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 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?