# Examples of non-linear field symmetries?

Consider a Lagrangian theory of fields $\phi^a(x)$. Sometime such a theory posseses a symmetry (let's talk about internal symmetries for simplicity), which means that the Lagrangian is invariant under replacement $\phi^a\to \phi'^a=\phi'^a(\phi,\epsilon)$. Here $\epsilon$ are some continuous transformation parameters. Usually one encounters symmetries that are linear in $\phi$, for example $\phi'=e^{i\epsilon}\phi$ for a single complex scalar or $\phi'^a=\epsilon^{ab}\phi^b$ with orthogonal matrix $\epsilon^{ab}$ for Lagrangian $\mathcal{L}=\frac12\left(\partial_\mu\phi^a\right)^2$.

My question is whether there are examples of non-linear symmetry transformations appearing in physical models? At the time I am mostly concerned with the classical fields but comments on the quantum extensions are surely welcome.

Clarification.

I appreciate references to various actual models. However I would like first to see an example as simple and explicit as possible, where the essence is not obstructed by technicalities. If there are some principle difficulties to construct a really simple example, there must be a reason for that?

Let me also narrow what I mean by a non-linear internal transformation. Assume that replacement $\phi^a(x)\to \phi^a_\epsilon(x)=f(\phi^b(x),\epsilon)$ with some function $f$ leaves Lagrangian invariant $L(\phi,\partial\phi)=L(\phi_\epsilon,\partial\phi_\epsilon)$. Parameter $\epsilon$ could be a vector. Then call such a transformation linear if $\frac{\partial \phi^a_\epsilon}{\partial\phi^b}$ is independent of co-ordinates, $\partial_\mu\frac{\partial \phi^a_\epsilon}{\partial\phi^b}=0$. In this sense, the shift transformation proposed by Andrew is also linear.

• This question (v2) seems like a list question. Here is one example of a non-linear gauge symmetry. – Qmechanic Oct 22 '15 at 8:39
• What does this mean? – Weather Report Oct 22 '15 at 8:43
• @Qmechanic the example seems too technical at the first glance. I believe that there must be either a simpler instance, or an argument for why it is not possible. Moreover, I would like a proper symmetry and not a gauge one to be addressed. – Weather Report Oct 22 '15 at 9:26

The simplest example I know of is a shift symmetry, $\phi\rightarrow \phi + c$ for constant $c$. It is not linear in the sense that is not of the form $\phi\rightarrow U\phi$ for some matrix $U$.

And example lagrangian with this symmetry is just a free massless scalar field, $\mathcal{L}=-1/2(\partial_\mu \phi)^2$.

A consequence of the fact that the symmetry is non-linearly realized (to use the jargon) is that we can think of this shift symmetry as a coming from spontaneously breaking another symmetry. Indeed, you can think of $\phi$ as being the Goldstone boson associated with this breaking.

Another related fact is that correlation functions are not invariant under this transformation. For example, the vacuum expectation value $\langle 0 | \phi | 0 \rangle$ is not invariant under $\phi\rightarrow\phi + c$.

• An interesting example, however not the one I had I mind. Depends on the conventions of course, but I would call this one linear inhomogeneous or something like that. Does not really suit for an application I've been thinking on. – Weather Report Oct 22 '15 at 13:08
• What application do you have in mind, out of curiosity? – Andrew Oct 22 '15 at 13:18
• It seems to me, that invariance of Lagrangian is equivalent to invariance of EOM only if this condition ('linearity' as stated in the appendix to the OP) is satisfied. I wish to check this at some particular examples. Maybe I'll make up a separate question on this. – Weather Report Oct 22 '15 at 13:59
• @WeatherReport Could you check this with my example? I expect that KPZ equation is invariant under the non-linear symmetry, but I never checked. – Steven Mathey Oct 22 '15 at 14:48
• @WeatherReport Hm I'm not sure I agree with that, if the actions are the same before and after the transformation then the EOMs derived from those actions will be the same too. A subtlety that may or may not be relevant is that when you have linearly realized symmetries involving multiple fields (such as an internal $SO(2)$ symmetry), the EOMs are technically covariant, not invariant, under the transformation. – Andrew Oct 22 '15 at 15:38

Consider any non-linear sigma-model with a (pseudo-)Riemannian target space(-time) $(X,g)$. For instance the relativistic particle propagating on a spacetime is an example. And any $p$-brane sigma model is an example. Also the scalar field sector in compactications of higher dimensional (super-)gravity theories are examples.

For these sigma-model field theories, the fields are smooth functions $\Sigma_{p+1} \longrightarrow X$ and the isometries of $(X,g)$ are (induce) non-gauge symmetries of the sigma-model. Any isometry whose underlying diffeomorphism $X \to X$ is, in any set of coordinates, not simply a linear function (and generically it won't) gives an example of a non-linear symmetry.

• Could you elaborate this explicitly at a simple enough example, say of a single particle? – Weather Report Oct 22 '15 at 11:49

The stochastic Kardar-Parisi-Zhang (KPZ) equation, $\partial_t h + \frac{\lambda}{2} \left[\vec{\nabla}h\right]^2 = \nu \nabla^2 h + \eta \, ,$ with $\langle \eta \rangle = 0$ and $\langle \eta(t,x) \eta(t',x') \rangle = D \, \delta(t-t') \, \delta(x-x') \, ,$ can be described as a field theory with the action $$S = \int_{t,x} \tilde{h}\left(\partial_t h + \frac{\lambda}{2} \left[\vec{\nabla}h\right]^2 - \nu \nabla^2 h\right) - D \, \tilde{h}^2 \, .$$ See e.g. this great book for details and further references. $h(t,x)$ is the height field as before and $\tilde{h}(t,x)$ is an auxiliary field that can be used to compute response functions.

This problem posses a non-linear symmetry. Indeed, in it's cole-hopf transformed version, $$h = \frac{2\nu}{\lambda} \, \log\left|w\right| \, , \qquad \tilde{h} = w \, \tilde{w} \, ,$$ the action is $$S = \int_{t,x} \tilde{w}\left(\partial_t w - \nabla^2 w\right) - \frac{1}{4} \frac{\lambda^2 D}{\nu^3} \left(w \tilde{w}\right)^2 \, .$$ The transformed problem is invariant under the transformation $w(t,x) \rightarrow \tilde{w}(-t,x)$, $\tilde{w}(t,x)\rightarrow w(-t,x)$. When translated back into the original variables, this becomes $$h'(t,x) = -h(-t,x) + \frac{2\nu}{\lambda} \log \left|\tilde{h}(-t,x)\right| \, , \qquad \tilde{h}'(t,x) = \tilde{h}(-t,x) \, ,$$ which is non-linear.

See e.g. this paper for details on the stochastic KPZ equation and further references.