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Let $G$ be a group. Suppose that the map $G \to G$, $g \mapsto g^r$ is a group homomorphism for every $r \in \mathbb N$. Then $G$ is abelian. Is this also true homotopy-theoretically?

(In the ordinary case, this need only hold for $r=2$, because if $(gh)^2 = g^2h^2$, then canceling a $g$ and an $h$ we find that $hg = gh$, i.e. $G$ is abelian).

More precisely, let $G$ be a group-like $E_n$-space. For each $r \in \mathbb N$, we may pick $m_r\in E_n(r)$ where $E_n$ is an $E_n$ operad (it doesn't matter which model we choose, nor which $m_r$ we choose), and look at the map

$G \to G$, $g \mapsto m_r(g,\dots,g)$

Questions:

  1. Is this map always a map of $E_{n-1}$-spaces?

  2. Supposing the answer to (1) is yes, if this $E_{n-1}$-map can be made into an $E_n$ map, does it follow that the $E_n$ structure on $G$ lifts to an $E_{n+1}$ structure?

I'm not sure I've chosen the right precisification of this question; I'm open to other interpretations.

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    $\begingroup$ 1. Yes (or, at least, in 'homotopy coherent' land- might be via a zig-zag or something in point-set land). This follows from the fact that an E_n algebra is an E_1 algebra in E_{n-1}-algebras (and that the diagonal map is a map of E_n-spaces). 2. I'd be surprised if the answer were yes. (Certainly, at the very least, one should ask for compatibility on the E_n-structures of the maps for each r). $\endgroup$ Jan 28, 2019 at 23:52

1 Answer 1

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Ok, here is a counterexample. Let $F$ be the fiber of the map $\Omega(K(\mathbb{F}_2, 2)\stackrel{i\cdot \mathrm{Sq}^1i}{\to} K(\mathbb{F}_2,5))$, with $\mathbb{E}_1$-structure as indicated. Then $F$ does not deloop further since $i\cdot\mathrm{Sq}^1i$ is not a loop map. On the other hand, each of the 'power maps' $K(\mathbb{F}_2, n) \to K(\mathbb{F}_2,n)$ is either 0 or 1 (depending on the parity of the power), and either way we can complete this to a commutative square with $i \cdot \mathrm{Sq}^1 i$ and then take loops and fibers to deduce that the power maps $F \to F$ are $\mathbb{E}_1$-maps.

The same argument does not work to show that $F \times F \to F$ is an $\mathbb{E}_1$-map precisely because the relevant diagram of EM-spaces no longer commutes.

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  • $\begingroup$ Thanks! I think I mostly follow. So $i$ is the fundamental class. I see that if $i \cdot Sq^1 i$ where a loop map, then $F$ would be $E_2$, but why does the converse hold? And I wonder if you could spell out the final comment a bit more? Also, what would happen if we did the whole arguent with the fiber of $iSq^1 i$ rather than $\Omega i Sq^1 i$? $\endgroup$
    – Tim Campion
    Jan 29, 2019 at 17:27
  • $\begingroup$ 1) If X is an n-fold loop space, the the postnikov tower of X is obtained from the Postnikov tower of B^nX by looping down- so all the k-invariants have to be n-fold loop maps. 2) If you did the argument for BF instead you’d learn that BF is a pointed space with some interesting self-maps and no other structure. $\endgroup$ Jan 29, 2019 at 18:51

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