One way to construct an $E_\infty$algebra is to consider the cochain complex $C^*(X;M)$ for $X$ a topological space and $M$ a module over some ring $\Lambda$. From what I can recall, the $E_\infty$algebra structure should contain the homotopy type of $X$. If $X$ is a $K(G,n)$, can the $E_\infty$structure on $$ C^*(K;\mathbb{Z}) $$ be made explicit?

$\begingroup$ I'm far from an expert but I think if you know all steenrod operations and all massey products on the cohomology then you know everything. $\endgroup$ – Saal Hardali Sep 7 '17 at 22:15

$\begingroup$ Perhaps this is only enough for the $A_\infty$structure though... $\endgroup$ – Saal Hardali Sep 7 '17 at 22:22

1$\begingroup$ @SaalHardali do you have a reference for this claim in the $A_\infty$ case? $\endgroup$ – 54321user Sep 7 '17 at 22:35

8$\begingroup$ I guess this depends on what you're looking for. An answer to this question probably involves three things: an explicit description of a model for $C^*(K(G,n);\Bbb Z)$, an explicit model for what an $E_\infty$ structure on a cochain complex is, and an explicit way that $C^*(K(G,n);\Bbb Z)$ has this structure. It would be easier to provide a helpful answer knowing if you have specific models in mind or if describing those would be part of what you need. $\endgroup$ – Tyler Lawson Sep 8 '17 at 1:21

1$\begingroup$ @54321user when G=Z/p and you take coefficients in a ring of characteristic p, there is actually a description by a single generator and single relation (described at the linked "Related question" mathoverflow.net/questions/48138/…) due to Mandell. You might be able to bootstrap that up to understanding other abelian pgroups when the coefficients are still in a ring (or preferably field) of char p. But Z coefficients will probably be sorta ugly, and so will more complicated G $\endgroup$ – Dylan Wilson Sep 17 '17 at 1:22
There is a very general framework that gives the $E_\infty$ structure on cochains in McClure and Smith's "Multivariable cochains and little $n$cubes", with formulas comparable to the AlexanderWhitney diagonal. Roughly, the $E_\infty$ structure is given by a collection of operations as follows:
(Warning: I am completely ignoring signs here. They are unpleasant, and McClureSmith do them more carefully.)
 Every operation takes a number $r$ of input cochains and has a degree $k$. It is multilinear in these $r$ variables.
 The operations of this type are denoted by $\langle n_1 n_2 \dots n_{k+r}\rangle$, where $1 \leq n_i \leq r$.
 If two adjacent $n_i$ are equal, or if one of the integers $1 \dots r$ is not among the $n_i$, then the operation is the zero operation.
 This is a $\Sigma_r$equivariant set of operations: permuting the inputs has the same effect as permuting $\{1,\dots,r\}$.
For example, the nonzero binary operations are: $\langle 12 \rangle, \langle 21 \rangle, \langle 121 \rangle, \langle 212 \rangle, \langle 1212 \rangle, \langle 2121 \rangle, \dots$, whereas $\langle 236154212\rangle$ is a degree3 operation of six inputs.
For any space $X$, these operations act on the singular cochain complex of $X$ roughly by taking a simplex $[v_0, \dots, v_n]$ and summing up over the ways to: break it up into $k+r$ overlapping partitions, reassemble the pieces corresponding to each number $1,\dots,r$ into a simplex, and feed those simplices to the input cochains.
For example, the operation $\langle 12\rangle$ applied to $(f,g)$ gives an operation that partition simplices into two pieces, gives the first half to the input number 1 ($f$), and the second half to input number 2 ($g$). This is the ordinary AlexanderWhitney cup product: if $f$ is a $p$cocycle and $g$ is a $q$cocycle, $\langle 12\rangle (f,g)$ is the cocycle that evaluates $f$ on simplex $[v_0,\dots,v_p]$ and evaluates $g$ on the simplex $[v_p,\dots,v_{p+q}]$, then multiplies them together. This is called $f \smile g$.
The operation $\langle 2 1\rangle(f,g)$ is an operation that partitions into two pieces, gives the first half to $g$, and the second half to $f$. This is the cup product of $g \smile f$.
The operation $\langle 1 2 1\rangle(f,g)$ is an operation that sums up the ways to divide into three pieces, give the first and last to $f$, and the middle to $g$. More explicitly, it sums up all the ways to divide $[v_0,\dots,v_{p+q1}]$ into $[v_0,\dots,v_i], [v_i,\dots,v_{i+q}],[v_{i+q},\dots,v_{p+q1}]$, evaluate $f$ on the simplex $[v_0,\dots,v_i,v_{i+q},\dots,v_{p+q1}]$, evaluate $g$ on the simplex $[v_i,\dots,v_{i+q}]$, and multiply them together. This is called the cup1 product $f \smile_1 g$ and it satisfies $$ \delta(\langle 121 \rangle (f,g)) = \langle 121 \rangle (\delta f, g) + \langle 121 \rangle (f, \delta g) + \langle 21 \rangle (f, g) + \langle 12 \rangle (f,g) $$ (again, up to signs that I'm not writing). There is a systematic formula for what $\delta$ applied to one of these operations gives: you sum up the ways to apply $\delta$ to the inputs, and then sum up the ways to first delete one of the numbers in $\langle n_1 \dots n_{k+r}\rangle$ and then apply it to the $r$ inputs.
There is also a systematic formula for how these operations compose, making them into an operad.
These operations work for the singular cochains of any space, or more generally for the cochains on a simplicial set. If you like, there are simplicial models for the $K(G,n)$'s, for example by doing symmetric products of simplicial models of Moore spaces, and at least some of these go back to classical work of Eilenberg and Mac Lane. These tend to look very similar to the "bar complex" of $G.$

1$\begingroup$ There is a very explicit simplicial model of $K(G,n)$ by taking $G\otimes \bar{\mathbb{Z}}[S^n]$ (and taking $S^n=\Delta^n/\partial \Delta^n$) $\endgroup$ – Denis Nardin Sep 21 '17 at 17:50

$\begingroup$ @DenisNardin Do you have a reference for this? $\endgroup$ – 54321user Sep 21 '17 at 20:29

$\begingroup$ @54321user It's just an easy consequence of the DoldKan correspondence. In particular, the homotopy groups of a simplicial abelian group are the homology groups of the associated complex (see C. Weibel, An introduction to homological algebra exercise 8.4.4). $\endgroup$ – Denis Nardin Sep 22 '17 at 5:57