Does an H-space have at most one delooping?
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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
New contributor
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show 2 more comments
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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
New contributor
$endgroup$
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.
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– Denis Nardin
2 days ago
4
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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.
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– Dylan Wilson
2 days ago
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Ah, Denis beat me to it.
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– Dylan Wilson
2 days ago
1
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Let me also make the stupid remark that you want $BX$ to be path-connected :)
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– Najib Idrissi
2 days ago
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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
2 days ago
|
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
New contributor
$endgroup$
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
New contributor
New contributor
New contributor
asked 2 days ago
AlexAlex
683
683
New contributor
New contributor
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.
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– Denis Nardin
2 days ago
4
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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.
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– Dylan Wilson
2 days ago
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Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
2 days ago
1
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Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
2 days ago
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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
2 days ago
|
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
2 days ago
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
2 days ago
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
2 days ago
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
2 days ago
$begingroup$
Ah, Denis beat me to it.
$endgroup$
– Dylan Wilson
2 days ago
1
1
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
2 days ago
$begingroup$
Let me also make the stupid remark that you want $BX$ to be path-connected :)
$endgroup$
– Najib Idrissi
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
|
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
yesterday
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@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
yesterday
add a comment |
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$begingroup$
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 yesterday
Lennart MeierLennart Meier
6,47924573
6,47924573
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 2 days ago
Dylan WilsonDylan Wilson
7,08773979
7,08773979
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
yesterday
$begingroup$
@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
yesterday
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
yesterday
$begingroup$
@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
yesterday
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 $S^7$.
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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 yesterday
Drew HeardDrew Heard
2,02211325
2,02211325
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But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
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– Dylan Wilson
yesterday
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@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)
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– Drew Heard
yesterday
add a comment |
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But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
yesterday
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@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
yesterday
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
yesterday
$begingroup$
But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?)
$endgroup$
– Dylan Wilson
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
add a comment |
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4
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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.
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– Denis Nardin
2 days ago
4
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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.
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– Dylan Wilson
2 days ago
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Ah, Denis beat me to it.
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– Dylan Wilson
2 days ago
1
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Let me also make the stupid remark that you want $BX$ to be path-connected :)
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– Najib Idrissi
2 days ago
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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.
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– Dylan Wilson
2 days ago