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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 $\delta_{ij}$?

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    $\begingroup$ Are you using the Einstein summation convention? The first equation you list is only true if you are using it. The third equation is only true if you're not (or if your index set only has one element). $\endgroup$ – user137731 Nov 10 '15 at 23:34
  • $\begingroup$ It seems that the author is using Einstein notation $\endgroup$ – Omnomnomnom Nov 10 '15 at 23:36
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    $\begingroup$ Given two matrices $A$ and $B$, $A_{ij}$ is one entry of $A$ and $B_{jk}$ is another entry of $B$. Matrix multiplication is defined entrywise as $$[AB]_{ik} = \sum_j A_{ij}B_{jk}$$ But using the Einstein summation convention, the RHS can just be written $A_{ij}B_{jk}$ where we see that we're supposed to sum over $j$ because there are two of them. If you see an expression like $A_{ii}$ then you need to realize that this is NOT one entry of $A$ -- this has an implied summation as well (over $i$). $A_{ii}= A_{11} + A_{22} + \cdots + A_{nn} = \operatorname{trace}(A)$. $\endgroup$ – user137731 Nov 10 '15 at 23:51
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    $\begingroup$ Well if you put repeated indices anywhere in a term -- for instance $\delta_{ij}v_i$ has a repeated $i$ -- then readers will assume that you mean to imply summation over the $i$ UNLESS you write something like "summation not implied" afterward. $\endgroup$ – user137731 Nov 11 '15 at 0:09
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    $\begingroup$ @Whyka: One thing I will say as a warning is that whether or not people assume S.C. is being used can depend on context. As a rule of thumb, applied mathematicians and physicists will assume it's being used; but pure mathematicians will commonly state clearly if it is being used, and otherwise will not use it. $\endgroup$ – Sharkos Nov 11 '15 at 0:23
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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.

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  • $\begingroup$ So, if I use Einstein notation, then $\delta_{ij}$ always equals $3^{0.5}$ in 3-D? $\endgroup$ – Whyka Nov 10 '15 at 23:39
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    $\begingroup$ Maybe it would help you to understand the $\delta_{ij}$ just means the entry in the $i$th row and $j$th column of the **identity matrix**. So of course $\delta_{ij}\ne 3^{1/2}$. $\endgroup$ – user137731 Nov 10 '15 at 23:42
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    $\begingroup$ @Whyka: I've responded in the question because it's much easier that way. $\endgroup$ – Sharkos Nov 10 '15 at 23:43
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    $\begingroup$ Because we are looking at an 1*1 matrix? $\endgroup$ – Whyka Nov 10 '15 at 23:48
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    $\begingroup$ @Whyka Exactly! Another way to look at it is that if $i,j\in \{1\}$ then $i=j$ always! $\endgroup$ – Sharkos Nov 11 '15 at 0:13

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