Does the normal ordered product on differential operators lift to $U\left(\mathfrak{gl}_n\right)$?

Question

Let $n\in\mathbb N$. Let $k$ be a commutative ring in which $1,2,3,\ldots$ are invertible. Let $\Omega$ denote the $k$-algebra of polynomial differential operators on $n$ variables $x_1$, $x_2$, ..., $x_n$ over $k$. (The multiplication in this algebra is the composition of differential operators.)

Let $M:k\left[x_1,x_2,...,x_n\right]\otimes k\left[\dfrac{\partial}{\partial x_1},\dfrac{\partial}{\partial x_2},...,\dfrac{\partial}{\partial x_n}\right]\to\Omega$ be the $k$-linear map which sends every $P\otimes Q$ to $P\cdot Q$. Clearly, $M$ is an isomorphism of $k$-modules, but not of $k$-algebras (unless $k=0$ or $n\leq 1$). This allows us to define a commutative multiplication $\boxdot$ on $\Omega$ by letting

$A\boxdot B = M\left(M^{-1}\left(A\right)\cdot M^{-1}\left(B\right)\right)$ for all $A\in\Omega$ and $B\in\Omega$.

This $\boxdot$ is called the normal(ly?) ordered product on $\Omega$. (When $A$ and $B$ are atomic terms, one often writes $:AB:$ for $A\boxdot B$, but when $A$ and $B$ are composite terms, $:AB:$ can mean something slightly different.)

The $k$-linear map $\ell:\mathfrak{gl} _ {n}\to \Omega$ which sends every elementary matrix $E_{i,j}$ to $x_i\dfrac{\partial}{\partial x_j}$ is a Lie algebra homomorphism. Thus, it gives rise to a $k$-algebra homomorphism $L:U\left(\mathfrak{gl} _ {n}\right)\to \Omega$. This homomorphism $L$ is generally not injective (I guess it's like a noncommutative Segre embedding: it sends $E_{i,j}E_{k,l}$ to the same differential operator as $E_{i,l}E_{j,k}$ if $i$, $j$, $k$, $l$ are pairwise distinct). Hence the following question: Is there a (commutative?) multiplication $\boxdot$ on $U\left(\mathfrak{gl} _ {n}\right)$ such that any $A\in U\left(\mathfrak{gl} _ {n}\right)$ and $B\in U\left(\mathfrak{gl} _ {n}\right)$ satisfy $L\left(A\right)\boxdot L\left(B\right) = L\left(A\boxdot B\right)$ ?

Note that (by a theorem of Sylvester from 1867, in P92 from the second tome of his Collected Works, but not the one I asked about in that thread) the image of $L$ is closed under $\boxdot$, and actually is the $k$-subalgebra of $\Omega$ generated by the image of $\ell$ under $\boxdot$-multiplication. This speaks in favor of the existence of a $\boxdot$ on $U\left(\mathfrak{gl} _ {n}\right)$ (but is not a proof yet, since $L$ is not injective).

Answer 1

Yes, the multiplication $\boxdot$ on $U\left(\mathfrak{gl}_n\right)$ exists. Moreover, Alexander Chervov's conjecture is true: There is a commutative multiplication $\boxdot$ on $U\left(\mathfrak{gl}_n\right)$ such that, for every $m \in \mathbb{N}$, the $k$-algebra homomorphism $$ U\left(\mathfrak{gl}_n\right) \to \mathcal{D}_{n,m},\ E_{i,j} \mapsto \sum\limits_{k=1}^{m} x_{i,k} \dfrac{\partial}{\partial x_{j,k}} $$ (where $\mathcal{D}_{n,m}$ is the ring of all differential forms in the $nm$ variables $x_{i,k}$ with $1 \leq i \leq n$ and $1 \leq k \leq m$) sends the multiplication $\boxdot$ to the normal ordered product $\boxdot$ on $\mathcal{D}_{n,m}$ (which normal ordered product is defined similarly to the one above on $\Omega$).

For a proof, see Theorem 4.6 in my note On the PBW theorem for pre-Lie algebras. Unfortunately, the proof is really long; chiefly responsible for this are folklore lemmas which I couldn't easily find in the literature and annoying computations with differential operators which I had to do in all their gory detail in order not to get wrong. (As the introduction says, ignore the proofs except maybe that of Theorem 4.7 (b).)

Note that the notations in my note differ from those in my OP above; in particular, $\Omega$ means something different.