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The PBW theorem states:

$\omega:\mathfrak {S} \mapsto \mathfrak {E} $ is an isomorphism of algebras.

Where $\mathfrak {S} $ is the symmetric tensor algebra of a Lie algebra $ L $.

Where $\mathfrak {E} $ is the graded algebra of the universal enveloping algebra of $ L $.

Does anyone have an intuitive reason why this is true?

I know this stems from the map of $\phi:\mathfrak {T} \mapsto \mathfrak {E} $, where $\mathfrak {T} $ is tensor algrbra of $ L $. The kernel of this map is suppose to be the same ideal of the tensor algebra that is used to make the symmetric algebra. In other words ker $\phi $ is suppose to equal the kernel of the map $\mathfrak {T} \mapsto \mathfrak {S} $, but I don't see why this true intuitively.

Thanks. All help us appreciated.

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  • $\begingroup$ Essentially, the idea is that, while there's no actual notion of degree in the Universal Enveloping Algerba, the commutator in the tensor algebra being "degree 1" allows you to take a basis $\{X_{1}, ..., X_{n}\}$ and then move terms such as $X_{1}X_{n}X_{1}$ to $X_{1}^{2}X_{n}$, while only picking up lower "degree" terms. Now, when you pass to the graded algebra, the degree thing not only makes sense but the lower "degree" terms become 0. $\endgroup$ – Siddharth Venkatesh Aug 26 '14 at 4:32
  • $\begingroup$ So all commutators become zero when you pass from $\mathfrak {T} \mapsto \mathfrak {E} $? Do the tensors generated by $ x\otimes y-y\otimes x $ become zero too? Thanks $\endgroup$ – dylan7 Aug 26 '14 at 15:40
  • $\begingroup$ I was a little too informal in what I was saying. There isn't actually a canonical map from a filtered algebra to its associated graded algebra. So there isn't a sense in which commutators in the enveloping algera become 0 when you pass to the graded algebra unless you choose them to become 0. On the other hand, it is true that the graded algebra is abelian, and you can use notion of degree to show that the commutators in the tensor algebra do go to 0 in the graded algebra (under the map you described). If you like, I can go into more detail in a full answer. $\endgroup$ – Siddharth Venkatesh Aug 27 '14 at 3:22
  • $\begingroup$ @Siddharth Venkatesh yes please do go into more detail. Thank you. $\endgroup$ – dylan7 Sep 12 '14 at 1:17

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