Talk:06-1350/Class Notes for Thursday September 28: Difference between revisions
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I'm putting this here but it really belongs to Tuesday Oct 3rd... I could'n figure out how to access that day. |
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This is about the exercise to show that for the tautological Z, |
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Z{Ribbon knots} = {<math>u\gamma</math>:<math>\gamma \in K(O-O-...-O)</math> and d<math>\gamma</math> = {O O ...O} |
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(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) |
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The left included in the right part is obvvious. |
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We want to show the reverse. |
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For this, suppose <math>\Psi \in A(O-O-...-O), d\Psi \in Z(O O...O)</math>, |
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and suppose further that <math>\Psi=Z\gamma</math> for some <math>\gamma \in K(O-...-O)</math> |
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We want to show that in this case, <math>d\gamma=O O ...O</math>. |
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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>, |
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<math>Z\delta=Z(O O...O)</math>, then <math>\delta=O O...O</math>. |
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To do this, we use two invariants. |
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The first one will be the product over all connected components of: 2 if the component is an unknotted circle; 0 otherwise. |
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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. |
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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>. |
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We have one assumption to eliminate, and I'm not quite sure how to do this: |
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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. |
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