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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