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

Let $n\in\mathbb N$. Let $k$ be a commutative ring. 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).

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 Have not you checked the following ? U(gl_n) can be EMBEDDED in differential operators E_ij = \sum_k x_ik d/dx_jk (this is left invariant operators on GL_n; in matrix terms E=XD^t - see Wiki on Capelli). So since this is embeding you can probably pushback the normally ordered product from differential operators to U(gl). May be it will satisfy the property you are interested in. Such a product is called "special symmetrization". It has been considered by Olashanski and Okounkov. There is some kind of Wick rule for it see arxiv.org/abs/q-alg/9602027 – Alexander Chervov Jul 22 at 16:23 By the way, sometimes this "normally ordered" product is called "Wick ordering". Since you put "x" left and differential operators on the right. In some sense the "sense" of Capelli identities is that "Wick ordering is good". See arxiv.org/abs/1203.5759 page 8 section 1.3.1 – Alexander Chervov Jul 22 at 16:28 I like the \sum_k generalization; thanks for that! I'd wish there would be an easy argument why (if at all) the pushback is independent on the number of k's though. – darij grinberg Jul 22 at 22:01 Good question about "easy argument". I feel it should be it true, but at the moment have no arguments... – Alexander Chervov Jul 23 at 11:41