# Relation between Symmetric algebra and Universal enveloping algebra as Lie algebras.

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.

• 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. – tessellation Apr 18 at 9:14
• 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. – tessellation Apr 18 at 10:15
• @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. – YCor Apr 18 at 10:16
• @tessellation Have you looked at explicit cases such as $L$ non-abelian 2-dimensional, or $L=\mathfrak{sl}_2$? – YCor Apr 18 at 10:21
• @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.. – tessellation Apr 18 at 11:35