I am very confused when building the Lie algebra of the Lorentz Group. In every books, they expand $\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu}$ at the origin and you end up with the condition that $\omega_{\mu\nu} = - \omega_{\nu\mu}$.

I've done exercises where I have done basically the same thing (at least i feel like they're the same). As instance for the group $SU(2)$ we've done it and we showed that the condition was that the generators must be hermitian and that if you had a basis of the 2x2 hermitian matrices (pauli matrices) then it means you had a basis of the your lie algebra.

But in the case of the Lorentz group we saw that for instance the boost $K^i$ are not antisymmetric and I can't see how the condition on $\omega$ we derived does not impose that. As for the case with the group $SU(2)$ I feel like having a basis of the antisymmetric 4x4 matrices would provide a basis of our Algebra. In the case of the Lorentz group we ask that it is the different generators that are indexed by $\rho$ and $\sigma$ as $\mathcal{(J^{\rho\sigma})^\mu_\nu}$ to be antisymmetric according to those indices and not according to the indices of the matrice $J^{\rho\sigma}$ ($\mu$ and $\nu$).

I obviously have a big confusion with Lie algebra and generators but I really can't put a finger on it.

  • $\begingroup$ Are you cool with the standard picture? $\endgroup$ Commented Jan 18, 2022 at 23:05
  • 1
    $\begingroup$ You are confusing representation-independent algebra/group indices with the quartet-representation on four-vectors indices. The ${\cal J}^{\rho \sigma}$ are hermitian antisymmetric for rotations and antihermitian symmetric for boosts, respectively, in the 4-vector (quartet) representation. $\endgroup$ Commented Jan 18, 2022 at 23:36

2 Answers 2


Well, let us consider the fundamental representation of $SO(1,3)_+$ made of matrices $$\Lambda = [{\Lambda^a}_b]_{a,b=0,1,2,3}\:.$$ The position of indices is of crucial relevance here. The Lorentz-group condition is $$\Lambda^t \eta \Lambda = \eta$$ namely $${\Lambda^a}_b \eta_{ac} {\Lambda^c}_d = \eta_{bd}\:.$$ Let us expand around the identity the matrices and let us drop second order terms: $$(\delta^a_b + {\omega^a}_b)\eta_{ac}(\delta^c_d+ {\omega^c}_d) = \eta_{bd}$$ obtaining $$\delta^a_b \eta_{ac} {\omega^c}_d+ {\omega^a}_b\eta_{ac} \delta^c_d=0\:.$$ In other words $$\eta_{bc} {\omega^c}_d+ {\omega^a}_b\eta_{ad} =0\:.\tag{1}$$ There are $6$ linearly independent real $4\times 4$ matrices $\omega = [{\omega^a}_b]_{a,b=0,1,2,3}$ satisfying these identities, they form a basis of the Lie algebra of $SO(1,3)_+$.

In particular, the three standard boost generators written as $4\times 4$ matrices $$K_k= [{(K_k)^a}_b]_{a,b=0,1,2,3}, \quad k=x,y,z$$ do satisfy this identity!

The same fact is true for the three remaining generators of spatial rotations $S_k$, $k=x,y,z$.

The $6$ matrices $K_k$ and $S_k$ are also linearly independent so that they form a basis of the considered Lie algebra.

However, defining $$\omega_{ab}:= \eta_{ac} {\omega^c}_b$$ (1) can be equivalently restated as $$\omega_{ab}+ \omega_{ba}=0\:.\tag{2}$$ Notice that, as $\eta\eta = I$, the components ${\omega^a}_b$, and $\omega_{ab}$ carry the same information and one passes from a type to another simply exploiting the standard procedure of rising and lowering indices (though this procedure should be justified properly viewing the Lorentz transformations as $(1,1)$ tensors).

From an elementary viewpoint (1) are the definition of the generators of the Lie algebra of $SO(1,3)_+$, (2) are equivalent statements which should be handled cum grano salis.

The same procedure for $SU(2)$, writing the matrices as $U = [{U^a}_b]_{a,b=1,2}$, the unitarity condition produces $$\overline{{\omega^a}_b}+ {\omega^b}_a =0\:,$$ in place of (1), where the bar denotes the complex conjugation. Here a basis of the Lie algebra is made of the standard Pauli matrices with immaginary coefficient $-i\sigma_k = -i[{(\sigma_k)^a}_b]_{a,b =1,2}$, where $k=1,2,3$.Here we have also to impose that the element of the Lie algebra are traceless to fulfill the requirement $\det U=1$.

  • $\begingroup$ Thanks for you help, I ended up realizing my mistake but your answer was still insightful $\endgroup$ Commented Jan 19, 2022 at 13:28

Thanks for your answer. I am not sure that's really where my confusion lies. I can try develop my analogy with the SU(2) group so that you can better understand my problem. In this case let's not $T^a$ the generators, expanding and applying the group constraint we get that matrices $T^a$ must be hermitian so $(T^a)^{\mu}_{\nu} = (conj(T^a))^{\nu}_{\mu}$. If I understand correctly $a$ here are playing the same role as the pair $\rho \sigma$ for the Lorentz group i.e indexes the different generators. Doing the same thing for the latter group and applying its constraint I feel like $\omega_{\mu\nu} = -\omega_{\nu\mu}$ means $(\mathcal{J}^{\rho\sigma})^{\mu}_{\nu} = -(\mathcal{J}^{\rho\sigma})^{\nu}_{\mu}$ and not $\mathcal{J}^{\rho\sigma} = -\mathcal{J}^{\sigma\rho}$. I can't see why it is the second option that is true That's were my confusion is.

EDIT: I think I am starting to see my problem. I think it is because I considered that $\omega_{\mu\nu} = -\omega_{\nu\mu}$ was equivalent to $\omega^\mu_\nu = - \omega^\nu_\mu$.

  • $\begingroup$ Look at the real Lorentz transformation 4x4 matrices. They are not unitary / orthogonal, as they shouldn’t be, for a finite dimensional rep of a non compact group! $\endgroup$ Commented Jan 19, 2022 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.