# How does vectorization affect $\nabla$?

The homework and exercise was to prove $$\nabla \times {A}$$ transform as a vector, and I've solved it thorough hard algebra.

However, something occurred to my mind and I have a hard time to resolve it. Notice that $$\tilde{\nabla} =\partial_{\tilde{\alpha}}e^{\tilde{\alpha}} =\partial_{i}(M^{-1})_{i\alpha }\delta^{\alpha\beta}(M)_{\beta j}e^{j} =\partial_{i}(M^{-1})_{i\alpha }(M)_{\alpha j}e^{j} =\partial_{i}\delta^i_je^{j} =\partial_{i} e^{i} =\nabla,$$ so it's easy to see that, in index naming notation, the component/"coefficient" of $$\nabla$$, i.e. $$\partial_{\tilde{a}}$$ transform as a covariant vector, but $$\nabla$$ it self was invariant and unique.

Thus, if one take into account of vectorization. $$\tilde{V} =\tilde{\nabla}\times\tilde{A} \Rightarrow\tilde{V}\vec{\tilde{e}} =(\tilde{\nabla}\vec{\tilde{e}})\times (\tilde{A} \vec{\tilde{e}} ),$$ notice we manually inserted $$\vec{\tilde{e}}$$ to complete the vectorization. Using the fact that both $$\nabla$$ and $$A$$ the tensor themselves were invariant, i.e. $$\nabla=\tilde{\nabla},\tilde{A}=A$$. It's easy to see that $$\tilde{V}\vec{\tilde{e}} =({\nabla}\vec{\tilde{e}})\times ({A} \vec{\tilde{{e}}} ) =({\nabla}R\vec{e})\times ({A} R\vec{{e}} ) =R({\nabla}\vec{e})\times ({A} \vec{{e}} ) =R(V\vec{{e}} ),$$ where $$R$$ being orthogonal matrix with determinant $$=1$$, i.e. a rotation.

However, somehow this doesn't make quite sense. Comparing to index naming notation, the vectorization seemed to have made the $$\nabla$$ a "contravariant component".

My question:

1. Is the above process correct? or have I had made any mistakes?

2. Is the procedure of curl $$(\nabla\times)$$ regarded as a contravariant tensor or covariant tensor?($$(\nabla\times)$$ was a pseudotensor arise from covariant tensor $$\nabla$$, but I'm wondering if the complete anti symmetric tensor $$\epsilon_{\alpha\beta\gamma}$$ may have changed covariant component into the contravariant component.)

3. How does vectorization affect $$\nabla$$?