N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

38 Chapter 3. The Linear Case
which gives a condition on ¯ h,
¯∂ ¯ h = ¯ h − Z
¯Z − h
, (3.6)
h(Z) being the complex conjugate of ¯ h( ¯ Z). Differentiating once more with respect to
¯Z yields
¯∂ 2¯ ∂¯ h¯
h =
¯Z − h − ¯ h − Z
( ¯ 2 Z − h)
= 0 ,
thus ¯ ∂ ¯ h is constant. Furthermore, the absolute value of the right-hand side of eq. (3.6)
equals one, hence
¯∂ ¯ h = e 2iϕ ⇒ ¯ h = e 2iϕ ¯ Z + c , (3.7)
where ϕ ∈ R and c ∈ C are constant. Inserting this expression back into eq. (3.6),
we conclude
e 2iϕ = e2iϕ ¯ Z + c − Z
¯Z − e −2iϕ Z − ¯c = e2iϕ ¯ Z − e −2iϕ Z + e −2iϕ c
¯Z − e −2iϕ Z − ¯c
which eventually leads to the solution
e −2iϕ c = −¯c ⇒ c = ir e iϕ , r ∈ R , (3.8)
A =
2 e −iϕ
e iϕ ¯ Z − e −iϕ Z + ir . (3.9)
Eq. (3.4) requires r = 0, while the parameter ϕ may be removed by a U(1) rotation
Having determined A, we continue with eq. 7),
Z ↦→ e iϕ Z , D i α ↦→ e −iϕ/2 D i α . (3.10)
∂E = 1
1
AE ⇒ E = 2 2A ¯ Z ¯ ∂¯g , (3.11)
where ¯g( ¯ Z) is independent of Z, and the peculiar form chosen for E will soon proove
beneficial. Eq. 2) fixes the L-dependence of C,
while eq. 25) implies
∂LC = 1
1
A ⇒ C = 2 2LA + v(Z, ¯ Z) , (3.12)
0 = Zv + ¯ Z¯v . (3.13)
It remains to solve eq. 3). When C and E are inserted, the L-dependent terms cancel
and we arrive at
0 = ∂v − 1
2 A(v + ¯ Z∂g) , (3.14)
which is readily solved by making an Ansatz v = 1
2 A ¯ Z u(Z, ¯ Z) leading to
0 = ∂u − ∂g ⇒ u = g(Z) + ¯ k( ¯ Z)
for some function ¯ k. Eq. (3.13) then requires ¯ k = ¯g, so the general solution is given by
v =
¯Z
¯Z
g(Z) + ¯g( Z) ¯ . (3.15)
− Z
,