What Is the Lie Algebra of the Lorentz Group?

[Silviu]

Hello! I read that the for the lie algebra of the Lorentz group we can parametrize the generators as an antisymmetric tensor $J^{\mu \nu}$ and the parameters as an another antisymmetric tensor $\omega_{\mu \nu}$ and a general transformation would be $\Lambda = exp(-\frac{i}{2} \omega_{\mu \nu} J^{\mu \nu})$. So based on Einstein notation, this would mean $\Lambda = exp(-\frac{i}{2} (\omega_{00} J^{00} + \omega_{01} J^{01} + ... ))$. But for example $\omega_{00}$ and $J^{00}$ are numbers, so in the end we will end up with a complex number in the exponential and now with a (4x4 most probably) matrix. Can someone explain to me what I do wrong in reading this notation? Thank you!

 

[Orodruin]

The $J^{\mu\nu}$ are not tensors, they are the basis of the Lie algebra.

 

[Silviu]

The $J^{\mu\nu}$ are not tensors, they are the basis of the Lie algebra.

So they are in the case of Minkowski space 4x4 matrices?

 

[Orodruin]

So they are in the case of Minkowski space 4x4 matrices?

If you use the (usual) matrix representation of the Lorentz group, yes.

 

[vanhees71]

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Sect. 1.7.

 

[dextercioby]

The $J^{\mu\nu}$ are not tensors, they are the basis of the Lie algebra.

They are tensor operators with respect to the Lorentz group.

 

[Orodruin]

They are tensor operators with respect to the Lorentz group.

They can be represented by tensor operators. This still does not make them tensors in the sense implied by the OP.

 

[samalkhaiat]

Hello! I read that the for the lie algebra of the Lorentz group we can parametrize the generators as an antisymmetric tensor $J^{\mu \nu}$ and the parameters as an another antisymmetric tensor $\omega_{\mu \nu}$ and a general transformation would be $\Lambda = exp(-\frac{i}{2} \omega_{\mu \nu} J^{\mu \nu})$. So based on Einstein notation, this would mean $\Lambda = exp(-\frac{i}{2} (\omega_{00} J^{00} + \omega_{01} J^{01} + ... ))$. But for example $\omega_{00}$ and $J^{00}$ are numbers, so in the end we will end up with a complex number in the exponential and now with a (4x4 most probably) matrix. Can someone explain to me what I do wrong in reading this notation? Thank you!

We do not ‘parameterize’ the generators. The generators of any Lie algebra are independent of the parameters of the corresponding Lie group. This $\omega_{\mu\nu}$ is not a tensor. It is a set of six real numbers, $\{ \omega_{a}, \ a = 1, ..., 6 \}$, (3 rotation angles $\theta_{i} \equiv \frac{1}{2}\epsilon_{ijk}\omega_{jk}$, and 3 boosts $\beta_{i} \equiv \omega_{0i}$) grouped together under the condition $\omega_{\mu\nu} = - \omega_{\nu\mu}$. So, when you use the symbol $\omega_{\mu\nu}$ to represent the (six) parameters, you should keep in mind that $\omega_{00} = \cdots = \omega_{33} = 0$. This means that there are only six generators ($M^{a}, \ a = 1, ... , 6$) which (also) can be grouped to satisfy $M^{\mu \nu} = - M^{\nu\mu}$. Since Lorentz group is a connected Lie group, then (almost all) its elements can be represented by the exponential map $$g = e^{\vec{\omega}} = e^{\omega_{a}M^{a}} \equiv e^{\frac{1}{2}\omega_{\mu\nu}M^{\mu\nu}} ,$$ where $\vec{\omega}$ is an element of 6-dimensional vector space (the Lorentz algebra) and $\omega_{\mu\nu}$ are its components (local coordinates on the group manifold) “along” the basis vectors $M^{\mu\nu}$. This should remind you with the school-days vector equation $\vec{x} = x_{i}\hat{e}^{i}$.
Now, the question whether the generators are Lorentz tensors or not is a different story. To answer it, we need to study the representation of the Lorentz group. Basically, in the representation theory of Lorentz group, you encounter three objects. These are: the coordinates transformation matrix $\Lambda$, with matrix elements denoted by $\Lambda^{\mu}{}_{\nu}$; the finite-dimensional representation matrix $D(\Lambda) = \exp \left(\frac{1}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right)$ which acts on the index space of geometrical objects (classical fields) on Minkowski space-time, i.e., under a Lorentz transformation $\Lambda$, the components of a given field $\varphi_{a}(x)$ get mixed by the matrix $D(\Lambda)$; and the third object is the infinite-dimensional unitary representation operator $U(\Lambda) = \exp \left( \frac{i}{2} \omega_{\mu\nu}J^{\mu\nu}\right)$ which acts on states and operators in the Hilbert space. In QFT, the above three objects appear in one fundamental (transformation) equation, that is $$U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) .$$ From this equation we understand that on the one hand the field transforms under the $SO(1,3)$ transformation like a classical geometrical object in the Minkowski space-time, $$\varphi_{a}(x) \to \bar{\varphi}_{a}(x) = D_{a}{}^{c}(\Lambda) \ \varphi_{c}(\Lambda^{-1}x) ,$$ on the other as an operator-valued distribution in the Hilbert space, $$\varphi_{a}(x) \to \bar{\varphi}_{a}(x) = U^{\dagger}(\Lambda) \varphi_{a}(x) U(\Lambda) .$$ Also, the generators $\Sigma^{\mu\nu}$ of the $D(\Lambda)$ transformation are given by constant numerical (spin) matrices (i.e., not a tensor). To see that consider the transformation law in the case of a vector field $A^{\mu}(x)$, $$\bar{A}^{\mu}(x) = D^{\mu}{}_{\nu}(\Lambda) \ A^{\nu}(\Lambda^{-1}x) = \Lambda^{\mu}{}_{\nu} A^{\nu}(\Lambda^{-1}x) .$$ This tells you that for a vector field, the finite-dimensional representation $D(\Lambda)$ is given by the Lorentz matrix $\Lambda$: $$D^{\mu}{}_{\nu}(\Lambda) = \Lambda^{\mu}{}_{\nu} .$$ To first order in the parameters, this becomes $$\delta^{\mu}_{\nu} + \frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}{}_{\nu} .$$ This leads to $$\frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \omega^{\mu}{}_{\nu} ,$$ which can be rewritten as $$\frac{1}{2} \omega_{\rho\sigma} \left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \eta^{\mu\rho} \delta^{\sigma}_{\nu} \omega_{\rho\sigma} = \frac{1}{2} \omega_{\rho\sigma} \left( \eta^{\mu\rho} \ \delta^{\sigma}_{\nu} - \eta^{\mu\sigma} \ \delta^{\rho}_{\nu} \right) .$$ Thus
$$\left( \Sigma^{\rho\sigma} \right)^{\mu}{}_{\nu} = \eta^{\mu\rho} \ \delta^{\sigma}_{\nu} - \eta^{\mu\sigma} \ \delta^{\rho}_{\nu} .$$ This tells you that the generator $\Sigma^{\rho\sigma}$ represents a set of six antisymmetric, $4 \times 4$ constant matrices with matrix elements given by $(\Sigma^{\rho\sigma})^{\mu}{}_{\nu}$. Of course, you may regard the object $\Sigma^{\rho\sigma\mu}{}_{\nu}$ as a rank-4 invariant tensor, but this does not mean that $\Sigma^{\rho\sigma}$ itself is a rank-2 tensor.
However, Hermitian generator $J^{\mu\nu}$ of the $U(\Lambda)$ transformation is a rank-2 Lorentz tensor operator. Indeed, under Lorentz transformation, it is easy to show that $$J^{\mu\nu} \to U^{\dagger}J^{\mu\nu}U = \Lambda^{\mu}{}_{\rho} \ \Lambda^{\nu}{}_{\sigma} \ J^{\rho \sigma} ,$$ which is the correct transformation law for rank-2 tensor operator.