[Physics] Divergence of cross product of transverse component

Question

If I define the vector as $V_i=V^T_i+V^L_i$ and the transverse part is defined by $$V^T_i=\Big(\delta_{ij}-\frac{\partial_i\partial_j}{\partial^2}\Big)V_j$$ then is is obvious that $\nabla.V^T=0$ as well as $\nabla \times V^L=0$. What happened if I took the the divergence of the cross product of two different vectors with only transverse component? Are they zero too?
For example is $\nabla.(A^T\times V^T)=0?$

Answer 1

The identity you're looking for is $$ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) $$ so indeed, if the curl of both factors in the cross product vanish, the divergence of the cross product vanishes, too.