Re: This Week's Finds in Mathematical Physics (Week 228)

on 21/03/2006 2:32 am, Tim S at Tim@xxxxxxxxxxxxxxxxxxxxxxxx wrote:

I have a few more notes to add to this, hopefully not throwing away too much
context...

on 20/03/2006 3:39 am, John Baez at baez@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
wrote:

<snip>

"rational tangles"

<snip>

The players, call them A and B, start by facing each other and holding
ropes in each hand connecting them together like this:

A A
| |
| |
B B

<snip>

"Add one":

A A
\ /
/
/ \
B B

"inverse reciprocal": move all four ends of the ropes a quarter-turn
clockwise, like this:

A A
\_/
_
/ \
B B


<stuff>


The basic operations on rational tangles are "adding
one" and "negative reciprocal", and these generate all the
fractional linear transformations

az + b
z |-> -------
cz + d

with a,b,c,d integer and ad-bc = 1. The group of these transformations
is PSL(2,Z). It acts on rational tangles, and Conway's theorem says
this action is isomorphic to the obvious action of PSL(2,Z) as fractional
linear transformations of the "rational projective line", meaning the
rational numbers together with a point at infinity.

Does anybody know [proof using topological properties of PSL(2,Z)]?

No, but I can at least draw a (very simple and obvious) picture of the
natural action of PSL(2, Z), to go alongside the more usual algebraic type
of proof... :-)

So, we take the plane R^2, and consider the lattice Z^2 in it of points with
integer coordinates.

OK, the matrix corresponding to the fractional linear transformation

z -> -1/z

is

( 0 -1)
( 1 0)

which actually does a 90 degree rotation of the plane (or lattice) when
acting 'naturally' as a member of PSL(2, Z) on R^2 or Z^2.

And the matrix corresponding to the fractional linear transformation

z -> z+1

is

( 1 1)
( 0 1)

sending (x) to (x+y)
(y) ( y )

which does a shear of the plane, keeping everything on the x axis fixed and
moving the line y = 1 to the right by 1 unit.

So lets quotient out the plane by the lattice and get a torus, and look at
an elementary cell with corners W = (0, 1), X = (1, 1), Y = (1, 0) and
Z = (0, 0), and lets also put our players A and B on the lattice, with

B on the line y = 1/2, with
his left hand at (0, 1/2) and
his right hand at (1/2, 1/2)

and A on the x axis, with
her left hand at (0, 0) and
her right hand at (0, 1/2)

<Snip stuff showing we generate tangles on our torus just like the tangles
in the game.>



So Conway's theorem says that the isotopy class of the tangle is determined
by the final location of point W.

To be marginally more precise, the x coordinate of W tracks vertical twists
in the tangle, and the y coordinate tracks horizontal twists (introduced by
conjugation with the 90 degree rotation).

Note 1: This paragraph is is nonsense, the horizontal and vertical twists
are encoded in the continued fraction and aren't obvious from the simple
rational fraction.

Note 2: the process of generating tangles on the torus appears to show a
very intimate relationship between rational tangles and Dehn twists, which
would be more useful if I knew anything about Dehn twists...

Note 3: The proof in http://www.math.uic.edu/~kauffman/RTang.pdf makes use
of the following identity, due to Lagrange:

1 1
a - - = (a - 1) + --------------
b 1
1 + -------
(b - 1)

If we re-express this as

1 -1
a - - = (a - 1) + --------------
b -1
-1 + -------
(b - 1)

then we get two recipes for sequences of moves in our game. Playing with,
e.g. a, b > 0, the LHS of the equation puts our point W in the bottom left
quadrant, while the RHS puts us in the top right quadrant, at the
diametrically opposite point. So they are projectively equivalent.

Note 4: The proof also makes use of the Jones polynomial, but specialised to
a particular value of its argument, namely sqrt(i). In an attempt to get
some insight into the deeper meaning of the Jones polynomial, I avoided
specialising, to see what I got. This is what happened:

To remind readers, we construct the Jones polynomial of a knot or tangle in
link representation as follows:

a) For each crossing

\ /
\
/ \

we replace it by

\_/ | |
A _ + A^{-1} | |
/ \ | |

where A is the indeterminate in our polynomial, and for a crossing going the
other way, we get the same, except with A and A^{-1} interchanged.

b) For each loop with no crossings, we take it away, and multiply what
remains by a factor -A^2 - A^{-2}.

c) A diagram consisting of a loop and nothing else gets replaced by the
number 1. However, we never reach this stage with rational tangles. Instead,
we end up with a polynomial times the || diagram (two vertical lines), and
another polynomial times the = diagram (two horizontal lines). The first
polynomial is called the numerator and the second the denominator. We divide
the numerator by the denominator to get a rational function which is
invariant under isotopies of the tangle (assuming we leave the ends of the
ropes fixed). Goldman and Kauffman set A = sqrt(i), multiply the resulting
fraction by -i, and get back the rational number corresponding to the
rational tangle. If A = sqrt(i), then A^2 + A^{-2} = 0, so all diagrams with
loops conveniently disappear.

Suppose we leave A indeterminate? What happens?

Let's pick an arbitrary rational tangle T, and apply our two operations to
it. Lets represent the numerator by N(T) and the denominator by D(T). We'll
start by considering what happens to the rational function N(T)/D(T).

The operation "take the inverse reciprocal" is simple. All crossings are
replaced by their reverses, so we end up with all vertical pairs
interchanged with horizontal pairs, so we interchange the numerator and
denominator, and the function gets replaced by its reciprocal.

The operator "add one" is more complicated. In fact, I'll replace it by a
slightly different operator which hopefully will not materially alter the
substance of the argument.

So, given our tangle T, we define the tangle T+1 to be

\___ /\ /
| | \ /
| T | /
|___| / \
/ \/ \

Rule a) for constructing the Jones polynomial then gives us

\___ /\ / \___ _______
| | | / | |
A | T | | | + A^{-1} | T |
|___| | \ |___|_______
/ \/ \ /


The diagram in the second term of this expression is of course just T again.

Now, from the definition of the numerator and the denominator, we know we
can replace

\___ /
| |
| T |
|___|
/ \

by

\ / \___ /
| |
N(T) | | + D(T)
| | ___
/ \ / \


Substituting this into our diagram for T+1 above gives

\ /\ / \___ /\ /
| | | / | /
A N(T) | | | | + A D(T) | |
| | | \ ___ | \
/ \/ \ / \/ \


\ / \___ /
| |
+ A^{-1} N(T) | | + A^{-1} D(T)
| | ___
/ \ / \


(Don'cha just love ascii art? No? Me neither.)

Replacing the loop in the first diagram by (-A^2 - A^{-2}), and gathering
together the terms with the || diagram (the first, second and third), and
the terms with the = diagram (the fourth), we get

N(T+1) = A (-A^2 - A^{-2}) N(T) + A D(T) + A^{-1} N(T)

= -A^3 N(T) + A D(T)


and

D(T+1) = A^{-1} D(T).

So the fraction F(T) = N(T)/D(T) goes to

A D(T) - A^3 N(T)
F(T+1) = -------------------
A^{-1} D(T)


= A^2 - A^4 F(T)

Introducing the definition

R(T) - -A^2 F(T), we get

R(T+1) = -A^2 F(T+1)

= -A^4 + A^4 A^2 F(T)

= -A^4 - A^4 R(T)

= -A^4 (R(T) + 1).

Now, if A = sqrt(i), then the factor of -A^2 translates into the factor of
-i introduced by Goldman and Kauffman, so that is alright...

Going back to the "minus reciprocal" operation, we have, as you recall,

F(-1/T) = 1/F(T),

so

R(-1/T) = -A^2 F(-1/T)

= -A^2 / F(T)

= A^4 / (-A^2 F(T))

= A^4 / R(T)

= -A^4 . -1/R(T)

So in place of "add one", we get "add one and multiply by -A^4, while in
place of "negative reciprocal" we get "negative reciprocal and multiply by
-A^4).

With A = sqrt(i), -A^4 = 1, so this works out.

Now let q = -A^4, and work out some examples:

0, add one:

(0 + 1) * q = q

Add one again:

(q + 1) * q = q^2 + q

Add one again

(q^2 + q + 1() * q = q^3 + q^2 + q

So defining the "q-deformed integer" [n] by

[n] = q + q^2 + ... + q^{n-1}

We get "add one n times" translating into q [n].

Note that, for q = 1, we have [n] = n.

Now take a negative reciprocal:

q [n] -> q . -1/(q [n])

= -1/[n]

Now "add one" some more:

q(-1/[n] + 1) = -q/[n] + q = -1/[n] + q [1]

q (-q/[n] + q + 1) = -q^2/[n] + q^2 + q = -q^2/[n] + q [2]

And eventually,

-q^m/[n] + q [m]

That pesky factor of q^m is a bit annoying, but it's needed for the
"q-addition" to work, e.g. to get [n] [+] [m] = [n+m], where we define [+]
by

a [+] [m] = q^m a + [m].

So we have some "q-deformed" continued fractions, which go to the
straightforward continued fractions when q -> 1 (which, recall, is A ->
sqrt(i) in the Goldman & Kauffman paper).

I'm not sure if this has given me any deep insight into the Jones
polynomial, but I hope it is mildly interesting to someone.

Tim

.