# Is the distinction between the Poincaré group and other internal symmetry groups artificial?

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?

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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. – José Figueroa-O'Farrill Nov 8 '11 at 12:02

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Thank you. Yes, that is trivial indeed. Sorry, didn't see that. – Arpan Saha Nov 8 '11 at 14:32
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. – Squark Nov 12 '11 at 12:13
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. . – user6818 Jan 22 '12 at 22:00
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! – user6818 Jan 22 '12 at 22:00
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## 2 Answers

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.

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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.

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