# This Week's Finds in Mathematical Physics (Week 228)

1. Mar 20, 2006

### John Baez

Also available at http://math.ucr.edu/home/baez/week228.html

March 18, 2006
This Week's Finds in Mathematical Physics (Week 228)
John Baez

Last week I showed you some pretty pictures of dunes on Mars.
This week I'll talk about dunes called "barchans" and their relation
to self-organized criticality. Then I'll say a bit about Lauscher
and Reuter's work on quantum gravity... and then I'll beg for help
on a topology problem involving so-called "rational tangles".

But first, a demonstration of my psychic powers.

Take any book off the shelf and look at its 10-digit ISBN number.
Multiply the first digit by 1, the second digit by 2, the third
digit by 3 and so on... up to the NEXT TO LAST DIGIT. Add them up.

Then take this sum and see what it equals mod 11. At the end of

Okay. Here's a photo of the icy dunes of northern Mars. I love it
because it shows that Mars is a lively place with wind and water:

1) North polar sand sea, Mars Odyssey Mission, THEMIS (Thermal
emission imaging system), http://themis.mars.asu.edu/features/polardunes

These dunes, occupying a region the size of Texas, have been sculpted by
wind into long lines with crests 500 meters apart. Their hollows are
covered with frost, which appears bluish-white in this infrared photograph.
The big white spot near the bottom is a hill 100 meters high.

Where the dunes become sparser - for example, near that icy hill - they
break apart into "barchans". These are crescent-shaped formations
whose horns point downwind. Barchans are also found on the deserts of
Earth, and surely on many other planets across the Universe. They are
one of several basic dune patterns, an inevitable consequence of the
laws of nature under fairly common conditions.

The upwind slope of a barchan is gentle, while the downwind slope is
between 32 and 34 degrees. This is the "angle of repose" for sand - the
maximum angle it can tolerate before it starts slipping down:

2) Wikipedia, Barchan, http://en.wikipedia.org/wiki/Barchan

Wind-blown sand accumulates on the front of the barchan, and then
slides down the "slip face" on the back.

Barchans gradually migrate in the direction of the wind at speeds of
about 1-20 meters per year, with small barchans moving faster than big
ones. In fact, when they collide, the smaller barchans pass right
through the big ones! So, they act like solitons in some ways.

It would be great to see one of these frosty barchans close up.
We almost can do it now! The European Space Agency's orbiter
called Mars Express took this wonderful closeup, already shown
in "week211":

3) ESA/DLR/FU Berlin (G. Neukum),
Glacial, volcanic and fluvial activity on Mars: latest images,
http://www.esa.int/SPECIALS/Mars_Express/SEMLF6D3M5E_1.html

However, this is not a barchan - it's a lot bigger. On top of
the picture we see dunes, but then there's a cliff almost 2 kilometers
high leading down into what may be a volcanic caldera. The white stuff
is ice, while the dark stuff could be volcanic ash.

It's actually a bit surprising that there's enough wind on Mars to
create dunes. After all, the air pressure there is about 1% what it
is here on Earth! But in fact the wind speed on Mars often exceeds
200 kilometers per hour, with gusts up to 600 kilometers per hour.
There are dust storms on Mars so big they were first seen from
telescopes on Earth long ago. So, wind is a big factor in Martian
geology:

4) NASA, Mars exploration program: dust storms,
http://mars.jpl.nasa.gov/gallery/duststorms/

The Mars rover Spirit even got its solar panels cleaned by
some dust devils, and it took some movies of them:

5) NASA, Exploration rover mission: dust devils at Gusev, Sol 525,
http://marsrovers.nasa.gov/gallery/press/spirit/20050819a.html

Turning to mathematical physics per se, I can't resist pointing out
that sand piles became very fashionable in this subject a while back.

Why? Well, for this I need to explain "self-organized criticality".

First, note that when a pile of sand is exactly at its angle of repose,
it will suffer lots of little landslides - and a few of these will become
big.

The theory of "critical phenomena" suggests that in this situation,
the probability that a landslide grows to size L should satisfy a
power law. In other, it should be proportional to

1/L^c

for some number c called the "critical exponent". At least, this
type of behavior is seen in many other situations where a physical
system is on the brink of some drastic change - or more precisely,
a "critical point".

When a system is not at a critical ponit, we typically see exponential
laws, where the probability of a disturbance of size L is proportional
to

exp(-L/L_0)

where L_0 is a fixed length scale. This means that our system will
look qualitatively different depending how much we zoom in with our
microscope. At length scales shorter than L_c, disturbances are
really common, while at large length scales they're incredibly rare.

When a system *is* at a critical point, it's self-similar: you can
zoom in or zoom out, and it looks qualitatively the same! It has
no specific length scale. This is what the power law says.

Here's a good place to learn the basics of power laws and self-similarity:

6) Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman,
New York, 1992.

What makes sand dunes interesting is that as they seem to *enjoy*
living on the brink of danger. As the wind blows, they heap up until
their slip face is right at the angle of repose... ready for landslides!

This is the idea of "self-organized criticality": some physical systems
seem to spontaneously bring themselves towards critical points, without
any need for us to tune their parameters to special values.

The paper that introduced this idea came out in 1987:

7) Per Bak, Chao Tang and Kurt Wiesenfeld, Self-organized criticality:
an explanation of 1/f noise, Phys. Rev. Lett. 59 (1987) 381-384.

They came up with a simple model of a sand pile that exhibits
self-organized criticality. In the words of Jos Thijssen:

Bak and co-workers modelled the sand pile as a regular array
of columns consisting of cubic sand grains. Addition of new
grains is simply performed by selecting a column at random and
increasing its height by one. If the column then exceeds its
neighbours in height more than some threshold, it will "collapse":
it will lose some grains which are distributed evenly over its
nearest neighbours. As this collapse alters the height differences
involving those neighbours, there is the possibility that they
collapse in turn. A cascade process sets in until all height
differences are below the threshold. The size of such an avalanche
is defined as the number of sand grains sliding as a result of
a single grain of sand being added to the pile.

What is so interesting about the sand pile model? It turns out
that the sides of the sand pile acquire a specific slope, which is
such that the distribution of avalanches as function of size scales
as a power law. Power laws indicate the absence of scale and indeed
avalanches on all scales are sustained for the equilibrium slope.
If the slope is changed artificially from its equilibrium value,
the distribution is no longer a power law, but it will have an
intrinsic scale (e.g. exponential). Power laws and absence of scale
are the signature of a system being critical. Because the sand pile
tends to adjust the slope of its sides until the power law scaling
sets in, the criticality is called "self-organised".

If your computer runs Java applets, you can play with Thijssen's
simulation sand pile and see the avalanches yourself:

8) Jos Thijssen, The sand pile model and self organised criticality,
http://www.tn.tudelft.nl/tn/People/Staff/Thijssen/sandexpl.html

And here's a cellular automaton sand pile you can play with:

9) Albert Schueller, Cellular automaton sand pile model,
http://schuelaw.whitman.edu/JavaApplets/SandPileApplet/

This one is only 2-dimensional, so the avalanches are less dramatic,
but you can have some fun using the mouse to build structures that
impede the motion of sand.

Like a speck of sand landing at the right place at the right time,
the original paper of Bak et al started a huge landslide of work on
self-organized criticality, some of which has been popularized here:

10) Per Bak, How Nature Works: The Science of Self-Organized Criticality,
Copernicus, New York, 1996.

As you can guess from the title "How Nature Works", some people got a
little carried away with the importance of self-organized criticality.
Then there was a kind of backlash, just as happened with fractals,
chaos, and catastrophe theory. These are all perfectly respectable and
interesting topics in mathematical physics that suffered from being
oversold. People are always eager to find the secret key that will
unlock all the mysteries of the universe. So, if some new idea seems
very general, people will run around trying to unlock all the mysteries
of the universe with it - and become sorely disappointed when it only
unlocks *some*.

I'd be interested to see how well mathematical physicists can model
actual sand dunes. These display an interesting complexity of behavior,
as the pictures here show:

11) US Army Corps of Engineers, Dunes,
http://www.tec.army.mil/research/products/desert_guide/lsmsheet/lsdune.htm

I've only looked at a few papers on the subject, all dealing with
barchans:

12) V. Schwaemmle and H. J. Herrmann, Solitary wave behaviour of sand
dunes, Nature 426 (Dec. 11, 2003), 619-620.

13) Klaus Kroy, Gerd Sauermann, and Hans J. Hermann, Minimal model for
sand dunes, Phys. Rev. Lett. 88 (2002), 054301. Also available at
cond-mat/0101380.

14) H. Elbelrhiti, P. Claudin, and B. Andreotti, Field evidence for
surface-wave-induced instability of sand dunes, Nature 437 (Sep. 29, 2005),
720-723.

The first paper describes how barchans pass through each other like
solitons, simulating them by an equation that's described in the second
one. (By the way, the term "minimal model" in the title of the second
paper is not being used in the sense familiar in conformal field theory!)

The third paper reports the results of a 3-year field study: in reality,
barchans are not stable, and big ones (called "megabarchans") can break
apart into smaller "elementary barchans".

If you're more interested in Mars than the mathematical physics of sand
dunes, you'll be happy to hear that Google has just moved to drastically
expand its customer base by introducing "Google Mars":

Using this you can explore many features of Mars, including its dunes.

I'm getting a little tired out, but there's one thing I've been
meaning to mention for a while. It's actually related to renormalization,
which is secretly the same subject as this "critical point" business
I just mentioned. But, it's not about sand piles - it's about quantum
gravity!

In "week222" I spoke about the work of Lauscher and Reuter, who claim to
have found evidence for an ultraviolet fixed point in quantum gravity
without matter. In other words, as you zoom in closer and closer, they
claim quantum gravity without matter acts more and more like some fixed
theory. This would be big news: it would suggest that gravity without
matter is a sensible theory, contrary to what everyone in string theory
says!

Not surprisingly, the string theorist Jacques Distler examined Lauscher
and Reuter's work with a critical eye. And, he wrote up a nice
explanation of the problems with their work:

16) Jacques Distler, Unpleasantness,
http://golem.ph.utexas.edu/~distler/blog/archives/000648.html

Briefly, the problem is that Lauscher and Reuter make a drastic
equation", which is a beautiful thing: it says how a Lagrangian
for a field theory at one length scale gives rise to an effective
Lagrangian for the same theory at a larger length scale. However,
then they truncate the incredibly complicated formula for a fully
general Lagrangian, restricting to Lagrangians with only an
Einstein-Hilbert term and a cosmological constant. Like Distler,
I see no reason to think this approximation is valid. So, their
claimed ultraviolet fixed point could be an artifact of their method.

Whether it's worth going further and checking this by considering
a slightly less brutal approximation, using Lagrangians with a few
more terms, is a matter of taste. Distler doesn't think so. I
hope Lauscher and Reuter do. If they don't, we may never know for
sure what happens. I think it's actually rather amazing that they
get an fixed point with their brutal approximation, instead of
coupling constants that run to infinity or zero, which is what
I would have naively expected. But who knows? Maybe this is
easily understood if you think hard enough.

Today I was also going to talk about the 3-strand braid group, the
group PSL(2,Z), and rational tangles, but now I don't have the energy.
So instead, I'll just put out a request for help!

There's a wonderful game invented by John Conway called "rational
tangles". Here's how it works. It involves two players and a referee.

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

This is called "position 0". The referee then cries out either "add one!"
or "take the negative reciprocal!". If the referee yells "add one!",
player B has to switch which hand he's using to hold which rope, making
sure to pass the right one over the left, like this:

A A
\ /
/
/ \
B B

This is called "position 1", since we started with "position 0" and
then did "add one!". But if the referee had said "take the inverse
reciprocal!" both players must cooperate to move all four ends of
the ropes a quarter-turn clockwise, like this:

A A
\_/
_
/ \
B B

This is called "position -1/0", since we started with 0 and then
did "take the negative reciprocal!".

The referee keeps crying "add one!" or "take the negative reciprocal!"
in whatever order she feels like, and players A and B keep doing the
same sort of thing: either player B switches the ropes right over left,
or both players rotate the whole tangle a quarter-turn clockwise. It's
reciprocal!", since some people refuse to divide by zero, for religious
reasons.

Anyway, after a while the ropes are all tangled up and the rope is
in "position p/q" for some complicated rational number p/q. It will
be all tangled up - but in a special way, called a "rational tangle".

Then the players have to *undo* the tangling and get back to "position 0".
They may not remember the exact sequence of moves that got them into
the mess they are in. In fact the game is much more fun if they *don't*
remember. It's best to do it at a party, possibly after a few drinks.

Luckily, any sequence of "add one!" and "take the negative reciprocal!"
moves that carry their number back to 0, will carry their tangle back
to "position 0". So they just need to figure out how to get their
number back to 0, and the tangle will automatically untangle itself.
That's the cool part! It's a highly nonobvious theorem due to Conway.

I'm vaguely aware of a few proofs of this fact. As far as I know,
Conway's original proof uses the Alexander-Conway polynomial:

16) John Horton Conway, An enumeration of knots and links and some
of their algebraic properties, in Computational Problems of Abstract
Algebra, ed. D. Welsh, Pergamon Press, New York, 1970, pp. 329-358.

There's also a proof by Goldman and Kauffman using the Jones polynomial:

17) Jay R. Goldman and Louis H. Kauffman, Rational tangles, Advances in
Applied Mathematics 18 (1997), 300-332. Also available at
http://www.math.uic.edu/~kauffman/RTang.pdf

There are also two proofs in here:

18) Louis H. Kauffman and Sofia Lambropoulou, On the classification of
rational tangles, available as math.GT/0311499.

But here's what I want to know: is there a proof that makes
extensive use of the group PSL(2,Z) and its relation to topology?

After all, 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. Since PSL(2,Z) has
lots of relations to topology, there should be some proof of Conway's
theorem that *uses* these relations to get the job done.

Does anybody know one?

Finally, the answer to the psychic powers puzzle: if you did the
calculation right, you got the last digit of the book's ISBN number -
unless your answer was 10, in which case the ISBN number should end
in the letter X.

This trick is called a "check sum" or "check digit": it's a way to spot
errors. The Universal Product Code, used in those bar codes you see
everywhere, also has a check digit. So do credit cards.

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2. Mar 21, 2006

### Tim S

on 20/03/2006 3:39 am, John Baez at baez@math.removethis.ucr.andthis.edu
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>

> If the referee yells "add one!",
> player B has to switch which hand he's using to hold which rope, making
> sure to pass the right one over the left, like this:
>
> A A
> \ /
> /
> / \
> B B
>
> This is called "position 1", since we started with "position 0" and
> then did "add one!". But if the referee had said "take the inverse
> reciprocal!" both players must cooperate to move all four ends of
> the ropes a quarter-turn clockwise, like this:
>
> A A
> \_/
> _
> / \
> B B
>
> This is called "position -1/0", since we started with 0 and then
> did "take the negative reciprocal!".
>

<stuff>

>
> But here's what I want to know: is there a proof that makes
> extensive use of the group PSL(2,Z) and its relation to topology?
>
> After all, 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. Since PSL(2,Z) has
> lots of relations to topology, there should be some proof of Conway's
> theorem that *uses* these relations to get the job done.
>
> Does anybody know one?

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)

| | |
| | |
--------y=1-------W------+------X------
| | |
| | |
--------y=1/2-----+------+------+------
|B_l |B_r |
| | |
--------y=0-------Z------+------Y------
|A_l |A_r |
| | |
x=0 x=1/2 x=1

and now let's act on them with the generators of PSL(2, Z) and mod out by
the lattice.

So z -> -1/z rotates anticlockwise 90 degrees, like in the original tangle
scenario, (or was it clockwise? It doesn't make any material difference, I
hope). This lands A and B in a different cell of the lattice, to the left of
ours, so we translate back to our own cell. This puts them in the bottom
right corner, but, in our torus the right edge _is_ the left edge, so they
end up back in the bottom left where they started. In fact, A and B change
places in the latter operation, so it turns out to be a clockwise rotation
after all...

OK, so now what do we get with the shear corresponding to

z -> z+1 ?

A is unaffected, since the x axis is left alone, while B slides 1/2 a unit
to the right. This puts B's left hand at (1/2, 1/2) and B's right hand at
(1, 1/2), on the right edge of the cell. But the right edge is the left
edge. So B's right hand ends up at the left. In other words, A has stayed
the same but B has switched hands. Which is exactly what happens in the
tangle game.

In our quotiented version of the plane, we basically have lost all the
'non-commutative' information about the number of twists (except whether
it's odd or even), but if we stay with the original full plane, then W will
actually move 1 place to the right for every vertical twist we put in.

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). We can imagine the ends of the
strings sitting in little wheels that roll along the lines of integer x or y
coordinate which join the lattice points, making one half-turn for each
increment of 1 in the coordinate.

Since W starts at (0, 1), and 0 and 1 are coprime (duh!), and since both the
generators of PSL(2, Z) preserve the gcd of the coordinates, the final
coordinates of W will be also coprime. So they will uniquely determine a ray
(through the origin), the x-coordinate of whose intersection with the line
y = 1 is precisely the fraction of the tangle. I think. Up to possible signs
and reciprocals, anyway.

The actual proof of Conway's theorem is left as an exercise for the

Tim

3. Mar 27, 2006

### Tim S

on 21/03/2006 2:32 am, Tim S at Tim@timsilverman.demon.co.uk 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@math.removethis.ucr.andthis.edu
> 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>
>

>> 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

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 + 1) * q = q

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

(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]

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