# Supersymmetry Generator Definition for ${\cal N }= 1$

I am studying SYM $$\mathcal{N}$$ = 1 in D = 10, and using the bimodular representations for the 32x32 gamma matrices $$\Gamma^a$$. This means that I work with the off-diagonal 16x16 matrices, which I call $$\gamma^a$$. This notation is not the most conventional, but was used in "Superspace formulation of ten-dimensional N=1 supergravity coupled to N=1 super Yang-Mills theory" by Atick, Dhar and Ratra, for example. The action for abelian SYM (in flat space) is

$$S = \int d^{10} x \{ -\frac{1}{4} F^{a b}F_{a b} - \frac{1}{2} \chi^\alpha (\gamma^a)_{\alpha \beta} \partial_a \chi^\beta \} \ \ .$$

which I have proven to be invariant under

$$\delta_\epsilon A_a = \epsilon^\alpha (\gamma_a)_{\alpha \beta} \chi^\beta \ \ , \ \ \delta_\epsilon \chi^\alpha = -\frac{1}{2} F_{a b} (\gamma^{a b})^\alpha_{\ \beta} \epsilon^\beta \ \ .$$

The problem is that when I calculate $$[\delta_1, \delta_2]A_a = [ \epsilon^\alpha Q_\alpha, \epsilon^\beta Q_\beta ]A_a$$ (same for $$\chi^\alpha$$), the algebra closes to $$\{ Q_\alpha, Q_\beta\} = -2 (\gamma^a)_{\alpha \beta} \partial_a$$ (whereas the usual is $$\{ Q_\alpha, Q_\beta\} = -2i (\gamma^a)_{\alpha \beta} \partial_a$$).

So far so good I guess, but when I have to define $$Q_\alpha$$ to satisfy that anticommutator, I have to define it as $$Q_\alpha = \partial_\alpha - (\gamma^a)_{\alpha \beta} \theta^\beta \partial_a$$ (and every book I checked defines it as proportional to $$Q_\alpha = \partial_\alpha - i(\gamma^a)_{\alpha \beta} \theta^\beta \partial_a$$).

My question is: Is it valid to define $$Q_\alpha$$ without the $$i$$? It seems consistent, but the fact that I haven't seem that convention anywhere troubles me.

• if an $i$ is mandatory in the definition between $\partial_\alpha$ and $\partial_a$, I would like to know the reason. Thanks! – Luke Jun 14 at 3:47
• Just a remark: There are potential factors of $\text{i}$ in the $\gamma$s depending on your sign convention for the metric and the Dirac algebra (i.e. whether $\Gamma^0$ is Hermitean or anit-Hermitean). – Toffomat Jun 14 at 11:48