It is well known that $S^n$ admits an H-space structure if and only if $n=0,1,3,7$. I'm interested in whether there are other suspensions $\Sigma X$ that admit H-space structures:

Question 1 For which $X$ (not a sphere) is $\Sigma X$ an H-space? And what about $\Sigma X$ that are associative H-spaces?

My motivation is that we have a construction (in the framework of homotopy type theory, and presumably portable to a wide range of model categories) that gives an H-space structure on the join $\Sigma X * \Sigma X$ whenever $\Sigma X$ has a homotopy-associative H-space structure (that is compatible with an involution on $X$ – for details, see these slides). Thus, it would be interesting to know some more examples where this construction applies.

This also leads to a follow-up question (mostly in case the answer to Q1 is “none”):

Question 2 If we go to a localization, do we get more answers to Q1? What about in other (non-stable) model categories?

Any references would be appreciated.

Finally, let me share a little scratch-work in trying to answer Q1 (feel free to ignore!): Any H-space structure $\mu : \Sigma X \times \Sigma X \to \Sigma X$ gives rise a Hopf map $H(\mu) : \Sigma X * \Sigma X \to \Sigma^2 X$. Assuming $X$ is pointed, we get a map $\Sigma^3(X \wedge X) \to \Sigma^2 X$. Applying $K$-cohomology to the cofiber sequence should yield information restricting $X$, but I failed to get much milage out of it without a notion of “bidegree” for $\mu$ (generalizing the case of $X$ a sphere).

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    $\begingroup$ Rationally you know that on the one hand cup products on a suspension vanish, and on the other hand the rational cohomology of a reasonable H-space is the free (super) commutative algebra on some generators. So I think the conclusion is that $\Sigma X$ is rationally an odd sphere. $\endgroup$ – Qiaochu Yuan Apr 10 '16 at 19:19
  • $\begingroup$ Thanks! I had a feeling there were very few choices, and this seems to confirm it. It is also in keeping with my feeling that there are probably no other non-trivial applications of this Cayley-Dickson-like construction (though I'd love to be proved wrong!). $\endgroup$ – Ulrik Buchholtz Apr 10 '16 at 19:33

If $Y$ is a connected CW-complex of finite type which is both an H-space and a co-H-space, then $Y$ has the homotopy type of $S^1$, $S^3$, $S^7$ or a point. This is a result of Robert West:

Robert W. West, $H$-spaces which are co-$H$-spaces, Proc. Amer. Math. Soc. 31 (1972), 580--582.

It follows that if $X$ is a finite type CW-complex such that $\Sigma X$ is an H-space, then $\Sigma X$ is homotopy equivalent to one of these spaces.

On the other hand, Adams and Walker give an example of a $4$-dimensional infinite CW-complex $Y$ which is both an Eilenberg--Mac Lane space and a Moore space of type $(\mathbb{Q},3)$:

J. F. Adams and G. Walker, An example in homotopy theory, Proc. Cambridge Philos. Soc. 60 (1964), 699--700.

This $Y$ is a suspension by construction, and an H-space by virtue of being an Eilenberg--Mac Lane space of an abelian group.

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  • $\begingroup$ This is great! Just to expand a bit on the Adams-Walker construction, $Y$ is the homotopy colimit of a sequence of $S^3$s connected by the degree $i$ maps for all $i$ (is $Y=S^3_{\mathbb Q}$?), and it is thus the suspension of the corresponding sequence of $S^2$s. We only need that $S^3$ is an H-space to show that $Y$ is a $K(\mathbb Z,3)$, so a similar thing happens for a sequence of $S^7$s. $\endgroup$ – Ulrik Buchholtz Apr 10 '16 at 21:42
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    $\begingroup$ Yes, I think $Y$ is actually the sphere $S^3$ rationalized. In fact, the rationalization of any odd sphere is an example of a suspension which is also an H-space (as per Qiaochu's comment). $\endgroup$ – Mark Grant Apr 11 '16 at 6:35
  • $\begingroup$ Good, I'll accept this answer which together with Qiaochu's comment goes a long way, even though there's perhaps room to say more about suspension H-spaces in the space of homotopy types rationally equivalent to odd spheres and in other homotopy theories (perhaps equivariantly). (BTW I of course meant $K(\mathbb Q,3)$ in the comment above.) $\endgroup$ – Ulrik Buchholtz Apr 11 '16 at 14:27

Here are some comments about the case where $X$ is not assumed to have finite type. Put $Y=\Sigma X$. For any field $K$, the groups $H_*(Y;K)$ form a Hopf algebra in which all elements of the augmentation ideal are primitive. If $u$ and $v$ lie in the augmentation ideal, then $u$, $v$ and $uv$ are all primitive, which gives $u\otimes v+(-1)^{|u||v|}v\otimes u=0$. If $u$ and $v$ are nonzero, it follows that $|u|=|v|$ and $Ku=Kv$, and $|u|$ is odd unless $K$ has characteristic $2$. Thus, $H_*(Y;K)$ is either $K$ or $K\oplus Ku$ for some element $u$, usually of odd degree. In particular, we see that $Y$ is rationally trivial or an odd-dimensional sphere.

Now consider the groups \begin{align*} A(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}_{(p)}) \\ B(p)_k &= A(p)_k/p \\ C(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}/p) \\ D(p)_k &= \text{ann}(p,A(p)_{k-1}) \end{align*} so that

  • $A(p)\otimes\mathbb{Q}$ has dimension $0$ or $1$ over $\mathbb{Q}$
  • $C(p)$ has dimension $0$ or $1$ over $\mathbb{Z}/p$
  • There is a short exact sequence $B(p)\to C(p)\to D(p)$, so $B(p)$ and $D(p)$ have dimension $0$ or $1$, and at least one of them is zero.

There are a number of different possibilities here.

  • If $D(p)=0$ then $A(p)$ is torsion-free and so injects in $A(p)\otimes\mathbb{Q}$. This means that $A(p)=0$ or $A(p)\simeq\mathbb{Z}_{(p)}$ or $A(p)\simeq\mathbb{Q}$.
  • If $B(p)=0$ then $A(p)$ is divisible, and so is injective as a $\mathbb{Z}_{(p)}$-module. If we let $T(p)$ denote the torsion part of $A(p)$ then we find that $T(p)$ is also divisible and therefore injective and therefore a summand in $A(p)$. We therefore have $A(p)=T(p)\oplus Q(p)$, where $Q(p)$ is a $\mathbb{Q}$-module. It follows that $A(p)\otimes\mathbb{Q}\simeq Q(p)$, so $Q(p)$ is $0$ or $\mathbb{Q}$. If $D(p)=0$ then $T(p)=0$. If $D(p)=\mathbb{Z}/p$ then I think it follows that $T(p)=\mathbb{Z}/p^\infty$.

The most interesting question arising from this analysis is as follows. Let $Y$ be the Moore space with $H_2(Y)=\mathbb{Z}/p^\infty$ for some prime $p$. To avoid trouble from the low-dimensional homotopy groups of spheres, we may want to take $p\geq 5$. Note that $\widetilde{H}_*(Y;K)=0$ unless $K$ has characteristic $p$, and that $\widetilde{H}_*(Y;\mathbb{Z}/p)$ is a copy of $\mathbb{Z}/p$ in dimension $3$. Does $Y$ have an $H$-space structure? I do not see an easy way to answer that. Note that $Y\wedge Y$ is a Moore space with $H_5(Y\wedge Y)=\mathbb{Z}/p^\infty$, but that does not immediately give a good hold on $[Y\times Y,Y]$.

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