One version of the PBW theorem states:
$\omega $:$\mathfrak {S} \mapsto \mathfrak {E} $ is an isomorphism of algebras.
I am curious if this is a possible proof for the PBW theorem, part is taken from Humphreys' book:
$ L $ is a finite-dimensional Lie algebra over $\mathbb {R} $. All the algebras stated below are "of" $ L $. $ T_m $ is a filtration of the tensor algebra of $ L$, which equals ${T^m}\oplus {T^{m-1}}...\oplus {T^{0}}$. $ U_m $ is its image under $\pi $ defined below. $ G^m $ is defined below. $ T^m$ is defined as ${T^m}L=L\otimes L...\otimes L $ $ m $ times where $\otimes $ is the tensor product.
Let $\pi $ define a map from $\mathfrak {T} $ to $\mathfrak {U} $. We define a filtration on the tensor algebra by $ T_m $ and under $\pi $ it maps to $ U_m $. We look to construct the graded algebra of $\mathfrak {U} $ denoted by $\mathfrak {E} $ containing $ G^m $=$ \frac{U_m}{U_{m-1}} $. We now define new map $\phi $ which maps $\mathfrak {T} $ to $ \mathfrak {E} $. Due to the way $\mathfrak {U} $ is constructed: $\pi (x\otimes y -y\otimes x) \in U_2$ in addition $ \pi([xy])\in U_1$ however due to the construction $ U_2$=$ U_1$ and thus these two arguments of $\pi$ are in the kernel of $ \phi $. Another conclusion is that the kernel of $\phi $ contains the kernel ($I$) of the canonical map, maping $\mathfrak {T} $ to $\mathfrak {S} $, the symmetric tensor algebra. Thus $\omega $ is onto. And we conclude $ I $ $\subseteq ker \phi $.
Next part is not from the book...
For $\omega $ to be an isomorphism the $ ker \phi $ must equal $ I $
This means that (*) $ x\otimes y-y\otimes x \in ker \phi$, iff $[xy]=0$, when this is true * is in $ I $. If * were true when $[xy] $ is not $0$, this would not make $\mathfrak {E} $ the graded algebra of $\mathfrak {U} $, and thus a contradiction by the definition of $\mathfrak {E} $. Thus $ ker \phi $=$I$, and $\omega $ is an isomorphism of algebras.
If this not a correct proof of the into part can someone give me a reason why * is in $\ker \phi $ iff $[xy]=0$? I just kind of thought if this. If this a true proof, I don't know of someone has already prooved it this way.
Hope this is clear and I appreciate the help.