Scalar expansion perturbation ($\kappa =\delta K$)

With negative extrinsic curvature convention, the background trace is $\bar{K}=-3H$. The scalar expansion perturbation is $\kappa =K+3H$. For lapse function $N=\bar{N}(1+\Psi )$, induced metric $h_{ij}=a^2[(1-2\Phi ){\delta }_{ij}+2E_{,ij}]$ and shift vector $N_i=-a^2{B}_{,i}$, it is $\kappa =3(\dot{\Phi }/\bar{N}+H\Psi )-\Delta \chi$, where $\chi$ is the scalar shear potential. The expansion of the future normal congruence is $-K$, so its perturbation has the opposite sign.

For $N_i=a^2{B}_{,i}$, ${\gamma }_{ij}=a^2[(1-2\Phi ){\delta }_{ij}+2E_{,ij}]$ and $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, put ${\chi }_{+}=a^2(B-\dot{E})/\bar{N}$. Then $K^i{}_j=-H{\delta }^i{}_j+(\dot{\Phi }/\bar{N}+H\Psi ){\delta }^i{}_j+{\partial }^i{\partial }_j{\chi }_{+}$ and $\kappa =3(\dot{\Phi }/\bar{N}+H\Psi )+\Delta {\chi }_{+}$. Defining $\chi =-{\chi }_{+}$ instead makes the trace-free tensor coefficient negative and the trace gradient term $-\Delta \chi$. A sign change must affect both pieces.