This question is motivated by the desire to understand better the interplay between Drinfeld associators, algebraic structures up to homotopy and related results.

Recall that a Lie bialgebra is a Lie algebra $\mathfrak g$ with a Lie algebra structure on $\mathfrak g^*$ (cobracket) plus some compatibility conditions. These are infinitesimal versions of compatible Poisson structures on Lie group. A quantization of $\mathfrak g$ is a $\mathbb C[[\hbar]]$ deformation of the envelopping algebra of $\mathfrak g$ the coproduct of which gives back the cobracket of $\mathfrak g$ at the quasi-classical limit. Etingof-Kazhdan theorem states that every Lie bialgebra can be quantized, in a functorial way. Tamarkin's proof of this results relies on the formality of the little disc operad, and in fact so does the original proof in a slighty different language.

The original proof use the statement "the choice of an associator leads to a quasi-triangular quasi-Hopf algebra (aka braided monidal category) from any metrizable Lie algebra (i.e. equipped with an invariant symmetric bilinear form)". Indeed, any metrizable Lie algebra leads to a representation of a certain operad in Lie algebra $\mathfrak t$ and an associator is nothing but an isomorphism from some completion of the group algebra of pure braid operad to $U(\mathfrak t)$.

Then the proof goes a follow:

  1. To any Lie bialgebra $\mathfrak g$ is asociated its double $\mathfrak d$ which quasi-triangular and in particular metrizable. $\mathfrak d=\mathfrak g\oplus \mathfrak g^*$ as vector space and the pairing is just the canonical one.
  2. Use the above statement to turns $U(\mathfrak d)[[\hbar]]$ into a quasi-triangular quasi-Hopf algebra
  3. Solve the so-called "twist equation" which say that the quasi-Hopf algebra $U(\mathfrak d)[[\hbar]]$ can be turned into an honest Hopf algebra $U_{\hbar}(\mathfrak d)$. An important fact is that the product is not modified, i.e. $U_{\hbar}(\mathfrak d)$ is isomorphic to $U(\mathfrak d)[[\hbar]]$ as an algebra.
  4. That we have a quasi-triangular quantization means roughly that we get a quantum version of the pairing (given by the quantum R-matrix) which can be used to identify a sub-Hopf algebra $U_{\hbar}(\mathfrak g)$ of $U_{\hbar}(\mathfrak d)$ which is the quantization we were looking for. One can also show that $U_{\hbar}(\mathfrak d)$ is the quantum Drinfeld double of $U_{\hbar}(\mathfrak g)$.

Note that thinking of $U_{\hbar}(\mathfrak g)$ as a subalgebra of $U_{\hbar}(\mathfrak d)$ is a bit misleading, and that it's somehow better to see $U(\mathfrak d)$ as an auxiliary space acting on $S(\mathfrak g)$ by differential operators. In this sense the "twist" really leads to a star product and a "star coproduct" on $S(\mathfrak g)$ whose coefficients are given by the action of some differential operators.

Now Tamarkin's proof goes as follow:

  1. Any Lie bialgebra $\mathfrak g$ leads to a Gerstenhaber structure on the free graded commutative algebra $H=S(\mathfrak g_{\hbar}[-1])$ where in $\mathfrak g_{\hbar}$ the cobracket is multiplied by $\hbar$.
  2. The operad of Gerstenhaber algebras is quasi-isomorphic to $BU(\mathfrak t)$ (where $B$ is the bar construction).
  3. The little disc operad is formal, which is roughly a refinement of the above statement on the relation between braids and $\mathfrak t$, applied at the level of chain.
  4. The operad of chain of the little disc operad is quasi-isomorphic to the operad of brace algebras (this is a version of the so-called Deligne conjecture).
  5. Therefore we get a brace structure on $H$, which induces a differential graded Hopf algebra structure on the cofree coalgebra $C(H[1])$.
  6. Finally, so far I understand, one can show that the 0th cohomolgoy group of the dg Hopf algebra obtained this way is a Hopf algebra quantizing $\mathfrak g$.

Admittedly I'm less confortable with this proof, though I can roughly understand how it goes, except maybe the last step. Anyway, I'm interested in understanding how thse proofs relate. Clearly the key fact is quite similar, though different: in the first case the (trivial deformation of) the category of modules over $\mathfrak d$ is an algebra over the operad $\mathfrak t$, while in the second case we construct a space which is an algebra over some operad of complex attached to $\mathfrak t$ through some combinatorial quasi-isomorphism. Then the same isomorphism coming from an associator is applied but in different worlds. So are related Etingof-Kahdan 2 and Tamarkin 3.

Now I'd like to understand the relations between the other steps. The relation between the pairing on $\mathfrak d$ and the complex associated to $\mathfrak g$ is probably well known, but I didn't find a reference. The second part is probably more tricky, and I would really be happy to learn that there is indeed some relations between the Deligne conjecture and the statement that "a quasi Hopf algebra obtained from a Drinfeld associator can be turned into an Hopf algebra". Since the role of the Drinfeld double is to somehow merge the (Lie) algebra and the (Lie) coalgebra structure, it's temptating to think that there is a sort of correspondance between coalgebra deformations of $U(\mathfrak d)$ and bialgebra deformations of $U(\mathfrak g)$, and Etingof-Kahdan's proof show that it is indeed the case, in a highly non obvious way. Can this fact also be related to the Deligne conjecture ?

Edit I just came across these very interesting notes By Pavel Safronov which explain Tamarkin's proof along the same lines as Kevin's answer below.

  • $\begingroup$ Great question. I have some thoughts and ideas on this (actually, I have begun a project joint with O. Gwilliam related to these questions), but nothing completed or answer-worthy. I'd be happy to discuss some of our ideas offline. Conversely, if you come across any discussion in the literature (or write anything yourself), I'd be very grateful if you sent me a pointers. $\endgroup$ Feb 18, 2013 at 5:32
  • $\begingroup$ Theo> I would be happy to discuss this ! I'll send you an e-mail soon. $\endgroup$
    – Adrien
    Feb 28, 2013 at 16:20

1 Answer 1


I'm not sure I can answer everything Adrien asked, but perhaps I can explain a little about part 6 in Tamarkin's proof, and about how the Drinfeld double appears in Tamarkin's story.

First, I'd like to explain a slightly different way to phrase Tamarkin's story. It's relies on Koszul duality, and throughout, there are some subtleties about completness, finiteness, etc. which always appear in Koszul duality. These can be dealt with, but I'll surpress the details.

The key thing we need to know is that any augmented $E_2$ algebra produces a Hopf algebra. (This is part 6 in Tamarkin's proof, as explained in the question). We can see this using Tanakian theory and Koszul duality. If $A$ is an augmented associative algebra, with a map $A \rightarrow k$ to the ground field, we can form a Koszul dual associative algebra $$ A^! = \mathrm{Hom}_k(A,A). $$ You can find (in good situations) an equivalence $$ A^!\mbox{-}\mathrm{mod} = A\mbox{-}\mathrm{mod}. $$ Actually, a precise version of this statement is pretty subtle; see e.g. Positselski's work. The coalgebra version of this statement (where we use comodules over $k \otimes_A k$) works better.

Let's assume for simplicity that $A^!$ has cohomology in degree $0$, so we don't have to worry about homotopical analogs of Hopf algebra structures.

Now, if $A$ is an $E_2$ algebra then left modules over $A$ form a monoidal category $A\mbox{-}\mathrm{mod}$ (this is explained in Lurie's higher algebra, for instance). Thus $A^!\mbox{-}\mathrm{mod}$ is a monoidal category, and you can check that the forgetul functor $$ A^!\mbox{-}\mathrm{mod} \rightarrow \mathrm{dgVect} $$ is monoidal. Thus by Tannakian theory $A^!$ has a Hopf algebra structure.

There's a similar story for $E_3$ algebras. Suppose $A$ is an $E_3$ algebra which is augmented as an $E_2$ algebra. Then $A\mbox{-}\mathrm{mod}$ has an $E_2$ (i.e. braided monoidal) structure, so that $A^!\mbox{-}\mathrm{mod}$ is a braided monoidal category such that the forgetful functor to vector spaces is monoidal. This implies, I believe, that $A^!$ has the structure of a quasi-triangular Hopf algebra.

Let $P_2$ be the homology operad of $E_2$. Formality of the $E_2$ operad gives an isomorphism between $E_2$ and $P_2$, and allows us to turn any $P_2$ algebra into an $E_2$ algebra. Lie bialgebras give rise to augmented $P_2$ algebras, as follows. If $\mathfrak{g}$ is a Lie bialgebra, then $C^\ast(\mathfrak{g})$ is a commutative algebra, with a Poisson bracket defined on generators $\mathfrak{g}^\vee$ by the Lie coalgebra structure on $\mathfrak{g}$.

This story (due to Tamarkin, of course) produces a Hopf algebra from any Lie bialgebra (subject to various caveats about completeness, etc. etc. which is why we find a formal quantization instead of an actual quantization)

The question asked about the Drinfeld double. I would guess that this works as follows. The Drinfeld double of a Hopf algebra is a quasi-triangular Hopf algebra. Quasi-triangular Hopf algebras are Koszul dual to $E_3$ algebras, and under Koszul duality, Drinfeld double should corresponds to $E_2$ Hochschild cohomology. By the higher analog of Deligne's conjecture, the $E_2$ Hochschild cohomology of an $E_2$ algebra is an $E_3$ algebra.

Some aspects of this story about $E_2$ algebras and Hopf are explained in detail in a paper I plan to put on the arxiv soon (which is closely related to Johnson-Freyd and Gwilliam's work). Of course, the picture is not due to me, but to Tarmarkin, but I don't think he showed that there's an equivalence of monoidal categories in this story.

  • $\begingroup$ That's amazing, thanks ! I really don't know much about Koszul duality, but if I undertsand well you are claiming that (whenever it make senses and under some assumption), if $A$ is a P-algebra over some operad P, then the category of modules over the Koszul dual of A is a P-algebra in Cat ?I'm a bit confused about your last statement: the category of modules over a quasi-triangular (quasi-)Hopf algebra H is braided monoidal, hence an algebra over $E_2$ in Cat, so it's temptating to thinks that H is koszul dual to an E_2 algebra, not necessarily E_3, but I'm probably missing something. $\endgroup$
    – Adrien
    Feb 28, 2013 at 16:30
  • 1
    $\begingroup$ If A is a P-algebra then A^! is a P^! algebra.. but A^!-mod is the same (up to size issues) as A-mod. Now here something more subtle is going on - we're taking an E_2 algebra and Koszul dualizing it only with respect to one of its two commuting (in fact equal) associative (E_1) algebra structures, getting an associative coalgebra in addition to the "other" associative structure, hence Hopf. Another version of the same trick (separating E_n as n different algebra structures and treating one separately) says that modules for an E_n algebra form an E_{n-1} category, eg E_2 algebra gives.. $\endgroup$ Feb 28, 2013 at 16:44
  • 1
    $\begingroup$ monoidal category and E_3 gives braided monoidal category, so when we apply the Koszul duality trick we get a braiding on our Hopf algebra. [BTW note that P=E_n for n>1 are Koszul self-dual P=P^!, maybe being careful about shifts, but that's not being used here AFAIK, only Koszul dualizing over one product structure.] $\endgroup$ Feb 28, 2013 at 16:47

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