Why is every deformation of the universal enveloping algebra of a complex semisimple Lie algebra trivial?

Question

I have read in these lecture notes that every deformation $U_h(\mathfrak{g})$ of $U(\mathfrak{g})$ is trivial, i.e. isomorphic to $U(\mathfrak{g})[[h]]$ as associative $\mathbb{C}[[h]]$-algebras. Why is this true? The reason they cite is that $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))=0$. Now, I know how to compute Hochschild cohomology of simpler algebras, but several problems arise for me in this setup:

1. I am unsure how to proceed in the computation of $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))$ (or more generally, replacing $U(\mathfrak{g})$ with $T(V)$, the tensor algebra of a vector space). How can I conclude that this is $0$?

2. Why does the "trivial" deformation look like a power series ring? They say that the multiplication in $U_h(\mathfrak{g})$ looks like "multiplication modulo $h$" in $U(\mathfrak{g})$; what does this mean precisely?

3. The authors instead deform $U(\mathfrak{g})$ as a Hopf algebra, but don't elaborate on what such a deformation should look like, what cohomology theory classifies such deformations, etc. Where can I find a resource that concretely answers these questions? An English translation of Drinfel'd's paper "Quasi-Hopf algebras" would be a good starting point, but I haven't been able to find one.

Answer 1

The article Deformation par quantification et rigidite des algebres enveloppantes by M. Bordemann, A. Makhlouf, T. Petit addresses these questions. They call Lie algebras $\mathfrak{g}$ with $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))=0$ strongly rigid, and show that then every formal associative deformation is equivalent to the trivial deformation. For semisimple Lie algebras over an algebraically closed field of characteristic zero one has $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))=H^2(\mathfrak{g},S\mathfrak{g})$, which is zero by Whitehead's second Lemma for Lie algebra cohomology.

Answer 2

2. Means that as a $\mathbb C[[\hbar]]$ associative algebra $U_\hbar(\mathfrak g)$ is isomorphic to $U(\mathfrak g)[[\hbar]]$. Each element can be understood as a formal power series $\sum_{n=0}^\infty \hbar^n X_n$ where each $X_n\in U(\mathfrak g)$ and the product is given by the usual product formula of formal power series $$\sum_{n=0}^\infty \hbar^n X_n \cdot \sum_{n=0}^\infty \hbar^n Y_n=\sum_{n=0}^\infty \hbar^n \sum_{j=0}^n X_j\cdot Y_{n-j}$$

3. Means that $U_\hbar(\mathfrak g)$ is a $\mathbb C[[\hbar]]$ Hopf algebra such that $U_\hbar(\mathfrak g)/\hbar U_\hbar(\mathfrak g)$ is again a Hopf algebra isomorphic (as Hopf algebra) to $U(\mathfrak g)$ . To see this as a deformation problem in a cohomological manner you should look for what is called Gerstenhaber-Shack cohomology. You can have a look at the original papers on the subject:

   • 1 Gerstenhaber and Shack, Bialgebra cohomology, deformations and quantum groups, Proc. Nat. Acad. Sci. USA 87 (1990).
   • [2] Gerstenhaber and Shack, Algebras, bialgebras, quantum groups, and algebraic deformations, Contemp. Math. 134 (1992).

I have the impression (I ask for specialists to correct me on this point) that in general computing GS cohomology is very hard and very few general results are known and that is why these works for quite some time were somewhat not really considered in the qg community.

As an aside, difficulties in finding the English version of the paper by Drinfel'd were already addressed here: English version of “Quasi-Hopf Algebras”. It is often difficult to find the very short lived Leningrad Journal of Math but with help from you library you should succeed.