# A question on supersymmetry variation of the Wilson loop in $\mathcal{N}=4$ SYM

The Wilson loop in $$\mathcal{N}=4$$ SYM is $$W=\frac{1}{N}tr P \exp \int ds (i A_\mu(x) \dot{x}^\mu+\Phi_i(x)\theta^i|\dot{x}|).\tag{2.3}$$

In order to check whether this operator is supersymmetric I will use a variation. We know that the supersymmetry transformations of the bosonic fields are $$\delta_\epsilon A^\mu=\bar{\Psi}\Gamma^\mu \epsilon\tag{2.2a}$$ $$\delta_\epsilon\Phi^i=\bar{\Psi}\Gamma^i \epsilon\tag{2.2b}$$

The supersymmetry variation of the Wilson loop is $$\delta_\epsilon W=\frac{1}{N}tr P \int ds\bar{\Psi}(i \Gamma_\mu \dot{x}^\mu+\Gamma_i\theta^i|\dot{x}|)\epsilon \exp \int ds' (i A_\mu(x) \dot{x}^\mu+\Phi_i(x)\theta^i|\dot{x}|).\tag{2.4}$$ The references I have checked (e.g. arXiv:hep-th/0205160) say that vanishing of the supersymmetry variation of the Wilson loop require $$(i \Gamma_\mu \dot{x}^\mu+\Gamma_i\theta^i|\dot{x}|)\epsilon=0.\tag{2.5}$$

Why not

$$\bar{\Psi}(i \Gamma_\mu \dot{x}^\mu+\Gamma_i\theta^i|\dot{x}|)\epsilon=0\tag{A}$$ or $$i \Gamma_\mu \dot{x}^\mu+\Gamma_i\theta^i|\dot{x}|=0~?\tag{B}$$

## 1 Answer

Briefly, eq. (2.5) is preferred over eq. (A) because it only depend on intrinsic Wilson loop data and not the fermion $$\bar{\Psi}$$. Eq. (B) is too strong and does not have non-trivial solutions.

• Thanks for the answer. Sorry, how can we be sure that there are no solutions for (B) ? – physics_teacher Jun 6 at 20:34
• Because the Gamma matrices are lin. indep. – Qmechanic Jun 6 at 22:58