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I was taught that vectors are one rank tensors, so under diffeomorphism they do not change (their components do change but not them as tensors) they are contravariant, i.e. \begin{equation} A = A^{\mu} e_{\mu} = \frac{\partial x^{\mu}}{\partial x^{\nu}} A^{\nu} \frac{\partial x^{\rho}}{\partial x^{\mu}} e_{\rho} = A^{\nu} e_{\nu} \end{equation} But, why do the basis vectors transform but the $A$ vector doesn't? Are they different types of vectors?

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  • $\begingroup$ The tensor $A$ doesn't change - only the components of the tensor - $A^{\mu}$ - change when you change coordinate systems. $\endgroup$ – Cinaed Simson Aug 13 at 23:07
  • $\begingroup$ This is linear algebra. Do you know the difference between a vector and a basis of a vector space? $\endgroup$ – DanielC Aug 13 at 23:19
  • $\begingroup$ DanielC: A basis of a vector space is a set of linearly independent vectors that generate the whole space, in this sense I don't see a wide difference. $\endgroup$ – Geovanny Alexander Rave Franco Aug 14 at 3:13
  • $\begingroup$ Cinaed: Yes, that's my question, but I asked to my Cosmology Professor and he told me that we have to think about the vectors as differential operators and, in that sense, $A$ and $e_{\mu}$ are different differential operators so the do not have to transform in the same way. That seems a good answer for me but I'd like to read more answers. $\endgroup$ – Geovanny Alexander Rave Franco Aug 14 at 3:18
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In terms of a differential operator - also known as the natural coordinate system, I would write $A=A^{i}\frac{\partial}{\partial u^{i}}$.

Then transforming the$\frac{\partial}{\partial u^{i}}$ basis to the $\frac{\partial}{\partial w^{j}}$ basis, for instance, I would use the rule $\frac{\partial}{\partial u^{i}}=\frac{\partial w^{j}}{\partial u^{i}} \frac{\partial}{\partial w^{j}}$.

So the bottom line is, when you change the basis from $\frac{\partial}{\partial u^{i}}$ to $\frac{\partial}{\partial w^{j}}$, the transformation rule changes the coefficents from $A^{i} \rightarrow A^{i}\frac{\partial w^{j}}{\partial u^{i}} $ and $A$ is unchanged.

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