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 $\nabla$?


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