0

Let $L$ be a Lie algeba over $\mathbb{C}$. It is well known that the universal enveloping algebra of $L$ is defined to be $T(L)/J$ where $T(L)$ is the tensor algebra of $L$ and $J$ is its ideal generated by $v\otimes w-w\otimes v-[v,w]$ for $v,w\in L$.

If $L$ has a basis $\{x_1,x_2,\cdots,x_n\}$ then consider the elements $$(*)\,\,\,\,\,\,\,\,\,\,\,\,x_i\otimes x_j-x_j\otimes x_i -[x_i,x_j] \,\,\,\, i\neq j$$

Q. Is it true that the set $$\{ (x_{i_1} \otimes \cdots \otimes x_{i_r})\otimes (x_i\otimes x_j-x_j\otimes x_i -[x_i,x_j])\otimes (x_{j_1} \otimes \cdots \otimes x_{j_s})\}$$ where $i's,j's$ are in $\{1,2,\ldots,n\}$; $i\neq j$; and $r,s\ge 0$ forms a basis of the vector space $J$?


Elaboration: I am considering the simple system of generators of the ideal $J$ - namely first take the elements (*); and then tensor these from left as well as right by the chosen basis elements of $L$; it is clear that the collection of all such expression in $T(L)$ is a generating set for vector space $J$. My question is whether it is basis of $J$?

Beginner
  • 10,836
  • There is of course the trivial answer no, because you have both $x_i\otimes x_j-x_j\otimes x_i-[x_i,x_j]$ and $x_j\otimes x_i-x_i\otimes x_j-[x_j,x_i]$ in the set. – user10354138 Oct 22 '18 at 11:30
  • You are right; thanks for the clarification. I must have to say $i<j$ than $i\neq j$, right? Should I edit the question? – Beginner Oct 22 '18 at 13:09

0 Answers0