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)?
$E_k$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. $\endgroup$ – Tyler Lawson Nov 7 '10 at 13:31