Non-abelian cohomologies The Next CEO of Stack OverflowAre the strata of Nakajima quiver varieties simply-connected? Do they have odd cohomology?Equivariant homology of $Omega X$/-space (references needed)?Finite generation of equivariant cohomology ringsIs there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?Equivariant and basic cohomologyIs there any “deep” relation between the localization theorem of equivariant cohomology and the localization theorem of equivariant K-theoryWeights on equivariant cohomology?Krull dimension in equivariant cohomologyIs there a geometric interpretation of the cohomology of an automorphism group acting on a universal deformation ring?Which rings are cohomology rings?

Non-abelian cohomologies



The Next CEO of Stack OverflowAre the strata of Nakajima quiver varieties simply-connected? Do they have odd cohomology?Equivariant homology of $Omega X$/-space (references needed)?Finite generation of equivariant cohomology ringsIs there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?Equivariant and basic cohomologyIs there any “deep” relation between the localization theorem of equivariant cohomology and the localization theorem of equivariant K-theoryWeights on equivariant cohomology?Krull dimension in equivariant cohomologyIs there a geometric interpretation of the cohomology of an automorphism group acting on a universal deformation ring?Which rings are cohomology rings?










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 $mathbbR$). 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
    Mar 23 at 3:48










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    Mar 23 at 4:10






  • 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
    Mar 24 at 2:25















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 $mathbbR$). 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
    Mar 23 at 3:48










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    Mar 23 at 4:10






  • 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
    Mar 24 at 2:25













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 $mathbbR$). 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 $mathbbR$). 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 Mar 24 at 11:07









Ali Taghavi

13352085




13352085










asked Mar 23 at 3:25









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
    Mar 23 at 3:48










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    Mar 23 at 4:10






  • 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
    Mar 24 at 2:25












  • 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
    Mar 23 at 3:48










  • $begingroup$
    Thanks for the comment.
    $endgroup$
    – Kiu
    Mar 23 at 4:10






  • 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
    Mar 24 at 2:25







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
Mar 23 at 3:48




$begingroup$
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
Mar 23 at 3:48












$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10




$begingroup$
Thanks for the comment.
$endgroup$
– Kiu
Mar 23 at 4:10




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
Mar 24 at 2:25




$begingroup$
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
Mar 24 at 2:25










1 Answer
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "504"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326110%2fnon-abelian-cohomologies%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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 Mar 23 at 7:14









        Qiaochu YuanQiaochu Yuan

        77.6k27318604




        77.6k27318604



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to MathOverflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326110%2fnon-abelian-cohomologies%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Adding axes to figuresAdding axes labels to LaTeX figuresLaTeX equivalent of ConTeXt buffersRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeNumerical conditional within tikz keys?adding axes to shapesAlign axes across subfiguresAdding figures with a certain orderLine up nested tikz enviroments or how to get rid of themAdding axes labels to LaTeX figures

            Image
            Fear of getting stuck on one programming language / technology that is not used in my country Is there a single word describing earning money through any means? Why do we read the Megillah by night and by day? Pre-mixing cryogenic fuels and using only one fuel tank Is the U.S. Code copyrighted by the Government? Should I outline or discovery write my stories? What is this called? Old film camera viewer? Does the expansion of the universe explain why the universe doesn't collapse? Removing files under particular conditions (number of files, file age) Argument list too long when zipping large list of certain files in a folder When a Cleric spontaneously casts a Cure Light Wounds spell, will a Pearl of Power recover the original spell or Cure Light Wounds? Non-trope happy ending? Is it better practice to read straight from sheet music rather than memorize it? Do Legal Documents Require Signing In Standard Pen Colors? Does an advisor owe his/her student anyt...
            Read more

            Tähtien Talli Jäsenet | Lähteet | NavigointivalikkoSuomen Hippos – Tähtien Talli

            Image
            RaviurheiluUrheilun kunniagalleriat Suomen HippoksenravihevostenLämminverisiäsuomenhevosiaSuomen raviurheilunBrad de Veluwe Tähtien Talli on vuonna 1995 perustettu Suomen Hippoksen ylläpitämä suomalaisten ravihevosten kunniagalleria. Tähtien Tallissa on toistaiseksi yhteensä 20 hevosta. Lämminverisiä galleriassa on kymmenen ja suomenhevosia kymmenen. Ulkomailla syntyneitä hevosia on mukana kolme. Vaikka Suomen raviurheilun perinteet ulottuvat jo 1800-luvun puolelle, painottuvat Tähtien Tallien valinnat viimeisten 20-30 vuoden aikana kilpailleisiin hevosiin. Viimeisin kunnianosoituksen saanut hevonen on Brad de Veluwe joka liitettiin Tähtien Talliin vuoden 2015 Ravigaalassa. Jäsenet | Reipas vuoden 1958 Kuninkuusraveissa Lahdessa. Passionate Kemp lämminveristen Suomen mestaruuden voittoloimi yllään. Hevonen Kilpailuvuodet Kasvattaja Valinta- vuosi Charme Asserdal 1975–1981 Erik Andersson 1995...
            Read more

            Do these cracks on my tires look bad? The Next CEO of Stack OverflowDry rot tire should I replace?Having to replace tiresFishtailed so easily? Bad tires? ABS?Filling the tires with something other than air, to avoid puncture hassles?Used Michelin tires safe to install?Do these tyre cracks necessitate replacement?Rumbling noise: tires or mechanicalIs it possible to fix noisy feathered tires?Are bad winter tires still better than summer tires in winter?Torque converter failure - Related to replacing only 2 tires?Why use snow tires on all 4 wheels on 2-wheel-drive cars?

            Image
            Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact Why do we say “un seul M” and not “une seule M” even though M is a “consonne”? Read/write a pipe-delimited file line by line with some simple text manipulation Free fall ellipse or parabola? Why did the Drakh emissary look so blurred in S04:E11 "Lines of Communication"? A hang glider, sudden unexpected lift to 25,000 feet altitude, what could do this? Planeswalker Ability and Death Timing Which acid/base does a strong base/acid react when added to a buffer solution? Is the 21st century's idea of "freedom of speech" based on precedent? Shortening a title without changing its meaning Man transported from Alternate World into ours by a Neutrino Detector Is it reasonable to ask other researchers to send me their previous grant applications? Find a path from s to t using as few red nodes as possible What difference does it make matching a w...
            Read more