# The universal enveloping algebra and the Poincaré–Birkhoff–Witt theorem

*Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).*

A Lie action gives operators whose products need not be Lie operators. The enveloping algebra collects those products while imposing precisely the prescribed commutators. The Poincaré–Birkhoff–Witt theorem explains why this construction introduces no additional linear relation among the original Lie elements, and why its leading-degree behavior is a polynomial algebra.

We begin over an arbitrary field \(k\), with no finite-dimensional assumption on the Lie algebra \(\mathfrak g\). In every characteristic the bracket is alternating, so \([x,x]=0\). Finite dimension will be imposed for Noetherianity. Characteristic zero will be imposed for symmetrization; the applications to semisimple algebras are over \(\mathbb C\).

The Lie definitions, tensor constructions and commutator identity used below are from [Lie algebras: definitions, examples and first constructions](RT-LIE-01.md). In the semisimple application we also use the proved decomposition in [The root space decomposition of a semisimple Lie algebra](RT-LIE-07.md).

## 1. From Lie actions to an associative algebra

Let \(T(\mathfrak g)=\bigoplus_{d\geq0}\mathfrak g^{\otimes d}\), with multiplication by concatenation and degree-zero part \(k\). Let \(J\) be its two-sided ideal generated by

\[
x\otimes y-y\otimes x-[x,y],\qquad x,y\in\mathfrak g.
\tag{1.1}
\]

Define \(U(\mathfrak g)=T(\mathfrak g)/J\). The natural map \(j:\mathfrak g\to U(\mathfrak g)\) satisfies \(j([x,y])=j(x)j(y)-j(y)j(x)\). Its injectivity is a theorem, not part of this definition.

If \(A\) is a unital associative \(k\)-algebra and \(a:\mathfrak g\to A\) is a Lie homomorphism for the commutator bracket on \(A\), its linear extension to tensors is the unique unital algebra homomorphism sending \(x_1\otimes\cdots\otimes x_d\) to \(a(x_1)\cdots a(x_d)\). It kills (1.1), and therefore factors uniquely through \(U(\mathfrak g)\). Conversely, every algebra homomorphism out of \(U(\mathfrak g)\) restricts to such a Lie homomorphism. This proves the universal property.

In particular, a \(\mathfrak g\)-module is exactly a unital \(U(\mathfrak g)\)-module: apply the property to \(A=\operatorname{End}_k(V)\). A linear map between two modules commutes with every Lie operator if and only if it commutes with their products, so the same maps are module homomorphisms in both descriptions.

Let \(F_dU\) be the image of tensors of lengths at most \(d\), and put \(F_{-1}U=0\). This is an exhaustive multiplicative filtration. Relation (1.1) lowers a commutator of two generators by one degree. More generally,

\[
[F_rU,F_sU]\subseteq F_{r+s-1}U.
\tag{1.2}
\]

For products this follows by repeatedly using \([ab,c]=a[b,c]+[a,c]b\) and its second-variable version; each resulting term replaces one pair of letters by one bracket. Thus the associated graded algebra \(\operatorname{gr}U=\bigoplus_{d\geq0}F_dU/F_{d-1}U\) is commutative. The degree-one map induces a surjective graded algebra map

\[
\eta:S(\mathfrak g)\longrightarrow\operatorname{gr}U(\mathfrak g).
\tag{1.3}
\]

Here \(S(\mathfrak g)\) is the tensor algebra modulo \(x\otimes y-y\otimes x\). Surjectivity follows because the graded algebra is generated by the degree-one symbols. Proving injectivity is the substantial step.

## 2. Ordered words and the complete PBW proof

Choose a basis \(B\) of \(\mathfrak g\) with any total order. A word means a finite sequence of basis letters, regarded as a tensor; the empty word is \(1\). A word is ordered when its letters are nondecreasing. For a word \(w=x_1\cdots x_d\), let

\[
\iota(w)=\#\{(r,s):r<s,\ x_r>x_s\}.
\tag{2.1}
\]

For adjacent letters \(x>y\), use the replacement

\[
AxyB\ \rightsquigarrow\ AyxB+A[x,y]B.
\tag{2.2}
\]

A bracket is expanded in the chosen basis. This is a finite sum, even when the basis is infinite. The swapped term has the same length and one fewer inversion; every bracket term has smaller length. Consequently every new word is smaller in the lexicographic pair \((d,\iota)\). These pairs are well-founded in \(\mathbb N^2\). Recursive reduction therefore terminates, with finite branching, without requiring the basis order itself to be a well-order.

We prove that the final linear combination of ordered words is independent of every choice in the reductions. The proof is simultaneous induction on \((d,\iota)\). All reduced values of smaller words are already defined; extend their normal-form map \(N\) linearly. Compare two possible first replacements in \(w\).

If they choose the same pair, they coincide. If the pairs are disjoint, performing both replacements gives the same four-term expansion: both pairs swapped, either one replaced by its bracket, and both replaced by brackets. Each term has smaller measure, so its further normal form is independent by induction. The two choices thus have the same normal form.

The only overlapping case consists of three consecutive letters \(x>y>z\). Sorting the three ordinary letters along the two paths gives

\[
\begin{aligned}
L&=zyx+[y,z]x+y[x,z]+[x,y]z,\\
R&=zyx+z[x,y]+[x,z]y+x[y,z].
\end{aligned}
\tag{2.3}
\]

These are identities of tensor expressions produced by (2.2), with the surrounding words \(A,B\) left in place. The difference has one fewer letter than \(AxyzB\). For all expressions of that smaller length, induction already gives

\[
N\bigl(A(uv-vu)B\bigr)=N\bigl(A[u,v]B\bigr),
\qquad u,v\in\mathfrak g.
\tag{2.4}
\]

Indeed, for basis letters in decreasing order this is the invariance of the lower-length normal form under its replacement; for increasing order it follows by reversing the pair, for equal letters by alternation, and for general \(u,v\) by bilinearity. Applying (2.4) to the three differences in \(L-R\), their normal form is the normal form of

\[
A\bigl([[y,z],x]+[y,[x,z]]+[[x,y],z]\bigr)B=0.
\tag{2.5}
\]

The equality is Jacobi. Hence the two overlapping choices also agree after normal reduction. This establishes the induction: define \(N(w)\) using any first replacement and the already-defined lower normal forms. The comparison proves this definition is independent of that first choice, and recursively of all subsequent choices. Ordered words are fixed by \(N\).

We now separate the relation ideal, using the normal form just proved. For any words \(A,B\) and any basis letters \(x,y\),

\[
N\bigl(A(xy-yx-[x,y])B\bigr)=0.
\tag{2.6}
\]

For \(x>y\) this is (2.2); the other cases follow as in (2.4). Bilinearity gives it for all Lie elements. These contextual relations span the ideal \(J\), so \(N(J)=0\). Conversely, each replacement changes a tensor by an element of \(J\), so \(t-N(t)\in J\) for every tensor \(t\). If \(N(t)=0\), then \(t\in J\). Thus \(\ker N=J\), and the tensor algebra is the direct vector-space sum of \(J\) and the span of the ordered words.

**Theorem 2.1 (Poincaré–Birkhoff–Witt).** For any Lie algebra over any field, the images of the ordered words in any totally ordered basis, including the empty word, form a basis of \(U(\mathfrak g)\). The map (1.3) is an isomorphism of graded algebras.

**Proof.** The kernel calculation gives the ordered-word basis. It also shows that \(F_dU\) has precisely those basis words of length at most \(d\), so \(F_0U=k1\). The length-\(d\) words give a basis of \(F_dU/F_{d-1}U\). In the symmetric algebra the same nondecreasing words are a basis: it is the polynomial algebra on the basis symbols, whose monomials have finite support. Map (1.3) sends these bases bijectively in each degree, and respects multiplication. \(\square\)

The proof does not deduce linear independence from congruence modulo \(J\). Equality of normal forms is established first, by the lower-length commutator calculation and Jacobi; only then is the ideal annihilated.

## 3. Leading symbols, domains and Noetherianity

**Corollary 3.1.** The map \(j:\mathfrak g\to U(\mathfrak g)\) is injective, and \(U(\mathfrak g)\) has no zero divisors.

**Proof.** Distinct basis letters remain independent by PBW. If \(a,b\ne0\), let \(r,s\) be their degrees in the filtration, and let \(\sigma(a),\sigma(b)\) be their nonzero leading symbols. The symmetric algebra is a domain: any two polynomials use finitely many variables, and in that polynomial ring the coefficient of the product of the leading monomials is the product of two nonzero field elements. Hence

\[
\sigma(a)\sigma(b)\ne0,\qquad
\deg(ab)=r+s,\qquad ab\ne0.
\tag{3.1}
\]

The first assertion uses \(\operatorname{gr}U\cong S(\mathfrak g)\), and proves the other two. \(\square\)

After this result we identify \(x\in\mathfrak g\) with \(j(x)\). It also proves that the only units of \(U(\mathfrak g)\) are \(k^\times\): if \(ab=1\), degree additivity forces both degrees to be zero.

We supply the finiteness argument needed for the next consequence. A ring is left Noetherian when every left ideal is finitely generated; right Noetherian is defined similarly.

**Lemma 3.2.** A polynomial ring \(k[t_1,\ldots,t_m]\), with finitely many variables, is Noetherian.

**Proof.** First, every set of exponent vectors in \(\mathbb N^m\) has only finitely many minimal vectors for the coordinatewise order. To see this, an infinite sequence in \(\mathbb N\) has an infinite nondecreasing subsequence: either a value repeats infinitely, or successive larger values can be selected since each bounded set then occurs only finitely often. Repeating this selection for each of the \(m\) coordinates gives an infinite coordinatewise nondecreasing subsequence from any infinite sequence in \(\mathbb N^m\). Infinitely many distinct minimal vectors would contradict this. Every exponent dominates a minimal one, since a strict coordinatewise descent lowers the sum of its coordinates. Thus every upward-closed set of exponents is generated by finitely many minimal exponents. The case \(m=0\) is immediate.

Use degree-lexicographic order on monomials. For an ideal \(I\), the exponents of leading monomials of its nonzero elements are upward-closed, because multiplication by a monomial adds its exponent. Select \(p_1,\ldots,p_r\in I\) whose leading monomials are its finitely many minimal ones. For any nonzero \(p\in I\), some leading monomial of \(p_i\) divides that of \(p\). Subtract a suitable monomial multiple of \(p_i\) to cancel the leading term. Repeat. The leading monomial strictly decreases in a well-order, so the process stops. A nonzero remainder would still belong to \(I\) and have a divisible leading monomial, contradicting termination. Therefore \(I=(p_1,\ldots,p_r)\). \(\square\)

**Proposition 3.3.** If \(\mathfrak g\) is finite-dimensional, \(U(\mathfrak g)\) is both left and right Noetherian.

**Proof.** For a left ideal \(I\subseteq U\), the leading symbols form the graded ideal

\[
\operatorname{gr}I=\bigoplus_{d\geq0}
(I\cap F_dU)/(I\cap F_{d-1}U)
\subseteq\operatorname{gr}U.
\tag{3.2}
\]

It is an ideal because the graded algebra is commutative. By Lemma 3.2 and PBW it has finitely many homogeneous generators: start with finite generators and take their homogeneous components, which remain in this graded ideal. Lift these generators to \(a_1,\ldots,a_r\in I\).

For \(a\in I\) of degree \(d\), express its leading symbol as \(\sum_i q_i\sigma(a_i)\), with homogeneous \(q_i\) of degree \(d-\deg(a_i)\); omit negative degrees. Lift \(q_i\) to \(b_i\in U\). Then \(a-\sum_i b_i a_i\in I\) has smaller degree. Induction on \(d\), starting at degree zero, writes \(a\) in the left ideal generated by the \(a_i\). The same argument with products \(a_i b_i\) proves finite generation of right ideals. \(\square\)

Finite dimension is essential here. For an abelian Lie algebra with basis \(x_1,x_2,\ldots\), its enveloping algebra is \(k[x_1,x_2,\ldots]\). The ideals \((x_1)\subsetneq(x_1,x_2)\subsetneq\cdots\) never stabilize: evaluation at \(x_1=\cdots=x_n=0\) leaves \(x_{n+1}\) nonzero. Their union cannot be finitely generated, since finitely many generators would all belong to one stage and would force the union to equal that stage. Its PBW basis and domain property nevertheless hold.

## 4. Subalgebras, direct sums and triangular multiplication

If \(\mathfrak a\subseteq\mathfrak g\) is a Lie subalgebra, the natural map \(U(\mathfrak a)\to U(\mathfrak g)\) is injective. Extend an ordered basis of \(\mathfrak a\) to one of \(\mathfrak g\), placing the new letters after the old ones. Its PBW words are then a subset of the PBW basis of \(U(\mathfrak g)\). We henceforth regard these enveloping algebras as subalgebras.

**Proposition 4.1.** There is a natural algebra isomorphism

\[
U(\mathfrak g\oplus\mathfrak g')\cong
U(\mathfrak g)\otimes_k U(\mathfrak g').
\tag{4.1}
\]

**Proof.** Send \((x,y)\) to \(x\otimes1+1\otimes y\). These operators obey the direct-sum bracket, so the universal property gives an algebra map from the left side to the right. In the other direction use multiplication of the two enveloping subalgebras. Their elements commute: the generators commute because the mixed Lie bracket is zero, and the commutator derivation identities extend this to all their products. Hence multiplication defines an algebra map from the tensor product. The two composites fix every generator and the unit, so both are identity maps. \(\square\)

Now let \(\mathfrak g\) be complex semisimple, choose a Cartan subalgebra and positive roots, and write its proved Lie decomposition as

\[
\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+.
\tag{4.2}
\]

Order a basis of \(\mathfrak n^-\) first, one of \(\mathfrak h\) next, and one of \(\mathfrak n^+\) last. Every ordered word in the combined basis has a unique negative, Cartan and positive block. Consequently:

**Proposition 4.2 (triangular decomposition).** Multiplication is a vector-space isomorphism

\[
U(\mathfrak n^-)\otimes U(\mathfrak h)\otimes U(\mathfrak n^+)
\xrightarrow{\ \sim\ }U(\mathfrak g).
\tag{4.3}
\]

**Proof.** The tensor products of the three PBW bases map bijectively to the combined PBW basis. \(\square\)

This multiplication map is generally not an algebra homomorphism for the ordinary tensor-product multiplication. For \(\mathfrak{sl}_2\), with \([h,e]=2e,[h,f]=-2f,[e,f]=h\), the basis is

\[
f^a h^b e^c,\qquad a,b,c\in\mathbb Z_{\geq0}.
\tag{4.4}
\]

The negative and positive factors commute in the tensor-product algebra, whereas \(ef-fe=h\ne0\) in \(U(\mathfrak{sl}_2)\). This distinction will matter when induced highest-weight modules are formed.

## 5. Symmetrization and the centre

Assume now that \(k\) has characteristic zero. For a commutative monomial define

\[
\operatorname{sym}(x_1\cdots x_d)=\frac1{d!}
\sum_{\pi\in S_d}x_{\pi(1)}\cdots x_{\pi(d)}\in U(\mathfrak g),
\qquad \operatorname{sym}(1)=1.
\tag{5.1}
\]

The expression is multilinear and symmetric, so it gives a well-defined linear map from \(S^d(\mathfrak g)\). It preserves the degree filtration. All products in (5.1) have the same leading symbol, so its associated graded map, under PBW, is the identity.

This proves that symmetrization is a filtered vector-space isomorphism. Injectivity follows by taking the highest nonzero homogeneous component of a polynomial: its symbol cannot vanish. For surjectivity, given \(u\in F_dU\), choose its leading polynomial \(p_d\in S^d(\mathfrak g)\); then \(u-\operatorname{sym}(p_d)\in F_{d-1}U\). Induction on \(d\) completes the construction of a preimage. Neither argument assumes finite dimension.

Let \(\mathfrak g\) act on itself by the adjoint action, on \(S(\mathfrak g)\) by its derivation extension, and on \(U(\mathfrak g)\) by commutators. Since a commutator acts on a product by the Leibniz rule,

\[
[x,\operatorname{sym}(x_1\cdots x_d)]
=\sum_{r=1}^d\operatorname{sym}
(x_1\cdots[x,x_r]\cdots x_d).
\tag{5.2}
\]

One obtains the equality by expanding each of the \(d!\) products: each replaced labeled factor occurs in exactly the corresponding permutation sum. Thus symmetrization is equivariant.

**Proposition 5.1.** In characteristic zero, symmetrization is an isomorphism of filtered \(\mathfrak g\)-modules and induces a vector-space isomorphism

\[
S(\mathfrak g)^{\mathfrak g}\xrightarrow{\ \sim\ }Z(U(\mathfrak g)).
\tag{5.3}
\]

**Proof.** Equivariance and bijectivity identify invariant vectors. The invariants of the commutator action on \(U\) are exactly its centre: commuting with every generator implies commuting with every product by the Leibniz rule, and those products span \(U\). \(\square\)

The map (5.1) is generally not multiplicative. In \(\mathfrak{sl}_2\),

\[
\operatorname{sym}(ef)=\tfrac12(ef+fe)=fe+\tfrac12h,\qquad
\operatorname{sym}(e)\operatorname{sym}(f)=ef=fe+h.
\tag{5.4}
\]

Their difference is \(\tfrac12h\ne0\). Formula (5.3) therefore asserts a vector-space identification, without asserting a ring identification by this particular map.

For a finite-dimensional Lie algebra with a nondegenerate invariant symmetric form \(B\), let \(v_i,v^i\) be dual bases. The quadratic tensor \(q=\sum_i v_i v^i\in S^2(\mathfrak g)\) is invariant. Indeed, if \([x,v_i]=\sum_j A_{ji}v_j\), invariance of \(B\) gives \([x,v^i]=-\sum_j A_{ij}v^j\), and the two sums in \(x\cdot q\) cancel after renaming indices. Hence

\[
\Omega_B=\operatorname{sym}(q)=\sum_i v_i v^i\in Z(U(\mathfrak g)).
\tag{5.5}
\]

For the second equality, the inverse-form tensor is symmetric under interchange of its factors, so the sum of their products equals the sum in reversed order. The same inverse-form tensor also shows that the definition is independent of the dual bases. This is the Casimir element associated with \(B\).

For the Killing form of \(\mathfrak{sl}_2\), the adjoint trace calculation in [The Killing form and Cartan's criteria](RT-LIE-03.md) gives \(B(h,h)=8\), \(B(e,f)=4\), with other relevant pairings zero. Thus

\[
\Omega_B=\frac18h^2+\frac14(ef+fe),\qquad
C:=8\Omega_B=h^2+2h+4fe.
\tag{5.6}
\]

The last expression is in the PBW order (4.4). In a highest-weight module with \(ev=0,hv=mv\), it gives \(Cv=m(m+2)v\); this calculation requires only the defining relations. The choice of scalar multiple in the notation \(C\) is explicit.

## 6. Exercises with complete solutions

### Exercise 6.1 — An abelian Lie algebra

Prove \(U(V)=S(V)\) when the bracket on \(V\) is zero, including when \(V\) is infinite-dimensional.

**Solution.** The defining ideal (1.1) is then exactly the two-sided ideal generated by \(xy-yx\), which is the defining ideal of \(S(V)\). The two quotients of \(T(V)\) are therefore canonically the same algebra, and their degree filtrations agree. Equivalently, their common basis consists of finite-support monomials in any chosen basis of \(V\). An infinite number of available variables changes neither the definition of a polynomial nor this quotient argument. For finite-dimensional \(V\) this is a polynomial ring in \(\dim V\) variables; the infinite-dimensional example following Proposition 3.3 explains why finite generation of ideals cannot be inferred in general.

### Exercise 6.2 — The ordered Casimir of \(\mathfrak{sl}_2\)

Write \(h^2+2(ef+fe)\) in the PBW order \(f,h,e\), and verify its centrality directly.

**Solution.** Since \(ef=fe+h\), the element is \(C=h^2+2h+4fe\). It commutes with \(h\), since \([h,fe]=-2fe+2fe=0\). For \(e\),

\[
[e,C]=-2eh-2he-4e+4he
=2(he-eh)-4e=0.
\tag{6.1}
\]

For \(f\),

\[
[f,C]=2fh+2hf+4f-4fh
=2(hf-fh)+4f=0.
\tag{6.2}
\]

We used \([e,h]=-2e\), \([f,h]=2f\), \([e,fe]=he\), and \([f,fe]=-fh\). Since \(e,f,h\) generate the enveloping algebra, these three vanishing commutators prove centrality. Formula (5.6) identifies \(C\) as eight times the Killing-form Casimir; arbitrary normalizations must not be conflated.

### Exercise 6.3 — Equivariance in every degree

Prove equivariance of symmetrization directly, and explain why it identifies adjoint invariants with the centre.

**Solution.** For a labeled tuple \(x_1,\ldots,x_d\), expand

\[
[x,x_{\pi(1)}\cdots x_{\pi(d)}]
=\sum_{s=1}^d x_{\pi(1)}\cdots[x,x_{\pi(s)}]\cdots x_{\pi(d)}.
\tag{6.3}
\]

Fix a label \(r\). Among all pairs \((\pi,s)\), those with \(\pi(s)=r\) contain each permutation of the tuple with \(x_r\) replaced by \([x,x_r]\) exactly once. Summing and dividing by \(d!\) gives (5.2). This works also when some letters are equal, because labels retain all multiplicities before passing to the symmetric product. Degree zero is fixed on both sides. Linearity now gives equivariance on the whole symmetric algebra. The filtered leading-symbol argument already proves symmetrization bijective, so it restricts bijectively to the kernels of all adjoint operators. In \(U\), an element in those kernels commutes with every Lie generator, hence with every word and every element. This is exactly the centre.

### Exercise 6.4 — Independence in PBW

Prove the full ordered-monomial theorem, explaining why the triple overlap suffices and why the resulting ordered words are independent, rather than merely spanning.

**Solution.** Induct on word length, and within a fixed length on the inversion number. Every replacement (2.2) lowers this pair, so it produces a finite linear combination of already-normalized lower words. Two different adjacent pairs either are disjoint or share one letter; no other possibility exists for pairs of length two. Disjoint replacements commute after expansion. Shared pairs can both be inverted only for \(x>y>z\), and the two sorting paths are precisely (2.3).

Their difference is

\[
([y,z]x-x[y,z])+(y[x,z]-[x,z]y)
+([x,y]z-z[x,y]).
\tag{6.4}
\]

With any surrounding words this has smaller length than the original triple word. By the completed lower-length induction, normalizing each commutator replaces it by its Lie bracket. Its normal form is therefore (2.5), zero by Jacobi. The induction defines a unique linear normal-form map on all tensors, fixing every ordered word.

Every contextual relation has zero normal form by the same replacement rule, including pairs in either order and equal pairs. Hence the map kills their span \(J\). Each tensor differs from its normal form by a sum of contextual relations, so the kernel is exactly \(J\). If a linear combination of distinct ordered words represented zero in \(U\), it would belong to \(J\); its normal form, which is that same linear combination of distinct tensor-basis words, would be zero. All coefficients must therefore be zero. This proves independence and spanning, for every field and every totally ordered basis. Grouping the ordered words by length yields the graded isomorphism (1.3).

## 7. Scope and references

All results used specifically in this lesson have been proved above or imported from the named earlier lessons. No additional theorem is left without proof. PBW and the domain assertion allow arbitrary dimension and characteristic; Noetherianity requires finite dimension; symmetrization requires characteristic zero. We have not classified the centre as a commutative algebra or identified its generators. Those questions belong to the later lesson on Harish-Chandra's theorem. The triangular decomposition proved here is also the algebraic input used in the planned compact-group course's treatment of highest weights.

Milne's [*Algebraic Groups: The Theory of Group Schemes of Finite Type over a Field*, corrected author edition](https://www.jmilne.org/math/Books/iAG2022.pdf), §10j, Theorem 10.36 and its following proof, is the principal comparison for the enveloping construction and PBW. The corrected 2021 revision has the theorem on printed page 200 and the proof on pages 201–204. Etingof's [*Lie Groups and Lie Algebras I*, MIT notes](https://ocw.mit.edu/courses/18-745-lie-groups-and-lie-algebras-i-fall-2020/mit18_745_f20_lec_full.pdf), §§13–14, provides an additional comparison for enveloping algebras, symmetrization and central elements. The normal-form argument here proves ideal separation after resolving the ambiguities; its full logic does not require any external diamond-lemma theorem.

The official [Stacks Project](https://stacks.math.columbia.edu/) is a separate reference collection. Unofficial AI drafts in AI Integrated Stacks Project are not an independent verification of this lesson.