Mixing
Relation between 2 different homology and cohomology pairings
$begingroup$
Consider $X$ $n-$dimensional smooth compact oriented manifold. Denote $H^i(X,Omega)$ as the cohomology of smooth differential form, $H_i(X)$ as the $i-$th cycles and $C_i(X)$ as the singular chains and $Z^i(X)$ as the $i-$the cocycles of differential forms.
There are 2 different pairings.
$C_i(X)times Z^i(X)to R$ by $int_comega$ where $cin C_i(X),omegain Z^i(X)$. This obviously descends to $H_i(X)times H^i(X,Omega)to R$ by Stokes Thm. Clearly I have embedding $H_i(X)to (H^i(X,Omega))^star$ by shrinking the target cycle I need.
$textbf{Q:}$ Is this isomorphism?
The other one is $C^i(X)times C^{n-i}(X)to R$ by $int_X omega_iwedgeomega_{n-i}$ with $omega_kin C^k(X)$. Clearly this also descends to $H^i(X)times H^{n-i}(X,Omega)to R$. This pairing is non degenerate.
$textbf{Q:}$ What is the relationship between the 2 pairings? By poincare duality, I have $H_i(X)cong H^{n-i}(X)$. So second pairing can be replaced by $H_i(X)times H^i(X,Omega)to R$. Should I say if $omega_2in H^{n-i}(X,Omega)$ is identified with a cycle $cin H_i(X)$, given $omega_1in H^i(X)$, I consider $int_comega_1=int_Xomega_1wedgeomega_2$? Is this even correct?
general-topology geometry differential-geometry algebraic-topology
asked Jan 9 at 18:09
user45765user45765
2,6942722
$endgroup$
|
show 4 more comments
$begingroup$
Consider $X$ $n-$dimensional smooth compact oriented manifold. Denote $H^i(X,Omega)$ as the cohomology of smooth differential form, $H_i(X)$ as the $i-$th cycles and $C_i(X)$ as the singular chains and $Z^i(X)$ as the $i-$the cocycles of differential forms.
There are 2 different pairings.
$C_i(X)times Z^i(X)to R$ by $int_comega$ where $cin C_i(X),omegain Z^i(X)$. This obviously descends to $H_i(X)times H^i(X,Omega)to R$ by Stokes Thm. Clearly I have embedding $H_i(X)to (H^i(X,Omega))^star$ by shrinking the target cycle I need.
$textbf{Q:}$ Is this isomorphism?
The other one is $C^i(X)times C^{n-i}(X)to R$ by $int_X omega_iwedgeomega_{n-i}$ with $omega_kin C^k(X)$. Clearly this also descends to $H^i(X)times H^{n-i}(X,Omega)to R$. This pairing is non degenerate.
$textbf{Q:}$ What is the relationship between the 2 pairings? By poincare duality, I have $H_i(X)cong H^{n-i}(X)$. So second pairing can be replaced by $H_i(X)times H^i(X,Omega)to R$. Should I say if $omega_2in H^{n-i}(X,Omega)$ is identified with a cycle $cin H_i(X)$, given $omega_1in H^i(X)$, I consider $int_comega_1=int_Xomega_1wedgeomega_2$? Is this even correct?
general-topology geometry differential-geometry algebraic-topology
asked Jan 9 at 18:09
user45765user45765
2,6942722
$endgroup$
$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
1
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49
|
show 4 more comments
$begingroup$
Consider $X$ $n-$dimensional smooth compact oriented manifold. Denote $H^i(X,Omega)$ as the cohomology of smooth differential form, $H_i(X)$ as the $i-$th cycles and $C_i(X)$ as the singular chains and $Z^i(X)$ as the $i-$the cocycles of differential forms.
There are 2 different pairings.
$C_i(X)times Z^i(X)to R$ by $int_comega$ where $cin C_i(X),omegain Z^i(X)$. This obviously descends to $H_i(X)times H^i(X,Omega)to R$ by Stokes Thm. Clearly I have embedding $H_i(X)to (H^i(X,Omega))^star$ by shrinking the target cycle I need.
$textbf{Q:}$ Is this isomorphism?
The other one is $C^i(X)times C^{n-i}(X)to R$ by $int_X omega_iwedgeomega_{n-i}$ with $omega_kin C^k(X)$. Clearly this also descends to $H^i(X)times H^{n-i}(X,Omega)to R$. This pairing is non degenerate.
$textbf{Q:}$ What is the relationship between the 2 pairings? By poincare duality, I have $H_i(X)cong H^{n-i}(X)$. So second pairing can be replaced by $H_i(X)times H^i(X,Omega)to R$. Should I say if $omega_2in H^{n-i}(X,Omega)$ is identified with a cycle $cin H_i(X)$, given $omega_1in H^i(X)$, I consider $int_comega_1=int_Xomega_1wedgeomega_2$? Is this even correct?
general-topology geometry differential-geometry algebraic-topology
asked Jan 9 at 18:09
user45765user45765
2,6942722
$endgroup$
Consider $X$ $n-$dimensional smooth compact oriented manifold. Denote $H^i(X,Omega)$ as the cohomology of smooth differential form, $H_i(X)$ as the $i-$th cycles and $C_i(X)$ as the singular chains and $Z^i(X)$ as the $i-$the cocycles of differential forms.
There are 2 different pairings.
$C_i(X)times Z^i(X)to R$ by $int_comega$ where $cin C_i(X),omegain Z^i(X)$. This obviously descends to $H_i(X)times H^i(X,Omega)to R$ by Stokes Thm. Clearly I have embedding $H_i(X)to (H^i(X,Omega))^star$ by shrinking the target cycle I need.
$textbf{Q:}$ Is this isomorphism?
The other one is $C^i(X)times C^{n-i}(X)to R$ by $int_X omega_iwedgeomega_{n-i}$ with $omega_kin C^k(X)$. Clearly this also descends to $H^i(X)times H^{n-i}(X,Omega)to R$. This pairing is non degenerate.
$textbf{Q:}$ What is the relationship between the 2 pairings? By poincare duality, I have $H_i(X)cong H^{n-i}(X)$. So second pairing can be replaced by $H_i(X)times H^i(X,Omega)to R$. Should I say if $omega_2in H^{n-i}(X,Omega)$ is identified with a cycle $cin H_i(X)$, given $omega_1in H^i(X)$, I consider $int_comega_1=int_Xomega_1wedgeomega_2$? Is this even correct?
general-topology geometry differential-geometry algebraic-topology
general-topology geometry differential-geometry algebraic-topology
asked Jan 9 at 18:09
user45765user45765
2,6942722
asked Jan 9 at 18:09
user45765user45765
2,6942722
edited Jan 9 at 19:14
user45765
asked Jan 9 at 18:09
user45765user45765
2,6942722
asked Jan 9 at 18:09
user45765user45765
2,6942722
asked Jan 9 at 18:09
user45765user45765
2,6942722
2,6942722
$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
1
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49
|
show 4 more comments
$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
1
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49
$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
1
1
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49
|
show 4 more comments
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$begingroup$
May be you want to say "Let $X$ be an $n$-dimensional manifold".. Is your $R$ the real numbers $mathbb{R}$?
$endgroup$
– Praphulla Koushik
Jan 9 at 18:22
$begingroup$
This is, more or less, the definition of the Poincare dual.
$endgroup$
– Amitai Yuval
Jan 9 at 18:52
$begingroup$
@AmitaiYuval I thought poincare duality is obtained by intersection with top form rather than by this identification.
$endgroup$
– user45765
Jan 9 at 19:05
$begingroup$
Do you see that you are writing $M$ in the first place and then $X$ there after... The real numbers are related by $mathbb{R}$..
$endgroup$
– Praphulla Koushik
Jan 9 at 19:09
1
$begingroup$
I'm saying that's the definition of Poincaré duality, the cohomology class $[omega_2]$ being the Poincaré dual of the cycle $c$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:49