Set of Jones polynomials as the knot varies












7














Is a characterization known for the set of Laurent polynomials arising as the Jones polynomial of some knot? More generally, is such a characterization known for any of the famous knot polynomials?










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  • 1




    "...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
    – David G. Stork
    Jan 5 at 18:39






  • 4




    As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
    – pre-kidney
    Jan 5 at 18:40


















7














Is a characterization known for the set of Laurent polynomials arising as the Jones polynomial of some knot? More generally, is such a characterization known for any of the famous knot polynomials?










share|cite|improve this question


















  • 1




    "...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
    – David G. Stork
    Jan 5 at 18:39






  • 4




    As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
    – pre-kidney
    Jan 5 at 18:40
















7












7








7







Is a characterization known for the set of Laurent polynomials arising as the Jones polynomial of some knot? More generally, is such a characterization known for any of the famous knot polynomials?










share|cite|improve this question













Is a characterization known for the set of Laurent polynomials arising as the Jones polynomial of some knot? More generally, is such a characterization known for any of the famous knot polynomials?







at.algebraic-topology knot-theory jones-polynomial






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asked Jan 5 at 18:32









pre-kidneypre-kidney

530215




530215








  • 1




    "...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
    – David G. Stork
    Jan 5 at 18:39






  • 4




    As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
    – pre-kidney
    Jan 5 at 18:40
















  • 1




    "...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
    – David G. Stork
    Jan 5 at 18:39






  • 4




    As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
    – pre-kidney
    Jan 5 at 18:40










1




1




"...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
– David G. Stork
Jan 5 at 18:39




"...as the knot varies"? Varies how? If the variation is due to an ambient isotopy or the Reidemeister moves, then invariant polynomials are... well... invariant.
– David G. Stork
Jan 5 at 18:39




4




4




As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
– pre-kidney
Jan 5 at 18:40






As the knot ranges over the set of all knots; I believe this is clear from the content of my question, if not from the title.
– pre-kidney
Jan 5 at 18:40












1 Answer
1






active

oldest

votes


















6














I believe your question is open for the Jones polynomial. However, it is solved for the Alexander polynomial. On page 171 of Rolfsen's Knots and Links, the following theorem appears.



Theorem. Let $p(t)$ be any Laurent polynomial satisfying:





  1. $p(1) = pm 1$, and


  2. $p(t)=p(t^{-1})$.


There exists a knot $K$ whose Alexander polynomial $Delta_K(t)$ is $p(t)$.



It is well-known that the Alexander polynomial satisfies the above conditions. In the proof, Rolfsen shows how to explicitly construct a knot $K$ whose Alexander polynomial is a given Laurent polynomial $p(t)$ satisfying the conditions above.



Rolfsen gives the original reference of this result as




  • Seifert, H.; Über das Geschlecht von Knoten. Math. Ann. 110 (1935), no. 1, 571–592.






share|cite|improve this answer























  • Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
    – pre-kidney
    Jan 5 at 20:35






  • 1




    I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
    – Adam Lowrance
    Jan 5 at 20:47













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1 Answer
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1 Answer
1






active

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6














I believe your question is open for the Jones polynomial. However, it is solved for the Alexander polynomial. On page 171 of Rolfsen's Knots and Links, the following theorem appears.



Theorem. Let $p(t)$ be any Laurent polynomial satisfying:





  1. $p(1) = pm 1$, and


  2. $p(t)=p(t^{-1})$.


There exists a knot $K$ whose Alexander polynomial $Delta_K(t)$ is $p(t)$.



It is well-known that the Alexander polynomial satisfies the above conditions. In the proof, Rolfsen shows how to explicitly construct a knot $K$ whose Alexander polynomial is a given Laurent polynomial $p(t)$ satisfying the conditions above.



Rolfsen gives the original reference of this result as




  • Seifert, H.; Über das Geschlecht von Knoten. Math. Ann. 110 (1935), no. 1, 571–592.






share|cite|improve this answer























  • Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
    – pre-kidney
    Jan 5 at 20:35






  • 1




    I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
    – Adam Lowrance
    Jan 5 at 20:47


















6














I believe your question is open for the Jones polynomial. However, it is solved for the Alexander polynomial. On page 171 of Rolfsen's Knots and Links, the following theorem appears.



Theorem. Let $p(t)$ be any Laurent polynomial satisfying:





  1. $p(1) = pm 1$, and


  2. $p(t)=p(t^{-1})$.


There exists a knot $K$ whose Alexander polynomial $Delta_K(t)$ is $p(t)$.



It is well-known that the Alexander polynomial satisfies the above conditions. In the proof, Rolfsen shows how to explicitly construct a knot $K$ whose Alexander polynomial is a given Laurent polynomial $p(t)$ satisfying the conditions above.



Rolfsen gives the original reference of this result as




  • Seifert, H.; Über das Geschlecht von Knoten. Math. Ann. 110 (1935), no. 1, 571–592.






share|cite|improve this answer























  • Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
    – pre-kidney
    Jan 5 at 20:35






  • 1




    I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
    – Adam Lowrance
    Jan 5 at 20:47
















6












6








6






I believe your question is open for the Jones polynomial. However, it is solved for the Alexander polynomial. On page 171 of Rolfsen's Knots and Links, the following theorem appears.



Theorem. Let $p(t)$ be any Laurent polynomial satisfying:





  1. $p(1) = pm 1$, and


  2. $p(t)=p(t^{-1})$.


There exists a knot $K$ whose Alexander polynomial $Delta_K(t)$ is $p(t)$.



It is well-known that the Alexander polynomial satisfies the above conditions. In the proof, Rolfsen shows how to explicitly construct a knot $K$ whose Alexander polynomial is a given Laurent polynomial $p(t)$ satisfying the conditions above.



Rolfsen gives the original reference of this result as




  • Seifert, H.; Über das Geschlecht von Knoten. Math. Ann. 110 (1935), no. 1, 571–592.






share|cite|improve this answer














I believe your question is open for the Jones polynomial. However, it is solved for the Alexander polynomial. On page 171 of Rolfsen's Knots and Links, the following theorem appears.



Theorem. Let $p(t)$ be any Laurent polynomial satisfying:





  1. $p(1) = pm 1$, and


  2. $p(t)=p(t^{-1})$.


There exists a knot $K$ whose Alexander polynomial $Delta_K(t)$ is $p(t)$.



It is well-known that the Alexander polynomial satisfies the above conditions. In the proof, Rolfsen shows how to explicitly construct a knot $K$ whose Alexander polynomial is a given Laurent polynomial $p(t)$ satisfying the conditions above.



Rolfsen gives the original reference of this result as




  • Seifert, H.; Über das Geschlecht von Knoten. Math. Ann. 110 (1935), no. 1, 571–592.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 19:20









Martin Sleziak

2,92032028




2,92032028










answered Jan 5 at 19:09









Adam LowranceAdam Lowrance

28122




28122












  • Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
    – pre-kidney
    Jan 5 at 20:35






  • 1




    I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
    – Adam Lowrance
    Jan 5 at 20:47




















  • Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
    – pre-kidney
    Jan 5 at 20:35






  • 1




    I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
    – Adam Lowrance
    Jan 5 at 20:47


















Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
– pre-kidney
Jan 5 at 20:35




Thank you - even if the Jones polynomial question remains open, are you aware of any variant of Seifert's construction that produces many (if not all) Jones polynomials?
– pre-kidney
Jan 5 at 20:35




1




1




I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
– Adam Lowrance
Jan 5 at 20:47






I think that not much is really known about how to produce knots with prescribed Jones polynomial. Manchon arXiv:0201160 used a construction of Morton and Bae arXiv:0012089 to construct prime knots whose Jones polynomials have prescribed extreme coefficients.
– Adam Lowrance
Jan 5 at 20:47




















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