorem_
due to [Cuntz](http://wwwmath.uni-muenster.de/u/cuntz/cuntz.html) and
[Quillen](http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Quillen.html). In more details : If A is a qurve, then a simple
$n $-dimensional representation corresponds to an epimorphism
$\pi~:~A \rightarrow S = M_n(\mathbb{C}) $ and if we take
$\mathfrak{m}=Ker(\pi) $, then
$M=\mathfrak{m}/\mathfrak{m}^2 $ is an $S $-bimodule and
the $\mathfrak{m} $-adic completion of A is isomorphic to the
completed tensor-algebra $\hat{T}_S(M) \simeq M_n(\mathbb{C}
\langle \langle x_1,\ldots,x_d \rangle \rangle) $ In contrast with
the commutative case however where the dimension remains constant over
all points, here the numbers n and d can change from simple to simple.
For n this is clear as it gives the dimension of the simple
representation, but also d changes (it is the local dimension of the
variety classifying simple representations of the same dimension). Here
an easy example : Consider the skew group algebra $A =
\mathbb{C}[x] \star C_2 $ with the action given by sending $x
\mapsto -x $. Then A is a qurve and its center is
$\mathbb{C}[y] $ with $y=x^2 $. Over any point $y
\not= 0 $ there is a unique simple 2-dimensional representation of A
giving the local algebra $M_2(\mathbb{C}[[y]]) $. If
$y=0 $ the situation is more complicated as the local structure
of A is given by the algebra $\begin{bmatrix} \mathbb{C}[[y]] &
\mathbb{C}[[y]] \\ (y) & \mathbb{C}[[y]] \end{bmatrix} $ So, over
this point there are precisely 2 one-dimensional simple representations
corresponding to the maximal ideals $\mathfrak{m}_1 =
\begin{bmatrix} (y) & \mathbb{C}[[y]] \\ (y) & \mathbb{C}[[y]]
\end{bmatrix}~\qquad \text{and}~\qquad \mathfrak{m}_2 = \begin{bmatrix}
\mathbb{C}[[y]] & \mathbb{C}[[y]] \\ (y) & (y) \end{bmatrix} $ and
both ideals are idempotent, that is $\mathfrak{m}_i^2 =
\mathfrak{m}_i $ whence the corresponding bimodule $M_i =
0 $ so the local algebra in either of these two points is just
$\mathbb{C} $. Ok, so the comleted local algebra at each point
is of the form $M_n(\mathbb{C}\langle \langle x_1,\ldots,x_d \rangle
\rangle) $, but what is the corresponding dual coalgebra. Well,
$\mathbb{C} \langle \langle x_1,\ldots,x_d \rangle \rangle $ is
the algebra dual to the _cofree coalgebra_ on $V = \mathbb{C} x_1 +
\ldots + \mathbb{C}x_d $. As a vectorspace this is the
tensor-algebra $T(V) = \mathbb{C} \langle x_1,\ldots,x_d
\rangle $ with the coalgebra structure induced by the bialgebra
structure defined by taking all varaibales to be primitives, that is
$\Delta(x_i) = x_i \otimes 1 + 1 \otimes x_i $. That is, the
coproduct on a monomial gives all different expressions $m_1 \otimes
m_2 $ such that $m_1m_2 = m $. For example,
$\Delta(x_1x_2) = x_1x_2 \otimes 1 + x_1 \otimes x_2 + 1 \otimes
x_1x_2 $. On the other hand, the dual coalgebra of
$M_n(\mathbb{C}) $ is the _matrix coalgebra_ which is the
$n^2 $-dimensional vectorspace $\mathbb{C}e_{11} + \ldots +
\mathbb{C}e_{nn} $ with comultiplication $\Delta(e_{ij}) =
\sum_k e_{ik} \otimes e_{kj} $ The coalgebra corresponding to the
local algebra $M_n(\mathbb{C}\langle \langle x_1,\ldots,x_d \rangle
\rangle) $ is then the tensor-coalgebra of the matrix coalgebra and
the cofree coalgebra. Having obtained the coalgebra at each point
(=simple representation) of our noncommutative manifold one might think
that the _coalgebra of non-commutative distributions_ should be the
direct sum of all this coalgebras, summed over all points, as in the
commutative case. But then we would forget about a major difference
between the commutative and the non-commutative world : distinct simples
can have non-trivial extensions! The mental picture one might have
about simples having non-trivial extensions is that these points lie
‘infinitesimally close’ together. In the $\mathbb{C}[x] \star
C_2 $ example above, the two one-dimensional simples have
non-trivial extensions so they should be thought of as a cluster of two
infinitesimally close points corresponding to the point $y=0 $
(that is, this commutative points splits into two non-commutative
points). Btw. this is the reason why non-commutative algebras can be
used to resolve commutative singularities (excessive tangents can be
split over several non-commutative points). While this is still pretty
harmless when the algebra is finite over its center (as in the above
example where only the two one-dimensionals have extensions), the
situation becomes weird over general qurves as ‘usually’ distinct
simples have non-trivial extensions. For example, for the free algebra
$\mathbb{C}\langle x,y \rangle $ this is true for all simples…
So, if we want to continue using this image of points lying closely
together this immediately means that non-commutative ‘affine’ manifolds
behave like compact ones (in fact, it turns out to be pretty difficult
to ‘glue’ together qurves into ‘bigger’ non-commutative manifolds, apart
from the quiver examples of [this old
paper](http://www.arxiv.org/abs/math.AG/9907136)). So, how to bring
this new information into our coalgebra of distributions? Well, let’s
repeat the previous argument not with just one point but with a set of
finitely many points. Then we have a _semi-simple algebra_ quotient
$\pi~:~A \rightarrow S = M_{n_1}(\mathbb{C}) \oplus \ldots \oplus
M_{n_k}(\mathb{C}) $ and taking again
$\mathfrak{m}=Ker(\pi) $ and
$M=\mathfrak{m}/\mathfrak{m}^2 $, then $M $ is again an
S-bimodule. Now, any S-bimodule can be encoded into a _quiver_ Q on k
points, the number of arrows from vertex i to vertex j being the number
of components in M of the form $M_{n_i \times
n_j}(\mathbb{C}) $. Again, it follows from the tubular neighborhood
theorem that the $\mathfrak{m} $-adic completion of A is
isomorphic to the completion of an algebra Morita equivalent to the
_path algebra_ $\mathbb{C} Q $ (being the tensor algebra
$T_S(M) $). As all the local algebras of the points are
quotients of this quiver-like completion, on the coalgebra level our
local coalgebras will be sub coalgebras of the coalgebra which is
co-Morita equivalent (and believe it or not but coalgebraists have a
name for this : _Takeuchi equivalence_) to the _quiver coalgebra_ which
is the vectorspace of the path algebra $\mathbb{C} Q $ with
multiplication induced by making all arrows from i to j skew-primitives,
that is, $\Delta(a) = e_i \otimes a + a \otimes e_j $ where the
$e_i $ are group-likes corresponding to the vertices. If all of
ths is a bit too much co to take in at once, I suggest the paper by Bill
Chin [A brief introduction to coalgebra representation
theory](http://condor.depaul.edu/~wchin/crt.pdf#search=%22%22A%20brief%20introduction%20to%20coalgebra%20representation%20theory%22%22). The
_coalgebra of noncommutative distributions_ we are after at is now the
union of all these Takeuchi-equivalent quiver coalgebras. In easy
examples such as the $\mathbb{C}[x] \star C_2 $-example this
coalgebra is still pretty small (the sum of the local coalgebras
corresponding to the local algebras $M_2(\mathbb{C}[[x]]) $
summed over all points $y \not= 0 $ summed with the quiver
coalgebra of the quiver $\xymatrix{\vtx{} \ar@/^/[rr] & & \vtx{}
\ar@/^/[ll]} $ In general though this is a huge object and we would
like to have a recipe to construct it from a manageable _blue-print_ and
that is what we will do next time.

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non-(commutative) geometry

Published June 21, 2006 by lievenlb

Now
that my non-geometry
post is linked via the comments in this
string-coffee-table post which in turn is available through a
trackback from the Kontsevich-Soibelman
paper it is perhaps useful to add a few links.

The little
I’ve learned from reading about Connes-style non-commutative geometry is
this : if you have a situation where a discrete group is acting with a
bad orbit-space (for example, $GL_2(\mathbb{Z})$ acting on the whole
complex-plane, rather than just the upper half plane) you can associate
to this a $C^*$-algebra and study invariants of it and interprete them
as topological information about this bad orbit space. An intruiging
example is the one I mentioned and where the additional noncommutative
points (coming from the orbits on the real axis) seem to contain a lot
of modular information as clarified by work of Manin&Marcolli and
Zagier. Probably the best introduction into Connes-style
non-commutative geometry from this perspective are the Lecture on
Arithmetic Noncommutative Geometry by Matilde Marcolli. To
algebraists : this trick is very similar to looking at the
skew-group algebra $\mathbb{C}[x_1,\ldots,x_n] * G$ if
you want to study the _orbifold_ for a finite group action on affine
space. But as algebraist we have to stick to affine varieties and
polynomials so we can only deal with the case of a finite group,
analysts can be sloppier in their functions, so they can also do
something when the group is infinite.

By the way, the
skew-group algebra idea is also why non-commutative algebraic
geometry enters string-theory via the link with orbifolds. The
easiest (and best understood) example is that of Kleinian singularities.
The best introduction to this idea is via the Representations
of quivers, preprojective algebras and deformations of quotient
singularities notes by Bill Crawley-Boevey.

Artin-style non-commutative geometry aka
non-commutative projective geometry originated from the
work of Artin-Tate-Van den Bergh (in the west) and Odeskii-Feigin (in
the east) to understand Sklyanin algebras associated to elliptic curves
and automorphisms via ‘geometric’ objects such as point- (and
fat-point-) modules, line-modules and the like. An excellent survey
paper on low dimensional non-commutative projective geometry is Non-commutative curves and surfaces by Toby
Stafford and
Michel Van den Bergh. The best introduction is the (also
neverending…) book-project Non-
commutative algebraic geometry by Paul Smith who
maintains a
noncommutative geometry and algebra resource page page (which is
also available from the header).

Non-geometry
started with the seminal paper ‘Algebra extensions and
nonsingularity’, J. Amer. Math. Soc. 8 (1995), 251-289 by Joachim
Cuntz and Daniel Quillen but which is not available online. An
online introduction is Noncommutative smooth
spaces by Kontsevich and Rosenberg. Surely, different people have
different motivations to study non-geometry. I assume Cuntz got
interested because inductive limits of separable algebras are quasi-free
(aka formally smooth aka qurves). Kontsevich and Soibelman want to study
morphisms and deformations of $A_{\infty}$-categories as they explain in
their recent
paper. My own motivation to be interested in non-geometry is the
hope that in the next decades one will discover new exciting connections
between finite groups, algebraic curves and arithmetic groups (monstrous
moonshine being the first, and still not entirely understood,
instance of this). Part of the problem is that these three topics seem
to be quite different, yet by taking group-algebras of finite or
arithmetic groups and coordinate rings of affine smooth curves they all
turn out to be quasi-free algebras, so perhaps non-geometry is the
unifying theory behind these seemingly unrelated topics.

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non-geometry

Published June 16, 2006 by lievenlb

Here’s
an appeal to the few people working in Cuntz-Quillen-Kontsevich-whoever
noncommutative geometry (the one where smooth affine varieties
correspond to quasi-free or formally smooth algebras) : let’s rename our
topic and call it non-geometry. I didn’t come up with
this term, I heard in from Maxim Kontsevich in a talk he gave a couple
of years ago in Antwerp. There are some good reasons for this name
change.

The term _non-commutative geometry_ is already taken by
much more popular subjects such as _Connes-style noncommutative
differential geometry_ and _Artin-style noncommutative algebraic
geometry_. Renaming our topic we no longer have to include footnotes
(such as the one in the recent Kontsevich-Soibelman
paper) :

We use “formal” non-commutative geometry
in tensor categories, which is different from the non-commutative
geometry in the sense of Alain Connes.

or to make a
distinction between _noncommutative geometry in the small_ (which is
Artin-style) and _noncommutative geometry in the large_ (which in
non-geometry) as in the Ginzburg notes.

Besides, the stress in _non-commutative geometry_ (both in Connes-
and Artin-style) in on _commutative_. Connes-style might also be called
‘K-theory of $C^*$-algebras’ and they use the topological
information of K-theoretic terms in the commutative case as guidance to
speak about geometrical terms in the nocommutative case. Similarly,
Artin-style might be called ‘graded homological algebra’ and they
use Serre’s homological interpretation of commutative geometry to define
similar concepts for noncommutative algebras. Hence, non-commutative
geometry is that sort of non-geometry which is almost
commutative…

But the main point of naming our subject
non-geometry is to remind us not to rely too heavily on our
(commutative) geometric intuition. For example, we would expect a
manifold to have a fixed dimension. One way to define the dimension is
as the trancendence degree of the functionfield. However, from the work
of Paul Cohn (I learned about it through Aidan Schofield) we know that
quasi-free algebras usually do’nt have a specific function ring of
fractions, rather they have infinitely many good candidates for it and
these candidates may look pretty unrelated. So, at best we can define a
_local dimension_ of a noncommutative manifold at a point, say given by
a simple representation. It follows from the Cunz-Quillen tubular
neighborhood result that the local ring in such a point is of the
form

$M_n(\mathbb{C} \langle \langle z_1,\ldots,z_m \rangle
\rangle) $

(this s a noncommutative version of the classical fact
than the local ring in a point of a d-dimensional manifold is formal
power series $\mathbb{C} [[ z_1,\ldots,z_d ]] $) but in non-geometry both
m (the _local_ dimension) and n (the dimension of the simple
representation) vary from point to point. Still, one can attach to the
quasi-free algebra A a finite amount of data (in fact, a _finite_ quiver
and dimension vector) containing enough information to compute the (n,m)
couples for _all_ simple points (follows from the one quiver to rule them
all paper or see this for more
details).

In fact, one can even extend this to points
corresponding to semi-simple representations in which case one has to
replace the matrix-ring above by a ring Morita equivalent to the
completion of the path algebra of a finite quiver, the _local quiver_ at
the point (which can also be computer from the one-quiver of A. The
local coalgebras of distributions at such points of
Kontsevich&Soibelman are just the dual coalgebras of these local
algebras (in math.RA/0606241 they
merely deal with the n=1 case but no doubt the general case will appear
in the second part of their paper).

The case of the semi-simple
point illustrates another major difference between commutative geometry
and non-geometry, whereas commutative simples only have self-extensions
(so the distribution coalgebra is just the direct sum of all the local
distributions) noncommutative simples usually have plenty of
non-isomorphic simples with which they have extensions, so to get at the
global distribution coalgebra of A one cannot simply add the locals but
have to embed them in more involved coalgebras.

The way to do it
is somewhat concealed in the
third version of my neverending book (the version that most people
found incomprehensible). Here is the idea : construct a huge uncountable
quiver by taking as its vertices the isomorphism classes of all simple
A-representations and with as many arrows between the simple vertices S
and T as the dimension of the ext-group between these simples (and
again, these dimensions follow from the knowledge of the one-quiver of
A). Then, the global coalgebra of distributions of A is the limit over
all cotensor coalgebras corresponding to finite subquivers). Maybe I’ll
revamp this old material in connection with the Kontsevich&Soibelman
paper(s) for the mini-course I’m supposed to give in september.

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