differential geometry - Difference between matrix representations of tensors and $delta^{i}_{j}$ and $delta_{ij}$?

My question basically is, is Kronecker delta $\delta_{ij}$ or $\delta^{i}_{j}$. Many tensor calculus books (including the one which I use) state it to be the latter, whereas I have also read many instances where they use the former. They cannot be the same as the don't have the same transformation laws. What I think is that since

$\delta_{j}^{i}$ = ($\delta_{j}^{i}$)$'$, but $\delta_{ij}$ doesn't, so the latter cannot be a tensor. But the problem is that both have the same value :- ($1,0$) depending upon the indices. So it makes me think that $\delta_{ij}$ is just the identity matrix $I$ and not a tensor, and $\delta^{i}_{j}$ is a function. But since $\delta_{j}^{i}$ also has the same output as $\delta_{ij}$, WHAT IS THE DIFFERENCE?

I think it could be matrix representations. IN GENERAL, is there a difference between matrix representations of $\delta_{ij}$, $\delta^{ij}$ and $\delta_{j}^{i}$ (or any other tensor for that matter). Please answer these (the difference between mixed indices AND matrix representations).

Answer

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.]

II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing

$$\tag{1} V \otimes V^{*}~\stackrel{\delta}{\longrightarrow}~ \mathbb{F}$$

$$ v\otimes f ~\stackrel{\delta}{\mapsto}~ f(v) , \quad v\in V, \quad f\in V^{*},$$

between an $\mathbb{F}$-vector space $V$ and its dual vector space $V^{*}$.

III) If we choose a basis $(e_i)_{i\in I}$ for $V$, there is an dual basis $(e^{j*})_{j\in I}$ for $V^{*}$ such that

$$\tag{2} e^{j*}(e_i)~=~\delta^j_i~:=~\left\{ \begin{array}{rcl} 1 &\text{for}& i=j, \\ \\ 0 &\text{for}& i\neq j. \end{array}\right. $$

[Here we distinguish between covariant and contravariant indices.] Then for a vector $v=\sum_{i\in I} v^i e_i\in V$ and a covector $ f=\sum_{j\in I} f_j e^{j*}\in V^{*}$, the contraction map (1) is

$$\tag{3} \delta(v,f)~=~ \sum_{i,j\in I} f_j \delta^j_i v^i~=~ \sum_{i\in I}f_i v^i. $$

In other words, $\delta^j_i$ is the matrix representation for the $\delta$ contraction map (1). It's an interesting fact that the matrix representation is independent of the choice of basis $(e_i)_{i\in I}$ for $V$, as long as we choose the corresponding dual basis for $V^{*}$ in the natural way. We often say that $\delta^j_i$ transforms as a tensor, or is a tensor.

IV) Now what about $\delta_{ij}$ with lower indices? Well, first we must introduce a symmetric bilinear form, or metric,

$$\tag{4} V\times V ~\stackrel{g}{\longrightarrow}~ \mathbb{F} $$ $$ g(v,w)=g(w,v) .$$

If we choose a basis $(e_i)_{i\in I}$ for $V$, then we can write

$$ \tag{5} g ~=~ \sum_{i,j\in I} g_{ij}~ e^{i*}\otimes e^{j*} .$$

Often we will choose a metric which is the unit matrix in a certain basis

$$\tag{6} g_{ij} ~=~\delta_{ij}~:=~\left\{ \begin{array}{rcl} 1 &\text{for}& i=j, \\ \\ 0 &\text{for}& i\neq j. \end{array}\right. $$

If we now choose another basis then the matrix representation $g_{ij}$ for the metric (4) will in general change. It will in general no longer be the unit matrix $\delta_{ij}$. We say that $\delta_{ij}$ does not transform as a tensor under general change of bases/coordinates.

In a nutshell, $\delta_{ij}$ with lower indices implicitly signals the presence of a metric (4), or in other words, a notion of length scale in the vector space $V$. It is important to realize that the choice of a metric (4) in $V$ is a non-canonical choice.

V) However, once we are given a metric $g$, it is natural to study changes of bases/coordinates that preserve this metric $g$. These correspond to orthogonal transformations and $\delta_{ij}$ behaves as a covariant tensor under such orthogonal transformations.