Showing Levi-Civita properties in 4 dimensions

[Phys pilot]

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
$$3(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})$$
$$3(\delta_{kk}\delta_{lm}-\delta_{km}\delta_{lk})$$
$$3(3\delta_{lm}-\delta_{lm})$$
$$3(2\delta_{lm})$$
$$6(\delta_{lm})=3!(\delta_{lm})$$

 

[Dr Transport]

Start with the matrix representation using Kronecker deltas then simplify.

https://en.wikipedia.org/wiki/Levi-Civita_symbol#Four_dimensions

 

[Phys pilot]

Start with the matrix representation using Kronecker deltas then simplify.

https://en.wikipedia.org/wiki/Levi-Civita_symbol#Four_dimensions

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

 

[Orodruin]

In 4 dimensions $\delta_{ii} = 4$.

 

[Phys pilot]

In 4 dimensions $\delta_{ii} = 4$.

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
$$\epsilon_{ijkl}\epsilon_{ijkl}=\delta_{ii}\delta_{jj}\delta_{kk}\delta_{ll}+\delta_{ij}\delta_{jk}\delta_{kl}\delta_{li}+\delta_{ik}\delta_{jl}\delta_{ki}\delta_{lj}+\delta_{il}\delta_{ji}\delta_{kj}\delta_{lk}-\delta_{ii}\delta_{jl}\delta_{kk}\delta_{lj}-\delta_{il}\delta_{jk}\delta_{kj}\delta_{li}-\delta_{ik}\delta_{jj}\delta_{ki}\delta_{ll}-\delta_{ij}\delta_{ji}\delta_{kl}\delta_{lk}=3^4+\delta_{lj}\delta_{jl}+\delta_{ii}\delta_{jj}+\delta_{jl}\delta_{lj}-\delta_{ii}\delta_{jj}\delta_{kk}-\delta_{ii}\delta_{jj}-\delta_{ii}\delta_{jj}\delta_{ll}-\delta_{ii}\delta_{kk}=3^4+\delta_{ll}+3^2+\delta_{jj}-3^3-3^2-3^3-3^2=3^4+3+3^2+3-3^3-3^2-3^3-3^2=24=4!$$

 

[Orodruin]

You are missing terms. The anti-symmetrisation over the 4 indices should have 4! = 24 terms, not 8.

 

[Phys pilot]

You are missing terms. The anti-symmetrisation over the 4 indices should have 4! = 24 terms, not 8.

Yeah, its true, i am sorry.

i am going to do the determinant 4x4

and then simplify.
Thanks!