# Why Lorentz algebra is not represented by the basis of antisymmetric $4\times 4$ tensors? Confusion building Lorentz Lie algebra

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.

• Are you cool with the standard picture? Jan 18, 2022 at 23:05
• 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. Jan 18, 2022 at 23:36

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 $$\det U=1$$.

• Thanks for you help, I ended up realizing my mistake but your answer was still insightful 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 $$\omega^\mu_\nu = - \omega^\nu_\mu$$.

• 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! Jan 19, 2022 at 11:51