I'm trying to understand better the idea of the standard model, where particle states are described within vector spaces corresponding to irreducible representations of the group of symmetry of physics (e.g. the Poincaré group, or some covering of it).

My question is simple:

  • Why do we only consider linear representations of the underlying symmetry group?

  • I know that the maths are much simpler when studying linear reps. because we know a lot about linear algebra, but why would the physical "reality" be described in a vector space?

  • Are we implicitly studying only a first order approximation of that reality?

  • 2
    $\begingroup$ Do you understand non-linear realization of a symmetry means the symmetry is spontaneously broken? So much of the internal symmetry of the EW interactions is non-linearly realized? $\endgroup$ – Cosmas Zachos Mar 23 '20 at 16:12
  • $\begingroup$ @CosmasZachos yeah, but those non-linearly realised symmetries become linear in the "right variables". They are not linear, but affine, which is pretty much the same thing. Quadratic, or wilder non-linear, is non-linear in any variable. My bet is that OP refers to those (but I could very well be wrong!) $\endgroup$ – AccidentalFourierTransform Mar 23 '20 at 16:19
  • $\begingroup$ @AccidentalFourierTransform Well, the SBroken axials in the nonlinear σ-model are as resoundingly nonlinear as they come, no? I guess I am inviting the OP to reveal if he is asking about Low-Manohar SSBroken Lorentz... $\endgroup$ – Cosmas Zachos Mar 23 '20 at 16:25
  • $\begingroup$ Well to be more precise, I don't have a background in theoretical physics and I'm just getting started with the standard model. So I'm not refering to any of the. Just wondering why there seems to be interest only for the linear representations ? $\endgroup$ – Weier Mar 23 '20 at 16:28
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    $\begingroup$ You all seem to be about SSB. However what I think OP is trying to ask is that when we setup the standard model we say fermions are j=1/2 represntation of the lorentz algebra, gauge bosons are j=1/vector representations etc which are all representation of the algebra over vector spaces, but why just vector spaces? Is there another algebraic space upon which it is possible to represent the lorentz group (I don't know the answer)? @Weier is this what you mean? $\endgroup$ – Toby Peterken Mar 23 '20 at 16:46

Two reasons that come to my mind:

  1. Physicists use Hilbert spaces to describe state spaces. So a model of a particle is set to a (inner product) linear space naturally in this stream. The theory has (group) symmetry. Encoding it gives us a group linear representation on Hilbert spaces.

  2. You might worry that this could be an over-simplification. But by Tannakian reconstruction, the collection of knowledge about all linear representations of a compact Lie group can reconstruct the group. So no information is lost about the group.

  • $\begingroup$ The Tannakian reconstruction seems to be a very good argument explaining why physical models tend to be described in Hilbert spaces on which we have a group action... how come nobody ever mentions that ? It also seems to be recent math things, compared to the standard model. Did the theoretical justification come after the standard model was established ? $\endgroup$ – Weier Jun 5 '20 at 19:58
  • $\begingroup$ I think it's the other way around, but I'm not certain: groups were originally defined as a collection of actions (on roots, linear spaces.. etc, for example), axiomatized only after a certain point of time. As an abstract object, people did not forget its actions, and linearized actions in particular. So in some sense, that you can reconstruct a group from its linear representations is not surprising.. TL;DR: a group = {all its possible actions}. $\endgroup$ – Student Jun 5 '20 at 21:53

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