PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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The Schwinger term obeys the Lie cocycle condition

Statement

For the finite-dimensional real Lie algebra $g$ , invariant symmetric form $κ$ , smooth loop algebra $L g$ , and bilinear form $ω$ defined in Loop current Lie algebra and central term, the form $ω$ is alternating and satisfies $ω ([X, Y], Z) + ω ([Y, Z], X) + ω ([Z, X], Y) = 0 (X, Y, Z \in L g) .$ Consequently the candidate bracket on $\hat{L g}$ is a Lie bracket and gives a one-dimensional central extension of $L g$ by $R$ .

Statement references

Loop current Lie algebra and central term

Exact sequence diagram

The arrows are the linear inclusion $i (a) = (0, a)$ and projection $p (X, a) = X$ . The proof verifies that these maps are Lie-algebra homomorphisms, that $\ker p = im i$ , and that $im i$ is central. The diagram records the resulting exact sequence; the proof does not rely on the diagram.

Facts & Assumptions

Given: The finite-dimensional real Lie algebra, periodic smooth loops, invariant symmetric form, cocycle formula, and candidate bracket of Loop current Lie algebra and central term.

[F1]

Loop current Lie algebra and central term: $L g = C^{\infty} (S^{1}, g)$ has the pointwise Lie bracket; $κ$ is symmetric and satisfies $κ ([U, V], W) = κ (U, [V, W])$ ; $ω (X, Y) = (2 π)^{- 1} \int_{0}^{2 π} κ (X, Y^{'}) d θ$ ; and the candidate bracket on $L g \oplus R$ is $[(X, a), (Y, b)] = ([X, Y], ω (X, Y))$ .

Proof

Given: [F1].

1.1givenF1algebra

Bilinearity of $κ$ and differentiation makes $ω$ bilinear. By symmetry and the product rule, $ω (X, Y) + ω (Y, X) = \frac{1}{2 π} \int_{0}^{2 π} (κ (X, Y^{'}) + κ (X^{'}, Y)) d θ = \frac{1}{2 π} [κ (X, Y)]_{0}^{2 π} = 0.$ The endpoint values agree because $X$ and $Y$ are smooth maps on $S^{1}$ , so $ω$ is skew-symmetric; over $R$ , this implies $ω (X, X) = 0$ .

1.2givenF1algebra

Invariance and symmetry give $κ ([X, Y], Z^{'}) = κ (X, [Y, Z^{'}]), κ ([Y, Z], X^{'}) = κ (X^{'}, [Y, Z]), κ ([Z, X], Y^{'}) = κ (X, [Y^{'}, Z]) .$ The product rule and $[Y, Z]^{'} = [Y^{'}, Z] + [Y, Z^{'}]$ therefore yield $κ ([X, Y], Z^{'}) + κ ([Y, Z], X^{'}) + κ ([Z, X], Y^{'}) = \partial_{θ} κ (X, [Y, Z]) .$ Integrating over one period makes the right-hand side zero. By the definition of $ω$ , this is the required cyclic cocycle identity.

2.1givenF1step 1.1step 1.2algebra

The displayed operation is bilinear and skew-symmetric by the pointwise Lie bracket and step 1.1. The vector component of its Jacobi sum is the pointwise Jacobi sum in $g$ , hence vanishes at every $θ$ . Its real component is the cyclic sum in step 1.2, hence also vanishes. Thus the operation satisfies Jacobi and is a Lie bracket.

3.1givenF1step 2.1algebra∎

The maps $i (a) = (0, a)$ and $p (X, a) = X$ are Lie-algebra homomorphisms: the bracket of any two elements in $im i$ is zero, and $p ([(X, a), (Y, b)]) = [X, Y]$ . The projection is surjective because $p (X, 0) = X$ , and $\ker p = {(0, a) : a \in R} = im i$ . Moreover $[(0, a), (Y, b)] = (0, ω (0, Y)) = (0, 0)$ by bilinearity of $ω$ , so the kernel is central. This proves the asserted central extension.

Depends on

Loop current Lie algebra and central term