By Rosellen M.

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4 Conformal Jacobi Identity We prove that the Leibniz identity for an algebra g is equivalent to the conformal Jacobi identity for g[[z ±1 ]]. The Leibniz identity for an algebra g is [[a, b], c] = [a, [b, c]] − ζ ab [b, [a, c]]. It is satisfied iff left multiplication g → gl(g), a → [a, ], is an algebra morphism iff [a, ] is a derivation. What we have defined is the left Leibniz identity. The right Leibniz identity is [a, [b, c]] = [[a, b], c] − ζ bc [[a, c], b]. If [ , ] is skew-symmetric then they are both equivalent to the Jacobi identity [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0.

The functor R → R1 , from N-graded conformal algebras that satisfy conformal skew-symmetry to skew-symmetric 1-truncated conformal algebras, has a left adjoint a → R(a). Proof. We construct the left adjoint a → R(a) as follows. Let R(a) := O⊕a[T ] be the K[T ]-module defined by T := d on O and T = T · on a[T ]. It is Ngraded with R0 = O and R1 = a. Define fλ g := 0, xλ f := xf, fλ x := −xf , and xλ y := [x, y] + (x, y)λ. Then it is clear that conformal skew-symmetry holds for any a, b ∈ O ⊕ a. We have (T f )λ g = −λfλ g since (df )g = 0.

Conversely, a Lie algebra morphism φ : g → g′ such that φF ⊂ F¯g′ yields a morphism (S, o) → (R(g′ ), o+ ), a → φa(z). The composition (S, o) → (R, o+ ) → (R(g(R)), o+ ), a → Y (a(z)), yields a morphism φ : g → g(R) such that at → a(z)t . On the other hand, by the universal property of g(R) the inclusion R ⊂ g[[z ±1 ]] induces a morphism φ′ : g(R) → g such that a(z)t → at for any a(z) ∈ R. In general, if R is a vertex Lie algebra and F ⊂ R a subset such that F¯ = R then the algebra g(R) is generated by {at | a ∈ F, t ∈ Z} because (ar b)t = (a(z)r b(z))t = i (−1)i ri [ar−i , bt+i ] and (T a)t = −tat−1 for a, b ∈ R and r ≥ 0, t ∈ Z.

### A Course in Vertex Algebra by Rosellen M.

by Anthony

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