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N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

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2.3. The Ansatz 33<br />

unknown coefficients in the Ansatz. This is quite obvious as the number of equations<br />

is not sufficient to fix the dependence of the coefficients on all three variables L, Z, ¯ Z.<br />

Our goal is to generalize the solutions found in the previous section. To this end we<br />

simplify the Ansatz by setting<br />

a = b = c = 0 ⇒ N ij<br />

α ˙α = 0 , (2.71)<br />

which can be justified a posteriori by showing that the resulting constraints do indeed<br />

yield the linear and nonlinear vector-tensor multiplet <strong>with</strong> gauged central charge. Note<br />

that we are still allowed to redefine<br />

L = κ ˆ L + f(Z) + ¯ f( ¯ Z) , κ ∈ R ∗ . (2.72)<br />

This is the general solution to the differential equations<br />

L ′′ = ∂L ′ = ¯ ∂∂L = 0 ,<br />

which according to eq. (2.54) guarantee the preservation of Ansatz (2.71).<br />

The simplification is motivated by the fact that now condition (C.4) may easily be<br />

evaluated. With the reduced Ansatz put in, it reads<br />

<br />

0 = δz ( 1<br />

2L + ZC) DiD j Z + (1 + ZA) D (i Z D j) L + (1 + ZB) ¯ D (i Z¯ D¯ j)<br />

L<br />

+ ZD D i Z D j Z + ZE ¯ D i <br />

Z¯ D¯ j<br />

Z ¯ i j<br />

+ ZF D L D L + ZG D¯ i<br />

L D¯ j<br />

L (2.73)<br />

+ c.c.<br />

from which we infer<br />

as well as<br />

23) 0 = 2 + ZA + ¯ Z ¯ B 24) 0 = ZF + ¯ Z ¯ G , (2.74)<br />

25) u(Z, ¯ Z) = L + ZC + ¯ Z ¯ C 26) v(Z, ¯ Z) = ZD + ¯ ZĒ , (2.75)<br />

u and v being arbitrary L-independent functions that contribute only to δz-invariant<br />

terms inside the square brackets in eq. (2.73). Equations 23 − 26) allow to eliminate<br />

B, D and G in the eqs. (2.46), (2.50) and (2.51), leaving the four unknown functions<br />

A, C, E and F , whereas u and v may be removed using the gauge freedom (2.72). To<br />

show this, let us first consider eq. 16) in (2.50). Replacing B and D according to the<br />

relations above, we obtain<br />

0 = ¯ ∂v − ¯ Z( ¯ ∂<br />

1 Ē − ĀĒ) . 2<br />

Eq. 7) then implies ¯ ∂v = 0. Next we consider eq. 15). Using eqs. 25), 3) and 26), we<br />

find<br />

0 = ∂u − Z(∂C − 1<br />

2AC) − ¯ ZĒ = ∂u − v .

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