I was reading my prof's notes and came across a passage that I didn't understand:

Consider a chiral SU(3) symmetry, under which the left-handed parts of the spin-1/2 fields of a fermion-number- conserving theory are valued in the fundamental representation, while the right-handed parts are all singlets.

I'm not very strong in group theory, so could someone please explain in simple terms what it means for the left handed parts to transform as triplets and right handed parts to transform as singlets? How would you go about writing down such terms in a Lagrangian?

Many thanks

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    $\begingroup$ It just means that the $SU(3)$ acts as identity on the right-handed fermions. I don't think "triplet" makes any sense here, your already said the left-handed fermions transform as the fundamental representation. $\endgroup$ – Meng Cheng Apr 24 '15 at 22:05

I'm not very strong in group theory, so could someone please explain in simple terms what it means for the left handed parts to transform as triplets and right handed parts to transform as singlets? How would you go about writing down such terms in a Lagrangian?

Groups are abstract. They have elements that can be "multiplied" and they have other properties (for example, every element has an inverse, etc). Representations of groups are concrete. For example a 3x3 matrix could be used to represent an abstract group element. Representations of groups can be multiplied, for example matrices can be multiplied by matrix multiplications.

Say, for example, that you have three abstract elements of a group called "a", "b", and "c". And say that $$ a\cdot b=c $$

Say, also, that you have a specific representation of those three above-mentioned group elements, the representation being 3x3 matrices called: R(a), R(b), and R(c). Then, you want the same multiplication rules to hold for the representation as for the group itself, namely: $$ R(a)\cdot R(b)=R(c)\;, $$ where the $\cdot$ now means matrix multiplication.

For every element in the group there should be some element in the representation of the group, and all the multiplication rules that hold for the elements of the group should hold for the elements of the representation.

The representation of the group does not have to be 3x3 matrices. There are an infinite number of different representations of a group.

One interesting representation is 1. I.e., the representation in which every element of the group is represented by 1. This is trivial, right? Yeah, that's why it is called the "trivial representation". But it works because, suppose that $a \to 1$, $b \to 1$, and $c \to 1$. Then $$ a\cdot b = c \to 1\cdot 1 = 1 $$ which is true.

So every group has a trivial representation.

When we say something "transforms under" some representation of the group, we mean that it gets multiplied by some representation of the group under the group transformation. If something transforms as a singlet it is transforming under the trivial representation. I.e., it gets multiplied by 1. I.e., it doesn't transform.

On the other hand, objects transforming under the fundamental representation of SU(3) get multiplied by 3x3 special unitary matrices. I.e., objects transforming under this representation are like 3x1 column vectors. I.e., triplets. There seems to be a commentator on your question that is confusing this issue.

There are other representations beside the trivial and the fundamental. As I said earlier there are an infinite number of representations. In the case of SU(3) there are also an infinite number of irreducible representations... but a discussion of this point is beyond the scope of this answer.

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    $\begingroup$ +1 Great answer at the right level! I have a slight pedantic quibble: "For every element in the group there should be one element in the representation of the group": you seem to be describing here a faithful representation, whereas a representation is in general simply a homomorphism rather than an isomorphism. You clearly already understand this because you give the trivial representation example: so what I'm saying is that your answer slightly contradicts itself and may thus be confusing to someone struggling to understand this stuff. $\endgroup$ – Selene Routley Apr 25 '15 at 3:48
  • $\begingroup$ PS: Does your cat like salami sandwiches? $\endgroup$ – Selene Routley Apr 25 '15 at 3:49
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    $\begingroup$ changed "one" to "some". $\endgroup$ – hft Apr 25 '15 at 16:55

A spinor has two components that describes its transformation under a lorentz transformation, classified by two indices $\Psi_{a\dot{b}}$, the first transforms under the left part of the lorentz group $SO(1,3)\approx SU(2)_L \otimes SU(2)_R$ (in more detail $\Psi_{a\dot{b}}\rightarrow L(\Lambda)_a^{a'} R(\Lambda)_\dot{b}^{\dot{b}'}\Psi_{a'\dot{b}'}$ where $\Lambda$ is the transformation and $L$ and $R$ are the representations).

Because we also have a global $SU(3)$, the spinor also has an additional index $I$ describing how it transforms under an $SU(3)$ group element $G$ with any representation $M$, i.e. $\Psi_I\rightarrow M(G)_I^J \Psi_J$

In your example we have two spinors, the left handed triplet $\chi_{a\dot{b}I}$ where $L(\Lambda)_a^{a'}$ acts as the $SU(2)_L$ doublet, $R(\Lambda)_{\dot{b}}^{\dot{b}'}=\delta _{\dot{b}}^{\dot{b}'}$ just teh $SU(2)_R$ singlet, and $I\in \{1,2,3\}$ so that $M(G)_I^J$ is just in the fundamental representation of $SU(3)$, i.e. a $3\times 3$ special unitary matrix. The second is the right handed singlet $\eta_{a\dot{b}I}$, for which $L_a^{a'}=\delta_a^{a'}$ and $R$ is an $SU(2)_R$ doublet, and $M(G)_I^J=\delta_I^J$ is just the singlet of $SU(3)$.

In other words they transform under a combined lorentz and $SU(3)$ transformation as \begin{align} \chi_{a\dot{b}I}&\rightarrow L_a^{a'}(\Lambda)M_I^J(G)\chi_{a' \dot{b} J}\\ \eta_{a\dot{b}I}&\rightarrow R_{\dot{b}}^{\dot{b}'}(\Lambda)\eta_{a' \dot{b} I} \end{align}

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