Set of Jones polynomials as the knot varies
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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
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
$endgroup$
add a comment |
$begingroup$
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
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
$endgroup$
1
$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
5
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40
add a comment |
$begingroup$
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
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
$endgroup$
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
at.algebraic-topology knot-theory jones-polynomial
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
asked Jan 5 at 18:32
pre-kidneypre-kidney
540215
540215
1
$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
5
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40
add a comment |
1
$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
5
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40
1
1
$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
5
5
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$p(1) = pm 1$, and
$p(t)=p(t^{-1})$.
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.
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
$endgroup$
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
add a comment |
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1 Answer
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$begingroup$
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:
$p(1) = pm 1$, and
$p(t)=p(t^{-1})$.
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.
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
$endgroup$
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
add a comment |
$begingroup$
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:
$p(1) = pm 1$, and
$p(t)=p(t^{-1})$.
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.
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
$endgroup$
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
add a comment |
$begingroup$
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:
$p(1) = pm 1$, and
$p(t)=p(t^{-1})$.
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.
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
$endgroup$
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:
$p(1) = pm 1$, and
$p(t)=p(t^{-1})$.
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.
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
edited Jan 5 at 19:20
Martin Sleziak
3,09032231
3,09032231
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
answered Jan 5 at 19:09
Adam LowranceAdam Lowrance
30622
30622
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
add a comment |
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
$begingroup$
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?
$endgroup$
– pre-kidney
Jan 5 at 20:35
1
1
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
$begingroup$
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.
$endgroup$
– Adam Lowrance
Jan 5 at 20:47
add a comment |
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$begingroup$
"...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.
$endgroup$
– David G. Stork
Jan 5 at 18:39
5
$begingroup$
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.
$endgroup$
– pre-kidney
Jan 5 at 18:40