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In a previous question of mine it was established that special relativity may be formulated as the theory of a (smooth) Lorentz manifold $(M,g)$ that is diffeomorphic to (the standard diff. structure of) $\mathbb{R}^4$ where $g$ is a flat metric of signature $(+---)$. In this case we have a representation of the group $SO(1,3)$ on each tangent space $T_pM$, always preserving $g$. Furthermore the group $$\text{Isom}(M)=\{\phi:M\to M|\ \phi \text{ is a diffeomorphism and } \phi_*g=g\}$$ of isometries of $M$ is isomorphic to the Poincaré group.

The continuation of this question is as follows:

If one defines a general field to be some section of a (possibly complex) vector bundle over $M$, why does one expect that vector bundle to carry a representation of the Poincaré group, or some subgroup? This as opposed to the obvious claim that the coordinate representation of such a bundle section can be transformed into the representation with respect to any other coordinate system. In particular I want to know if there is a meaningful difference between considering the Poincaré transformations as coordinate changes and considering them as more meaningful transformations, under the actions of which we would expect things like actions and fundamental equations of motion to be invariant. To illustrate what I mean: If $x,y:M\to\mathbb{R}^4$ are global charts of $M$ such that $x\circ y^{-1}$ is actually a Poincaré transformation then transforming the coordinate representations of the same field with respect to $x$ to the representation with respect to $y$ will involve Lorentz matrices.

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    Sorry to trouble you again, but I think these three questions are not focused enough to be answered in a single question. There are aspects there which deserve separate questions: Why every classical field is expected to carry a representation of the Poincaré group, and what the motivation for gauge groups is. Your second questions just sounds weird to me: The whole point of irreducible representations is that (except for some very strange groups) every other representation is a direct sum of them, so demanding that the field transform in an irrep is not actually a constraint. – ACuriousMind Nov 20 '16 at 23:38
  • I agree. I removed all but the first part, which is a free-standing question on its own; I'd encourage you to post the other parts as their own individual questions. – David Z Nov 21 '16 at 02:25

2 Answers2

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Consider a vector bundle $E \overset{\pi}{\rightarrow} M$. In your question, you construct a diffeomorphism $\phi=x\circ y^{-1}:M\rightarrow M$ out of two global charts $x, y:M\rightarrow \mathbb{R}^4$. To me, it seems that you say that $\phi$ defines a transformation of the space of sections: \begin{align} \Phi:\rm{Diff}(M)\times\Gamma(M,E)&\rightarrow \Gamma(M,E) \\ (\phi, X) &\mapsto (p \mapsto X_{\phi(p)}) \end{align} asuming that we are identifying of all the fibers $\pi^{-1}(p)$ for $p\in M$ using the flat connection associated to the flat metric $g$. Of course this defines some action of the Poincaré group.

However, for some suitable bundle a non-equivalent action of the Poincaré group can be defined. Consider a representation of the Lorentz group $R:SO(1,3)\rightarrow \rm{End}(V)$ and suppose that the fibers of our bundle are isomorphic to (and identified with) $V$. The action over $\Gamma(M,E)$ of the Poincaré group that we want is \begin{align} \Psi_R: \rm{Iso}(M)\times\Gamma(M,E )&\rightarrow \Gamma(M,E) \\ (\phi, X) &\mapsto (p \mapsto R(d\phi) X_{\phi(p)}) \end{align}

This is the action that is used in physics. It includes as a particular case the restriction of $\Phi$ from $\rm{Diff}(M)$ to $\rm{Iso}(M)$. This can be seen substituting $R$ by the tensor product of the trivial representation with itself. $\Psi_R$ is the action that involves the Lorentz matrices, but only when $R$ is the fundamental representation.

Now, it is true that for every Poincaré transformation $\phi$ there's an associated coordinate transformation $C^\infty(M,\mathbb{R}^4)\rightarrow C^\infty(M,\mathbb{R}^4):x\mapsto \phi\circ x$. There's also a "trivial" action of every Poincaré transformation over the space of sections: $\Phi$, but it's not the one used in physics and it doesn't involve Lorentz matrices. For the action used in physics we need some extra information, namely the representation $R$.

As for the reason why such a non-trivial action of the Poincaré group is used in physics, in special relativity the space-time symmetry group is $\rm{Iso}(M)\simeq ISO(1,3)$ (the reason for this is, if you want, experimental evidence) and we want to know how the relevant objects in the theory transform under it. In relativistic field theory, those objects are the fields: sections of the vector bundles equipped with some $\Psi_R$.


In general, the statement that the action $S:\Gamma(M,E)\rightarrow \mathbb{R}$ is invariant under some transformation $T:\Gamma(M,E)\rightarrow\Gamma(M,E)$ means that $S[X]=S[T(X)]$ for every section $X$.

The integral that defines the action can be written in coordinates. Thus the action may be seen as a functional $S_{\mathbb{R}^4}$ over sections of a bundle over $\mathbb{R}^4$ instead of the original over $M$. Then the trivial statement that a coordinate transformation $\phi:\mathbb{R}^4\rightarrow \mathbb{R}^4$ induces another action $S'_{\mathbb{R}^4}$ (a different functional) given by $S'_{\mathbb{R}^4}[X]=S_{\mathbb{R}^4}[X\circ \phi]$ is not equivalent to the invariance of the action under the transformation $\phi$ (it would be so only if $S'=S$). Every coordinate transformation induces a new action, but not all of them leave the action invariant (only the Poincaré transformations do).

coconut
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  • Thank you very much for your comment! I might be coming closer to understanding something. I have two questions about the above:
    1. Why does one choose the action on the bundle as defined by a representation on a single fibre? Why is it that precise action you gave the only one worth considering on sections?
    2. What does it mean physically to say "The space-time symmetry group is ..."? (I thought it was Poincaré?) Is it correct to think of transforming between the observations of observers as coordinate changes? From this point of view I still don't see what's special about isometries.
    – Adomas Baliuka Nov 21 '16 at 19:38
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    I'm glad it helped. About you second question, by ISO(1,3) I meant the Poincaré group, so you are totally right. It is correct in some sense to think about transformations changing between observers as coordinate transformations. The special thing about isometries is that the action (the lagrangian, the equations of motion...) is invariant under them (but not under general diffeomorphisms). – coconut Nov 21 '16 at 21:11
  • Suppose one has an action functional $S:\Gamma(M,E)\to\mathbb{R}$. Then to say that $S$ is invariant under $\text{Iso}(M)$ is (?) to say that for all sections $X$ and $\Lambda\in \text{Iso}(M)$ we have $S[\Psi_R(\Lambda,X)]=S[X]$. Usually one writes these actions as integrals of chart representations. The integral expression can of course be written using any chart. Applying a symetry transformation in this formalism looks to me exactly like converting expressions to the integral of the same quantity using a different chart. How does one clarify the difference between these "transformations"? – Adomas Baliuka Nov 21 '16 at 22:10
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    I have added a brief text to my answer trying to clarify this point (if I'm understanding it correctly). I have done so because I needed more space than a comment and I thought that the answer would me more complete with it. – coconut Nov 22 '16 at 09:13
  • Thank you! I'll try to play around with this and maybe make another question dealing with a specific example, if I'm still confused. – Adomas Baliuka Nov 22 '16 at 09:16
  • Of course I would still like to know why one chooses the action on sections $\Psi_R$ induced by a representation of the symmetry group in that precise way and not something else. – Adomas Baliuka Nov 22 '16 at 09:28
  • I'm not sure about that one. I would say that fields encode "local information" (at each point). When applying a transformation $\phi$ taking a point $p$ to $\phi(p)$ you want to translate that information from the fiber at $p$ to the fiber at $\phi(p)$ in a way that is consistent with $\phi$ (using $d\phi$) – coconut Nov 22 '16 at 10:00
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A particle shall have an identity that is kept invariant under symmetry transformations. Irreducible representations provide such an identity, since if you transform a state of a particle you will end up in a state in the same space. On the contrary, if we had not defined particles as irreducible representations, you could apply a symmetry transformation and alter the identity of a particle. You could obtain, for example, a neutrino state by applying a Poincare transformation on an electron state, which is nonsense. Since irreducible representations are indexed by the eigenvalues of the Casimir invariants, the possible values of those invariants index the possible identities of particles. Poincarre symmetry inherits particles with mass and spin, which partially define the identity of a particle and correspond to the two Casimirs of the Poincare group (squared translation generator and squared Pauli-Lubanski vector). The rest of the particle’s identity is given by the Casimirs of other, internal, symmetries. U(1) will give charge, SU(2) isospin and so forth.

  • Sorry, but I don't really see how that answers my question(s). You say that "Poincaré symmetry gives ..." but I'm asking about what Poincaré symmetry is and why we postulate it. My interest in particle physics being zero, I would have to know why $U(1)$ "gives charge" and what exactly that statement means as in why that invariant should be interpreted as charge (which is not part of the question above). Also I don't know what "a neutrino state" or "the identity of a particle"should be in the context of my question (which is not concerning quantum mechanics at all yet). – Adomas Baliuka Nov 20 '16 at 23:29
  • Why we postulate Poincare symmetry is a philosophical question. You should consult the teachings of the great minds of the last century. And about why this gives charge and the other gives spin, well, first of all, forget about all these names. Qualities are attributed to physical entities acording to symmetries, the last being abelian groups. Casimir invariants of Lie groups index their irreducible representations. An irreducible representation is then defined as one-particle state. – user3257624 Nov 21 '16 at 03:57
  • Accepting a set of symmetries, due to philosophical reasons, we attribute qualities to the particles. A set of such qualitites defines the identity of the particle. We then define the corresponding fields as global sections of a twisted vector bundle, associated to the G-principal bundle, where G is the internal symmetries group (internal meaning non-spacetime symmtries) – user3257624 Nov 21 '16 at 04:06
  • The question has been radically modified by an admin. In any case the modification is justified in focusing on my main question: Are Poincaré transformations just coordinate changes? All of physics can be formulated in a way which works with arbitrary coordinates. Why do we focus on Poincaré transformations? What special physical meaning do they gain as isometries? In your answers you talk about concepts such as particles, qualities and identity. My idea of what those things are is extremely vague. Your answers seem to concern questions I would ask once I know the answer to this one. – Adomas Baliuka Nov 21 '16 at 07:28