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This is perhaps somewhat related to this question. Fix a field $k$ of characteristic $p>0$. Suppose that $A$ is an $E_\infty$-algebra over $k$. Then $A$ also has an $A_\infty$-algebra structure, and therefore so does its homology $HA$. Its homology is also graded commutative.

I'm looking for an extended version of graded commutativity, a result like this: if $m_n$ is any of the higher multiplications in the $A_\infty$ structure on $HA$, then for any $1 \leq j \leq n$, $$ x m_n(a_1 \otimes \cdots \otimes a_n) = \pm m_n(a_1 \otimes \cdots \otimes x a_j \otimes \cdots \otimes a_n). $$ This is certainly true if $n=2$ by graded commutativity. What about for larger values?

(I'm tempted to tag any question about $E_\infty$-algebras as "commutative algebra", but I suppose that would be misleading...)

Edit: as Fernando points out in his comment, this is too much to expect in general. The $A_\infty$-algebra structure on $HA$ is not unique, so perhaps the right question is, are there conditions on $x$ and the $a_i$ so that, for some choice of $m_n$, $x m_n(\dots) = \dots$?

Along with Fernando's example, another one to consider is the mod $p$ cohomology of a cyclic group of order $p$, with $p$ odd. If $x$ is the generator of $H^1$ and $y$ is the generator of $H^2$, then $m_p(x^{\otimes p}) = \pm y$, so $x m_p(x^{\otimes p}) = \pm xy \neq 0$ while I think $m_p(x^2 \otimes x^{\otimes p-1}) = 0$, since $x^2=0$.

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Is $x\in HA$ any element? If so then $m_n(a_1\otimes\cdots\otimes a_n)= \pm a_1\cdots a_n\cdot m_n(1\otimes\cdots\otimes 1)=0$ if the $A$-infinity structure is normalized (you can always assume this). – Fernando Muro Mar 13 '11 at 14:24

as Fernando points out in his comment, this is too much to expect in general. The A∞-algebra structure on HA is not unique

There is a unique $A_\infty$-algebra structure on $H(A)$ such that $H(A)$ and $A$ are weakly equivalent (i.e. $A_\infty$-quasi-isomorphic).

About your question, in characteristic 0 you could replace your $E_\infty$-algebra by a weakly equivalent DG commutative algebra... then on its cohomology you would get a $C_\infty$-structure (which is a strictly commutative $A_\infty$-structure).

In positive characteristic this is more difficult. But there is still the possibility to deal with divided power algebras (see e.g. on page 18 for a hint).

I know this is not really an answer to your question. But I hope to can help.

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Damien, the structure is not unique. Perhapes you are mislead by the fact that it is essentially unique, but there are tons precisely because of that. – Fernando Muro Apr 26 '11 at 12:11
By essentieally unique, do you mean that it is unique up to a unique $A_\infty$-isomorphism ? If so, then we agree :-) – DamienC Apr 26 '11 at 12:52
@Damien: Yes, exactly. – Fernando Muro Apr 26 '11 at 19:59
The non-uniqueness is important, because $m_n(a_1 \otimes \dots \otimes a_n)$ may have very different values for two different (but isomorphic) $A_\infty$ structures. Anyway, I will look at the citation you mentioned -- I mainly care about the positive characteristic case. – John Palmieri Apr 27 '11 at 20:42

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