Values of ##k## for which ##A_{ij}A_{ij} = |\vec a|^2##?

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The discussion centers on determining the values of k for which the equation A_{ij}A_{ij} = |\vec a|^2 holds, given the antisymmetric tensor A_{ij} = kε_{ijk}a_k. The attempt at a solution leads to A_{ij}A_{ij} resulting in zero, indicating a potential error in the calculations. Participants highlight the importance of correctly applying the summation convention, particularly with the Kronecker delta identity. The conversation emphasizes the need to reassess the numerical factors involved in the calculations to avoid discrepancies. The resolution of this problem hinges on accurately interpreting the summation convention and its implications for the tensor product.
Incand
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Homework Statement


The antisymmetric tensor is constructed from a vector ##\vec a## according to ##A_{ij} = k\varepsilon_{ijk}a_k##.
For which values of ##k## is ##A_{ij}A_{ij} = |\vec a|^2##?

Homework Equations


Identity
##\varepsilon_{ijk}\varepsilon_{klm} = \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}##

The Attempt at a Solution


##A_{ij}A_{ij} = k^2\varepsilon_{ijk}\varepsilon_{ijm}a_ka_m = k^2\varepsilon_{jki}\varepsilon_{ijm}a_ka_m = k^2(\delta_{jj}\delta_{km}-\delta_{jm}\delta_{kj})a_ka_m = k^2(\delta_{km}-\delta_{km})a_ka_m =0##
Which I obviously shouldn't get but I can't see where I'm making an error.
 
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You are using the summation convention, so
$$\delta_{jj} \equiv \sum_{j=1}^3 \delta_{jj}.$$
This gives a different numerical factor in front of the first ##\delta_{km}## term.
 
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fzero said:
You are using the summation convention, so
$$\delta_{jj} \equiv \sum_{j=1}^3 \delta_{jj}.$$
This gives a different numerical factor in front of the first ##\delta_{km}## term.
Right thanks!
 
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