User:Sankaran/06-1350-HW4

The Generators

Our generators are [math]\displaystyle{ T }[/math], [math]\displaystyle{ R }[/math], [math]\displaystyle{ \Phi }[/math] and [math]\displaystyle{ B^{\pm} }[/math]:

Picture
Generator	[math]\displaystyle{ T }[/math]	[math]\displaystyle{ R }[/math]	[math]\displaystyle{ \Phi }[/math]	[math]\displaystyle{ B^+ }[/math]	[math]\displaystyle{ B^- }[/math]
Perturbation	[math]\displaystyle{ t }[/math]	[math]\displaystyle{ r }[/math]	[math]\displaystyle{ \varphi }[/math]	[math]\displaystyle{ b^+ }[/math]	[math]\displaystyle{ b^- }[/math]

(Thanks Zavosh for the nice picture)

The Relations

The Reidemeister Move R2

(Courtesy of Andy)

In formulas, this is

[math]\displaystyle{ 1 = (123)^\star B^- (132)^\star B^+. }[/math]

Linearized and written in functional form, this becomes

[math]\displaystyle{ \rho_2(x_1,x_2,x_3) = - b^-(x_1,x_2,x_3) - b^+(x_1,x_3,x_2). }[/math]

The Reidemeister Move R3

(Picture and first example courtesy of Dror)

There are eight of these (each crossing in the picture can be + or - ). For example, if all the crossings are positive, the picture (with three sides of the shielding removed) is

In formulas, this is

[math]\displaystyle{ (1230)^\star B^+ (1213)^\star B^+ (1023)^\star B^+ = (1123)^\star B^+ (1203)^\star B^+ (1231)^\star B^+ }[/math].

Linearized and written in functional form, this becomes

[math]\displaystyle{ \rho_3[+++](x_1, x_2, x_3, x_4) = }[/math]	[math]\displaystyle{ b^+(x_1,x_2,x_3) + b^+(x_1+x_3,x_2,x_4) + b^+(x_1,x_3,x_4) }[/math]
	[math]\displaystyle{ - b^+(x_1+x_2,x_3,x_4) - b^+(x_1,x_2,x_4) - b^+(x_1+x_4,x_2,x_3). }[/math]

Here are the rest of them, linearized and in functional form - I think this is too many, but it's probably easier to write these out than to figure the relationships between them. Also, some better notation is needed.

[math]\displaystyle{ \rho_3[++-](x_1,x_2,x_3,x_4) = b^+(x_1,x_2,x_3) + b^+(x_1+x_3,x_2,x_4) + b^-(x_1,x_3,x_4) - b^-(x_1+x_2,x_3,x_4) - b^+(x_1,x_2,x_4) - b^+(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[+-+](x_1,x_2,x_3,x_4) = b^+(x_1,x_2,x_3) + b^-(x_1+x_3,x_2,x_4) + b^+(x_1,x_3,x_4)- b^+(x_1+x_2,x_3,x_4) - b^-(x_1,x_2,x_4) - b^+(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[-++](x_1,x_2,x_3,x_4) = b^-(x_1,x_2,x_3) + b^+(x_1+x_3,x_2,x_4) + b^+(x_1,x_3,x_4)- b^+(x_1+x_2,x_3,x_4) - b^+(x_1,x_2,x_4) - b^-(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[+--](x_1,x_2,x_3,x_4) = b^+(x_1,x_2,x_3) + b^-(x_1+x_3,x_2,x_4) + b^-(x_1,x_3,x_4)- b^-(x_1+x_2,x_3,x_4) - b^-(x_1,x_2,x_4) - b^+(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[-+-](x_1,x_2,x_3,x_4) = b^-(x_1,x_2,x_3) + b^+(x_1+x_3,x_2,x_4) + b^-(x_1,x_3,x_4)- b^-(x_1+x_2,x_3,x_4) - b^+(x_1,x_2,x_4) - b^-(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[--+](x_1,x_2,x_3,x_4) = b^-(x_1,x_2,x_3) + b^-(x_1+x_3,x_2,x_4) + b^+(x_1,x_3,x_4)- b^+(x_1+x_2,x_3,x_4) - b^-(x_1,x_2,x_4) - b^-(x_1+x_4,x_2,x_3). }[/math]

[math]\displaystyle{ \rho_3[---](x_1,x_2,x_3,x_4) = b^-(x_1,x_2,x_3) + b^-(x_1+x_3,x_2,x_4) + b^-(x_1,x_3,x_4)- b^-(x_1+x_2,x_3,x_4) - b^-(x_1,x_2,x_4) - b^-(x_1+x_4,x_2,x_3). }[/math]

The Reidemeister Move R4

(Courtesy of Andy)

There are two (ostensibly) different versions:

In formulas, this is

[math]\displaystyle{ (1230)^\star B^+ (1213)^\star B^+ (1023)^\star \Phi = (1123)^\star \Phi (1233)^\star B^+ }[/math].

Linearized and written in functional form, this becomes

[math]\displaystyle{ \rho_{4a}(x_1,x_2,x_3,x_4) = b^+(x_1,x_2,x_3) + b^+(x_1+x_3,x_2,x_4) + \phi(x_1,x_3,x_4) - \phi(x_1+x_2,x_3,x_4) - b^+(x_1,x_2,x_3+x_4). }[/math]

Second:

In formulas, this is

[math]\displaystyle{ (1123)^\star B^+ (1203)^\star B^+ (1231)^\star \Phi = (1230)^\star \Phi (1223)^\star B^+ }[/math].

Linearized and written in functional form, this becomes

[math]\displaystyle{ \rho_{4b}(x_1,x_2,x_3,x_4) = b^+(x_1+x_2,x_3,x_4) + b^+(x_1,x_2,x_4) + \phi(x_1+x_4,x_2,x_3) - \phi(x_1,x_2,x_3) - b^+(x_1,x_2+x_3,x_4). }[/math]

The Syzygies

The "B around B" Syzygy

The picture, with all shielding removed, is

(Drawn with Inkscape) (note that lower quality pictures are also acceptable)

The functional form of this syzygy is

[math]\displaystyle{ BB(x_1,x_2,x_3,x_4,x_5) = }[/math]	[math]\displaystyle{ \rho_3(x_1, x_2, x_3, x_5) + \rho_3(x_1 + x_5, x_2, x_3, x_4) - \rho_3(x_1 + x_2, x_3, x_4, x_5) }[/math]
	[math]\displaystyle{ - \rho_3(x_1, x_2, x_4, x_5) - \rho_3(x_1 + x_4, x_2, x_3, x_5) - \rho_3(x_1, x_2, x_3, x_4) }[/math]
	[math]\displaystyle{ + \rho_3(x_1, x_3, x_4, x_5) + \rho_3(x_1 + x_3, x_2, x_4, x_5). }[/math]

A Mathematica Verification

The following simulated Mathematica session proves that for our single relation and single syzygy, [math]\displaystyle{ d^2=0 }[/math]. Copy paste it into a live Mathematica session to see that it's right!

In[1]:=

d1 = { rho3[x1_, x2_, x3_, x4_] :> bp[x1, x2, x3] + bp[x1 + x3, x2, x4] + bp[x1, x3, x4] - bp[x1 + x2, x3, x4] - bp[x1, x2, x4] - bp[x1 + x4, x2, x3] }; d2 = { BAroundB[x1_, x2_, x3_, x4_, x5_] :> rho3[x1, x2, x3, x5] + rho3[x1 + x5, x2, x3, x4] - rho3[x1 + x2, x3, x4, x5] - rho3[x1, x2, x4, x5] - rho3[x1 + x4, x2, x3, x5] - rho3[x1, x2, x3, x4] + rho3[x1, x3, x4, x5] + rho3[x1 + x3, x2, x4, x5] };

In[3]:=	BAroundB[x1, x2, x3, x4, x5] /. d2
Out[3]=	- rho3[x1, x2, x3, x4] + rho3[x1, x2, x3, x5] - rho3[x1, x2, x4, x5] + rho3[x1, x3, x4, x5] - rho3[x1 + x2, x3, x4, x5] + rho3[x1 + x3, x2, x4, x5] - rho3[x1 + x4, x2, x3, x5] + rho3[x1 + x5, x2, x3, x4]

In[4]:=	BAroundB[x1, x2, x3, x4, x5] /. d2 /. d1
Out[4]=	0