How to represent tensors in a base? And some questions about indices

Question

I was reading some notes about tensor algebra. And they say that you can imagine vectors like objects independent of the coordinates. And once you choose a base for your space $V$, you can write vectors like $$\vec A= A^\mu \vec e_\mu$$ where $\{ \vec e_\mu\}$ is your base for $V$.

And in a similar way you can write a one-form like $$\widetilde A = A_\mu \widetilde e^\mu$$ where $\{ \widetilde e^\mu\}$ is the base of $V^*$ given by the base of $V$.

And I they also say that you can write tensors like $$ R_1= R_\mu \, ^\nu \, _\rho (\widetilde e^\mu \otimes \vec e_\nu \otimes \widetilde e^\rho),$$ and $$R_2 = R^\mu \, _\nu \, _\rho (\vec e_\mu \otimes \widetilde e^\nu \otimes \widetilde e^\rho). $$

In all these examples $\vec A ,\widetilde A, R_1, R_2$ are invariant under change of coordinates.

As far as I understand $\vec A \in V$, $\widetilde A \in V^*$, $R_1 \in V^* \otimes V \otimes V^*$ and $R_2 \in V \otimes V^* \otimes V^*$. But I thought that tensors are elements of the set $V \otimes ...\otimes V \otimes V^* \otimes ... V^* $ in that specific order. Now I am very confused.

Can a tensor can be an element of the tensor product between $V$ and $V^*$ in any order or is there a particular order? Also, what is the importance of the rank, since both $R_1$ and $R_2$ have rank $(1,2)$ but they are from different sets, and have indices in different places, some are contravariant while others are covariant.

And last I understand that you can change contravariant indices for covariant indices with the metric tensor $g$. Does that mean that you can interpret the metric as a function between the tensor product of some copies of $V$ and some of $V^*$. For example $$g_{\rho\pi} R^{\mu~~\pi}_{~\nu} = R^{\mu}_{~\nu\rho}$$ Could you interpret that equation as sending elements of $V \otimes V^* \otimes V$ to elements of $V \otimes V^* \otimes V^*$?

Answer 1

Usually books present the tensor products with $V$ first and $V*$ later for simplicity, but it's perfectly possible to have a tensor that takes any number of vectors and covectors in any order. For example, some people define the Riemann tensor with its upper index last: $R_{abc}^{\quad d}$. This also shows that technically the rank is not enough to identify the set of tensors you're using.

However, given any tensor with its indices in any order, you can always push all the upper indices to the left and get a uniquely defined tensor with its indices in canonical order. There is no real difference between these two tensors except for the index order, that's why books simply consider the indices to be in the standard order.

Things get tricky when you introduce a metric, because now you can move indices up and down. The metric itself is a tensor, not a function between tensor spaces; rather, it induces a function (an isomorphism) between any two spaces with the same total number of indices, by raising and lowering.

Now the problem is that there are two ways of putting indices in the standard order: by moving them around, or by raising/lowering with the metric, and these two ways produce different results. Since the latter is much more common (at least in physics), I would say that it should be the standard interpretation.