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Past this question in MO have raised the following questions for me.

Question

In characteristic $0$, it is well-known that a Kadeishvili‘s $C_{\infty}$-algebra is an $E_{\infty}$-algebra.

However, do we have an example of $E_{\infty}$-algebras which is not a $C_{\infty}$-algebra? (We always only consider characteristic 0)

Any comment is welcome. Thank you!

See also comments below.

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    $\begingroup$ Any $C_\infty$-algebra is weakly equivalent to a cdga, namely the cobar complex of its bar complex. For $E_\infty$-algebras, Dyer-Lashov operations are an obstruction to the existence of such a weak equivalence. $\endgroup$
    – Bertram Arnold
    Commented Oct 8, 2023 at 13:23
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    $\begingroup$ You should maybe clarify what you mean by $E_\infty$ and $C_\infty$. I’ve only seen the latter considered in characteristic zero, where both notions coincide for almost all possible definitions. $\endgroup$
    – Fernando Muro
    Commented Oct 8, 2023 at 13:28
  • $\begingroup$ @FernandoMuro Thank you. I only consider characteristic 0 and the definitions of $C_{\infty} $ and $E_{\infty}$ algebras in Section 13.1.8~13.1.10 of Loday-Vallette's book. $\endgroup$
    – YkMz
    Commented Oct 8, 2023 at 13:35
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    $\begingroup$ @Walterfield then you can find an answer to your question in the first paragraph of 13.1.10. $\endgroup$
    – Fernando Muro
    Commented Oct 8, 2023 at 21:54
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    $\begingroup$ @FernandoMuro I misunderstood. You mean that $Com_\infty$ is just one model of $E_\infty$-operad. So it is possible to have an $E_\infty$-algebra that does not appear as an algebra over $Com_\infty$. $\endgroup$
    – YkMz
    Commented Oct 9, 2023 at 9:01

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