Poincare dual simplicial structure of complexes homotopy equivalent to manifolds Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)References Request (Poincare Duality via Unimodular pairing, de Rham isomorphism)Invariance of combinatorial/geometric euler characteristic(References) Relationship between skeletons of simplicial complexes and manifolds?Poincare duality in group (co)homologyClosed manifolds with isomorphic cohomology rings, but different cohomology modules over the Steenrod algebraBeginner question: Poincare Duality for Simplicial ComplexesHomotopy type of simplicial complexesA surface of genus g is not homotopy equivalent to a wedge sum of CW-complexesGeneralising dual triangulation of manifoldsDo contractible homology manifolds have one end?
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Poincare dual simplicial structure of complexes homotopy equivalent to manifolds
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)References Request (Poincare Duality via Unimodular pairing, de Rham isomorphism)Invariance of combinatorial/geometric euler characteristic(References) Relationship between skeletons of simplicial complexes and manifolds?Poincare duality in group (co)homologyClosed manifolds with isomorphic cohomology rings, but different cohomology modules over the Steenrod algebraBeginner question: Poincare Duality for Simplicial ComplexesHomotopy type of simplicial complexesA surface of genus g is not homotopy equivalent to a wedge sum of CW-complexesGeneralising dual triangulation of manifoldsDo contractible homology manifolds have one end?
$begingroup$
Given a closed $n$-manifold $M$, Poincare duality equips us with an isomorphism: $$H_k(M)cong H^n-k(M)$$
Here I'm speaking of singular homology with coeffecient in $mathbbZ_2$.
Suppose now $M$ has a triangulation $K$. Then clearly, we have a similar Poincare isomorphism as above for $K$ as well. In fact, even more can be said: there exists a polyhedral dual structure $K^vee$ of $K$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^vee)cong C^n-k(K)$$
(in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)
My question is:
Is there an isomorphism of the form $C_k(K^vee)cong C^n-k(K)$
even when $K$ is not necessarily a triangulation of $M$, but instead
only homotopy equivalent $M$
I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $K$ is the nerve of some "good cover" of $M$ (and thus homotopy equivalent from the nerve lemma)
algebraic-topology homology-cohomology triangulation simplicial-complex poincare-duality
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
$endgroup$
add a comment |
$begingroup$
Given a closed $n$-manifold $M$, Poincare duality equips us with an isomorphism: $$H_k(M)cong H^n-k(M)$$
Here I'm speaking of singular homology with coeffecient in $mathbbZ_2$.
Suppose now $M$ has a triangulation $K$. Then clearly, we have a similar Poincare isomorphism as above for $K$ as well. In fact, even more can be said: there exists a polyhedral dual structure $K^vee$ of $K$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^vee)cong C^n-k(K)$$
(in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)
My question is:
Is there an isomorphism of the form $C_k(K^vee)cong C^n-k(K)$
even when $K$ is not necessarily a triangulation of $M$, but instead
only homotopy equivalent $M$
I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $K$ is the nerve of some "good cover" of $M$ (and thus homotopy equivalent from the nerve lemma)
algebraic-topology homology-cohomology triangulation simplicial-complex poincare-duality
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
$endgroup$
add a comment |
$begingroup$
Given a closed $n$-manifold $M$, Poincare duality equips us with an isomorphism: $$H_k(M)cong H^n-k(M)$$
Here I'm speaking of singular homology with coeffecient in $mathbbZ_2$.
Suppose now $M$ has a triangulation $K$. Then clearly, we have a similar Poincare isomorphism as above for $K$ as well. In fact, even more can be said: there exists a polyhedral dual structure $K^vee$ of $K$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^vee)cong C^n-k(K)$$
(in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)
My question is:
Is there an isomorphism of the form $C_k(K^vee)cong C^n-k(K)$
even when $K$ is not necessarily a triangulation of $M$, but instead
only homotopy equivalent $M$
I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $K$ is the nerve of some "good cover" of $M$ (and thus homotopy equivalent from the nerve lemma)
algebraic-topology homology-cohomology triangulation simplicial-complex poincare-duality
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
$endgroup$
Given a closed $n$-manifold $M$, Poincare duality equips us with an isomorphism: $$H_k(M)cong H^n-k(M)$$
Here I'm speaking of singular homology with coeffecient in $mathbbZ_2$.
Suppose now $M$ has a triangulation $K$. Then clearly, we have a similar Poincare isomorphism as above for $K$ as well. In fact, even more can be said: there exists a polyhedral dual structure $K^vee$ of $K$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^vee)cong C^n-k(K)$$
(in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)
My question is:
Is there an isomorphism of the form $C_k(K^vee)cong C^n-k(K)$
even when $K$ is not necessarily a triangulation of $M$, but instead
only homotopy equivalent $M$
I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $K$ is the nerve of some "good cover" of $M$ (and thus homotopy equivalent from the nerve lemma)
algebraic-topology homology-cohomology triangulation simplicial-complex poincare-duality
algebraic-topology homology-cohomology triangulation simplicial-complex poincare-duality
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
edited Apr 1 at 14:54
Itamar Vigi
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
asked Apr 1 at 13:42
Itamar VigiItamar Vigi
326110
326110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
$endgroup$
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
$endgroup$
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
add a comment |
$begingroup$
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
$endgroup$
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
add a comment |
$begingroup$
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
$endgroup$
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
answered Apr 2 at 12:52
Najib IdrissiNajib Idrissi
42k473143
42k473143
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
add a comment |
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
$begingroup$
I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality?
$endgroup$
– Itamar Vigi
Apr 2 at 13:49
1
1
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
@ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) cong C^n-*(K^vee, partial K^vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^-1(X) = mathbbF_2$. There is never anything in negative degrees.
$endgroup$
– Najib Idrissi
Apr 2 at 13:51
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
$begingroup$
Thank you for pointing out the absurdity of what I'm looking for.
$endgroup$
– Itamar Vigi
Apr 4 at 8:12
add a comment |
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