Reference request for K-Theory linearization

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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:



$$A(X)simeq mathbbZtimes BwidehatGl(Omega^inftySigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehatGl(Omega^inftySigma^infty |G|)$ to be $A^infty$, but since $Omega^inftySigma^infty |G|$ 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.










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    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    yesterday










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    yesterday














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:



$$A(X)simeq mathbbZtimes BwidehatGl(Omega^inftySigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehatGl(Omega^inftySigma^infty |G|)$ to be $A^infty$, but since $Omega^inftySigma^infty |G|$ 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
    yesterday










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    yesterday












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:



$$A(X)simeq mathbbZtimes BwidehatGl(Omega^inftySigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehatGl(Omega^inftySigma^infty |G|)$ to be $A^infty$, but since $Omega^inftySigma^infty |G|$ 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:



$$A(X)simeq mathbbZtimes BwidehatGl(Omega^inftySigma^infty |G|)$$



Where $|G|$ is a loop group of $X$. My problem is that to apply $B$ we need $widehatGl(Omega^inftySigma^infty |G|)$ to be $A^infty$, but since $Omega^inftySigma^infty |G|$ 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






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Check out our Code of Conduct.











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









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edited yesterday





















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asked yesterday









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






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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
    yesterday










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    yesterday












  • 1




    Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
    – Denis Nardin
    yesterday










  • @DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
    – Noah Riggenbach
    yesterday







1




1




Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
– Denis Nardin
yesterday




Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
– Denis Nardin
yesterday












@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
– Noah Riggenbach
yesterday




@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
– Noah Riggenbach
yesterday










1 Answer
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up vote
8
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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).






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

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    up vote
    8
    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
      8
      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
        8
        down vote



        accepted







        up vote
        8
        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 yesterday









        Denis Nardin

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