In condensed matter physics or quantum field theory we often write down terms in our Lagrangian which are invariant under given symmetries. The standard model for example is invariant under $SU(3)\times SU(2) \times U(1)$ (spare spontaneous symmetry breaking) - whilst a typical free energy in the Heisenberg model is: $$H=\int d^3\vec r\left(\frac{1}{2} \nabla_i M_j \nabla_i M_j+a_2 \vec M \cdot \vec M+a_4(\vec M \cdot \vec M)^2+\cdots\right)$$

It is clear in this case that is invariant under $SO(3)$ and $O(3)$.

My question is - in general what terms can we add to Lagrangian (in CPM and QFT) to make it invariant under solely $SO(n)$ ($SU(n)$) rather then $O(n)$ ($U(n)$)?

  • 2
    $\begingroup$ Terms that are not invariant under parity transformations, such as terms involving cross products. $\endgroup$ – Slereah May 11 '18 at 9:13

Take a field theory with a scalar field $\phi_i$ as a multiplet of $O(N)$, where $i$ is the $O(N)$ index.

You can construct two scalar combinations:

  • $\phi_i \phi^i$ that is both $O(N)$ and $SO(N)$ invariant

  • $\epsilon^{i j k \dots}\phi_i \phi_j \phi_k\dots$ that is $SO(N)$ invariant but not $O(N)$ invariant.

So a Lagrangian that contains some combination of the second term will be only $SO(N)$ invariant.

EDIT for clarification: The term $\epsilon^{i j k \dots}\phi_i \phi_j \phi_k\dots$ is zero written like this, but adding derivatives makes it non zero without changing its transformation properties. For example

$$\partial_\mu \phi_i \partial_\nu\phi_j \partial_\sigma \phi_k \epsilon^{ijk}\epsilon^{\mu\nu\sigma\rho}V_\rho$$

is $SO(3)$ invariant and non zero (where $V_\rho$ is some Lorentz vector).

  • $\begingroup$ I take it that we are assuming in your second expression that $\phi_i$, $\phi_j$ and $\phi_k$ are different fields - else this term is just zero. $\endgroup$ – Quantum spaghettification May 11 '18 at 9:45

The answer is fundamentally different for global vs. gauge symmetries. For global symmetries FrodCube's answer is correct, but for gauge symmetries in a continuum field theory (like the one you mention in your question), the gauge group must be connected because the gauge field must be continuous. So it doesn't make sense to have a disconnected gauge group like $\mathrm{O}(N)$ in a continuum gauge field theory.

  • $\begingroup$ I don't think this is correct -- gauge theories with disconnected groups are the bread-and-butter of several active research areas. Say, theories with gauge groups $\mathrm O(N)$ or even $\mathbb Z_n$. $\endgroup$ – AccidentalFourierTransform Jun 4 '18 at 3:36
  • $\begingroup$ @AccidentalFourierTransform You may be correct, I'm not an expert - do you have any references? People talk about $\mathbb{Z}_n$ lattice gauge theories all the time, but I've never heard of a $\mathbb{Z}_n$ continuum gauge theory. How would one reach parts of the gauge group disconnected from the identity via a spatially smooth gauge transformation? $\endgroup$ – tparker Jun 4 '18 at 3:40
  • $\begingroup$ I'm far from an expert either, so take what I say with a grain of salt. For a gauge theory over $\mathrm{O}(N)$ I found arxiv.org/abs/1710.06069, while for $\mathbb Z_n$ I think you need something of the for of Dijkgraaf-Witten theory. $\endgroup$ – AccidentalFourierTransform Jun 4 '18 at 3:43
  • $\begingroup$ @AccidentalFourierTransform Some quick Googling suggests that Dijkgraaf-Witten theory may only be rigorously defined on a lattice. But I would definitely welcome comments from experts on how disconnected gauge groups work. $\endgroup$ – tparker Jun 4 '18 at 3:53
  • $\begingroup$ BTW, you may find this PSE post interesting. $\endgroup$ – AccidentalFourierTransform Jun 4 '18 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.