# Can you have more than one $D$ term for each gauge group?

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I'm reading this note in chapter 6 which discusses ${\cal N} = 2$ supermultiplets, though the discussion is quite general and applies to ${\cal N} = 1$ fields in the adjoint representation. I summarize the discussion given in the paper here with a slightly simplified situation and notation for clarity. Suppose we have some field, $\Phi$, in the adjoint representation of a gauge group $SU(N)$. The Kahler term for the field is

\begin{equation}\int d ^2 \theta \Phi ^\dagger e ^{ 2 g V ^a T ^a} \Phi =g D ^a \phi ^\dagger T ^a \phi + ...\end{equation}

and the kinetic gauge term gives,

\begin{equation}W^\alpha _a W_\alpha^a=\frac{1}{2} D ^a D ^a+...\end{equation}

The paper then goes on to integrate out the $D^a$ fields which gives,

\begin{equation}D^a =\Phi ^\dagger T ^a \Phi\end{equation}

Then the authors go on to introduce more ${\cal N} =1$ field, $H$. The $H$ field Lagrangian takes the form,

\begin{align}{\cal L} &= \int d^4\theta H ^\dagger e^{2q V ^a t ^a} H+ \int d^2\theta W^\alpha _a W_\alpha^a + ...\\&= g D ^a \Phi ^\dagger T ^a \Phi + \frac{1}{2}D_aD_a...\end{align}

where $t_a$ are the generators in the representation of the $H$ fields.

So far I have no issues, however then do something strange. They go on to again integrate out the $D$ fields, but they do so independently of the fields introduced earlier in the adjoint representation. However, don't we have a single $\vec{D}$ field for every gauge group, not every representation? So how come we are able to consider the sectors independently of one another?

edited May 27, 2014

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There is indeed one $D$ field for each gauge group, it is simply one field in the vector supermultiplet. I think that the way the paper presents things can be explained as follows. The Lagrangian for the $D$-field will always be of the form

$\frac{1}{2} D^{2} + (a+b+...)D$

where $a$,$b$,... are the contributions of the various charged fields and the equation of motion for $D$ is $D = -a-b-...$. In particular, each charged field gives a well-defined contribution to $D$ and the total $D$ is just  the sum of these contributions, there is no interaction between the various terms. When they consider one special field and integrate out $D$, I think they mean to compute the contribution of this field to the $D$-term.

answered May 28, 2014 by (5,140 points)

Thanks for your response. That makes sense. Just to make sure I understand correctly. Is it then indeed true that we will also have cross terms of the form,

\begin{equation} (\phi ^\dagger T ^a \phi ) H ^\dagger t ^a H \end{equation}

arising from the $D^a D^a$ term?

Yes, I think so.

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