Friday, March 10, 2017

homework and exercises - How to calculate 3otimes3 and 3otimes3otimes3 in SU(3)?



EDIT: I have boiled my question down to


How many independent components does a rank three totally symmetric tensor have in n dimensions?


A derivation would be nice too.



OP: I know that I can represent 33 by a rank two tensor. Then the symmetric part is 3(3+1)/2=6 and the antisymmetric part is 3(31)/2=3. Thus 33=6ˉ3 I'm not quite sure about the bar though...It is not clear to me why there is a bar when I do this diagrammatically, because the ˉ3 has the same isospin and hypercharge states as a quark triplet. So that's my first question.


So then 333=3(6ˉ3)=(36)(3ˉ3) I can represent 3ˉ3 by a mixed tensor and separate out the trace. Thus 3ˉ3=81 I get this just fine.


But I'm not sure how to go about doing 36. Since 6 is a symmetric tensor, 36 should be a rank three tensor, symmetric over two indices. Then I could form a totally symmetric tensor and a tensor antisymmetric over two indices. But I have no clue which one is 10 and which one is 8 and obviously no clue how to prove it.


Any help would be greatly appreciated.


EDIT I: I once heard that in three dimensions an antisymmetric tensor is a vector. Is this the reason for writing the second rank antisymmetric tensor as ˉ3? But how does one identify an antisymmetric tensor with the conjugate representation?


EDIT II: Strike the first question. If φkl is the antisymmetric tensor, then I can use the Levi-Civita symbol to write φi=εiklφkl, which is the ˉ3.


EDIT III: I know how to get the 8. Represent 6 by the symmetric Sij and 3 by the vector φk. Then define φlm=εlmkφk and Tijlm=Sijφlm. The tensor Tijlj is traceless and is thus a 321=8. From this I see that this 8 has mixed symmetry properties, as is should, because Tijlj=Tjilj=Tjijl=Tijjl Now I know that 63=108, but how do I figure out that the 10 is an irrep and totally symmetric?


EDIT IV: Now I tried writing Tijk=S(ikφk). This should be the 10. I wrote down the 27 components and verified there are 10 degrees of freedom, as expected. But this is kinda laborious. So my question now becomes: How many independent components does a rank three totally symmetric tensor have in n dimensions? If the answer is 10 for n=3, then my question will be answered.




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