## One universal enveloping algebra

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 .

Proof:

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 .

## No maverick elements in a free space

This keeps cropping up, and every time it takes me some effort to rediscover the proof. It happened first with free groups, then with free vector spaces and again with free Lie algebras.

I’ll formulate the issue with groups, but it is easy to carry it over to other cases.

A free group on set is defined by its universal property; furthermore, a particular specimen can be constructed. This construction is essential for proving the existence of a free group, but that should be its only role.

The question is: Is the subgroup of generated by necessarily itself? Can’t there be any “maverick” elements in , that are not in that subgroup? It seems that there shouldn’t be any, because they would be “too free”; when applying the universal property, the value given to them by the supposedly unique group morphism from to the “test group” might not be uniquely determined. But how do we prove this?

If we examine the constructed specimen ‑ in the case of free groups, the set of “strings” of elements of considered as “letters” plus their “primed versions”, with a certain composition rule ‑ the answer is that indeed, the whole of is generated by . And since all free groups on a given set are isomorphic, this answers our question.

But having to make use of a particular specimen doesn’t seem right to me. I want a proof with the universal property alone! Here it is, for my own repeated reference.

Let be a set, and any free group on . Let be the subgroup of generated by .

Now since, by definition of a generated subgroup, , is also a mapping from to ; or rather, we will call it as the codomain has changed. Thus is the mapping .

We thus have a group and a mapping ; the universal property of free groups satisfied by implies the existence of a group morphism such that .

Now can also be seen as a mapping from to ; or rather, since again we have changed the codomain, we define the new mapping . Since is a group morphism, so is .

Our equality  can be rewritten with the larger codomain , yielding . So is a group morphism from to such that . We know that there is only one such group morphism. Since we also have , necessarily .

For any , we have . Hence .