Difference between $T^{i}_{\;\;j}$ and $T_i^{\;\;j}$?

Question

Let $T_{ij}$ be a $(0,2)$ tensor. Then we defined $T^{i}_{\;\;j}=g^{ik}T_{kj}$.

What is $T_j^{\;\;i}$? Are $T_j^{\;\;i}$ and $T^{i}_{\;\;j}$ equal?

I understand that the vertical placement of indices indicates the type of the tensor ($(0,2)$ versus $(1,1)$ in my example), but what does the horizontal placement of the indices represent?

Some sources I have read simply write $T_j^i$ (as if they are disregarding the horizontal placement of indices).

What is the reason for this?

Answer 1

A $(0, 2)$-tensor $T$ on a real vector space $V$ is a multilinear map $T : V \times V \to \mathbb{R}$. Given a non-degenerate $(0, 2)$-tensor $g$ on $V$, we can define two new tensors $\widehat{T} : V^*\times V\to\mathbb{R}$ and $\widetilde{T} : V\times V^* \to \mathbb{R}$ which are determined by the identities $\widehat{T}(g(v, \cdot), w) = T(v, w)$ and $\widetilde{T}(v, g(w, \cdot)) = T(v, w)$. If $\{e_i\}$ is a basis for $V$, then $\widehat{T}^i_{\; j} = g^{ik}T_{kj}$ and $\widetilde{T}_i^{\;\; j} = g^{kj}T_{ik}$; by relabelling the indicies, we also have $\widetilde{T}_j^{\;\; i} = g^{ki}T_{jk}$.

Suppose now that $T$ is symmetric, i.e. $T(v, w) = T(w, v)$. Then, as $V\times V^*$ and $V^*\times V$ are naturally isomorphic via the map $(v, \alpha) \mapsto (\alpha, v)$, one often identifies $\widehat{T}$ and $\widetilde{T}$, and refers to this $(1, 1)$-tensor as $T$. Because the ordering of the inputs is being disregarded, we write $T^i_j$.

In general, $\widehat{T}^i_{\: j} = \widetilde{T}_j^{\:\; i}$ if and only if $T_{ij} = T_{ji}$.

Answer 2

They are the components of $$T=T^i{}_jb_i\otimes\beta^j,$$ and $$U=T_i{}^j\beta^i\otimes b_j,$$ where the base $\{b_i\}$ span the vectorspace and $\{\beta^j\}$ span its dual.