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Endow $S^7$ with a structure of an $H$-space induced from multiplication in octonions $\mathbb{O}=\mathbb{R}^8$. It is not associative as octonion multiplication is not associative.

Is it associative up to homotopy, i.e. are maps $m(m(-,-),-):S^7\times S^7\times S^7\to S^7$ and $m(-,m(-,-)):S^7\times S^7\times S^7\to S^7 $ homotopic?

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It is not. See Theorem 1.4 of this paper by I.M. James (Trans. AMS 84 (1957), 545-558).

In particular, there exists no homotopy associative multiplication on $S^n$ unless $n=1$ or $n=3$.

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    $\begingroup$ Or n=0, of course. $\endgroup$
    – Theo Johnson-Freyd
    Commented Dec 4, 2016 at 3:25
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    $\begingroup$ Multiplications on $S^7$ are in 1-1 correspondence with the elements of $\pi_{14}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{5}$. Given a product $\mu:S^7\times S^7\rightarrow S^7$, the obstruction it being homotopy associative lies in $\pi_{21}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{4}\oplus\mathbb{Z}_{3}$. As Jon says, there are no homotopy associative multiplications on $S^7$. In fact $S^7$ has no 2-local or 3-local associative products. One must invert both the primes 2 and 3 before $S^7$ admits a homotopy associative multiplication (and then any multiplication will be HA). $\endgroup$
    – Tyrone
    Commented Dec 4, 2016 at 15:59
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    $\begingroup$ @Tyrone A question arises which I wanted to formulate rigorously but failed - is it still "homotopy something"? That is, is the existing multiplication completely generic in some sense? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 5, 2016 at 5:47
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    $\begingroup$ @მამუკაჯიბლაძე The octonions form an alternative algebra, i.e. $x(xy) = (xx)y$ and $(yx)x = y(xx)$ (and $S^7$ is the unit sphere of the algebra of octonions as far as I can tell). $\endgroup$
    – Najib Idrissi
    Commented Dec 5, 2016 at 8:42
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    $\begingroup$ @ConnorMalin, correct. You need a multiplication $\mu$ to begin with, then you get a new one by defining $\mu'=\mu+\alpha\circ q$, (as elements of the algebraic loop $[X\times X,X]$) where $\alpha:X\wedge X\rightarrow X$ and $q$ is the quotient to the smash. Since the only spheres that admit (integral) multiplications are $S^1,S^3,S^7$, among spheres the statement is quite special to these chaps. $\endgroup$
    – Tyrone
    Commented Jul 25, 2019 at 18:03
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There is a Proof due to Stasheff in "H-space from homotopy point of view" (Theorem 6.7). The argument is fairly simple to describe. $\def\OP{{\mathbb O\mathbf P}}$

If $S^7$ admits a homotopy associative multiplication, then one should be able to construct $\OP^3$. It would follow that $\tilde{H}^*(\OP^3;\mathbb{Z}/3)$ is generated by $u_8$ in degree $8$ with the relation that $u_8^4 =0$. It follows that, $$ P^4(u_8) = u_8^3. $$ However, $P^4 = P^1P^3$, which means the $u_8^3 = P^1x$, where $x \in \tilde{H}^{20}(\OP^3;\mathbb{Z}/3) = 0$. Thus $u_8^3 = 0$ which contradicts the existence of $\OP^3$ and consequently the existence of homotopy associative multiplication on $S^7$.

More interestingly this proof suggests that the obstruction to homotopy associative multiplication is $3$-local. This leads to the following question.

Question: Is $S^7_{(2)}$ homotopy associative? Does the reference to James work in the answer due to Jon Beardsley, addresses this question? (Tyrone's comment made me think of this question.)

(Sorry for asking a question inside an "answer".)

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    $\begingroup$ The answer to your question seems to be no: link.springer.com/chapter/10.1007%2FBFb0069236#page-1 $\endgroup$
    – Drew Heard
    Commented Dec 5, 2016 at 10:40
  • $\begingroup$ Isn't the associativity itself required to construct the classifying space which turns out to have the same cohomology as $\mathbb{OP}^{\infty}$? (assuming I got the argument right) $\endgroup$
    – SashaP
    Commented Dec 5, 2016 at 18:37
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    $\begingroup$ What you need is $A_{\infty}$-structure (slightly weaker than strict associativity) to construct classifying space/bar complex. Stasheff showed that you can constructed an $n$-truncated bar-complex for an $H$-space which is $A_n$. In particular homotopy associativity is equivalent to $A_3$-structure. If $S^7$ admits $A_3$-structure then you construct the $3$-truncated bar-complex aka $\mathbb{OP}^{3}$, and that is all is required for the proof above. $\endgroup$
    – Prasit
    Commented Dec 5, 2016 at 19:20
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This response is to the question raised above whether S^7_(2), the 2-localization of the 7-sphere, can admit a homotopy associative multiplication. The answer is no. This fact was established by I.M. James. A later proof by Daciberg Goncalves, "Mod 2 homotopy associative H-spaces" in Geometric Applications of Homotopy Theory I 198-216, Edited by M.G.Barratt and M.E. Mahowald, Lecture Notes in Mathematics 657, Springer-Verlag (1978), revealed the universal nature of the obstruction. The proof involves a secondary cohomology operation and has wider implications. In the same volume, John Hubbock gave a K-theory proof. The calculation of the secondary operation is also presented in the book "Secondary Cohomology Operations" Graduate Studies in Mathematics 49 AMS, by the undersigned. The details are on pages 192-3, and 217-8.

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