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Recently, Ching and Salvatore have proven that the $E_n$ operad is Koszul self dual. While thinking about the analogous question for the framed $E_n$ operad, I realized there is an obvious first question: does the spectrum $\Sigma^\infty_+ O(n)^\vee$ have an $A_\infty$-structure? Perhaps worth noting, since $\Sigma^\infty_+ O(n)$ is self dual, it suffices to show $\Sigma^{-n(n-1)/2}_+ O(n)$ has an $A_\infty$ structure.

Edit: I realized this question is very basic; the Spanier-Whitehead dual of any space is canonically $E_\infty$ via the diagonal.

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    $\begingroup$ Isn't it a split square zero extension of the sphere? $\endgroup$
    – Tim Campion
    Commented Dec 3, 2022 at 18:03
  • $\begingroup$ @TimCampion I don't know what that actually means. Does the multiplication from dualizing $O(n) \times O(n) \xrightarrow{\mu \wedge 1} O(n) \times O(n)$ and composing one of the factors with $O(n)_+ ^\vee \rightarrow S^0$? $\endgroup$
    – Connor Malin
    Commented Dec 3, 2022 at 19:26
  • $\begingroup$ I just mean that $\Sigma^\infty_+ O(n)^\vee = \mathbb S \oplus \Sigma^\infty O(n)^\vee$. Since $\Sigma^\infty O(n)^\vee$ is a $\mathbb S$-bimodule, there's an algebra structure on $\mathbb S \oplus \Sigma^\infty O(n)^\vee$ where the multiplication is zero on $\Sigma^\infty O(n)^\vee \otimes \Sigma^\infty O(n)^\vee$. (so no, the multiplication on $O(n)$ is not involved here) $\endgroup$
    – Tim Campion
    Commented Dec 3, 2022 at 19:46
  • $\begingroup$ Ching and Salvatore's proof is O(n) equivariant- doesn't that imply the result for all the 'framed' variants? $\endgroup$
    – Dylan Wilson
    Commented Dec 3, 2022 at 21:57
  • $\begingroup$ @DylanWilson Well there is not even a definition yet of the Koszul dual of an unreduced operad that coincides with Ching and Salvatore's for reduced operads (as far as I know). $\endgroup$
    – Connor Malin
    Commented Dec 3, 2022 at 23:27

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The question as stated probably requires clarification. If

$X$ is a space, then the S-dual $D_+(X)$ (i.e., functions from $X_+$ to the sphere) is always an $E_\infty$-ring spectrum. In particular, it will also be an $A_\infty$-ring.

Perhaps what is being asked is whether $D_+(G)$ is an $A_\infty$-coalgebra when $G$ is a topological group. The answer is yes, because $G$ is an $A_\infty$-space.

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