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Fields are one of the following: scalars, vectors, spinors or some Lie algebra elements, right? And it's often said that scalars are spin-0 and vectors are spin-1. So, what's idea of correspondence between nature of field and it's spin value?

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  • $\begingroup$ I see a lot of votes to close, but no explanatory comments. What is going on? $\endgroup$
    – S. Carnahan
    May 21, 2012 at 9:01

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A classical field is a section of a vector bundle on the space-time manifold $M$.

That vector bundle is typically obtained by using the associated bundle construction applied to the frame bundle of $M$, and some irreducible representation of SO(d) (you've noted that I'm in Eucledian signature).

Sometimes, this is not enough, and one has to start with a double cover of the frame bundle of $M$, and use some irreducible representation of the double cover Spin(d) of SO(d).

So, the types of field (scalar, vector, ...) are in one-to-one correspondence with types of irreducible representations of the spin group Spin(d).

Those irreducible representations are classified.
For d=3, there is exactly one representation of each dimension. The irreducible representations of the group Spin(3)= SU(2) are:

spin-0 = 1-dimensional trivial irrep
spin-1/2 = 2-dimensional irrep
spin-1 = 3-dimensional irrep
etc.

For d=4, we have an isomorphism between Spin(4) and SU(2)xSU(2), and so irreducible representations are classified by pairs of non-negative half-integers.

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  • $\begingroup$ It's worth mentioning that there's an exceptional isomorphism Spin(3,1) = SL(2,C) as real Lie groups. I don't know what reasonable "physical" restrictions there are to place on the representations of Spin(3,1) — unitarity is probably to strong, and with no restriction at all the complex representation theory extends (and is determined by its extension to) the complexification, so there's no difference. Is it possible that certain physical restrictions do impose that SL(2,C) act holomorphicly? Then you still get the classification in d=3+1 that you had in d=3. $\endgroup$ May 21, 2012 at 7:29

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