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?

  • $\begingroup$ I'd say this is a complicated question, since I think you cannot work with chain complexes (for lack of a model structure), so you have to work with modules over the Eilenberg-MacLane commutative ring spectrum of the field. $\endgroup$ – Fernando Muro Sep 18 '15 at 8:59
  • $\begingroup$ I see. Ok then this question needs some elaborate setup like homotopy operads (where formality would be an oo-morphism, not just a zig-zag, like defined in the paper of Dehling&Vallette arxiv.org/abs/1503.02701) $\endgroup$ – Mark.Neuhaus Sep 18 '15 at 9:29
  • $\begingroup$ Mark, not really, zig-zags are fine. $\endgroup$ – Fernando Muro Sep 18 '15 at 10:14
  • $\begingroup$ I meant that the theory of Dehling&Vallette established a homotopy theory for operads over arbitrary rings. And in that framework zig-zags are 'the same' as actuall oo-morphisms $\endgroup$ – Mark.Neuhaus Sep 18 '15 at 10:36
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    $\begingroup$ Not sure what precisely formal means in this context, but if it means any kind of equivalence between an operad and its homology, it is surely false in characteristic p, by the original examples and calculations in iterated loop spaces. The difference gives the Araki-Kudo-Dyer-Lashof operations. $\endgroup$ – Peter May Sep 18 '15 at 13:57

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.


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