Symmetrization map over the polynomial ring of a vector space.

Question

Let $V$ be a finite-dimensional complex vector space. Is the morphism \begin{gather*} \mathrm{Sym}^{\bullet}(V \oplus V^*) \to D(V) \cong \frac{\mathrm{T}^{\bullet}(V\oplus V^*)}{I} \,, \\[0.5em] (v_1, w_1) \cdots (v_n, w_n) \mapsto \frac{1}{n!} \sum_{\sigma\in S_n}(v_{\sigma(1)}, w_{\sigma(1)}) \otimes \cdots\otimes (v_{\sigma(n)}, w_{\sigma(n)}) \bmod I \,, \end{gather*} where $I := \langle [x,y] = y(x) : x \in V, y \in V^* \rangle$, a morphism of $\mathbb{C}[V]$-modules? Clearly, it is a linear map over $\mathbb{C}$.

Answer 1

This doesn't look to me to be true even if $V$ is $1$-dimensional. In this case the algebra of differential operators $D(V)$ is the Weyl algebra $k[x, \partial]/([\partial, x] = 1)$ while the LHS is the polynomial algebra $k[x, y]$ in two variables, and the symmetrization map, if I have understood correctly, sends

$$x \mapsto x, y \mapsto \partial$$ $$xy \mapsto \frac{x \partial + \partial x}{2}$$

which shows that it does not respect either the left multiplication $k[x]$-module structure or the right multiplication $k[x]$-module structure on the Weyl algebra (not surprising since it "mixes" the two through symmetrization).