Notes for AKT-090917-1/0:23:37: Difference between revisions

Latest revision as of 21:38, 4 September 2011

Let ${\mathcal {K}}_{m}=\{m$ -singular knots $\}$

Given $V$ of type $m$ , We have $V^{(m)}:{\mathcal {K}}_{m}\rightarrow A$ .

Since $V^{(m+1)}=0$ , $V^{(m)}$ does not distinguish over crossing and under crossings in ${\mathcal {K}}_{m}$ .

Let Failed to parse (unknown function "\overcrossing"): {\displaystyle \mathcal{D}_m = \mathcal{K}_m / (\overcrossing=\undercrossing)} .

Hence, the weight system ${\mathcal {D}}_{m}\rightarrow A$ given by $W_{V}=V^{(m)}$ is well-defined.

@@ Line 1: / Line 1: @@
 Let <math>\mathcal{K}_m = \{ m</math>-singular knots <math>\}</math>
-We have <math>V^{(m)}: \mathcal{K}_m \rightarrow A</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>.
-Let <math>\mathcal{D}_m = \mathcal{K}_m / ( \mbox{over crossing}=\mbox{under crossing})</math>.
+Given <math>V</math> of type <math>m</math>, We have <math>V^{(m)}: \mathcal{K}_m \rightarrow A</math>.
-Hence the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined.
+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>.
+Let <math>\mathcal{D}_m = \mathcal{K}_m / (\overcrossing=\undercrossing)</math>.
+Hence, the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined.