Tensor multiplication problem

[DuckAmuck]

Homework Statement

Show that
epsilon_{ijkl} ( M^{ij} N^{kl} + N^{ij} M^{kl}) = 0

Relevant Equations

epsilon is the 4D anti-symmetric Levi-Cevita tensor. M and N are also anti-symmetric tensors.

ep_{ijkl} M^{ij} N^{kl} + ep_{ijkl}N^{ij} M^{kl}
The second term can be rewritten with indices swapped
ep_{klij} N^{kl}M^{ij}
Shuffle indices around in epsilon
ep{klij} = ep{ijkl}
Therefore the expression becomes
2ep_{ijkl}M^{ij}N^{kl}
Not zero.
What is wrong here?

 

[malawi_glenn]

I tried, but got the same result as you did. Are you sure its not supposed to be
$\epsilon_{ijkl} ( M^{ij} N^{kl} - N^{ij} M^{kl}) = 0$?
What I did was to write everything out, using all even permutations of 1,2,3,4:
{1,2,3,4}, {1,3,4,2}, {1,4,2,3}, {2,1,4,3}, {2,3,1,4}, {2,4,3,1}, {3,1,2,4}, {3,2,4,1}, {3,4,1,2}, {4,1,3,2}, {4,2,1,3}, {4,3,2,1}
and all odd ones:
{1,2,4,3}, {1,3,2,4}, {1,4,3,2}, {2,1,3,4}, {3,2,1,4}, {4,2,3,1}, {2,3,4,1}, {2,4,1,3}, {3,1,4,2}, {3,4,2,1}, {4,1,2,3}, {4,3,1,2}
and the fact that $M$ and $N$ are anti-symmetrical, i.e. $M^{12}= - M^{21}$ etc.

 

[DuckAmuck]

ok i think i have solid reasoning here:

Suppose $C^{ij} = M^{ij} + N^{ij}$

From symmetry and antisymmetry we have:

$\epsilon_{ijkl} C^{ij}C^{kl} = 0$

Also if you foil the CC product in terms of M and N you get $C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}$

The MM and NN terms are zero for the same reason the CC product is when multiplied by epsilon.

So this demands that

$\epsilon_{ijkl} (M^{ij}N^{kl} + N^{ij}M^{kl}) = 0$

Can someone please verify?

 

[Orodruin]

ok i think i have solid reasoning here:

Suppose $C^{ij} = M^{ij} + N^{ij}$

From symmetry and antisymmetry we have:

$\epsilon_{ijkl} C^{ij}C^{kl} = 0$

Also if you foil the CC product in terms of M and N you get $C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}$

The MM and NN terms are zero for the same reason the CC product is when multiplied by epsilon.

So this demands that

$\epsilon_{ijkl} (M^{ij}N^{kl} + N^{ij}M^{kl}) = 0$

Can someone please verify?

No, the statement as it stands seems false to me. It is not generally the case thar $\epsilon_{ijkl} C^{ij} C^{kl} = 0$.

 

[DuckAmuck]

No, the statement as it stands seems false to me. It is not generally the case thar $\epsilon_{ijkl} C^{ij} C^{kl} = 0$.

You’re right. I am just trying to figure out *how* this could be zero at this point, as in what conditions. Otherwise I’m stumped.

 

[malawi_glenn]

I am just trying to figure out *how* this could be zero at this point, as in what conditions.

You should have been given all the conditions already, that M and N are antisymmetric rank-2 tensors.

There is always the possibility that whoever gave you this problem, is wrong / made a typo. I have been tearing my hair off several times doing excersices in general relativity books... to find out there was some typo in the problem as written.

Here is my "expanded" calculation that I did btw:

The underlined terms I will collect at the end.

$\underline{M^{12}N^{34}} + M^{13}N^{42} + M^{14}N^{23} + \underline{M^{21}N^{43}} + M^{23}N^{14} + M^{24}N^{31} + M^{31}N^{24} + M^{32}N^{41} + M^{34}N^{12} + M^{41}N^{32} + M^{42}N^{13} + M^{43}N^{21}$
$- ( \underline{M^{12}N^{43}} + M^{13}N^{24} + M^{14}N^{32} + \underline{M^{21}N^{34}} + M^{32}N^{14} + M^{42}N^{31} + M^{23}N^{41} + M^{24}N^{13} + M^{31}N^{42} + M^{34}N^{21} + M^{41}N^{23} + M^{43}N^{12} )$
$+ N^{12}M^{34} + N^{13}M^{42} + N^{14}M^{23} + N^{21}M^{43} + N^{23}M^{14} + N^{24}M^{31} + N^{31}M^{24} + N^{32}M^{41} + \underline{N^{34}M^{12}} + N^{41}M^{32} + N^{42}M^{13} + \underline{N^{43}M^{21}}$
$- ( N^{12}M^{43} + N^{13}M^{24} + N^{14}M^{32} + N^{21}M^{34} + N^{32}M^{14} + N^{42}M^{31} + N^{23}M^{41} + N^{24}M^{13} + N^{31}M^{42} + \underline{N^{34}M^{21}} + N^{41}M^{23} + \underline{N^{43}M^{12}} \: )$

The stuff I underlined:
$M^{12}N^{34} + M^{21}N^{43} - M^{12}N^{43} - M^{21}N^{34} + N^{34}M^{12} +N^{43}M^{21} -N^{34}M^{21} - N^{43}M^{12}$

($M^{21}= - M^{12}$ and $N^{43}= - N^{34}$)

$M^{12}N^{34} + (-1)^2 M^{12}N^{34} - (-1)M^{12}N^{34} - (-1)M^{12}N^{34} + N^{34}M^{12} +(-1)^2N^{34}M^{12} - (-1)N^{34}M^{12} - (-1)N^{34}M^{12} = 8M^{12}N^{34}$

Well that was fun.