singular cohomology and Poincaré duality
$begingroup$
Suppose $M$ is a n-dimensional, finite type, oriented, smooth manifold. A $k$-dimensional cycle in $M$ is a pair ($S$,$phi$), where $S$ is a compact, oriented $k$-dimensional manifold without boundary, and $phi$: $S$ → $M$ is a smooth map. A $k$-cycle ($S$, $phi$) defines a linear map $H^k(M)rightarrow R$ given by $w mapsto int_S phi^* w$ where $w in H^k(M)$ and $phi^*$ is the pullback. In other words, that each cycle defines an element in $(H^k(M))^*$. Via the Poincaré duality we identify the vector space $(H^k(M))^∗$ with $H^{n−k}_{cpt}(M)$ where $H^{n-k}_{cpt}(M)$ denotes the compactly supported cohomology group. Thus there exists $delta_S in H^{n−k}_{cpt}(M)$, such that
begin{align}
int_M w wedge delta_S=int_S phi^*w.
end{align}
The compactly supported cohomology class $delta_S$ is called the Poincare dual of the pair $(S, phi)$.
My question is that: since in the above formula uses the wedge product, it is more or less related to De Rham cohomology(not?). Can the above formula be rephrased into the language of singular cohomology? More explicitly, consider a triangulation $Gamma$ of $M$, and a $k$-cycle ($S,phi)$ can be represented by a singular $k$-cycle, denoted as $C_S$ and the cohomology class $w$ can also defined in the singular cohomology sense, so we still use $w$ to denote it. Then I try to rephrase the above formula as
begin{align}
C_S cap w= Gamma cap (wcup w_S)
end{align}
or equivalently
begin{align}
int_{C_S} w= int_{Gamma} wcup w_S
end{align}
where $w_S in H^{n-k}(Gamma)$ is the Poincaré dual of $C_S$ in $Gamma$, so that $w cup w_S in H^n(Gamma)$, parallel to $delta_S$ defined above. $cap$ is the cap product and $cup$ means the cup product. Do this rephrase make sense? Is this $w_S$ right to be Poincaré dual of $C_S$?
algebraic-topology homology-cohomology de-rham-cohomology poincare-duality
edited Jan 2 at 0:59
Bernard
122k740116
asked Jan 2 at 0:48
user630944user630944
162
$endgroup$
add a comment |
$begingroup$
Suppose $M$ is a n-dimensional, finite type, oriented, smooth manifold. A $k$-dimensional cycle in $M$ is a pair ($S$,$phi$), where $S$ is a compact, oriented $k$-dimensional manifold without boundary, and $phi$: $S$ → $M$ is a smooth map. A $k$-cycle ($S$, $phi$) defines a linear map $H^k(M)rightarrow R$ given by $w mapsto int_S phi^* w$ where $w in H^k(M)$ and $phi^*$ is the pullback. In other words, that each cycle defines an element in $(H^k(M))^*$. Via the Poincaré duality we identify the vector space $(H^k(M))^∗$ with $H^{n−k}_{cpt}(M)$ where $H^{n-k}_{cpt}(M)$ denotes the compactly supported cohomology group. Thus there exists $delta_S in H^{n−k}_{cpt}(M)$, such that
begin{align}
int_M w wedge delta_S=int_S phi^*w.
end{align}
The compactly supported cohomology class $delta_S$ is called the Poincare dual of the pair $(S, phi)$.
My question is that: since in the above formula uses the wedge product, it is more or less related to De Rham cohomology(not?). Can the above formula be rephrased into the language of singular cohomology? More explicitly, consider a triangulation $Gamma$ of $M$, and a $k$-cycle ($S,phi)$ can be represented by a singular $k$-cycle, denoted as $C_S$ and the cohomology class $w$ can also defined in the singular cohomology sense, so we still use $w$ to denote it. Then I try to rephrase the above formula as
begin{align}
C_S cap w= Gamma cap (wcup w_S)
end{align}
or equivalently
begin{align}
int_{C_S} w= int_{Gamma} wcup w_S
end{align}
where $w_S in H^{n-k}(Gamma)$ is the Poincaré dual of $C_S$ in $Gamma$, so that $w cup w_S in H^n(Gamma)$, parallel to $delta_S$ defined above. $cap$ is the cap product and $cup$ means the cup product. Do this rephrase make sense? Is this $w_S$ right to be Poincaré dual of $C_S$?
algebraic-topology homology-cohomology de-rham-cohomology poincare-duality
edited Jan 2 at 0:59
Bernard
122k740116
asked Jan 2 at 0:48
user630944user630944
162
$endgroup$
add a comment |
$begingroup$
Suppose $M$ is a n-dimensional, finite type, oriented, smooth manifold. A $k$-dimensional cycle in $M$ is a pair ($S$,$phi$), where $S$ is a compact, oriented $k$-dimensional manifold without boundary, and $phi$: $S$ → $M$ is a smooth map. A $k$-cycle ($S$, $phi$) defines a linear map $H^k(M)rightarrow R$ given by $w mapsto int_S phi^* w$ where $w in H^k(M)$ and $phi^*$ is the pullback. In other words, that each cycle defines an element in $(H^k(M))^*$. Via the Poincaré duality we identify the vector space $(H^k(M))^∗$ with $H^{n−k}_{cpt}(M)$ where $H^{n-k}_{cpt}(M)$ denotes the compactly supported cohomology group. Thus there exists $delta_S in H^{n−k}_{cpt}(M)$, such that
begin{align}
int_M w wedge delta_S=int_S phi^*w.
end{align}
The compactly supported cohomology class $delta_S$ is called the Poincare dual of the pair $(S, phi)$.
My question is that: since in the above formula uses the wedge product, it is more or less related to De Rham cohomology(not?). Can the above formula be rephrased into the language of singular cohomology? More explicitly, consider a triangulation $Gamma$ of $M$, and a $k$-cycle ($S,phi)$ can be represented by a singular $k$-cycle, denoted as $C_S$ and the cohomology class $w$ can also defined in the singular cohomology sense, so we still use $w$ to denote it. Then I try to rephrase the above formula as
begin{align}
C_S cap w= Gamma cap (wcup w_S)
end{align}
or equivalently
begin{align}
int_{C_S} w= int_{Gamma} wcup w_S
end{align}
where $w_S in H^{n-k}(Gamma)$ is the Poincaré dual of $C_S$ in $Gamma$, so that $w cup w_S in H^n(Gamma)$, parallel to $delta_S$ defined above. $cap$ is the cap product and $cup$ means the cup product. Do this rephrase make sense? Is this $w_S$ right to be Poincaré dual of $C_S$?
algebraic-topology homology-cohomology de-rham-cohomology poincare-duality
edited Jan 2 at 0:59
Bernard
122k740116
asked Jan 2 at 0:48
user630944user630944
162
$endgroup$
Suppose $M$ is a n-dimensional, finite type, oriented, smooth manifold. A $k$-dimensional cycle in $M$ is a pair ($S$,$phi$), where $S$ is a compact, oriented $k$-dimensional manifold without boundary, and $phi$: $S$ → $M$ is a smooth map. A $k$-cycle ($S$, $phi$) defines a linear map $H^k(M)rightarrow R$ given by $w mapsto int_S phi^* w$ where $w in H^k(M)$ and $phi^*$ is the pullback. In other words, that each cycle defines an element in $(H^k(M))^*$. Via the Poincaré duality we identify the vector space $(H^k(M))^∗$ with $H^{n−k}_{cpt}(M)$ where $H^{n-k}_{cpt}(M)$ denotes the compactly supported cohomology group. Thus there exists $delta_S in H^{n−k}_{cpt}(M)$, such that
begin{align}
int_M w wedge delta_S=int_S phi^*w.
end{align}
The compactly supported cohomology class $delta_S$ is called the Poincare dual of the pair $(S, phi)$.
My question is that: since in the above formula uses the wedge product, it is more or less related to De Rham cohomology(not?). Can the above formula be rephrased into the language of singular cohomology? More explicitly, consider a triangulation $Gamma$ of $M$, and a $k$-cycle ($S,phi)$ can be represented by a singular $k$-cycle, denoted as $C_S$ and the cohomology class $w$ can also defined in the singular cohomology sense, so we still use $w$ to denote it. Then I try to rephrase the above formula as
begin{align}
C_S cap w= Gamma cap (wcup w_S)
end{align}
or equivalently
begin{align}
int_{C_S} w= int_{Gamma} wcup w_S
end{align}
where $w_S in H^{n-k}(Gamma)$ is the Poincaré dual of $C_S$ in $Gamma$, so that $w cup w_S in H^n(Gamma)$, parallel to $delta_S$ defined above. $cap$ is the cap product and $cup$ means the cup product. Do this rephrase make sense? Is this $w_S$ right to be Poincaré dual of $C_S$?
algebraic-topology homology-cohomology de-rham-cohomology poincare-duality
algebraic-topology homology-cohomology de-rham-cohomology poincare-duality
edited Jan 2 at 0:59
Bernard
122k740116
asked Jan 2 at 0:48
user630944user630944
162
edited Jan 2 at 0:59
Bernard
122k740116
asked Jan 2 at 0:48
user630944user630944
162
edited Jan 2 at 0:59
Bernard
122k740116
edited Jan 2 at 0:59
Bernard
122k740116
edited Jan 2 at 0:59
Bernard
122k740116
122k740116
asked Jan 2 at 0:48
user630944user630944
162
asked Jan 2 at 0:48
user630944user630944
162
asked Jan 2 at 0:48
user630944user630944
162
162
add a comment |
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