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Let $\mathfrak{g}$ and $\mathfrak{g}'$ be Lie algebras. It is known that if $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as associative algebras, then it is not necessarily true that $\mathfrak{g}\cong \mathfrak{g}'$ as Lie algebras.

I am looking for examples such that $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as algebras but $\mathfrak{g}\not\cong \mathfrak{g}'$ as Lie algebras (over an algebraically closed field). Moreover, are there examples such that the categories $U(\mathfrak{g})-\text{Mod}$ and $U(\mathfrak{g}')-\text{Mod}$ are not monoidally equivalent?

I'm not very familiar with the isomorphism problem for enveloping algebras, a quick google search only gave me counterexamples in positive characteristic. I'd be very happy with examples in characteristic zero (infinite dimensions are allowed). I'm more into the monoidal stuff and might figure out myself whether the representation categories are monoidally equivalent.

Edit: I'm asking this because I naturally encountered a quantized version of this problem. Obviously the categories $U(\mathfrak{g})-\text{Mod}$ and $U(\mathfrak{g}')-\text{Mod}$ are Morita equivalent but there is more information here. First of all $U(\mathfrak{g})\cong U(\mathfrak{g}')$ as algebras which clearly is stronger but they are also enveloping algebras of Lie algebras, further restricting possibilities. In the quantized version I'm looking at, I suspect the representation rings of both categories to be the same making the difference in the monoidal structure very subtle. So I'm wondering whether anything on this subject is known in the non-quantized world.

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In the recent paper Lie, associative, and commutative quasi-isomorphism, R. Campos, D. Petersen, F. Wierstra, and I settled the question above for nilpotent Lie algebras: if two nilpotent Lie algebras have universal enveloping algebras that are isomorphic as unital associative algebras, then the two Lie algebras also are isomorphic.

In fact, we proved a more general result in the differential graded context:

Theorem B: Let $\mathfrak{g}, \mathfrak{h}$ be two dg Lie algebras. If $U\mathfrak{g}$ and $U\mathfrak{h}$ are quasi-isomorphic as unital associative dg algebras, then the homotopy completions $\mathfrak{g}^{\wedge h}$ and $\mathfrak{h}^{\wedge h}$ are quasi-isomorphic as dg Lie algebras.

This has the statement above as a corollary, since one can show that a Lie algebra that is either strictly positively graded or non-negatively gradedand nilpotent is always quasi-isomorphic to its homotopy completion (in the language of the paper, it is homotopy complete). There are other interesting implications of this result in rational homotopy theory.

In my view (my coauthors might disagree) the spirit of the proof is mostly deformation theoretical, but operad theory play a big supporting role. For those who are interested in the structure of the proof without the technical details, we give a sketch of the arguments in paragraphs 0.27-0.31.


In a previous version of this answer and of the paper, we claimed that the more general statement that if two dg Lie algebras have universal enveloping algebras that are quasi-isomorphic as associative dg algebras, then the two dg Lie algebras are themselves quasi-isomorphic. Unfortunately, the proof had a gap that we were not able to fix. This more general statement remains open.

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  • $\begingroup$ Would you edit the post so as to make it easier to read and updated (emphasizing what you prove, rather than being the concatenation of a result and a preceding paragraph saying it has a mistake)? You could state what's proved in v3 of your arxiv paper and then say at the end that in a first version of the post/the paper a stronger version was claimed, which is still an open problem? $\endgroup$ – YCor Feb 3 '20 at 1:42
  • $\begingroup$ @YCor Good point, thank you. Done. $\endgroup$ – Daniel Robert-Nicoud Feb 4 '20 at 23:06
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FWIW, a ten year old article states: "We stress that, in spite of all this, the characteristic zero case of the isomorphism problem remains entirely open."

(https://link.springer.com/article/10.1007/s10468-007-9083-0)

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  • $\begingroup$ What a shame, this probably also explains why I was completely stuck on the quantized problem I encountered. The quantized version shouldn't be too much different as the representation theory of $\mathfrak{g}$ and $U_v(\mathfrak{g})$ are often very similar. Anyway, in the near future we will publish an article on something completely different where all of the sudden this question pops up. Maybe smarter people can use our example to book some progress. $\endgroup$ – Mathematician 42 Feb 6 '18 at 8:59
  • $\begingroup$ Thanks btw, I encountered this paper but completely missed the sentence saying that the characteristic zero case is still open. Thank you for spotting this. $\endgroup$ – Mathematician 42 Feb 6 '18 at 9:12

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