# State counting in the d = 1+2, $\cal{N} = 2$ vector multiplet

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The question is from Box 8.2, page 282 of the book "Gauge Gravity Duality" by Ammon and Erdmenger. The link to the specific page from Google Books is here.

According to the authors, a $\mathcal{N} = 2$ vector superfield includes a vector potential $A_\mu$, a real scalar field $\sigma$, two real (Majorana) gauginos, and an auxiliary real scalar field $D$, all in the adjoint representation of the gauge group.

I am not sure how the counting works:

• Vector potential $A_\mu$ has $(3-2) = 1$ (bosonic) degree of freedom, as a gauge field in $d = 1 + 2$ dimensions.

• A real scalar field has $1$ (bosonic) degree of freedom.

• Two real Majorana gauginos have $2 \times 2^{(3-1)/2}$ (real) fermionic degrees of freedom, i.e. $4$ fermionic degrees of freedom.

• An auxiliary real field has $1$ bosonic degree of freedom.

The number of fermionic and bosonic components do not match.

This post imported from StackExchange Physics at 2015-11-13 21:41 (UTC), posted by SE-user leastaction
Why do you claim the gauge field has only one d.o.f. in 3+1D? A gauge field has $d-2$ d.o.f. for $d$ the number of spacetime dimensions.

This post imported from StackExchange Physics at 2015-11-13 21:41 (UTC), posted by SE-user ACuriousMind
The spacetime is 3 dimensional..d=1+2. So d-2=1.

This post imported from StackExchange Physics at 2015-11-13 21:41 (UTC), posted by SE-user leastaction
Sorry I corrected the typos. What does 0 bosonic degrees of freedom mean?

This post imported from StackExchange Physics at 2015-11-13 21:41 (UTC), posted by SE-user leastaction
Okay, if you are in 2+1D, a gauge field has indeed only one degree of freedom (your post currently says 0, you introduced another typo). The case of 1+1 dimensions would mean the gauge theory is perturbatively trivial, and the only non-trivial states are topological/global.

This post imported from StackExchange Physics at 2015-11-13 21:41 (UTC), posted by SE-user ACuriousMind

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