Notes for AKT-090915/0:03:24 - Revision history

Iva at 17:36, 30 August 2011

2011-08-30T17:36:00Z

← Older revision		Revision as of 13:36, 30 August 2011
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	Review and additions to last class, corrections:		Review and additions to last class, corrections:
	# Jones polynomial is usually normalized by diving ~~the~~ '<>' of an unknot (i.e. ~~divide~~ by an additional factor of <math>d</math>).		# The Jones polynomial is usually normalized by diving by <math>\left\langle \bigcirc \right\rangle</math>, the bracket of the unknot (i.e. dividing by an additional factor of <math>d</math>).
	# We can prove ~~all~~ knot <math>K</math> ~~has~~ <math>J(K)</math> ~~being~~ a polynomial of <math>A^4</math>, ~~hence~~ we substitute <math>A=q^{1/4}</math>.		# We can prove that for any knot <math>K</math>, <math>J(K)</math> is a polynomial of <math>A^4</math>. Hence, we can substitute <math>A=q^{1/4}</math> to get a Laurent polynomial in <math>q</math>.

Drorbn at 10:21, 16 September 2009

2009-09-16T10:21:11Z

← Older revision		Revision as of 06:21, 16 September 2009
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			Review and additions to last class, corrections:
	~~Review and additions to last class, corrections: 1.~~ Jones polynomial is usually normalized by diving the '<>' of an unknot (i.e. divide by an additional factor of d) 2.~~We can prove all knot K has J(K) being a polynomial of A^4, hence we substitute A=q^{1/4}~~		# Jones polynomial is usually normalized by diving the '<>' of an unknot (i.e. divide by an additional factor of <math>d</math>).
			# We can prove all knot <math>K</math> has <math>J(K)</math> being a polynomial of <math>A^4</math>, hence we substitute <math>A=q^{1/4}</math>.

Conan777 at 02:18, 16 September 2009

2009-09-16T02:18:29Z

New page

Review and additions to last class, corrections: 1. Jones polynomial is usually normalized by diving the '<>' of an unknot (i.e. divide by an additional factor of d) 2.We can prove all knot K has J(K) being a polynomial of A^4, hence we substitute A=q^{1/4}