Product structure in Cohomology Spectral sequence
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In the Serre Spectral sequence, we know, the cup product structure induces a canonical product in all $E_r$ pages which is compatible with respect to the differential. I am trying to understand this product in $E_2$ page. $E_2^p,q= H^p(B,H^q(F,R)) times H^r(B,H^s(F,R))=E_2^r,s rightarrow H^p+r(B,H^q+s(F,R))=E_2^p+r,q+s $ has the description as follows, if $phi, psi$ are two cocycles , the we define the product as $phi cup psi$ and the coefficients are multiplied by the cup product structure in $H^*(F,R)$.
I believe, on the singular level, it means, for any ,p+r-singular simplex $sigma$, $ phi cup psi (sigma)= phi(sigma|[v_0,v_1,...,v_p]) cup psi(sigma|[v_p,..,v_p+r])$. Is my interpretation correct? Second question what does product mean exactly? I have seen, in some computation, Hatcher uses the fact product of two generator is again a generator? Why can the product not be some multiple of generator in the codomain? Am I missing something ? Kindly say some few words on the confusion. Thank you.
algebraic-topology homology-cohomology spectral-sequences
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up vote
2
down vote
favorite
In the Serre Spectral sequence, we know, the cup product structure induces a canonical product in all $E_r$ pages which is compatible with respect to the differential. I am trying to understand this product in $E_2$ page. $E_2^p,q= H^p(B,H^q(F,R)) times H^r(B,H^s(F,R))=E_2^r,s rightarrow H^p+r(B,H^q+s(F,R))=E_2^p+r,q+s $ has the description as follows, if $phi, psi$ are two cocycles , the we define the product as $phi cup psi$ and the coefficients are multiplied by the cup product structure in $H^*(F,R)$.
I believe, on the singular level, it means, for any ,p+r-singular simplex $sigma$, $ phi cup psi (sigma)= phi(sigma|[v_0,v_1,...,v_p]) cup psi(sigma|[v_p,..,v_p+r])$. Is my interpretation correct? Second question what does product mean exactly? I have seen, in some computation, Hatcher uses the fact product of two generator is again a generator? Why can the product not be some multiple of generator in the codomain? Am I missing something ? Kindly say some few words on the confusion. Thank you.
algebraic-topology homology-cohomology spectral-sequences
I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the Serre Spectral sequence, we know, the cup product structure induces a canonical product in all $E_r$ pages which is compatible with respect to the differential. I am trying to understand this product in $E_2$ page. $E_2^p,q= H^p(B,H^q(F,R)) times H^r(B,H^s(F,R))=E_2^r,s rightarrow H^p+r(B,H^q+s(F,R))=E_2^p+r,q+s $ has the description as follows, if $phi, psi$ are two cocycles , the we define the product as $phi cup psi$ and the coefficients are multiplied by the cup product structure in $H^*(F,R)$.
I believe, on the singular level, it means, for any ,p+r-singular simplex $sigma$, $ phi cup psi (sigma)= phi(sigma|[v_0,v_1,...,v_p]) cup psi(sigma|[v_p,..,v_p+r])$. Is my interpretation correct? Second question what does product mean exactly? I have seen, in some computation, Hatcher uses the fact product of two generator is again a generator? Why can the product not be some multiple of generator in the codomain? Am I missing something ? Kindly say some few words on the confusion. Thank you.
algebraic-topology homology-cohomology spectral-sequences
In the Serre Spectral sequence, we know, the cup product structure induces a canonical product in all $E_r$ pages which is compatible with respect to the differential. I am trying to understand this product in $E_2$ page. $E_2^p,q= H^p(B,H^q(F,R)) times H^r(B,H^s(F,R))=E_2^r,s rightarrow H^p+r(B,H^q+s(F,R))=E_2^p+r,q+s $ has the description as follows, if $phi, psi$ are two cocycles , the we define the product as $phi cup psi$ and the coefficients are multiplied by the cup product structure in $H^*(F,R)$.
I believe, on the singular level, it means, for any ,p+r-singular simplex $sigma$, $ phi cup psi (sigma)= phi(sigma|[v_0,v_1,...,v_p]) cup psi(sigma|[v_p,..,v_p+r])$. Is my interpretation correct? Second question what does product mean exactly? I have seen, in some computation, Hatcher uses the fact product of two generator is again a generator? Why can the product not be some multiple of generator in the codomain? Am I missing something ? Kindly say some few words on the confusion. Thank you.
algebraic-topology homology-cohomology spectral-sequences
algebraic-topology homology-cohomology spectral-sequences
asked Sep 11 at 3:13
Monkey.D.Luffy
234
234
I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32
add a comment |
I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32
I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32
I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32
add a comment |
1 Answer
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For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)otimes H^m(X;G')to
H^n+m(Xtimes X;Gotimes G')xrightarrowDelta^*H^n+m(X;Gotimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(Xtimes X)simeq C_*(X)otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $phi:C_n(X)to G$ and $phi':C_m(X)to G'$ to the cochain given by the composition $$C_n+m(Xtimes X)xrightarrowf_n+mbigopluslimits_i+j=n+mC_i(X)otimes C_j(X)xrightarrowtextprojC_n(X)otimes C_m(X)xrightarrowphiotimesphi'Gotimes G'$$Now suppose there is a group homomorphism $mu:Gotimes G'to G''$, for example the cup product $$H^i(Y;R)otimes H^j(Y;R)to H^i+j(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^n+m(X;Gotimes G')to H^n+m(X;G'')$$ via
$$bigg(phi:C_n+m(X)to Gotimes G'bigg)mapstobigg(phi:C_n+m(X)to Gotimes G'xrightarrowmuG''bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $mu$ the multiplication in $R$, this is the ordinary cup product.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)otimes H^m(X;G')to
H^n+m(Xtimes X;Gotimes G')xrightarrowDelta^*H^n+m(X;Gotimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(Xtimes X)simeq C_*(X)otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $phi:C_n(X)to G$ and $phi':C_m(X)to G'$ to the cochain given by the composition $$C_n+m(Xtimes X)xrightarrowf_n+mbigopluslimits_i+j=n+mC_i(X)otimes C_j(X)xrightarrowtextprojC_n(X)otimes C_m(X)xrightarrowphiotimesphi'Gotimes G'$$Now suppose there is a group homomorphism $mu:Gotimes G'to G''$, for example the cup product $$H^i(Y;R)otimes H^j(Y;R)to H^i+j(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^n+m(X;Gotimes G')to H^n+m(X;G'')$$ via
$$bigg(phi:C_n+m(X)to Gotimes G'bigg)mapstobigg(phi:C_n+m(X)to Gotimes G'xrightarrowmuG''bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $mu$ the multiplication in $R$, this is the ordinary cup product.
add a comment |
up vote
2
down vote
accepted
For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)otimes H^m(X;G')to
H^n+m(Xtimes X;Gotimes G')xrightarrowDelta^*H^n+m(X;Gotimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(Xtimes X)simeq C_*(X)otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $phi:C_n(X)to G$ and $phi':C_m(X)to G'$ to the cochain given by the composition $$C_n+m(Xtimes X)xrightarrowf_n+mbigopluslimits_i+j=n+mC_i(X)otimes C_j(X)xrightarrowtextprojC_n(X)otimes C_m(X)xrightarrowphiotimesphi'Gotimes G'$$Now suppose there is a group homomorphism $mu:Gotimes G'to G''$, for example the cup product $$H^i(Y;R)otimes H^j(Y;R)to H^i+j(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^n+m(X;Gotimes G')to H^n+m(X;G'')$$ via
$$bigg(phi:C_n+m(X)to Gotimes G'bigg)mapstobigg(phi:C_n+m(X)to Gotimes G'xrightarrowmuG''bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $mu$ the multiplication in $R$, this is the ordinary cup product.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)otimes H^m(X;G')to
H^n+m(Xtimes X;Gotimes G')xrightarrowDelta^*H^n+m(X;Gotimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(Xtimes X)simeq C_*(X)otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $phi:C_n(X)to G$ and $phi':C_m(X)to G'$ to the cochain given by the composition $$C_n+m(Xtimes X)xrightarrowf_n+mbigopluslimits_i+j=n+mC_i(X)otimes C_j(X)xrightarrowtextprojC_n(X)otimes C_m(X)xrightarrowphiotimesphi'Gotimes G'$$Now suppose there is a group homomorphism $mu:Gotimes G'to G''$, for example the cup product $$H^i(Y;R)otimes H^j(Y;R)to H^i+j(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^n+m(X;Gotimes G')to H^n+m(X;G'')$$ via
$$bigg(phi:C_n+m(X)to Gotimes G'bigg)mapstobigg(phi:C_n+m(X)to Gotimes G'xrightarrowmuG''bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $mu$ the multiplication in $R$, this is the ordinary cup product.
For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)otimes H^m(X;G')to
H^n+m(Xtimes X;Gotimes G')xrightarrowDelta^*H^n+m(X;Gotimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(Xtimes X)simeq C_*(X)otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $phi:C_n(X)to G$ and $phi':C_m(X)to G'$ to the cochain given by the composition $$C_n+m(Xtimes X)xrightarrowf_n+mbigopluslimits_i+j=n+mC_i(X)otimes C_j(X)xrightarrowtextprojC_n(X)otimes C_m(X)xrightarrowphiotimesphi'Gotimes G'$$Now suppose there is a group homomorphism $mu:Gotimes G'to G''$, for example the cup product $$H^i(Y;R)otimes H^j(Y;R)to H^i+j(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^n+m(X;Gotimes G')to H^n+m(X;G'')$$ via
$$bigg(phi:C_n+m(X)to Gotimes G'bigg)mapstobigg(phi:C_n+m(X)to Gotimes G'xrightarrowmuG''bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $mu$ the multiplication in $R$, this is the ordinary cup product.
edited Sep 14 at 19:45
answered Sep 14 at 6:15
Christian Carrick
912
912
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I think working with cellular cohomology would be more fruitful, given the construction of the LSSS.
– Tyrone
Sep 12 at 9:32