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.

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