Notes for AKT-170113/0:50:48 - Revision history

https://drorbn.net/index.php?action=history&feed=atom&title=Notes_for_AKT-170113%2F0%3A50%3A48 Notes for AKT-170113/0:50:48 - Revision history 2026-05-08T01:45:52Z Revision history for this page on the wiki MediaWiki 1.39.6 https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15913&oldid=prev Rolandvdv at 13:33, 14 January 2017 2017-01-14T13:33:52Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:33, 14 January 2017</td> </tr><tr> <td colspan="2" class="diff-lineno">Line 1:</td> <td colspan="2" class="diff-lineno">Line 1:</td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math><del style="font-weight: bold; text-decoration: none;">R</del> = 1 + <del style="font-weight: bold; text-decoration: none;">hr</del> + \frac{1}{2!}h^<del style="font-weight: bold; text-decoration: none;">2r</del>^2</math> and working modulo <math >h^3 </math>.</div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math><ins style="font-weight: bold; text-decoration: none;">R_{ij}</ins> = 1 + <ins style="font-weight: bold; text-decoration: none;">hr_{ij}</ins> + \frac{1}{2!}h^<ins style="font-weight: bold; text-decoration: none;">2r_{ij}</ins>^2</math> and working modulo <math >h^3 </math>.</div></td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice I added the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just for fun. </div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Notice I added the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just for fun. </div></td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I wanted to test this idea in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I wanted to test this idea in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> </tr>  </table>

Rolandvdv https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15912&oldid=prev Rolandvdv at 13:18, 14 January 2017 2017-01-14T13:18:25Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:18, 14 January 2017</td> </tr><tr> <td colspan="2" class="diff-lineno">Line 1:</td> <td colspan="2" class="diff-lineno">Line 1:</td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </math>.</div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </math>.</div></td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>I <del style="font-weight: bold; text-decoration: none;">put</del> the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just <del style="font-weight: bold; text-decoration: none;">a hint of more to</del> <del style="font-weight: bold; text-decoration: none;">come</del>. </div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Notice </ins>I <ins style="font-weight: bold; text-decoration: none;">added</ins> the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just <ins style="font-weight: bold; text-decoration: none;">for</ins> <ins style="font-weight: bold; text-decoration: none;">fun</ins>. </div></td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>I <del style="font-weight: bold; text-decoration: none;">thought it was fun</del> to <del style="font-weight: bold; text-decoration: none;">have</del> <del style="font-weight: bold; text-decoration: none;">an</del> <del style="font-weight: bold; text-decoration: none;">example of this</del> in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>I <ins style="font-weight: bold; text-decoration: none;">wanted</ins> to <ins style="font-weight: bold; text-decoration: none;">test</ins> <ins style="font-weight: bold; text-decoration: none;">this</ins> <ins style="font-weight: bold; text-decoration: none;">idea</ins> in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to get an invariant in this case.</div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to get an invariant in this case.<ins style="font-weight: bold; text-decoration: none;"> </ins></div></td> </tr> <tr> <td colspan="2" class="diff-empty diff-side-deleted"></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Taking <math>\tilde{r}_{ij} = r_{ij}+r_{ji}</math> we may do a little better in that now the curl yields <math>1+ h(EF+FE+\frac{1}{2}H^2)+\mathcal{O}(h^2)</math></div></td> </tr> <tr> <td colspan="2" class="diff-empty diff-side-deleted"></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>indicating our ambiguity may now be a central element (the Casimir at order h).</div></td> </tr>  </table>

Rolandvdv https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15911&oldid=prev Rolandvdv at 13:09, 14 January 2017 2017-01-14T13:09:07Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:09, 14 January 2017</td> </tr><tr> <td colspan="2" class="diff-lineno">Line 2:</td> <td colspan="2" class="diff-lineno">Line 2:</td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to <del style="font-weight: bold; text-decoration: none;">fix</del> <del style="font-weight: bold; text-decoration: none;">it</del>.</div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to <ins style="font-weight: bold; text-decoration: none;">get</ins> <ins style="font-weight: bold; text-decoration: none;">an invariant in this case</ins>.</div></td> </tr>  </table>

Rolandvdv https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15910&oldid=prev Rolandvdv at 13:07, 14 January 2017 2017-01-14T13:07:37Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 09:07, 14 January 2017</td> </tr><tr> <td colspan="2" class="diff-lineno">Line 2:</td> <td colspan="2" class="diff-lineno">Line 2:</td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{ij} = E_iF_j + \frac{1}{4} H_iH_j</math> is a solution to CYBE.</div></td> </tr> <tr> <td colspan="2" class="diff-empty diff-side-deleted"></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>If I got it right the positive Reidemeister 1 curl yields the value <math>1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})</math> bad news, we need the element <math>S</math> to fix it.</div></td> </tr>  </table>

Rolandvdv https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15909&oldid=prev Rolandvdv at 12:22, 14 January 2017 2017-01-14T12:22:01Z

<p></p> <table style="background-color: #fff; color: #202122;" data-mw="interface"> <col class="diff-marker" /> <col class="diff-content" /> <col class="diff-marker" /> <col class="diff-content" /> <tr class="diff-title" lang="en"> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">← Older revision</td> <td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 08:22, 14 January 2017</td> </tr><tr> <td colspan="2" class="diff-lineno">Line 1:</td> <td colspan="2" class="diff-lineno">Line 1:</td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </math>.</div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </math>.</div></td> </tr> <tr> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> <td class="diff-marker"></td> <td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. </div></td> </tr> <tr> <td class="diff-marker" data-marker="−"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{<del style="font-weight: bold; text-decoration: none;">12</del>} = <del style="font-weight: bold; text-decoration: none;">E_1F_2</del> + \frac{1}{4} <del style="font-weight: bold; text-decoration: none;">H_1H_2</del></math> is a solution to CYBE.</div></td> <td class="diff-marker" data-marker="+"></td> <td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{<ins style="font-weight: bold; text-decoration: none;">ij</ins>} = <ins style="font-weight: bold; text-decoration: none;">E_iF_j</ins> + \frac{1}{4} <ins style="font-weight: bold; text-decoration: none;">H_iH_j</ins></math> is a solution to CYBE.</div></td> </tr>  </table>

Rolandvdv https://drorbn.net/index.php?title=Notes_for_AKT-170113/0:50:48&diff=15908&oldid=prev Rolandvdv: Created page with "{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </mat..." 2017-01-14T12:21:28Z

<p>Created page with "{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </mat..."</p> <p><b>New page</b></p><div>{{Roland}} At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting <math>R = 1 + hr + \frac{1}{2!}h^2r^2</math> and working modulo <math >h^3 </math>.<br /> I put the <math>h^2</math> term to make the inverse <math>R^{-1}</math> be identical but with negative <math>h</math>, the factorial is just a hint of more to come. <br /> I thought it was fun to have an example of this in <math>U(sl_2)</math> where you can check that <math>r_{12} = E_1F_2 + \frac{1}{4} H_1H_2</math> is a solution to CYBE.</div>

Rolandvdv