# Does an H-space have at most one delooping?

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?

• 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. – Denis Nardin Jan 15 at 16:12
• 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. – Dylan Wilson Jan 15 at 16:12
• Ah, Denis beat me to it. – Dylan Wilson Jan 15 at 16:12
• Let me also make the stupid remark that you want $BX$ to be path-connected :) – Najib Idrissi Jan 15 at 16:17
• 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. – 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 $$S^7$$.

• But what about producing even more loop space structures? (Isn’t S^7 not a loop space, for example?) – Dylan Wilson Jan 16 at 12:12
• @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) – Drew Heard Jan 16 at 13:05