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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.

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    $\begingroup$ It would be helpful to explain your notation. $\endgroup$
    – YCor
    Aug 24, 2014 at 21:09
  • $\begingroup$ Sorry thought it was universal: $\mathfrak {S} $= symmetric tensor algebra, $\mathfrak {E} $=graded algebra of $\mathfrak {U} $ the universal enveloping algebra. $\mathfrak {T} $ is the tensor algebra of a Lie algebra $ L $. $\times $ is tensor product. $ x, y $ are elements of the Lie algebra. $ T_m $ is a filtration of the temsor algebra (=${T^{m}}\oplus {T^{m-1}}\oplus ... {T^{0}}$). $ U_m$ is its image under $\pi $. $ G^m $ is defined above. $\endgroup$
    – dylan7
    Aug 24, 2014 at 21:16
  • $\begingroup$ algebra of what? a Lie algebra? over a field? which field? finite-dimensional?... $\endgroup$
    – YCor
    Aug 24, 2014 at 21:18
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    $\begingroup$ This is a sketch of a standard proof, and one does not have to restrict to characteristic zero. See Section 22.2 of ``More Concise Algebraic Topology" by Kate Ponto and myself for a recent version, which deals with graded Lie algebras. $\endgroup$
    – Peter May
    Aug 24, 2014 at 22:30
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    $\begingroup$ If you want me to pinpoint one place which definitely looks wrong: in proving that $\operatorname{ker} \phi = I$, you cannot just restrict yourself to considering tensors of the form $x \otimes y - y \otimes x$. $\endgroup$ Aug 25, 2014 at 8:43

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