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)
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?
