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 W be a \K-vector space and (V, \alpha) a free associative algebra on W. The mapping \alpha is a vector space morphism W \to V.

V can also be viewed as a Lie algebra. \alpha^\rightarrow(W) is a part of V; let U be the sub-Lie algebra of V generated by \alpha^\rightarrow(W).

(It appears that, except in trivial cases, U is strictly smaller than V; for instance, if we take V as the free associative algebra on W constructed in chapter 18, V = W \oplus (W \otimes W) \oplus (W \otimes W \otimes W) \oplus \ldots, in U it appears that all the W \otimes W components are antisymmetric.)

We have a natural Lie algebra morphism \beta: U \to V, \pmb v \mapsto \pmb v. That is, \beta is the canonical injection from U to V.

Proposition: The associative algebra V is the universal enveloping algebra of the Lie algebra U.

Proof:

Let us first note that since U is the Lie subalgebra of V generated by \alpha^\rightarrow(W), we have \alpha^\rightarrow(W) \subseteq U. Let us define \delta: W \to U, \pmb w \mapsto \alpha(\pmb w), that is, \delta is \alpha with its codomain restricted to U. We can then write \alpha = \beta \circ \delta.

We must prove that given any associative algebra V' and any Lie algebra morphism \beta': U \to V', there exists one and only one associative algebra morphism \gamma: V \to V' such that \gamma \circ \beta = \beta'.

So let V' be an associative algebra and \beta' a Lie algebra morphism U \to V'; that is, a linear mapping U \to V' such that \beta'(\pmb v_1 \pmb v_2 - \pmb v_2 \pmb v_1) = \beta'(\pmb v_1) \beta'(\pmb v_2) - \beta'(\pmb v_2) \beta'(\pmb v_1).

Then we have \beta' \circ \delta vector space morphism W \to V'. Since (V, \alpha) is a free associative algebra on the vector space W, there exists a unique associative algebra morphism \gamma_0: V \to V' such that \gamma_0 \circ \alpha = \beta' \circ \delta, that is, \gamma_0 \circ \beta \circ \delta = \beta' \circ \delta.

Thus \gamma_0 \circ \beta and \beta' agree on all elements of \delta^\rightarrow(W), which is also \alpha^\rightarrow(W), which generates the sub Lie algebra U of V. We wish to show that they actually agree on all elements of U.

Let A be the set of all elements of U on which \gamma_0 \circ \beta and \beta' agree. Since \gamma_0 \circ \beta and \beta' are both linear mappings, \gamma_0 \circ \beta - \beta' is a linear mapping too, even if it may not be an associative algebra or Lie algebra morphism. A is its kernel, and is a sub vector space of U. Furthermore, if \gamma_0 \circ \beta and \beta' agree on vectors \pmb v_1 and \pmb v_2 of U, then (\gamma_0 \circ \beta - \beta')([\pmb v_1, \pmb v_2]) = (\gamma_0 \circ \beta)([\pmb v_1, \pmb v_2]) - \beta'([\pmb v_1, \pmb v_2]). Since both \gamma_0 \circ \beta and \beta' are Lie algebra morphisms, this is equal to [(\gamma_0 \circ \beta - \beta')(\pmb v_1), (\gamma_0 \circ \beta - \beta')(\pmb v_2)] - [\beta'(\pmb v_1), \beta'(\pmb v_2)], and, since \gamma_0 \circ \beta and \beta' agree on \pmb v_1 and \pmb v_2, this expression reduces to zero. Thus [\pmb v_1, \pmb v_2] too is an element of A. This completes the proof that A is a sub Lie algebra of U. Since A includes \alpha^\rightarrow(W), and since U is the smallest sub Lie algebra of itself including \alpha^\rightarrow(W), it follows that A = U, that is, that \gamma_0 \circ \beta and \beta' agree on all U. Since U is their common domain, they are equal: \gamma_0 \circ \beta = \beta'.

Hence there exists at least one associative algebra \gamma: V \to V', namely \gamma = \gamma_0, such that \gamma \circ \beta = \beta'.

Let us show that such a \gamma is unique.

Let \gamma be an associative algebra morphism V \to V' such that \gamma \circ \beta = \beta'.

Then \gamma \circ \beta \circ \delta = \beta' \circ \delta, that is, \gamma \circ \alpha = \beta' \circ \delta.

But, because (V, \alpha) is a free associative algebra on the vector space W, we know that there is a unique associative algebra morphism \gamma having such a property, namely \gamma = \gamma_0.

Hence there is one and only one associative algebra morphism V \to V' such that \gamma \circ \beta = \beta'.

This shows that (V, \beta) is the free associative algebra on the Lie algebra U.

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 (G, \alpha) on set S 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 G generated by \alpha^\rightarrow(S) necessarily G itself? Can’t there be any “maverick” elements in G, 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 G 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 S considered as “letters” plus their “primed versions”, with a certain composition rule ‑ the answer is that indeed, the whole of G is generated by \alpha^\rightarrow(S). 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 S be a set, and (G, \alpha) any free group on S. Let H be the subgroup of G generated by \alpha^\rightarrow(S).

Now since, by definition of a generated subgroup, \alpha^\rightarrow(S) \subseteq H, \alpha is also a mapping from S to H; or rather, we will call it \beta as the codomain has changed. Thus \beta is the mapping S \to H, s \mapsto \alpha(s).

We thus have a group H and a mapping \beta: S \to H; the universal property of free groups satisfied by (G, \alpha) implies the existence of a group morphism \gamma: G \to H such that \gamma \circ \alpha = \beta.

Now \gamma can also be seen as a mapping from G to G; or rather, since again we have changed the codomain, we define the new mapping \delta: G \to G, g \mapsto \gamma(g). Since \gamma is a group morphism, so is \delta.

Our equality \gamma \circ \alpha = \beta can be rewritten with the larger codomain G, yielding \delta \circ \alpha = \alpha. So \delta is a group morphism from G to G such that \delta \circ \alpha = \alpha. We know that there is only one such group morphism. Since we also have \Id_G \circ \alpha = \alpha, necessarily \delta = \Id_G.

For any g \in G, we have g = \Id_G(g) = \delta(g) = \gamma(g) \in H. Hence H = G.