Notes for AKT-090917-1/0:23:37 - Revision history

Iva at 02:38, 5 September 2011

2011-09-05T02:38:06Z

← Older revision		Revision as of 22:38, 4 September 2011
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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>.

	Let <math>\mathcal{D}_m = \mathcal{K}_m / ( \~~mbox{over crossing}~~=\~~mbox{under crossing}~~)</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.		Hence, the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined.

Conan777 at 01:00, 19 September 2009

2009-09-19T01:00:01Z

← Older revision		Revision as of 21:00, 18 September 2009
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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>

⚫	~~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>.
			Given <math>V</math> of type <math>m</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>.		Let <math>\mathcal{D}_m = \mathcal{K}_m / ( \mbox{over crossing}=\mbox{under crossing})</math>.

	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.

Conan777 at 00:47, 19 September 2009

2009-09-19T00:47:59Z

New page

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>.
Hence the '''weight system''' <math> \mathcal{D}_m \rightarrow A</math> given by <math>W_V = V^{(m)}</math> is well-defined.