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N=2 Supersymmetric Gauge Theories with Nonpolynomial Interactions

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42 Chapter 3. The Linear Case<br />

Again the terms in parentheses will turn out to be the covariant derivative of the<br />

potential Vµ. For Z = i the equation reduces to Gµν = Vµν.<br />

We are not done yet, as the equations for W µ and Gµν are still coupled. To simplify<br />

the following calculations let us introduce the abbreviations<br />

H µ ≡ 1<br />

Vµν ≡ I<br />

|Z| 2<br />

2εµνρσ∂νBρσ + Λ µ<br />

<br />

Vµν + ˜ Σµν<br />

Then we first solve eq. (3.31) for Gµν,<br />

and insert this into eq. (3.30),<br />

(3.32)<br />

R<br />

+<br />

|Z| 2<br />

<br />

˜Vµν − Σµν . (3.33)<br />

Gµν = Vµν − 2I<br />

|Z| 2 A[µWν] − R<br />

|Z| 2 εµνρσA ρ W σ , (3.34)<br />

IW µ = H µ + V µν Aν − 2I<br />

|Z| 2 A[µ W ν] Aν . (3.35)<br />

Collecting the terms <strong>with</strong> W µ , this can be written as<br />

where the field dependent matrix K µν is given by<br />

IK µν Wν = |Z| 2 (H µ + V µν Aν) , (3.36)<br />

K µν = η µν E + A µ A ν , (3.37)<br />

E being the expression (1.55) that we have already encountered in the central charge<br />

transformation of the spinors ψ i . To solve for W µ , we need to invert K µν . It can be<br />

easily checked that<br />

(K −1 )µν = 1<br />

E<br />

<br />

ηµν − |Z| −2 <br />

AµAν . (3.38)<br />

Due to the appearance of E in the denominator, this expression is nonpolynomial in<br />

the gauge field Aµ.<br />

Having determined W µ , we obtain Gµν from eq. (3.34). The final solution to the Bianchi<br />

identities then reads<br />

W µ = |Z|2 µ µν<br />

H + V Aν − |Z|<br />

IE<br />

−2 A µ AνH ν<br />

(3.39)<br />

Gµν = Vµν − 2<br />

E A[µ<br />

<br />

Hν] + Vν]ρA ρ − R<br />

IE εµνρσA ρ H σ + V σλ <br />

Aλ . (3.40)<br />

The fundamental fields of course are the gauge potentials Vµ and Bµν rather than<br />

W µ and Gµν, so we now have to determine their supersymmetry and central charge<br />

transformations as well. These can be obtained most easily from eqs. (3.30) and (3.31).<br />

Applying ∆ z to the former we have<br />

∆ z (C) (IW µ − Λ µ ) = 1<br />

2εµνρσ z µν<br />

∂ν ∆ (C) Bρσ − G ∂νC + CAν δzG µν .

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