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Tag: modular

devilish symmetries

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}) $!

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the Manin-Marcolli cave

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…

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the taxicab curve

(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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