Non-abelian cohomologies












3












$begingroup$


Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?



If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?






at.algebraic-topology cohomology







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$endgroup$








  • 1




    $begingroup$
    Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
    $endgroup$
    – LSpice
    yesterday










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    yesterday






  • 1




    $begingroup$
    I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
    $endgroup$
    – user51223
    yesterday
















3












$begingroup$


Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?



If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?






at.algebraic-topology cohomology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
    $endgroup$
    – LSpice
    yesterday










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    yesterday






  • 1




    $begingroup$
    I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
    $endgroup$
    – user51223
    yesterday














3












3








3


1



$begingroup$


Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?



If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?






at.algebraic-topology cohomology







share|cite|improve this question











$endgroup$




Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?



If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?






at.algebraic-topology cohomology




at.algebraic-topology cohomology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 15 hours ago









Ali Taghavi

8052085




8052085










asked yesterday









KiuKiu

393112




393112








  • 1




    $begingroup$
    Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
    $endgroup$
    – LSpice
    yesterday










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    yesterday






  • 1




    $begingroup$
    I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
    $endgroup$
    – user51223
    yesterday














  • 1




    $begingroup$
    Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
    $endgroup$
    – LSpice
    yesterday










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    yesterday






  • 1




    $begingroup$
    I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
    $endgroup$
    – user51223
    yesterday








1




1




$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
yesterday




$begingroup$
Although it's not what you're asking for, the Galois cohomology sets $operatorname H^1(E/F, mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $mathbb G$ make sense even if $mathbb G$ is non-Abelian.
$endgroup$
– LSpice
yesterday












$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday




$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
yesterday




1




1




$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
yesterday




$begingroup$
I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative?
$endgroup$
– user51223
yesterday










1 Answer
1






active

oldest

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8












$begingroup$


Is there a notion of cohomology ring of X with coefficients in A?




Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map



$$H^n(X, A) times H^m(X, A) to H^{n+m}(X, A)$$



in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.






share|cite|improve this answer









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    active

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    8












    $begingroup$


    Is there a notion of cohomology ring of X with coefficients in A?




    Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map



    $$H^n(X, A) times H^m(X, A) to H^{n+m}(X, A)$$



    in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$


      Is there a notion of cohomology ring of X with coefficients in A?




      Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map



      $$H^n(X, A) times H^m(X, A) to H^{n+m}(X, A)$$



      in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$


        Is there a notion of cohomology ring of X with coefficients in A?




        Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map



        $$H^n(X, A) times H^m(X, A) to H^{n+m}(X, A)$$



        in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.






        share|cite|improve this answer









        $endgroup$




        Is there a notion of cohomology ring of X with coefficients in A?




        Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A times A to A$ which induces a map



        $$H^n(X, A) times H^m(X, A) to H^{n+m}(X, A)$$



        in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Qiaochu YuanQiaochu Yuan

        77.5k27317603




        77.5k27317603






























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