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|align=right|(Drawn with [http://asymptote.sf.net/ Asymptote]) |
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|align=right|(Drawn with [http://asymptote.sf.net/ Asymptote], [[06-1350/Syzygies in Asymptote|Syzygies in Asymptote]]) |
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Revision as of 14:29, 4 December 2006
The Generators
Our generators are 
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, 
and 
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| Generator
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| Perturbation
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The Relations
The Reidemeister Move R3
The picture (with three sides of the shielding removed) is
In formulas, this is
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Linearized and written in functional form, this becomes
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The Reidemeister Move R4
To establish the syzygy below, I needed two versions of R4. First:
In formulas, this is
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Linearized and written in functional form, this becomes
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Second:
In formulas, this is
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Linearized and written in functional form, this becomes
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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
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(Drawn with Inkscape)
(note that lower quality pictures are also acceptable)
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The functional form of this syzygy is
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The " around B" Syzygy
The picture, with all shielding (and any other helpful notations) removed, is
The functional form of this syzygy is
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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]:=
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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]
};
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In[3]:=
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BAroundB[x1, x2, x3, x4, x5] /. d2
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Out[3]=
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- 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]
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In[4]:=
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BAroundB[x1, x2, x3, x4, x5] /. d2 /. d1
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Out[4]=
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0
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