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

Karenechu: /* The Operation unzip */

2007-01-12T01:55:11Z

The Operation unzip

← Older revision		Revision as of 21:55, 11 January 2007
Line 166:		Line 166:
	Linearized and in functional form:		Linearized and in functional form:

	<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2,x_3,x_2)+\phi(x_1,x_2,x_3)+\phi+(x_1+x_3,x_3,x_2)==0 </math>		<math>\phi(x_1,x_2,x_3)+\phi(x_1+x_2,x_3,x_2)+\phi(x_1,x_2,x_3)+\phi+(x_1+x_3,x_3,x_2)==0 </math>

	The part I don't quite understand:		The part I don't quite understand:

Karenechu: /* The Operation delete */

2007-01-12T01:52:35Z

The Operation delete

← Older revision		Revision as of 21:52, 11 January 2007
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	====The Operation delete====		====The Operation delete====

			Warning: sorry, I must be doing something wrong to obtain 0 on the LHS--since there are chords ending on the edge to be deleted, but non-zero on the RHS.


	As above, <math> d^A_e (Z (\gamma))= Z( d_e (\gamma))</math>:		As above, <math> d^A_e (Z (\gamma))= Z( d_e (\gamma))</math>:

128.100.59.68: /* The Relations */

2007-01-12T01:49:29Z

The Relations

← Older revision		Revision as of 21:49, 11 January 2007
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	The part I don't quite understand:		The part I don't quite understand:
	When we unzip an edge with chords ending on it, we need to sum up all combinations of individual chords ending on the 2 unzipped edges; but here we have been treating the chords ending on an edge as just one single collection, so I only summed up the 2 ways of the collection ending on the unzipped edges		When we unzip an edge with chords ending on it, we need to sum up all combinations of individual chords ending on the 2 unzipped edges; but here we have been treating the chords ending on an edge as just one single collection, so I only summed up the 2 ways of the collection ending on the unzipped edges

			====The Operation delete====

			As above, <math> d^A_e (Z (\gamma))= Z( d_e (\gamma))</math>:



			[[Image:del.jpg]]

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

128.100.59.68: /* The Operation unzip */

2007-01-12T01:46:58Z

The Operation unzip

← Older revision		Revision as of 21:46, 11 January 2007
Line 168:		Line 168:
	<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2,x_3,x_2)+\phi(x_1,x_2,x_3)+\phi+(x_1+x_3,x_3,x_2)==0 </math>		<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2,x_3,x_2)+\phi(x_1,x_2,x_3)+\phi+(x_1+x_3,x_3,x_2)==0 </math>

			The part I don't quite understand:
	When we unzip an edge with chords ending on it, we need to sum up all combinations of chords ending on the 2 unzipped edges, but we ~~look~~ at the chords ending on a edge of a ~~shield~~ ~~collectively~~, so I only summed up the 2 ways of the collection ~~of chords~~ ending on the unzipped edges.		When we unzip an edge with chords ending on it, we need to sum up all combinations of individual chords ending on the 2 unzipped edges; but here we have been treating the chords ending on an edge as just one single collection, so I only summed up the 2 ways of the collection ending on the unzipped edges

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

128.100.59.68: /* The Operation unzip */

2007-01-12T01:44:34Z

The Operation unzip

← Older revision		Revision as of 21:44, 11 January 2007
Line 162:		Line 162:
	[[Image:unzip.jpg]]		[[Image:unzip.jpg]]

	(I chose escapes routes for the chords ending on a shield first, then unzip in the space A, and this should equal to Z of the unzipped knot projection.)		(on the LHS,I chose escapes routes for the chords ending on any shield first, then unzip in the space A (here I am not sure if I handled it correctly), and this should equal to the LHS: Z of the unzipped knot projection.)

	Linearized and in functional form:		Linearized and in functional form:

	<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2+x_3,x_3,x_2)=0 </math>		<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2,x_3,x_2)+\phi(x_1,x_2,x_3)+\phi+(x_1+x_3,x_3,x_2)==0 </math>

			When we unzip an edge with chords ending on it, we need to sum up all combinations of chords ending on the 2 unzipped edges, but we look at the chords ending on a edge of a shield collectively, so I only summed up the 2 ways of the collection of chords ending on the unzipped edges.

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

128.100.59.68: /* The Operation unzip */

2007-01-12T01:38:43Z

The Operation unzip

← Older revision		Revision as of 21:38, 11 January 2007
Line 156:		Line 156:
	Warning: I am not sure if this is correct, but thought it better to post than not post.		Warning: I am not sure if this is correct, but thought it better to post than not post.

	We have found the <math> u^A </math>, <math> d^A </math>,<math> #^A </math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u</math>,<math> d</math>, <math> #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>		We have found the <math> u^A </math>, <math> d^A </math>,<math> \#^A </math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u</math>,<math> d</math>, <math> \#</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>:


Line 162:		Line 162:
	[[Image:unzip.jpg]]		[[Image:unzip.jpg]]

			(I chose escapes routes for the chords ending on a shield first, then unzip in the space A, and this should equal to Z of the unzipped knot projection.)


	This means:
	<math>(123)^\star B^+ (132)^\star B^- = 1</math>

	Linearized and in functional form:		Linearized and in functional form:

	<math>\~~rho_2~~(x_1,x_2,x_3)~~=b^~~+(x_1,x_2,x_3~~)+b^-(x_1~~,x_3,x_2) </math>		<math>\phi(x_1,x_2,x_3)+\phi+(x_1+x_2+x_3,x_3,x_2)=0 </math>

	And we get the other R2 by switching both crossings, i.e. switching b^+ and b^-:

	<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 Syzygies===		===The Syzygies===

128.100.59.68: /* The Operation unzip */

2007-01-12T01:32:05Z

The Operation unzip

← Older revision		Revision as of 21:32, 11 January 2007
Line 156:		Line 156:
	Warning: I am not sure if this is correct, but thought it better to post than not post.		Warning: I am not sure if this is correct, but thought it better to post than not post.

	We have found the <math> u^A , d^A ,#^A </math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u, d, #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>		We have found the <math> u^A </math>, <math> d^A </math>,<math> #^A </math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u</math>,<math> d</math>, <math> #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>

128.100.59.68: /* The Operation unzip */

2007-01-12T01:31:02Z

The Operation unzip

← Older revision		Revision as of 21:31, 11 January 2007
Line 156:		Line 156:
	Warning: I am not sure if this is correct, but thought it better to post than not post.		Warning: I am not sure if this is correct, but thought it better to post than not post.

	We have found the <math>u^A,d^A,#^A</math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u, d, #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>		We have found the <math> u^A , d^A ,#^A </math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u, d, #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>

128.100.59.68: /* The Operation unzip */

2007-01-12T01:30:06Z

The Operation unzip

← Older revision		Revision as of 21:30, 11 January 2007
Line 154:		Line 154:
	====The Operation unzip====		====The Operation unzip====

	~~warning~~: I am not sure if this is correct, but thought it better to post it ~~anyways~~.		Warning: I am not sure if this is correct, but thought it better to post than not post.

			We have found the <math>u^A,d^A,#^A</math> operations in the space <math>A(\Tau)</math> corresponding to the <math>u, d, #</math>, in the space of <math>K(\Tau) </math>, so that for the instance <math> u^A_e (Z (\gamma))= Z( u_e (\gamma))</math>

128.100.59.68: /* The Operation unzip */

2007-01-12T01:13:34Z

The Operation unzip

@@ Line 153: / Line 153: @@
 ====The Operation unzip====
 [[Image:unzip.jpg]]
 ===The Syzygies===