How Do You Prove \sum_{j,k} \epsilon_{ijk} \epsilon_{ljk} = 2\delta_{il}?

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Homework Help Overview

The problem involves proving the identity \(\sum_{j,k} \epsilon_{ijk} \epsilon_{ljk} = 2\delta_{il}\), which relates to the properties of the Levi-Civita symbol and Kronecker delta in tensor calculus.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of substituting equal subscripts in the Kronecker delta and how this affects the overall expression. Questions arise about the interpretation of these substitutions and their role in achieving the desired result.

Discussion Status

Some participants have offered guidance on simplifying the expression by replacing equal subscripts with 1. There is an ongoing exploration of the relationships between different Kronecker delta expressions, but no consensus has been reached on the overall proof.

Contextual Notes

Participants are navigating the complexities of tensor notation and the properties of the Levi-Civita symbol, with some assumptions about the indices being equal or not being questioned.

cashmerelc
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Homework Statement


Prove \sum_{j,k} \epsilon_{ijk} \epsilon_{ljk} = 2\delta_{il}

Homework Equations


\epsilon_{ijk} \epsilon_{ljk} = \delta_{il}(\delta_{jj}\delta_{kk} - \delta_{jk}\delta_{kj}) + \delta_{ij}(\delta_{jk}\delta_{kl} - \delta_{jl}\delta_{kk}) + \delta_{ik}(\delta_{jl}\delta_{kk} - \delta_{jj}\delta_{kl})<h2>The Attempt at a Solution</h2><br /> <br /> Okay, in cases where subscripts of the Kronecker delta are equal, then \delta_{jj} = 1. <br /> <br /> If the subscripts are not equal, then \delta_{il} = 0. <br /> <br /> So plugging those into the parenthesis of the above equation gives me:<br /> <br /> \delta_{il}(\delta_{jj}\delta_{kk}) ?<br /> <br /> If that is the case, then how could the two inside the parenthesis equal 2? I know I must be missing something.
 
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In your formula, replace the \delta_{jj}, \delta_{kk} etc... where the variables are the same... with 1.

Also, \delta_{ij}\delta_{jk} = \delta_{ik}
 
learningphysics said:
In your formula, replace the \delta_{jj}, \delta_{kk} etc... where the variables are the same... with 1.

Also, \delta_{ij}\delta_{jk} = \delta_{ik}

If \delta_{ij}\delta_{jk} = \delta_{ik} does that mean that \delta_{lk}\delta_{kj} = \delta_{lj} and so on?
 
cashmerelc said:
If \delta_{ij}\delta_{jk} = \delta_{ik} does that mean that \delta_{lk}\delta_{kj} = \delta_{lj} and so on?

Yes, exactly.
 
Okay, I think one more question will help me get it.

\delta_{jk}\delta_{kj} = ?
 
cashmerelc said:
Okay, I think one more question will help me get it.

\delta_{jk}\delta_{kj} = ?

= \delta_{jj} = 1
 

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