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

+ 2 like - 0 dislike
138 views

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?

asked May 23, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited May 27, 2014 by JeffDror

1 Answer

+ 3 like - 0 dislike

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 40227 (4,660 points) [ revision history ]

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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