Drorbn - User contributions [en]

Notes for Dancso-120602/0:00:56

2012-06-04T03:25:04Z

Zsuzsi:

My handout is [http://www.math.toronto.edu/zsuzsi/research/hyperplanes.pdf here].

[[image:hyperplanes.jpg|center]]

File:Hyperplanes.jpg

2012-06-04T03:24:19Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-05T23:48:57Z

Zsuzsi:

===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>
|}

A low-tech completed version of this chart:

[[Image:chart.jpg]]

===The Relations===

====The Symmetry of B====

To eliminate the choice involved in placing a B at a crossing, it has to have 180 degrees rotational symmetry. This yields the following picture:

[[Image:Symm1.jpg]]

The relation cannot be written in the first notation, as on the right side the chords ending on different red lines could end up on the same pink line.

In the linearized functional notation though we can express this:

<math>s_1(x_1,x_2,x_3)=b^+(x_1,x_2,x_3)-b^+(-x_1-x_2-x_3,-x_3,-x_2)</math>

Explanation: on the right side, chords on the first red line can drop off on either the third, second or the first strand, morover, the orders are reversed, hence the minus signs.

The same picture for B^- yields:

<math>s_2(x_1,x_2,x_3)=b^-(x_1,x_2,x_3)-b^-(-x_1-x_2-x_3,-x_3,-x_2)</math>

====The symmetry of <math>\Phi</math>====

<math>\Phi</math> has to have A(4)-symmetry. For example, rotation around the "top" vertex yields the following picture and relation:

[[Image:Symm3.jpg]]

The same explanation goes here, and we get the relation:

<math>s_3(x_1,x_2,x_3)=\varphi(x_1,x_2,x_3)-\varphi(-x_1-x_2,-x_2-x_3,x_2)</math>

And here is the full picture and the relations (I suppose it would be enough to take a few that generate A(4), but we're on the safe side writing all these up... and I kind of got into drawing tetrahedrons.):

[[Image:Tetrahedrons.jpg]]

<math>s_3(x_1,x_2,x_3)=\varphi(x_1,x_2,x_3)-\varphi(-x_1-x_2,-x_2-x_3,x_2)</math>

<math>s_4(x_1,x_2,x_3)=\varphi(-x_1-x_2,-x_2-x_3,x_2)-\varphi(x_1+x_2,x_3,-x_2-x_3)</math>

<math>s_5(x_1,x_2,x_3)=\varphi(x_1+x_2,x_3,-x_2-x_3)-\varphi(x_2,-x_1-x_2,x_1+x_2+x_3)</math>

<math>s_6(x_1,x_2,x_3)=\varphi(x_2,-x_1-x_2,x_1+x_2+x_3)-\varphi(-x_1-x_2,x_1,x_2+x_3)</math>

<math>s_7(x_1,x_2,x_3)=\varphi(-x_1-x_2,x_1,x_2+x_3)-\varphi(x_2+x_3,-x_3,-x_1-x_2)</math>

<math>s_8(x_1,x_2,x_3)=\varphi(x_2+x_3,-x_3,-x_1-x_2)-\varphi(-x_1-x_2-x_3,-x_2,x_1+x_2)</math>

<math>s_9(x_1,x_2,x_3)=\varphi(-x_1-x_2-x_3,-x_2,x_1+x_2)-\varphi(-x_2,x_2+x_3,-x_1-x_2-x_3)</math>

<math>s_{10}(x_1,x_2,x_3)=\varphi(-x_2,x_2+x_3,-x_1-x_2-x_3)-\varphi(-x_2-x_3,-x_1,x_1+x_2)</math>

<math>s_{11}(x_1,x_2,x_3)=\varphi(-x_2-x_3,-x_1,x_1+x_2)-\varphi(-x_1,x_1+x_2+x_3,-x_3)</math>

<math>s_{12}(x_1,x_2,x_3)=\varphi(-x_1,x_1+x_2+x_3,-x_3)-\varphi(x_3,-x_1-x_2-x_3,x_1)</math>

<math>s_{13}(x_1,x_2,x_3)=\varphi(x_3,-x_1-x_2-x_3,x_1)-\varphi(-x_3,-x_2,-x_1)</math>

====The Reidemeister move R1====

[[Image:Reidemeister1.jpg]]

As with the symmetry relations, we cannot write this one in the first notation either.

In the linearized functional notation, it looks like this:

<math> \rho_1(x_1,x_2)=b^-(x_1,x_2,-x_2)</math>

Where the negative sign is because the order of the chords is reversed as we slide them along the little loop.

====The Reidemeister move R2====

With three sides of the shielding removed, the picture is:
[[Image:Reidemeister2.jpg]]

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

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>

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 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 Reidemeister Move R4, source:Andy====
First version of R4:
[[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 version:
[[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>
|}

===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>
|}

====The "<math>\Phi</math> around B" Syzygy- I copy-pasted this from Andy, as well as R4====

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>
|}

====The "<math>\Phi</math> around <math>\Phi</math>" Syzygy -also taken from Andy====

note: I've changed Andy's notation to fit my version of R2.

The picture is
{| align=center
|- align=center
|[[Image:06-1350-PhiAroundPhi.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\Phi(x_1,x_2,x_3,x_4,x_5) = </math>
|<math>-\rho_2'(x_1+x_2,x_3,x_4) - \rho_2'(x_1+x_2+x_4,x_3,x_5) + \rho_{4b}(x_1+x_2,x_4,x_5,x_3)</math>
|-
|
|<math>- \rho_2'(x_1,x_2,x_4) - \rho_2'(x_1+x_4,x_2,x_5) + \rho_{4b}(x_1,x_4,x_5,x_2)</math>
|-
|
|<math>+ \rho_{4a}(x_1,x_4+x_5,x_2,x_3) - \rho_{4b}(x_1,x_4,x_5,x_2+x_3) - \rho_{4a}(x_1+x_4,x_5,x_2,x_3)</math>
|-
|
|<math>+ \rho_2'(x_1+x_4,x_2,x_5) + \rho_2'(x_1+x_2+x_4,x_3,x_5) - \rho_{4a}(x_1,x_4,x_2,x_3)</math>
|-
|
|<math>+ \rho_2'(x_1,x_2,x_4) + \rho_2'(x_1+x_2,x_3,x_4)</math>
|}

Note that the first and last terms cancel, as the two steps at the top of the diagram are opposites.

===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>}}

File:Tetrahedrons.jpg

2006-12-05T23:22:44Z

Zsuzsi:

06-1350/Homework Assignment 4

2006-12-05T19:01:02Z

Zsuzsi:

{{06-1350/Navigation}}

'''This assignment is due on Tuesday, December 5 2006.'''

This is an unusual assignment; the task at hand is to do some real research, stuff that to the best of my knowledge had never been done before and most definitely was never written up. Thus the rules will also be a bit different - your work (or at least the accumulation of work on this topic by everyone in class) is meant to be used and useful. So it must be presented in a very readable form (i.e., typed up and with figures) and it must be reliable; in fact, it will be computer verifiable. But some rules will be relaxed, as well.

The task is a bit technical. But hey, it is a homework assignment, after all!

==The Task==

Write all the relations between <math>T</math>, <math>R</math>, <math>\Phi</math> and <math>B^{\pm}</math> in a completely explicit way, both as formulas and as illuminating figures, and then do the same to all the syzygies between these relations. Finally, enter everything you have written into a Mathemmatica script that will verify that for the complex you have created, <math>d^2=0</math>.

Note that when I write "all relations" or "all syzygies" above I really mean "a complete independent set of relations/syzygies". And while this cannot be formalized, ''the prettier your representatives are, the better!''

==The Rules==

The first relation and the first syzygy were written by {{Dror}} (see below or visit [[User:Drorbn/06-1350-HW4]]). You are to copy his work into your user space and complete it there.
* On this wiki create a page named "User:YourUsernameHere/06-1350-HW4" (or simply "User:YourUsernameHere/HW4"). If necessary, go to [[Help:Contents]] to see how this is done.
* Copy [[User:Drorbn/06-1350-HW4]] into your page. The easiest way to do that is to edit [[User:Drorbn/06-1350-HW4]] and copy the source code into your page using copy-paste on your windowing system. Then "preview" or "save" your copy but "cancel" the edit to [[User:Drorbn/06-1350-HW4]].
* Now work on your page...
* Copying is legal! You are allowed, indeed encouraged, to collaborate with others or to simply copy results from other people's pages into yours. The goal is to get something complete. If one of you will start with something incomplete and somebody else will do some other incomplete thing and yet another person will merge the two, we may achieve the goal.
* If you copy, always credit the original source! Likewise, if I will ever use any of the material that will be first produced here, I am committed to crediting the source(s).
* You will get a good though not perfect grade on this assignment for doing anything at all, or even for doing nothing at all but copying on the understanding that by submitting your work you are testifying that you understand it. Perfect grades will go to the people who will make substantial contributions.
* Elegance counts! Beauty counts! A systematic approach counts!

==A Bonus Question==

Find the definitive completion of the silliest proof for the existence of exponentials. In other words, find the definitive proof that if <math>M(u,v)</math> is a two-variable power series for which <math>M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)=0</math>, the there exists a single-variable power series <math>\epsilon(t)</math> for which <math>M(u,v)=\epsilon(v)-\epsilon(u+v)+\epsilon(u)</math>.

==User:Drorbn/06-1350-HW4==

{{User:Drorbn/06-1350-HW4}}

==Links==
In order to make it easier for us to see each others work, and not all work on the same parts of the assignment I suggest that you can link here to your assignment page. You can also say what you have worked out there and what you are planing on working on.

My page is [[User:Jana/06-1350-HW4]]

[[User:Shawkm/06-1350-HW4]]

[[User:Andy/06-1350-HW4]]

[[User:Zak/06-1350-HW4]]

[[User:zsuzsi/HW4]] (worked out some relations and copied Andy's work)

06-1350/Homework Assignment 4

2006-12-05T19:00:23Z

Zsuzsi:

{{06-1350/Navigation}}

'''This assignment is due on Tuesday, December 5 2006.'''

This is an unusual assignment; the task at hand is to do some real research, stuff that to the best of my knowledge had never been done before and most definitely was never written up. Thus the rules will also be a bit different - your work (or at least the accumulation of work on this topic by everyone in class) is meant to be used and useful. So it must be presented in a very readable form (i.e., typed up and with figures) and it must be reliable; in fact, it will be computer verifiable. But some rules will be relaxed, as well.

The task is a bit technical. But hey, it is a homework assignment, after all!

==The Task==

Write all the relations between <math>T</math>, <math>R</math>, <math>\Phi</math> and <math>B^{\pm}</math> in a completely explicit way, both as formulas and as illuminating figures, and then do the same to all the syzygies between these relations. Finally, enter everything you have written into a Mathemmatica script that will verify that for the complex you have created, <math>d^2=0</math>.

Note that when I write "all relations" or "all syzygies" above I really mean "a complete independent set of relations/syzygies". And while this cannot be formalized, ''the prettier your representatives are, the better!''

==The Rules==

The first relation and the first syzygy were written by {{Dror}} (see below or visit [[User:Drorbn/06-1350-HW4]]). You are to copy his work into your user space and complete it there.
* On this wiki create a page named "User:YourUsernameHere/06-1350-HW4" (or simply "User:YourUsernameHere/HW4"). If necessary, go to [[Help:Contents]] to see how this is done.
* Copy [[User:Drorbn/06-1350-HW4]] into your page. The easiest way to do that is to edit [[User:Drorbn/06-1350-HW4]] and copy the source code into your page using copy-paste on your windowing system. Then "preview" or "save" your copy but "cancel" the edit to [[User:Drorbn/06-1350-HW4]].
* Now work on your page...
* Copying is legal! You are allowed, indeed encouraged, to collaborate with others or to simply copy results from other people's pages into yours. The goal is to get something complete. If one of you will start with something incomplete and somebody else will do some other incomplete thing and yet another person will merge the two, we may achieve the goal.
* If you copy, always credit the original source! Likewise, if I will ever use any of the material that will be first produced here, I am committed to crediting the source(s).
* You will get a good though not perfect grade on this assignment for doing anything at all, or even for doing nothing at all but copying on the understanding that by submitting your work you are testifying that you understand it. Perfect grades will go to the people who will make substantial contributions.
* Elegance counts! Beauty counts! A systematic approach counts!

==A Bonus Question==

Find the definitive completion of the silliest proof for the existence of exponentials. In other words, find the definitive proof that if <math>M(u,v)</math> is a two-variable power series for which <math>M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)=0</math>, the there exists a single-variable power series <math>\epsilon(t)</math> for which <math>M(u,v)=\epsilon(v)-\epsilon(u+v)+\epsilon(u)</math>.

==User:Drorbn/06-1350-HW4==

{{User:Drorbn/06-1350-HW4}}

==Links==
In order to make it easier for us to see each others work, and not all work on the same parts of the assignment I suggest that you can link here to your assignment page. You can also say what you have worked out there and what you are planing on working on.

My page is [[User:Jana/06-1350-HW4]]

[[User:Shawkm/06-1350-HW4]]

[[User:Andy/06-1350-HW4]]

[[User:Zak/06-1350-HW4]]

[[User:zsuzsi/HW4]]
(worked out some relations and copied Andy's work)

06-1350/Homework Assignment 4

2006-12-05T18:58:26Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-05T18:55:56Z

Zsuzsi:

===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>
|}

A low-tech completed version of this chart:

[[Image:chart.jpg]]

===The Relations===

====The Symmetry of B====

To eliminate the choice involved in placing a B at a crossing, it has to have 180 degrees rotational symmetry. This yields the following picture:

[[Image:Symm1.jpg]]

The relation cannot be written in the first notation, as on the right side the chords ending on different red lines could end up on the same pink line.

In the linearized functional notation though we can express this:

<math>s_1(x_1,x_2,x_3)=b^+(x_1,x_2,x_3)-b^+(-x_1-x_2-x_3,-x_3,-x_2)</math>

Explanation: on the right side, chords on the first red line can drop off on either the third, second or the first strand, morover, the orders are reversed, hence the minus signs.

The same picture for B^- yields:

<math>s_2(x_1,x_2,x_3)=b^-(x_1,x_2,x_3)-b^-(-x_1-x_2-x_3,-x_3,-x_2)</math>

====The symmetry of <math>\Phi</math>====

<math>\Phi</math> has to have A(4)-symmetry. A(4) is generated by 120 degree rotations around the vertices of the tetrahedron. For example, rotation around the "top" vertex yields the following picture and relation:

[[Image:Symm3.jpg]]

The same explanation goes here, and we get the relation:

<math>s_3(x_1,x_2,x_3)=\varphi(x_1,x_2,x_3)-\varphi(-x_1-x_2,-x_2-x_3,x_2)</math>

====The Reidemeister move R1====

[[Image:Reidemeister1.jpg]]

As with the symmetry relations, we cannot write this one in the first notation either.

In the linearized functional notation, it looks like this:

<math> \rho_1(x_1,x_2)=b^-(x_1,x_2,-x_2)</math>

Where the negative sign is because the order of the chords is reversed as we slide them along the little loop.

====The Reidemeister move R2====

With three sides of the shielding removed, the picture is:
[[Image:Reidemeister2.jpg]]

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

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>

====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 Reidemeister Move R4, source:Andy====
First version of R4:
[[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 version:
[[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>
|}

===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>
|}

====The "<math>\Phi</math> around B" Syzygy- I copy-pasted this from Andy, as well as R4====

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>
|}

===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>}}

User:Zsuzsi/HW4

2006-12-05T18:51:00Z

Zsuzsi:

File:Symm3.jpg

2006-12-05T18:31:51Z

Zsuzsi:

File:Symm1.jpg

2006-12-05T18:31:13Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-04T21:43:44Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-04T21:15:43Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-04T21:13:51Z

Zsuzsi:

File:Reidemeister2.jpg

2006-12-04T21:07:57Z

Zsuzsi:

File:Reidemeister1.jpg

2006-12-04T21:07:29Z

Zsuzsi:

File:Chart.jpg

2006-12-04T21:06:42Z

Zsuzsi:

User:Zsuzsi/HW4

2006-12-04T18:44:23Z

Zsuzsi:

06-1350/Class Photo

2006-10-04T22:37:48Z

Zsuzsi: /* Who We Are */

{{06-1350/Navigation}}

Our class on September 28, 2006:

[[Image:06-1350-ClassPhoto.jpg|thumb|center|500px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
|}

{| align=center border=1
|-
{{Photo Entry|last=Dancso|first=Zsuzsi|userid=Zsuzsi|email=zsuzsi@ math.toronto.edu|location=first row, second from the right|comments=Oops. I can't align this with the first row.}}
|}

06-1350/Class Notes for Tuesday October 3

2006-10-04T20:29:11Z

Zsuzsi:

=== Attempted solution of the exercise ===

I'm talking about the one stating that

Z{Ribbon knots} = {<math>u\Psi: \Psi \in A(O-O-...-O), d\Psi = Z(O O...O)</math>}

''my dumbbel didn't come out perfectly, and there might be further ugliness due to my lack of latex ability- I apologize''

The inclusion of the left in the right is obvious. We're trying to do the reverse one.

Assume that we have a <math>\Psi \in A(O-...-O)</math> such that <math>d\Psi=Z(O...O)</math>.

What we need to show is that <math>\Psi=Z(\gamma)</math> for some <math>\gamma \in K(O-...-O),~ d\gamma=(O...O)</math>.

To do this, we first prove that <math>d:A(O-...-O) \to A(O...O)</math> is (almost) injective.

Let's just do this for a simple (two circles) dumbbell, it will generalize to longer dumbbells easily.

Injectivity means that for some <math>\Psi \in A(O-O)</math>, if <math>d\Psi \in A(O O)</math> is zero on all
invariants <math>\varphi \in Inv ~ K(O O)</math>, then <math>\Psi</math> has to be zero on all invariants in <math>Inv ~ K(O-O)</math>.

We use that <math>d \Psi ( \varphi)=\Psi ( \varphi \circ d)</math>

The key here is that the edge connecting the two circles carries no topological information, it's contractible. Therefore, an invariant of the knotted dumbbell cannot depend on how the edge wiggles- we can contract the edge to a point and get a vedge of two knotted circles, and there's only one way to vedge circles together. Therefore, all invariants of the knottings of the dumbbell are of the form <math>\varphi \circ d</math>, where <math>\varphi</math> is an invariant of the knottings of two circles. And hence we're done.

Of course there's a lie here: we are talking about framed graphs and so the edge does carry some information: which side it's attached to on each circle, and whether it's twisted or not. So there are <math>2^3=8</math> preimages of each value of d. (Still a lie: If the circles are linked, I think it's countably many, given by the sides and how many times the edge is twisted, but we only care about the unlinked case, and there two twists equal nothing.)

Now, still thinking about a simple dumbbell, we can show that <math>\Psi=Z\gamma</math> for some knotted dumbbell <math>\gamma</math>.

We have <math>d:A(O-O) \to A(O O), ~ d(\Psi)=Z(O O)</math>.
We know that <math>Z(O O)</math> has 8 preimages, and all of those are Z-images of the dumbbells we get from the 8 different ways of attaching d (think connected sum). Therefore, one of those has to be the <math>\gamma</math> we are searching for.

Clearly, this works just as well for the dumbbell with k circles, with <math>2^{3(k-1)}</math> in place of 8.

And obviously, for the <math>\gamma</math> we've found, <math>d\gamma=(O...O)</math>.

And we are done.

''Yesterday I posted something stupid about this same problem, then realized and took it off, so I'm not sure anymore that this really works... any comments welcome!
''

Talk:06-1350/Class Notes for Thursday September 28

2006-10-03T22:36:24Z

Zsuzsi:

Talk:06-1350/Class Notes for Thursday September 28

2006-10-03T22:25:25Z

Zsuzsi:

I'm putting this here but it really belongs to Tuesday Oct 3rd... I could'n figure out how to access that day.

This is about the exercise to show that for the tautological Z,
Z{Ribbon knots} = {<math>u\gamma</math>:<math>\gamma \in K(O-O-...-O)</math> and d<math>\gamma</math> = {O O ...O}

''(the dumbbell didn't come out perfectly but you know what I mean...I apologize for any further ugliness resulting from my un-knowledge of latex)''

The left included in the right part is obvvious.
We want to show the reverse.
For this, suppose <math>\Psi \in A(O-O-...-O), d\Psi \in Z(O O...O)</math>,
and suppose further that <math>\Psi=Z\gamma</math> for some <math>\gamma \in K(O-...-O)</math>
We want to show that in this case, <math>d\gamma=O O ...O</math>.

Now <math>d\Psi=dZ\gamma=Zd\gamma=Z(O O...O)</math> so we need to show that if for some knotted garph <math>\delta</math>,
<math>Z\delta=Z(O O...O)</math>, then <math>\delta=O O...O</math>.
To do this, we use two invariants.

The first one will be the product over all connected components of: 2 if the component is an unknotted circle; 0 otherwise.
Z(O O...O) of this invariant is <math>2^k</math>, and for <math>\delta</math> to have the same, all components of it have to be unknotted circles, and it has to have the same number of components.

What is left is to make sure those circles are not linked. For this we can take our second invariant to be the sum over all pairs of connected componets the absolute value of their linking number. By playing the same game as above, we have now proved that <math>\delta=O O ...O</math>.

We have one assumption still to eliminate, and I'm not quite sure how to do this:
We have assumed that <math>\Psi</math> was the Z-image of some knotted version of the dumbbell graph. <math>\Psi</math> is a set of values of all imaginable knot invariants on knottings of the dumbbell, and we would have to find an actual knotting on which the invariants take exactly these values- does anyone know how to do this? It surprises me that it can be done.

Talk:06-1350/Class Notes for Thursday September 28

2006-10-03T22:24:45Z

Zsuzsi:

I'm putting this here but it really belongs to Tuesday Oct 3rd... I could'n figure out how to access that day.

This is about the exercise to show that for the tautological Z,
Z{Ribbon knots} = {<math>u\gamma</math>:<math>\gamma \in K(O-O-...-O)</math> and d<math>\gamma</math> = {O O ...O}

''(the dumbbell didn't come out perfectly but you know what I mean...I apologize for any further ugliness resulting from my un-knowledge of latex)''

The left included in the right part is obvvious.
We want to show the reverse.
For this, suppose <math>\Psi \in A(O-O-...-O), d\Psi \in Z(O O...O)</math>,
and suppose further that <math>\Psi=Z\gamma</math> for some <math>\gamma \in K(O-...-O)</math>
We want to show that in this case, <math>d\gamma=O O ...O</math>.

Now <math>d\Psi=dZ\gamma=Zd\gamma=Z(O O...O)</math> so we need to show that if for some knotted garph <math>\delta</math>,
<math>Z\delta=Z(O O...O)</math>, then <math>\delta=O O...O</math>.
To do this, we use two invariants.

The first one will be the product over all connected components of: 2 if the component is an unknotted circle; 0 otherwise.
Z(O O...O) of this invariant is <math>2^k</math>, and for <math>\delta</math> to have the same, all components of it have to be unknotted circles, and it has to have the same number of components.

What is left is to make sure those circles are not linked. For this we can take our second invariant to be the sum over all pairs of connected componets the absolute value of their linking number. By playing the same game as above, we have now proved that <math>\delta=O O ...O</math>.

We have one assumption to eliminate, and I'm not quite sure how to do this:
We have assumed that <math>\Psi</math> was the Z-image of some knotted version of the dumbbell graph. <math>\Psi</math> is a set of values of all imaginable knot invariants on knottings of the dumbbell, and we would have to find an actual knotting on which the invariants take exactly these values- does anyone know how to do this? It surprises me that it can be done.

Talk:06-1350/Class Notes for Thursday September 28

2006-10-03T22:24:08Z

Zsuzsi:

I'm putting this here but it really belongs to Tuesday Oct 3rd... I could'n figure out how to access that day.

This is about the exercise to show that for the tautological Z,
Z{Ribbon knots} = {<math>u\gamma</math>:<math>\gamma \in K(O-O-...-O)</math> and d<math>\gamma</math> = {O O ...O}

(the dumbbell didn't come out perfectly but you know what I mean...I apologize for any further ugliness resulting from my un-knowledge of latex)

The left included in the right part is obvvious.
We want to show the reverse.
For this, suppose <math>\Psi \in A(O-O-...-O), d\Psi \in Z(O O...O)</math>,
and suppose further that <math>\Psi=Z\gamma</math> for some <math>\gamma \in K(O-...-O)</math>
We want to show that in this case, <math>d\gamma=O O ...O</math>.

Now <math>d\Psi=dZ\gamma=Zd\gamma=Z(O O...O)</math> so we need to show that if for some knotted garph <math>\delta</math>,
<math>Z\delta=Z(O O...O)</math>, then <math>\delta=O O...O</math>.
To do this, we use two invariants.

The first one will be the product over all connected components of: 2 if the component is an unknotted circle; 0 otherwise.
Z(O O...O) of this invariant is <math>2^k</math>, and for <math>\delta</math> to have the same, all components of it have to be unknotted circles, and it has to have the same number of components.

What is left is to make sure those circles are not linked. For this we can take our second invariant to be the sum over all pairs of connected componets the absolute value of their linking number. By playing the same game as above, we have now proved that <math>\delta=O O ...O</math>.

We have one assumption to eliminate, and I'm not quite sure how to do this:
We have assumed that <math>\Psi</math> was the Z-image of some knotted version of the dumbbell graph. <math>\Psi</math> is a set of values of all imaginable knot invariants on knottings of the dumbbell, and we would have to find an actual knotting on which the invariants take exactly these values- does anyone know how to do this? It surprises me that it can be done.