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By A. Jaffe (Chief Editor)

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A) Given a k-tuple of positive numbers (c1 , . . , ck ) := (c(α¯ 1 ), . . , ¯ for all c(α¯ k )) satisfying the Assumption, Eq. 6) uniquely defines numbers c(α) ¯ α ∧ E−α satisfies positive quasiroots α¯ = α¯ i such that the bivector v = c(α)E the condition [[v, v]] = K 2 ϕM . 8) for all quasiroots α. ¯ Proof. a) Since by assumption the denominator is never zero, Eq. 5) for all positive β. 4) becomes c(α¯ + β)(c( α) ¯ + c(β)) ¯ β) Eq. 6) implies that c(α) ¯ = 0 for all quasiroots. So setting λ(α) ¯ = 1/c(α), ¯ we find ¯ = λ(α) ¯ Thus λ is a linear that Eq.

15), but this time we consider the series which terminate at r = − 21 : −1/2 ψ(η) = r=0 arα (η − eα )2 −r They are parametrized by zeros of the polynomial a α + 21 ψ(η; Eiα ). 24) since Uα and τˆ (u) commute. 1. 2. Assume that ω2 = τ is pure imaginary and that parameters zn are all real numbers. Then all Eiα are real and Eiα = Ejα for distinct i, j . The solutions ψ(η; Eiα ) and Uα ψ(η; Eiα ) span the space V ( ) . In particular Eiα (i = 1, . . , + 21 ) for α = 1, 2, 3 coincide up to order, and a 1 1 (E) = a 2 1 (E) = a 3 1 (E).

Using the Killing form, it is easy to see that representations mβ¯ and m−β¯ are dual. such that β¯1 + β¯2 ∈ the representation Question. Is it true that for β¯1 , β¯2 ∈ 2 mβ¯1 +β¯2 is contained in mβ¯1 ∧ mβ¯2 ⊂ ∧ m with multiplicity one? Since g contains the Cartan subalgebra h, each g invariant tensor over m has to be of weight zero. It follows that there are no invariant vectors in m. Hence, there are no invariant vector fields on M. Consider the invariant bivector fields on M. From the above, such fields correspond to the g invariant bivectors from ∧2 m.

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