Let $L$ be a Lie algebra over $\mathbb{C}$. Assume $L$ satisfies PBW theorem. We can associate two Lie algebras with $L$:

1) $U(L):$ the universal enveloping algebra. Here the Lie bracket is defined by $[x,y]=x*y-y*x$ where $*$ is the product induced from tensor product.

2)$S(L):$ the symmetric algebra. Here the Lie bracket is defined by extending the Leibniz rule $[xy,z]=x[y,z]+[x,z]y.$

Q) Are theses two Lie algebras isomorphic as Lie algebras.

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    $\begingroup$ I have defined the Lie bracket. With this bracket $S(L)$ is a Lie algebra. There is no grading involved to define this Lie algebra. $\endgroup$ – tessellation Apr 18 at 9:14
  • $\begingroup$ Yes. Surely you can define Lie product in the Symmetric Lie algebra in many ways. Given a vector space, there may be more than one way to define Lie bracket. My question is regarding the Lie bracket defined in the question. $\endgroup$ – tessellation Apr 18 at 10:15
  • $\begingroup$ @DietrichBurde Exercise: prove that there is a unique Lie bracket on $S(L)$ that restricts to the original Lie bracket on $L$ and that satisfies the Leibniz rule as given by the OP. $\endgroup$ – YCor Apr 18 at 10:16
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    $\begingroup$ @tessellation Have you looked at explicit cases such as $L$ non-abelian 2-dimensional, or $L=\mathfrak{sl}_2$? $\endgroup$ – YCor Apr 18 at 10:21
  • $\begingroup$ @YCor Hi. I am quite new to this subject and my knowledge is limited to wiki entries and some basic lecture notes. Therefore I am not even capable of understanding $\mathfrak{sl}_2$. Therefore it would be really helpful if you could give some hints regarding whether this is even true or not.. $\endgroup$ – tessellation Apr 18 at 11:35

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