# Is there something similar to Noether's theorem for discrete symmetries?

Noether's theorem states that, for every continuous symmetry of a system, there exists a conserved quantity, e.g. energy conservation for time invariance, charge conservation for $U(1)$. Is there any similar statement for discrete symmetries?

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discrete symmetries of a lagrangian or just anything? – anon Aug 24 '10 at 8:53
I guess $t \mapsto -t$ is an interesting discrete symmetry. – anon Aug 24 '10 at 9:02

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– Tobias Kienzler Aug 24 '10 at 12:17

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I am curious if there is a conservation law associated with symmetries of the form psi(x)=psi(x+2*Pi*R) ( a 4-sphere) or psi(x,y)=psi(x+k,y-q) (klein bottle?) – Holowitz Apr 13 '11 at 5:33
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## 7 Answers

For continuous global symmetries, Noether theorem gives you a locally conserved charge density (and an associates current), whose integral over all of space is conserved (time independent).

For global discrete symmetries you have to distinguish between the cases where the conserved charge is continuous or discrete. For infinite symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one. So in such case momentum is conserved modulo vectors in the reciprocal lattice. The conservation is local just as in the case of continuous symmetries.

In the case of finite group of symmetries the conserved quantity is itself discrete. You then don't have local conservation laws because the conserved quantity cannot vary continuously in space. Nevertheless for such symmetries you still have a conserved charge which gives constraints (selection rules) on allowed processes. For example, for parity invariant theories you can give each state of a particle a "parity charge" which is simply a sign, and the total charge has to be conserved for any process, otherwise the amplitude for it is zero.

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Isn't this called Pontryagin duality or something? – Keenan Pepper Apr 13 '11 at 20:24
– Tobias Kienzler Apr 14 '11 at 9:58

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can you provide references on this? – Tobias Kienzler Apr 18 '11 at 9:10

Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not.

Now the question asks if continuous can be replace by discrete?

It should immediately be stressed that Noether Theorem is a machine that for each input in form of an appropriate symmetry produces an output in form of a conservation law. To claim that a Noether Theorem is behind, it is not enough to just list a couple of pairs (symmetry, conservation law).

Now, where could a discrete version of Noether's Theorem live? A good bet is in a discrete lattice world, if one uses finite differences instead of differentiation. Let us investigate the situation.

Our intuitive idea is that finite symmetries, e.g., time reversal symmetry, etc, can not be used in a Noether Theorem in a lattice world because they don't work in a continuous world. Instead we pin our hopes to that discrete infinite symmetries that become continuous symmetries when the lattice spacings go to zero, can be used.

Imagine for simplicity a 1D point particle that can only be at discrete positions $q_t\in\mathbb{Z}a$ on a 1D lattice $\mathbb{Z}a$ with lattice spacing $a$, and that time $t\in\mathbb{Z}$ is discrete as well. (This was, e.g., studied in J.C. Baez and J.M. Gilliam, Lett. Math. Phys. 31 (1994) 205; hat tip: Edward.) The velocity is the finite difference

$$v_{t+\frac{1}{2}}:=q_{t+1}-q_t\in\mathbb{Z}a,$$

and is discrete as well. The action $S$ is

$$S[q]=\sum_t L_t$$

with Lagrangian $L_t$ on the form

$$L_t=L_t(q_t,v_{t+\frac{1}{2}}).$$

Define momentum $p_{t+\frac{1}{2}}$ as

$$p_{t+\frac{1}{2}} := \frac{\partial L_t}{\partial v_{t+\frac{1}{2}}}.$$

Naively, the action $S$ should be extremized wrt. neighboring virtual discrete paths $q:\mathbb{Z} \to\mathbb{Z}a$ to find the equation of motion. However, it does not seem feasible to extract a discrete Euler-Lagrange equation in this way, basically because it is not enough to Taylor expand to the first order in the variation $\Delta q$ when the variation $\Delta q\in\mathbb{Z}a$ is not infinitesimal. At this point, we throw our hands in the air, and declare that the virtual path $q+\Delta q$ (as opposed to the stationary path $q$) does not have to lie in the lattice, but that it is free to take continuous values in $\mathbb{R}$. We can now perform an infinitesimal variation without worrying about higher order contributions,

$$0 =\delta S := S[q+\delta q] - S[q] = \sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_t + p_{t+\frac{1}{2}}\delta v_{t+\frac{1}{2}} \right]$$ $$=\sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_{t} + p_{t+\frac{1}{2}}(\delta q_{t+1}- \delta q_t)\right]$$ $$=\sum_t \left[\frac{\partial L_t}{\partial q_t} - p_{t+\frac{1}{2}} + p_{t-\frac{1}{2}}\right]\delta q_t + \sum_t \left[p_{t+\frac{1}{2}}\delta q_{t+1}-p_{t-\frac{1}{2}}\delta q_t \right].$$

Note that the last sum is telescopic. This implies (with suitable boundary conditions) the discrete Euler-Lagrange equation

$$\frac{\partial L_t}{\partial q_t} = p_{t+\frac{1}{2}}-p_{t-\frac{1}{2}}.$$

This is the evolution equation. At this point it is not clear whether a solution for $q:\mathbb{Z}\to\mathbb{R}$ will remain on the lattice $\mathbb{Z}a$ if we specify two initial values on the lattice. We shall from now on restrict our considerations to such systems for consistency.

As an example, one may imagine that $q_t$ is a cyclic variable, i.e., that $L_t$ does not depend on $q_t$. We therefore have a discrete global translation symmetry $\Delta q_t=a$. The Noether current is the momentum $p_{t+\frac{1}{2}}$, and the Noether conservation law is that momentum $p_{t+\frac{1}{2}}$ is conserved. This is certainly a nice observation. But this does not necessarily mean that a Noether Theorem is behind.

Imagine that the enemy has given us a global vertical symmetry $\Delta q_t = Y(q_t)\in\mathbb{Z}a$, where $Y$ is an arbitrary function. (The words vertical and horizontal refer to translation in the $q$ direction and the $t$ direction, respectively. We will for simplicity not discuss symmetries with horizontal components.) The obvious candidate for the bare Noether current is

$$j_t = p_{t-\frac{1}{2}}Y(q_t).$$

But it is unlikely that we would be able to prove that $j_t$ is conserved merely from the symmetry $0=S[q+\Delta q] - S[q]$, which would now unavoidably involve higher order contributions. So while we stop short of declaring a no-go theorem, it certainly does not look promising.

Perhaps, we would be more successful if we only discretize time, and leave the coordinate space continuous? I might return with an update about this in the future.

An example from the continuous world that may be good to keep in mind: Consider a simple gravity pendulum with Lagrangian

$$L(\varphi,\dot{\varphi}) = \frac{m}{2}\ell^2 \dot{\varphi}^2 + mg\ell\cos(\varphi).$$

It has a global discrete periodic symmetry $\varphi\to\varphi+2\pi$, but the (angular) momentum $p_{\varphi}:=\frac{\partial L}{\partial\dot{\varphi}}= m\ell^2\dot{\varphi}$ is not conserved if $g\neq 0$.

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This paper may be useful for the discrete action ideas you suggest: arxiv.org/abs/nlin.CG/0611058 A "No-Go" Theorem for the Existence of an Action Principle for Discrete Invertible Dynamical Systems. I haven't read through it yet, but it sounds interesting. – Edward Apr 13 '11 at 10:14
If you solve the simple gravity pendulum problem, you can construct two independent conserved quantities. They can be combined in a quantity known as the total energy in this case. – Vladimir Kalitvianski Apr 13 '11 at 22:09

You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector.

There is an additional result: Suppose the Hamiltonian itself is time independent, and suppose the symmetry is related to an operator $\hat S$. An example would be the parity operator $\hat P|x\rangle = |-x\rangle$. If this operator is a symmetry, then $[H,P] = 0$. But since the commutator of an operator with the Hamiltonian also gives you the derivative, you have $\dot P = 0$.

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No, because discrete symmetries have no infinitesimal form which would give rise to the (characteristic of) conservation law. See also this article for a more detailed discussion.

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– Tobias Kienzler Aug 26 '10 at 8:32

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Who says that conservation laws can arise onyly from infinitesimal forms? – Lagerbaer Apr 12 '11 at 19:37

Maybe,

http://www.technologyreview.com/blog/arxiv/26580/

I am by no means an expert, but I read this a few weeks ago. In that paper they consider a 2d lattice and construct an energy analogue. They show it behaves as energy should, and then conclude that for this energy to be conserved space-time would need to be invariant.

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As was said before, this depends on what kind of 'discrete' symmetry you have: if you have a bona fide discrete symmetry, as e.g. $\mathbb{Z}_n$, then the answer is in the negative in the context of Nöther's theorem(s) — even though there are conclusions that you can draw, as Moshe R. explained.

However, if you're talking about a discretized symmetry, i.e. a continuous symmetry (global or local) that has been somehow discretized, then you do have an analogue to Nöther's theorem(s) à la Regge calculus. A good talk introducing some of these concepts is Discrete Differential Forms, Gauge Theory, and Regge Calculus (PDF): the bottom line is that you have to find a Finite Difference Scheme that preserves your differential (and/or gauge) structure.

There's a big literature on Finite Difference Schemes for Differential Equations (ordinary and partial).

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Sobering thoughts:

Conservation laws are not related to any symmetry, to tell the truth. For a mechanical system with N degrees of freedom there always are N conserved quantities. They are complicated combinations of the dynamical variables. Their existence is provided with existence of the problem solutions.

When there is a symmetry, the conserved quantities get just a simpler look.

EDIT: I do not know how they teach you but the conservation laws are not related to Noether theorem. The latter just shows how to construct some of conserved quantities from the problem Lagrangian and the problem solutions. Any combination of conserved quantities is also a conserved quantity. So what Noether gives is not unique at all.

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– anna v Apr 14 '11 at 6:31

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General comment: there must be something missing in the current generation's education. The past three yeasr I have been following scientific blogs, I find that most difficulties and misunderstandings arise because people cannot understand or see the difference between necessary and sufficient conditions. How is mathematics taught at present bemuses me. – anna v Apr 14 '11 at 6:34

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– kakaz Apr 14 '11 at 7:53

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Cont. Then we have theory that for Hamiltonian systems if N integrals of motion exists - system is "integrable" So Vladimir statement in a case of Hamiltonian dynamics is wrong. Of course that there exists constants of motion not related to symmetry. But they are not related to structure of phase space and there is no foliation so in certain meaning they are particular, accidental one. And the may be represented ( after mathematical transformation) as initial conditions of well defined system. – kakaz Apr 14 '11 at 7:55

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– Tobias Kienzler Apr 14 '11 at 8:36
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