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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.

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  • $\begingroup$ Are you cool with the standard picture? $\endgroup$ Commented Jan 18, 2022 at 23:05
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    $\begingroup$ 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

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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 $\det U=1$.

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  • $\begingroup$ Thanks for you help, I ended up realizing my mistake but your answer was still insightful $\endgroup$ Commented Jan 19, 2022 at 13:28
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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 $\omega^\mu_\nu = - \omega^\nu_\mu$.

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  • $\begingroup$ 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

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