A three rank Levi-Civita tensor in four dimensional spacetime

Question

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.

Answer 1

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.

Answer 2

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] .)