Can Two Levi-Civita Symbols be Reduced to One with Indices?

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SUMMARY

The discussion centers on the reduction of two Levi-Civita symbols, specifically ## \epsilon^{ckl} \epsilon_{ibj} ##, to a single Levi-Civita symbol ## \epsilon_{kij} ## in the context of matrix multiplication involving antisymmetric matrices A and B. The indices b and c range from 4 to 5, while the other indices run from 1 to 3. The participants explore the properties of the Levi-Civita symbol and its contraction with tensors, questioning the feasibility of simplifying the expression while maintaining the integrity of the indices involved.

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  • Understanding of Levi-Civita symbols and their properties
  • Familiarity with tensor algebra and matrix multiplication
  • Knowledge of antisymmetric matrices and their characteristics
  • Basic concepts of index notation in mathematical physics
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This discussion is beneficial for physicists, mathematicians, and students engaged in advanced topics in tensor analysis and those working with antisymmetric matrices in theoretical frameworks.

Safinaz
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Hi there,

Can two Levi-Civia symbols ## \epsilon^{ckl} \epsilon_{ibj} ## reduced to one with indices ## \epsilon_{kij} ## ?

WHERE b and c run from 4 to 5, the other indices run from 1 to 3 and both symbols are multiplied by the following matrices:

## A_{ib}^c ~ B^{kl} ~ C_j ~\epsilon^{ckl} \epsilon_{ibj} ~ [1] ##,

## B^{kl}## is antisymmetric in k and l, and A is antisymmetric in b and c ..

So I hope my question is clear enough .. in summary, can expression [1] written in the forum : ## A_i ~ B_k ~ C_j ~ \epsilon_{ikj} ## ?

I think doing this needs with using Levi-Civita symbol properties , to use some direct products of matrices, as: ## 3 \times 3 = 3^*_{Antisymm}+6_{Symm} ##, and ## 2 \times 2 = 1_A +3_S ## .

Any ideas ?
Thanx.
 
Last edited:
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Are you sure that this is even possible? you are dropping away free indices,..
 
mmmm .. which one ? in [1] all the indices (i,b,j,k,l,c) are contacted .

Can we use here Levi-Civita symbol contraction with second or third rank tensor ? if there any relation like in the metric tensor ## A^\alpha = g^{\alpha\beta} A_\beta ## ..
 
Last edited:

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