Notes for AKT-090917-1/0:46:11: Difference between revisions
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Let <math>K</math> be a knot, <math>J_K(q)</math> be its Jones polynomial. Substitute <math>q = e^x</math> and expand <math>J_k(e^x)</math> into power series. We have <math>J_K(e^x) = \sum_n j_{n,K} \ x^n</math> where the coefficients <math>j_{n,\cdot}: \{knots \} \rightarrow \mathbb{Z}</math> are knot invariants. |
Let <math>K</math> be a knot, <math>J_K(q)</math> be its Jones polynomial. Substitute <math>q = e^x</math> and expand <math>J_k(e^x)</math> into power series. We have <math>J_K(e^x) = \sum_n j_{(n,K)} \ x^n</math> where the coefficients <math>j_{(n,\cdot)}: \{knots \} \rightarrow \mathbb{Z}</math> are knot invariants. |
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Thm: <math>j_{n, \cdot}</math> is of type <math>n</math>. |
Thm: <math>j_{(n, \cdot)}</math> is of type <math>n</math>. |
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Latest revision as of 20:38, 18 September 2009
Let [math]\displaystyle{ K }[/math] be a knot, [math]\displaystyle{ J_K(q) }[/math] be its Jones polynomial. Substitute [math]\displaystyle{ q = e^x }[/math] and expand [math]\displaystyle{ J_k(e^x) }[/math] into power series. We have [math]\displaystyle{ J_K(e^x) = \sum_n j_{(n,K)} \ x^n }[/math] where the coefficients [math]\displaystyle{ j_{(n,\cdot)}: \{knots \} \rightarrow \mathbb{Z} }[/math] are knot invariants.
Thm: [math]\displaystyle{ j_{(n, \cdot)} }[/math] is of type [math]\displaystyle{ n }[/math].
