In my search for less abstract or trivial examples of a universal enveloping algebra, I have come upon this one.
Let be a vector space and a free associative algebra on . The mapping is a vector space morphism .
can also be viewed as a Lie algebra. is a part of ; let be the sub-Lie algebra of generated by .
(It appears that, except in trivial cases, is strictly smaller than ; for instance, if we take as the free associative algebra on constructed in chapter 18, , in it appears that all the components are antisymmetric.)
We have a natural Lie algebra morphism . That is, is the canonical injection from to .
Proposition: The associative algebra is the universal enveloping algebra of the Lie algebra .
Let us first note that since is the Lie subalgebra of generated by , we have . Let us define , that is, is with its codomain restricted to . We can then write .
We must prove that given any associative algebra and any Lie algebra morphism , there exists one and only one associative algebra morphism such that .
So let be an associative algebra and a Lie algebra morphism ; that is, a linear mapping such that .
Then we have vector space morphism . Since is a free associative algebra on the vector space , there exists a unique associative algebra morphism such that , that is, .
Thus and agree on all elements of , which is also , which generates the sub Lie algebra of . We wish to show that they actually agree on all elements of .
Let be the set of all elements of on which and agree. Since and are both linear mappings, is a linear mapping too, even if it may not be an associative algebra or Lie algebra morphism. is its kernel, and is a sub vector space of . Furthermore, if and agree on vectors and of , then . Since both and are Lie algebra morphisms, this is equal to , and, since and agree on and , this expression reduces to zero. Thus too is an element of . This completes the proof that is a sub Lie algebra of . Since includes , and since is the smallest sub Lie algebra of itself including , it follows that , that is, that and agree on all . Since is their common domain, they are equal: .
Hence there exists at least one associative algebra , namely , such that .
Let us show that such a is unique.
Let be an associative algebra morphism such that .
Then , that is, .
But, because is a free associative algebra on the vector space , we know that there is a unique associative algebra morphism having such a property, namely .
Hence there is one and only one associative algebra morphism such that .
This shows that is the free associative algebra on the Lie algebra .