devilish symmetries

By lieven

continued fractions

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 $B_3 \simeq SL2(\mathbb{Z})$ 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 $\sigma_1 \sigma2 \sigma_1 = \sigma2 \sigma_1 \sigma2$ as can be seen because the 3-strings below can be transformed into each other

But from this relation it follows that $c=(\sigma_1 \sigma2 \sigma_1)^2$ is a central element in B3

and it is not difficult to verify that indeed $B_3/ \langle c \rangle \simeq PSL2(\mathbb{Z})$ and $B_3/ \langle c^2 \rangle \simeq SL2(\mathbb{Z})$ 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 $PSL2(\mathbb{Z}) = C_2 \ast C3$ (and similarly, there are only 12 one-dimensional $SL_2(\mathbb{Z})$ -representations). Btw. for those still interested in noncommutative geometry : $(P)SL2(\mathbb{Z})$ 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 $C2 \ast \mathbb{N}$ 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 $a = \langle a0;a_1,a2,\hdots \rangle$ is the continued fraction of a (see this post for details) then if we look at the n-th approximations $\frac{p_n}{qn}$ (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,

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 $S \rightarrow GL2(\mathbb{C})~\qquad r \mapsto \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix}~\qquad g \mapsto \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$ 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 $rg^{i1}rg^{i_2}r \hdots rg^{il}r$ can be the identity in S. We will prove this by computing its action on the continued fraction expression of $a = \langle 0;a_0,a1,\hdots \rangle$ . It is a pleasant exercise to show that $g. \langle 0;a_1,a2,\hdots \rangle = \langle 0;a_1+1,a2,\hdots \rangle$ whence by induction $g^n. \langle 0;a_1,a2,\hdots \rangle = \langle 0;a_1+n,a2,\hdots \rangle$ Moreover, the action on r is given by $r. \langle 0;a_1,a2,\hdots \rangle = \langle 0;1,a_1-1,a2,\hdots \rangle$ if $a_1 \not= 1$ whereas $r. \langle 0;1,a2,a_3,\hdots \rangle = \langle 0;a2+1,a_3,\hdots \rangle$ But then, as a consequence we have that $g^{n-1}rg . \langle 0;a1,a_2,\hdots \rangle = \langle 0;n,a1,a_2,\hdots \rangle$ and iterating this procedure gives us finally that an expression $g^{j-1} r g^k r g^l r \hdots g^z r g = (g^{j-1} r g)(g^{k-1} r g)(g^{l-1} r g) \hdots (g^{z-1} r g)$ acts on $a = \langle 0;a1,a_2,\hdots \rangle$ by sending it to $\langle 0;j,k,l,\hdots,z,a1,a_2,\hdots \rangle$ whence such an expression can never act as the identity element, proving that indeed $S \simeq C2 \ast \mathbb{N}$ . 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 $GL2(\mathbb{Z})$ !

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Artin, braid group, Conway, games, geometry, modular, noncommutative, representations

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Posted in groups
Written on Mon, 02 April 2007 at 10:29 am
Tags: Artin, braid group, Conway, games, geometry, modular, noncommutative, representations
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devilish symmetries

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