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Is it possible to construct a Lorentz invariant, three rank Levi-Civita tensor in Minkowski Spacetime? If not, why so? I am talking about something like this $\epsilon_{\alpha\beta\gamma}$ or $\epsilon^{\alpha\beta\gamma}$, where each indices run from $0$ to $3$. As in this answer here, which proves the Lorentz co-variance of the Levi-Civita tensor by using the determinant formula, I guess one would run into trouble if we have three rank Levi-Civita tensors. Kindly elaborate on that.

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  • $\begingroup$ The levi civita symbol is defined to be this: $\epsilon_{0123}=1$ and every odd permutation of 0123 would have value of $\epsilon$ to be zero. So, how would we define a 3 ranked levi civita symbols in 4 dimensions? $\endgroup$ – Naman Agarwal Jun 6 '18 at 14:14
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You can use Young tableaux/diagrams and the permutation group to figure out the symmetries of the general rank-3 tensor. The spaces correspond to the partitions of the rank:

3=3:

One 20 dimensional total symmetric subspace.

3=2+1:

Two 20 dimensional mixed symmetry subspaces.

3=1+1+1:

One 4 dimensional totally antisymmetric subspace:

$$ A_{\alpha\beta\gamma} = \frac 1 6 [T_{\alpha\beta\gamma} + T_{\beta\gamma\alpha} + T_{\gamma\alpha\beta} - T_{\gamma\beta\alpha} - T_{\beta\alpha\gamma} - T_{\alpha\gamma\beta}] $$

That is the only antisymmetric thing you can make according to Schurl-Weyl theory.

To find the dimensions, I used the Hook Length Formula (summing of the boxes $x$ in a diagram $Y(\lambda)$) for the Young diagram corresponding to the integer partition:

$$ {\rm dim}\pi_{\lambda} = \frac {n!}{\prod_{x\in Y}{\rm hook}(x)}$$

If you consider 3 dimensions ($n=3$), you get ${\rm dim} = 1$, that is the standard Levi-Civita symbol $\epsilon_{ijk}$.

If you set $n=4$, the result is ${\rm dim} = 4$.

That means $A_{\alpha\beta\gamma}$ transforms like a 4-vector.

So, the only antisymmetric part of a rank-3 tensor in Minkowski space rotates like a 4-vector, which means it is not invariant and is not a candidate to be Levi-Civita like.

Meanwhile, the dimensions of the 3 other irreducible spaces are all 20--which are certainly not scalars, and thus not candidates to be Levi-Civita like.

Note that if you consider rank-4 tensors, the partitions are as follows:

4=4:

35 dimensional and symmetric.

4=3+1:

Three 45-dimensional mixed symmetry spaces.

4=2+2:

Two 20-dimensional mixed symmetry spaces.

4=2+1+1:

Three 15-dimensional mixed symmetry spaces.

4=1+1+1+1:

One total antisymmetric 1 dimensional space, which is proportional to the Levi-Civita symbol $\epsilon_{\mu\nu\sigma\lambda}$.

In summary, the answer is "No", and the reason why has to do with the representations of the symmetric group on 3-letters. You partition the rank=3, use the Robinson-Schensted correspondence to associate that partition with irreducible representations of the permutation group. (The Young Diagrams make this step a snap). Then, Schur-Weyl duality associates those with irreducible subspaces of and rank-N tensor (signed permutations of indices). Finally, the Hook Length formula tells you the dimensions of those subspaces.

The Levi-Civita symbol needs to be invariant (e.g., dimension 1, like a scalar) and it need to be totally antisymmetric in all indices--and that simply did not exist for rank 3 in 4 dimensions.

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  • $\begingroup$ The answer is much more engaged than I thought it would be. It will take me a bit of time to understand the whole thing. Can you please refer something where I can read about Young tableaux, partitions and Hook Length formula etc.? Thank you. $\endgroup$ – fogof mylife Jun 5 '18 at 23:25
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    $\begingroup$ @fogofmylife I had to cull many references. Most are too mathy for us. Finite group theory is abstract, but a starting point is yufeizhao.com/research/youngtab-hcmr.pdf which covers the permutation group and Young tableaux. That the Young symmetrizer applied to indices yields the irreducible subspaces for all ranks in any dimension over all fields (Schur-Weyl Duality) is not clear and I know of no physics oriented reference. It's best to try it by hand for 0=0 (scalars), 1=1 (vectors), 2=2,1+1 (tensors), 3=3, 2+1, 1+1+1 (the 1st nontrivial one) by hand and see what's happening. Goodluck $\endgroup$ – JEB Jun 6 '18 at 3:02
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    $\begingroup$ @fogofmylife: The best physics-oriented reference for the Schur-Weyl duality is probably Hamermesh's Group Theory and Its Application to Physical Problems, but even that one isn't the clearest. $\endgroup$ – Michael Seifert Jun 6 '18 at 14:47
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Does this satisfy the requirements?

In Minkowski spacetime (signature: -,+,+,+), let $\epsilon_{abcd}$ be the alternating tensor satisfying $\epsilon_{abcd}=\epsilon_{[abcd]}$ and $\epsilon_{abcd}\epsilon^{abcd}=-24$ and let $v^a$ be the 4-velocity of an observer ($v^av_a=-1$).

Define the "spatial alternating tensor seen by the observer $v^a$" $$\epsilon_{abc}=\epsilon_{abcd}v^d,$$ which satisfies $\epsilon_{abc}=\epsilon_{[abc]}$, $\epsilon_{abc}\epsilon^{abc}=6$, and $v^a\epsilon_{abc}=0$.

(This is extracted from Robert Geroch's "General Relativity, 1972 Lecture Notes" [ISBN 978-0987987174] .)

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  • $\begingroup$ This wouldn't be Lorentz-invariant, since its definition depends on the four-vector $v^a$ chosen, and this vector will transform non-trivially between reference frames. $\endgroup$ – Michael Seifert Jun 6 '18 at 14:45
  • $\begingroup$ @MichaelSeifert , I agree. By construction, this certainly depends on the choice of four-vector. $\endgroup$ – robphy Jun 6 '18 at 15:03

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