Here a list of saved pdf-files of previous NeverEndingBooks-posts on geometry in reverse chronological order.

Leave a Comment# Tag: Klein

Today we will explain how curves defined over

$\overline{\mathbb{Q}} $ determine permutation representations

of the carthographic groups. We have seen that any smooth projective

curve $C $ (a Riemann surface) defined over the algebraic

closure $\overline{\mathbb{Q}} $ of the rationals, defines a

_Belyi map_ $\xymatrix{C \ar[rr]^{\pi} & & \mathbb{P}^1} $ which is only ramified over the three points

$\\{ 0,1,\infty \\} $. By this we mean that there are

exactly $d $ points of $C $ lying over any other point

of $\mathbb{P}^1 $ (we call $d $ the degree of

$\pi $) and that the number of points over $~0,1~ $ and

$~\infty $ is smaller than $~d $. To such a map we

associate a _dessin d\’enfant_, a drawing on $C $ linking the

pre-images of $~0 $ and $~1 $ with exactly $d $

edges (the preimages of the open unit-interval). Next, we look at

the preimages of $~0 $ and associate a permutation

$\tau_0 $ of $~d $ letters to it by cycling

counter-clockwise around these preimages and recording the edges we

meet. We repeat this procedure for the preimages of $~1 $ and

get another permutation $~\tau_1 $. That is, we obtain a

subgroup of the symmetric group $ \langle \tau_0,\tau_1

\rangle \subset S_d $ which is called the **monodromy
group** of the covering $\pi $.

For example, the

dessin on the right is

associated to a degree $8 $ map $\mathbb{P}^1 \rightarrow

\mathbb{P}^1 $ and if we let the black (resp. starred) vertices be

the preimages of $~0 $ (respectively of $~1 $), then the

corresponding partitions are $\tau_0 = (2,3)(1,4,5,6) $

and $\tau_1 = (1,2,3)(5,7,8) $ and the monodromy group

is the alternating group $A_8 $ (use

GAP ).

But wait! The map is also

ramified in $\infty $ so why don\’t we record also a

permutation $\tau_{\infty} $ and are able to compute it from

the dessin? (Note that all three partitions are needed if we want to

reconstruct $C $ from the $~d $ sheets as they encode in

which order the sheets fit together around the preimages). Well,

the monodromy group of a $\mathbb{P}^1 $ covering ramified only

in three points is an epimorphic image of the fundamental

group of the sphere

minus three points $\pi_1(\mathbb{P}^1 – { 0,1,\infty

}) $ That is, the group of all loops beginning and

ending in a basepoint upto homotopy (that is, two such loops are the

same if they can be transformed into each other in a continuous way

while avoiding the three points).

This group is generated by loops

$\sigma_i $ running from the basepoint to nearby the i-th

point, doing a counter-clockwise walk around it and going back to be

basepoint $Q_0 $ and the epimorphism to the monodromy group is given by sending

$\sigma_1 \mapsto \tau_0~\quad~\sigma_2 \mapsto

\tau_1~\quad~\sigma_3 \mapsto \tau_{\infty} $

Now,

these three generators are not independent. In fact, this fundamental

group is

$\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =

\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2

\sigma_3 = 1 \rangle $

To understand this, let us begin

with an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is

$~\mathbb{Z} $ as it encodes how many times we walk around the

point. However, on the sphere the situation is different as we can make

our walk around the point longer and longer until the whole walk is done

at the backside of the sphere and then we can just contract our walk to

the basepoint. So, there is just one type of walk on a sphere minus one

point (upto homotopy) whence this fundamental group is trivial. Next,

let us consider the sphere minus two points

Repeat the foregoing to the walk $\sigma_2 $, that

is, strech the upper part of the circular tour all over the backside of

the sphere and then we see that we can move it to fit with the walk

$\sigma_1$ BUT for the orientation of the walk! That is, if we do this

modified walk $\sigma_1 \sigma_2^{\’} $ we just made the

trivial walk. So, this fundamental group is $\langle

\sigma_1,\sigma_2~\mid~\sigma_1 \sigma_2 = 1 \rangle =

\mathbb{Z} $ This is also the proof of the above claim. For,

we can modify the third walk $\sigma_3 $ continuously so that

it becomes the walk $\sigma_1 \sigma_2 $ but

with the **reversed orientation** ! As $\sigma_3 =

(\sigma_1 \sigma_2)^{-1} $ this allows us to compute the

\’missing\’ permutation $\tau_{\infty} = (\tau_0

\tau_1)^{-1} $ In the example above, we obtain

$\tau_{\infty}= (1,2,6,5,8,7,4)(3) $ so it has two cycles

corresponding to the fact that the dessin has two regions (remember we

should draw ths on the sphere) : the head and the outer-region. Hence,

the pre-images of $\infty$ correspond to the different regions of the

dessin on the curve $C $. For another example,

consider the degree 168 map

$K \rightarrow \mathbb{P}^1 $

which is the modified orbit map for the action of

$PSL_2(\mathbb{F}_7) $ on the Klein quartic.

The corresponding dessin is the heptagonal construction of the Klein

quartic

Here, the pre-images of 1 correspond to the midpoints of the

84 edges of the polytope whereas the pre-images of 0 correspond to the

56 vertices. We can label the 168 half-edges by numbers such that

$\tau_0 $ and $\tau_1 $ are the standard generators b

resp. a of the 168-dimensional regular representation (see the atlas

page ).

Calculating with GAP the element $\tau_{\infty} = (\tau_0

\tau_1)^{-1} = (ba)^{-1} $ one finds that this permutation

consists of 24 cycles of length 7, so again, the pre-images of

$\infty $ lie one in each of the 24 heptagonal regions of the

Klein quartic. Now, we are in a position to relate curves defined

over $\overline{Q} $ via their Belyi-maps and corresponding

dessins to Grothendiecks carthographic groups $\Gamma(2) $,

$\Gamma_0(2) $ and $SL_2(\mathbb{Z}) $. The

dessin gives a permutation representation of the monodromy group and

because the fundamental group of the sphere minus three

points $\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =

\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2

\sigma_3 = 1 \rangle = \langle \sigma_1,\sigma_2

\rangle $ is the free group op two generators, we see that

any dessin determines a permutation representation of the congruence

subgroup $\Gamma(2) $ (see this

post where we proved that this

group is free). A **clean dessin** is one for which one type of

vertex has all its valancies (the number of edges in the dessin meeting

the vertex) equal to one or two. (for example, the pre-images of 1 in

the Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu

example ) The corresponding

permutation $\tau_1 $ then consists of 2-cycles and hence the

monodromy group gives a permutation representation of the free

product $C_{\infty} \ast C_2 =

\Gamma_0(2) $ Finally, a clean dessin is said to be a

**quilt dessin** if also the other type of vertex has all its valancies

equal to one or three (as in the Klein quartic or Mathieu examples).

Then, the corresponding permutation has order 3 and for these

quilt-dessins the monodromy group gives a permutation representation of

the free product $C_2 \ast C_3 =

PSL_2(\mathbb{Z}) $ Next time we will see how this lead

Grothendieck to his anabelian geometric approach to the absolute Galois

group.

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}) $…

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.

The Klein Four Group is an a

capella group from the maths department of Northwestern. Below a link to

one of their songs (grabbed from P.P. Cook’s Tangent Space

).

**Finite
Simple Group (of order two)**

A Klein Four original by

Matt Salomone

The path of love is never

smooth

But mine’s continuous for you

You’re the upper bound in the chains of my heart

You’re my Axiom of Choice, you know it’s true

But lately our relation’s not so well-defined

And

I just can’t function without you

I’ll prove my

proposition and I’m sure you’ll find

We’re a

finite simple group of order two

I’m losing my

identity

I’m getting tensor every day

And

without loss of generality

I will assume that you feel the same

way

Since every time I see you, you just quotient out

The faithful image that I map into

But when we’re

one-to-one you’ll see what I’m about

‘Cause

we’re a finite simple group of order two

Our equivalence

was stable,

A principal love bundle sitting deep inside

But then you drove a wedge between our two-forms

Now

everything is so complexified

When we first met, we simply

connected

My heart was open but too dense

Our system

was already directed

To have a finite limit, in some sense

I’m living in the kernel of a rank-one map

From my

domain, its image looks so blue,

‘Cause all I see are

zeroes, it’s a cruel trap

But we’re a finite simple

group of order two

I’m not the smoothest operator in my

class,

But we’re a mirror pair, me and you,

So

let’s apply forgetful functors to the past

And be a

finite simple group, a finite simple group,

Let’s be a

finite simple group of order two

(Oughter: “Why not

three?”)

I’ve proved my proposition now, as you

can see,

So let’s both be associative and free

And by corollary, this shows you and I to be

Purely

inseparable. Q. E. D.

If you go

to Oberwolfach and the weather

predictions are as good as last

weeks, try to bring your mountain-bike along! Here is a nice

1hr30 to 2hrs tour : from the institute to Walke (height 300m), follow

the road north to Rankach and at the Romanes Hof turn left to Hackerhof.

Next, off-road along the Hacker lochweg until you hit the road

Haaghutte-Mooshutte at an height of 855m (this should be doable in under

one hour). A few metres further, you have a view at the highest

mountains in the vicinity of the Institute : the Grosser Hundskopf

(947m) and Kleiner Hundskopf (926m) as on the left. Then, descend along

the Kirchhofweg over Moosbauerhof all the way down to the

Dohlenbacherhof where you hit the main road which brings you back to the

institute going SW. Please take a pump and repair material along, I

had 2 flat tires in 4 days. If you happen to have a GPS, you can

download the gpx-file.

You can feed this to Tom Carden’s Google Map

GPX Viewer and study it in more detail (I made one wrong turn in the

descent and also briefly lost GPS reception in the forest near the top

causing the top waypoint (the lower waypoint is the Institute)).

If you

were not present and still want to see some of the talks or if you are

just curious in the outcome of Paul’s

frantic typing on his PowerBook, he has put his (selection of

talks)-notes

online. Perhaps I’ll write down some of my own recollections of this

meeting later.

Klein’s

quartic $X$is the smooth plane projective curve defined by

$x^3y+y^3z+z^3x=0$ and is one of the most remarkable mathematical

objects around. For example, it is a Hurwitz curve meaning that the

finite group of symmetries (when the genus is at least two this group

can have at most $84(g-1)$ elements) is as large as possible, which in

the case of the quartic is $168$ and the group itself is the unique

simple group of that order, $G = PSL_2(\mathbb{F}_7)$ also known as

Klein\’s group. John Baez has written a [beautiful page](http://math.ucr.edu/home/baez/klein.html) on the Klein quartic and

its symmetries. Another useful source of information is a paper by Noam

Elkies [The Klein quartic in number theory](www.msri.org/publications/books/Book35/files/elkies.pd).

The quotient map $X \rightarrow X/G \simeq \mathbb{P}^1$ has three

branch points of orders $2,3,7$ in the points on $\mathbb{P}^1$ with

coordinates $1728,0,\infty$. These points correspond to the three

non-free $G$-orbits consisting resp. of $84,56$ and $24$ points.

Now, remove from $X$ a couple of $G$-orbits to obtain an affine open

subset $Y$ such that $G$ acts on its cordinate ring $\mathbb{C}[Y]$ and

form the Klein stack (or hereditary order) $\mathbb{C}[Y] \bigstar G$,

the skew group algebra. In case the open subset $Y$ contains all

non-free orbits, the [one quiver](www.matrix.ua.ac.be/master/coursenotes/onequiver.pdf) of this

qurve has the following shape $\xymatrix{\vtx{} \ar@/^/[dd] \\

\\ \vtx{} \ar@/^/[uu]} $ $\xymatrix{& \vtx{} \ar[ddl] & \\

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

\vtx{} \ar[dll] & & \\ \vtx{} \ar[d] & & & & \vtx{} \ar[ull] \\ \vtx{}

\ar[dr] & & & & \vtx{} \ar[u] \\ & \vtx{} \ar[rr] & & \vtx{} \ar[ur]

&} $ Here, the three components correspond to the three

non-free orbits and the vertices correspond to the isoclasses of simple

$\mathbb{C}[Y] \bigstar G$ of dimension smaller than $168$. There are

two such of dimension $84$, three of dimension $56$ and seven of

dimension $24$ which I gave the non-imaginative names \’twins\’,

\’trinity\’ and \’the dwarfs\’. As we want to spice up later this

Klein stack to a larger group, we need to know the structure of these

exceptional simples as $G$-representations. Surely, someone must have

written a paper on the general problem of finding the $G$-structure of

simples of skew-group algebras $A \bigstar G$, so if you know a

reference please let me know. I used an old paper by Idun Reiten and

Christine Riedtmann to do this case (which is easier as the stabilizer

subgroups are cyclic and hence the induced representations of their

one-dimensionals correspond to the exceptional simples).

Here the

story of an idea to construct new examples of non-commutative compact

manifolds, the computational difficulties one runs into and, when they

are solved, the white noise one gets. But, perhaps, someone else can

spot a gem among all gibberish…

[Qurves](http://www.neverendingbooks.org/toolkit/pdffile.php?pdf=/TheLibrary/papers/qaq.pdf) (aka quasi-free algebras, aka formally smooth

algebras) are the \’affine\’ pieces of non-commutative manifolds. Basic

examples of qurves are : semi-simple algebras (e.g. group algebras of

finite groups), [path algebras of

quivers](http://www.lns.cornell.edu/spr/2001-06/msg0033251.html) and

coordinate rings of affine smooth curves. So, let us start with an

affine smooth curve $X$ and spice it up to get a very non-commutative

qurve. First, we bring in finite groups. Let $G$ be a finite group

acting on $X$, then we can form the skew-group algebra $A = \mathbfk[X]

\bigstar G$. These are examples of prime Noetherian qurves (aka

hereditary orders). A more pompous way to phrase this is that these are

precisely the [one-dimensional smooth Deligne-Mumford

stacks](http://www.math.lsa.umich.edu/~danielch/paper/stacks.pdf).

As the 21-st century will turn out to be the time we discovered the

importance of non-Noetherian algebras, let us make a jump into the

wilderness and consider the amalgamated free algebra product $A =

(\mathbf k[X] \bigstar G) \ast_{\mathbf k G} \mathbfk H$ where $G

\subset H$ is an interesting extension of finite groups. Then, $A$ is

again a qurve on which $H$ acts in a way compatible with the $G$-action

on $X$ and $A$ is hugely non-commutative… A very basic example :

let $\mathbb{Z}/2\mathbb{Z}$ act on the affine line $\mathbfk[x]$ by

sending $x \mapsto -x$ and consider a finite [simple

group](http://mathworld.wolfram.com/SimpleGroup.html) $M$. As every

simple group has an involution, we have an embedding

$\mathbb{Z}/2\mathbb{Z} \subset M$ and can construct the qurve

$A=(\mathbfk[x] \bigstar \mathbb{Z}/2\mathbb{Z}) \ast_{\mathbfk

\mathbb{Z}/2\mathbb{Z}} \mathbfk M$ on which the simple group $M$ acts

compatible with the involution on the affine line. To study the

corresponding non-commutative manifold, that is the Abelian category

$\mathbf{rep}~A$ of all finite dimensional representations of $A$ we have

to compute the [one quiver to rule them

all](http://www.matrix.ua.ac.be/master/coursenotes/onequiver.pdf) for

$A$. Because $A$ is a qurve, all its representation varieties

$\mathbf{rep}_n~A$ are smooth affine varieties, but they may have several

connected components. The direct sum of representations turns the set of

all these components into an Abelian semigroup and the vertices of the

\’one quiver\’ correspond to the generators of this semigroup whereas

the number of arrows between two such generators is given by the

dimension of $Ext^1_A(S_i,S_j)$ where $S_i,S_j$ are simple

$A$-representations lying in the respective components. All this

may seem hard to compute but it can be reduced to the study of another

quiver, the Zariski quiver associated to $A$ which is a bipartite quiver

with on the left the \’one quiver\’ for $\mathbfk[x] \bigstar

\mathbb{Z}/2\mathbb{Z}$ which is just $\xymatrix{\vtx{}

\ar@/^/[rr] & & \vtx{} \ar@/^/[ll]} $ (where the two vertices

correspond to the two simples of $\mathbb{Z}/2\mathbb{Z}$) and on the

right the \’one quiver\’ for $\mathbf k M$ (which just consists of as

many verticers as there are simple representations for $M$) and where

the number of arrows from a left- to a right-vertex is the number of

$\mathbb{Z}/2\mathbb{Z}$-morphisms between the respective simples. To

make matters even more concrete, let us consider the easiest example

when $M = A_5$ the alternating group on $5$ letters. The corresponding

Zariski quiver then turns out to be $\xymatrix{& & \vtx{1} \\\

\vtx{}\ar[urr] \ar@{=>}[rr] \ar@3[drr] \ar[ddrr] \ar[dddrr] \ar@/^/[dd]

& & \vtx{4} \\\ & & \vtx{5} \\\ \vtx{} \ar@{=>}[uurr] \ar@{=>}[urr]

\ar@{=>}[rr] \ar@{=>}[drr] \ar@/^/[uu] & & \vtx{3} \\\ & &

\vtx{3}} $ The Euler-form of this quiver can then be used to

calculate the dimensions of the EXt-spaces giving the number of arrows

in the \’one quiver\’ for $A$. To find the vertices, that is, the

generators of the component semigroup we have to find the minimal

integral solutions to the pair of equations saying that the number of

simple $\mathbb{Z}/2\mathbb{Z}$ components based on the left-vertices is

equal to that one the right-vertices. In this case it is easy to see

that there are as many generators as simple $M$ representations. For

$A_5$ they correspond to the dimension vectors (for the Zariski quiver

having the first two components on the left) $\begin{cases}

(1,2,0,0,0,0,1) \\ (1,2,0,0,0,1,0) \\ (3,2,0,0,1,0,0) \\

(2,2,0,1,0,0,0) \\ (1,0,1,0,0,0,0) \end{cases}$ We now have all

info to determine the \’one quiver\’ for $A$ and one would expect a nice

result. Instead one obtains a complete graph on all vertices with plenty

of arrows. More precisely one obtains as the one quiver for $A_5$

$\xymatrix{& & \vtx{} \ar@{=}[dll] \ar@{=}[dddl] \ar@{=}[dddr]

\ar@{=}[drr] & & \\\ \vtx{} \ar@(ul,dl)|{4} \ar@{=}[rrrr]|{6}

\ar@{=}[ddrrr]|{8} \ar@{=}[ddr]|{4} & & & & \vtx{} \ar@(ur,dr)|{8}

\ar@{=}[ddlll]|{6} \ar@{=}[ddl]|{10} \\\ & & & & & \\\ & \vtx{}

\ar@(dr,dl)|{4} \ar@{=}[rr]|{8} & & \vtx{} \ar@(dr,dl)|{11} & } $

with the number of arrows (in each direction) indicated. Not very

illuminating, I find. Still, as the one quiver is symmetric it follows

that all quotient varieties $\mathbf{iss}_n~A$ have a local Poisson

structure. Clearly, the above method can be generalized easily and all

examples I did compute so far have this \’nearly complete graph\’

feature. One might hope that if one would start with very special

curves and groups, one might obtain something more interesting. Another

time I\’ll tell what I got starting from Klein\’s quartic (on which the

simple group $PSL_2(\mathbb{F}_7)$ acts) when the situation was sexed-up

to the sporadic simple Mathieu group $M_{24}$ (of which

$PSL_2(\mathbb{F}_7)$ is a maximal subgroup).