Category: featured

the taxicab curve

Published March 26, 2007 by lievenlb

(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!

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anabelian geometry

Published March 22, 2007 by lievenlb

Last 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…

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.

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