Tensor contractions in the transformation coefficients of a primed tensor

Question

Consider some arbitrary primed tensor of rank (2,2) $${T'}^{\mu\nu}_{\rho\sigma},$$ where $${T'}^{\mu\nu}_{\rho\sigma}\equiv T^{\mu'\nu'}_{\rho'\sigma'}.$$ The tensor $T$ transforms as

$${T'}^{\mu\nu}_{\rho\sigma}=\frac{\partial {x'}^\mu}{\partial x^\delta}\frac{\partial {x'}^\nu}{\partial x^\epsilon}\frac{\partial x^\omega}{\partial x'^\rho}\frac{\partial x^\lambda}{\partial {x'}^\sigma}\,T^{\delta\epsilon}_{\omega\lambda}.$$

Like wise for some arbitrary (1,1) tensor

$${T'}^\Sigma_{\Sigma}=\frac{\partial {x'}^\Sigma}{\partial x^\alpha}\frac{\partial x^\beta}{\partial {x'}^\Sigma}\, T^\alpha_{\beta} = \delta^\beta_{\alpha}\,T^\alpha_{\beta}=T^\beta_{\beta}.$$

$$ I understand this process for the tensor ${T'}^\Sigma_{\Sigma}$; however, I do not understand how to do the same process for ${T'}^{\mu\nu}_{\rho\sigma}$. How can I manipulate the expression

$${T'}^{\mu\nu}_{\rho\sigma}=\frac{\partial {x'}^\mu} {\partial x^\delta}\frac{\partial {x'}^\nu}{\partial x^\epsilon}\frac{\partial x^\omega}{\partial {x'}^\rho}\frac{\partial x^\lambda}{\partial {x'}^\sigma}\,T^{\delta\epsilon}_{\omega\lambda}$$

such that I obtain deltas from inverse transformation coefficients, such that the resulting tensor contraction will simplify my expression for how ${T'}^{\mu\nu}_{\rho\sigma}$ transforms?

Answer 1

You can't. $T'^{\Sigma}_{\Sigma}$ has all indices contracted, and hence it is now a scalar. That is why its transformation law is so simple. $T'^{\mu\nu}_{\alpha\beta}$ is a full $(2,2)$ tensor, and hence each of its indices will transform separately. Notice that a different object, such as $S^{\mu}_\nu$, will also have a complicated transformation law, but contracting the indices to $S^{\mu}_{\mu}$ will make it simple. Similarly, if some of the indices on $T'^{\mu\nu}_{\alpha\beta}$ were contracted (e.g. $T'^{\mu\nu}_{\mu\beta}$) the transformation would be simplified, because the rank of the tensor would be smaller.