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Topic review (newest first)

Zhylliolom
2007-02-21 19:42:54

I realize the poster is probably long gone by now, but I am in a knot theory class currently and might as well post some content in this thread. I can try to give more knot theory formulas if there is any request (it doesn't seem like a hugely popular field, however).

The Conway Polynomial

The Conway polynomial is a polynomial invariant of knots and links described by the following three axioms:

Axiom 1: For each oriented knot or link K there is an associated polynomial ∇K(z) ∈ Z[z] (Z[z] is the ring of polynomials in z with integer coefficients). If one knot K is ambient isotopic to another knot K' ( K ~ K'), then ∇K = ∇K'.

Axiom 2: If K is ambient isotopic to the unknot (K ~ O), then ∇K = 1.

Axiom 3: Suppose that three knots or links K+, K-, and L differ at one crossing in the manner shown below:

http://i4.photobucket.com/albums/y144/Zhylliolom/K.jpg K+
http://i4.photobucket.com/albums/y144/Zhylliolom/Kbar.jpg K-
http://i4.photobucket.com/albums/y144/Zhylliolom/L.jpg L

Then ∇K+ - ∇K- = z∇L.


Axiom 1 tells us that for any knot or link there exists a Conway polynomial; Axioms 2 and 3 give us a way to find it. Tomorrow I shall post an example of how to use these axioms to find the Conway polynomial of a knot (using a specific example, most likely the trefoil, but maybe some others), and perhaps I shall also describe the Jones polynomial, the HOMFLY polynomial, the chromatic polynomial, and more.

Edit: Wow, sorry about that, I could have sworn this topic had been replied to very recently, and I didn't realize it had been moved from the formulas section.

Devantè
2006-11-26 05:17:30

Here's a good explanation from Wolfram's MathWorld: http://mathworld.wolfram.com/ConwayPolynomial.html

rosana
2006-11-12 05:58:46

hi,
How are you ?
I want any information , any topic ,website or any definition can help me to understand and explain these searching ( THE CONWAY POLYNOMIAL for louis H.Kauffman )because I do not understand it

thank you for your help

Moved to Help Me! - Ricky

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