Notes for AKT-090917-1/0:23:37: Difference between revisions
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Let <math>\mathcal{K}_m = \{ m</math>-singular knots <math>\}</math> |
Let <math>\mathcal{K}_m = \{ m</math>-singular knots <math>\}</math> |
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Given <math>V</math> of type <math>m</math>, We have <math>V^{(m)}: \mathcal{K}_m \rightarrow A</math>. |
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Let <math>\mathcal{D}_m = \mathcal{K}_m / ( \mbox{over crossing}=\mbox{under crossing})</math>. |
Let <math>\mathcal{D}_m = \mathcal{K}_m / ( \mbox{over crossing}=\mbox{under crossing})</math>. |
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Hence the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined. |
Hence the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined. |
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Revision as of 20:00, 18 September 2009
Let [math]\displaystyle{ \mathcal{K}_m = \{ m }[/math]-singular knots [math]\displaystyle{ \} }[/math]
Given [math]\displaystyle{ V }[/math] of type [math]\displaystyle{ m }[/math], We have [math]\displaystyle{ V^{(m)}: \mathcal{K}_m \rightarrow A }[/math].
Since [math]\displaystyle{ V^{(m+1)}=0 }[/math], [math]\displaystyle{ V^{(m)} }[/math] does not distinguish over crossing and under crossings in [math]\displaystyle{ \mathcal{K}_m }[/math].
Let [math]\displaystyle{ \mathcal{D}_m = \mathcal{K}_m / ( \mbox{over crossing}=\mbox{under crossing}) }[/math].
Hence the weight system [math]\displaystyle{ \mathcal{D}_m \rightarrow A }[/math] given by [math]\displaystyle{ W_V = V^{(m)} }[/math] is well-defined.
