Why the Universal Enveloping Algebra–Commutator Adjunction Exists

Why the Universal Enveloping Algebra–Commutator Adjunction Exists

The universal enveloping algebra is usually introduced by a construction:

$$U(\mathfrak{g})=T(\mathfrak{g})/⟨x\otimes y-y\otimes x-[x,y]⟩.$$

This is correct, but it hides the reason why the construction exists.

The real reason is categorical:

the commutator bracket is a morphism of algebraic theories, and the universal enveloping algebra is the left adjoint to the induced pullback functor.

Let us make this precise.

A Lawvere theory is a category encoding a kind of algebraic structure. Its models are finite-product-preserving functors into $\mathbf{Set}$$\mathbf{Set}$. For example, groups, rings, associative algebras, and Lie algebras can all be described by algebraic theories.

For $k$$k$-linear structures, we include the operations of addition and scalar multiplication in the theory. Thus there is a theory

$${\mathbb{T}}_{\mathrm{Lie}}$$

for Lie algebras over $k$$k$, and a theory

$${\mathbb{T}}_{\mathrm{Ass}}$$

for associative $k$$k$-algebras.

Now observe the basic fact:

Every associative algebra $A$$A$ has an underlying Lie algebra

$$A_{\mathrm{Lie}}$$

with bracket

$$[a,b]=ab-ba.$$

This is not an accident. It comes from a morphism of theories

$$F:{\mathbb{T}}_{\mathrm{Lie}}\to {\mathbb{T}}_{\mathrm{Ass}}.$$

The morphism $F$$F$ interprets the Lie bracket operation as the commutator operation in associative algebras:

$$[-,-]\mapsto m-m\circ \tau ,$$

where $m$$m$ is multiplication and $\tau (a,b)=(b,a)$$\tau(a,b)=(b,a)$.

This interpretation respects the Lie identities. The commutator in any associative algebra is antisymmetric and satisfies the Jacobi identity. Therefore it defines a genuine morphism of theories.

Every theory morphism

$$F:\mathbb{T}\to \mathbb{S}$$

induces a pullback functor on models:

$$F^{\ast }:\mathrm{Mod}(\mathbb{S})\to \mathrm{Mod}(\mathbb{T}).$$

It sends an $\mathbb{S}$$\mathbb S$-model to a $\mathbb{T}$$\mathbb T$-model by restricting its operations along $F$$F$.

In our case,

$$F^{\ast }:{\mathrm{AssAlg}}_k\to {\mathrm{LieAlg}}_k$$

is exactly the commutator functor:

$$A\mapsto A_{\mathrm{Lie}}.$$

So the familiar operation “take an associative algebra and form its commutator Lie algebra” is simply pullback along a morphism of theories.

The universal enveloping algebra is the left adjoint to this pullback functor.

That is, there is an adjunction

$$U:{\mathrm{LieAlg}}_k\rightleftarrows {\mathrm{AssAlg}}_k:F^{\ast }$$

such that

$${\mathrm{Hom}}_{{\mathrm{AssAlg}}_k}(U(\mathfrak{g}),A)\cong {\mathrm{Hom}}_{{\mathrm{LieAlg}}_k}(\mathfrak{g},A_{\mathrm{Lie}}).$$

This is the universal property of $U(\mathfrak{g})$$U(\mathfrak g)$.

It says:

To give an associative algebra map from $U(\mathfrak{g})$$U(\mathfrak g)$ to $A$$A$ is the same as giving a Lie algebra map from $\mathfrak{g}$$\mathfrak g$ to the commutator Lie algebra of $A$$A$.

The concrete construction realizes this adjoint explicitly. Start with the tensor algebra

$$T(\mathfrak{g})=\underset{n\ge 0}{⨁}{\mathfrak{g}}^{\otimes n}.$$

This is the free associative algebra generated by the vector space $\mathfrak{g}$$\mathfrak g$. But it does not yet know the Lie bracket on $\mathfrak{g}$$\mathfrak g$. To force the commutator in the associative algebra to agree with the given Lie bracket, quotient by the relations

$$xy-yx=[x,y].$$

Thus

$$U(\mathfrak{g})=T(\mathfrak{g})/⟨x\otimes y-y\otimes x-[x,y]⟩.$$

The canonical map

$$i:\mathfrak{g}\to U(\mathfrak{g})$$

is a Lie algebra homomorphism, where $U(\mathfrak{g})$$U(\mathfrak g)$ is viewed as a Lie algebra under commutator.

The universal property is now immediate. Given any associative algebra $A$$A$ and any Lie algebra map

$$f:\mathfrak{g}\to A_{\mathrm{Lie}},$$

the freeness of $T(\mathfrak{g})$$T(\mathfrak g)$ gives an algebra map

$$T(\mathfrak{g})\to A.$$

The condition that $f$$f$ is a Lie algebra map says precisely that the relations

$$xy-yx-[x,y]$$

are sent to zero. Therefore the map factors uniquely through

$$U(\mathfrak{g}).$$

So the quotient construction is not ad hoc. It is the explicit construction of the left adjoint forced by the theory morphism

$${\mathbb{T}}_{\mathrm{Lie}}\to {\mathbb{T}}_{\mathrm{Ass}}.$$

The same perspective also explains why this kind of construction is common. Whenever we have a morphism of algebraic theories

$$F:\mathbb{T}\to \mathbb{S},$$

we get a restriction functor

$$F^{\ast }:\mathrm{Mod}(\mathbb{S})\to \mathrm{Mod}(\mathbb{T}).$$

Under standard finitary algebraic hypotheses, this restriction functor has a left adjoint

$$F_{!}:\mathrm{Mod}(\mathbb{T})\to \mathrm{Mod}(\mathbb{S}).$$

This left adjoint freely equips a $\mathbb{T}$$\mathbb T$-model with the extra structure needed to become an $\mathbb{S}$$\mathbb S$-model, subject to the equations imposed by the theory morphism.

For the morphism

$${\mathbb{T}}_{\mathrm{Lie}}\to {\mathbb{T}}_{\mathrm{Ass}},$$

this left adjoint is precisely the universal enveloping algebra.

Finally, $U(\mathfrak{g})$$U(\mathfrak g)$ also carries a natural Hopf algebra structure. On generators,

$$\mathrm{\Delta }(x)=x\otimes 1+1\otimes x,\epsilon (x)=0,S(x)=-x.$$

These maps extend to $U(\mathfrak{g})$$U(\mathfrak g)$ and make it a Hopf algebra. This is why representations of $\mathfrak{g}$$\mathfrak g$ can be studied as modules over $U(\mathfrak{g})$$U(\mathfrak g)$, and why tensor products of representations are controlled by the coproduct.

The conclusion is simple:

The universal enveloping algebra exists because the commutator bracket is a morphism of theories.

The commutator functor is pullback along this morphism.

The universal enveloping algebra is its left adjoint.

So the adjunction

$$U⊣(-{)}_{\mathrm{Lie}}$$

is not a coincidence. It is an instance of the general principle:

$$\text{morphisms of theories}⟹\text{restriction functors}⟹\text{free left adjoints}.$$