I think it has to do with a counting of degrees of freedom, which is explained in every set of lecture notes on higher spin theory, e.g. the ones by Sorokin, "Introduction to the Classical Theory of Higher Spins". It's been a while since I've been into this stuff, but I strongly recommend to take a specificif reference and be more explicit if you want people to help you ;)
Ok there is s-covariant totally symmetric tensor T with indices m1,m2,...,ms, 1<=mi<=4 there is (4+s-1)!/((s!)*(4+s-1-4) independent components! This result is non equal to 2s+1 degrees of freedom(independent components, why ????
Yes, I read this. But problem is transversality and traceless condition. Is transversality and traceless condition in conection with irreducibility condition ? If yes, then where I can find irreducibility conditions for tensors ?
For any given s, the first member of the sets (2.1) and (2.2) defines the mass–shell, the second eliminates
unwanted “time” components, and finally the last confines the available excitations to irreducible multiplets.
So I guess the divergenceless condition removes ghosts (negative norm states), and the tracelessness condition gives you an irrep. Heuristically, this is to be expected, since for rotations every representation can be written as the direct sum of three irreps: the trace, the antisymmetric part and the traceless-symmetric part.
For some representation theory, maybe Zee's QFT-book is nice. But I must admit I've never seen really clear expositions of this kind of stuff. I only know it handwavingly since I've never had to use it in my own research.
If I google, I find often references to the original papers of Fierz, Pauli and Fronsdal. To be honest, I've never seen clear expositions of this representation-stuff aimed at physicists without an overload of mathematical jargon or notation. As I said, Zee has a nice appendix about group theory and building irreducible representations. I also like the exposure of Srednicki's QFT-book.
A very heuristic explanation as I remember it from the top of my head (someone correct me if I'm wrong!): in four dimensions, the (complexified) Lorentz algebra is isomorphic to the product of two SU(2) (complexified) algebra's. That's why for instance the vector representation can be written as the product of two fundamental SU(2) representations. However, these products as such are often not irreducible and also contain other spin-representations. With the tracelessness condition you eliminate those lower spin parts.
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