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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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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 $$[S^4\vee S^4, S^3] \to [S^6, S^3] \to [S^3\times S^3, S^3] \to [S^3\vee S^3, S^3]$$ 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 $[S^3\times S^3, S^3]$. The group $[S^6, S^3] = \pi_6S^3$ is $\mathbb{Z}/12$. Moreover, the map $S^6 \to S^4\vee S^4$ is null as it is the suspension of the Whitehead product of the two inclusions $S^3 \to S^3\vee S^3$. 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.


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 $S^3\to K(\mathbb{Z},7)$ 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 $A_{k\pm \varepsilon}$-structures.


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 $S^7$.

  • $\begingroup$ 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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