Proving isomorphism between the symmetric algebra $\mathfrak{S}$ and the associated graded algebra $\mathfrak{G}$

Question

I am studying universal enveloping algebras following Humphrey's book (https://www.math.uci.edu/~brusso/humphreys.pdf - especially pages $91,92$).

In the book it is proved that $\omega: \mathfrak{S}\rightarrow \mathfrak{G}$ is a isomorphism of algebras (where $\mathfrak{S}$ denotes the symmetric algebra and $\mathfrak{G}$ denotes the associated graded algebra of the universal enveloping algebra $\mathfrak{U(L)}$) and this result is then utilized to prove the coordinate version of PBW-Theorem:

Let $(x_1,x_2,\cdots)$ be any ordered basis of $L$.

Then the elements $x_{i(1)}\cdots x_{i(m)} = \pi(x_{i(1)}\otimes\cdots \otimes x_{i(m)}) ,m \in Z^+, i(1)\leq\cdots\leq i(m),$ along with $1$, form a basis of $\mathfrak{U(L)}$.

But I would like to prove the opposite way. I want to assume the result above and prove that $\mathfrak{S}$ is isomorphic to $\mathfrak{G}$ (coordinate free version of PBW-Theorem).

Following the arguments given in the book i really could see that the map $\phi$ is an isomorphism of algebras and that it is surjective. But i can't prove that the induced map $\omega$ is injective using the theorem above.

Any hints on how to use the fact that $x_{i(1)}\cdots x_{i(m)} $ forms a basis of $\mathfrak{U(L)}$ in order to prove that $\omega $ is injective?

Thanks in advance!

Answer 1

Here is the outline of the proof:

1. The images of words of length $d$ $$\{x_{i_1} \cdots x_{i_d} \ | \ i_1 \leq i_2 \leq \cdots \}$$ in the degree $d$ piece of the associated graded $\mathfrak{G}_d$ are a basis of $\mathfrak{G}_d$.

2. The same words are a basis of the degree $d$ piece $\mathfrak{S}_d$ of the symmetric algebra.

3. The map $\omega$ takes the second set of words to the first.