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.