Poincare duality in Bott Tu Differential Forms in Algebraic Topology
$begingroup$
This is related to Sec 5, Chpt 1 of Bott Tu Differential Forms in Algebraic Topology.
Given $Ssubset M$ with $M$ a oriented manifold and $S$ a closed oriented submanifold, one takes $omegain Z_c^k(M)$ where $Z_c^k$ denotes $k-$th de Rham cocycle defined by $Ker(Omega^k_c(M)xrightarrow{d}Omega^{k+1}_c(M))$ where $Omega_c^star$ are compactly supported differential forms. Now $i:Sto M$ as the inclusion map gives rise to $i^staromega$. Since $S$ is closed, $supp(i^star(omega))subset S$ is closed and compact by $supp(i^star(omega))subset supp(omega)$ and $S$ closed. Hence $int_S i^star:Z_c^k(M)to R$ is defined.
$textbf{Q:}$ Then the book says by Stokes Thm, $int_Si^starin Hom(H_c^k(M),R)$. There is no good reason to say $int_Si^star$ descends to cohomology map to $R$ unless Stokes Thm guarantees $partial S=emptyset$. Then in this case, it does descend to cohomology level map. Do I have to assume $partial S=emptyset$ here? If I assume simplicial cohomology and simplicial homology here, I do have $partial S=emptyset$ by standard algebraic topology version Poincare duality via cap product.
general-topology geometry differential-geometry algebraic-topology
asked Jan 16 at 0:17
user45765user45765
2,6882724
$endgroup$
add a comment |
$begingroup$
This is related to Sec 5, Chpt 1 of Bott Tu Differential Forms in Algebraic Topology.
Given $Ssubset M$ with $M$ a oriented manifold and $S$ a closed oriented submanifold, one takes $omegain Z_c^k(M)$ where $Z_c^k$ denotes $k-$th de Rham cocycle defined by $Ker(Omega^k_c(M)xrightarrow{d}Omega^{k+1}_c(M))$ where $Omega_c^star$ are compactly supported differential forms. Now $i:Sto M$ as the inclusion map gives rise to $i^staromega$. Since $S$ is closed, $supp(i^star(omega))subset S$ is closed and compact by $supp(i^star(omega))subset supp(omega)$ and $S$ closed. Hence $int_S i^star:Z_c^k(M)to R$ is defined.
$textbf{Q:}$ Then the book says by Stokes Thm, $int_Si^starin Hom(H_c^k(M),R)$. There is no good reason to say $int_Si^star$ descends to cohomology map to $R$ unless Stokes Thm guarantees $partial S=emptyset$. Then in this case, it does descend to cohomology level map. Do I have to assume $partial S=emptyset$ here? If I assume simplicial cohomology and simplicial homology here, I do have $partial S=emptyset$ by standard algebraic topology version Poincare duality via cap product.
general-topology geometry differential-geometry algebraic-topology
asked Jan 16 at 0:17
user45765user45765
2,6882724
$endgroup$
1
$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40
add a comment |
$begingroup$
This is related to Sec 5, Chpt 1 of Bott Tu Differential Forms in Algebraic Topology.
Given $Ssubset M$ with $M$ a oriented manifold and $S$ a closed oriented submanifold, one takes $omegain Z_c^k(M)$ where $Z_c^k$ denotes $k-$th de Rham cocycle defined by $Ker(Omega^k_c(M)xrightarrow{d}Omega^{k+1}_c(M))$ where $Omega_c^star$ are compactly supported differential forms. Now $i:Sto M$ as the inclusion map gives rise to $i^staromega$. Since $S$ is closed, $supp(i^star(omega))subset S$ is closed and compact by $supp(i^star(omega))subset supp(omega)$ and $S$ closed. Hence $int_S i^star:Z_c^k(M)to R$ is defined.
$textbf{Q:}$ Then the book says by Stokes Thm, $int_Si^starin Hom(H_c^k(M),R)$. There is no good reason to say $int_Si^star$ descends to cohomology map to $R$ unless Stokes Thm guarantees $partial S=emptyset$. Then in this case, it does descend to cohomology level map. Do I have to assume $partial S=emptyset$ here? If I assume simplicial cohomology and simplicial homology here, I do have $partial S=emptyset$ by standard algebraic topology version Poincare duality via cap product.
general-topology geometry differential-geometry algebraic-topology
asked Jan 16 at 0:17
user45765user45765
2,6882724
$endgroup$
This is related to Sec 5, Chpt 1 of Bott Tu Differential Forms in Algebraic Topology.
Given $Ssubset M$ with $M$ a oriented manifold and $S$ a closed oriented submanifold, one takes $omegain Z_c^k(M)$ where $Z_c^k$ denotes $k-$th de Rham cocycle defined by $Ker(Omega^k_c(M)xrightarrow{d}Omega^{k+1}_c(M))$ where $Omega_c^star$ are compactly supported differential forms. Now $i:Sto M$ as the inclusion map gives rise to $i^staromega$. Since $S$ is closed, $supp(i^star(omega))subset S$ is closed and compact by $supp(i^star(omega))subset supp(omega)$ and $S$ closed. Hence $int_S i^star:Z_c^k(M)to R$ is defined.
$textbf{Q:}$ Then the book says by Stokes Thm, $int_Si^starin Hom(H_c^k(M),R)$. There is no good reason to say $int_Si^star$ descends to cohomology map to $R$ unless Stokes Thm guarantees $partial S=emptyset$. Then in this case, it does descend to cohomology level map. Do I have to assume $partial S=emptyset$ here? If I assume simplicial cohomology and simplicial homology here, I do have $partial S=emptyset$ by standard algebraic topology version Poincare duality via cap product.
general-topology geometry differential-geometry algebraic-topology
general-topology geometry differential-geometry algebraic-topology
asked Jan 16 at 0:17
user45765user45765
2,6882724
asked Jan 16 at 0:17
user45765user45765
2,6882724
asked Jan 16 at 0:17
user45765user45765
2,6882724
asked Jan 16 at 0:17
user45765user45765
2,6882724
asked Jan 16 at 0:17
user45765user45765
2,6882724
2,6882724
1
$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40
add a comment |
1
$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40
1
1
$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40
add a comment |
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$begingroup$
Two comments. (1) Topologists use the word "closed" in the context of manifolds to mean "compact and without boundary." (2) You need to say that $S$ is a $k$-dimensional closed oriented submanifold.
$endgroup$
– Ted Shifrin
Jan 16 at 22:59
$begingroup$
@TedShifrin Ah. Thanks a lot. That is what I need.
$endgroup$
– user45765
Jan 17 at 2:40