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This might be more of a soft question, since I don't While learning about representations of the Lorentz group, I found in Maggiore's book (Chapter 2) that massive particles of spin $j$ have $2j+1$ degrees of freedom, whereas massless particles only have one degree of freedom, the helicity. That's why a photon, which we often call a massless spin-1 particle, has two polarizations and its "spin" (helicity really) can only take two values.

The argument follows from having two Casmir operators for the Lorentz group, $P^\mu P_\mu, W^\mu W_\mu$, with $W^\mu$ being the Pauli-Lubanski 4-vector, which commute. After that, we make the distiction between $P^\mu P_\mu=m=0$ and $m\neq 0$. So, then $m=0$ reduces the degrees of freedom of the field that transforms under the Lorentz group. Please, do correct me if I'm wrong.

This might be a bit generic but, in the case of an internal symmetry (like a gauge symmetry), can there be an analogous restriction to the degrees of freedom? If so, are there any examples of models that do that?

As an example, I'm thinking about a theory with internal symmetry under a group $G$, with its Casimirs, $C_1, C_2, .. C_n$, with one $C_j$ (analogous to $P^\mu P_\mu$ for the Lorentz group) such that $C_j=0$ (with $C_j$ not being trivial).

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What you should really be thinking of is a group and representations.

In the massive case, the little group is $SU(2)$ whose representations are labelled by a half-integer $j$ and has dimension $2j+1$.

In the massless case, the little group os $U(1)$ whose representations are labelled by a half-integer $h$ and have dimension $1$. However, we want our Hilbert space to be CPT invariant so for every helicity $h$ representation, we also include a helicity $-h$ representation giving 2 d.o.f. for each $|h|$.

There is no "restriction" of any sense happening anywhere. The two types of particles are described by two completely different groups so their structures are different.

Internal symmetry groups can also have such properties. It's definitely possible. For instance $SL(2,{\mathbb R})$ has the usual highest weight representations (which has real and discrete scaling dimensions), but they also have a continuous series representation (which has complex and continuous scaling dimensions) which are totally different in structure.

We do not generically see such things in QFT since internal symmetry groups are usually compact (The Poincare group and $SL(2,{\mathbb R})$ are not!)

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  • $\begingroup$ Oh, I didn't know about the concept of a little group. So, generalizing this argument for gauge symmetries, it's again all about the little group, meaning that depending on which the little group of the gauge group is (which I guess is dependent on the value of the aforementioned Casimir that's the analogue of $P^\mu P^\mu$, it can give different results? $\endgroup$ Jul 12 at 19:20
  • $\begingroup$ Little groups are simply a method of constructing representations. At the end of the day, the only thing that matters is the group representations. I was just telling you that what you are thinking of as "restrictions" are not that at all - they are simply two completely different types of representations of the Poincar\'e group. $\endgroup$
    – Prahar
    Jul 12 at 19:24
  • $\begingroup$ Yes, thanks for clearing up the confusion. So, returning to what I asked (apart from the "restriction" mistake), I see no reason why this should not be the case for internal group symmetries, right? $\endgroup$ Jul 12 at 20:08
  • $\begingroup$ I suppose I really don't understand what you are trying to get to here. It's definitely possible that some representations have some vanishing Casimirs. $\endgroup$
    – Prahar
    Jul 12 at 20:10
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    $\begingroup$ It's definitely possible. For instance $SL(2,{\mathbb R})$ has the usual highest weight representations, but they also have a continuous series representation which are totally different in structure. We do not generically see such things in QFT since internal symmetry groups are usually compact (The Poincare group and $SL(2,{\mathbb R})$ are not!) $\endgroup$
    – Prahar
    Jul 12 at 20:14

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