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This is a partial duplicate to this MO question, I apologize for that. I'm asking since the answers there still do not allow me to work out an answer to my question, which is a bit more specific.

Informally speaking, an $E_\infty$-algebra is an algebra which is associative and commutative up to coherent homotopies, a notion that can be properly formalized via the language of operads. However, in characteristic zero, one can can obtain an equivalent category to $E_\infty$ algebras by taking commutative differential graded algebras and inverting the quasi-isomorphisms. This is the same mechanism by which one can define the category of $L_\infty$ algebras in characteristic zero by starting with differential graded Lie algebras and inverting quasi-isomorphisms.

However, in the $L_\infty$-case it is well known that $L_\infty$-morphisms between differential graded Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ can be also given an "explicit" definition as sequences of multilinear maps $\varphi_k\colon \wedge^k \mathfrak{g}\to \mathfrak{h}[1-k]$ which satisfy a set of quadratic equations. This explicit definition obscures the theory behind it but has nice "practical" consequences. For instance one immediately sees that an $L_\infty$-morphism between two Lie algebras (seen as differential graded Lie algebras concentrated in degree zero) is precisely the same thing as a Lie algebra morphism between them, or that an $L_\infty$-morphism from a Lie algebra $\mathfrak{g}$ to a differential graded Lie algebra $\mathfrak{h}$ concentrated in nonegative degree is the same thing as a morphism of differential graded Lie algebras from $\mathfrak{g}$ to $\mathfrak{h}$. Or even, if $\mathfrak{g}$ is a Lie algebra and $\mathfrak{h}=(\mathfrak{h}^{-1}\to \mathfrak{h}^0)$ is a differential graded Lie algebra concentrated in degrees -1 and 0, one has a complete, simple and explicit control on what an $L_\infty$ morphism from $\mathfrak{g}$ to $\mathfrak{h}$ would be.

I'm wondering whether something similar can be said for $E_\infty$-morphisms between differential garded commutative algebras.

For instance: can it be given an explicit description of the $E_\infty$-morphisms between a commutative algebra $A$ (seen as a differential graded commutative algebra concentrated in degree zero) and the differential graded commutative algebra $\mathbb{K}[1]\oplus \mathbb{K}$, where $\mathbb{K}$ is the base field? (the differential on $\mathbb{K}[1]\oplus \mathbb{K}$ is the zero differential and the multiplication $(\mathbb{K}[1]\oplus \mathbb{K})^0\otimes (\mathbb{K}[1]\oplus \mathbb{K})^{-1}\to (\mathbb{K}[1]\oplus \mathbb{K})^{-1}$ is the multiplication in $\mathbb{K}$).

I guess such an $E_\infty$-moprhism should reduce to the datum of two $\mathbb{K}$-linear maps $\varphi_0\colon A\to \mathbb{K}$ and $\varphi_1\colon A\otimes A\to \mathbb{K}$ satisfying suitable equations which shuld precisely encode the fact that $\varphi_0\colon A\to \mathbb{K}$ is a morphism of commutative algebras up to a homotopy given by $\varphi_1$. But I'm not able to find a rigorous statement along these lines anywhere in the literature.

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  • $\begingroup$ Not sure I completely understand the answer: if $E_\infty$-morphisms between dgca's are the same things as $A_\infty$-morphisms between them, then this already solves the question: $A_\infty$-morphisms have a well known explicit descriprion in terms of multilinear maps of the kind I would like. But why should $E_\infty$-alg $\hookrightarrow$ $A_\infty$-alg be fully faithful? Any reference? $\endgroup$
    – domenico fiorenza
    Commented Mar 6, 2015 at 13:31
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    $\begingroup$ I would be a bit surprised if the inclusion of $E_{\infty}$-algebras in $A_{\infty}$-algebras was fully faithful. I believe it is known that not every $A_{\infty}$ map is $E_{\infty}$. Unless you are in Characteristic 0. $\endgroup$
    – Sean Tilson
    Commented Mar 6, 2015 at 14:08
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    $\begingroup$ It is not true that the inclusion of $E_\infty$-algebras into $A_\infty$-algebras is fully faithful in characteristic zero. As an example, let $A$ be the (discrete) $E_\infty$-algebra $\mathbb{Q}[x,y]$ and let $B$ be an $E_\infty$-algebra with nontrivial $\pi_1$. As an $E_\infty$-algebra, $A$ is free on two generators, so homotopy classes of maps $A \to B$ give $\pi_0 B \oplus \pi_0 B$. As an $A_\infty$-algebra, $A$ is free on two generators $x,y$ together with a homotopy $xy \simeq yx$... $\endgroup$
    – Akhil Mathew
    Commented Mar 6, 2015 at 18:07
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    $\begingroup$ In characteristic zero, an $E_\infty$-algebra is an $A_\infty$-algebra such that the operations $m_n$ vanish on shuffles, and similarly for morphisms. This is kind of well known. A good reference is Loday-Vallette's book, which deals with this example explicitly in the last chapter. Positive characteristic (or arbitrary ground rings) is a totally different world. $\endgroup$
    – Fernando Muro
    Commented Mar 7, 2015 at 13:18
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    $\begingroup$ ok, perfect, this solves everything! thanks a lot. Actually I was suspecting that with the definition in Loady-Vallette also $Comm$ would have been $E_\infty$, but since they do not say this and instead mention the fact for $Comm_\infty$ I got confused and found a wrong reason to justify that $Comm$ was not good (but indeed it is, as you say). The cofibrancy condition comes later in what they write and this did not help me in understanding what was going on here. But you did! Thanks a lot! $\endgroup$
    – domenico fiorenza
    Commented Mar 8, 2015 at 17:15

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For future readers' convenience, I'm summarizing the comments to the original question in the form of an answer. Many thanks to Adeel, Sean Tilson, Akhil Mathew and Fernando Muro for their inputs. I'm obviously the only one to blame for eventual inaccuracies I'll be including in this answer. So, the answer is as follows:

Let $A$ and $B$ two differential graded commutative $\mathbb{K}$-algebras, where $\mathbb{K}$ is a characteristic zero field. An $E_\infty$-morphism $\varphi\colon A\to B$ is precisely a morphism of $C_\infty$-algebras between $A$ and $B$. Equivalently, it is a morphism of $A_\infty$-algebras form $A$ to $B$ such that all of its Taylor coefficients $\varphi_k\colon A^{\otimes k}\to B[1-k]$ vanish on the shuffle products.

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    $\begingroup$ Hi, equivalently one can define a morphism between $C_{\infty}$-algebras $A$ and $B$ as a map of DG Lie coalgebras $L^cA[1] \to L^cB[1]$ where $L^cV$ is the free Lie coalgebra generated by the vector space $V$. The DG Lie coalgebra structures on $L^cA[1]$ and $L^cB[1]$ are defined by the CDGA structures on $A$ and $B$ respectively. Moreover, one can define a $C_{\infty}$-algebra structure on a vector space $V$ to be a coderivation $D: L^cV \to L^cV$ such that $D^2=0$... and the definition of $C_{\infty}$-morphism also makes sense in this general case. $\endgroup$
    – Manuel Rivera
    Commented Jul 2, 2015 at 16:14
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    $\begingroup$ Working over characteristic zero "misses the point" in that there's no difference between $E_\infty$ and $C_\infty$ there. One of the features of an $E_\infty$ operad (over $\mathbb Z$, say) is that it resolves the trivial representation of the symmetric group in each arity $n$ in the catgory of $S_n$-modules. In characteristic zero, the homological algebra of finite groups is trivialized. $\endgroup$
    – Pedro
    Commented Jan 14, 2020 at 14:09

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