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The Helmholtz decomposition separates the peculiar velocity into a gradient field and a transverse vector field:With decaying or periodic spatial boundary conditions this is equivalently the Fourier transform projection, for $k\ne 0$,The transverse projection operator annihilates a gradient, because $({\delta }_{ij}-k_i{k}_j/k^2)k_j=0$. Both forces in the linear cosmological Euler equation are gradients, with spatially homogeneous coefficients. Applying the transverse projection operator therefore givesSince $H=\dot{a}/a$, this is $d(a{\mathbf{v}}_{\perp })/dt=0$, and henceAt this order the scalar pressure and gravitational forces cannot generate vorticity; an initially vanishing transverse vector field remains zero. Its nonzero initial value is damped by cosmic expansion.

For $E_{\mu \nu }={\eta }_{\mu \nu }$, transversality would require $k^{\mu }{\eta }_{\mu \nu }=k_{\nu }=0$. Hence the operator has a third-order stress-tensor pole and is not a primary operator for any nonzero $k$.

Forthe two transversality conditions coincide and reduce toIf $k^2=0$, this says $k\cdot \xi =-1$, with arbitrary additional component transverse to $k$. If $k^2\ne 0$, it forcesso $E_{\mu \nu }={\eta }_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ is the transverse projection operator. Once this condition removes the third-order poles, the conformal weights are again $(1+{\alpha }^{\prime }{k}^2/4,1+{\alpha }^{\prime }{k}^2/4)$; in the null case they are $(1,1)$.