I have come across this phrase

left handed fermions transform under $SU(3)\times{}SU(2)\times{}U(1)$ differently from the way right handed fermions do.

I am just beginning to learn about how the particles of the standard model are related to irreps of the gauge group. So, I would like to know what the phrase transforming under a gauge group exactly means, and what's the meaning of an action of a gauge group on whatever it acts on.

  • $\begingroup$ @JamalS perhaps you can explain? ;) In light of our recent discussion of group theory terminology $\endgroup$
    – Danu
    Jun 1, 2014 at 15:13
  • $\begingroup$ @Danu which discussion is that? $\endgroup$
    – Yossarian
    Jun 1, 2014 at 15:17
  • $\begingroup$ @silvrfück: It was a private discussion, I'm answering your question as we speak now. $\endgroup$
    – JamalS
    Jun 1, 2014 at 15:20
  • $\begingroup$ Related: physics.stackexchange.com/q/19751/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 1, 2014 at 15:26

1 Answer 1


Basic Lie Facts

A group is a general collection of abstract entities; in physics we concern ourselves with those of unitary operators which act on vector spaces of quantum states. Continuously generated groups, for which any element is arbitrarily close to the identity, by the repeated application of infinitesimal elements, are denoted Lie groups. Any infinitesimal element may be written as,

$$g=1+i\alpha_a T^a + \dots$$

where $T^a$ are hermitian generators of the group $G$. These must satisfy a relation,


where $f^{ab}_c$ are the structure constants of the group $G$. For a theory, once the symmetry group is known, the fields transform under a representation of the group, which is unitary and finite-dimensional. A representation is simply a set of matrices $t^a$ which satisfy the aforementioned commutation relation, which is referred to as a Lie algebra.

Group Action

In general, if a group $G$ acts on a set $X$, there must exist a map, $\phi : G \times X \to X$, such that for all elements of the set, $x\in X$, we have:

  • $\phi(I,x)=x$ for the identity $I$
  • $\phi(g,\phi(h,x))=\phi(gh,x) \,$ for all $g,h \in G$

We denote $\phi$ the group action; essentially it's what we do to the set we act on. For the case of the $U(1)$ group, the structure constants vanish, and when we act on the field, it takes the form,

$$A_{\mu}\to e^{i\alpha} A_{\mu}$$

if we consider global symmetries, then $\alpha \in \mathbb{R}$. If you expand the $\exp(i\alpha)$ as a Taylor series, you see it reproduces the expression in the first equation; unsurprisingly, it's called the exponential map.

Role of handedness

Consider the Dirac fermion; it transforms under a reducible representation; we can decompose it into two irreducible representations which act only one two-component spinors, $u_{\pm}$ Weyl spinors. They transform the same way under rotations, but oppositely under boosts - one is left-handed, the other right-handed.

A consequence of their different handedness is that they lie in different representations;the $u_{+}$ in $(1/2,0)$ whilst the other within $(0,1/2)$. A Dirac spinor lies within $(1/2,0) \otimes (0,1/2)$, which is treated as a representation of the double cover of the Lorentz group $SL(2,\mathbb{C})$.

  • 1
    $\begingroup$ But regarding to the problem i suggested, LR handed fermions, that the action is different, does it just mean that the irreps of different handedness particles are different? $\endgroup$
    – Yossarian
    Jun 1, 2014 at 15:43
  • 1
    $\begingroup$ @silvrfück: I've explained the role of handedness with the simplest example: Dirac and Weyl spinors. So yes, they transform under different representations. $\endgroup$
    – JamalS
    Jun 1, 2014 at 15:49

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