Tensor Notation Upper and Lower Indices I want to ask what the difference between the tensors $T_i^{\; j}$ , $T_j^{\; i}$ , $T_{\; i}^{ j}$ , and $T_{\;i}^{j}$ are. In particular I am asking about the matrix representations of these tensors and their relationships. 
 A: First of all, $T_{i}^{\; j}$, $T_{j}^{\; i}$, $T^{i}_{\; j}$ and $T^{j}_{\; i}$ are not tensors, strictly speaking. They are the coordinates of the same second-order tensor $\boldsymbol{T}$ on the chosen basis $\lbrace\boldsymbol{e}_i\rbrace$  of the vector space (e.g. $\mathbb{R}^n$). More precisely, they are all mixed co- and contravariant coordinates of $\boldsymbol{T}$, but they don't have all the same variance (see e.g. [1, 2]). Indeed, if the dual basis of $\lbrace\boldsymbol{e}_i\rbrace$ is denoted by $\lbrace\boldsymbol{e}^j\rbrace$, then


*

*$T_{i}^{\; j} = \boldsymbol{T}(\boldsymbol{e}_i,\boldsymbol{e}^j)$ and $T_{j}^{\; i} = \boldsymbol{T}(\boldsymbol{e}_j,\boldsymbol{e}^i)$ are both 1-covariant 2-contravariant coordinates of $\boldsymbol{T}$. The only difference between them is the notation used for sub- and superscripts;

*$T^{i}_{\; j} = \boldsymbol{T}(\boldsymbol{e}^i,\boldsymbol{e}_j)$ and $T^{j}_{\; i} = \boldsymbol{T}(\boldsymbol{e}^j,\boldsymbol{e}_i)$ are both 1-contravariant 2-covariant coordinates of $\boldsymbol{T}$. The only difference between them is the notation used for sub- and superscripts.


Let us introduce the metric tensor $\boldsymbol{g}$. Its covariant coordinates are $g_{j\ell} = \boldsymbol{e}_j \cdot \boldsymbol{e}_\ell$, and its contravariant coordinates are $g^{ik} = \boldsymbol{e}^i \cdot \boldsymbol{e}^k$. On the one hand, using the Einstein summation convention [4] and the rule of raising and lowering indices [3], we write
\begin{aligned}
\boldsymbol{T} &= T_{i}^{\; j} \left(\boldsymbol{e}^i \otimes \boldsymbol{e}_j\right) \\
&= T_{i}^{\; j} \left( (g^{ik}\, \boldsymbol{e}_k) \otimes (g_{j\ell }\, \boldsymbol{e}^\ell )\right) .
\end{aligned}
On the other hand, we write $\boldsymbol{T} = T^{k}_{\; \ell} \left(\boldsymbol{e}_k \otimes \boldsymbol{e}^\ell\right)$. By unicity of the 1-contravariant 2-covariant coordinates of $\boldsymbol{T}$, we obtain the relationship
$$
T^{k}_{\; \ell} = g^{ik}\, g_{j\ell }\, T_{i}^{\; j}  \, .
$$

Notes:


*

*the dual basis $\lbrace\boldsymbol{e}^j\rbrace$ of $\lbrace\boldsymbol{e}_i\rbrace$ is defined by $\boldsymbol{e}^j \cdot \boldsymbol{e}_i = \delta_i^j$, where $\delta_i^j$ is the Kronecker delta.

*low and high indices are useful when working with a non-orthonormal basis. Indeed, an orthonormal basis is its own dual.

A: I've been wondering about this also, my conclusion from reading up in this pdf is that when using matrix representations, we are used to having the first component represent which row and the second which column. 
Therefore if we are multiplying $T^\mu _\nu v^\nu$ this can be visualised by a m,n matrix multiplying with a n dimentional column vector. 
As with the scalar product it is sometimes convenient to visualise convectors as row vectors, to maintain rules of matrix multiplication. Then one could write $w_\nu {A ^\nu} _\mu$ as the product of an n dimensional row vector with A a n,m matrix. But for example in some contexts one prefers to consider transposes multiplying with column vectors instead of using row vectors. Then one could write ${A_\mu} ^\nu w_\nu$ where one has essentially the transpose of an n,m matrix (m,n) post multiplied by a n dimentional column vector
Hence when writing these indeces, the first one from left to right represents the row and the second the column of the matrix, and they are up or down so that the latter one can be contracting with a corresponding lower or upper index of whatever comes after.
The difference between ${T^\mu} _\nu$ and ${T_\mu} ^\nu$ is that the first would be used to postmultiply by a contravariant vector $v^\nu$ and the second would be used to postmultiply by a covariant vector $v_\nu$ if one cared about maintaining representation with matrices and their products more visual. Then in this sense switching the order from left to right is equivalent to doing the transpose. 
If however one doesn't care about matrix representation and just works with the indeces then 
Disclaimer: I have never had any course on this and am just learning online, so please correct me if I misunderstood something
A: If the indicies are at the top or at the bottom depends how you are allowd to contract them. ( See covariance and contravariance)
For example you have two vectors $v_i$ and $u_i$ and a metrix $g^{ij}$ then you are not allowd to contract $$\sum_i v_i u_i$$ to get a meaning full scalar product, because you would get the wrong result respect to the metric. You easly check this by noting: Never contract two upper or two lower indicies. Thus you have to find a 'workaround' using the metric:
$$\sum_{ij}v_iu_jg^{ij} = \sum_{i}v_iu^i$$
As you can see at the right hand side you on lower and one upper index, which are allowd to be contracted.
A tensor $T^{ij}$ is nothing else as linear function, which pictures two vectors on a scalar (also called linear form). Similar to the example with the vectors you are not allowed to contract two upper or two lower indicies.
