One universal enveloping algebra

olivierd

Jul 2, 2017 4 min read

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 $U$ .