Exterior Algebra as Universal Enveloping Algebra? 1. Motivation:
Let $V$ denote a vector space over a field.

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*The symmetric algebra $S(V)$ can be realized as a a universal enveloping algebra:
Consider $V$ as an abelian Lie algebra with Lie bracket given by the zero map. Then the universal enveloping algebra $U(V)$ is isomorphic (as an algebra) to the symmetric algebra $S(V)$.


*Similarly, the tensor algebra $T(V)$ can be realized as a universal enveloping algebra:
Consider the free Lie algebra $\operatorname{FreeLieAlg}(V)$ on $V$. Then the universal enveloping algebra $U(\operatorname{FreeLieAlg}(V))$ is isomorphic (as an algebra) to the tensor algebra $T(V)$.
2. Question:

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*What about the exterior algebra $\bigwedge (V)$? Can it be realized as the universal enveloping algebra of a Lie algebra?

 A: Let $$ be a Lie algebra.

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*It follows from the PBW-theorem that $\operatorname{U}()$ is one-dimensional if $$ is zero, and otherwise infinite-dimensional.

*One version of the PBW-theorem asserts that the associated graded algebra of $\operatorname{U}()$ is the symmetric algebra $\operatorname{S}()$.
The symmetric algebra $\operatorname{S}()$ has no zero divisors, so it follows that $\operatorname{U}()$ also has no zero-divisors.

Let now $V$ be a vector space.

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*If $V$ is zero-dimensional, then $⋀(V)$ is just the ground field, and thus the enveloping algebra of the zero Lie algebra.


*If $V$ is non-zero but still finite-dimensional, then $⋀(V)$ is again of finite-dimension, namely of dimension $2^{\dim(V)} > 1$.
Therefore, $⋀(V)$ cannot be isomorphic to a universal enveloping algebra.


*Suppose more generally that $V$ is non-zero but of arbitrary dimension.
The exterior algebra $⋀(V)$ contains zero divisors (e.g., all elements of $V$).
Therefore, $⋀(V)$ cannot be isomorphic to a universal enveloping algebra.
We see that $⋀(V)$ is never a universal enveloping algebra of a Lie algebra, except for when $V$ is the zero vector space.

As mentioned in the comments, this answer changes if we extend our notion of Lie algebra.

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*We may regard the vector space $V$ as a $ℤ$-graded Lie algebra $$ concentrated in degree $1$.
The resulting universal enveloping algebra $\operatorname{U}()$ (which is a $ℤ$-graded algebra satisfying a certain universal property) is precisely the exterior algebra $⋀(V)$, including the grading.

*Instead of $ℤ$ we could also use $ℤ/2$.
It is common to use the prefix “super” for $ℤ/2$-graded structures.
We have thus a Lie superalgebra $$, whose universal enveloping algebra $\operatorname{U}()$ is precisely given by $⋀(V)$ as superalgebras.

It should be noted that a graded Lie algebra $$ is typically not a Lie algebra in the usual, non-graded sense.
(But every graded algebra is also an algebra in the non-graded sense.)
And even if the Lie bracket of $$ also makes it into a non-graded Lie algebra, then the universal enveloping algebra of $$ as a graded Lie algebra will typically still be different from its universal enveloping algebra as a non-graded Lie algebra.
That is, the two notions of “universal enveloping algebra” need not be compatible.
(This is actually what happens in the above example:
if we regard $V$ as a graded Lie algebra $$ concentrated in degree $1$, then $$ is abelian, and therefore also a non-graded Lie algebra.
But the universal enveloping algebra of $$ as a graded Lie algebra is $⋀(V)$, whereas its universal enveloping algebra as a non-graded Lie algebra is $\operatorname{S}(V)$.)
We have therefore no contradiction between “$⋀(V)$ is the universal enveloping algebra of a graded Lie algebra” and “$⋀(V)$ is never the universal enveloping algebra of a Lie algebra”.
