Poincaré-Birkoff-Witt Theorem

In this section we prove the Poincaré-Birkoff-Witt Theorem. The treatment here essentially follows [1]. Let $\mathfrak{g}$ be a Lie algebra over a field $k$ . To prove the theorem we consider $S_k (\mathfrak{g})$ , the symmetric algebra of $\mathfrak{g}$ over $k$ . Let us denote the multiplication of $S_k (\mathfrak{g})$ by $(x,y) \mapsto x\circ y$ . We note that each element $x$ of $S_k (\mathfrak{g})$ has its degree $\deg(x)$ . (as a polynomial in elements of $\mathfrak{g}$ .)

LEMMA 3.1   We choose a ordered basis $(x_\lambda ; \lambda \in \Omega)$ . (That means, a basis with a totally ordered index set $\Omega$ .) Then there exists a linear action of $\mathfrak{g}$ on $S_k (\mathfrak{g})$ which obeys the following rules:

1. For any $x\in \mathfrak{g}$ and for any $y \in S_k(\mathfrak{g})$ ,
   $$\deg(x. y - x \circ y) \leq \deg(y)$$

2. If $\lambda_0 \leq \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n$ , then we have
   $$x_{\lambda_0} . (x_{\lambda_1} \circ x_{\lambda_2} \circ x_{\lamb... ...da_1} \circ x_{\lambda_2} \circ x_{\lambda_3} \circ \dots\circ x_{\lambda_n} .$$

3. For any $x,y\in \mathfrak{g}$ and for any $z\in S_k(\mathfrak{g})$ , we have
   $$x.(y.z)-y.(x.z)=[x,y].z$$

The proof is done by a careful use of induction. Namely,

SUBLEMMA 3.2   We employ the same assumption of the above Lemma. Then for each $m\in \mathbb{Z}_{>0}$ , there exists a unique $k$ -bilinear map

$$f_m: \mathfrak{g}\times S_k(\mathfrak{g})_{\leq m} \to S_k(\mathfrak{g})_{\leq m+1}$$

which obeys the following rules:

1. For any $x\in \mathfrak{g}$ and for any $y \in S_k(\mathfrak{g})_{\leq m}$ ,
   $$\deg(f_m(x, y) - x \circ y) \leq \deg(y)$$

2. If $\lambda_0 \leq \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n$ and $n\leq m$ , then we have
   $$f_m(x_{\lambda_0} , x_{\lambda_1} \circ x_{\lambda_2} \circ x_{\l... ...da_1} \circ x_{\lambda_2} \circ x_{\lambda_3} \circ \dots\circ x_{\lambda_n} .$$

3. For any $x,y\in \mathfrak{g}$ and for any $z\in S_k(\mathfrak{g})_{\leq m-1}$ , we have
   $$f_m(x,f_m(y,z))=f_m(y,f_m(x,z))+f_m([x,y],z)$$

PROOF.. We note first that $S_k (\mathfrak{g})$ has the set of monomials

$$\{x_{\lambda_1} \circ x_{\lambda_2} \circ x_{\lambda_3} \circ \d... ...\lambda_n} ;\lambda_1\leq \lambda_2\leq \lambda_3 \leq \dots \leq \lambda_n\}$$

as a $k$ -basis. For monomial $w=x_{\lambda_1} \circ x_{\lambda_2} \circ x_{\lambda_3} \circ \dots\circ x_{\lambda_n}$ such that $\lambda_1\leq \lambda_2\leq \lambda_3 \leq \dots \leq \lambda_n$ , we put $z=\lambda_2\leq \lambda_3 \leq \dots \leq \lambda_n$ . Then

$$w=f_{m-1}(x_{\lambda_1},z).$$

We define inductively the action of $x_{\lambda_0}$ on it by the following equations.

$$f_m(x_{\lambda_0} , x_{\lambda_1} \circ z) = \begin{cases} x _... ...ligned}\right ) &(\text{if } \lambda_0>\lambda_1)\\ \end{cases}$$

We first note that the above definition is necessary to meet our conditions. Indeed, by (2) we necessarily define as above for $\lambda_0\leq \lambda_1$ . When $\lambda_0>\lambda_1$ , we compute

$$x_{\lambda_0}.(x_{\lambda_1}\circ z)$$

$$\overset{(3)}{=} x_{\lambda_1}.x_{\lambda_0}.z+[x_{\lambda_0},x_{\lambda_1}].z$$

$$= x_{\lambda_1}.(x_{\lambda_0}.z-x_{\lambda_0}\circ z) + x_{\lambda_1}.(x_{\lambda_0}\circ z) +[x_{\lambda_0},x_{\lambda_1}].z$$

$$\overset{(2)}{=} x_{\lambda_1}.(x_{\lambda_0}.z-x_{\lambda_0}\circ z) + x_{\lambda_1}\circ x_{\lambda_0}\circ z +[x_{\lambda_0},x_{\lambda_1}].z$$

and take a careful look at degrees of each monomials using (1). From this argument we see in particular that the action is uniquely determined by conditions (1),(2),(3).

It is easy to see that the conditions (1),(2) are satisfied by $f_m$ defined as above.. Let us proceed to verify that the $f_m$ so defined also satisfies (3). Let us consider $x_\lambda, x_\mu$ $z=x_{\mu_1}\circ x_{\mu_2} \circ \dots \circ x_{\mu_n}$ with $\mu_1 \leq \mu_2 \leq \dots \leq \mu_n$ , $n\leq m-1$ . We need to prove

$$\flat x_\lambda.x_\mu. z-x_\mu.x_\lambda.z=[x_\lambda,x_\mu].z.$$

Since the equation above is antisymmetric in $\mu,\nu$ , we may assume that $\lambda\leq \mu$ .

(i) Case where $\lambda \leq \mu_1$ .

$$x_\lambda. x_\mu.z$$

$$= x_\lambda.(x_\mu \circ z)+x_\lambda.(x_\mu. z-x_\mu\circ z)$$

$$\overset{(1)}{=} x_\lambda\circ x_\mu \circ z +x_\lambda.(x_\mu.z-x_\mu\circ z)$$

In other words,

$$f_m(x_\lambda, f_m(x_\mu,z)) =x_\lambda\circ x_\mu \circ z + f_{m-1}(x_\lambda,(f_{m-1}(x_\lambda,z)-x_\mu \circ z)).$$

On the other hand we have

$$x_\mu.x_\lambda.z$$

$$= x_\mu.(x_\lambda\circ z)$$

$$\overset{\text{by def}}{=} x_\lambda\circ x_\mu \circ z+ f_{m-1}(x_\lambda,f_{m-1}(x_\mu,z)-x_\mu\circ z)+f_{m-1}([x_\mu,x_\lambda], z)$$

So the equation $&flat#flat;$ surely holds in this case.

(ii) Case where $\lambda ,\mu >\mu_1$ .

In this case we need to ``decompose'' $z$ further:

$$z=x_\nu .w .$$

We first forget about the hypothesis $\lambda\leq \mu$ and prove

$$x_\lambda. (x_\mu. (x_\nu. w)) \qquad (\heartsuit)$$

$$= x_\nu.( x_\lambda.( x_\mu.w)) + [x_\lambda,x_\nu].( x_\mu.w) + [x_\mu,x_\nu]. (x_\lambda.w) +[x_\lambda,[x_\mu,x_\nu]].w$$

(Since we are doing induction, we need to pay a special attention on degrees on operands. That means, we should use $f_m$ 's rather than the above ``lazy'' notation. But that is fairly cumbersome, so we keep on being lazy here.)

Let us now admit that the above equation $\heartsuit$ is true and prove the rest of the equation (3). By interchanging $\lambda$ and $\mu$ in the equation ( $\heartsuit$ ), we obtain

$$x_\mu. (x_\lambda. (x_\nu. w)) \qquad (\diamondsuit)$$

$$= x_\nu.( x_\mu.( x_\lambda.w)) + [x_\mu,x_\nu].( x_\lambda.w) + [x_\lambda,x_\nu]. (x_\mu.w) +[x_\mu,[x_\lambda,x_\nu]].w$$

Then by subtracting $(\diamondsuit)$ from $(\heartsuit)$ , we obtain

$$x_\lambda.(x_\mu. (x_\nu .w))-x_\mu. (x_\lambda. (x_\nu. w))$$

$$= x_\nu.(x_\lambda.(x_\mu. w)-x_\mu. (x_\lambda. w))$$

$$+ ([x_\lambda,[x_\mu,x_\nu]] - [x_\mu,[x_\lambda,x_\nu]]).w.$$

Since $\deg(w)$ is smaller than $\deg(z)$ , by induction hypothesis the first term in the right hand side may be replaced by $x_\nu. ([x_\lambda,x_\mu].w)$ . The second term may be replaced, by the Jacobian identity, by $[[x_\lambda,x_\mu],x_\nu]$ . So the equation ($&flat#flat;$) holds in this case too.

It remains to prove the equation ( $\heartsuit$ ). By the induction hypothesis we have

$$x_\mu. (x_\nu.w)=x_\nu.(x_\mu.w)+[x_\mu,x_\nu].w.$$

Also by the induction hypothesis we have

$$x_\lambda.([x_\mu,x_\nu].w) = [x_\mu,x_\nu].(x_\lambda.w) +[x_\lambda,[x_\mu,x_\nu]].w$$

Lastly, we decompose $x_\mu.w$ as

$$x_\mu.w=(x_\mu\circ w)+(x_\mu.w-x_\mu\circ w). =(x_\mu\circ w)+y$$

Then the second term $y$ has degree smaller than $\deg(z)=\deg(w)+1$ . The case (i) applies to the first term and we obtain:

$$x_\lambda.(x_\nu.(x_\mu.w))= x_\nu.(x_\lambda.(x_\mu.w)) +[x_\lambda,x_\nu].(x_\mu.w).$$

These altogether complete the proof. $\qedsymbol$

THEOREM 3.3 (Poincaré, Birkoff, Witt(PBW))   Let $\mathfrak{g}$ be a Lie algebra over a field $k$ . Then we have a $k$ -algebra isomorphism

$$\Psi:\operatorname{Gr}(U(\mathfrak{g}))\cong S(\mathfrak{g}).$$

PROOF.. Let

$$\iota_0: \mathfrak{g}\to \operatorname{Gr}(U(\mathfrak{g}))$$

be the obvious $k$ -linear map.

Using the universality of symmetric algebra, there exists a unique $k$ -algebra homomorphism

$$\Phi: S(\mathfrak{g})\to \operatorname{Gr}(U(\mathfrak{g}))$$

which extends $\iota_0$ . On the other hand the action defined in the Lemma 1.3 gives us a linear map

$$\Psi_0:U(\mathfrak{g}) \ni x \mapsto x.1 \in S(\mathfrak{g})$$

which is clearly degree-decreasing. So it defines a $k$ -linear map

$$\Psi: \operatorname{Gr}(U(\mathfrak{g}))\to\operatorname{Gr}(S(\mathfrak{g}))\cong S(\mathfrak{g}).$$

Now the composition we obtain

$$\Psi\circ \Phi: S(\mathfrak{g})\overset{\Phi}{\to} \operatorname{Gr}(U(\mathfrak{g}))\overset{\Psi}{\to} S(\mathfrak{g})$$

coincides with the identity map. Indeed, it coincides with the identity on monomials of the form

$$x_{\lambda_1}\circ x_{\lambda_2}\circ x_{\lambda_3}\circ \dots \circ x_{\lambda_{n-1}}\circ x_{\lambda_n}.$$

The map $\Phi$ is easily verified to be surjective. So we conclude that $\Phi$ and $\Psi$ are both bijective and are inverse to each other. $\qedsymbol$