Symmetrische Aufteilung für Tensor dritten Ranges?

$$\begin{eqnarray} T_{[{\mu\nu}\alpha]}&=& T_{{\rho\sigma}\beta}\delta^\rho_{[\mu}\delta^\sigma_\nu\delta^\beta_{\alpha]}\;,\\ &=& \frac 1 6 T_{{\rho\sigma}\beta} \bigg[ \delta^\rho_{\mu}\delta^\sigma_\nu\delta^\beta_{\alpha}- \delta^\rho_{\mu}\delta^\sigma_\alpha\delta^\beta_{\nu}+\delta^\rho_{\alpha}\delta^\sigma_\mu\delta^\beta_{\nu} -\delta^\rho_{\alpha}\delta^\sigma_\nu\delta^\beta_{\mu} +\delta^\rho_{\nu}\delta^\sigma_\alpha\delta^\beta_{\mu} -\delta^\rho_{\nu}\delta^\sigma_\mu\delta^\beta_{\alpha} \bigg]\;,\\ &=& \begin{cases} &\frac 1 3 \big\{ T_{\mu[\nu\alpha]} +T_{\nu[\alpha\mu]} +T_{\alpha[\mu\nu]} \big\} \\ &\frac 1 3 \big\{ T_{[\mu\nu]\alpha} +T_{[\nu\alpha]\mu} +T_{[\alpha\mu]\nu} \big\} \\ &\frac 1 3 \big\{ T_{[\mu|\nu|\alpha]} +T_{[\nu|\alpha|\mu]} +T_{[\alpha|\mu|\nu]} \big\} \end{cases}\;,\quad (3)\\ &=& \frac 1 6 \bigg[ T_{\mu\nu\alpha}-T_{\mu\alpha\nu} +T_{\alpha\mu\nu}-T_{\alpha\nu\mu} +T_{\nu\alpha\mu}-T_{\nu\mu\alpha} \bigg]\quad (1) \end{eqnarray}$$

$$\begin{eqnarray} T_{({\mu\nu}\alpha)}&=& T_{{\rho\sigma}\beta}\delta^\rho_{(\mu}\delta^\sigma_\nu\delta^\beta_{\alpha)}\;,\\ &=& \frac 1 6 T_{{\rho\sigma}\beta} \bigg[ \delta^\rho_{\mu}\delta^\sigma_\nu\delta^\beta_{\alpha}+ \delta^\rho_{\mu}\delta^\sigma_\alpha\delta^\beta_{\nu}+\delta^\rho_{\alpha}\delta^\sigma_\mu\delta^\beta_{\nu} +\delta^\rho_{\alpha}\delta^\sigma_\nu\delta^\beta_{\mu} +\delta^\rho_{\nu}\delta^\sigma_\alpha\delta^\beta_{\mu} +\delta^\rho_{\nu}\delta^\sigma_\mu\delta^\beta_{\alpha} \bigg]\;,\\ &=& \begin{cases} &\frac 1 3 \big\{ T_{\mu(\nu\alpha)} +T_{\nu(\alpha\mu)} +T_{\alpha(\mu\nu)} \big\} \\ &\frac 1 3 \big\{ T_{(\mu\nu)\alpha} +T_{(\nu\alpha)\mu} +T_{(\alpha\mu)\nu} \big\} \\ &\frac 1 3 \big\{ T_{(\mu|\nu|\alpha)} +T_{(\nu|\alpha|\mu)} +T_{(\alpha|\mu|\nu)} \big\} \end{cases}\;,\quad (4)\\ &=& \frac 1 6 \bigg[ T_{\mu\nu\alpha}+T_{\mu\alpha\nu} +T_{\alpha\mu\nu}+T_{\alpha\nu\mu} +T_{\nu\alpha\mu}+T_{\nu\mu\alpha} \bigg]\quad (2) \end{eqnarray}$$ Aus (1)(2) haben wir $$\begin{equation} T_{[{\mu\nu}\alpha]}+T_{({\mu\nu}\alpha)} = \frac 1 3 \big[ T_{\mu\nu\alpha} +T_{\alpha\mu\nu} +T_{\nu\alpha\mu} \big] \end{equation}$$ Außerdem verwenden wir die erste Zeile von (3) und die zweite Zeile von (4) für diese Beziehung, die wir haben $$\begin{equation} T_{\mu[\nu\alpha]} +T_{\nu[\alpha\mu]} +T_{\alpha[\mu\nu]} + T_{(\mu\nu)\alpha} +T_{(\nu\alpha)\mu} +T_{(\alpha\mu)\nu} = T_{\mu\nu\alpha} +T_{\alpha\mu\nu} +T_{\nu\alpha\mu}\;, \end{equation}$$ impliziert $$\begin{equation}\boxed{ T_{\mu\nu\alpha} = T_{\mu[\nu\alpha]} +T_{\nu[\alpha\mu]} +T_{\alpha[\mu\nu]} + T_{(\mu\nu)\alpha} +T_{(\nu\alpha)\mu} +T_{(\alpha\mu)\nu} -T_{\alpha\mu\nu} -T_{\nu\alpha\mu}}\;. \end{equation}$$ Tatsächlich gibt es acht weitere Möglichkeiten, dies zu schreiben. Aber beginnen Sie damit, dass wir schreiben können $$\begin{equation}\boxed{ T_{\mu\nu\alpha} = T_{\mu[\nu\alpha]} +T_{\nu[\alpha\mu]} -T_{\alpha[\mu\nu]} + T_{(\mu\nu)\alpha} -T_{(\nu\alpha)\mu} +T_{(\alpha\mu)\nu}} \;.\label{ssp5} \end{equation}$$ Die andere äquivalente Form von symmetrischen Splits sind $$\begin{equation} T_{\mu\nu\alpha} \begin{cases} & = T_{\mu[\nu\alpha]} -T_{\nu[\alpha\mu]} +T_{\alpha[\mu\nu]} + T_{(\mu|\nu|\alpha)} +T_{(\nu|\alpha|\mu)} -T_{(\alpha|\mu|\nu)}\\ & = T_{[\mu\nu]\alpha} -T_{[\nu\alpha]\mu} +T_{[\alpha\mu]\nu} + T_{\mu(\nu\alpha)} +T_{\nu(\alpha\mu)} -T_{\alpha(\mu\nu)}\\ & = T_{[\mu\nu]\alpha} +T_{[\nu\alpha]\mu} -T_{[\alpha\mu]\nu} + T_{(\mu|\nu|\alpha)} -T_{(\nu|\alpha|\mu)} +T_{(\alpha|\mu|\nu)}\\ & = T_{[\mu|\nu|\alpha]} +T_{[\nu|\alpha|\mu]} -T_{[\alpha|\mu|\nu]} + T_{\mu(\nu\alpha)} -T_{\nu(\alpha\mu)} +T_{\alpha(\mu\nu)}\\ & = T_{[\mu|\nu|\alpha]} -T_{[\nu|\alpha|\mu]} +T_{[\alpha|\mu|\nu]} + T_{(\mu\nu)\alpha} +T_{(\nu\alpha)\mu} -T_{(\alpha\mu)\nu} \end{cases}\label{fivet} \end{equation}$$