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







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$endgroup$












  • $begingroup$
    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




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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






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oldest

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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











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1 Answer
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active

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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


















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