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?










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$endgroup$








  • 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
















14












$begingroup$


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?










share|cite|improve this question









$endgroup$








  • 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














14












14








14


7



$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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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














  • 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










3 Answers
3






active

oldest

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18












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    15












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      10












      $begingroup$

      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 $S^7$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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












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      3 Answers
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      3 Answers
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      active

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      18












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        18












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          18












          18








          18





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 9:03









          Lennart MeierLennart Meier

          6,55424675




          6,55424675























              15












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                15












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  15












                  15








                  15





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 17:08









                  Dylan WilsonDylan Wilson

                  7,39073982




                  7,39073982























                      10












                      $begingroup$

                      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 $S^7$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        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
















                      10












                      $begingroup$

                      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 $S^7$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        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














                      10












                      10








                      10





                      $begingroup$

                      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 $S^7$.






                      share|cite|improve this answer









                      $endgroup$



                      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 $S^7$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      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


















                      • $begingroup$
                        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


















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