This is a short question:
Is it just unproven folklore (yet), or is it definitively known that $E_n$-operads are not formal, if the characteristic of the underlying field is not equal to zero?
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Sign up to join this communityThis is a short question:
Is it just unproven folklore (yet), or is it definitively known that $E_n$-operads are not formal, if the characteristic of the underlying field is not equal to zero?
Ok, Sean. I'll write in terms of homology operations. Let $\mathcal C$ be any $\Sigma$-free operad, in spaces or in chain complexes, makes no real difference to the answer. For definiteness, take chain complexes over a field $K$; $\Sigma$-free means that $\mathcal C(j)$ is a free $K[\Sigma_j]$-module for each $j$. For any $K$-chain complex $X$, $\mathcal C(j)\otimes X^j$ is $K[\Sigma_j]$-chain homotopy equivalent to $\mathcal C(j)\otimes H_*(X)^j$ (exercise or see Lemma 1.1 in http://www.math.uchicago.edu/~may/PAPERS/10.pdf). Therefore, if $\mathcal C$ is formal then $$ H_*(\mathcal C[j]\otimes_{\Sigma_j} X^j) \cong H_*(\mathcal C[j])) \otimes_{\Sigma_j} H_*(X)^j $$ so the only homology operations come from $H_*(\mathcal C[j]))$. But $H_*(\mathcal C[j]))$ is just way too small if $\mathcal C$ is an $E_n$-operad and $K$ has characteristic $p$. You can see this most obviously if you let $n$ go to $\infty$, when $H_*(\mathcal C(j))$ is zero in positive degrees. For finite $n$ the algebras over $H_*(\mathcal C)$ are $n-1$-braid algebras (if $p\neq 2$ or $3$), but the homology of algebras over $\mathcal C$, such as $H_*(\Omega^n Y)$ for a space $Y$, have a much richer structure. A brief discussion of this is given in Section 5 of http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf.
Paolo Salvatore put a paper on the arXiv last week (1807.11671) proving that $E_2$ is not formal over $\mathbb{F}_2$ as a non-symmetric operad (what he calls planar operad), i.e. you cannot find a zigzag of quasi-isomorphisms of operads (not necessarily $\Sigma$-equivariant) between $H_*(E_2; \mathbb{F}_2)$ and $C_*(E_2; \mathbb{F}_2)$. He uses obstruction theory for that.