# Does all symmetry breaking have corresponding unitary group?

In high energy physics. Symmetry breaking like electroweak's has corresponding $$SU(2)\times U(1)$$ unitary gauge group broken down to $$U(1)$$. Does it mean all kinds of symmetry breaking (even low temperature) like balancing a pencil with tip of finger and removing finger has corresponding unitary group? What is the unitary group for the pencil symmetry breaking then? If it is unrelated. How do you know what kind of symmetry breaking will have a gauge group like $$U(1)$$, $$SU(2)$$ or none at all?

• The left over symmetry is always a subgroup as far as I know. Commented Jul 13 at 23:42

As in this SE post, you can have the Lagrangian with $$SO(N)$$ symmetry $${\mathcal{L}} = \frac{1}{2}(\partial_\mu \Phi)^T (\partial^\mu \Phi) - \left(\frac{1}{2}\mu^2 \Phi^T \Phi + \frac{1}{4}\lambda (\Phi^ T \Phi)^2 \right). \tag{1}$$

where $$\Phi =\begin{bmatrix} \phi_1 \\ \phi_2\\ \vdots \\ \phi_N \end{bmatrix}.$$

Now, imagine that there is a vev of $$\Phi$$ $$\left<{\Phi}\right> =\begin{bmatrix} 0 \\ 0\\ \vdots \\ v \end{bmatrix}, \tag{2}$$ which spontaneously breaks the $$SO(N)$$ symmetry. Just by looking at (2) we can see that there is still an $$SO(N-1)$$. Indeed, we can further rewrite the Lagrangian to make this explicit by parametrising $$\Phi =\begin{bmatrix} \pi_1 \\ \pi_2\\ \vdots \\ v+\sigma \end{bmatrix} = \begin{bmatrix} \vec{\pi} \\ v+\sigma \end{bmatrix},$$ which can be plugged into (1) to find $${\mathcal{L}} = \frac{1}{2}(\partial_\mu \vec{\pi})^T (\partial^\mu \vec{\pi}) - \left(\frac{1}{2}\mu^2 \vec{\pi}^T \vec{\pi} + \frac{1}{4}\lambda (\vec{\pi}^ T \vec{\pi})^2 \right) + \text{Terms dependent on }\sigma,v,$$ which has a $$SO(N-1)$$ symmetry.

PD: I don't think there's a $$U(1)$$ symmetry in this case, but I'm not 100% sure. I don't see it in (2), but it might be prudent to fully expand the Lagrangian to be sure!

• There is a $U(1)\simeq SO(2)$ subgroup of the remaining symmetry in this case for N>2. But N=2 does constitute an example of no remaining symmetry. Indeed you can introduce scalars to just spontaneously break all the symmetries. Commented Jul 14 at 0:28
• I asked all this because let say there were magnetic monopoles that would make Maxwell Equation symmetric with the presence of single magnetic charges. What would be the corresponding gauge group then of the magnetic monopoles. Let's say there is a low temperature symmetry breaking such that in one mode is the symmetric Maxwell Equation then in one mode it breaks the symmetry producing only electric charges without the presence of magnetic monopoles.
– Jtl
Commented Jul 14 at 22:49

In general you can have broken symmetry groups different from $$U(n)$$. For example the quantum Ising model has a discrete symmetry (enacted by $$P=\prod_j \sigma^z_j$$) that breaks spontaneously in the ferromagnetic phase. Since $$P^2=1$$ the symmetry group is $$\mathbb{Z}_2$$. The Hamiltonian is

$$H = \sum_j \sigma^x_j \sigma^x_{j+1} +h\sigma^z_j \ .$$

But other symmetry group (that breaks spontaneously) are possible. For example in some situations the translation group can break spontaneously.