Here a collection of pdf-files of NeverEndingBooks-posts on groups, in reverse chronological order.
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After a lengthy spring-break, let us continue with our course on noncommutative geometry and $SL_2(\mathbb{Z}) $-representations. Last time, we have explained Grothendiecks mantra that all algebraic curves defined over number fields are contained in the profinite compactification
$\widehat{SL_2(\mathbb{Z})} = \underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $ of the modular group $SL_2(\mathbb{Z}) $ and in the knowledge of a certain subgroup G of its group of outer automorphisms. In particular we have seen that many curves defined over the algebraic numbers $\overline{\mathbb{Q}} $ correspond to permutation representations of $SL_2(\mathbb{Z}) $. The profinite compactification $\widehat{SL_2}=\widehat{SL_2(\mathbb{Z})} $ is a continuous group, so it makes sense to consider its continuous n-dimensional representations $\mathbf{rep}_n^c~\widehat{SL_2} $ Such representations are known to have a finite image in $GL_n(\mathbb{C}) $ and therefore we get an embedding $\mathbf{rep}_n^c~\widehat{SL_2} \hookrightarrow \mathbf{rep}_n^{ss}~SL_n(\mathbb{C}) $ into all n-dimensional (semi-simple) representations of $SL_2(\mathbb{Z}) $. We consider such semi-simple points as classical objects as they are determined by – curves defined over $\overline{Q} $ – representations of (sporadic) finite groups – modlart data of fusion rings in RCTF – etc… To get a feel for the distinction between these continuous representations of the cofinite completion and all representations, consider the case of $\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n \mathbb{Z} $. Its one-dimensional continuous representations are determined by roots of unity, whereas all one-dimensional (necessarily simple) representations of $\mathbb{Z}=C_{\infty} $ are determined by all elements of $\mathbb{C} $. Hence, the image of $\mathbf{rep}_1^c~\hat{\mathbb{Z}} \hookrightarrow \mathbf{rep}_1~C_{\infty} $ is contained in the unit circle
and though these points are very special there are enough of them (technically, they form a Zariski dense subset of all representations). Our aim will be twofold : (1) when viewing a classical object as a representation of $SL_2(\mathbb{Z}) $ we can define its modular content (which will be the noncommutative tangent space in this classical point to the noncommutative manifold of $SL_2(\mathbb{Z}) $). In this way we will associate noncommutative gadgets to our classical object (such as orders in central simple algebras, infinite dimensional Lie algebras, noncommutative potentials etc. etc.) which give us new tools to study these objects. (2) conversely, as we control the tangentspaces in these special points, they will allow us to determine other $SL_2(\mathbb{Z}) $-representations and as we vary over all classical objects, we hope to get ALL finite dimensional modular representations. I agree this may all sound rather vague, so let me give one example we will work out in full detail later on. Remember that one can reconstruct the sporadic simple Mathieu group $M_{24} $ from the dessin d’enfant
- This
- dessin determines a 24-dimensional permutation representation (of
- $M_{24} $ as well of $SL_2(\mathbb{Z}) $) which
- decomposes as the direct sum of the trivial representation and a simple
- 23-dimensional representation. We will see that the noncommutative
- tangent space in a semi-simple representation of
- $SL_2(\mathbb{Z}) $ is determined by a quiver (that is, an
- oriented graph) on as many vertices as there are non-isomorphic simple
- components. In this special case we get the quiver on two points
- $\xymatrix{\vtx{} \ar@/^2ex/[rr] & & \vtx{} \ar@/^2ex/[ll]
- \ar@{=>}@(ur,dr)^{96} } $ with just one arrow in each direction
- between the vertices and 96 loops in the second vertex. To the
- experienced tangent space-reader this picture (and in particular that
- there is a unique cycle between the two vertices) tells the remarkable
- fact that there is **a distinguished one-parameter family of
- 24-dimensional simple modular representations degenerating to the
- permutation representation of the largest Mathieu-group**. Phrased
- differently, there is a specific noncommutative modular Riemann surface
- associated to $M_{24} $, which is a new object (at least as far
- as I’m aware) associated to this most remarkable of sporadic groups.
- Conversely, from the matrix-representation of the 24-dimensional
- permutation representation of $M_{24} $ we obtain representants
- of all of this one-parameter family of simple
- $SL_2(\mathbb{Z}) $-representations to which we can then perform
- noncommutative flow-tricks to get a Zariski dense set of all
- 24-dimensional simples lying in the same component. (Btw. there are
- also such noncommutative Riemann surfaces associated to the other
- sporadic Mathieu groups, though not to the other sporadics…) So this
- is what we will be doing in the upcoming posts (10) : explain what a
- noncommutative tangent space is and what it has to do with quivers (11)
- what is the noncommutative manifold of $SL_2(\mathbb{Z}) $? and what is its connection with the Kontsevich-Soibelman coalgebra? (12)
- is there a noncommutative compactification of $SL_2(\mathbb{Z}) $? (and other arithmetical groups) (13) : how does one calculate the noncommutative curves associated to the Mathieu groups? (14) : whatever comes next… (if anything).
In another post we introduced
Minkowski’s question-mark function, aka the devil’s straircase
and related it to
Conways game of _contorted fractions_. Side remark : over at Good Math, Bad Math Mark Chu-Carroll is running
a mini-series on numbers&games, so far there is a post on surreal numbers,
surreal arithmetic and the connection with
games but
probably this series will go on for some time.
About a year ago I had
an email-exchange with Linas Vepstas because I was
intrigued by one of his online publications linking the fractal
symmetries of the devil’s staircase to the modular group. Unfortunately,
his paper contained some inaccuracies and I’m happy some of my comments
made it into his rewrite The Minkowski question mark, GL(2,Z) and the
modular group. Still, several
mistakes remain so read this paper only modulo his own caveat
XXXX This paper is unfinished. Although this version
corrects a number of serious errors in the previous drafts, it is still
misleading and confusing in many ways. The second half, in particular
must surely contain errors and mis-statements! Caveat emptor! XXXX
For example, on page 15 of the march 24-version he claims
that the third braid group $B_3 \simeq SL_2(\mathbb{Z}) $ which
would make life, mathematics and even physics a lot easier, but
unfortunately is not true. Recall that Artin’s defining relation for the
3-string braid group is $\sigma_1 \sigma_2 \sigma_1 = \sigma_2
\sigma_1 \sigma_2 $ as can be seen because the 3-strings below can
be transformed into each other
But from this
relation it follows that $c=(\sigma_1 \sigma_2 \sigma_1)^2 $ is
a central element in $B_3 $ and it is not difficult to verify
that indeed $B_3/ \langle c \rangle \simeq PSL_2(\mathbb{Z}) $
and $B_3/ \langle c^2 \rangle \simeq SL_2(\mathbb{Z}) $ An easy
way to see that the third braid group and the modular group are quite
different is to look at their one-dimensional representations. Any
group-map $B_3 \rightarrow \mathbb{C}^_ $ is determined by
non-zero complex numbers x and y satisfying $x^2y=y^2x $ so are
parametrized by the torus $\mathbb{C}^_ $ whereas there are only
6 one-dimensional representations of $PSL_2(\mathbb{Z}) = C_2 \ast
C_3 $ (and similarly, there are only 12 one-dimensional
$SL_2(\mathbb{Z}) $-representations). Btw. for those still
interested in noncommutative geometry : $(P)SL_2(\mathbb{Z}) $
are noncommutative manifolds whereas $B_3 $ is definitely
singular, if I ever get to the definitions of all of this… Still,
there is a gem contained in Linas’ paper and here’s my reading of it :
the fractal symmetries of the devil’s staircase form a generating
sub-semigroup $C_2 \ast \mathbb{N} $ of
$GL_2(\mathbb{Z}) $ . To begin, let us recall that the
question-mark function is defined in terms of continued fraction
expressions. So, what group of symmetries may be around the corner?
Well, if $a = \langle a_0;a_1,a_2,\ldots \rangle $ is the
continued fraction of a (see this
post for details) then if we
look at the n-th approximations $\frac{p_n}{q_n} $ (that is, the
rational numbers obtained after breaking off the continued fraction at
step n) it is failrly easy to show that $\begin{bmatrix} p_n &
p_{n-1} \\ q_n & q_{n-1} \end{bmatrix} \in GL_2(\mathbb{Z}) $ and
recall (again) that this group acts on
$\mathbb{P}^1_{\mathbb{C}} $ via Moebius transformations
$\begin{bmatrix} a & b \ c & d \end{bmatrix} $ via $z
\mapsto \frac{az+b}{cz+d} $ One of the symmetries is easy to spot
(reflexion along the 1/2-axis) 
That is, $?(x-1) = 1 – ?(x) $ Observe that the left-hand
side transformation is given by the Moebius transformation determined by
the matrix $r = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \in
GL_2(\mathbb{Z}) $ Other symmetries are harder to see as they are
_fractal symmetries_, that is they are self-symmetries but at different
scales. For example, let us blow-up the ?-function at the interval
[1/3,1/2] and compare it with the function at the interval [1/2,1] 
which has the same graph, while halving the function value. More
generally, substituting the ?-function definition using continued
fraction expressions one verifies that $?(\frac{x}{x+1}) =
\frac{1}{2} ?(x) $ and this time the left-hand transformation is
determined by the matrix $g = \begin{bmatrix} 1 & 0 \\ 1 & 1
\end{bmatrix} \in GL_2(\mathbb{Z}) $ We obtain a semi-group $S
= \langle r,g \rangle $ of fractal symmetries which are induced (the
right hand sides of the above expressions) via a 2-dimensional
representation of S $S \rightarrow GL_2(\mathbb{C})~\qquad r
\mapsto \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix}~\qquad g \mapsto
\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} $ acting
via left-multiplication on the two-dimensional vectorspace
$\mathbb{C}1+\mathbb{C}x $. We claim that S is the free
semi-group $C_2 \ast \mathbb{N} $. Clearly, $r^2=1 $ and
g is of infinite order, but we have to show that no expression of the
form $rg^{i_1}rg^{i_2}r \ldots rg^{i_l}r $ can be the identity
in S. We will prove this by computing its action on the continued
fraction expression of $a = \langle 0;a_0,a_1,\ldots \rangle $.
It is a pleasant exercise to show that $g. \langle 0;a_1,a_2,\ldots
\rangle = \langle 0;a_1+1,a_2,\ldots \rangle $ whence by induction
$g^n. \langle 0;a_1,a_2,\ldots \rangle = \langle 0;a_1+n,a_2,\ldots
\rangle $ Moreover, the action on r is given by $r. \langle
0;a_1,a_2,\ldots \rangle = \langle 0;1,a_1-1,a_2,\ldots \rangle $ if
$a_1 \not= 1 $ whereas $r. \langle 0;1,a_2,a_3,\ldots
\rangle = \langle 0;a_2+1,a_3,\ldots \rangle $ But then, as a
consequence we have that $g^{n-1}rg . \langle 0;a_1,a_2,\ldots
\rangle = \langle 0;n,a_1,a_2,\ldots \rangle $ and iterating this
procedure gives us finally that an expression $g^{j-1} r g^k r g^l
r \ldots g^z r g = (g^{j-1} r g)(g^{k-1} r g)(g^{l-1} r g) \ldots
(g^{z-1} r g) $ acts on $a = \langle 0;a_1,a_2,\ldots
\rangle $ by sending it to $\langle
0;j,k,l,\ldots,z,a_1,a_2,\ldots \rangle $ whence such an expression
can never act as the identity element, proving that indeed $S \simeq
C_2 \ast \mathbb{N} $. As for the second claim, recall from this
post that
$GL_2(\mathbb{Z}) $ is generated by the matrices $U =
\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}~\quad V = \begin{bmatrix}
0 & 1 \ -1 & 1 \end{bmatrix}~\quad R = \begin{bmatrix} 0 & 1 \ 1 & 0
\end{bmatrix} $ and a straightforward verification shows that
$r = RV,~\quad g = VU $ and $R = g^{-1}rg,~\quad
V=g^{-1}rgr,\quad U=rg^{-1}rg^2 $ whence, indeed, the semi-group S
generates the whole of $GL_2(\mathbb{Z}) $!
Yesterday, Yuri Manin and Matilde Marcolli arXived their paper
Modular shadows and the Levy-Mellin infinity-adic transform which is a
follow-up of their previous paper Continued fractions, modular symbols, and non-commutative geometry.
They motivate the title of the recent paper by :
In
[MaMar2](http://www.arxiv.org/abs/hep-th/0201036), these and similar
results were put in connection with the so called “holography”
principle in modern theoretical physics. According to this principle,
quantum field theory on a space may be faithfully reflected by an
appropriate theory on the boundary of this space. When this boundary,
rather than the interior, is interpreted as our observable
space‚Äìtime, one can proclaim that the ancient Plato’s cave metaphor
is resuscitated in this sophisticated guise. This metaphor motivated
the title of the present paper.
Here’s a layout of
Plato’s cave
Imagine prisoners, who have been chained since childhood deep inside an
cave: not only are their limbs immobilized by the chains; their heads
are chained as well, so that their gaze is fixed on a wall.
Behind
the prisoners is an enormous fire, and between the fire and the
prisoners is a raised walkway, along which statues of various animals,
plants, and other things are carried by people. The statues cast shadows
on the wall, and the prisoners watch these shadows. When one of the
statue-carriers speaks, an echo against the wall causes the prisoners to
believe that the words come from the shadows.
The prisoners
engage in what appears to us to be a game: naming the shapes as they
come by. This, however, is the only reality that they know, even though
they are seeing merely shadows of images. They are thus conditioned to
judge the quality of one another by their skill in quickly naming the
shapes and dislike those who begin to play poorly.
Suppose a
prisoner is released and compelled to stand up and turn around. At that
moment his eyes will be blinded by the firelight, and the shapes passing
will appear less real than their shadows.
Right, now how
does the Manin-Marcolli cave look? My best guess is : like this
picture, taken from Curt McMullen’s Gallery
Imagine
this as the top view of a spherical cave. M&M are imprisoned in the
cave, their heads chained preventing them from looking up and see the
ceiling (where $PSL_2(\mathbb{Z}) $ (or a cofinite subgroup of
it) is acting on the upper-half plane via
Moebius-transformations ). All they can see is the circular exit of the
cave. They want to understand the complex picture going on over their
heads from the only things they can observe, that is the action of
(subgroups of) the modular group on the cave-exit
$\mathbb{P}^1(\mathbb{R}) $. Now, the part of it consisting
of orbits of cusps
$\mathbb{P}^1(\mathbb{Q}) $ has a nice algebraic geometric
description, but orbits of irrational points cannot be handled by
algebraic geometry as the action of $PSL_2(\mathbb{Z}) $ is
highly non-discrete as illustrated by another picture from McMullen’s
gallery
depicting the ill behaved topology of the action on the bottom real
axis. Still, noncommutative _differential_ geometry is pretty good at
handling such ill behaved quotient spaces and it turns out that as a
noncommutative space, this quotient
$\mathbb{P}^1(\mathbb{R})/PSL_2(\mathbb{Z}) $ is rich enough
to recover many important aspects of the classical theory of modular
curves. Hence, they reverse the usual NCG-picture of interpreting
commutative objects as shadows of noncommutative ones. They study the
_noncommutative shadow_
$\mathbb{P}^1(\mathbb{R})/PSL_2(\mathbb{Z}) $ of a classical
commutative object, the quotient of the action of the modular group (or
a cofinite subgroup of it) on the upper half-plane.
In our
noncommutative geometry course we have already
seen this noncommutative shadow in action (though at a very basic
level). Remember that we first described the group-structure of the
modular group $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $ via the
classical method of groups acting on trees. In particular, we
considered the tree
and
calculated the stabilizers of the end points of its fundamental domain
(the thick circular edge). But
later we were able to give a
much shorter proof (due to Roger Alperin) by looking only at the action
of $PSL_2(\mathbb{Z}) $ on the irrational real numbers (the
noncommutative shadow). Needless to say that the results obtained by
Manin and Marcolli from staring at their noncommutative shadow are a lot
more intriguing…
(After-math of last week’s second year lecture on elliptic
curves.)
We all know the story of Ramanujan and the taxicab, immortalized by Hardy
“I remember once going to see him when he was lying ill at Putney. I had ridden in taxicab no. 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. ‘No,’ he replied, ‘it’s a very interesting number; it is the smallest number expressible as a sum of two cubes in two different ways’.”
When I was ten, I wanted to become an archeologist and even today I can get pretty worked-up about historical facts. So, when I was re-telling this story last week I just had to find out things like :
the type of taxicab and how numbers were displayed on them and, related to this, exactly when and where did this happen, etc. etc. Half an hour free-surfing further I know a bit more than I wanted.
Let’s start with the date of this taxicab-ride, even the year changes from source to source, from 1917 in the dullness of 1729 (arguing that Hardy could never have made this claim as 1729 is among other things the third Carmichael Number, i.e., a pseudoprime relative to EVERY base) to ‘late in WW-1’ here…
Between 1917 and his return to India on march 13th 1919, Ramanujan was in and out a number of hospitals and nursing homes. Here’s an attempt to summarize these dates&places (based on the excellent paper Ramanujan’s Illness by D.A.B. Young).
(may 1917 -september 20th 1917) : Nursing Hostel, Thompson’s Lane in Cambridge.
(first 2 a 3 weeks of october 1917) : Mendip Hills Senatorium, near Wells in Somerset. (november 1917) : Matlock House Senatorium atMatlock in Derbyshire.
(june 1918 – november 1918) : Fitzroy House, a hospital in Fitzroy square in central London. (december 1918 – march 1919) : Colinette House, a private nursing home in Putney, south-west London. So, “he was lying ill at Putney” must have meant that Ramanujan was at Colinette House which was located 2, Colinette Road and a quick look with Google Earth
shows that the The British Society for the History of Mathematics Gazetteer is correct in asserting that “The house is no longer used as a nursing home and its name has vanished” as well as.”
“It was in 1919 (possibly January), when Hardy made the famous visit in the taxicab numbered 1729.”
Hence, we are looking for a London-cab early 1919. Fortunately, the London Vintage Taxi Association has a website including a taxi history page.
“At the outbreak of the First World War there was just one make available to buy, the Unic. The First World War devastated the taxi trade.
Production of the Unic ceased for the duration as the company turned to producing munitions. The majority of younger cabmen were called up to fight and those that remained had to drive worn-out cabs.
By 1918 these remnant vehicles were sold at highly inflated prices, often beyond the pockets of the returning servicemen, and the trade deteriorated.”
As the first post-war taxicab type was introduced in 1919 (which became known as the ‘Rolls-Royce of cabs’) more than likely the taxicab Hardy took was a Unic,
and the number 1729 was not a taxicab-number but part of its license plate. I still dont know whether there actually was a 1729-taxicab around at the time, but let us return to mathematics.
Clearly, my purpose to re-tell the story in class was to illustrate the use of addition on an elliptic curve as a mean to construct more rational solutions to the equation $x^3+y^3 = 1729 $ starting from the Ramanujan-points (the two solutions he was referring to) : P=(1,12) and Q=(9,10). Because the symmetry between x and y, the (real part of) curve looks like
and if we take 0 to be the point at infinity corresponding to the asymptotic line, the negative of a point is just reflexion along the main diagonal. The geometric picture of addition of points on the curve is then summarized
in
and sure enough we found the points $P+Q=(\frac{453}{26},-\frac{397}{26})$ and $(\frac{2472830}{187953},-\frac{1538423}{187953}) $ and so on by hand, but afterwards I had the nagging feeling that a lot more could have been said about this example. Oh, if Im allowed another historical side remark :
I learned of this example from the excellent book by Alf Van der Poorten Notes on Fermat’s last theorem page 56-57.
Alf acknowledges that he borrowed this material from a lecture by Frits Beukers ‘Oefeningen rond Fermat’ at the National Fermat Day in Utrecht, November 6th 1993.
Perhaps a more accurate reference might be the paper Taxicabs and sums of two cubes by Joseph Silverman which appeared in the april 1993 issue of The American Mathematical Monthly.
The above drawings and some material to follow is taken from that paper (which I didnt know last week). I could have proved that the Ramanujan points (and their reflexions) are the ONLY integer points on $x^3+y^3=1729 $.
In fact, Silverman gives a nice argument that there can only be finitely many integer points on any curve $x^3+y^3=A $ with $A \in \mathbb{Z} $ using the decomposition $x^3+y^3=(x+y)(x^2-xy+y^2) $.
So, take any factorization A=B.C and let $B=x+y $ and $C=x^2-xy+y^2 $, then substituting $y=B-x $ in the second one obtains that x must be an integer solution to the equation $3x^2-3Bx+(B^2-C)=0 $.
Hence, any of the finite number of factorizations of A gives at most two x-values (each giving one y-value). Checking this for A=1729=7.13.19 one observes that the only possibilities giving a square discriminant of the quadratic equation are those where $B=13, C=133 $ and $B=19, C=91 $ leading exactly to the Ramanujan points and their reflexions!
Sure, I mentioned in class the Mordell-Weil theorem stating that the group of rational solutions of an elliptic curve is always finitely generated, but wouldnt it be fun to determine the actual group in this example?
Surely, someone must have worked this out. Indeed, I did find a posting to sci.math.numberthy by Robert L. Ward : (in fact, there is a nice page on elliptic curves made from clippings to this newsgroup).
The Mordell-Weil group of the taxicab-curve is isomorphic to $\mathbb{Z} \oplus \mathbb{Z} $ and the only difference with Robert Wards posting was that I found besides his generator
$P=(273,409) $ (corresponding to the Ramanujan point (9,10)) as a second generator the point
$Q=(1729,71753) $ (note again the appearance of 1729…) corresponding to the rational solution $( -\frac{37}{3},\frac{46}{3}) $ on the taxicab-curve.
Clearly, there are several sets of generators (in fact that’s what $GL_2(\mathbb{Z}) $ is all about) and as our first generators were the same all I needed to see was that the point corresponding to the second Ramanujan point (399,6583) was of the form $\pm Q + a P $ for some integer a. Points and their addition is also easy to do with sage :
sage: P=T([273,409])
sage: Q=T([1729,71753])
sage: -P-Q
(399 : 6583 : 1)
and we see that the second Ramanujan point is indeed of the required form!
2 CommentsLast time we saw
that a curve defined over $\overline{\mathbb{Q}} $ gives rise
to a permutation representation of $PSL_2(\mathbb{Z}) $ or one
of its subgroups $\Gamma_0(2) $ (of index 2) or
$\Gamma(2) $ (of index 6). As the corresponding
monodromy group is finite, this representation factors through a normal
subgroup of finite index, so it makes sense to look at the profinite
completion of $SL_2(\mathbb{Z}) $, which is the inverse limit
of finite
groups $\underset{\leftarrow}{lim}~SL_2(\mathbb{Z})/N $
where N ranges over all normalsubgroups of finite index. These
profinte completions are horrible beasts even for easy groups such as
$\mathbb{Z} $. Its profinite completion
is
$\underset{\leftarrow}{lim}~\mathbb{Z}/n\mathbb{Z} =
\prod_p \hat{\mathbb{Z}}_p $
where the right hand side
product of p-adic integers ranges over all prime numbers! The
_absolute Galois group_
$G=Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ acts on all curves
defined over $\overline{\mathbb{Q}} $ and hence (via the Belyi
maps ans the corresponding monodromy permutation representation) there
is an action of $G $ on the profinite completions of the
carthographic groups.
This is what Grothendieck calls anabelian
algebraic geometry
Returning to the general
case, since finite maps can be interpreted as coverings over
$\overline{\mathbb{Q}} $ of an algebraic curve defined over
the prime field $~\mathbb{Q} $ itself, it follows that the
Galois group $G $ of $\overline{\mathbb{Q}} $ over
$~\mathbb{Q} $ acts on the category of these maps in a
natural way.
For instance, the operation of an automorphism
$~\gamma \in G $ on a spherical map given by the rational
function above is obtained by applying $~\gamma $ to the
coefficients of the polynomials P , Q. Here, then, is that
mysterious group $G $ intervening as a transforming agent on
topologico- combinatorial forms of the most elementary possible
nature, leading us to ask questions like: are such and such oriented
maps ‚conjugate or: exactly which are the conjugates of a given
oriented map? (Visibly, there is only a finite number of these).
I considered some concrete cases (for coverings of low degree) by
various methods, J. Malgoire considered some others ‚ I doubt that
there is a uniform method for solving the problem by computer. My
reflection quickly took a more conceptual path, attempting to
apprehend the nature of this action of G.
One sees immediately
that roughly speaking, this action is expressed by a certain
outer action of G on the profinite com- pactification of the
oriented cartographic group $C_+^2 = \Gamma_0(2) $ , and this
action in its turn is deduced by passage to the quotient of the
canonical outer action of G on the profinite fundamental group
$\hat{\pi}_{0,3} $ of
$(U_{0,3})_{\overline{\mathbb{Q}}} $ where
$U_{0,3} $ denotes the typical curve of genus 0 over the
prime field Q, with three points re- moved.
This is how my
attention was drawn to what I have since termed anabelian
algebraic geometry, whose starting point was exactly a study
(limited for the moment to characteristic zero) of the action of
absolute Galois groups (particularly the groups Gal(K/K),
where K is an extension of finite type of the prime field) on
(profinite) geometric fundamental groups of algebraic varieties
(defined over K), and more particularly (break- ing with a
well-established tradition) fundamental groups which are very far
from abelian groups (and which for this reason I call
anabelian).
Among these groups, and very close to
the group $\hat{\pi}_{0,3} $ , there is the profinite
compactification of the modular group $Sl_2(\mathbb{Z}) $,
whose quotient by its centre ±1 contains the former as congruence
subgroup mod 2, and can also be interpreted as an oriented
cartographic group, namely the one classifying triangulated
oriented maps (i.e. those whose faces are all triangles or
monogons).
and a bit further, on page
250
I would like to conclude this rapid outline
with a few words of commentary on the truly unimaginable richness
of a typical anabelian group such as $SL_2(\mathbb{Z}) $
doubtless the most remarkable discrete infinite group ever
encountered, which appears in a multiplicity of avatars (of which
certain have been briefly touched on in the present report), and which
from the point of view of Galois-Teichmuller theory can be
considered as the fundamental ‚building block‚ of the
Teichmuller tower
The element of the structure of
$Sl_2(\mathbb{Z}) $ which fascinates me above all is of course
the outer action of G on its profinite compactification. By
Bielyi’s theorem, taking the profinite compactifications of subgroups
of finite index of $Sl_2(\mathbb{Z}) $, and the induced
outer action (up to also passing to an open subgroup of G), we
essentially find the fundamental groups of all algebraic curves (not
necessarily compact) defined over number fields K, and the outer
action of $Gal(\overline{K}/K) $ on them at least it is
true that every such fundamental group appears as a quotient of one
of the first groups.
Taking the anabelian yoga
(which remains conjectural) into account, which says that an anabelian
algebraic curve over a number field K (finite extension of Q) is
known up to isomorphism when we know its mixed fundamental group (or
what comes to the same thing, the outer action of
$Gal(\overline{K}/K) $ on its profinite geometric
fundamental group), we can thus say that
all algebraic
curves defined over number fields are contained in the profinite
compactification $\widehat{SL_2(\mathbb{Z})} $ and in the
knowledge of a certain subgroup G of its group of outer
automorphisms!
To study the absolute
Galois group $Gal(\overline{\mathbb{\mathbb{Q}}}/\mathbb{Q}) $ one
investigates its action on dessins denfants. Each dessin will be part of
a finite family of dessins which form one orbit under the Galois action
and one needs to find invarians to see whether two dessins might belong
to the same orbit. Such invariants are called _Galois invariants_ and
quite a few of them are known.
Among these the easiest to compute
are
- the valency list of a dessin : that is the valencies of all
vertices of the same type in a dessin - the monodromy group of a dessin : the subgroup of the symmetric group $S_d $ where d is
the number of edges in the dessin generated by the partitions $\tau_0 $
and $\tau_1 $ For example, we have seen
before that the two
Mathieu-dessins
form a Galois orbit. As graphs (remeber we have to devide each
of the edges into two and the midpoints of these halfedges form one type
of vertex, the other type are the black vertices in the graphs) these
are isomorphic, but NOT as dessins as we have to take the embedding of
them on the curve into account. However, for both dessins the valency
lists are (white) : (2,2,2,2,2,2) and (black) :
(3,3,3,1,1,1) and one verifies that both monodromy groups are
isomorphic to the Mathieu simple group $M_{12} $ though they are
not conjugated as subgroups of $S_{12} $.
Recently, new
Galois invariants were obtained from physics. In Children’s drawings
from Seiberg-Witten curves
the authors argue that there is a close connection between Grothendiecks
programme of classifying dessins into Galois orbits and the physics
problem of classifying phases of N=1 gauge theories…
Apart
from curves defined over $\overline{\mathbb{Q}} $ there are
other sources of semi-simple $SL_2(\mathbb{Z}) $
representations. We will just mention two of them and may return to them
in more detail later in the course.
Sporadic simple groups and
their representations There are 26 exceptional finite simple groups
and as all of them are generated by two elements, there are epimorphisms
$\Gamma(2) \rightarrow S $ and hence all their representations
are also semi-simple $\Gamma(2) $-representations. In fact,
looking at the list of ‘standard generators’ of the sporadic
simples
(here the conjugacy classes of the generators follow the
notation of the Atlas project) we see that all but
possibly one are epimorphic images of $\Gamma_0(2) = C_2 \ast
C_{\infty} $ and that at least 12 of then are epimorphic images
of $PSL_2(\mathbb{Z}) = C_2 \ast
C_3 $.
Rational conformal field theories Another
source of $SL_2(\mathbb{Z}) $ representations is given by the
modular data associated to rational conformal field theories.
These
representations also factor through a quotient by a finite index normal
subgroup and are therefore again semi-simple
$SL_2(\mathbb{Z}) $-representations. For a readable
introduction to all of this see chapter 6 \”Modular group
representations throughout the realm\” of the
book Moonshine beyond the monster the bridge connecting algebra, modular forms and physics by Terry
Gannon. In fact, the whole book
is a good read. It introduces a completely new type of scientific text,
that of a neverending survey paper…
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.
Fortunately,
there is a drastic shortcut to the general tree-argument of last time, due to
Roger Alperin. Recall that the Moebius
transformations corresponding to u resp. v send z resp. to
$-\frac{1}{z} $ and $\frac{1}{1-z} $
whence the Moebius transformation
corresponding to $v^{-1} $ send z to $1-\frac{1}{z} $.
Consider
the set $\mathcal{P} $ of all positive irrational real numbers and the
set $\mathcal{N} $ of all negative irrational real numbers and observe
that
$u(\mathcal{P}) \subset \mathcal{N} $ and
$v^{\pm}(\mathcal{N}) \subset \mathcal{P} $
We have to show
that no alternating word $w=(u)v^{\pm}uv^{\pm}u \dots v^{\pm}(u) $ in
u and $v^{\pm} $ can be the identity in $PSL_2(\mathbb{Z}) $.
If the
length of w is odd then either $w(\mathcal{P}) \subset
\mathcal{N} $ or $w(\mathcal{N}) \subset
\mathcal{P} $ depending on whether w starts with a u or with
a $v^{\pm} $ term. Either way, this proves that no odd-length word can
be the identity element in $PSL_2(\mathbb{Z}) $.
If the length of
the word w is even we can assume that $w = v^{\pm}uv^{\pm}u \dots
v^{\pm}u $ (if necessary, after conjugating with u we get to this form).
There are two subcases, either $w=v^{-1}uv^{\pm}u \dots
v^{\pm}u $ in which case $w(\mathcal{P}) \subset v^{-1}(\mathcal{N}) $
and this latter set is contained in the set of all positive irrational
real numbers which are strictly larger than one .
Or, $w=vuv^{\pm}u \dots v^{\pm}u $ in which case
$w(\mathcal{P}) \subset v(\mathcal{N}) $ and this set is contained in
the set of all positive irrational real numbers strictly smaller than
one .
Either way, this shows that w cannot be the identity
morphism on $\mathcal{P} $ so cannot be the identity element in
$PSL_2(\mathbb{Z}) $.
Hence we have proved that
$PSL_2(\mathbb{Z}) = C_2 \ast C_3 = \langle u,v : u^2=1=v^3 \rangle $
A
description of $SL_2(\mathbb{Z}) $ in terms of generators and relations
follows
$SL_2(\mathbb{Z}) = \langle U,V : U^4=1=V^6, U^2=V^3 \rangle $
It is not true that $SL_2(\mathbb{Z}) $ is the free
product $C_4 \ast C_6 $ as there is the extra relation $U^2=V^3 $.
This relation says that the cyclic groups $C_4 = \langle U \rangle $
and $C_6 = \langle V \rangle $ share a common subgroup $C_2 = \langle
U^2=V^3 \rangle $ and this extra condition is expressed by saying that
$SL_2(\mathbb{Z}) $ is the amalgamated free product of $C_4 $ with
$C_6 $, amalgamated over the common subgroup $C_2 $ and denoted
as
$SL_2(\mathbb{Z}) = C_4 \ast_{C_2} C_6 $
More
generally, if G and H are finite groups, then the free product $G
\ast H $ consists of all words of the form $~(g_1)h_1g_2h_2g_3
\dots g_nh_n(g_{n-1}) $ (so alternating between non-identity
elements of G and H) and the group-law is induced by concatenation
of words (and group-laws in either G or H when end terms are
elements in the same group).
For example, take the dihedral groups $D_4
= \langle U,R : U^4=1=R^2,(RU)^2=1 \rangle $ and $D_6 = \langle V,S :
V^6=1=S^2,(SV)^2=1 \rangle $ then the free product can be expressed
as
$D_4 \ast D_6 = \langle U,V,R,S :
U^4=1=V^6=R^2=S^2=(RV)^2=(RU)^2 \rangle $
This almost fits in with
our obtained description of
$GL_2(\mathbb{Z}) $
$GL_2(\mathbb{Z}) = \langle U,V,R :
U^4=1=V^6=R^2=(RU)^2=(RV)^2, U^2=V^3 \rangle $
except for the
extra relations $R=S $ and $U^2=V^3 $ which express the fact that we
demand that $D_4 $ and $D_6 $ have the same subgroup
$D_2 = \langle U^2=V^3,S=R \rangle $
So, again we can express these relations by
saying that $GL_2(\mathbb{Z}) $ is the amalgamated free product of
the subgroups $D_4 = \langle U,R \rangle $ and $D_6 = \langle V,R
\rangle $, amalgamated over the common subgroup $D_2 = C_2 \times C_2 =
\langle U^2=V^3,R \rangle $. We write
$GL_2(\mathbb{Z}) = D_4
\ast_{D_2} D_6 $
Similarly (but a bit easier) for
$PGL_2(\mathbb{Z}) $ we have
- $PGL_2(\mathbb{Z}) = \langle u,v,R
- u^2=v^3=1=R^2 = (Ru)^2 = (Rv)^2 \rangle $
which can be seen as
the amalgamated free product of $D_2 = \langle u,R \rangle $ with $D_3
= \langle v,R \rangle $, amalgamated over the common subgroup $C_2 =
\langle R \rangle $ and therefore
$PGL_2(\mathbb{Z}) = D_2
\ast_{C_2} D_3 $
Now let us turn to congruence subgroups of
the modular group.
With $\Gamma(n) $ one denotes the kernel of the natural
surjection
$PSL_2(\mathbb{Z}) \rightarrow
PSL_2(\mathbb{Z}/n\mathbb{Z}) $
that is all elements represented by a matrix
$\begin{bmatrix} a
& b \\ c & d \end{bmatrix} $
such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand $\Gamma_0(n) $ consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of $PSL_2(\mathbb{Z}) $.
As we have seen that
$PSL_2(\mathbb{Z}) = C_2 \ast C_3 $ it follows from general facts
on free products that any finite index subgroup is of the
form
$C_2 \ast C_2 \ast \dots \ast C_2 \ast
C_3 \ast C_3 \ast \dots \ast C_3 \ast C_{\infty}
\ast C_{\infty} \dots \ast C_{\infty} $
that is the
free product of k copies of $C_2 $, l copies of $C_3 $ and m copies
of $C_{\infty} $ where it should be noted that k,l and m are allowed
to be zero. There is an elegant way to calculate explicit generators of
these factors for congruence subgroups, due to Ravi S. Kulkarni (An
Arithmetic-Geometric Method in the Study of the Subgroups of the Modular
Group , American Journal of Mathematics, Vol. 113, No. 6. (Dec.,
1991), pp. 1053-1133) which deserves another (non-course) post.
Using this method one finds that $\Gamma_0(2) $ is generated by
the Moebius transformations corresponding to the
matrices
$X=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $ and
$Y=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} $
and that
generators for $\Gamma(2) $ are given by the
matrices
$A=\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} $
and $B=\begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix} $
Next,
one has to write these generators in terms of the generating matrices
u and v of $PSL_2(\mathbb{Z}) $ and as we know all relations between
u and v the relations of these congruence subgroups will follow.
We
will give the details for $\Gamma_0(2) $ and leave you to figure out
that $\Gamma(2) = C_{\infty} \ast C_{\infty} $ (that is that
there are no relations between the matrices A and
B).
Calculate that $X=v^2u $ and that $Y=vuv^2 $. Because the
only relations between u and v are $v^3=1=u^2 $ we see that Y is an
element of order two as $Y^2 = vuv^3uv^2= v^3 = 1 $ and that no power of
X can be the identity transformation.
But then also none of the
elements $~(Y)X^{i_1}YX^{i_2}Y \dots YX^{i_n}(Y) $ can be the identity
(write it out as a word in u and v) whence,
indeed
$\Gamma_0(2) = C_{\infty} \ast C_2 $
In fact,
the group $\Gamma_0(2) $ is staring you in the face whenever you come to
this site. I fear I’ve never added proper acknowledgements for the
beautiful header-picture
so I’d better do it now. The picture is due to Helena
Verrill and she has a
page with
more pictures. The header-picture depicts a way to get a fundamental
domain for the action of $\Gamma_0(2) $ on the upper half plane. Such a
fundamental domain consists of any choice of 6 tiles with different
colours (note that there are two shades of blue and green). Helena also
has a
Java-applet
to draw fundamental domains of more congruence subgroups.