Using definition of SE tensor as a response to the infinitesimal coordinate change $$\delta_{\epsilon} S=\int\partial_{\mu}\epsilon_{\nu}T_{\mu\nu}d^Dx;\quad \partial_{\mu}T_{\mu\nu}=0;\quad (7)$$ We can prove that for $SO(D)$ invariant theories $T_{\mu \nu}$ can always be made symmetric. It can be seen as follows. Consider $\epsilon_\nu=\omega_{\nu \lambda} x_\lambda$, where $\omega_{\mu \nu}=-\omega_{\nu \mu}$, then from (7) we have $$ \delta_\epsilon S=\frac{1}{2} \int_{\mathbb{R}^D}\left[\partial_\mu \omega_{\nu \lambda}\left(x_\lambda T_{\mu \nu}-x_\nu T_{\mu \lambda}\right)-\omega_{\mu \nu}\left(T_{\mu \nu}-T_{\nu \mu}\right)\right] d^D \boldsymbol{x} . $$ For constant $\omega_{\mu \nu}$ this variation should vanish for rotationally invariant theories, which implies $$ T_{\mu \nu}-T_{\nu \mu}=\partial_\lambda f_{\lambda \mu \nu}, \quad f_{\lambda \mu \nu}=-f_{\lambda \nu \mu} , \tag{i}$$ cf. my related Phys.SE question here.
We can build Belinfante-Rosenfeld SEM tensor by: $$ \tilde{T}_{\mu \nu} \stackrel{\text { def }}{=} T_{\mu \nu}-\partial_\lambda B_{\lambda \mu \nu} \tag{ii}$$ where $$ B_{\lambda \mu \nu}=\frac{1}{2}\left(f_{\lambda \mu \nu}-f_{\mu \lambda \nu}-f_{\nu \lambda \mu}\right) \text {. } \tag{iii} $$ But nobody actually says how to find this tensor, do I have to just find noether current under rotations or what?
