How are characteristic classes morphisms of infinite loop spaces? (if they are)Group Completions and Infinite-Loop SpacesCommutativity of a diagram of boundary morphisms from the long exact sequence of homotopy groups of a fibration and its loop spacesWhat are the algebras over $Omega^kSigma^k$ ?$mathcalI$-functors and infinite loop spacesInfinite loop of a p-completed specta vs p-completion of infinite loop of the spectraComonadicity of spaces over spectra?Is there a fibration sequence of spectra $KmathbbF_qto KUto KU$?Homology of spectra vs homology of infinite loop spacesEquivalent definitions of Thom spectra
How are characteristic classes morphisms of infinite loop spaces? (if they are)
Group Completions and Infinite-Loop SpacesCommutativity of a diagram of boundary morphisms from the long exact sequence of homotopy groups of a fibration and its loop spacesWhat are the algebras over $Omega^kSigma^k$ ?$mathcalI$-functors and infinite loop spacesInfinite loop of a p-completed specta vs p-completion of infinite loop of the spectraComonadicity of spaces over spectra?Is there a fibration sequence of spectra $KmathbbF_qto KUto KU$?Homology of spectra vs homology of infinite loop spacesEquivalent definitions of Thom spectra
$begingroup$
The direct sum of real vector bundles endows $BO=mathrmcolim BO(n)$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $O$, i.e., one has a natural structure of abelian group up to homotopy on $BSO$, $BSpin$, $BString$, etc.
As $w(Eoplus F)=w(E)cup w(F)$ one sees that
$$
w_1colon BO to K(mathbbZ/2mathbbZ,1)
$$
is a morphism of abelian groups up to homotopy, and that similarly
$$
w_2colon BSO to K(mathbbZ/2mathbbZ,2)
$$
is a morphism of abelian groups up to homotopy. One can even make a step further and see
$$
BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)
$$
and
$$
BSpin to BSOxrightarroww_2 K(mathbbZ/2mathbbZ,2)
$$
as ``short exact sequences of abelian groups up to homotopy''.
One can be more ambitious here. Not only $BO$ is an abelian group up to homotopy, but it is an $infty$-loop space, i.e. $BO=Omega^infty bo$ for a certain connective spectrum $bo$. The same applies to $BSO$, $BSpin$,etc., and it also applies to $K(mathbbZ/2mathbbZ,n)$ as $K(mathbbZ/2mathbbZ,n)=Omega^inftySigma^n HmathbbZ/2$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations
$$
bso to boxrightarrowOmega^-inftyw_1 Sigma HmathbbZ/2
$$
and
$$
bspin to bsoxrightarrowOmega^-inftyw_2 Sigma^2 HmathbbZ/2
$$
of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $BSpin^c$ participates in a fibration of infinite loop spaces $K(mathbbZ,2)to BSpin^cto BSOxrightarrowbw_2K(mathbbZ,3)$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF.
My question is:
Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $Omega^infty$?
Where can I find a rigorous proof of this statement?
at.algebraic-topology loop-spaces
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
$endgroup$
add a comment
|
$begingroup$
The direct sum of real vector bundles endows $BO=mathrmcolim BO(n)$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $O$, i.e., one has a natural structure of abelian group up to homotopy on $BSO$, $BSpin$, $BString$, etc.
As $w(Eoplus F)=w(E)cup w(F)$ one sees that
$$
w_1colon BO to K(mathbbZ/2mathbbZ,1)
$$
is a morphism of abelian groups up to homotopy, and that similarly
$$
w_2colon BSO to K(mathbbZ/2mathbbZ,2)
$$
is a morphism of abelian groups up to homotopy. One can even make a step further and see
$$
BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)
$$
and
$$
BSpin to BSOxrightarroww_2 K(mathbbZ/2mathbbZ,2)
$$
as ``short exact sequences of abelian groups up to homotopy''.
One can be more ambitious here. Not only $BO$ is an abelian group up to homotopy, but it is an $infty$-loop space, i.e. $BO=Omega^infty bo$ for a certain connective spectrum $bo$. The same applies to $BSO$, $BSpin$,etc., and it also applies to $K(mathbbZ/2mathbbZ,n)$ as $K(mathbbZ/2mathbbZ,n)=Omega^inftySigma^n HmathbbZ/2$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations
$$
bso to boxrightarrowOmega^-inftyw_1 Sigma HmathbbZ/2
$$
and
$$
bspin to bsoxrightarrowOmega^-inftyw_2 Sigma^2 HmathbbZ/2
$$
of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $BSpin^c$ participates in a fibration of infinite loop spaces $K(mathbbZ,2)to BSpin^cto BSOxrightarrowbw_2K(mathbbZ,3)$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF.
My question is:
Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $Omega^infty$?
Where can I find a rigorous proof of this statement?
at.algebraic-topology loop-spaces
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
$endgroup$
add a comment
|
$begingroup$
The direct sum of real vector bundles endows $BO=mathrmcolim BO(n)$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $O$, i.e., one has a natural structure of abelian group up to homotopy on $BSO$, $BSpin$, $BString$, etc.
As $w(Eoplus F)=w(E)cup w(F)$ one sees that
$$
w_1colon BO to K(mathbbZ/2mathbbZ,1)
$$
is a morphism of abelian groups up to homotopy, and that similarly
$$
w_2colon BSO to K(mathbbZ/2mathbbZ,2)
$$
is a morphism of abelian groups up to homotopy. One can even make a step further and see
$$
BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)
$$
and
$$
BSpin to BSOxrightarroww_2 K(mathbbZ/2mathbbZ,2)
$$
as ``short exact sequences of abelian groups up to homotopy''.
One can be more ambitious here. Not only $BO$ is an abelian group up to homotopy, but it is an $infty$-loop space, i.e. $BO=Omega^infty bo$ for a certain connective spectrum $bo$. The same applies to $BSO$, $BSpin$,etc., and it also applies to $K(mathbbZ/2mathbbZ,n)$ as $K(mathbbZ/2mathbbZ,n)=Omega^inftySigma^n HmathbbZ/2$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations
$$
bso to boxrightarrowOmega^-inftyw_1 Sigma HmathbbZ/2
$$
and
$$
bspin to bsoxrightarrowOmega^-inftyw_2 Sigma^2 HmathbbZ/2
$$
of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $BSpin^c$ participates in a fibration of infinite loop spaces $K(mathbbZ,2)to BSpin^cto BSOxrightarrowbw_2K(mathbbZ,3)$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF.
My question is:
Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $Omega^infty$?
Where can I find a rigorous proof of this statement?
at.algebraic-topology loop-spaces
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
$endgroup$
The direct sum of real vector bundles endows $BO=mathrmcolim BO(n)$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $O$, i.e., one has a natural structure of abelian group up to homotopy on $BSO$, $BSpin$, $BString$, etc.
As $w(Eoplus F)=w(E)cup w(F)$ one sees that
$$
w_1colon BO to K(mathbbZ/2mathbbZ,1)
$$
is a morphism of abelian groups up to homotopy, and that similarly
$$
w_2colon BSO to K(mathbbZ/2mathbbZ,2)
$$
is a morphism of abelian groups up to homotopy. One can even make a step further and see
$$
BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)
$$
and
$$
BSpin to BSOxrightarroww_2 K(mathbbZ/2mathbbZ,2)
$$
as ``short exact sequences of abelian groups up to homotopy''.
One can be more ambitious here. Not only $BO$ is an abelian group up to homotopy, but it is an $infty$-loop space, i.e. $BO=Omega^infty bo$ for a certain connective spectrum $bo$. The same applies to $BSO$, $BSpin$,etc., and it also applies to $K(mathbbZ/2mathbbZ,n)$ as $K(mathbbZ/2mathbbZ,n)=Omega^inftySigma^n HmathbbZ/2$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations
$$
bso to boxrightarrowOmega^-inftyw_1 Sigma HmathbbZ/2
$$
and
$$
bspin to bsoxrightarrowOmega^-inftyw_2 Sigma^2 HmathbbZ/2
$$
of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$BSO to BOxrightarroww_1 K(mathbbZ/2mathbbZ,1)$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $BSpin^c$ participates in a fibration of infinite loop spaces $K(mathbbZ,2)to BSpin^cto BSOxrightarrowbw_2K(mathbbZ,3)$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF.
My question is:
Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $Omega^infty$?
Where can I find a rigorous proof of this statement?
at.algebraic-topology loop-spaces
at.algebraic-topology loop-spaces
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
asked Sep 27 at 21:37
domenico fiorenzadomenico fiorenza
5,33518 silver badges35 bronze badges
5,33518 silver badges35 bronze badges
add a comment
|
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = Omega^infty x$. One always has a
fibration sequence
$$y rightarrow x rightarrow Sigma^n Hpi_n(X)$$
and applying $Omega^infty$ to this yields a fibration sequence of spaces
$$Y rightarrow X rightarrow K(pi_n(X),n).$$
$Y$ is the $n$--connected cover of $X$.
(In your situation, one has successive covers $bspin rightarrow bso rightarrow bo$.)
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
$endgroup$
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
add a comment
|
$begingroup$
Yes it is true. You have correctly interpreted the intended meaning of the phrase ``fibration of infinite loop spaces''. One early reference is chapter I of $E_infty$ ring spaces and $E_infty$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
$endgroup$
add a comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = Omega^infty x$. One always has a
fibration sequence
$$y rightarrow x rightarrow Sigma^n Hpi_n(X)$$
and applying $Omega^infty$ to this yields a fibration sequence of spaces
$$Y rightarrow X rightarrow K(pi_n(X),n).$$
$Y$ is the $n$--connected cover of $X$.
(In your situation, one has successive covers $bspin rightarrow bso rightarrow bo$.)
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
$endgroup$
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
add a comment
|
$begingroup$
Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = Omega^infty x$. One always has a
fibration sequence
$$y rightarrow x rightarrow Sigma^n Hpi_n(X)$$
and applying $Omega^infty$ to this yields a fibration sequence of spaces
$$Y rightarrow X rightarrow K(pi_n(X),n).$$
$Y$ is the $n$--connected cover of $X$.
(In your situation, one has successive covers $bspin rightarrow bso rightarrow bo$.)
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
$endgroup$
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
add a comment
|
$begingroup$
Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = Omega^infty x$. One always has a
fibration sequence
$$y rightarrow x rightarrow Sigma^n Hpi_n(X)$$
and applying $Omega^infty$ to this yields a fibration sequence of spaces
$$Y rightarrow X rightarrow K(pi_n(X),n).$$
$Y$ is the $n$--connected cover of $X$.
(In your situation, one has successive covers $bspin rightarrow bso rightarrow bo$.)
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
$endgroup$
Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = Omega^infty x$. One always has a
fibration sequence
$$y rightarrow x rightarrow Sigma^n Hpi_n(X)$$
and applying $Omega^infty$ to this yields a fibration sequence of spaces
$$Y rightarrow X rightarrow K(pi_n(X),n).$$
$Y$ is the $n$--connected cover of $X$.
(In your situation, one has successive covers $bspin rightarrow bso rightarrow bo$.)
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
answered Sep 28 at 3:30
Nicholas KuhnNicholas Kuhn
5,23013 silver badges29 bronze badges
5,23013 silver badges29 bronze badges
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
add a comment
|
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
1
1
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
$begingroup$
Ah, sure! For any spectrum $x$ one has the fiber sequence $x_>n to x to x_leq n$ given by the standard ($n$-connected,$n$-truncated) $t$-structure on spectra. If $x$ is $(n-1)$-connected then $x_leq n=Sigma^n Hpi_n(x) = Sigma^n Hpi_n(Omega^infty x)$. Then one applies $Omega^infty$. I was missing the obvious here.
$endgroup$
– domenico fiorenza
Sep 28 at 5:52
add a comment
|
$begingroup$
Yes it is true. You have correctly interpreted the intended meaning of the phrase ``fibration of infinite loop spaces''. One early reference is chapter I of $E_infty$ ring spaces and $E_infty$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes it is true. You have correctly interpreted the intended meaning of the phrase ``fibration of infinite loop spaces''. One early reference is chapter I of $E_infty$ ring spaces and $E_infty$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes it is true. You have correctly interpreted the intended meaning of the phrase ``fibration of infinite loop spaces''. One early reference is chapter I of $E_infty$ ring spaces and $E_infty$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
$endgroup$
Yes it is true. You have correctly interpreted the intended meaning of the phrase ``fibration of infinite loop spaces''. One early reference is chapter I of $E_infty$ ring spaces and $E_infty$ ring spectra, available at http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
answered Sep 28 at 3:11
Peter MayPeter May
26.1k3 gold badges79 silver badges121 bronze badges
26.1k3 gold badges79 silver badges121 bronze badges
add a comment
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add a comment
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