N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions
N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions
N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions
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2.3. The Ansatz 29<br />
18) ′<br />
19) ′<br />
20) ′<br />
21) ′<br />
22) ′<br />
0 = ∂(A − 2a) − 1<br />
2 (A − 2a)2 + 2D(F − c) − 2ĒG 0 = ∂L(A − 2a) − ¯ BG + 2a(F − c) (2.51)<br />
0 = ¯ ∂(A − 2a) − 1<br />
2 B ¯ B + 2bF<br />
0 = ¯ ∂G + 1<br />
2<br />
ĀG + 1<br />
2 B(2c − F − ¯ F )<br />
0 = ¯ ∂B − 2E(F − c) − 1<br />
2 B(Ā − 2ā) + 2 ¯ DG .<br />
Before we are trying to solve these equations, it is important to note that the solutions to<br />
the consistency conditions decompose into mutually disjoint equivalence classes, where<br />
two sets of constraints are deemed equivalent if they are related by a local superfield<br />
redefinition<br />
L → L = L( ˆ L, Z, ¯ Z) . (2.52)<br />
Each representative of such a class effectively describes the same physics. We may<br />
employ this to choose representatives which simplify the subsequent calculations as<br />
much as possible. Using<br />
D i αL = L ′ D i α ˆ L + ∂L D i αZ (2.53)<br />
and similar for ¯ D i ˙αL, where L ′ ≡ ∂ ˆ L L = 0, we rewrite the Ansatz (2.40), (2.41) in terms<br />
of ˆ L. The coefficients as functions of ˆ L are then given by<br />
â = a + c ∂L − ∂L ′ /L ′<br />
ˆ b = (b + a ¯ ∂L + ā ∂L + c ∂L ¯ ∂L − ∂ ¯ ∂L)/L ′<br />
ĉ = c L ′ − L ′′ /L ′<br />
 = A + 2F ∂L − 2∂L ′ /L ′<br />
ˆB = B + 2G ¯ ∂L (2.54)<br />
Ĉ = (C − ∂L)/L ′<br />
ˆD = (D + A ∂L + F ∂L ∂L − ∂ 2 L)/L ′<br />
Ê = (E + B ¯ ∂L + G ¯ ∂L ¯ ∂L)/L ′<br />
ˆF = F L ′ − L ′′ /L ′<br />
ˆG = GL ′ .<br />
The differential equations (2.46), (2.48) and (2.50) are invariant under the above transformations<br />
in the sense that if A, B, etc. are solutions to the consistency conditions,<br />
then Â, ˆ B, etc. satisfy the same equations <strong>with</strong> ∂L replaced by ∂ ˆ L . Consider for instance<br />
eq. 1) in (2.46),<br />
∂ ˆ F − 1<br />
2 ∂ ˆ L Â = ∂ (F L ′ − L ′′ /L ′ ) − 1<br />
2 ∂ ˆ L (A + 2F ∂L − 2∂L ′ /L ′ )<br />
= ∂F L ′ + ∂L ∂LF L ′ + F ∂L ′ − ∂L ′′ /L ′ + L ′′ ∂L ′ /L ′2<br />
− 1<br />
2 (L′ ∂LA + 2L ′ ∂LF ∂L + 2F ∂L ′ − 2∂L ′′ /L ′ + 2L ′′ ∂L ′ /L ′2 )<br />
= L ′ (∂F − 1<br />
2 ∂LA) = 0 .