We are over some field $k$ of characteristic $0$. The general linear group $\mathrm{GL}_n$ canonically acts from the left and from the right on the space $\mathrm{M}_n$ of all $n\times n$-matrices, and thus also acts from the left and from the right on the ring $k\left[\mathrm{M}_n\right]$ of polynomial functions on $\mathrm{M}_n$. Differentiating these two actions yields a left and a right action of the Lie algebra $\mathfrak{gl}_n$ on $k\left[\mathrm{M}_n\right]$. These actions are Lie algebra actions, and thus can be lifted to algebra actions (again, a left and a right one) of the universal envelopping algebra $\mathfrak U\left(\mathfrak{gl}_n\right)$ on the space $k\left[\mathrm{M}_n\right]$. Explicitly, these are given by

$L\left(E_{i,j}\right)=-\sum\limits_{l=1}^n x_{j,l}\dfrac{\partial}{\partial x_{i,l}}$;

$R\left(E_{i,j}\right)=\sum\limits_{l=1}^n x_{l,i}\dfrac{\partial}{\partial x_{l,j}}$,

where $E_{i,j}$ denotes the elementary matrix with $1$ in cell $\left(i,j\right)$ and $0$ in all other cells, and $x_{u,v}$ are the coordinate functions on the space $\mathrm{M}_n$ (so that $\sum\limits_{i=1}^n\sum\limits_{j=1}^n E_{i,j}\cdot x_{i,j}=\mathrm{id}$). Of course, $L$ stands for left action and $R$ for right action.

So we have two maps $L$ and $R$ from the algebra $\mathfrak U\left(\mathfrak{gl}_n\right)$ to the algebra of polynomial differential operators on $k\left[\mathrm{M}_n\right]$.

**Question:** Why are $L$ and $R$ injective?

Actually I am not sure they are, since...

**Motivation:** ... this question comes from reading

Roger Howe, Tôru Umeda, *The Capelli identity, the double commutant theorem, and multiplicity-free actions*, Math. Ann. 290, 565-619 (1991)

(I can send you the paper should you wish), where the authors claim (on page 567) that

$L\left(\mathfrak Z\mathfrak U\left(\mathfrak{gl}_n\right)\right) = L\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right) \cap R\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right) = R\left(\mathfrak Z\mathfrak U\left(\mathfrak{gl}_n\right)\right)$

(where $\mathfrak Z\mathfrak U\left(\mathfrak{gl}_n\right)$ denotes the center of $\mathfrak U\left(\mathfrak{gl}_n\right)$) **because** the images of $L\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right)$ and $R\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right)$ in the algebra of polynomial differential operators commute with each other. It's this "because" that makes me think that $L$ and $R$ are injective, as otherwise I don't know how to prove that the intersection $L\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right) \cap R\left(\mathfrak U\left(\mathfrak{gl}_n\right)\right)$ is inside $L\left(\mathfrak Z\mathfrak U\left(\mathfrak{gl}_n\right)\right)$. Maybe I am just blind?