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Let $L$ be a Lie algebra of finite dimension and $U(L)$ be its universal enveloping algebra. For every $x\in L$, define $ad_x:U(L)\rightarrow U(L)$ by $ad_x(t)=xt-tx$ for $t\in U(L)$. Following is an exercise in Humphreys' Lie algebra:

Prove that every $t\in U(L)$ lies in finite dimensional $L$-submodule of $U(L)$.

My justification: Fix a basis $\{x_1,x_2,\ldots,x_n\}$ of $L$.

For example, if $t$ is a monomial of degree $2$, say $t=x_{i_1}x_{i_2}$ then $ad_{x_j}(t)=x_j(x_{i_1}x_{i_2})-(x_{i_1}x_{i_2})x_j$. This can be written as $$x_j(x_{i_1}x_{i_2})-x_{i_1}x_jx_{i_2} + x_{i_1}x_jx_{i_2}-x_{i_1}x_{i_2}x_j=[x_{j},x_{i_1}]x_{i_2} + x_{i_1}[x_{j},x_{i_2}]$$ and this expression is again a combination of monomials of degree $\le 2$. So the upshot is - we can, in general, say in similar way:

The subspace of $U(L)$ consisting of all monomials in $x_1,x_2,\ldots,x_n$ of degree $\le k$ is $L$-submodule of $U(L)$.

Q. Is the last assertion correct?

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1 Answers1

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Since the OP did not say over which field the Lie algebra is defined, I let $F$ be the base field. By the Poincaré–Birkhoff–Witt (PBW) theorem, each $t\in U(L)$ can be written as a linear combination $$\sum_{\alpha}c_\alpha x^\alpha$$ where $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_n)$ runs through all multi-indices with $\alpha_i\in\mathbb{N}_0$, $c_\alpha \in F$ is nonzero for only finitely many values of $\alpha$, and $x^\alpha$ denotes $$x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}.$$ The maximum of $|\alpha|=\alpha_1+\alpha_2+\ldots+\alpha_n$ among nonzero terms $x^\alpha$ that appear in $t$ is called the degree of $t$, provided $t\neq 0$. (You can define the degree of $0$ to be $0$, $-1$, or $-\infty$, but it is not important.)

First, note that $\operatorname{ad}_{x_j}:U(L)\to U(L)$ does not increase the degree of $t$ (this requires a bit of work to verify, but not too difficult). Therefore, $\operatorname{ad}_{x_j}(t)$ lies within $U(L,d)$ if $U(L,d)$ is the subspace of $U(L)$ consisting of all PBW polynomials of degree at most $d$. Consequently, the $L$-submodule $M_t$ of $U(L)$ generated by $t$ is a subspace of $U(L,d)$. However, $$\dim U(L,d)=\#\big\{\alpha:|\alpha|\leq d\big\}=\binom{n+d}{n}.$$ Hence, $$\dim M_t\leq \binom{n+d}{n}<\infty.$$

Here is how to prove that $\operatorname{ad}_x$ does not increase the degree. First, show that $$\operatorname{ad}_x(y_1y_2\cdots y_d)=\sum_{i=1}^dy_1y_2\cdots y_{i-1}\operatorname{ad}_x(y_i)y_{i+1}y_{i+2}\cdots y_d.$$ If $y_1,y_2,\ldots,y_d$ are of degree at most $1$, then verify that $\operatorname{ad}_x(y_i)$ is of degree at most $1$ for each $i$. So, indeed, $U(L,d)$ is an $L$-submodule of $U(L)$.