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?



Is the space of polynomial functions on M_n a faithful U(gl_n)-module?
The usual way to prove something like this is by taking associated graded; if you have a filtered map between two filtered rings, then if the induced map of associated gradeds is injective, the original map was injective (the other way is wildly false).


So what is this associated graded of this map for the PBW filtration on the UEA and the order filtration on differential operators? The universal enveloping algebra becomes the symmetric algebra on $\mathfrak{gl}\_n=M\_n$ and the differential operators become functions on $M\_n\times M\_n$ (the $x\_{j,k}$ and the $\frac{d}{dx\_{j,k}}$ are commuting polynomial generators now). So I want to know if this map of polynomial rings is injective.


Of course, by the usual tricks of algebraic geometry, this is the same as asking if the corresponding map of affine spaces gotten by taking Spec is dominant (has dense image). So, one interprets for formulae for $L$ and $R$ you wrote above as formulae in commuting coordinates for a map $M\_n\times M\_n\to M\_n$.


What map is it? For $L$ you get $(A,B)\mapsto AB$ and for $R$ you get $(A,B)\mapsto BA$, both of which are obviously surjective. Thus, the original maps are injective.


