# Are the elements of a Lie algebra separated by its finite-dimensional representations?

Let $$\mathbb{k}$$ be some field and let $$\mathfrak{g}$$ be a $$\mathbb{k}$$-Lie algebra.

Does there exist for every nonzero element $$x$$ of $$\mathfrak{g}$$ a finite-dimensional representation of $$\mathfrak{g}$$ on which $$x$$ acts nonzero?

In other words, do the finite-dimensional representations of $$\mathfrak{g}$$ separate the elements of $$\mathfrak{g}$$?

• If $$\mathfrak{g}$$ is finite-dimensional, then this is true by Ado’s theorem.

• As pointed out in an answer to a similar question, the finite-dimensional representations of $$\mathfrak{g}$$ even separate the points in the universal enveloping algebra $$\operatorname{U}(\mathfrak{g})$$ if $$\mathfrak{g}$$ is finite-dimensional and $$\mathbb{k}$$ is of characteristic zero. (This seems to be Theorem 2.5.7 in Dixmier’s Enveloping Algebras.)

Both of the above arguments show even stronger assertions, but also need some additional assumptions and quite a bit of work.

• An infinite-dimensional simple Lie algebra wouldn't have this property (which one could call "residually finite-dimensional"), and presumably these exist. – Qiaochu Yuan Dec 7 '20 at 0:33

The Wikipedia says there are no finite-dimensional representations of (non-trivial) affine Lie algebras, which implies the desired result is false: https://en.wikipedia.org/wiki/Affine_Lie_algebra

There are many possible answers. Here's one Edit: Here are a few:

Consider the Lie algebra (over a fixed field of characteristic zero) with presentation $$\mathfrak{g}=\langle x,y,z\mid [x,y]=y,\;[y,z]=z\rangle.$$

(1) It's easy to check that $$f(z)=0$$ for every finite-dimensional representation.

(2) However, $$z\neq 0$$. This is because this is by definition amalgam of two 2-dimensional Lie algebras $$\langle x,y\mid [x,y]=y\rangle$$ and $$\langle y,z\mid [y,z]=z\rangle$$ the common 1-dimensional subalgebra $$Ky$$, and it's known (see Encyclopedia of Math.) that subalgebras embed into their amalgam.

For (1), it's a simple consequence of the study of finite-dimensional representations of the 2-dimensional non-abelian Lie algebra $$\langle x,y\mid [x,y]=y\rangle$$, which we can assume to be over an algebraically closed field. Every such representation maps $$y$$ to a nilpotent matrix. Now consider a finite-dimensional representation of $$\mathfrak{g}$$, mapping $$x,y,z$$ to $$X,Y,Z$$.. Using the first subalgebra, $$Y$$ is nilpotent. Also, we can make the second subalgebra act as upper triangular matrices, and $$Z$$ is nilpotent. So both $$Y,Z$$ are strictly upper triangular, and $$[Y,Z]=Z$$ forces $$Z=0$$.

Similarly we can deduce that the analogue of Higman group, the Lie algebra $$\langle x_0,x_1,x_2,x_3\mid [x_{i-1},x_i]=x_i: i\in\mathbf{Z}/4\mathbf{Z}\rangle$$ has no non-trivial finite-dimensional representation. I guess one can elaborate using amalgams (but haven't checked details) that it's not trivial (hence infinite-dimensional).

$$\DeclareMathOperator\h{\mathfrak{h}}$$Here's now an example that is completely self-contained.

Consider the Lie algebra $$\h$$ with basis $$u$$, $$(e_n)_{n\in\mathbf{Z}}$$, law $$[e_i,e_j]=(i-j)e_{i+j}$$, $$[u,e_i]=ie_i$$, over a field $$K$$ of characteristic zero.

I claim that every finite-dimensional representation of $$\h$$ kills all $$e_i$$. Indeed, consider operators $$U$$, $$E_n$$ of a finite-dimensional vector space satisfying the same relations. Since $$[U,E_n]=nE_n$$, the $$E_n$$ are in distinct eigenspaces for $$\mathrm{ad}(U)$$, and hence the $$KE_n$$ generate their direct sum. Since the dimension is finite, there exists $$n$$ such that $$E_n=0$$. Then for $$m\neq 2n$$, $$E_m=\frac{1}{2n-m}[E_n,E_{m-n}]=0$$. In turn $$E_{2n}=\frac{1}{2-2n}[E_1,E_{2n-1}]=0$$, so $$E_m=0$$ for all $$m\in\mathbf{Z}$$.

Actually, in this example the subalgebra $$\mathfrak{r}$$ already has the property that every finite-dimensional representation is trivial, but using a slightly more elaborate argument, which however works in arbitrary characteristic $$\neq 2$$.

Let $$W_n$$ be the subspace generated by $$\{E_k:k\ge n\}$$, and $$W_\infty=\bigcap_n W_n$$, so $$W_\infty=W_n$$ for large enough $$n$$, say $$n\ge n_0$$. Then $$[E_n,W_\infty]=W_{\infty}$$ for all $$n$$.

Suppose by contradiction that $$W_\infty\neq 0$$. Choose $$n\ge n_0$$. Take a block-diagonal decomposition of $$E_n$$. Then the sum $$M$$ of characteristic subspaces for nonzero eigenvalues of $$\mathrm{ad}(E_n)$$ consists of those matrices in this block decomposition all of whose diagonal blocks are zero. The condition $$[E_n,W_\infty]=W_\infty$$ forces $$W_\infty\subset M$$. In particular, $$E_n$$ has this form. But by definition $$E_n$$ is block-diagonal. So $$E_n=0$$, and this works for all $$n\ge n_0$$.

So $$W_{\infty}=0$$, that is, $$E_n=0$$ for all large $$n$$. Similarly $$E_{-n}=0$$ for all large $$n$$. Using commutators we deduce that $$E_n=\frac{1}{n+2q}[E_{n+q},E_{-q}]=0$$ (choosing $$q$$ such that $$n+2q\neq 0$$ in $$K$$).