Inverting homotopy groups of spectra
$begingroup$
Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$
To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?
at.algebraic-topology stable-homotopy
edited Dec 22 '18 at 18:24
David White
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
$endgroup$
add a comment |
$begingroup$
Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$
To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?
at.algebraic-topology stable-homotopy
edited Dec 22 '18 at 18:24
David White
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
$endgroup$
add a comment |
$begingroup$
Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$
To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?
at.algebraic-topology stable-homotopy
edited Dec 22 '18 at 18:24
David White
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
$endgroup$
Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$pi_*(X)cong pi_{-*}(X')?$$
To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?
at.algebraic-topology stable-homotopy
at.algebraic-topology stable-homotopy
edited Dec 22 '18 at 18:24
David White
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
edited Dec 22 '18 at 18:24
David White
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
edited Dec 22 '18 at 18:24
David White
11.5k460101
edited Dec 22 '18 at 18:24
David White
11.5k460101
edited Dec 22 '18 at 18:24
David White
11.5k460101
11.5k460101
asked Dec 22 '18 at 16:56
user09127user09127
3888
asked Dec 22 '18 at 16:56
user09127user09127
3888
asked Dec 22 '18 at 16:56
user09127user09127
3888
3888
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):
$pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$
So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.
See here for more details
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
$endgroup$
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319293%2finverting-homotopy-groups-of-spectra%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
$begingroup$
Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):
$pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$
So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.
See here for more details
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
$endgroup$
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
add a comment |
$begingroup$
Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):
$pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$
So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.
See here for more details
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
$endgroup$
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
add a comment |
$begingroup$
Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):
$pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$
So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.
See here for more details
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
$endgroup$
Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{mathbb{Q}/mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):
$pi_{-ast}(I_{mathbb{Q}/mathbb{Z}}X) simeq Hom(pi_ast(X),mathbb{Q}/mathbb{Z})$
So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.
See here for more details
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
edited Dec 23 '18 at 12:15
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
answered Dec 22 '18 at 17:37
Sam GunninghamSam Gunningham
4,3922029
4,3922029
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
add a comment |
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
1
1
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
$begingroup$
There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$? More precisely, is the assignment $Xmapsto I_{mathbb Q/mathbb Z}X$ a functor from the $infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $infty$-groupoid level?
$endgroup$
– André Henriques
Dec 23 '18 at 21:42
1
1
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
$begingroup$
@AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_mathbf{Q/Z} := I_mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_mathbf{Q/Z}(X) = F(X, I_mathbf{Q/Z})$, and this is functorial in $X$.
$endgroup$
– skd
Dec 24 '18 at 5:47
1
1
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
$begingroup$
@skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{mathbb Q/mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices?
$endgroup$
– André Henriques
Dec 24 '18 at 16:18
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319293%2finverting-homotopy-groups-of-spectra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
