Let $A$ be an $E_1$-algebra in chain complexes over $\mathbb Q$.

Is there an easy way to check if $A$ admits the structure of an $E_2$-algebra (or $E_\infty$-algebra)?

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Let $A$ be an $E_1$-algebra in chain complexes over $\mathbb Q$.

Is there an easy way to check if $A$ admits the structure of an $E_2$-algebra (or $E_\infty$-algebra)?

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Since I am most familiar with the (co)chain complex version (using the surjection operad (cf. McClure-Smith)), I will give an example in this setting. The surjection operad $S$ is also filtered by the chain operads $S_k$ consisting of all surjections of complexity $\le k$ (as define in the paper above).

Playing a bit around with the definition of complexity shows that all operations of complexity $\le 1$ are just iterated applications of the $\cup$-product (which is also called $\langle 12\rangle$) and possibly permuting the arguments. So a cochain complex with the structure of an $S_{1}$-algebra is just a DGA.

Now the operation $\cup_1=\langle 121\rangle$ lives in $S_{2}$. It is a reason why the cohomology ring of our DGA is graded commutative, i.e. for two cocycles $x,y$ we have $$d(x\cup_1 y)=x \cup y \pm y\cup x.$$ So given a DGA whose cohomology ring is not graded commutative, cannot be quasiisomorphic to a DGA with the additional structure of an $S_2$-algebra, i.e. there is no map of DGAs which induces an isomorphism in cohomology.

Similarly, in the case of $\mathbb{F}_2$-coefficients, one can do the following to get obstructions to extend an $S_2$-algebra structure to an $S_3$-algebra. Suppose we have an $S_2$-algebra $C^*$. Especially we have $\cup$ and $\cup_1$-products so we can define for $[x]\in H^n(C^*)$ the elements $Sq^n([x])=[x\cup x]\in H^{2n}(C^*)$ and $Sq^{n-1}(x)=[x\cup_1 x]\in H^{2n-1}(C^*)$. Now if we have an Adem- or a Cartan relation that only involves these Operations, we can try to find a proof of those relations via the surjection operad, i.e. find a linear combination of operations whose boundary is the difference of the representatives of both sides (analogous to the case of graded commutativity above). Now those operations typically live in $S_k$ for some $k> 2$, and we end up with an obstruction to lifting the $S_2$-algebra structure to an $S_k$-algebra structure. The easiest example of this is probably that $Sq^1Sq^1=0:H^1\rightarrow H^3.$

So given a cochain complex over $\mathbb{F}_2$ with the structure of an $S_2$-algebra not satisfying that $Sq^1Sq^1=0:H^1\rightarrow H^3$, then the $S_2$-algebra structure cannot be lifted to an $S_3$-algebra structure. These are of course just a couple of examples of rather explicit obstructions. I have no idea whether this strategy is in any sense exhaustive, i.e. whether all obstructions can be constructed in a similar way.

EDIT:

So what both examples have in common is that I am looking for a linear combination of operations with $m$ arguments of complexity $k$ such that its boundary has smaller complexity. So maybe we are really interested in $H^*((S_{k,r}/S_{m,r})\otimes_{RG}R)$, where $S_{k,r}$ is the cochain complex of all operations of complexity $\le k$ with $r$ arguments and $G$ is a subgroup of $\Sigma_r$ fixing a decomposition of $\{1,\ldots,r\}$ corresponding to plugging in the same argument multiple times. In the second example $G=\Sigma_4$. If we would look at a Cartan-relation, we would look at operations with four arguments and plug in something like $x\otimes x\otimes y\otimes y$ and get $G=\Sigma_2\times \Sigma_2\subset \Sigma_4$.

It also turns out that if we pick a boundary in the upper complex, then its boundary is strictly zero, i.e. the obstruction that we were hoping for holds on the nose in any $S_m$-algebra.

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This isn't a complete answer, as it would fill a whole paper; and I haven't checked everything. But I think you can use a strategy similar to what is done in the paper *The Intrinsic Formality of $E_n$-operads* by Fresse and Willwacher. I'm not saying it's easy, but it's systematic.

The situation is the following. You have the diagram: $$\require{AMScd} \begin{CD} E_1 @>>> \mathsf{End}_A \\ @V{\iota}VV \\ E_2 \end{CD}$$ and you want to know whether there exists a lift. (Of course there is nothing special about $\mathsf{End}_A$ here, it could be any operad.) This is a two-pronged question:

- Does there exist a map $E_2 \to \mathsf{End}_A$ at all?
- Can you find one that makes the diagram commute?

(The problem I see with this approach is that the second part depends on the first, but the first could have many different answers...)

You can use Bousfield's obstruction theory to answer these question. You want to find a path-component in the mapping space $\operatorname{Map}(E_2, \mathsf{End}_A)$, i.e. you're looking at $\pi_0$. Of course, the very first obstruction is to find a Poisson $2$-algebra structure on $H_*(A)$. If you deal with the resolutions correctly and everything, I believe that such a structure would correspond to a class in the first page of the Bousfield spectral sequence. You want to know whether it survives until the end. Bousfield's obstruction theory should tell you that obstructions live in the cohomotopy groups $\pi^{s+1} \pi_s X$ where $X$ is a certain bicosimplicial space. You want all these obstructions to vanish. Based on the work of Fresse–Willwacher, my (perhaps wild) guess is that you get obstructions indexed by classes in Kontsevich's graph complex $\mathrm{GC}_2$.

Now you have your morphism and you want to know if the diagram commutes, at least up to homotopy I guess. In other words, you want to know if the two maps $E_1 \to \mathsf{End}_A$ (the first one is given by the $E_1$-structure, the second by restricting the $E_2$-structure you found through $\iota$) are in the same path component, i.e. they are joined by a path. Relative Bousfield obstruction theory should tell you that obstructions to finding such a path live in $\pi^s \pi_s X$. If you check that they all vanish, this means that your two morphism are equal up to homotopy. Again, my guess is that obstructions should be indexed by the hairy graph complex $\mathrm{HGC}_{12}$.