Kronecker Delta summation (easy)

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    Delta Summation
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Discussion Overview

The discussion revolves around the properties and summation rules of the Kronecker delta, specifically the expression ##\delta_{ij} \delta_{jk} = \delta_{ik}##. Participants explore the implications of summing over repeated indices and clarify the conditions under which such summation applies.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why summation occurs over the repeated index ##j## in the expression ##\delta_{ij} \delta_{jk}##.
  • Another participant explains that the summation over ##j## results in a compact form, leading to ##\delta_{ik}## when ##i=k##.
  • A different participant presents a counterexample using specific values for ##i,j,k##, suggesting that the sum does not yield ##\delta_{ik}## and questions the reasoning.
  • Further responses reiterate the summation process and clarify that only the index ##j## is summed, while ##i## and ##k## are not summed in the original expression.
  • One participant acknowledges their misunderstanding after reviewing the responses.

Areas of Agreement / Disagreement

Participants express differing views on the summation of indices, with some supporting the standard interpretation while others challenge it with specific examples. The discussion remains unresolved regarding the implications of the summation in the context provided.

Contextual Notes

There are assumptions about the ranges of indices and the specific conditions under which the Kronecker delta operates. The discussion does not resolve the mathematical steps or the implications of the counterexamples presented.

member 428835
Hi PF!

As outlined in my book ##\delta_{ij} \delta_{jk} = \delta_{ik}## but don't we sum over repeated indices (and the ##j## is repeated)? Can someone explain why we do not sum in this situation?

Thanks!
 
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Yes it is summed over ##j##. Written explicitly, the sum looks like
$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$
You see that if ##i\neq k##, all terms above will vanish. If ##i=k## there will be only one term surviving, therefore that sum can be written compactly as ##\delta_{ik}##.
 
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Thanks for responding blue_leaf77! Let's assume ##i,j,k## vary from ##1,2,3##. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?
 
joshmccraney said:
Thanks for responding blue_leaf77! Let's assume ##i,j,k## vary from ##1,2,3##. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?
Why would ##i## and ##k## be summed?

And, if you sum them on the right hand side you would get equality.
 
Last edited:
PeroK said:
Why would ##i## and ##k## be summed?
That's what blue_leaf wrote.
blue_leaf77 said:
$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$

Since ##i## and ##k## can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.
 
joshmccraney said:
That's what blue_leaf wrote.Since ##i## and ##k## can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.
blue leaf summed only ##j##, which is a repeated index.
 
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PeroK said:
blue leaf summed only ##j##, which is a repeated index.
Ohhhhh I see now! Shoot, yea I was doing it all wrong! Thank you both!
 

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