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Showing Levi-Civita properties in 4 dimensions

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  1. Apr 23, 2017 #1
    first of all english is not my mother tongue sorry. I want to ask if you can help me with some of the properties of the levi-civita symbol.

    I am showing that

    $$\epsilon_{ijkl}\epsilon_{ijmn}=2!(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$

    so i have this...

    $$\epsilon_{ijkl}\epsilon_{ijmn}=\delta_{ii}\delta_{jj}\delta_{km}\delta_{ln}+\delta_{ij}\delta_{jm}\delta_{kn}\delta_{li}+\delta_{im}\delta_{jn}\delta_{ki}\delta_{lj}+\delta_{in}\delta_{ji}\delta_{kj}\delta_{lm}-\delta_{ii}\delta_{jn}\delta_{km}\delta_{lj}-\delta_{in}\delta_{jm}\delta_{kj}\delta_{li}-\delta_{im}\delta_{jj}\delta_{ki}\delta_{ln}-\delta_{ij}\delta_{ji}\delta_{kn}\delta_{lm}=3^2\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}+\delta_{km}\delta_{ln}+\delta_{kn}\delta_{lm}-3\delta_{km}\delta_{ln}-\delta_{km}\delta_{ln}-3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3\delta_{km}\delta_{ln}-3\delta_{kn}\delta_{lm}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$

    which is not equal to the supposedly correct answer. can the statement be wrong?
    if not i can't see my error.

    Also i need to prove that:

    $$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
    i know and i proved that
    $$\epsilon_{ijkl}\epsilon_{ijkl}=24=4!$$
    so if $l=m$ i have this
    $$\epsilon_{ijkl}\epsilon_{ijkl}=3!\delta_{ll}=3!\cdot 3=18$$
    which is not equal to 4!=24 so its the statement wrong again? it must be
    $$\epsilon_{ijkl}\epsilon_{ijkm}=8\delta_{lm} $$
    Actually i proved that:
    $$\epsilon_{ijkl}\epsilon_{ijkm}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{lm}+\delta_{ij}\delta_{jk}\delta_{km}\delta_{li}+\delta_{ik}\delta_{jm}\delta_{ki}\delta_{lj}+\delta_{im}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jm}\delta_{kk}\delta_{lj}-\delta_{im}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{lm}-\delta_{ij}\delta_{ji}\delta_{km}\delta_{lk}=3^3\delta_{lm}+\delta_{ik}\delta_{km}\delta_{li}+\delta_{ii}\delta_{lj}\delta_{jm}+\delta_{jm}\delta_{kj}\delta_{lk}-3^2\delta_{lm}-\delta_{jj}\delta_{li}\delta_{im}-3\delta_{ii}\delta_{lm}-\delta_{ii}\delta_{lm}=3^3\delta_{lm}+\delta_{lk}\delta_{km}+\delta_{ii}\delta_{lm}+\delta_{lj}\delta_{jm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=3^3\delta_{lm}+\delta_{lm}+3\delta_{lm}+\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}-3^2\delta_{lm}-3\delta_{lm}=8\delta_{lm} $$
    Even more if i use my solution:

    $$\epsilon_{ijkl}\epsilon_{ijmn}=3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
    to prove
    $$\epsilon_{ijkl}\epsilon_{ijkm}=3!\delta_{lm}$$
    using $m=k$ and $n=m$ i have this
    $$3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
    $$3(\delta_{kk}\delta_{lm}-\delta_{km}\delta_{lk})$$
    $$3(3\delta_{lm}-\delta_{lm})$$
    $$3(2\delta_{lm})$$
    $$6(\delta_{lm})=3!(\delta_{lm})$$
     
  2. jcsd
  3. Apr 23, 2017 #2

    Dr Transport

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  4. Apr 23, 2017 #3
    That's what i did but when i simplify i get the error and i don't know where i think that the statement is false because i repeat the product too many times and i always get the same
     
  5. Apr 23, 2017 #4

    Orodruin

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    In 4 dimensions ##\delta_{ii} = 4##.
     
  6. Apr 23, 2017 #5
    that makes sense but if $\delta_{ii}=4$
    $$\epsilon_{ijkl}\epsilon_{ijkl}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{ll}+\delta_{ij}\delta_{jk}\delta_{kl}\delta_{li}+\delta_{ik}\delta_{jl}\delta_{ki}\delta_{lj}+\delta_{il}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jl}\delta_{kk}\delta_{lj}-\delta_{il}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{ll}-\delta_{ij}\delta_{ji}\delta_{kl}\delta_{lk}=4^4+\delta_{lj}\delta_{jl}+\delta_{ii}\delta_{jj}+\delta_{jl}\delta_{lj}-\delta_{ii}\delta_{jj}\delta_{kk}-\delta_{ii}\delta_{jj}-\delta_{ii}\delta_{jj}\delta_{ll}-\delta_{ii}\delta_{kk}=4^4+\delta_{ll}+4^2+\delta_{jj}-4^3-4^2-4^3-4^2=4^4+4+4^2+4-4^3-4^2-4^3-4^2=120=5!$$

    And it should be 4!=24, actually if $\delta_{ii}=3$ it works
    $$\epsilon_{ijkl}\epsilon_{ijkl}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{ll}+\delta_{ij}\delta_{jk}\delta_{kl}\delta_{li}+\delta_{ik}\delta_{jl}\delta_{ki}\delta_{lj}+\delta_{il}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jl}\delta_{kk}\delta_{lj}-\delta_{il}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{ll}-\delta_{ij}\delta_{ji}\delta_{kl}\delta_{lk}=3^4+\delta_{lj}\delta_{jl}+\delta_{ii}\delta_{jj}+\delta_{jl}\delta_{lj}-\delta_{ii}\delta_{jj}\delta_{kk}-\delta_{ii}\delta_{jj}-\delta_{ii}\delta_{jj}\delta_{ll}-\delta_{ii}\delta_{kk}=3^4+\delta_{ll}+3^2+\delta_{jj}-3^3-3^2-3^3-3^2=3^4+3+3^2+3-3^3-3^2-3^3-3^2=24=4!$$
     
  7. Apr 23, 2017 #6

    Orodruin

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    You are missing terms. The anti-symmetrisation over the 4 indices should have 4! = 24 terms, not 8.
     
  8. Apr 23, 2017 #7
    Yeah, its true, i am sorry.

    i am going to do the determinant 4x4
    gif&s=2&w=82.&h=76..gif gif&s=2&w=446.&h=57..gif
    and then simplify.
    Thanks!!!
     
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