This is more a mathematical question but in my string theory lecture we always divide in the Polyakov path integral by $$\mathrm{Diff}\ltimes \mathrm{Weyl}$$ and I was wondering why there is the semidirect product. Is there a simple argument for the semidirect product like for the Poincare group by the composition of two transformations?


1 Answer 1

  1. A group element $$(f,\Omega)~\in~\mathrm{Diff}\ltimes \mathrm{Weyl}$$ acts from the right on a metric tensor $g$ as $$g.(f,\Omega)~=~f^{\ast}g\cdot \Omega$$ via pullback and multiplication.

  2. This leads to an outer semidirect product rule: $$(f_1,\Omega_1)\bullet (f_2,\Omega_2)~=~(f_1\circ f_2,\underbrace{\Omega_1\circ f_2}_{\equiv f_2^{\ast}\Omega_1}\cdot\Omega_2).$$

  • $\begingroup$ Qmechanic, this is your turf, maybe you know a quick answer. I just had a fleeting wild thought, I can't post a Q, it's too... fuzzy. Is the OP's semidirect product isomorphic to $SU(2)\ltimes\mathbb{R}^3$? $\endgroup$ Commented Dec 16, 2023 at 20:23
  • $\begingroup$ No, for starters the latter is only 6 dimensional while OP's group is infinite dimensional. $\endgroup$
    – Qmechanic
    Commented Dec 16, 2023 at 20:25
  • $\begingroup$ Thanks, I suspected I'm on the wrong path. I love Physics.SO for this: while thinking on the answer I suddenly find myself on a possibly unexplored sidetrack. Even tho I much more often find myself in a roadside ditch, like this time :-)) $\endgroup$ Commented Dec 16, 2023 at 20:31

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