Reference request for K-Theory linearization











up vote
7
down vote

favorite
1












I posted this question on math.se here:https://math.stackexchange.com/q/2996787/482732, but I think it may be more appropriate here, sorry if I am wrong about that.



In Waldhausen's paper Algebraic K theory of Spaces(the long one) he proves the following:



$$A(X)simeq mathbb{Z}times Bwidehat{Gl}(Omega^{infty}Sigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehat{Gl}(Omega^{infty}Sigma^infty |G|)$ to be $A^infty$, but since $Omega^{infty}Sigma^infty |G|$ is only a ring up to homotopy this isn't obvious to me. Waldhausen just brushes past this point, so I was hoping to get a reference to somewhere that shows this in detail. Thanks.










share|cite|improve this question









New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    Nov 13 at 15:49










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    Nov 13 at 15:57















up vote
7
down vote

favorite
1












I posted this question on math.se here:https://math.stackexchange.com/q/2996787/482732, but I think it may be more appropriate here, sorry if I am wrong about that.



In Waldhausen's paper Algebraic K theory of Spaces(the long one) he proves the following:



$$A(X)simeq mathbb{Z}times Bwidehat{Gl}(Omega^{infty}Sigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehat{Gl}(Omega^{infty}Sigma^infty |G|)$ to be $A^infty$, but since $Omega^{infty}Sigma^infty |G|$ is only a ring up to homotopy this isn't obvious to me. Waldhausen just brushes past this point, so I was hoping to get a reference to somewhere that shows this in detail. Thanks.










share|cite|improve this question









New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    Nov 13 at 15:49










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    Nov 13 at 15:57













up vote
7
down vote

favorite
1









up vote
7
down vote

favorite
1






1





I posted this question on math.se here:https://math.stackexchange.com/q/2996787/482732, but I think it may be more appropriate here, sorry if I am wrong about that.



In Waldhausen's paper Algebraic K theory of Spaces(the long one) he proves the following:



$$A(X)simeq mathbb{Z}times Bwidehat{Gl}(Omega^{infty}Sigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehat{Gl}(Omega^{infty}Sigma^infty |G|)$ to be $A^infty$, but since $Omega^{infty}Sigma^infty |G|$ is only a ring up to homotopy this isn't obvious to me. Waldhausen just brushes past this point, so I was hoping to get a reference to somewhere that shows this in detail. Thanks.










share|cite|improve this question









New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I posted this question on math.se here:https://math.stackexchange.com/q/2996787/482732, but I think it may be more appropriate here, sorry if I am wrong about that.



In Waldhausen's paper Algebraic K theory of Spaces(the long one) he proves the following:



$$A(X)simeq mathbb{Z}times Bwidehat{Gl}(Omega^{infty}Sigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehat{Gl}(Omega^{infty}Sigma^infty |G|)$ to be $A^infty$, but since $Omega^{infty}Sigma^infty |G|$ is only a ring up to homotopy this isn't obvious to me. Waldhausen just brushes past this point, so I was hoping to get a reference to somewhere that shows this in detail. Thanks.







reference-request homotopy-theory kt.k-theory-and-homology






share|cite|improve this question









New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Nov 13 at 15:54





















New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Nov 13 at 15:41









Noah Riggenbach

1385




1385




New contributor




Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Noah Riggenbach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    Nov 13 at 15:49










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    Nov 13 at 15:57














  • 1




    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    Nov 13 at 15:49










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    Nov 13 at 15:57








1




1




Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
– Denis Nardin
Nov 13 at 15:49




Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
– Denis Nardin
Nov 13 at 15:49












@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
– Noah Riggenbach
Nov 13 at 15:57




@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
– Noah Riggenbach
Nov 13 at 15:57










1 Answer
1






active

oldest

votes

















up vote
9
down vote



accepted










I claim that for every $A_infty$-space $A$, there is a canonical $A_infty$-ring structure on $Omega^inftySigma^infty_+A$.



First, $Sigma^infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_infty$-space $A$ to an $A_infty$-algebra in spectra $Sigma^infty_+A$, that is an $A_infty$-ring spectrum. The fact that $Sigma^infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra).



Secondly I claim that if $E$ is an $A_infty$-ring spectrum, then $Omega^infty E$ has a canonical $A_infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_infty$-ring spaces and $E_infty$-ring spectra).





If all you care for is a construction of the $A_infty$-structure on $GL_1(Sigma^infty_+A)$, I particularly like the approach in




Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology




where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_infty$-structure, since all automorphism groups in an ∞-category "trivially" do).






share|cite|improve this answer





















    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',
    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
    });


    }
    });






    Noah Riggenbach is a new contributor. Be nice, and check out our Code of Conduct.










     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315239%2freference-request-for-k-theory-linearization%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








    up vote
    9
    down vote



    accepted










    I claim that for every $A_infty$-space $A$, there is a canonical $A_infty$-ring structure on $Omega^inftySigma^infty_+A$.



    First, $Sigma^infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_infty$-space $A$ to an $A_infty$-algebra in spectra $Sigma^infty_+A$, that is an $A_infty$-ring spectrum. The fact that $Sigma^infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra).



    Secondly I claim that if $E$ is an $A_infty$-ring spectrum, then $Omega^infty E$ has a canonical $A_infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_infty$-ring spaces and $E_infty$-ring spectra).





    If all you care for is a construction of the $A_infty$-structure on $GL_1(Sigma^infty_+A)$, I particularly like the approach in




    Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology




    where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_infty$-structure, since all automorphism groups in an ∞-category "trivially" do).






    share|cite|improve this answer

























      up vote
      9
      down vote



      accepted










      I claim that for every $A_infty$-space $A$, there is a canonical $A_infty$-ring structure on $Omega^inftySigma^infty_+A$.



      First, $Sigma^infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_infty$-space $A$ to an $A_infty$-algebra in spectra $Sigma^infty_+A$, that is an $A_infty$-ring spectrum. The fact that $Sigma^infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra).



      Secondly I claim that if $E$ is an $A_infty$-ring spectrum, then $Omega^infty E$ has a canonical $A_infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_infty$-ring spaces and $E_infty$-ring spectra).





      If all you care for is a construction of the $A_infty$-structure on $GL_1(Sigma^infty_+A)$, I particularly like the approach in




      Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology




      where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_infty$-structure, since all automorphism groups in an ∞-category "trivially" do).






      share|cite|improve this answer























        up vote
        9
        down vote



        accepted







        up vote
        9
        down vote



        accepted






        I claim that for every $A_infty$-space $A$, there is a canonical $A_infty$-ring structure on $Omega^inftySigma^infty_+A$.



        First, $Sigma^infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_infty$-space $A$ to an $A_infty$-algebra in spectra $Sigma^infty_+A$, that is an $A_infty$-ring spectrum. The fact that $Sigma^infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra).



        Secondly I claim that if $E$ is an $A_infty$-ring spectrum, then $Omega^infty E$ has a canonical $A_infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_infty$-ring spaces and $E_infty$-ring spectra).





        If all you care for is a construction of the $A_infty$-structure on $GL_1(Sigma^infty_+A)$, I particularly like the approach in




        Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology




        where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_infty$-structure, since all automorphism groups in an ∞-category "trivially" do).






        share|cite|improve this answer












        I claim that for every $A_infty$-space $A$, there is a canonical $A_infty$-ring structure on $Omega^inftySigma^infty_+A$.



        First, $Sigma^infty_+$ from spaces to spectra is symmetric monoidal. So it sends an $A_infty$-space $A$ to an $A_infty$-algebra in spectra $Sigma^infty_+A$, that is an $A_infty$-ring spectrum. The fact that $Sigma^infty_+$ is symmetric monoidal (at the model category/∞-category level) can be found in any modern book treating the smash product of spectra (e.g. it is proposition 4.7 in Elmendorf-May-Kriz-Mandell and corollary 4.8.2.19 in Lurie's Higher Algebra).



        Secondly I claim that if $E$ is an $A_infty$-ring spectrum, then $Omega^infty E$ has a canonical $A_infty$-ring space structure. Exactly how this works will depend on your preferred definition of $A_infty$-ring space, but it can be proven, e.g., with the same technique that May uses to prove the analogous statement for $E_infty$-ring spaces (Corollary 7.5 in May's What precisely are $E_infty$-ring spaces and $E_infty$-ring spectra).





        If all you care for is a construction of the $A_infty$-structure on $GL_1(Sigma^infty_+A)$, I particularly like the approach in




        Matthew Ando, Andrew J. Blumberg, David Gepner, Michael J. Hopkins, Charles Rezk An ∞-categorical approach to R-line bundles, R-module Thom spectra, and twisted R-homology




        where they identify $GL_1(R)$ with the automorphism group of $R$ as an $R$-module (and so it has an $A_infty$-structure, since all automorphism groups in an ∞-category "trivially" do).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 13 at 16:48









        Denis Nardin

        7,09512552




        7,09512552






















            Noah Riggenbach is a new contributor. Be nice, and check out our Code of Conduct.










             

            draft saved


            draft discarded


















            Noah Riggenbach is a new contributor. Be nice, and check out our Code of Conduct.













            Noah Riggenbach is a new contributor. Be nice, and check out our Code of Conduct.












            Noah Riggenbach is a new contributor. Be nice, and check out our Code of Conduct.















             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315239%2freference-request-for-k-theory-linearization%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

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            How to change which sound is reproduced for terminal bell?

            Can I use Tabulator js library in my java Spring + Thymeleaf project?