Higher homology groups of knots and links
I know that $H_1$ of the complement of a knot or a link can be obtained by taking the commutative quotient group which can be computed by Wirtinger presentation theorem. My questions are following
- I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?
- When I consider the higher homology groups, I tried to calculate them by M-V sequence, but it seems not work, so I wonder if there are some ways to compute them.
Thanks advanced!
algebraic-topology homology-cohomology
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
add a comment |
I know that $H_1$ of the complement of a knot or a link can be obtained by taking the commutative quotient group which can be computed by Wirtinger presentation theorem. My questions are following
- I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?
- When I consider the higher homology groups, I tried to calculate them by M-V sequence, but it seems not work, so I wonder if there are some ways to compute them.
Thanks advanced!
algebraic-topology homology-cohomology
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
1
The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31
add a comment |
I know that $H_1$ of the complement of a knot or a link can be obtained by taking the commutative quotient group which can be computed by Wirtinger presentation theorem. My questions are following
- I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?
- When I consider the higher homology groups, I tried to calculate them by M-V sequence, but it seems not work, so I wonder if there are some ways to compute them.
Thanks advanced!
algebraic-topology homology-cohomology
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
I know that $H_1$ of the complement of a knot or a link can be obtained by taking the commutative quotient group which can be computed by Wirtinger presentation theorem. My questions are following
- I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?
- When I consider the higher homology groups, I tried to calculate them by M-V sequence, but it seems not work, so I wonder if there are some ways to compute them.
Thanks advanced!
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
edited Nov 27 '18 at 1:09
Kyle Miller
8,452928
8,452928
asked Jun 20 '18 at 5:11
user365833
1097
asked Jun 20 '18 at 5:11
user365833
1097
asked Jun 20 '18 at 5:11
user365833
1097
1097
1
The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31
add a comment |
1
The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31
1
1
The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31
add a comment |
1 Answer
1
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Alexander Duality is you friend here. It relates the homology of the complement with the cohomology of the link itself, which is just a disjoint union of circles. In particular, $H_1$ has one $mathbb Z$ for every link component. $H_2$ has one fewer copy of $mathbb Z$, essentially because you need to use reduced homology in the statement of the Alexander Duality isomorphism.
MV should provide an alternate proof of this, if you take $mathbb R^3=Ucup V$ with $U$ being a tubular neighborhood of the link, and $V$ being the link complement.
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
add a comment |
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Alexander Duality is you friend here. It relates the homology of the complement with the cohomology of the link itself, which is just a disjoint union of circles. In particular, $H_1$ has one $mathbb Z$ for every link component. $H_2$ has one fewer copy of $mathbb Z$, essentially because you need to use reduced homology in the statement of the Alexander Duality isomorphism.
MV should provide an alternate proof of this, if you take $mathbb R^3=Ucup V$ with $U$ being a tubular neighborhood of the link, and $V$ being the link complement.
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
add a comment |
Alexander Duality is you friend here. It relates the homology of the complement with the cohomology of the link itself, which is just a disjoint union of circles. In particular, $H_1$ has one $mathbb Z$ for every link component. $H_2$ has one fewer copy of $mathbb Z$, essentially because you need to use reduced homology in the statement of the Alexander Duality isomorphism.
MV should provide an alternate proof of this, if you take $mathbb R^3=Ucup V$ with $U$ being a tubular neighborhood of the link, and $V$ being the link complement.
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
add a comment |
Alexander Duality is you friend here. It relates the homology of the complement with the cohomology of the link itself, which is just a disjoint union of circles. In particular, $H_1$ has one $mathbb Z$ for every link component. $H_2$ has one fewer copy of $mathbb Z$, essentially because you need to use reduced homology in the statement of the Alexander Duality isomorphism.
MV should provide an alternate proof of this, if you take $mathbb R^3=Ucup V$ with $U$ being a tubular neighborhood of the link, and $V$ being the link complement.
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
Alexander Duality is you friend here. It relates the homology of the complement with the cohomology of the link itself, which is just a disjoint union of circles. In particular, $H_1$ has one $mathbb Z$ for every link component. $H_2$ has one fewer copy of $mathbb Z$, essentially because you need to use reduced homology in the statement of the Alexander Duality isomorphism.
MV should provide an alternate proof of this, if you take $mathbb R^3=Ucup V$ with $U$ being a tubular neighborhood of the link, and $V$ being the link complement.
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
answered Jun 20 '18 at 5:53
Cheerful Parsnip
20.9k23396
20.9k23396
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
add a comment |
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
thx for your answer, but i still have some problems when using M-V sequebce, if we take u,v as you said, then homology of u may be computed seeing it as some torus (is this right?), but how to compute the homology of the intersection of U and V?(maybe i should use cutting lemma?)
– user365833
Jun 20 '18 at 6:31
1
1
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
$U$ is homotopy equivalent to a circle. The intersection is homotopy equivalent to a 2D torus.
– Cheerful Parsnip
Jun 20 '18 at 14:18
add a comment |
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The usual way (for knots) is to use the sphere theorem. I think this is probably in any beginning knot theory book (and I want to say it's almost certainly in the book by Lickorish). Essentially, the higher homology for a knot is trivial.
– Steve D
Jun 20 '18 at 5:55
I guess you could also use Alexander duality to get $H_1$ and $H_2$.
– Steve D
Jun 20 '18 at 5:58
@SteveD so is it true that H_i(R3-K) is zero if i ≥2 ? sorry i haven't studied cohomology...
– user365833
Jun 20 '18 at 7:10
Yes that's true. Here's maybe the easiest way to see it. You have a knot in $S^3$. If you assume the knot contains the point at infinity, then you complement is $mathbb{R^3}$ minus a squiggly line. Now take $U$ to be a nbhd of that line, $V$ to be the complement of that line. Then $Ucup V=mathbb{R^3}$ is contractible, and so is $U$. So by M-V, the homology of $V$ (your complement) is the same as $Ucap V$, which is an infinite cylinder, homotopic to a circle. So your complement has the homology of a circle.
– Steve D
Jun 20 '18 at 15:31