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

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>

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
reader... :-(

Tim

.

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