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Let $k$ be a field of characteristic zero, and $\mathcal{C}$ be a $k$-linear additive symmetric monoidal category. A braided deformation of $\mathcal{C}$ over a local artin ring $R$ with residue field $k$ is an $R$-linear braided monoidal category $\mathcal{C}'$, whose hom-sets are free $R$-modules, together with an equivalence of braided monoidal categories $\mathcal{C}' \otimes_{R} k \simeq \mathcal{C}$.

If I've understood correctly, there is a deep result of Drinfeld-Cartier that states that braided deformations satisfy a degree of "unobstructedness." More specifically, any deformation over $k[\epsilon]/\epsilon^2$, which is trivial as a deformation of monoidal categories, can be lifted to a formal deformation over $k[[\epsilon]]$. The proof starts by considering the "cocycle" that a braided deformation over the dual numbers yields, and then exponentiates it to get the braiding over $k[[\epsilon]]$. Unfortunately, that will not satisfy the hexagon axiom (because the exponential of matrices is not a homomorphism in general), so one has to modify the associativity constraint via "Drinfeld associators."

I'm curious if this result can be understood from a more homotopy-theoretic standpoint. For example, I'd like to know if there is an analogous picture in higher category theory.

  • A possible analog of $\mathcal{C}$ would be a $k$-linear presentable symmetric monoidal, stable $\infty$-category in the sense of Lurie. One benefit of working with these is that there is a good theory of tensor products and base change.
  • The deformation problem could be the following: over a local artin ring $R$ with residue field $k$, consider all $R$-linear $E_2$-categories $\mathcal{C}'$ (presentable, stable) $\mathcal{C}$ equipped with an equivalence of $E_2$-categories $\mathcal{C}' \otimes_R k \simeq \mathcal{C}$. This defines a deformation problem $R \mapsto \mathrm{Def}(R)$.

  • Then a natural question would be: is this deformation problem unobstructed in general? Or, at least, is every first-order deformation trivialized in the $E_1$-direction liftable to a formal deformation?

  • Are there generalizations of this to $E_n$-monoidal (presentable, stable) $\infty$-categories?
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    $\begingroup$ I don't see where the assumption that $\mathcal{C}$ is symmetric is used. Is it necessary? $\endgroup$ – S. Carnahan Jun 16 '14 at 11:22
  • $\begingroup$ @S.Carnahan : I don't know. I'll let an expert answer your question. $\endgroup$ – Akhil Mathew Jun 17 '14 at 14:30
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I'm not very familiar with $\infty$-categories, but it's probably worth mentionning that the Drinfeld-Cartier result you quote is essentially equivalent to the formality of the $E_2$ operad. Every infinitesimal deformation of the kind you mention turns the trivial deformation of your category into an algebra over the $P_2$ (aka Gerstenhaber) operad, and the braided monoidal structure you get by plugging an associator the way you described is exactly the one you get using the formality morphism constructed from the same associator.

There are probably subtleties I'm unaware of in defining this carefully in the $\infty$-world but I've no doubt it's known and written somewhere that all of this extends to this setting. Also, regarding your last question, the $E_n$-operad is formal for $n\geq 2$.

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    $\begingroup$ This is the type of answer I was hoping for, but why does a deformation over the dual numbers make the trivial deformation over $\mathbb{C}[[\epsilon]]$ into an algebra over the $P_2$-operad? $\endgroup$ – Akhil Mathew Jun 16 '14 at 14:22
  • $\begingroup$ That's because it is, more or less by construction, an algebra over the operad of chord diagrams. The relation between the latter and the P2 operad is explained in Tamarkin's paper (arxiv.org/abs/math/9809164) Prop. 4.1. $\endgroup$ – Adrien Jun 17 '14 at 11:37
  • $\begingroup$ Interesting. Thanks for pointing me to this. $\endgroup$ – Akhil Mathew Jun 17 '14 at 14:30

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