In the vector representation of the Lorentz group its generators are given by -

$$(J^{\mu\nu})_{\alpha\beta} = i(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)$$

It can be shown also that $$\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\rho\sigma}=-2(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)$$

Implying that $$(J^{\mu\nu})_{\alpha\beta}=-\frac{i}{2} \epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\rho\sigma}$$

Can the first formula be derived from the second (or vice versa)? Is there some intuition behind this relation?

  • 1
    $\begingroup$ I think the first is just a property of the Lorentz group you stated (so specific case) and the second is a property of the Levi-Civita pseudotensor (this holds in general). $\endgroup$ Nov 10, 2018 at 13:02
  • $\begingroup$ @Dani this relation doesn't hold in other dimensions? $\endgroup$
    – proton
    Nov 10, 2018 at 13:25
  • $\begingroup$ I don't think the first equation changes in a different number of dimensions; the factor of 2 in front of your second equation would change in different dimensions (I think it would be $(d-2)!$ in general.) So all that would change in your final equation would be that the $2$ would be replaced by $(d-2)!$. $\endgroup$ Nov 10, 2018 at 14:25
  • $\begingroup$ Also, my guess is that it's just a coincidence. Both tensors have to be antisymmetric under exchange of $\{\mu, \nu\}$ and $\{\rho, \sigma\}$, and there are only so many tensors you can construct with those symmetries. $\endgroup$ Nov 10, 2018 at 14:49
  • 4
    $\begingroup$ Check the indices of $\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\mu\nu}=-2(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)$. I guess that the indices of the second $\epsilon$ should be $\alpha\beta\rho\sigma$ instead of $\alpha\beta\mu\nu.$ $\endgroup$
    – md2perpe
    Nov 11, 2018 at 22:45

1 Answer 1


No, your first equation is not a consequence of your second one, nor vice-versa. The third one combines the first two.

The antisymmetric permutation symbols in n (spacetime) dimensions obey $$ \epsilon^{j_1 \dots j_n}\epsilon_{i_1 \dots i_n} = n! \delta_{[ i_1}^{j_1} \dots \delta_{i_n ]}^{j_n} ,$$ where [...] denotes complete antisymmetrization of the lower indices.

As a result, contracting n-2 pairs, that is all indices but the leading two pairs, yields $$ \epsilon^{\mu\nu j_3 \dots j_n}\epsilon_{\alpha\beta i_3 \dots i_n} = -(n-2)! (\delta_{ \alpha}^{\mu} \delta_{ \beta}^{\nu}- \delta_{ \beta}^{\mu} \delta_{ \alpha}^{\nu} ), $$ antisymmetric in all up and down free index pairs.

Your n=4 has a coefficient -2, while n=3 and n=2 a coefficient -1, and n=5 a coefficient -6, etc. This is all pure combinatorics, logically independent of the Lorentz group.

Now, the Lorentz group has an n(n-1)/2 -dimensional Lie algebra, (so 6d for n=4) indexed by the corresponding antisymmetric pairs of indices $\mu\nu$, etc, $$ [J^{\mu\nu},J^{\rho \sigma}]=i(J^{\mu\rho} g^{\nu\sigma} +J^{\nu\sigma} g^{\mu\rho}- J^{\nu\rho} g^{\mu\sigma}-J^{\mu\sigma} g^{\nu\rho}), $$ a unique form bearing all single and pairwise antisymmetries of the (pairwise antisymmetric) commutator on the left.

It is represented by all types of m × m matrices. in general, whose indices are suppresed here. But when it is represented by a "fundamental" set of n × n matrices, 4 × 4 ones in your case, with indices $\alpha\beta$ (antisymmetric, but symmetrized by the action of the metric if one of the two is 0--see below) it is straightforward, albeit tedious brute force, to check that your first and third equations in fact satisfy it. I'd like to speculate that the 3rd eqn form is easier than the first, but I'd be lying... it's subjective.

In any case, this is only one of several bases in that dimension used, with the SO(3) subroup of rotations represented by sparse antisymmetric matrices and the three boosts in the coset by sparse symmetric ones. In the mixed rotation/boost basis much of the compact structure you are observing is gone.

For n=3, your first formula still works for the 3 × 3 generators, but the third one is supplanted by $$(J^{\mu\nu})_{\alpha\beta}=- i \epsilon^{\mu\nu\rho }\epsilon_{\alpha\beta\rho },\\ (M^\lambda)_{\alpha\beta}\equiv g^{\lambda\kappa}\epsilon_{\kappa\mu\nu} (J^{\mu\nu})_{\alpha\beta}=-2ig^{\lambda\kappa}\epsilon_{\kappa\alpha\beta} ,$$ one rotation (antisymmetric matrix $(M^0)^\alpha_{~~\beta}$); and two boosts ($(M^1)^\alpha_{~~\beta}, ~ (M^2)^\alpha_{~~\beta}$, now symmetric matrices, on account of the uneven action of the metric in raising the last index!).

*Intuition? I'm nonlinear at that... If you think of SO(n) instead of its noncompact brother the homogeneous Lorentz group SO(1,n-1), then the metric is the identity, and co/contra technically equivalent, so all is antisymmetric matrices and your 3rd equation is arguably more direct. It is yet another reformulation for your toolbox... Does it expedite Pauli-Lubanski for you?


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