Computing the de Rham cohomology of $S^2$












1












$begingroup$


I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $Ucap V cong S^1 times [0,1] cong S^1$. Thus, with this information we can start considering the following long exact sequence.



$$ 0 xrightarrow{p} H^0(S^2) xrightarrow{alpha_0} H^0(U) oplus H^0(V)cong mathbb{R} oplus mathbb{R} xrightarrow{beta_0} H^0(Ucap V)cong mathbb{R} xrightarrow{d_0} H^1(S^2) xrightarrow{alpha_1} H^1(U) oplus H^1(V) cong 0 xrightarrow{beta_1} H^1(U cap V) cong mathbb{R} xrightarrow{d_1} H^2(S^2) xrightarrow{alpha_2} 0 xrightarrow{beta_2}0$$



I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = text{im } p = ker alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/ker(alpha_0) cong text{im } alpha_0 = _{text{(by exactness)}} ker beta_0 $. But then, I know how to figure out the kernel of $beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) cong mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.



Thanks for your help!






differential-topology de-rham-cohomology







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
    $endgroup$
    – Prototank
    Dec 12 '18 at 18:55
















1












$begingroup$


I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $Ucap V cong S^1 times [0,1] cong S^1$. Thus, with this information we can start considering the following long exact sequence.



$$ 0 xrightarrow{p} H^0(S^2) xrightarrow{alpha_0} H^0(U) oplus H^0(V)cong mathbb{R} oplus mathbb{R} xrightarrow{beta_0} H^0(Ucap V)cong mathbb{R} xrightarrow{d_0} H^1(S^2) xrightarrow{alpha_1} H^1(U) oplus H^1(V) cong 0 xrightarrow{beta_1} H^1(U cap V) cong mathbb{R} xrightarrow{d_1} H^2(S^2) xrightarrow{alpha_2} 0 xrightarrow{beta_2}0$$



I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = text{im } p = ker alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/ker(alpha_0) cong text{im } alpha_0 = _{text{(by exactness)}} ker beta_0 $. But then, I know how to figure out the kernel of $beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) cong mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.



Thanks for your help!






differential-topology de-rham-cohomology







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
    $endgroup$
    – Prototank
    Dec 12 '18 at 18:55














1












1








1





$begingroup$


I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $Ucap V cong S^1 times [0,1] cong S^1$. Thus, with this information we can start considering the following long exact sequence.



$$ 0 xrightarrow{p} H^0(S^2) xrightarrow{alpha_0} H^0(U) oplus H^0(V)cong mathbb{R} oplus mathbb{R} xrightarrow{beta_0} H^0(Ucap V)cong mathbb{R} xrightarrow{d_0} H^1(S^2) xrightarrow{alpha_1} H^1(U) oplus H^1(V) cong 0 xrightarrow{beta_1} H^1(U cap V) cong mathbb{R} xrightarrow{d_1} H^2(S^2) xrightarrow{alpha_2} 0 xrightarrow{beta_2}0$$



I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = text{im } p = ker alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/ker(alpha_0) cong text{im } alpha_0 = _{text{(by exactness)}} ker beta_0 $. But then, I know how to figure out the kernel of $beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) cong mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.



Thanks for your help!






differential-topology de-rham-cohomology







share|cite|improve this question









$endgroup$




I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $Ucap V cong S^1 times [0,1] cong S^1$. Thus, with this information we can start considering the following long exact sequence.



$$ 0 xrightarrow{p} H^0(S^2) xrightarrow{alpha_0} H^0(U) oplus H^0(V)cong mathbb{R} oplus mathbb{R} xrightarrow{beta_0} H^0(Ucap V)cong mathbb{R} xrightarrow{d_0} H^1(S^2) xrightarrow{alpha_1} H^1(U) oplus H^1(V) cong 0 xrightarrow{beta_1} H^1(U cap V) cong mathbb{R} xrightarrow{d_1} H^2(S^2) xrightarrow{alpha_2} 0 xrightarrow{beta_2}0$$



I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = text{im } p = ker alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/ker(alpha_0) cong text{im } alpha_0 = _{text{(by exactness)}} ker beta_0 $. But then, I know how to figure out the kernel of $beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) cong mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.



Thanks for your help!






differential-topology de-rham-cohomology




differential-topology de-rham-cohomology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 18:52









BOlivianoperuano84BOlivianoperuano84

1778




1778












  • $begingroup$
    en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
    $endgroup$
    – Prototank
    Dec 12 '18 at 18:55


















  • $begingroup$
    en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
    $endgroup$
    – Prototank
    Dec 12 '18 at 18:55
















$begingroup$
en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
$endgroup$
– Prototank
Dec 12 '18 at 18:55




$begingroup$
en.wikipedia.org/wiki/De_Rham_cohomology#The_n-sphere
$endgroup$
– Prototank
Dec 12 '18 at 18:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

Since
$$ 0 to H^1(Ucap V)cong mathbb{R} xrightarrow{d_1}H^2(S)to 0$$
is exact then $d_1$ is an isomorphism, so $H^2(S)cong mathbb{R}.$



Now note that if $M$ is any manifold, then
$$ H^0(M)={fin C^infty(M) text{ | } df=0}={fin C^{infty}(M) text{ | }f text{ is locally constant}},$$
so if $M$ has $n$ connected components, then $H^0(M)cong mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=mathbb{R}.$



Finally, look at the exact sequence
$$ 0to H^0(S^2)cong mathbb{R} to H^0(U)oplus H^0(V)cong mathbb{R}oplus mathbb{R}to H^0(Ucap V) cong mathbb{R}to H^1(S^2) to 0$$
and use the following:



Proposition. If $0to A_1to A_2to cdots to A_n to 0$ is an exact sequence of vector spaces then
$$ sum_{i=1}^n(-1)^idim A_i=0.$$



Then you find that $1-2+1-dim H^1(S^2)=0,$ so $H^1(S^2)=0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
    $endgroup$
    – BOlivianoperuano84
    Dec 12 '18 at 20:46











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%2f3037083%2fcomputing-the-de-rham-cohomology-of-s2%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









2












$begingroup$

Since
$$ 0 to H^1(Ucap V)cong mathbb{R} xrightarrow{d_1}H^2(S)to 0$$
is exact then $d_1$ is an isomorphism, so $H^2(S)cong mathbb{R}.$



Now note that if $M$ is any manifold, then
$$ H^0(M)={fin C^infty(M) text{ | } df=0}={fin C^{infty}(M) text{ | }f text{ is locally constant}},$$
so if $M$ has $n$ connected components, then $H^0(M)cong mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=mathbb{R}.$



Finally, look at the exact sequence
$$ 0to H^0(S^2)cong mathbb{R} to H^0(U)oplus H^0(V)cong mathbb{R}oplus mathbb{R}to H^0(Ucap V) cong mathbb{R}to H^1(S^2) to 0$$
and use the following:



Proposition. If $0to A_1to A_2to cdots to A_n to 0$ is an exact sequence of vector spaces then
$$ sum_{i=1}^n(-1)^idim A_i=0.$$



Then you find that $1-2+1-dim H^1(S^2)=0,$ so $H^1(S^2)=0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
    $endgroup$
    – BOlivianoperuano84
    Dec 12 '18 at 20:46
















2












$begingroup$

Since
$$ 0 to H^1(Ucap V)cong mathbb{R} xrightarrow{d_1}H^2(S)to 0$$
is exact then $d_1$ is an isomorphism, so $H^2(S)cong mathbb{R}.$



Now note that if $M$ is any manifold, then
$$ H^0(M)={fin C^infty(M) text{ | } df=0}={fin C^{infty}(M) text{ | }f text{ is locally constant}},$$
so if $M$ has $n$ connected components, then $H^0(M)cong mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=mathbb{R}.$



Finally, look at the exact sequence
$$ 0to H^0(S^2)cong mathbb{R} to H^0(U)oplus H^0(V)cong mathbb{R}oplus mathbb{R}to H^0(Ucap V) cong mathbb{R}to H^1(S^2) to 0$$
and use the following:



Proposition. If $0to A_1to A_2to cdots to A_n to 0$ is an exact sequence of vector spaces then
$$ sum_{i=1}^n(-1)^idim A_i=0.$$



Then you find that $1-2+1-dim H^1(S^2)=0,$ so $H^1(S^2)=0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
    $endgroup$
    – BOlivianoperuano84
    Dec 12 '18 at 20:46














2












2








2





$begingroup$

Since
$$ 0 to H^1(Ucap V)cong mathbb{R} xrightarrow{d_1}H^2(S)to 0$$
is exact then $d_1$ is an isomorphism, so $H^2(S)cong mathbb{R}.$



Now note that if $M$ is any manifold, then
$$ H^0(M)={fin C^infty(M) text{ | } df=0}={fin C^{infty}(M) text{ | }f text{ is locally constant}},$$
so if $M$ has $n$ connected components, then $H^0(M)cong mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=mathbb{R}.$



Finally, look at the exact sequence
$$ 0to H^0(S^2)cong mathbb{R} to H^0(U)oplus H^0(V)cong mathbb{R}oplus mathbb{R}to H^0(Ucap V) cong mathbb{R}to H^1(S^2) to 0$$
and use the following:



Proposition. If $0to A_1to A_2to cdots to A_n to 0$ is an exact sequence of vector spaces then
$$ sum_{i=1}^n(-1)^idim A_i=0.$$



Then you find that $1-2+1-dim H^1(S^2)=0,$ so $H^1(S^2)=0.$






share|cite|improve this answer









$endgroup$



Since
$$ 0 to H^1(Ucap V)cong mathbb{R} xrightarrow{d_1}H^2(S)to 0$$
is exact then $d_1$ is an isomorphism, so $H^2(S)cong mathbb{R}.$



Now note that if $M$ is any manifold, then
$$ H^0(M)={fin C^infty(M) text{ | } df=0}={fin C^{infty}(M) text{ | }f text{ is locally constant}},$$
so if $M$ has $n$ connected components, then $H^0(M)cong mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=mathbb{R}.$



Finally, look at the exact sequence
$$ 0to H^0(S^2)cong mathbb{R} to H^0(U)oplus H^0(V)cong mathbb{R}oplus mathbb{R}to H^0(Ucap V) cong mathbb{R}to H^1(S^2) to 0$$
and use the following:



Proposition. If $0to A_1to A_2to cdots to A_n to 0$ is an exact sequence of vector spaces then
$$ sum_{i=1}^n(-1)^idim A_i=0.$$



Then you find that $1-2+1-dim H^1(S^2)=0,$ so $H^1(S^2)=0.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 20:36









positrón0802positrón0802

4,353520




4,353520












  • $begingroup$
    Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
    $endgroup$
    – BOlivianoperuano84
    Dec 12 '18 at 20:46


















  • $begingroup$
    Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
    $endgroup$
    – BOlivianoperuano84
    Dec 12 '18 at 20:46
















$begingroup$
Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
$endgroup$
– BOlivianoperuano84
Dec 12 '18 at 20:46




$begingroup$
Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism!
$endgroup$
– BOlivianoperuano84
Dec 12 '18 at 20:46


















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037083%2fcomputing-the-de-rham-cohomology-of-s2%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

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more