18.747: Problem set 4; due Thursday, March 9

1. Let 𝔤 be a finite dimensional simple Lie algebra.

a) Show that the Verma module M(−ρ) is irreducible, and, moreover, it splits off as a direct summand in every module, i.e. every module M in category O has the form M ≅ M(−ρ)^{n} ⊕ N, where N does not have subquotients isomorphic to M(−ρ).

b) Let 𝔤̂ = 𝔤[t, t^{−1}]⊕ℂK be the (untwisted) affine Kac-Moody algebra. Here we normalize the invariant bilinear form on 𝔤 used to define 𝔤̂ so that the induced form on 𝔥^{∗} has the property that (α, α) = 2 for a long root α. For a simple root α_{i} define a number ǎ_{i} from the relation α̌ = ∑_{i} ǎ_{i}h_{i} where α̌ is the coroot corresponding to the maximal root (thus it is a coroot of minimal length). The dual Coxeter number is defined by ȟ = 1 + ∑_{i} ǎ_{i}. Show that the level (i.e., the constant by which the element K acts) of the Verma module M(−ρ) equals −ȟ. This is called the critical level.

c) (Optional) Now equip 𝔤 with the Killing form κ(x, y) = T r(ad(x)ad(y)) and consider the corresponding central extension of the current algebra 𝔤̂_{κ} = 𝔤[t, t^{−1}] ⊕ ℂK_{κ}. Show that K_{κ} acts in M(−ρ) by −\frac{1}{2} times the identity. [Hint: use Proposition 6.9(f) on p.63 in Feigin-Zelevinsky paper].

2. The Weyl-Kac denominator formula for affine Lie algebras states that

∑_{w∈W} sign(w)e^{(wρ,h)} = ∏_{γ}(1 − e^{−(γ,h)})^{mult(γ)},

where the product is taken over positive roots (see Remark 11.3 in Kac-Raina). Derive it in the case of affine sl(2) from the Jacobi triple product identity:

∏_{k=1}^{∞}(1 − q^{2k})(1 − q^{2k−1}z)(1 − q^{2k−1}z^{−1}) = ∑_{m∈ℤ}(−1)^{m}q^{m^{2}}z^{m}.

3. Let M(λ)^{+} be the highest weight Verma module over a finite dimensional simple Lie algebra g, and M(λ)^{−} be the lowest weight Verma module. Let V be any h-diagonalizable module over this Lie algebra. Show that Hom_{g}(M(λ)^{+} ⊗M(μ)^{−}, V) (as representations) is the space V [λ + μ] of vectors in V of weight λ + μ.

4. Show that the adjoint representation of an affine Kac-Moody Lie algebra is not completely reducible. [Thus not every integrable representation is completely reducible].

5. Do problem 10.1 in Kac’s book, p. 183.