Proof that generators of Lorentz algebra transform as tensors

Question

At page 60 of Weinberg book "Quantum Theory of Fields" (vol 1) it is proved that $J^{\mu\nu}$ transform as a tensor, but I can't understand the step from eq (2.4.7) to (2.4.8). It seems to me that he assumes the following relation:

$(\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} = \omega_{\rho\sigma}\Lambda_\mu^{\,\,\rho}\Lambda_\nu^{\,\,\sigma}J^{\mu\nu}$

where $(\Lambda^{-1})^{\rho}_{\,\,\mu} = \Lambda_\mu^{\,\,\rho}$

but I can't find a way to prove that relation, cause in my calculation I get

$(\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu} = \omega_{\rho\sigma}(\Lambda^T)_{\phantom.\mu}^{\rho}\Lambda_\nu^{\phantom.\sigma}J^{\mu\nu}$

Note the different position of the indexes and the transpose in the first $\Lambda$.

In the Wu-Ki Tung book he proves that $J_{\mu\nu}$ transform as a tensor using the following theorem (page 185, 10.2-14):

$\Omega\Lambda(\omega)\Omega^{-1} = \Lambda(\omega^\prime)\,\,,$ where $\omega^{\prime\mu\nu}=\Omega^{\mu}_{\,\,\lambda}\Omega^{\nu}_{\,\,\sigma}\omega^{\lambda\sigma}$

$\Omega$ is a generic Lorentz transformation and $\Lambda(\omega) = e^{-i/2\omega^{\mu\nu}J_{\mu\nu}}$, but again there's no explicit proof of that theorem.

Answer 1

Is there a typo in the specific relation you're wondering about?

By definition $$ \Lambda=({\Lambda^\rho}_\mu)\qquad\text{and}\qquad\Lambda^{-1}=({\Lambda_\nu}^\sigma). $$ So by definition $$(\Lambda\omega\Lambda^{-1})_{\mu\nu}={\Lambda^\rho}_\mu\omega_{\rho\sigma}{\Lambda_\nu}^\sigma=\omega_{\rho\sigma}{\Lambda^\rho}_\mu{\Lambda_\nu}^\sigma $$ and should the relation be $$ (\Lambda\omega\Lambda^{-1})_{\mu\nu}J^{\mu\nu}={\Lambda^\rho}_\mu\omega_{\rho\sigma}{\Lambda_\nu}^\sigma J^{\mu\nu}=\omega_{\rho\sigma}{\Lambda^\rho}_\mu{\Lambda_\nu}^\sigma J^{\mu\nu}\,? $$ Specifically, note the indices in ${\Lambda^\rho}_\mu$, which is where the typo seems to be? And this is all just the definitions of the tensors in use?

Answer 2

Weinberg is correct. Let us interpret the action of the Lorentz group actively after having chosen a Minkowskian reference frame. In that case the Lorentz transformation is a tensor ${\Lambda^a}_b$ of type $(1,1)$, it being a liner transformation of vectors. We can therefore use the standard procedure of lowering and raising indices by means of the Minkowski metric $\eta$.

First of all notice that the coefficients $\omega$ are introduced in the book with the same type of indices as the ones of the Lorentz matrices in the fundamental representation: $${\omega^\mu}_\nu\:.$$

That is the way one has to interpret the matrix in the parenthesis of $(\Lambda \omega \Lambda^{-1})_{\mu\nu}$. So,

first one computes ${(\Lambda \omega \Lambda^{-1})^\mu}_\nu$,

next he/she lowers the index $\mu$ using the metric.

We also take advantage of the identity $${(\Lambda^{-1})^a}_b= {\Lambda_b}^a$$ (see below why it is correct from two viewpoints).

$${(\Lambda \omega \Lambda^{-1})^\mu}_\nu = {\Lambda^\mu}_\rho {\omega^\rho}_\sigma {(\Lambda^{-1})^\sigma}_\nu= {\Lambda^\mu}_\rho {\omega^\rho}_\sigma {\Lambda_\nu}^\sigma= {\Lambda^\mu}^\rho {\omega_\rho}_\sigma {\Lambda_\nu}^\sigma\:.$$ Hence $$(\Lambda \omega \Lambda^{-1})_{\mu\nu} = {\Lambda_\mu}^\rho {\omega_\rho}_\sigma {\Lambda_\nu}^\sigma = \omega_{\rho\sigma} {\Lambda_\mu}^\rho {\Lambda_\nu}^\sigma\:.$$ That is the wanted identity.

REMARK

The identity $${(\Lambda^{-1})^a}_b= {\Lambda_b}^a \tag{1}$$ has a correct meaning in this formalism. Indeed, it is equivalent to another true identity as all the following passages are reversible. From (1), $${(\Lambda^{-1T})^b}_a= {\Lambda_a}^b\tag{2}$$ and thus $$\eta_{cb}{(\Lambda^{-1T})^b}_a\eta^{ad}= \eta_{cb}{\Lambda_a}^b\eta^{ad} = {\Lambda^d}_c$$ In matrix form $$\eta (\Lambda^{-1T}) \eta = \Lambda$$ which is true because it immediately arises from the known identities for the elements of the Lorentz group $$\Lambda^{-1} = \eta \Lambda^T \eta\:, \quad \eta= \eta^T = \eta^{-1}$$ where $$\eta := diag(-1,1,1,1)\:.$$

Finally, (2) is also coherent with the fact that ${\Lambda_a}^b$ is the standard notation for the (active) action of the Lorentz group on covariant vectors (the fundamental representation ${\Lambda^a}_b$ acts on contravariant vectors). Let us prove it. If $$G \ni g \mapsto L_g \in GL(V)$$ is a representation of a group, referred to a basis $e_1,\ldots, e_n$ of the vector space $V$ of dimension $n$, the dual action acting on the dual space $V^*$ is just $$G \ni g \mapsto L'_g := (L_g)^{-1T} \in GL(V^*)$$ referred to the dual basis $e^{*1}, \ldots, e^{*n}$. This is nothing but (2) when $G$ is the Lorentz group.

Answer 3

I believe a cleaner way to prove it is the following.

It is not difficult to verify that the 4-vectors matrix representation of the generators of the Lorentz algebra can be written as follow (see Wu-Ki Tung "Group Theory" page 185 or Maggiore "A modern introduction to Quantum Field Theory" page 19):

$(J^{\mu\nu})^\sigma_{\phantom.\alpha} = i(\eta^{\mu\sigma}\delta^\nu_\alpha - \eta^{\nu\sigma}\delta^\mu_\alpha)$

where $\mu\nu$ identifies the generator and $\sigma\alpha$ the raw,column matrix element of that generator.

Remind also the formula for the inverse of the Lorentz trasformation matrix:

$(\Lambda^{-1})^\rho_{\phantom.\sigma} = \Lambda^\chi_{\phantom.\phi}\eta_{\sigma\chi}\eta^{\rho\phi}$

Using the two equations above is not difficult to get the following two equations:

$(\Lambda^{-1})^\rho_{\phantom.\sigma}(J^{\mu\nu})^\sigma_{\phantom.\alpha}\Lambda^\alpha_{\phantom.\beta} = i(\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\beta} - \Lambda^\mu_{\phantom.\beta}\Lambda^\nu_{\phantom.\phi})\eta^{\rho\phi}$

$(J^{\phi\sigma})^\rho_{\phantom.\beta}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma} = i(\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\beta} - \Lambda^\mu_{\phantom.\beta}\Lambda^\nu_{\phantom.\phi})\eta^{\rho\phi}$

Since the RHS are equal then also the LHS are equal, and you get:

$(\Lambda^{-1})^\rho_{\phantom.\sigma}(J^{\mu\nu})^\sigma_{\phantom.\alpha}\Lambda^\alpha_{\phantom.\beta} = (J^{\phi\sigma})^\rho_{\phantom.\beta}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma}$

Note that in the LHS you have the product of the three matrices $(\Lambda^{-1}J^{\mu\nu}\Lambda)$, while in the RHS you have a linear combination of the matrices $J^{\phi\sigma}$, and on both side $(\rho,\beta)$ is the (raw,column) that identify the element in the resultant matrix: $(M^{\mu\nu})^\rho_{\phantom.\beta}$.

So as a matrix equation It can be written as:

$\Lambda^{-1}J^{\mu\nu}\Lambda = J^{\phi\sigma}\Lambda^\mu_{\phantom.\phi}\Lambda^\nu_{\phantom.\sigma}$

Which proves that the generators $J^{\mu\nu}$ transform as a tensor under similarity Lorentz transformation.

The proof is much more easier when the generators of the Lorentz algebra are represented by differential operators instead of matrices (for the definition of $J^{\mu\nu}$ with differential operator see Schwartz "Quantum Filed Theory" page 161, or Zee "Group Theory" page 434, or Maggiore "A modern introduction to Quantum Field Theory" page 30):

$ J^{\mu\nu} = i(x^\mu\partial^\nu - x^\nu\partial^\mu) $

$ x^\mu \xrightarrow[\text{}]{\text{Lorentz}} \Lambda^\mu_{\phantom.\rho} x^{\rho}\,\,,\,\, \partial^\mu \xrightarrow[\text{}]{\text{Lorentz}} \Lambda^\mu_{\phantom.\rho} \partial^{\rho}$

$ J^{\mu\nu} \xrightarrow[\text{}]{\text{Lorentz}} i(\Lambda^\mu_{\phantom.\rho} x^{\rho}\Lambda^\nu_{\phantom.\sigma}\partial^\sigma - \Lambda^\nu_{\phantom.\rho} x^{\rho}\Lambda^\mu_{\phantom.\sigma}\partial^\sigma) = i(\Lambda^\mu_{\phantom.\rho}\Lambda^\nu_{\phantom.\sigma}x^\rho\partial^\sigma - \Lambda^\nu_{\phantom.\sigma} \Lambda^\mu_{\phantom.\rho}x^\sigma\partial^\rho) = \Lambda^\mu_{\phantom.\rho}\Lambda^\nu_{\phantom.\sigma} J^{\rho\sigma} $