Let $X$ be a smooth scheme, then an infinity enchancement of $QCoh(X)$ has an $E_\infty$ structure and in particular an $E_n$ structure for any $n$. In this paper, http://arxiv.org/abs/0805.0157 Ben-Zvi, Francis, and Nadler compute $E_n$ Hochschild cohomology of this category as $T_X[-n]$. I believe there is also separate work of Francis explaining how the $E_n$ Hochschild cohomology of $QCoh(X)$ gives deformations of these categories. I've been trying to get some kind of idea as to what this result means. The case $n=1$, I understand pretty well, but I am curious about all of the other odd cases in particular. The thing that I find strange is that the vector space and Lie-structure of the deformation space are very similar for all of the odd $n$ and yet I believe from degree considerations, it must be the case that more often than not, deformations of the $E_{2k+1}$ structure are trivial as deformations of the $E_{2k-1}$ structure (the notable exception coming from commutative deformations of the underlying scheme).

Just to keep things concrete (I hope), let's just assume $X$ is $Spec(A)$ and $n=3$. Can these deformations be made reasonably explicit in the same way as in the $n=1$ case (e.g. by deforming some sort of associator map)? What about in some special case such as that of a function $f$ (these will be some kind of derived deformations in the sense that will break the $\mathbb{Z}$-grading)?

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    $\begingroup$ $E_k$ deformations require you to be able to impose more structure, and this makes it less likely that a random $E_{k-1}$-deformation will extend. E.g. in the case n=3, an $E_3$-deformation of A should come equipped with a braided monoidal structure on its category of modules, and this forces (at a minimum) the cohomology ring to be commutative-associative, whereas a general $E_1$-deformation is just an associative deformation. $\endgroup$ Nov 7, 2010 at 13:31
  • $\begingroup$ That much is intuitive, what I don't know how to do is use a section of the tangent bundle to deform the braided monoidal structure other than by trying to deform the linear structure(which as you say doesn't even really work). My hope was that this could be made reasonably concrete by deforming some associator... $\endgroup$ Nov 7, 2010 at 20:55
  • $\begingroup$ Deforming the associator seems likely, or possibly deforming the chain homotopy expressing the braid relation. Seems like a polynomial algebra k[x,y], with a square-zero extension k[x,y] + M[2], might be a decent place to start. $\endgroup$ Nov 8, 2010 at 3:15
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    $\begingroup$ I thought that $E_n$-Hochschild cohomology of $QCOh(X)$ was $S(T_X[-n])$ ? $\endgroup$
    – DamienC
    Apr 26, 2011 at 19:15


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