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?



Are $E_n$-operads not formal in characteristic 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>. 


