I am trying to understand Poincare duality at a concrete example. Consider X as the boundary of the embedded standard 2-simplex, i.e. X looks like this a / \ / \ v v b --- > c where the opposite edge of a,b,c gets the name A,B,C respectively. This is an orientated 1-manifold without boundary which is homeomorphic to S^1. To calculate the homology with integer coefficients one has to consider the chain complex 0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0 where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is for example (A,-B,C) and H_1(X)=Z. Now one obtains the dual cell structure (which is a simplicial complex too in this case) by taking the center of the edge A as a new 0- simplex a', b' and c' the same way and an edge connecting for example a' and b' is named C' (because it runs over the vertex c). The new simplicial complex (let's call it X' even X and X' are the same topological spaces) yields a canonical orientation a' > b' > c'. Now I like to show that there is a map f_* of cell complexes 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 | | f_1 | f_0 | v v v v 0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0 which means that f_0D=D'f_1. From this I can go further to show that the two complexes are chain-homotopic.
But here is my problem: What is the map D'? Every time I have seen cohomology the co-chain complex which have occurred has been of the form hom(C_*,Z) where C_* has been a normal chain complex. What can I doo here? I can't see how to interpretate the elements a' or A' (for example) as a map ?-->Z...? Can anybody help me to see how the relation to the cohomology occurs in my above example?
> I am trying to understand Poincare duality at a concrete example. > Consider X as the boundary of the embedded standard 2-simplex, i.e. X > looks like this > a > / \ > / \ > v v > b --- > c > where the opposite edge of a,b,c gets the name A,B,C respectively. > This is an orientated 1-manifold without boundary which is > homeomorphic to S^1. To calculate the homology with integer > coefficients one has to consider the chain complex > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0 > where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is > for example (A,-B,C) and H_1(X)=Z. > Now one obtains the dual cell structure (which is a simplicial complex > too in this case) by taking the center of the edge A as a new 0- > simplex a', b' and c' the same way and an edge connecting for example > a' and b' is named C' (because it runs over the vertex c). The new > simplicial complex (let's call it X' even X and X' are the same > topological spaces) yields a canonical orientation a' > b' > c'. > Now I like to show that there is a map f_* of cell complexes > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 > | | f_1 | f_0 | > v v v v > 0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0 > which means that f_0D=D'f_1. From this I can go further to show that > the two complexes are chain-homotopic.
> But here is my problem: What is the map D'? Every time I have seen > cohomology the co-chain complex which have occurred has been of the > form hom(C_*,Z) where C_* has been a normal chain complex. What can I > doo here? I can't see how to interpretate the elements a' or A' (for > example) as a map ?-->Z...? Can anybody help me to see how the > relation to the cohomology occurs in my above example?
> Thanks, > S.
Poincaré duality is an isomorphism from homology to cohomology. What you are doing is constructing a map between the complexes which compute *homology* for two related triangulations.
I doubt I'd be able to explain how Poincaré duality works in ASCII. The very besy you can do is get a textbook and follow the construction. Greenberg's book has a very nice discussion, for example.
<mariano.suarezalva...@gmail.com> wrote: > On 5 jul, 14:24, sanchopanch...@web.de wrote:
> > Hello,
> > I am trying to understand Poincare duality at a concrete example. > > Consider X as the boundary of the embedded standard 2-simplex, i.e. X > > looks like this > > a > > / \ > > / \ > > v v > > b --- > c > > where the opposite edge of a,b,c gets the name A,B,C respectively. > > This is an orientated 1-manifold without boundary which is > > homeomorphic to S^1. To calculate the homology with integer > > coefficients one has to consider the chain complex > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0 > > where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is > > for example (A,-B,C) and H_1(X)=Z. > > Now one obtains the dual cell structure (which is a simplicial complex > > too in this case) by taking the center of the edge A as a new 0- > > simplex a', b' and c' the same way and an edge connecting for example > > a' and b' is named C' (because it runs over the vertex c). The new > > simplicial complex (let's call it X' even X and X' are the same > > topological spaces) yields a canonical orientation a' > b' > c'. > > Now I like to show that there is a map f_* of cell complexes > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 > > | | f_1 | f_0 | > > v v v v > > 0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0 > > which means that f_0D=D'f_1. From this I can go further to show that > > the two complexes are chain-homotopic.
> > But here is my problem: What is the map D'? Every time I have seen > > cohomology the co-chain complex which have occurred has been of the > > form hom(C_*,Z) where C_* has been a normal chain complex. What can I > > doo here? I can't see how to interpretate the elements a' or A' (for > > example) as a map ?-->Z...? Can anybody help me to see how the > > relation to the cohomology occurs in my above example?
> > Thanks, > > S.
> Poincaré duality is an isomorphism from homology > to cohomology. What you are doing is constructing > a map between the complexes which compute *homology* > for two related triangulations.
> I doubt I'd be able to explain how Poincaré duality > works in ASCII. The very besy you can do is get > a textbook and follow the construction. > Greenberg's book has a very nice discussion, for > example.
> -- m
Thank you for your answer. Anyway, I don't think that my construction from above is very far from what is done in the Poincare duality construction. Some textbook of mine gives a similar example with S^2 and wikipedia (http://en.wikipedia.org/wiki/ Poincaré_duality#Dual_cell_structures) starts the same way as I have done as far as I have understood that. My problem from above rises in the wikipedia article here:
"Then one can show that the boundary operator C_k(X) \to C_{k-1}(X) is the transpose of the boundary operator C_{n-k+1}(Y)=Hom_Z(C_{k-1}(X),Z) \to C_{n-k}(Y)=Hom_Z(C_k(X),Z)."
If I can answer that problem and get to know what the boundary map D' is, I can show that the two complexes from above are chain homotopic because they are both chain homotopic to the complex of the barycentric subdivision. I am pretty confident, that the second complex my be easily seen as a simplicial cochain complex.
I don't want to proof anything of the Poincare duality theorem, I'd just like to understand this 'nearly obvious' (like the statement fom wikipedia) things written everywhere which are sadly not obvious to me.
> > > I am trying to understand Poincare duality at a concrete example. > > > Consider X as the boundary of the embedded standard 2-simplex, i.e. X > > > looks like this > > > a > > > / \ > > > / \ > > > v v > > > b --- > c > > > where the opposite edge of a,b,c gets the name A,B,C respectively. > > > This is an orientated 1-manifold without boundary which is > > > homeomorphic to S^1. To calculate the homology with integer > > > coefficients one has to consider the chain complex > > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0 > > > where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is > > > for example (A,-B,C) and H_1(X)=Z. > > > Now one obtains the dual cell structure (which is a simplicial complex > > > too in this case) by taking the center of the edge A as a new 0- > > > simplex a', b' and c' the same way and an edge connecting for example > > > a' and b' is named C' (because it runs over the vertex c). The new > > > simplicial complex (let's call it X' even X and X' are the same > > > topological spaces) yields a canonical orientation a' > b' > c'. > > > Now I like to show that there is a map f_* of cell complexes > > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 > > > | | f_1 | f_0 | > > > v v v v > > > 0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0 > > > which means that f_0D=D'f_1. From this I can go further to show that > > > the two complexes are chain-homotopic.
> > > But here is my problem: What is the map D'? Every time I have seen > > > cohomology the co-chain complex which have occurred has been of the > > > form hom(C_*,Z) where C_* has been a normal chain complex. What can I > > > doo here? I can't see how to interpretate the elements a' or A' (for > > > example) as a map ?-->Z...? Can anybody help me to see how the > > > relation to the cohomology occurs in my above example?
> > > Thanks, > > > S.
> > Poincaré duality is an isomorphism from homology > > to cohomology. What you are doing is constructing > > a map between the complexes which compute *homology* > > for two related triangulations.
> > I doubt I'd be able to explain how Poincaré duality > > works in ASCII. The very besy you can do is get > > a textbook and follow the construction. > > Greenberg's book has a very nice discussion, for > > example.
> > -- m
> Thank you for your answer. Anyway, I don't think that my construction > from above is very far from what is done in the Poincare duality > construction. Some textbook of mine gives a similar example with S^2 > and wikipedia (http://en.wikipedia.org/wiki/ > Poincaré_duality#Dual_cell_structures) starts the same way as I have > done as far as I have understood that. My problem from above rises in > the wikipedia article here:
> "Then one can show that the boundary operator > C_k(X) \to C_{k-1}(X) > is the transpose of the boundary operator > C_{n-k+1}(Y)=Hom_Z(C_{k-1}(X),Z) \to C_{n-k}(Y)=Hom_Z(C_k(X),Z)."
> If I can answer that problem and get to know what the boundary map D' > is, I can show that the two complexes from above are chain homotopic > because they are both chain homotopic to the complex of the > barycentric subdivision. I am pretty confident, that the second > complex my be easily seen as a simplicial cochain complex.
> I don't want to proof anything of the Poincare duality theorem, I'd > just like to understand this 'nearly obvious' (like the statement fom > wikipedia) things written everywhere which are sadly not obvious to > me.
The boundary map for the complex arising from the dual triangulation is computed in exactly the same way as one compuex the boundary map arising from any triangulation.
> > > > I am trying to understand Poincare duality at a concrete example. > > > > Consider X as the boundary of the embedded standard 2-simplex, i.e. X > > > > looks like this > > > > a > > > > / \ > > > > / \ > > > > v v > > > > b --- > c > > > > where the opposite edge of a,b,c gets the name A,B,C respectively. > > > > This is an orientated 1-manifold without boundary which is > > > > homeomorphic to S^1. To calculate the homology with integer > > > > coefficients one has to consider the chain complex > > > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0 > > > > where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is > > > > for example (A,-B,C) and H_1(X)=Z. > > > > Now one obtains the dual cell structure (which is a simplicial complex > > > > too in this case) by taking the center of the edge A as a new 0- > > > > simplex a', b' and c' the same way and an edge connecting for example > > > > a' and b' is named C' (because it runs over the vertex c). The new > > > > simplicial complex (let's call it X' even X and X' are the same > > > > topological spaces) yields a canonical orientation a' > b' > c'. > > > > Now I like to show that there is a map f_* of cell complexes > > > > 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 > > > > | | f_1 | f_0 | > > > > v v v v > > > > 0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0 > > > > which means that f_0D=D'f_1. From this I can go further to show that > > > > the two complexes are chain-homotopic.
> > > > But here is my problem: What is the map D'? Every time I have seen > > > > cohomology the co-chain complex which have occurred has been of the > > > > form hom(C_*,Z) where C_* has been a normal chain complex. What can I > > > > doo here? I can't see how to interpretate the elements a' or A' (for > > > > example) as a map ?-->Z...? Can anybody help me to see how the > > > > relation to the cohomology occurs in my above example?
> > > > Thanks, > > > > S.
> > > Poincaré duality is an isomorphism from homology > > > to cohomology. What you are doing is constructing > > > a map between the complexes which compute *homology* > > > for two related triangulations.
> > > I doubt I'd be able to explain how Poincaré duality > > > works in ASCII. The very besy you can do is get > > > a textbook and follow the construction. > > > Greenberg's book has a very nice discussion, for > > > example.
> > > -- m
> > Thank you for your answer. Anyway, I don't think that my construction > > from above is very far from what is done in the Poincare duality > > construction. Some textbook of mine gives a similar example with S^2 > > and wikipedia (http://en.wikipedia.org/wiki/ > > Poincaré_duality#Dual_cell_structures) starts the same way as I have > > done as far as I have understood that. My problem from above rises in > > the wikipedia article here:
> > "Then one can show that the boundary operator > > C_k(X) \to C_{k-1}(X) > > is the transpose of the boundary operator > > C_{n-k+1}(Y)=Hom_Z(C_{k-1}(X),Z) \to C_{n-k}(Y)=Hom_Z(C_k(X),Z)."
> > If I can answer that problem and get to know what the boundary map D' > > is, I can show that the two complexes from above are chain homotopic > > because they are both chain homotopic to the complex of the > > barycentric subdivision. I am pretty confident, that the second > > complex my be easily seen as a simplicial cochain complex.
> > I don't want to proof anything of the Poincare duality theorem, I'd > > just like to understand this 'nearly obvious' (like the statement fom > > wikipedia) things written everywhere which are sadly not obvious to > > me.
> The boundary map for the complex arising from the dual > triangulation is computed in exactly the same way as one > compuex the boundary map arising from any triangulation.
> -- m
Ok. here is my problem more exactly: This 0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0 is the chain complex arising from the original traingulation and D is the matrix [[0,-1,1],[-1,0,1],[-1,1,0]]. If I apply hom(-,Z) I get 0 <---- Z[A,B,C] <--M-- Z[a,b,c] <------0 where M is the transposed matrix D. Calculating the homology of the last complex gives me the cohomology of the space X. Now wikipedia says in other symbols that
"Then one can show that the boundary operator D: C_1(X) \to C_0(X) is the transpose of the boundary operator D' C_1(X')=Hom_Z(C_0(X),Z) \to C_0(X')=Hom_Z(C_1(X),Z)."
Where X' denotes the space X with the dual simplicial structure as considered above. D' should be the boundary map which you get by concerning this dual simplicial structure X' and forming the ordinary simplicial complex to calculate homology of X'. I can't see that D' equals M. More precise: To get this second complex I have to choose an orientation of the vertices of X'. Every orientation I can imagine does not give D' as the boundary map in the related complex. Can you help me?
