As we saw in the first post, a representation of a finite group ${G}$ can be thought of simply as a module over a certain ring: the group ring. The analog for Lie algebras is the enveloping algebra. That’s the topic of this post.

Definition

The basic idea is as follows. Just as a representation of a finite group ${G}$ was a group-homomorphism ${G \rightarrow Aut(V)}$ for a vector space, a representation of a Lie algebra ${\mathfrak{g}}$ is a Lie-algebra homomorphism ${\mathfrak{g} \rightarrow \mathfrak{g}l(V)}$. Now, ${\mathfrak{g}l(V)}$ is the Lie algebra constructed from an associative algebra, ${End(V)}$—just as ${Aut(V)}$ is the group constructed from ${End(V)}$ taking invertible elements.

So, in general, it follows that what we’re really looking at here is maps ${f: \mathfrak{g} \rightarrow A}$ for an associative algebra ${A}$, such that ${f}$ is a Lie-algebra homomorphism, i.e. $\displaystyle f([x,y]) = f(x)f(y) - f(y)f(x) = [f(x),f(y)].$

Call such a function ${f}$ an ${\alpha}$-map. Note that ${f}$ makes any ${A}$-module into a representation of ${\mathfrak{g}}$, so these are interesting objects to consider.

The universal enveloping algebra has the universal ${\alpha}$-map. In detail, if ${\mathfrak{g}}$ is fixed, we can consider the category ${\mathfrak{C}}$ of pairs ${(f, A)}$, where ${A}$ is an associative algebra and ${f: \mathfrak{g} \rightarrow A}$ an ${\alpha}$-map. A morphism between ${(f,A)}$ and ${(g,B)}$ is an algebra-homomorphism ${h:A \rightarrow B}$ such that ${h \circ f = g}$.

Definition 1 In this category, the universal enveloping algebra ${\mathcal{U}(\mathfrak{g})}$ is the initial object: thus for any ${\alpha}$-map ${f: \mathfrak{g} \rightarrow A}$, we have a unique algebra-homomorphism ${\mathcal{U}(\mathfrak{g}) \rightarrow A}$ making the usual diagram commutative.

By the uniqueness of universal objects, the enveloping algebra is unique if it exists.

Proposition 2 To give a ${\mathfrak{g}}$-representation is the same as to give a ${\mathcal{U}(\mathfrak{g})}$-module. Alternatively, the categories of ${\mathfrak{g}}$-representations and ${\mathcal{U}(\mathfrak{g})}$-modules are equivalent.

Proof: Indeed, a ${\mathcal{U}(\mathfrak{g})}$-module is a vector space ${V}$ with an algebra homomorphism ${\mathcal{U}(\mathfrak{g}) \rightarrow End(V)}$. The set of such algebra-homomorphisms is functorially isomorphic to the set of ${\alpha}$-mappings ${\mathfrak{g} \rightarrow End(V)}$, or ${\mathfrak{g}}$-representations. $\Box$

Of course, we have to still show ${\mathcal{U}(\mathfrak{g})}$ exists.

Tensor Algebras

Given a vector space ${V}$, we can form its tensor powers ${V^{\otimes n}}$ for all ${n \mathfrak{g}eq 0}$ (say ${V^{\otimes 0} = {\mathbb C}}$ if we are working over the complex numbers).

We can thus form the graded vector space $\displaystyle TV := \bigoplus_n V^{\otimes n},$

which we can make into a noncommutative ring. Indeed, we have a bilinear map $\displaystyle V^{\otimes i} \times V^{\otimes j} \rightarrow V^{\otimes i+j} ,$

and we can extend by linearity to the direct sum to get a graded ring. Basically, the structure is described simply: if ${e_v, 1 \leq v \leq m}$, form a basis for ${V}$, then multiplication sends $\displaystyle (e_{v_1} \otimes e_{v_2} \otimes \dots \otimes e_{v_a} )( e_{w_1} \otimes \dots \otimes e_{w_b}) = e_{v_1} \otimes \dots \otimes e_{v_a} \otimes e_{w_1} \otimes \dots \otimes e_{w_b}.$

The ordering of the ${e}$‘s does matter, which explains the noncommutativity; if we demand that the ${e}$‘s do commute, we end up with a polynomial algebra.

Anyway, the importance of the tensor algebra is the universal property:

Proposition 3 Given a linear map ${f: V \rightarrow A}$ for a ring ${A}$, there is a unique algebra-homomorphism ${TV \rightarrow A}$ such that the map ${V \rightarrow TV \rightarrow A}$ is just ${f}$.

Proof: The proof follows by sending ${e_{v_1} \otimes \dots \otimes e_{v_a}}$ to ${f(e_{v_1}) \dots f(e_{v_a})}$; this gives a ring-homomorphism ${TV \rightarrow A}$. The rest is straightforward to check. $\Box$

The Enveloping Algebra

It is often useful to quotient out the tensor algebra ${TV}$ by some relations to still get a useful object. For instance, if one abelianizes, one gets the symmetric algebra; if one quotients out by the ideal generated by ${x \otimes x, x \in V}$, one gets the exterior algebra.

In our case, we have a Lie algebra ${\mathfrak{g}}$, and I claim that

Proposition 4 ${\mathcal{U}(\mathfrak{g})}$ is isomorphic to ${ T\mathfrak{g} / I }$ where ${I}$ is the two-sided ideal generated by $\displaystyle x \otimes y - y \otimes x - [x,y], \quad x,y \in \mathfrak{g}.$

Proof: We can check this by looking at universal properties. So, first, there’s a map ${\mathfrak{g} \rightarrow T\mathfrak{g}}$ (this is true for any vector space) and a quotient map ${T\mathfrak{g} \rightarrow T\mathfrak{g}/I}$. Thus there’s a map ${f:\mathfrak{g} \rightarrow T\mathfrak{g}/I}$. Now ${f(x)f(y)-f(y)f(x) = [f(x),f(y)]}$ by the relations defining ${I}$. From this it follows that ${f}$ is an ${\alpha}$-map.

Conversely, if ${\mathfrak{g} \rightarrow A}$ is an ${\alpha}$-map, then there is a unique algebra-homomorphism ${T \mathfrak{g} \rightarrow A}$. By the definition of ${\alpha}$-maps, it must factor through ${I}$, so it follows that ${\mathcal{U}(\mathfrak{g})}$ is universal. $\Box$

So, next we should probably look at some of the properties of the enveloping algebra. For instance, it’s Noetherian. It also has a nice basis by the PBW theorem.