de Rham cohomology of doubly punctured torus
$begingroup$
Let $T^2=S^1times S^1$. I'd like to know all de Rham cohomology groups of $M=T^2-{a,b}$ but I couldn't find a result. So I want to compute it and I'm thinking of using Mayer Vietoris sequence. I need two open sets whose union covers $M$. I'm having difficulty choosing these open sets. Any help is appreciated.
algebraic-topology differential-topology homology-cohomology de-rham-cohomology
edited Dec 8 '18 at 0:30
klirk
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
$endgroup$
add a comment |
$begingroup$
Let $T^2=S^1times S^1$. I'd like to know all de Rham cohomology groups of $M=T^2-{a,b}$ but I couldn't find a result. So I want to compute it and I'm thinking of using Mayer Vietoris sequence. I need two open sets whose union covers $M$. I'm having difficulty choosing these open sets. Any help is appreciated.
algebraic-topology differential-topology homology-cohomology de-rham-cohomology
edited Dec 8 '18 at 0:30
klirk
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
$endgroup$
1
$begingroup$
The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39
add a comment |
$begingroup$
Let $T^2=S^1times S^1$. I'd like to know all de Rham cohomology groups of $M=T^2-{a,b}$ but I couldn't find a result. So I want to compute it and I'm thinking of using Mayer Vietoris sequence. I need two open sets whose union covers $M$. I'm having difficulty choosing these open sets. Any help is appreciated.
algebraic-topology differential-topology homology-cohomology de-rham-cohomology
edited Dec 8 '18 at 0:30
klirk
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
$endgroup$
Let $T^2=S^1times S^1$. I'd like to know all de Rham cohomology groups of $M=T^2-{a,b}$ but I couldn't find a result. So I want to compute it and I'm thinking of using Mayer Vietoris sequence. I need two open sets whose union covers $M$. I'm having difficulty choosing these open sets. Any help is appreciated.
algebraic-topology differential-topology homology-cohomology de-rham-cohomology
algebraic-topology differential-topology homology-cohomology de-rham-cohomology
edited Dec 8 '18 at 0:30
klirk
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
edited Dec 8 '18 at 0:30
klirk
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
edited Dec 8 '18 at 0:30
klirk
2,591530
edited Dec 8 '18 at 0:30
klirk
2,591530
edited Dec 8 '18 at 0:30
klirk
2,591530
2,591530
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
asked Dec 7 '18 at 23:29
tangentbundletangentbundle
410211
410211
1
$begingroup$
The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39
add a comment |
1
$begingroup$
The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39
1
1
$begingroup$
The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
The most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal.
But some parts of the boundary are identified with each other.
Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
$endgroup$
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
add a comment |
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
});
}
});
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%2f3030496%2fde-rham-cohomology-of-doubly-punctured-torus%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
$begingroup$
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal.
But some parts of the boundary are identified with each other.
Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
$endgroup$
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal.
But some parts of the boundary are identified with each other.
Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
$endgroup$
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal.
But some parts of the boundary are identified with each other.
Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
$endgroup$
Its easy to visualize when looking at the fundamental domain of the torus:
$U$ is a neighbourhood of $x$, $V$ is a neighbourhood of $y$. Removing $x$ and $y$ from the torus is homotopy equivalent to removing the whole neighbourhoods $U$ and $V$. Further we can choose $U,V$ so large, that they fill the entire triangle they lie in.
So the fundamental domain is homotopy equivalent to the union of the boundary of the square with the diagonal.
But some parts of the boundary are identified with each other.
Doing the identifications, we obtain the wedge sum of three circles (their common point is $A$). One circle corresponds to the left side = right side of the bundary, one circle corresponds to the top side = bottom side and one circle corresponds to the diagonal.
More precisely, first identifying top with bottom, we get two distinct edges connecting the bottom/top left vertex with the bottom/top right vertex (corresponding to the bottom/top edge and the diagonal) and two circles (corresponding to the left and right edges).
Then we identify the left and right edges. The two circles we already had get identified, and as the bottom/top left edge gets identified with the bottom/top right edge, the edges which previously connected these two vetices become circles.
So in the end we get three circles which are connected in their common point $A$, the bottom/top now left/right vertex.
I hope the detailled description did not make things more confusing. I probably should've also drawn this.
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
edited Dec 8 '18 at 0:02
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
answered Dec 7 '18 at 23:53
klirkklirk
2,591530
2,591530
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
add a comment |
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
$begingroup$
Thanks for the elaborate explanation.
$endgroup$
– tangentbundle
Dec 8 '18 at 0:19
add a comment |
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.
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%2f3030496%2fde-rham-cohomology-of-doubly-punctured-torus%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 most easiest argument is probably to show that $M$ is homotopy equivalent to the wedge sum of three circles. Do you want an explicit calculation of the differential forms, or is an argument using homotopy equivalences enough for you?
$endgroup$
– klirk
Dec 7 '18 at 23:37
$begingroup$
How did you know it contracts to the wedge sum of 3 circles? I can't visualize it. An argument in homotopy equivalence works. Thanks.
$endgroup$
– tangentbundle
Dec 7 '18 at 23:39