Confusion about the Kronecker $\delta$ Something disturbs me, concerning the Kronecker $\delta$.
Assuming these hold: $$\delta_{ij}\delta_{jk}=\delta_{ik}$$ $$\delta_{ij}=\delta_{ji}$$ $$\delta_{ii}=1$$ does it follow that for every  $\delta_{ij}$ we have $(\delta_{ij})^2=\delta_{ij}\delta_{ji}=\delta_{ii}=1$?
This makes no sense, as $\delta_{ij}$ can also be equal $0$.
Can anyone clear the confusion?

Edit: I am using Einstein notation. Do Kronecker deltas in Einstein notation always equal something different then zero? For example, if $\delta_{ii}=n$, does it imply $\delta_{ij}=n^{0.5}$ for all $\delta_{ij}$?
 A: You're confused about this notation, most likely because of what I suspect is the inconsistent use of summation convention.
The first statement is only correct if implicitly summed over $j$:
$$\sum_j \delta_{ij}\delta_{jk} = \delta_{ik}$$
The last statement however is only true if there is no summation over $i$, since generically if $i=1,2,\ldots,n$ then
$$\sum_i \delta_{ii} = n$$
This inconsistency is what is causing you problems.
(Notice for example that if $n=1$ then there is no such ambiguity but indeed it is not possible for $\delta_{ij}$ to be anything other than 1!)

Edit in response to comment: To respond to your comment, no. $\delta_{ij}$ takes different values depending on what $i,j$ are. For example in 3D,
$$\delta_{ij} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}_{ij}$$
Hence using summation convention, e.g. $\delta_{ii} \equiv \delta_{11} + \delta_{22} + \delta_{33} = 3$
Generally, $\delta_{ij}$ is just a thing that tells you whether or not $i = j$. If yes, then it is 1, if no, then it is 0. You can also think of these as the components of the identity matrix as suggested in the comments by Bye_World.
