Computing de Rham cohomology group $H^1(mathbb{RP}^2#mathbb{RP}^2)$
$begingroup$
I am trying to compute the de Rham cohomology group $H^p(mathbb{RP}^{n+1}#mathbb{RP}^{n+1})$ and I am stuck at computing $H^1(mathbb{RP}^2#mathbb{RP}^2)$. ($#$ stand for the connected sum)
Let $Usimeq mathbb{RP^2setminus {p}}$ and $Vsimeq mathbb{RP}^2setminus {q}$ such that $M=mathbb{RP}^2#mathbb{RP}^2=Ucup V$ and $Ucap Vsimeq S^1$. Then by using Mayer Vietoris, I get
begin{align*}
0&xrightarrow{a} H^0(M)=mathbb{R}xrightarrow{b} H^0(U)oplus H^0(V)=H^0(mathbb{RP}^1)oplus H^0(mathbb{RP}^1)=mathbb{R}oplus mathbb{R}xrightarrow{c} H^0(S^1)=mathbb{R} \& xrightarrow{d} H^1(M)xrightarrow{e} H^1(U)oplus H^1(V)=H^1(mathbb{RP}^1)oplus H^1(mathbb{RP}^1)=mathbb{R}oplus mathbb{R} xrightarrow{f} H^1(S^1)=mathbb{R}\&xrightarrow{g} H^2(M)xrightarrow{h} H^2(U)oplus H^2(V)=H^2(mathbb{RP}^1)oplus H^2(mathbb{RP}^1)=0.
end{align*}
But the fact that it is exact sequence does not give me specific feature of $H^1(M)$ and $H^2(M)$. I have computed two possibilities.
Note that $d$ is zero map, so $e$ is injective. So only possible choice of $H^1(M)$ is $0$, $mathbb{R}$ and $mathbb{R}oplus mathbb{R}$. But $H^1(M)$ cannot be $0$ since otherwise $f$ is injective and it is impossible.
Then observe the following two cases.
(1) If $H^1(M)$ is $mathbb{R}$, $g$ is surjective zero map which means $H^2(M)=0$.
(2) If $H^1(M)$ is $mathbb{R}oplus mathbb{R}$, $f$ is zero map so $g$ is isomorphism. Thus, $H^2(M)=mathbb{R}$.
As we can see, either ways does not make any contradiction.
I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)
projective-space de-rham-cohomology
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to compute the de Rham cohomology group $H^p(mathbb{RP}^{n+1}#mathbb{RP}^{n+1})$ and I am stuck at computing $H^1(mathbb{RP}^2#mathbb{RP}^2)$. ($#$ stand for the connected sum)
Let $Usimeq mathbb{RP^2setminus {p}}$ and $Vsimeq mathbb{RP}^2setminus {q}$ such that $M=mathbb{RP}^2#mathbb{RP}^2=Ucup V$ and $Ucap Vsimeq S^1$. Then by using Mayer Vietoris, I get
begin{align*}
0&xrightarrow{a} H^0(M)=mathbb{R}xrightarrow{b} H^0(U)oplus H^0(V)=H^0(mathbb{RP}^1)oplus H^0(mathbb{RP}^1)=mathbb{R}oplus mathbb{R}xrightarrow{c} H^0(S^1)=mathbb{R} \& xrightarrow{d} H^1(M)xrightarrow{e} H^1(U)oplus H^1(V)=H^1(mathbb{RP}^1)oplus H^1(mathbb{RP}^1)=mathbb{R}oplus mathbb{R} xrightarrow{f} H^1(S^1)=mathbb{R}\&xrightarrow{g} H^2(M)xrightarrow{h} H^2(U)oplus H^2(V)=H^2(mathbb{RP}^1)oplus H^2(mathbb{RP}^1)=0.
end{align*}
But the fact that it is exact sequence does not give me specific feature of $H^1(M)$ and $H^2(M)$. I have computed two possibilities.
Note that $d$ is zero map, so $e$ is injective. So only possible choice of $H^1(M)$ is $0$, $mathbb{R}$ and $mathbb{R}oplus mathbb{R}$. But $H^1(M)$ cannot be $0$ since otherwise $f$ is injective and it is impossible.
Then observe the following two cases.
(1) If $H^1(M)$ is $mathbb{R}$, $g$ is surjective zero map which means $H^2(M)=0$.
(2) If $H^1(M)$ is $mathbb{R}oplus mathbb{R}$, $f$ is zero map so $g$ is isomorphism. Thus, $H^2(M)=mathbb{R}$.
As we can see, either ways does not make any contradiction.
I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)
projective-space de-rham-cohomology
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
$endgroup$
2
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
2
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
2
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08
|
show 1 more comment
$begingroup$
I am trying to compute the de Rham cohomology group $H^p(mathbb{RP}^{n+1}#mathbb{RP}^{n+1})$ and I am stuck at computing $H^1(mathbb{RP}^2#mathbb{RP}^2)$. ($#$ stand for the connected sum)
Let $Usimeq mathbb{RP^2setminus {p}}$ and $Vsimeq mathbb{RP}^2setminus {q}$ such that $M=mathbb{RP}^2#mathbb{RP}^2=Ucup V$ and $Ucap Vsimeq S^1$. Then by using Mayer Vietoris, I get
begin{align*}
0&xrightarrow{a} H^0(M)=mathbb{R}xrightarrow{b} H^0(U)oplus H^0(V)=H^0(mathbb{RP}^1)oplus H^0(mathbb{RP}^1)=mathbb{R}oplus mathbb{R}xrightarrow{c} H^0(S^1)=mathbb{R} \& xrightarrow{d} H^1(M)xrightarrow{e} H^1(U)oplus H^1(V)=H^1(mathbb{RP}^1)oplus H^1(mathbb{RP}^1)=mathbb{R}oplus mathbb{R} xrightarrow{f} H^1(S^1)=mathbb{R}\&xrightarrow{g} H^2(M)xrightarrow{h} H^2(U)oplus H^2(V)=H^2(mathbb{RP}^1)oplus H^2(mathbb{RP}^1)=0.
end{align*}
But the fact that it is exact sequence does not give me specific feature of $H^1(M)$ and $H^2(M)$. I have computed two possibilities.
Note that $d$ is zero map, so $e$ is injective. So only possible choice of $H^1(M)$ is $0$, $mathbb{R}$ and $mathbb{R}oplus mathbb{R}$. But $H^1(M)$ cannot be $0$ since otherwise $f$ is injective and it is impossible.
Then observe the following two cases.
(1) If $H^1(M)$ is $mathbb{R}$, $g$ is surjective zero map which means $H^2(M)=0$.
(2) If $H^1(M)$ is $mathbb{R}oplus mathbb{R}$, $f$ is zero map so $g$ is isomorphism. Thus, $H^2(M)=mathbb{R}$.
As we can see, either ways does not make any contradiction.
I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)
projective-space de-rham-cohomology
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
$endgroup$
I am trying to compute the de Rham cohomology group $H^p(mathbb{RP}^{n+1}#mathbb{RP}^{n+1})$ and I am stuck at computing $H^1(mathbb{RP}^2#mathbb{RP}^2)$. ($#$ stand for the connected sum)
Let $Usimeq mathbb{RP^2setminus {p}}$ and $Vsimeq mathbb{RP}^2setminus {q}$ such that $M=mathbb{RP}^2#mathbb{RP}^2=Ucup V$ and $Ucap Vsimeq S^1$. Then by using Mayer Vietoris, I get
begin{align*}
0&xrightarrow{a} H^0(M)=mathbb{R}xrightarrow{b} H^0(U)oplus H^0(V)=H^0(mathbb{RP}^1)oplus H^0(mathbb{RP}^1)=mathbb{R}oplus mathbb{R}xrightarrow{c} H^0(S^1)=mathbb{R} \& xrightarrow{d} H^1(M)xrightarrow{e} H^1(U)oplus H^1(V)=H^1(mathbb{RP}^1)oplus H^1(mathbb{RP}^1)=mathbb{R}oplus mathbb{R} xrightarrow{f} H^1(S^1)=mathbb{R}\&xrightarrow{g} H^2(M)xrightarrow{h} H^2(U)oplus H^2(V)=H^2(mathbb{RP}^1)oplus H^2(mathbb{RP}^1)=0.
end{align*}
But the fact that it is exact sequence does not give me specific feature of $H^1(M)$ and $H^2(M)$. I have computed two possibilities.
Note that $d$ is zero map, so $e$ is injective. So only possible choice of $H^1(M)$ is $0$, $mathbb{R}$ and $mathbb{R}oplus mathbb{R}$. But $H^1(M)$ cannot be $0$ since otherwise $f$ is injective and it is impossible.
Then observe the following two cases.
(1) If $H^1(M)$ is $mathbb{R}$, $g$ is surjective zero map which means $H^2(M)=0$.
(2) If $H^1(M)$ is $mathbb{R}oplus mathbb{R}$, $f$ is zero map so $g$ is isomorphism. Thus, $H^2(M)=mathbb{R}$.
As we can see, either ways does not make any contradiction.
I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)
projective-space de-rham-cohomology
projective-space de-rham-cohomology
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
edited Dec 25 '18 at 15:56
Lev Ban
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
asked Dec 24 '18 at 22:58
Lev BanLev Ban
1,0821317
1,0821317
2
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
2
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
2
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08
|
show 1 more comment
2
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
2
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
2
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08
2
2
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
2
2
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
2
2
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08
|
show 1 more comment
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2
$begingroup$
The top cohomology of a non-orientable manifold must be $0$.
$endgroup$
– Cheerful Parsnip
Dec 24 '18 at 23:41
$begingroup$
@CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from?
$endgroup$
– Lev Ban
Dec 24 '18 at 23:42
$begingroup$
See the computation here: math.stackexchange.com/questions/187413/…
$endgroup$
– anomaly
Dec 24 '18 at 23:43
2
$begingroup$
@LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:04
2
$begingroup$
Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle.
$endgroup$
– Cheerful Parsnip
Dec 25 '18 at 0:08