# Singlets and gauge invariance

## Main Question or Discussion Point

Hi folks!

Another stupid question: Consider a Yukawa coupling $\lambda \bar{\psi}_1 \psi_2 \phi$ where $\phi$ is a scalar field in the $(2,-\frac{1}{2})$ representation and $\psi_1$ and $\psi_2$ are lh. Weyl fields in the $(2,-\frac{1}{2})$ and $(1,1)$ representation of $\mathrm{SU}(2) \times \mathrm{U}(1)$. Why does the occurrence of the singlet $(1,0)$ on the rhs of

$$(2,-\frac{1}{2}) \otimes (2,-\frac{1}{2}) \otimes (1,1) = (1,0) \oplus (3,0)$$

imply that this term is gauge-invariant? What about the $(3,0)$ part? I just can't see it.

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Interesting question. I'm still learning this stuff so take this explanation with a grain of salt.

$\varphi,\psi_1$ live in two-dimensional field vector spaces, so when we juxtapose them we must end up with a 4-dimensional space according to the tensor product rule for multiplication of vector spaces. We can ignore $\psi_2$ because it is 1-dimensional.

Intuitively it makes sense that a doublet $\varphi$ multiplied by the Dirac conjugate of another doublet is singlet, so $\bar{\psi}_1\varphi$ must form a 1-dimensional subspace of our 4-dimensional tensor product space. The fact that we have a 3-dimensional subspace left over should not concern us much because $\bar{\psi}_1\varphi$ doesn't live in there.

Yes, I think so. The appearance of a singlet in the above equation is just a necessary but not a sufficient condition for the term to be gauge invariant, right?

I'd say that's correct yes.

samalkhaiat
The appearance of a singlet in the above equation is just a necessary but not a sufficient condition for the term to be gauge invariant,
Do not confuse the REDUCIBLE tensor

$$\bar{\psi}_{2} \otimes \phi \otimes \psi_{1} \equiv (2,1) \otimes (2,1) \otimes (1,-2) = (1,0) \oplus (3,0)$$

with the local product of the fields

$$\bar{\psi}_{2} \phi \psi_{1} \in (1,0)$$

which forms the su(2)Xu(1)-invariant Yukawa coupling; anything belongs to a singlet is a scalar (invariant).
Indeed, using the transformations

$$\psi_{1} \rightarrow u_{1} \psi_{1}, \ \phi \rightarrow u_{2} \phi$$

$$\psi_{2} \rightarrow u_{1}u_{2} \psi_{2} = u_{2}u_{1} \psi_{2}$$

where $u_{1} \in u(1)$ and $u_{2} \in su(2)$, it is easy to show that

$$\bar{\psi}_{2} \phi \psi_{1} \rightarrow \bar{\psi}_{2} \phi \psi_{1}$$

We build up higher-dimensional irreducible representations by taking tensor pruducts of the fundamental representations and projecting out the invariant subspaces. Take the simple case of su(3), the fundamental representations are $q \equiv [3]$ and $\bar{q} \equiv [\bar{3}]$;

$$\bar{q} \otimes q \equiv [\bar{3}] \otimes [3] = [8] \oplus [1]$$

This corresponds to the tensor identity

$$\bar{q}_{i} q_{j} = \left( \bar{q}_{i} q_{j} - \frac{1}{3} \delta_{ij} \bar{q}_{k}q_{k} \right) + \frac{1}{3} \delta_{ij} \bar{q}_{k}q_{k}$$

So, while

$$\{\bar{q} \otimes q \equiv \bar{q}_{i}q_{j}\} \in [\bar{3}]\otimes [3]$$

is a reducible su(3)-tensor,

$$\bar{q}_{k}q_{k} \in [1]$$

is an invariant ( su(3)-scalar).

The other object in the bracket(...) is an irreducible, traceless tensor with 8 components, i.e., it belongs to the octet [8] ( mesons or gluons matrix).

regards

sam

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Enlightenment! :)

A very nice explanation by samalkhaiat there.

Something I've been wondering is that given tensor components, e.g.,

$\left( \bar{q}_{i} q_{j} - \frac{1}{3} \delta_{ij} \bar{q}_{k}q_{k} \right)$

how does one deduce that the corresponding subspace is irreducibly invariant?

I believe a necessary and sufficient condition for this to hold is that it must not be possible to form a tensor of lower rank by contracting with the Kronecker delta and permutation symbol, although I don't quite understand why this works.

Suppose that I want to decompose $\mathbf{3} \otimes \mathbf{6}$. Is it sufficient to conclude that the symmetric tensors of the form $T^{ijk}\equiv A^{ij}B^k+A^{ik}B^j+A^{kj}B^i$ form an irreducible subspace and that the remaining orthogonal subspace must be the adjoint rep because it has dimension 8?

Thus $\mathbf{3}\otimes\mathbf{6} = \mathbf{10} \oplus \mathbf{8}$.