5
$\begingroup$

I am trying to find canonical references and the history of trivial symmetries. The earliest text book reference I can find is on page 69 of Quantization of Gauge Systems by Henneaux and Teitelboim.

A trivial symmetry is a symmetry transformation of a classical mechanical or field theory system that reduces to the trivial transformation on-shell* (OS), i.e. in the classical mechanical system with action $S(q,\dot q, \dots)$ $$ q \to q'=q'(q,\dot q,\dots) \quad\mathrm{st}\quad S(q',\dot q'\dots)=S(q,\dot q, \dots) \quad\mathrm{and}\quad q' \xrightarrow{OS} q $$ For infinitesimal symmetries $q \to q'=q+\delta q$ the above is written as (introducing indices $i$ for the coordinates $q$ and the summation convention) $$ q^i \to q^i+\delta q^i \quad\mathrm{st}\quad \delta q^i \frac{\delta S}{\delta q^i}=0 \quad\mathrm{and}\quad \delta q^i \xrightarrow{OS} 0 $$ Theorem (3.1) of Henneaux and Teitelboim says that such a transformation must be proportional* to the equation of motion $$ \delta q^i = \varepsilon^{ij}\frac{\delta S}{\delta q^j} $$ where $\varepsilon^{ij}=-\varepsilon^{ji}$. This all generalises to field theories with both commuting and anticommuting fields. It is also proved in the article Symmetries and physical functions in general gauge theory by Gitman and Tyutin.

The above result means that infinitesimal trivial symmetries form an ideal in the algebra of gauge symmetries and can be basically ignored. In fact, they apparently weren't even really noticed in modern field theory until they turned up as the commutator of some non-trivial symmetries in some supergravity calculations. (This is stated without reference in Remarks on Gauge Invariance and First-Class Constraints.)

A more mathsy discussion can be found in (eg) Rigid Symmetries and Conservation Laws in Non-Lagrangian Field Theory.


So, to summarise, my questions are:

  1. What is a canonical reference to be given when introducing trivial symmetries?
  2. Where is theorem (3.1) of Henneaux and Teitelboimm first given?
  3. What are the historical references for trivial gauge symmetries?*

EDIT - Footnotes:

  • On-shell (OS) means that $q$ satisfies its equation of motion $\delta S/\delta q = 0$.

  • Actually, it is normally written using a DeWitt-like condensed notation, so the index contraction actually includes an integration over time (or spacetime in field theories) - this is because there can also be terms with time derivatives of the equations of motion.

  • Where were trivial symmetries first discussed in classical mechanics? Where were they first discussed in field theories? etc...

$\endgroup$
6
  • 2
    $\begingroup$ A slightly earlier, by 3 years, ref is from Henneaux slac.stanford.edu/spires/find/hep?key=2070154 I do not have it in hand; so, I do not know if it references an earlier paper. $\endgroup$ Nov 2, 2010 at 10:48
  • $\begingroup$ @Kelly Thanks for the reference. Unfortunately, once again, the discussion seems to be given with no citations of earlier works. It might have been "common knowledge" in those communities but not put in the literature until about that time... $\endgroup$
    – Simon
    Nov 2, 2010 at 12:35
  • $\begingroup$ Yvette Kosmann-Schwarzbach has a book about the history of symmetries of Lagrangian systems and the Noether Theorems springer.com/mathematics/history+of+mathematics/book/… I don't know what it says about trivial symmetries but maybe you can email her and ask. $\endgroup$ Nov 2, 2010 at 14:26
  • 1
    $\begingroup$ I like this question. I recommend, however, modifying the title to include the actual question, not just the topic. For example, "What are the canonical and earliest references to trivial symmetries in gauge systems?" will make it clearer to the readers (and to Google!) what question you mean to ask. (I tend to skim posts for question marks on first reading, and if I like the sentences ending with question marks, I read the post. The title counts. In that direction, you may also want to mildly rewrite your "looking for" list into entirely question form.) $\endgroup$ Nov 2, 2010 at 16:06
  • $\begingroup$ @Theo Thanks. I made the changes you suggested. $\endgroup$
    – Simon
    Nov 2, 2010 at 21:39

1 Answer 1

1
$\begingroup$

Remark: I think my answer should be ignored. (Apparently I did not understand the problem properly. Probably $S$ should be a linear functional on $C^\infty(M)$ (time-dependent?) etc. I still believe that the question ultimately boils down to something elementary once it is formulated in the right way. I just don't understand how the objects are defined.)

I think the question would be easier to answer for mathematicians if formulated in standard math language. I am not sure I am able to translate it, but let me try: You have some configuration space (probably a manifold, maybe infinite-dimensional) $M$, and a one-parameter group $\{g_t\}$ of $M$ (probably a diffeomorphism) and I guess that you assume that this is contained in some nice Lie group $G$ so that $g_t = \exp(tX)$ with $X \in \mathfrak g$, the Lie algebra of $G$. Now you have a (smooth?) function S on M and you want it to be invariant under $\{g_t\}$. This just means $XS = 0$. Now what does OS mean? Maybe you could explain this. It seems to me that you want to deduce something about $X$ and the derivative of $S$, but I am not quite sure I understand your notation there. I believe if you reformulate your question along these lines, more people can help.

$\endgroup$
3
  • $\begingroup$ The action is a functional, not a function! $\endgroup$ Nov 2, 2010 at 10:37
  • $\begingroup$ Well, a functional is also a function, and the notation $S(q, \dot q, ...)$ seems to indicate that it depends only on input from the configuration space. $\endgroup$ Nov 2, 2010 at 12:08
  • $\begingroup$ @Tobias Thanks for reminding me that on-shell (OS) is not standard maths terminology. I've added a footnote. I used this notation, because my question is a reference request, and as far as I know, most of the early work was done by physicists using this type of notation. (Also I wouldn't be entirely comfortable translating it) $\endgroup$
    – Simon
    Nov 2, 2010 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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