Exterior Algebra as Universal Enveloping Algebra

Question

1. Motivation:

Let $V$ denote a vector space over a field.

• 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:

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

Answer 1

Let $𝔤$ be a Lie algebra.

• It follows from the PBW-theorem that $\operatorname{U}(𝔤)$ is one-dimensional if $𝔤$ is zero, and otherwise infinite-dimensional.
• One version of the PBW-theorem asserts that the associated graded algebra of $\operatorname{U}(𝔤)$ is the symmetric algebra $\operatorname{S}(𝔤)$. The symmetric algebra $\operatorname{S}(𝔤)$ has no zero divisors, so it follows that $\operatorname{U}(𝔤)$ also has no zero-divisors.

Let now $V$ be a vector space.

• 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.

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

• We may regard the vector space $V$ as a $ℤ$-graded Lie algebra $𝔤$ concentrated in degree $1$. The resulting universal enveloping algebra $\operatorname{U}(𝔤)$ (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 $𝔤$, whose universal enveloping algebra $\operatorname{U}(𝔤)$ is precisely given by $⋀(V)$ as superalgebras.

It should be noted that a graded Lie algebra $𝔤$ 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 $𝔤$ also makes it into a non-graded Lie algebra, then the universal enveloping algebra of $𝔤$ 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 $𝔤$ concentrated in degree $1$, then $𝔤$ is abelian, and therefore also a non-graded Lie algebra. But the universal enveloping algebra of $𝔤$ 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”.