Does the union of all finite groups yield a complete knot invariant for prime knots?
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It is established in Whitten - Knot complements and groups together with the Gordon-Luecke theorem (that knot complements determine knot type) that the type of a prime knot is determined by the isomorphy type of its knot group.
In the book Charles Livingston - Knot Theory, the author uses surjective homomorphisms from knot groups into finite groups as knot invariants (i.e., two knot groups are nonisomorphic if one of them can be mapped onto a certain finite group and the other one can't). My question is:
If two prime knots are distinct, does that mean there is a finite group such that exactly one of them can be mapped surjectively into it?
or:
Do all finite groups combined yield (as described above) a complete knot invariant for prime knots?
finite-groups knot-theory
edited Dec 6 at 22:10
YCor
27k380132
asked Dec 6 at 20:39
Simon1729
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up vote
9
down vote
favorite
It is established in Whitten - Knot complements and groups together with the Gordon-Luecke theorem (that knot complements determine knot type) that the type of a prime knot is determined by the isomorphy type of its knot group.
In the book Charles Livingston - Knot Theory, the author uses surjective homomorphisms from knot groups into finite groups as knot invariants (i.e., two knot groups are nonisomorphic if one of them can be mapped onto a certain finite group and the other one can't). My question is:
If two prime knots are distinct, does that mean there is a finite group such that exactly one of them can be mapped surjectively into it?
or:
Do all finite groups combined yield (as described above) a complete knot invariant for prime knots?
finite-groups knot-theory
edited Dec 6 at 22:10
YCor
27k380132
asked Dec 6 at 20:39
Simon1729
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
It is established in Whitten - Knot complements and groups together with the Gordon-Luecke theorem (that knot complements determine knot type) that the type of a prime knot is determined by the isomorphy type of its knot group.
In the book Charles Livingston - Knot Theory, the author uses surjective homomorphisms from knot groups into finite groups as knot invariants (i.e., two knot groups are nonisomorphic if one of them can be mapped onto a certain finite group and the other one can't). My question is:
If two prime knots are distinct, does that mean there is a finite group such that exactly one of them can be mapped surjectively into it?
or:
Do all finite groups combined yield (as described above) a complete knot invariant for prime knots?
finite-groups knot-theory
edited Dec 6 at 22:10
YCor
27k380132
asked Dec 6 at 20:39
Simon1729
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It is established in Whitten - Knot complements and groups together with the Gordon-Luecke theorem (that knot complements determine knot type) that the type of a prime knot is determined by the isomorphy type of its knot group.
In the book Charles Livingston - Knot Theory, the author uses surjective homomorphisms from knot groups into finite groups as knot invariants (i.e., two knot groups are nonisomorphic if one of them can be mapped onto a certain finite group and the other one can't). My question is:
If two prime knots are distinct, does that mean there is a finite group such that exactly one of them can be mapped surjectively into it?
or:
Do all finite groups combined yield (as described above) a complete knot invariant for prime knots?
finite-groups knot-theory
finite-groups knot-theory
edited Dec 6 at 22:10
YCor
27k380132
asked Dec 6 at 20:39
Simon1729
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 6 at 22:10
YCor
27k380132
asked Dec 6 at 20:39
Simon1729
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 6 at 22:10
YCor
27k380132
edited Dec 6 at 22:10
YCor
27k380132
edited Dec 6 at 22:10
YCor
27k380132
27k380132
asked Dec 6 at 20:39
Simon1729
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 6 at 20:39
Simon1729
462
asked Dec 6 at 20:39
Simon1729
462
462
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Simon1729 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
11
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Though it is not completely obvious, it turns out that if $G_1$ and $G_2$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $G_1$ and $G_2$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.
I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
Though it is not completely obvious, it turns out that if $G_1$ and $G_2$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $G_1$ and $G_2$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.
I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
add a comment |
up vote
11
down vote
Though it is not completely obvious, it turns out that if $G_1$ and $G_2$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $G_1$ and $G_2$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.
I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
add a comment |
up vote
11
down vote
up vote
11
down vote
Though it is not completely obvious, it turns out that if $G_1$ and $G_2$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $G_1$ and $G_2$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.
I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
Though it is not completely obvious, it turns out that if $G_1$ and $G_2$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $G_1$ and $G_2$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.
I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
answered Dec 6 at 21:33
Andy Putman
31.1k5132212
31.1k5132212
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
add a comment |
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
1
1
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots.
– Mike Miller
Dec 6 at 22:24
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic.
– Kapil
Dec 7 at 3:01
1
1
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
@Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions.
– Andy Putman
Dec 7 at 3:24
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $mathbb{C}$. That is much more than what is required here.
– Kapil
2 days ago
add a comment |
Simon1729 is a new contributor. Be nice, and check out our Code of Conduct.
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