For instance, given a theory and a formulation thereof in terms of a principal bundle with a Lie group $G$ as its fiber and spacetime as its base manifold, would a principle bundle with the Poincaré group as its fiber and $\mathcal{M}$ as its base manifold, where $\mathcal{M}$ is a manifold the group of whose isometries is $G$, lead to an equivalent formulation? Why? Why not?

On a related note, can any Lie group be realized as the group of isometries of some manifold?

  • 5
    $\begingroup$ On the related note, the answer is trivially true and not very interesting. Any finite-dimensional Lie group is a group of isometries of any left-invariant metric on its underlying manifold. $\endgroup$ – José Figueroa-O'Farrill Nov 8 '11 at 12:02
  • 1
    $\begingroup$ Thank you. Yes, that is trivial indeed. Sorry, didn't see that. $\endgroup$ – Arpan Saha Nov 8 '11 at 14:32
  • $\begingroup$ I don't understand why you expect this sort of duality to exist. Normally, switching spacetime and the target space in a non-linear sigma model leads to something completely differently. Mappings from M to N are one thing, mappings from N to M completely other. $\endgroup$ – Squark Nov 12 '11 at 12:13
  • $\begingroup$ I am just wondering - isn't trying to write a theory with the Poincare group as the fibre on the space-time the "same" as doing Einstein's gravity? Of course the later part of the question doesn't make sense to me - I mean in the usual theory of connections on some G-bundle (i.e Yang-Mill's theory!) I don't see how the gauge group is acting as isometry on the space-time!? That doesn't look right at all. . $\endgroup$ – user6818 Jan 22 '12 at 22:00
  • $\begingroup$ I would think that the vielbein language of gravity is sort of the correct formulation in which these two pictures are manifest that in some sense $G\times Poincare\text{ }Group$ is the local gauge group of a Yang-Mill's theory with the gauge group $G$ on a space-time. I am not sure. I would love to be corrected! $\endgroup$ – user6818 Jan 22 '12 at 22:00

I think no. In a local trivialization of your G-bundle $\mathcal{M}\times G$, we have a right action $g(x,h)=(x,gh)$ - in other words, $G$ does not act on $\mathcal{M}$ in that sense. So moving to the second case must change the base space to some manifold $\mathcal{M}$ whose Poincare group is isomorphic to $G$ (see what Squark said). They have equivalent fibers but inequivalent base spaces.


I know only very basic things about fiber bundles, but I think that this "swapping" you propose fails even in simplest cases.

I mean, lets take $\mathbb{R}$ as a base space and $S^1$ as a fiber space -- we will get a cylinder, right? Now lets exchange them: $S^1$ is a base space and $\mathbb{R}$ is a fiber space -- that way we can get either cylinder or Möbius strip.

The two things seem to be inequivalent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy