Lie Algebras

3 Poincaré-Birkhoff-Witt

We would in general like to understand the structure of $U (g)$ .

To do this, we will use a general tool due to George Bergman for finding canonical forms for elements of associative $k$ -algebras.

Say that we have a set $X$ , and we consider the tensor algebra $T_{R} (X)$ on $X$ over a commutative ring $R$ . Say we also have some relations $σ \in S$ of the form $W_{σ} = f_{σ}$ , where $W$ is a word in $X$ . For any other words $A, B$ , we can consider $r_{A σ B}$ , the $R$ -linear endomorphism on $T_{R} (X)$ replacing $A W_{σ} B$ with $A f_{σ} B$ . We call applying this to an element of $T_{R} (X)$ a reduction. If all reductions on an element are trivial, then that element is irreducible. The submodule of irreducible elements is called $T_{R} (X)_{i r r}$ . We say that an element is reduction finite if for any infinite sequence of reductions, only finitely many act nontrivially. The reduction finite elements form an $R$ -submodule of $T_{R} (X)$ called $T_{R} (X)_{f i n}$ . A sequence of reductions is final if it results in an irreducible element. An element $a$ is reduction unique if it is reduction finite and any final sequence results in the same irreducible element. The unique result will be denoted $r (a)$ .

Lemma 3.1. The reduction unique elements form a submodule denoted $T_{R} (X)_{u n}$ , and $r : T_{R} (X)_{u n} \to T_{R} (X)_{i r r}$ is $R$ -linear.

Proof. Suppose $a, b \in T_{R} (X)_{u n}, k \in R$ . $a + k b$ is reduction finite, and if we have two reduction sequences of it to an irreducible, we can consider doing the same sequence on $a, b$ and extend them to one that makes each irreducible (note that this won’t change the result, only the sequence). Then by uniqueness for $a, b$ we see that the two reductions are the same, and that $r$ is $R$ -linear. □

Lemma 3.2. Let $a, b, c \in T_{R} (X)$ have the property that if $A, B, C$ are nonzero monomials in $a, b, c$ , then $A B C$ is reduction unique. Then if $r_{Σ} (b)$ is the result of some finite reductions on $b$ , then $a r_{Σ} (b) c$ is reduction unique.

Proof. Note in particular that the hypotheses imply that if $a b c$ is reduction unique, and that if the conclusion holds, $r (a r_{Σ} (b) c) = r (a b c)$ . By linearity we only need to show this when $a, b, c$ are monomials, and when $r_{Σ} (b)$ is a single reduction $r_{d σ e}$ . But $a r_{d σ e} (b) c = r_{a d σ e c} (a b c)$ , so this follows since $a b c$ is reduction unique. □

We say that a overlap ambiguity is a pair $σ, σ^{'} \in S$ and a triple $A, B, C$ of nonempty words such that $A B = W_{σ}, B C = W_{σ}^{'}$ (they overlap). It is resolvable if there is are sequences of reductions $r, r^{'}$ such that $r \circ r_{σ} (A B C) = r^{'} \circ r_{σ^{'}} (A B C)$ . An inclusion ambiguity is the same, except when $W_{σ} = B, W_{σ}^{'} = A B C$ , and has the same conditions for resolvability (except $A, C$ can be empty).

Say that a compatible partial ordering $\leq$ on words of $X$ is one such that $A \leq B ⟹ C A D \leq C B D$ and such that the monomials of the $f_{σ}$ are smaller than the monomials of $W_{σ}$ . Let $I$ be the two-sided ideal generated by the relations in $S$ , and $I_{A}$ be the $R$ -submodule of $T_{R} (X)$ generated by $B σ C < A, σ \in S$ (every monomial is smaller than $A$ ). An ambiguity is resolvable relative to $\leq$ if $r_{σ} (A B C) - r_{σ}^{'} (A B C) \in I_{A B C}$ . Note that this is an easy condition to check.

Finally we arrive at this theorem, which can be considered a Diamond Lemma for rings:

Theorem 3.3. Suppose we have $S, X, \leq, I$ as above where $\leq$ is compatible with $S$ and satisfies the descending chain condition. Then the following are equivalent:
- 1. Every ambiguity is resolvable.
- 2. Every ambiguity is resolvable with respect to $\leq$ .
- 3. Every element is reduction unique.
- 4. The natural quotient identifies the submodule spanned by irreducible monomials with $T_{R} (X) / I$ , which is $T_{R} (X)_{i r r}$ .

Proof. Clearly $(3) ⟹ (1) ⟹ (2)$ . If $(3)$ is true, $r$ defines a projection onto $T_{R} (X)_{i r r}$ . The kernel is contained in $I$ by definition of $r$ and contains $I$ , as for any $A B$ , $r (A σ B) = 0$ by Lemma 3.2. Thus as an $R$ -module, $T_{R} (X) = I \oplus T_{R} (X)_{i r r}$ , giving $(4)$ . Conversely if $(4)$ is true, Then since reductions are equal in the quotient, they must be unique.

Thus it suffices to prove $(2) ⟹ (3)$ , and by linearity, we need only show this for a monomial $A$ . From the descending chain condition, we can assume that any smaller monomial is reduction unique. Now suppose that we have two reductions of a monomial $A$ .

If there is no ambiguity, they commute, so we can create two more reductions to an irreducible where the first two steps are the first step of these reductions but in different orders. They will give the same element, and by induction will show that the two original reductions are also the same.

If there is ambiguity, then since it is resolvable relative to $\leq$ by induction the difference can be resolved to $0$ , so by linearity the two resolutions must agree. □

Theorem 3.4 (Poincaré-Birkhoff-Witt). Let $g$ be a Lie algebra over $R$ where the underlying module is free. Choose a well-ordered basis of $g$ , $x_{α}, α \in I$ . Then $x_{a_{1}}^{e_{1}} \dots x_{a_{n}}^{e_{n}}$ for $x_{a_{1}}$ in increasing order, over all possible $a_{i}$ and $e_{i}$ form a basis of $U (g)$ .

Proof. Construct an ordering on monomials in the basis where monomials of smaller degree are smaller, and if two monomials have the same degree, we compare them lexicographically using the order on our basis. This is a well-ordering on monomials, and a sufficient set of relations for $U (g)$ can be written as $x y = y x + [x, y]$ for $x > y$ in the basis. The order is compatible with this, and we can check that every ambiguity is resolvable with respect to our order. Here, the only nontrivial kind of ambiguity that can occur is an overlap ambiguity, when we have something of the form $A (y x z + [x, y] z) B = A x y z B = A (x z y + x [y, z]) B$ . The difference is:

$A (y x z + [x, y] z - (x z y + x [y, z])) B = A (y z x + y [x, z] + z [x, y] + [[x, y], z]$

$- (z x y + [x, z] y + [y, z] x + [x, [y, z]])) B = A (z y x + [y, z] x + y [x, z] + z [x, y] + [[x, y], z]$

$- (z y x + z [x, y] + y [x, z] + [[x, z], y] + [y, z] x + [x, [y, z]]) B$

$= A ([[x, y], z] - [[x, z], y] - [x, [y, z]]) B = 0$

Where at the last step, we have used the Jacobi identity. This shows that the ambiguity is resolvable under the ordering, so that by Theorem 3.3 we are done. □

Note that as a corollary, any Lie algebra $g$ (not necessarily finite-dimensional) is faithfully represented in $U (g)$ .