Does an H-space have at most one delooping?
$begingroup$
I am new to H-spaces, delooping, etc. I know that not every $H$-space has a delooping (e.g. Stasheff's theorem, one needs a group-like $A_infty$ space). I also know that the same space can have inequivalent deloopings corresponding to different $H$-space structures. An example is $BU$, which has two $H$-space structures (coming from direct sum, respectively tensor product, of vector bundles).
Once you fix an $H$-space structure on a space $X$, is there at most one (up to homotopy) delooping $BX$ such that $X$ and $Omega BX$ are equivalent as $H$-spaces, in some appropriate sense?
homotopy-theory stable-homotopy
$endgroup$
|
show 2 more comments
$begingroup$
I am new to H-spaces, delooping, etc. I know that not every $H$-space has a delooping (e.g. Stasheff's theorem, one needs a group-like $A_infty$ space). I also know that the same space can have inequivalent deloopings corresponding to different $H$-space structures. An example is $BU$, which has two $H$-space structures (coming from direct sum, respectively tensor product, of vector bundles).
Once you fix an $H$-space structure on a space $X$, is there at most one (up to homotopy) delooping $BX$ such that $X$ and $Omega BX$ are equivalent as $H$-spaces, in some appropriate sense?
homotopy-theory stable-homotopy
$endgroup$
4
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
4
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18
|
show 2 more comments
$begingroup$
I am new to H-spaces, delooping, etc. I know that not every $H$-space has a delooping (e.g. Stasheff's theorem, one needs a group-like $A_infty$ space). I also know that the same space can have inequivalent deloopings corresponding to different $H$-space structures. An example is $BU$, which has two $H$-space structures (coming from direct sum, respectively tensor product, of vector bundles).
Once you fix an $H$-space structure on a space $X$, is there at most one (up to homotopy) delooping $BX$ such that $X$ and $Omega BX$ are equivalent as $H$-spaces, in some appropriate sense?
homotopy-theory stable-homotopy
$endgroup$
I am new to H-spaces, delooping, etc. I know that not every $H$-space has a delooping (e.g. Stasheff's theorem, one needs a group-like $A_infty$ space). I also know that the same space can have inequivalent deloopings corresponding to different $H$-space structures. An example is $BU$, which has two $H$-space structures (coming from direct sum, respectively tensor product, of vector bundles).
Once you fix an $H$-space structure on a space $X$, is there at most one (up to homotopy) delooping $BX$ such that $X$ and $Omega BX$ are equivalent as $H$-spaces, in some appropriate sense?
homotopy-theory stable-homotopy
homotopy-theory stable-homotopy
asked Jan 15 at 16:02
AlexAlex
733
733
4
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
4
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18
|
show 2 more comments
4
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
4
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18
4
4
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
4
4
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
1
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Another example is $S^3$. If I am not mistaken, there are exactly $12$ H-space structures on $S^3$. Indeed, we can consider the long exact sequence
$$[S^4vee S^4, S^3] to [S^6, S^3] to [S^3times S^3, S^3] to [S^3vee S^3, S^3]$$
of groups (using an arbitrary loop structure on $S^3$). We have to count the preimages of the folding map $S^3 vee S^3 to S^3$ in $[S^3times S^3, S^3]$. The group $[S^6, S^3] = pi_6S^3$ is $mathbb{Z}/12$. Moreover, the map $S^6 to S^4vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 to S^3vee S^3$. Thus, the number of $H$-space structures (up to homotopy) is in bijection with $mathbb{Z}/12$.
On the other hand, there are uncountably many loop structures on $S^3$. This was proven by Rector in Loop structures on the homotopy type of $S^3$. What he does is roughly the following: He considers spaces that are $p$-locally equivalent to $mathbb{HP}^infty$ for every prime $p$, but not necessarily equivalent to $mathbb{HP}^infty$ themselves, and whose loop space is equivalent to $S^3$. He shows that there is a ${0,1}^{infty}$ worth of them, i.e. uncountably many. In particular, there are $H$-space structures on $S^3$ with infinitely many deloopings.
This technique having certain fixed pieces at all the primes, but mixing them in a new way to obtain a new (integral) space is sometimes called Zabrodsky mixing. There is an amusing description on p.79 of Adams's Infinite loop spaces.
$endgroup$
add a comment |
$begingroup$
Alright, here's an example (assuming I didn't mess up). Let $S^3=Omega mathbb{H}P^{infty}$ have the standard loop-space structure. Let $X = K(mathbb{Z}, 6) times S^3$ with the product $A_{infty}$-structure, and let $Y$ be the fiber of the map $S^3 to K(mathbb{Z},7)$ obtained by looping down a nontrivial map $mathbb{H}P^{infty}to K(mathbb{Z},8)$. Then $X$ and $Y$ are have inequivalent deloopings, by design, but they are equivalent as $H$-spaces because there is a unique $H$-space map $S^3to K(mathbb{Z},7)$ up to homotopy through $A_2$-maps (namely, zero). (This is because an $A_k$-map to an Eilenberg-MacLane space from an $A_{infty}$-space is determined by a cohomology class in a skeleton of the delooping).
More generally, this trick should produce inequivalent $A_{infty}$-structures on $K(mathbb{Z}, 4k-2) times S^3$ which are equivalent as $A_{kpm varepsilon}$-structures.
$endgroup$
add a comment |
$begingroup$
The real projective spaces $mathbb{R}P^3 cong SO(3)$ and $mathbb{R}P^7$ also give fun examples. Naylor proved that there exist 768 $H$-space structures on $SO(3)$, while Rees shows that there exist 30,720 (!) $H$-space structures on $mathbb{R}P^7$.
For the spheres, James proved that $H$-space structures on $S^n$ (when they exist) are in 1-1 correspondence with elements of $pi_{2n}S^n$. When $n = 3$, this gives Lennart's calculation that there exist 12 $H$-space structures on $S^3$. When $n = 7$, we have $pi_{14}S^7 cong mathbb{Z}/120$, so there exist 120 $H$-space structures on $S^7$.
$endgroup$
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
add a comment |
Your Answer
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%2f320950%2fdoes-an-h-space-have-at-most-one-delooping%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another example is $S^3$. If I am not mistaken, there are exactly $12$ H-space structures on $S^3$. Indeed, we can consider the long exact sequence
$$[S^4vee S^4, S^3] to [S^6, S^3] to [S^3times S^3, S^3] to [S^3vee S^3, S^3]$$
of groups (using an arbitrary loop structure on $S^3$). We have to count the preimages of the folding map $S^3 vee S^3 to S^3$ in $[S^3times S^3, S^3]$. The group $[S^6, S^3] = pi_6S^3$ is $mathbb{Z}/12$. Moreover, the map $S^6 to S^4vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 to S^3vee S^3$. Thus, the number of $H$-space structures (up to homotopy) is in bijection with $mathbb{Z}/12$.
On the other hand, there are uncountably many loop structures on $S^3$. This was proven by Rector in Loop structures on the homotopy type of $S^3$. What he does is roughly the following: He considers spaces that are $p$-locally equivalent to $mathbb{HP}^infty$ for every prime $p$, but not necessarily equivalent to $mathbb{HP}^infty$ themselves, and whose loop space is equivalent to $S^3$. He shows that there is a ${0,1}^{infty}$ worth of them, i.e. uncountably many. In particular, there are $H$-space structures on $S^3$ with infinitely many deloopings.
This technique having certain fixed pieces at all the primes, but mixing them in a new way to obtain a new (integral) space is sometimes called Zabrodsky mixing. There is an amusing description on p.79 of Adams's Infinite loop spaces.
$endgroup$
add a comment |
$begingroup$
Another example is $S^3$. If I am not mistaken, there are exactly $12$ H-space structures on $S^3$. Indeed, we can consider the long exact sequence
$$[S^4vee S^4, S^3] to [S^6, S^3] to [S^3times S^3, S^3] to [S^3vee S^3, S^3]$$
of groups (using an arbitrary loop structure on $S^3$). We have to count the preimages of the folding map $S^3 vee S^3 to S^3$ in $[S^3times S^3, S^3]$. The group $[S^6, S^3] = pi_6S^3$ is $mathbb{Z}/12$. Moreover, the map $S^6 to S^4vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 to S^3vee S^3$. Thus, the number of $H$-space structures (up to homotopy) is in bijection with $mathbb{Z}/12$.
On the other hand, there are uncountably many loop structures on $S^3$. This was proven by Rector in Loop structures on the homotopy type of $S^3$. What he does is roughly the following: He considers spaces that are $p$-locally equivalent to $mathbb{HP}^infty$ for every prime $p$, but not necessarily equivalent to $mathbb{HP}^infty$ themselves, and whose loop space is equivalent to $S^3$. He shows that there is a ${0,1}^{infty}$ worth of them, i.e. uncountably many. In particular, there are $H$-space structures on $S^3$ with infinitely many deloopings.
This technique having certain fixed pieces at all the primes, but mixing them in a new way to obtain a new (integral) space is sometimes called Zabrodsky mixing. There is an amusing description on p.79 of Adams's Infinite loop spaces.
$endgroup$
add a comment |
$begingroup$
Another example is $S^3$. If I am not mistaken, there are exactly $12$ H-space structures on $S^3$. Indeed, we can consider the long exact sequence
$$[S^4vee S^4, S^3] to [S^6, S^3] to [S^3times S^3, S^3] to [S^3vee S^3, S^3]$$
of groups (using an arbitrary loop structure on $S^3$). We have to count the preimages of the folding map $S^3 vee S^3 to S^3$ in $[S^3times S^3, S^3]$. The group $[S^6, S^3] = pi_6S^3$ is $mathbb{Z}/12$. Moreover, the map $S^6 to S^4vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 to S^3vee S^3$. Thus, the number of $H$-space structures (up to homotopy) is in bijection with $mathbb{Z}/12$.
On the other hand, there are uncountably many loop structures on $S^3$. This was proven by Rector in Loop structures on the homotopy type of $S^3$. What he does is roughly the following: He considers spaces that are $p$-locally equivalent to $mathbb{HP}^infty$ for every prime $p$, but not necessarily equivalent to $mathbb{HP}^infty$ themselves, and whose loop space is equivalent to $S^3$. He shows that there is a ${0,1}^{infty}$ worth of them, i.e. uncountably many. In particular, there are $H$-space structures on $S^3$ with infinitely many deloopings.
This technique having certain fixed pieces at all the primes, but mixing them in a new way to obtain a new (integral) space is sometimes called Zabrodsky mixing. There is an amusing description on p.79 of Adams's Infinite loop spaces.
$endgroup$
Another example is $S^3$. If I am not mistaken, there are exactly $12$ H-space structures on $S^3$. Indeed, we can consider the long exact sequence
$$[S^4vee S^4, S^3] to [S^6, S^3] to [S^3times S^3, S^3] to [S^3vee S^3, S^3]$$
of groups (using an arbitrary loop structure on $S^3$). We have to count the preimages of the folding map $S^3 vee S^3 to S^3$ in $[S^3times S^3, S^3]$. The group $[S^6, S^3] = pi_6S^3$ is $mathbb{Z}/12$. Moreover, the map $S^6 to S^4vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 to S^3vee S^3$. Thus, the number of $H$-space structures (up to homotopy) is in bijection with $mathbb{Z}/12$.
On the other hand, there are uncountably many loop structures on $S^3$. This was proven by Rector in Loop structures on the homotopy type of $S^3$. What he does is roughly the following: He considers spaces that are $p$-locally equivalent to $mathbb{HP}^infty$ for every prime $p$, but not necessarily equivalent to $mathbb{HP}^infty$ themselves, and whose loop space is equivalent to $S^3$. He shows that there is a ${0,1}^{infty}$ worth of them, i.e. uncountably many. In particular, there are $H$-space structures on $S^3$ with infinitely many deloopings.
This technique having certain fixed pieces at all the primes, but mixing them in a new way to obtain a new (integral) space is sometimes called Zabrodsky mixing. There is an amusing description on p.79 of Adams's Infinite loop spaces.
answered Jan 16 at 9:03
Lennart MeierLennart Meier
6,55424675
6,55424675
add a comment |
add a comment |
$begingroup$
Alright, here's an example (assuming I didn't mess up). Let $S^3=Omega mathbb{H}P^{infty}$ have the standard loop-space structure. Let $X = K(mathbb{Z}, 6) times S^3$ with the product $A_{infty}$-structure, and let $Y$ be the fiber of the map $S^3 to K(mathbb{Z},7)$ obtained by looping down a nontrivial map $mathbb{H}P^{infty}to K(mathbb{Z},8)$. Then $X$ and $Y$ are have inequivalent deloopings, by design, but they are equivalent as $H$-spaces because there is a unique $H$-space map $S^3to K(mathbb{Z},7)$ up to homotopy through $A_2$-maps (namely, zero). (This is because an $A_k$-map to an Eilenberg-MacLane space from an $A_{infty}$-space is determined by a cohomology class in a skeleton of the delooping).
More generally, this trick should produce inequivalent $A_{infty}$-structures on $K(mathbb{Z}, 4k-2) times S^3$ which are equivalent as $A_{kpm varepsilon}$-structures.
$endgroup$
add a comment |
$begingroup$
Alright, here's an example (assuming I didn't mess up). Let $S^3=Omega mathbb{H}P^{infty}$ have the standard loop-space structure. Let $X = K(mathbb{Z}, 6) times S^3$ with the product $A_{infty}$-structure, and let $Y$ be the fiber of the map $S^3 to K(mathbb{Z},7)$ obtained by looping down a nontrivial map $mathbb{H}P^{infty}to K(mathbb{Z},8)$. Then $X$ and $Y$ are have inequivalent deloopings, by design, but they are equivalent as $H$-spaces because there is a unique $H$-space map $S^3to K(mathbb{Z},7)$ up to homotopy through $A_2$-maps (namely, zero). (This is because an $A_k$-map to an Eilenberg-MacLane space from an $A_{infty}$-space is determined by a cohomology class in a skeleton of the delooping).
More generally, this trick should produce inequivalent $A_{infty}$-structures on $K(mathbb{Z}, 4k-2) times S^3$ which are equivalent as $A_{kpm varepsilon}$-structures.
$endgroup$
add a comment |
$begingroup$
Alright, here's an example (assuming I didn't mess up). Let $S^3=Omega mathbb{H}P^{infty}$ have the standard loop-space structure. Let $X = K(mathbb{Z}, 6) times S^3$ with the product $A_{infty}$-structure, and let $Y$ be the fiber of the map $S^3 to K(mathbb{Z},7)$ obtained by looping down a nontrivial map $mathbb{H}P^{infty}to K(mathbb{Z},8)$. Then $X$ and $Y$ are have inequivalent deloopings, by design, but they are equivalent as $H$-spaces because there is a unique $H$-space map $S^3to K(mathbb{Z},7)$ up to homotopy through $A_2$-maps (namely, zero). (This is because an $A_k$-map to an Eilenberg-MacLane space from an $A_{infty}$-space is determined by a cohomology class in a skeleton of the delooping).
More generally, this trick should produce inequivalent $A_{infty}$-structures on $K(mathbb{Z}, 4k-2) times S^3$ which are equivalent as $A_{kpm varepsilon}$-structures.
$endgroup$
Alright, here's an example (assuming I didn't mess up). Let $S^3=Omega mathbb{H}P^{infty}$ have the standard loop-space structure. Let $X = K(mathbb{Z}, 6) times S^3$ with the product $A_{infty}$-structure, and let $Y$ be the fiber of the map $S^3 to K(mathbb{Z},7)$ obtained by looping down a nontrivial map $mathbb{H}P^{infty}to K(mathbb{Z},8)$. Then $X$ and $Y$ are have inequivalent deloopings, by design, but they are equivalent as $H$-spaces because there is a unique $H$-space map $S^3to K(mathbb{Z},7)$ up to homotopy through $A_2$-maps (namely, zero). (This is because an $A_k$-map to an Eilenberg-MacLane space from an $A_{infty}$-space is determined by a cohomology class in a skeleton of the delooping).
More generally, this trick should produce inequivalent $A_{infty}$-structures on $K(mathbb{Z}, 4k-2) times S^3$ which are equivalent as $A_{kpm varepsilon}$-structures.
answered Jan 15 at 17:08
Dylan WilsonDylan Wilson
7,39073982
7,39073982
add a comment |
add a comment |
$begingroup$
The real projective spaces $mathbb{R}P^3 cong SO(3)$ and $mathbb{R}P^7$ also give fun examples. Naylor proved that there exist 768 $H$-space structures on $SO(3)$, while Rees shows that there exist 30,720 (!) $H$-space structures on $mathbb{R}P^7$.
For the spheres, James proved that $H$-space structures on $S^n$ (when they exist) are in 1-1 correspondence with elements of $pi_{2n}S^n$. When $n = 3$, this gives Lennart's calculation that there exist 12 $H$-space structures on $S^3$. When $n = 7$, we have $pi_{14}S^7 cong mathbb{Z}/120$, so there exist 120 $H$-space structures on $S^7$.
$endgroup$
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
add a comment |
$begingroup$
The real projective spaces $mathbb{R}P^3 cong SO(3)$ and $mathbb{R}P^7$ also give fun examples. Naylor proved that there exist 768 $H$-space structures on $SO(3)$, while Rees shows that there exist 30,720 (!) $H$-space structures on $mathbb{R}P^7$.
For the spheres, James proved that $H$-space structures on $S^n$ (when they exist) are in 1-1 correspondence with elements of $pi_{2n}S^n$. When $n = 3$, this gives Lennart's calculation that there exist 12 $H$-space structures on $S^3$. When $n = 7$, we have $pi_{14}S^7 cong mathbb{Z}/120$, so there exist 120 $H$-space structures on $S^7$.
$endgroup$
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
add a comment |
$begingroup$
The real projective spaces $mathbb{R}P^3 cong SO(3)$ and $mathbb{R}P^7$ also give fun examples. Naylor proved that there exist 768 $H$-space structures on $SO(3)$, while Rees shows that there exist 30,720 (!) $H$-space structures on $mathbb{R}P^7$.
For the spheres, James proved that $H$-space structures on $S^n$ (when they exist) are in 1-1 correspondence with elements of $pi_{2n}S^n$. When $n = 3$, this gives Lennart's calculation that there exist 12 $H$-space structures on $S^3$. When $n = 7$, we have $pi_{14}S^7 cong mathbb{Z}/120$, so there exist 120 $H$-space structures on $S^7$.
$endgroup$
The real projective spaces $mathbb{R}P^3 cong SO(3)$ and $mathbb{R}P^7$ also give fun examples. Naylor proved that there exist 768 $H$-space structures on $SO(3)$, while Rees shows that there exist 30,720 (!) $H$-space structures on $mathbb{R}P^7$.
For the spheres, James proved that $H$-space structures on $S^n$ (when they exist) are in 1-1 correspondence with elements of $pi_{2n}S^n$. When $n = 3$, this gives Lennart's calculation that there exist 12 $H$-space structures on $S^3$. When $n = 7$, we have $pi_{14}S^7 cong mathbb{Z}/120$, so there exist 120 $H$-space structures on $S^7$.
answered Jan 16 at 10:50
Drew HeardDrew Heard
2,06211326
2,06211326
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
add a comment |
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
Jan 16 at 12:12
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
$begingroup$
@DylanWilson Thanks! I got over excited, and answered a different question. You're right, $S^7$ is not a loop space (presumably $mathbb{R}P^7$ is not either)
$endgroup$
– Drew Heard
Jan 16 at 13:05
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%2f320950%2fdoes-an-h-space-have-at-most-one-delooping%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
4
$begingroup$
You are essentially asking if an h-space structure can extended at most uniquely at an $A_∞$-structure. I'm pretty sure the answer is no, although I don't have a counterexample in my head right now.
$endgroup$
– Denis Nardin
Jan 15 at 16:12
4
$begingroup$
Nope. This is equivalent to asking whether an $A_2$-structure has at most one refinement to an $A_{infty}$-structure, but this need not hold.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
Jan 15 at 16:12
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
Jan 15 at 16:17
$begingroup$
Ok, let's try an example: $Omega S^2$ is equivalent to $S^1 times Omega S^3$ but they have different deloopings. We'd be good if we can show they are equivalent as $H$-spaces.
$endgroup$
– Dylan Wilson
Jan 15 at 16:18