$\nabla \varphi \overset{?}{=} \nabla \cdot \varphi \bar{\bar{I}}$ where $\varphi$ is scalar, $\bar{\bar{I}}$ is identity tensor

Question

I am trying to determine if these two are equivalent. I have a function written with both terms, and this is the only discrepancy. The gradient increases the rank of the scalar to a vector, while the divergence reduces the rank of a tensor to a vector, but is this statement correct?

Answer 1

Yes. Maybe this is easiest to see in coordinates: $$g^{ij}\nabla_i(\varphi\delta)_{jk}=g^{ij}\nabla_i\varphi\delta_{jk}+g^{ij}\varphi\nabla_i\delta_{jk}=g^{ij}\nabla_i\varphi\delta_{jk}=\nabla_k\varphi.$$

Without coordinates, this would be $$\nabla\cdot\varphi\overline{I}=\operatorname{tr}_{12}\nabla\varphi\otimes \overline{I}+\varphi\nabla\overline{I}=\operatorname{tr}_{12}\nabla\varphi\otimes\overline{I}=\nabla\varphi.$$