This is related to this earlier question I had asked.

I am using the so called ``Majorana" representation of gamma matrices in $2+1$ dimensions in which everything is real. After doing the dimensional reduction of the $\cal{N}=1$ supersymmetry transformations of the components of the vector superfield in $3+1$ dimensions the supersymmetry transformations of the resulting $\cal{N}=2$ components of the vector superfield in $2+1$ dimensions are,

$$\delta F_A = i \bar{\alpha}^a\lambda_{Aa}$$ $$\delta D_A = i \bar{\alpha}^a\gamma_3^\mu D_\mu \lambda_{Aa}$$ $$\delta V_{A\mu} = i \bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}$$ $$\delta \lambda_{Aa} = -\frac{1}{2}f^3_{A\mu \nu}\gamma_3^{\mu \nu}\alpha_a + D_A\alpha^a + \gamma_3^\mu D_\mu F_A\alpha ^a$$

where $A,B,..$ are the gauge group indices, $f^3_{A\mu \nu}$ is the non-Abelian field strength and $\alpha$ is a spinor parameter whose components are raised and lowered as, $\alpha^1 = \alpha_2$ and $\alpha^2 = -\alpha_1$.

Using the above one can derive the following transformations for the possible terms in the intended super-Chern-Simons' theory,

$$\delta(Tr[FD]) = Tr[t_At_B] \{ i\bar{\alpha}^a \lambda _{Aa}D_B - i \bar{\alpha}^a \gamma_3^\mu\lambda_{Ba}\partial_\mu F_A + i\bar{\alpha}^a \gamma_3^\mu \lambda_{Ca} C_{BB'C}V_{B'\mu}F_A\}$$

$$\delta(Tr[\bar{\lambda}_a\lambda_a]) = 2Tr[t_A t_B]\{ \frac{1}{2}\bar{\alpha}_a\gamma_{3\rho}\lambda_{Ba}f^3_{A\mu\nu}\epsilon ^{\mu \nu \rho} + \bar{\alpha}^a\lambda_{Ba}D_A - \bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}\partial_\mu F_A$$

$$- \bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB'C}V_{B'\mu}F_C \}$$

$$\delta(Tr[\epsilon^{\mu \nu \rho}(V_\mu \partial_\nu V_\rho)]) = -i\epsilon^{\mu \nu \rho}Tr[t_At_B]\bar{\alpha}^a\gamma_{3\rho}\lambda_{Aa}(\partial_\mu V_{B\nu} - \partial_\nu V_{B\mu})$$

$$\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho]) = -\frac{3}{2}\epsilon^{\mu \nu \rho}Tr[t_At_D]C_{DBC}\bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}V_{B\nu}V_{C\rho}$$

(where $t_A$ are a chosen basis in the lie algebra of the gauge group such that the structure constants are defined as, $[t_A,t_B]=iC_{DAB}t_D$)

It is clear that by choosing a coefficient of $-2$ for the $Tr[FD]$ and $i$ for the $Tr[\bar{\lambda}_a\lambda_a]$ some of the terms the variation of the auxiliary fields can be cancelled and some of the remaining terms of the variation of the fermionic term totally cancel the supersymmetric variation of the kinetic term of the gauge fields.

What remains are,

$$\delta(Tr[\epsilon^{\mu \nu \rho}V_\mu\partial _ \nu V_\rho + i \bar{\lambda}_a\lambda_a - 2FD]) = Tr[t_At_B]\{i\bar{\alpha}_a\gamma_{3\rho}\lambda_{Aa}C_{BCD}V_{C\mu}V_{D\nu}\epsilon^{\mu \nu \rho}$$ $$ - 2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB'C}V_{B'\mu}F_C - 2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ca}C_{BB'C}V_{B'\mu}F_A\}$$


$$\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho]) = -\frac{3}{2}Tr[t_At_B]\bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}C_{BCD}V_{C\nu}V_{D\rho}\epsilon^{\mu \nu \rho}$$

  • Its not clear that a coefficient can be chosen for the last term so that the supersymmetric variation of the sum of the LHSs go to zero.

The above terms seem to be structurally very different and hence its not clear how they will cancel. Like the variation of the fermionic self-coupling term produces a coupling of the fermionic component, gauge field and auxiliary field. Such a term is not produced by the variation of the gauge field cubed term!

One expects the lagrangian should look something like,

$$Tr[\epsilon^{\mu \nu \rho}(V_\mu\partial _\nu V_\rho -i\frac{2}{3}V_\mu V_\nu V_\rho ) + i \bar{\lambda}_a\lambda_a - 2FD ]$$

I would like to get some help in establishing the above!

  • One progress would be if the two terms with the structure constant actually cancel i.e,


$$Tr[t_At_B]\{C_{AB'C}\lambda_{Ba} V_{B'\mu}F_C + C_{BB'C}\lambda_{Ca}V_{B'\mu}F_A\} = 0$$

But the above is not clear!

NB. My structure constants are defined as $[t_A,t_B]=iC_{DAB}t_D$


2 Answers 2


Without going through and doing the calculation myself, I can only make some general comments.

Your index contractions seem a little weird. In the first term on the RHS of the $$\begin{align} &\delta(Tr[\epsilon^{\mu \nu \rho}V_\mu\partial _\nu V_\rho + i \bar{\lambda}_a\lambda_a - 2FD]) \\ = &Tr[t_At_B]\{i\bar{\alpha}_a\gamma_{3\rho}\lambda_{Aa}C_{ABC}V_{B\mu}V_{C\nu} - 2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB'C}V_{B'\mu}F_C \} \,, \end{align} $$ the Lorentz indices are not contracted, are you missing a $\epsilon^{\mu \nu \rho}$? Also, in the same term you have triply repeated gauge indices (i.e. the same index appears 3 times), which is unpleasant.

If you fix the above, and maybe use the symmetry of the $Tr[t_At_D]$ term to move the $D$ index in the RHS of the cubic term $\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho])$, then maybe you can get it cancel the first term talked about above.

Finally, the second term on the RHS of the above displayed equation can not be canceled by anything else. So check your result for $\delta(Tr[\bar{\lambda}_a\lambda_a])$ - maybe the trouble term is meant to vanish... Are you sure that your Susy variations are correct? Have you got a reference (e.g. http://arxiv.org/abs/hep-th/9506170) you can check against?

  • $\begingroup$ you have triply repeated gauge indices, which is unpleasant ... @Simon this is the Einstein summation convention and is used all the time. Not sure what you mean by "unpleasant". $\endgroup$
    – user346
    Mar 16, 2011 at 0:47
  • $\begingroup$ @Deepak: Normally you only sum over doubly repeated indices - sticking to this convention avoids strange errors. In some situations you can bend that rule, such as when you have a sign factor that depends on the index... $\endgroup$
    – Simon
    Mar 16, 2011 at 0:53
  • $\begingroup$ @Deepak: In the 2nd term you have B' indices, but in the first you simply repeat the A and B indices. I thought perhaps it was by mistake. You don't seem to have chosen a basis for your gauge group (or at least the metric for your basis), so there's no reason to assume that they are trace orthogonal. Maybe you should choose $tr(t_a t_b)\propto\delta_{ab}$ and simplify the expressions. $\endgroup$
    – Simon
    Mar 16, 2011 at 0:55
  • $\begingroup$ @Simon let me give a simple example. Given three vectors, the expression $\textbf{A}\cdot\textbf{B}\times\textbf{C}$ gives the volume of the parallelepiped defined by these vectors. One can also write this as $\epsilon^{abc} A_a B_b C_c$ in index notation. $\endgroup$
    – user346
    Mar 16, 2011 at 0:57
  • $\begingroup$ @Deepak: I'm not stupid. $\endgroup$
    – Simon
    Mar 16, 2011 at 0:57

@Deepak: triply repeated is operationally well-defined as part of Einstein summation convention, but is meaningless in general: contractions are supposed to be group invariants, and in that sense such triple "contractions" are meaningless.


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