I have a particular kind of algebraic structure that's come up in my work. It's basically a chain complex equipped with a multiplication which is commutative and associative up to homotopy in a particularly nice way. So one might hope that it's some variation of an $E_\infty$-algebra. Unfortunately I cannot see a direct relation between the two notions.
Fix a ground ring $R$. (Everything below will make sense more generally than for $R$-modules, though.) Let $A$ be a left module over the commutative operad, a.k.a. a twisted commutative algebra, in the category of dg $R$-modules. We suppose that $A(0)=0$.
To be explicit, this means that I have a sequence $\{A(n)\}_{n \geq 1}$ of chain complexes over $R$ with actions of the symmetric groups $S_n$, and a multiplication map $A(m) \otimes A(n) \to A(m+n)$ which is $S_m \times S_n$-equivariant. I impose also the condition that the multiplication is associative, and that it's commutative in a "twisted" sense, meaning that the diagram $$ A(m) \otimes A(n) \to A(m+n) $$ $$ \cong \hspace{6em}\cong$$ $$ A(n) \otimes A(m) \to A(n+m) $$ commutes. Here the left vertical isomorphism is the "flip" map, and the right isomorphism is acting by the "box" permutation $(n+1,n+2,...,n+m,1,2,\ldots,n)$.
So far I'm only repeating standard definitions. Let me define $A$ to be a homotopy twisted commutative algebra if it is moreover equipped with an element $\mathbf 1 \in A(1)$ such that multiplication with $\mathbf 1$ induces a quasi-isomorphism $A(n) \to A(n+1)$ for all $n$.
Example. If $C$ is a unital commutative algebra over $R$ in the usual sense, then if I put $A(n)=C$ with trivial $S_n$-action, with multiplication given by the usual multiplication in $C$, then I obtain a rather trivial example of a homotopy twisted commutative algebra.
Claim. If $A$ has vanishing differential, then it is necessarily of the form of the previous example. Indeed multiplication by $\mathbf 1$ gives me isomorphisms $A(n) \cong A(n+1)$ which I can use the to identify all the different components; one can prove that under these isomorphisms, the multiplication $A(1) \otimes A(1) \to A(2) \cong A(1)$ becomes on-the-nose commutative and associative.
So in general a homotopy twisted commutative algebra will on the level of cohomology give me a strictly commutative multiplication. But on the chain level I have something weaker. If I choose a quasi-inverse $A(2) \stackrel \sim \to A(1)$ I get a multiplication on $A(1)$, but now it is only commutative and associative up to homotopy.
Question: Is there a relationship between homotopy twisted commutative algebras and $E_\infty$-algebras? For example, is there a functor from one of them to the other? A Quillen equivalence?
