Higher homology groups of knots and links












1














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




  1. I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?

  2. 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







share|cite|improve this question




















  • 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














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




  1. I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?

  2. 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







share|cite|improve this question




















  • 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








1







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




  1. I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?

  2. 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







share|cite|improve this question















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




  1. I have shown that for a knot, the first homology group is the infinitely cyclic group, is this also true for a link?

  2. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 1:09









Kyle Miller

8,452928




8,452928










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














  • 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










1 Answer
1






active

oldest

votes


















4














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.






share|cite|improve this answer





















  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2825676%2fhigher-homology-groups-of-knots-and-links%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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.






share|cite|improve this answer





















  • 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
















4














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.






share|cite|improve this answer





















  • 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














4












4








4






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2825676%2fhigher-homology-groups-of-knots-and-links%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more