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

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===The Relations===
===The Relations===

====The Reidemeister Move R2 (Andy's)====
The following version of R2 was the easiest to use to build my [[media:06-1350-PhiAroundPhi.png|original <math>\Phi</math> around <math>\Phi</math> syzygy]]:
[[Image:06-1350-R2-weird.png|center]]
In formulas, this is
<center><math>1 = (123)^\star B^- (132)^\star B^+.</math></center>
Linearized and written in functional form, this becomes
{| align=center
|-
|<math>\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====
====The Reidemeister Move R3====
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====R4====
====R4====


This Reidemeister move has a number of forms. I will put two here, both in linearized functional form. Pictures to come I hope.
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.




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R4d
R4d

[[Image:R4d.JPG]]
{| align=center
|-
|<math>\rho_4d(x_1, x_2, x_3, x_4) = </math>
|<math>b^-(x_1,x_2,x_3) + \phi(x_1+x_2,x_3+x_4,x_4)</math>
|-
⚫
|
|<math>- \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).</math>
|}

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===
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|<math>+ \rho_3(x_1, x_3, x_4, x_5) + \rho_3(x_1 + x_3, x_2, x_4, x_5).</math>
|<math>+ \rho_3(x_1, x_3, x_4, x_5) + \rho_3(x_1 + x_3, x_2, x_4, x_5).</math>
|}
|}

====The "<math>\Phi</math> around B" Syzygy (By Andy)====

The picture, with all shielding (and any other helpful notations) removed, is
{| align=center
|- align=center
|[[Image:06-1350-PhiAroundB.png|center]]
|-
|align=right|(Drawn with [http://asymptote.sf.net/ Asymptote], [[06-1350/Syzygies in Asymptote|Syzygies in Asymptote]])
|}

The functional form of this syzygy is

{| align=center
|-
|<math>\Phi B(x_1,x_2,x_3,x_4,x_5) = </math>
|<math>\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)</math>
|-
|
|<math>- \rho_3(x_1,x_2,x_3+x_4,x_5) - \rho_{4a}(x_1,x_2,x_3,x_4)</math>
|-
|
|<math>- \rho_{4b}(x_1,x_3,x_4,x_5) + \rho_3(x_1+x_3,x_2,x_4,x_5).</math>
|}

====Phi around Phi====


[[Image:Phiphi.svg]]



===A Mathematica Verification===
===A Mathematica Verification===

Latest revision as of 23:14, 5 December 2006

The Generators

Our generators are , , and :

Picture Phi.PNG 06-1350-BPlus.svg
Generator
Perturbation

The Relations

The Reidemeister Move R2 (Andy's)

The following version of R2 was the easiest to use to build my original around syzygy:

06-1350-R2-weird.png

In formulas, this is

Linearized and written in functional form, this becomes


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

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

R4c.JPG



R4d

R4d.JPG

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 (By Andy)

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

06-1350-PhiAroundB.png
(Drawn with Asymptote, Syzygies in Asymptote)

The functional form of this syzygy is

Phi around Phi

Phiphi.svg


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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