# User:Jana/06-1350-HW4

### The Generators

Our generators are ${\displaystyle T}$, ${\displaystyle R}$, ${\displaystyle \Phi }$ and ${\displaystyle B^{\pm }}$:

 Picture Generator ${\displaystyle T}$ ${\displaystyle R}$ ${\displaystyle \Phi }$ ${\displaystyle B^{+}}$ ${\displaystyle B^{-}}$ Perturbation ${\displaystyle t}$ ${\displaystyle r}$ ${\displaystyle \varphi }$ ${\displaystyle b^{+}}$ ${\displaystyle b^{-}}$

### The Relations

#### The Reidemeister Move R2 (Andy's)

The following version of R2 was the easiest to use to build my original ${\displaystyle \Phi }$ around ${\displaystyle \Phi }$ syzygy:

In formulas, this is

${\displaystyle 1=(123)^{\star }B^{-}(132)^{\star }B^{+}.}$

Linearized and written in functional form, this becomes

 ${\displaystyle \rho _{2}(x_{1},x_{2},x_{3})=-b^{-}(x_{1},x_{2},x_{3})-b^{+}(x_{1},x_{3},x_{2}).}$

#### The Reidemeister Move R3

The picture (with three sides of the shielding removed) is

In formulas, this is

${\displaystyle (1230)^{\star }B^{+}(1213)^{\star }B^{+}(1023)^{\star }B^{+}=(1123)^{\star }B^{+}(1203)^{\star }B^{+}(1231)^{\star }B^{+}}$.

Linearized and written in functional form, this becomes

 ${\displaystyle \rho _{3}(x_{1},x_{2},x_{3},x_{4})=}$ ${\displaystyle b^{+}(x_{1},x_{2},x_{3})+b^{+}(x_{1}+x_{3},x_{2},x_{4})+b^{+}(x_{1},x_{3},x_{4})}$ ${\displaystyle -b^{+}(x_{1}+x_{2},x_{3},x_{4})-b^{+}(x_{1},x_{2},x_{4})-b^{+}(x_{1}+x_{4},x_{2},x_{3}).}$

#### R4

This Reidemeister move has a number of forms. I will put two here, both in linearized functional form. The two following were copied from Andy.

R4c

 ${\displaystyle \rho _{4}c(x_{1},x_{2},x_{3},x_{4})=}$ ${\displaystyle b^{+}(x_{1},x_{2},x_{3})+\phi (x_{1}+x_{2},x_{3}+x_{4},x_{4})}$ ${\displaystyle -\phi (x_{1}+x_{2}+x_{3},x_{3}+x_{4},x_{4})-b^{+}(x_{1},x_{2},x_{4})-b^{+}(x_{1}+x_{4},x_{3}+x_{4},x_{4}).}$

R4d

 ${\displaystyle \rho _{4}d(x_{1},x_{2},x_{3},x_{4})=}$ ${\displaystyle b^{-}(x_{1},x_{2},x_{3})+\phi (x_{1}+x_{2},x_{3}+x_{4},x_{4})}$ ${\displaystyle -\phi (x_{1}+x_{2}+x_{3},x_{3}+x_{4},x_{4})-b^{-}(x_{1},x_{2},x_{4})-b^{-}(x_{1}+x_{4},x_{3}+x_{4},x_{4}).}$

To establish the syzygy below, I needed two versions of R4. First:

In formulas, this is

${\displaystyle (1230)^{\star }B^{+}(1213)^{\star }B^{+}(1023)^{\star }\Phi =(1123)^{\star }\Phi (1233)^{\star }B^{+}}$.

Linearized and written in functional form, this becomes

 ${\displaystyle \rho _{4a}(x_{1},x_{2},x_{3},x_{4})=b^{+}(x_{1},x_{2},x_{3})+b^{+}(x_{1}+x_{3},x_{2},x_{4})+\phi (x_{1},x_{3},x_{4})-\phi (x_{1}+x_{2},x_{3},x_{4})-b^{+}(x_{1},x_{2},x_{3}+x_{4}).}$

Second:

In formulas, this is

${\displaystyle (1123)^{\star }B^{+}(1203)^{\star }B^{+}(1231)^{\star }\Phi =(1230)^{\star }\Phi (1223)^{\star }B^{+}}$.

Linearized and written in functional form, this becomes

 ${\displaystyle \rho _{4b}(x_{1},x_{2},x_{3},x_{4})=b^{+}(x_{1}+x_{2},x_{3},x_{4})+b^{+}(x_{1},x_{2},x_{4})+\phi (x_{1}+x_{4},x_{2},x_{3})-\phi (x_{1},x_{2},x_{3})-b^{+}(x_{1},x_{2}+x_{3},x_{4}).}$

Are these independent, or can they be shown to be equivalent using other relations?

### 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

 ${\displaystyle BB(x_{1},x_{2},x_{3},x_{4},x_{5})=}$ ${\displaystyle \rho _{3}(x_{1},x_{2},x_{3},x_{5})+\rho _{3}(x_{1}+x_{5},x_{2},x_{3},x_{4})-\rho _{3}(x_{1}+x_{2},x_{3},x_{4},x_{5})}$ ${\displaystyle -\rho _{3}(x_{1},x_{2},x_{4},x_{5})-\rho _{3}(x_{1}+x_{4},x_{2},x_{3},x_{5})-\rho _{3}(x_{1},x_{2},x_{3},x_{4})}$ ${\displaystyle +\rho _{3}(x_{1},x_{3},x_{4},x_{5})+\rho _{3}(x_{1}+x_{3},x_{2},x_{4},x_{5}).}$

#### The "${\displaystyle \Phi }$ around B" Syzygy (By Andy)

The picture, with all shielding (and any other helpful notations) removed, is

 (Drawn with Asymptote, Syzygies in Asymptote)

The functional form of this syzygy is

 ${\displaystyle \Phi B(x_{1},x_{2},x_{3},x_{4},x_{5})=}$ ${\displaystyle \rho _{3}(x_{1},x_{2},x_{3},x_{5})+\rho _{4a}(x_{1}+x_{5},x_{2},x_{3},x_{4})+\rho _{4b}(x_{1}+x_{2},x_{3},x_{4},x_{5})}$ ${\displaystyle -\rho _{3}(x_{1},x_{2},x_{3}+x_{4},x_{5})-\rho _{4a}(x_{1},x_{2},x_{3},x_{4})}$ ${\displaystyle -\rho _{4b}(x_{1},x_{3},x_{4},x_{5})+\rho _{3}(x_{1}+x_{3},x_{2},x_{4},x_{5}).}$

### A Mathematica Verification

The following simulated Mathematica session proves that for our single relation and single syzygy, ${\displaystyle d^{2}=0}$. 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