Deriving symmetries of a Gauge theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:31:02Z http://mathoverflow.net/feeds/question/17997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17997/deriving-symmetries-of-a-gauge-theory Deriving symmetries of a Gauge theory Pedro 2010-03-12T17:28:14Z 2012-08-12T23:58:19Z <p>Hello, I don't know if this is a good place for exposing my problem but I'll try...</p> <p>I have a gauge theory with action:</p> <p><code>$S=\int\;dt L=\int d^4 x \;\epsilon^{\mu\nu\rho\sigma} B_{\mu\nu\;IJ} F_{\mu\nu}^{\;\;IJ}$</code></p> <p>Where <code>$B$</code> is an antisymmetric tensor of rank two and <code>$F$</code> is the curvature of a connection <code>$A$</code> i.e: <code>$F=dA+A\wedge A$</code>, <code>$\mu,\nu...$</code> are space-time indices and <code>$I,J...$</code> are Lie Algebra indices (internal indices) I would like to find its symmetries. So I rewrite the Lagrangian by splitting time and space indices <code>$\{\mu,\nu...=0..3\}\equiv \{O; i,j,...=1..3\}$</code> I find:</p> <p><code>$L = \int d^3 x\;(P^i_{\;IJ}\dot{A}_i+B_i^{\,IJ}\Pi^i_{\,IJ}+A_0^{\;IJ}\Pi_{IJ})$</code></p> <p>Where <code>$\dot{A}_i = \partial_0 A_i$</code>, <code>$P^i_{\;IJ} = 2\epsilon^{ijk}B_{jk\,IJ}$</code> is hence the conjugate momentum of <code>$A_i^{\,IJ}$</code></p> <p><code>$B_i^{\,IJ}$</code> and <code>$A_0^{\;IJ}$</code> being Lagrange multipliers we obtain respectively two primary and two secondary constraints:</p> <p><code>$\Phi_{IJ} = P^0_{\;IJ} \approx0$</code></p> <p><code>$\Phi_{\;\;IJ}^{\mu\nu} = P^{\mu\nu}_{\;\;IJ} \approx0$</code></p> <p><code>$\Pi^i_{\,IJ} = 2\epsilon^{ijk}F_{jk\,IJ} \approx0$</code></p> <p><code>$\Pi_{IJ}=(D_i P^i)_{IJ} \approx0$</code></p> <p>Where <code>$P^0_{\;IJ}$</code> are the conjugate momentums of <code>$A_0^{\,IJ}$</code> and <code>$P^{\mu\nu}_{\;\;IJ}$</code> those of <code>$B_{\mu\nu}^{\;\;IJ}$</code>. Making these constraints constant in time produces no further constraints.</p> <p>Whiche gives us a general constraint:</p> <p><code>$\Phi = \int d^3 x \;(\epsilon^{IJ}P^0_{\,IJ}+\epsilon_{\mu\nu}^{IJ}\;P^{\mu\nu}_{\;\;IJ}+\eta^{IJ}\Pi_{IJ}+\eta_i^{IJ}\Pi^i_{\;IJ})$</code></p> <p>Each quantity <code>$F$</code> have thus a Gauge transformation <code>$\delta F = \{F,\Phi\}$</code> where <code>$\{...\}$</code> denotes the Poisson bracket.</p> <p>Knowing that this theory have the following Gauge symmetry:</p> <p><code>$\delta A = D\omega$</code></p> <p><code>$\delta B = [B,\omega]$</code> </p> <p>Where <code>$\omega$</code> is a 0-form, I would like to retrieve these transformations using the relation below. (where <code>$\Phi$</code> is considered as the generator of the Gauge symmetry) but my problem is that I don't know how to proceed, I already did this with a Yang-Mills theory and it worked... but for this theory it seems to le intractable! Someone to guide me?</p> http://mathoverflow.net/questions/17997/deriving-symmetries-of-a-gauge-theory/18082#18082 Answer by mathphysicist for Deriving symmetries of a Gauge theory mathphysicist 2010-03-13T18:29:54Z 2010-03-16T01:09:47Z <p>Unless for some reason you absolutely must work within the Hamiltonian approach, you can just directly look for the complete set of (infinitesimal Lie point) symmetries of the Euler--Lagrange equations or of the action itself. The procedure is standard and described in many good books. For instance, you can look into those by <a href="http://books.google.com/books?id=sI2bAxgLMXYC" rel="nofollow">Olver</a> (more math-y) or <a href="http://books.google.com/books?id=nFSJn7dIYysC" rel="nofollow">Stephani</a> (somewhat closer to physics). Using the theory from these books you can also verify whether the transformation at the end of your question is indeed a symmetry.</p> http://mathoverflow.net/questions/17997/deriving-symmetries-of-a-gauge-theory/63388#63388 Answer by Amitabha Lahiri for Deriving symmetries of a Gauge theory Amitabha Lahiri 2011-04-29T08:41:32Z 2011-04-29T08:41:32Z <p>Although this question is over a year old, some readers may be interested in the following comments. </p> <ol> <li><p>The symmetries of the Lagrangian and the action are not identical.</p></li> <li><p>I don't know why you cannot recover the gauge transformations from the constraints, perhaps it has to do with not including the $B_{0i}$ fields in the second expression for the Lagrangian?</p></li> <li><p>The constraints give infinitesimal transformations leading to the usual gauge transformations, but there seem to be symmetries which are non-trivial extensions of these.</p></li> <li><p>For example, for a gauge group SU(n), the transformations $$A' = UAU^\dagger + \phi$$</p></li> </ol> <p>$$B' = UBU^\dagger + UAU^\dagger\wedge\phi + \phi\wedge UAU^\dagger + \phi\wedge\phi<br>$$</p> <p>where $U(x) \in$ SU(n), and $\phi = -dUU^\dagger$, leaves the action invariant. </p>