User:Andy/06-1350-HW4: Difference between revisions

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(Added Phi around B picture.)
(Adding R4 relations.)
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|<math>- 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>- 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 R3====
To establish the syzygy below, I needed two versions of R4. First:
[[Image:06-1350-R4a.png|center]]
In formulas, this is
<center><math>(1230)^\star B^+ (1213)^\star B^+ (1023)^\star \Phi = (1123)^\star \Phi (1233)^\star B^+</math>.</center>
Linearized and written in functional form, this becomes
{| align=center
|-
|<math>\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:
[[Image:06-1350-R4b.png|center]]
In formulas, this is
<center><math>(1123)^\star B^+ (1203)^\star B^+ (1231)^\star \Phi = (1230)^\star \Phi (1223)^\star B^+</math>.</center>
Linearized and written in functional form, this becomes
{| align=center
|-
|<math>\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>
|}

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


===The Syzygies===
===The Syzygies===

Revision as of 18:12, 3 December 2006

The Generators

Our generators are , , and :

Picture 06-1350-BPlus.svg
Generator
Perturbation

The Relations

The Reidemeister Move R3

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

06-1350-R4.svg

In formulas, this is

.

Linearized and written in functional form, this becomes

The Reidemeister Move R3

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

06-1350-R4a.png

In formulas, this is

.

Linearized and written in functional form, this becomes

Second:

06-1350-R4b.png

In formulas, this is

.

Linearized and written in functional form, this becomes

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

06-1350-BAroundB.svg
(Drawn with Inkscape)
(note that lower quality pictures are also acceptable)

The functional form of this syzygy is

The " around B" Syzygy

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

06-1350-PhiAroundB.png
(Drawn with Asymptote)

A Mathematica Verification

The following simulated Mathematica session proves that for our single relation and single syzygy, . 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
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