2
$\begingroup$

I am performing an irreducible decomposition of a tensor of rank 4, where it is symmetric in the first two indices: $T_{abmn} = T_{bamn}$. In English notation, the Young tableaux I need to evaluate correspond to the Young diagrams (4), (3,1), and (2,2). The others [(2,1,1) and (1,1,1,1)] are not relevant to the symmetry of the tensor: they contain antisymmetrization between three or more indices, and hence the projectors will be trivial.

Now to my question:

In my projection operators on the corresponding subspaces I encounter terms like $S_{134}A_{12} T_{abmn} = -2T_{ambn} - 2T_{anbm} + 2T_{bman} + 2T_{bnam} \not= 0$.

On the other hand, $A_{12}T_{abmn} = 0$ because of the symmetry property of $T_{abmn} = T_{bamn}$.

How are these two statements not in contradiction with each other?

$\endgroup$
  • $\begingroup$ I suppose my question is: why is this not associative? I have to evaluate $S_{134} A_{12}$ first, and only then apply it onto the tensor. That is, I should write it as $\left( S_{134} \circ A_{12} \right) T_{abmn}$. Make sense? $\endgroup$ – Jens Jul 4 '17 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.