Poincaré-Birkhoff-Witt theorem

Let $𝔤$ be a Lie algebra over a field $k$, and let$B$ be a $k$-basis of $𝔤$ equipped with a linearorder $\le$. The Poincaré-Birkhoff-Witt-theorem (oftenabbreviated to PBW-theorem) states that the monomials

$$x_1{x}_2\mathrm{\cdots }{x}_n\text{ with }{x}_1\le x_2\le \mathrm{\cdots }\le x_n\text{ elements of }B$$

constitute a $k$-basis of the universal enveloping algebra$U(𝔤)$ of $𝔤$. Such monomials are often calledordered monomials or PBW-monomials.

It is easy to see that they span $U(𝔤)$: for all $n\in \mathbb{N}$, let $M_n$ denote the set

$$M_n=\{(x_1,\mathrm{\dots },x_n)\mid x_1\le \mathrm{\cdots }\le x_n\}\subset B^n,$$

and denote by $\pi :{\bigcup }_{n=0}^{\mathrm{\infty }}{B}^n\to U(𝔤)$ themultiplication map. Clearly it suffices to prove that

$$\pi (B^n)\subseteq \sum _{i=0}^n\pi (M_i)$$

for all $n\in \mathbb{N}$; to this end, we proceed by induction. For $n=0$the statement is clear. Assume that it holds for $n-1\ge 0$, and consider alist $(x_1,\mathrm{\dots },x_n)\in B^n$. If it is an element of $M_n$, then we aredone. Otherwise, there exists an index $i$ such that $x_i>x_{i+1}$.Now we have

$\pi (x_1,\mathrm{\dots },x_n)$$=\pi (x_1,\mathrm{\dots },x_{i-1},x_{i+1},x_i,x_{i+2},\mathrm{\dots },x_n)$

$+x_1\mathrm{\cdots }{x}_{i-1}[x_i,x_{i+1}]x_{i+1}\mathrm{\cdots }{x}_n.$

As $B$ is a basis of $𝔨$, $[x_i,x_{i+1}]$ is a linearcombination of $B$. Using this to expand the second term above, we findthat it is in ${\sum }_{i=0}^{n-1}\pi (M_i)$ by the induction hypothesis.The argument of $\pi$ in the first term, on the other hand, islexicographically smaller than $(x_1,\mathrm{\dots },x_n)$, but contains thesame entries. Clearly this rewriting proces must end, and thisconcludes the induction step.

The proof of linear independence of the PBW-monomials is slightly moredifficult, but can be found in most introductory texts on Lie algebras, such as the classic below.