Mathematics Lecture Notes

TheoremPoincaré-Birkhoff-Witt (PBW) Theorem

Let $\mathfrak{g}$ be a Lie algebra with ordered basis $\{X_1, \ldots, X_n\}$ . Then the monomials: $X_1^{k_1} X_2^{k_2} \cdots X_n^{k_n}, \quad k_i \geq 0$ form a basis for $U(\mathfrak{g})$ as a vector space.

ProofMain Proof

Step 1: Define filtrations

Introduce a filtration on $U(\mathfrak{g})$ : $U_0 \subseteq U_1 \subseteq U_2 \subseteq \cdots \subseteq U(\mathfrak{g})$ where $U_k$ is spanned by products of at most $k$ elements from $\mathfrak{g}$ .

Similarly, filter the symmetric algebra $S(\mathfrak{g})$ by degree: $S_k =$ polynomials of degree $\leq k$ .

Step 2: Construct the associated graded

The associated graded algebra is: $\text{gr}(U(\mathfrak{g})) = \bigoplus_{k \geq 0} U_k / U_{k-1}$

with multiplication induced from $U(\mathfrak{g})$ . The canonical map $\mathfrak{g} \to U(\mathfrak{g}) \to \text{gr}(U(\mathfrak{g}))$ extends to a surjection: $\psi: S(\mathfrak{g}) \twoheadrightarrow \text{gr}(U(\mathfrak{g}))$

Step 3: Show $\psi$ is injective

Consider the PBW monomials $X_1^{k_1} \cdots X_n^{k_n}$ . Their images in $\text{gr}(U(\mathfrak{g}))$ are: $\overline{X_1^{k_1} \cdots X_n^{k_n}} \in U_{k_1 + \cdots + k_n} / U_{k_1 + \cdots + k_n - 1}$

To show linear independence, suppose: $\sum_{(k_1, \ldots, k_n)} c_{k_1, \ldots, k_n} X_1^{k_1} \cdots X_n^{k_n} = 0$ in $U(\mathfrak{g})$ with not all $c$ zero. Let $d = \max\{k_1 + \cdots + k_n : c_{k_1, \ldots, k_n} \neq 0\}$ .

Passing to $\text{gr}(U(\mathfrak{g}))$ , the degree $d$ part gives: $\sum_{k_1 + \cdots + k_n = d} c_{k_1, \ldots, k_n} \overline{X_1^{k_1} \cdots X_n^{k_n}} = 0$

Step 4: Use symmetry

In $S(\mathfrak{g})$ , the monomials $X_1^{k_1} \cdots X_n^{k_n}$ are linearly independent (they form a basis for $S(\mathfrak{g})$ as a polynomial ring).

Since $\psi: S(\mathfrak{g}) \to \text{gr}(U(\mathfrak{g}))$ is surjective and preserves the monomial structure, it must be injective (dimension count).

Therefore $\psi$ is an isomorphism: $S(\mathfrak{g}) \cong \text{gr}(U(\mathfrak{g}))$ .

Step 5: Lift to $U(\mathfrak{g})$

Since $\text{gr}(U(\mathfrak{g})) \cong S(\mathfrak{g})$ and both have the same Hilbert series (dimension in each degree), the PBW monomials in $U(\mathfrak{g})$ are linearly independent.

Moreover, they span $U(\mathfrak{g})$ by construction (every element can be written as a linear combination using the commutation relations).

Conclusion: The PBW monomials form a basis for $U(\mathfrak{g})$ .

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Remark

The PBW theorem is fundamental for:

Structure theory: $U(\mathfrak{g})$ is a filtered deformation of $S(\mathfrak{g})$
Quantization: Connection to Poisson geometry and deformation quantization
Homological algebra: Computing cohomology via Koszul resolution
Quantum groups: q-deformed versions $U_q(\mathfrak{g})$ satisfy q-PBW theorems

The theorem shows that the "non-commutativity" of $U(\mathfrak{g})$ is completely controlled by the Lie bracket—there are no additional relations beyond the Lie algebra structure.

ExamplePBW for $\mathfrak{sl}_2$

For $\mathfrak{sl}_2$ with basis $\{H, E, F\}$ and relations $[H,E] = 2E$ , $[H,F] = -2F$ , $[E,F] = H$ :

A PBW basis is $\{H^a E^b F^c : a, b, c \geq 0\}$ .

Any element like $EFE$ can be rewritten: $EFE = E(FE) = E(EF - H) = E^2 F - EH$ $= E^2 F - (HE - 2E) = E^2 F - HE + 2E$ expressing it in PBW form $E^2 F - HE + 2E$ .

Math Notes

Proof of the PBW Theorem