Bowling balls

I'm quixotically trying to prove the volume conjecture using my bowling ball representation of braid groups.

N is a number.
q is e^(2 pi i / N)
[k] is 1 + ... + q^(k-1).
[k]! is [1]*...*[k]
[k choose r] is [k]! / [r]![k-r]!

We have a positive braid, made of lanes of a bowling alley. Each lane can hold up to N-1 balls. The chance c balls fall when a pass over b is f^a_b(c) =

[a choose c] * [N-1-b choose c] * [c]! * (1-q)^c * q^(a-c)(N-1-b-c).

I want to let N go to infinity, and have "liquid bowling balls".

So, what's the argument of f^a_b(c)?

• [k] = sqrt(q)^k-1 |[k]|
• [k]! = sqrt(q)^kchoose2 |[k]!|
• [k choose r] = sqrt(q)^r(k-r) |[k choose r]|
• 1-q = sqrt(-q) |1-q| = sqrt(q)^(1 - N/2) |1-q|

Altogether I get the fourth root of q to the power of

c(2b + c + 3) - a(4b + 2c + 4) + cN    (mod 4N)

Now let's figure out the change of argument when you move one strand from the right to the left side of a whole region of the braid diagram.

For a crossing on the right, a and c go down

2a + 2b + 1 - N 

On the left, b goes up and c goes down

-2a - 2b - 3 - N

On the top, a and c go up, b goes down

2a - 2b - 2c - 1 + N

On the bottom, c goes up

2pi(-c/N - 1/4)

-2a + 2b + 2c + 3 + N

--------------------------------------------------------------------
c(a-c)theta + c(N-b-c)theta + c-choose-2 theta + (theta-pi/2)c + 2(a-c)(N-b-c)theta
= pi/2N times -4ab - 2ac + 4aN + 2bc + c^2 - 3cN + c

Now let's figure out the change of argument when you move one strand from the right to the left side of a whole region of the braid diagram.

For a crossing on the right of the region, a and c goes down and b stays constant. So let's take a kind of negative derivative:

pi times -1/2 + a/N + b/N - 1/2N

For a crossing on the left side of the region, b goes up and c goes down.

pi times 3/2 - a/N - b/N - 1/2N

For the crossing on the bottom, c goes up.

pi times -3/2 - a/N + b/N + c/N + 1/2N

For the crossing on the top, a goes up, b goes down, c goes up.

pi times 1/2 + a/N - b/N - c/N + 1/2N

We want the derivative on each face to be zero. Ignore the 1/N terms.


For sigma1 sigma2 sigma1, with a,b,c at the bottom, d,e,f at top, x,y,z in the middle reading from bottom to top, I get

x = -3/2 + b + e + f
y = 3/2 + d - b - c
z = -3/2 + b + c + e

For sigma2 sigma1 sigma2

x = -1/2 + b + d + e
y = 1/2 + f - a - b
z = -1/2 + a + b + e

The fudges seem to be wrong. Call them L,R,T,B.

sigma1 sigma2 sigma1:
x = b+e+f + (R+T+B)
y = a-e-f - (R+T+B)
z = e+b+c + (R+T+B)

sigma2 sigma1 sigma2:
x = b+d+e - (T+L+B)
y = f-a-b + (T+L+B)
z = a+b+e - (T+L+B)

It all seems to work if T+B = LR = 0.

There must be a mistake.

Let [N]=0, and a'=a-c, b'=b+c. I get
f^a_b(c) = [a choose a'] [b' - 1]! (1-q)^(b'-b) q^(-a'b') / [b-1]!
Does that help?