TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Universal property of the enveloping superalgebra

Statement

Let $g$ be a complex Lie superalgebra, $A$ a unital associative superalgebra, and $ϕ : g \to A$ a parity-preserving linear map that preserves brackets, where the bracket on $A$ is its supercommutator. There is a unique unital associative-superalgebra homomorphism $\hat{ϕ} : U (g) \to A$ with $\hat{ϕ} \circ ι_{g} = ϕ$ .

Facts & Assumptions

Given: $g, A, ϕ$ as in the Statement.

[F1]

Universal enveloping superalgebra: $U (g) = T (g) / I$ , where $I$ is generated by $x \otimes y - (- 1)^{∣ x ∣ ∣ y ∣} y \otimes x - [x, y]$ .

[F2]

Associative superalgebra over the complex numbers: homogeneous products add parity, and the supercommutator is $[a, b]_{s} = a b - (- 1)^{∣ a ∣ ∣ b ∣} b a$ .

[F3]

Universal property of the tensor algebra: a linear map from $g$ into a unital associative algebra extends uniquely to a unital algebra homomorphism from $T (g)$ , acting on pure tensors by multiplying the images of their entries.

Proof

1.1givenF2F3construct

The tensor-algebra extension $\tilde{ϕ} : T (g) \to A$ exists uniquely by [F3]; on a homogeneous pure tensor it is $ϕ (x_{1}) \dots ϕ (x_{n})$ , whose parity is $\sum_{i} ∣ x_{i} ∣$ , so it preserves the supergrading.

2.1givenF1F2step 1.1algebra

For each homogeneous pair $x, y$ , bracket preservation gives $\tilde{ϕ} (x \otimes y - (- 1)^{∣ x ∣ ∣ y ∣} y \otimes x - [x, y]) = [ϕ (x), ϕ (y)]_{s} - ϕ ([x, y]) = 0$ . Since $\tilde{ϕ}$ is an algebra homomorphism, it kills the two-sided ideal $I$ generated by these elements.

3.1F1F3step 1.1step 2.1∎

Therefore $\tilde{ϕ}$ factors through a unique homomorphism $\hat{ϕ} : T (g) / I \to A$ , which has $\hat{ϕ} ι_{g} = ϕ$ . Any other such homomorphism would pull back to an extension of $ϕ$ to $T (g)$ ; uniqueness in [F3] makes it equal to $\hat{ϕ}$ .

Depends on

Used by

Lie supermodules are graded enveloping-superalgebra modules Theorem

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sigiswald Barbier, A minimal representation for the orthosymplectic Lie-superalgebra, §2.6.2 (standard reference, not scraped)
Lauren Caston and Rita Fioresi, Mathematical Foundations of Supersymmetry, §§1.2–1.3 (standard reference, not scraped)