User:Shawkm/06-1350-HW4 - Revision history

Shawkm at 00:38, 3 December 2006

2006-12-03T00:38:06Z

← Older revision		Revision as of 20:38, 2 December 2006
Line 111:		Line 111:


	What I will do now is linearize and write down in functional form the relation R2		What I will do now is linearize and write down in functional form the relation R2. We know R2 correspnds to <math>(123)^\star B^\pm\cdot(132)^\star B^\mp=1_3</math>. Rewriting it gives,

	{\| align=center		{\| align=center

Shawkm at 00:34, 3 December 2006

2006-12-03T00:34:05Z

← Older revision		Revision as of 20:34, 2 December 2006
Line 109:		Line 109:

	This is probably a completely trivial remark, and it is also trivial to see that mathematica will deal with this in the same way as for <math> b^+ </math> so I won't verify that <math> d^2 =0 </math>.		This is probably a completely trivial remark, and it is also trivial to see that mathematica will deal with this in the same way as for <math> b^+ </math> so I won't verify that <math> d^2 =0 </math>.


			What I will do now is linearize and write down in functional form the relation R2

			{\| align=center
			\|-
			\|<math>\rho_2(x_1, x_2, x_3) = </math>
			\|<math>b^+(x_1,x_2,x_3) + b^-(x_1,x_3,x_2)</math>
			\|}

			or

			{\| align=center
			\|-
			\|<math>\rho_2(x_1, x_2, x_3) = </math>
			\|<math>b^-(x_1,x_2,x_3) + b^+(x_1,x_3,x_2)</math>
			\|}

Shawkm at 00:29, 3 December 2006

2006-12-03T00:29:43Z

← Older revision		Revision as of 20:29, 2 December 2006
Line 97:		Line 97:
	===Here Goes Something New(?)===		===Here Goes Something New(?)===

	The first thing to notice is that the relation <math>\rho_3 </math> holds for <math> b^+ <\math> and <math> b^- <\math> so we have another version:		The first thing to notice is that the relation <math>\rho_3 </math> holds for <math> b^+ </math> and <math> b^- </math> so we have another version:

	{\| align=center		{\| align=center
	\|-		\|-
Line 107:		Line 108:
	\|}		\|}

	This is probably a completely trivial remark, and it is also trivial to see that mathematica will deal with this in the same way as for <math> b^+ <\math> so I won't verify that <math> d^2 =0 <\math>.~~</math>~~		This is probably a completely trivial remark, and it is also trivial to see that mathematica will deal with this in the same way as for <math> b^+ </math> so I won't verify that <math> d^2 =0 </math>.

Shawkm at 00:27, 3 December 2006

2006-12-03T00:27:18Z

← Older revision		Revision as of 20:27, 2 December 2006
Line 92:		Line 92:

	{{InOut\|n=4\|in=<nowiki>BAroundB[x1, x2, x3, x4, x5] /. d2 /. d1</nowiki>\|out=<nowiki>0</nowiki>}}		{{InOut\|n=4\|in=<nowiki>BAroundB[x1, x2, x3, x4, x5] /. d2 /. d1</nowiki>\|out=<nowiki>0</nowiki>}}



			===Here Goes Something New(?)===

			The first thing to notice is that the relation <math>\rho_3 </math> holds for <math> b^+ <\math> and <math> b^- <\math> so we have another version:
			{\| align=center
			\|-
			\|<math>\rho_3(x_1, x_2, x_3, x_4) = </math>
			\|<math>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>- 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>
			\|}

			This is probably a completely trivial remark, and it is also trivial to see that mathematica will deal with this in the same way as for <math> b^+ <\math> so I won't verify that <math> d^2 =0 <\math>.</math>

Shawkm at 00:19, 3 December 2006

2006-12-03T00:19:09Z

New page

===The Generators===

Our generators are <math>T</math>, <math>R</math>, <math>\Phi</math> and <math>B^{\pm}</math>:
{| align=center cellpadding=10 style="border: solid orange 1px"
|- align=center valign=middle
|align=left|Picture
|
|
|
|[[Image:06-1350-BPlus.svg|100px]]
|
|- align=center valign=middle
|align=left|Generator
|<math>T</math>
|<math>R</math>
|<math>\Phi</math>
|<math>B^+</math>
|<math>B^-</math>
|- align=center valign=middle
|align=left|Perturbation
|<math>t</math>
|<math>r</math>
|<math>\varphi</math>
|<math>b^+</math>
|<math>b^-</math>
|}

===The Relations===

====The Reidemeister Move R3====
The picture (with three sides of the shielding removed) is
[[Image:06-1350-R4.svg|400px|center]]
In formulas, this is
<center><math>(1230)^\star B^+ (1213)^\star B^+ (1023)^\star B^+ = (1123)^\star B^+ (1203)^\star B^+ (1231)^\star B^+</math>.</center>
Linearized and written in functional form, this becomes
{| align=center
|-
|<math>\rho_3(x_1, x_2, x_3, x_4) = </math>
|<math>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>- 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 Syzygies===

====The "B around B" Syzygy====

The picture, with all shielding removed, is
{| align=center
|- align=center
|[[Image:06-1350-BAroundB.svg|center]]
|-
|align=right|(Drawn with [http://www.inkscape.org/ Inkscape])<br>(note that lower quality pictures are also acceptable)
|}

The functional form of this syzygy is

{| align=center
|-
|<math>BB(x_1,x_2,x_3,x_4,x_5) = </math>
|<math>\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>- \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>+ \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>d^2=0</math>. Copy paste it into a live Mathematica session to see that it's right!

{{In|n=1|in=<nowiki>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]
};</nowiki>}}

{{InOut|n=3|in=<nowiki>BAroundB[x1, x2, x3, x4, x5] /. d2</nowiki>|out=<nowiki>- 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]</nowiki>}}

{{InOut|n=4|in=<nowiki>BAroundB[x1, x2, x3, x4, x5] /. d2 /. d1</nowiki>|out=<nowiki>0</nowiki>}}