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N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

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3.2. Transformations and Bianchi Identities 41<br />

whereas the determination of δzW µ is somewhat involved. According to eq. (2.39),<br />

2 µ i<br />

Iδz |Z| W + 2L(Z∂µ Z¯ − Z∂ ¯ µ i Z) + 2 (Zψiσ µ¯ λi − ¯ Zλ i σ µ ψi) ¯ =<br />

= I 2 DνG µν − IR Dν ˜ G µν + 1<br />

6IR Diσ µ DjM ¯ ij<br />

= I 2 DνG µν − G µν R∂νR − R Dν(I ˜ G µν + Σ µν ) + ˜ F µν <br />

1<br />

Wν + 2UR ∂µ |Z| 2<br />

+ 1<br />

2R δz(Zψ i σµ ¯ λi + ¯ Zλ i σµ ¯ ψi) .<br />

The expression in square brackets can be rewritten by means of the Bianchi identity<br />

(3.27),<br />

2 µ i<br />

Iδz |Z| W + 2L(Z∂µ Z¯ − Z∂ ¯ µ i Z) + 2 (Zψiσ µ¯ λi − ¯ Zλ i σ µ ψi) ¯ =<br />

= |Z| 2 DνG µν + 1<br />

2UR ∂µ |Z| 2 + 1<br />

2R δz(Zψ i σµ ¯ λi + ¯ Zλ i σµ ¯ ψi) ,<br />

from which follows<br />

µ µ<br />

δz IW − Λ = DνG µν . (3.29)<br />

One can now easily check that the Bianchi identities, together <strong>with</strong> the central charge<br />

transformations just obtained, satisfy the integrability condition<br />

D[µDν] = 1<br />

2 Fµνδz ,<br />

which is the covariant analogue of d 2 = 0.<br />

We first solve the constraint on W µ . Let us split the covariant derivative in eq. (3.25)<br />

into the partial derivative and the central charge transformation, which we then replace<br />

using eq. (3.29),<br />

µ µ<br />

∂µ IW − Λ = 1<br />

2 FµνG µν − Aµδz<br />

= 1<br />

2 Fµν − AµDν<br />

= ∂µ(G µν Aν) .<br />

IW µ − Λ µ <br />

G µν<br />

In the last step we employed the identity A[µDν] = A[µ∂ν]. We conclude that the terms<br />

in parentheses equal the dual field strength of an antisymmetric tensor gauge field Bµν,<br />

IW µ = 1<br />

2 εµνρσ (∂νBρσ − Aν ˜ Gρσ) + Λ µ . (3.30)<br />

Note how the first two terms on the right-hand side resemble a covariant derivative,<br />

and indeed we shall find that Bµν transforms into − ˜ Gµν under δz. In the limit Z = i<br />

we recover the relation W µ = H µ just as in the free case.<br />

To solve the Bianchi identity (3.27) we proceed along the same lines,<br />

<br />

∂ν IG ˜µν µν µν<br />

+ RG + Σ = −ε µνρσ ( 1<br />

2FνρWσ − AνDρWσ)<br />

Hence there is a vector field Vµ such that<br />

= −ε µνρσ ∂ν(AρWσ) .<br />

I ˜ G µν + RG µν = ε µνρσ (∂ρVσ − AρWσ) − Σ µν . (3.31)

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