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Tag: Grothendieck

permutation representations of monodromy groups

Published March 20, 2007 by lievenlb

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.

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.

4 Comments

The best rejected proposal ever

Published March 7, 2007 by lievenlb


The Oscar in
the category The Best Rejected Research Proposal in Mathematics
(ever) goes to … Alexander Grothendieck
for his proposal Esquisse d’un Programme, Grothendieck\’s research program from 1983, written as
part of his application for a position at the CNRS, the French
equivalent of the NSF. An English translation is
available.

Here is one of the problems discussed :
Give TWO non-trivial elements of
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ the _absolute_
Galois group of the algebraic closure of the rational numbers
$\overline{\mathbb{Q}} $, that is the group of all
$\mathbb{Q} $-automorphisms of $\overline{\mathbb{Q}} $. One element
most of us can give (complex-conjugation) but to find any other
element turns out to be an extremely difficult task.

To get a handle on
this problem, Grothendieck introduced his _’Dessins d’enfants’_
(Children’s drawings). Recall from last session the pictures of the
left and right handed Monsieur Mathieu

The left hand side drawing was associated to a map
$\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}} $ which was
defined over the field $\mathbb{Q} \sqrt{-11} $ whereas the right side
drawing was associated to the map given when one applies to all
coefficients the unique non-trivial automorphism in the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ (which is
complex-conjugation). Hence, the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ acts _faithfully_ on the
drawings associated to maps $\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} \rightarrow
\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} $ which are ramified only over
the points ${ 0,1,\infty } $.

Grothendieck’s idea was to
extend this to more general maps. Assume that a projective smooth curve
(a Riemann surface) X is defined over the algebraic numbers
$\overline{\mathbb{Q}} $ and assume that there is a map $X
\rightarrow \mathbb{P}^1_{\mathbb{C}} $ ramified only over the points
${ 0,1,\infty } $, then we can repeat the procedure of last time and
draw a picture on X consisting of d edges (where d is the degree
of the map, that is the number of points lying over another point of
$\mathbb{P}^1_{\mathbb{C}} $) between white resp. black points (the
points of X lying over 1 (resp. over 0)).

Call such a drawing a
‘dessin d\’enfant’ and look at the collection of ALL dessins
d’enfants associated to ALL such maps where X runs over ALL curves
defined over $\overline{\mathbb{Q}} $. On this set, there is an action
of the absolute Galois group
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ and if this action would be
faithful, then this would give us insight into this
group. However, at that time even the existence of a map $X \rightarrow
\mathbb{P}^1 $ ramified in the three points ${ 0,1,\infty } $
seemed troublesome to prove, as Grothendieck recalls in his proposal

In more erudite terms, could it be true that
every projective non-singular algebraic curve defined over a number
field occurs as a possible ‚ modular curve‚ parametrising
elliptic curves equipped with a suitable rigidification? Such a
supposition seemed so crazy that I was almost embarrassed to submit
it to the competent people in the domain. Deligne when I consulted
him found it crazy indeed, but didn’t have any counterexample up
his sleeve. Less than a year later, at the International Congress in
Helsinki, the Soviet mathematician Bielyi announced exactly that result,
with a proof of disconcerting simplicity which fit into two little
pages of a letter of Deligne ‚ never, without a doubt, was such a
deep and disconcerting result proved in so few lines!

In
the form in which Bielyi states it, his result essentially says that
every algebraic curve defined over a number field can be obtained as
a covering of the projective line ramified only over the points 0,
1 and infinity. This result seems to have remained more or less
unobserved. Yet, it appears to me to have considerable importance. To
me, its essential message is that there is a profound identity
between the combinatorics of finite maps on the one hand, and the
geometry of algebraic curves defined over number fields on the
other. This deep result, together with the algebraic- geometric
interpretation of maps, opens the door onto a new, unexplored world within reach of all, who pass by without seeing it.

Belyi’s proof is indeed relatively easy
(full details can be found in the paper Dessins d’enfants on the
Riemann sphere by Leila
Schneps). Roughly it goes as follows : as both X and the map are
defined over $\overline{\mathbb{Q}} $ the map is only ramified over
(finitely many) $\overline{\mathbb{Q}} $-points. Let S be the finite
set of all Galois-conjugates of these points and consider the polynomial

$f_0(z_0) = \prod_{s \in S} (z_0 -s) \in
\mathbb{Q}[z_0] $

Now, do a
resultant trick. Consider the
polynomial $f_1(z_1) = Res_{z_0}(\frac{d f_0}{d
z_0},f_0(z_0)-z_1) $ then the roots of $f_1(z_1) $ are exactly the
finite critical values of $f_0 $, $f_1 $ is again defined over
$\mathbb{Q} $ and has lower degree (in $z_1 $) than $f_0 $ in $z_1 $.
Continue this trick a finite number of times untill you have constructed
a polynomial $f_n(z_n) \in \mathbb{Q}[z_n] $ of degree zero.

Composing
the original map with the maps $f_j $ in succession yields that all
ramified points of this composition are
$\mathbb{Q} $-points! Now, we only have to limit the number of
these ramified $\mathbb{Q} $-points (let us call this set T) to three.

Take any three elements of T, then there always exist integers $m,n
\in \mathbb{Z} $ such that the three points go under a linear
fractional transformation (a Moebius-function associated to a matrix in
$PGL_2(\mathbb{Q}) $) to ${ 0,\frac{m}{m+n},1 } $. Under the
transformation $z \rightarrow \frac{(m+n)^{m+n}}{m^m
n^n}z^m(1-z)^n $ the points 0 and 1 go to 0 and
$\frac{m}{m+n} $ goes to 1 whence the ramified points of the
composition are one less in number than T. Continuing in this way we
can get the set of ramified $\mathbb{Q} $-points of a composition at
most having three elements and then a final Moebius transformation gets
them to ${ 0,1,\infty } $, done!

As a tribute for this clever
argument, maps $X \rightarrow \mathbb{P}^1 $ ramified only in 0,1 and
$\infty $ are now called Belyi morphisms. Here is an example of
a Belyi-morphism (and the corresponding dessin d’enfants) associated to
one of the most famous higher genus curves around : the Klein
quartic (if you haven’t done
so yet, take your time to go through this marvelous pre-blog post by
John Baez).

One can define the Klein quartic as the plane projective
curve K with defining equation in
$\mathbb{P}^2_{\\mathbb{C}} $ given by $X^3Y+Y^3Z+Z^3X = 0 $ K has
a large group of automorphism, namely the simple group of order
168 $G = PSL_2(\mathbb{F}_7) =
SL_3(\mathbb{F}_2) $ It is a classical fact (see for example
the excellent paper by Noam Elkies The Klein quartic in number theory) that the quotient map $K \rightarrow K/G =
\mathbb{P}^1_{\mathbb{C}} $ is ramified only in the points
0,1728 and $\infty $ and the number of points of K lying over them
are resp. 56, 84 and 24. Now, compose this map with the Moebius
transormation taking ${ 0,1728,\infty } \rightarrow { 0,1,\infty } $
then the resulting map is a Belyi-map for the Klein quartic. A
topological construction of the Klein quartic is fitting 24 heptagons
together so that three meet in each vertex, see below for the gluing
data-picture in the hyperbolic plane : the different heptagons are given
a number but they appear several times telling how they must fit
together)

The resulting figure has exactly $\frac{7 \times 24}{2} =
84 $ edges and the 84 points of K lying over 1 (the white points in
the dessin) correspond to the midpoints of the edges. There are exactly
$\frac{7 \times 24}{3}=56 $ vertices corresponding to the 56 points
lying over 0 (the black points in the dessin). Hence, the dessin
d\’enfant associated to the Klein quartic is the figure traced out by
the edges on K. Giving each of the 168 half-edges a
different number one assigns to the white points a permutation of order
two and to the three-valent black-points a permutation of order three,
whence to the Belyi map of the Klein quartic corresponds a
168-dimensional permutation representation of $SL_2(\mathbb{Z}) $,
which is not so surprising as the group of automorphisms is
$PSL_2(\mathbb{F}_7) $ and the permutation representation is just the
regular representation of this group.

Next time we will see how
one can always associate to a curve defined over
$\overline{\mathbb{Q}} $ a permutation representation (via the Belyi
map and its dessin) of one of the congruence subgroups $\Gamma(2) $ or
$\Gamma_0(2) $ or of $SL_2(\mathbb{Z}) $ itself.

6 Comments

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) $.

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way too ambitious

Published January 30, 2007 by lievenlb

Student-evaluation sneak preview : I am friendly and
extremely helpful but have a somewhat chaotic teaching style and am way
too ambitious as regards content… I was about to deny vehemently
all assertions (except for the chaotic bit) but may have to change my
mind after reading this report on
Mark Rowan’s book ‘Symmetry and the monster’ (see also
my post
)

Oxford University Press considers this book
“a must-read for all fans of popular science”. In his blog,
Lieven le Bruyn, professor of algebra and geometry at the University of
Antwerp, suggests that “Mark Ronan has written a beautiful book
intended for the general public”. However, he goes on to say:
“this year I’ve tried to explain to an exceptionally
good second year of undergraduates, but failed miserably Perhaps
I’ll give it another (downkeyed) try using Symmetry and the
Monster as reading material”.

As an erstwhile
mathematician, I found the book more suited to exceptional maths
undergraduates than to the general public and would strongly encourage
authors and/or publishers to pass such works before a few fans of
popular science before going to press.

Peggie Rimmer,
Satigny.

Well, this ‘exceptionally good
year’ has moved on and I had to teach a course ‘Elementary
Algebraic Geometry’ to them last semester. I had the crazy idea to
approach this in a historical perspective : first I did the
Hilbert-Noether period (translating geometry to ideal theory of
polynomial rings), then the Krull-Weil-Zariski period (defining
everything in terms of coordinate rings) to finish off with the
Serre-Grothendieck period (introducing scheme theory)… Not
surprisingly, I lost everyone after 1920. Once again there were
complaints that I was expecting way too much from them etc. etc. and I
was about to apologize and promise I’ll stick to a doable course
next year (something along the lines of Miles Reid’s
‘Undergraduate Algebraic Geometry’) when one of the students
(admittedly, probably the best of this ‘exceptional year’)
decided to do all exercises of the first two chapters of Fulton’s
‘Algebraic Curves’ to become more accustomed to the subject.
Afterwards he told me “You know, I wouldn’t change the
course too much, now that I did all these exercises I realize that your
course notes are not that bad after all…”. Yeah, thanks!

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2006 paper nominees

Published December 29, 2006 by lievenlb

Here are
my nominees for the 2006 paper of the year award in mathematics &
mathematical physics : in math.RA : math.RA/0606241
: Notes on A-infinity
algebras, A-infinity categories and non-commutative geometry. I by

Maxim Kontsevich and
Yan Soibelman. Here is the abstract :

We develop
geometric approach to A-infinity algebras and A-infinity categories
based on the notion of formal scheme in the category of graded vector
spaces. Geometric approach clarifies several questions, e.g. the notion
of homological unit or A-infinity structure on A-infinity functors. We
discuss Hochschild complexes of A-infinity algebras from geometric point
of view. The paper contains homological versions of the notions of
properness and smoothness of projective varieties as well as the
non-commutative version of Hodge-to-de Rham degeneration conjecture. We
also discuss a generalization of Deligne’s conjecture which includes
both Hochschild chains and cochains. We conclude the paper with the
description of an action of the PROP of singular chains of the
topological PROP of 2-dimensional surfaces on the Hochschild chain
complex of an A-infinity algebra with the scalar product (this action is
more or less equivalent to the structure of 2-dimensional Topological
Field Theory associated with an “abstract” Calabi-Yau
manifold).

why ? : Because this paper
probably gives the correct geometric object associated to a
non-commutative algebra (a huge coalgebra) and consequently the right
definition of a map between noncommutative affine schemes. In a previous post (and its predecessors) I’ve
tried to explain how this links up with my own interpretation and since
then I’ve thought more about this, but that will have to wait for
another time. in hep-th : hep-th/0611082 : Children’s Drawings From
Seiberg-Witten Curves by Sujay K. Ashok, Freddy Cachazo, Eleonora
Dell’Aquila. Here is the abstract :

We consider N=2
supersymmetric gauge theories perturbed by tree level superpotential
terms near isolated singular points in the Coulomb moduli space. We
identify the Seiberg-Witten curve at these points with polynomial
equations used to construct what Grothendieck called “dessins
d’enfants” or “children’s drawings” on the Riemann
sphere. From a mathematical point of view, the dessins are important
because the absolute Galois group Gal(\bar{Q}/Q) acts faithfully on
them. We argue that the relation between the dessins and Seiberg-Witten
theory is useful because gauge theory criteria used to distinguish
branches of N=1 vacua can lead to mathematical invariants that help to
distinguish dessins belonging to different Galois orbits. For instance,
we show that the confinement index defined in hep-th/0301006 is a Galois
invariant. We further make some conjectures on the relation between
Grothendieck’s programme of classifying dessins into Galois orbits and
the physics problem of classifying phases of N=1 gauge theories.

why ? : Because this paper gives the
best introduction I’ve seen to Grothendieck’s dessins d’enfants
(slightly overdoing it by giving a crash course on elementary Galois
theory in appendix A) and kept me thinking about dessins and their
Galois invariants ever since (again, I’ll come back to this later).

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coalgebras and non-geometry

Published September 6, 2006 by lievenlb

In this
series of posts I’ll try to make at least part of the recent
[Kontsevich-Soibelman paper](http://www.arxiv.org/abs/math.RA/0606241) a
bit more accessible to algebraists. In non-geometry, the algebras
corresponding to *smooth affine varieties* I’ll call **qurves** (note
that they are called **quasi-free algebras** by Cuntz & Quillen and
**formally smooth** by Kontsevich). By definition, a qurve in an affine
$\mathbb{C} $-algebra A having the lifting property for algebra
maps through nilpotent ideals (extending Grothendieck’s characterization
of smooth affine algebras in the commutative case). Examples of qurves
are : finite dimensional semi-simple algebras (for example, group
algebras $\mathbb{C} G $ of finite groups), coordinate rings of
smooth affine curves or a noncommutative mixture of both, skew-group
algebras $\mathbb{C}[X] \ast G $ whenever G is a finite group of
automorphisms of the affine curve X. These are Noetherian examples but
in general a qurve is quite far from being Noetherian. More typical
examples of qurves are : free algebras $\mathbb{C} \langle
x_1,\ldots,x_k \rangle $ and path algebras of finite quivers
$~\mathbb{C} Q $. Recall that a finite quiver Q s just a
directed graph and its path algebra is the vectorspace spanned by all
directed paths in Q with multiplication induced by concatenation of
paths. Out of these building blocks one readily constructs more
involved qurves via universal algebra operations such as (amalgamated)
free products, universal localizations etc. In this way, the
groupalgebra of the modular group $SL_2(\mathbb{Z}) $ (as well
as that of a congruence subgroup) is a qurve and one can mix groups with
finite groupactions on curves to get qurves like $ (\mathbb{C}[X]
\ast G) \ast_{\mathbb{C} H} \mathbb{C} M $ whenever H is a common
subgroup of the finite groups G and M. So we have a huge class of
qurve-examples obtained from mixing finite and arithmetic groups with
curves and quivers. Qurves can we used as *machines* generating
interesting $A_{\infty} $-categories. Let us start by recalling
some facts about finite closed subschemes of an affine smooth variety Y
in the commutative case. Let **fdcom** be the category of all finite
dimensional commutative $\mathbb{C} $-algebras with morphisms
being onto algebra morphisms, then the study of finite closed subschemes
of Y is essentially the study of the covariant functor **fdcom** –>
**sets** assigning to a f.d. commutative algebra S the set of all onto
algebra maps from $\mathbb{C}[Y] $ to S. S being a f.d.
commutative semilocal algebra is the direct sum of local factors $S
\simeq S_1 \oplus \ldots \oplus S_k $ where each factor has a
unique maximal ideal (a unique point in Y). Hence, our study reduces to
f.d. commutative images with support in a fixed point p of Y. But all
such quotients are also quotients of the completion of the local ring of
Y at p which (because Y is a smooth variety, say of dimension n) is
isomorphic to formal power series
$~\mathbb{C}[[x_1,\ldots,x_n]] $. So the local question, at any
point p of Y, reduces to finding all settings
$\mathbb{C}[[x_1,\ldots,x_n]] \twoheadrightarrow S
\twoheadrightarrow \mathbb{C} $ Now, we are going to do something
strange (at least to an algebraist), we’re going to take duals and
translate the above sequence into a coalgebra statement. Clearly, the
dual $S^{\ast} $ of any finite dimensional commutative algebra
is a finite dimensional cocommutative coalgebra. In particular
$\mathbb{C}^{\ast} \simeq \mathbb{C} $ where the
comultiplication makes 1 into a grouplike element, that is
$\Delta(1) = 1 \otimes 1 $. As long as the (co)algebra is
finite dimensional this duality works as expected : onto maps correspond
to inclusions, an ideal corresponds to a sub-coalgebra a sub-algebra
corresponds to a co-ideal, so in particular a local commutative algebra
corresponds to an pointed irreducible cocommutative coalgebra (a
coalgebra is said to be irreducible if any two non-zero subcoalgebras
have non-zero intersection, it is called simple if it has no non-zero
proper subcoalgebras and is called pointed if all its simple
subcoalgebras are one-dimensional. But what about infinite dimensional
algebras such as formal power series? Well, here the trick is not to
take all dual functions but only those linear functions whose kernel
contains a cofinite ideal (which brings us back to the good finite
dimensional setting). If one takes only those good linear functionals,
the ‘fancy’-dual $A^o $of an algebra A is indeed a coalgebra. On
the other hand, the full-dual of a coalgebra is always an algebra. So,
between commutative algebras and cocommutative coalgebras we have a
duality by associating to an algebra its fancy-dual and to a coalgebra
its full-dual (all this is explained in full detail in chapter VI of
Moss Sweedler’s book ‘Hopf algebras’). So, we can dualize the above pair
of onto maps to get coalgebra inclusions $\mathbb{C} \subset
S^{\ast} \subset U(\mathfrak{a}) $ where the rightmost coalgebra is
the coalgebra structure on the enveloping algebra of the Abelian Lie
algebra of dimension n (in which all Lie-elements are primitive, that is
$\Delta(x) = x \otimes 1 + 1 \otimes x $ and indeed we have that
$U(\mathfrak{a})^{\ast} \simeq \mathbb{C}[[x_1,\ldots,x_n]] $.
We have translated our local problem to finding all f.d. subcoalgebras
(containing the unique simple) of the enveloping algebra. But what is
the point of this translation? Well, we are not interested in the local
problem, but in the global problem, so we somehow have to **sum over all
points**. Now, on the algebra level that is a problem because the sum of
all local power series rings over all points is no longer an algebra,
whereas the direct sum of all pointed irreducible coalgebras $~B_Y
= \oplus_{p \in Y} U(\mathfrak{a}_p) $ is again a coalgebra! That
is, we have found a huge coalgebra (which we call the coalgebra of
‘distributions’ on Y) such that for every f.d. commutative algebra S we
have $Hom_{comm alg}(\mathbb{C}[Y],S) \simeq Hom_{cocomm
coalg}(S^{\ast},B_Y) $ Can we get Y back from this coalgebra of
districutions? Well, in a way, the points of Y correspond to the
group-like elements, and if g is the group-like corresponding to a point
p, we can recover the tangent-space at p back as the g-primitive
elements of the coalgebra of distributions, that is the elements such
that $\Delta(x) = x \otimes g + g \otimes x $. Observe that in
this commutative case, there are no **skew-primitives**, that is
elements such that $\Delta(x) = x \otimes g + h \otimes x $ for
different group-likes g and h. This is the coalgebra translation of the
fact that a f.d. semilocal commutative algebra is the direct sum of
local components. This is something that will definitely change if we
try to extend the above to the case of qurves (to be continued).

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Krull & Paris

Published September 5, 2006 by lievenlb

The
Category-Cafe ran an interesting post The history of n-categories
claiming that “mathematicians’ histories are largely
‘Royal-road-to-me’ accounts”

To my mind a key
difference is the historians’ emphasis in their histories that things
could have turned out very differently, while the mathematicians tend to
tell a story where we learn how the present has emerged out of the past,
giving the impression that things were always going to turn out not very
dissimilarly to the way they have, even if in retrospect the course was
quite tortuous.

Over the last weeks I’ve been writing up
the notes of a course on ‘Elementary Algebraic Geometry’ that I’ll
be teaching this year in Bach3. These notes are split into three
historical periods more or less corresponding to major conceptual leaps
in the subject : (1890-1920) ideals in polynomial rings (1920-1950)
intrinsic definitions using the coordinate ring (1950-1970) scheme
theory. Whereas it is clear to take Hilbert&Noether as the leading
figures of the first period and Serre&Grothendieck as those of the
last, the situation for the middle period is less clear to me. At
first I went for the widely accepted story, as for example phrased by Miles Reid in the
Final Comments to his Undergraduate Algebraic Geometry course.

…
rigorous foundations for algebraic geometry were laid in the 1920s and
1930s by van der Waerden, Zariski and Weil (van der Waerden’s
contribution is often suppressed, apparently because a number of
mathematicians of the immediate post-war period, including some of the
leading algebraic geometers, considered him a Nazi collaborator).

But then I read The Rising Sea: Grothendieck
on simplicity and generality I by Colin McLarty and stumbled upon
the following paragraph

From Emmy Noether’s viewpoint,
then, it was natural to look at prime ideals instead of classical and
generic points—or, as we would more likely say today, to identify
points with prime ideals. Her associate Wolfgang Krull did this. He gave
a lecture in Paris before the Second World War on algebraic geometry
taking all prime ideals as points, and using a Zariski topology (for
which see any current textbook on algebraic geometry). He did this over
any ring, not only polynomial rings like C[x, y]. The generality was
obvious from the Noether viewpoint, since all the properties needed for
the definition are common to all rings. The expert audience laughed at
him and he abandoned the idea.

The story seems to be
due to Jurgen Neukirch’s ‘Erinnerungen an Wolfgang Krull’
published in ‘Wolfgang Krull : Gesammelte Abhandlungen’ (P.
Ribenboim, editor) but as our library does not have this book I would
welcome any additional information such as : when did Krull give this
talk in Paris? what was its precise content? did he introduce the prime
spectrum in it? and related to this : when and where did Zariski
introduce ‘his’ topology? Answers anyone?

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a noncommutative topology 2

Published February 1, 2006 by lievenlb

A *qurve*
is an affine algebra such that $~\Omega^1~A$ is a projective
$~A~$-bimodule. Alternatively, it is an affine algebra allowing lifts of
algebra morphisms through nilpotent ideals and as such it is the ‘right’
noncommutative generalization of Grothendieck’s smoothness criterium.
Examples of qurves include : semi-simple algebras, coordinate rings of
affine smooth curves, hereditary orders over curves, group algebras of
virtually free groups, path algebras of quivers etc.

Hence, qurves
behave a lot like curves and as such one might hope to obtain one day a
‘birational’ classification of them, if we only knew what we mean
by this. Whereas the etale classification of them is understood (see for
example One quiver to
rule them all or Qurves and quivers )
we don’t know what the Zariski topology of a qurve might be.

Usually, one assigns to a qurve $~A~$ the Abelian category of all its
finite dimensional representations $\mathbf{rep}~A$ and we would like to
equip this set with a topology of sorts. Because $~A~$ is a qurve, its
scheme of n-dimensional representations $\mathbf{rep}_n~A$ is a smooth
affine variety for each n, so clearly $\mathbf{rep}~A$ being the disjoint
union of these acquires a trivial but nice commutative topology.
However, we would like open sets to hit several of the components
$\mathbf{rep}_n~A$ thereby ‘connecting’ them to form a noncommutative
topological space associated to $~A~$.

In a noncommutative topology on
rep A I proposed a way to do this and though the main idea remains a
good one, I’ll ammend the construction next time. Whereas we don’t know
of a topology on the whole of $\mathbf{rep}~A$, there is an obvious
ordinary topology on the subset $\mathbf{simp}~A$ of all simple finite
dimensional representations, namely the induced topology of the Zariski
topology on $~\mathbf{spec}~A$, the prime spectrum of twosided prime ideals
of $~A~$. As in commutative algebraic geometry the closed subsets of the
prime spectrum consist of all prime ideals containing a given twosided
ideal. A typical open subset of the induced topology on $\mathbf{simp}~A$
hits many of the components $\mathbf{rep}_n~A$, but how can we extend it to
a topology on the whole of the category $\mathbf{rep}~A$ ?

Every
finite dimensional representation has (usually several) Jordan-Holder
filtrations with simple successive quotients, so a natural idea is to
use these filtrations to extend the topology on the simples to all
representations by restricting the top (or bottom) of the Jordan-Holder
sequence. Let W be the set of all words w such as $U_1U_2 \ldots U_k$
where each $U_i$ is an open subset of $\mathbf{simp}~A$. We can now define
the *left basic open set* $\mathcal{O}_w^l$ consisting of all finite
dimensional representations M having a Jordan-Holder sequence such that
the i-th simple factor (counted from the bottom) belongs to $U_i$.
(Similarly, we can define a *right basic open set* by counting from the
top or a *symmetric basic open set* by merely requiring that the simples
appear in order in the sequence). One final technical (but important)
detail is that we should really consider equivalence classes of left
basic opens. If w and w’ are two words we will denote by $\mathbf{rep}(w
\cup w’)$ the set of all finite dimensional representations having a
Jordan-Holder filtration with enough simple factors to have one for each
letter in w and w’. We then define $\mathcal{O}^l_w \equiv
\mathcal{O}^l_{w’}$ iff $\mathcal{O}^l_w \cap \mathbf{rep}(w \cup w’) =
\mathcal{O}^l_{w’} \cap \mathbf{rep}(w \cup w’)$. Equivalence classes of
these left basic opens form a partially ordered set (induced by
set-theoretic inclusion) with a unique minimal element 0 (the empty set
corresponding to the empty word) and a uunique maximal element 1 (the
set $\mathbf{rep}~A$ corresponding to the letter $w=\mathbf{simp}~A$).
Set-theoretic union induces an operation $\vee$ and the operation
$~\wedge$ is induced by concatenation of words, that is,
$\mathcal{O}^l_w \wedge \mathcal{O}^l_{w’} \equiv \mathcal{O}^l_{ww’}$.
This then defines a **left noncommutative topology** on $\mathbf{rep}~A$ in
the sense of Van Oystaeyen (see [part
1](http://www.neverendingbooks.org/index.php/noncommutative-topology-1 $
). To be precise, it satisfies the axioms in the left and middle column
of the following picture and
similarly, the right basic opens give a right noncommutative topology
(satisfying the axioms of the middle and right columns) whereas the
symmetric opens satisfy all axioms giving the basis of a noncommutative
topology. Even for very simple finite dimensional qurves such as
$\begin{bmatrix} \mathbb{C} & \mathbb{C} \\ 0 & \mathbb{C}
\end{bmatrix}$ this defines a properly noncommutative topology on the
Abelian category of all finite dimensional representations which
obviously respect isomorphisms so is really a noncommutative topology on
the orbits. Still, while this may give a satisfactory local definition,
in gluing qurves together one would like to relax simple representations
to *Schurian* representations. This can be done but one has to replace
the topology coming from the Zariski topology on the prime spectrum by
the partial ordering on the *bricks* of the qurve, but that will have to
wait until next time…

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noncommutative topology (1)

Published January 29, 2006 by lievenlb

A couple of days ago Ars Mathematica had a post Cuntz on noncommutative topology pointing to a (new, for me) paper by Joachim Cuntz

A couple of years ago, the Notices of the AMS featured an article on noncommutative geometry a la Connes: Quantum Spaces and Their Noncommutative Topology by Joachim Cuntz. The hallmark of this approach is the heavy reliance on K theory. The first few pages of the article are fairly elementary (and full of intriguing pictures), before the K theory takes over.

A few comments are in order. To begin, the paper is **not** really about noncommutative geometry a la Connes, but rather about noncommutative geometry a la Cuntz&Quillen (based on quasi-free algebras) or, equivalently, a la Kontsevich (formally smooth algebras) or if I may be so bold a la moi (qurves).

About the **intruiging pictures** : it seems to be a recent trend in noncommutative geometry research papers to include meaningless pictures to lure the attention of the reader. But, unlike aberrations such as the recent pastiche by Alain Connes and Mathilde Marcolli A Walk in the Noncommutative Garden, Cuntz is honest about their true meaning

I am indebted to my sons, Nicolas and Michael,
for the illustrations to the examples above. Since
these pictures have no technical meaning, they
are only meant to provide a kind of suggestive
visualization of the corresponding quantum spaces.

As one of these pictures made it to the cover of the **Notices** an explanation was included by the cover-editor

About the Cover :

The image on this month’s cover arose from
Joachim Cuntz’s effort to render into visible art
his own internal vision of a noncommutative
torus, an object otherwise quite abstract. His
original idea was then implemented by his son
Michael in a program written in Pascal. More
explicitly, he says that the construction started
out with a triangle in a square, then translated
the triangle by integers times a unit along a line
with irrational slope; plotted the images thus
obtained in a periodic manner; and stopped
just before the figure started to seem cluttered.
Many mathematicians carry around inside
their heads mental images of the abstractions
they work with, and manipulate these objects
somehow in conformity with their mental imagery. They probably also make aesthetic judgements of the value of their work according to
the visual qualities of the images. These presumably common phenomena remain a rarely
explored domain in either art or psychology.

—Bill Casselman(covers@ams.org)

There can be no technical meaning to the pictures as in the Connes and Cuntz&Quillen approach there is only a noncommutative algebra and _not_ an underlying geometric space, so there is no topology, let alone a noncommutative topology. Of course, I do understand why Cuntz&others name it as such. They view the noncommutative algebra as the ring of functions on some virtual noncommutative space and they compute topological invariants (such as K-groups) of the algebras and interprete them as information about the noncommutative topology of these virtual and unspecified spaces.

Still, it is perfectly possible to associate to a qurve (aka quasi-free algebra or formally smooth algebra) a genuine noncommutative topological space. In this series of posts I’ll explain the little I know of the history of this topic, the thing I posted about it a couple of years ago, why I abandoned the project and the changes I made to it since and the applications I have in mind, both to new problems (such as the birational_classification of qurves) as well as classical problems (such as rationality problems for $PGL_n $ quotient spaces).

Although others have tried to define noncommutative topologies before, I learned about them from Fred Van Oystaeyen. Fred spend the better part of his career constructing structure sheaves associated to noncommutative algebras, mainly to prime Noetherian algebras (the algebras of preference for the majority of non-commutative algebraists). So, suppose you have an ordinary (meaning, the usual commutative definition) topological space X associated to this algebra R, he wants to define an algebra of sections on every open subset $X(\sigma) $ by taking a suitable localization of the algebra $Q_{\sigma}(R) $. This localization is taken with respect to a suitable filter of left ideals $\mathcal{L}(\sigma) $ of R and is defined to be the subalgebra of the classiocal quotient ring $Q(R) $ (which exists because $R$ is prime Noetherian in which case it is a simple Artinian algebra)

$Q_{\sigma}(R) = { q \in Q(R)~|~\exists L \in \mathcal{L}(\sigma)~:~L q \subset R } $

(so these localizations are generalizations of the usual Ore-type rings of fractions). But now we come to an essential point : if we want to glue this rings of sections together on an intersection $X(\sigma) \cap X(\tau) $ we want to do this by ‘localizing further’. However, there are two ways to do this, either considering $~Q_{\sigma}(Q_{\tau}(R)) $ or considering $Q_{\tau}(Q_{\sigma}(R)) $ and these two algebras are only the same if we impose fairly heavy restrictions on the filters (or on the algebra) such as being compatible.

As this gluing property is essential to get a sheaf of noncommutative algebras we seem to get stuck in the general (non compatible) case. Fred’s way out was to make a distinction between the intersection $X_{\sigma} \cap X_{\tau} $ (on which he put the former ring as its ring of sections) and the intersection $X_{\tau} \cap X_{\sigma} $ (on which he puts the latter one). So, the crucial new ingredient in a noncommutative topology is that the order of intersections of opens matter !!!

Of course, this is just the germ of an idea. He then went on to properly define what a noncommutative topology (and even more generally a noncommutative Grothendieck topology) should be by using this localization-example as guidance. I will not state the precise definition here (as I will have to change it slightly later on) but early version of it can be found in the Antwerp Ph.D. thesis by Luc Willaert (1995) and in Fred’s book Algebraic geometry for associative algebras.

Although _qurves_ are decidedly non-Noetherian (apart from trivial cases), one can use Fred’s idea to associate a noncommutative topological space to a qurve as I will explain next time. The quick and impatient may already sneak at my old note a non-commutative topology on rep A but please bear in mind that I changed my mind since on several issues…

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