Koebe 1/4-theorem and Inequalities in N=2 Super-qcd

The critical curve C on which Im τ̂ = 0, τ̂ = aD/a, determines hyperbolic domains whose Poincaré metric is constructed in terms of aD and a. We describe C in a parametric form related to a Schwarzian equation and prove new relations for N = 2 Super SU(2) Yang-Mills. In particular, using the Koebe 1/4-theorem and Schwarz’s lemma, we obtain inequalities involving u, aD and a, which seem related to the Renormalization Group. Furthermore, we obtain a closed form for the prepotential as function of a. Finally, we show that ∂τ̂ 〈trφ〉τ̂ = 1 8πib1 〈φ〉τ̂ , where b1 is the one-loop coefficient of the beta function. Partly supported by the European Community Research Programme Gauge Theories, applied supersymmetry and quantum gravity, contract SC1-CT92-0789 1. The effective action of the low-energy limit of N = 2 super Yang-Mills, solved exactly in [1], is described in terms of the prepotential F [2] Seff = 1 4π Im (∫ dθdθ̄ΦDΦi + 1 2 ∫ dθτ WiWj ) , (1) where ΦD ≡ ∂F/∂Φi and τ ij ≡ ∂F/∂Φi∂Φj . Let us denote by ai ≡ 〈φi〉 and aD ≡ 〈φD〉 the vevs of the scalar component of the chiral superfield. For gauge group SU(2) the moduli space of quantum vacua, parameterized by u ≡ 〈trφ2〉, is Σ3 = C\{−Λ2,Λ2}, the Riemann sphere Ĉ = C ∪ {∞} with punctures at ±Λ2 and ∞, where Λ is the dynamically generated scale. It turns out that [1] (we set Λ = 1) aD = ∂aF = √ 2 π ∫ u 1 dx √ x− u √ x2 − 1 , a = √ 2 π ∫ 1 −1 dx √ x− u √ x2 − 1 . (2) A crucial property of aD and a is that they satisfy the equation [3] (see also [4]) [ 4(u − 1)∂ u + 1 ] aD = 0 = [ 4(u − 1)∂ u + 1 ] a. (3) This equation is the “reduction” of the uniformizing equation for Σ3 [5][3][4] [ 4(1− u)∂ u + u + 3 ] ψ = 0, (4) which is satisfied by √ 1− u∂uaD and √ 1− u∂ua. Let us summarize the main results in [4]. First of all it has been shown that u = πi(F − a∂aF/2), (5) that is F (〈φ〉) = 1 πi 〈trφ〉+ 1 2 〈φ〉〈φD〉. (6) In order to specify the functional dependence of u we set u = G1(a), u = G2(τ̂ ) and u = G3(τ) where τ̂ = aD/a and τ = ∂ 2 aF . By Eq.(3) we have (1− G 1)∂ aG1 + a 4 (∂aG1) = 0, (7) that by (5) provides recursion relations for the instanton contribution and implies ∂ aF = π (a∂ aF − ∂aF) 3 16 [ 1 + π2 (F − a∂aF/2) ] . (8) By (2) we have a(u = −1) = −i4/π and a(u = 1) = 4/π so that the initial conditions for the second-order equation (7) are G1(−i4/π) = −1 and G1(4/π) = 1.