# Singlets and gauge invariance

1. Jul 6, 2008

### ledamage

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.

2. Jul 8, 2008

### jdstokes

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.

3. Jul 9, 2008

### ledamage

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?

4. Jul 9, 2008

### jdstokes

I'd say that's correct yes.

5. Jul 17, 2008

### samalkhaiat

Last edited: Jul 17, 2008
6. Jul 18, 2008

### ledamage

Enlightenment! :)

7. Jul 19, 2008

### jdstokes

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.

8. Jul 19, 2008

### jdstokes

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