A question of BRST symmetry of bosonic string theory

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I tried to verify Eq. (4.2.7) in Polchinski's string theory book vol I p. 127 but I miserably miss a sign

$$\delta_B (b_A F^A) = i \epsilon (S_2 + S_3) \tag{4.2.7}$$

Necessary equations: $$S_2 =-iB_A F^A (\phi) \tag{4.2.4}$$ $$S_3 = b_A c^{\alpha} \delta_{\alpha} F^A(\phi) \tag{4.2.5}$$ $$\delta_B \phi_i = -i \epsilon c^{\alpha} \delta_{\alpha} \phi_i \tag{4.2.6a}$$ $$\delta_B B_A=0 \tag{4.2.6b}$$ $$\delta_B b_A = \epsilon B_A \tag{4.2.6c}$$ $$\delta_B c^{\alpha} = \frac{i}{2} \epsilon f^{\alpha}_{\beta \gamma} c^{\beta} c^{\gamma} \tag{4.2.6d}$$

My attempt to verify Eq (4.2.7), $$\delta_B b_A F^A = (\delta_B b_A) F^A + b_A (\delta_B F^A)$$ $$= \epsilon B_A F^A + b_A (\partial^i F^A) \delta_B \Phi_i = \epsilon B_A F^A + b_A (\partial^i F^A) (-i \epsilon c^{\alpha} \delta_{\alpha} \phi_i)$$ $$= \epsilon B_A F^A - i \epsilon b_A c^{\alpha} \delta_\alpha F^A = i \epsilon (S_2 - S_3)$$

Why did I get $-S_3$?

1 Answer

I got it. $\epsilon$ anticommutes with $b_A$.