PBW symmetrization in characteristic zero

By [L1] the coefficient $1\mathrm{/}n!$ exists. Reindexing the finite sum by $\pi \mapsto \pi \tau$ for any $\tau \in S_n$ shows by [L2] that the displayed multilinear expression is invariant under permuting the inputs, so it descends to a linear map on $S^n(\mathfrak{g})$; for $n=0$ it sends $1$ to $1$. Taking the graded direct sum defines $\mathrm{sym}$, and degree $n$ maps into $F_{n}U(\mathfrak{g})$.

In degree zero, $\mathrm{sym}:S^0(\mathfrak{g})=k\to F_{0}U(\mathfrak{g})$ has associated-graded map ${\sigma }_0$, hence is bijective by [L3].

Assume that every element of $F_{n-1}U(\mathfrak{g})$ has a unique preimage in ${⨁}_{r<n}{S}^r(\mathfrak{g})$.

In $F_n\mathrm{/}{F}_{n-1}$ all reordered products of $v_1,\dots ,v_n$ have the same symbol, because interchanging adjacent factors changes a word by a bracket term of degree $n-1$. Thus the leading symbol of $\mathrm{sym}(v_1\cdots v_n)$ is the average of $n!$ identical symbols, namely $\sigma (v_1\cdots v_n)$. Hence $\mathrm{gr}(\mathrm{sym})=\sigma$.

Given $u\in F_n$, use the surjectivity of ${\sigma }_n$ and step 2.1 to choose $s_n\in S^n(\mathfrak{g})$ whose symmetrization has the same class as $u$ in $F_n\mathrm{/}{F}_{n-1}$. Then $u-\mathrm{sym}(s_n)\in F_{n-1}$ and step 1.3 supplies a preimage, proving surjectivity through degree $n$.

If $s={\sum }_{r\le n}{s}_r$ and $\mathrm{sym}(s)=0$, its top filtration class is ${\sigma }_n(s_n)$ by step 2.1, so $s_n=0$ by injectivity of [L3]. Descending in degree, or using the uniqueness clause in step 1.3, gives every $s_r=0$. Thus symmetrization is injective through degree $n$.

Steps 1.2–3.2 complete the filtration induction. Every element of either algebra has finite degree, so $\mathrm{sym}$ is a filtered vector-space isomorphism on the full direct sums. The proof asserts no multiplicativity.