# An automorphism of the Lorentz algebra

Considering the Lorentz algebra: $$[J_{\mu\nu},J_{\rho\sigma}] = -i(\eta_{\mu\rho}J_{\nu\sigma} - \eta_{\nu\rho}J_{\mu\sigma} + \eta_{\nu\sigma}J_{\mu\rho} - \eta_{\mu\sigma}J_{\nu\rho})$$ with $$J_{0i} = K_i$$, $$J_{ij} = \epsilon_{ijk}J_k$$ and $$J_{\mu\nu} = -J_{\nu\mu}$$.

For each $$\Lambda \in O(1,3)$$ I have that the following map is an automorphism of Lorentz algebra: $$f_\Lambda: J_{\mu\nu} \rightarrow \tilde{J}_{\mu\nu} = {\Lambda_\mu}^\rho{\Lambda_\nu}^\sigma J_{\rho\sigma}$$

Calling $$\pi(J_{\mu\nu})$$ an irreducible representation of the Lorentz algebra, I would like to show that if $$\Lambda$$ is in the connected part of Lorentz group $$SO(1,3)^+$$, then the representation $$\pi(\tilde{J}_{\mu\nu})$$ is equivalent to $$\pi(J_{\mu\nu})$$, that is: $$\pi(\tilde{J}_{\mu\nu}) = U^{-1}(\Lambda)\pi(J_{\mu\nu})U(\Lambda)$$

I really don't know where to start to prove it. Can anyone give me some advice?

• 1. Since you did not give any physical context for this, might this question be more appropriate for Mathematics? 2. Your notation doesn't really make sense to me: What's the $U$ in the equation you want to show - why is that not $\pi$, too? How is $\pi(\tilde{J})$ supposed to be a different representation than $\pi(J)$? The $\pi$ is the representation map in standard notation and defines the representation, not its argument! Sep 24, 2022 at 13:38
• You do not need representations to show that JTILDE also satisfy the commutation relations. Sep 24, 2022 at 15:48

As Daniel mentioned in the comment section, you can prove the equivalence by showing $$\pi(\tilde{J}_{\mu\nu})$$'s have the same Lie bracket as $$\pi(J_{\mu\nu})$$'s, but if you want to find the exact form of $$U \in \text{GL}(V)$$ where $$\pi(J_{\mu\nu}) \in gl(V)$$, a solution is provided in the content below.
Let $$\Lambda(\theta,\hat{m})=e^{-{i \over 2}\theta^{\mu\nu} \ M_{\mu\nu}}$$ where $$\hat{m}$$ is the axis of rotation, $$\theta$$ is the degree of rotation, and $$M_{\mu\nu}$$'s are the generators of $$\text{SO}(1,3)^+$$ and have the same commutation relationship as $$J_{\mu\nu}$$'s. We write $$\Lambda(\theta,\hat{m})$$ as $$\Lambda(\theta)$$ for notation convenience and define the unitary operator $$U(\Lambda(\theta))=e^{-{i \over 2}\theta^{\mu\nu} \ \pi(J_{\mu\nu})}=\lim_{n \rightarrow +\infty}{\big(e^{-{i \over 2}\big({1 \over n}\big)\theta^{\mu\nu} \ \pi(J_{\mu\nu})}\big)^n}=\lim_{n \rightarrow +\infty}{U\big(\Lambda\big({\theta \over n}\big)\big)^n}$$ And we are going to show $$U(\Lambda(\theta))$$ satisfies $$\pi(\tilde{J}_{\mu\nu})=U^{-1}(\Lambda(\theta))\pi(J_{\mu\nu})U(\Lambda(\theta))$$.
With the Lie bracket of $$J_{\mu\nu}$$'s and the definition of the representation, we know $$[\pi(J_{\mu\nu}),\pi(J_{\rho\sigma})]=-i(\eta_{\mu\rho}\pi(J_{\nu\sigma})-\eta_{\nu\rho}\pi(J_{\mu\sigma})+\eta_{\nu\sigma}\pi(J_{\mu\rho})-\eta_{\mu\sigma}\pi(J_{\nu\rho}))$$ and when $$n \rightarrow +\infty$$, we have $$\big(\Lambda\big({\theta \over n}\big)\big)_{\mu}^{\rho}=\delta_{\mu}^{\rho}-{i \over 2n}\theta^{\kappa\lambda}(M_{\kappa\lambda})_{\mu}^{\rho}+O\big({1 \over n^2}\big)=\delta_{\mu}^{\rho}+{1 \over 2n}\theta^{\kappa\lambda}(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho})+O\big({1 \over n^2}\big)$$ Therefore, up to the first order (the terms proportional to $${1 \over n^2}$$ or smaller are discarded), \begin{align} \pi\big(\big(\Lambda\big({\theta \over n}\big)\big)_{\mu}^{\rho}\big(\Lambda\big({\theta \over n}\big)\big)_{\nu}^{\sigma}J_{\rho\sigma}\big) & \sim \pi(J_{\mu\nu})+{1 \over 2n}\big(\delta_{\mu}^{\rho}\theta^{\kappa\lambda}(\eta_{\kappa\nu}\delta_{\lambda}^{\sigma}-\eta_{\lambda\nu}\delta_{\kappa}^{\sigma})+\theta^{\kappa\lambda}(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho})\delta_{\nu}^{\sigma}\big)\pi(J_{\rho\sigma}) \\ & = \pi(J_{\mu\nu})+{1 \over 2n}\theta^{\kappa\lambda}\big(\eta_{\kappa\nu}\delta_{\lambda}^{\sigma}-\eta_{\lambda\nu}\delta_{\kappa}^{\sigma}\big)\pi(J_{\mu\sigma})+{1 \over 2n}\theta^{\kappa\lambda}\big(\eta_{\kappa\mu}\delta_{\lambda}^{\rho}-\eta_{\lambda\mu}\delta_{\kappa}^{\rho}\big)\pi(J_{\rho\nu}) \\ & = \pi(J_{\mu\nu})+{1 \over 2n}\big(\eta_{\kappa\nu}\theta^{\kappa\sigma}\pi(J_{\mu\sigma})-\eta_{\lambda\nu}\theta^{\sigma\lambda}\pi(J_{\mu\sigma})+\eta_{\kappa\mu}\theta^{\kappa\rho}\pi(J_{\rho\nu})-\eta_{\lambda\mu}\theta^{\rho\lambda}\pi(J_{\rho\nu})\big) \\ & = \pi(J_{\mu\nu})-{1 \over 2n}\big(-\eta_{\nu\rho}\theta^{\rho\sigma}\pi(J_{\mu\sigma})+\eta_{\nu\sigma}\theta^{\rho\sigma}\pi(J_{\mu\rho})-\eta_{\mu\sigma}\theta^{\rho\sigma}\pi(J_{\nu\rho})+\eta_{\mu\rho}\theta^{\rho\sigma}\pi(J_{\nu\sigma})\big) \\ & = \pi(J_{\mu\nu})-{i \over 2n}\theta^{\rho\sigma}[\pi(J_{\mu\nu}),\pi(J_{\rho\sigma})] \\ & = U^{-1}\big(\Lambda\big({\theta \over n}\big)\big)\pi(J_{\mu\nu})U\big(\Lambda\big({\theta \over n}\big)\big) \end{align} It results in \begin{align} \pi\big(\big(\Lambda(\theta)\big)_{\mu}^{\rho}\big(\Lambda(\theta)\big)_{\nu}^{\sigma}J_{\rho\sigma}\big) & = \lim_{n \rightarrow +\infty}{\pi\bigg(\Pi_{k=1}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}J_{\rho_{1}\sigma_{1}}\bigg)} \ \ \ (\rho_{n+1}=\mu,\sigma_{n+1}=\nu) \\ & = \lim_{n \rightarrow +\infty}{\Pi_{k=1}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}\pi\big(J_{\rho_{1}\sigma_{1}}\big)} \\ & = \lim_{n \rightarrow +\infty}{U^{-1}\big(\Lambda\big({\theta \over n}\big)\big)\bigg(\Pi_{k=2}^{n}{\bigg(\big(\Lambda\big({\theta \over n}\big)\big)_{\rho_{k+1}}^{\rho_{k}}\big(\Lambda\big({\theta \over n}\big)\big)_{\sigma_{k+1}}^{\sigma_{k}}\bigg)}\pi\big(J_{\rho_{2}\sigma_{2}}\big)\bigg)U\big(\Lambda\big({\theta \over n}\big)\big)} \\ & = \cdots = U^{-1}(\Lambda(\theta))\pi(J_{\mu\nu})U(\Lambda(\theta)) \end{align}