MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Tags changed (lie-groups -> algebraic-groups) so it doesn't look like it's a geometry question. – darij grinberg Dec 25 '10 at 16:46
up vote 7 down vote accepted

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.

share|cite|improve this answer
Nice one! I should have thought about this - it's the way it is shown that a Lie algebra injection $\mathfrak g\to \mathfrak h$ gives rise to an injective map $\mathfrak U\left(\mathfrak g\right)\to \mathfrak U\left(\mathfrak h\right)$. – darij grinberg Dec 25 '10 at 17:38

Yes, the map is injective.

In general, suppose I have a Lie algebra $\mathfrak g$ acting on a space $X$; then I get a Lie algebra structure on the vector bundle $\mathfrak g \times X$ with anchor map $\rho: (\mathfrak g \times X) \to {\rm T}X$. I can form the universal enveloping algebroid for each: $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$. The universal enveloping algebroid of a Lie algebroid $A \to X$ is the $\mathcal O(X)$-algebra generated by $\Gamma(A)$ with commutation relations coming from the bracket $[,]: \Gamma(A) \otimes_k \Gamma(A)$ — note that $\Gamma(A)$ is a $k$-Lie algebra but not an $\mathcal O(X)$-Lie algebra.

Anyway, the trivialization gives a map $\mathfrak U \mathfrak g \hookrightarrow \mathfrak U(\mathfrak g \times X)$ by sending each element of $\mathfrak g$ to the corresponding constant section (in fact, it determines an isomorphism of $\mathcal O(X)$-modules but not of algebras $\mathfrak U(\mathfrak g \times X) \cong \mathfrak U \mathfrak g \otimes \mathcal O(X)$, with each tensorand embedding as a subalgebra).

On the other hand, $\mathfrak U({\rm T}X)$ is the algebra of differential operators on $X$, and your map $\mathfrak U \mathfrak g \to \mathfrak U({\rm T}X)$ factors through $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$.

Now suppose that for densely many $x \in X$, the action map $\mathfrak g \to {\rm T}_x X$ is injective. Then $\Gamma\rho: \Gamma(\mathfrak g \times X) \to \Gamma({\rm T}X)$ is injective. In particular, any relation imposed in the construction of $\mathfrak U({\rm T}X)$ pulls back to a relation in $\mathfrak U(\mathfrak g \times X)$. Letting $X = k^n$ and $\mathfrak g = \mathfrak{gl}_n$ finishes your question. Edit: No it doesn't: fiberwise, the map $\mathfrak{gl} \to {\rm T}k^n$ cannot be injective, just by dimensions! So my argument is broken. It might be salvageable, so I'll leave it up in case someone has a fix (if so, let me know and I'll switch the answer to CW), but I won't be fixing it today.

(I was going to claim that as soon as $\mathfrak g \to \Gamma({\rm T}X)$ was injective, then the answer would be yes, but in fact the "densely many $x$" requirement is important. For example, let $\mathfrak g = k^2 = \operatorname{span}(e_1,e_2)$ be the two-dimensional abelian Lie algebra, and pick $X$ with opens $U_1,U_2$ so that there are nonzero functions supported on one and not the other. Let the basis vector $e_i$ of $\mathfrak g$ act by some non-zero vector field supported entirely in $U_i$. This is certainly a representation. But at the level of universal enveloping algebra/oids, the product $e_1e_2$ is non-zero in $\mathfrak U\mathfrak g$ and $\mathfrak U(\mathfrak g \times X)$ but acts as the zero differential operator.)

Oh, a final comment: Of course, you should probably sprinkle the words "sheaf of" throughout the above exposition.

share|cite|improve this answer
Also, there very well might be a much more direct argument. This was the first that came to mind. – Theo Johnson-Freyd Dec 25 '10 at 4:26
Thanks, I've +1ed this although I don't understand much. Do you happen to know where one can learn about sheaves of Lie algebras / algebroids without having to know geometry? – darij grinberg Dec 25 '10 at 9:40
Thanks, but don't +1 too quickly, I realized there's a bit of an error. I do think that a "universal enveloping algebroid" style answer should work, but there might be better ones. In any case, I don't think very sheafily, but I do think about Lie algebroids sometimes. For definition and examples, they're locally all actions of Lie groups, but the idea is to forget which vector fields are the "constant" ones. For global results, you'll have to ask a geometer. – Theo Johnson-Freyd Dec 25 '10 at 16:44

Let me remark.

L, R define the left and right invariant vector fields on GL_n. And hence corresponding maps of U(g) are surjective ( this is true for any Lie algebra and Lie group).

I mean that we identified GL as an open subspace in Mat_n. So this gives us explicit coordinates and formulas above are writting invariant objects in coordinates.

It is easy to see that L and R are invariant vector fields. This can be done like this: rewrite formulas in the matrix form: L(E) = XD^t where E is matrix with E_ij element and X is matrix with x_ij matrix , D is matrix with partial derivatives.

Now natural action of GL is X-> Xg, D-> Dg^{-1}^t, so when we multiply XD^t you see that "g" cancels and you see that L(E) is invariant.

My argument is a little informal but it can be made formal.

Actually sometimes left invariant is called right and vice versa.

This stuff is popular in Capelli identities story as You know since you read Howe Umeda, I wrote something on it on wikipedia - see "Capelli identities"...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.