I was reading up Humphrey's Introduction to Lie Algebras and Representation Theory and have a confusion regarding a consequence of PBW.
First some notations:
Let $L$ be a Lie algebra over $\mathbb{C}$, define $T^{\le m}(L) = \bigoplus_{n = 0}^m T^n(L)$ and $T(L) = \bigoplus_{m = 0}^\infty T^m(L)$. Let $S(L)$ be the symmetric algebra of $L$ define as $T(L)/I$, where $I$ is the ideal generated by elements of the form $x\otimes y-y\otimes x$. Let $\tilde{S}(L)$ be the symmetric algebra of $L$ defined as a subspace of $T(L)$, i.e the image of $S:T(L)\to T(L)$ where $$S(x_1\otimes \cdots\otimes x_n) = \frac{1}{n!}\sum_{\sigma\in S_n} x_{\sigma(1)}\otimes\cdots\otimes x_{\sigma(n)}.$$ We know $\tilde{S}(L)\cong S(L)$ as graded commutative algebras with 1.
Let $J$ be the ideal generated by elements of the form $x\otimes y-y\otimes x-[xy]$ inside $T(L)$. Define the universal enveloping algebra $U(L) = T(L)/J$ and let $\pi:T(L)\to U(L)$ be the canonical map. Define a filtration on $U^m = \pi(T^{\le m}(L))$ with $U^{-1} = 0$ and the layers $G^m = U^m/U^{m-1}$. Define bilinear multiplication $G^m\times G^p\to G^{m+p}$ by the multiplication in $T(L)$, namely $$((a+J)+U_{m-1})\cdot((b+J)+U_{p-1}) = (ab+J)+U_{m+p-1}.$$ This multiplication is well defined. Set $\mathfrak{S} = \bigoplus_{m = 0}^\infty G^m$, and is it a graded associative algebra with 1.
Let $\phi_n: T^m(L)\to U^m\to G^m$ be the composition map, observe that it is surjective because $\phi_n(T^m(L)-T^{m-1}(L)) = U^m-U^{m-1}$. Therefore $\phi_n$ extends to a surjective map $\phi: T(L)\to \mathfrak{S}$. Further observe that this map is an algebra homomorphism and $I\subset \ker \phi$, thus it factors through $S(L)$. We have $$\overline{\phi}: S(L)\to \mathfrak{S}(L). $$
PBW theorem states that $\ker \phi = I$ and $\overline{\phi}:S(L) \cong \mathfrak{S}(L)$ as graded associative algebras.
The question is the following:
Consider $\pi: T^m\to U^m$ and $p:U^m\to G^m$ the canonical maps, Then $p\circ \pi: T^m\to G^m$ factors through $S^m$. From this, we can deduce that if $W$ is a subspace of $T^m$ such that it maps isomorphically onto $S^m$, it is mapped isomorphically onto $G^m$. Thus we must have $W$ mapping isomorphically onto $G^m$ by $p\circ \pi$. As a result, $\pi(W)$ is a direct complement to $U^{m-1}$ in $U^m$. If we set $W = T^1 = L$ and consider $m = 1$ then we see that $L$ injects into $U^1$ by $\pi$ and $U^1 = \mathbb{C}\oplus L$. Now take $W = \tilde{S}^m$, which is mapped isomorphically onto $S^m$. Thus we will have that $\pi$ injects $\tilde{S}^m$ into $U^m$ as a direct complement to $U^{m-1}$. Then wouldn't it mean by induction that $U \cong \mathbb{C}\oplus L\oplus \tilde{S}^1(L)\oplus\cdots = \tilde{S}(L)$?
Yes, it is indeed true. In fact there is $S(L)$ and $U(L)$ are isomorphic as $L$-modules, in particular as vector spaces.