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I have consulted this topic several books, but I am still not completely clear on how the index upload and download operation works.

For example, if $T_{k}^{s r}$ are the components of a 2-contravariant 1-covariant tensor and we want to lower the two contravariant indices $s$ and $r$, we can do it using the metric tensor $g_{ij}$ in the following way:

$$ g_{i s} g_{j r} T_{k}^{s r} = g_{i s} T_{j k}^{s} = T_{i j k} $$

My doubts are these ones:

  1. Must the index we want to move always be placed in the second position of the metric tensor's components? That is, if we want to lower the index $r$ from $T_{k}^{s r}$, this wouldn't make any sense, right?

$$g_{r j} T_{k}^{s r} \overset{?}{=} T_{j k}^{s} \tag{1}$$

  1. Can we move any index of the tensor or only the rightmost one? That is, if we want to lower the index $s$ from $T_{k}^{s r}$, would it be possible to do this?

$$g_{j s} T_{k}^{s r} \overset{?}{=} T_{j k}^{r} \tag{2}$$

  1. Must the moved index always be placed at the left? That is, would this be correct?

$$g_{j r} T_{k}^{s r} \overset{?}{=} T_{k j}^{r} \tag{3}$$

  1. I have seen that in some texts it is used a notation with indents, as in equation $(4)$. What is its purpose? Does it allow to move indices that are not the rightmost one?

$$g_{j r} T^{s r k} = T^{s}{ }_{j}{}^{k} \tag{4}$$

  1. Finally, is there a straightforward way to demonstrate why this operation works?
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    $\begingroup$ It’s called “raising and lowering” for a reason. It doesn’t move indices left or right. You aren’t keeping track of which is first, which is second, etc. $\endgroup$ – G. Smith Jan 9 at 18:59
  • $\begingroup$ But, if we don't keep track of the order of the indices, in $(3)$ we could write both $T_{k j}^{r}$ and $T_{j k}^{r}$? Wouldn't we be assuming that those indices $j$ and $k$ are symmetric? $\endgroup$ – Invenietis Jan 9 at 19:48
  • $\begingroup$ In general, tensors are not symmetric or antisymmetric on any pairs of indices, so you cannot change their index order. If you know symmetry or antisymmetry exists, you can, but you should not confuse this with raising and lowering indices. $\endgroup$ – G. Smith Jan 9 at 20:19
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    $\begingroup$ One thing should also be mentioned that G.Smith sort of hints at here, many are sloppy about their placement of indices, but one should place them with appropriate spaces to see that there is only one spot you can drop an upper index to so there is less ambiguity. For instance $T_k^{sr}$ can refer to the three different tensors $T^{\; s r}_{ k}$, $T^{s\; r}_{\; k}$, or $T^{s r\; }_{\; \; k}$ $\endgroup$ – Triatticus Jan 10 at 1:41
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As already noted, it is really important too keep each index in each own "slot" which then should answer most of your questions. So you should always use the notation used in eq. 4, because then there is no ambiguity.

To answer your point 5:

Remember that at each point $p$ of your manifold $M$, the metric is a map $$ g_p \colon T_pM \times T_p M \to \mathbb{R}, $$ so you can use this to map vectors $v$, i.e. elements of $T_pM$ to covectors, i.e. elements of $T^*_pM$ via $$ v \mapsto g(v, .)=: v^\flat,$$ which acts for $w\in T_pM$ as $v^\flat(w) = g(v, w)$, which in coordinates gives you $(v^\flat)_i = g_{ij}v^j$. So in physics one drops $\flat$ and just writes $v_i = g_{ij}v^j$.
By non-degeneracy of the metric, the $\flat$-map is even an isomorphism, allowing us to identify $T_p M$ with $T^*_pM$, which is ultimately the reason we can freely lower and raise indices.
The same logic applies to covectors and the inverse metric $g^{ij}$, which then allows us to extend this action to arbitrary tensors.

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