0

Let $L$ be a Lie algebra and $U(L)$ be the corresponding universal enveloping algebra. If $L$ is finite-dimensional then by the virtue of PBW theorem we know that $L$ is embedded in $U(L).$ Can we conclude the same thing for infinite dimensional Lie algebra as well? In order to figure it out I started to look at the construction of $U(L)$ which is obtained by quotienting the tensor algebra $T(L)$ modulo the two sided ideal $J$ of $T(L)$ generated by the elements of the form $[x,y] - (xy - yx),$ $x,y \in L.$ For finite-dimensional case the embedding is precisely given by the quotient map $x \mapsto x + J.$ But I firmly believe that it continues to be an embedding for infinite dimensional case as well. For that what we nee to argue is that $L \cap J = (0)$ But I couldn't quite able to prove it. Could anyone please help me in this regard?

Thanks in advance.

Anacardium
  • 2,357
  • The infinite dimensional case also follows from PBW theorem. – luxerhia Apr 27 '23 at 15:25
  • @luxerhia$:$ I only know finite-dimensional version of PBW theorem. Could you please tell me about the infinite dimensional version of PBW theorem? Can't we prove it directly by showing that $L \cap J = (0)\ $? Thanks for your prompt response and hoping to learn more on this. – Anacardium Apr 27 '23 at 15:32
  • Not so different from the finite-dimensional version. Let ${x_i }{ i \in I } $ be a basis of $L$ with a totally ordered index set $I$. Then the elements $x{i_1} x_{i_2} \cdots x_{i_n} $ form a basis of $U(L)$, where $i_1 \leq i_2 \leq \cdots i_n$ and $n \in \mathbb{Z}_{\geq 0}$. Another way of saying this is that the associated graded algebra $\mbox{gr} ~ U(L)$ is canocically isomorphic to the symmetric algebra $S(L)$ as an algebra. – luxerhia Apr 27 '23 at 18:19
  • I'm not sure one can establish the embedding $L \rightarrow U(L)$ without PBW, but it is possible in some special cases. See this and this. – luxerhia Apr 27 '23 at 18:27
  • @luxerhia$:$ But can $L$ always have a Hamel basis with totally ordered index set? – Anacardium Apr 27 '23 at 19:11
  • Yes, because every set can be well-ordered, it can be totally ordered. But this result is independent of ZF. – luxerhia Apr 28 '23 at 04:10

0 Answers0