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Higher cup-1 product of coboundaries is also a coboundary?



The 2019 Stack Overflow Developer Survey Results Are InPontrjagin square (Mosher and Tangora Question)The complement of a Cartesian product for characterizing the closed sets of a product spaceGroup cohomology: cup product for cyclic groupsCohomology ring of a direct sum via Poincare dualityDoes the cup product on de Rham cohomology induce a nondegenerate bilinear form?Reference Request: Equivariant cup product in singular cohomologyNotational issues with group cohomologyLinking number and cup productDoes $H^bullet(G, mathbbZ)$ have a coalgebra structure?singular cohomology and Poincaré duality










4












$begingroup$


In the cohomology or the group cohomology theory, suppose $mu_1$ and $mu_2$ are coboundaries of arbitrary dimensions,
$$
mu_1=delta eta_1
$$

$$
mu_2=delta eta_2
$$

where $eta_1$ and $eta_2$ are their lower dimensional split cochains.




Could we prove that the higher cup 1 product is also a coboundary?
$$
mu_1 cup_1 mu_2=(delta eta_1)cup_1 (delta eta_2)=delta(beta)?
$$



If so, how do we write this $beta$ explicitly?



Is a Higher cup-1 product of coboundaries also a coboundary?











share|cite|improve this question









$endgroup$











  • $begingroup$
    See this post for higher cup product mathoverflow.net/questions/268181/…
    $endgroup$
    – annie heart
    Mar 23 at 16:05










  • $begingroup$
    I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
    $endgroup$
    – annie heart
    Mar 23 at 16:44















4












$begingroup$


In the cohomology or the group cohomology theory, suppose $mu_1$ and $mu_2$ are coboundaries of arbitrary dimensions,
$$
mu_1=delta eta_1
$$

$$
mu_2=delta eta_2
$$

where $eta_1$ and $eta_2$ are their lower dimensional split cochains.




Could we prove that the higher cup 1 product is also a coboundary?
$$
mu_1 cup_1 mu_2=(delta eta_1)cup_1 (delta eta_2)=delta(beta)?
$$



If so, how do we write this $beta$ explicitly?



Is a Higher cup-1 product of coboundaries also a coboundary?











share|cite|improve this question









$endgroup$











  • $begingroup$
    See this post for higher cup product mathoverflow.net/questions/268181/…
    $endgroup$
    – annie heart
    Mar 23 at 16:05










  • $begingroup$
    I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
    $endgroup$
    – annie heart
    Mar 23 at 16:44













4












4








4


2



$begingroup$


In the cohomology or the group cohomology theory, suppose $mu_1$ and $mu_2$ are coboundaries of arbitrary dimensions,
$$
mu_1=delta eta_1
$$

$$
mu_2=delta eta_2
$$

where $eta_1$ and $eta_2$ are their lower dimensional split cochains.




Could we prove that the higher cup 1 product is also a coboundary?
$$
mu_1 cup_1 mu_2=(delta eta_1)cup_1 (delta eta_2)=delta(beta)?
$$



If so, how do we write this $beta$ explicitly?



Is a Higher cup-1 product of coboundaries also a coboundary?











share|cite|improve this question









$endgroup$




In the cohomology or the group cohomology theory, suppose $mu_1$ and $mu_2$ are coboundaries of arbitrary dimensions,
$$
mu_1=delta eta_1
$$

$$
mu_2=delta eta_2
$$

where $eta_1$ and $eta_2$ are their lower dimensional split cochains.




Could we prove that the higher cup 1 product is also a coboundary?
$$
mu_1 cup_1 mu_2=(delta eta_1)cup_1 (delta eta_2)=delta(beta)?
$$



If so, how do we write this $beta$ explicitly?



Is a Higher cup-1 product of coboundaries also a coboundary?








general-topology algebraic-topology homology-cohomology group-cohomology simplicial-complex






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 23 at 15:55









annie heartannie heart

673721




673721











  • $begingroup$
    See this post for higher cup product mathoverflow.net/questions/268181/…
    $endgroup$
    – annie heart
    Mar 23 at 16:05










  • $begingroup$
    I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
    $endgroup$
    – annie heart
    Mar 23 at 16:44
















  • $begingroup$
    See this post for higher cup product mathoverflow.net/questions/268181/…
    $endgroup$
    – annie heart
    Mar 23 at 16:05










  • $begingroup$
    I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
    $endgroup$
    – annie heart
    Mar 23 at 16:44















$begingroup$
See this post for higher cup product mathoverflow.net/questions/268181/…
$endgroup$
– annie heart
Mar 23 at 16:05




$begingroup$
See this post for higher cup product mathoverflow.net/questions/268181/…
$endgroup$
– annie heart
Mar 23 at 16:05












$begingroup$
I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
$endgroup$
– annie heart
Mar 23 at 16:44




$begingroup$
I also ask a related question: mathoverflow.net/q/326155/106497 in Use of Steenrod's Higher Cup product and the graded-commutativity
$endgroup$
– annie heart
Mar 23 at 16:44










0






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