Notes for AKT-090917-1/0:23:37: Difference between revisions
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Since <math>V^{(m+1)}=0</math>, <math>V^{(m)}</math> does not distinguish over crossing and under crossings in <math>\mathcal{K}_m</math>. |
Since <math>V^{(m+1)}=0</math>, <math>V^{(m)}</math> does not distinguish over crossing and under crossings in <math>\mathcal{K}_m</math>. |
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Let <math>\mathcal{D}_m = \mathcal{K}_m / ( |
Let <math>\mathcal{D}_m = \mathcal{K}_m / (\overcrossing=\undercrossing)</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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Latest revision as of 21:38, 4 September 2011
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 / (\overcrossing=\undercrossing) }[/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.
