Reference request for K-Theory linearization
$begingroup$
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
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
$endgroup$
add a comment |
$begingroup$
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
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
$endgroup$
1
$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57
add a comment |
$begingroup$
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
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
$endgroup$
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
reference-request homotopy-theory kt.k-theory-and-homology
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
edited Nov 13 '18 at 15:54
Noah Riggenbach
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
asked Nov 13 '18 at 15:41
Noah RiggenbachNoah Riggenbach
1385
1385
1
$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57
add a comment |
1
$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57
1
1
$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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).
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
$endgroup$
add a comment |
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$begingroup$
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).
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
$endgroup$
add a comment |
$begingroup$
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).
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
$endgroup$
add a comment |
$begingroup$
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).
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
$endgroup$
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).
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
answered Nov 13 '18 at 16:48
Denis NardinDenis Nardin
8,23223359
8,23223359
add a comment |
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$begingroup$
Uh, I might be a bit confused, but $Omega^inftySigma^infty_+ |G|$ is certainly an $A_infty$-ring
$endgroup$
– Denis Nardin
Nov 13 '18 at 15:49
$begingroup$
@DenisNardin I understand why both operations separately are $A^infty$, I just don't understand the distributivity I suppose.
$endgroup$
– Noah Riggenbach
Nov 13 '18 at 15:57