Transformation law of Lie algebra of Lorentz group $SO^+(1,3)$

Question

I'd like to understand why the Lie algebra of Lorentz group is described as $(2,0)$ tensor.

One can take a basis of the Lie algeba $\mathfrak{so}(1,3)$ as

\begin{align} J^{(23)}=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}\ \ \ J^{(31)}=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}\ \ \ J^{(12)}=\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}\\ J^{(01)}=\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ J^{(02)}=\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ J^{(03)}=\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}.\tag1 \end{align}

and each of them makes $\theta$ rotation with respect to $\mu$ -$\nu $ plane by $\exp(\theta J^{(\mu\nu)})$ where $\{e_\mu \}_{\mu=0}^{3}$ is canonical basis of $\mathbb{R}^{1,3}$. I use parenthesis for indices because it's not sure that they transform as tensors. In my text book, it is written that $J^{(\mu\nu)}$ transforms as $(2,0)$ tensor, that is,

$$J'^{(\alpha\beta)}=\Lambda^\alpha_\mu\Lambda^\beta_\nu J^{(\mu\nu)}\tag{2}$$

where $\{e'_\alpha \}_{\alpha=0}^3$ is another basis for $\mathbb{R}^{1,3}$ such that for the dual basis $\{e'^\alpha \}_{\alpha=0}^3\,$, $e'^\alpha (e_\mu)=\Lambda^\alpha_\mu$ and, $J^{(32)}$ is defined as $-J^{(23)}$ and $J^{(10)}$ is defined as $J^{(01)}$ and so on. I don't understand $(2)$. I write the reason. Each $J'^{(\alpha\beta)}$ should be defined by $\theta$ rotation with respect to $\alpha$ -$\beta $ plane for basis $\{e'_\alpha \}_{\alpha=0}^3$;

$$\tag{3}\exp(\theta J'^{(\alpha\beta)})=\cos\theta\left(e'_\alpha\otimes e'^\alpha+e'_\beta\otimes e'^\beta\right)+\sin\theta \left(-e'_\alpha\otimes e'^\beta+e'_\beta\otimes e'^\alpha\right) $$ for $(\alpha\beta)=(23),(31),(12)$,

$$\tag{4}\exp(\theta J'^{(\alpha\beta)})=\cosh\theta\left(e'_\alpha\otimes e'^\alpha+e'_\beta\otimes e'^\beta\right)+\sinh\theta\left(-e'_\alpha\otimes e'^\beta-e'_\beta\otimes e'^\alpha\right) $$ for $(\alpha\beta)=(01),(02),(03)$ where $\alpha$ and $\beta$ are not summed up.

Then, differentiating with respect to $\theta$ at $\theta=0$ and using $\sum_{\mu=0}^{3}e_\mu\otimes e^\mu=identity$,

\begin{align} J'^{(\alpha\beta)}&=-e'_\alpha\otimes e'^\beta+e'_\beta\otimes e'^\alpha\\ &=-\left(\sum_{\mu=0}^{3}e_\mu\otimes e^\mu\left(e'_\alpha\right)\right)\otimes \left(\sum_{\nu=0}^{3}e'^\beta \left(e_\nu\right)\otimes e^\nu\right)+\left(\sum_{\mu=0}^{3}e_\mu\otimes e^\mu\left(e'_\beta\right)\right)\otimes \left(\sum_{\nu=0}^{3}e'^\alpha \left(e_\nu\right)\otimes e^\nu\right)\\ &=\sum_{\mu,\nu=0}^{3}-\left(\Lambda^{-1}\right)^\mu_\alpha \Lambda^\beta_\nu \,e_\mu\otimes e^\nu+\sum_{\mu,\nu=0}^{3}\left(\Lambda^{-1}\right)^\mu_\beta \Lambda^\alpha_\nu \,e_\mu\otimes e^\nu\\ &=\sum_{\mu,\nu=0}^{3}\left(-\left(\Lambda^{-1}\right)^\mu_\alpha \Lambda^\beta_\nu +\left(\Lambda^{-1}\right)^\mu_\beta \Lambda^\alpha_\nu\right)\,e_\mu\otimes e^\nu \tag{5} \end{align} for $(\alpha\beta)=(23),(31),(12)$ and $$J'^{(\alpha\beta)}=\sum_{\mu,\nu=0}^{3}\left(\left(\Lambda^{-1}\right)^\mu_\alpha \Lambda^\beta_\nu +\left(\Lambda^{-1}\right)^\mu_\beta \Lambda^\alpha_\nu\right)\,e_\mu\otimes e^\nu \tag{6}$$ for $(\alpha\beta)=(01),(02),(03)$.

In conclusion of $(5)$ and $(6)$, $J'^{(\alpha\beta)}$ isn't subject to $(2)$.

How is $(2)$ correct? I'd like you to share mathematical way to prove (2).

Answer 1

$J^{(\mu\nu)}$ are just a fixed set of matrices -- you should not take the indices to denote components of a tensor. In particular, this object does not transform under Lorentz transformations: the components of these matrices are the same in every frame of reference, just like $\eta_{\mu\nu}$ which (despite apparently having two vector indices) does not transform, it takes the same values in every frame of reference. Same goes for Dirac matrices $\gamma^\mu$.

The equation $$ J^{(\alpha\beta)}=\Lambda^\alpha_\mu\Lambda^\beta_\nu J^{(\mu\nu)}\tag{2} $$ is true, but note that there is no prime in the left hand side: $J^{(\mu\nu)}$ is frame independent and does not transform. This equation is not in fact making a statement about $J^{(\alpha\beta)}$; instead, this equation is a constraint on $\Lambda$.

Specifically, the matrices $J$ have the following components (by definition!): $$ (J^{(\alpha\beta)})^{\sigma}{}_\rho:=-i(\eta^{\rho\beta}\delta^{\alpha}_\rho-\eta^{\alpha\sigma}\delta^\beta_\rho) $$ and equation $(2)$ follows from the well-known equation $$ \Lambda \eta \Lambda^T=\eta $$ which is simply the definition of a Lorentz transformation. This proves $(2)$.

A posteriori, one recognizes that $(2)$ is formally identical to the transformation law of a rank-2 tensor, and therefore one might pretend that the indices in $J^{(\alpha\beta)}$ are true vector indices. They are not, when initially defining this object, but they can be taken to be, since they satisfy the same equation a true tensor would (except that, again, we do not write a prime on $J$: it is an invariant tensor and its components are the same in every frame of reference).

See also this PSE post for some more details on Lorentz covariance, invariant symbols, etc.

Footnote: I decided to include a factor of $i$ in the generators $J$ because that is the convention I like. Feel free to drop them to match the convention in the OP. I like rotations to be hermitian instead of anti-hermitian.