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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 $\delta_{ij}$?

Whyka
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    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). –  Nov 10 '15 at 23:34
  • It seems that the author is using Einstein notation – Ben Grossmann Nov 10 '15 at 23:36
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    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 $i$). $A_{ii}= A_{11} + A_{22} + \cdots + A_{nn} = \operatorname{trace}(A)$. –  Nov 10 '15 at 23:51
  • @Bye_World Thank you for a good comment. So, how do I know which meaning does the notation have when I see it? – Whyka Nov 10 '15 at 23:52
  • If there are any repeated indices then a summation is implied. Otherwise there is no summation implied. So $A_{ij}$ is one entry of $A$ because there are no repeated indices but $A_{ii} = \sum_i A_{ii}$ because $i$ is a repeated index. –  Nov 10 '15 at 23:54
  • @Bye_World But if I do want to refer to the regular form of the Kronecker delta ($\delta_{ij}=1$ if $i=j$, $0$ else), how do I make sure it won't be understood as in the summation? – Whyka Nov 11 '15 at 00:06
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    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. –  Nov 11 '15 at 00:09
  • @Bye_World I see. So the Einstein notation is the "defualt", unless said otherwise. Thanks! – Whyka Nov 11 '15 at 00:13
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    @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. – not all wrong Nov 11 '15 at 00:23

1 Answers1

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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.

not all wrong
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