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Question

If an infinitesimal symmetry transformation parametrized by Killing field $f^\mu(x)$ $$ \delta_f\phi=\phi'(x)-\phi(x)=f^\mu\partial_\mu\phi\tag1 $$ can be generated via Poisson bracket $$ \delta_f\phi=\{\phi,Q_f\}\tag2 $$ Is the following equation correct? $$ \delta_f\delta_g\phi=\{\{\phi,Q_g\},Q_f\}\tag3 $$ Motivation

I would like to study the connection between the algebra of generators and transformations and possible central extensions in 2D conformal classical field theories. Equation (3) gives the clear connection and allows to find the connection (using the Jacobi identity) $$ [\delta_f\delta_g]\phi=\{\phi,\{Q_g,Q_f\}\} = \delta_{\{Q_g,Q_f\}}\phi. $$ Problem

The point is if I use (1) twice I get $$ \delta_f\delta_g\phi=\delta_f(g^\mu\partial_\mu\phi) = f^\nu\partial_\nu(g^\mu\partial_\mu\phi) $$ If I use (2) twice I get $$ \{\{\phi,Q_g\},Q_f\} = \{g^\mu\partial_\mu\phi,Q_f\}=g^\mu\{\partial_\mu\phi,Q_f\} $$ In the last equation I assumed that Poisson bracket ignores $g^\mu$ as it is independent of field $\phi$. Therefore this result is independent of $\partial_\nu g^\mu$ term and contrary may obtain $\partial_\mu f^\nu$.

That is the source of discrepancy, but I am not sure what is the correct approach - adjust Poisson bracket so that they contain variation with respect to $g^\mu$ or to forget about (3), but then I am not sure how to find the connection between the two algebras.

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  • $\begingroup$ I'm pretty sure the answer comes from the fact that the Poisson bracket in field space has an integration and there's an integration by parts that needs to happen when computing the functional derivative. If I remember and have time later I'll check this and maybe make it an answer. $\endgroup$ – Richard Myers Feb 5 at 19:57
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I spent some time double checking this, and actually, unlike my comment indicated, the resolution to your issue has nothing to do with integrations by parts, but rather with a minus sign you seem to have missed somewhere in your calculation. Show that there is indeed a missing minus sign, let me first reproduce the argument for identifying the charge algebra with (a possible central extension of) the Killing vector algebra.

To begin, let me make a point on notation. There is a well-defined notion of a differential on the phase space, and so the notation $\delta_\xi F$ where $\xi$ is one of your Killing vectors (not actually important that it's a Killing vector, only that it generates a symmetry) and $F$ is any function on the phase space is really a shorthand for the contraction of the phase space vector field associated to $\xi$ contracted on the phase space differential of $F$. This is, almost by definition, precisely equivalent to the action of the spacetime Lie derivative of $F$ with respect to $\xi$. So with this in mind, we know, by a matter of definition, that these $\delta_f$ and $\delta_g$ operators satisfy the same Lie bracket algebra as the Killing vectors we started with. The task then is to show that the Noether charges also obey the same algebra.

Hence, I will assume that I have some collection of vector fields $\xi_a$ over spacetime where $a$ indexes which vector field I'm talking about. These vector fields should satisfy some algebra $$ [\xi_a,\xi_b]=f^{c}_{\ ab}\xi_c. $$ I denote the Noether charge associated to each of these by $Q_a$. Letting $F$ be any function on the phase space, it then follows that $$ [\delta_{\xi_a},\delta_{\xi_b}]F=\delta_{\xi_a}\delta_{\xi_b}F-\delta_{\xi_b}\delta_{\xi_a}F=\{\{F,Q_b\},Q_a\}-\{\{F,Q_a\},Q_b\}. $$ Using the Jacobi identity written in the form (and note that we haven't missed a minus sign because all these terms can be put on the same side of the equality with cyclic orders $a,b,F$) $$ \{\{F,Q_b\},Q_a\}-\{\{F,Q_a\},Q_b\}=-\{F,\{Q_a,Q_b\}\} $$ we immediately obtain $$ [\delta_{\xi_a},\delta_{\xi_b}]F=-\{F,\{Q_a,Q_b\}\} $$ so the action of the commutator of the Killing fields is minus the action of the Poisson bracket of the Noether charges. If we were to put the structure constants I defined earlier into here, we would find that the algebra of Noether charges is isomorphic to that of the Killing vectors, up to possible central extension, but with negated structure constants.

There's an important point to be made here by this minus sign. Vector fields are naturally defined to act from the left on functions. However if we look at the action of the Noether charges via the Poisson bracket, $\{\cdot,Q_a\}$, this is actually defining an action from the right. Hence the difference in the minus sign.

So, now that we have taken the time to find this minus sign and point out that the Poisson bracket is actually defining a right action, how does this answer your question? Well let's do the calculation you're asking about: We know that $$ \delta_f\delta_g\phi=f^\nu\partial_\nu(g^\mu\partial_\mu\phi). $$ Let's do the calculation in terms of the Poisson brackets: $$ \{\{\phi,Q_g\},Q_f\}=\{g^\mu\partial_\mu\phi,Q_f\}=g^\mu\partial_\mu\{\phi,Q_f\}=g^\mu\partial_\mu(f^\nu\partial_\nu\phi), $$ so we see that the Poisson bracket calculation actually gives the correct result, but ordered differently. This is what I mean by the bracket defining the right action as opposed to a right action. The ordering we supposed should give the result we wanted was based on thinking about actions from the left, not right. It's easy to see from here that noting the bracket gives a right action will immediately tell us that the algebra should have negated structure constants.

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  • $\begingroup$ Thank you @Richard Myers for the response. But I am not sure the minus sign was the problem (I used antisymmetry of Poisson bracket and changed the order of charges). Rather, I have problem with putting the last two equations of yours into the second one - On LHS I have $(f^\mu\partial_\mu g^\nu - g^\mu\partial_\mu f^\nu)\partial_\nu\phi$, while on RHS I have the opposite order, which clearly is not equal. $\endgroup$ – AccidentalThought Feb 8 at 11:08
  • $\begingroup$ Also one other thought, if I just change the ordering $\{\cdot,Q_a\} = -\{Q_a,\cdot\}$, does it make Poisson bracket an action from the left? In two consecutive actions the minus sign vanish, but antisymmetry of PB gives the same results. Then the different order based on left or right action seems incorrect. Do you have any reference for that right and left action? $\endgroup$ – AccidentalThought Feb 8 at 11:25
  • $\begingroup$ @AccidentalThought The point I was looking to get at is more susinctly made by this MSE post. Note that here we are dealing with generators so there is a sign and the order of composition is reversed. However you are correct that there's a contradiction remaining and the LHS and RHS disagree by a sign. I think I know where this sign appears, but need to check. $\endgroup$ – Richard Myers Feb 8 at 21:15

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