Timeline for Gauge invariance of Faddeev-Popov determinant in bosonic string theory
Current License: CC BY-SA 4.0
7 events
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| Apr 26, 2020 at 9:40 | comment | added | Leonard | I don't think the determinant should be anything else, it is just the general rule that $\delta(f(g(x)))=\sum\frac{\delta(g(x)-g_0)}{|\det df(g_0)|}$. | |
| Apr 25, 2020 at 17:17 | comment | added | pseudo-goldstone | @Leonard Therefore you perhaps don't want to consider $\det|\frac{\delta h^\epsilon}{\delta h}|_{h=0}$ but rather something like $\det|\frac{\delta(\zeta\circ \epsilon)}{\delta \zeta}|$, i.e. the change in group element near the identity. I'm certainly not an expert on these things and that statement was very schematic so hopefully someone who knows more can say something else about it but I do think that's potentially where the invariance of the delta function comes from. | |
| Apr 25, 2020 at 16:53 | comment | added | pseudo-goldstone | @Leonard sorry yes you're absolutely right. I've actually made a larger error in the Taylor expansion I wrote. For example, if your gauge group has a Lie Algebra structure then the new metric is given by $h' = he^{i\omega_a t_a}$ and so even if $\omega$ is small all the terms are only first order in $h$. I removed my edit since I don't believe I know enough on this to say something definitely but I think possibly the thing to look at is the determinant you've written down. When you make a "change of variables" in the integral, it's not the metrics you integrate over but the gauge group. | |
| Apr 25, 2020 at 16:50 | history | edited | pseudo-goldstone | CC BY-SA 4.0 |
deleted 868 characters in body
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| Apr 25, 2020 at 6:33 | comment | added | Leonard | Hello goldmans, for a 'larger' gauge transformation, why can't the leading term be non unity and thus have a non unit determinant? Especially: if the $\epsilon$ is a pure Weyl transformation, then $h^\epsilon=\phi*\epsilon$, and multiplying by phi has certainly non unit determinant. | |
| Apr 25, 2020 at 5:59 | history | edited | pseudo-goldstone | CC BY-SA 4.0 |
Added an explanation of why the delta function is invariant.
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| Apr 25, 2020 at 3:48 | history | answered | pseudo-goldstone | CC BY-SA 4.0 |