How to use two number to form a Jones polynomial
1
$begingroup$
According to the Wikipedia article on Knots,
The number of crossing (rule $1$) and a line crossing the triangle (rule $2$) form a number such as $3,1$. With these two numbers, how do you form a Jones polynomial?
update
Where does $t^{3/2}$ come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)?
How do notation come out from skein relation?
Where do $V_{O_2}(t)$ come from and equal to what number?
How does $V_{sigma^3}$ become a combination of $V_sigma$ and $V_{sigma^2}$
polynomials knot-theory
edited Dec 24 '18 at 10:28
amWhy
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
$endgroup$
add a comment |
1
$begingroup$
According to the Wikipedia article on Knots,
The number of crossing (rule $1$) and a line crossing the triangle (rule $2$) form a number such as $3,1$. With these two numbers, how do you form a Jones polynomial?
update
Where does $t^{3/2}$ come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)?
How do notation come out from skein relation?
Where do $V_{O_2}(t)$ come from and equal to what number?
How does $V_{sigma^3}$ become a combination of $V_sigma$ and $V_{sigma^2}$
polynomials knot-theory
edited Dec 24 '18 at 10:28
amWhy
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
$endgroup$
1
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50
add a comment |
1
1
1
$begingroup$
According to the Wikipedia article on Knots,
The number of crossing (rule $1$) and a line crossing the triangle (rule $2$) form a number such as $3,1$. With these two numbers, how do you form a Jones polynomial?
update
Where does $t^{3/2}$ come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)?
How do notation come out from skein relation?
Where do $V_{O_2}(t)$ come from and equal to what number?
How does $V_{sigma^3}$ become a combination of $V_sigma$ and $V_{sigma^2}$
polynomials knot-theory
edited Dec 24 '18 at 10:28
amWhy
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
$endgroup$
According to the Wikipedia article on Knots,
The number of crossing (rule $1$) and a line crossing the triangle (rule $2$) form a number such as $3,1$. With these two numbers, how do you form a Jones polynomial?
update
Where does $t^{3/2}$ come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)?
How do notation come out from skein relation?
Where do $V_{O_2}(t)$ come from and equal to what number?
How does $V_{sigma^3}$ become a combination of $V_sigma$ and $V_{sigma^2}$
polynomials knot-theory
polynomials knot-theory
edited Dec 24 '18 at 10:28
amWhy
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
edited Dec 24 '18 at 10:28
amWhy
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
edited Dec 24 '18 at 10:28
amWhy
1
edited Dec 24 '18 at 10:28
amWhy
1
edited Dec 24 '18 at 10:28
amWhy
1
1
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
asked May 18 '13 at 3:50
FermionGhostFermionGhost
1815
1815
1
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50
add a comment |
1
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50
1
1
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50
add a comment |
0
active
oldest
votes
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f395181%2fhow-to-use-two-number-to-form-a-jones-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f395181%2fhow-to-use-two-number-to-form-a-jones-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
The linked article mentions nothing about this. What are you trying to ask?
$endgroup$
– Henry T. Horton
May 18 '13 at 3:56
$begingroup$
i add a picture which i do not understand V(3,2) form a polynomial
$endgroup$
– FermionGhost
May 18 '13 at 4:32
$begingroup$
In your picture, Equations (6.7)-(6.9) give a recipe to recursively determine the Jones polynomial by Skein relations (see here for the picture of $K_pm$ and $K_0$: en.wikipedia.org/wiki/Jones_polynomial#Properties). $V_{T(3,2)}$ denotes the Jones polynomial of the trefoil knot $T(2,3)$ (the notation comes from the fact that the trefoil is the (2,3)-torus knot). I can't make sense of what you are trying to say about "rule 1," "rule 2," and what the numbers you mention are supposed to be. The Wikipedia article you link still has no relation to the question other than being about knots.
$endgroup$
– Henry T. Horton
May 18 '13 at 4:45
$begingroup$
could you demonstrate the final two steps in diagram, when i see the properties, i just know crossing is separated in two kind of crossing, one is on top of another. i do not understand how do notation come out from skein relation? and how do t^(3/2) come from, where is a variable for 3 to substitute in (6.7)(6.8)(6.9)
$endgroup$
– FermionGhost
May 18 '13 at 7:08
$begingroup$
Draw the knot, with the braid $sigma^3$. Now switch the top crossing either from an overcrossing to an undercrossing or from an undercrossing to an overcrossing. Then by Reidemeister's 2nd move, 2 crossings will be out, so we are left with $sigma$. Then change the top crossing to a "no-crossing", then two crossings are left, so we have $sigma^2$. Now put into the skein relation, and solve.
$endgroup$
– wilsonw
Dec 24 '18 at 13:50