Emphasis is on the irreducible. I get what's special about them. But is there some principle that I'm missing, that says it can only be irreducible representations? Or is it just 'more beautiful' and usually the first thing people tried?

Whenever I'm reading about some GUT ($SU(5)$, $SO(10)$, you name it) people usually consider some irreducible rep as a candidate field. Also, the SM Lagrangian is constructed in this way. (Here, experimental evidence of course suggests it.)

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    $\begingroup$ The Dirac spinor is a reducible representation of the Lorentz group. Also, the adjoint of $\mathrm{SO}(n)$ is not always irreducible (e.g. $\mathrm{SO}(4)$), so a gluon on that group would belong to a reducible representation. $\endgroup$
    – MannyC
    Aug 23, 2019 at 16:09

4 Answers 4


Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means that interaction terms that treat irreducible parts of a reducible field representation differently are allowed and generically expected. This in turn means that we will instead reclassify/perceive any reducible field in terms of their irreducible constituents.

  • $\begingroup$ A really great answer! $\endgroup$
    – andypea
    Aug 22, 2019 at 22:13
  • $\begingroup$ Let me try to make an example. Say I have two fields that transform under representations of the same group. There is nothing to prevent me from writing down different mass terms for both fields as the two dont mix (And you say Gell-Manns's totalitarian principle should compel us to do so?). It would therefore not be particularly useful to write the two fields as a direct sum of some 'unified' field, forming a reducible representation. $\endgroup$
    – BeneIT
    Aug 23, 2019 at 11:53
  • $\begingroup$ However, I might digress and come to the conclusion (maybe supported by experimental evidence) that my model would be simpler if the masses were the same, in fact if I wouldn't write down any terms which distinguish the two fields. $\endgroup$
    – BeneIT
    Aug 23, 2019 at 11:53
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    $\begingroup$ @BeneIT In that case, your “simpler” model presumably has a larger symmetry group than you started with, and the reducible representation is in fact an irrep of the larger group. $\endgroup$ Aug 23, 2019 at 16:58

This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. A field that transforms as $\mathbf R$ is the same thing as $N$ fields, which transform as $\mathbf R_1, \dots, \mathbf R_N$. When talking about fundamental fields, we can therefore assume that they transform as irreducible representations of the symmetry group.

  • $\begingroup$ I think this answer is incorrect (or at least, incomplete). $\endgroup$ Aug 23, 2019 at 7:58
  • $\begingroup$ Take an $n$-component massive field transforming as an irreducible representation of the group $\mathcal G$. Symmetry requires that the $n$ components should have the same mass, because $\mathcal G$ mixes the components. If I take two such fields $\varphi_1$ and $\varphi_2$ transforming under irreducible representations $\mathbf R_1$ and $\mathbf R_2$, then $\varphi_1\oplus\varphi_2$ transforms under the reducible representation $\mathbf R_1\oplus\mathbf R_2$. But there is no requirement that $\varphi_1$ and $\varphi_2$ have the same mass. $\endgroup$ Aug 23, 2019 at 7:58
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    $\begingroup$ @StephenPowell After seeing Qmechanic's answer (+1), I might agree that mine is not quite complete. I do not think that it is wrong though, and I don't quite understand your point. I agree with everything you wrote, but I do not see a contradiction to my answer. $\endgroup$
    – Noiralef
    Aug 23, 2019 at 11:28
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    $\begingroup$ There is nothing wrong with this answer. In fact, IMHO, it is the correct answer, and it is complete. I don't understand @StephenPowell's comment. Nowhere in this answer it is said that all the irreps of $\boldsymbol R$ have the same mass. $\endgroup$ Aug 23, 2019 at 12:30
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    $\begingroup$ Note that, although non-reductive groups do have non-completely reducible representations (that's one definition of reductivity), this can't happen for unitary representations (of any group)—at least, not for finite-dimensional unitary representations. $\endgroup$
    – LSpice
    Aug 23, 2019 at 16:50

Irreducible representations are always determined by some numbers, labeling the representation, which correspond to the eigenvalues of some observables which are invariant under the (unitary) action of the Lie group.

If the group represents physical transformations connecting different reference frames (Lorentz, Poincare',...), these numbers are therefore viewed as observables which do not depend on the reference frame so that they define some intrinsic property of the elementary physical system one is considering.

If the group represents gauge transformations, these numbers correspond to quantities which are gauge invariant. In this sense they are physical quantities.

Finally, it turns out that in many cases (always if the Lie group is compact), generic unitary representations are constructed as direct sums of irreducible representations. This mathematical fact reflects the physical idea that physical objects are made of elementary physical objects (described by irreducible representations)


The premise of the question is just false. When doing phenomenology it is useful to split a field into its irreducible components, essentially because each irrep carries its own coupling constant. But when analysing QFT from a theoretical point of view, it is convenient to consider a single "big" field in a reducible representation. So it is just not true that in QFT fields are irreducible: sometimes they are not.

For the representations that are relevant to conventional QFT, all reducible representations are completely reducible, so thinking of a single reducible rep, or a collection of individual irreps, is nothing but a matter of convenience: both descriptions carry the exact same information.

Take for example the beta function of Yang-Mills plus matter. The first coefficient is of the form $$ b_0\sim C_2(G)-T(R) $$ where $T(R)$ is the index of the representation for the matter fields. If $R$ is reducible, $R=R_1\oplus R_2\oplus\cdots\oplus R_n$, one has $T(R)=T(R_1)+T(R_2)+\cdots+T(R_n)$. Therefore, if there are $N_F$ copies of a certain irrep, $R=R_1^{\oplus N_F}$, one would write $$ b_0\sim C_2(G)-N_F T(R_1) $$ which is the formula one often finds in textbooks. Both formulas are identical, and one may or may not want to explicitly split $R$ into its irreps. The general case is the same: one can think of a single field in a rep $R$, or a collection of fields into the irreps of $R$. Both conventions are valid, and sometimes one is more useful than the other. But it is emphatically wrong to claim that all fields are irreducible.


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