# I Lie Algebra of Lorentz group

1. May 17, 2017

### 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!

2. May 17, 2017

### Orodruin

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

3. May 17, 2017

### Silviu

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

4. May 17, 2017

### Orodruin

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

5. May 18, 2017

### vanhees71

6. May 18, 2017

### dextercioby

They are tensor operators with respect to the Lorentz group.

7. May 18, 2017

### Orodruin

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

8. May 20, 2017

### samalkhaiat

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.