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In the chapter IV.1 "Reducible or Irreducible?" of Zee's Group Theory book (p. 188-), the author breaks a 2nd rank tensor $T^{ij}$ into invariant subspaces with respect to the action of $\mathrm{SO(3)}$ group. The tensor $T^{ij}$ breaks into a five-dimensional (symmetric traceless), three-dimensional (antisymmetric) and one-dimensional invariant subspaces. The author then claims that each of the subspaces corresponds to an irreducible representation and proceeds to use this as a fact in the further discussions. I do not see a reason to be making claims about irreducibility of the five- and three-dimensional representations because it is not clear that they do not have their own non-trivial invariant subspaces. I wonder if someone could help me here. Why are the three- and five-dimensional representations are irreducible?

Pavlo. B.
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    Do you know what an irreducible representation is? It is a representation with no non-trivial invariant subspaces. It is not very clear to me what you're trying to ask. – Vincent Thacker Oct 03 '21 at 04:05
  • Zee does not prove that the three- and five-dimensional subspaces do not have non-trivial invariant subspaces in themselves (or, may be I just don't see where he proves it). – Pavlo. B. Oct 03 '21 at 04:08
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    You can refer to the answer provided to another question https://physics.stackexchange.com/questions/18228/irreducible-tensors-concept to get clarity on why 3,5-dimensional representations are irreducible. – PRAJWAL H.P. Oct 03 '21 at 09:58
  • @PRAJWAL thank you for the reference. It is helpful, but leaves a question, why $\delta_{ab}$ and $\epsilon_{abc}$ are the only invariant tensors? – Pavlo. B. Oct 03 '21 at 16:19
  • @PRAJWALH.P. I understand why the two tensors are invariant. But is there a proof that there is no other invariant tensors? – Pavlo. B. Oct 04 '21 at 23:55

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