As the gauge group of the Standard Model is $SU(3) \times SU(2) \times U(1)$, would the associated fermions fields be the product of a triplet, a doublet and a singlet, for all particles, or is that incorrect (I know it is not usually written that way, but I wonder if it's just for notational brevity)? For instance, would an up quark be written as something like

$$(u) \otimes (u, d) \otimes (u_r, u_g, u_b)$$

or somesuch, where each component itself is a spinor, or am I missing something. If that is the case, would a particle that does not interact with it have a null multiplet, for instance the electron being

$$(e) \otimes (e, \nu_e) \otimes (0,0,0)$$

Though I suppose that if I write each fermion thusly, there will be a lot of fields written several times, so I don't know.


$\newcommand{\bs}{\boldsymbol}$First off the Standard Model (SM) is chrial, so left and right handed fermions are in different representations of the gauge group.

The rep of $SU(3)$ is determined by the color charge. Gluons are in the adjoint of $SU(3)$, which is the $\bs 8$ of $SU(3)$. Both left and right handed quarks are in the fundamental rep, which is a $\bs 3$ (or a $\bs{\bar{ 3}}$) (for example, the up quark field transforms as a $\bs 3$ of $SU(3)$; stated in more physics-y language, an up quark comes in three colors). All other fundamental fields in the SM are colorless, meaning they transform in the $\bs 1$ rep.

The rep of $SU(2)_L$ is determined by the transformation properties under the weak charge.The $W^\pm$ and $Z$ bosons transform under the adjoint, which is a $\bs 3$ rep. The left handed fermions transform. Within each generation, the different quarks mix and the different leptons mix. So $u_L$ and $d_L$ together form a $\bs 2$ rep of $SU(2)$, similarly $e_L$ and $\nu_{e,L}$ form a $\bs 2$ of $SU(2)$. The higgs forms a doublet (another $\bs 2$) of $SU(2)$, although the vev of the higgs breaks the $SU(2)_L\times U(1)$ down to the $U(1)$ of electromagnetism. The rest of the particles in the SM (in particular the gluons and the right handed fermions) are in the singlet of $SU(2)_L$.

Finally the rep of $U(1)$ is the hypercharge of the object. The hypercharge assignments are chosen ultimately to give the right electric charge, you can find the hypercharge assignments at Wikipedia.

So, putting this together, as an example a left handed up quark transforms as \begin{equation} {\rm left\ handed\ up\ quark}\sim( 3, 2,1/3) \end{equation}

The full list of reps for each particle is a bit annoying to write out, although you can also construct this from the rules above.


Based on the comment, let me give a quick example of how group theory translates into the Lagrangian.

Let's just do a simple case. Let's say we had a set of 2 complex scalar fields transforming in the $\bs 2$ of a global $SU(2)$ symmetry. The representation $\bs 2$ (not the $2$ in $SU(2)$) means that the scalar fields should be arranged in a 2d column vector, so instead of writing $\Phi$ and $\Psi$ to represent the fields, we write $\Phi_a$, where $a$ is an index. In other words, the two scalar fields become linked because of the existence of the symmetry.

Under an $SU(2)$ transformations, \begin{equation} \Phi_a \rightarrow U_{a}^{\ \ b} \Phi_b, \end{equation} where $U$ is unitary, $U U^\dagger = U^\dagger U = 1$.

Similarly, it is useful to define $\Phi^a = (\Phi_a)^\dagger$ (that is, $\Phi^a$ is a conjugate transpose of $\Phi_a$, so it is like a row vector). Then $\Phi^a$ transforms as \begin{equation} \Phi^a \rightarrow \Phi^b(U^\dagger)^{\ \ a}_{b}. \end{equation} Note that $\Phi^a \Phi_a$ is invariant under these $SU(2)$ transformations by unitarity.

Thus, you need to write lagrangians invariant under this transformation. This can by done by writing for example \begin{equation} \mathcal{L} = -\partial_\mu \Phi^a \partial^\mu \Phi_a - V(\Phi^a \Phi_a). \end{equation} If you want the theory to be renormalizable, then $V$ should only contain terms up to $O(\Phi^4)$ in 4 space-time dimensions.

The SM has lots of different reps, which at a pragmatic level shows up as lots of different kinds of indices on fields, each with their own transformation rule. The symmetries are also local, which involves coupling in the gauge bosons. I think if you want more detail you need to ask a separate question, the original question was about the representations present in the SM, the full Lagrangian of the SM is a completely different question.

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  • $\begingroup$ What would this translate as in the Lagrangian? $\endgroup$ – Slereah Jun 20 '15 at 10:40
  • $\begingroup$ I updated the answer to give a bit of detail on this. A full answer to that question would be too broad for this forum I think--you're basically asking to go through the entire SM lagrangian and explain every term. However, there's some toy examples you can understand first, and the SM is essentially just lots of copies of the basic toy examples. $\endgroup$ – Andrew Jun 20 '15 at 15:15

It has quite a nice representation if you consider it a subgroup of $SU(5)$. Then you can think of fermion particles as having 5 charges $(R,G,B,W,H)$ which can be each be $\pm1$.

The other charges can be written as:

Color $C = R + \omega G + \omega^2 B $ with $\omega = \frac{1}{2}(1 + \sqrt{-3})$

Electric charge $Q = \frac{1}{6}(R+G+B-3W)$

Weak Isospin $T_3 = \frac{1}{4}(W-H)$ (says if it is left or right handed.)

Weak Hypercharge $Y_W \equiv 2(Q-T_3)$

Six of the gluons $SU(3)$ carry charges of the form $(\pm2,\mp2,0,0,0)$, $(\pm2,0,\mp2,0,0)$ and $(0,\pm2,\mp2,0,0)$. There are also two neutral gluons to complete $SU(3)$ which act on the two components of the colour charge $C$.

Two of the Weak particles of $SU(2)$, $W^{\pm}$ carry charges of the form $(0,0,0,\pm2,\mp2)$. There is also a netural weak-vector boson to complete $SU(2)$ which acts on $T_3$.

Finally there is the neural photon acting on charge $Q$.

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